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These roughly 100 lesson plans are related to mathematics technology and are provided by the non-profit Education Development Center. The lesson plans are designed to help students develop both deep conceptual...
These roughly 200 lesson plans are related to mathematical problem solving and are provided by the non-profit Education Development Center. The lesson plans are designed to help students develop both deep conceptual...
A problem (with solution) sent by Prof. W. McWorter's: Byzantine Basketball is like regular basketball except that foul shots are worth a points instead of two points and field shots are worth b points instead of three...
A tutorial on wavelet filters aimed at engineers. Focusses on "lifting," a technique for creating a general framework to design filters for every possible wavelet transform. May be read online or downloaded in...
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Careers for Math Majors
Careers for Math Majors
A good mathematics education can lead to a good and rewarding job. Statistics and data analysis skills are particularly well-suited to jobs in industry.
Programming experience is also important. In particular, know standard packages such as the Microsoft Office Package, MatLab, and Maple (Mathematica is not as useful here and is more used in academic circles).
Job ads may not specifically mention math. Look for terms like "analyst," "data analysis," and "operations research." Here are some websites with job ads:
The subject of the first exam is calculus-based probability. BU courses that cover the material on this exam are 53.241 (Probability and Statistics) and 53.462 (Mathematical Statistics). Since 53.462 is only offered in the spring semester of odd-numbered years, you need to plan ahead!
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MATLAB: An Introduction with Applications 4th Edition
MATLAB®: An Introduction with Applications is
used by more college students than any other MATLAB® text or
reference. This concise book is known for its just-in-time learning
approach, giving students the information when they need it. The
new edition presents the latest MATLAB® functionality
gradually and in detail. Equally effective as a freshmen-level
text, self-study tool, or course reference, the book is generously
illustrated through computer screen shots and step-by-step
tutorials, with abundant and motivating applications to problems in
mathematics, science, and engineering.
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11.Reading Graphs Reading Graphs Abstract This lesson is designed to introduce students to graphing functions and to reading simple functions from graphs. Many of the examples are motivated by a situation described by the graph. Objectives Upon completion of this lesson, students will: have practiced plotting functions on the Cartesian coordinate plane seen several catego...
14.Function-al Machines & Spaghetti Graphs Summary: Students will learn to identify the relationship between patterns and functions. Main Curriculum Tie: Mathematics Grade 6 Represent and analyze quantitative relationships between dependent and independent variables. 9. Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of a...
16.Algebra II: Quadratic Functions ...s a Clarifying Lesson? A model lesson teachers can implement in their classroom. Clarifying Lessons combine multiple TEKS statements and may use several Clarifying Activities in one lesson. Clarifying Lessons help to answer the question "What does a complete lesson look like that addresses a set of related TEKS statements, and how can these TEKS statements be connected to other parts of the TEKS?" TEKS Addressed in This Lesson Foundations for functions: 2A.1.A, B Quadratic and square root functions: 2A.6.A, B, C; 2A.8.A, B, C, D Materials Graphing calculator Student worksheet (pdf 60kb) home » instructional materials » algebra 2 » clarifyin...
19.Functions and the Vertical Line Test Functions and the Vertical Line Test Abstract The following discussions and activities are designed to lead the students to explore the the vertical line test for functions. Plotting points and drawing simple piecewise functions are practiced along the way. Objectives Upon completion of this lesson, students will: be able to recognize functions f...
User Rating:
Grade Level: 3-5
20.Impossible Graphs Impossible Graphs Abstract This lesson is devoted to impossible graphs. Users of the module can learn to distinguish between possible and impossible graphs of functions, and to learn why some graphs are impossible. Objectives Upon completion of this lesson, students will: have practiced plotting functions on the Cartesian coordinate plane be able to read a...
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Lead your child into advanced math with the new Horizons Pre-Algebra! This highly anticipated course for junior high students reviews basic math operations and walks your child into the intriguing world of algebra, trigonometry, geometry, and real-life applications with a colorful student workbook, a user-friendly teacher's guide, and a tests and resources book.
College Test Prep Prepare your child for the rigors of standardized math testing! Because many colleges base admission and scholarships on test scores, this unique feature provides practice with multiple choice problems that include the same style, format, and difficulty of pre-algebra questions on today's standardized tests.
Math Minute Interviews Before each set of ten lessons, students read insightful interviews with people who use algebra in their daily vocations. Adding a human interest to algebra, these interviews make math come to life and are the basis for word problems within that section.
Hands-on Learning Acquiring an understanding of new concepts is more fun with hands-on materials that assist both visual and kinesthetic learners in mastering challenging algebraic concepts with formula strips, full-color diagrams of 3-D shapes, and color-coded algebra tiles.
Classwork Section Reinforce new concepts with additional practice! This handy layout feature cements new material with extra exercises that can be completed in homeschool groups, with older siblings, or between parent and child.
What Others Have Said about Horizons Math "Your Horizons Math books are fantastic! The spiral teaching continually reinforces the skills, and the colorful and fun way it is presented keeps my daughters interested. Horizons Math is truly the best!" - Debbie D., North Carolina
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Horizons Pre-Algebra Placement Tests Download and complete the free Horizons Pre-Algebra Readiness Test to see if your child is ready to begin studying math at the pre-algebra level.
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This book takes a fresh, student-oriented approach to teaching the material covered in the senior- and first-year graduate-level matrix structural analysis course. Unlike traditional texts for this course that are difficult to read, Kassimali takes special care to provide understandable and exceptionally clear explanations of concepts, step-by-step procedures for analysis, flowcharts, and interesting and modern examples, producing a technically and mathematically accurate presentation of the subject.
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Calculus
This course is an introduction to the fundamental concepts of calculus. The first semester consists of a review of analytic geometry and trigonometry, and the study of the derivative, continuity and limits, and differentials. The second semester includes a study of integration, logarithmic and exponential functions, techniques of integration, and applications of integration. A graphing calculator is required. Prerequisite: Precalculus
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9780534495015
ISBN:
053449501X
Pub Date: 2005 Publisher: Brooks/Cole
Summary: An increasing number of computer scientists from diverse areas are using discrete mathematical structures to explain concepts and problems. Based on their teaching experiences, the authors offer an accessible text that emphasizes the fundamentals of discrete mathematics and its advanced topics. This text shows how to express precise ideas in clear mathematical language. Students discover the importance of discrete ma...thematics in describing computer science structures and problem solving. They also learn how mastering discrete mathematics will help them develop important reasoning skills that will continue to be useful throughout their careers.
Schlipf, John is the author of Discrete Mathematics For Computer Science With Student Solutions Manual on CDROM, published 2005 under ISBN 9780534495015 and 053449501X. Four hundred forty eight Discrete Mathematics For Computer Science With Student Solutions Manual on CDROM textbooks are available for sale on ValoreBooks.com, sixty used from the cheapest price of $0.01, or buy new starting at $30
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Algebra and Trigonometry presents the essentials of algebra and trigonometry with some applications. The emphasis is on practical skills, problem solving, and computational techniques. Topics covered range from equations and inequalities to functions and graphs, polynomial and rational functions, and exponentials and logarithms. Trigonometric functions... more...
Algebra for College Students, Revised and Expanded Edition is a complete and self-contained presentation of the fundamentals of algebra which has been designed for use by the student. The book provides sufficient materials for use in many courses in college algebra. It contains chapters that are devoted to various mathematical concepts, such as the... more...
College Algebra, Second Edition is a comprehensive presentation of the fundamental concepts and techniques of algebra. The book incorporates some improvements from the previous edition to provide a better learning experience. It provides sufficient materials for use in the study of college algebra. It contains chapters that are devoted to various... more...
College Algebra and Trigonometry, Second Edition provides a comprehensive approach to the fundamental concepts and techniques of college algebra and trigonometry. The book incorporates improvements from the previous edition to provide a better learning experience. It contains chapters that are devoted to various mathematical concepts, such as the... more...
First Course in Algebra and Number Theory presents the basic concepts, tools, and techniques of modern algebra and number theory. It is designed for a full year course at the freshman or sophomore college level. The text is organized into four chapters. The first chapter is concerned with the set of all integers - positive, negative, and zero. It... more...
Algebra, Topology, and Category Theory: A Collection of Papers in Honor of Samuel Eilenberg is a collection of papers dealing with algebra, topology, and category theory in honor of Samuel Eilenberg. Topics covered range from large modules over artin algebras to two-dimensional Poincaré duality groups, along with the homology of certain H-spaces as... more...
Test Bank for Precalculus: Functions & Graphs is a supplementary material for the text, Precalculus: Functions & Graphs. The book is intended for use by mathematics teachers. The book contains standard tests for each chapter in the textbook. Each set of test focuses on gauging the level of knowledge the student has achieved during the course. The... more...
Intermediate Algebra focuses on the principles, operations, and approaches involved in intermediate algebra. The book first elaborates on basic properties and definitions, first-degree equations and inequalities, and exponents and polynomials. Discussions focus on the greatest common factor and factoring by grouping, factoring trinomials, special... more...
Elementary Algebra, Third Edition focuses on the basic principles, operations, and approaches involved in elementary algebra. The book first ponders on the basics, linear equations and inequalities, and graphing and linear systems. Discussions focus on the elimination method, solving linear systems by graphing, word problems, addition property of... more...
The History of Modern Mathematics, Volume I: Ideas and their Reception documents the proceedings of the Symposium on the History of Modern Mathematics held at Vassar College in Poughkeepsie, New York on June 20-24, 1989. This book is concerned with the emergence and reception of major ideas in fields that range from foundations and set theory, algebra... more...
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books.google.com - This... Number Theory
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Find a Fayville Algebra fluent in the many advanced formalisms of special and general relativity and quantum mechanics, including classical Hamiltonian and Lagrangian dynamics, quantum mechanical Hilbert spaces, Riemannian geometry in general relativity, and the geometry of gauges (connections on fiber bundles) in cLearning English for ESL/ESOL students is about developing the necessary written and communication skills to be able to function effectively in daily life in an English-speaking society. Conversational and pronunciation skills are critical for students so that they can converse with and understa...
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Algeba Acrostics
0.54MB. 0 audio & 122 images. Updated 2014-08-22.
Description
Update Aug 4 2014
Connection words are in bold
Many interference fixes
First word list has been shortened
The instructions are in the deck. tag: "guide"
Acrostics have a particular advantage over other mnemonic methods. Memory connections for English, math, science, foreign languages, and other subjects can be created well in advance of learning the material. The process can be described as a sort of "pre-loading" of the material into memory. Anyone of reading age can utilize this advantage.
This system was created to replace traditional forms of memorization and fill in a number of missing elements in the memory recall process.
First word recall (use this list while reviewing)
A list of the potential choices for the first word of the sentence is narrowed to about 7+/- choices. Using this list while reviewing will help to categorize the initial recalls.
problem: The student remembers the formula/procedure/steps but does not remember its name or application.
solution: The topic ID is at the head of the acrostic. Here is an example.
Adding and subtracting terms- [for fractions- same denominator/for polynomials- same exponent and variable/for radicals- same index and radicand]
Verification
problem: The student remembers the formula/procedure/definition but is unsure of the accuracy of the recall.
solution: A weakened or an interference-ladened recall can be verified via interconnections to the retrieval cue, material, or the acrostic.
problem: The student has forgotten that a formula/procedure/definition exists.
solution: The category lists are maintained in memory via the spaced repetition system.
problem: The student does not know what to do or where to begin solving the math problem.
solution: Parts of the math problem can be identified and matched to their category.
problem: The student creates his or her own rule or procedure thinking that the rule or procedure exists and is valid.
solution: Every action in the example has an ID and is maintained in memory by the spaced repetition system.
problem: The student has partially or completely forgotten a formula/procedure/definition.
solution: The acrostic is interconnected with the material and is maintained in the spaced repetition system. The topic ID and subcategories are used as retrieval cues to help recall the acrostic and the material. Here is an example.
(The connection words TF are in bold)
A S T [ F... P... R... ]
Adding and subtracting terms- [for fractions- same denominator/for polynomials- same exponent and variable/for radicals- same index and radicand]
Assisted initial repetition
problem: New material requires a large amount of initial drill.
solution: For new material, the acrostic letters and knowledge of the material are used to assist the initial recall of new material.
example of the process
------------------------------------
3 on the answer button graph is the number of retrieval cue successes.
2 on the answer button graph is the number of retrieval cue failures.
Learning stage stage (Workload starts to decrease)
In this stage, over-learning and the spacing effect pushes the cards into maturity.
[35]
[10] mature stage (Workload decreases to very small numbers)
You should wait until the cards have reached this stage before you use the system for a test. You should also be able to recall the category lists.
[8]
[5]
Here are scans of my preparation for two tests.
Sample (from 164Front
Function reflections
Back
The frontline royal navy forces' x-planes were fielded with new x-weapons for their operation.Function reflections- [for negative f- x-axis/for f negative x- opposite]
Acrostic
frnfxfnxo
Tags
Front
Sequence methods (list)
Back
The social message bar on the cable news newscaster's program permitted the nation's participation in the show.Binomial theoremCircular permutationsN choose rNth of an arithmetic sequencePermutations of n, n at a timePermutations of n, r at a timeNth of a geometric sequencePermutations with repetitionsSum of the first n terms of an arithmetic sequence
Acrostic
smbcnnppnps
Tags
alist category
Front
Standard equation of an ellipse or hyperbola
Back
The scrambling of the entire encampment to hastily handle the x-planes that managed to hide and slip into their territory with the aim of striking their production and manufacturing yards was mobilized to knock out the sudden threat to their bases and to stabilize the volatile state of the nation.Standard equation of an ellipse or hyperbola- [for horizontal- x minus h squared through a squared plus or minus y minus k squared through b squared/for vertical- switch numerators]
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Created by David Lane of Rice University, this applet demonstrates how a histogram is affected by bin width and starting point of first bin. It also illustrates cross-validation criterion for assessing histograms. The...
Exercises posted on this web site offer an opportunity for students to evaluate how much they have retained in various subjects of Algebra. Topics covered include geometry, functions, vectors, and statistics. There are...
This webpage from Johns Hopkins University provides an introduction to Fourier series. The site describes Fourier series in the context of audio tones, and interactive applets that illustrate the concepts involved are...
Created by David Smith for the Connected Curriculum Project, the purpose of this module is to study data that may be modeled by sinusoidal functions; in particular, to determine average level, period, frequency,...
Created by Nathan Kahl for the Connected Curriculum Project, the purpose of this module is to study properties of the graphs of the basic trigonometric functions, sine and cosine. This is one within a much larger set...
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In the first paragraph of the preface of Mathematical Tools for Data Mining, the authors state that "we emphasize that this book is about mathematical tools for data mining, not data mining itself." Mathematicians will be reassured to know that the authors hold true to their word. In addition to the set theory, partial orders, and combinatorics mentioned in the subtitle, readers will find solid presentations of topologies and measures, linear spaces, norms and inner products, lattices and Boolean algebras, and more.
The authors also hold true to their word when they note that the book "is intended as a reference for the working data miner," though this reviewer would add one caveat. Data miners who intend to use the book as a reference must have a facility with mathematical notation prior to using the book. As long as they do, they will find that the chapters and their subsections are well labeled and organized. Further, the prose is clear and the comprehensive index facilitates ease of use.
Mathematics faculty may consider using the book as more than simply a reference. That is, this textbook is appropriate for an advanced undergraduate or graduate mathematics elective class. All theorems are proved, notation is standard, and ample exercise sets are included at the end of every chapter. The treatment of topics progresses quickly from the introductory — for example rank, multilinear forms, and determinants are covered in the chapter titled "Linear Spaces" — to less common topics such as topological linear spaces. Significant sections of the book present the math on its own terms, though readers will also discover applications, such as those to databases and data mining.
At 800+ pages, Mathematical Tools for Data Mining: Set Theory, Partial Orders, Combinatorics is more than just the data-mining reference book. It is highly readable textbook that successfully connects classic, theoretical mathematics to an enormously popular current application in modern society.
Susan D'Agostino is an Associate Professor of Mathematics at Southern New Hampshire University. She also serves on NH Governor Maggie Hassan's STEM Education Task Force.
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Math Skills Program - Information for All Students
Program Description
The Math Skills Program consists of three non-credit courses: M102F, M103F and M104F (expressed collectively as M10XF) and is intended for students who either have a weak background in mathematics or are returning to the subject after some years.
These courses are offered on campus, daytime and evenings, and through distance education, in the Fall, Winter and Spring semesters.
Registration Fees
Each individual Math 10XF course carries the same tuition fee as other Memorial University courses. Students pay for whatever course they register for; if students manage to complete two or all three M10XF courses in the one semester they will not be charged additional tuition fees. If necessary, course registration will be changed, at the end of the semester, to reflect the highest M10XF level completed.
Please note that students completing more than one course per semester may have to purchase an additional set of textbooks.
Methodology
The Math 10XF courses are designed to be self-study with tutorial assistance; they are not lecture courses. Students write a set of diagnostic surveys at the beginning of each level of the program (M102F, M103F, M104F) to assess their math skills pertinent to that level. Answers to the survey questions are analyzed and an individualized program is designed for each student based on the results of the diagnostics. Students are then given their personal programs outlining the work to be completed during the semester. Because each student's program is individualized, the content of personal programs will vary.
Please note that these courses are not self-paced; all assigned work must be completed and all tests passed by the end of the semester.
Evaluation
The Math 10XF courses do not have midterm or final examinations. The method of evaluation used is continuous assessment. Students write a series of module tests throughout the semester. The tests are divided into sections or parts and students are required to demonstrate their mastery of topics by passing all tests sections assigned to them. Each test section requires that a minimum number of questions be answered correctly to pass the section. The pass standard for each section varies but can be as high as 80 to 90%. In fact, some sections actually require a mark of 100% to pass.
Answers to test questions are marked right or wrong only; no part marks are given. All questions are worth one mark each regardless of the number of steps involved in finding the solution. Normally marks will not be given for answers only. Students must show how they arrived at their answers unless the questions can easily be done mentally or with the aid of a calculator (for those test sections where calculator usage is permitted). Certain test sections require that students demonstrate the use of a particular method to arrive at their answers. For these test sections, answers that are numerically correct will still be marked wrong if the method specified is not used.
If a student fails to meet the minimum requirements needed to pass a section(s) of a module test, s/he will have the opportunity to rewrite those sections. The student will not have to rewrite the entire test, just a different version of those sections of the test s/he failed. Follow-up work will be given to the student to help him/her prepare to rewrite those sections.
Students will be allowed a maximum of three attempts at each module test they write. Certain restrictions apply and will be explained to the student during the semester. Results of tests written in the last week of the semester and all third attempts of module tests will not be given back to students. Course results can be obtained when final grades are released via Memorial's Self-Service system.
A final grade of P (pass) or F (fail) in the course is assigned at the end of the semester, based on whether or not the student passed all assigned sections of all required module tests, as specified on their personal program.
Since the Math 10XF courses are non-credit, they cannot affect a student's grade point average but the grade P or F will appear on the student's transcript.
Effort Required
It takes an average of four hours to complete one unit of work from our textbooks. Students are expected to spend at least eight hours per week working on their personal program. If a student works at this pace; attends classes regularly (if completing the course on-campus) so as to receive additional help, valuable tips and reinforcing materials; and writes his/her module tests in a timely fashion, any one of the Math 10XF courses should be passable.
While the majority of M10XF students complete one course per semester, students who are highly motivated and have a good work ethic may be able to complete two or even all three courses in the same semester. Of course the student's work pace would have to be increased accordingly.
It is extremely important that students organize their time carefully to ensure all work is completed and all tests passed by the end of the semester.
Calculator Usage
Calculator usage is monitored and controlled throughout the Math 10XF program. Students are only allowed to use calculators for specific parts of the program. The Math 10XF course manual lists, in detail, the sections of the textbook and module tests where calculator usage is permissible. At no time, will students be allowed to use graphics or programmable calculators when writing module tests. For this reason, it is also advisable that students not use graphics or programmable calculators while working on their personal programs. A regular scientific calculator will suffice for all Math 10XF work.
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Statistics for the Behavioral Sciences 6e to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Essentials of Statistics for the Behavioral Sciences 6e
Select this link to open drop down to add material Essentials of Statistics for the Behavioral Sciences 6 Statistics for the Behavioral Sciences 6e to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Fundamental Statistics for the Behavioral Sciences 6e
Select this link to open drop down to add material Fundamental Statistics for the Behavioral Sciences 6e to your Bookmark Collection or Course ePortfolio
'Geometry is concerned with the various aspects of size, shape and space. In this unit, you will explore the concepts of...
see more
'Geometry is concerned with the various aspects of size, shape and space. In this unit, you will explore the concepts of angles, shapes, symmetry, area and volume through interactive activities.After studying this unit you should be able to:understand geometrical terminology for angles, triangles, quadrilaterals and circles;measure angles using a protractor;use geometrical results to determine unknown angles;recognise line and rotational symmetries;find the areas of triangles, quadrilaterals and circles and shapes based on these;find the volume of boxes, cyclinders and objects with constant cross-sectional areas.Geometry is concerned with the various aspects of size, shape and space. In this unit, you will explore the concepts of angles, shapes, symmetry, area and volume through interactive activities. After studying this unit you should be able to: understand geometrical terminology for angles, triangles, quadrilaterals and circles; measure angles using a protractor; use geometrical results to determine unknown angles; recognise line and rotational symmetries; find the areas of triangles, quadrilaterals and circles and shapes based on these; find the volume of boxes, cyclinders and objects with constant cross-sectional areas or Course ePortfolio for material Geometry
Select this link to open drop down to add material Geometry
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Synopses & Reviews
Please note that used books may not include additional media (study guides, CDs, DVDs, solutions manuals, etc.) as described in the publisher comments.
Publisher Comments:
Just the facts (and figures) to understanding algebra.
The Complete Idiot's Guide® to Algebra has been updated to include easier-to-read graphs and additional practice problems. It covers variations of standard problems that will assist students with their algebra courses, along with all the basic concepts, including linear equations and inequalities, polynomials, exponents and logarithms, conic sections, discrete math, word problems and more.
—Written in an easy-to-comprehend style to make math concepts approachable
—Award- winning math teacher and author of The Complete Idiot's Guide® to Calculus and the bestselling advanced placement book in ARCO's "Master" series
Synopsis:
About the Author
W. Michael Kelley is an award-winning calculus teacher and the author of bestselling The Complete Idiot's Guide to Calculus and Master the AP Calculus AB and BC Tests-by far the bestselling advanced placement book in ARCO's Master series. He has been recognized as an Outstanding High School Mathematics Teacher by the Maryland Council of Teachers of Mathematics and has been named "Most Popular Teacher" for four years running in his home
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MATH FOR LIBERAL ARTS I
Prerequisites are a
satisfactory score on an appropriate proficiency examination and Algebra
II and Geometry, or the equivalent of these two courses. Presents topics
in sets, logic, numeration systems, geometric systems, and elementary
computer concepts. Lecture 3 hours per week.
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What knowledge of mathematics do secondary school math teachers need to facilitate understanding, competency, and interest in mathematics for all of their students? This unique text and resource bridges the gap between the mathematics learned in college and the mathematics taught in secondary...
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US Education - charts and graphs ged worksheets
50 Tagged Resources - "charts and graphs ged worksheets"andFree educational programs, games, and printable worksheets in Math and English for grades K-8. Fulfills supplemental educational needs of both homeschooled and elementary school kids... Practice spelling words with fun software andworksheets. Fill in the missing letters game and colorful word se...Online tutorials for using the TI-83 or TI-84 graphing calculator in mathematics courses. Appropriate for the high school and college level... ... for using the TI-83 or TI-84 graphing calculator in a trigonometry course. ... Trigonometry. Precalculus ... a precalculus level trigonometry course, ...Universal Class is the leading provider of online, instructor-led, professional development courses... Dozens of writing classes at levels from beginning to advanced, developed by professional and published writers... GED Science... GED Science ... 2, 2008 You Are Here: Home @amp;gt;School of Sci...
High school level mathematics and computer programming lessons and activities that are fun and challenging. Materials for students, parents and teachers... Solving Equations... Solving Equations ... Enter the right side of the equation into Y2. Graph (you may need to adjust your window to see whe...
This site is dedicated to providing free algebra 1 worksheetsand the the piano, woodwinds and brass, music composer biographies andworksheets, music theory games andworksheets, printable award certificates, and more!
ClassBrain.com is an educational resource site that has specially designed resources for K-12 students and extensive resources for parents and teachers. Explore ClassBrain.com today!... ... taking the new SAT exam, graphing calculators are also an excellent study aid ... the library, students canWolfram Research, makers of Mathematica, the only fully integrated technical computing software... ... college and professional settings, graphing calculators are still common in high ... Key Advantages of Mathematica Compared to Graphing Calculators: ... Free Seminars from the Wolfram Education ...
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free Algebra eBooks
Algebra, Second Edition, by Michael Artin, provides
comprehensive coverage at the level of an honors-undergraduate or
introductory-graduate course. The second edition of this classic
text incorporates twenty years of feedback plus the author's own
teaching experience. This book discusses concrete topics of
algebra in greater detail than others, preparing readers for the
more abstract concepts; linear algebra is tightly integrated... more...
Most topics in near-ring and near-field theory are treated here,
along with an extensive introduction to the theory. There are two
invited lectures: ``Non-Commutative Geometry, Near-Rings and
Near-Fields'' which indicates the relevance of near-rings and
near-fields for geometry, while ``Pseudo-Finite Near-Fields'' shows
the impressive power of model theoretic methods. The remaining
papers cover such topics as D.G. near-rings, radical theory,... more...
This volume is dedicated to Bill Helton on the occasion of his
sixty fifth birthday. It contains biographical material, a list of
Bill's publications, a detailed survey of Bill's contributions to
operator theory, optimization and control and 19 technical
articles. Most of the technical articles are expository and should
serve as useful introductions to many of the areas which Bill's
highly original contributions have helped to shape over the last
forty... more...
Integral transforms, such as the Laplace and Fourier transforms,
have been major tools in mathematics for at least two centuries. In
the last three decades the development of a number of novel ideas
in algebraic geometry, category theory, gauge theory, and string
theory has been closely related to generalizations of integral
transforms of a more geometric character. "Fourier–Mukai and Nahm
Transforms in Geometry and Mathematical Physics" examines... more...
Algebraic K-theory encodes important invariants for several
mathematical disciplines, spanning from geometric topology and
functional analysis to number theory and algebraic geometry. As is
commonly encountered, this powerful mathematical object is very
hard to calculate. Apart from Quillen's calculations of finite
fields and Suslin's calculation of algebraically closed fields, few
complete calculations were available before the discovery of... more...
Offering a uniquely modern, balanced approach,
Tussy/Gustafson/Koenig's BASIC MATHEMATICS FOR COLLEGE STUDENTS,
Fourth Edition, integrates the best of traditional drill and
practice with the best elements of the reform movement. To many
developmental math students, mathematics is like a foreign
language. They have difficulty translating the words, their
meanings, and how they apply to problem solving. Emphasizing the
"language of mathematics," the... more...
This market-leading text continues to provide students and
instructors with sound, consistently structured explanations of the
mathematical concepts. Designed for a one-term course that prepares
students for further study in mathematics, the new Eighth Edition
retains the features that have always made College Algebra a
complete solution for both students and instructors: interesting
applications, pedagogically effective design, and innovative... more...
This volume contains selected refereed papers based on lectures
presented at the ´;Fifth International Fez Conference on
Commutative Algebra and Applications´ that was held in Fez, Morocco
in June 2008. The volume represents new trends and areas of
classical research within the field, with contributions from many
different countries. In addition, the volume has as a special focus
the research and influence of Alain Bouvier on commutative algebra
over... more...
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In the transition from students to mathematicians, many find it
difficult to move from equations to proofs. Here Sally (mathematics, U.
of Chicago) gives students and their instructors an easier passage,
working through an extraordinary number of exercises embedded in the
text. Designed to cover two semesters or three quarters, this starts
with a review of sets, functions and other basic ideas and moves on to
linear algebra, the construction of the real and complex numbers, metric
and Euclidean spaces, complete metric spaces and the p-adic completion
of Q. The exercises lend themselves well to classroom use, independent
study, or inquiry based learning projects students can work and present
to the class. Sally includes a set of challenge problems for the
particularly motivated and/or gifted.
([c]20082005 Book News, Inc., Portland, OR)
COPYRIGHT 2008 Book News, Inc.
No portion of this article can be reproduced without the express written permission from the copyright holder.
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This math problem demonstrates the concept of geometric progression, through an example of a million dollar contract between an employee and an employer. Application of the concept of geometric progression to social cause activism is addressed. This...(View More) resource is from PUMAS - Practical Uses of Math and Science - a collection of brief examples created by scientists and engineers showing how math and science topics taught in K-12 classes have real world applications.(View Less)
Using the simple example of calculating the probability of reaching a traffic light while green, students are shown how to build a mathematical model using a very commonly-taught formula (sum of first n integers) to solve a rather practical problemThis exercise shows a practical application of trigonometry in the aviation environment, where student pilots consider the relationship between altitude and distance to complete a landing. It requires a scientific calculator. simple example shows how algebra can be useful in the real world by exploring the question: Should Grandpa start receiving his Social Security benefits at age 62 or should he wait until age 65? This resource is from PUMAS - Practical Uses of...(View More) Math and Science - a collection of brief examples created by scientists and engineers showing how math and science topics taught in K-12 classes have real world applications.(View Less)
This series of example calculations applies basic trigonometry to to calculate the altitude of satellites and Iridium satellite flares. This resource is from PUMAS - Practical Uses of Math and Science - a collection of brief examples created by...(View More) scientists and engineers showing how math and science topics taught in K-12 classes have real world applications.(View Less)
In this exercise, students learn about the historical development of the Julian and the Gregorian Calendars and design a reasonable calendar for an imaginary planet, considering the cycle period and making design tradeoffs, math example shows how to calculate the distance one can see from different heights using trigonometry. This resource is from PUMAS - Practical Uses of Math and Science - a collection of brief examples created by scientists and engineers...(View More) showing how math and science topics taught in K-12 classes have real world applications
| 677.169 | 1 |
Word Problems? NO PROBLEM!
Now anyone, even those whose palms begin to sweat at the first sight of math problems that begin "A train left the station going 65 mph..." can overcome anxiety and learn to solve word problems. In Math Word Problems Demystified, experienced math instructor Allan G. Bluman provides an effective, tension-free, approach to conquering the word problems on the SATs and many other standardized tests, in algebra, and in other mathematics and science classes.
With Math Word Problems Demystified, you master the subject one simple step at a time -- at your own speed. This unique self-teaching guide offers practice problems, a quiz at the end of each chapter to pinpoint weaknesses, and a 40 question final exam to reinforce the methods and material presented in the book.
If you want to master math word problems, here's the self-teaching course that will get it done. Get ready to --
Transform word problems into solvable equations
Master a 4-step strategy that empowers you to understand and solve the most common word problems
A fast, effective, and fun way to master word problems, Math Word Problems Demystified is the perfect shortcut to gain the confidence and develop the necessary skills to solve these tough test questions.
| 677.169 | 1 |
This substantially illustrated manual describes how to use Maple as an investigative tool to explore calculus concepts numerically, graphically, symbolically and verbally. Every chapter begins with Maple commands employed in the chapter, an introduction to the mathematical concepts being covered, worked examples in Maple worksheet format, followed by thought-provoking exercises and extensive discovery projects to encourage readers to investigate ideas
| 677.169 | 1 |
Mathematical Concepts and Methods in Modern Biology
Using Modern Discrete Models
Mathematical Concepts and Methods in Modern Biology offers a quantitative framework for analyzing, predicting, and modulating the behavior of complex biological systems. The book presents important mathematical concepts, methods and tools in the context of essential questions raised in modern biology.
Designed around the principles of project-based learning and problem-solving, the book considers biological topics such as neuronal networks, plant population growth, metabolic pathways, and phylogenetic tree reconstruction. The mathematical modeling tools brought to bear on these topics include Boolean and ordinary differential equations, projection matrices, agent-based modeling and several algebraic approaches. Heavy computation in some of the examples is eased by the use of freely available open-source software.
Audience
Researchers, educators, and students engaged in Biological Research and Mathematics
Reviews
"Contributors in biology, in mathematics, and in bioinformatics introduce undergraduate students and their instructors to more applications of discrete mathematics to biology than can be found in standard textbooks. The goal is not to be comprehensive, but to open the door to more advanced and specialized resources."--Reference and Research Book News, August 2013
| 677.169 | 1 |
Robert P Hostetler See less
My order was canceled
Calculus: Early Transcendental Functions
It's a textbook
Not sure what is there to review. I'm taking calculus in college, this is the required textbook.
It's pretty good at explaining things and the there are interesting historical context stories
| 677.169 | 1 |
Universal Math Solver 10.0.2.9
The latest version of Universal Math Solver is UMS Online 10.0.2.9 (October 20, 2011). UMS is designed to solve and explain mathematical problems entered by the user. This program provides step by step solutions to most problems in arithmetic, algebra, trigonometry and introductory/intermediate calculus for middle- to high-school students and first year university students. All solutions are accompanied by step by step verbal and written commentaries, graphs and tables.
Universal Math Solver Similar Software
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... area where elementary students overwhelmingly display difficulties. Master Math Word Problems can help sharpen skills through practice. ... watch for key words and translate those into mathematical operations. Students can learn new math skills, practice logic, get extended practice with word ... With regular practice your students may become master math word problem solvers. Download and try out .... Free download of Master Math Word Problems 1.9j
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Browse related Subjects ...
Read More mathematics. Attention to detail, an exceptionally clear writing style, and continuous review and reinforcement are McKeague hallmarks that constitute the solid foundation of the text, while new pedagogy help students 'bridge the concepts.' These 'bridges' guide students and help them make successful connections from concept to concept-and from this course to the next. "Elementary Algebra, 9th edition, International Edition" is one of the most current and reliable texts you will find for the course, and is ideally structured and organized for a lecture-format. Each section can be discussed in a 45- to 50-minute class session, allowing you to easily construct your course to fit your needs.
Read Less
New. 0840064217
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Synopses & Reviews
Publisher Comments:
Was plane geometry your favourite math course in high school? Did you like proving theorems? Are you sick of memorising studies of the subject, by stressing the importance of pictures in mathematics and hard problems. The exposition is informal and relaxed, with many helpful asides, examples and occasional comments from mathematicians like Dieudonne, Littlewood and Osserman. The author has taught the subject many times over the last 35 years at Berkeley and this book is based on the honours version of this course. The book contains an excellent selection of more than 500 exercises.
Synopsis:
Synopsis:
Real Functions In this new introduction to undergradute real analysis the author takes a different approach from past studies of the subject by introducing the importance of pictures in mathematics and hard problems. The exposition is informal and relaxed
"Synopsis"
by Springer,"Synopsis"
by Springer,
| 677.169 | 1 |
If you are a currently enrolled FSU student and have a math question,
please visit the Math Help Center. If you are not an FSU student and
have a math question, please try visiting our Math Virtual Library.
Good luck!
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Job Type
Job Status
TEST DESIGN AND FRAMEWORK
below is the material i need to know to past the test. i will supply sample test notes and question to help me go over.
TEST FRAMEWORK
Mathematics
NUMBER CONCEPTS AND OPERATIONS
0001 Understand number operations and basic principles of number theory.
For example:
• analyzing and applying properties of
positive and negative num
bers (e.g., absolute value)
• analyzing number operations
related to rational numbers
• demonstrating knowledge of ...
| 677.169 | 1 |
Discrete Mathematics for Computer Science243.70.
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About the Book
Master the fundamentals of discrete mathematics with DISCRETE MATHEMATICS FOR COMPUTER SCIENCE with Student Solutions Manual CD-ROM! An increasing number of computer scientists from diverse areas are using discrete mathematical structures to explain concepts and problems and this mathematics text shows you how to express precise ideas in clear mathematical language. Through a wealth of exercises and examples, you will learn how mastering discrete mathematics will help you develop important reasoning skills that will continue to be useful throughout your career.
| 677.169 | 1 |
Project Laboratory in Mathematics is a course designed to give students a sense of what it's like to do mathematical...
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Project Laboratory in Mathematics is a course designed to give students a sense of what it's like to do mathematical research. In teams, students explore puzzling and complex mathematical situations, search for regularities, and attempt to explain them mathematically. Students share their results through professional-style papers and presentations. This course site was created specifically for educators interested in offering students a taste of mathematical research. This site features extensive description and commentary from the instructors about why the course was created and how it operates821 Project Laboratory in Mathematics (MIT) to your Bookmark Collection or Course ePortfolio
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This course asks students to consider the ways in which social theorists, institutional reformers, and political...
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This course asks students to consider the ways in which social theorists, institutional reformers, and political revolutionaries in the 17th through 19th centuries seized upon insights developed in the natural sciences and mathematics to change themselves and the society in which they lived. Students study trials, art, literature and music to understand developments in Europe and its colonies in these two centuries. Covers works by Newton, Locke, Voltaire, Rousseau, Marx, and Darwin433 The Age of Reason: Europe from the 17th to the Early 19th Centuries (MIT) to your Bookmark Collection or Course ePortfolio
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William Shakespeare didn't go to college. If he time-traveled like Dr. Who, he would be stunned to find his words on a...
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William Shakespeare didn't go to college. If he time-traveled like Dr. Who, he would be stunned to find his words on a university syllabus. However, he would not be surprised at the way we will be using those words in this class, because the study of rhetoric was essential to all education in his day. At Oxford, William Gager argued that drama allowed undergraduates "to try their voices and confirm their memories, and to frame their speech and conform it to convenient action": in other words, drama was useful. Shakespeare's fellow playwright Thomas Heywood similarly recalled: In the time of my residence in Cambridge, I have seen Tragedies, Comedies, Histories, Pastorals and Shows, publicly acted…: this is held necessary for the emboldening of their Junior scholars, to arm them with audacity, against they come to be employed in any public exercise, as in the reading of Dialectic, Rhetoric, Ethic, Mathematic, the Physic, or Metaphysic Lectures. Such practice made a student able to "frame a sufficient argument to prove his questions, or defend any axioma, to distinguish of any Dilemma and be able to moderate in any Argumentation whatsoever" (Apology for Actors, 1612). In this class, we will use Shakespeare's own words to arm you "with audacity" and a similar ability to make logical, compelling arguments, in speech and in writing. Shakespeare used his ears and eyes to learn the craft of telling stories to the public in the popular form of theater. He also published two long narrative poems, which he dedicated to an aristocrat, and wrote sonnets to share "among his private friends" (so wrote Francis Meres in his Palladis Tamia, 1598). Varying his style to suit different audiences and occasions, and borrowing copiously from what he read, Shakespeare nevertheless found a voice all his own–so much so that his words are now, as his fellow playwright Ben Jonson foretold, "not of an age, but for all time." Reading, listening, analyzing, appreciating, criticizing, remembering: we will engage with these words in many ways, and will see how words can become ideas, habits of thought, indicators of emotion, and a means to transform010 Writing with Shakespeare (MIT) to your Bookmark Collection or Course ePortfolio
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"The Art of the Probable" addresses the history of scientific ideas, in particular the emergence and development of...
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"The Art of the Probable" addresses the history of scientific ideas, in particular the emergence and development of mathematical probability. But it is neither meant to be a history of the exact sciences per se nor an annex to, say, the Course 6 curriculum in probability and statistics. Rather, our objective is to focus on the formal, thematic, and rhetorical features that imaginative literature shares with texts in the history of probability. These shared issues include (but are not limited to): the attempt to quantify or otherwise explain the presence of chance, risk, and contingency in everyday life; the deduction of causes for phenomena that are knowable only in their effects; and, above all, the question of what it means to think and act rationally in an uncertain world. Our course therefore aims to broaden students' appreciation for and understanding of how literature interacts with – both reflecting upon and contributing to – the scientific understanding of the world. We are just as centrally committed to encouraging students to regard imaginative literature as a unique contribution to knowledge in its own right, and to see literary works of art as objects that demand and richly repay close critical analysis. It is our hope that the course will serve students well if they elect to pursue further work in Literature or other discipline in SHASS, and also enrich or complement their understanding of probability and statistics in other scientific and engineering subjects they elect to take017 The Art of the Probable: Literature and Probability to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material 21L.017 The Art of the Probable: Literature and ProbabilityBeginning with the decline of the Roman Empire, this course discusses German, Muslim, Viking and Magyar invasions, the...
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Beginning with the decline of the Roman Empire, this course discusses German, Muslim, Viking and Magyar invasions, the development of Catholicism in Western Europe and of Eastern Orthodoxy in the Byzantine Empire, the Arabic contribution to mathematics, science, and philsophy and the institutions of feudalism and manorialism. The course concludes with the economic, demographic and urban revival which began around 1000 ADIST 304 - The Dark Ages, Summer 2008 to your Bookmark Collection or Course ePortfolio
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The STEM Concept Videos are designed to help students learn a pivotal concept in science, technology, engineering, and/or...
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The STEM Concept Videos are designed to help students learn a pivotal concept in science, technology, engineering, and/or mathematics (STEM). These ideas are the building blocks of many engineering curricula, and learning them will help students master more difficult material. The STEM Concept Videos were produced by the Teaching and Learning Lab (TLL) at MIT for the Singapore University of Technology and Design (SUTDTLL-004 STEM Concept Videos (MIT) to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material RES.TLL-004 STEM Concept Videos341 History and Philosophy of Mechanics: Newton's Principia Mathematica (MIT) to your Bookmark Collection or Course ePortfolio
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Synopses & Reviews
Publisher Comments:
Algebraic Geometry provides an impressive theory targeting the understanding of geometric objects defined algebraically. Geometric Modeling uses every day, in order to solve practical and difficult problems, digital shapes based on algebraic models. In this book, we have collected articles bridging these two areas. The confrontation of the different points of view results in a better analysis of what the key challenges are and how they can be met. We focus on the following important classes of problems: implicitization, classification, and intersection. The combination of illustrative pictures, explicit computations and review articles will help the reader to handle these subjects.
Synopsis:
"Synopsis"
by Springer,
| 677.169 | 1 |
More About
This Textbook
Overview
This volume provides a basic understanding of Fourier series, Fourier transforms, and Laplace transforms. It is an expanded and polished version of the authors' notes for a one-semester course intended for students of mathematics, electrical engineering, physics and computer science. Prerequisites for readers of this book are a basic course in both calculus and linear algebra. The material is self contained with numerous exercises and various examples of
| 677.169 | 1 |
Math & Statistics
Mathematics is the study of numbers, sets of points and various other abstract elements and deals with the size, order, shape and various relationships among these features. Statistics is a branch of Mathematics that includes the study of methods for data collection, analysis, interpretation and principles of experimental design. Mathematics contributes to the formulation and solution of problems in diverse fields such as Medicine, Economics and the Social Sciences in addition to being the foundation of the field of Computer Science and the "language" of Science and Engineering.
| 677.169 | 1 |
More About
This Textbook
Overview
For more than two thousand years a familiarity with mathematics has been regarded as an indispensable part of the intellectual equipment of every cultured person. Today, unfortunately, the traditional place of mathematics in education is in grave danger. The teaching and learning of mathematics has degenerated into the realm of rote memorization, the outcome of which leads to satisfactory formal ability but does not lead to real understanding or to greater intellectual independence. This new edition of Richard Courant's and Herbert Robbins's classic work seeks to address this problem. Its goal is to put the meaning back into mathematics.
Written for beginners and scholars, for students and teachers, for philosophers and engineers, What is Mathematics?, Second Edition is a sparkling collection of mathematical gems that offers an entertaining and accessible portrait of the mathematical world. Covering everything from natural numbers and the number system to geometrical constructions and projective geometry, from topology and calculus to matters of principle and the Continuum Hypothesis, this fascinating survey allows readers to delve into mathematics as an organic whole rather than an empty drill in problem solving. With chapters largely independent of one another and sections that lead upward from basic to more advanced discussions, readers can easily pick and choose areas of particular interest without impairing their understanding of subsequent parts.
Brought up to date with a new chapter by Ian Stewart, What is Mathematics?, Second Edition offers new insights into recent mathematical developments and describes proofs of the Four-Color Theorem and Fermat's Last Theorem, problems that were still open when Courant and Robbins wrote this masterpiece, but ones that have since been solved.
Formal mathematics is like spelling and grammar—a matter of the correct application of local rules. Meaningful mathematics is like journalism—it tells an interesting story. But unlike some journalism, the story has to be true. The best mathematics is like literature—it brings a story to life before your eyes and involves you in it, intellectually and emotionally. What is Mathematics is like a fine piece of literature—it opens a window onto the world of mathematics for anyone interested to view.
Editorial Reviews
From the Publisher
"Without doubt, the work will have great influence. It should be in the hands of everyone, professional or otherwise, who is interested in scientific thinking."—The New York Times
"Should prove a source of great pleasure and satisfaction."—Journal of Applied Physics
"Succeeds brilliantly in conveying the intellectual excitement of mathematical inquiry and in communicating the essential ideas and methods."Journal of Philosophy
"It is a work of high perfection, whether judged by aesthetic, pedagogical or scientific standards. It is astonishing to what extent What is Mathematics? has succeeded in making clear by means of the simplest examples all the fundamental ideas and methods which we mathematicians consider the life blood of our science."—Herman Weyl
Related Subjects
Meet the Author
The late Richard Courant, headed the Department of Mathematicas at New York University and was Director of the Institute of Mathematical Sciences—which has subsequently renamed the Courant Institute of Mathematical Sciences. His book Mathematical Physics is familiar to every physicist, and his book Differential and Integral Calculus is acknowledged to be one of the best presentations of the subject written in modern times. Herbert Robbins is New Jersey Professor of Mathematical Statistics at Rutgers University. Ian Stewart is Professor of Mathematics at the University of Warwick, and author of Nature's Numbers and Does God Play Dice?. He also writes the "Mathematical Recreations" column in Scientific 15, 2003
a superb book
I love this book. The treatment is masterly and the topics are interesting. You can find things discussed here that will never be mentioned in standard treatments of these topics. Moreover they are discussed here in a self contained, if brief, style. The treatment omits some details, and proceeds quickly. As a young math major I felt embarrassed by how difficult it could be to read it, even about elementary topics. I was also surprized at finding things I did not know in a book intended for the lay public. As a mature mathematician, I have learned to read more slowly, and appreciate that many of the topics simply do not occur elsewhere at a similarly accessible level. If the book is hard for a true beginner, perhaps it is a good choice for a teacher of an introductory course, even an advanced introduction to the ideas of mathematics. For example, last night while preparing my "proofs" class, I wondered why in 2,000 years no one had given a proof of unique factorization of integers into primes, different from Euclid's original proof. As soon as I turned to the appropriate section in Courant-Robbins I found a more recent proof, clever in its avoidance of the concept of gcd, which is used in every other proof I had seen. I am now motivated to present this argument in my course, and to recommend this book. On another occasion I fielded a phone call with an interesting question about Appolonius' problem on circles tangent to a given set of circles. I struggled with an answer, then found a beautiful, succint discussion in Courant - Robbins. This is a work by a real master, both of the mathematics and of exposition. It is not easy, and not perfect, but I know of no substitute for it.
2 out of 2 people found this review helpful.
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Anonymous
Posted April 16, 2003
Timeless.
Einstein writes...'Easily understandable.' And Herman Weyl,...'It is a work of high perfection.' It is both for beginners and for scholars. The first edition by Courant and Robbins, has been revised, with love and care, by Ian Stewart. Of the sciences, math stands out in the way some central ideas and tools are timeless. Key math ideas from our first mathematical experiences, perhaps early in life, often have more permanence this way. While the fads do change in math, there are some landmarks that remain, and which inspire generations. And they are as useful now as they were at their inception, the fundamentals of numbers, of geometry, of calculus and differential equations, and more. Much of it is presenterd with an eye to applications. The authors are ambitious (and remarkably sucessful)in trying to cover the essetials within the span of 500 plus pages. You find the facts, presented in clear and engaging prose, and with lots of illustrationsclassic
I give this book 5 stars because it is a classic. I believe, however, that it is too sketchy to be useful for the beginner as it is advertised. For chapter 1, for example, on number theory, I recommend Hardy's 'Introduction to the Theory of Numbers.' For the second chapter, on the number systems, I recommend a book like Birkhoff and MacLane's 'Modern Algebra.' It's difficult to write a survey of mathematics textbook without being sketchy and Courant isn't up to the task. In addition, the bibliography at the end of the book is fairly outdated, although the two books I mentioned above are included there. I also wish Courant would have provided more information on the evolution of mathematical concepts and ideas. This is something Kline does in his 'History of Mathematical Thought.' I find this information vital in answering the question 'what is mathematics?' If you really want to get a good idea of what mathematics is you should start with a general history of mathematics like Kline's book and quickly move on to Greek mathematics. Even a small understanding of Euclid's axiomatic method will help you understand modern day mathematics and why mathemticians do what they do the way they do it. Having said that, I plan on making more use of Courant's book later on in my mathematics career.
1 out of 1 people found this review helpful.
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3.
i i i
STUDYTEXT
Acknowledgment
We gratefully acknowledge permission to quote from the past examination papers of the
following bodies: Kenya Accountants and Secretaries National Examination Board
(KASNEB); Chartered Association of Certified Accountants (ACCA).
9.
3
STUDYTEXT
MATHEMATICAL TECHNIQUES
CHAPTER ONE
MATHEMATICAL TECHNIQUES
Objectives
At the end of this chapter, one should be able to:
Interpret data in a Venn diagram••
Apply matrix algebra to input-output analysis and Markovian process••
Apply calculus to economic models••
Introduction
Linear algebra has extensive applications innatural sciences and social sciences, since nonlinear
models are often approximated by linear ones.
One of the applications of linear algebra is in finding solutions of simultaneous linear equations.
The simplest case is when the number of unknowns is the same as the number of equations.
One could begin with the problem of solving n simultaneous linear equations for n unknowns.
Fast Forward: Linear algebra is the branch of mathematics concerned with the study of vectors,
vector spaces, linear maps and systems of linear equations.
Definition of key terms
Set theory - Set theory is the branch of mathematics that studies sets, which are collections of
objects.
Function - The mathematical concept of function expresses a relationship between two variables
i.e the dependence between two quantities, one of which is known (theindipendent variable,
argument of the function, or its 'input') and the other produced (the dependent variable, value of
the function, or 'output').
Matrix - In mathematics, a matrix (plural matrices) is a rectangular table of elements (or entries),
which may be numbers or, any abstract quantities that can beadded and multiplied. Matrices
are used to describe linear equations, keep track of the coefficients of linear transformations
and to record data that depend on multiple parameters. Matrices can be added, multiplied, and
decomposed in various ways.
Determinant – This is a characteristic of a matrix and is obtained from the elements of a matrix
by specified calculation. The determinant is only specified for a square matrix.
Equilibrium state – This is a situation in a Markov process when there is no further gain in
market share. It is also known as the steady state or the long-run state.
10.
quantitative techniques4
STUDYTEXT
INDUSTRY CONTEXT
In practice, Business Calculus presents some of the mathematical tools that are useful in the
analysis of business and economic problems. A few topics are compound interest, annuities,
differential and integral calculus.
Exam context
The algebra and calculus has been emphasized by examiners in various sittings as shown
below:
1.1 Need for quantitative techniques in the
business world
Mathematics is logical and precise as applied to measurable phenomena. Measurable quantities
include:
-Output
-Revenue
-Commissions
-Costs
-Profits etc
Rationale: Management needs to be able to influence factors which can be manipulated or
controlled to achieve objectives e.g. sales level is determined by a sales function which has
determinants such as level of advertising, price, income etc.
Optimization theory i.e. maximising or minimising some measure of revenue or costs calls for the
use of mathematical analysis.
Maximise: Revenue, profits, productivity, and motivation
Minimise: Costs, risks, lateness, fatigue
Mathematical analysis also helps in marginal analysis i.e. the conversion of marginal function to
a total function.
Drawback: It is not applicable to non-measurable phenomena even if such a factor is crucial to
the success of an organization.
11.
5
STUDYTEXT
MATHEMATICAL TECHNIQUES
1.2 Functions
Definitions
1. A constant – This is a quantity whose value remains unchanged throughout a particular
analysis e.g fixed cost, rent, and salary.
2. A variate (variable) – This is a quantity which takes various values in a particular problem
Illustration 1.1.
Suppose an item is sold at Sh 11 per unit. Let S represent sales rate revenue in shillings and let
Q represent quantity sold.
Then the function representing these two variables is given as:
S = 11Q
S and Q are variables whereas the price - Sh 11 - is a constant.
Variables
Types of variables
Independent variable – this is a variable which determines the quantity or the value of some other
variable referred to as the dependent variable. In Illustration 1.1, Q is the independent variable
while S is the dependent variable.
An independent variable is also called a predictor variable while the dependent variable is also
known as the response variable i.e Q predicts S and S responds to Q.
3. A function – This is a relationship in which values of a dependent variable are determined by
the values of one or more independent variables. In illustration 1.1 sales is a function of quantity,
written as S = f(Q)
Demand = f( price, prices of substitutes and complements, income levels,….)
Savings = f(investment, interest rates, income levels,….)
Note that the dependent variable is always one while the independent variable can be more than
one.
14.
quantitative techniques8
STUDYTEXT
Diagram 1.1
Discrete function – Is a function in which the dependent variable is constant for particular values of
the independent variable. It might change to another level for different values of the independent
variable and so on e.g fixed cost for various production, capacities or levels.
Fixed cost is Ksh 1000 for 0 – 100 units
Ksh 1,500 for 101 – 150 units
Ksh 3,000 for 101 – 150 units
Ksh 4,000 there after
0 x
y
y = x2
16.
quantitative techniques1 0
STUDYTEXT
Characteristics of the straight line
1) It has only one solution i.e can cross the x axis only once, y = α + bx
Therefore 0 = a+bx
x = −
a
b
2) It has no turning point. The turning point is also known as critical or stationary point
3) It is completely defined when critical
i) Either two points on the line are given, or
ii) One point and the slope are specified.
Illustration 1.2
1. Determine the linear function that goes through the following points
X 2 5
Y 5 17
2. What is the straight line which has slope b = -0.5 and goes through the point (x,y) = (10,18)
Solution
Slope =
∆Y
=
17− 5
=
12
= 4
∆X 5−2 3
Y = a+ bx
5 = a+ 4x2
5 = a +8
a = -3
x
(3)
b > 0 (-ve)
e.g price and
quantity
y
x
(4)
b is undefined
or infinite e.g.
prices of addictive
goods in relation
to quantity
consumed
y
17.
11
STUDYTEXT
MATHEMATICAL TECHNIQUES
Therefore y = -3 +4x
y = a +bx
b = -0.5
at (x, y) = (10,18)
18 = a + (- 0.5) (10)
18 = a -5
a = 23
y = 23 – 0.5x
Applications of linear functions
Analysis of commission and wages1.
Demand and supply function (market equilibrium)2.
Fixed assets accounting – depreciation – straight line method3.
Cost volume profit ( c-v-p) analysis (profit planning)4.
Profit = f (price, output, cost)
Profit is a function of costs, prices, volume etc. The problem is how does management manipulate
factors which determine profit in order to maximize that profit. For linear C-V-P analysis, we make
the following assumptions:
Linearity1. – all functional (mathematical) relationships are linear with respect to the
activity level such as quantity produced and so on. Revenue, cost and profit functions
are linear.
Price per unit is constant e.g. there are no quantity discounts.2.
Unit variable cost is constant e.g. - direct material costs do not change3.
- Wages rates are constant
Fixed costs do not change4.
All costs can be categorised as either fixed or varied i.e. there are no semi variables 5.
cost.
All units produced are sold i.e. inter period changes are negligible6.
The only factor that influences revenues, costs and profits is the level of activity7.
There are no demand or other restrictions i.e. no constraints8.
All factors under consideration are known with certainty.9.
23.
17
STUDYTEXT
MATHEMATICAL TECHNIQUES
Quadratic functions
General formula: y =a + b1x + b2x2
Where x = independent variable
y = dependent variable
a, b1, b2 = constants
Note b2 ≠ 0
Properties of quadratic functions
1). Number of solutions (roots) is 2, i.e. it can cross the x-axis twice.
Recall: if ax2
+bx+c = 0
2). A quadratic function has a single turning point.
3). A quadratic function is completely specified once any three points which lie on this curve are
given.
Quadratic sketches
Cubic functions
These are 3rd degree polynomials.
General form: y = a + b1
x + b2
x2 + b3
x3
Where b3
≠ 0
Properties of a cubic function
i. It will have at least one and at most three real roots i.e. the number of times it crosses the
x – axis
ii. It has either two turning points (one max , the other min) or a point of inflexion
iii. It is completely defined once we have four points which lie on the curve
Exponential functions
Exponential functions
These are functions whose, at least, one term as independent variable is part of an exponent or
power.
y y
xx
b2
>4ac
Max
Min
b2
<4ac
25.
19
STUDYTEXT
MATHEMATICAL TECHNIQUES
Illustration 1.5
Population growth
The population growth of a country was 100 million in 1990. It has since grown at 4% per annum.
The growth rate is exponential of the form p = aekt
p = population
k = growth rate
t = time in years
a = a constant
Required:
a). What is the function which describes population growth through time?
b). What is the population in the year 2010?
Solution
It is of the form p = aekt
Let 1990 be t = 0
Constant % growth rate is the coefficient of t which is k = 4%
For t = 0 (1990) the population, P = 100
Replacing these values in our equation we get.
100 = ae0.04×0
a = 100 million since e0
= 1
Therefore, P = 100e0.04t
For year 2010, t = 20
P = 100e0.04t
P = 100e0.04×20
P = 222.5541 million (4 d. p)
Asset valuation
The resale value v, of a certain piece of industrial equipment has been found to be described by
the function
V = 100 × 4+0.1t
or 100(4+0.1t
) where t = time
depreciation rate = 10%
Required :
a) What was the purchase cost of the equipment?
b) What is the expected re-sale value after: i) 5 years, ii) 10 years
Solutions
a) Purchase cost – it is the value when t = 0 which is 100
27.
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STUDYTEXT
MATHEMATICAL TECHNIQUES
Diagram 1.7
1). Compound interest
If a certain amount of money Sh p is deposited in an account which earns a compound interest at
a rate of I % per annum, the amount after n years ( interest + principal) is given as n
I)(1PS +=
, if P = sh 1,000 and I = 8%
What is the sum after 25 yrs if interest is compounded annually?
Solution
S = 1000(1+0.08)25
S = sh. 6,848
Logarithmic functions
A logarithmic is a power to which a base must be raised in order to give a certain number i.e. a
logarithm is an exponent. Consider the following equation.
23
= 8
The table below shows the logarithmic and exponential form
Exponential equations Logarithmic equation
104
= 10,000 4 = log10
10, 000
24
= 16 4 = log2
16
52
= 25 2 = log5
25
102
= 100 2 = log10
100
100%
95%
P(%)
y = 0.95(1-e0.7t
)
t0
28.
quantitative techniques2 2
STUDYTEXT
1.3 Matrices
Fast Forward: Due to their widespread use, considerable effort has been made to develop
efficient methods of matrix computing.
A matrix is a rectangular array of numbers in rows and columns enclosed in brackets. A matrix
that has m rows and n columns is said to be an m × n matrix. For example
A =
1 3
is a 2 × 2 matrix.
-2 1
2 3 4
B =
7 8 2
is a 2×3 matrix.
Matrix names are usually represented by capital letters while elements of a matrix are represented
by a lower case subscripted letters : aij
stands for real numbers where aij
is the element in the ith
row and jth
column.
Types of matrices
Row matrix – is a matrix which has only one row.
For example (2, 2), (2, 3, -2, 5)
Column matrix – this is a matrix which has only one column.
2
2
3For example,
2
-2
Square matrix – this is a matrix in which the number of rows is equal to the number of columns.
A =
2 3
7 8
Diagonal matrix – this is a square matrix that has zeros everywhere except on the main diagonal
- that is the diagonal running from the upper left to the lower right.
29.
23
STUDYTEXT
MATHEMATICAL TECHNIQUES
For example:
2 0 0
A = 0 8 0
0 0 4
1 0 0
B = 0 0 0
0 0 0
A matrix s diagonal if aij
is equal to zero for all elements when i≠j and aij
is not equal to zero and
at least i = j
1). Identity matrix – this is a square matrix with the leading diagonal elements all equal to one
and all other elements equal to zero i.e. it is a diagonal matrix whose diagonal matrix is equal to
one.
1 0
I =
0 1
If you multiply a matrix by an identity matrix, you will get the same matrix regardless if you pre-
multiply or post-multiply.
Amm
Imm
= Amm
= Imm
Amm
2). Zero or null matrix – this is a square matrix where every element is zero.
Note:
a) When null matrix is added or subtracted from another matrix that matrix remains
unchanged.
b) Pre or post-multiplying a matrix with a null matrix results in another matrix.
3). Scalar matrix – is a diagonal matrix whose diagonal elements are equal.
A =
10 0
0 10
1 0 0
B = 0 1 0
0 0 1
Triangular matrix – A square matrix whose element aij
is equal to zero. Whenever i<j, it is called
a lower triangular matrix, whenever i>j, it is called an upper triangular matrix.
1 0 0
A = 3 2 0
6 8 3
10 2 4
B = 0 4 3
0 0 3
30.
quantitative techniques2 4
STUDYTEXT
Matrices are used because they are able to summarise data. Through matrices, operations,
formulation and solution of data are simplified which would almost be impossible or complicated
in conventional or algebraic operations. Knowledge of matrices can be used in solving problems
that arise in various fields of operations such as:
1. Simultaneous equations
2. Markov processes
3. Input-output analysis
4. Linear programming
5. Game theory
Operations of matrices
The following operations can be carried out in matrices:
1. Addition
2. Subtraction
3. Multiplication
4. Determinant
5. Transposition
6. Matrix Inversion
Matrix addition and subtraction
If A and B are two matrices of the same order then the addition of A and B is defined to be the
matrix obtained by adding the corresponding elements of A and B.
Illustration 1.6
3 2 1
A = 4 3 9
5 6 8
0 1 8
B = 3 7 6
2 6 4
3 2 10
A + B = C = 7 10 15
7 12 12
To subtract one matrix from another, the rule is to subtract corresponding elements just like
matrix addition. (Note: order of subtraction is important).
Multiplication of two matrices
The main rule is that, for it to be possible to multiply two matrices, the number of columns in the
first matrix should equal to the number of rows in the second. For instance, 2×2 and 1×2 cannot
be multiplied because there are two columns in first and one row in the second. On the other
hand, 2×2 and 2×1 can be multiplied and it should give 2×1 matrix.
32.
quantitative techniques2 6
STUDYTEXT
Note:
The transpose of a diagonal matrix is the same diagonal matrix.
If a square matrix and its transpose are equal, that is, if aij
= aji
for all i and j the matrix is said to
be symmetric about its main diagonal.
-1 5 -3
A = 5 0 4
-3 4 9
-1 5 -3
A' = 5 0 4
-3 4 9
Hence matrix A is a symmetric matrix.
A symmetric matrix that reproduces itself when multiplied by itself is said to be idempotent; that
is, A is idempotent if
A' = A
AA = A
Determinant of a matrix
This is a scalar obtained from the elements of a matrix by a specified operation, which is a
characteristic of a matrix. The determinant is only specified for a square matrix.
For a 2×2 matrix:
a11
a12
A =
a21
a2
2
|A| or Adet
= a11
a22
– a12
a21
Illustration 1.8
A =
8 9
|A| = 8×-5 - 9×6
6 -5
= -40-54
= -94
B =
1 2
|B| = (1×-10)-(2×1)
1 -10
= -10-2
= -12
For a 3×3 matrix.
There are two methods to get a determinant:
a) Artistic method
b) Co-factor method/ Laplace method
34.
quantitative techniques2 8
STUDYTEXT
Cofactor (Laplace method)
For any square matrix there can be found a matrix of cofactors, 'Ac
'. The matrix of cofactors will
have dimensions of A and will consist of elements Cij which are known as cofactors of each
element aij obtained in A. The corresponding cofactors Cij will be determined as follows:
Either mentally or with a pencil cross off row i and column j in the original matrix. Focus on the
remaining elements. The remaining elements will form a sub-matrix of the original matrix.
Find the determinant of the remaining sub-matrix. This determinant is called the minor of element
aij
.
The cofactor aij
is found by multiplying minor by either positive one or negative one depending on
the position of element aij
that is (-1)i+j(minor).
a11
a12
a13
A = a21
a22
a23
a31
a32
a33
|A| = a11
a22 a23
− a12
a21 a23
+ a13
a21 a22
a32 a33 a31 a33 a31 a32
Illustration 1.10
3 1 2 + − +
A = -1 2 3 signs = − + −
3 -2 1 + − +
2 4
-1
-1 4
+2
-1 2
A =
-2 1 3 1 3 -2
A = 3│(2×1) − (-2×4)│ -1│(-1×1) − (3×4)│ +2│(-1×-2) − (3×2)│
A = 3│2+8│-1│-1−12│+2│2−6│
A = 30+13-8
A = 35
35.
29
STUDYTEXT
MATHEMATICAL TECHNIQUES
+
2 4
−
-1 4
+
-1 2
-2 1 3 1 3 -2
Ac
=
−
1 2
+
3 2
−
3 1
-2 1 3 1 3 -2
+
1 2
−
3 2
+
3 1
2 4 -1 4 -1 2
Therefore
│A│= (1×3) + (2×-3) + (-2×-14)
│A│=35
Inverse of a matrix
If for an n×n matrix, A there is another n× n matrix, B such that their product is the identity matrix
of product n× n.
A n× n
× B n× n
= I n× n
Then B is said to be inverse or reciprocal of A. A matrix which has an inverse is known as a non-
singular matrix. A matrix which has no inverse is said to be a singular matrix.
The inverse of a 2×2 matrix is calculated by interchanging the elements in the leading diagonal
and multiplying the elements in the other diagonal by -1, the resultant matrix is then multipliead
by the inverse of the original matrix's determinant A
A =
a11
a12
a21
a22
A-1
=
1 a22 -a12
│A│ -a21 a11
Illustration 1.11
2 3
B =
1 4
1 4 -3
B-1
=
5 -1 2
4
/5 -3
/5
B-1
=
-1
/5 2
/5
36.
quantitative techniques3 0
STUDYTEXT
For a 3×3 matrix
Procedure:
(1) Determine the matrix of cofactors Ac
from matrix A.
The cofactor of any element aij
(known as cij
) is the signed minor associated with that element.
The sign is not changed if (i+j) is even and it is changed if (i+j) is odd. Thus the sign alternated
whether vertically or horizontally, beginning with a plus in the upper left hand corner.
+ − +
i.e. 3 x 3 signed matrix will have signs − + −
+ − +
1 2 0
For matrix A = 1 0 -1
-1 3 2
Hence the cofactor of element a11
is m11
= 3, cofactor of a12
is –m12
= -1 the cofactor of element
a13
is +m13
= 3 and so on.
3 -1 3
Matrix of cofactors of A: Ac
= -4 2 -5
-2 1 -2
a b c
in general for a matrix M = e d f
g h i
Cofactor of a is written as A, cofactor of b is written as B and so on.
Hence matrix of cofactors of M is written as
A B C
= D E F
G H I
The determinant of a n×n matrix can be calculated by adding the products of the element in any
row (or column) multiplied by their cofactors. If we use the symbol ∆ for determinant.
Then ∆ = aA + bB + cC
or
= dD + eE + fF e.t.c
( )
( )
( )
42.
quantitative techniques3 6
STUDYTEXT
2. Markov processes
Markov analysis is a technique that deals with probabilities of future occurrences by analysing
presently known probabilities. The technique has numerous applications in business including
market share analysis, bad debt prediction, university enrolment prediction, determining whether
a machine will break down in future, and so on.
Markov analysis makes the assumption that the system starts at initial state of condition for
instance two competing manufacturers might have 40% and 60% of the market sales respectively
as initial state. Perhaps in two months the market shares of the two companies will change to
45% and 55% of the market share respectively. Predicting these future states involves knowing
the systems likelihood or probability of changing from one state to another. For a particular
problem these probabilities can be collected and placed in a matrix table. This matrix of transition
probability shows the likelihood of the system would change from one period to the next. This is
the Markov process and enables us to predict future states and conditions.
Assumptions
There are a lim1. ited or finite number of states.
The states in the system are collectively exhaustive. This means that an object can only2.
belong to or subscribe to a state within that system but not outside.
The states in the system are mutually exclusive that is an object in the system can be3.
found in one and only one state.
The probability that an object will shift from one state to another during any specified4.
period of time remains constant from period to period or transition to transition.
We can predict any future state from the previous state (initial state) and the matrix of5.
transition probability.
Transition matrix, T
TO
S1
S2
SM
S1
P11
P12
P1m
FROM S2
P21
P22
P2m
SM
P1m
P2m
Pmm
Where: Si/Sj represent state of a given phenomena
Pij represent the probability that the object shift from state i to state j (transition matrix)
The transition probabilities for Markov processes are normally organised in a matrix called
transition matrix.
Properties of a transition matrix
(1) The sum of probabilities across a row is one.
(2) The row indicates one source of object to all its destinations.
(3) Transition matrix is a square matrix.
(4) Probabilities Pij are obtained empirically that is by observation and data collection.
43.
37
STUDYTEXT
MATHEMATICAL TECHNIQUES
Illustration 1.14: For 2×2 matrix
In a certain country there are two daily newspapers: The Citizen and The Mirror. A researcher
interested in the reading habit of this country found the following:
Of the readers who read Citizen on a given day 50% do so following day while the rest change
to the Mirror. Of those who read Mirror on a given day 40% change to the Citizen the following
day.
Yesterday the readership levels were 30% Citizen and 70% Mirror. Assume all conditions hold.
Required:
(a) Determine the readership levels of both dailies:-
(i). Today
(ii). Tomorrow
(b) If this process persists long enough, what will be the eventual readership?
Solution
Initial state
C M
(0.3 0.7)
Transition matrix
To
C M
From C 0.5 0.5
M 0.4 0.6
(a) (i) Today
0.5 0.5
(0.3 0.7)
0.4 0.6
= (0.43 0.57)
C = 0.43
M = 0.57
(ii) Tomorrow
(0.43 0.57)
0.5 0.5
=(0.443 0.557)
0.4 0.6
44.
quantitative techniques3 8
STUDYTEXT
Equilibrium state
In a Markov process a situation reaches when there is no further gain in market share. This
situation is known as equilibrium state or steady state or the long-run state.
By definition equilibrium condition exist if state probabilities do not change after a large number of
periods. At equilibrium, state probability from next period equals state probability of this period.
p = long-run readership level of Citizen
q = long-run readership level of Mirror.
(p q)T = (pq)
0.5 0.5
(p 1-p)
0.4 0.6
= (p 1-p)
0.5p + 0.4 − 0.4p = p
0.1 + 0.4 = p
0.9p = 0.4
p = 4/9 = 0.444
Citizen = 44.45
Mirror = 55.6%
Applications
(1) Marketing to future market shares.
(2) Finance to predict share prices in stock exchange market.
(3) Human resource management to analyse shifts of personnel among various organisation
units like departments, divisions and so on.
(4) Financial accounting to estimate provision for bad debts.
3. Input-output analysis
The sectors in the economy are interdependent for continuity. In an economy the sectors depend
on each other for the continued production of output.
In input-output analysis the main task is to determine the output required from each sector in
the economy in order to satisfy both inter-sectorial and external demand. The technique was
developed by Wassily Leontief in 1961.
Types of input-output models
Input-output models are of two types:-
The closed in which the entire production is consumed by those participating in1.
production.
Open in which some of the products are consumed by an external body.2.
In the closed model, one seeks the income of each participant in the system. In the open model,
one seeks the amount of production needed to achieve a future demand when the amount of
production needed to achieve current demand is known.
The most useful application of input-output analysis in the economy or the common broker is the
ability to tell how the change in demand for one industry affects the entire economy.
47.
41
STUDYTEXT
MATHEMATICAL TECHNIQUES
Illustration 1.16
Three clients of Disrup Ltd P, Q and R are direct competitors in the retail business. In the first week
of the year P had 300 customers Q had 250 customers and R had 200 customers. During the
second week, 60 of the original customers of P transferred to Q and 30 of the original customers
of P transferred to R. similarly in the second week 50 of the original customers of Q transferred
to P with no transfers to R and 20 of the original customers of R transferred to P with no transfers
to Q.
Required
a) Display in a matrix the pattern of retention and transfers of customers from the first to
the second week
(4 marks)
b) Re-express the matrix that you have obtained in part (a) showing the elements as
decimal fractions of the original numbers of customers of P, Q and R
(2 marks) Refer to this re-expressed matrix as B
c) Multiply matrix B by itself to determine the proportions of the original customers that
have been retained or transferred to P, Q and R from the second to the third week.
(4 marks)
d) Solve the matrix equation (xyz)B = (xyz) given that x + y + z = 1 (8 marks)
e) Interpret the result that you obtain in part (d) in relation to the movement of customers
between P, Q and R (2marks)
(Total 20 marks)
Solution
a) Think of each row element as being the point from which the customer originated and each
column element as being the destination e.g. 210 customers move from P to P, 60 move
from P to Q and 30 move from P to R. The sum of the elements of the first row totalling 300,
that is the number of customers originally with P.
Hence required matrix is
P Q R To
P 210 60 30 row total 300
From Q 50 200 0 row total 250
R 20 0 180 row total 200
b). The requirement of this part is to express each element as a decimal fraction of its
corresponding row total. The second row, first element is therefore 50/250, that is 0.2 and the
second element is therefore 200/250 that is 0.8.
0.7 0.2 0.1
Hence B = 0.2 0.8 0
0.1 0 0.9
0.7 0.2 0.1 0.7 0.2 0.1 0.54 0.30 0.16
c). 0.2 0.8 0 0.2 0.8 0 = 0.30 0.68 0.20
0.1 0 0.9 0.1 0 0.9 0.16 0.02 0.82
( )
( )
( ) ( ) ( )
( )
48.
quantitative techniques4 2
STUDYTEXT
The result can be checked by the normal rules of matrix multiplication.
0.7 0.2 0.1
d) (x y x) X 0.2 0.8 0 = (x y z)
0.1 0 0.9
This produces from the first row
0.7x + 0.2y + 0.1z = x
Which reduces to 0.2y + 0.1z = 0.3x
Or 2y + z = 3x ………………………...(i)
Or
The second row produces, 0.2x + 0.8y = y
Reducing to 0.2x = 0.2 y
X = y ………………………..(ii)
Or
The third row produces 0.1x + 0.9z = z
Reducing to 0.1x = 0.1z
X = z ………. (iii)
At this point you will notice that condition h (ii) and condition (iii) produce 2x + x = 3x when
substituted into condition (i), we therefore need extra condition x + y + z = 1 to solve the
problem.
Thus x + x + x = 1
Or 3x = 1
That is x = 1/3
Leading to x = 1/3, y = 1/3, z = 1/3
e). In proportion terms this solution means that P, Q, and R will in the long term each have one
third of the total customers
To apply the input-output model the sectors in the economy should be (assumptions):-
Finite.1.
Mutually exclusive2.
Output from all sectors in the economy can be expressed in the same monetary unit.3.
Input requirement per unit for each sector remains constant irrespective of the number4.
of units produced from the sector and are known and are estimated in reliable degree off
accuracy that is A = technology matrix.
External demand from each sector is known.5.
( )
49.
43
STUDYTEXT
MATHEMATICAL TECHNIQUES
Technology matrix
The technology matrix (technical coefficient matrix) is usually denoted by A and will describe the
relation a sector has with others. The technology matrix A will be a matrix such that each column
vector represents a different industry and each corresponding row vector represents what that
industry inputs as a commodity in the column industry for example the technology matrix A below
represents the relationship between industries of construction and farming and clothing.
Inputs
Farming construction clothing
Farming a11
a12
a13
Outputs Construction a21
a22
a23
Clothing a31
a32
a33
The relationships between these industries in the example are as follows:
The entry a11
holds the number of units the farmer uses from his own products for producing 1
unit.
The entry a21
holds the number of units the farmer needs for construction to produce one more
unit of farming.
The entry a31
holds the number of units the farmer needs for clothing to produce 1 more unit of
farming.
Input-output model derivation
Let X1
= total output required from sector 1
X2
= total output required from sector 2
aij
= number of units required in sector i as input to sector j.
Table 1.3
To Total output
= total input to
each sector
External demand
From
S1
S2
S1
a11
x1
a12
x2
X1
d1
S2
a21
x1
a22
x2
X2
d2
50.
quantitative techniques4 4
STUDYTEXT
Let the external demand from sector 1 be D1
and from sector 2 be D2
.
Sector 1 x1
= a11
x1
+ a12
x2
+d1
Sector 2 x2
= a21
x1
+ a22
x2
+d2
x
a11
a12
x1
d1
x2
=
a21
a22
x2
+
d2
X = AX + D
X-AX = D
(I-A)X = D
X = (I-A)-1
D
Where A is the technological matrix
D is the external demand
(I-A) is called Leontief matrix,
(I-A)-1
is called Leontief inverse matrix
Note: The primary purpose of input-output analysis is to calculate the level required at a particular
level of intermediate and final demand.
Input-output analysis can also be used to study sectors of an economy either using the closed
model or the open model.
In a closed economy the sum of the elements in a column will always total 1.
In general each entry in the technological matrix is represented as aij
which is equal to xij/xj
where xj represent the physical output of sector j.
1.4 Calculus
Calculus is concerned with the mathematical analysis of change or movement. There are two
basic operations in calculus:
i) Differentiation
ii) Integration
These two basic operations are reverse of one another in the same way as addition and subtraction
or multiplication and division.
Utilities of calculus in Business
Often we may be interested in optimisation e.g. maximisation of revenue, profit,1.
productivity.
Minimisation of cost, waste
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STUDYTEXT
Comment: the two lines are parallel i.e. the rate of change of y with respect to x for the two is the
same. As x changes by one unit y changes by 3 units in both cases.
Mathematical sign for change is the Greek letter , delta, ∆
Therefore
∆y
= 3 ∆
∆x
=
y2
− y1
e.g. for the 1st
line
∆y
=
7−1
=3
x2
− x1
∆x 1− (-1)
∆y
is also known as the slope or gradient
∆x
For any linear function the slope is constant and is equal to b in the form: y = a + bx
Forms of slope of linear functions
Exercise
What is the slope of line joining the points (x,y) = (2,5) and (x,y) = (5,18)?
∆y
=
18 − 5
=
13
= 4
1
∆x 5 − 2 3 3
Slope is -ve
y
x
b < 0
Slope is Zero
y
x
b = 0
Slope is Infinite
or undefined
b = ∞
Slope is +ve
x
y
b > 0
53.
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STUDYTEXT
MATHEMATICAL TECHNIQUES
Non-Linear Functions
A slope of straight line is constant however slopes of other (non-linear) functions are different at
different sections/ points of the function
Consider the general function, y = f(x)
Diagram 1.8
Slope of Y f(x) varies at different points along the curve e.g. to the left of point A, slope is
positive
Between points A and B slope is negative
Between points B and C slope is positive
Slope at point D is greater than at point E
Problem: to calculate the slope of non –linear functions at any point of interest given the
function
Integration
It is the reversal of differentiation.
An integral can either be indefinite (when it has no numerical value) or definite (have specific
numerical values).
It is represented by the sign ʃf(x)dx.
Rules of integration
i. The integral of a constant
∫adx = ax +c where a = constant
D
E
y
A
B
C
x
Y = f(x)
57.
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STUDYTEXT
MATHEMATICAL TECHNIQUES
Find its total cost function.
Solution
The total cost C is given by
C = ∫MC.dq
= ∫a + a1
q + a2
q2
).dq
Example 4.
Your company manufacturers large scale units. It has been shown that the marginal (or variable)
cost, which is the gradient of the total cost curve, is (92 – 2x) Shs. thousands, where x is the
number of units of output per annum. The fixed costs are Shs. 800,000 per annum. It has also
been shown that the marginal revenue which is the gradient of the total revenue is (112 – 2x)
Shs. thousands.
Required
i. Establish by integration the equation of the total cost curve
ii. Establish by integration the equation of the total revenue curve
iii. Establish the break even situation for your company
iv. Determine the number of units of output that would
a) Maximize the total revenue and
b) Maximize the total costs, together with the maximum total revenue and total costs
Solution
i) First find the indefinite integral limit points of the marginal cost as the first step to
obtaining the total cost curve
Thus ∫(92 – 2x) dx = 92x – x2
+ c
Where c is constant
Since the total costs are the sum of variable costs and fixed costs, the constant term in
the integral represents the fixed costs, thus if Tc are the total costs then,
Tc = 92x – x2
+ 800
or Tc = 800 + 92x - x2
ii) As in the above case, the first step in determining the total revenue is to form the
indefinite integral of the marginal revenue
Thus ∫(112 - 2x) dx = 112x – x2
+ c
Where c is a constant
The total revenue is zero if no items are sold, thus the constant is zero and if Tr represents
the total revenue, then
Tr = 112x – x2
iii) At break even the total revenue is equal to the total costs
Thus 112x – x2
= 800 + 92x - x2
20x = 800
x = 40 units per annum
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STUDYTEXT
1st order and 2nd order conditions
In most situations one is faced with the process of differentiation for example what is the maximum
level of profit, at what points you have max. Revenue applied in NSE statistics
Diagram 1.10
Maxima and minima are found at the turning point of the curve, when the curve is parallel to x-
axis, at this point the gradient = 0 i.e the derivative = 0
2nd
Order derivatives
2nd
order derivative is a result of differentiating a function twice. It is usually donated as y″ or
d2
y
or f ″ (x)
dx2
y = 3x + x2
y ′ = 3 + 2x
y ″ = 2
Example.
Determine the maxima or minima of the graph y = 3x2
- 2x
Solution
We use 1st
and 2nd
order derivatives to determine if a point is min. or max.
1st
derivative
dy
= 6x + 2,
dy
= 0 but at min or max point:
dx dx
therefore 6x + 2 = 0 ; at max or min
Which gives x = - 1/3
2nd
derivative
d2
y
= 6
dx2
Local maxima
Local minima
Global minima
Global maxima
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STUDYTEXT
Second order condition (SOC)
d2
y
=
dy dy
dx1
2
dx1
dx1
dy
=
dy dy
dx2
2
dx2
dx2
Example
Y = 3x1
2
+x2
3
+3x1
x2
+6
Solution
F.O.C
dy
= 6x1
+ 3x2
dy
= 3x2
2
+ 3x1 dx1
dx2
S.O.C
d2
y
= 6
d2
y
= 6x2 dx1
2
dx2
2
Applications of partial differentiation
In marginal costs – if costs in a function are of two or more elements.1.
Related commodities be it substitutes or complimentary where demand is not only2.
influenced by product price but the price of other commodities.
Partial elasticity of demand.3.
Profit maximization, revenue optimization and cost optimization.4.
Set Theory
Introduction
Study of sets is popular in economic and business world since the basic understanding of concept
in sets algebra provides a form of logical language through which the business specialists can
communicate important concepts and ideas.
It is also used in solving counting problems of a logical nature
A further study of set algebra provides a solid background to understanding probability and
statistics which are important business decision-making tools
Any well defined collection of group of objects is a set e.g. set of all courses offered in Strathmore
University, all first year students at Egerton University, student studying medicine in Kenya.
Requirement of a set
A set must be well defined i.e. it must not leave any room for ambiguities e.g sets of all1.
students- which? Where? When?
A set must be defined in terms of space and time
( )
( )
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STUDYTEXT
MATHEMATICAL TECHNIQUES
The objective (elements or members) from a given set must be distinct i.e each object2.
must appear once and only once, Must appear but not more than once
The order of the presentation of elements of a given set is immaterial3.
e.g 1,2,3 = 1,3,2 = 3,2,14.
Specifying (Naming) sets
By convention sets are specified or named using a capital letter. Further the elements of a
given set are designed by either listing all the elements or by using descriptive characteristics or
patterns e.g.
A = (0, 1, 2, 3, 4, 5, 6) : Listing all
A = (whole No's from zero to six) : descriptive characteristics
A = (0, 1, 2….6) : Pattern
Set membership- set membership is expressed by using the Greek letter epsilon,1. ∈ e.g.
3 ∈ A; 0, 5 ∈ A
A finite set consist of a limited or countable number of elements e.g. A has 7 members2.
therefore A is a finite set. An infinite set consists of unlimited or uncountable number of
members e.g. set of all odd numbers.
A subset S of some other set A is such that all elements in S are members of A e.g. if S3.
= [1, 5] then S is a subset of A denoted: S ⊂ A equally A is a superset to S denoted A ⊃
S.
Equality of sets – if all elements in set D4. 1
are also in D2
and all elements in D2
are also in
D1
then D1
and D2
are equal e.g. let D1
= (a,
c,
f) D2
= (c, a, f) Then D1
= D2
.Further,
D1
⊂
D2
and D2
⊂ D1
i.e. each set is subset to itself.
Members of universal set are denoted by U or5. ξ. The universal set is that which contains
all elements under consideration by the analyst or the researcher.
Let U = set of all students at universities in Kenya in the year 2005
S1 = Student of Strathmore University
S2 = Engineering Students
S3 = Students above 25years at Kenyatta University
S4 = All female students below 20years of age.
The null or the empty set- this is the set with no elements denoted:6.
E = ( ) or E = Ø
Compliment of a set. If U = Universal set and A is a subset of the universal set, then the7.
compliment of A denoted A' or Ac
represents all elements in the universal set which are
not members of A e.g
A = (whole No's from 0-6)
U = (whole No's from 0-10)
Then A' = (whole No's from 7-10)
Pictorial or Diagrammatic representation of sets- This is done using Venn diagrams8.
(named after the 18th
C. English Logician, John Venn)
= a single set/ordinary set (Not a universal set)
= universal set
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STUDYTEXT
Diagram 1.11
Set Algebra
This consists of ways or operations whereby sets are combined in order to obtain other sets of
interest.
Basic Set Operations
1. Let P = ( 1, 2, 3) and Q = ( 1, 3, 5,6)
Union of sets, denoted U .Therefore PUQ represents all elements in P or Q.
Note: "Or" could be used in place of U
PUQ = (1,3,2,5,6)
2. Intersection of sets denoted ∩ This consists of elements in both P and Q. It is the common
area.
P∩Q = (1,3)
Note: "and" could be used in place of ∩
3. Set difference or set disjunction denoted by (-) e.g P-Q consist of elements in P but not Q i.e
(2). Whereas Q – P = (5, 6)
P
1, 3
2
Q
5
6
U = { 0, 1, 2, ….10}
4 1 6
2 3
5
10
9
8
7
A
71.
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STUDYTEXT
MATHEMATICAL TECHNIQUES
We have to subtract the No of the elements with the intersection to avoid counting them twice.
However if S1 and S2 are disjoint sets then n (S1
US2
) = n (S1) +n (S2
)
S1 and S2 have an intersection i.e they share some elements
S1 and S2 are disjoint i.e they lack an intersection.
For symmetry difference number of elements n(S1
∆ S2
) = n(S1US2
) – n(S1
∩S2
)
Illustration (2-sets)
In a recent survey of 400 students in a college 100 were listed as studying typing (T) and (150)
were listed as doing accountancy (A), 75 registered as doing both courses.
Required: -
Find the number of students in the college who are not registered in either course.
How many students were registered for typing only.
Solution
N(T) = 100 n(A) = 150 n(T∩A) = 75
Diagram 1.13
Venn diagram
S1 S2
S1 S2
S1 S2
b=25 d=75
c
75
a=225
n (U) = 400
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STUDYTEXT
MATHEMATICAL TECHNIQUES
(i) Daily nation only b = 25
S and K only = 10+20+10=40
Total (U) = a+b+c+d+e+f+g+h = 100
1.5 Chapter summary
Functions of a single independent variable may either be linear or non linear.
Linear functions can be represented by:
y = a + bx
Whereas non – linear functions can be represented by functions such as:
y = α0
+ α1
3
x + α2
x3
y2
= 3x + 18
y = 2x2
+ 5x + 7
ax2
+ bx + cy + d = 0
xy = k
y = ax
The solution of a system of linear simultaneous equations is a set of values of the variables which
simultaneously satisfy all the equations of the system.
The determinant of a square matrix det (A) or A is a number associated to that matrix. If the
determinant of a matrix is equal to zero, the matrix is called singular matrix otherwise it is called
non-singular matrix.
Differentiation deals with the determination of the rates of change of business activities or simply
the process of finding the derivative of a function.
Integration deals with the summation or totality of items produced over a given period of time or
simply the reverse of differentiation
An Implicit function is one of the y = x2
y + 3x2
+ 50. It is a function in which the dependent
variable (y) appears also on the right hand side.
To differentiate the above equation we use the differentiation method for a product, quotient or
function of a function.
A set can be classified as a finite or infinite set, depending upon the number of elements it has.
A finite set has a finite number of elements whereas an infinite set has an infinite number of
elements.
Where α, a, b, c, d, k = constants
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STUDYTEXT
Chapter Quiz
A ………….associates a single output to each input element drawn from a fixed set,1.
such as the real numbers, although different inputs may have the same output.
Which one of the following is not a function:2.
(i) Polynomials
(ii) Multivariate
(iii) Logarithmic
(iv)Exponential
(v) Calculus
3. Which of the following operations cannot be applied to a matrix equation:
(i) Multiplication
(ii) Addition
(iii) Division
(iv) Subtraction
4. List the basic operations in calculus.
75.
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STUDYTEXT
MATHEMATICAL TECHNIQUES
Answers to Chapter quiz
Function1.
(v) Calculus2.
(iii) Division3.
Differentiation and integration4.
Questions from previous exams
JUNE 2000 Question 1
a) Define the following terms:
(i) Stochastic process
(ii) Transition matrix
(iii) Recurrent state
(iv) Steady state
b) UC Limited specialises in selling small electrical appliances. A considerable portion of
the company's sales is on installment basis. Although most of the company's customers
make their installment payments on time, a certain percentage of their accounts are always
overdue and some are never paid at all. The Company's experience with overdue accounts
has been that if a customer falls two or more installments behind schedule, then this account
is generally not going to be paid; hence it is the company's policy of discontinuing credit
to such customers and to write these accounts off as bad debts. At the beginning of each
month, the company reviews each account and classifies them as either "paid-up", "current"
(being paid on time), "overdue (one payment past due) or "bad debt". To investigate this
problem, the analysts at UC Limited have constructed matrix representing the various states
that each shilling in the accounts receivable can take on at the beginning of two consecutive
months.
Transition Matrix of UC Limited
State of each Sh. at next month
Paid Current Overdue Bad debt
(P1
) (P2
) (P3
) (P4
)
State of each Sh.Paid (P1
) 1.00 0.00 0.00 0.00
First Month Current (P2
) 0.30 0.50 0.20 0.00
Overdue (P3
) 0.50 0.30 0.10 0.10
Bad debt (P4
) 0.00 0.00 0.00 1.00
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STUDYTEXT
Required:
(i) Interpret P22
and P23
(2 marks)
(ii) "Paid" and "bad debts" states have values of 1. interpret (3 marks)
(iii) If the original amount of money outstanding was Sh. 100,000, determine how much UC
Limited expected to be paid back if the records indicate that for every Sh. 100 in payments
due Sh. 70 are classified as "current" and Sh. 30 are classified as "overdue".
(7 marks)
(Total: 20 marks)
Note:
1.00 0 0 0
0.95 0 0 0.05
Pn
= 0.87 0 0 0.13
0 0 0 1.00
June 2002: Question 2
a) Define the following terms as used in Markovian analysis
i) Transition matrix (2 marks)
ii) Initial probability vector (1 marks)
iii) Equilibrium (1 marks)
iv) Absorbing state (2 marks)
b) A company employs four classes of machine operators (A, B, C and D), all new employees
are hired as class D, through a system of promotion, may work up to a higher class. Currently,
there are 200 class D, 150 class C, 90 class B and 60 class A employees. The company
has signed an agreement with the union specifying that 20 percent of all employees in each
class be promoted, one class in each year. Statistics show that each year 25 percent of
the class D employees are separated from the company by reasons such as retirement,
resignation and death. Similarly 15 percent of class C, 10 per cent of class B and 5 per cent
of class A employees are also separated. For each employee lost, the company hires a new
class D employee.
Required
i) The transition matrix (7 marks)
ii) The number of employees in each class two years after the agreement with the union.
(7 marks)
(Total: 20 marks)
June 2003: Question 2
The break-even point is that level of output at which revenue equals cost.
Required:
Graphically show the break-even point using the cost, revenue and profit functions.
(6 months)
77.
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STUDYTEXT
MATHEMATICAL TECHNIQUES
a) ABC Manufacturers Ltd. Produces spare part for motor vehicles. The demand for this spare
part is given by;
D = 15 – 0.5p; where d = demand and p = price
Required:
i) Explain clearly why the coefficient to the price is negative.
ii) Determine the total revenue function
iii) Determine the price elasticity at a price of Sh. 12. Clearly interpret your answer.
(4 marks)
b) The revenue of Better Option Pink Mobile Phone Company is related to advertising (a) and
phone sold (q). specifically, the relationship can be expressed as
R = q2
+ 3qa + a2
However, the budget constraint on advertising and production is given by
q + a = 100
Required:
i) Determine the maximum revenue with the advertising and production constraints.
(4 marks)
ii) What does a represent economically? (2 marks)
Total: 20 marks)
December 2003: Question 2
a) Explain the purpose of Venn diagrams (3 marks)
b) A market study taken at a local sporting goods store, Maua Wahome Store, showed that of
the 200 people interviewed, 60 owned tents, 100 owned sleeping bags, 80 owned stoves,
40 owned both tents and camping stoves and 40 owned both sleeping bags and camping
stoves.
Required:
c) If 20 people interviewed owned a tent, a sleeping bag and a camping stove, determine how
many people owned only a camping stove. In this case, is it possible for 30 people to own
both a tent and a sleeping bag, but not a camping stove? (6 marks)
d) "Under One Thousand Shillings" Corner Stone is planning to open a new store on the
corner of Main and Crescent Streets. It has asked the Tomorrow's Marketing Company to
do a market study of randomly selected families within a five kilometers radius of the store.
Among the questions it wishes Tomorrow's Marketing Company to ask each homeowner
are:
i) Family income.
ii) Family size
iii) Distance from home to the store site
iv) Whether or not the family owns a car or uses public transport.
Required:
"One Thousand Shilling" Corner Store. Denote which of these are discrete and which are
continuous random variables. (11 marks)
(Total: 20 marks)
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STUDYTEXT
December 2004: Question 2
Ujumi Industries Ltd. Has two subsidiary companies; apex and Maxima. Apex is solely involved
in manufacturing of Ujumi's products. Apex operates as a cost centre and supplies all its products
to Maxima. The total cost of production for Apex is given by the equation C = 3Q3
– 30Q2
+ 50Q
+ 300 where Q is the number of units (in hundreds of thousands) produced. Currently, apex does
not charge Maxima for the transfer of the goods.
Maxima is involved in the distribution and marketing of the products. The total cost associated
with maxima's activities is given by the equation
C = 2Q3
– 10Q2
+ 25Q + 100.
The revenue generated by Maxima on selling Q units is, R = 400Q – 250Q2
Ujumi's management has accused apex managers for not controlling the production cost which
has caused the company's profit to fall. In response apex managers argued that much of the
overhead cost is incurred in the transfer of the products to maxima. Apex Management have
argued that Ujumi's management should allow them to charge maxima a transfer fee of Sh. 100
per unit, so as to cut down on the overheads.
Required
a) The optimal number of units that Maxima should receive from apex if no fee is
charged. (6 marks)
b) The optimal number of units that apex can transfer without charging any fee, to minimize
total cost. (4 marks)
c) The optimal policy that Ujumi Industies Ltd. as a whole should adopt to maximise profits
when no transfer fee is charged. (6 marks)
d) Explain clearly whether allowing apex to charge the transfer fee, would help Ujumi
Industries Ltd. as a whole to improve their optimal policy.
(4 marks)
(Total: 20 marks)
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STUDYTEXT
Descriptive Statistics
CHAPTER TWO
Descriptive Statistics
Objectives
At the end of this chapter, you should be able to::••
Define statistics;••
Mention uses of statistics in business and other areas;••
Explain what descriptive statistics and inferential statistics are;••
Construct unweighted and weighted indices;••
Construct a price, quantity, value and special purpose index;••
Explain how the consumer price index is constructed and used;••
Compute various measures of dispersion for data organised in a frequency••
distribution;
Compute and explain the uses of the coefficient of variation and the coefficient of••
skewness.
Fast Forward: Descriptive Statistics are used to describe the basic features of data gathered
from an experimental study in various ways.
Introduction
Statistics is the art and science of getting information from data or numbers to help in decision
making.
As a science, statistics follows a systematic procedure to reach objective decisions or solutions
to problems.
As an art statistics utilises personal judgment and intuition to reach a solution. It depends on
experience of the individual involved. It is more subjective.
Statistics provides us with tools that aid decision making. For example, using statistics we can
estimate that expected returns and associated risks of a given investment opportunity.
Definitions of key terms
Statistical inference is deduction about a population based on information obtained from a
sample drawn from it. It includes:
point estimation••
interval estimation••
hypothesis testing (or statistical significance testing)••
prediction••
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STUDYTEXT
Measures of central tendency are single numbers that are used to summarise a larger set
of data in a distribution of scores. The three measures of central tendency are mean, median,
and mode.
Measures of dispersion – These are important for describing the spread of the data, or its
variation around a central value. Such measures of dispersion include: standard deviation,
interquartile range, range, mean difference, median absolute deviation, avarage absolute
deviation (or simply average deviation)
Variance is the sum of squared deviations divided by the number of observations. It is the average
of the squares of the deviation of the individual values from their means.
Skewness describes the degree of symmetry in a distribution. When data are uni-modal and
symmetrical, the mean, mode and median will be almost the same value.
Kurtosis describes the degree of peakedness or steepness in a distribution.
EXAM CONTEXT
Statistics as an art and science of getting information provides us with tolls that aid decision-
making, for example, with the aid of statistics we can estimate the expected returns and
associated risks of a given investment opportunity. It has been a wealthy area for examiners as
outlined below:
12/06, 12/06, 12/05, 12/04, 6/04, 6/01, 12/00, 6/00
Statistics has two broad meanings
Statistics1. refers to the mass of components or measures such as the mean, mode,
standard deviation etc. that help in describing the characteristics of a given set of data
or distribution. In this respect, statistics is simply, numerical facts about a given situation.
This is the original meaning of statistics. Governments were the first organisations to
collect vital statistics such as death rates, birth rates, economic growth etc.
Statistics2. refers to a method of study. This is commonly known as the scientific method.
Statistics is the process that gives thorough systematic steps to aid problem solving or
decision making.
The steps are:
a) Problem identification and definition
Once the problem is identified, it should be formulated as clearly and as unambiguously as
possible. The questions to be answered will clearly be defined at this point. This is necessary
because it enables us to make appropriate conclusion from the study.
b) Research methodology design
After formulating the problem, we need to come up with a defined plan on how the study will be
conducted to solve the problem.
Research design entails the following:
i. Determine the population of study. Population of study refers to the entire collection of
objects or subjects for which we want to make a decision e.g. the number of customers
of a supermarket; the number of students in public universities, the number of matatus
in Nairobi, the number of employees in campus, the number of hospitals in Nyeri.
ii. Decide whether to use a census or a sample study. A census entails studying each
and every element in a population. A sample study entails studying a portion of the
population.
85.
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STUDYTEXT
Descriptive Statistics
iii. Decide the data collection techniques to use e.g. observation, questionnaires,
interviews, Internet pop-ups etc.
iv. Identify the data analysis technique to be used.
Actual data collection/fieldwork•
Data analysis•
After data is collected, it is analysed so as to enable us make decisions. Data analysis involves
the following:
Organisation of data•• - This enables us to check for completeness, accuracy and
reference. Organising data can also involve coding the data for ease of analysis.
Data presentation• – Diagrams, graphs, tables etc. are used to present data so as
to highlight the visual impression of the data. The kinds of graphs and charts used
will depend on the audience. Type of graphs and charts may also be determined
by type of information being presented.
2.1 Statistical inference
The four steps above are referred to as descriptive statistics. Descriptive statistics helps us to
organise, summarise and present data in a convenient and informative way.
Inferential statistics, on the other hand, enables us to draw conclusions about the characteristics
of a population. Because most studies depend on samples, we use the sample results to draw
conclusions about the population from which the sample was drawn. The process of making
conclusions about the population based on sample statistics is known as sample inference.
Use sample statistics conclude on population parameters
Based on the statistical inference, then we answer the question or make the decision or solve the
problem that we set out to do.
Case Study
Statistics is applied in various fields. In the pension industry, actuarial methods are used to
measure the costs of alternative strategies with regard to the design, maintenance or redesign
of pension plans.
The strategies are greatly influenced bycollective bargaining; the employer's old, new and foreign
competitors; the changing demographics of the workforce; changes in the internal revenue code;
changes in the attitude of the internal revenue service regarding the calculation of surpluses; and
equally importantly, both the short and long term financial and economic trends.
It is common with mergers and acquisitions that several pension plans have to be combined or at
least administered on an equitable basis. When benefit changes occur, old and new benefit plans
have to be blended, satisfying new social demands and various government discrimination test
calculations, and providing employees and retirees with understandable choices and transition
paths. Benefit plans liabilities have to be properly valued, reflecting both earned benefits for past
service, and the benefits for future service. Finally, funding schemes have to be developed that
are manageable and satisfy the Financial Accounting Standards Board (FABS)
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STUDYTEXT
Summary measures
Fast Forward: In descriptive statistics, summary statistics are used to summarise a set of
observations, in order to communicate the largest amount as simply as possible.
The purpose of computing summary measures is to have single values that describe
characteristics of samples or population of interest. For example, we can compute measure of
central tendency or averages that represent the typical value of the distribution.
Measures computed from population are known as parameters while those computed from
samples are known on statistics e.g. population mean (µ), population std deviation ( ) proportion
(π ) the sample statistics are used to estimate the population parameters because most studies
use samples. There are various types of summary measures including averages and measures
of dispersion.
Measures of central tendency
a) Mean
It is the most popular measure of central tendency. It is the sum of all the values divided by the
number of values. When the mean is computed from a sample, it is represented by the symbol (
) pronounced x-bar. When computed from a population it is represented by the symbol µ (mu)
i) Arithmetic mean
Arithmetic mean represents the sum of all observations divided by the number of observations.
x =
x1
+ x2
+x3
+... +xn
= ∑
xi
n n
Illustration
Mr. Ochieng would like to know the average amount of money students carry when they go
to class. From a random sample of his closest friends, he has collected the following values
Shs250, 2000, 650, 4000,2000, 1050, 90, 8000, 1500,264
What is the mean amount of money carried by these students?
x =
250 + 2,000 + 650 + 4,000 + 2,000 + 1,050 + 90 + 8,000 + 1,500 +264
10
x = Shs 1,980.40
Weighted mean
x =
f1
x1
+ f2
x2
... + fn
xn
f1
+ f2
... + fn
Where f is / are the respective frequencies or weights of the observed values.
n
i−1
_
_
_
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We may also get the mean of grouped data using the assumed mean method. The method
simplifies calculation by reducing the magnitude of numbers used in the calculation. We start by
making one of the class mid-points, the assumed mean. The actual mean is then found by using
the formula.
Where d is the diff.6th
the assumed mean and the respective class midpoints.
i.e. d+xi-4
A is the assumed mean.
Example 4
Let's take A in the previous example to be 55
Therefore x
_
= 55+
370
=55 + 5 74
x
_
= 60
(b) Median
The median is a positional measure. A median is the middle value in an ordered set of data
values. There are as many values above the median as there are below it. Given a set of data,
the first step is to arrange the data in order. When we have an odd number. of observations the
median is the value of the middle observation. Median is denoted as X0.5
.To locate the middle value of a distribution, we use the formula
X0.5
=
n + 1
Value in the distribution 2
Example1
The salaries of a sample of CEOs of companies in Westlands area are:
Kshs, 90,000, 150 000, 1.2m, 500 000, 600 000, 1.5m
What is the median salary for this sample of CEOs?
90 000, 150 000, 500 000, 600 000, 800 000, 1.2m, 1.5m
X0.5
=
7 + 1
2
= 4th observation which is Ksh 600,000
For an even number of observations, the median is the arithmetic mean of the two middle
values.
Example 2
What is the median in the following values?
1200, 18, 30, 17, 60, 90, 48, 65, 120
17, 18, 30, 48, 49, 60, 65, 90, 120, 1200
X 0.5 =
49+60
= 54.5 2
Median for grouped data
In grouped data, just like in ungrouped data, the median is the value or point occupying the
middle position. It divides the data into 2 equal parts.
x = A +
∑ fi
di
∑ fi
_
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Descriptive Statistics
X0.5
= L0.5
+
n/2 − y
f
i where L0.5
= The lower limit or lower boundary of the median class
cf is the cumulative frequency of the distribution upto the class immediately preceding the median
class.
F= The frequency of the median class
i =The interval or class width of the median class
n= The total no. of observations
Steps
Identify the median class as it contains the median. The median class is the one that contains the
observation slightly above n/2 observation.
Example 3
Compute the median score in the business statistics class
Class f cf
30-40 3 3 0-2
40-50 10 13 3-12
50-60 19 32 13-31
60-70 31 63 31-62 (37th
)
70-80 11 74 62-73
The median class contains the observation slightly above
2
74
observation i.e 37th
Therefore is the 60-70 class
74
− 32
X0.5
= 60 + 10 2
31
= 60 + 1.61
X0.5
= 61.61
c) Mode
The word originated from the French word 'La mode' meaning mostly fashionable. It represents
the value that occurs most often in a distribution. "The value with the highest frequency."
In grouped data, the mode will fall into the modal class. The modal class is the class with the
highest frequency. The mode can be computed for any level of data.
It is most for normal data because for this level data, we cannot compute the other level of central
tendency.
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The mode can be used when considering the popularity of given attributes e.g. the most popular
car in a given town.
A distribution can have more than one mode. When there are two modes, it is said to be bi-modal
denoted Xm
Example
What is the mode in following data set 2, 11, 25, 11, 2, 5, 17, 38, 25, 17, 25, 13
= 25 (since it appears 3 times)
Mode for grouped data
The mode for grouped data falls in the modal class. After the modal class is determined, the
mode is found as:
Xm = Lm
+
d1
i
d1 + d2
Where Lm
= The lower limit of the modal class
d1
= The difference between the frequency of the modal class and that of the
class before it
d2
= The difference between the frequencies of the modal class and that of the class
just after it.
i = The class mdth of the modal class.
Example
Compute the modal mark in the business statistics class last semester
Mark Frequency
30 – 40 3
40 – 50 10
50 – 60 19
60 – 70 31
70 – 80 11
Modal class – 60 -70 with a frequency of 31
Xm
= 60 +
(31 − 19)
× 10
(31 − 19) + (13 − 11)
Xm
= 63.75
ii) Measures of dispersion
The measures of central tendencies give us values that may be considered to be typical values
samples of population from which they are computed.
Measures of dispersion enable us to know how far or how near observed values are spread
from the averages. They show the extent to which such values differ from the average value
(usually the mean). When observed values are close to the mean, we say there is low dispersion.
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STUDYTEXT
Descriptive Statistics
Dispersion is also known as spread, scatter or variation. Some of the most commonly used
measures of dispersion include: - range, variance and standard deviation.
Range
It is the difference between the highest and the lowest values in a data set
Therefore, R= Xmax
– Xmin
The range is the simplest measure of dispersion because it only uses two values. It is most useful
in cases where there are erratic changes.
Example
What is the range in the following exchange rates of the shilling to the US dollar? 75, 74,77, 68,
69, 70, 73, 74, 68.5, 75.5, 69, 78.5, 70
Range = max – min
= 78.5 – 68
= 10.5
Example
The following data shows salaries earned by the top management of Kabete International Ltd.
105,000; 2,000,000; 300,000; 250,000; 120,000; 350,000, 130,000
Range = Max – Min
= 2,000,000 – 105 000
= 1,895,000
Weakness
Range depends on only two values. This means that it can be influenced by extreme values that
may be considered to be outliers.
It doesn't not give an indication as to how the values are spread in a distribution
To overcome this weakness, we use the inter-quartile range (IR).
The inter-quartile range is the difference between the top quartile and the lower quartile
IR = Q3
– Q1
Q3
– The value in the observation below which ¾ or 75% of the observation lie and above in the
remaining ¼ or 25% of the observations
Q1
- Will have 25% of observation are less than and 75% above it.
Quartiles, Deciles, percentiles
Another way to describe variation in data is to determine the location of what divides a set
of observation into equal parts. These values include the median, quartile, deciles and
percentiles.
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Quartiles
They divide an ordered set into four equal parts. The first quartile Q1
is the value withih which
25% of the observation lie and Q3
is the value below which 75% of the observations lie. When
computing quartiles, the first step is to locate the quartile class. The location of the quartile is
found as:
Qj
= (n+1) j/4= Where Qj
quartile and is 1, 2, 3, 4.
The quartile value is then found as;
Qj
= Lj
+ (jt/4-cf) i
F
Where Lj
- the lower limit of the quartile class
cf - Cumulative frequency up to the class before the quartile class
f - the frequency of the quartile class
i - width of the quartile class
Example
Compute the values of Q1
and Q3
for the scores in the Business Statistics course last semester
Marks f cumulative cf
30-40 3 3 0-2
40-50 0 13 3-12
50-60 19 32 13-31 (18)
60-70 31 63 31-62
70-80 11 74 62-73
Solution
Find the location of Q1
Q1
falls in the class width (74+1) ¼ the observation is 18.75th
observation. This is the class 50-
60
Q1
will therefore be
50 +
(¼ (74) − 13)
×10
19
= 52.89
This means that 25% of the students scored less than 52.89% in the course.
Q3
= Location = (n+1)3/4
= (74+1)3/4
75 x 3/4 = 56.25
Value = 60 +
(¾ (74) − 32
× 10
31
= 67.58
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STUDYTEXT
Descriptive Statistics
This means that 75% of the students scored less than 67.58% marks
Percentiles
They divide an ordered data set into 100 equal parts
Given a set distribution X1
, X2
, X3
, ……………Xn
then pth
percentile is the value of X such that P%
of the observation are less than P and (100-p) % of the observation are greater than P.
Example 1
P40 means 40% of the observations are less than P and 60% of the observation are greater than
P
P40 is the 40th
percentile
The location of the percentile has to be determined before we get the percentile. The percentile
class is the one that contains the (n+1) k/100th
observation where k-1, 2,3,……100
The percentile will then be found as;
Pk
= Lk +
(kn/100 – cf)
f
Where Lk - The lower limit of the percentile class
Cf - Cumulative frequency upto class just before the percentile
f - Frequency of P class
j - Width of P class
Example 2
Compute the following percentiles for the scores in the Business Statistics course last
semester.
P10, p25, p50, p75, p60, p50
P10, location (74+1) 10/100 = 75x0.1 = 7.5
We look for the class with 7.5th
observation
40+
(10 x 74 -3)
10 = 44.4
100
10
Variance and standard deviation
The two common measures of dispersion are variance and standard deviation.
Adata set that is more variable will have a larger variance than one that is relatively homogeneous.
The variance is the sum of the square deviations divided by the number of observations. It is the
average of the squares of the deviation of the individual values from their means. For any set of
values, the sum of square deviations from the mean is smaller than the sum of square from any
other point.
Population variance is denoted as δ2
→ parameter
Sample variance is denoted as S2
→ statistics
95.
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STUDYTEXT
Descriptive Statistics
S2
=
328
5
= 65.6
Sample variance, S2
=
∑ (Xi
− X)2
n − 1
=
328
4
= 82
For group data, we only get an approximation (estimate) of the variance.
S2
=
∑ fi
(Xi
− X)2
∑ fi
− 1
The standard deviation is the square root of the variance. It is expressed in the same units as
the original data.
Population variance,σ2
=
∑ (xi
− m)2
N
Sample standard deviation, S =
∑ (xi
− x)2
√ n − 1
For grouped data, S = ∑ (xi
− x)2
√ ∑fi − 1
Coefficient of variation
Coefficient of variation is useful when comparing the levels of variability in sets of data. It is a
relative measure of variability. It is especially useful when comparing sets that are not measured
in the same units e.g. in weights of people vs. income, or when comparing data with means that
are of different magnitudes, or risk of projects. The coefficient of variation is dimensionless (free
of units). It is generally expressed in percentage or in decimal form.
CV −
s
or =
s
× 100%
x x
The higher the coefficient of variation, the higher the variability.
i =1
n
i =1
n
i =i
N
−
i =1
n
i =1
n
−
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Example
Which of these 2 sets of data has greater variability?
A B
χ = 150kgs χ = 0.85cm
S = 30.5kgs S = 0.015cm
CV = 30.5 CV = 0.015
150 0.85
= 0.203 = 0.018
Set A has greater variability than set B
Measures of normality/shape
A normal distribution is data that forms a symmetrical bell curve. Measures of normality tells us
more about the way data is distributed e.g. figures A, B and C below appear to be symmetrical.
Distributions may have same averages and measures of dispersion but have different shapes.
Measures of normality give us an idea of how the data is distributed. Measures of normality
include coefficient of skewness and coefficient of kurtosis.
Skewness
Skewness describes the degree of symmetry in a distribution. When data are uni-modal and
symmetrical, the mean, mode and median will be almost the same value. In a skewed distribution,
we have higher frequencies occurring to one end of the distribution e.g.
A
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STUDYTEXT
Graphs A and C represent skewed distributions;
Skewed to the right (negatively skewed)
Skewed to the left (positively skewed)
In a positively skewed distribution the mean > median >mode
In a negatively skewed distribution the mean < median < mode
When data are skewed, the mean will be pulled towards the skew. The degree of skewness is
composed by using either the 1st
Pearsonian coefficient of skewness or the 2nd
First coefficient of skewness, SK1 =
X − Xm
where
S
Second coefficient of skewness, SK2 = 3
X − X0.5
S
xm
= mode
S = standard deviation
X0.5
= median
If SK1 or SK2 = 0, the distribution is normally distributed or is symmetrical.
If SK>0, the distribution is positively skewed.
If SK<0, the distribution is negatively skewed.
Kurtosis
Kurtosis describes the degree of peakedness or steepness in a distribution.
Example
Leptokurtic
(Highly peaked)
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STUDYTEXT
Empirical Rule
i. The empirical rule says that if a sample or population of measurement has a normal
distribution
ii. Approximately 68% of the observations lie within one standard deviation of the mean
iii. Approximately 95% of the observations lie within two standard deviations of the mean
iv. Approximately 99.7% of the observations lie within three standard deviations of the
mean.
Diagram 1.1
Chapter Summary
Statistics is the art and science of getting information from data or numbers to help in decision
making.
The following are some characteristics of index numbers
1. They are specialised averages to obtain a typical measure of central tendency like an
average. The items must both be comparable and the unit of measurement must be the
same
2. Measure the change in the level of a phenomenon3. Measure the effect of changes over a
period of time
Counting techniques may be classified into:
i. Probability trees
ii. Permutations
iii. Combinations
68%
95%
99.7%
-3δ -2δ -1δ μ δ 2δ 3δ
101.
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STUDYTEXT
Descriptive Statistics
Chapter Quiz
1. Define Mean
2. Which of the following is the odd one out?
i. Mean
ii. Mode
iii. Median
iv. Range
3. What is the importance of Kurtosis?
4. ………… describes the degree of symmetry in a distribution when data are uni-modal
and symmetrical, the mean, mode and median will be almost the same value.
5. List three counting techniques.
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STUDYTEXT
Answers to Chapter Quiz
1. It is the sum of all the values divided by the number of values.
2. Range – it is not a central tendancy measure
3. Kurtosis describes the degree of peakedness or steepness in a distribution.
4. Skewness
5. i) Probability trees
ii) Permutations
iii)Combinations
Questions from previous exams
1. The weights of 15 parcels recorded at the GPO were as follows:
16.2, 17, 20, 25(Q1) 29, 32.2, 35.8, 36.8(Q2) 40, 41, 42, 44(Q3) 49, 52, 55 (in kgs)
Required
Determine the semi interquartile range for the above data
2. The following table shows the levels of retirement benefits given to a group of workers in a
given establishment.
Retirement benefits £ '000 No of retirees (f) UCB cf
20 – 29 50 29.5 50
30 – 39 69 39.5 119
40 – 49 70 49.5 189
50 – 59 90 59.5 279
60 – 69 52 69.5 331
70 – 79 40 79.5 371
80 – 89 11 89.5 382
Required
a) Determine the semi interquartile range for the above data
b) Determine the minimum value for the top ten per cent.(10%)
c) Determine the maximum value for the lower 40% of the retirees
3. The following information was obtained from an NGO which was giving small loans to some
small scale business enterprises in 1996. the loans are in the form of thousands of Kshs.
107.
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STUDYTEXT
Probability
CHAPTER THREE
Probability
Objectives
At the end of this chapter, you should be able to:
i. Define probability.
ii. Describe the classical, the relative frequency, and the subjective approaches to
probability.
iii. Calculate probabilities, applying the rules of addition and multiplication.
iv. Determine the number of possible permutations and combinations.
v. Calculate a probability using Baye's Theorem.
vi. Define a probability distribution.
vii. Distinguish between a discrete probability distribution and a continuous probability
distribution.
viii Calculate the mean, variance and standard deviation of a probability distribution.
ix. Construct binomial and poisson distribution.
x. Determine which probability distribution to use in a given situation.
Fast Forward: Probability, or chane, is a way of expressing knowledge or belief of the possibility
or the strength of the possibility that an event will occur.
Introduction
Probability is a measure of likelihood, the possibility or chance that an event will happen in future.
It can be considered as a quantification of uncertainty.
Uncertainty may also be expressed as likelihood, chance or risk theory. It is a branch of
mathematics concerned with the concept and measurement of uncertainty.
Much in life is characterised by uncertainty in actual decision making.
Probability can only assume a value between 0 and 1 inclusive. The closer a probability is to zero
the more improbable that something will happen. The closer the probability is to one the more
likely it will happen.
Definitions of key terms
Random experiment results in one of a number of possible outcome e.g. tossing a coin
Outcome is the result of an experiment e.g. head up, gain, loss, etc. Specific outcomes are
known as events.
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Trial- Each repetition of an experiment can be thought of as a trial which has an observable
outcome e.g. in tossing a coin, a single toss is a trial which has an outcome as either head or
tail
Sample space is the set of all possible outcomes in an experiment e.g. a single toss of a coin,
S=(H,T). The sample space can be finite or infinite. A finite sample space has a finite number of
possible outcomes e.g. in tossing a coin only 2 outcomes are possible.
An infinite sample space has an infinite number of possible outcomes e.g. time between arrival
of telephone calls and telephone exchange.
An Event of an experiment is a subset of a sample space e.g in tossing a coin twice S= (HH, HT,
TH, TT) HH is a subset of a sample space.
Mutually exclusive event - A set of events is said to be mutually exclusive if the occurrence of
any one of the events precludes the occurrence of other events i.e the occurrence of any one
event means none of the others can occur at the same time e.g. the events head and tail are
mutually exclusive
Collectively exclusive event - A set of events is said to be collectively exclusive if their union
accounts for all possible outcomes i.e. one of their events must occur when an experiment is
conducted.
Favourable events refers to the number of possible occurrences of a given event in an experiment
e.g. if we pick a card from a deck of 52 cards the number favorable to a red card is 26, in tossing
a coin the number favourable to a head is one.
Independent events – Events are independent if the happening or non-happening of one has no
effect on the future happening of another event. E.g. in tossing two times of a coin, the outcome
of 1st
toss does not affect 2nd
toss.
Equally likely events – Events are equally likely if the happening of one is not favoured over the
happening of others. In tossing a coin the tail and head are equally likely.
OTHER CONCEPTS
Unconditional and conditional probabilities – with unconditional probability we express
probability of an event as a ratio of favourable outcomes to the number of all possible
outcomes.
A conditional probability is the probability that a second event occurs if the first event has
already occurred.
Joint probability – joint probability gives the probability of the joint or simultaneous occurrence
of two or more characteristics.
Marginal probability – is the sum of two or more joint probabilities taken over all values of one
or more variables. It is the probability that the results when we ignore one or more criteria of
classification when computing probability.
109.
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STUDYTEXT
Probability
Industry context
Probability is used throughout business to evaluate financial risks and decision-making.
Every decision made by management carries some chance for failure, so probabiity
analysis is conducted formally.
In many natural processes, random variation conforms to a particular probability
distribution known as the normal distribution, which is the most commonly observed
probability distribution.
EXAM CONTEXT
The probability topic has been examined previously:
6/06, 12/05, 6/05, 12/04, 6/04, 6/03, 12/02, 6/02, 12/01, 12/00, 6/00
Laws of Probability
1. Rules of Addition
a) Special rule of addition
If two events A and B are mutually exclusive the probability of one or other occurring is equal to
the sum of their probability
P (A or B) = P (A) + P (B)
P (A or B or C) = P (A) + P (B) + P(C)
Illustration
An automatic plastic bag - a mixture of beans, broccoli and other vegetables, most of the
bags contain the correct weight but because of slight variations in the size of beans and other
vegetables. A package may be slightly under or overweight. A check of 4,000 packages of past
reveals the following:
Weight Event No of packages
Underweight A 100
Satisfactory B 3600
Overweight C 300
What is the probability that a particular package will be either underweight or overweight?
The two events are mutually exclusive
P (A or C) = P (A) + P (C)
P(A) =100/4000
P(C) =300/4000
P(A or C) =400/4000 = 1/10 =0.1
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STUDYTEXT
Note: The probability that a bag of mixed vegetable is selected underweight P(A) plus the
probability that it is not underweight P(NA) must logically be equal to one. This is referred to as
Complement rule.
b) General rule of Addition
The outcome of an experiment may not be mutually exclusive. This rule is used to combine
events that are not mutually exclusive.
P(A or B) = P(A) + P(B)- P(A and B)
Example
a) What is the probability that a card chosen at random from a pack of well shuffled deck will
either be a king or a heart
P(king or heart) = P(king) + P(heart) - P(king and heart)
4 + 13 + 1 = 16
52 52 52 52
b) Routine physical examination is conducted annually as part of the programme of a particular
organisation. It was discovered that 8% needed correcting shoes, 15% needed major dented
work and 3% needed for both correcting shoes and major dental work.
What's the probability that an employee selected at random will either need correcting shoes or
major dental work?
2. Rule of Multiplication
a) Special Rule of multiplication
For two events A and B the probability that A and B will both occur is found by multiplying the
probability
P(A and B)=P(A) × P(B)
P(A and B and C) = P(A) × P(B) × P(C)
This rule is applicable for independent events i.e. the occurrence of one does not depend on the
occurrence of the other.
Example
i) Two coins are tossed. What is the probability that both will land tails up
The two events are independent
P (T and T) = P (T) × P (T)
=1/2 × 1/2 = ¼ = 0.25
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STUDYTEXT
Probability
ii) From long experience firestone tyres have a 0.8 probability that their xB.70 will last 40,000
miles before it becomes bald and adjustments made. If you purchase four xB.70
What's the probability that all tyres will last 40,000 miles?
What's the probability that at least two will last 40,000 miles?
Solution
P(L and L and L and L) =P(L) × P(L) × P(L) × P(L)
= 0.8 × 0.8 × 0.8 × 0.8
= 0.4096
Probability of lasting, P(L) = 0.8; Probability of not lasting, P(N)= 0.2
Probability of at least two tyres lasting 40,000 miles is given by the following combinations:
b) General rules of Multiplication
It states that of two events A and B the joint probability that both events will happen is found by
multiplying the probability that A will happen by the conditional probability of event B happening
P(A and B) = P(A) × P(B|)
Where P(B/A) stand for probability that B will occur given that A has already occurred.
Example
1. Assume that there are 10 rolls of film in a box 3 of which are defective. Two rolls are to be
selected one after another. What's the probability of selecting a defective roll followed by
another defective roll?
P (D and D) = P (D) × P (D |D)
3/10 × 2/9 =6/90 =0.07
2. Three effective toothbrushes were accidentally shipped to a chemist by Clean Brand Product
along with 17 non-effective ones
What's the probability that of the first 2 toothbrushes sold one will be effective and the other one
will not.
Solution
P (One defective = P (D and N) or P (N and D)
= 2/20 x 17/19 + 17/20 x 3/19
Contingency Table
A two dimensional contingency table is formed by classifying two factors. One factor determines
the row categories and the other determines the column categories. Each element in the table
belongs to the two classifications known as a cell e.g. classifying subjects by gender (M and F)
and smoking status (Current/Former/Never). Such categories are said to be mutually exclusive
and collective. Exhaustive mutually exclusive means the categories include all possibilities and
so there is a category for everyone.
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STUDYTEXT
Table 3.1
Smoking Status
Gender
Current Former Never
Male a b c
Female d e f
The two factors here are smoking status and gender and the cells are a,b,c,d,e and f.
Example
A survey of executives dealt with their loyalty to the company. Out of the questions asked "If you
were given an offer by another Company equal or slightly better than the present will you remain
in the Company or take another position. The responses were classified with their level of service
within the company.
Length of Service
Loyalty Less than 1
year
1-5 years 6-10 years More than
10 years
Total
Remain
Not Remain
10
25
30
15
5
10
75
30
120
80
200
What's the probability of randomly selecting an executive who is loyal to the company (who will
remain) and who has been in the service in more than 10 years?
P(R and >10) = P(R) × P (>10|R)
=
120
×
75
200 120
=
75
200
What's the probability of selecting at random an executive who will remain loyal to the company
and has less than 6 years of service
= 10 + 30
200
=
40
200
= 0.2
Tree Diagram (Probability Trees)
Is very useful for portraying conditional and joint probabilities. It is particularly useful for analysing
business decisions where there are several stages of problems.
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Probability
Bayes Theorem
In the 18th Century Rev Thomas Bayes, an English Presbyterian Minister, ordered this question,
"Does God really exists" Being interested in mathematics he attempted to develop a formula to
arrive at the probability that God really exists based on the evidence available on earth.
Later, Laplace refined Bayes work and gave it the name Bayes Theorem in a workable form, the
Bayes Theorem is
P(A/B) =
P(A)P(B)
P(B)
=
P(A1
) × P(B/A1
)
P(A) × P(B/A) + P(A2
) × P(B/A2
)
Note: P(PA and B) = P(A) × P(B/A)
Example 1
Suppose 5% of the population of a third world country has a disease peculiar to that country.
Let A1
refer to the event that a person has the disease and A2
refer to the event that a person
does not have the disease. Therefore, if a person is chosen at random, the probability that the
individual chosen will have a disease will be P(A1
) = 0.05 thus P(A2
) = 0.95.
These probabilities are called prior probabilities (they are assigned before any empirical data is
obtained). It's the probability based on the present level of information. Assume further that there
is a diagnostic technique to detect the disease but it is not accurate. Let B denote the event test
shows disease is present. Assume that historical evidence shows that if a person actually has the
disease the probability that the test will indicate presence of a disease is 0.9.
P (B|A1
) =0.9
Assume that there is probability of 0.15 that a person does not actually have the disease but the
test indicates that the disease is present = (P (B/A2
) = 0.15.
Let us randomly select a person from the country and perform a test. The test indicates that the
disease is present. What's the probability that the person actually has the disease?
P(A1
| B) =
P(A1
) × P(B/A1
)
P(A1
) × P(B/A1
) + P(A2
) × P(B/A2
)
Note: P(B) = P(A1
) × P(B/A1
) + P(A2
) × P(B/A2
)
Example 2
An electronic firm purchases its supplies from 4 different suppliers. A Ltd supplies 20%, B Ltd
30% C Ltd 25% and D Ltd 25%.
A Ltd tends to have the best quality. Only 3% of their supplies are defective. B Ltd supplies are
4% defective, C Ltd 7.0% and D Ltd 6.5% defective
Required:
a) What is the probability of selecting a defective item?
b) A defective supply was discovered in two shipments. What's the probability that it came from
A Ltd
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What's the probability that the defective supply came from A ltd, C ltd and D ltd
Solution
a) What is the probability of selecting a defective item
P (D) = P( A and D) + P(B and D) + P(C and D)+P(D and D)
0.006+0.012+0.0175+0.01625
=0.05175
b)
P(A/D) =
P(A and D)
P(D)
=
0.006
0.05157
c) P(A/D) = P(A) × P(D/A)
=
P(A) × P(D/A) + P(B) × P(D/B)) + P(C) × P(D/C)) + P(D)) × P(D/D))
P(D)
Principle of Counting
If the number of possible outcomes in an experiment is relatively easy to count and list all the
possible events e.g. there are six possible events resulting from a roll of a dice. If however,
there a large number of possible outcomes such as number of boys and girls for families with
10 children it would be tedious to list and count all the possibilities: we could have two boys and
eight girls, 1 boy and 9 girls etc. This situation is further complicated when we consider the order
of arrangements.
To facilitate counting, counting rules can be employed. These include multiplication rule,
permutation rule, and combination rule.
Multiplication Rule
If there are M ways of doing one thing and N ways of doing another thing, there are
M x N ways of doing both; that is, the total number of arrangements equals M×N. This can
be extended to more than two events e.g. for three events M, N and O the total number of
arrangements equal M × N × O.
Example I
An automobile dealer wants to advertise that for Kshs 500,000 you can buy a convertible, a two-
door or a four-door model with your choice as either a saloon or a station wagon. How many
different arrangements of models and car types can the dealer offer?
Solution
Convertible Saloon Convertible S.W
Two-door saloon 2-door S.W
Four-door Saloon 4-door S.W
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Probability
We can employ multiplication rule as a check when M- No of models
N- No of car types
Solution = 6
Example 2
The marketing manager of ABC Ltd would like to send a representative to the branches of a
company in Central Kenya. The company has three regions in Central Kenya, each with
5-branches. The sales representative will be based in one of the sub-branches. How many ways
can this be done?
M – No. of regions
N – No. of branches
M x N = 3x5 = 15
Solution =15 ways
Example 3
Pioneer manufactures 3 models of stereo receivers, 2 cassettes decks, 4 speakers and 3CD-
carousels. When the 4 types of compatible are sold together they form a system. How many
different systems can electronic firm offer?
2 x 4 x3=24syatems
Permutation Rule
The multiplication formula is applied in finding the number of possible arrangements for two or
more groups. The permutation formula is applied to find the number of possible arrangements
where there is only one group of objects.
Permutation is thus any arrangement of r objects from n possible objects. However this can be
calculated using the permutation formula:Example 1
In how many ways can you arrange A, B and C?
Solution
In this case, note that the order in which the elements appear matters i.e ABC is different from
ACB.
Therefore
ABC BAC CAB
ACB BCA CBA
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Answer: Six ways.
Using the permutation formulaIn this example
3
P3
=
3!
(3 − 3)!
= 3 × 2 × 1 = 6 ways
Note that 0! = 1
ii) 3 electronic points are to be assembled in a plug-in unit for a TV set. The parts can be
assembled in any order. In how many different ways can the 3 parts be assembled?
All the objects (electronic parts) are to be assembled and therefore r=3. .
Solution
n =3, r =3
n
Pr
=
n! = 3!
(n − r)! (3 − 3)!
=
6
1
= 6
Example 2
There are 10 numbers; 0 through to 9, which are to be used in code group of four to identify an
item of clothing e.g. code 1083 is to identify blue blouse, code 1030 is identify a pair of socks and
so on. Repetition of numbers is not permitted i.e. same number can't be used twice or more in a
total sequence e.g. 2256 or 2872
How many different code groups can be designed?
n = 10 r = 4
n
Pr
= 5.040
Permutation allowing repetition
If repetitions are permitted the permutation formula n
Pr
= nr
e.g assume a 2-letter code where 'a'
and 'b' are to be taken each at a time with repetitions such as aa allowed.
In this case n = 2, r = 2
The no. of permutations 2
P2
= 22
= 4
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Probability
Combination Rule
In determining the number of permutations of n-different things taken at a time, the order of the
items is important.
Combination determines the number of ways to choose r objects from group of n objects without
regard to order.
Combination Formula=
n
Cr
=
n!
(n − r)!
Example
The sales department has been asked to design 42 colour codes for 42 different parts. A colour
code should consist of three colours and if 3 colours are used for one part, the same cannot be
used to identify a different part. Would 7 different colours be adequate to generate the 42 parts
colour codes?
Solution
n=7 r=3
7
C3
= 35
35 colour codes would be inadequate. Thus 7 colours would not be adequate.
As an alternative to the use of three colour combination it has been suggested that only two
colours (r) be used for one colour code. Would 10 colours be adequate to colour code the 42
different parts?
n = 10 r = 2
10
C2
= 45
This number would be adequate.
Probability Distributions
Statistical inference has the objective of making inferences or statement about a population
based on a sample selected from a population.
Probability distribution – Is a listing of all the outcomes of an experiment and the probability
associated with each outcome.
Example:
Suppose we are interested in determining the number of heads showing upon tossing a coin
three times.
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Table 3.2: Enumerating the samples
Options 1st
coin
toss
Second Third No. of
heads
Prob
1 T T T 0 1/8
2 T T H 1 1/8
3 T H T 1 1/8
4 T H H 2 1/8
5 H H H 3 1/8
6 H T H 2 1/8
7 H H T 2 1/8
8 H T T 1 1/8
x = outcome P(x) = Probability of outcome
0 1⁄8
1 2⁄8
2 3⁄8
3 1⁄8
Note:
Two important characteristics of probability distribution
The probability of a particular outcome must always be between zero and one.
The sum of probabilities of all mutually exclusive probabilities in a distribution is one
Exercise
Assuming you roll a die and observe the possible outcome of an experiment involving a six side
die;
Develop a probability distribution for this outcome
Portray the probability distribution graphically
Random variable
In any experiment of chance the outcomes occur randomly e.g. rolling a single dice is an
experiment in which any of the six possible outcomes can occur. Some experiments result in
outcomes that are quantitative such as shs, weight or number of children while others result in
outcomes that are qualitative e.g. colour, or religious preference.
A random variable is a quantity resulting from a random experiment that by chance can assume
different values, for example:
1. If we count the number of students absent (the random variable) from class on a
Monday morning we can have 0, 1, 20, 30 etc.
2. If we toss two coins and count the number of heads we could have 0,1 or 3 heads.
Since the exact number of heads resulting from this experiment is due to chance, it will
be called a random variable.
Other random variables might be the number of defective light bulbs produced during a week, the
weight of members of DMS 201 regular class.
A random variable can either be discrete or continuous.
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Probability
Discrete random variable
A discrete random variable can assume only a certain number of separate values of an interval;
for example, if there are 100 employees, the count of number absent on Friday can be 0, 2, and
100
In most cases a discrete random variable is usually a result of counting something.
Note: A discrete variable can assume fractional or decimal values in some cases. These values
must however be separated or countable.
Continuous random variable
A continuous random variable can assume any value in an interval. It is usually the result of
measuring something e.g. height of a person. It can assume one of an infinitely large number of
values within certain limitations.
E.g. the weight of a cloth can be 67, 67.2, 67.24kg, 67.241kg depending on the accuracy of
measuring instrument
Logically, if we organise a set of discrete random in a probability distribution the distribution is
called a discrete probability distribution.
How to determine mean, variance, standard deviation of probability distribution
The mean represents the central location of the data
The variance describes the spread in the data
In a similar way, a probability distribution is summarised by its mean and its variance
Mean (µ)
The mean is a typical value to represent a probability distribution. It is also the long-run average
value of the random variable. It is also referred to as expected value E(x).
E (x) can be computed as
E (x) = µ = Σx.P(x)
For a continuous variable
E (x) = µ = ∫ x.P(x)dx
Where P (x) is the probability of various outcomes of x
Variance and standard deviation
The mean does not describe the amount of spread (variation) of a distribution. The variance does
this.
s2
= ∑ (x − µ)2
P(x)
The standard deviation = square root of Variance
Example
Bill sells new cars for GM. Bill usually sells the largest number of cars on Saturday. He has
established the following probability distribution for the cars he expects to sell on a particular
Saturday.
∞
- ∞
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Probability
More examples
A product is either classified as acceptable by the quality control department or not acceptable.
A sales call results in a customer either purchasing or not purchasing a product. Frequently we
classify the possible two outcomes as "success and failure" however, this classification does not
imply one outcome is good and the other is bad.
The information obtained in a binomial distribution is as a result of count i.e. we count the number
of success in the total number Of trials e.g. if we flip a coin five times, we could count the number
of times a head appears.
The probability of a success remains the same from one trail to another e.g. the probability that
you will give the first question of a true /false test correctly is one half (1/2). This is the 1st
trial.
The probability that you will guess in the 2nd
question (2nd
trial) is also ½ and so on.
One trial in a binomial distribution is independent of another trial. In fact this is the same as saying
that there is no rhythmic pattern with respect to the outcomes. As an example, the answers to the
true/false question are not arranged in any order e.g. FFFF TTTTTTTTT FFFF and so on
The number of trials is fixed at a certain value
Formula in binomial distribution
To construct a binomial Probability Distribution we must know
1. The number of trials
2. The probability of success on each trial
P(r) =
n!
pr
qn − r
r!(n − r)
= n
Cr
pr
qn − r
Where n - Number of trials
r - Number of observed outcomes
p - Probability of success of each trial
q - Probability of failure = 1-p
Example
The answer to a true/false question is either true or false. Assume that;:
1. An examination consists of four true/false questions
2. A student has no knowledge of subject matter. The chance (probability) that the student will
guess the correct answer to the first question is ½. Likewise the probability of guessing right
the remaining number of question is 0.5 for each question. What's the probability that:
a) The student will get none out of the four questions correct;
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STUDYTEXT
b) The students will get exactly one out of the four correct?
n= 4
r = ?
p = 0.5
q = 0.5
r = 0
P(0) =
4!×0.5×0.54
0!(4 − 0)
= 0.0625
Poisson Probability Distribution
Fast Forward
The binomial probability distribution for probabilities of success P less than 0.05 can be computed
but the calculations would be quite time consuming especially for a large n say 100 or more. The
distribution of probabilities would become more and more skewed as the probability of success
becomes smaller. Where the probability of success is very small and n is larger, we refer to it as
as Poisson probability distribution named after Simeon Poisson. The probability of a particular
event happening is quite small.
The poisson distribution is also a discrete probability distribution because it is formed by counting
something. An important characteristic of such a distribution is that we can count only the
occurrences, we cannot count the non-occurrences. For example, we can only count the number
of telephone calls coming to a telephone exchange (switch board) each minute. We cannot count
the number of calls that we did not make. We can count accidents within a given time but we
cannot count the number of accidents that did not occur within a given period.
A poisson situation can be recognised by the following characteristics:
a) The existence of events that
- occur randomly
- are rare
b) An interval of time, space or distance is defined within which events can occur
Assumptions
The probability of occurrence is the same everywhere in the interval
The probability of multiple occurrences at precisely same point is negligible
The occurrence rate is a random variance
Formula
p(x) =
e−x
|x
x!
Where x equals the arithmetic mean number of occurrences (success) in a particular interval. It's
the specific value of the random variable.
e - is an exponential.
x - is the number of occurrences.
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Probability
Continuous Probability Distribution
In continuous probability distribution, random variable can take any value within a given range of
values by chance. The distribution of continuous random variable is characterised by a probability
density function which represents all the possible outcomes in an experiment. The probability
density function can be a graph, an equation or a curve, which represents these possible values.
A probability of an experiment = 1.
Given a probability density function of a random variable x we can determine the probability that x
has a value between two given points A and B. This probability is equal to the total area enclosed
by the curve of the function, two perpendicular lines erected at points A and B and the x-axis as
a proportion of the total area enclosed by the function curve and x-axis.
Daigram 3.3
y F(x)
A B x
There are several types of continuous probability distribution:
Normal distribution
The normal distribution curve is symmetrical and bell-shaped. Most observations in the
distribution are close to the mean with fewer observations further away. A normal distribution can
be determined by the values of the mean and the standard deviation.
The normal curve is asymptotic to the x-axis i.e. its tails continuously approach the x-axis but
never quite touch it to infinity.
In a normal distribution the following observations are true:
68.26% of all observations lie within 1 standard deviation of the mean
95.44% of all observations lie within 2 standard deviations of the mean
99.73% of all observations lie within 3 standard deviations of the mean
Diagram 3.4
68.26%
95.44%
99.73%
-3δ -2δ -1δ μ δ 2δ 3δ
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The normal distribution is a family of distribution e.g. standard normal distribution. Standard
normal distribution is a special case of the normal distribution. It has a mean of zero and a
standard deviation of 1 given any normal distribution. We may convert it to the standard normal
distribution by simply converting its μ to o and ó equal to 1. We convert each observation to the
standard normal variation as follows:
Z =
c − m
σ
Where Z is the standardised random variable. Also known as standard normal deviate
orZ-score
x is the individual observed value of the random variable
μ is the mean
σ is standard deviation
By standardising any normally distributed random variable we can use the normal distribution to
estimate the area enclosed by the normal curve and true z-value. In the normal table, the values
of z only go to about z=3
If we have a list of observations of a normaly distributed random variable, what proportion of the
observation will be expected to be?
Example
If x is a continuous random variable with a mean of 50 and standard deviation of 2, what is the
probability that an observation picked at random is
1. Less than 52
2. Less than 46
3. Between 44 and 48
Solution
1. μ = 50
σ= 2
when x = 52
z =
x-μ
σ
=
52 – 50
2
Calculate z = 1
= 0.3413
= 0.5 + 0.3413
= 0.8413
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Probability
Less than 46
x = 46
x =
46-50
= -4/2 = -2
2
Calculate 2 = -2
From table 2 = 0.4772
Less than 46
= 0.5 -0.4772
= 0.0228
2. Between 44 and 48
z =
44-50
2
=6/2 = -3
z = 0.49865
z = 48-50
2
= 2/2 = 1
z = 0.3413
3. Between 44 and 48
0.49865 – 0.3413 = 0.15735
Chapter Summary
A set of events is said to be mutually exclusive if the occurance of any one of the events precludes
the occurrence of any of the other events e.g. when tossing a coin, the events are a head or a tail
these are said to be mutually exclusive since the occurrence of heads for instance implies that
tails cannot and has not occurred.
Binomialprobabilitydistributionisasetofprobabilitiesfordiscreteevents.Discreteeventsarethose
whose results or outcomes can be counted. Binomial probabilities are commonly encountered in
business situations e.g. in quality control activities when determining the probability of having a
certain number of defective items in a given consignment.
General rules of Multiplication – It states that of two events, A and B, the joint probability that
both events will happen is found by multiplying the probability that A will happen by the conditional
probability of event B happening
P(A and B) = P(A) x P(B/A)
Where P(B/A) stand for probability that B will occur given that A has already occurred.
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Chapter Quiz
1. The mean does not describe the amount of spread (variation) of a distribution. The
………… does this.
2. The probability of multiple occurrences at precisely same point is negligible
True
False
3. The ………………. curve is symmetrical and bell-shaped.
4. Which distribution is represented by the formula
n!
pr
qn − r
r!(n − r)
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Probability
Answers to Chapter Quiz
1. Variance
2. True
3. Normal distribution
4. Binomial distribution
Questions from previous exams
June 2000: Question 7
i) Define probability as used in Quantitative Techniques. (3 marks)
ii) What is Bayes Theorem? Explain how Bayes Theorem can be utilised practically
(5 marks)
iii) KK accounting firm has noticed that of the companies it audits, 85% show no inventory
shortages, 10% show small inventory shortages and 5% show large inventory
shortages. KK firm has devised a new accounting test for which it believes the following
probabilities hold:
P (company will pass test/no shortage) = 0.90
P (company will pass test/small shortage) = 0.50
P (company will pass test/large shortage) = 0.20
Required:
(i) Determine the probability if a company being audited fails this test has large or small
inventory shortage. (7 marks)
(ii) If a company being audited passes this test, what is the probability of no inventory
shortage? (5 marks)
(Total: 20 marks)
December 2001 Question 4
a) The past records of Salama Industries indicate that about 4 out of 10 of the company's
orders are for export. Further, their records indicate that 48 per cent of all orders are for
export in one particular financial quarter. They expect to satisfy about 80 orders in the
next financial quarter.
Required:
(i) Determine the probability that they will break their previous export record. (7 marks)
(ii) Explain why you have used the approach you have chosen to solve part (i) above
(2 marks)
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b) Gear Tyre Company has just developed a new steel-belted radial tyre that will be sold
through a national chain of discount stores. Because the tyre is a new product, the
company's management believes that the mileage guarantee offered with the tyre will be
an important factor in the consumer acceptance of the product. Before finalising the tyre
mileage guarantee policy, the actual road test with the tyres shows that the mean tyre
mileage is 36,500 kilometres and the standard deviation is 5,000 kilometres. In addition,
the data collected indicate that a normal distribution is a reasonable assumption.
Required:
(i) Gear Tyre Company will distribute the tyres if 20 per cent of the tyres manufactured can
be expected to last more than 40,000 kilometres. Should the company distribute the
tyres? (4 marks)
(ii) The company will provide a discount on a new set of tyres if the mileage on the original
tyres does not exceed the mileage stated on the guarantee.
What should the guarantee mileage be if the company wants no more that 10% of the
tyres to be eligible for the discount. (4 marks)
c) Explain briefly some of the advantages of the standard normal distribution. (3 marks)
(Total: 20 marks)
June 2002: Question 3
a) State clearly what is meant by two events being statistically independent. (2 marks)
b) In a certain factory which employs 500 men, 20% of all employees have a minor accident
in a given year. Of these, 30% had safety instructions whereas 80% of all employees
had no safety instructions.
Required:
Find the probability of an employee being accident-free given that he had:
(i) No safety instructions (5 marks)
(ii) Safety instructions (5 marks)
c) An electric utility company has found that the weekly number of occurrences of lightning
striking the transformers is a Poisson distribution with mean 0.4.
Required:
(i) The probability that no transformer will be struck in a week. (3 marks)
(ii) The probability that at most two transformers will be struck in a week. (5 marks)
(Total: 20 marks)
June 2003: Question 7
a) The J.R Muchemi Computer Company is considering a plant expansion that will enable
the company to begin production of a new computer product. The company's executive
director must determine whether to make the expansion a medium or large scale project.
Uncertainly exists in the demand for the new product, which for planning purposes
may be low, medium or high demand. The probability estimates for demand are 0.20,
0.50 and 0.30 respectively. The firm's accountants have developed the following annual
profit (in thousands of shillings) forecast for the medium and large scale expansion
projects:
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Probability
Medium
scale profit
Expansion
profitability
Large
scale profit
Expansion
profitability
Low 50 0.20 0 0.20
Demand Medium 150 0.50 100 0.50
High 200 0.30 300 0.30
Required:
(i) Which decision is preferred for the objective of maximising the expected profit
(4 marks)
(ii) Which decision is preferred for the objective of minimising the risk or uncertainty?
(4 marks)
From your answers in (i) and (ii) above, should the company go for medium scale expansion or
the large scale expansion? Explain. (2 marks)
b) A manufacturing firm based in Nairobi Kenya receives shipment of parts from two
different suppliers from UK and Japan. Currently, 65% of the parts purchased by the
company are from suppliers 1 (UK) and the remaining 35% are from supplier 2 (Japan).
The quality of the purchased parts varies with the source of supply. Historical data
suggest that the quality rating of the two suppliers are as shown below:
Percent of good parts Percent of bad
parts
Supplier 1 98 2
Supplier 2 95 5
The parts from the two suppliers are used in the firm's manufacturing process and during the
processing, a machine breaks down because it attempts to process a bad part.
Required:
(i) Given the information that a part is bad, determine the probability that it came from
Supplier 1 and it came from Supplier 2. (5 marks)
(ii) Show the above information in a probability tree. (5 marks)
(Total: 20 marks)
December 2004: Question 7
a) Indicate which of the following statements you agree with and which you disagree with and
defend your opinion.
(i) When performing a Bayesian decision analysis, the prior probabilities must inevitably
be subjective probabilities (3 marks)
(ii) The Bayes action will be the same regardless of whether it is selected using expected
monetary pay-offs or expected utilities. (3 marks)
(iii) For a given decision situation, the maximin criterion, on one hand, and the maximax
criterion on, on the other, will always point towards different actions.
(3 marks)
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c) In a flow-chart diagram, show the steps involved in a standard posterior analysis.
(6 marks)
With reference to probability theory, briefly but clearly, explain the statement "there is only
one thing certain and that is that nothing is certain" (5 marks)
(Total: 20 marks)
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Sampling and Estimation
CHAPTER FOUR
Sampling and Estimation
Objectives
At the end of this chapter, you should be able to:
Explain why in many situations a sample is the only feasible way to learn something••
about a population.
Explain the various methods of selecting a sample.••
Distinguish between probability sampling and non-probability sampling.••
Define and construct a sampling distribution of sample means.••
Explain the Central Limit theorem and its importance in statistical inference.••
Calculate confidence intervals for means and proportions.••
Determine how large a sample should be for both means and proportions.••
Fast Forward: Sampling is that part of statistical practice concerned with the selection of individual
observations intended to yield some knowledge about a population of concern, especially for the
purposes of statistical inference.
Introduction
A population consists of all the items with which a particular study is concerned. A sample is a
much smaller number chosen from this population. The sample must be chosen randomly. The
data collected in the sample is used to draw inferences about the corresponding population
parameter.
The three types of distributions:
1. Population distribution
Population distribution is the distribution of the individual values of population. Its mean is
denoted by "μ".
2. Sample distribution
It is the distribution of the individual values of a single sample. Its mean is generally written
as 'm.' It is extremely unlikely that it will be the same as "μ".
3 Distribution of sample means
A sample of size n is taken from the parent population and mean of the sample is calculated.
This is repeated for a number of samples so that a population of sample means is obtained.
This population approaches a normal distribution as n increases and is the distribution of
sample means.
Definition of key terms
1. Estimate – an approximate calculation of quantity or degree or worth; an estimate of what
it would cost; a rough idea how long it would take.
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2. Sample - a small part of something intended as representative of the whole. In statistics, a
sample is a subset of a population.
3. Probability – Probability is the likelihood or chance that something is the case or will happen.
Probability theory is used extensively in areas such as statistics, mathematics, science and
philosophy to draw conclusions about the likelihood of potential events and the underlying
mechanics of complex systems.
4. Proportion – The quotient obtained when the magnitude of a part is divided by the magnitude
of the whole; a quantity of something that is part of the whole amount or number.
5. Null hypothesis – describes some aspect of the statistical behaviour of a set of data.
This description is treated as valid unless the actual behaviour of the data contradicts this
assumption.
6. An alternative hypothesis is one that specifies that the null hypothesis is not true. The
alternative hypothesis is false when the null hypothesis is true, and true when the null
hypothesis is false. The symbol H1
is used for the alternative hypothesis.
Industry Context
With the realisation of the fact that in business time is money, dynamic technologies for
forecasting
have been a necessary toolin a wide range of managerial decisions.. In making strategic
decisions
under uncertainty, we all make forecasts. We may not think that we are forecasting, but our
choices will be directed by our anticipation of results of our actions or inactions.
Indecision and delays are the parents of failure. For instance, budgets are intended to help
managers and administrators do a better job of anticipating, and hence a better job of
managing
uncertainty, by using effective forecasting and other predictive techniques.
EXAM CONTEXT
Sampling and estimation has been a popular field for examiners. The student must understand
the formulae for the previous tests to avoid confusion during an exam. Previous exam papers
where the topic has featured are:
12/02, 6/06, 12/05, 6/05, 12/04, 6/04, 6/03, 12/02, 12/00, 6/00
Fast forward: Sampling is that part of statistical practice concerned with the selection of individual
observations intended to yield knowledge about a population of concerns, especially for the
purpose of statistical inference.
137.
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STUDYTEXT
Sampling and Estimation
4.1 Methods of probability sampling
There is no one 'best' method of selecting a probability sample from a population of interest. A
method used to select a sample of invoices in a file drawer might not be the most appropriate
method to use when choosing a national sample of voters. However, all probability sampling
methods have a similar goal, namely, to allow chance to determine the items or persons to be
included in the sample.
These sampling methods include:
1. Simple random sampling
A sample is formulated in such a manner that each item or person in the population has the same
chance of being included. For instance suppose a population consists of 576 employees of Yana
Tires. A sample of 63 employees is to be selected from that population. One way is to first write all
their names, put the names in a box, mix them thoroughly then make the first selection. Repeat
this process until the 63 employees are selected.
The other method which is convenient is to use the identification number of each employee and
a table of random numbers. As the name implies, these numbers have been generated by a
random process (in this case by a computer). Bias is therefore completely eliminated from the
selection process.
2. Systematic random sampling
The items or individuals of the population are arranged in some way- alphabetically, in a file
drawer by date received, or some other method. A random starting point is selected and then
every kth number of the population is selected for the sample. A systematic sample should not
be used, however, if there is a predetermined pattern to the population.
3. Stratified random sampling
A population is first divided into subgroups called strata and a sample is selected from each
stratum.After the population has been divided into strata, either a proportional or non-proportional
sample can be selected. A proportional sampling procedure requires that the number of items in
each stratum be in the same proportion as found in the population.
In a non-proportional stratified sample, the number of items studied in each stratum is
disproportionate to their number in the population. We then weight the sample results according
to the stratum's proportion of the total population.
4. Cluster sampling
It is often employed to reduce the cost of sampling a population scattered over a large geographic
area. Suppose you want to conduct a survey to determine the views of industrialists in a state.
Selecting a random sample of industrialists in the state and personally contacting each one
would be time-consuming and very expensive. Cluster sampling would be useful by subdividing
the state into small units of regions often called primary units.
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STUDYTEXT
Standard error of the mean.
The standard deviation of the sample mean from the overall mean μ is called the standard error
(Se
)
For large samples, Se
=
s
√ n
Where s is the standard deviation of the population and n is the size of the population.
Note: In general, the standard deviation of the population is not known. In such cases, the
standard deviation of the sample (provided it is large) is a good estimate of the population
standard deviation (s ).
The standard error of the mean then becomes
Standard DeviationOf Sample
√ n
Proportions
There are times when information cannot be given as a mean or as a measure but only as a
fraction or percentage.
Examples:
Percentage of female in a certain population.
Proportion of defective production in total production.
In these cases, we are faced with estimating the population proportion from a single sample.
The sampling theory states that if repeated large random samples are taken from a population,
the sample proportion 'p' will be normally distributed with mean equal to the population proportion
and standard error equal to
p(1 − p)
√ n
where n is sample size.
The procedure for estimating a proportion is similar to that for estimating a mean.
The Sampling distribution of sample proportions
Population proportions are used in business particularly in market research, where we might
investigate the proportions of populations displaying a particular characteristic.
Provided p, (1-p) are both greater than 5, we may use
p(1 − p)
√ n
to represent the standard error of the sampling distribution of sample proportions
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Sampling and Estimation
Confident intervals
Usually we use a single sample, to produce an estimate of population parameter. It is important
to know how reliable this estimate is based on sample results.
The standard error of the sampling distribution (or proportions) will give us an indication of
reliability – the smaller the standard error, the less variable the sample statistic.
However it is simpler to appreciate the reliability of an estimate if we set up a range of values
within which we can be reasonably sure that the population parameters lie. This range of values
is called confidence interval.
Confidence limits
Confidence limits are the outer limits to a confidence interval. This is a zone of values within
which we may be confident that the true population mean (or the parameter being considered)
does lie.
Example: the 95% confidence interval for the population proportion is:
p ± 1.96 p(1 − p)
√ n
Note: It is usual to write 1-p = q
Test of significance
In practice, sizes of population parameters such as mean and proportions, are generally unknown.
However a claim may be made that the size of a given parameter is equal to 'x' (say). Such a
claim may be true or false. There is need for such a claim to be put to test.
To test this claim we take a sample. It will be a miracle if the sample taken from the population
will have the same mean or standard deviation as the corresponding population parameters. The
difference could be:
a) because the original belief was wrong, or
b) because the difference was purely due to ordinary chance.
If the difference cannot be explained solely due to ordinary chance, then the difference is said to
be statistically significant.
4.2 The Null hypothesis
Fast Forward
A common convention is to use the symbol H0
to denote the null hypothesis.
It is an assumption that nothing has changed i.e. what we have assumed to be true (H0
), is actually
true. In other words, there is no contradiction between the believed mean and the sampled mean
and the difference is solely due to chance.
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Testing the null hypothesis
We know that 95% of the means of all samples, will fall within 1.96 standard errors of the
population mean (or believed mean). If the sample mean lies more than 1.96 standard error from
the believed mean, one can reject the null hypothesis at the 95% level of confidence and say
that there is a contradiction between the believed population mean and the sample. If the null
hypothesis is not rejected, then we say there is not enough evidence to prove that the true mean
is not as we believe.
Sampling theory for small samples (t-Distribution)
In applying sampling theories to small samples (n ≤30), we assume:
a) the parent population is normal or near normal
b) In this case the sample means do not follow a normal distribution.
This distribution is similar to the standard normal distributions in that, it is symmetrical and
continuous. The major difference between the two is that in the case of normal distribution there
is only one defined distribution; there are however many t-distributions depending upon the
sample size.
The value of t is obtained from the t- tables depending upon the degrees of freedom and the level
of confidence.
Type I and Type II Errors
If we reject a hypothesis when it should be accepted, we say that a type I error has been made.
If on the other hand, we accept a hypothesis when it should be rejected, we say that a type II
error has been made.
In either case a wrong decision or error in judgment has occurred. For a given sample size, an
attempt to decrease one type of error is accompanied in general by an increase in another type
of error.
While testing hypothesis (H0
) and deciding to either accept or reject a null hypothesis, there are
four possible occurrences.
a) Acceptance of a true hypothesis (correct decision) – accepting the null hypothesis
and it happens to be the correct decision. Note that statistics does not give absolute
information, thus its conclusion could be wrong only that the probability of it being right
are high.
b) Rejection of a false hypothesis (correct decision).
c) Rejection of a true hypothesis – (incorrect decision) – this is called type I error, with
probability = α.
d) Acceptance of a false hypothesis – (incorrect decision) – this is called type II error, with
probability = β.
In practice, type I error is considered more serious than the type II errors. The maximum probability
with which we would be willing to risk a type I error is called the level of significance of the test.
141.
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Sampling and Estimation
One-tailed and two –tailed test
When we are interested in extreme values of the statistics S or its corresponding Z score on
both sides of the statistic, we perform a test called a two-tailed test (or two-sided test). Similarly
we may be interested only in extreme values to one side of the mean, such tests are called one-
tailed tests (or one-sided tests)
Levels of significance
A level of significance is a probability value which is used when conducting tests of hypothesis.
A level of significance is basically the probability of one making an incorrect decision after the
statistical testing has been done. Usually such probability used are very small e.g. 1% or 5%
NB: If the standardised value of the mean is less than –1.65 we reject the null hypothesis (H0
)
and accept the alternative hypothesis (H1
) but if the standardised value of the mean is more than
–1.65 we accept the null hypothesis and reject the alternative hypothesis
The above sketch graph and level of significance are applicable when the sample mean is less
than the population mean)
The following is used when sample mean > population mean
0.45
0
5% = 0.05
CriticalRegion
Critical Value = −1.65
0.5000
0.4900
1% provision for errors
0
Critical Value
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STUDYTEXT
NB: If the sample mean standardized value < 1.65, we accept the null hypothesis but reject
the alternative. If the sample mean value > 1.65 we reject the null hypothesis and accept the
alternative hypothesis
The above sketch is normally used when the sample mean given is greater than the population
mean
NB: If the standardized value of the sample mean is between –2.58 and +2.58 accept the null
hypothesis but otherwise reject it and therefore accept the alternative hypothesis
Two-tailed tests
Atwo-tailed test is normally used in statistical work (tests of significance) e.g. if a complaint lodged
by the client is about a product not meeting certain specifications i.e. the item will generate a
complaint if its measurements are below the lower tolerance limit or above the upper tolerance
limit
Acceptance region
Critical region (rejection region)
5% = 0.05
0 Z = 1.65 (critical value)
Accept null hyp (reject alternative hyp)
Reject null hyp (accept alt hyp) Reject null hyp (accept alt hyp)
0.05% = 0.05 0.495 0.495 0.5% = 0.05
-2.58 +2.58
143.
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STUDYTEXT
Sampling and Estimation
NB: Alternative hypothesis is usually rejected if the standardized value of the sample mean lies
beyond the tolerance limits (15cm and 17 ½ cm).
One-tailed test
This is a test where the alternative hypothesis (H1
:) is only concerned with one of the tails of the
distribution e.g. to test a business complaint if the complaint is above the measurements of an
item being shorter than is required.
E.g. a manufacturer of a given brand of bread may state that the average weight of the bread is
500gms but if a consumer takes a sample and weighs each of the pieces of bread and happens
to have a mean of 450 gms he will definitely complain about the bread which is underweight. The
statistical analysis to be done will concentrate on the left tail of the normal distribution in which
one will have to establish whether 450gms being less than 500g is statistically significant. Such
a test is referred to as one-tailed test.
On the other hand, the test may tend towards the right hand tail of the normal distribution. When
this happens, the major complaint is likely to do with oversize items bought. The test is known as
one-tailed as the focus is on one end of the normal distribution.
Region of acceptance for
HO
Critical RegionCritical Region
15cm 17 ½ cm
Left
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STUDYTEXT
Number of standard errors
Two-tailed
test
One-tailed
test
5% level of
significance
1.96 1.65
1% level of
significance
2.58 2.33
Hypothesis Testing Procedure
Whenever a business complaint comes up there is a recommended procedure for conducting a
statistical test. The purpose of such a test is to establish whether the null hypothesis or alternative
hypothesis is to be accepted.
The following are steps normally adopted:
Statement of the null and alternative hypothesis1.
Statement of the level of significance to be used.2.
Statement about the test statistic i.e. what is to be tested e.g. the sample mean, sample3.
proportion, difference between sample means or sample proportions
Type of test whether two tailed or one tailed.4.
Statement on critical values using the appropriate level of significance5.
Standardising the test statistic6.
Conclusion showing whether to accept or reject the null hypothesis7.
The F-Distribution (The Variance Ratio Test)
The F-test is based on the ratio of the variances of two samples.
The F-distribution has the following shape
Diagram 4.1
Critcal Piont F - Value
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STUDYTEXT
Sampling and Estimation
The null-hypothesis is that there is no significant difference between the variances of the two
samples.
How to use the F-test
Step I: Calculate the variance for each sample and use these to estimate the population
variance.
Step II: Obtain the F-value which is
Larger variance
Smaller variance
Step III: Find F – value from F-tables using degrees of freedom and level of confidence.
4.3 THE CHI –SQUARE DISTRIBUTION X2
This distribution can be used to test whether an observed series of values differs significantly
from what is expected.
Formula:
x2
= ∑
(observed value − expected value)2
= ∑
(o − E)2
expected value E
Testing the difference between two sample means (large samples)
A large sample is defined as one which contains 30 or more items (n≥30 where n is the sample
size)
In a business those involved are constantly observant about the standards or specifications
of the item which they sell e.g. a trader may receive a batch of items at one time and another
batch at a later time. At the end, he may have concluded that the two samples are different in
certain specifications e.g. mean weight, mean lifespan, mean length e.t.c. Further it may become
necessary to establish whether the observed differences are statistically significant or not. If the
differences are statistically significant then it means that such differences must be explained
i.e. there are known causes but if they are not statistically significant then it means that the
differences observed have no known causes and are mainly due to chance.
If the differences are established to be statistically significant then it implies that the complaints,
which necessitated that kind of test, are justified.
Let X1
and X2
be any two samples whose sizes are n1
and n2
and mean X
1
and X
2
. Standard
deviation S1
and S2
respectively. In order to test the difference between the two sample means,
we apply the following formulas
Z =
X1
− X2
where = S(X1
− X2
=
S1
2
+ S2
2
S (X1
− X2
) √ n1
n2
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STUDYTEXT
Example 1
An agronomist was interested in a particular fertilizer yield output. He planted maize on 50 equal
pieces of land and the mean harvest obtained later was 60 bags per plot with a standard deviation
of 1.5 bags. The crops grew under natural circumstances and conditions without the soil being
treated with any fertilizer. The same agronomist carried out an alternative experiment where he
picked 60 plots in the same area and planted the same plant of maize but a fertilizer was applied
on these plots. After the harvest it was established that the mean harvest was 63 bags per plot
with a standard deviation of 1.3 bags
Required
Conduct a statistical test in order to establish whether there was a significant difference between
the mean harvest under the two types of field conditions. Use 5% level of significance.
Solution
H0
: µ1
= µ2
H1
: µ1
≠ µ2
Critical values of the two-tailed test at 5% level of significance are 1.96
The standardised value of the difference between sample means is given by Z where
Z =
X1
− X2
where = S(X1
− X2
=
1.52
+ 1.32
S (X1
− X2
) √ 50
60
Z =
(60 − 63)
√ 50
60
= 11.11
Since 11.11 < -1.96, we reject the null hypothesis but accept the alternative hypothesis at 5%
level of significance i.e. the difference between the sample mean harvest is statistically significant.
This implies that the fertilizer had a positive effect on the harvest of maize
Note: You don't have to illustrate your solution with a diagram.
−− −
− −
- 1.96 0 +1.96
147.
141
STUDYTEXT
Sampling and Estimation
Example 2
An observation was made about reading abilities of males and females. The observation led
to a conclusion that females are faster readers than males. The observation was based on the
times taken by both females and males when reading out a list of names during graduation
ceremonies.
In order to investigate the observation and the consequent conclusion a sample of 200 men
were given lists to read. On average, each man took 63 seconds with a standard deviation of 4
seconds. A sample of 250 women was also taken and asked to read the same list of names. It
was found that they took 62 seconds on average with a standard deviation of 1 second.
Required
By conducting a statistical hypothesis testing at 1% level of significance establish whether the
sample data obtained supports the earlier observation.
Solution
H0
: µ1
= µ2
H1
: µ1
≠ µ2
Critical values of the two tailed test is at 1% level of significance is 2.58.
Z =
X1
− X2
S(X1
−X2
)
63 − 62
Z = 42
+
12
√ 200 250
Since 3.45 > 2.33 reject the null hypothesis but accept the alternative hypothesis at 1% level of
significance i.e. there is a significant difference between the reading speed of Males and females,
thus females are actually faster readers.
Acceptance region
Rejection Region
- 2.58 0 +2.58 +3.45
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STUDYTEXT
Test of hypothesis on proportions
This follows a similar method to the one for means except that the standard error used in this
case is:
Sp =
Pq
√ n
Z score is calculated as, Z =
P − �
Where P = Proportion found in the sample.
Sp
Π – the hypothetical proportion.
Example
Amember of parliament (MP) claims that in his constituency only 50% of the total youth population
lacks university education. A local media company wanted to ascertain that claim thus they
conducted a survey taking a sample of 400 youths, of these 54% lacked university education.
Required:
At 5% level of significance, confirm if the MP's claim is wrong.
Solution
Note: This is a two-tailed test since we wish to test that the hypothesis is different (≠) and not
against a specific alternative hypothesis e.g. less than or more than.
H0
: π = 50% of all youth in the constituency lack university education.
H1
: π ≠ 50% of all youth in the constituency lack university education.
Sp =
pq
=
0.5 × 0.5
= 0.025
√ n √ 400
Z =
0.54 − 0.50
= 1.6
0.025
at 5% level of significance for a two-tailed test the critical value is 1.96 since calculated Z value
< tabulated value (1.96).
i.e. 1.6 < 1.96 we accept the null hypothesis.
Thus the MP's claim is accurate.
Hypothesis testing of the difference between proportions
Example
Ken industrial manufacturers have produced a perfume known as "fianchetto." In order to test
its popularity in the market, the manufacturer carried a random survey in Back rank city where
10,000 consumers were interviewed after which 7,200 showed preference. The manufacturer
also moved to area Rook town where he interviewed 12,000 consumers out of which 10,000
showed preference for the product.
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STUDYTEXT
( )
( ) ( )
1 2
0.78 0.22 0.78 0.22
10,000 12,000
S P P- = +
= 0.00894
Z =
0.72 0.83
0.00894
-
= 12.3
Since 12.3 > 1.96, we reject the null hypothesis but accept the alternative. The differences
between the proportions are statistically significant. This implies that the perfume is much more
popular in Rook town than in Back rank city.
Hypothesis testing about the difference between two proportions
Is used to test the difference between the proportions of a given attribute found in two random
samples.
The null hypothesis is that there is no difference between the population proportions. It means
two samples are from the same population.
Hence
H0
: π1
= π2
The best estimate of the standard error of the difference of P1 and P2 is given by pooling the
samples and finding the pooled sample proportions (P) thus
P = 1 1 2 2
1 2
p n p n
n n
+
+
Standard error of difference between proportions
( )1 2
1 2
pq pq
S p p
n n
- = +
And Z =
( )
1 2
1 2
P P
S p p
-
-
Example
In a random sample of 100 persons taken from village A, 60 are found to be consuming tea. In
another sample of 200 persons taken from a village B, 100 persons are found to be consuming
tea. Do the data reveal significant difference between the two villages so far as the habit of taking
tea is concerned?
Solution
Let us take the hypothesis that there is no significant difference between the two villages as far
as the habit of taking tea is concerned i.e. π1
= π2
We are given
P1
= 0.6; n1
= 100
P2
= 0.5; n2
= 200
151.
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STUDYTEXT
Sampling and Estimation
Appropriate statistic to be used here is given by
P = 1 1 2 2
1 2
p n p n
n n
+
+
=
( )( ) ( )( )0.6 100 0.5 200 60 100
100 200 300
+ +
=
+
= 0.53
q = 1 – 0.53
= 0.47
( )1 2S P P-
=
1 2
pq pq
n n
+
=
( )( ) ( )( )0.53 0.47 0.53 0.47
100 200
+
= 0.0608
Z =
0.6 0.5
0.0608
-
= 1.64
Since the computed value of Z is less than the critical value of Z = 1.96 at 5% level of significance;
therefore we accept the hypothesis and conclude that there is no significant difference in the
habit of taking tea in the two villages A and B
t-distribution (student's t distribution) tests of hypothesis (test for small samples
n < 30)
For small samples n < 30, the method used in hypothesis testing is similar to the one for large
samples except that t values are used from t distribution at a given degree of freedom v, instead
of z score, the standard error Se statistic used is also different.
Note that v = n – 1 for a single sample and n1
+ n2
– 2 where two samples are involved.
a) Test of hypothesis about the population mean
When the population standard deviation (S) is known then the t statistic is defined as
t =
X
X
S
m-
where X
S
S
n
=
Follows the students t distribution with (n-1) d.f. where:
X = Sample mean
μ = Hypothesis population mean
n = sample size
and S is the standard deviation of the sample calculated by the formula:
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quantitative techniques14 6
STUDYTEXT
S =
( )
2
1
X X
n
-
-
∑ for n < 30
If the calculated value of t exceeds the table value of t at a specified level of significance, the null
hypothesis is rejected.
Example
Ten oil tins are taken at random from an automatic filling machine. The mean weight of the tins is
15.8 kg and the standard deviation is 0.5kg. Does the sample mean differ significantly from the
intended weight of 16kgs. Use 5% level of significance.
Solution
Given that n = 10; x = 15.8; S = 0.50; μ = 16; v = 9
H0
: μ = 16
H1
: μ ≠ 16
=
0.5
10
X
S =
t = 0.5
10
15.8 16-
=
0.2
0.16
= -1.25
The table value for t for 9 d.f. at 5% level of significance is 2.26. The computed value of t is
smaller than the table value of t. therefore, difference is insignificant and the null hypothesis is
accepted.
b) Test of hypothesis about the difference between two means
The t test can be used under two assumptions when testing hypothesis concerning the difference
between the two means; that the two are normally distributed (or near normally distributed)
populations and that the standard deviation of the two is the same or at any rate not significantly
different.
Appropriate test statistic to be used is
t =
( )21
1 2
X X
X X
S -
-
at (n1
+ n2
– 2) d.f.
The standard deviation is obtained by pooling the two sample standard deviation as shown
below.
Sp
=
( ) ( )2 2
1 1 2 2
1 2
1 1
2
n S n S
n n
- + -
+ -
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STUDYTEXT
Sampling and Estimation
Where: 1X =170, 2X = 205, n1
= 20, n2
= 18, S1
= 20, S2
= 25, V = 36
Sp
=
( )( ) ( )( )2 2
19 20 17 25
20 18 2
+
+ -
= 22.5
( )1 2
38
22.5
360X X
S -
=
= 7.31
t =
170 205
7.31
-
= 4.79
t0.05
(36) = 1.9 (Since d.f > 30 we use the normal tables)
The table value of t at 5% level of significance for 36 d.f. when d.f. >30, that t distribution is the
same as normal distribution, 1.9. Since the computed value of t is more than the table value, we
reject the null hypothesis. Thus, we conclude that there is significant difference in the average
sales between the two salesmen.
Testing the hypothesis equality of two variances
The test for equality of two population variances is based on the variances in two independently
selected random samples drawn from two normal populations
Under the null hypothesis 2
2
2
1 óó =
F =
2
2
2
2
2
1
2
1
ó
s
ó
s
Now under the H0
: 2
2
2
1 óó = it follows that
F =
2
1
2
2
S
S
which is the test statistic.
Which follows F – distribution with V1
and V2
degrees of freedom. The larger sample variance is
placed in the numerator and the smaller one in the denominator.
If the computed value of F exceeds the table value of F, we reject the null hypothesis i.e. the
alternate hypothesis is accepted
Example
In one sample of observations the sum of the squares of the deviations of the sample values from
sample mean was 120 and in the other sample of 12 observations it was 314; test whether the
difference is significant at 5% level of significance
157.
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STUDYTEXT
Sampling and Estimation
The computed value of χ2
is compared with that of tabulated χ2
for a given significance level and
degrees of freedom.
i) Test for goodness of fit
These tests are used when we want to determine whether an actual sample distribution matches
a known theoretical distribution
The null hypothesis usually states that the sample is drawn from the theoretical population
distribution and the alternate hypothesis usually states that it is not.
Example
Mr Nguku carried out a survey of 320 families in Ateka district, each family had 5 children and
they revealed the following distribution
No. of boys 5 4 3 2 1 0
No. of girls 0 1 2 3 4 5
No. of families 14 56 110 88 40 12
Is the result consistent with the hypothesis that male and female births are equally probable at
5% level of significance?
Solution
If the distribution of gender is equally probable then the distribution conforms to a binomial
distribution with probability P(X) = ½.
Therefore
H0
= the observed number of boys conforms to a binomial distribution with P = ½
H1
= the observations do not conform to a binomial distribution.
On the assumption that male and female births are equally probable, the probability of a male
birth is P = ½ . The expected number of families can be calculated by the use of binomial
distribution. The probability of male births in a family of 5 is given by
P(x) = 5
cX
Px
q5-x
(for x = 0, 1, 2, 3, 4, 5,)
= 5
cX
( ½ )5
(Since P = q = ½ )
To get the expected frequencies, multiply P(x) by the total number N = 320. The calculations are
159.
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Sampling and Estimation
ii) Test of independence of attributes
This test discloses whether there is any association or relationship between two or more attributes.
The following steps are required to perform the test of hypothesis.
1. The null and alternative hypothesis are set as follows
H0
: No association exists between the attributes
H1
: An association exists between the attributes
2. Under H0
an expected frequency E corresponding to each cell in the contingency table is
found by using the formula
E =
R C
n
×
Where R = a row total, C = a column total and n = sample size
3. Based upon the observed values and corresponding expected frequencies the χ2
statistic is
obtained using the formular
χ2
=
( )
2
O E
E
-
∑
4. The characteristics of this distribution are defined by the number of degrees of freedom
(d.f.) which is given by
d.f. = (r-1) (c-1),
Where r is the number of rows and c is number of columns corresponding to a chosen level
of significance, the critical value is found from the chi squared table
5. The calculated value of χ2
is compared with the tabulated value χ2
for (r-1) (c-1) degrees
of freedom at a certain level of significance. If the computed value of χ2
is greater than the
tabulated value, the null hypothesis of independence is rejected. Otherwise we accept it.
Example
In a sample of 200 people where a particular device was selected, 100 were given a drug and
the others were not given any drug. The results are as follows
Drug No drug Total
Cured 65 55 120
Not cured 35 45 80
Total 100 100 200
Test whether the drug will be effective or not, at 5% level of significance.
Solution
Let us take the null hypothesis that the drug is not effective in curing the disease.
Applying the χ2
test
The expected cell frequencies are computed as follows
E11
=
1 1R C
n =
120 100
200
×
= 60
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Sampling and Estimation
The analytical procedure is the same as that given for the test of independence.
Example
A random sample of 400 persons was selected from each of three age groups and each person
was asked to specify which types of TV programmes they preferred. The results are shown in
the following table
Type of programme
Age group A B C Total
Under 30 120 30 50 200
30 – 44 10 75 15 100
45 and above 10 30 60 100
Total 140 135 125 400
Test the hypothesis that the populations are homogenous with respect to the types of television
programmes they prefer, at 5% level of significance.
Solution
Let us take the hypothesis that the populations are homogenous with respect to different types
of television programmes they prefer
Applying χ2
test
O E (O – E) 2
(O – E) 2
/E
120 70.00 2500.00 35.7143
10 35.00 625.00 17.8571
10 35.00 625.00 17.8571
30 67.50 1406.25 20.8333
75 33.75 1701.56 50.4166
30 33.75 14.06 0.4166
50 62.50 156.25 2.500
15 31.25 264.06 8.4499
60 31.25 826.56 26.449
Σ(O – E) 2
/E = 180.4948
χ2
=
( )
2
O E
E
-
∑
The table value of χ2
for 4 d.f. at 5% level of significance is 9.488.
The calculated value of χ2
is greater than the table value. We reject the hypothesis and conclude
that the populations are not homogenous with respect to the type of TV programmes preferred,
thus the different age groups vary in choice of TV programmes.
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Chapter Summary
Methods of sampling
a. Random or probability sampling methods
These include
Simple random sampling
Stratified sampling
Systematic sampling
Multi stage sampling
b. Non random probability sampling methods
These consist of
Judgment sampling
Quota sampling
Cluster sampling
- A hypothesis is a claim or an opinion about an item or issue. Therefore it has to be tested
statistically in order to establish whether it is correct
- When testing a hypothesis, one must fully understand the 2 basic hypothesis to be tested
namely
The null hypothesis (H0
)
The alternative hypothesis(H1
)
Standard hypothesis tests
Normal test
Test a sample mean ( X ) against a population mean (µ) (where samples size n > 30 and population
variance σ2
is known) and sample proportion, P (where sample size np >5 and nq >5 since in this
case the normal distribution can be used to approximate the binomial distribution).
t test
Tests a sample mean ( X ) against a population mean, especially where the population variance
is unknown and n < 30.
Variance ratio test or f test
It is used to compare population variances with samples of any size drawn from normal
populations.
Chi squared test
It can be used to test the association between attributes or the goodness of fit of an observed
frequency distribution to a standard distribution
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Sampling and Estimation
Chapter Quiz
1. Which is the odd one out?
(a) Simple random sampling
(b) Stratified sampling
(c) Systematic sampling
(d) Continuous sampling
(e) Multi stage sampling
2. If the difference cannot be explained solely due to ordinary chance, then the difference
is said to be ____________
3 __________ is an assumption that nothing has changed.
4. What is the formula for T- Distribution?
5. Which one is not in this category?
a) Judgment sampling
b) Quota sampling
c) Systematic sampling
d) Cluster sampling
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Answers to chapter quiz
1. (d) Continuous sampling
2. Statistically significant.
3. Null Hypothesis
4. t =
x − µ
σ
√ n
5. (c) Systematic Sampling
questions from previous exams
December 2000 Question 4
a) Briefly explain each of the following distributions indicating whether it is a discrete or a
continuous distribution.
(i) Binomial distribution. (3 marks)
(ii) Poisson distribution (3 marks)
(iii) Normal distribution (3 marks)
(iv) Chi-square distribution (3 marks)
(v) Fisher (F) distribution (3 marks)
b) Give one example in the accounting profession where each of the above distributions
can be applied. (5 marks)
(Total: 20 marks)
December 2002 Question 2
a) Transparency and Certified Public Accountants (CPAs) have been appointed to
audit accounts of Health National Hospital. Due to the large number of accounts, the
consultants have decided to audit a random sample of the accounts.
Required
(i) State the advantages of auditing a sample of the accounts. (3 marks)
(ii) Describe briefly the sampling technique you would recommend. (5 marks)
(iii) What are the advantages and disadvantages of the sampling technique recommended
in (ii) above? (4 marks)
b) State and briefly explain the qualities of a good point estimator. (8 marks)
(Total: 20 marks)
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Sampling and Estimation
June 2003 Question 4
a) Nation Standard Newspaper poll for the year 2002 presidential campaign in Kenya
sampled 491 potential voters in October 2002. A primary purpose of the poll was to
obtain an estimate of the promotion of potential voters who favour each candidate.
Close to December 2002 elections, better precision and smaller margins of error were
desired. Assume a planning value for the population proportion of p = 0.50 and that a
95% confidence level is desired.
Required:
Determine the recommended sample size for each of the following surveys:
Survey Margin of error
Early December 0.02
Pre-election day 0.01
(8 marks)
b) Future Computer Company has developed a new computer accounting software
package to help accountancy analysts reduce the time required to design, develop
and implement an accounting system. To evaluate the benefits of the software
package, a random sample of 24 accountancy analysts is selected. Each analyst is
given specifications for a hypothetical accounting system. Then 12 of the analysts are
instructed to produce the accounting system by using the current technology. The other
12 analysts are trained in the use of the software package and then instructed to use it
to produce the accounting system. The 24 analysts complete the study and the results
are shown below:
Completion Time and Summary Statistics for the Software Testing Study
Current technology New technology
300 276
280 222
344 310
385 338
372 200
360 302
288 317
321 260
376 320
290 312
301 334
283 265
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Regression, Time Series and Forecasting
CHAPTER FIVE
Regression, Time Series and Forecasting
Objectives
At the end of this chapter, you should be able to:
Establish relationship between two or more variables.••
Understand a particular situation, explain it and then analyse it.••
Discuss assumptions underlying analysis of the linear regression model.••
To build a model qualitatively about which factors are likely to influence the dependent••
variable.
Fast Forward:Time series forecasting is the use of a model to forecast future events based on
known past events to forecast future data points before they are measured.
Introduction
All businesses have to plan their future activities, both short and long term. Managers will have
to make forecasts of the future values of important variables such as sales, interest rates, costs
etc. In this chapter, we will look at ways of using past information to make these forecasts.
For example, we may wish to explain the variability in sales by looking at the ways in which they
have changed with time, ignoring other factors. If we can explain the past pattern, we can use it
to forecast future values. A data set, in which the independent variable is time, is referred to as
time series.
Care is required since the historical pattern is not always relevant for particular forecasts. A
company may deliberately plan to change its pattern if, for example it has been making a loss.
There may be large external factors which completely modify the pattern. There may be a major
change in raw material prices, world inflation may suddenly increase or a natural disaster may
affect the business unpredictably.
In this section we begin by looking at time series which contain components such as trend,
seasonal variation and cyclical variation. The components can be combined in a number of
ways. We will look at two specific models: the additive components model and the multiplicative
component model. As the names imply, the components are added or multiplied, respectively.
For each of these models, there are different ways of calculating the trend component. We will
use a combination of moving averages and linear regression.
You should note that the techniques described in this chapter are not the only ones, nor necessarily
the best, forecasting methods for any particular forecasting situation. There are many more
sophisticated statistical techniques; there are qualitative methods which must be used when
there is little or no past data. The Delphi technique and the scenario writing method are examples
of these.
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Definitions of key terms
1. Time series - A time series is a sequence of data points, measured typically at
successive times, spaced at (often uniform) time intervals.
2. Trend – This is a pattern manifested by collected information over a period of time
oftenly used to predict future events. It could also be used to estimate uncertain events
in the past.
3. Forecast - Forecasting is the process of estimation in unknown situations. Prediction
is a similar, but more general term. Both can refer to estimation of time series, cross-
sectional or longitudinal data. Usage can differ between areas of application: for
example in hydrology, the terms "forecast" and "forecasting" are sometimes reserved
for estimates of values at certain specific future times.
Industry context
All businesses have to plan their future activities. When making both short and long term plans,
managers will have to make forecasts of the future values of important variables such as sales,
interest rates, costs, etc. Care should be taken since historical pattern is not always relevant for
particular forecasts. A company may deliberately plan to change its pattern if it has been making
a loss.
Exam context
Time is money. Students should clearly have an indepth understanding of time
series. The topic has been previously examined as follows:
12/06, 6/06, 12/05, 6/05, 12/04, 6/04, 12/03, 6/03, 6/02, 12/01, 6/01, 12/00, 6/00
5.1 Linear Regression
Introduction
Consider a company which regularly places advertisements for one of its products in a local
newspaper. The company keeps records, on a monthly basis, of the amount of money spent on
advertising and the corresponding sales of this product. If advertising is effective at all, then we
can see intuitively that there is likely to be a relationship between the amount of money spent on
advertising and the corresponding monthly sales. We would expect that the larger the sum spent
on advertising, the greater the sales, at least within certain limits. There are a number of factors
which will work together to determine the exact value of sales each month such as the price of
a competitor's product, perhaps the time of the year, or the weather conditions, Nevertheless, if
the expenditure on advertising is thought to be a major factor in determining sales, knowledge of
the relationship between the two variables would be of greater use in the estimation of sales and
related budgeting and planning activities.
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Regression, Time Series and Forecasting
The term association is used to refer to the relationship between variables. For the purposes of
the statistical analysis two aspects of the problem are defined. The term regression is used to
describe the nature of the relationship, while the term correlation is used to describe the strength
of the relationship.
We need to know, for example, whether the monthly advertising expenditure is strongly related to
the monthly sales and therefore will provide a reliable estimate of sales, or whether the relationship
is weak and will provide a general indicator only.
The general procedure in the analysis of the relationship between variables is to use a sample
of corresponding values of the variables to establish the nature of the relationship. We can then
develop a mathematical equation, or model, to describe this relationship.
From the mathematical point of view, linear equations are the simplest to set up and analyse.
Consequently, unless a linear relationship is clearly out of the question, we would normally try to
describe the relationship between the variables by means of a linear model. This procedure is
called Linear Regression.
A measure of the fit of the linear model to the data is an indicator of the strength of the linear
relationship between the variables and, hence, the reliability of any estimates made using this
model. For Example:
Diagram 5.1
A linear relationship
This graph indicates that a linear regression model will be a suitable way of assesing
the relationship between sales and advertising expenditure.
Diagram 5.2
A non-Linear relationship
Sales/month
Advertising expenditure/Month
********
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STUDYTEXT
This graph indicates that a linear model would not be suitable in describing the relationship
between sales and advertising expenditure.
Linear regression is our first example of the use of mathematical models. The purpose of a model
is to help us to understand a particular situation, possibly to explain it and then analyse it. We
may use the model to make forecasts or expeditions. A model is usually a simplification of the
real situation. We have to make assumptions so that we can construct a manageable model.
Models range from the simple twovariable type to the complicated multivariable models. These
models are widely used because there are many easy-to-use computer programmes available to
carry out the required calculations.
This chapter will cover the steps in the analysis of a simple linear regression model. We will take
one sample of data and use it to illustrate each of the steps.
This chapter concludes with sections on multiple regression models, examples of non-linear
relationships and finally a non-parametric measure of correlation, Spearman's rank correlation
coefficient.
Simple linear Regression Model
We are interested in whether there is any linear relationship between the two variables. For
example, we may be interested in the heights and weights of a number of people, the price
and quantity of a product sold, employees age and salary, chicken's age and weight, weekly
departmental costs and hours worked or distance traveled and the time taken.
As an example, let us say that a poultry farmer wishes to predict the weights of the chicken he
is rearing. Weight is the variable which we wish to predict, therefore weight is the dependent
variable. We will plot the dependent variable on the y axis. It is suggested to the farmer that
weight depends on the chicken's age.Age is said to be an independent variable. The independent
variable will be plotted on the x-axis. If we can establish the nature of the relationship between
the age and the weight of the chickens, then we can predict the weight of a chicken by looking
at the age.
Sales/month
Advertising expenditure/Month
****
*
*
*
*
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Regression, Time Series and Forecasting
Setting up of a simple linear regression model: Illustration
We are running a special delivery service for short distances in a city. We wish to cost the service
and to do this we must estimate the time for deliveries of any given distance.
There are factors, other than the distance travelled, which will affect the times taken –traffic
congestion, time of day, road works, the weather,, the road system, the driver, his transport.
However the initial investigation will be as simple as possible. We consider distance only,
measured as the shortest practical route in miles and the time taken in minutes.
The relevant population is all of the possible journeys, with their times, which could be made
in the city. It is an infinite population and we require a random sample from this population. For
simplicity, let us use a systematic sample design for this preliminary sample. We will measure the
time and distance for every tenth journey starting from randomly selected day and a randomly
selected hour of next week. The firm works a 6-day week, excluding Sundays. The random
number, chosen by throwing a dice is 2, so next Tuesday is the chosen day. The service runs
from 8am to 6pm. A random number, between 0-9 is chosen from random number tables to
select the starting time. The number chosen is 6, so the first journey chosen is the first one after
1pm (i.e. this is the sixth hour, beginning at 8am), then we take every tenth delivery after that.
The sample data for the first 10 deliveries will be used for the analysis.
Table 5.1 Sample data for delivery distances and times
Distance, mile Time, Minutes
2.5 16
3.4 13
1.9 19
t.2 18
3.0 12
α.3 11
3.0 8
3.0 14
1.5 9
4.1 16
We wish to explain variations in the time taken, the dependent variable (y) by introducing distance
as the independent variable(x). Generally we would expect the time taken to increase as distance
increases. If this data was plotted it would not appear on a straight line. It also looks through
the plotted points cluster around a straight line. This means that we could use a linear model to
describe the relationship between these two variables. The points are not exactly on a line. It
would be surprising if they were, in view of all the other factors which we know can affect journey
time. A linear model will be an approximation only to the true relationship between journey time
and distance, but the evidence of the plot is that is the best available.
In the population from which we have taken our sample, for each distance, there are numerous
different journeys and numerous different times for each of the journeys. In fact, for any distance
there is a distribution of possible delivery times. Our sample of 10 journeys is, in effect, a number
of different samples, each taken from these different distributions. We have taken a sample of
size 1 from 1.0 mile deliveries and the 1.3 mile deliveries, a sample of size 2 from the 3.0 mile
deliveries and we have taken no samples from the distributions for distance not on our list.
We now require a method of finding the most suitable line to fit through our sample of points. The
line is referred to as the line of best fit.
The line of best fit is called the least squares regression line.
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The equations for the slope and the intercept of the least squares regression line are:
Slope, b =
n∑xy
−
∑x∑y
n∑x3
(∑x)2
Where n is the sample size.
Intercept, α =
∑y2
−
b∑x
n
n
The calculations for our sample of size n=10 are given below. The linear model is:
y = x + ab
Table 5.2 calculations for the regression line
Distance Time
X miles Y miles xy x2
y2
3.5 16 56.0 12.25 256
2.4 13 31.2 5.76 169
4.9 19 93.1 24.01 361
4.2 18 75.6 17.64 324
3.0 12 36.0 9.0 144
1.3 11 14.3 1.69 121
1.0 8 8.0 1.0 64
3.0 14 42.0 9.0 196
1.5 9 13.5 2.25 81
4.1 16 65.6 16.81 256
Totals 28.9 136 435.3 99.41 1972
The final column is used in later calculations
The slope, b =
10 × 435.3 − 28.9 × 136
10 × 99.41 − 28.92
=
422.6
158.9
= 2.66
We now insert these vales in the linear model, giving
y =5.91 +2.66x
Or
Delivery time (Min) = 5.91+2.66×delivery distance (miles)
The slope of the regression line (2.66 minutes per mile) is the estimated number of minutes per
mile needed for a delivery. The intercept (5.91 minutes) is the estimated time to prepare for the
journey and to deliver the goods, that is, the time needed for each journey other than the actual
travelling time. The intercept gives the average effect on journey time of all of the influential
factors except distance which is explicitly included in the model. It is important to remember
that these values are based on a small sample of data. We must determine the reliability of the
estimates, that is, we must calculate the confidence intervals for the population parameters.
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Regression, Time Series and Forecasting
Strength of the linear relationship- the correlation coefficient, r
Assessment of how well the linear model fits the data. Let us consider two variables, x and y. We
consider that a relationship is likely between these variables.
The ratio of the explained variation to the total variation is used as a measure of the strength of
the linear relationship. The stronger the linear relationship, the closer this ratio will be to one. The
ratio is called the coefficient determination and is given the symbol r2
where
r2
=
∑ (y − y)2
∑ (y − y
The coefficient of determination is frequently expressed as a percentage and tells us the amount
of the variation in y which is explained by the introduction of x into the model. A perfect linear
relationship at all means r2
= 0 or 0%. The coefficient of determination does not indicate whether
y increases or decreases as x increases. This information may be obtained from the Pearson
product moment correlation coefficient, r. This coefficient is the square root of the coefficient
of determination:
r =
(∑ y − y)
2
√ (∑ y − y )
2
For the purpose of calculations it is useful to re-arrange this expression algebraically to giveThis is the sample correlation coefficient.
The value of r always lies between-1 and +1. The sign of r is the same as the sign of the slope,
b. If b is positive, showing a positive relationship between the variables, then the correlation
coefficient, r, will also be positive. If the regression coefficient, b, is negative then the correlation
coefficient, r, is also negative.
As the strength of the linear relationship between the variables increases, the plotted points will
lie more closely along a straight line and the magnitude of r will be closer to 1. As the strength of
the linear relationship diminishes, the value of r is closer to zero. When r is zero, there is no linear
relationship between the variables. This does not necessarily means that there is no relationship
of any kind. Figure 8 and 9 below will both give values of the correlation coefficient which are
close to zero.
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Return to the example above, in which a model was set up to predict delivery times for journeys
of a given distance within a city. The correlation coefficient, r is calculated as followsNotice that we have already evaluated the top line and part of the bottom line in the calculation
for the slope of the regression line, b
r =
10 × 435.2 − 28.9 × 136
=
422.6
√ (10 × 99.41 − 28.92
)(10 × 1972 − 1362 158.9−1224
r = 0.958
This value of the correlation coefficient is very close to +1 which indicates a very strong linear
relationship between the delivery distance and the time taken. This conclusion confirms the
subjective assessment made from the scatter diagram.
The coefficient of determination (r2
x 100%) gives the percentage of the total variability in delivery
time which we have explained in terms of the linear relationship with the delivery distance. In this
example, the coefficient of determination is high:
r2
= 0.958 x 100 =91.8%
The sample model: time (minutes) = 5.91 + 2.66 × distance in miles has explained 91.8% of the
variability in the observed times. It has not explained 8.2% of the variability in the journey time
but have not included in the model.
Diagram 5.8
No relationship of any kind between the
variable r ≈0
y
x
Diagram 5.9
A very strong linear relationship
relationship between the variables
r≈1
x
y
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Regression, Time Series and Forecasting
Prediction and estimation using the linear regression model
Predictions within the range of the sample data
We can use the model to predict the mean journey time for any given distance. If the distance is
4.0 miles, then our estimated mean journey time is:
A word of caution: It is good practice to use the model to make predictions for values of the
independent variables which are outside the range of the data used.
The relationship between time and distance may change as distance increases. For example, a
longer journey might include the use of a high speed motorway, but the sample data was drawn
only from slower city journeys. In the same way, longer journeys may have to include meal or rest
stops which will considerably distort the time taken.
The information from which we have been working is a sample from the population of journey
times within the distance range 1 to 4.9 miles. If we wish to extrapolate to distance outside this
range, we should collect more data. If we are unable to do this, we must be very careful using the
model to predict the journey times. These predictions are likely to be unreliable.
Estimation, error and residuals
How accurate are our predictions likely to be? In the next section, we will consider this question
in terms of the familiar idea of confidence intervals. However, it is also to assess the reliability
of the predictions by mean of the differences between the observed value of the dependent
variable, y, and the predicted value yˆ , for each value of the independent variable x. These
errors or residuals, e, are the unexplained part of each observation and are important for two
reasons. Firstly, they allow us to check that the model and its underlying assumptions are sound.
Secondly, we can use them to give a crude estimate of the likely errors in the predictions made
using the line.
The table below gives the residuals for the example above.
Table 5.3 Calculation of residuals (y- yˆ )
Observed Estimated times
Distance in Time, =5.91+2.66x Residual
Miles x Mins, y Mins, yˆ e=(y- yˆ )
3.5 16 15.22 0.78
2.4 13 12.29 +0.71
4.9 19 18.94 +0.06
4.2 18 17.08 +0.92
3.0 12 13.89 -1.89
1.3 11 9.37 +1.63
1.0 8 8.57 -0.57
3.0 14 13.89 +0.11
1.5 9 9.90 -0.90
4.1 16 16.82 -0.82
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We can examine the suitability of the model by plotting the residuals on the y-axis, against either
the calculated values of
^
y or, in bivariate problems, the x-values. This procedure is particularly
important in multiple regressions, when the original data cannot be plotted initially on a scatter
diagram so that the linearity of the proposed relationship may be assessed. If the linear model is
a good fit, the residuals will be randomly and closely scattered about zero. There should be no
pattern apparent in the plot.
If the underlying relationship had in fact been a curve, then the residual pattern would have
shown this very clearly. An example of the effect of fitting a linear model when the relationship
between the variables is actually curvilinear. The residential also allow us to assess the spread
of the errors. One of the basic assumptions behind the last method is that the spread of data
about the line is the same for all of the values of x, that is, the amount of variability in the data is
the same across the range of x.
The residual pattern for the journey time example shows two large residuals. This may indicate
that the data used do not conform to the assumptions of uniform spread. The consequences of
this will be that the confidence limits, described in the next section, will be unusable.
The only way to continue with the statistical analysis of confidence intervals and hypothesis
testing in this case, is to transform the data (often by taking logs of the x values) until the residual
plot gives a random scatter of points about e = 0, with no larger values.
Computer generated solution
The setting up and evaluation of a linear regression model can be a lengthy task, especially if
the data set is moderately large. Fortunately there are many computer packages available which
will take drudgery out of the work by performing the entire arithmetic task to produce the required
statistics. Unfortunately, it is all too easy to collect data and, without any further thought, to enter
it into a computer. The programme will produce a linear model, no matter how unsuitable that
may be.
Illustration
Leisure Publishers Ltd. recently published 20 romantic novels by 20 different authors. Sales
ranged from just over 5,000 copies for one novel to about 24,000 copies for another novel.
Before publishing, each novel had been assessed by a reader who had given it a rating between
1 and 10. The managing director suspects that the main influence on sales is the cover of the
book. The illustrations on the front covers were drawn either by artist A or artist B. The short
description on the back cover of the novel was written by either editor C or editor D.
A multiple regression analysis was done using the following variables:
Y sales (millions of shillings).
X1
1 if front cover is by artist A.
2 if front cover is by artist B.
X2
readers' rating
X3
1 if the short description of the novel is by editor C.
2 if the short description of the novel is by editor D.
The computer analysis produced the following results:
Correlation coefficient r = 0.921265
Standard error of estimate = 2.04485
Analysis of variance
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STUDYTEXT
95% confidence interval = - 6.25485 ± 2.12 × 0.961897
= - 6.25485 ± 2.03922
= - 8.29407 < B1
< - 4.21563
We are 95% confident that a cover by artist B instead of artist A reduces sales by between
4216 and 8294 copies.
(e) 95% confidence interval = 5.86599 ± 2112 × 0.922233
= 5.86599 ± 1.95513
= 3.911 < B3
< 7.821
We are 95% confident that using short descriptions by Editor D instead of Editor C will increase
sales by 3911 and 7821 copies.
Statistical influence in linear regression analysis
The underlying assumptions
In this section, we will discuss some of the necessary assumptions underlying the further analysis
of the linear regression model. The data from which a linear regression model is constructed is a
sample of the population of pairs of x and y values. Essentially we are using the sample to build a
model which we hope will represent the relationship in the population as a whole. The relationship
between the dependent variable, y and the independent variable, x, is described by:
y = a + bx + e
where ε is the deviation of the actual value of y from the line:
y = a + bx
for given value of x.
y = a + bx
is the linear model we would set up if we had all the population data.
For any given x, the population of y value is assumed to be normally distributed about the
population line, with the same variance for all x's
yˆ is the mean of all the y values for a given x. As before, Greek letters refer to the population
parameters such as µ and alfa. E is the error, or residual, the difference between the actual y
value and the mean value from the line. If the least squares method is used to determine the line
of best fit, then we are minimising summation of ∑e2
the linear model which we calculate from
the sample is:
Where
^
y is an estimate of the population mean y for a given value of x, and a and b are the
sample statistic used to estimate the population parameters α and .β
As in sampling situation, if we take a second sample, different values of a and b will arise. There
is an exact analogy between the use of x to estimate µ and the use of a to estimate beta. By
making assumptions about the sampling distribution of x, we can find confidence intervals for the
value of the population mean µ. Exactly the same procedure may be used for alpha and beta by
making inference from the sample values of a and b. Our basic model is;
eba ++= xy
183.
177
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Regression, Time Series and Forecasting
Assumptions
1. The underlying relationship is linear
2. The independence
3. The errors or residuals, ε are normally distributed
4. For any given x, the expected value of e is zero, i.e E(ε) =0
5. The variance of ε is constant for all values of x, ie the variance ε=
2
s
6. The errors re independent.
These assumptions hold if the population of y values, for a given x, is normal, with mean;
µy/x
= α + βx
Where µ denotes the mean of y for a given x, and variance = δ2
The line, set up from the sample data, is the estimate of this population line, with a as the best
estimate of α and b as the best estimate of β. Since there are many possible samples which
could be drawn from a given population, it is not possible to be sure that a particular set of
sample data is actually drawn from the given population. Hypothesis tests show how confident
we may be on the sample used to asses the compatibility of the sample with population. The
tests show how confident we may be about the linearity of the parent population. If there is no
linear relationship in the population r , the population correlation coefficient, will be zero, and β,
the slope of the regression line, will also be zero. Once we have tested the overall linearity, we
may wish to calculate confidence intervals for the slope β, for the intercept α, for the mean value
of y given a value of x,or for individual values of y for a given x. We will use a random sample to
calculate sample statistics and to estimate the corresponding population parameters.
Hypothesis tests to assess the overall linearity of the relationship
We are using sample data which has been drawn at random from a population in order to estimate
a suitable linear relationship for the population. We do not actually know that the underlying
relationship in the population is linear. The random sampling process could, quite legitimately
result in a sample which exhibits linear properties but which was actually drawn from a population
in which the underlying relationship is not linear.
We require some means of assessing the likelihood that a linear relationship in the sample
implies a linear relationship in the population. Hypothesis tests help with this assessment. As in
any situation in which hypothesis tests are used, we can never prove beyond all doubt that the
population relationship is compatible with the relationship derived from the sample. We simply
determine the consistency, or otherwise, of the sample evidence with the given null hypothesis.
Linear regression generates several statistical figures and it is possible to perform separate
hypothesis tests on these. We therefore build up an accumulative picture of the evidence for or
against the basic hypothesis of linear relationship in the population.
We will now look at this hypothesis test in turn. The null hypothesis is essentially the same for
all of the tests: That there is no linear relationship between the dependent and independent
variables in the population.
184.
quantitative techniques17 8
STUDYTEXT
Testing the population correlation coefficient
The evaluation of the Pearson product moment correlation coefficient, r, depends on the size of
the sample. The interpretation of the value of r is independent of size from the point of view of the
sample, but the implications for the population relationship are different sample sizes. A different
inference will be drawn when considering a correlation coefficient of, 0.90, which arises from a
sample of 6 items, compared to the same value which arises from a sample of 20 items. We can
feel more confident that the underlying relationship is linear in the second case, since the chance
of obtaining a sample which exhibits linearity, from a population which does not decrease as the
size of the sample increases. The correlation coefficient is assessed using a t test:
Ho
; there is no linear relationship between the y and x variables. The independent variable does
not help in predicting the values of y, i.e. 0=r
H1
: 0≠r there is some linear relationship between the x and y variables. X does help to predict
the y values.
Using this alternative hypothesis, we have a two-sided test. If we had decided that only positive
value for r would be sensible, then H1
: 0>r and we would now use a one-sided test. The test
statistic is:
( )
( )2
2
1
2
r
nr
t
-
-
=
The number of degrees of freedom is (n-2), because we have calculated x and y to find r, using
up two degrees of freedom, n is the number of pairs of values in the sample. If we wish to test
at the 5% level using a two-tail test statistic would be compared with t0.025,(n-2)
found from the
tables.
To illustrate the procedure, we will return to an example which was concerned with the estimation
of journey times from the journey distance. Previously we have found that r=0.958. Therefore the
test statistic is:
( ) 082.0
342.7
958.01
8958.0
2
2
=
-
×
=t
= 9.45
The number of degrees of freedom is: (10-2) =8
From the tables
t0.025, 8
=2.306
The test statistic (9.45) is greater than 2.306; therefore, we reject H0
at the 5% level of significance
and choose to accept H1
. The evidence is not consistent with the null hypothesis at this level.
We assume that the correlation coefficient in the population is not zero and that there is a linear
relationship between journey time and distance. This is the result we would expect with such a
high value of the sample correlation coefficient, r.
185.
179
STUDYTEXT
Regression, Time Series and Forecasting
Hypothesis test on the slope of the simple regression line
In simple linear regression, the hypothesis test on the slope of the line does exactly the same
job as the test on the correlation coefficient. We do either one test or the other, but not both. In
multiple regressions, however, where we have a regression coefficient for each of the independent
variables, the two tests fulfill different functions and both are needed.
H0
: There is no linear relationship between the variables, x does not help predicting y, ie β=0
H1
: β≠ 0 i.e. there is a linear relationship and x does help to predict the y values.
In this case a two sided test is used. However, as in the test on r , we can alter this to a one sided
test if we think that β>0 or B<0 is a more appropriate alternative hypothesis. When the population
variance is unknown, the test statistic for a sample mean is:
t =
(sample statistics parameter assumed in H0
best estimate of standard error of statistic
=
(x − µ)
s/ √ n − 1
The test statistic for the regression coefficient, b, is:
t =
(b − o)
Estimated standard error of b
The estimated standard error of b is:
seb
=
σe
√∑(x − x)2
Where is the variance of the distribution of the residuals about the population regression line.
Remember, we are assuming that this variance is the same for all values of x. σe
2
is the best
estimate of this population variance σe
2
σe
2
=
∑e2
=
∑ (y − y)2
(n − 2) (n − 2)
For computational purposes, this expression may be re-arranged algebraically as:
Again to illustrate the procedure, refer to the example about the journey times and distances.
Using the first expression for
Therefore: = 1.12minutes
−
−
>
>
2
ˆ
2
2
n
xybyay
e
25.1
8
01.10
8
82.0...71.078.0
ˆ
222
2
e
2
ˆe
2
ˆe
186.
quantitative techniques18 0
STUDYTEXT
And:
Therefore:
The test statistic for β is:
If allowance is made for rounding errors, this value of t is the same as the t statistic obtained in
the test on the correlation coefficient, 9.47 compared with 9.45.
To test at the 5% level using a two-sided test, we compare the test statistic with the boundary
value found from the tables.
t 0.025, 8
=2.306
Since 9.47>2.306, we reject H0
and choose to accept H1
. At the 5% decision level, the evidence
is not consistent with the null hypothesis. This is the same conclusion as before. We choose to
assume that there is a linear relationship between the journey time and distance. i.e. β≠0, x does
help to explain the variability in y.
Confidence intervals in linear regression analysis
Confidence interval for the slope of the population regression line β
The (1-p) 100% confidence interval for the slope, β, defines a range of values about b, the
sample estimate of β, within which we may be (1-p) 100% confident that the actual value of β lies.
Put another way, for (1-p) 100% of the sample, the true value of β will lie within the confidence
interval.
From the above we know that:
Let us calculate the 95% confidence interval for the slope of the regression model derived in the
example about journey times and distance
b±t0.025, 8
seb
= 2.66±2.31×0.281=2.66±0.65
We are 95% confident that the population slope, β lies between 2.01 and 3.31 minutes per mile.
There is a 5% chance that β lies outside the range.
99.3889.15
2
2
2
n
x
xxx
281.0
99.3
12.1
bse
47.9
281.0
66.2
t
bnp setb 2,2
2
^
xx
se e
b
187.
181
STUDYTEXT
Regression, Time Series and Forecasting
Confidence for the mean value of y for a given value of x
We now return to the basic assumption in regression analysis that for a given value of x, which
we will call x0, the possible values of y are normally distributed. The mean value of these normal
distributions is the value of y on the population regression line. We will call this mean value μy/x.
The (1-p) 100% confidence interval for μy/x
..is :
Where
^
y is the estimated value, calculated from the sample regression, bxay +=
^
Notice that this confidence interval depends on the value of x0
. The width of the interval about
the regression line, therefore, varies as x varies. The interval is at its narrowest when
xx =0 , the
sample mean of the x's. The interval is then the more familiar:
As x0
moves further away from x , in either direction, the width increases.
In the example which we have been following, the 95% confidence interval for μy/x
.is
Values for this interval will be calculated in the next section.
Confidence interval for individual values of y for given value of x
A further assumption of the model is that y values are distributed about the regression line with
variance,
2
es , which is the same for all values of x, since we are using a sample, there are two
elements of variability for individual y values. One arises from the estimated position of the mean,
μy/x
. and the other arises from the availability of the individual values about this mean.
The two elements are different in that the first is due to the fluctuations inherent in sampling
and the effect can be reduced if the sample is increased. The second is due to the nature of the
variables and is unavoidable. It can be argued, therefore, that the confidence interval for individual
values of y is not like other confidence intervals which arise entirely due to sampling fluctuation.
Some books refer to this as 'prediction' interval rather than confidence interval. Whatever name
is used, it is important to understand the distinction between the (1-p) 100% interval for μy/x
and
that for the individual y's, when x is given. The confidence interval expressions look similar; the
only difference is that the variance for individual y's given x is increased by
2
es . Similarly, the
(1-p) 100% confidence interval for individual y values, given x = x0
is
Where:
2
2
0
^
2,2/
^ 1
,
xx
xx
n
ty np
nty enp /
^
2,2/
^
89.15
89.2
10
1
12.131.2
2
0
^
x
y
0
^
66.291.5 xy
2
2
0
2,2/
1
1ˆˆ
xx
xx
n
ty enp
0
ˆ bxay
188.
quantitative techniques18 2
STUDYTEXT
The 95% confidence interval in example above is:
The table below illustrates the behaviour of the two confidence intervals, as x0
changes
Table 5.4 Calculation of confidence intervals for .μy/x
and y given x0
, for example above
Estimated time
yˆ =5.91+2.66x095%confidence intervals for
y given x0±
± mins
Distance x0
Miles
Minutes μy/x
± mins
1.0 8.57 1.47 2.98
2.0 11.23 1.00 2.77
2.89( x )
13.60 0.82 2.71
3.0 13.89 0.82 2.71
4.0 16.55 1.09 2.81
4.9 18.94 1.54 3.01
These are long and complicated calculations for a sample of only 20 values. Much of the work
can be taken away by the computer package but it is important to understand what the package
is doing and how to interpret its output. Unfortunately different packages use slightly different
terms and symbols. If you have a regression analysis package available to you, it will help if you
work through a simple example, like the one in this book, by hand/manually, and then use the
package. Compare the computer output with the hand/manual calculation until you thoroughly
understand it. A clear understanding of the two variable linear models will also help considerably
when we tackle multiple regression, the calculation for which are always done by computer.
Multiple linear regression models
In the previous section it was mentioned that the chosen independent variable is unlikely to be
the only factor which affects the dependent variable. There will be many situations in which we
can identify more than one factor we feel must influence the dependent variable. For example,
we wish to predict cost per week for a production department. It is reasonable to suppose that
departmental costs will be affected by production hours worked, raw material used, and number
of items produced and maintenance hours worked.
It seems sensible to use all of the factors we have identified to predict the departmental costs
for a sample week, we can collect the data on costs, production hours, raw material usage, etc,
but we will no longer be able to investigate the nature of the relationship between costs and
other variables by means of a scatter diagram. In the absence of any evidence to the contrary,
we begin by assuming a linear relationship and only if this proves to be unsuitable, will we try to
establish a non-linear model. The linear model for multiple linear regressions is:
89.15
89.2
10
1
112.131.2ˆ
2
0x
y
189.
183
STUDYTEXT
Regression, Time Series and Forecasting
The variation in y is explained in terms of the variation in a number of independent variables which,
ideally, should be independent of each other. For example, if we decide to use 5 independent
variables, the model is:
As with simple linear regression, the sample data is used to estimate α, β1,
β2.
etc. The line of best
fit for the sample data is:
y = a + b1
x1
+ b2
x2
+…bn
xn
These are the same as those for the simple regression case. However, here they lead to very
complex calculations. Fortunately, a computer package performs these calculations leaving us
free to concentrate on the interpretation and evaluation of the multiple linear regression models.
In the following section, we will indicate the steps which should be taken when setting up and
using a multiple regression model, but at all stages, we assume that the actual calculations will
be done by computer.
The time series components
The value of a variable, such as sales will change over time due to a number of factors. For
example, a company may be expanding hence there will be an upward movement in the sales
of the product as time progresses. The general change in the value of a variable over time is
referred to as the trend T. In time series analysis the trend is the line of best fit. The model is then
used to forecast future trend. In practice, there may be no trend at all, with demand fluctuating
about a fixed value, or more likely, there may be a non-linear trend. The graphs below represent
the trend demand at different stages of a product's life cycle.
There is underlying upward trend for the newly launched product and a dying curve of the old
product reaching the end of its economic life. It is difficult to fit the equation to these trend
curves.
The moving average technique, described in the following sections, can be used to separate
the trend from the seasonal pattern. The technique uncovers the historical trend by smoothing
away the seasonal fluctuations. However, the moving average trend is not used to forecast future
trend values because there is too much uncertainty involved in extending such a series.
Diagram 5.10 Sales of a successful new product
nn xxxxy ...332211
5544332211 xxxxxy
Sales
Time
190.
quantitative techniques18 4
STUDYTEXT
Diagram 5.12 Sales of product nearing the end of the life
Generally, it is found that values do not indicate a trend only. If these regular fluctuations occur
in a short term, they are referred to as seasonal variation,'S.' Longer term fluctuation is called
cyclical variation.
The seasonal pattern in examples used in this chapter refers to the traditional seasons, but
in forecasting generally, the term 'season" is applied to any systematic pattern. It may be the
pattern of retail sales during the week, in which case the 'season' is a day. We may be interested
in a seasonal pattern of traffic flow during the day and during the week. This will give us an hourly
'seasonal' pattern, superimposed on a daily 'seasonal' pattern, which both fluctuate about a daily
trend. If we use annual data, we cannot identify a seasonal pattern. Any fluctuation about the
annual trend data would be described by the cyclical component. This cyclical factor will not be
included in our examples. It is seen only in long-term data, covering 10, 15 or 20 years, where
large scale economic factors cause additional fluctuations about the trend.
These cyclical factors were apparent economic data from about 1960 to 1975. This was the
time when many of these forecasting ideas were being developed, but since then, the overall
economic pattern has changed. We will concentrate on short-term models which exclude the
cyclical component.
The final term in our model also arises in the linear regression model. It is error or residual, the
part of the actual observation that we cannot explain using the model. We can use the errors to
give us measure of how well our model fits the data. Two measures are usually used. These are
the mean absolute deviation.
This is the sum of all the errors, ignoring their sign, divided by number of forecasts, and the mean
square error
Sales
Sales
Time
n
E
n
ForecastActual
MAD
t
n
E
MSE
t
2
191.
185
STUDYTEXT
Regression, Time Series and Forecasting
This is the sum of the squares of the errors, divided by the number of forecasts. This second
measure emphasises the large errors.
In the analysis of a time series, we attempt to identify the factors which are present and to build
a model which combines them in an appropriate way.
Example: To illustrate the choice of an appropriate time series model.
The data below are quantities of a product sold by Lewplan Plc during the last 13-three monthly
periods.
Table 5.11 Quantities of product sold over last 13-three monthly periods.
Date Quantity sold 000
Jan – May 19 x 6 239
Apr – Jun 201
Jul – Sep 182
Oct – Dec 297
Jan – Mar 19 x 7 324
Apr – Jun 278
Jul – Sep 257
Oct – Dec 384
Jan – Mar 19 x 8 401
Apr – Jun 360
Jul – Sep 335
Oct – Dec 462
Jan – Mar 19 x 9 481
We wish to analyzse this data set to see if we can find historical pattern. If there is a consistent
pattern, we will use it to forecast the quantity sold in subsequent three monthly periods.
Solution: The figures are plotted below. In time series diagrams it is customary to join the points
with straight lines so that any pattern can be seen more clearly.
Diagram 5.10 Lewplan plc, sales per 3 months.
Quantity sold per 3
months, 000s
500
400
300
200
100
Approximately equal seasonal variation
indicating an additive model
1 2 3 4 1 2 3 4 1 2 3 4 1 Time Quarter
19x6 19x7 19x8
192.
quantitative techniques18 6
STUDYTEXT
The diagram suggests that there may be an increasing trend, overlaid by seasonal fluctuations.
The sales in the winter seasons, 1 and 4 are consistently higher than those in the summer
seasons 2 and 3. The seasonal component appears to be fairly constant over the 3 years. The
trend is for the sales to increase overall from around 230 in 19x6 to 390, but the seasonal
fluctuations have not increased. This indicates that the additive component model should be
more suitable. See section 9.3.
The analysis of an additive component model A = T + S + E
The addictive component model is one in which the variation of the value of the variable overtime
can be described by adding the relevant components. Assuming that cyclical variation is not
included the actual value of the variables A may be modeled by:
Actual value = trend + seasonal variation + error
That is
A = T + S + E
In both the addictive and multiplicative components modeled, the general analysis procedure is
the same:
Step 1: Calculate the seasonal components
Step 2: Remove the seasonal component from the actual values. This is called deseasonalising
the data.
Calculate the trend from these deseasonalised figures.
Step 3: Deduct the trend figures from deaseasonalised figures to leave the errors
Step 4: Calculate the mean average deviation (MAD) or the mean square error (MSE) to judge
whether the model is reasonable, or to select the best from different models.
Calculate the seasonal components for the addictive model
Example: Setting up the addictive component model for a time series
Refer to the example in the previous section which relates to the quarterly sales of Lewplan plc.
We have already decided that an addictive model is appropriate for these data; therefore the
actual sales may be modeled by:
A = T + S +E
To eliminate the seasonal components we will use the method of moving averages. If we add
together the first 4 data points, we obtain the total sales for 19x6; dividing this by 4 gives the
quarterly average for 19 x 6, i.e.
(239 + 201 + 183 + 297)/4 = 229.75
This figure contains no seasonal component because we have averaged them out over the
year, that is, for mid-point of quarters 2 and 3. If we move on for 3 months, we can calculate the
average quarterly figure for April 19x6 – March 19x 7 (251), for July 19x6 –June 19 x 7 (270.5)
and so on. This process generates the 4 point moving averages for this set of data. The set of
moving averages represent the best estimate of the trend in the demand.
We now use these trend figures to produce estimates of the seasonal components. We
calculate:
A – T = S + E
193.
187
STUDYTEXT
Regression, Time Series and Forecasting
Unfortunately, the estimated values of the trend given by the 4 point moving averages are for
points in time which are different from those for the actual data. The first value, 229.75, represents
a point in the middle of 19 x x, exactly between the April June and the July- September quarters.
The second value, 251, falls between the July-September and the October-December actual
figures. We require a deseasonalised average which corresponds with the figure for an actual
quarter. The position of the deseasonalised averages is moved by reaveraging pairs of values.
The first and second values are averaged, centring them on July-September 19 x 6, ie:
(229.75 + 251)/2 = 240.4
This is the deseasonalised average for July-September 19 x 6. This deseasonalised value, called
the centred moving average can be compared directly with the actual value for July-September
19 x 6 of 182. Notice that this means that we have no estimated trend figures for the first 2 or last
2 quarters of the time series. The results of these calculations are shown in the table below:
Table 5.12 Calculations of the centered 4 point moving average trend values for the model A = T + S= E
Date Quantity
000's A
4 Quarter Total 4quarter
moving
average
Centered
moving
average
Estimated
seasonal
component
A – T = S + E
Jan-Mar
19x6
239
Apr – Jun 201
919 229.75
Jul – Sep 182 240.4 - 58.4
1004 251
Oct –Dec 297 260.6 + 36.4
1081 270.25
Jan -Mar
19x7
324 279.6 +44.4
1156 289
Apr-Jun 278 299.9 -21.9
1243 310.75
Jul-Sep 257 320.4 -63.4
1320 330
194.
quantitative techniques18 8
STUDYTEXT
Oct – Dec 384 340.3 +43.8
1402 350.5
Jan-Mar
19x8
401 360.2 +40.8
1480 370
Apr-Jun 360 379.8 - 19.8
1558 389.5
Jul-Sep 335 399.5 -64.5
1638 409.5
Oct –Dec 462
481
For each quarter of the year, we have estimates of the seasonal components, which include
some error or residual. There are two further stages in the calculations before we have usable
seasonal components. We average the seasonal estimates for each season of the year. This
should remove some of the errors. Finally we adjust the averages, moving them all up or down by
the same amount, until their total zero. This is done because we require the seasonal components
to average out over the year. The correction factor required is: (the sum of the estimated seasonal
values)/4. The estimates from the last column in Table.2 are shown under their corresponding
quarter numbers. The procedure is shown in the table below.
Table 5.13 Calculations of the average seasonal components
1 2 3 4
19X6 - - - 58.4 + 36.4
19X7 +44.4 -21.9 -63.4 + 43.8
19X8 +40.8 -19.8 -64.5
Total + 85.2 -41.7 -186.3 + 80.2
Average 85.2÷ 2 -41.7 ÷2 -186.3÷3 80.2÷2
Estimated
seasonal
component
+ 42.6 -20.8 -62.1
Sum
= -0.2
Adjusted
seasonal
component
+ 42.6 -20.7 -62.0
Sum
= 0
In this example, two of the seasonal components have been rounded up, and two have been
rounded down, so that the sum equals zero.
The seasonal components confirm our comments on the diagram at the end of the section
above. Both winter quarters are above the trend by about 40 thousand units and the two summer
quarters are below the trend by 21 and 62 thousand units, respectively.
196.
quantitative techniques19 0
STUDYTEXT
Diagram 5.11 Lewplan Plc, actual and deseasonalise sales per 3 months
The equation of the trend line is:
T = a + bx quarter number
Where a and b represent the intercept and slope of the line. The least squares method can be
used to determine the line of best fit, therefore the equation for a and b, from the previous chapter
on linear regression are:
Where X is the quarter number and y is (T + E) in the above table. Using a calculator, we find:
∑x = 91 ∑x2
=819 ∑y = 4158.7 ∑xy = 32747.1 n = 13
It follows by substitution that:
b = 19.978 and a = 180.046
Hence the question of the trend model may be written.
Trend quantity (000s) = 180.0 + 20.0 x quarter number.
Calculate the errors
Step 3 in the procedure, before using a model to forecast, is to calculate the errors or residuals.
The model is:
A = T + S + E
We have calculated S and T. We can now deduct each of these from A the actual quantity, to find
the errors in the model.
Qty sold per
3 months,
000s
500
400
300
200
100
actuals
Estimated trend
22
xxn
yxxyn
b
n
xb
n
y
a
and
198.
quantitative techniques19 2
STUDYTEXT
The quarter number for the next three monthly periods, April-June 19 x9, is 14, therefore the
forecast trend is:
T14
= 180 + 20 x 14 = 460 (000 units per quarter)
The appropriate seasonal component is – 20.7 (000 units). Therefore the forecast for this quarter
is:
F (April – Jun 19 x9) = 460 – 20.7 = 439.3 (00 units)
It is important to remember that the further ahead the forecast, the more unreliable it becomes.
We are assuming that the historical pattern continues uninterrupted. This assumption may hold
for short periods but is less and less likely to be true the further we go into the future.
The analysis of multiplicative component model: A = T ×S × E
In some time series, the seasonal component is not a fixed amount each year. Instead it is a
percentage of the trend values. As the trend increases, so does the seasonal variation.
Example: Setting up a multiplicative component model for a time series.
CD Plc sells a range of products. The quarterly sales of one product for the last 13 quarters are
given below
Table 5.16 CD Plc quarterly sales.
Date Quarter number Quantity sold A
Jan-Mar 19x6 1 70
Apr – Jun 2 66
July – Sep 3 65
Oct-Sep 4 71
Oct – Dec 5 79
Jan- Mar 19x8 6 66
Jul- Sep 7 67
Oct – Dec 8 82
Jan-Mar 19x8 9 84
Apr – Jun 10 69
July-Sep 11 72
Oct-Dec 12 87
Jan-Mar 19x9 13 94
The scatter diagram for these data is:
This product has a similar seasonal pattern to the previous example, with high winter values and
low summer values, but the size of variations about the trend line are increasing. A multiplicative
component model should be suitable for these data.
Actual values = trend x seasonal variation x error
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Regression, Time Series and Forecasting
That is: A = T X S X E
In this example, the trend looks linear but this will become clearer when we have smoothed the
series.
Calculations of the seasonal components
Initially the same procedure is followed as for the addictive model. The centred moving average
trend values are calculated but the estimated seasonal components are ratios derived from A/T
= S X E. The calculations are shown in the table below:
The seasonal components ratios are derived from the quarterly estimates in a similar way to
those for the addictive model. Since the seasonal values are ratios and there are 4 seasons,
we require the seasonal components to total to 4 rather than zero. (If the data comprised 7 daily
seasons in each week, then the seasonal components would be required to total to 7). If the total
is not 4, the values are adjusted as before. The estimates from the last column above are shown
under their corresponding quarter numbers below.
Qty sold/3
months
100
90
80
70
60
1 2 3 4 1 2 3 4 1 2 3 4 1 Time quarter year
19x6 19x7 19x8
Increasing seasonal variations as
the trend increases indicates a
multiplication
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The trend values are superimposed on the original scatter diagram:
Diagram 5.12 CD Plc actual and deseasonalised sales per 3 months
The trend which emerges is erratic. The sales values in this time series are not consistent like the
ones in the first example for Lewplan Plc. CD plc is probably a more realistic example.
We now have to decide how to model the trend. It is not a curve but looks roughly linear even
though the values are erratic, particularly in 19 x 6. For simplicity, we will assume that the trend
is linear and use the least squares method to fit the best-line to the data. The trend line, using
the same procedure is:
T = 64.6 + 1.36 X quarter number (000 units per 3 months)
We use this equation to estimate the value of the trend sales for each of the periods.
Calculation of the errors: A /(T X S) = E OR A – (T XS) = E
We have now calculated the trend and seasonal components. We can use these to find the errors
between the observed sales, A, and the sales which are forecast by the model, T x S. The table
below gives the errors, both as a proportion, E = A/(T x S), and as absolute values, A – (T x S)
Qty sold/3
months
100
90
80
70
60
1 2 3 4 1 2 3 4 1 2 3 4 1 Time quarter year
19x6 19x7 19x8
actuals
Estimated trend
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And the seasonal component ratios are 1.116 for quarter 2, 0.922 for quarter 3, and 1.055 for
quarter 4. The next quarter is April- June19 x 9, which is quarter 14 in the series and quarter 2 in
the year. The forecast sales are:
F = T X S
= (64.6 + 1.36 X 14) X 0.907
= 83.64 X 0.907 = 75.9 (000 units per quarter)
Given the errors for the model, we would hope that this estimate will be within 2% or 3% actual
value. Similarly, the forecast for October-December 19 x 9 is found using quarter number is 16
and the seasonal component for quarter 4:
F= T X S
= (64.6 + 1.36 X16) X 1.055
= 86.64 X 1.055 = 91.1 (000 units per quarter)
We expect the error on this forecast to be larger than previous one because it is further into the
future.
CHAPTER SUMMARY
A time series is any set of data which is recorded over time. It may, for example, be annual,
quarterly, monthly or weekly data. Models use the historical pattern of the time series to forecast
how the variable will behave in the future. The short term forecast will be more accurate than the
longer term ones. The further ahead we forecast, the less likely it is that the historical pattern will
remain unchanged, and the larger will be the errors.
There are two basic models. In both cases, it is assumed that the value of the variable is made
up of a number of components. The series may contain trend –general movement in the value of
the variable, seasonal variation – short term periodic fluctuations in the variable values: cyclical
variation – long term periodic fluctuations in the variable values, and error components – residual
term. Data sets enough to include the cyclical component were not considered in this text.
The component models are:
Addictive A = T + S + E
Multiplicative A = T × S × E
In both cases, moving averages are used to deseasonalise the time series. These deseasonalised
data are used to set up a model to describe the trend. The model is used to forecast the best
model. Two measures give guidance on how well a model fits the past data. These are:
Mean absolute deviation (MAD) =
n
Et∑
Mean square error (MSE) =
( )
n
Et∑
2
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Regression, Time Series and Forecasting
Chapter Quiz
1. A data set, in which the independent variable is time, is referred to as…………..
2. …………. is the process of estimation in unknown situations.
3. Variance is that part of the actual observation that we cannot explain using the model.
(a) True
(b) False
4. Is the following component model right? A + T= S + E
5. The formula for mean absolute deviation is
(a) True
(b) False
n
Et
2
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Answers
1. Time series
2. Forecasting
3. (b) False. This is called Error or residual
4. Wrong. It is A – T = S + E
5. False. MAD =
n
Et∑
Questions from previous exams
December 2000 Question 5
An increasing number of organisations operate sophisticated corporate planning models. Quite
often these models feature integrated models for the production, marketing and financing sub-
systems. While linkage between these models is important, forecasts of the external environment
are also needed and econometric models are frequently used to provide these forecasts.
Required
a) Under production, marketing and finance sub-systems, what information about the
environment might be provided by the econometric models? (9 marks)
b) The monthly electricity bill at the Chez Paul Restaurant over the past 12 months has
been as follows:
Month Amount (Sh.)
December 30,660
January 27,190
February 30,570
March 30,640
April 29,730
May 31,530
June 29,720
July 33,070
August 30,010
September 27,550
October 30,130
November 27,940
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Regression, Time Series and Forecasting
Paul is considering using exponential α = 0.5 to forecast future electricity sales (index α
smoothing with.)
Required:
Determine next January's forecast. (11 marks)
(Total: 20 marks)
June 2002 Question 5
The following regression equation was calculated for a class of 24 CPA II students.
Ŷ = 3.1 + 0.021X1
+ 0.075X2
+ 0.043X3
Standard error (0.019) (0.034) (0.018)
Whereby: Y = Standard score on a theory examination
X1
= Students rank (from the bottom) in high school
X2
= Students verbal aptitude score
X3
= a measure of the students character
Required:
a) Calculate the t ratio and the 95% confidence interval for each regression coefficient.
(8 marks)
b) What assumptions did you make in (a) above? How reasonable are they?
(4 marks)
c) Which regressor gives the strongest evidence of being statistically discernible?
(2 marks)
d) In writing up a final report, should one keep the first regressor in the equation, or drop
it? Why? (6 marks)
(Total 20 marks)
June 2003 Question Five
a) For an additional time series model, what does the term "residual variation" mean?
Describe briefly its main constituents. (4 marks)
b) Explain the moving average centering and why it is employed. (4 marks)
c) Given a time series with trend figures already calculated, describe in words only, the
method for calculating seasonal variation values using the additive model. (4 marks)
d) When projecting a moving average trend, what basis would make the choice of the
following appropriate?
(i) Projecting 'by eye'. (2 marks)
(ii) Using the method of semi-averages (1 mark)
(iii) Using the average change in trend per period from the range (1 mark)
d) Explain how management of an organisation might use seasonal variation figures and
seasonally adjusted data. (4 marks)
(Total: 20 marks)
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Decision Theory
CHAPTER SIX
Decision Theory
Objectives
At the end of this chapter, you should be able to:••
Define such terms as state of nature, event, act and payoff.••
Organise data into a payoff table.••
Determine the expected payoff of an act.••
Compute opportunity loss and expected opportunity loss.••
Assess the value of perfect information.••
Fast Forward: Decision theory in mathematics and statistics is concerned with identifying the
values, uncertainities and other issues relevant in a given decision and the resulting optimal
decision.
Introduction
Decision making can be regarded as an outcome of mental processes (cognitive processes)
leading to the selection of a course of action among several alternatives. Every decision making
process produces a finalchoice. The output can be an action or an opinion of choice.
Definitions of key terms
Decision – is a commitment to irrevocably allocate valuable resources. It is a commitment to act
and action is the irrevocable allocation of valuable resources.
Outcome/consequences – A consequence is a result of a course of action or decision taken by
a decision maker.
Decision theory – is a body of knowledge related analytical techniques of different degrees of
formalities designed to help a decision maker choose a set of alternatives in light of possible
consequences. It is a theory that applies in conditions of uncertainty, risks and certainty.
Choice of judgement – it is a tough decision e.g. I will take Maria as my wife. It can be defined
simply as firmness of conviction. These are results arrived at by judges.
Decision makers – A group of persons who make final choice among alternatives.
Alternative – option – this is one of the mutually exclusive courses of action in attaining the
objective.
Course of action – strategy or means available to a decision maker by which the objectives may
be attained.
Objective – something that a decision maker seeks to accomplish or to obtain by means of his
decision. It is short term.
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Goal – a general objective. It is long term. Goal is used to denote a very general objective
Cyclic variation - Any change in economic activity that is due to some regular and/or recurring
cause, such as the business cycle or season.
Industry context
We apply decision theory models to political science with the intent of creating a positive theory
of politics. We use models of self-interested individuals to consider various voting mechanisms
used to generate group decisions.
Several important applications of decision theory with uncertainty, applying decision theory to
investments, insurance, and search.
Exam Context
For successful organizations and institutions, good ideas have to be made. In this case the
examiners have previously examined the students as outlined below:
12/06, 6/06, 6/06, 12/05, 12/05, 6/05, 12/03, 12/02, 6/02, 12/01, 12/00
Decision Theory
Almost everything that a human being does involves decisions. We make choices all the time.
Some decisions are easy others are not.
Consider the following and the problem they give:
Shall I bring an umbrella today? - This decision depends on something I don't know,1.
namely whether it will rain.
I am looking for a house, shall I buy this house? This house looks fine, but perhaps I'll2.
find a better house if I go searching
Shall I smoke the next cigarette? One single cigarette is no problem, but if I make the3.
same decision sufficiently many times it may kill me.
A decision maker may have more than one objective. The term goal is sometimes used to denote
a very general objective, usually long term.
Decision making process.
Is the process of choosing among alternative courses of actions which are feasible.
Process:
Identify the objectives••
Search for alternative courses of action••
Gather data about the alternatives••
Select alternative courses of action••
Compare actual and planned outcome••
Respond to divergencies for the plan.••
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Decision Theory
Decision theory generally involve 4 steps
Example: Consider a manufacturing company that is thinking of several methods to increase its
production.
Steps:
1. List all the viable alternatives for the company considered above. There may be the
following options.
a) Expand the present plant
b) Construct a new plant
c) Subcontract the plan for extra demand.
2. Identify the expected future events. Often it's possible to identify most of the events that
can occur i.e. states of nature.
The difficulty is to identify which particular event will occur. For the manufacturing company,
the greatest uncertainty will be about product demand. The future events related to a demand
will be
i) high demand
ii) Moderate demand
iii) Low demand
iv) No demand
3. Construct a payoff table – The decision maker makes a payoff table representation
profits or benefits for each combination course of action of a data and states of nature.
The possible payoffs for the manufacturing company expansion.
Table 6.1
States of nature
Alternatives High Moderate Low Nil
Expand 50,000 25,000 (25,000) (45,000)
Construct 70,000 30,000 (40,000) (80,000)
Subcontract 30,000 15,00 (1,000) (10,000)
4. Select optimum decisions criteria - Decision makers will choose criteria which result in
target profit. The criteria may be economic, qualitative or quantitative.
Decision making environment
1. Certainty – In this environment, there exists only one state of nature i.e there is complete
certainty about the future. Complete information is also available as to which state of nature
is going to occur. It is thus easy to analyse the situation and make good decisions. The
decision making process is just picking the best alternative.
Unfortunately certainty environment is a very simplistic environment which is rarely applicable
in real life situation.
Examples: Linear programming, transportation and assignment techniques, I/O analysis
and EOQ
Few complex managerial problems ever enjoy the luxury of complete information about the
future and thus decision making under certainty is of little consequential interest.
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2. Uncertainty – More than one state of nature exists but the decision maker lacks sufficient
knowledge to allow him assign probabilities with the various states of nature. Uncertain
events are those events that cannot be predicted with statistical confidence i.e the decision
maker does not know the relevant variables nor their probability therefore decision making
in this environment depends on the risk attitude of the decision maker i.e. risk averse or a
risk seeker or risk neutral.
A risk averse avoids risk at all cost; the individual is conservative and assumes the worst
situation will occur.
A risk seeker takes high risks in expectation of high returns. He assumes that the best
outcome will occur.
A risk neutral person is not affected by risks. Such people will make decisions based on
something else but not risk.
3. Risks – Here more than one state of nature exists and the decision maker has sufficient
information to allow him probabilities to each of these states of nature. Therefore it involves
situations or events which may/may not occur but its probability of occurrence can be
calculated statistically. There are frequency of occurrence predicted from first records.
4. Competition – In this environment, the decisions of the firm are affected by the decision of
other firms with opposing interests.
Decision making under uncertainty
When the probability of occurrence of each state of nature can be assessed the expected monetary
value (EMV) of the expected opportunity loss (EOL) decision criteria are usually appropriate.
When a manager can't assess the outcome probability with confidence or virtually no probability
data are available other decision criteria may be applied. These include:-
i. Maxmax
ii. Maxmin
iii. Laplace (equally likely/ Criteria of rationality)
iv. Hurcwiz (criteria of realism)
v. Minimax regret criteria (savage criterion)
1. Maxmax
Finds alternatives that maximise the max outcome or consequence of every alternative. The
decision maker first locates the maximum number since this decision criterion locates the
alternatives with the highest possible gain it has been called optimistic decision criterion. The
criterion appeals to risk takers for optimism who are ready to make huge profits if they occur.
Table 6.2
Alternative high medium low No Action Max. of row
Expand 50,000 25,000 -25,000 -45,000 50,000
Contract 70,000 30,000 -40,000 -80,000 70,000
Subcontract 30,000 15,000 -10,000 -10,000 30,000
Maxmax payoff is sh70,000 corresponding to contract alternative.
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Decision Theory
2. Maxmin
Finds alternatives that maximise the min or consequences of each alternative. Under this rule,
the decision maker picks the worst possible outcome under each alternative and then the best
of these worst outcomes. Since this decision maker locates the alternatives that has the least
possible loss it has been called pessimist decision criteria. The criterion appeals to risk averse
decision makers because it's a criterion of extreme caution. It assumes that the worst outcome
will occur.
Table 6.3
Alternative high medium low No Action min. of row
Expand 50,000 25,000 -25,000 -45,000 -45,000
Contract 70,000 30,000 -40,000 -80,000 -80,000
Subcontract 30,000 15,000 -10,000 -10,000 -10,000
The maxmin payoff is sh (10,000) to the company. Corresponding to the decision subcontract.
3. Laplace (criterion of rationality)
It holds that if a decision maker does not know the probability of the various states of nature
and has reason to think otherwise then the states of nature will be considered equally likely.
This criterion is based upon the principle of insufficient reason. The criterion assigns equal
probabilities to all the events of each alternative decision and selects the alternatives with the
max expected payoff.
Symbolically if n denotes the no of events and P1 donates the payoffs the expected values of
strategy say S1
= 1/n (p1
+p2
…. +pn
)
Table 6.4
Alternative high medium low No action expected payoff
Expand 50 25 -25 -45 ¼ (50 + 25 -45)=1250
Construct 70 30 -40 -80 (500)
Sub construct 30 15 -10 -10 8,500
NB: Units in '000'
The alternative subcontract will be the best to take since it results in max average payoff of sh
8,500.
4. Hurcwicz criterion (criterion of realism / weighted average criterion)
Is a compromise between maxmax and maxmin decision criteria. The criterion is based on
Hurwicz concept of optimism or pessimism. This concept allows the decision maker to take into
account both the maximum and minimum of each alternative and assigns the weight according
to the degree of pessimism or optimism
The alternative which maximises the sum of these payoffs is then selected.
The criterion of realism consists the following steps:
i. A coefficient of realism (alpha) is selected. The coefficient is between zero and one.
When alpha is close to 1 the decision maker is optimistic about the future and when is
it close to zero he is pessimistic about the future.
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ii. Determine the max as well as min of each alternative and obtain the following:
iii. P = α (maximum) + 1-α (minimum)
iv. Choose the alternative that yields the maximum value P using example above
Table 6.5
Alternative high medium low No Action min. of row max p=α(max)+(1-α)min
Expand 50 25 -25 -45 -45 50 31
Contract 70 30 -40 -80 -80 70 40
Subcontract 30 15 -10 -10 -10 30 22
NB: Units in '000'
Assume that coefficient of realism, α = 0.8
The best decision under Hurwicz criterion would be to construct since it has the highest weighted
payoff of Ksh 40,000.
Minimax regret criterion (Savage criterion)
Was developed by L.J Savage. He pointed out that the decision has been made and state of
nature has occurred. Thus, the decision maker should attempt to minimise regret by minimising
the maximum opportunity loss of each alternative.
Steps
i. Develop opportunity loss table and then find the max opportunity loss within each
alternative.
ii. Select the alternative with the min or smallest opportunity loss
Table 6.6
Alternative High Moderate Low Nil Max
Expand 20,000 5,000 24,000 35,000 35,000
Construct 0 0 39,000 70,000 70,000
Subcontract 40,000 15,000 0 0 40,000
The company should expand because this decision will minimise its regret which is Sh. 35,000.
Decision making under risk
In this environment, it is possible to attach some probabilities in the various states of nature.
Several states of nature may occur each at a given probability. The decision maker would either
choose the expected monetary value or the expected opportunity loss (EOL)
These two are similar in the sense that the alternatives that maximise the EMV will minimise EOL
at the same time.
Expected Monetary Value
Given a decision payoff table with conditional value (payoffs) and probability assessments for all
states of nature, it is possible to determine the expected monetary value for each alternative if the
decision would be expected a large number of times. The EMV of an alternative is just the sum
of possible payoffs of the alternative each weighted by the probability of that pay off occurring so
that EMV (alternatives) =
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Decision Theory
Payoff of 1st
state of nature ×probability of 1st
state of nature + Payoff of 2nd
state of nature
× probability of 2nd
state of nature + payoff of last state of nature ×probability of last state of
nature.
Steps
i. Construct a payoff table listing the alternative decisions and various states of nature.
Enter conditional profit for each decision event combination along with the associated
probability.
ii. Calculate the EMV for each alternative decision by multiplying the conditional profit by
assigned probability and adding the resulting values
iii. Select the alternative that yields the highest EMV.
Example
Consider the following table for Thomson Manufacturing Company
States of nature Favourable mkt unfavourable mkt
Alternative $ $
Construct a large plant 200,000 -180,000
Construct a small plant 100,000 -20,000
Do nothing 0 0
Probability 0.5 0.5
Which alternative will give the best EMV?
EMV (large) = 0.5 (200,000) + 0.5 (-180,000) = 10,000
EMV (small) = 0.5 (100,000) + 0.5 (-20,000) =40,000
EMV (Nothing) = 0
The largest expected monetary value results from the 2nd
alternative which is to construct a small
plant.
Expected opportunity loss
An alternative for maximising EMV is to minimise EOL. EOL represents the amounts by which
the maximum possible profit will be reduced under various possible stock action. The cost of
action that minimise these losses is the optimal decisions alternative.
Procedures
i. Prepare the conditional profit table for each decision taken and the associated profit
ii For each event determine the conditional opportunity loss by subtracting the pay off of
the max pay off of that event.
iii. Calculate the EOLfor each decision alternative by multiplying the conditional opportunity
loss by the associated probabilities and then adding the values.
iv. Select the alternative that yields the lowest value
Alternative favourable unfavourable
Large 0 180,000
Small 100,000 20,000
Do nothing 200,000 0
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EOL (large) = 0.5 0) +0.5(180,000) = 90,000
EOL (small) = 0.5 (100,00o) + 0.5(20,000) = 60,000
EOL (nothing) = 0.5 (200,000) + 0.5 (0) = 100,000
The decision to construct the small plant will be selected since it has the lowest EOL
Note: i) The minimum EOL will always result in the same decision as EMV
ii) The minimum EOL= Expected value of project information ( EVPI)
Expected value of perfect information (EVPI)
EV with PI is the expected or average return in the long run. If we have perfect information before
a decision has been made. To calculate this value we choose the best alternative for each state
of nature and multiply its payoff by the probability of occurrence of that state of nature
The EVPI is the expected outcome with perfect information minus the expected outcome without
perfect information i.e. the maximum EMV.
EVPI = EV, PI – EMU (max) EVPI = Min (EOL)
EVPI = (200,000 × 0.5) + ( 0 x 0.5) – 40,000
EVPI = $60,000 = MIN. EOL
EV PI = is the information which guarantees the future with 100% degree of accuracy.
Multistage decision making
Decision trees
It is a graphic representation of the decision alternatives, states of nature, probabilities attached
to the state of nature and conditional losses.
It consists of a network of nodes. Two types of nodes are used; decision node represented by a
square and states of nature (chance of event) node represented by a circle.
Alternative courses (strategies) originate from the decision node as main branches (decision
branches)
At the end of each decision branch there's a state of nature mode from which emanates chance
events in the form of sub-branches (chance branches)
The respective pay offs and probability associated with alternative course and chance events
are shown alongside those branches. At the terminal of those branches are shown the expected
value of outcomes.
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Decision Theory
Diagram 6.1
The general approach used in decision tree analysis is to work backward through the tree from
right to left compiling the expected value of each chance node and then choosing the particular
branch leaving a decision node which leads to the chance node with the highest expected
value.
Steps in decision tree analysis
1. Identify the decision point and the alternative courses of action at each decision point
systematically.
2. At each decision point determine the probability in payoff associated with each course
of action.
3. Commencing from the extreme right, compute the expected pay off (EMV) from each
course of action
4. Choose the course that gives the best payoff for each of the decision.
5. Proceed backwards to the next stage of decision point.
6. Repeat above steps until the 1st
decision point is reached.
7. Identify the courses of action to be adopted from the beginning to the end under the
different possible outcomes for the situation as a whole.
Advantages of Decision Tree Approach
i. It structures the decision process and helps decision making in orderly and systematic
and consequent manner.
ii. It requires the decision maker to examine all possible outcomes whether desirable or
undesirable.
iii. It communicates the decision making to others in an easy clear manner illustrating each
assumption about the future.
iv. It displays the logical relationship between the parts of a complex decision and identifies
the time sequence in which various actions and subsequent events occur.
Limitations
i. Decision tree diagram becomes more complex as the number of decision alternatives
increases and more variables are introduced.
ii. It becomes highly complicated when interdependent alternatives and dependent
variable are present in the problem.
Action A1
Oucome X1
Oucome X2
Oucome X1
Oucome X2
Action A2
Action B1
Action B1
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iii. It analyses the problem in terms of expected value and thus yields an average value
solution
iv. There is often inconsistency in assigning probabilities of different events
Perfect and imperfect information
Uncertainty about the future can sometimes be reduced by first obtaining more information
about what is likely to occur. Such information can be obtained from various sources like market
research, pilot testing, building a prototype model, etc. Information can be categorised into:
i. Perfect information – is information which guarantees the future with 100% surety.
ii. Imperfect information – is information which, although good and hence better than
having no information at all, could be wrong in its prediction. Also referred to as sample
information.
Expected value of sample information therefore is the maximum that the decision maker should
be willing to spend in order to obtain sampled information.
Decision Trees and Bayes Theorem
There are many ways of getting probability data. The numbers can be assessed by a manager
based on past experience and intuition. They can be derived from historical data or they can be
computed from other available data using Bayes theorem approach. It recognises that a decision
maker does not know with certainty what state of nature will occur. It allows the manager to revise
his /her initial or prior probability assessment. These are called posterior probability.
This rule or theorem is given by
P(A|B) =
( ) ( )
P(B)
ABPAP ×
It is used frequently in decision making where information is given in the form of conditional
probabilities and the reverse of these probabilities must be found.
Example 1
In a class of 100 students, 36 are male and studying accounting, 9 are male but not studying
accounting, 42 are female and studying accounting, 13 are female and are not studying
accounting.
Use these data to deduce probabilities concerning a student drawn at random.
Solution:
Accounting A
Not accounting
A
Total
Male M 36 9 45
Female F 42 13 55
Total 78 22 100
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It is used frequently in decision making where information is given in the form of conditional
probabilities and the reverse of these probabilities must be found.
Example 2
Analysis of questionnaire completed by holiday makers showed that 0.75 classified their holiday
as good at Malindi. The probability of hot weather in the resort is 0.6. If the probability of
regarding holiday as good given hot weather is 0.9, what is the probability that there was hot
weather if a holiday maker considers his holiday good?
Solution
P(A|B) =
( ) ( )
P(B)
ABPAP ×
Let H = hot weather
G = Good
P(G) = 0.75 P(H) = 0.6 and P(G|H) = 0.9 (Probability of regard holiday as good given hot
weather)
Now the question requires us to get
P(H|G) = Probability of (there was) hot weather given that the holiday has been rated as good).
=
( ) ( ) ( )( )
0.75
0.90.6
P(G)
HGPHP
=
= 0.72.
Decision Making Under Competition
Game theory
Competition is an important economic factor. The strategy taken by individuals or organisations
can dramatically affect the outcome of our decision. In the automobile industry, for example, the
strategies of competitors to introduce certain models can dramatically affect the profitability of
car makers.
Today's business cannot make important decisions without considering what other organisation
or individuals are doing or might do. A game theory is one way to consider the impact of others
on our strategies and outcome. A game is a context involving two or more decision makers each
of whom wants to win.
Game theory is a study of how optimal studies are formulated in conflict situations. In the business
context, the term game refers to conflicts through time. It is to do with business situations where
usually the success of one person is at the expense of the other.
Competitive game
Properties
i. There's a finite number of participants. If the number of participants is two, the game is
referred to as a two-person game. If we have M number of participants it is called an
M-person game.
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Decision Theory
ii. Each participant has a finite number of possible courses of action
iii. Each participant must know all the courses of action available for the others but must
not know which of these will be chosen.
iv. A play of the game is said to occur when each player chooses one of its courses of
action. The choices are assumed to be made simultaneously so that no participant
knows the choice of the other until he has decided on his own.
v. After all participants have chosen their courses of action their respective game options
are finite.
vi The game of participants depends on his action as well as those of others
Useful terminology
1. Player – each participant (interested party) is called a player
2. A play of the game results when each player has chosen a course of action. After each
play of the game one player pays the other an amount determined by the courses of
action chosen.
3. Strategy – the decision/role which a player determines his/her course of action is called
strategy. To reach a decision of which strategy to use neither player needs to know the
other's strategy.
4. Pure strategy – if a player decides to use only one particular course of action during
every play he is said to use a pure strategy. A pure strategy is usually represented by a
number with which a course of action is associated.
5. Mixed strategy – if a player decides in advance to use all or some of his available
courses of action in a fixed proportion he is said to use a mixed strategy. Thus a mixed
strategy is a selection among pure strategies with some fixed proportions.
6. Zero sum game – A game where a gain of one equals the loss of the other is known
as a two person zero-sum game. In such a game, the interest of the two players are
opposed so that the sum of their net gain (sum of the game) is zero. If there are n-
players and the sum of the game is zero, it is an n-person zero sum game.
Two person zero sum game
Characteristics of two-person zero – sum game
i. Only two players participate
ii. Each player has finite number of strategies to use
iii. Each specific strategy results in a payoff.
iv. Total payoff to the two players at the end of each play is zero
v. Payoff is the outcome of playing the game. A payoff (game matrix) is a table showing
the amounts received by the players named at the left hand side after all possible plays
of the game. The payment is made by the player at the top of the table.
Fast Forward: Game theory attempts to mathematically capture behaviour in strategic situations,
in which an individual's success in making choices depends on the choices of others.
Assumptions of Game theory
i. Economic theory – Each player is assumed to be economically rational; each player
desires to win i.e. sometimes referred to as requirements
ii. Information – Both players have the same information set i.e. each player knows the
strategies of the opponent and the consequential payoffs.
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iii. Zero Sum of the game – The gain of a given player equals the loss of the other player
so that the game is zero sum in nature
iv. No indifference – Each player has to play even if they are losing
v. Repeatedness – The game is played expectedly over a long period of time
Example
Consider the following:
Strategy Y1
strategy Y2
min. of row
X1
3 5 3
X2
1 -2 -2
Max. of column 3 5
Determine the strategy that each player will employ.
Both players have dominant or pure strategies therefore the game has a saddle point. Numerical
value of saddle point is the game outcome. Saddle point exists only in pure strategy game. In
this example the saddle point is 3
Note:
In reality player X and Y may not see the saddle point.After the game is played for some••
time, however, each player will realise that there is only one strategy to be played.
From then on these players will only play one strategy which corresponds to the saddle
point.
The value of the game is the average of expected game outcome. If the game is played••
in an infinite number of times the value of this game is 3. If a game has a saddle point
the value of the game is equal to its nominal value.
The saddle point in this example is the largest number in its column and the smallest••
number in its row. This is true for all saddle points.
The minimax procedure is accomplished as follows
1. Find the smallest number in each row
2. Pick the largest of them. The number is called the lower (maxmin) and the row is the X's
maxmin strategy.
3. If the upper value and lower value are the same there is a saddle point which is equal
to the lower and upper value; if not equal, there is no saddle point and if the value of the
game lies between two values such a strategy is called a mixed strategy game.
Note:
The game is said to be fair if the maxmin value = minmax value =0 and it is said to be••
strictly determinable if the maxmin value = minmax value
Always look for a saddle point before attempting to solve a game.••
Mixed Strategy Game
When there is no saddle point, players will play each strategy for a certain percentage of the time.
This is called a mixed strategy game.
Two methods
1. Algebraic method
2. Arithmetic method
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Decision Theory
Algebraic method.
For a 2×2 game an algebraic approach can be used to solve for the percentage of the time each
strategy is played.
Principle of rational expectation
X wants to decide his/her time so that he wins as much when Y is playing Y1
as when Y is playing
Y2
. The following diagram can be used:
Table 6.7
Y
Y1
(p) Y2
(1-p)
X
X1
(Q)
X2
(1-Q)
Where Q, 1-Q equals the proportion time when X plays strategy X1
and X2
respectively P, 1-P
equals the percentage of time when y plays Y1
and Y2
respectively. The overall objective of
each player is to determine the fraction of time that each strategy is to be played to maximise
winnings.
Steps
1. To find X`s best strategy multiply Q and 1-Q times the appropriate game outcome and
solve for Q and 1-Q by setting column 1 equal column 2 in the game.
2. To find Y`s best strategy multiply P and 1-P times the appropriate outcome numbers
and solve for p and 1-p by setting row 1 equals to tow 2 in the game
Arithmetic Method
Steps
1. Subtract the two digits in column 1 and write them under column 2 ignoring the sign
2. Subtract the two digits in column 2 and write them under column 1 sign
3. Similarly proceed to the two rows. These values are called oddments. They are the
frequencies with which the players must use their courses of action in their optimum
strategies.
Table 6.8
Y
Y1
(p) Y2
(1-p)
X X1
(Q) 4 2 9/11
X2
(1-Q) 1 10 2/11
8/11 3/11
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Value of the game= 9/11×2 + 2/11×10
=35
/11
Note: The arithmetic and algebraic method can only be employed if we are dealing with a 2x2
game matrix i.e. this is a situation where each of the two players has two possible courses of
action.
Principle of dominance
The principle of dominance can be used to reduce the size of the games by eliminating strategies
that will never be played. A strategy or a play can be eliminated if the player can always do as will
or better than another strategy. In other words, a strategy can be eliminated if all of its outcomes
are the same or worse than the corresponding game outcomes of another strategy.
Note: The objective is to reduce the game to 2x2 so that to use algebraic or arithmetic method.
Example
Using the principle of dominance reduce the size of the following game
y1
y2
y3
y4
min of row
x1
-5 -4 6 -3 -5
x2
-2 6 2 -20 -20
Max of col. -2 6 6 -3
It has no saddle point.
Y will never play Y2
and Y3
as he will lose. He will play Y1
and Y4
.
Reduce the following by dominance and find the value of the game.
Non zero-sum game
An assumption of zero-sum game is that the gain of a given player equals the loss of the other
player. This may be possible in practice for some situations e.g. many games or sports like
football; athletics etc. Also a fixed make competition can be modeled as a zero-sum game.
However, in the business world many situations cannot realistically be modeled as zero-sum. In
many cases the players will all gain e.g. cartels in the oil industry. They may all lose e.g. fierce
competition in the matatu industry. Some may gain while others lose, not necessarily by the
same amount.
Chapter Summary
There are many types of decision making
Decision making under uncertainty
This refers to situations where more than one outcome can result from any single decision.
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Decision Theory
Decision making under certainty
When only one outcome exists for a decision, we are dealing with this category e.g. linear
programming, transportation assignment and sequencing.
Decision making using prior data
It occurs whenever it is possible to use past experience (prior data) to develop probabilities for
the occurrence of each data.
Decision making without prior data
No past experience exists that can be used to derive outcome probabilities. In this case, the
decision maker uses his/her subjective estimates of probabilities for various outcomes.
Decision making under uncertainty
Several methods are used to make decision in circumstances where only the payoffs are known
and the likelihood of each state of nature are known:
Maximin Method••
Maximax method••
The Laplace method••
The Hurwicz method••
Expected Opportunity Loss (EOL) method••
The Expected Monetary Value method••
Minimax regret method••
Chapter Quiz
1. …………..theorem recognises that a decision maker does not know with certainty
what state of nature will occur.
2. …………. is also called criterion of rationality.
3. …………. is also called criterion of realism.
4. Payoff is the outcome of playing the game is one of the characteristics of two-persons
zero-sum game.
(a) True
(b) False
5. An assumption of ……………. is that the gain of a given player exactly equals the loss
of the other player.
6. Which one of the following is not a characteristic of game theory?
(a) Economic theory
(b) Information
(c) Infinite players
(d) Zero-sum of the game
(e) No indifference
(f) Repeatedness
7. Which principle can be used to reduce the size of the games by eliminating strategies
that will never be played?
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Decision Theory
(b) What would be the company's optimal decision under the optimistic approach and the
minimax regret approach? (8 marks)
(c) Suppose that the sales department has determined that the number of subscribers will
be a function of the pricing plan. The probability for the pricing plans are given below:
Probability under pricing plan
Number of subscribers I II III IV
10,000 0 0.05 0.10 0.20
20,000 0.05 0.10 0.20 0.25
30,000 0.05 0.20 0.20 0.25
40,000 0.40 0.30 0.20 0.15
50,000 0.30 0.20 0.20 0.10
60,000 0.20 0.15 0.10 0.05
Which pricing plan is optimal?
(d) Briefly explain the main difference between the approaches used in part (b) and (c)
above.
June 2002 Question 7
(a) Define the following terms used game theory:
(i) Dominance (2 marks)
(ii) Saddle point (2 marks)
(iii) Mixed strategy (2 marks)
(iv) Value of the game (2 marks)
(b) Consider the two-person zero-sum game between playersAand B given by the following
payoff table
Player B Strategies
1 2 3 4
Player A
strategies
1 2 2 3 -1
2 4 3 2 6
Required:
(i) Using the maximin and minimax values, is it possible to determine the value of the
game? Give reasons. (3 marks)
(ii) Use graphical methods to determine optimal mixed strategy for player A and determine
the value of the game. (9 marks)
(Total: 20 marks)
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December 2002 Question 7
(a) The probability distribution showing the number of hours a specially trained accounts
clerk is required to work daily is as follows:
Hours Frequency
0 0.4
1 0.3
2 0.2
3 0.1
The accounts clerk is assigned to this work daily based on the expected daily requirement. In
addition, customers pay Sh. 2,500 per hour for this service. The accounts clerk contracted wage
is Sh. 1,500 per hour. If more hours are required than are specified in the contract, the cost is
Sh. 3,000 per hour.
Required:
(i) Determine the hours the accounts clerk should be contracted. (4 marks)
(ii) Calculate the net gain or loss to the company for each possible number of hours if the
contract is written for the expected number of hours. (6 marks)
(b) A student team was planning strategy for a management game being played in one of
their classes. They had to decide whether to price their firm's products high or low. They
know that their subsequent profits in either case depended on whether the economy
moved up or down – a variable controlled by the instructor. They estimated that a high
price in an upward economy would net Sh. 5 million in profit but a low price would yield
Sh. 3 million profit. If the economy went down, a high price would net Sh. 2 million and
a low price Sh. 1 million in profit.
Required:
Formulate the above information as a game and determine the students' best strategy.
(4 marks)
(c) A charity is about to launch an appeal for a children's ward in a national hospital and
has to choose between the following fund raising strategies. Strategy I involves an initial
outlay of Sh. 40,000 and administrative costs of 5 cents for each Sh. 1.00 collected.
Strategy II has no initial outlay but instead involves an extensive door-to-door campaign
with administrative costs of 30cents for each Sh. 1.00 collected. It is estimated that,
whichever strategy is used, the amount collected will be as follows:
Amount (Sh.) 50,000 100,000 150,000 200,000
Probability 0.2 0.4 0.3 0.1
Required:
(i) Construct a payoff table (2 marks)
(ii) Determine which strategy the charity should adopt (4 marks)
(Total: 20 marks)
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Decision Theory
December 2003 Question 7
a) In the context of game theory, explain the following statement "winning isn't everything;
it is the only thing" (3 marks)
b) The optimal solution of a two-person zero-sum game always represents a saddle point
regardless of whether the players use pure or mixed strategies. Explain.
(5 marks)
c) Joseph Njau and Anne Wairimu can use one strategy some of the time and the other
strategy the rest of the time. Consider the following case of player one (Joseph Njau)
and the opponent (Ann Wairimu). Joseph Njau has two strategies A and B whereas
the opponent, Ann Wairimu has strategies X and Y. Utilities have been assigned as
indicated below:
Ann Wairimu
X Y
A 1 -2
Joseph Njau
B -15 2
Required:
Determine the percentage time Joseph Njau plays strategy A and strategy B. Also determine the
percentage time that Ann Wairimu should play strategy X and Strategy Y. (12 marks)
(Total: 20 marks)
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Mathematical Programming
CHAPTER SEVEN
Mathematical Programming
Objectives
i. At the end of this chapter, you should be able to:
ii. Define terms such as decision variables, objective and constraint function.
iii. Formulate a linear programming problem.
iv. Interpret a computer assisted solution.
v. Cite applications to transport and assignment problems.
Fast Forward: A business model is a framework for creating economic, social, and/or other
forms of value.
Introduction
There are many activities in an organisation which involve the allocation of resources. These
resources include labour, raw materials, machinery and money. The allocation of these resources
is sometimes referred to as programming. Problems arise because the resources are usually in
limited supply. If a company makes several products, using the same machine and labour force,
management must decide how many of each product to produce. The decision will be made so
that management's objective is satisfied. Management may wish to plan production in such a
way as to maximise the total contribution made each month, or to maximise the utilisation of the
machinery each week, or to minimise the cost of labour each week. The decision variable in this
case is the amount of each product to be made in a given time period.
Similarly, if the company has an amount of capital to invest in a number of projects, the money
allocated to each project will be governed by some objective. It may be necessary to minimise
risk, or to maximise capital growth. The decision variable in this case is the amount of money
allocated to each project.
In general, the objective is to determine the most efficient method of allocating these resources
to the variables so that some measure of performance is optimised. Modelling methods can often
be used to aid in this allocation process.
Mathematicalprogrammingistheuseofmathematicalmodelsandtechniquestosolveprogramming
problems. There are a number of techniques which fall under mathematical programming but we
will only consider the one which is most commonly used: linear programming.
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7.1 LINEAR PROGRAMMING
Linear programming is a suitable method for modelling an allocation problem if the objective and
the constraints on the resources can all be expressed as linear relationships of the variables. The
techniques have a number of distinct steps.
The linear programme must first be formulated mathematically. This means that the variables over
which we have control and the objective must be identified. The objective and the constraints on
the resources are then written down as linear relationships in terms of the variables.
Once the linear programme is complete, all of the feasible combinations of the variables are
identified. The combination which optimises the objective may then be selected. If only two
variables are involved, a graphical solution is possible. If, however, we have a multivariable
problem, we must resort to an algebraic method for which a computer package will be used.
Once the optimum solution has been identified, it must be evaluated. This will include a sensitivity
analysis.
As with any other mathematical aid to decision making, the solution from the linear programme is
just one piece of management information which contributes to the final decision.
Definition of key terms
1. Linear programming - A branch of mathematics that uses linear inequalities to solve
decision-making problems involving maximums and minimums; or, a mathematical procedure
for minimising or maximising a linear function of several variables, subject to a finite number of
linear restrictions.
2. Slack variable - In linear programming a slack variable is a variable which is added to a
constraint to turn the inequality into an equation. This is required to turn an inequality into an
equality where a linear combination of variables is less than or equal to a given constant in the
former.
3. Surplus variable - In linear programming a surplus variable is a variable which is subtracted
from a constraint to turn the inequality into an equation.
This is required to turn an inequality into an equality where a linear combination of variables is
greater than or equal to a given constant in the former.
4. Basic variables – These are variables associated with unit columns of the Simplex Tableau.
All other variables are called non-basic variables.
5. Feasible region - The set of all values which satisfy all constraints; otherwise it is an infeasible
region
6. Sensitivity analysis - A technique of assessing the extent to which changes in assumptions
or input variables will affect the ranking of alternatives.
7. Shadow price - is the maximum price that management is willing to pay for an extra unit of a
given limited resource.
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Mathematical Programming
Industry Context
Linear programming models are used to allocate scarce resources in a way which meets a
business objective. The objective might be to maximize weekly profit or to minimize daily costs.
The steps in formulating a linear-programming model are:
Step 1: Identify the decision variables
Step 2: Identify the linear objective function and the constraints
Step 3: Express the objective function in terms of the variables
Step 4: Express the constraints in terms of the variables
Exam Context
Student should be very prepared to tackle business modeling questions. In most cases the
questions are not straightforward and this will require in-depth analytical skills. The past papers
have, however, been outlined below for the student to practice adequately:
Past Paper Analysis
12/06, 6/06, 12/04, 6/04, 12/03, 6/03, 12/02, 6/02, 12/02, 6/02, 12/01, 12/00, 6/00
Problem formulation
The basic procedure is the same for the formulation of all linear programmes:
Step 1: Identify the variables in the problem for which the values can be chosen, within the limits
of the constraints.
Step 2: Identify the objective and the constraints on the allocation.
Step 3: Write down the objective in terms of variables.
Step 4: Write down the constraints in terms of the variables.
The same procedure is used no matter how many variables there are, but we will look first at two
variable problems.
To formulate a two variable linear programme
Example 1
A small family firm produces two old-fashioned non-alcoholic drinks. 'Pink fizz' and 'Mint pop'.
The firm can sell all that it produces but production is limited by the supply of a major ingredient
and by the amount of the machine capacity available. The production of 1 litre of 'Pink fizz'
requires 0.02 hours of machine time, whereas the production of 1 litre of 'Mint pop' requires 0.04
hours of machine time. 0.01 kg of a special ingredient is required for 1 litre of 'Pink fizz'. 'Mint pop'
requires 0.04 kg of this ingredient per litre. Each day the firm has 24 machine hours available
and 16kg of the special ingredient. The contribution is £0.10 on 1 litre of 'Pink fizz' and £0.30 on
1 litre of 'Mint pop'. How much of each product should be made each day, if the firm wishes to
maximise the daily contribution?
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Solution
Step 1: Identify the variables within the limits of the constraints. The firm can decide how much
of each type of drink to make. Let p be the number of litres of 'Pink Fizz' produced per day. Let m
be the number of litres of 'mint pop' produced per day.
Step 2: Identify the objective and the constraints. The objective is to maximise the daily
contribution
Let £P per day be the contribution. This is maximised within constraints on the amounts of
machine time and the special ingredients available.
Step 3: Express the objectives in terms of the variables:
P = 0.01 + 0.30m (£/day)
This is the objective function – the quantity we wish to optimise.
Step 4: Express the constraints in terms of the variables. The contribution is maximised subject
to the following constraints on production:
Machine time: to produce p litres of 'Pint fizz' and m litres of 'Mint pop' requires (0.02p + 0.04m)
hours of machine time each day. There is a maximum of 24 machine hours available each day,
therefore the production must be such that the number of machine hours required is less than or
equal to 24 hours per day. Therefore:
0.02p + 0.04 m < 16 24 hours/day
Special ingredient: to produce p litres of 'pink fizz' and m litres of 'Mint pop' requires (0.01p +
0.04m) kg of the ingredient each day. There is a maximum of 16 kg available each day, therefore
the production must be such that the amount of the special ingredient required is at most 16kg
per day. Therefore:
0.01p + 0.04m < 16 kg/day
There are no further constraints on the production, but it is sensible to assume that the firm
cannot make negative amounts of drink, therefore.
Non negativity:
P >0, m >0
The complete linear programme is as follows.
Maximise:
P = 0.10p + 0.30 m (£/day)
Subject to:
Machine time: 0.02 p + 0.04m < 24 hours/day
Special ingredient: 0.01p + 0.04m < 16 kg/day
p,m >0
Example 2
A manufacturer of high precision machined components produces two different types, X and
Y. In any given week, there are 4,000 man-hours of skilled labour available. Each component
X requires one man-hour for its production and each component Y requires 2 man-hours. The
manufacturing plant has the capacity to produce a maximum of 2,250 components of type X each
week as well as 1,750 components of type Y. Each component X requires 2kg of plate. Each week
there are 10,000 kg each of rod and plate available. The company supplies a car manufacturer
with the unions that at least 1,500 components will be produced each week in total.
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Mathematical Programming
If the unit contribution for component X is £30 and for component Y is £40, how many of each
type should be made in order to maximise the total contribution per week?
Solution:
First the linear programme must be formulated.
Step 1: Choose the variables: produce X components of type X per week and Y components of
type Y per week.
Step 2: What is the objective? What are the constraints on the production? The objective is to
maximise the total weekly contribution. The production is constrained by the amount of:
a) Labour – maximum available is 4000 hours per week
b) Machine capacity – there is a separate limit for each product. The machines can make
at most 2,250 components of type X each week and at most 1750 components of type
Y each week.
c) Rod – maximum available is 10,000kg per week
d) Plate – maximum available is 10,000kg per week
In addition, minimum amounts of each product are required:
a) Regular orders – the number of component X made must be at least enough to satisfy
the regular orders
b) Union agreement – the total number of components (x +y) must at least satisfy the
agreement.
Step 3: The objective function. Let £P be the total contribution per week, where:
P = 30 x + 40 y (V/week)
Step 4: The constraints on production. For each resource, the amount of resource required each
week to produce X components of type X and Y components of type Y is given below, together
with the maximum amount of the resource available.
Labour required: 1 x + 2y ≤ 4000 hours/week
Machine capacity required: x ≤ 2250 components/week
Y ≤1750 components /week
Rod required: 2x + 5y ≤ 10,000 kg/week
Plate required: 5x + 2y ≤ 10,000 kg/week
In addition
Regular orders: x ≥ 600 components/week
Union agreement: x + y ≥1500 components/week
Non-negativity: x, y ≥ 0
The complete linear programme is:
Produce x components of type X and Y components of type Y each week. Maximise:
P = 30 x + 40 y (£ /week)
Subject to the constraints:
Labour: 1x + 2y ≤ 4000 hours/week
Machine capacity: x ≤ 2250 components/week
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Y ≤1750 components/week
Rod: 2x + 5y ≤ 10,000 kg/week
Plate: 5x + 2y ≤ 10,000 kg/week
Regular orders: x ≥ 600 components/week
Union agreement: x + y ≥1500 components/week
Non-negativity: x, y ≥ 0
We will look next at a problem involving more than two variables. The procedure is identical.
To formulate a multivariable linear programme
Example 1
Electra Plc produces personal computers and word processors. Currently four models are being
produced.
The Jupiter – 512K memory, single disk drive;
The Venus – 512K memory, double disk drive;
The Mars – 640K memory, double disk drive;
The Saturn – 640K memory, hard disk
Each computer passes through three departments in the factory – sub, assembly, and test. The
details of the times required in each department for each model are given in Table 1 together
with the maximum capacities of the departments. The marketing department has assessed the
demand of each model. The maximum forecast demands are also given in the table, together
with the unit contribution for each model:
Table 7.1 Times required in each department for each model
Unit production time, hours Maximum available hours/month
Department Jupiter Venus Mars Saturn
Sub-assembly 5 8 20 25 800
Assembly 2 3 8 14 420
Test 0.1 0.2 2 4 150
Maximum forecast
Demand/month 100 45 25 20
Contribution 15 30 120 130
Construct the linear programme for this product mix problem, if the objective is to maximise the
total contribution per month.
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Solution
Step 1: Choose the variables produce:
Let
J units of the Jupiter per month
V units of the Venus per month
M units of the Mars per month and
S units of the Saturn per month
Step 2: What is the objective? What are the constraints on the production? The objective is to
maximise the total contribution each month. The production is constrained by the number of the
labour hours available in the three departments and by the number of each model which can be
sold.
Step 3: The objective function. Let £p be the total contribution per month, where:
P = 15j + 30v + 120m + 130s ( £/month
Step 4: The constraints on production. For each department, the amount of time required to
produce j, v, m and s units of the respective models is linked to the maximum time available.
Sub – assembly: 5j + 8v+ 20m +24s ≤ 800 hours/month
Assembly: 2j + 3v + 8m + 14s ≤ 420 hours/month
Test: 0.1 j + 0.2v + 2m + 4s ≤ 150 hours/month
Demand for Jupiter: j ≤ 100 units/ month
Demand for Venus: v ≤ 45 units/month
Demand for Mars: m ≤ 25 units/month
Demand for Saturn: s ≤ 20 units/month
Non-negativity: j, .v, m, .s ≥0
The complete linear programme is each month produce j, v, m and s models, respectively of the
Jupiter , Venus, Mars and Saturn. Maximize:
P= 15 j + 30 v + 120m + 130 s (£/month)
Subject to the constraints given above.
Example 2:
A portfolio manager wishes to invest up to £100,000 to maximise the total annual interest income.
She has narrowed her choices to four possible investments, A, B, C and D. Investment A yields
6% per annum, investment B yields 8% per annum, investment C 10% per annum and investment
D 9% per annum. The four investments have varying risks and conditions attached to them. For
liquidity, at least 25% of the funds must be placed in investment D. Volatile government policies
indicate that no more than 20% of the investments should be in C, whereas tax considerations
require at least 30% of the funds to be placed in A. Formulate a linear programming model for
this investment problem.
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STUDYTEXT
Solution:
Invest £a in investments, A, £b in investment B, £c in investment C and £d in investment D. The
objective is to maximise the total interest income per year. The investment is constrained by the
considerations of safety, liquidity, government policy and taxation. Let £R be the total interest
income per year, where:
R = 0.06a + 0.08b + 0.10 c+ 0.09d (£/year)
Maximise subject to:
Total investment: a +b + c + d ≤ 100,000 £ available
Safety: a +b ≥ 0.5 (a +b + c +d) £
Liquidity: d ≥ 0.25 (a + b + c + d) £
Government policy: c≤ 0.2 (a + b +c +d) £
Tax: a≥ 0.3 (a + b + c + d) £
Non-negativity: a, b, c, d ≥ 0
To solve a linear programme, it is conventional to arrange the constraints so that the variables
appear only on the left hand side of the inequalities. This is done below. The complete linear
programme is:
Invest £a in investment A,
£b in investment B
£c in investment C
And £d in investment D.
Maximise the total interest income per year, where:
R = 0.06a + 0.08 b + 0.10 c + 0.09 d (£ /year)
Subject to the constraints:
Total investment: a + b + c + d ≤ 100.000 £
Safety: 0.5 a + 0.5 b – 0.5c + 0.5 d ≥ 0 £
Liquidity: - 0.25 a – 0.25b – 0.5 c – 0.5 d ≥ 0 £
Government policy: - 0.2 a – 0.2 b + 0.8 c – 0.2 d ≤ 0 £
Tax: 0.7 a – 0.3 b – 0.3 c- 0.3 d ≥0 £
Non-negativity: a, b, c, d ≥ 0
The objectives in the four examples above have all required a quantity to be maximised. The
procedure is identical at this stage if the objective is to minimise some measure.
Solving the linear programme
We are now at the stage of considering how to find the values of the variables which will satisfy
all of the constraints simultaneously and optimise the objective. We are more accustomed to
dealing with equations rather than inequalities. It is a simple matter to change the inequalities into
equations. We add an additional variable to the left hand side of the inequality.
245.
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Mathematical Programming
To demonstrate this procedure, let us refer to example 2 which concerned the manufacturer of
machine components X and Y. We will include an additional variable in each constraint to produce
a set of equations. This variable is denoted by s hence s1 is included in the first constraint, s2
in the second constraint and so on. We will also impose the condition that the values of these
variables cannot be negative i.e. s1 ≥ 0 all ≥. This means that the variable is added to the
left hand side for all ≤ constraints and subtracted from all ≥ constraints. The linear programme
becomes: produce x components of type x and y components of type Y each week. The objective
is to maximise the total weekly contribution.
Maximise
P = 30x + 40y (£/week)
Subject to the constraints:
Labour: 1x + 2y + s1
= 4000 hours/week
Machine capacity: x + s2
= 2250 components/week
Y + s 3
= 1750 components/week
Rod: 2x + 5y + s4
= 10,000 kg/week
Plate: 5x + 2y + s5
= 10,000 kg/week
Regular orders: x - s6
= 600 components/week
Union agreement: x + y –s7
= 1500 components/week
Non-negativity: x, y ≥ 0
These additional variables are called Slack variables. In the ≤ constraints. They represent
the amount of the resource not used, that is, the difference between the resource used and
the maximum available. For example, look at the labour constraint above. Suppose 1,000
components of hours. Since 4,000 hours are available, the spare capacity, or slack is (4,000-
3,000) = 1,000 hours. For this combination, the negative slack variables are referred to as Surplus
variables since they represent the amount of resource being used over and above the minimum
requirement. For example, look at the 'regular orders' constraint, when 1,000 components of type
X are being produced. The minimum number of type X required by this constraint is 600, hence
a production level of 1,000 gives a surplus of 400 components above the minimum. Therefore,
s6 takes the value 400.
Start
We now have a set of simultaneous equations. However, the number of variables is greater than
the number of equations. A unique set of solutions will arise only if the number of variables and
the number of equations are the same. The best we can do is to identify a set of feasible solutions
to the equations. This set of feasible solutions gives all combinations of the variables which
satisfy all of the constraints. We will then select from this set the particular solution or solutions
which optimise the objective.
How do we set about identifying the set of feasible solutions? This can be done graphically if
the problem involves only two variables. However, we must resort to an algebraic method if the
problem is multivariate.
246.
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The algebraic solution of a linear programme
Alinear equation represents a set of points which lie on a straight line.Alinear inequality represents
an area of a graph. For example, x ≤ 7 says that X takes a value which is less than 7 or is equal to
7. The situation can be illustrated graphically as follows. Draw the line x = 7, see left hand graph
in figure 12.1. This divides the graph into three sets of points for which x = 7, the line itself; those
for which x < 7, the area to the left of the line; and those for which x > 7, the area to the right of
the line. We do not require this last set. It is usual to shade the area not required. See, the right
hand graph in Diagram 7.1 below
Diagram 7.1 Graphical representation of the inequality x ≤ 7
Suppose x + y ≤ 10, which area does this represent? The procedure is the same as in the previous
example. First of all we draw the line x + y = 10. See the left hand graph below. Again the line
divides the graph into three sets of points: those for which x +y = 10, the line; those for which x +
y < 10, the area below the line; and those for which x + y > 10, the area above the line.
A useful technique for deciding which is the rejected area on the graph is to take any point on the
graph away from the line and substitute its values into the inequality. If the inequality still makes
sense, then that point is feasible solution. If the inequality is untrue, then the point is infeasible
and lies in the rejected region. The origin is a convenient point to use. Substitute x = y = 0 into
the inequality x +y ≤ 10, we have 0 + 0 ≤ 10 which is a true statement, therefore the origin is a
feasible solution and we should reject the other side of the line. See the right hand graph I figure
2
Diagram 7.2 Graphical representation of the inequality x +y ≤ 10
y
LineX=7
X=7
Area X>7Area X<7 Area X<7 Area not required
y
X = 7
7 7x x
y
X + y=1010
X + y>10
10
x
X+ y = 10
X+ y<0
Rejected area
y
10
10
x
247.
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Mathematical Programming
Each constraint in the linear programme can be drawn in this way and the rejected area shaded.
If all of the constraints are drawn on the same graph, the area which remains unshaded is the
set of points which satisfies all of the constraints simultaneously. This area is called the feasible
region. For a linear programme, it does not matter which variable is plotted on which axis. The
origin should always be included on the graph. False zeros must not be used. Let us now apply
this procedure to the linear programme for example about the production of the two types of soft
drink. We can illustrate the constraints graphically.
Machine time; 0.02p + 0.04 m≤ 24 hours/day.
Plot the line 0.2p + 0.04 m = 24. An easy way of plotting the line is to find the points where the
line crosses the p and the m axes. Put p = 0 into the equation and calculate p, ie when m = 0, p =
1200. Plot these two points and join them to give the line. This method always works unless the
line passes through the origin. In that case revert to the alternative procedure of substituting any
other value of p and finding the equivalent value of m.
To find which side of the line to shade put p = 0 and m = 0 in the inequality:
0.02 x 0 + 0.04 x 0 < 24
This statement is true, so the origin is included in the feasible area
Special ingredient: 0.01p + 0.04m ≤ 16
Then you plot the line:
0.01p + 0.04 m = 16
Again the origin is included in the feasible region so we shade out the area above the line
Shade out negative values of each variable
Putting these four constraints together on one graph gives:
Diagram 7.3 Graphical representations of the constraints for example 1
Special ingredient
0.01p+0.04m=16kg/day
nFeasible
region
200 400 600
248.
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STUDYTEXT
The area left unshaded by all of the constraints is the feasible region and this contains all of the
possible combinations of productions which will satisfy the given constraints. The co-ordination
of any point within the feasible region represents a possible combination of soft drink production
for this firm.
We must now consider how to choose the production which will maximise the firm's daily
contribution. The objective function is:
P= 0.01p + 0.30m (£/day)
If we like p = 100 £ per day, then we can illustrate the objective function graphically. If we then
give p another value, the new line will be parallel to the one for p = 100 £ per day.
We can generate the entire family of possible contribution lines by drawing one in particular, then
moving across the feasible region parallel to it. The further from the origin we move, the larger is
the contribution.
If we draw a contribution line on the graph of the linear programme, as in figure 4 we can move
parallel to this line across the feasible region in the direction of increasing contribution until we
reach the last feasible solution(s), before the line moves into the infeasible region.
Diagram 7.4 Linear Programme For Example 1
We can see the point A is the last feasible solution. The co-ordinates of point A give the optimum
combination of production for the two drinks. The approximate co-ordinates of point A can be
read from the graph, but, for precision, the co-ordinates are calculated by solving simultaneously
the equations of the two constraints which form point A.
These two constraints are called the binding or limiting constraints. They are the resources
which are being used fully and therefore prevent the daily contributions from increasing further.
The optimum solution is the intersection of:
0.02p + 0.04 m = 24 (1)
0.01p + 0.04m = 16 (2)
Subtract (2) from (1)
0.01p = 8
Special ingredient
0.01p+0.04m=16kg/day
pA point giving max.
contribution
Typical contribution
0.1p+0.3m=£100
200 400 600
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Mathematical Programming
Therefore:
P = 800 litres/day'
Substitute into (2) and find m:
x 800 + 0.04 m = 16
Therefore: m = 200 litres/day
To maximise the daily contribution, the firm should produce 800 litres of 'Pink Fizz' and 200 litres
of 'Mint Pop' each day. This will yield a maximum contribution of:
0.10 x 800 + 0.30 x 200 = £ 140/day
This combination utilises all of the machine time and special ingredient available each day. There
is no spare capacity or slack on either of the constraints.
This method of identifying the optimum corner depends on a suitable profit line being drawn.
The following is a practical note which will help to obtain a suitable profit line from which to
identify the optimum corner. Choose any inconvenient point near the middle of the feasible region.
Suppose in the above example the point m = 200, p = 200 is chosen. The daily contribution from
this product mix is:
P = 0.10 p + 0.30 m = 0.10 × 200 + 0.30 x 200 = £80 /day
The other entire product mix which give daily contribution of £80 lie on the line:
80 = 0.10p + 0.30 m (£/day)
One point on this line is already known, i.e. m = 200, p = 200. A second point might be m =
0, hence, p = 800. This particular daily contribution line is now drawn on the graph and the
procedure described above is followed to identify the optimum solution (s). It is clear from the
procedure that the optimum will always be at a corner of the feasible region, or, if the objective
function is parallel to one of the constraints, at any point on the line joining two corners.
We have assumed that the variables in the linear programme are continuous or, if not, then
fractions are acceptable. It will often be the case that part units are allowed the time period of the
problem. For example, if two models of a car are being produced and the objective of the linear
programme is to maximise the machine usage per week. For such a product 'work in progress'
is allowable on weekly basis.
If, however, we are allocating workers to task, part workers are not acceptable. In this case the
optimum solution must produce integer values. The feasible solutions are all the points in the
feasible region where the variables are integers. The last point within the feasible region which
has integer coordinates is selected and this may no longer be at a corner of the feasible region.
For two variable linear programmes, it does not make much difference to the solution procedure
if the variables must be integers. The feasible region is placed by the set of feasible points within
the constraint boundaries. The typical objectives function is moved through these points, rather
than through the feasible region as a whole. In the multivariable case, however, the method of
integer programming is used.
Refer to Example 2 which is concerned with the production of two machined components. We
wish to know the product mix which will achieve maximum total contribution per week.
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STUDYTEXT
Solution:
The feasible region for each constraint is as follows:
The feasible region, containing all of the possible product mixes for this problem, is shown
unshaded.
We wish to identify the optimum product mix which will maximise the weekly contribution. The
objective function is:
P = 30 x + 40 y (£/week)
To plot this function for a typical value of the weekly contribution, we select the point x = 1000, y
= 1000 which is in the feasible region. The weekly contribution for this product mix is:
P = 30 x 1000 + 40 x 1000 = £ 70,000/week
We will use the contribution line:
70,000 = 30x + 40y (£/week) as the trial line. This line also passes through the point x = 0,y =
1750. It is shown by the broken line on figure 5. In the direction of increasing contribution leads
us to point A as the last feasible solution.
Diagram 7.5 A linear programme for the weekly production of machined components of type x and y
Plate
5x+2y=10 000
W e e k l y
production
of type y in
000s
Orders x=600
m/c capacity
y=2250
Point of max
contribution
Rod 2x+5y=10 000
m/c capacity
y= 1750
Labour x+2y=4000
5
4
3
2
1
0
0 1 2 3 4 5
Agreement
x+y=1500
Typical
contribution
30x+40y=70000
x
y
251.
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Mathematical Programming
The binding constraints are therefore:
Labour: x + 2y ≤ 4000 hours/week
Plate: 5x + 2y ≤ 10,000 kg/week
Solving the corresponding equations simultaneously gives
X + 2y = 4000 (1)
5x + 2y = 10,000 (2)
(2) – (1) 4x = 6,000
Therefore x = 1250, and y = 1500 by substitution.
The optimum product mix is 1,500 of component x and 1,250 of component y each week. The
maximum contribution per week will then be:
P max = 30 x 1500 + 40 x 1250 = £95,000/week
This product mix uses all of the labour hours available and all plate. These are the binding
constraints. However, there will be spare capacity on machine time for both components and
spare rod capacity. The production will also exceed the minimum required by the regular orders
and the minimum required by the union agreement.
We find the value of the slack variables in the machine time constraints are 750 of component x
and 500 of machine tool y, i.e.
1500 + s2 = 2250, therefore s2 = 750 components/week, and
1250 + s3= 1750, therefore s3 = 500 components/week
The slack in the rod constraint is:
2x 1500 + 5 x 1250 + s = 10,000
Therefore s = 750 kg/week
The surplus on the regular orders constraints is:
1500 – s6 = 600
Therefore: s6 = 900 components/week
Above the minimum needed for the regular orders. The surplus on the agreement is:
1500 + 1250 – s7 = 1500
Therefore s7 = 1250 components/week
Above the minimum required by the union agreement.
As we have already said, the optimum solution will normally be at a corner of the feasible region.
It is possible, therefore, once the graph has been drawn to identify the optimum corner by
evaluating the objective function at each corner of the feasible region. A basic solution is the
name given to the set of variable values at a corner of the feasible region. The basic variables
are those variables which have non-zero values at a particular corner.
Occasional problems arise when solving a linear programme. The problem may be infeasible.
In this case, there is no feasible region. No combination of the variables satisfies all of the
constraints simultaneously and the linear programme is unbounded. In this case, the solution
can be increased indefinitely without violating constraint. This usually means that the linear
programme is formulated incorrectly, with some constraints missing.
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STUDYTEXT
The issue of non-unique solutions was mentioned earlier. These arise when objective function is
parallel to a binding constraint. Any point on that constraint between the two optimum corners,
will give the optimum value of the objective function. Any one of these points forms an optimum
solution to the model. This can be useful situation since it gives the decision maker some
flexibility.
Sensitivity Analysis
In most decision making activities it is prudent to maximise the preferred course of action to
see what effect changes in the problem will have on the decision. Linear programming is no
exception. There are three aspects of the problem which we need to consider.
The effect of additional supplies of the limiting resources;
The effect of changes in the non-limiting resources;
The effect of changes in the coefficients of the objective function.
How do changes in the non-limiting resource affect the optimum solution?
In example 2.6 we considered the two limiting constraints of labour and plate. The other constraints
are not binding at the original optimum solution. These constraints are:
-Machine time to produce component X
-Machine time to produce component Y
-Rod
-Regular orders
-Union agreement
What happens as each of these constraints is charged?
The first three are less-than-or – equal – to constraints. Any of their availabilities will not affect
the optimum solution. However, any decrease in these three constraints can affect the solution.
The tightening of one of the non-limiting constraints will cause it to move towards the origin. At
first, the only change will be a reduction in the size of the feasible region. When, however, the
particular constraint passes through the original optimum corner. It will itself become limiting and
a new optimum solution will emerge.
It is useful to know what the lower limits are on these constraints. The machine capacity for
X can be reduced by 750 hours, from 2,250 to 1,500 hours, before it affects the solution. The
machine capacity for Y can be reduced by 500, from 1,750 to 1,250 hours. The supply of rod
can be reduced by 750 per week, from 10,000kg to 9,250kg. These reductions are the values of
the slack variables mentioned earlier. The greater- than- or-equal to constraints, for the regular
orders and the union agreement, act in the opposite way. Any reduction in these requirements
will increase the feasible region but will not affect the optimum solution.
Any increase in these constraints will first reduce the feasible region but will not affect the optimum
solution. If the regular orders for X increase by at least 90 to 1,500, the optimum will begin to
change. If the union's agreement was increased by at least 1,250, to more than 2,750, there
would be no feasible region and no solution. These increases are the surplus referred to earlier.
253.
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Mathematical Programming
How do changes in the coefficients of the objective function affect the optimum
solution?
It is inevitable that the circumstances under which a linear programme is formulated will change.
Major changes will probably mean that the work will have to be done again but may be possible
to identify the effect of minor changes from the solution to the original problem. In this section, we
consider changes to the objective function. If the objective is to maximise weekly contribution, a
change in the cost of raw material will alter the coefficients in the objective function.
In an investment portfolio problem, if the objective is to maximise the annual return on the
investments, a change in the interest rate earned on one of the investment will change that
coefficient in the objective function.
We will consider situations when the coefficient change is one at a time.
Suppose:
P = ax + 4y (£/week) represents the objective function for a profit maximising linear programme,
where £4 per unit is the profit on product Y and £a per unit is the profit on product X. The profit
on product X is liable to change. Suppose this linear programme has been graphed with X and
Y in the conventional directions. It is helpful to re-arrange the objective function so that y is the
subject:
Y = p/4 – (a/4) x
The profit line cuts the Y axis at p/4 and the slope of the profit line is – (a/4). The intercept on
the Y axis is independent of the value of a, but the slope of the line increases as an increase,
and decreases as a decrease. In other words, as the value of slope change, the profit lines
rotate. Small rotations in either direction will not usually alter the optimum corner. However, larger
rotations will result in different corners emerging as the optimum. It is useful to know the range
which 'a' can take before a particular corner ceases to be the optimum. A similar argument would
apply if the coefficient of x was fixed and the coefficient of y was liable to change.
The simplex solution of multi-variable linear programmes.
An algebraic solution method is required if a linear programme contains more than two variables.
The basic principle of solution of multi-variable model is very simple. It is assumed that the
optimum solution is at one of the 'corner' of the feasible region. Therefore, we systematically
evaluate the objective function for each corner until we find the one which gives the optimum
value of the objective function. We employ the techniques of matrix algebra and an algorithm
for moving from corner to corner of the feasible region in such a way that a move is made only if
it improves the value of the objective function. If, at a particular basic solution, no further move
is recommended then we know that the optimum solution has been reached. This algorithm is
called the Simplex method. A detailed explanation of the simplex method is not necessary,
since multi-variable linear programming models are normally solved by using one of the many
computer packages which are readily available for this purpose. However, an understanding of
the basic principles of the method is helpful for fully interpreting and evaluating the solution to a
linear programme which has been obtained by computer package.
The basic simplex method assumes that the linear programming model is a maximising one,
subject to a set of ≤ constraints. This means that the algorithm can take the origin as the initial
corner. The search for the optimum always starts from a zero value for the objective function.
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STUDYTEXT
The simplex method can be adopted for minimising problems and for problems with ≥or =
constraints. This involves the introduction of artificial, as well as slack and surplus, variables. We
have omitted these complications, since the problems are usually solved by a computer package
which automatically introduces these variables into the model.
The basic model with which we will work may be formally written as:
Maximise Z = c1 x1 + c2 x 2 + …… + c n x n
The C1 are constants. This is maximised subject to a set of m linear constraints:
a 11
x1
+ a 12
x2
+a13
x3
+ …… a1n
x n
≤ b1
a21
x1
+ a22
x2
+a 23
x3
+ …… a2n
x n
≤ b2
a31
x1
+ a32
x2
+a33
x3
+ …… a3n
x n
≤ b3
a m1
x1
+ a m2
x2
+a m3
x3
+ …… a mn
x n
≤ b m
x ≥ 0
There are n variables and m constraints. The double subscripts for the coefficients in the left
hand side of the constraints refer first to the constraints, then to the variable. For example, + a32
is in constraint 3 and is the coefficient of the variable x2. We will illustrate the use of the simplex
method by considering a simple two variable problem which we will first solve graphically. This
will enable us to compare the graphical and algebraic solutions.
Interpretation of computer generated solution
Example
Maximise 25x1
+ 20x2
+ 24x2
where: x1
= Xtrgrow, x2
= Youngrow, x3
= Zupergrow
Subject 0.3x1
+0.2x3
≤ 500
0.5x2
+ 0.4x3
≤ 1000
0.2x1
+ 0.1x2
+0.1x3
≤ 800
0.4x2
0.3x3
≤ 600
x1 ≥ 1500
x1
≥ 0, x2
≥ 0, x3
≥ 0
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STUDYTEXT
Value: this is value that the variables assume at optimal solution (to optimise the••
objective function one needs to produce this amounts of the variables). In our example
we are required to produce 1,666.67 of x1 and 1,750 of x2 and none of x3
Objective coefficients: these are the coefficients of the objective function••
Objective value contribution: this is the value contributed by each variable to the••
objective function (for x1=25×1,666.67), the total of this is equal to our objective value
(i.e 41,666.67+35,000=76,666.67).
The 3 columns of the second part of table1 can be interpreted as follows;
Constraints: this is constraints of the model representing the limited resources.••
RHS: the Right hand side value is the limiting value of the constraint. e.g for the first••
constraint the maximum amount of material A is 500 tons.
Slack/surplus: at optimal production not all the materials for some of the constraints••
will be fully utilised, slack is the amount of material that is left over after production. For
constraint 1 and 4 no material remained; this also implies that these are the binding
constraints i.e their adjustment directly affects the objective solution
To illustrate the use of the simplex method.
A firm manufactures two products, X and Y, subject to constraint on three raw materials, RM1,
RM2, and RM3. The objective of the firm is to select a product mix which will maximise weekly
profit.
The linear programme for the problem is:
Produce x units of product x per week and y units of product y per week.
Maximise the weekly profit, £P, where P = 2x =Y (£/week)
Maximise subject to:
RM1: 3x≤ 27 kg/week
RM2: 2y≤ 30 kg/week
RM3: x +y≤ 20 0kg/week
x, y ≥ 0
Determine the optimum product mix and the maximum value of the weekly profit. State the spare
capacity on each resource.
Solution
Simplex method arranges the coefficients in the left hand side of the constraints equations in a
matrix format. Label the columns with the name of the variables to which they refer. Put the right
hand side values of the constraints in a separate column on the right of the matrix label the rows
with the names of the variables which are basic (have --- values) at the initial corner (the origin).
Finally, add the objective function as an addition row to the table. There are several slightly
different of the objective function to be entered as negative values. The resulting matrix is called
coefficients of the objective function to be entered as negative values. The resulting matrix is
called the first simplex tableau. The above procedure is step 1 in the logarithm
259.
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STUDYTEXT
Mathematical Programming
The basic variables are those which are non-zero at the optimum corner. The values of the basic
variables are in the corresponding row of the B column. Therefore:
X = 9 units/week
Y = 11 units/week
And the slack for raw material 2 is 8kg per week.
All other variables are zero that is the slack on constraint 1, s1 and constraint 3, s3 are
zero. This means that these constraints are binding and the available raw material 1 and 3 are
fully used. The optimum value of the objective functions is in the 'b' column of the objective
function row. The maximum value of weekly profit is £ 29. This solution corresponds exactly with
the graphical solution. The figures in the objective function row and the slack variable columns
shown in Table above give the shadow prices. The shadow price on constraint 1, RMI, is £ 1/3
per kg and the shadow price for constraint 3 is £1 per kilogramme of RM1 becomes available;
the weekly profit will increase by 33 pence (less any additional costs above the normal cost of
RM1). Similarly if an extra kilogram of RM3 becomes available, the weekly profit will increase
by £ 1 (less any additional costs). The figures for the shadow prices may be checked from the
graphical solution.
We will look at constraint 1 only to illustrate the point.
Constraint 1, for raw material 1 is 3x = 27 kg per week, if this constraint is relaxed by one
kilogramme, 3x = 28. The optimum corner will still be the intersection of constrains 1 and 3. Look
back at the graph to check this. The new optimum corner has the co-ordinates:
X = 28/3 = 9 1/3
And 28/3 + y = 20
Giving:
Y = 32/3 = 10 2/3
The new value of the maximum weekly profit is:
2 x (28/3) + (32/3) = 88/3 = £ 29.33/ week
This is again 33 pence for one kilogramme
The shadow price for RM1 is 33 pence kilogram. The remaining values on the page on the edge
of the final tableau, are those in the objective function row and the variable columns. In this
example, the value in both the x and the y columns are zero. These numbers will be non- zero if
any of the variables is non-basic in the optimum solution. For example, if the optimum solution
had said that we should produce only product X, then Y would be non-basic, is y = 0. In that case,
the figure in the Y column of the objective function row tells us by how much the maximum value
of the objective function would decrease if we insisted on producing one unit of Y.
Suppose the following had been the final tableau for the current problem:
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Table 7.3 Modified final tableau
Basic variables
Variables
X y s1
s2
s3
Right hand side , b
x
s2
s3
9
8
11
Objective function, p 0 0.5 1/3 0 0
18 = value of maximum
profit
In this solution, the optimum produces 9 units of x and none of y. If we feel that we must produce
some units of y, the value of the objective function will decrease by £0.5 for each unit of y which
is produced.
Shadow or dual prices
Definition: A shadow price or a dual price is the amount increase (or decrease) of the objective
function when one more (or one less) of the binding constraints is made available.
Maximise 4X1 + 3 X2
Subject to: 0.5X1 + 0.33X2 ≤ 12 (Machine hours)
0.5X1 + 0.5X2 ≤ 14 (labor hours)
Starting with machine hours; let's assume that one more machine hour is available (with labour
hours remaining constant)
We get:
0.5X1 + 0.33X2 = 13
0.5X1 + 0.5X2 = 14
Solving this simultaneously we get the values of X1 and X2 as
0.17 X2 = 1
X2 = 5.88
X1 = 22.12
Thus the contribution is
4(22.12) + 3(5.88) = Sh.106.12
Comparing this with its original contribution of Sh100.24 (see example 1) we see increasing
machine hours by one unit has increased contribution by Sh5.88, which is the shadow price per
machine hour.
Note: This figure is also arrived at if we assume that machine hours are reduced by 1 unit ie
12-1.
Similarly assuming that one more labour hour is made available, then contribution change is:
0.5X1 + 0.33X2 = 12
0.5X1 + 0.5X2 = 15
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Mathematical Programming
Solving this simultaneously gives:
0.17 X2 =3
X2 = 17.65
X1 = 12.35
Which give a contribution of:
4(12.35) + 3(17.65) = Sh102.35
The contribution change is Sh2.11 which is the shadow price per labour hour.
Note:
The shadow prices apply in so far as the constraint is binding; for example, if more and more
labour hours are available, it will reach a point where labour hours are no longer scarce thus
labour hours cease to be a binding constraint and its shadow price becomes a zero. (All non-
binding constraints have zero shadow price). Logically, it is senseless to pay more to increase a
resource, which is already abundant.
Interpretation of shadow prices
A shadow price of a binding constraint indicates to management how much extra contribution will
be gained by increasing a unit of the scarce resource.
In the example above, Sh2.11 is the shadow price for labour hours. This implies that management
is ready to pay up to Sh2.11 extra per hour for the extra hours i.e. say an employee is paid Sh5
per hour and one day he works for two hours extra (overtime), the management is prepared to
pay up to Sh7.11 per hour for the two hours overtime worked.
Sensitivity analysis and the simplex method
The final tableau of the simplex algorithm can be used to carry out a sensitivity analysis of the
solution of a linear programming model. For limiting constraints, the values in the corresponding
slack variable column represent the change in the values of the basic variables if one additional
unit of the limiting resource is available.
Example
To illustrate the use of the final simplex tableau for a sensitivity analysis of the limiting constraints,
we will use the final simplex tableau of Example above to determine:
The effect on the optimum solution if one additional kilogramme of RM1 becomes available
The effect on the optimum solution if two additional kilogrammes of RM1 become available
The effect on the optimum solution if five additional kilogrammes of RM3 become available
The maximum number of additional kilogrammes of RM3 which can be used without spare
capacity being created
The effect on the optimum solution if two fewer kilogrammes of RM1 are available
Solution
The linear programme and the final simplex tableau are now produced
Maximise the weekly profit, £P, where P = 2x + y (£/week)
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Subject to RM1: 3x ≤ 27 kg/week
RM2: 2y ≤ 30 kg/week
RM3: x +y ≤ 20 kg/week
x, y ≥ 0
Table 7.4 Final simplex tableau
Basic variables
Variables
X y s1 s2 s3
Right hand side , b
x
s2
y
1 0 1/3 0 0
0 0 2/3 1 -2
0 1 -1/3 0 1
9
8
11
Objective function, p 0 0 1/3 0 1 29
If one additional kilogramme of RM1 is available, this limiting constraint is relaxed by one
kilogramme. The values in the s1 column are the changes in the basic variables which result from
this relaxation. The final tableau is re-written below with only the relevant values and calculations
shown.
Table 7.5 partial modified final simplex tableau (1kg extra of RM1)
Basic variables
Variables
X y s1 s2 s3
Right hand side , b
X
s2
y
1/3
2/3
- 1/3
9 + (1/3) = 9 1/3
8 + (2/3) = 8 2/3
11 – (1/3) = 10 2/3
Objective function, p 1/3 29 + (1/3) = 29 1/3
One additional kilogramme of RM1 causes the value of x to increase by 1/3 of a unit, the slack
for RM2 to increase by 2/3 of a kilogramme, the value of y to decrease by 1/3 of a unit and
the maximum value of the weekly profit to increase by £1/3 i.e. by the shadow price. The new
optimum solution requires 9 1/3 of product x and 10 2/3 of product y to be produced each week.
The slack on constraint 2, the amount of raw materials not used, is 82/3 kg. The other variables
are zero. The slack on constraints 1 and 3 is zero; therefore all RM1 and RM3 are used.
If two additional kilogrammes of RM1 are available, this limiting constraint is relaxed by two
kilogrammes. The values in the s1 column are multiplied by two. The resulting values are then
the changes in the values of the basic variables which arise from the additional two kilogrammes.
The final tableau is shown in table below with only the relevant values and calculations shown.
263.
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STUDYTEXT
Mathematical Programming
Table 7.5 Partial modified final simplex tableau (2kg extra of RM1)
Basic variables
Variables
X y C s1 s2 s3
Right hand side , b
x
s2
y
1/3 x2
2/3 x2
- 1/3 x2
9 + (2/3) = 9 2/3
8 + (4/3) = 8 1/3
11 – (2/3) = 10 1/3
Objective function, p 1/3×2 29+ (2/3) = 29 2/3
The new optimum solution requires 9 2/3 of product x and 10 1/3 of product y to be produced
each week. The slack on constraint 2 is 9 1/2kg. The other variables s2 and s3 are zero. This
means that these constraints are binding. The maximum value of weekly profit is £29.67. This
solution can be illustrated in the same way as 1 above.
If five additional kilogrammes of RM3 are available, this limiting constraint is relaxed by five
kilogrammes. The values in the s3 column multiplied by five kilogrammes are shown in the
modified final tableau below
Table 7.6
Basic variables
Variables
X y s1 s2 s3
Right hand side , b
x
s2
y
0 x5
-2 x5
- 1 x5
9 + (0) = 9
8 + (-10) = - 2
11 + (5) = 16
Objective function, P 1×5 29 + (5) = 34
A problem has now arisen. The value of the slack variable, s2 for raw materials 2, has become
negative. This is not allowed, since variables must always be positive or zero. If we consult the
graphical solution we can see at once what has happened. The RM3 constraint has been relaxed
so far that it is no longer limiting. The tableau gives a point outside the feasible region. We are
not able to use all the extra five kilogrammes of RM3. This problem is discussed further in section
4.
The RM3 constraint is represented by the s column of the final tableau. The only negative value
in the s column is the marginal value of s2 which is –2. As RM3 is relaxed, the value of s2
decreases by 2, but it cannot be negative. When s2 reaches zero, the limiting position for the
RM3 constraint will occur. Suppose this limiting position is reached when the RM3 constraint has
been relaxed by r kg, the value of s2 will then be zero, therefore:
8 + ( -2 xr) = 0
Which means that r = 4kg. The RM3 constraint may be relaxed by 4kg from 20 to 24kg before it
ceases to be binding. This takes it to intersection of the RM1 and RM2 constraints where x = 9,
y =15 and the RM3 constraint is x +y =24
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STUDYTEXT
If fewer kilogrammes of RM1 are available, this limiting constraint is tightened by two kilogrammes.
The values in the s column are multiplied by two. The resulting values are deducted from the
values of the basic variables. The final tableau is re-written below.
Table 7.7
Basic variables
Variables
X y s1
s2
s3
Right hand side , b
X
s2
y
0 x 5
-2 x 5
- 1 x 5
9 - (2/3) = 8 1/3
8 - (4/3) = 6 2/3
11 – (- 2/3) = 11 2/3
Objective function, p 1/3×2 29 - (2/3) = 28 1/3
The new optimum solution requires 8 1/3 of product x and 11 2/3 of products y to be produced
each week. The slack on constraint 2 is 6 2/3kg while the slack on constraints 1 and 3 is zero.
This means that these constraints are binding. The maximum value of weekly profit is £ 28.33.
This analysis is tedious to do by hand using the simplex method, even with the simplest two
variable models. All commercial standard linear programming computer packages will provide
the information. This is how multivariable sensitivity analysis is done in practice. The principles
are exactly the same as those for the two variable models we have just completed.
The dual linear programming model
The dual linear programming is used to investigate a problem from a different perspective to the
one obtained from the usual primal model. The primal and dual models give the same solution
and the same sensitivity information. The only reason for using one rather than the other is that
computationally one may be easier to solve. With increasingly powerful computer packages the
need for the primal/dual switch is becoming less relevant. The variables in the dual model are
the shadow prices of the original, or primal, model. The structures of dual and primal models are
similar. If the primal model has been built, the corresponding dual model is derived from it. In
general, a linear programming problem can be described by:
Maximise Z = c1 x1 + c x + …. + cn xn
Subjects to a set of m linear constraints
a11
x1
+ a12
x2
+ a13
x3
+ …. a 1n
xn
≥ b1
a21
x1
+ a22
x2
+ a23
x3
+ …. a2n
xn
≥b2
a31
x1
+ a32
x2
+ a33
x3
…. A3n
xn
≥b3
am1
x1
+ am2
x2
+ am3
x3
+ …. amn
xn
≥ bm
xi
≥0
265.
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STUDYTEXT
Mathematical Programming
The above linear programme maximises and has all ≤ constraints. Any linear programming model
can be put into this form and converted into its dual, as we show below. The dual model is:
Minimise G = b1 y1 + b2 y2 + …. bm ym
Subject to a set of n linear constraints
a11
y1
+ a21
y2
+ a31
y3
+ …. a m1
xm
≤ c1
a12
y1
+ a22
y2
+ a32
y3
+ …. am2
xm
≤c2
a13
y1
+ a23
y2
+ a33
y3
…. am3
xm
≤c3
a1n
y1
+ a2n
y2
+ a3n
y3
+ …. amn
xm
≤ cn
yi
≥0
There are m dual variables y, one for each of the m primal constraints, and n constraints, one
for each of the x variables in the primal. The coefficients c, in the primal objective function and
the right hand side values, b, of the constraints in the primal constraints are interchanged row to
column on the dual. The dual variables, y are the shadow prices in the minimal problem and vice
versa. In this case the dual minimises, the primal maximises. The primal has ≤ constraints and
the dual ≥ constraints.
To set and interpret the dual linear programme
A firm makes two products, R and Q both of which require two raw materials RM1 and RM2. Each
kilogramme of product R requires 2kg of RM1 and 3.5kg of RM2. Each kilogramme of product Q
requires 3kg of RM1 and 1.5kg of RM2. Each week 10kg of RM1 and 12kg of RM2 are available.
There is an unlimited supply of labour and machine time and the firm can sell all its production.
The unit profit on product R is £5 and on product Q is £8.
Required
1. Set up a profit maximising linear programming model for this problem
2. Set up the dual linear programming model
3. Explain the relationship between the two models in 1 and 2
4. Find the optimum solution for the two models graphically.
Solution:
Produce x1 kg of product R and x2 kg of product Q each week. Maximise weekly profit, £p
where:
P = 5x1 + 8x2 (£/week)
Subject to; RM1: 2x1 + 3x2≤10
3.5x1+1.5x2≤12
Using the primal model, the dual model is:
minimise : G = 10 y1 + 12y2
subject to:
product R: 3y1 + 1.5 y2 ≥ 8 £/ unit
Primal model: The variables are the amount of each product to be produced each week. The
objective function is the total profit per week from the production of R and Q. Each constraint
refers to one raw material. The left hand side of the constraint gives the total requirement for that
raw material. The left hand side of the constraint gives the total requirement for that raw material
by both of the products. The right hand side gives the total raw material available each week.
266.
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STUDYTEXT
Dual model: the variables are the primal shadow price, that is, the amount which would be
added to the value of the objective function if one more unit of the raw material was available.
The shadow prices represent the value of one unit of the raw material. The objective function is
the total value per week of the raw materials used in the production of R and Q. each constraint
refers to one product. The left hand side of the constraint gives the total value of both raw
materials used to make one kilogramme of that product. The right hand side gives the unit profit
generated by that product. Let us look at the dual model again and try to identify the individual
components
Minimise G:
10y1 + 12y2
Product R:
2y1 + 3.5 y2 ≥ £5 profit per R
Product Q:
3y1 + 1.5y2 ≥ £8 profit per Q
Each constraint says that the total value of the raw materials used in that product must be more
than or equal to the unit profit on that product. The solution of either the primal or the dual model
enables us to solve the other model.
The graphical solution for the primal problem is given below.
Diagram 7.6 Primal model
a11
x1
+ a12
x2
+ a13
x3
+ …. a 1n
xn
≤ b1
a21
x1
+ a22
x2
+ a23
x3
+ …. a2n
xn
≤b2
a31
x1
+ a32
x2
+ a33
x3
…. A3n
xn
≤b3
am1
x1
+ am2
x2
+ am3
x3
+ …. amn
xn
≤ bm
Primal model
The optimum solution is at point A, the intersection of the raw material 1 constraint and the Q
Product
Q, kg
per
week
X2
8
6
4
2
0
RM2 3.5x1
+ 1.5x2
= 12
Point giving max. profit (0, 3.5)
RM 2x1
+ 3x2
= 10
0 1 2 3 4 5 x1
Product R, kg
per week
267.
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STUDYTEXT
Mathematical Programming
axis. To maximise profit we should produce 31/2 kg of Q and zero kilogrammes of R. all raw
material 1 will be used, but not all of RM2. The maximum profit is 31/2 x 8 = £ 26.67 per week.
The graphical solution for the dual is shown below.
Dual model
This is a minimisation problem. We wish to make the objective function as small as possible,
hence we move towards the origin, parallel to trial objective function. The last point in the feasible
region is given by Z. this is the optimum solution for the dual. Z is the intersection of the product
Q constraint and the y-axis i.e.
Y2 =0 and 3y1 + 1.5y2 = 8
Therefore y2 = 0 and y1 = 2 2/3
The minimum value for the dual problem is:
G = 10 x 2 2/3 + 12 x 0 = £26.67
Diagram 7.7
This is the same value of the objective function as the one for the primal model. The two solutions
combined tell us that the maximum profit is £26.67 per week, when we produce 31/2 kg of Q and
none of R, the value of the raw materials, the shadow price, is £2.67 per kg of RM1 and zero for
RM2. This is the same as the information which we would derive from a full analysis of the primal
problem alone.
Product Q
3y1
+1.5y2
= 8
Feasible
region
Point giving min. value
used (3.5, 0)
Typical value line
10y1
+12y2
=36
Product R
2y1
+3.5y2
= 5
Value per
unit of
RM2 5
4
3
2
1
0
0 1 2 3 Value per unit of
RM1
Y1
Y1
268.
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STUDYTEXT
Chapter summary
The same method of formulation applies to two variable and multivariable models.The two variable
models, however, can be solved graphically. The constraints, which are usually inequalities either
≤ or ≥ are presented on the graph by lines and areas. Each constraint divides the graph into a
rejected area and an acceptable area. The area in which all of the constraints are satisfied is
called the feasible region. This feasible region contains all possible solutions to the problem.
The optimum point which is always at a corner of the feasible region is found by plotting typical
objective function on the graph. The objective function is moved, parallel to this trial line away
from the origin if the objective is to maximise, or towards the origin if the objective is to minimise.
The last point that this line touches before it completely leaves the feasible region gives the
values of the variables which will optimize the objective function.
Sensitivity analysis is very important in linear programming since the values used in the model
may be subject to uncertainty. The procedure allows us to consider variation and uncertainty in
the objective function coefficients and in the right hand side values of the constraints.
Multivariable linear programming models are solved by computer using the simplex algorithm. It
provides the optimum value of the objective function, the values of the decision variables and the
values of the slack or surplus variables. In addition, it gives the shadow prices of the resources.
The final tableau can also be used in sensitivity analysis to show the full effect of variation in the
scarce resources, on the objective function, and on each of the constraints.
Each primal linear programme has dual formulation. The solutions to the primal and the dual
are identical. The dual may be derived from the primal model by interchanging the role of the
coefficients in the model. There are sometimes advantages in solving a simpler dual rather than
a complex primal, formulation.
Chapter quiz
1. In Linear programming a ………… is a variable which is added to a constraint to turn
the inequality into an equation
2. In Linear programming a ………….. is a variable which is subtracted from a constraint
to turn the inequality into an equation.
3. The area which remains unshaded is the set of points which satisfies all of the constraints
simultaneously. This area is called?
4. A …………. is the name given to the set of variable values at a corner of the feasible
region.
5. Each …….. linear programme has dual formulation. The solutions to the ……… and the
dual are identical. The dual may be derived from the ………… model by interchanging
the role of the coefficients in the model.
269.
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STUDYTEXT
Mathematical Programming
Answers to chapter quiz
1. Slack variable
2. Surplus variable
3. Feasible region
4. Basic solution
5. Primal
questions from previous exams
June 2000 Question 6
Regal Investments has just received instructions from a client to invest in three stocks: airline,
insurance and information technology. The client has Sh10,000 available for investment. She
has instructed that her money be invested in the three stocks so that no more than Sh5,000
is invested in any one stock but at least 1,000 is invested in each stock. She further instructed
Regal Investments to use its current data and invest in a manner that maximises her expected
overall gain during a one-year period. The stocks, the current price per share, and the projected
stock price, a year from now, are summarized as follows:
Stock Current price (Sh.)
Projected price (Sh)
1 year hence
Airline 25 35
Insurance 50 60
Information Technology (IT) 100 125
271.
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STUDYTEXT
Mathematical Programming
Required:
a) Formulate the above problem (6½ marks)
b) Determine the values of the columns with blanks, that is, coefficient sensitivity, slack/
surplus, objective coefficient ranges current value and right hand side ranges current
value columns. For slack/surplus column indicate whether it is a slack or a surplus.
c) What is the optimal solution including the optimal value of the objective function?
1½ marks)
d) If the client had an additional Sh1,000 available for investing, how much would the
expected over-all one-year gain increase? (1½ marks)
e) If the client increased the allowed maximum investment amount to Sh6,000 for just one
stock, should it be IT stock? Why? (1/2 mark)
f) Based on your stock choice in (e) above, how much would the objective function
increase? (1/2 mark)
g) For your choice in (e) above, how much could be allowed maximum investment amount
be raised before optimal investment mix might change? (1 ½ marks)
(Total: 20 marks)
June 2002 Question 6
a) Define the following terms as used in linear programming:
(i) Feasible solution (2 marks)
(ii) Transportation problem (2 marks)
(iii) Assignment problem (2 marks)
b) The Tamu-Tamu products Company Ltd. is considering an expansion into five new sales
districts. The company has been able to hire four new experienced salespersons. Upon
analysing the new salesperson's past experience in combination with a personality test
which was given to them, the company assigned a rating to each of the salespersons
for each of the districts. These rating are as follows:
District
1 2 3 4 5
Salesperson
A 92 90 94 91 83
B 84 88 96 82 81
C 90 90 93 86 93
D 78 94 89 84 88
The company knows that with four salespersons, only four of the five potential districts can be
covered.
Required:
(i) The four districts that the salespersons should be assigned to in order to maximize the
total of the ratings. (6 marks)
(ii) Maximum total rating. (8 marks)
(Total: 20 marks)
273.
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STUDYTEXT
Mathematical Programming
Required
(a) Formulate the above problem (6 marks)
(b) What is the optimal daily production plan (2 marks)
(c) What is the maximum selling price for the standard model that will keep the same
optimal solution as in (b) above? (1 mark)
(d) Suppose C1 was changed from its current value of Sh5.50. Would the optimal solution
change? Why? (1 mark)
(e) Suppose C2 was changed from its current value to Sh4.00. Would the optimal solution
change? Why? (1 mark)
(f) Suppose simultaneously C1 changed to Sh5.50 and C2 changed to Sh.4.00. Would the
optimal solution change? Why? (2 marks)
(g) What is the shadow price for man-hours? Interpret. (2 marks)
(h) What is the shadow price of the packing? Interpret. (2 marks)
(i) Suppose simultaneously the amount of material available increased from 350 to 500,
the number of boxes available increased from 300 to 310 and the number of man-hours
increased from 80 to 84. What conclusion can be drawn regarding the shadow prices?
Why? (3 marks)
(Total: 20 marks)
December 2003 Question 6
a) One of the properties (assumptions) of Linear programming is that fractional values of
the decision variables are permitted that is divisibility. However, fractional values might
sometimes be meaningless.
Explain one case where fractional values might be meaningless. What approach should
be used to over-come this assumption? (3 marks)
b) Explain the connection between the following:
(i) Reduced costs column and the range of optimality (2 marks)
(ii) Dual prices and the range feasibility (2 marks)
c) Lake Naivasha Floatway Tours has Sh4 million that may be used to purchase new rental
boats for hire during the coming December holidays. The boats can be purchased from
the two different manufacturers. Pertinent data concerning the boats are summarised
below:
Boat type Manufacturer Cost (Sh.)
Maximum seating
capacity
Expected Daily
profit (Sh.)
Speedhawk Sleekboat 60,000 3 7,000
Silverbird Sleekboat 70,000 5 8,000
Chui Racer 50,000 2 5,000
Simba Racer 90,000 6 11,000
Floatway Tours would like to purchase at least 50 boats and would like to purchase the same
number from Sleekboat as from Racer to maintain goodwill. At the same time, Floatway Tours
wishes to have a total seating capacity of at least 200.
279.
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STUDYTEXT
Network Planning and Analysis
CHAPTER EIGHT
Network Planning and Analysis
Objectives
i. At the end of this chapter, you should be able to:
ii. Define network analysis
iii. Cite importance and application of network analysis in real business world
iv. Crash a project
v. Find an optimum solution for transportation and assignment models
vi. Discuss various special cases of the assignment problem
Introduction
Network analysis is a family of related techniques developed to aid management in the planning,
co-ordination and controlling of large complex projects using limited resources like personnel,
material, money, time etc in order to achieve some objective.
Fast Forward: A project is a finite endeavour (having specific start and completion dates)
undertaken to create a unique product or service which brings about beneficial change or added
value.
Definition of key terms
Project: It is a human undertaking which consists of a series of interrelated tasks geared toward
a definite objective. It involves a considerable amount of time, human and financial resources for
completion.
Task/activity: This is an individual identifiable activity in a project which has a definite beginning
and end. A task requires time and other resources for its completion. An activity or task is
considered non divisible. An activity is represented in a network by an arrow.
Event: This denotes the beginning or the ending of an activity. It is just a moment in time hence
does not require any resources or time. An event is represented in a network by a circle or
node.
Industry context
Business modelling has traditionally been done by operations research (OR) and management
science professionals. It involves the use of quantitative and computer methods for planning the
efficient allocation of resources in business, industry and the agencies of government. It overlaps
economics as a discipline but it primarily focuses on the internal operations of organisations
emphasises planning rather than market mechanisms for allocating resources makes more
extensive use of quantitative methods.
280.
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STUDYTEXT
Operations management often presents complex problems that can be modelled by linear
functions. The mathematical technique of linear programming is instrumental in solving a wide
range of operations management problems.
Linear programming is used to solve problems in many aspects of business administration
including:
product mix planning••
distribution networks••
truck routing••
staff scheduling••
financial portfolios••
corporate restructuring••
Exam context
The most recognisable feature of PERT is the "PERT Networks", a chart of interconnecting
timelines. PERT is intended for very large-scale, one-time, complex, non-routine projects.
Students should be keen on the flow of scheduling as well as flow of events while coming up with
the ultimate allocation pattern to avoid misallocation.
A student will be required to tackle questions from network analysis step by step. Questions are
easily raised as outlined below:
Past Paper Analysis
12/06, 6/06, 12/05, 6/05, 12/04, 6/04, 12/03, 6/03, 12/02, 6/02, 12/01, 6/01, 12/00, 6/00
8.1 NETWORK ANALYSIS
This is a system of interrelationships between jobs and tasks for planning and control of resources
of a project by identifying critical path of the project.
In Network analysis attention should be directed to some activities. These include:
Dummy activity
This is used in a network for
i. improving clarity of the network
ii. To facilitate a logical flow of activities in the network.
A dummy activity consumes no time or resources. Its duration is zero. A dummy activity is
represented by a dotted arrow/line.
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Network Planning and Analysis
Activity floats
Floats give a measure of the flexibility which exists for either individual activities or a path of
activities in a network in relation to whether we can extend project completion time as measured
by the critical path. These are the total, free and independent floats.
a) Total float(TFij)
This indicates the amount of time by which a non-critical activity can be delayed without affecting
the project duration dates. It is equal to the difference between the total time allowed on a
performance of an activity and the actual time required for its performance. For critical activities,
the total float is zero. For non-critical activities, the total float can be computed using the following
formula:
TFij = LCj – Esi – Dij
Where
TFij = total float
LCj= latest completion time at the head of the activity
ESi = earliest start time at the tail of the activity
Dij = means the duration of the activity.
b) Free float (FFij)
This indicates how far a non-critical activity can be delayed beyond its earliest start time without
affecting the earliest start time of the activities immediately following it. Critical activities have
zero free float. For non-critical activities, free float can be determined as follows:
FFij = ESj – Esi – Dij
Where
FFij = free float
ESj = earliest start time at the head of the activity.
ESi = earliest start time at the tail of the activity.
Dij = duration of the activity.
c) Independent float (IF)
This is the amount of time an activity can be delayed without delaying the project completion time
if all preceding activities are completed as early as possible and subsequent activities are started
as soon as possible.
It is given as follows:
IFij = ESj –LCi – Dij
Where
IFij = independent float
ESj = earliest start time at the head of the activity.
LCi = latest completion time at the tail of the activity
Dij = duration of the activity
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STUDYTEXT
Event slack
An event slack is the maximum time an event can be delayed without delaying the overall project
completion time. For an event i ,it given as LCi – Esi
Note:
Total float ≥ free float ≥ independent float
When asked to calculate float or slack without specifying, take that to mean the total float.
Role of network analysis
Network analysis in project management will help:
Define the job to be done, breaking it into individual activities;
Integrate them in a logical time sequence;
Afford a system of dynamic control over the progress of the plan
How would you use network analysis to solve a management
problem?
Phase I: Formulation
Identify the project
Determine and list all the activities in the project.
Estimate the duration and cost of all activities listed above listing all resources required.
Establish a logical flow of all the activities – first, second, and so on.
Phase II: Scheduling
Draw the network diagram of the project indicating various events and their duration.
Determine the earliest and the latest occurrence of each event.
Calculate activity floats.
Determine the critical path for network – the critical path is defined as an unbroken continuity
of activities sequence from start to finish which is the longest path (and) which represents the
minimum time the project can be completed. A delay of an activity on the critical path inevitably
delays the whole project.
Note: There can be more than one critical path in one network.
Phase III: Implementation and control
i) Analyse the effects of delays in the project time, reduction of costs (activity crashing).
ii) Check off the progress of implementation with the plans.
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Network Planning and Analysis
iii) Reassign or reschedule the resources that are used by the project as appropriate.
iv) Revise the network and set a new schedule where necessary.
There are two approaches to pictorially represent a network:
Activity on arrow approach
Activity on node approach
Rules to construct network diagrams
1. As a rule, no two activities can begin and end at the same event node as they would not be
uniquely identified. (Use a dummy activity to ensure uniqueness).
Diagram 8.1
A complete network can have only one beginning event node and one ending event node.
Diagram 8.2
2. Every activity must have one preceding or "tail" event and one succeeding or "head" event.
Note that many activities may use the same tail event and many may use the head event.
Diagram 8.3
3. "Loops" are a series of activities which lead back to the same event and are NOT allowed
because a network is a progression of activities always moving onwards in time.
A
B
Dummy
Start Stop
Tail
Event
Haed
Event
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STUDYTEXT
Diagram 8.4
4. All activities must be tied into the network i.e. they must contribute to the progression or
be discarded as irrelevant. Activities which do not link into the overall project are termed
"danglers". Danglers should not be used.
Diagram 8.5
5. Networks proceed from left to right.
6. Networks are not drawn to scale
7. Use straight lines, not bent or curved ones.
8. The arrows should not cross each other unless it is completely necessary.
9. The length of the arrow is not proportional to its duration.
10. When the drawing is complete, use code number to number the events. The code number at
the beginning of an activity must be smaller than the code number at the end of an activity.
These code numbers ensure that each activity can be uniquely identified in the diagramOther initial differences:
Dangler
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Network Planning and Analysis
i) Mechanic of drawing network
PERT used AOA approach where AOA = Activity-on-Arrow approach.
Diagram 8.6
CPM used AON approach where AON = Activity-on–Node approach.
Diagram 8.7
ii) How to estimate activity time
CPM used a single time estimate therefore it was deterministic in the estimate.
PERT on the other hand used multiple time estimates allowing for any doubt to be included in the
estimates therefore probabilistic.
These estimates were reduced to three for examination purposes.
a) Optimistic time estimate = a
This is the shortest time an activity can take to be complete. It represents uder real estimate such
that the probability is small that the activity can be completed in less time.
b) Pessimistic time estimate = b
This is the longest time an activity can take to be completed. It is the worst time estimate
representing bad luck, such that the probability is small that the activity will take longer.
MostCritical activity
1. It must be critical for the completion of the project.
2. It must be started and completed as originally scheduled otherwise the project will be late.
3. It must have the first priority in the use of resources.
4. Requires keen management attention and control.
A B C
A B C
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STUDYTEXT
Three conditions to be satisfied simultaneously
1. ESi = LCi
2. ESj = LCj
3. ESj – Esi = Dij or LCj – Lci = Dij
Critical path calculations
Fast Forward: Today, it is commonly used with all forms of projects, including construction,
software development, research projects, product development, engineering, and plant
maintenance, among others.
Two phases are computed:
a) Phase I is the Forward Pass computations. It is used to determine the ESi for each
activity.
b) Phase II is the Backward pass computations. It used to determine the LCj for each
activity.
Phase I – Forward Pass = ESi
Computation begins from the initial event. Move forward until you reach the terminal event.
The initial event time is zero which is base time.
Earliest start time of the activity beginning from the immediate next event which is base time plus
the duration of the activity.
If there is more than one activity going into an event, for the ESi from that event, take the maximum
allowable time using the following formual:-
ESj = max ESi + Dij for all (ij) acitivities.
Phase II – Backward pass = LCj
Start from the terminal event and move backwards to reach the initial event.
Terminal event LCij= ESi
For the next event it is the difference between the maximum allowable time and the duration of
the activity.
Take minimum allowable time, LCi = min LCj – Dij for (ij) all activities
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Network Planning and Analysis
Crashing the project
Crashing means trying to perform an activity in a time shorter than the activity required i.e.
putting in extra effort to deliberately reduce the activity time. Consequently, the duration of some
activities can be reduced if some additional resources are employed but by introducing such
additional resources the cost of performing such activities increases and the normal time by
which an activity is required to be reduced the greater the amount of resources required to be
employed.
Normal duration (Dn)
This is the time required to perform an activity under normal circumstances and with minimum
direct costs.
Crash duration (Dc)
This is the time taken to perform an activity if the duration is reduced or shortened.
Normal costs (Cn)
This is the absolute minimum direct costs required to perform an activity within the normal
duration.
Crash costs (Cc)
This is the cost incurred to achieve the reduced performance time. It is generally more than the
normal costs because of introduction of additional resources.
Incremental costs (Ic)
This is the increase in cost incurred per unit of time that is used. It is determined by the following
formula:
Incremental costs (Ic) =
crashcost − normalcost
=
∆ cost
Normalduration − Crashduration ∆ duration
Guideline for crashing
Identify the critical activities and the critical path(s)
Crash the durations of the critical activities only because they are the ones that determine the
project duration.
Determine the incremental cost of all the critical activities.
If there are several critical activities to be crashed select the critical activity with the least
incremental cost to be crashed first.
For each crashing step a unit of time is the maximum amount that can be used because there is
a possibility of new critical activities and critical path emerging.
Check to be sure that the critical path being crashed is still critical. This is because a reduction
in activity time along the critical path may cause non-critical activities and paths to become
critical.
If a project has more than one critical path then all critical paths must be reduced simultaneously
by an equal amount.
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STUDYTEXT
When to stop crashing
For technical reasons the duration of an activity cannot be reduced indefinitely. If crash time
represents the minimum duration that is allowable consequently stop crashing activity time or
duration when:
The management wish has been achieved.
When all crashable activities have reached their climax crash limits.
When it is no longer economical to continue crashing.
Resource scheduling and profiling
So far we have assumed that an activity can begin as soon as all the activities which must
precede it have been completed. This assumes that there are sufficient resources to perform all
work defined by the activity even if concurrently. Such an assumption may not always be true in
practice. Further, resource utilisation needs to be planned so that it is as steady (smooth/even)
as possible.
Planning resource requirements is done through the use of a time (Gantt) chart levelling of
resources usage which will lead to the construction of a resource profile. A resource profile is
used as a calendar of time schedule by the personnel who will implement the project.
We perform resource scheduling and profiling because:
We accept that concurrent activities may not be possible due to resource constraints.
For other resources such as manpower, the application should be planned for in advance to
minimize losses due to shortages at certain time or idleness at other times.
Smoothing a profile
This is the process of attempting to reduce the peaks and troughs in the resource allocation
so that we have a more even usage of personnel. Smoothing requires that we make use of
activity floats since we generally cannot extend completion time. We cannot reschedule critical
activities. We can normally reschedule only non-critical activities because they have floats. Such
rescheduling should be for as little as possible since it is prudent to delay an activity for as short
a time as possible. Smoothing is conducted on the basis of trial and error because there is no
analytical method for it.
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STUDYTEXT
Network Planning and Analysis
Illustration
Consider the following project network and activity times (in weeks).
Activity
Activity Times
(weeks)
A
B
C
D
E
F
G
H
5
3
7
6
7
3
10
8
Required:
(i) How long will it take to finish this project?
(ii) Can activity D be delayed without delaying the entire project? If so, by how many
weeks?
(iii) What is the schedule for activity E?
Solution
i) We can determine the project duration by drawing a schedule to ascertain the critical
activities.
C
C
D
B E G
F Finish
Start
H
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Network Planning and Analysis
The critical path is A – G – H – J.
Clearly, the project cannot be completed within the remaining time of 1 week. There is need to
increase the duration of critical path by ½ week.
The assumption of PERT as used in network analysis is that the activity time duration can be best
described by a beta probability distribution.
Illustration
Lilian Wambugu is the project manager of Jokete Construction Company. The company is bidding
on a contract to install telephone lines in a small town. It has identified the following activities
along with their predecessor restrictions, expected times and worker requirements.
Activity Predecessors Duration Weeks Crew Size Workers
A - 4 4
B - 7 2
C A 3 2
D A 3 4
E B 2 6
F B 2 3
G D, E 2 3
H F, G 3 4
Lilian Wambugu has agreed with the client that the project should be completed in the
shortest duration.
Required:
(i) Draw a network for the project.
(ii) Determine the critical path and the shortest project duration.
(iii) Lilian Wambugu will assign a fixed number of workers to the project for its entire duration
and so she would like to ensure that the minimum number of workers is assigned and
that the project will be completed in 14 weeks.
Draw a schedule showing how the project will be completed in 14 weeks.
4
8
8
B
A
0 0
3
9 9 12 12
G 3
3
E D
6
258
J5
17 17 25
34
13
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STUDYTEXT
Critical path = A – C – G – H - I
To have at most 6 technicians Mr. Mutiso should reschedule activity F to start on the 6th week
i.e. delay it by 5 weeks.
8.2 Transportation and Assignment Models
The linear programming methods discussed in Chapter Seven are suitable for a range of allocation
problems. The work involved in solving the model can be drastically reduced by use of a computer
package. This leaves the decision maker free to concentrate on the interpretation and evaluation
of final solution. However, the package still requires the formulation of the linear programming
model. This can be a major task for large problems. The variables must be identified and the
constraints formulated.
For certain types of allocation problems, the use of specially designed algorithm simplifies the
building of the initial model. In this chapter, we will look at two related examples of this type of
algorithm which are suitable for solving transportation and assignment problems.
In both cases the allocation concerns items which are transferred from a number of origins
to a set of destinations according to a particular objective. The objective is often one which
minimises the total cost of the transfer. For example, a company has three factories and five
6
6
1
6
2
6
4
4
2
2
F
E
D
D
Personnel 2
Activities
Critical activities ACGHI
Now critical activities BDEF
4
C G H I
2 2 2
T/F
D
E
F
G
H
I
T/F
3
3
5
0
0
0
3
6
4
10 10 10 10
6
2
65
4
2
7
4
6
2
8
6
2
9
6
2
10
6
2 Duration (wks)
rescheduled
Unbalanced resource profile
11 12
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Network Planning and Analysis
regional distribution centres. The management requires the transfer of the finished goods from
the factories to the distribution centres to be achieved at minimum cost. This is a situation in
which the transportation method would be appropriate.
Assignment is a particular case of the transportation problem. For each combination of origin
and destination, the transfer involves one item only. For example, a machine shop has six lathes
of varying ages and design. On a given morning, the machine shop manager has six jobs to
allocate. The jobs will take different lengths of time on different lathes. The manager wishes to
allocate one job to each lathe to minimise the total working time. The assignment algorithm may
be used to solve problems such as this.
In this chapter we will describe the application of these two algorithms using small problems. It
should be borne in mind that, in practice, the problems will he much larger and are solved using
computer packages. In addition, transportation models often involve several stages, for example,
factory-to depot-to-retail outlet. In these cases the basic algorithm has to be modified and more
sophisticated methods used.
Transportation Problem and Algorithm
This problem is concerned with the allocation of items between suppliers (called origins) and
consumers (called destinations) so that the total cost of the allocation is minimised. The problem
can he solved using either linear programming methods or the special transportation algorithm.
The linear programming method is illustrated in Example 1.
The transportation problem
Example 1: To Illustrate a basic transportation problem
Ace Foods Ltd manufacture soft drinks at two plants, A and B. Bottles for the two plants are
supplied by two firms, P and Q. For the month of November, plant A requires 5,000 bottles and
plant B requires 3,500 bottles. Firm P is able to supply a maximum of 7,500 bottles and firm Q
is able to supply a maximum of 4,000 bottles. The cost per bottle of transport between each
supplier and each plant is shown in the
Table 1: costs, requirements and availabilities of bottle supply
Transport cost, pence/bottle to plant available bottles
A B
From supplier P 4 4 7500
Q 3 2 4000
Bottles required 5000 3500
How should the bottles be supplied to the plants to minimise the total transport cost?
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STUDYTEXT
Solution
It is always useful with transportation to see if there is an obvious solution. Ideally, we would like
to use only the cheapest routes. Supplier Q will be preferred by both plants, since it is cheaper
than P. Unfortunately Q has only 4,000 bottles, compared with the total requirement of 8,500.
The cheapest solution will probably be to use the 2p per bottle route from Q to plant B, to supply
all the requirement at B (3,500). The balance from Q (500) should be sent to A at 3p per bottle.
The rest of the demand at plant A will come from P at a cost of 4p per bottle. The total cost of this
allocation is:
0.02 x 3500 + 0.03 x 500 + 0.04 x 4500 = £265/month
We have no proof that this is the most economic allocation. One of the important aspects of
the model which we are about to consider, is that it provides a solution, demonstrates that it is
the optimum solution and shows the effect on the solution of any changes which arise in the
problem.
We will now solve the above problem using a conventional linear programming model with a
graphical solution.
Suppose firm P supplies x bottles to plant A and y bottles to plant B. Firm Q must supply the
remaining (5000 — x) bottles to A and the remaining (3500 — y) bottles to B. The objective is to
minimise the total transport cost, C, pence, where:
C = 4x + 4y + 3(5000 — x) + 2(3500 — y)
Therefore:
C = x + 2y + 22,000
and:
Z = C - 22,000 = x + 2y
Z will take its minimum value, when C takes its minimum value. The values of x and y which
minimise Z = x + 2y, will also minimise C. The objective function is minimised subject to:
Requirement at A: x <5000 bottles
Requirement at B: y <3500 bottles
Supply from P: x + y <7500 bottles
Supply from Q: (5000 — x) + (3500y) <4000 bottles
i.e.:
x + y >4500 bottles
x, y> 0
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Network Planning and Analysis
These constraints are illustrated in Figure 1.
The point x = 4000, y = 2000 is a feasible solution. At this point:
Z = 4000 + 2 x 2000 = 8000 p
The trial objective function is 8000=x+2y. This is shown in figure.1. Moving in the direction of
decreasing values of Z, corner A is the optimum. At this corner, x = 4500 and y = 0. Therefore the
optimum solution is for P to supply 4500 bottles to A and none to B, while 0 supplies 500 bottles
to A and 3500 bottles B. At this solution, the minimum cost is:
Cmin = 4500 + 2 x 0 + 22,000 26,500 p = £265
The only spare capacity is at firm P which retains 3,000 bottles. This is the solution that we
thought would be the minimum. We have now shown that this is the case.
The transportation algorithm
The problem in section1 may be solved using the transportation a1gorithm. To use this algorithm,
a number of conditions must be satisfied:
1. The cost per item for each combination of origin and destination must be specified.
2. The supply of items at each origin must be known.
3. The requirement of items at each destination must be known.
4. The total supply must equal the total demand.
Example.1 satisfies the first three conditions but not the last one. However, a dummy plant can
be included for which the requirement is the difference between the total available and the total
required. In Example 1 the dummy plant would have a requirement for (11,500 - 8500) =3,000
bottles.
Any items allocated to a dummy destination represent items which do not leave the supplier. In
a similar way, if the total supply is less than the total demand, a dummy supplier is included to
supply the shortfall. Any items allocated from this dummy represent items not supplied.
x=5000
x + y=7500
000
No. of
bottles
supplied
from P
to B
y
8
6
4
2
0
0 1 2 4 6 8
Typical objective function
x+2y=8000
y=3500
No. of bottles supplied from P to AA
x + y=4500
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quantitative techniques29 2
STUDYTEXT
The transportation algorithm has four stages:
Stage 1:Arrange the data in tableau format and find any feasible allocation.Afeasible allocation is
one in which all demand at the destinations is satisfied and all supply at the origins is allocated.
Stage 2: Test the allocation to see if it is the optimal solution.
Stage 3: If the first allocation is not optimal, re-allocate in order to move to a better, lower cost,
solution.
Stage 4: Test again for optimality.
Repeat this iterative process until the optimum allocation is found.
Finding an initial allocation
The initial allocation can be made using any method which will produce a feasible solution.
However, a systematic approach tends to produce more useful solutions. We will look at two
methods of finding an initial allocation, the minimum cost method and Vogel's method. The
procedure is explained in Example 2.
Example 2: To Illustrate methods of finding an Initial allocation
Three warehouses P, Q and R can supply 9, 4 and 8 items respectively. Three stores at A, B and
C require 3, 5 and 6 items respectively. What is the minimum cost of allocating the items from the
warehouses to the stores if the unit transportation costs are as shown in table below?
Table.2 Costs, requirements and availabilities for Example 2
Transport costs £/item, to
stores
A B C
Total available
From warehouses P
Q
R
10 20 5 9
4
8
2 10 8
1 20 7
Total required 3 5 6
Solution
The information on costs, availability and requirements are given but total supply is bigger than
demand. The warehouses have 21 items available but the stores require only 14, a dummy store
is needed to absorb the 7 items which are surplus to requirements. These 7 items will never
actually leave the warehouse, and therefore their transportation costs are assumed to be zero.
The first transportation tableau is given below:
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STUDYTEXT
Network Planning and Analysis
Table.3 Balanced transportation tableau
Transport costs £/item, to
stores
A B C dummy
Total available
From warehouses P
Q
R
10 20 5 0 9
4
8
2 10 8 0
1 20 7 0
Total required 3 5 6 21
To find the first feasible allocation we will use the minimum cost method and Vogel's method in
turn. However, it should be remembered that only one method is actually required.
Method 1: Minimum Cost Method
1. We allocate as much as possible to the cell with the minimum unit cost.
2. We adjust the remaining availabilities and requirements.
3. We choose the next smallest cost and allocate as much as possible to this cell, and so
on, until supply and demand are all zero.
4. If more than one cell has the smallest value of unit cost, then we choose at random.
In table.4, the transport costs are placed in the separate boxes in the upper right of each cell. The
subscripts indicate the order in which the allocations are made and should help you to follow the
explanation. The dash in a cell indicates that this cell is no longer available.
Table 4 First allocation using minimum cost method
To store
A B C Dummy
Total available
P
From
warehouse Q
R
10 20 5 0
9 2 0
- - 23 71
2 10 8 0
4 0
- 45 - -
1 20 7 0
8 5 1 0
32 16 44 -
Total required
3
0
5
1
0
6
4
0
7
0
21
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quantitative techniques29 4
STUDYTEXT
Key:
The smallest cost is zero. We choose any one of cells (P, Dummy), (0, Dummy) or (R, Dummy).
Cell (P, Dummy) is chosen and we allocate the maximum amount, 7 units, to the cell. We reduce
by 7 the amount available at P and the amount required by the dummy. Cross out the cells which
now cannot be used, that is, (0, Dummy) and (R, Dummy).
Neither of the other zero costs are available, so the next smallest cost is 1 in cell (RA). We
allocate as many units as possible, 3, to this cell. Adjust the row and column totals and cross out
cells which are no longer available, that is (P, A) and (Q, A).
The smallest cost, still available, is 5 in cell (P, C). The remaining 2 items at P are allocated to this
cell. The row and column totals are adjusted and we cross out the remaining cell in row P.
The final allocations are, in order, (R, C), (Q, B) and (RB).
If the allocation is feasible, the total available at each warehouse and the total required at each
store should now be zero. The above allocation is feasible.
Cost £ ((3 x 1) + (4 x 10) + (1 x 20) + (2 x 5) + (4 x 7) + (7 x 0) = £101
We do not know yet whether this allocation is the cheapest but it should give a reasonable cost.
Method 2: Vogel's Method
Thismethodusespenaltycosts.Foreachrowandcolumnthepenaltycostisthedifferencebetween
the cheapest available route and the next cheapest. We try to minimise these penalties.
To calculate the penalty cost for each row and column, we look at the least cost cell and next
smallest cost cell. For each row and column subtract the smallest cost from the next smallest.
This gives the penalty cost of not allocating into the cell with the cheapest cost.
We choose the row or column with the largest penalty cost and allocate as much as possible
to the cell with the smallest cost in that row or column. In this way, the high penalty costs are
avoided as far as possible.
As with the previous method, we adjust the row and column totals.
We cross out the remaining cells in any row or column for which the supply or demand is now
zero, since these cells are no longer available.
Return to 1 and re-calculate the penalty costs, ignoring the cells which have been used or crossed
out.
We repeat these steps until all demand is satisfied. The subscripts in table.5 show the order of
choosing the penalty costs and making the allocations.
Unit transport cost
Units
allocated
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STUDYTEXT
Network Planning and Analysis
Table 5 First allocation using Vogel's method
To store
A B C Dummy
Total available
Penalty
1 2 3
P
From
warehouse
Q
R
10 20 5 0
9,8,2,0 5 5 5
- 1 6 2
2 10 8 0
4,0 2 - -
- 41 - -
1 20 7 0
8,5,0 1 1 73
32 - - 53
Total required
3
0
5
1,0
6
0
7
2,0
21
1st penalty
2nd penalty
3rd penalty
1
92
-
101
0
0
2
2
2
0
0
0
After the third allocation, there is only one way of completing the solution. The remaining items
are allocated as follows—(P, B), (P, C) and (P, Dummy).
Cost = (1 x20+6x5+2x0+4x 10+3x 1+5x0)'93
Again, we do not yet know whether this is optimal but we do know that it is a cheaper allocation
than the £101 for the minimum cost method.
Testing for optimality
To test for optimality, we must first determine whether the initial allocation is basic, that is, whether
it is a solution at a corner of the feasible region. The tableau shown in table 4 gives a feasible
solution, that is, a solution inside or on the edge of the feasible region. If the allocation is basic,
there should be one basic variable for every constraint. In a problem with m warehouses and n
stores (including the dummy), there are (m + n - 1) independent constraints. A basic solution will
therefore have (m + n - 1) allocated cells. These (m + n - 1) variables must be in independent
positions. It is not necessary to worry about independence at this stage since any problems will
emerge during the test for optimality.
If the allocation has (m + n - I) independent variables, the methods for testing for optimality may
be applied directly. If there are fewer variables, the tests have to be modified, as will be illustrated
in table 6. However, if there are more than (m + n - 1) variables, then the allocation procedure has
been used incorrectly. It should be possible to modify the allocation to give a cheaper cost with
the correct number of variables.
302.
quantitative techniques29 6
STUDYTEXT
Refer to Example.2. We will test each of the allocations for basicness. The tableau has 3 rows
and 4 columns, therefore a basic solution will have (3 + 4 -1) = 6 allocated cells. We can see that
this is the case for both allocation methods. The two methods have given solutions at different
corners of the feasible region. The testing procedures may be used without modification.
The initial allocation is tested to determine whether it is the cheapest solution and, if it is not, it
should be changed. We will illustrate two methods of testing for optimality.
In the stepping stone method, the costs of using the unallocated cells - the shadow costs - are
calculated. The procedure is long and rather clumsy but the physical meaning is clear.
The MODI (modified distribution) method is a mathematical procedure which gives the same
shadow costs much more quickly, although the physical meaning is not so obvious.
In both methods, if the allocation is not optimal, a stepping stone procedure is used to move to
the next basic allocation. Once a basic solution has been found, the algorithm enables us to
move from corner to corner of the feasible region, until the optimum solution is found.
Example.3: To illustrate the test for optimality using the stepping stone method
We will use the allocation produced by the minimum cost method to illustrate the procedure. The
allocation is repeated in table 6 below
Table 6 First allocation using minimum cost method
To store
A B C Dummy
Total available
P
From warehouse
Q
R
10 20 5 0
9
- - 2 7
2 10 8 0
4
- 4 - -
1 20 7 0
8
3 1 4 -
Total required 3 5 6 7 21
Key:
The stepping stones are the cells which have allocations in them - (P, C), (P, dummy), (0 B), (A,
A), (R, B) and (A, C). We take one of the empty cells and pretend that we move one item into
it. This move upsets the totals for the row or column in which the empty cell lies. The amounts
in some of the allocated cells are then adjusted to restore the balance. We use these allocated
cells, the stepping stones, to calculate the cost of the transfer of this one item into the empty cell.
If the cost is positive, using the empty cell will increase total costs and we do not want to do this.
If the cost is negative, using the cell will reduce costs. This means that the present allocation is
not optimal and we can find a better solution using this cell.
P
R
P
R
Unit transport cost
Units
allocated
303.
297
STUDYTEXT
Network Planning and Analysis
It does not matter which empty cell is chosen as the start. We will choose (P, A). We add 1 item
to (P, A). The allocation is no longer correct. Store A is receiving 4 items, when it wants only 3.
Warehouse P is supplying 10 when it has only 9 items. We must adjust the A column and the
P row. To balance the A column, we must deduct 1 item from the stepping stone (R, A). This
corrects column A, but unbalances row R, reducing its supply from 8 to 7.
We can re-balance row P by subtracting 1 item from either (P, C) or (P, Dummy). If we choose (P,
Dummy), there is no other allocated cell in the dummy column which could be used to re-adjust
the dummy column, therefore we do not make this choice. Adjustments can be made using only
those cells which have allocations already. We must use (P, C). We deduct 1 item from (P, C).
This corrects row P, but unbalances column C. We now have problems with row A and column
C. Both can be adjusted simultaneously by adding 1 item into (R, C). The physical effect of using
the empty cell (P, A) and returning to a balanced allocation is shown in table 7 below.
The net cost effect of moving 1 item into (P, A) is:
+1 x (P, A) cost—1 x (R, A) cost +1 x (R, C) cost—i x (P, C) cost
= + (1 x 10) – (1 x 1) + (1 x 7) – (1 x 5) = +£11/item
Table 7 Testing empty cell (P, A)
Physical change- items
Using (P, A) would cost an extra £11 for each item sent from P to A. The shadow price is positive
therefore we do not choose to use this empty cell.
We return to the original allocation and repeat the procedure for the other empty cells in turn.
Look next at the cell (R, Dummy) and use the stepping stones (P, Dummy), (P. C), and (A, C) to
show the physical and cost changes of moving 1 unit into (P, Dummy):
Table 8 Testing empty cell (R, Dummy) Table9 testing empty cell (R,dummy)
Physical change—Items
cost change, £
P
R
Test cell
+1
Allocated cell
-1
Allocated
cell
-1
Allocated
cell
+1
A C
Test cell
+1
Allocated cell
-1
Test cell
+1
Allocated cell
-1
Allocated
cell
-1
Allocated
cell
+1
Allocated
cell
+1
Allocated
cell
+1
P
R
304.
quantitative techniques29 8
STUDYTEXT
The net cost change of adding 1 item to (R, Dummy) is:
+0—0 + 5—7 = —2 per item
By allocating into cell (R, Dummy), it is possible to reduce the costs, therefore, the present
allocation is not optimal. We can find a cheaper allocation, saving £2 per item, by using (R,
Dummy) and this stepping stone route. We must, however, complete the testing of all the empty
cells, since there may be a cell which gives an even better saving.
Let us next construct the stepping stone path for the empty cell (0, Dummy). We must remember
that to re-balance rows and columns we can step on allocated cells only. A four step circuit is not
possible this time. We must look for a more complex route. We allocate 1 item to cell (0, Dummy).
There is only 1 allocated cell in row Q and only 1 allocated cell in the Dummy column. Suppose
we choose to move from (Q, Dummy) to (Q, B). We deduct 1 item from this cell which balances
row Q. Column B can be balanced by (R, B) only, therefore we add 1 item to this cell. We can
balance row R via (R, A) or (R, C) but (R, A) is the only allocation in column A, so we do not use
this cell. If we did, we would not be able to balance column A. We deduct 1 item from (R, C). The
route home should now be clear. We balance column C by adding 1 item to (P, C) and balance
row P by deducting one item from (P, Dummy). This last move also balances the Dummy column
and the circuit is complete. We should remember that as long as the initial allocation is basic, it is
possible to find a suitable route round the tableau which starts and ends with the chosen empty
cell. The physical effects and cost changes are summarised in Tables.10 and 11.
Table 10 testing empty cell (Q, dummy)
Physical change - items
B C Dummy
Empty
Allocated
+1
allocated
-1
Allocated
-1
empty
test
+1
Allocated
+1
Allocated -1 Empty
Table 11 testing empty cell (Q, dummy)
Cost change £
B C Dummy
Empty Allocated +5 allocated
-0
Allocated
-10
empty
Test +0
Allocated
+20
Allocated -7 empty
305.
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STUDYTEXT
Network Planning and Analysis
The net cost affect of adding 1 item to the empty cell (Q, Dummy) is:
+0—0+5--7+20—10= +8/item
The total cost of allocation will increase by £8 per item if we allocate into this empty cell. We do
not choose to make this change. The shadow costs for the remaining empty cells are calculated
in a similar way to that described above. The full set of values is shown in table 12:
Table 12
To store
Total
available
P
From warehouse
Q
R
10 20 5 0
9
+11 +2 2 7
2 10 8 0
4
+11 4 +11 +8
1 20 7 0
8
3 1 4 -2
Total required 3 5 6 7 21
Key:
This is not the optimal allocation because cell (R, Dummy) has a negative shadow cost of -.E2.
The present cost of the allocation, £101, can be reduced by using this cell and the stepping stone
circuit which gave the net saving of £2 per item.
We will continue with this example, to find the optimum allocation, in section.5, but first we will
discuss the MOD1 method of calculating the shadow costs. The stepping stone procedure is
a clumsy method and it is easy to make errors. It is more sensible to use the mathematical
approach of the MODI method to test for optimality. The procedure does not give an insight into
the physical problem but it does produce the same shadow prices with much less effort.
Shadow cost
of empty cost
Unit transport
cost
Units
allocated
m
306.
quantitative techniques30 0
STUDYTEXT
To begin with, consider only the allocated cells. Each unit cost, Cjj, for these cells is split into
two components, ui for the row and vj for the column. For example, cell (R,B), which is in row 3
and column 2, has a unit cost c32 = £20. This is split into the row component u3 and the column
component v2, i.e.
c32 = 20 = U3 + v2
For each empty (non-basic) cell, we calculate the shadow cost from:
Sij = cij – (ui + vj)
This shadow cost is the extra cost of transporting one item by the route i to j. If all of the shadow
costs are positive or zero, that is, Sij > 0, then the solution is optimal. In this case, if an allocation
is moved into an empty cell for which the shadow cost is positive, the total costs will increase, if
the shadow cost is zero, the total costs remains unchanged.
Example.4: To test a basic allocation (or optimality using the MODI method
Refer again to the initial allocation obtained using the minimum cost method. We will test this
allocation for optimality using the MODI method. The initial allocation is repeated below:
Table 13 First allocation using minimum cost method
P
From warehouse
Q
R
To store
A B C Dummy
Total
available
10 20 5 0
9
- - 2 7
2 10 8 0
4
- 4 - -
1 20 7 0
8
3 1 4 -
Total required 3 5 6 7 21
The row components, u and the column components, v,are calculated using the allocated cells.
The allocated cells are (P, C), (P, Dummy), (0, B), (R, A), (A, B), and (A, C), which give the
following six simultaneous equations. These six equations contain seven variables, hence there
is no unique solution. The actual values given to the components are not important as long as
the set of values is consistent,
c13 = 5 = U1 + v3 for allocated cell P. C)
c14 = 0 = U1 + v4 for allocated cell (P, Dummy)
c33 = 7 = u3 + v3 for allocated cell (A, C)
c31 = 1 = u3 + v1 for allocated cell (A, A)
c32 = 20 = u3 = v2 for allocated cell (R, B)
c22 = 10= u2 + v2 for allocated cell (0, B)
308.
quantitative techniques30 2
STUDYTEXT
Identify those cells from which items are to be deducted and determine the amount which could
be deducted from each cell, without any of the allocations becoming negative. The minimum
value of these figures gives the maximum amount that can be allocated to the chosen cell.
Reallocate around the circuit.
There is no guarantee that further improvements cannot be made. The new allocation must
be checked for optimality using the MODI method. The minimum cost is found when all of the
shadow costs are positive or zero.
We will continue with Example.4.
Example.4 continued: To test a basic allocation for optimality using the MODI method
Cell (R, Dummy) is the only one with a negative shadow cost, -—2. We wish to allocate as much
as possible into this cell.
The stepping stone circuit for (R, Dummy), which gives the —2, is shown below with the existing
allocations and Unit costs.
Table 15 Stepping stone circuit for (R, Dummy)
C Dummy
P
R
+ denotes items are to be added to this cell. – denotes that items are to be deducted from this
cell
The - cells are (P, Dummy) and (R, C), which contain allocations of 7 and 4 items. The minimum
value in the — cells is 4, which means that 4 items can be moved round the circuit, into the +
cells and out of the — cells. The total saving in cost is (2 x 4) = £8. The revised tableau is given
in table 16.
Table 16 Revised allocations
P
From warehouse
Q
R
To store
A B C Dummy
Total available
10 20 5 0
9
- - 2+4 7-4
2 10 8 0
4
- 4 - -
1 20 7 0
8
3 1 4-4 0+4
Total required 3 5 6 7 21
+ 5 - 0
2 7
- 7 + 0
4 -2
310.
quantitative techniques30 4
STUDYTEXT
If this second allocation is not optimal, the re-allocation procedure is repeated as many times as
is necessary.
It should be noted that the minimum cost was achieved on the first allocation using Vogel's
method. This will frequently happen for small scale problems. Vogel's method tends to produce
a better first tableau but there is no guarantee that it will give the optimum immediately. It should
also be noted that the allocation produced by Vogel's method is different to the one above (see
Example.2). There is an alternative optimum solution:
• Warehouse P sends 1 item to Store B, 6 items to Store C and retains 2 items.
• Warehouse Q sends 4 items to Store B.
• Warehouse R sends 3 items to Store A and retains 5 items.
The existence of an alternative optimum solution was signaled to us by the zero shadow cost for
cell (P, B). These zero shadow costs are associated with alternative allocations which generate
the same total cost.
Sensitivity analysis
The final allocation, together with the shadow costs of the empty cells, can be used for sensitivity
analysis. The shadow cost tells us by how much the total cost will increase, if we are forced to
allocate one item to that empty cell. If we are forced to send one item from warehouse Q to store
C, the extra cost will be £13, much more than the unit cost of £18 for the (Q C) route itself. The
extra cost arises because we have to re-balance the allocation using the following stepping stone
circuit.
Table 18 Stepping stone circuits for (Q, C)
Physical change - items
B C Dummy
Table 19 Stepping stone circuit for (Q, C)
B C Dummy
empty Allocated
-5
Allocated +0
Allocated
-10
Test +8 Empty
Allocated
+20
empty Allocated
-0
P
Q
R
empty Allocated
-1
Allocated
+1
Allocated
-1
Test +1 empty
Allocated
+1
empty Allocated
-1
P
Q
R
311.
305
STUDYTEXT
Network Planning and Analysis
The net change in cost is:
+8 - 5+0 + 0 + 20 -10= +£l3/item
The maximum number of items which would be moved round this circuit is the minimum quantity
in the - cell that is:
(P, C) = 6, (R, Dummy) = 4 or (Q, B) = 4
Four items is the maximum number which could be moved.
The zero shadow cost in cell (P, B) was mentioned in the previous section. The stepping stone
circuit for this empty cell is:
Table.20 stepping stone Circuit for (P, B) Table.21 stepping stone circuit for (P, B)
Items can be allocated to cell (P, B) and the net effect on the cost is zero. This means that there is
another allocation which will give the same minimum cost of £93. The maximum number of items
which can be added to (P, B) is the minimum quantity in the — cells, (R, B) = 1 and (P, Dummy)
= 3. Only one item can, therefore, be moved around the circuit into (P, B).
The shadow costs may also be used to indicate how the cost for an empty cell must change
before the optimum allocation is affected; the shadow cost for the empty cell (R, C) is + £2, and
the actual cost of transfer is £7 per item. The actual cost would have to reduce to at most (7 - 2)
= £5 per item, before we would use this cell to reduce the overall costs.
It is more difficult to determine the effect of cost changes in the allocated cells. If the costs reduce,
we are encouraged to put more items into that cell. If the costs of an allocated cell increase, at
some point we will wish to stop using that cell and transfer to another.
If we look at the allocated cell, (P, C), it has an actual cost of £5 per item. If this cost is reduced,
it will not affect the physical allocation since this cell already has the full requirement for Store C
allocated to it.
If the cell cost is increased from £5, we must look at the stepping stone circuits which use (P,
C). These are the circuits which give the shadow costs of £3 for (Q, C) and £2 for (R, C). In
both of these circuits (P, C) is a — cell and any increase in the £5 cost will reduce the shadow
cost of these empty cells. The physical allocation will change when the unit transfer cost of (P,
C) increases by more than £2, from £5 to more than £7. The shadow cost of (R, C) will then
become negative. At this point, it will be advantageous to use this empty cell, changing the (P,
C) allocation.
For the current optimum allocation, the cost of (P, C) has an upper limit of £7 and a lower limit of
£0. Between these limits the physical allocation is the same but the total costs will change.
test
+20
allocated
-0
allocated
-20
allocated
+0
test
+1
allocated
-1
allocated
—1
allocated
+1
312.
quantitative techniques30 6
STUDYTEXT
Variations in the transportation problem
Forbidden Allocations
If an allocation from a particular origin to a particular destination is impossible for some reason,
the algorithm can be forced to avoid this allocation by assigning a large cost to the cell. The exact
value is unimportant but it must be much larger than the other costs in the tableau. The algorithm
will then automatically avoid this cell.
Example 5: Forbidden allocations
This example also illustrates how the transportation algorithm can be used to solve problems
which are not the straight forward transfer of goods from origins to destinations. In this example,
we will deal with the transfer of goods through time. We have a four-month production schedule
to meet. The demand and production capacities are given below.
Table 22 Demand and production capacity
Month Production capacity Demand items
1 300 300
2 350 275
3 325 400
4 375 300
There is an initial stock of 50 items held at the beginning of month 1. Items can be made to meet
immediate demand or for stock to meet future demand. If orders are not met during the required
month, the sales are lost. The variable costs are £100 per item. The stockholding costs are £2
per item per month. What is the optimum production schedule?
Solution
The situation can be modelled using a transportation tableau with the rows representing the
initial stock and monthly production, and the columns representing monthly requirements. The
forbidden cells are those which involve meeting past orders from future production. These cells
are given an infinite cost in table.23.
Table 23 Data for production schedule for months 1 - 4
Cost £ per item months
M1 M2 M3 M4
Total Available
stock M1 2 4 6 8 50
M1
Production
100 102 104 106 300
M2 ∞ 100 102 104 350
M3 ∞ ∞ 100 102 325
M4 ∞ ∞ ∞ 100 375
Total demand 300 275 400 300
The normal procedure is now followed to solve this transportation problem to minimise the cost
of meeting the schedule (See Example 8).
313.
307
STUDYTEXT
Network Planning and Analysis
Degeneracy
A solution is degenerate when there are fewer than (m + n - 1) allocations in the tableau. This
problem can be overcome by allocating a very small amount, essentially zero, to an independent
cell. The number of allocations is then increased to (m + n — 1). The procedure in the MODI test
for optimality will show us which empty cells to use.
Example 6: Degenerate solutions
Three warehouses (X, Y and Z) can supply 6, 3 and 4 items to 3 shops (L, M and N), which
require 4, 5 and 1 items respectively. The unit costs of transport are given in the tableau.
Table 24 Data for example.6
To shop (£/item)
L M N
Total Available
X
From Warehouse Y
Z
6 4 9 6
5 3 2 3
2 3 6 4
Total Required 4 5 1
How should the items be allocated in order to minimise the total cost of transport?
Solution
There is a total of 13 items available which is more than the total requirement for 10 items,
therefore, we include a dummy shop which absorbs the surplus from the warehouses. We will
use Vogel's method to find an initial allocation:
Table.25 initial allocation for example 8 using Vogel's method
The cost of the allocation is:
4 x 3+0 x 3+3 x 2+2 x l +2 x4=28
To shop
L M N Dummy Total available
Penalty cost
1 2 3
6 4 9 0 41
2 2
X - 3 - 31
6 3 0
From warehouse 5 3 2 0 2 1 2
Y - 2 12
- 3 2 0
2 3 6 0 2 1 1
Z 43
- - - 4 0
Total required 4
0
5
3 0
1
0
3
0 13
1st
penalty
2nd
penalty
3rd
penalty
3
3
33
0
0
0
4
42
-
0
-
-
314.
quantitative techniques30 8
STUDYTEXT
For a basic solution, there should be (3 + 4 — 1) = 6 allocations, but here there are only 5.
The allocation is degenerate. As the MODI method proceeds, we will have to make one zero
allocation, to convert an empty cell into a pretend allocated cell. This gives the required total of 6
allocated cells. It will then be possible to calculate all of the u and v components and hence the
shadow costs.
We begin the MODI procedure using the initial 5 allocated cells. The extra zero allocation is
made when we can go no further. See table.26.
The allocated cells are used to find the row and column components from cij = ui + vj with = 0.
We can calculate v2, v4, u2 and v3 without any problem but we cannot find u3 or v1. We need an
additional allocated cell. We can put the zero allocation into any empty cell in the v1 column or
the u3 row. It does not matter which of these cells is chosen. Cell (Z, N) is used. The procedure
can be completed and the shadow costs found for the empty cells from sij cij - (ui + vi). The
figures are given in the table:
Table 26 Testing a degenerate allocation—MODI method
Two of the shadow costs are negative. The allocation is not optimal. It is necessary to re-allocate
into cell (Z, M) or (Z, Dummy). We will start with (Z, M), since this has the larger negative shadow
cost. The stepping stone circuit for (Z, M), showing the items allocated to each cell is:
Table 27 Stepping stone circuit for (Z, M)
M N
Allocated -2 Allocated +1
Test + Zero allocation 0
To shop
L M N Dummy Total available
6 4 9 0 u1
=0
X +7 3 +6 3 6
From warehouse 5 3 2 0 u2
=-1
Y +7 2 1 +1 3
2 3 6 0 u3
=3
Z 4 -4 0 -3 4
Total required 4 5 1 3 13
v1
=-1 v2
=4 v3
=3 v4
=0
315.
309
STUDYTEXT
Network Planning and Analysis
To find the number of items to move round the circuit, we look at the — cells, (Y, M) and (Z, N),
which contain 2 and zero units. This means we must move the zero allocation round the circuit
so that cell (Z, N) becomes empty again, cell (Z, M) takes the zero allocation and becomes the
pretend allocated cell. The other allocations remain unchanged. When we test for optimality this
time, we find that all of the shadow costs are positive. The allocation is optimal. This means that
the initial solution with the 5 allocations is actually the optimal solution. See the tableau in table
28:
Table 28 Testing the optimal allocation – MODI method
To shop
L M N Dummy Total available
6 4 9 0 u1=0
X +3 3 +6 3 6
From warehouse 5 3 2 0 u2=-1
Y +3 2 1 +1 3
2 3 6 0 u3=-1
Z 4 0 +4 +1 4
Total required 4 5 1 3 13
v1=3 v2=4 v3=3 v4=0
This type of result does occur in some problems in which then an ordinary allocation is degenerate.
There will be some problems in which the re-allocation process adds items to the zero allocation
cells and it becomes an ordinary allocated cell. The degeneracy of the solution then disappears.
The usual procedures will then lead to a better allocation.
Maximisation
The transportation algorithm assumes that the objective function is to be minimised. However, if
a suitable problem requires the objective function to be maximised, the algorithm can be modified
slightly to deal with this. For example, we may wish to transfer the items in Example 5 in such
a way that the total contribution is maximised. In such a case the data required are the unit
contributions between each origin and destination. The procedure is to multiply all of the unit
contributions by (-1_ and then to proceed in the usual way.
The Assignment Problem
The assignment problem is a special case of the transportation problem, in which the number
of origins must equal the number of destinations, that is, the tableau is square. Also, at each
destination, the 'demand' = 1 and at each origin the 'supply' = 1. Any assignment problem may be
solved using either linear programming or the transportation algorithm. However, the particular
structure of this problem has resulted in the development of a specially designed solution
procedure called the Hungarian algorithm
316.
quantitative techniques31 0
STUDYTEXT
The assignment algorithm
The algorithm has three stages to it.
Stage 1:
1. Set out the problem in tableau format, as in the transportation algorithm.
2. For each row in the tableau, find the smallest row element and subtract it from every
element in the row
3. Repeat for the columns.
There is now at least one zero in every row and every column. The assignment problem
represented by this 'reduced' tableau is equivalent to the original problem and the optimum
allocation will be the same for both, the objective of the Hungarian algorithm is to continue to
reduce the matrix until all of the items to be assigned, can be allocated to a cell with a zero value.
This means that the total value of the reduced objective function will be zero. Since negative
values are not allowed, an objective function values of zero is the optimum.
Stage 2: For a feasible solution, there must be exactly one assignment in every row and every
column.
1. If the assignments are made only to cells with zero values, this will give us the minimum
value of the objective function.
2. Find a row with only one zero in it, and make an assignment to this zero. If no such row
exists, begin with any zero.
3. Cross out all other zeros in the same column.
Repeat 1 and 2 until no further progress can be made.
If, at this stage, there are still zeros which are not either assigned or deleted, then:
4. Find a column with only one zero and assign to it.
5. Cross out all other zeros in the same row.
6. Repeat 4 and 5 until no further movement is possible.
If all zeros are still not accounted for, repeat I to 6. If the solution is feasible, that is, all of the
allocations have been made to zeros, and then the solution must also be optimal. If the solution
is not feasible, go on to Stage 3.
Stage 3:
1. Draw the minimum number of straight lines through the rows and columns (not diagonals)
that all zeros in the tableau are covered.
2. Find the smallest element without a line through it.
3. Subtract this number from every element without a line through it.
4. Add the chosen number to every element with two lines through it.
5. Leave alone all elements with one line through them.
This procedure has now created at least one new zero. Return to Stage 2 and repeat the procedure
until the optimum solution is reached.
Example 7: To illustrate the application of the assignment algorithm
A company has 4 distribution depots and 4 orders to be delivered to separate customers. Each
depot has one lorry available which is large enough to carry one of these orders. The distances
between each depot and each customer are given in table.29:
317.
311
STUDYTEXT
Network Planning and Analysis
Table 29 Distance from depots to customers
Distance in miles
Customers
I II III IV
A 68 72 75 83
Depot B 56 60 58 63
C 38 40 35 45
D 47 42 40 45
How should the orders be assigned to the depots in order to minimise the total distance
travelled?
Solution
It will help you to understand the problem if you try to find a solution to it using a familiar technique,
before applying the mechanics of the Hungarian algorithm. Try using Vogel's penalty cost method.
See how close you are to the optimum given at the end of the section. The availabilities and
requirements are one for each row and column.
Stage 1 of the Hungarian algorithm: Find the smallest row elements.
Table 30 To find the smallest row elements
Customers Smallest row element
I II III IV
A 68 72 75 83 68
Depot B 56 60 58 63 56
C 38 40 35 45 35
D 47 42 40 45 40
Subtract the smallest element from each element in its row
Table 31
0 4 7 15
0 4 2 7
3 5 0 10
7 2 0 5
0 2 0 5
Smallest column
element
318.
quantitative techniques31 2
STUDYTEXT
Subtract the smallest column element from each element in its column
Table 32
0 2 7 10
0 2 2 2
3 3 0 5
7 0 0 0
Make assignments as described in Stage 2 above; (0) denotes an assignment
Table 33
(0) 2 7 10
0 2 2 2
3 3 (0) 5
7 (0) 0 0
We can make only three zero assignments and we require four. This is not a feasible solution. Go
on to Stage 3. We draw the least number of lines to cover all of the zeros.
Table 34
(0) 2 7 10
0 2 2 2
3 3 (0) 5
7 (0) 0 0
The smallest number without a line through it is 2. Adjust the tableau as described in Stage 3
above, that is, deduct 2 from any number without a line through it, add 2 to any number at the
intersection of two lines and leave all the remaining numbers, which are cut by one line. We then
re-allocate depots to customers.
Table 35 Adjusted tableaus with assignments to zeros
I II III IV
A (0) 0 7 8
B 0 (0) 2 0
C 3 1 (0) 3
D 9 0 2 (0)
319.
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STUDYTEXT
Network Planning and Analysis
We have now made the required four allocations to zeros and therefore the solution is optimal.
We allocate depot A to customer I, depot B to customer II, depot C to customer Ill and depot
D to customer IV. The solution, though optimal, is not unique. (C, Ill) must always be allocated
because it is the only zero in the C row. There are two other optimum allocations.
Table 36 First alternative optimum allocation
I II III IV
A (0) 0 7 8
B 0 0 2 (0)
C 3 1 (0) 3
D 9 (0) 2 0
Table 37 Second alternative optimum allocation
I II III IV
A 0 (0) 7 8
B (0) 0 2 0
C 3 1 (0) 3
D 9 0 2 (0)
The minimum mileage for each of these three allocations can be calculated from the original
tableau:
Allocation 1 = 68 + 60 + 35 + 45 = 208 miles
Allocation 2 = 68 + 63 + 35 + 42 = 208 miles
Allocation 3 = 72 + 56 + 35 + 45 = 208 miles
All three solutions give the same total mileage.
Note: For larger problems than the one in Example.7, it may be more difficult to be sure that,
at Stage 3, step 1, the minimum number of lines has been drawn to cover all of the zeros. The
following rule of thumb' may be helpful:
i. Choose any row or column which has a single zero in it.
ii. If a row has been chosen, draw a line through the column in which the zero lies.
iii. If a column is chosen, draw a line through the row in which the zero lies.
iv. Repeat steps 1 to 3 until all of the zeros have been covered.
Special cases of the assignment problem
Maximise the objective function
The assignment algorithm is designed to minimise the objective function. If we have an assignment
problem, but we wish to maximise the objective function, we deal with this as we would for the
transportation algorithm. We set up the first tableau and multiply all the values in the cells by -1.
321.
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STUDYTEXT
Network Planning and Analysis
Table 40 Subtract row elements and find smallest column elements
Area
I II III IV V VI
A 15 11 8 0 8 14
B 7 3 5 0 2 4
C 10 7 5 0 20 18
D 13 11 6 8 0 17
E 8 0 2 3 1 0
F 7 9 3 2 0 4
7 0 2 0 0 0
Smallest element
We subtract the smallest column element from each element in the column.
Table 41 Subtract smallest column element
Area
I II III IV V VI
A 8 11 6 0 8 14
B 0 3 3 0 2 4
C 3 7 3 0 20 18
D 6 11 4 8 0 17
E 1 0 0 3 1 0
F 0 9 1 2 0 4
Forbidden Allocations
Again this problem is solved in the same way as it was in the transportation algorithm. If a
particular assignment is impossible for some reason, we insert a value into the relevant cell which
is much bigger than any other value. The algorithm will then automatically avoid this allocation.
Unequal numbers of origins and destinations
If the tableau is not square, then additional dummy rows or columns must be included to make
it square. The values assigned to these dummy cells will usually be zero. Destinations which
receive allocations from dummy rows (origins) are the ones which, in practice, will not receive
an allocation. Allocations which are made to dummy columns represent items which are not
allocated.
322.
quantitative techniques31 6
STUDYTEXT
Exercise 1
In the Kingdom of the Republic of Idion there are five coal mines which have the following outputs
and production costs:
Mine output (tonnes/day production cost (£/tonne)
1 120 25
2 150 29
3 80 34
4 160 26
5 140 28
Before the coal can be sold, it must be 'cleaned' and graded at one of three coal preparation
plants. The capacities and operating costs of these three plants are as follows:
Plant capacity (tones/day) operating cost (£/tonne)
A 300 2
B 200 3
C 200 3
All coal is transported by rail at cost of £0.5 per tonne kilometer, and the distances (in kilometers)
from each mine to the three preparation plants are:
Preparation
plant
Mines (Distance)
1 2 3 4 5
A 22 44 26 52 24
B 18 16 24 42 48
C 44 32 16 16 22
Required
a) Using a transportation model, determine how the output of each mine should be
allocated to the three preparation plants.
b) Following the installation of new equipment at coal mine number 3, the production cost
is expected to fall to £30 per tonne. What effect, if any, will this have on the allocation
of coal to the preparation plants?
c) It is planned to increase the output of coal mine number 5 to 180 tonnes per day which
can be achieved without any increase in production cost per tonne. How will this affect
the allocation of coal to the preparation plants?
(ACCA, June 1986)
324.
quantitative techniques31 8
STUDYTEXT
Preparation plant A has 50 tonnes per day spare capacity even though it has the cheapest
operating costs. The total costs of the above allocation are:
70 x 38 + 50 x 37 + 150 x 40 + 40 x 49 + 40 x 45 + 160 x 37 + 140 x 42 = $26,070/day
a) Production costs at Mine 3 fall from $34 to $30 per tonne. All mine output is already
taken by the plants and production costs are like a fixed cost and do not affect the
allocation, therefore total cost will be reduced by 80 x 4 = $320 per day
b) Mine 5 plans to increase output by 40 tonnes per day from 140 to 180. All of Mine 5's
output is allocated to Plant A which has 50 tonnes per day spare capacity. The extra
40 tonnes per day output will go from Mine 5 TO Plant A, increasing costs by 40 x 42 =
$1,680 per day
Exercise 2
a) Briefly describe and contrast two methods of finding an initial feasible solution to a
transportation problem
b) The Braintree Electronics Company produces video cassette tapes for purchase by the
general public. The demand (in hundreds) and production capacities (in hundreds) for
the three months in the forth quarter of the year are shown below
Month Normal time Overtime Demand capacity
October 300 400 150
November 450 400 150
December 800 400 150
Note that capacity exists to produce the video cassette tapes by working normal and overtime
hours, where capacity remains constant over time but demand increases for the Christmas sales.
The company does not have any initial inventory and does not wish to have any inventory on
hand after December.
The costs for production of the video cassette tapes are £150 (per hundred) if produced during
normal working hours and £180 (per hundred) if produced during overtime. It has been determined
that inventory costs are £20 (per hundred) per month. You should assume that all orders are
satisfied on time and that all demands and outputs occur at the midpoint of each month.
Required
i) Formulate this production scheduling situation as a transportation problem with six
'sources' and three 'destinations', showing the unit cost associated with each source
destination combination
ii) Use the transportation algorithm to find the optimum production schedule over this
period. State the total production cost of your solution
(ACCA, June 1988)
a) Describe two methods of finding an initial feasible solution. We describe two methods
these are the minimum cost and Vogel's penalty cost methods
b) Braintree Electronics – video tapes. There are six sources of production: October
normal (400) and December overtime (150). Total capacity is 1650 ('00 tapes) and total
demand is 1550 ('00 tapes). Therefore, we need a dummy demand column. The costs
in the tableau are production and inventory costs.
326.
quantitative techniques32 0
STUDYTEXT
Product Order units
Mugs 4000
Cups 2400
Bowls 1000
There are three machines available for the manufacturing operations, and all three can produce
each of the products at the same production rate. However, the unit costs of these products
vary depending upon the machine used. The unit costs (in £) of each machine are given in the
following table
Mugs cups bowls
A 1.20 1.30 1.10
Machine B 1.40 1.30 1.50
C 1.10 1.00 1.30
Furthermore, it is known that capacity for next week for machines B and C is 3000 units and for
machine A is 2000 units
Required
i) Use the transportation model to find the minimum cost production schedule for the
products and machines. Determine this minimum cost
ii) If this optimal solution is not unique, describe all other production schedule with the
minimum cost. If the production manager would like the minimum cost schedule to
have the smallest number of changeovers of production on machines, recommend the
optimal solution
Answers
(a) (i) Degeneracy:
In the transportation algorithm, an allocation is degenerate if the number of routes used
is less than (number of rows + number of columns - l). The difficulty may be overcome
by the use of dummy routes. For example, if the actual number of routes used is one
too few, then an empty cell is selected and treated as if it were an allocated cell, if we
wish, we can allocate a very small amount to this cell. The amount allocated should be
so small that the associated transport costs can be ignored. If the transport allocation
is degenerate by two routes then we use two empty cells in this way, and so on.
(ii) Inequality of supply and demand:
Thetransportationalgorithmrequiresthetotalamountdemandedatallofthedestinations
to be equal to the total amount available at all of the origins. If this is not the case, then
the algorithm cannot be used until a dummy origin or destination has been added to
satisfy this condition.
If the total supply exceeds the total demand, then a dummy destination is added which
has a demand equal to the excess supply. If the total supply is less than the total
demand, then a dummy origin added which has a supply equal to the shortfall. The
transportation costs for all routes involving the dummy are zero.
(iii) Non-unique optimal solution:
The optimal solution is non-unique if there is more than one allocation of routes which
produce the minimum cost. The existence of other optimal allocations can be identified
by examining the shadow cost. If any of the shadow costs are zero, then alternative
optima exist.
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STUDYTEXT
Network Planning and Analysis
A zero shadow cost means that items may be moved into that cell without increasing
the total cost of the allocation. Non-unique optimal solutions are not a difficulty in the
transportation problem, but it does mean that the decision maker must use some
criterion, other than cost to choose the allocation of routes to be used.
(b) (i) Find the minimum cost production schedule using the transportation method:
Total capacity of the machines is 8,000 units.
Total requirement is 7,400 units.
Therefore (here is excess capacity of 600 units. A dummy product must be included in
the transportation tableau, for which the demand is 600 units next week.
Set up a transportation tableau and use Vogel's method to make the initial allocation.
Product capacity penalty
mugs cups bowls dummy
A 1.20 1.30 1.10 0.00 2000 1.10,0.10
0.1010004 _ 10003 _ 1000
0
B 1.40 1.30 1.50 0.00 3000
2400
0
1.30,0.10
0.10Machine 24004 _ _ 6001
C 1.10 1.00 1.30 0.00 3000
600
0
1.00,0.10
0.206004 24002 _ _
Demand 4000
0
2400
0
1000
0
600
0
Penalty 0.10
0.10
0.10
0.30
0.302
0.20
0.20
0.203
0.00
For a basic allocation we require:
No of allocated cells = no of rows + no of columns -1
In this case, no of rows + no of columns 3 + 4-1= 6, which is the same as the number of allocated
cells. The allocation is basic.
Test for optimality
For each allocated cell, we split the unit cost, cij, into a row component, ui, and a column
component, vj
For each empty cell, we calculate the shadow cost, sij, where suj = cij – (ui + vj)
329.
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STUDYTEXT
Network Planning and Analysis
Product
mugs cups bowls dummy
A 1.20 1.30 1.10 0.00
1000 _ 1000 _
B 1.40 1.30 1.50 0.00
Machine 2400 _ 600
C 1.10 1.00 1.30 0.00
3000 _ _ _
This solution requires only one changeover (machine A) and therefore is the preferred choice.
The preferred minimum cost production schedule is:
Machine A makes 1,000 mugs and l000 bowls and is filly used;
Machine B makes 2,400 cups and is under-used by 600 units;
Machine C makes 3,000 mugs and is fully used.
Exercise 4
a) Explain briefly how the transportation algorithm can be modified for profit maximisation
rather than the minimisation of costs.
b) The Orange Computer Company manufactures one product, a dot-matrix printer, which
is currently in short supply. Four of Orange's main outlets, large specialty computer
shops at Abbotstown, Beswich, Carlic and Denstone, already have requirements which
in total exceed the combined capacity of its three production plants at Rexford, Eadon
and Tristron. The company needs to know how to allocate its production capacity to
maximize profits
Distribution costs (£) per unit from each production plant to each speciality shop are given in the
following table.
From
T o
Abbotstown
To Beswich To Carlic To Denstone
£ £ £ £
Rexford 22 24 22 30
Seadon 24 20 18 28
Tristron 26 20 26 24
Since the four specialty shops are in different parts of the country, and as there are differing
transportation costs between the production plants and the specialty shops, along with slightly
three of its different production costs at each of the production plants, there is a pricing structure
that enables different prices to be charged at the four shops. Currently the price per unit charged
is £230 at Abbotstown, £235 at Beswich, £225 at Carlic, and £240 at Denstone. The variable unit
production costs arc £150 at plants Rexford and Tristron, and £155 at plant Seadon.
330.
quantitative techniques32 4
STUDYTEXT
Required:
i) Set up a matrix showing the unit contribution to profit associated with each production
plant/speciality shop allocation.
ii) The demands at Abbotstown, Beswich, Carlic, and Denstone are 850, 640, 380, and
230 to produce respectively. The plant capacity at Rexford is 625, at Seadon is 825,
and at Tristron is 450. Denstone vary. Use the transportation algorithm to determine the
optimal allocation.
iii) Determine the contribution to profit for the optimal allocation.
(ACCA, June 1990)
Answers
(a) How can the transportation algorithm be modified to maximise rather than minimise?
Instead of minimising the positive unit costs of all of the cells, calculate the unit profits,
make them negative and put these in each cell. Use the transportation algorithm as
usual to minimise these negative profits.
Alternatively, load the cells with the largest profits (instead of smallest costs) to give an
initial allocation. Test the empty cells as usual, but use any cell which has a price. If all
the shadow prices are negative or zero, then allocation give the maximum profit.
(b) Orange produces printers in three factories and sells them to four main outlets. How
many of the available printers should be supplied from each factory to each outlet in
order to maximise profit?
Factories R, S and T supply shops A, B, C and D.
The contribution per printer = selling price at the shop for a particular cell
- variable cost at the factory
- Factory to shop transport cost
For example the contribution per printer supplied A = 230 – 50 = £180 per printer.
The matrix of contributions is given below:
Contribution £ per printer
To Shop
From
Factory
A B C D
R 58 61 53 60
S 51 60 52 57
T 54 65 49 66
The total demand from the four shops is:
850 - 640+380+ 230 = 2100 printers
The total supply t the three Factories is:
625+825+450=1900 printers
332.
quantitative techniques32 6
STUDYTEXT
This gives the following table:
To shop U
A B C D
R 58 61 53 60 u1=0
625 -6 -6 -8
S
From
factory
51 60 52 57 u2=-7
25 420 380 -4
T 54 65 49 66 u3=-2
-2 220 -8 230
Dummy 0 0 0 0 u4=-58
200 -9 -1 -10
V v1=58 v2=67 v3=59 v4=68
All of the shadow prices are negative, therefore any change would reduce the contribution,
consequently this is the optimum allocation.
The Rexford factory supplies all of its 625 printers to Abbotstown.
Seadon supplies 25 to Abbotstown, 420 to Beswich and 380 to Garlic.
Triston supplies 220 to Beswich and 230 to Ocustonc.
Abbotstown does not receive 200 oldie printers it asked for.
(iii) The contribution to profit of this allocation is:
58 x 625 +51 x 25 + 60 x 420 + others
Exercise 5
(a) The assignment problem can be regarded as a special case of a transportation
problem.
Describe these special features of the assignment problem and explain why the
transportation algorithm tends not to be used to solve such problems.
(b) The Midland Research Association has recently been notified that it has received
government grant for the research grants to undertake four major projects. The
managing director has to assign a research officer to each of these projects. Currently
there are five research officers - Adams, Brown, Carr, Day, Evans who are available to
carry out these duties. The amount of time required to complete each of the research
projects is dependent on the experience and ability of the research optimal officer who
is assigned to the project. The managing director has been provided with an estimate
of the project completion time (in days) for each officer and each project.
333.
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STUDYTEXT
Network Planning and Analysis
R e s e a r c h
officer
Project
1 2 3 4
Adams 80 120 60 104
Brown 72 144 48 110
Carr 96 148 72 120
Day 60 108 52 92
Evans 64 140 60 96
As the four projects have equal priority, the managing director would like to assign research
officers in a way that would minimise the total time (in days) necessary to complete all four shops
at projects.
Required:
(i) Determine an optimal assignment of research officers to projects, and hence determine
the minimum total number of days allocated to these four projects.
(ii) State any further allocations that would result in the same total number of days. If
research officers Brown, Carr and Day express a preference for projects 2 or 3, whilst
officers Adams and Evans express their preference for projects 1 or 4, which of the
optimal allocations seems to be the most sensible for the managing director to make?
iii) What feature of this particular project duration matrix could be exploited to simplify the
problem?
(ACCA, June 1989)
Answers
(a) The special features of the assignment problem:
The assignment problem must have an equal number of origins and destinations. The
supply is one unit at each origin and the demand is one unit at each destination
The transportation algorithm can not to be used to solve transportation problems under
the above conditions because of difficulties due to degenerate allocations. There
will usually be several degenerate stages which increases the number of iterations
required.
(b) (i) Determine the optimal allocation of people to projects and the minimum total time
required for the four projects
Since there are 5 research officers and 4 projects, we must include a dummy project
which takes zero time no matter which officer is assigned to it. However, if we
look at the data in more detail, we can see that Carr is estimated to require the longest
time to complete each of the projects. It follows, therefore, that the algorithm will allocate
the dummy project to Carr. Knowing this we can reduce the problem to the allocation of
4 projects to 4 people.
335.
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STUDYTEXT
Network Planning and Analysis
There are three allocations only, therefore we know we have not reached the optimum solution.
Step 4: Draw the minimum number of lines through the cells with zeros. Find the smallest number
in a cell without a line through it. Subtract this number from the values in the unlined' cells and
add it to the values in the cell, with two lines through them.
Research
Officer
Project time, days
1 2 3 4
Adams 16 4 0 8
Brown 20 40 0 26
Day 4 0 0 4
Evans 0 24 0 0
Research
Officer
Project time, days
1 2 3 4
Adams 12 0 0 4
Brown 16 36 0 22
Day 4 0 4 4
Evans 0 24 4 0
Step 5: Repeat the allocation procedure with the new tableau
Research
Officer
Project time, days
1 2 3 4
Adams 12 0 0 4
Brown 16 36 0 22
Day 4 0 4 4
Evans 0 24 4 0
The allocation is still not optimal. Repeat steps 4 and 5 above
Research
officer
Project time, days
1 2 3 4
Adams 12 0 0 4
Brown 16 36 0 22
Day 4 0 4 4
Evans 0 24 4 0
The smallest
Number in an
Uncovered cell
Is 4.
The smallest number in a cell,
not covered by a line is 4.
336.
quantitative techniques33 0
STUDYTEXT
Research
Officer
Project time, days
1 2 3 4
Adams 8 0 0 0
Brown 12 36 0 18
Day 0 0 4 0
Evans 0 28 8 0
Repeat the allocation procedure
Research
Officer
Project time, days
1 2 3 4
Adams 8 0 0 0
Brown 12 36 0 18
Day 0 0 4 0
Evans 0 28 8 0
We now find that the solution is optimal but not unique. Brown must be allocated to project 3, but
Adams, Day and Evans can be allocated to more than one of the other projects. The total time
used will not be affected.
We will take the option in which Adams does project 2, Day does project I and Evans does project
4. The total time used is 120+48+60-,-96=324 days.
(ii) State the alternative optimum and select the most sensible one:
The alternative optimum allocations are:
Officer Project Project Project
Adams 2 2 4
Brown 3 3 3
Day 1 4 2
Evans 4 1 1
Officer Brown must have his preference for project 3 which means that Officer can have his
preference for project 2. We must, therefore, choose alternaiivenumba3, which gives project 4 to
Adams and project I to Evans.
(iii) How can the problem be simplified?
The elimination of Officer Carr from the project has already been exploited in
part (b)(i).
337.
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STUDYTEXT
Network Planning and Analysis
Chapter SummaryTime Estimates
(a) Optimistic time estimate = a
This is the shortest time an activity can take to be complete. It represents real estimate such that
the probability is small that the activity can be completed in less time.
(b) Pessimistic time estimate = b
This is the longest time an activity can take to be completed. It is the worst time estimate
representing bad luck. Such that the probability is small that the activity will take longer.
(c) MostChapter quiz
1. What denotes the beginning or the ending of an activity?
2. These desriptions relate to a certain activity. Which activity is it?
(a) used to improve clarity of the network
b) Used to facilitate a logical flow of activities in the network.
3. This is the longest time an activity can take to be completed
4. This is the process of attempting to reduce the peaks and troughs in the resource
allocation so that we have a more even usage of personnel.
5. A solution is ……….. when there are fewer than (m + n - 1) allocations in the tableau.
338.
quantitative techniques33 2
STUDYTEXT
Answers to chapter quiz
1. Event
2. Dummy activity
3. Pessimistic time estimate
4. Smoothing a profile
5. Degenerate
questions from previous exams
June 2000 Question 8
a) Explain the following terms as used in network analysis:
(i) Backward pass (2 marks)
(ii) Crashing (2 marks)
(iii) Stock (2 marks)
(iv) Earliest start times (2 marks)
(v) Critical-path activities (2 marks)
b) XYZ Construction Company is building a 250-unit apartment complex in Embakasi,
Kenya. The project consists of hundreds of activities involving excavating, framing,
wiring, plastering, painting, landscaping and more. Some of the activities must be done
sequentially and others can be done simultaneously. Also, some of the activities can be
completed faster than normal by acquiring additional resources.
Required:
(i) How would Quantitative Techniques be used to solve this problem? (2 marks)
(ii) What would be the uncontrollable inputs? (2 marks)
(iii) What would be the decision variables of the model? The objective function? The
constraints? (3 marks)
(iv) Is the model deterministic or stochastic? (1 mark)
v) Suggest assumptions that could be made to simplify the model. (2 marks)
(Total: 20 marks)
340.
quantitative techniques33 4
STUDYTEXT
Required:
a) Determine the project's expected completion time and its critical path. (12 marks)
b) Can activities E and G be performed at the same time without delaying the completion
of the project? (2 marks)
c) Can one person perform A, G and I without delaying project? (2 marks)
d) By how much time can activities G and L be delayed without delaying the entire
project? (2 marks)
e) By how much time would the project be delayed if activity G were delayed by 3 hours
and activity L by 4 hours? (2 marks)
(Total: 20 marks)
June 2004 Question 8
Mrs. Mwangi wants to open a cafeteria in Kisumu. A small business enterprise adviser whom she
approached, listed for her six major activities to be carried out. The table below gives a summary
of the normal time estimates of each activity, crash time and the cost reduction per day.
Activity Predecessor
Normal time
(weeks)
Crash time
(weeks)
A Procurement of materials - 3 3
B Plumbing A 6 4
C Masonry - 5 3
D Electrical works C 8 7
E Carpentry C 6 4
F Finishing B,D,E 4 2
Activity Cost Slope (Sh.)
A -
B 45,000
C 30,000
D 60,000
E 22,500
F 75,000
Required
(a) The normal completion time of the project and the critical activities (6 marks)
(b) (i) The shortest time the project can be completed. (8 marks)
(ii) The additional cost to be incurred if the project is crashed. (2 marks)
(c) Explain the meaning of the cost slope and how it is computed. (2 marks)
(d) Assumption made when crashing (2 marks)
341.
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STUDYTEXT
Network Planning and Analysis
December 2004 Question 8
The following data relate to activities in the development of a two-stage audit software by Mwangi
and Onyango associates for Mwitu Ltd.
Activity Description Immediate
predecessor
Expected time
(days)
A Evaluate the current system - 7
B Adjust the current system A 3
C Develop a new system - 6
D Dry-run the new system C 3
E Test the system B,D 2
The recommended crashed activity times and total costs are shown below:
Activity Expected time (days) Cost (Sh.'000')
A 6 600
B 3 200
C 6 500
D 3 200
E 1 550
At the start of day 8, the following activity status report is received:
Activity Actual cost (AC, sh. '000) Percent completion
A 800 100
B 100 67
C 450 100
D 250 50
E 0 0
Required:
a) (i) In terms of cost, is the project on schedule? Make a recommendation on the action
to be taken. (10 marks)
(ii) Clearly explain the assumptions that are made in calculating the 'crash' costs for the
activities. (2 marks)
(b) Explain how the crash cost per unit is computed. Why is it important to compute
them? (3 marks)
(c) Explain the purpose of a cost status report. (5 marks)
(Total: 20 marks)
345.
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STUDYTEXT
Simulation and Queuing Theory
CHAPTER NINE
Simulation and Queuing Theory
Objectives
At the end of this chapter, you should be able to:
Cite various applications of simulation in the business world••
Discuss merits and demerits of simulation••
Cite various features of queuing theory••
Mention various problems associated with queuing systems••
Define optimum order quantity, re-order level and re-order interval••
Interpret the total cost equation••
Explain the effect of different quantity discount levels••
Introduction
Simulation is a technique used to make decisions under conditions of uncertainty whereby a
model of the real system is used and then a chain of repeated trial and error experiments are
conducted to forecast the behaviour of the system over a period. It is an imitation of the real
system; for instance, model of an aircraft, a windmill.
Fast Forward: The act of simulating something generally entails representing certain key
characteristics or behaviours of a selected physical or abstract system.
Definition of key terms
Random numbers – A set of numbers arranged in random order. Used in statistics to refer to
numbers drawn without bias or conscious choice to ensure equal chances for each item drawn.
Queue – A group of customers waiting for service in a system rendering some service.
Service facility – this is the facility that provides service to customers on the queue.
Waiting time – this is time taken in a queue before being provided with a service.
Time in the system – is equal to waiting time plus service time.
Balking – this is a case when a member of the population refuses to join the queue due to the
size of the queue.
FIFO queue discipline – this system that gives priority to those who arrive first.
Priority queue discipline – this system allows different priorities entitling them to get service on
a pre-emptive basis or a non-pre-emptive basis.
Pre-emptive basis allows some customers to interrupt customers already receiving service
while non-pre-emptive basis does not allow interruption of those already receiving the service.
346.
quantitative techniques34 0
STUDYTEXT
Single phase system – service is received from only one station.
Multi-phase system – service must be received from more than one station sequentially.
Industry Context
Most practical inventory systems deal with hundreds or thousands of items. A large manufacturer
or a supermarket are examples. Not all of the stock items should be dealt with in the same way. It
is sensible to concentrate efforts on the items which have high annual value, rather than on those
which have a small annual value. Hence, the use of Pareto effect. For instance the airline uses a
large amount of fuel costing pence per litre but with a high annual value due to volume.
It is important to note that most of the questions from simulation are run by computers and
the student will be expected to analyse and provide solutions. It is of essence for a student to
understand the various formulae for queuing to avoid interchanging them during exam.
Exam Context
The student will be expected to apply the imaginary theory of what happens in a manufacturing
industry or wholesale businesses and retail businesses and apply to exam questions.
Past Paper Analysis
12/05, 6/06, 6/05, 6/01, 6/00, 12/04
9.1 Simulation
Reasons for adopting Simulation
Simulation may be the only method available because it is difficult to observe the actual
environment.
It may be infeasible to develop a mathematical solution.
Actual observation of a real system may be too expensive.
Time may be insufficient to allow the real system operate extensively, for instance, study of sales
in a firm for a number of years.
It may be disruptive to adopt actual operation and observation of a real system.
Monte Carlo method is one of the most common simulation techniques. It uses random numbers
and yields a solution that is close to the optimal but not necessarily exact solution. It should be
noted that only models under uncertainty could be studied with the help of Monte Carlo method
because Monte Carlo simulation method is based on the continued observation of the system
over a long period and experience.
347.
341
STUDYTEXT
Simulation and Queuing Theory
Generating Random Numbers
To simulate a sample will involve the use of random numbers. Random numbers refers to numbers
selected in such a way that each number has an equal chance of selection. Once it is selected,
it is transformed into an observation drawn from the probability distribution specified in the model
under study.
The following are the methods used to generate random numbers:
1. They may be obtained from random number table in the memory of the computer.
2. An electronic device may be constructed as part of a digital computer to generate true
random numbers.
3. Mid-square method:involves the following procedures:
a) Take a four digited number and square it e.g 15372
=2362369
b) From the result above take the four digits starting with the third from the right e.g
2362369
c) To get the 2nd
random start with the new four-digited number obtained in (b) above and
repeat the procedure.
Advantages of Simulation
It is less costly and less risky than if the actual system was put in experiment.
Currently, simulation is done with computers hence less time is required.
Additionally,computersimulationcanberepeatedmakingusershavecontroloverthedevelopment
of the model.
Simulation does not require simplifications and assumptions as is the case in analytical
solutions.
It is easier to explain a simulation model to managers since it describes characteristic of a real
system process.
Simulation may still be used to check the correctness of the analytical solution.
In situations where it is difficult to predict bottlenecks, simulation may be employed.
Since simulation calls for breaking down of a system into subsystems, it allows one to gain
increased knowledge of the operating system.
Simulation is of magnitude importance in cases where complex relationships of a predictable
nature exist.
Since simulation keeps mathematics aside, it is easily understood by the operating personnel.
Simulation models are flexible hence modifiable to accommodate the changing environments of
the real situation.
It is easier to use than mathematical models and it is quite superior to mathematical models.
Simulation involves personnel into the exercise making the trainee gain confidence and familiarise
himself with data processing.
Disadvantages of Simulation
It does not produce optimum solutions since each simulation is run like a single experiment.
It is time consuming since it involves repetition of the experiment especially when done
manually.
348.
quantitative techniques34 2
STUDYTEXT
The difficulty in finding the optimum number of parameters increases as the number of parameters
increase.
There is a high chance of over reliance on simulation technique even in cases where mathematical
models would be appropriate.
Business applications of Simulation
i. Solution of queuing problems.
ii. To study inventory systems under uncertainties.
iii. Planning for the new product introduction into the market.
iv. Developing a number of network simulation models.
v. Testing of decision rules for Hospital operating policies.
Example
1. Describe the advantages and disadvantages of using simulation to investigate queuing
situations compared with the use of queuing theory formulae.
2. The time between arrivals at a complaints counter in a large department store has been
observed to follow the distribution shown below:
Time between arrivals
(minutes)
Probability
0 – 4 0.25
4 – 8 0.45
8 – 12 0.20
12 – 16 0.10
Customers' complaints are handled by a single complaints officer but all customers who consider
their complaints to be 'serious' or who have to wait 5 minutes or more before being seen by the
complains officer demand to see the store manager, who deals with them separately. The time to
deal with complaints by the complaints officer has a normal
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books.google.com - David Cohen's PRECALCULUS: A PROBLEMS-ORIENTED APPROACH, Sixth Edition, focuses on teaching mathematics by using a graphical perspective throughout to provide a visual understanding of college algebra and trigonometry. The author is known for his clear writing style and the numerous quality exercises... A Problems-Oriented Approach, Enhanced Edition
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Mathematical Ideas
9780321361462
ISBN:
0321361466
Pub Date: 2007 Publisher: Addison-Wesley
Summary: One of the biggest issues college math instructors face is capturing and keeping student interest. Over the years, John Hornsby has refined a creative solution--bringing the best of Hollywood into his mathematics classroom. Mathematical Ideas applies this same strategy of engaging students through descriptions of video clips from popular cinema and television to the textbook. Alongside fresh data and tools, this Elev...enth Edition uses up-to-the-minute images as well as old favorites of math being done in Hollywood. In addition, examples are clarified with additional annotations, the margin notes have been freshened, and Chapter 14: Personal Financial Management has been updated to meet the needs of today's students. With great care and effort, Vern Heeren and John Hornsby have crafted this new edition to serve the needs of today's students and instructors.
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DALLAS, March 1 /PRNewswire/ -- Texas Instruments (TI) has released a new operating system update for its TI-84 Plus family of graphing calculators that helps educators and students make a stronger visual connection between the math in textbooks and on their handhelds.
The TI-84 Plus OS version 2.53MP with MathPrint(TM) mode enables users to input and view math symbols and formulas, including stacked fractions, in their handhelds exactly as the equations appear in textbooks.
DALLAS, July 28 /PRNewswire/ -- As the back-to-school season approaches, parents want their children to start the school year on the right foot. It's important for parents to prepare their children by offering advice and tools that will further their academic careers.
By Steele, Marcee M Introduction High school algebra and geometry teachers use graphing calculators extensively in their classes today. National and state mathematics curriculum standards recommend their use for class work, tests and assignments (Dion, Jackson, Klag, Liu, & Wright, 2001).
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An unusual and attractive edition of Euclid was published in 1847 in England, edited by an otherwise unknown mathematician named Oliver Byrne. It covers the first 6 books of Euclid, which range through most of...
Overview of a movie on medical imaging. Imagine that you are given an image, say a medical (MRI or CT) scan. Suppose you want to extract the important feature within the image of a digital subtraction angiogram (DSA);...
This research investigated the process of argument co-construction in 14 cooperative problem-solving groups in an algebra-based, college level, introductory physics course. The results of the research provide a rich...
The Open Door Web Site is a reference source for both students and teachers containing separate sections for physics, biology, technology, history, electronics, and chemistry. Within each of these sections are tutorials...
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College Algebra - 5th edition reinf
orce what you've learned. In addition, the book includes many real-world examples that show you how mathematics is used to model in fields like engineering, business, physics, chemistry, and biology. reinforce what you've learned. In addition, the book includes many real-world examples that show you how mathematics is used to model in fields like engineering, business, physics, chemistry, and
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Elementary and Inter. Algebra : Graphs and Models - 4th edition
TheBittinger Graphs and Models Serieshelps readers learn algebra by making connections between mathematical concepts and their real-world applications. Abundant applications, many of which use real data, offer students a context for learning the math. The authors use a variety of tools and techniques-including graphing calculators, multiple approaches to problem solving, and interactive features-to engage and motivate all types of00 +$3.99 s/h
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what do you want to known about how to learn multiplication and division in china , you may look this .Just a multiplication table. It will make your ability improve in math .You will find calculation is geting easier.
This book contains the fundamental trigonometric and hyperbolic functions, 25 challenging problems, allong with their solutions and analysis.
The readers of this book should be familiar with Trigonometry, Algebra, Equations and Complex Numbers, and will be able to challenge and evaluate their knowledge and understanding of the subject.
Vedic Mathematics is a system of reasoning and mathematical working based on ancient Indian teachings called Veda. It is fast, efficient and easy to learn and use.
This E-book "Learn And Teach Vedic Mathematics" will explain in brief all the 16 Sutras with practical examples so that you can get rid of your maths phobia and become an expert in this subject.
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Gifts by Price
Pre Algebra Student Text
Product description
Presents concepts in numerous examples worked out with clear step-by-step explanations and provides additional exercises. (Note: About half of all students will be prepared to take this course during the 7th grade.)
Type: Hardcover (Student/Stdy Gde)Category: > Home SchoolingISBN / UPC: 9780890847626/0890847622Publish Date: 1/1/1994Item No: 84204Vendor: Bob Jones University Press
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Don't just tell students "Algebra helps you to think in the abstract". That's a great tool, but many people need an application. They may be creative, but not as good with abstract concepts. Give them examples of what Algebra does and what it's used for. That would keep a lot of them from dropping the class, out of school and falling through the cracks. I really, really needed to be great with Algebra for my college major, even if it wouldn't have been needed after graduation.
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Advanced Calculus (Rea's Problem Solvers)
Synopses & Reviews
Publisher Comments:
Each Problem Solver is an insightful and essential study and solution guide chock-full of clear, concise problem-solving gems. All your questions can be found in one convenient source from one of the most trusted names in reference solution guides. More useful, more practical, and more informative, these study aids are the best review books and textbook companions available. Nothing remotely as comprehensive or as helpful exists in their subject anywhere. Perfect for undergraduate and graduate studies.
Here in thisDETAILS
- The PROBLEM SOLVERS are unique - the ultimate in study guides.
- They are ideal for helping students cope with the toughest subjects.
- They greatly simplify study and learning tasks.
- They enable students to come to grips with difficult problems by showing them the way, step-by-step, toward solving problems. As a result, they save hours of frustration and time spent on groping for answers and understanding.
- They cover material ranging from the elementary to the advanced in each subject.
- They work exceptionally well with any text in its field.
- PROBLEM SOLVERS are available in popular subjects.
- Each PROBLEM SOLVER is prepared by supremely knowledgeable experts.
- Most are over 1000 pages.
- PROBLEM SOLVERS are not meant to be read cover to cover. They offer whatever may be needed at a given time. An excellent index helps to locate specific problems rapidly.
- Educators consider the PROBLEM SOLVERS the most effective and valuable study aids; students describe them as "fantastic" - the best books on the market.
TABLE OF CONTENTS
Introduction
Chapter 1: Point Set Theory
Sets and Sequences
Closed and Open Sets and Norms
Metric Spaces
Chapter 2: Vector Spaces
Definitions
Properties
Invertibility
Diagonalization
Orthogonality
Chapter 3: Continuity
Showing that a Function is Continuous
Discontinuous Functions
Uniform Continuity and Related Topics
Paradoxes of Continuity
Chapter 4: Elements of Partial Differentiation
Partial Derivatives
Differentials and the Jacobian
The Chain Rule
Gradients and Tangent Planes
Directional Derivatives
Potential Functions
Chapter 5: Theorems of Differentiation
Mean Value Theorems
Taylor's Theorem
Implicit Function Theorem
Chapter 6: Maxima and Minima
Relative Maximum and Relative Minimum
Extremes Subject to a Constraint
Extremes in a Region
Method of Lagrange Multipliers
Functions of Three Variables
Extreme Value in Rn
Chapter 7: Theory of Integration
Riemann Integrals
Stieltjes Integrals
Chapter 8: Line Integrals
Method of Parametrization
Method of Finding Potential Function (Exact Differential)
Independence of Path
Green's Theorem
Chapter 9: Surface Integrals
Change of Variables Formula
Area
Integral Function over a Surface
Integral Vector Field over a Surface
Invergence Theorem
Stoke's Theorem
Differential Form
Chapter 10: Improper Integrals
Improper Integrals of the 1st, 2nd, and 3rd Kind
Absolute and Uniform Convergence
Evaluation of Improper Integrals
Gamma and Beta Functions
Chapter 11: Infinite Sequences
Convergence of Sequences
Limit Superior and Limit Inferior
Sequence of Functions
Chapter 12: Infinite Series
Tests for Convergence and Divergence
Series of Functions
Operations on Series
Differentiation and Integration of Series
Estimates of Error and Sums
Cesaro Summability
Infinite Products
Chapter 13: Power Series
Interval of Convergence
Operations on Power Series
Chapter 14: Fourier Series
Definitions and Examples
Convergence Questions
Further Representations
Applications
Chapter 15: Complex Variables
Complex Numbers
Complex Functions and Differentiation
Series
Integration
Chapter 16: Laplace Transforms
Definitions and Simple Examples
Basic Properties of Laplace Transforms
Step Functions and Periodic Functions
The Inversion Problem
Applications
Chapter 17: Fourier Transforms
Definition of Fourier Transforms
Properties of Fourier Transforms
Applications of Fourier Transforms
Chapter 18: Differential Geometry
Curves
Surfaces
Chapter 19: Miscellaneous Problems and Applications
Miscellaneous Applications
Elliptic Integrals
Physical Applications
Index
WHAT THIS BOOK IS FOR
Students have generally found calculus a difficult subject to understand and learn. Despite the publication of hundreds of textbooks in this field, each one intended to provide an improvement over previous textbooks, students of advanced calculus continue to remain perplexed as a result of numerous subject areas that must be remembered and correlated when solving problems. Various interpretations of advanced calculus terms also contribute to the difficulties of mastering the subject.
In a study of calculus, REA found the following basic reasons underlying the inherent difficulties of advanced calculus:
No systematic rules of analysis were ever developed to follow in a step-by-step manner to solve typically encountered problems. This results from numerous different conditions and principles involved in a problem that leads to many possible different solution methods. To prescribe a set of rules for each of the possible variations would involve an enormous number of additional steps, making this task more burdensome than solving the problem directly due to the expectation of much trial and error.
Current textbooks normally explain a given principle in a few pages written by a mathematician who has insight into the subject matter not shared by others. These explanations are often written in an abstract manner that causes confusion as to the principle's use and application. Explanations then are often not sufficiently detailed or extensive enough to make the reader aware of the wide range of applications and different aspects of the principle being studied. The numerous possible variations of principles and their applications are usually not discussed, and it is left to the reader to discover this while doing exercises. Accordingly, the average student is expected to rediscover that which has long been established and practiced, but not always published or adequately explained.
The examples typically following the explanation of a topic are too few in number and too simple to enable the student to obtain a thorough grasp of the involved principles. The explanations do not provide sufficient basis to solve problems that may be assigned for homework or given on examinations.
Poorly solved examples such as these can be presented in abbreviated form which leaves out much explanatory material between steps, and as a result requires the reader to figure out the missing information. This leaves the reader with an impression that the problems and even the subject are hard to learn - completely the opposite of what an example is supposed to do.
Poor examples are often worded in a confusing or obscure way. They might not state the nature of the problem or they present a solution, which appears to have no direct relation to the problem. These problems usually offer an overly general discussion - never revealing how or what is to be solved.
Many examples do not include accompanying diagrams or graphs, denying the reader the exposure necessary for drawing good diagrams and graphs. Such practice only strengthens understanding by simplifying and organizing calculus processes.
Students can learn the subject only by doing the exercises themselves and reviewing them in class, obtaining experience in applying the principles with their different ramifications.
In doing the exercises by themselves, students find that they are required to devote considerable more time to advanced calculus than to other subjects, because they are uncertain with regard to the selection and application of the theorems and principles involved. It is also often necessary for students to discover those "tricks" not revealed in their texts (or review books) that make it possible to solve problems easily. Students must usually resort to methods of trial and error to discover these "tricks," therefore finding out that they may sometimes spend several hours to solve a single problem.
When reviewing the exercises in classrooms, instructors usually request students to take turns in writing solutions on the boards and explaining them to the class. Students often find it difficult to explain in a manner that holds the interest of the class, and enables the remaining students to follow the material written on the boards. The remaining students in the class are thus too occupied with copying the material off the boards to follow the professor's explanations.
This book is intended to aid students in advanced calculus overcome the difficulties described by supplying detailed illustrations of the solution methods that are usually not apparent to students. Solution methods are illustrated by problems that have been selected from those most often assigned for class work and given on examinations. The problems are arranged in order of complexity to enable students to learn and understand a particular topic by reviewing the problems in sequence. The problems are illustrated with detailed, step-by-step explanations, to save the students large amounts of time that is often needed to fill in the gaps that are usually found between steps of illustrations in textbooks or review/outline books.
The staff of REA considers advanced calculus a subject that is best learned by allowing students to view the methods of analysis and solution techniques. This learning approach is similar to that practiced in various scientific laboratories, particularly in the medical fields.
In using this book, students may review and study the illustrated problems at their own pace; students are not limited to the time such problems receive in the classroom.
When students want to look up a particular type of problem and solution, they can readily locate it in the book by referring to the index that has been extensively prepared. It is also possible to locate a particular type of problem by glancing at just the material within the boxed portions. Each problem is numbered and surrounded by a heavy black border for speedy identification.
Synopsis:
REAs Advanced Calculus Problem SolverSynopsis:"Synopsis"
by Firebrand,
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Signal Hill Calculus-2 is the door to real life related math. From Algebra-2 students start to deal with real life problems and learn how to solve them. What is calculus?
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The authors help students ''see the math'' through their focus on functions; visual emphasis; side-by-side algebraic and graphical solutions; real-data applications; and examples and exercises. By remaining focused on today's students and their needs, the authors lead students to mathematical understanding and, ultimately, success in class.
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Barron's College Review volumes make excellent supplements to college survey courses. Their broad overviews of subject material also make them fine brush-up texts for students preparing for standardized tests. Books in this series contain succinct explanations of subject matter along with examples, exercises, and solutions.
This general review covers equations, functions, and graphs; limits, derivatives; integrals and antiderivatives; word problems; applications of integrals to geometry; and much more. Additional features make this volume especially helpful to students working on their own. They include worked-out examples, a summary of the main points of each chapter, exercises, and where needed, background material on algebra, geometry, and reading comprehension.
Table of Contents:
Introduction
Equations, Functions, and Graphs
Change, and the Idea of the Derivative
The Idea of Limits
Computing Some Derivatives
Formulas for Derivatives
Extreme Values, the Mean Value Theorem, and Curve Sketching
Word Problems
The Idea of the Integral
Computing Some Integrals
Formulas for Integrals: Integrals, Antiderivatives and the Fundamental Theorem of Calculus
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"A very stimulating book ... in a class by itself." ? American Mathematical Monthly Advanced students, mathematicians and number theorists will welcome this stimulating treatment of advanced number theory, which approaches the complex topic of algebraic number theory from a historical standpoint, taking pains to show the reader how concepts, definitions... more...
This excellent textbook introduces the basics of number theory, incorporating the language of abstract algebra. A knowledge of such algebraic concepts as group, ring, field, and domain is not assumed, however; all terms are defined and examples are given ? making the book self-contained in this respect. The author begins with an introductory chapter... more...
An advanced monograph on Galois representation theory by one of the world's leading algebraists, this volume is directed at mathematics students who have completed a graduate course in introductory algebraic topology. Topics include abelian and nonabelian cohomology of groups, characteristic classes of forms and algebras, explicit Brauer induction... more...
This volume contains the two most important essays on the logical foundations of the number system by the famous German mathematician J. W. R. Dedekind. The first presents Dedekind's theory of the irrational number-the Dedekind cut idea-perhaps the most famous of several such theories created in the 19th century to give a precise meaning to irrational... more...
The teaching of mathematics has undergone extensive changes in approach, with a shift in emphasis from rote memorization to acquiring an understanding of the logical foundations and methodology of problem solving. This book offers guidance in that direction, exploring arithmetic's underlying concepts and their logical development. This volume's great... more...
The three-volume series History of the Theory of Numbers is the work of the distinguished mathematician Leonard Eugene Dickson, who taught at the University of Chicago for four decades and is celebrated for his many contributions to number theory and group theory. This second volume in the series, which is suitable for upper-level undergraduates... more...
Careful organization and clear, detailed proofs characterize this methodical, self-contained exposition of basic results of classical algebraic number theory from a relatively modem point of view. This volume presents most of the number-theoretic prerequisites for a study of either class field theory (as formulated by Artin and Tate) or the contemporary... more...
This and a particularly helpful collection of exercises,... more...
An excellent introduction to the basics of algebraic number theory, this concise, well-written volume examines Gaussian primes; polynomials over a field; algebraic number fields; and algebraic integers and integral bases. After establishing a firm introductory foundation, the text explores the uses of arithmetic in algebraic number fields; the fundamental... more...
An imaginative introduction to number theory, this unique approach employs a pair of fictional characters, Ant and Gnam. Ant leads Gnam through a variety of theories, and together, they put the theories into action?applying linear diophantine equations to football scoring, using a black-magic device to simplify problems in modular structures, and... more...
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Description
Numerical Methods provides a clear and concise exploration of standard numerical analysis topics, as well as nontraditional ones, including mathematical modeling, Monte Carlo methods, Markov chains, and fractals. Filled with appealing examples that will motivate students, the textbook considers modern application areas, such as information retrieval and animation, and classical topics from physics and engineering. Exercises use MATLAB and promote understanding of computational results.
The book gives instructors the flexibility to emphasize different aspects--design, analysis, or computer implementation--of numerical algorithms, depending on the background and interests of students. Designed for upper-division undergraduates in mathematics or computer science classes, the textbook assumes that students have prior knowledge of linear algebra and calculus, although these topics are reviewed in the text. Short discussions of the history of numerical methods are interspersed throughout the chapters. The book also includes polynomial interpolation at Chebyshev points, use of the MATLAB package Chebfun, and a section on the fast Fourier transform. Supplementary materials are available online.
Provides flexibility so instructors can emphasize mathematical or applied/computational aspects of numerical methods or a combination
Includes recent results on polynomial interpolation at Chebyshev points and use of the MATLAB package Chebfun
Short discussions of the history of numerical methods interspersed throughout
Supplementary materials available online
About the authors
Anne Greenbaum is professor of applied mathematics at the University of Washington. She is the author of Iterative Methods for Solving Linear Systems. Timothy P. Chartier is associate professor of mathematics at Davidson frequentlyThe Second Edition of the highly regarded An Introduction to Numerical Methods and Analysis provides a fully revised guide to numerical approximation. The book continues to be accessible and expertly guides readers through the many available techniques of numerical methods and analysis. Second An appendix that contains proofs of various theorems and other material
The book is an ideal textbook for students in advanced undergraduate mathematics and engineering courses who are interested in gaining an understanding of numerical methods and numerical analysis.
This book provides an extensive introduction to numerical computing from the viewpoint of backward error analysis. The intended audience includes students and researchers in science, engineering and mathematics. The approach taken is somewhat informal owing to the wide variety of backgrounds of the readers, but the central ideas of backward error and sensitivity (conditioning) are systematically emphasized. The book is divided into four parts: Part I provides the background preliminaries including floating-point arithmetic, polynomials and computer evaluation of functions; Part II covers numerical linear algebra; Part III covers interpolation, the FFT and quadrature; and Part IV covers numerical solutions of differential equations including initial-value problems, boundary-value problems, delay differential equations and a brief chapter on partial differential equations.
The book contains detailed illustrations, chapter summaries and a variety of exercises as well some Matlab codes provided online as supplementary material.
"I really like the focus on backward error analysis and condition. This is novel in a textbook and a practical approach that will bring welcome attention." Lawrence F. Shampine
A Graduate Introduction to Numerical Methods and Backward Error Analysis" has been selected by Computing Reviews as a notable book in computing in 2013. Computing Reviews Best of 2013 list consists of book and article nominations from reviewers, CR category editors, the editors-in-chief of journals, and others in the computing community.
A solutions manual to accompany An Introduction to Numerical Methods and Analysis, Second EditionSecond® An appendix that contains proofs of various theorems and other material
This book gathers threads that have evolved across different mathematical disciplines into seamless narrative. It deals with condition as a main aspect in the understanding of the performance ---regarding both stability and complexity--- of numerical algorithms. While the role of condition was shaped in the last half-century, so far there has not been a monograph treating this subject in a uniform and systematic way. The book puts special emphasis on the probabilistic analysis of numerical algorithms via the analysis of the corresponding condition. The exposition's level increases along the book, starting in the context of linear algebra at an undergraduate level and reaching in its third part the recent developments and partial solutions for Smale's 17th problem which can be explained within a graduate course. Its middle part contains a condition-based course on linear programming that fills a gap between the current elementary expositions of the subject based on the simplex method and those focusing on convex programming.
This book gives an introduction to the finite element method as a general computational method for solving partial differential equations approximately. Our approach is mathematical in nature with a strong focus on the underlying mathematical principles, such as approximation properties of piecewise polynomial spaces, and variational formulations of partial differential equations, but with a minimum level of advanced mathematical machinery from functional analysis and partial differential equations. In principle, the material should be accessible to students with only knowledge of calculus of several variables, basic partial differential equations, and linear algebra, as the necessary concepts from more advanced analysis are introduced when needed. Throughout the text we emphasize implementation of the involved algorithms, and have therefore mixed mathematical theory with concrete computer code using the numerical software MATLAB is and its PDE-Toolbox. We have also had the ambition to cover some of the most important applications of finite elements and the basic finite element methods developed for those applications, including diffusion and transport phenomena, solid and fluid mechanics, and also electromagnetics.
Matrix algorithms are at the core of scientific computing and are indispensable tools in most applications in engineering. This book offers a comprehensive and up-to-date treatment of modern methods in matrix computation. It uses a unified approach to direct and iterative methods for linear systems, least squares and eigenvalue problems. A thorough analysis of the stability, accuracy, and complexity of the treated methods is given.
Numerical Methods in Matrix Computations is suitable for use in courses on scientific computing and applied technical areas at advanced undergraduate and graduate level. A large bibliography is provided, which includes both historical and review papers as well as recent research papers. This makes the book useful also as a reference and guide to further study and research work.
Åke Björck is a professor emeritus at the Department of Mathematics, Linköping University. He is a Fellow of the Society of Industrial and Applied Mathematics.
Computational science is fundamentally changing how technological questions are addressed. The design of aircraft, automobiles, and even racing sailboats is now done by computational simulation. The mathematical foundation of this new approach is numerical analysis, which studies algorithms for computing expressions defined with real numbers. Emphasizing the theory behind the computation, this book provides a rigorous and self-contained introduction to numerical analysis and presents the advanced mathematics that underpin industrial software, including complete details that are missing from most textbooks.
Using an inquiry-based learning approach, Numerical Analysis is written in a narrative style, provides historical background, and includes many of the proofs and technical details in exercises. Students will be able to go beyond an elementary understanding of numerical simulation and develop deep insights into the foundations of the subject. They will no longer have to accept the mathematical gaps that exist in current textbooks. For example, both necessary and sufficient conditions for convergence of basic iterative methods are covered, and proofs are given in full generality, not just based on special cases.
The book is accessible to undergraduate mathematics majors as well as computational scientists wanting to learn the foundations of the subject.
Presents the mathematical foundations of numerical analysis Explains the mathematical details behind simulation software Introduces many advanced concepts in modern analysis Self-contained and mathematically rigorous Contains problems and solutions in each chapter Excellent follow-up course to Principles of Mathematical Analysis by Rudin
Linear algebra is one of the central disciplines in mathematics. A student of pure mathematics must know linear algebra if he is to continue with modern algebra or functional analysis. Much of the mathematics now taught to engineers and physicists requires it. This well-known and highly regarded text makes the subject accessible to undergraduates with little mathematical experience. Written mainly for students in physics, engineering, economics, and other fields outside mathematics, the book gives the theory of matrices and applications to systems of linear equations, as well as many related topics such as determinants, eigenvalues, and differential equations. Table of Contents: l. The Algebra of Matrices 2. Linear Equations 3. Vector Spaces 4. Determinants 5. Linear Transformations 6. Eigenvalues and Eigenvectors 7. Inner Product Spaces 8. Applications to Differential Equations For the second edition, the authors added several exercises in each chapter and a brand new section in Chapter 7. The exercises, which are both true-false and multiple-choice, will enable the student to test his grasp of the definitions and theorems in the chapter. The new section in Chapter 7 illustrates the geometric content of Sylvester's Theorem by means of conic sections and quadric surfaces. 6 line drawings. lndex. Two prefaces. Answer section.
". . .recommended for the teacher and researcher as well as for graduate students. In fact, [it] has a place on every mathematician's bookshelf." —American Mathematical Monthly
Linear Algebra and Its Applications, Second Edition presents linear algebra as the theory and practice of linear spaces and linear maps with a unique focus on the analytical aspects as well as the numerous applications of the subject. In addition to thorough coverage of linear equations, matrices, vector spaces, game theory, and numerical analysis, the Second Edition features student-friendly additions that enhance the book's accessibility, including expanded topical coverage in the early chapters, additional exercises, and solutions to selected problems.
Beginning chapters are devoted to the abstract structure of finite dimensional vector spaces, and subsequent chapters address convexity and the duality theorem as well as describe the basics of normed linear spaces and linear maps between normed spaces.
Further updates and revisions have been included to reflect the most up-to-date coverage of the topic, including:
The QR algorithm for finding the eigenvalues of a self-adjoint matrix
The Householder algorithm for turning self-adjoint matrices into tridiagonal form
The compactness of the unit ball as a criterion of finite dimensionality of a normed linear space
Additionally, eight new appendices have been added and cover topics such as: the Fast Fourier Transform; the spectral radius theorem; the Lorentz group; the compactness criterion for finite dimensionality; the characterization of commentators; proof of Liapunov's stability criterion; the construction of the Jordan Canonical form of matrices; and Carl Pearcy's elegant proof of Halmos' conjecture about the numerical range of matrices.
Clear, concise, and superbly organized, Linear Algebra and Its Applications, Second Edition serves as an excellent text for advanced undergraduate- and graduate-level courses in linear algebra. Its comprehensive treatment of the subject also makes it an ideal reference or self-study for industry professionals. frequentlyMuch recent research has concentrated on the efficient solution of large sparse or structured linear systems using iterative methods. A language loaded with acronyms for a thousand different algorithms has developed, and it is often difficult even for specialists to identify the basic principles involved. Here is a book that focuses on the analysis of iterative methods. The author includes the most useful algorithms from a practical point of view and discusses the mathematical principles behind their derivation and analysis. Several questions are emphasized throughout: Does the method converge? If so, how fast? Is it optimal, among a certain class? If not, can it be shown to be near-optimal? The answers are presented clearly, when they are known, and remaining important open questions are laid out for further study. Greenbaum includes important material on the effect of rounding errors on iterative methods that has not appeared in other books on this subject. Additional important topics include a discussion of the open problem of finding a provably near-optimal short recurrence for non-Hermitian linear systems; the relation of matrix properties such as the field of values and the pseudospectrum to the convergence rate of iterative methods; comparison theorems for preconditioners and discussion of optimal preconditioners of specified forms; introductory material on the analysis of incomplete Cholesky, multigrid, and domain decomposition preconditioners, using the diffusion equation and the neutron transport equation as example problems. A small set of recommended algorithms and implementations is
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A Beginner's Guide to Graph Theory
Browse related Subjects
This beginner's textbook is intended for a first course in graph theory. It strikes a balance between a theoretical and practical approach, consisting of carefully chosen topics to develop graph-theoretic reasoning for a mixed audience.
Read More
This beginner's textbook is intended for a first course in graph theory. It strikes a balance between a theoretical and practical approach, consisting of carefully chosen topics to develop graph-theoretic reasoning for a mixed audience.
Read Less
Good. Paperback. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9780817644840-4-0-3 Orders ship the same or next business day. Expedited shipping within U.S. will arrive in 3-5 days. Hassle free 14 day return policy. Contact Customer Service for questions. ISBN: 9780817644840
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This learning object from Wisc-Online covers the triangle, examining the properties and components of the shape. The lesson uses the geometric formulas for finding the perimeter and area of the shape. Practice questions...
Gilbert Strang, of the Massachusetts Institute of Technology, highlights calculus in a series of short videos that introduces the basic ideas of calculus ? how it works and why it is important. The intended audience i...
This is a basic course, produced by Gilbert Strang of the Massachusetts Institute of Technology, on matrix theory and linear algebra. Emphasis is given to topics that will be useful in other disciplines, including...
This course, authored by Denis Auroux of Massachusetts Institute of Technology, covers vector and multi-variable calculus. It is the second semester in the freshman calculus sequence. Topics include vectors and matrices...
This learning object from Wisc-Online covers the sphere, examining the properties and components of the shape. The lesson uses the geometric formulas for finding the volume and surface area of the shape. Practice...
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I once took my class outdoors and positioned the students at various points on a hillside. I asked them to point in the direction of greatest slope away from the point where they stood, extending their arms far from their bodies or near, according as they thought the maximum slope was large or small. To convey the idea of the gradient vector, I've also drawn arrows on basketballs, used transparencies provided by book publishers, and sketched elaborate diagrams in colored chalk on the blackboard. Like this one, several concepts in multivariable calculus present a teaching challenge because a student's ability to understand them is likely to have more to do with his or her facility with spatial relations than with mathematics.
Maple's Release 4 permits what previous releases did not: displaying in the same window both an animation and a background, or "dead," plot. When I learned that Maple could this, I thought of several animated demonstrations that I might write for multivariable calculus showing, for example, a plane moving through a fixed surface. In this paper, I share demonstrations for the topics that I've found to be most illuminated by animations. Nothing else has worked as well.
One way to understand why quadric surfaces have the shapes they do, is to consider their traces (cross-sections) in various planes parallel to the coordinate planes. The difficult part is putting the traces together in space to form the surface. The hyperboloid of two sheets
is a good first example because, with a little analysis, it's easy to see that certain horizontal planes will not intersect the surface at all (Animation 1), and that all vertical ones will meet the surface in hyperbolas (Animation 2).
is a little harder to visualize because, in vertical planes, the hyperbolic traces change orientation when the plane gets farther from the origin than 2 in the direction of the y-axis or 3 in the direction of the x-axis. Animation 4 shows the surface and the plane x = k for varying values of k, each trace having equation
.
The most difficult of the quadric surfaces for students to visualize, and certainly to draw, is the hyperbolic paraboloid. Like the hyperboloid of one sheet, the traces in certain planes are hyperbolas that change orientation after a point. As an example, I use horizontal planes intersecting the surface
.
(See Animation 5.) I constructed the frames to include the plane that produces the degenerate hyperbola (two intersecting lines) lying between the two families of hyperbolas. The animation shows this reasonably well.
Because it is a function of x and y, the previous example leads naturally to the topic of level curves (orthogonal projections onto the xy-plane of traces in horizontal planes). Given a real-valued function of two real variables, one way to understand the nature of its surface is to make a contour map, a plot in the plane of various level curves. The first animation I use shows the surface
and the moving plane that creates the level curves (see Animation 6). The second shows the traces on the surface, then rotates into a view straight down the z-axis, which is the same as a contour map. (See Animation 7.) This shows very clearly how the contour map corresponds to the traces as they sit in space.
To step up one dimension is to consider the level surfaces of a real-valued function of three real variables. We can try to imagine these surfaces as stacked, somehow, in four dimensions--a challenge for us all. I use the example
whose level surface, f (x, y, z) = k, is a hyperboloid of two sheets when k < 0, a hyperboloid of one sheet when k > 0, and an elliptic cone when k = 0. The animation that I use shows these surfaces "stacked" in time. (See Animation 8.) This is a very nice analogue to the previous example of a hyperbolic paraboloid. There, the level curves are hyperbolas with branches in one direction when k < 0, hyperbolas with branches in the other direction when k > 0, and two intersecting lines when k = 0.
To demonstrate the vector concepts of velocity and acceleration, I created an animation of a particle moving along a curve ( r(t ) = [(sin 3t ) (cos t ), (sin 3t ) (sin t )] ) at varying speed. At the point in the course where I use this, the students know that, given a vector-valued function of a real variable which is to be viewed as a position function, the velocity and acceleration vectors are the first and second derivatives, respectively, and that the speed is the magnitude of the velocity vector. At the chalkboard, I explain that the velocity vector, always tangent to the path, points in the direction of motion, lengthening as the particle gains speed and shortening as it slows down. This concept has not been particularly difficult to teach using the chalkboard. Harder to communicate, though, is the dual role of the acceleration vector: that it acts to some degree in a direction orthogonal to the velocity vector to move the particle off its course; and that it acts to some degree in the direction of the velocity vector when the particle is gaining speed and in a direction opposite to the velocity vector when the particle is slowing down. Animation 9 illustrates this behavior very well; the velocity vector shown in red, the acceleration vector in green.
The most difficult idea to convey in the entire course, for me, is that of the directional derivative and the gradient vector. The animation that I've created to help me, uses the function
The animation shows:
the surface
a unit vector rotating about the point (1, 1, 0)
a rotating plane parallel to the unit vector
the traces of the planes in the surface
the tangent lines to the traces at (1, 1, f (1, 1))
the gradient vector (shown in green)
(See Animation 10.) As the traces and tangents rotate along with the planes and unit vectors, the effect of a changing unit vector is evident. Moreover, the animation makes very clear what is so difficult to show by other means: that the maximum slope away from a point occurs when the unit vector has the direction of the gradient vector.
These demonstrations represent a significant leap in my ability to communicate these ideas. They tie the abstract to the concrete, bind a concept to an image that we can call up in our mind's eye.
When I was writing Animation 10, I couldn't quite make it work before it was time that I teach the topic, so I taught directional derivatives and gradient vectors the old way, using a chalkboard and prepared transparency. That night, I found a way to get what I was after (by foiling the hidden-line algorithm a little) and used the demonstration on the following day. After class, two people thanked me for writing it. "We didn't understand before," they said. "But we do now."
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Math 220. Calculus
Lecture Syllabus
This syllabus assumes MWF lectures and Tuesday-Thursday discussion sections, with 43 lecture hours in the semester. It includes 36 lectures, leaving 7 hours for leeway and exams. Note that a minimum of 4 one-hour exams is recommended for Math 220, but one can give 5 if there is time.
Math 220 is intended for students who have NOT had a year of calculus in high school.
This is primarily a course on calculation and problem solving; proofs should not be emphasized. Chapter 1 is optional since it represents review material. Instructors may choose to eliminate it or cover it at a pace more rapid than that suggested by the 4 lecture provision. However, many Math 220 students do need review of this material.
It is assumed that the Teaching Assistants in this course may need to do some lecturing in their discussion sections so as to keep the timeline for the syllabus on track.
Chapter 1: Functions and Models (4 lectures)
1.1 Four Ways to Represent a Function
1.2 Mathematical Models: A Catalog of Essential Functions
1.3 New Functions from Old Functions
1.5 Exponential Functions
1.6 Inverse Functions and Logarithms
Chapter 2: Limits and Derivatives (5 lectures)
2.1 The Tangent and Velocity Problems
2.2 The Limit of a Function
2.3 Calculating Limits Using the Limit Laws
2.4 The Precise Definition of a Limit (optional)
2.5 Continuity
2.6 Limits at Infinity; Horizontal Asymptotes
2.7 Derivatives and Rates of Change
2.8 The Derivative as a Function
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Chapter 8: Exponential and Logarithmic Functions
Introduction
In this chapter, we will analyze two new types of functions, exponents and logarithms. Up until now, the variable has been the base, with numbers in the exponent; linear, quadratic, cubic, etc. Exponential functions have the variable in the exponent. Logarithmic functions are the inverse of exponential functions. We will graph these functions, solve equations, and use them for modeling real-life situations.
Chapter Summary
Summary
In this chapter you'll learn about exponential and logarithmic functions. We will graph these functions and solve equations. We will also learn about the properties of logarithms and the natural number, e.
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The text presents a circular function approach to trigonometry by demonstrating connections between the familiar algebra and the new language of trigonometry. This methodalong with foreshadowingis used throughout the text to provide students with a comfortable base for learning something new from something old or familiar. With just a few connections to algebra, students have the tools to understand the circular functions, their domains and ranges, and the relationship between the circular functions and the functional values. The approach immediately launches the student into the concept of periodic functions, their applications, graphs, and use in modeling many periodic phenomena. Beginning with this approach provides the student with a common thread that can be used to discover, connect and understand the remaining concepts of trigonometryAHA-BUCH GmbH via Germany
ISBN 0201771748 Publisher: Cram101,Dez , 2010 Cram101,Dez 2010; NEUBUCH! 280x210x4201771749. This item is printed on demand. (EDUCATION / General)
Hardcover, ISBN 0201771748 Publisher: Pearson, 2003 Used - Acceptable, Usually ships in 1-2 business days, Orders ship the same or next business day. This book is used in acceptable condition and may have curling of the front or back cover, and some markings or highlights throughout. May have torn cover. All text is legible. May or may not include supplemental materials such as CDs or access codes. Still a good book for class and will save you a ton!
Hardcover, ISBN 0201771748 Publisher: Pearson, 2003 Used - Acceptable, Usually ships in 1-2 business days, This Book is in acceptable Condition. Used Copy With medium amount of Wear from use.some pages has moister damage. back cover on binding small tear.but still very readable . 100% Guaranteed. plus a no-questions-asked return policy. inv a1.
Hardcover, ISBN 0201771748 Publisher: Pearson, 2003 Used - Acceptable, Usually ships in 24 hours, The cover of this book is somewhat worn, and it even has been repaired with a scotch tape on the inside, but it's pretty nice and clean on the inside.. Shipped from Amazon. Amazon Prime. Eligible for FREE Super Saver Shipping201771748 Publisher: Addison Wesley201771748 Publisher: Pearson, 2003 Used - Good, Usually ships in 1-2 business days, Item in good condition. Textbooks may not include supplemental items i.e. CDs, access codes etc... All Day Low Prices!
Hardcover, ISBN 0201771748 Publisher: Pearson, 2003 Used - Good, Usually ships in 1-2 business days, Choose EXPEDITED SHIPPING for FASTEST delivery! SHIPS SAME or NEXT business day. Covers and corners may show shelf wear. Pages may contain some highlighting/underlining and/or notes in the margins. USED stickers on cover. Used books do not include valid access codes or other supplements. Delivery confirmation standard. Satisfaction guaranteed
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4.1 Introduction Information and communication technologies (ICTs) systems now dominate our everyday lives. This unit will explain what constitutes such a system and how ICT systems work. You will also look at how ICT systems convey, store and manipulate data, and how they process data. Finally you will learn how these systems are used. Author(s): The Open University
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How to Determine the Value of a Coin This clip shows how to determine the value of a coin. An antique store owner explains that it depends on several factors, including what the coin is made from, if the coin has a mint mistake and if the coin is rare. he gives tips for understand the grading system of coins.
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The Real Number Set - Part 2 In this Language of Mathematics video the instructor continues his demonstration of the real number system and how it is broken down. He starts off by speaking of the number zero and the impact of the breakthrough of the 'discovery' of this number. Author(s): No creator set
Coordinate Geometry Proof (Part 1) The instructor uses a chalkboard to model the Cartesian coordinate system. Discussed is the midpoint of two points and how to find it using the formula for midpoint. This video is part of The Language of Mathematics series - Part 32. Author(s): No creator set
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The Language of Mathematics (33): Coordinate Geometry Proof, Part 2 In this segment, the instructor continues discussing the Cartesian coordinate system. Instructor uses a small chalkboard for demonstration. The viewer may want to open the screen to 'Full View' as the room the instructor is in is a little dark and his chalkboard has a lot if information. Author(s): No creator set
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Crohns Disease Module 10: Research Learn about research currently focusing on abnormalities in the immune system, bacteria in the intestine, and genetics. This module will also address how you can help with research to find better ways to treat Crohns, how this will this help develop better medications, and what this research will provide. Author(s): No creator set
Musculoskeletal System Using still images, this is a brief overview of the musculoskeletal system (02:40). Author(s): No creator set
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English Grammar Training- Articles This clip is a lesson on articles a, an, and the. Rules for using articles are reviewed on a chalk board. The sound quality in this video is poor due to an echo. Author(s): No creator set
Atmosphere The atmosphere is a critical system that helps to regulate Earth's climate and distribute heat around the globe. In this unit, discover the fundamental processes that cause atmospheric circulation and create climate zones and weather patterns, and learn how carbon cycling between atmosphere, la Author(s): No creator set
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Workshop 6: The Mind's Intelligences This workshop considers Howard Gardner's theory of multiple intelligences and shows his theory being applied in a range of classrooms. As Gardner shares his thoughts on educational reform, you will learn how to create learning environments that support the full spectrum of students' abilities. Author(s): No creator set
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Microbial Control In certain situations, microbial control is a necessity. For instance, our food system requires sanitary conditions and hospitals require sterilization techniques. Here we see the options available for various levels of microbial control. Author(s): No creator set
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Angular Momentum An old momentum with a new twist. Kepler's second law of planetary motion, which is rooted here in a much deeper principle, imagined a line from the sun to a planet that sweeps out equal areas in equal times. Angular momentum is a twist on momentum -- the cross product of the radius vector and momentum. A force with twist is torque. When no torque acts on a system, the angular momentum Author(s): No creator set
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How To Completely Format A Hard Drive Watch How To Completely Format A Hard Drive. How To Completely Format A Hard Drive. Sometimes your hard drive crashes and you need to reformat it to get it running again. Other times you might reformat it because you have a virus or you want to upgrade your operating system. Author(s): No creator set
The class engages in a variety of activities and keeps journals of data and their thoughts while examining the five senses. Ms. Collier assesses learning by listening to conversations as students work and discuss what they are learning in small and large groups.
Perception: Inverted Vision The peculiar image inversion process that takes place in the normal visual system is examined in this module. The program traces the experiences of an art student who volunteers to wear lenses that invert her visual world, connecting the adaptation process she undergoes with how the visual system functions. Graphic animations reinforce understanding of the mechanism invo Author(s): No creator set
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Pre-Writing Skills Writers are more effective when they first develop a writing plan, such
as organizing their thoughts into main ideas and supporting details.
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Elayn Martin-Gay firmly believes that every student can succeed, and her developmental math textbooks and video resources are motivated by this belief. Intermediate Algebra, Fourth Edition was written to provide students with a solid foundation in algebra and to help them transition to their next mathematics course. The new edition offers new resources like the Student Organizer and now includes Student Resources in the back of the book to help students on their quest for success.
This is the standalone book, if you want the Book/Access Card order the ISBN listed below:
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Elayn is the author of 12 published textbooks and numerous multimedia interactive products, all specializing in developmental mathematics courses. She has participated as an author across a broad range of educational materials: textbooks, videos, tutorial software, and courseware. This offers an opportunity for multiple combinations for an integrated teaching and learning package, offering great consistency for the student.
I bought this book, advertised as new, with access card. The access card was inside the front page, and certainly looked legitimate to me. I scraped the gray covering over the code to reveal it and typed it in the first time I used it. It would not work after several tries, and I got a message that I would have temporary access. After 3 weeks, this temporary access apparently expired. It took me a full week and 10 hours trying to contact Pearson by phone, chat and email before they finally called me back. The one time I actually got a person via chat, I spent 3 hours on hold and then he couldn't offer any help. Finally, because a college administrator and my college professor contacted Pearson, they reactivated my account and I now have access. I am disappointed because the access code did not work, I was told it was because it was bought on Amazon a "third-party vendor". The book was sealed in plastic, and appeared brand new. Just a HUGE hassle, that at least turned out ok so far.
My wife has decided to go back to school for radiography and needed this book for algebra. Its more than likely something she wont be reading again so why pay over 120 dollars for a book she wont ever use again when I can just rent it for her for a quarter of the cost. The book arrived and is in better shape than any used book I have ever gotten from a student bookstore!
Save a ton of money!!!!!!!! The book was stated as being in good condition- was in like new condition when recieved. The delivery was fast too. Like i said saved a bunch of money compared to buying a new book or even a used book at the collage book store. Very happy with this purchase.
I got this book because i had to for school definatly not beacuse i wanted it. cd was in tack as well as access code needed for online course. online course in pain if you make the tiniest mistakes. good shipping and perfect condition
Seems to be a good self learning product. Comes with book and online access code which gives step by step instructions via video on how to complete equations. Seems to also give plenty of opportunity to get additional help.
More About the Author
An award-winning instructor and best-selling author, El developed an acclaimed series of lecture videos to support developmental mathematics students in their quest for success. These highly successful videos originally served as the foundation material for her texts. Today, the videos are specific to each book in the Martin-Gay series. Elayn also pioneered the Chapter Test Prep Video to help students as they prepare for a test--their most "teachable moment!"
Elayn's experience has made her aware of how busy instructors are and what a difference quality support makes. For this reason, she created the Instructor-to-Instructor video series. These videos provide instructors with suggestions for presenting specific math topics and concepts in basic mathematics, prealgebra, beginning algebra, and intermediate algebra. Seasoned instructors can use them as a source for alternate approaches in the classroom. New or adjunct faculty may find the videos useful for review. Her textbooks and acclaimed video program support Elayn's passion of helping every student to succeed.
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Probability is an area of mathematics of tremendous contemporary importance across all aspects of human endeavour. This book is a compact account of the basic features of probability and random processes at the level of first and second year mathematics undergraduates and Masters' students in cognate fields. It is suitable for a first course in...
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(Meets GE requirements only for Elementary Education majors who have taken Math 305) This course is for Elementary Education majors. The content is enrichment or extended study of the concepts covered in Math 305. Geometry, probability, statistics and informal logic for elementary education teachers.
DESCRIPTION:
TOPICS:
Geometry with translations, networks, and coordinant geometry; permutations and combinations; probability and statistics, including its misuse;geometry as shape, transformation and measurement; Geometer's Sketchpad; Logo
OBJECTIVES:
Students in Math 306 will learn, model, and apply concepts from probability, statistics, and geometry to a wide variety of problems and situations. This knowledge will enable them to accurately teach concepts from the grade K-8 mathematics curriculum.
REQUIREMENTS:
Attendance, a copy of the text and activities manual, and a calculator
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0817639144
9780817639143
Details about Trigonometry:
Covers the basics of trigonometry through illustrations and examples. This title includes the definitions of the trigonometric functions that are geometrically motivated. It presents geometric relationships that are rewritten in trigonometric form and extended. It studies algebraic and analytic properties of trigonometric functions.
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Rent Trigonometry 1st edition today, or search our site for I. M. textbooks. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Birkhauser Boston.
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try to keep it simple because precalculus can be confusing if students don't understand what they will be using it for in calculus yet. I have a degree in math, discrete was one of the first math classes I took in college. I have a lot of experience with proofs and Boolean logic.
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Discovering Mathematics from the UK within 24 hours. Your purchase supports authors through the Book Author Resale Right. No Dustjacket.
Bookbarn International
Somerset, GBR
$27.79
FREE
About the Book
One of the most striking characteristics of mathematics is that thoughtful and persistent mathematical analysis often provokes totally unexpected insights into what may at first have looked like an uninteresting or intractable problem. This book gives students an opportunity to discover the nature and process of mathematics by developing their ability to investigate problems without relying on the standardized methods usually taught. The techniques required are elementary, allowing the student to concentrate on the way the material is explored and developed, and on the strategies for addressing questions whose answers are not immediately obvious. The book will challenge high school and college mathematics students as well as interested general readers.
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MathWorks at MIT IAP 2014
MathWorks is hosting six sessions during MIT's Independent Activities Period (IAP) 2014. Join us to learn how you can use MATLAB and Simulink for technical computing and application development in engineering, math, and science.
Tuesday, January 28
Explore, Visualize, and Analyze Your Data with MATLAB
10:00 a.m.–12:00 p.m.
Room 4-163
In this session, you will learn how to use MATLAB to gain insight into your engineering and scientific data. With the MATLAB language, interactive tools, and built-in math functions, you can explore and model your data, build customized analyses, and share your discoveries with others.
Through product demonstrations, you will see how to:
Access data from files and spreadsheets
Manage complex and messy data
Plot data and customize figures
Perform statistical analysis and fitting
Generate reports and build apps
This session is for students, faculty, and researchers who are new to MATLAB. Experienced MATLAB users may also benefit from the session, which features capabilities from recent releases of MATLAB.
About the Presenter
Vipresh Gangwal (vipresh.gangwal@mathworks.com) works in the Engineering Development Group at MathWorks. He has an M.S. in electrical engineering from University of Southern California (USC) and a B.E. in electrical engineering from University of Pune, India. His research at USC focused on probabilistic robotics involving computer vision for perception and planning for robots.
Introduction to MATLAB: Problem Solving and Programming
1:00–5:00 p.m.
Room 4-163
MATLAB is a high-level language that allows you to quickly perform computation and visualization through easy-to-use programming constructs. This hands-on lab presents the essentials you need to use MATLAB for your classes or research.
In this session, we import historical temperature data collected in the Northern Hemisphere from an external file, plot the data over time, then perform some analysis to view the data trend to determine if global warming is happening. You'll learn how to write a MATLAB script and publish it to a format for sharing, such as HTML. You'll also learn how to write your own MATLAB functions, use flow control, and create loops.
By the end of the session, you'll have learned to create an application in MATLAB.
Key topics include:
Navigating the MATLAB desktop
Working with variables in MATLAB
Calling MATLAB functions
Importing and extracting data
Visualizing data
Conducting computational analysis
Fitting data to a curve
Automating analysis with scripts
Publishing MATLAB programs
Programming in MATLAB
Note: Attendees should bring a laptop to this hands-on lab.
About the Presenters
Laura Proctor (laura.proctor@mathworks.com) develops and delivers training targeted to academic users and creates web-based content for online learning. She joined MathWorks in 2009 as an application support engineer and in 2010 transferred to the Customer Training Group. Laura has a B.S. in mathematics and a B.S. in mechanical and aerospace engineering with a minor in physics from the University of Missouri. She also has an M.S. in mechanical engineering and an M.S. in computation for design and optimization from MIT. Her graduate research included simulating and optimizing satellite trajectories, electronic circuit design and implementation, biomimetic design and prototyping, computational methods applied to nanostructural ion flow, and course material design with an emphasis on visual learning.
Eoin Moore (eoin.moore@mathworks.com) delivers MATLAB training to academic and corporate audiences and also develops online self-paced MATLAB courses. He holds a B.S. in physics from the University of Massachusetts and an M.S. in physics from the University of California San Diego. His graduate research involved the analysis of plasmas, including turbulent flow and nonlinear dynamics.
Wednesday, January 29
Introduction to Simulink for Dynamic Systems Modeling and Simulation
9:00 a.m.–12:00 p.m.
Room 35-225
Simulink is an environment for multidomain simulation and Model-Based Design for dynamic and embedded systems. It provides an interactive graphical environment and a customizable set of block libraries that let you design, simulate, implement, and test a variety of time-varying systems in multiple domains, including communications, controls, and signal, video, and image processing.
Using specific, real-world examples, we show you how to model and simulate dynamic systems as part of a top-down design workflow. We start by modeling and simulating the differential equations governing the system dynamics using an approach based on first principles. We then show how to use additional domain-specific libraries for system design and architectural modeling to create more detailed models as physical networks of components. We present examples from a variety of domains, including electrical and mechanical, and for a variety of applications, including both control systems and digital signal processing.
Highlights include:
Creating a new model from scratch
Using libraries of predefined blocks
Creating your own reusable subsystems
Incorporating MATLAB code into Simulink models
Combining Simulink with physical modeling libraries
About the Presenters
Carlos Osorio (carlos.osorio@mathworks.com) is a principal applications engineer at MathWorks focusing on control system design and implementation. Carlos received a B.S. from the Pontificia Universidad Catolica del Peru and an M.S. from the University of California at Berkeley, both in mechanical engineering. He specializes in automatic control systems and vehicle dynamics. Before joining MathWorks in October 2007, he worked in the automotive industry in the Advanced Chassis Technology Department at Visteon Corporation, where he was involved in the development and implementation of prototype electronic active and semi-active suspensions and steer-by-wire and brake-by-wire systems for passenger vehicles.
Rick Rosson (rick.rosson@mathworks.com) is a senior applications engineer specializing in signal processing and communications systems. He supports a wide variety of MATLAB and Simulink products, including Signal Processing Toolbox, DSP System Toolbox, and MATLAB Coder, in industries such as communications, electronics, semiconductors, aerospace, and automotive. His current professional interests include digital signal processing, digital communications, analog and mixed-signal system design, and embedded C code generation. Prior to joining MathWorks, Rick worked for 12 years in manufacturing management and process improvement, primarily in the automotive and electronics industries. Rick has an M.S. in electrical engineering from Boston University and an M.S. in management from the MIT Sloan School of Management.
MATLAB and Simulink with Raspberry Pi: A Hands-On Workshop on Hardware Support
1:00–4:00 p.m.
Room 35-225
Addressing the growing need in curriculum and research for low-cost, easy-to-use hardware and software environments, this session describes the built-in support in MATLAB and Simulink for prototyping, testing, and running Simulink models on Raspberry Pi.
Simulink includes the capability to program Arduino, Raspberry Pi, LEGO MINDSTORMS NXT, and other low-cost hardware platforms. This hands-on workshop introduces the hardware support capabilities in Simulink. Participants develop, simulate, and test custom algorithms and implement the code on an embedded system from within the Simulink environment. Lab modules include examples of video and image processing algorithms, from very simple video in/out handling to more sophisticated processing such as object recognition and edge detection. The workshop provides practical hands-on experience and gives attendees an understanding of the potential for use in the classroom, research, and student projects.
Participants will:
Design, simulate, and test custom algorithms in Simulink
Implement these algorithms on embedded hardware
Discover the ease of using Simulink to program
Note:
Necessary software and Raspberry Pi Kits will be made available to attendees for the duration of the workshop.
We have a limited class size for this workshop. Register now and we will contact you to confirm your seat.
Faculty, staff and graduate students will be given preference as attendees.
About the Presenter
Sumit Tandon (sumit.tandon@mathworks.com) joined MathWorks in 2007 and worked on Simulink and code generation products before transferring to application engineering in 2011. Previous work experience includes a year as a software developer in COBOL and mainframe technology. He has a B.E. in electrical engineering from Jadavpur University, India, and an M.S. in electrical engineering from University of Texas at Arlington. His master's thesis focused on cancer research, and he developed a simulator for biopsy using concepts of computer graphics, computer vision, and haptics.
Thursday, January 30
Machine Learning with MATLAB
10:00 a.m.–12:00 p.m.
Room 4-231
Machine learning techniques are often used for data analysis and decision-making tasks such as forecasting, classification of risk, estimating probabilities of default, and data mining. However, implementing and comparing machine learning techniques to choose the best approach can be challenging. In this session, you will learn about several machine learning techniques available in MATLAB and how to quickly explore your data, evaluate machine learning algorithms, compare the results, and apply the best technique to your problem.
Highlights include unsupervised and supervised learning techniques such as:
K-means and other clustering tools
Neural networks
Decision trees and ensemble learning
Naïve Bayes classification
Linear, logistic, and nonlinear regression
About the Presenter
Abhishek Gupta (abhishek.gupta@mathworks.com) joined Technical Support at MathWorks in 2007 and later moved to the Application Engineering Group. He primarily focuses on academia and the financial services industry. He has an M.S. in mechanical engineering from Texas A&M University. His research involved the development of mathematical models of heat exchangers using MATLAB and Simulink.
Accelerating MATLAB Algorithms and Applications
1:00–3:30 p.m.
Room 4-231
Analyze data, develop algorithms, and create models and applications – all more quickly. In this session we will present strategies and techniques to accelerate your MATLAB computations, and highlight ways that you can use MATLAB with HPC environments without needing to be an expert in parallel programming with CUDA or MPI.
The acceleration topics covered include:
Parallel computing on multicore processors and GPUs
Scaling computations to clusters and clouds
Generating and incorporating C-based functions that can be scaled with your code base
We will describe the underlying acceleration technology, and explain when it is most applicable.
About the Presenter
Adam Filion (adam.filion@mathworks.com) joined the MathWorks Engineering Development Group in 2010 and moved to Application Engineering in 2012. He holds a B.S. and M.S. in aerospace engineering from Virginia Tech. His research involved nonlinear controls of spacecraft and periodic orbits in the three-body problem.
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MERLOT Search - materialType=Open%20Textbook&category=2513&sort.property=overallRating
A search of MERLOT materialsCopyright 1997-2015 MERLOT. All rights reserved.Tue, 21 Apr 2015 01:56:29 PDTTue, 21 Apr 2015 01:56:29 PDTMERLOT Search - materialType=Open%20Textbook&category=2513&sort.property=overallRating
4434Rice Virtual Lab in Statistics IntroduDimensions (geometry)
This is a free, online textbook that provides information on dimensions, from longtitude and latitude to the proof of a theorem of geometry. There are 9 chapters, each 13 minutes long. The book contains a total of 117 minutes of video, but can also be read as an ordinary textbook.The film can be enjoyed by anyone, provided the chapters are well-chosen. There are 9 chapters, each 13 minutes long. Chapters 3-4, 5-6 and 7-8 are double chapters, but apart from that, they are more or less independent of each other.A First Course in Linear Algebra
A First Course in Linear Algebra is an introductory textbook aimed at college-level sophomores and juniors. Typically such a student will have taken calculus, but this is not a prerequisite. The book begins with systems of linear equations, then covers matrix algebra, before taking up finite-dimensional vector spaces in full generality. The final chapter covers matrix representations of linear transformations, through diagonalization, change of basis and Jordan canonical form.PDF versions are available to download for printing or on-screen viewing, an online version is available, and physical copies may be purchased from the print-on-demand service at Lulu.com. GNU Free Documentation LicenseA ProblemText in Advanced Calculus
Advanced Calculus open textbook. Download LaTeX source or PDF. Creative Commons BY-NC-SA.Algebra: Abstract and Concrete
The book provides a thorough introduction to "modern'' or "abstract'' algebra at a level suitable for upper-level undergraduates and beginning graduate students. The book addresses the conventional topics: groups, rings, fields, and linear algebra, with symmetry as a unifying theme.Applied Finite Mathematics Textbook
Applied Finite Mathematics Textbook onlineCalculus
OCW is pleased to make this textbook available online. Published in 1991 and still in print from Wellesley-Cambridge Press, the book is a useful resource for educators and self-learners alike. It is well organized, covers single variable and multivariable calculus in depth, and is rich with applications. There is also an online Instructor's Manual and a student Study Guide.Formal Logic
This is an 'undergraduate college level textbook covering first order predicate logic with identity but omitting metalogical proofs.'Brief ContentsPreliminariesSentential LogicPredicate LogicIdentity Theory Intermediate Algebra
Intermediate Algebra textbook in PDF form
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,...
Show More, modeling, communication, and technology skills to develop students' fluency in the "language of algebra". Tussy and Gustafson understand the challenges of teaching developmental students, and this book reflects a holistic approach to teaching mathematics that includes developing study skills, problem solving, and critical thinking alongside mathematical concepts. New features in this edition include a pretest for students to gauge their understanding of prerequisite concepts, problems that make correlations between student life and the mathematical concepts, and study skills information designed to give students the best chance to succeed in the course. Additionally, the text's widely acclaimed Study Sets at the end of every section are tailored to improve students' ability to read, write and communicate mathematical ideas
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Math Textbooks On-line!
You must sign up for an account and use the activations codes provided. Look in the mathematics locker on the left for the full instructions.
Math Course Sequence
Math Deparment D Policy
At California Aerospace Academy, the policy for moving on to the next level of mathematics is defined as follows:
Students earning a grade of "A", "B" or "C" may move on to the next course.
Students earning a grade of "D" must repeat the semester they received that D in and may move on only when they earn a "C" or better for that semester.
Students earning a grade of "F" must repeat the class.
Rationale:
I know of no classroom in America in which a "D" represents anything other than the failure of the student to demonstrate proficiency and the failure of the teacher to acknowledge it. The availability of a "D" is simply the policy option that allows a school to explicitly acknowledge that a student failed to demonstrate proficiency in the subject, while refusing to require the student to do so. In a genuinely standards-based school-system, the grade of "D" should not exist. Either students are proficient (usually a grade of at least an "A" or "B" and, sometimes a "C") or they are not. The failure to be proficient should, in most circumstances, result in a grade of "incomplete" while the student is afforded more opportunities to learn and demonstrate proficiency. Should the student refuse to do so, a failing grade, not a "D" is the only accurate grade.
Indeed, most teachers would agree that students to whom they have given a "D" grade do not meet the standards for that class, and the teacher would have regarded the "D" as an unsatisfactory grade. Nevertheless, for the purposes of awarding a high school diploma the "D" grade is regarded as satisfactory.
If standards mean anything, they mean that students must demonstrate proficiency in order to obtain credit for classes and, ultimately, in order to obtain a high school diploma from that school system.
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books.google.com - For...
Analysis: With an Introduction to Proof
For abstract mathematics by its emphasis on proofs. Student oriented and instructor friendly, it features clear expositions and examples, helpful practice problems, many drawings that illustrate key ideas, and hints/answers for selected exercises. *NEW - True/False questions - (More than 250 total) located at the beginning of the exercises for each section and relating directly to the reading. *NEW - 8 new illustrations of key concepts make this the most visually compelling analysis text. *Straightforward discussion of logic - As it applies to the proving of theorems in analysis (Ch. 1). Can be covered briefly or in depth, depending on the needs of students. *Practice problems - Scattered throughout the narrative (more than 140 total). These problems relate directly to what has just been presented. Includes complete answers at the end of each section. *Fill-in-the-blank proofs. Helps stude
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Introduction to Real Analysis
9780471433316
ISBN:
0471433314
Edition: 4 Pub Date: 2011 Publisher: Wiley
Summary: This book provides the fundamental concepts and techniques of real analysis for readers in all of these areas. It helps one develop the ability to think deductively, analyze mathematical situations and extend ideas to a new context. Like the first three editions, this edition maintains the same spirit and user-friendly approach with addition examples and expansion on Logical Operations and Set Theory. There is also c...ontent revision in the following areas: introducing point-set topology before discussing continuity, including a more thorough discussion of limsup and limimf, covering series directly following sequences, adding coverage of Lebesgue Integral and the construction of the reals, and drawing reader attention to possible applications wherever possible.
Bartle, Robert G. is the author of Introduction to Real Analysis, published 2011 under ISBN 9780471433316 and 0471433314. Seven hundred seven Introduction to Real Analysis textbooks are available for sale on ValoreBooks.com, eight used from the cheapest price of $36.75, or buy new starting at $73.50
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La Costa, CA Precalculus. This program challenges students to apply basic math and science skills to solve simple real world problems
...Algebra 2 reviews the important topics from Algebra 1 before introducing more complex ideas. A variety of important mathematical concepts are covered. These include families of functions, absolute value functions, polynomial functions, rational exponents, radical functions, exponential and logarithmic functions, sequences and series, probability and counting methods, and rational functions.
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A stimulating excursion into pure mathematics aimed at "the mathematically traumatized," but great fun for mathematical hobbyists and serious mathematicians as well. Requiring only high school algebra as mathematical background, the book leads the reader from simple graphs through planar graphs, Euler's formula, Platonic graphs, coloring, the genus of a graph, Euler walks, Hamilton walks, and a discussion of The Seven Bridges of Konigsberg. Exercises are included at the end of each chapter. "The topics are so well motivated, the exposition so lucid and delightful, that the book's appeal should be virtually universal . . . Every library should have several copies" — Choice. 1976 edition.
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Mr. Trudeau has done a fabulous job of introducing graph theory in a way which is understandable and intellectually provocative. He mentions that some of the problems are easy, and that some have been unsolved. In both cases, they both are fully illustrative of the subject matter. If you want to begin exploring graph theory, this book is for you!
This book make you want to know more about graph theory. The concepts are first intuitively explained and then formally stated. The numerous examples are completely treated and then easy to follow. R. Trudeau devoted a large part of the book to the puzzling problems of planar graphs and coloring and explains them in a very pleasant manner. As a result, these problems almost appear as trivial (which of course is not the case). The main criticism I would make is the following. This book is a corrected and enlarged version of another book. Unfortunately, the updating is not very convincing when the "four color problem" is a conjecture in the body of the book and a theorem in footnotes and afterwords.
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Singapore is number one in Mathematics and number two in Science worldwide in the Third International Mathematics And Science Study (TIMSS) 1999. 93% and 80% of our students are in the international top half for Mathematics and Science respectively.
The mathematics and science curriculum in Singapore has been found to be more comprehensive than that of many countries. Singapore's rigorous curriculum is continually reviewed to ensure that it remains relevant for our students.
Editor's picks
This is a comprehensive curriculum that will give your child a solid foundation in mathematics, build up their confidence and give them a head start on their peers.
"Initially I thought of buying just the text and work books as used by the Singapore students, but the recommended workbooks in the respective Singapore Mathematics and Science grade packages are very well selected. They include solutions that are helpful for me to grade my children's work. The Mathematics teacher's Guide is colourful, but sadly the Science Teacher's Guide comes in Black & white. C. H."
This is a comprehensive curriculum that will give your child a solid foundation in mathematics, build up their confidence and give them a head start on their peers.
"Dear SGBox The books arrived last Friday. The parcel arrived in perfect condition and we are very satisfied with the contents. My daughter has started work in the maths books and finds that they are very well laid out with clear examples and instruction. We have had a quick look through the science books and they promise to be as good as the maths! So far we find that these are excellent products and look forward to ordering from you in the future. Best wishes Sandie McDonnell"
This is a comprehensive curriculum that will give your child a solid foundation in mathematics, build up their confidence and give them a head start on their peers.
"While I have used the Singapore Primary Math books for over 10 years (pre-2001 syllabus), I had not realized that there were more contemporary textbooks available for US based customers. As a homeschool parent, these new texts have far exceeded my expectations and are a significant improvement on the Primary Mathematics texts currently available from US based retailers. The graphics are much more engaging and the supporting materials are very thorough and provide a significant amount of extra practice. I will say, making the transition from 5B old text to 6A new text, the 6A (except for algebra) will predominantly be review (3 of the six chapters) of 5B with some additional challenging problems and twists. That said, for the student having previously worked with Singapore Math, the transition in texts should be a confidence booster because the problems build in complexity at a nice pace. I also can not emphasize enough how pleased I am with the supplementary texts, which include answer keys. The explanations are clear and the additional practice problems are excellent for building skill, speed, accuracy, and fundamental understanding. I would highly recommend these products - not just the textbooks and workbooks but also the additional guides. Karen D."
This is a comprehensive curriculum that will give your child a solid foundation in mathematics, build up their confidence and give them a head start on their peers.
"This is an email to let you know that we have recieved Our order. Thank you for the speedy delivery, the packages arrived two days before scheduled. After reviewing the order, we are very impressed with the quality of both the books and the service. again thank you SGBox And Thanks Mom and Dad! Joe B."
Friends at SGBox.com, Thank you for efficiently processing and sending my order. I was out of town yesterday, and when I returned today your order was waiting for me. It came either on the 22nd or 23rd. Yes, you may print my previous response to your service as well as the exciting news that I am about to share. I had the fortunate opportunity to attend a two-day workshop on why and how to teach Singapore math. I shared my first grade package from you with the presenter and fellow teachers. They were very impressed with the quality and quantity of books I received for my money. They all wanted your website, which I shared with them. Thank you again, Marla O'Keefe"
Your child will develop lifelong learning skills, social skills such as effective communication, cooperation and team spirit, and IT skills through participation in these hands-on and IT activities.
"We received our order of books today. They arrived in perfect shape and both our children have begun devouring them. The books look to have excellent content on our initial review. We are all excited to have these resources to work through this summer! S D"
This is a comprehensive curriculum that will give your child a solid foundation in mathematics, build up their confidence and give them a head start on their peers.
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This is a comprehensive curriculum that will give your child a solid foundation in mathematicsThis is a comprehensive curriculum that will give your child a solid foundation in mathematics, build up their confidence and give them a head start on their peers.
"This is the third order from SGBox. All purchases arrived on time in v. good condition. The kids enjoy their math and the curriculum package provides lots of practice, growth potential and review opportunities through out the year. Karen"
This is a comprehensive curriculum that will give your child a solid foundation in mathematics, build up their confidence and give them a head start on their peers.
"Hi, I am extremely happy with the SGBox set I ordered from you. First of all your customer support is fantastic and I was very happy to recieve the order in a shorter time than you had promised. I have been using these books with my 5 year old son and he loves it. He cannot wait to get back from school and start with "maks" as he calls it. I particularly love the way the books are designed in very simple and clear pictures , that my son can work on it even without my help sometimes. The books have a lot of simple skill practice interspersed with some thought provoking word questions. It gives the child a perspective different from just regular drilling. On the whole, I love it and it has given my son a lot of confidence in maths. I hope to follow up on the sets as and when he masters them. Thank you SGBox for the fantastic service and product. Chandrika, Dubai, United Arab Emirates"
The questions in this Singapore Mathematics workbook are designed to develop and enhance your child's problem-solving skills, stimulate their creative thinking and build up their interest in Mathematics.
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More About
This Textbook
Overview
This best-selling textbook meets the needs of students who will be directly or indirectly involved in the activities of merchandising and buying at the retail level. Mathematics for Retail Buying explains the essential concepts, practices, procedures, calculations, and interpretations of figures that relate to producing profitable retail buying and selling operations. Now in its seventh edition, the text has been reorganized and expanded to provide real world examples that reflect current industry practices and trends. A companion CD-ROM, now containing all practice problems from the text, allows hands-on practice computing retail buying functions and setting up formulas in spreadsheet format.
Editorial Reviews
Booknews
This straightforward textbook/workbook introduces retail pricing and repricing of merchandise, the relationship of markup to profit, the retail method of inventory, dollar planning and control, and terms of sale. The fifth edition adds spreadsheet problems, store forms for practice problems, and a glossary. Annotation c. Book News, Inc., Portland, OR (booknews.com)
Product Details
ISBN-13: 9781563670886
Publisher: Fairchild Books
Publication date: 3/28/1996
Edition description: Older Edition
Edition number: 4
Pages: 333
Age range: 16 - 17 Years
Meet the Author
Bette K. Tepper is a former faculty member of the Fashion Management Department at the Fashion Institute of Technology (FIT), USA, where she taught for more than 30 years, and served as Assistant Chairperson of Marketing Fashion and Related Industries Department
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THE STORIES BEHIND OUR ICONIC NUMBERS Rogerson's Book of Numbers is based on a numerical array of virtues, spiritual attributes, gods, devils, sacred cities, powers, calendars, heroes, saints, icons, and cultural symbols. It provides a dazzling mass of information for those intrigued by the many roles numbers play in folklore and popular culture,... more...
? the most outstanding aspect of this book is its excellent structure: it is as though we have been given a map to help us move around this technology from the base to the summit ? I highly recommend this book ? ?Jose Lloret, Computing Reviews , March 2010 more...
This book originates as an essential underlying component of a modern, imaginative three-semester honors program (six undergraduate courses) in Mathematical Studies. In its entirety, it covers Algebra, Geometry and Analysis in One Variable. The book is intended to provide a comprehensive and rigorous account of the concepts on sets, mapping, family,... more...
INTRODUCTION About the software MATLAB An Introduction to MATLAB Taylor Series NUMBER SYSTEM AND ERRORS Floating-Point Arithmetic Round-Off Errors Truncation Error Interval Arithmetic ROOTS OF EQUATIONS The Bisection Method The Method of False Position Fixed-Point Iteration The Secant method Newton's Method Convergence of the Newton and Secant Methods... more...
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Call us now on
01745 832211
Our GCSE mathematics home study course is suitable for all three main examination boards (AQA, Edexcel and OCR). Whatever your reason - whether you want to apply to university, start a college course, do teacher training (PGCE or ITT), become a nurse, a social worker or generally improve your job prospects it is ideal for you. Covering a wide range of topics, it is divided into easy to follow modules. Each module contains figures, diagrams, examples and in-text questions to help you progress steadily. Particularly attention has been paid to difficult topics like algebra. Everything you need to start studying right away is provided - workbooks, two resource CDs and tutor support and tutor marked assignments. Also included is apen, scientific calculator, geometry set, GCSE exam papers and carry case.
Flexibility is the Key to Success
The main advantage of home study is flexibility. This means you are in control. Simply fit in your studying whenever and wherever you want. Study at home, in your lunch break at work, when the children have gone to bed, literally anytime, it is entirely up to you. Study for ten minutes, half an hour, a couple of hours or more, remember you set the pace. This is very important for a subject like maths.
Topics Covered
Five specially written workbooks guide you through all the main subjects in the latest GCSE mathematics syllabus. They are:
Core unit - Looking at basic number skills.
Statistics (and number) - Data. analysis.
Algebra.- Use of letters and formulae.
Further Algebra - Graphs and practical aspects of algebra in everyday situations.
Geometry - Looking at shapes
Course Content
This is a more detailed breakdown of the contents of the five workbooks (syllabus):
Course Fees and How to Order
The cost of the full course, which includes tutor support and FREE UK postage, is ONLY £199.
Remember you are not alone, once you have enrolled, you can always contact our tutors for friendly help and advice. To purchase this course on-line with your credit/ debit card, click the button below and follow the instructions.
Or you can order over the phone on 01745 832211. Lines are open from 10.30am - 9pm, 7 days a week.
How to Enter for the GCSE Exam
Unless you are already enrolled on a college or school course, you will have to enter yourself for the examination. To do this you need register with an approved centre near to you. This is normally a school or college. The examinations officer at the centre should help you, but if you have any problems, get in touch with us.
You will be entering as an external or private candidate. Make sure you enter for the exam in plenty of time. Exams are usually held twice a year. For further details click on the link below:
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... more...
Part I of this coherent, well-organized text deals with formal principles of inference and definition. Part II explores elementary intuitive set theory, with separate chapters on sets, relations, and functions. Ideal for undergraduates. more...
.... more...
Computer-Assisted Instruction at Stanford, 1966-68: Data, Models, and Evaluation of the Arithmetic Programs provides an analysis and assessment of the arithmetic programs in computer-assisted instruction at Stanford for the years 1966-68. This book focuses on behavioral data, the application of models to these data, and an assessment of the effectiveness... more...
In modern mathematics, both the theory of proof and the derivation of theorems from axioms bear an unquestioned importance. The necessary skills behind these methods, however, are frequently underdeveloped. This book counters that neglect with a rigorous introduction that is simple enough in presentation and context to permit relatively easy comprehension.... more...
Additive and Polynomial Representations deals with major representation theorems in which the qualitative structure is reflected as some polynomial function of one or more numerical functions defined on the basic entities. Examples are additive expressions of a single measure (such as the probability of disjoint events being the sum of their probabilities),... more...
Presents a chronology of Nelson Goodman's writings. This book delineates the constraints imposed upon the aesthetics by both the metaphysics and the epistemology. It is suitable for those who want to understand Goodman's disparate writings within their proper context. more...
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Browse related Subjects ...
Read More just the right level for precalculus students like you. The book also provides calculator examples, including specific keystrokes that show you how to use various graphing calculators to solve problems more quickly. Perhaps most important-this book effectively prepares you for further courses in mathematics.
Read Less
Fair. Hardcover. All text is legible, may contain markings, cover wear, loose/torn pages or staining and much writing. SKU: 9780840068576-5 Hardcover. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9780840068576-4Very good. Hardcover. Has minor wear and/or markings. SKU: 9780840068576-3Hardcover. New Condition. SKU: 9780840068576-1Not enough examples
This book moves too fast with poor examples. I also purchased the solutions guide which was also poor. The first example of a set of problems is shown with all stepps required to complete it but then the more difficult problems do not show the important steps so that I could see where I went wrong
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Math in Our World - 2nd edition
ISBN13:978-0077356651 ISBN10: 0077356659 This edition has also been released as: ISBN13: 978-0072982534 ISBN10: 0072982535 Ste
p Approach. The result is an exceptionally engaging text that is able to both effectively and creatively convey the basic concepts fundamental to a liberal arts math curriculum for even the most hesitant student. Step Approach. The result is an exceptionally engaging text that is able to both effectively and creatively convey the basic concepts fundamental to a liberal arts math curriculum for even the most hesitant student. ...show less900 +$3.99 s/h
Acceptable
Nettextstore Lincoln, NE
2010 Hardcover Fair CONTAINS SLIGHT WATER DAMAGE / STAIN, STILL VERY READABLE, SAVE! This item may not include any CDs, Infotracs, Access cards or other supplementary material.
$156651-5-0
$15.99 +$3.99 s/h
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HPB-North Olmsted North OlmstedColumbus Reynoldsburg, OH75 +$3.99 s/h
New
Books Pole Skokie, IL
2010-01-20 Hardcover New New. There is a slight shelf or time wear. Otherwise new. We Ship Every Day! Free Tracking Number Included! International Buyers Are Welcome! Satisfaction Guaranteed!
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Introductory Algebra-Text - 8th edition exe
rcises and applications, and increased Summary Exercises to enhance comprehension and challenge students' knowledge of the subject matter. exercises and applications, and increased Summary Exercises to enhance comprehension and challenge students' knowledge of the subject matter. ...show less
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...
more mathematics program.
Please provide a link to your portfolio
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Math Fundamentals of Mathematics II (Math 0308-0030)
Fundamentals of Mathematics II: Topics include real numbers, basic geometry, polynomials, factoring, linear equations, inequalities, quadratic equations and rational expressions. A departmental final examination must be passed with a score of 60% or more in order to pass the course. Prerequisite: MATH 0306 or equivalent test score.
This course is intended for students who have either never been exposed to algebra or who have been away from the subject for quite some time. Particularly, this course is intended to prepare students for the study of Intermediate Algebra.
Course Student Learning Outcomes (SLO):
Identify and apply properties of real numbers, and perform accurate arithmetic operations with numbers in various formats and number systems.
2.2 factor polynomials using the techniques of the greatest common factor, grouping, difference of two squares and special trinomials.
2.3 multiply and divide, and simplify rational expressions
3.1 plot ordered pairs and graph linear equations.
4.1 find the perimeter and area of rectangles, squares, parallelograms, triangles, trapezoids and circles; volume and surface area, relations between angle measures, congruent and similar triangles, and properties of parallelograms.
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Introduction. Teaching multivariable calculus, which is the study of curves and surfaces in space, can definitely be enhanced by using the 3-D graphics, and the spacial vector capabilities of Maple. This book provides a new look at vector calculus by integrating the use of Maple with the traditional vector calculus curriculum. Close examination of this text will show that while the method of presenting the topics makes extensive use of Maple, this innovation has been introduced without significantly revising the traditional topics. We have deliberately decided to guide instructors and students through material which is familiar to the instructors, to investigate and compare the capabilities of Maple with the more traditional presentation of vector calculus. As a result, this text can be used in place of a traditional vector calculus text without causing any radical changes in the subject matter.
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The Basically Math Study Guide is designed to help students learn the basic facts. There is a Manual that coordinates with the Study Guide. The Basically Math Manual contains tests and graphs for student use. Basically Math is being used in schools in western Wisconsin and Mesa, Arizona. Basically Math was developed by an experienced teacher after a futile search for materials that would help students.
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More About
This Textbook
Overview
Features an introduction to advanced calculus and highlights its inherent concepts from linear algebra
Advanced Calculus reflects the unifying role of linear algebra in an effort to smooth readers' transition to advanced mathematics. The book fosters the development of complete theorem-proving skills through abundant exercises while also promoting a sound approach to the study. The traditional theorems of elementary differential and integral calculus are rigorously established, presenting the foundations of calculus in a way that reorients thinking toward modern analysis.
Following an introduction dedicated to writing proofs, the book is divided into three parts:
Part One explores foundational one-variable calculus topics from the viewpoint of linear spaces, norms, completeness, and linear functionals.
Part Two covers Fourier series and Stieltjes integration, which are advanced one-variable topics.
Part Three is dedicated to multivariable advanced calculus, including inverse and implicit function theorems and Jacobian theorems for multiple integrals.
Numerous exercises guide readers through the creation of their own proofs, and they also put newly learned methods into practice. In addition, a "Test Yourself" section at the end of each chapter consists of short questions that reinforce the understanding of basic concepts and theorems. The answers to these questions and other selected exercises can be found at the end of the book along with an appendix that outlines key terms and symbols from set theory.
Guiding readers from the study of the topology of the real line to the beginning theorems and concepts of graduate analysis, Advanced Calculus is an ideal text for courses in advanced calculus and introductory analysis at the upper-undergraduate and beginning-graduate levels. It also serves as a valuable reference for engineers, scientists, and mathematicians.
Related Subjects
Meet the Author
Leonard F. Richardson, PhD, is Herbert Huey McElveen Professor and Assistant Chair of the Department of Mathematics at Louisiana State University, where he is also Director of Graduate Studies in Mathematics. Dr. Richardson's research interests include harmonic analysis and representationAllen1
Posted April 6, 2013
This is another example of someone who either refuses to or simp
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Embedding Mathematics in Context
An interview with Chris Arney, United States Military Academy
These days biology is the most popular science on most campuses. Based
on your years of experience in interdisciplinary curriculum
development, what do you see as the main opportunities and primary
challenges for mathematics and biology faculty--both in curriculum and in
pedagogy?
Biology is the science of life and of living organisms, including their
structure, function, growth, origin, evolution, and distribution.
Biologists try to find relationships that help us understand and explain
the structure of life, which is obviously of great importance to the
advancement of society. However, these relationships can be very complex.
To facilitate their analysis, biologists often transform these
relationships into mathematical constructs. The process of explaining
real-world behaviors using mathematical constructs is known as
mathematical modeling. Thus biology can be viewed as a form of applied
mathematics.
The biologist and the mathematician have different bases of knowledge,
however, they have the same goals. Both want to find relationships that
provide a better understanding of the universe.
For example, effective modeling of the population of a particular species
requires both the knowledge of a biologist and the capabilities of a
mathematician. Biologists can provide insight on a species' birth rate
and death rates, environmental capacities, migration, and epidemics.
Furthermore, biologists can identify the model's simplifying assumptions
and understand their implications. However, the processes generated by a
mathematical model can be most effectively examined and verified by a
mathematician. Biologists are concerned about the relationships involved
in population growth, whereas mathematicians are concerned with the
process that validates these relations. Both groups, however, are
concerned about gaining a better understanding of real-world behavior.
Mathematics and biology faculty have many opportunities to contribute to
curriculum and pedagogy. There are many recent textbooks in the areas of
biostatistics and mathematical modeling for biology and environmental
sciences. Nevertheless, additional opportunities for students to learn
statistical principles within a setting of biology would be a positive
change. In addition, discrete dynamical models can help students
understand population growth (e.g., "predator-prey" models) under many
conditions, thus leading to insight into the complex dynamics of
nature.
I see this kind of interdisciplinary setting as beneficial. The challenge
lies in the dynamics of making it work with faculty and for students. I
would recommend an interdisciplinary course in "Mathematical Modeling with
Biology," taught by both a biology instructor and a mathematics
instructor. Such a course could introduce biological concepts by having
students employ scientific experiments, collect data, and perform
mathematical analyses. The biggest challenge is time. The
departments must provide the extra time necessary for instructors to
build, plan, and teach a course that functions as a single course. Course
preparation is often much longer for interdisciplinary courses, especially
when they are team taught.
Has the widespread use of computers for modeling and graphical analysis
changed the kinds of mathematics (in school or college) students need
in order to pursue the study of science?
The advent of computers has altered the skills used to apply
mathematics in the analysis of a problem. No longer is it necessary to
have the skill (or patience) to manipulate numbers in order to transform
data into a useful form. Because of the computer, real problems with
large data sets are now accessible to everyone. The computer has given us
more time and more opportunities to analyze problems and discover
relationships. In essence the computer has allowed mathematics,
especially at the undergraduate level, to become more relevant.
The widespread use of technology for mathematical modeling and data
analysis has provided mathematics students with more visual power over the
subject. Thus today's mathematics students have the tools to be more
competent problem-solvers. Some students may be less capable of by-hand
manipulations, but all students can use computer tools to gain valuable
insights into problem. Moreover, today's students are often asked to
describe verbally what they see and how to model the data. Students
frequently need help in translating the results back into the subject
area. Instructors must be prepared to bridge the gap between verbal
descriptions and the ability to "make it happen."
Mathematics teachers often worry that the distinctive nature of
mathematics will be lost unless it is taught as a separate subject rather
than as part of an integrated interdisciplinary program. Do you share this
concern?
I do not know the distinctive nature of mathematics. I do know that
mathematics is a framework for analysis. This framework can be used
effectively without understanding its underpinning. However, we do need
to continue to increase and improve the current mathematical framework.
Often in interdisciplinary programs where the framework is used, it is not
examined and critiqued.
Mathematics teachers also worry that the context-rich environment of
an interdisciplinary course will impede rather than enhance learning since
it may be harder for students to sort out the mathematics principles from
the surrounding context. What has been your experience in this regard?
Interdisciplinary courses motive the study of mathematics! We do not want
to sort out the mathematics from the surrounding context. We want the
mathematics woven within the context. However, we also want students
to understand the mathematics and not apply it blindly.
One must articulate course objectives prior to designing a course. An
interdisciplinary course needs to be a blend of science and mathematics
designed to enlighten the science. New mathematics can be taught and
learned if there is a need. But the scientific subject area should create
the motivation to learn the mathematics. The blend must not be weighted
more heavily to either science or mathematics. Moreover, you do not want
to blend a 3-hour biology course and 3-hour statistics course into a
single 3-hour bio-statistics course covering all the former material.
This simply cannot be done. An interdisciplinary course works best as a
sequel that builds on basic building blocks from previous courses.
Colonel Chris Arney is a member of the Department of Mathematical
Sciences at the United States Military Academy at West Point and editor of
Interdisciplinary Lively Application Projects (ILAP) published in
1997 by the Mathematical Association of America. He can be reached via
e-mail at ad6819@exmail.usma.army.mil.
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books.google.com - Written for liberal arts students and based on the belief that learning to solve problems is the principal reason for studying mathematics, Karl Smith introduces students to Polya's problem-solving techniques and shows them how to use these techniques to solve unfamiliar problems that they encounter... of Mathematics
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Introduction to MATLAB for Engineers, 3e
Written for use either in a freshman engineering course, or as a self-study text, or as a supplementary reference, this book provides a comprehensive introduction to MATLAB. The book begins with the essentials of MATLAB, and later focuses on advanced topics such as programming in MATLAB, plotting and model building, linear algebraic equations, probability, calculus, and differential equations.
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Tutorials
The tutorials presented here provide a general introduction to various
software packages, and are not intended to be a substitute for a full
semester course, nor are they intended to replace the software's original
documentation. Complete manuals for most programs are available through
the University of Texas Perry Castaneda Library; you can check their availability
through UTNETCAT.
Most materials presented in the tutorials were developed for use in three
hour lecture-format workshops taught every semester to interested UT students,
faculty, and staff. Outside tutorials can be found in our Outside
Resources section.
Mathematical Software Tutorials
Matlab
An overview of commands available in Matlab can be accessed through
an online tutorial.
In addition, the following is a collection of Matlab m files that
should be useful in providing some practice in working with Matlab.
Matlab short course: The two files course.m
and the squarex.m are ASCII
files, contained within course.m are instructions for its use.You
will need to download both, since course.m calls squarex.m. They
must be downloaded into a directory in Matlab's search path, then
accessed by typing 'course' at the Matlab command prompt or calling
squarex.
No prior knowledge of Matlab or any mathematical software is assumed.
This course will cover the basic aspects of Matlab such as statement
syntax, mathematical operations and graphics, as well as some system
interactions such as saving and printing your files. This course
assumes a basic familiarity with matrices.
On-line tutorials are also available from The Mathworks, Inc.,
the developer of Matlab:
Maple
Maple is a mathematical software package for symbolic computation.
Conventional mathematical software packages usually require numerical
values for all variables. In contrast, Maple can evaluate both symbolic
and numerical expressions.
This UT
Maple tutorial is designed for beginning Maple users. No prior
knowledge of Maple or any symbolic mathematical software is assumed.
It will cover the basic aspects of Maple such as statement syntax,
mathematical operations and graphics. While this tutorial will deal
with some calculus related material, it is designed so that anyone
with a basic algebra background will benefit from it.
Mathematica
An extensive documentation of Mathematica commands can be accessed
through the help browser provided along with Mathematica.
tut.nb
is a Mathematica notebook that has been found to be useful as a
mathematica tutorial. The tutorial in it will cover the basic aspects
of Mathematica such as statement syntax, mathematical operations
and graphics, as well as some system interactions such as saving
and printing your files.
While this tutorial will deal with some calculus related material,
it is designed so that anyone with a basic algebra background will
benefit from it. No prior knowledge of Mathematica or any symbolic
mathematical software is assumed.
Statistical Software Tutorials
AMOS
The AMOS (Analysis of Moment Structures) software program features
a powerful, yet easy to use graphical interface. It is designed
primarily for structural equation modeling and similar analyses
(e.g., path analysis, confirmatory factor analysis), though it can
also be used to fit MANOVA, MANCOVA, ANOVA, ANCOVA, and regression
models.
HLM
HLM (Hierarchical Linear Models) are used for analyzing data in
a clustered or "nested" structure, in which lower-level units of
analysis are nested within higher-level units of analysis. For example,
students are nested within classrooms, which are nested within schools.
While experimenters are often not interested in the effects of a
particular classroom or school when they are examining the effects
of a classroom intervention, these units potentially have an effect
on the outcome of the study that should be accounted for in a statistical
model. The program can be used to analyze a variety of questions
using either categorical or continuous dependent variables.
PRELIS
and LISREL
PRELIS and LISREL are designed primarily for structural equation
modeling and similar analyses (e.g., confirmatory factor analysis
and path analysis), though it can also be used to fit ANOVA, ANCOVA,
MANOVA, and MANCOVA models. Also, it be used to perform regression
analysis and some multilevel or hierarchical linear modeling (HLMs). Many of the statistical methods are also now available for the analysis of complex sampling designs.
Mplus
Mplus is primarily designed for conducting exploratory factor
analysis, confirmatory factor analysis, and structural equation
modeling. The program can also handle multiple group analysis and
multilevel SEM.
SAS
Over the years SAS has developed a reputation of being a powerful
and full-featured package for general statistical analysis. The
new release of SAS (version 8) has a number of new features that
promise to make SAS more user-friendly. In particular, the current
version of SAS has a substantially enhanced windows-driven interface
which allows you to point and click your way through many tasks
that previously required knowledge of SAS programming syntax.
Getting Started: introduces readers to the SPSS for Windows
environment, and discusses how to create or import a dataset, transform
variables, manipulate data, and perform descriptive statistics.
Descriptive and Inferential Statistics: describes the use of
SPSS to obtain descriptive and inferential statistics. In this module,
you will be introduced to procedures used to obtain several descriptive
statistics, frequency tables, and crosstabulations in the first
section. In the second section, the Chi-square test of independence,
independent and paired sample t tests, bivariate and partial correlations,
regression, and the general linear model will be covered.
Displaying Data: describes the use of SPSS to create and modify
tables which can be exported to other applications. Graphical displays
of data are also discussed, including bar graphs and scatterplots
as well as a discussions on how to modify graphs using the SPSS
Chart Editor and Interactive Graphs.
Data Manipulation and Advanced Topics: describes the use of
SPSS to do advanced data manipulation such as splitting files for
analyses, merging two files, aggregating datasets, and combining
multiple tables in a database into an SPSS dataset. Several advanced
topics are also included, such as the use of SPSS syntax, the SPSS
Visual Basic editor, and SPSS Macros.
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Abstract. This chapter provides a general introduction to common mathematical
concepts used in engineering applications. Simple algebraic applications, includingexponents and logarithms, are reviewed. Other useful topics include: interpolation,graphical presentations, the rate equation, and the concept of mass and energybalances.
Knowledge of basic problem-solving principles is essential for understanding manyof the engineering applications in food engineering. This unit includes a review ofseveral basic problem-solving topics needed to understand subjects covered insubsequent units. Guidelines for presentation of information are also included. Thematerial is presented as a review only. If you understand the topics, you do not need toread this unit. If you do not understand them, you should read the unit, study theexamples, and work the problems given at the end of the unit.
1.2 Significant Digits
The writing of any number indicates a certain degree of precision for that number.
This is best shown by the use of examples as shown below.
Table 1.01. Significant digit examples.
Number
Tolerance
Significant Digits
2.2
2.2 \u00b1 0.05
2
2.20
2.20 \u00b1 0.005
3
3.1416
3.1416 \u00b1 0.00005
5
2
2 \u00b1 0.5
1
0.0030
0.0030 \u00b1 0.00005
2
2200
2200 \u00b1 50
2
2200
2200 \u00b1 5
3
2200
2200 \u00b1 0.5
4
Note the significance of the zeros in the examples above. The numbers 2.2 and 2.20represent different degrees of precision because the added zero is a significant digit.On the other hand, zeros following a decimal but preceding the first non-zero digit are
not significant digits since they only serve to place the decimal. The last number(2200) is ambiguous since it may have been rounded to the nearest whole number (4significant digits) or the nearest 10 or the nearest 100. This ambiguity can be elimi-nated by the use of scientific notation (e.g., 2.200\u00d7 103 and 2.2\u00d7 103 show four andtwo significant digits respectively.)
The precision of results from any arithmetic calculation depends upon the precisionof the least precise number used in the calculations. Thus if two numbers are multi-plied (i.e., 2.2\u00d7 3.1416) the result, 6.9115, implies false precision because it containsmore significant digits than the least accurate factor in the multiplication. Thus theresult in this example should be rounded to 6.9. This procedure applies identically fordivision. For addition and subtraction, the results should be rounded to the number ofdecimal points included in the least precise number.
The above comments represent the official guidelines for significant digits. Inactual practice, these rules are not always followed. For example, a length may beexpressed as 1 m when it is actually 1.0 m or 1.00 m. Because numbers are oftenpresented in less than the true number of significant digits, it is common practice topresent calculated results as 3 significant digits\u2014even though the rules may call foronly one or two significant digits.
You should also be cautious in rounding numbers. A series of numerical calcula-tions should be carried out with several significant digits and the final answer roundedto an appropriate number of significant digits. For example, 10/3 equals 3.33333\u2026,not 3 or 3.3.
1.3 Unit Factors
Numerical quantities without units generally carry little meaning in engineeringproblems. Consequently, including the correct units with the numerical answer to aproblem is just as important as arriving at the correct numerical value. A briefdiscussion of units and a list of many common unit conversion factors are included in
the appendix.
Unit analysis affords a valuable aid in solving many physical problems. Someproblems require only the conversion of units for a solution. Others may require amore thorough analysis of the problem before finding the number and units for thesolution. Virtually all problems involving food engineering applications involvenumbers with units. Thus, close monitoring of units is essential.
The conversion of units is accomplished by the use of unit factors, defined as
ratios whose actual value is unity, or one. Consider the following:
Equation
Ratio
Unit Factor
1 hr = 3600 s
1
3600 s
hr
1
=
s
3600
hr
1
1 hr = 3600 s
1
hr
1
s
3600=
hr
1
s
3600
1m3 = 1000 L
1
m
1
1000 L
3
=
3
m
1
L
1000
Chapter 1 Introduction: Problem-Solving Tools
3
The combination of numbers and units in the numerator of a unit factor is equal tothe combination of numbers and units in the denominator. Thus, the actual value of theratio is one, or unity. The numerical values alone are usually not equal to one sincethey serve as conversion multipliers for the units involved.
The following example shows how a unit factor problem can be set up for a sys-tematic solution. The steps may seem unnecessary for this simple problem; however,you can avoid difficulties with more complicated problems by following this proce-dure.
Example 1.1
A conveyor belt moves 3 ft in 15 s. What is the belt speed in meters per hour?
Solution:
First enter a blank followed by the final desired units.An equality sign should
follow this, and then enter the pertinent information needed for the solution:
s
15ft
3
hr
m=
Once the pertinent information is included, the solution consists of multiplyingthe basic data on the right hand side of the equation by appropriate unit factorsto obtain the desired final results.Many different unit factors can usually beused to obtain the desired results; however, the better solutions will be those thatproduce a logical solution with a minimum number of unit factors.The follow-ing is one such \u201cbetter\u201d solution:
219.4
hr
1
s
3600
ft
3.281
m
1
s
15
ft
3
hr
m
=
\u00d7
\u00d7
=
The units on the right side of the equation are canceled to produce the resultingunits on the left side of the equation.The numeric answer is then written in theblank on the left side of the equation.
1.4 Algebraic Equations
An equation is a statement of equality between one or more expressions involvingvariables and constants. The equation for a straight line (y = ax + b) is a simpleexample. We solve equations by manipulating them such that the equality of theequation is not affected. Such changes are: addition or subtraction of the same numberor variable to each side (y \u2013 y = ax + b \u2013 y and b + y = ax + b + b); multiplication ofeach side by the same number or variable (ky = kax + kb); or dividing each side by thesame non-zero number or variable (y/a = x + b/a).
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Intermediate Algebra : Graphs and Models - 3rd edition
The Third Edition of the Bittinger Graphs and Models series helps students succeed in algebra by emphasizing a visual understanding of concepts. This latest edition incorporates a new Visualizing the Graph feature that helps students make intuitive connections between graphs and functions without the aid of a graphing calculator.
3.1 Systems of Equations in Two Variables 3.2 Solving by Substitution or Elimination 3.3 Solving Applications: Systems of Two Equations 3.4 Systems of Equations in Three Variables 3.5 Solving Applications: Systems of Three Equations 3.6 Elimination Using Matrices 3.7 Determinants and Cramer's Rule 3.8 Business and Economics Applications39 +$3.99 s/h
Good
PenguinXpress Youngstown, OH
2007
| 677.169 | 1 |
Main navigation
The aim of this 200 page book is to enable talented students to tackle the sort of problems on number theory which are set in mathematics competitions. Topics include primes and divisibility, congruence arithmetic and the representation of real numbers by decimals. A useful summary of techniques and hints is included. This is a fully revised and extended edition of a book which was initially published as part of the composite volume 'Introductions to Number Theory and Inequalities'.
The author is a former Oxford University lecturer and teacher at Clifton College, Bristol.
For ages 16+ competition preparation.
For delivery addresses outside the UK please remember to add extra postage.
See 'Non-UK postage charges' at the top of the 'UKMT Books' page.
Latest News
The UKMT runs 1-day professional development seminars for maths teachers. Our Outreach section has full details of course content at each venue, and a booking form.
Results of the Intermediate Olympiad and Kangaroo will be sent to schools in late April/early May.
The Junior Maths Challenge is held on Thursday 30th April. Buy downloadable past papers for all challenges here.
The downloadable UKMT resources on this website are copyright UKMT, but may be used by teachers/pupils for educational, non-profit purposes
About the Trust
The UK Mathematics Trust (UKMT) is a registered charity whose aim is to advance the education of children and young people in mathematics. The UKMT organises national mathematics competitions and other mathematical enrichment activities for 11-18 year old UK school pupils. We were established in 1996 and last academic year over 600,000 pupils from 4,000 schools took part in the three individual challenges, the UK's biggest national maths competitions. Each challenge leads into a follow-on Olympiad round and we run mentoring schemes and summer schools for high performing students as well as training the team of six to represent the UK in the International Mathematical Olympiad. We also run team maths competitions for two age ranges, publish books and organise enrichment seminars for teachers.
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$105.00Modern Computer Arithmetic focuses on arbitrary-precision algorithms for efficiently performing arithmetic operations such as addition, multiplication and division, and their connections to topics such as modular arithmetic, greatest common divisors, the Fast Fourier Transform (FFT), and the computation of elementary and special functions. Brent and Zimmermann present algorithms that are ready to implement in your favorite language, while keeping a high-level description and avoiding too low-level or machine-dependent details. The book is intended for anyone interested in the design and implementation of efficient high-precision algorithms for computer arithmetic, and more generally efficient multiple-precision numerical algorithms. It may also be used in a graduate course in mathematics or computer science, for which exercises are included. These vary considerably in difficulty, from easy to small research projects, and expand on topics discussed in the text. Solutions are available from the authors.
Contains all the important algorithms relevant to integers, modular arithmetic, floating-point numbers and special functions
Reviews & endorsements
"Very few books do justice to material that is suitable for both professional software engineers and graduate students. This book does just that, without losing its focus or stressing one audience over the other."
Marlin Thomas, Computing Reviews
"This is a concise, well-written and beautiful book in modern computer arithmetic. I found the book very pleasant to read."
Song Yan for SIGACT News
Resources for
Modern Computer Arithmetic
Richard P. Brent, Paul Zimmermann
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Introduction to Modern Algebra and Matrix Theory: Second Edition
This unique text provides students with a single-volume treatment of the basics of calculus and analytic geometry. It reflects the teaching methods and philosophy of Otto Schreier, an influential mathematician and professor. The order of its presentation promotes an intuitive approach to calculus, and it offers a strong emphasis on algebra with minimal prerequisites. Starting with affine space and linear equations, the text proceeds to considerations of Euclidean space and the theory of determinants, field theory and the fundamental theorem of algebra, elements of group theory, and linear transformations and matrices. Numerous exercises at the end of each section form important supplements to the text.
| 677.169 | 1 |
About this item
Essential math concepts for professional chefs and culinary students
Ideal for students and working professionals, Math for the Professional Kitchen explains all the essential mathematical skills needed to run a successful, profitable operation. From scaling recipes and converting units of measure, to costing ingredients and setting menu prices, it covers crucial information that will benefit every foodservice provider.
Written by three veteran math instructors from The Culinary Institute of America, the book utilizes a teaching methodology based on daily in-classroom practice. The entirety of the standard culinary math curriculum is covered, including conversions, determining yields, purchasing, portioning, and more.
This is a thorough, comprehensive main text for culinary students as well as a great kitchen reference for working professionals
Math for the Professional Kitchen will be an invaluable resource not only in the classroom but also in the kitchen as students embark on their professional careers, where math skills play a crucial role in the ever-important bottom
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Elementary Algebra -Text Only - 6th edition
ISBN13:978-0077224790 ISBN10: 0077224795 This edition has also been released as: ISBN13: 978-0073533506 ISBN10: 0073533505
Elementary Algebra, 6/e is part of the latest offerings in the successful Dugopolski series in mathematics. Given the importance of examples within a math book, the author has paid close attention to the most important details for solving the given topic. Dugopolski includes a double cross-referencing system between the examples and exercise sets, so no matter which one the students start with, they will see the connection to the other.
Mark Dugopolski was born and raised in Menominee, Michigan. He received a degree in mathematics education from Michigan State University and then taught high school mathematics in the Chicago area. While teaching high school, he received a master's degree in mathematics from Northern Illinois University. He then entered a doctoral program in mathematics at the University of Illinois in Champaign, where he earned his doctorate in topology in 1977. He was then appointed to the faculty at Southeastern Louisiana University, where he now holds the position of professor of mathematics. He has taught high school and college mathematics for over 30 years. He is a member of the MAA, the AMS, and the AMATYC. He has written many articles and mathematics textbooks. He has a wife and two daughters. When he is not working, he enjoys hiking, bicycling, jogging, tennis, fishing, and motorcycling58 +$3.99 s/h
Good
Campus_Bookstore Fayetteville, AR
Used - Good Hardcover. TEXTBOOK ONLY 6
| 677.169 | 1 |
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