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Lecture NotesEdition April 28, 2010

D.M. Tulett

c© April 28, 2010 D.M. Tulett i

This work is c©D.M. Tulett, 2010. Some of the problems at the end of the doc-ument were created by A.R. Redlack, and are used with his permission. Excel R© isa registered trademark of Microsoft, OpenOffice.org 3 R© is a registered trademarkof OpenOffice.org, and LINDO R© is a registered trademark of LINDO SystemsInc. The Solver used by Excel was made by Frontline Systems, Inc. The graph onthe cover is a modified version of the one that appears near the end of Lecture 4.

http://www.microsoft.com

http://www.openoffice.org

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ii c© April 28, 2010 D.M. Tulett

PrefaceIn this e-book, a word, phrase, or page number in red indicates a link to somewhereelse in this document; a word or phrase in pink indicates a link to the web. Inparticular, both the Table of Contents and the Index are linked to the appropriateplaces in the main body of the material. The table of contents is divided into 22sections. Each of these corresponds with a lecture given at a rate of two per week.

Each lecture (except the very last one) ends with an exercise, which consistsof one or more problems. The learning which comes from doing these exerciseswill prepare you for the tests and the final exam. Your professor may require thatsome of these be done for credit. It would be beneficial, of course, to do all theother exercises as well.

It is assumed that the reader has learn the basics of spreadsheets. In this book,the syntax of Microsoft R©Excel R©is used, but there are other programs which couldbe used such as the free OpenOffice.org 3 OpenOffice R© software.

A companion volume called Mathematics for Management Science: A Bridg-ing Course (hereafter simply called Mathematics for Management Science) ad-dresses in detail some of the material which is covered here: special applicationsof spreadsheets; probability; probability trees; binomial distribution; Bayesian re-vision; and graphing. In this e-book, we concentrate on management science as-suming that the details of the mathematics have been seen either in Mathematicsfor Management Science or elsewhere.

In addition to the above, this e-book should be read in conjunction with anyother materials which have been required for your course.

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mailto:[email protected]

c© April 28, 2010 D.M. Tulett iii

Contents1 Introduction 1

1.1 The Paradigm of Management Science . . . . . . . . . . . . . . . 11.2 Example – Cellular Phone Plans . . . . . . . . . . . . . . . . . . 1

1.2.1 Problem Identification . . . . . . . . . . . . . . . . . . . 21.2.2 Building the Model . . . . . . . . . . . . . . . . . . . . . 31.2.3 Model Solution . . . . . . . . . . . . . . . . . . . . . . . 41.2.4 Implementation . . . . . . . . . . . . . . . . . . . . . . . 61.2.5 Commentary . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Uncertainty 1 102.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 Problem Description . . . . . . . . . . . . . . . . . . . . 102.2.2 Model Formulation . . . . . . . . . . . . . . . . . . . . . 112.2.3 Model Solution . . . . . . . . . . . . . . . . . . . . . . . 132.2.4 Recommendation . . . . . . . . . . . . . . . . . . . . . . 19

2.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Uncertainty 2 213.1 Salvage Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1.1 Theatre Example with Salvage Value . . . . . . . . . . . 213.2 Expected Value of Perfect Information . . . . . . . . . . . . . . . 25

3.2.1 Direct Calculation of the EVPI . . . . . . . . . . . . . . . 263.2.2 Indirect Calculation of the EVPI . . . . . . . . . . . . . . 27

3.3 Decision Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3.1 Pessimism . . . . . . . . . . . . . . . . . . . . . . . . . 283.3.2 Optimism . . . . . . . . . . . . . . . . . . . . . . . . . . 293.3.3 Hurwicz . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.3.4 Laplace . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.3.5 The Regret Matrix . . . . . . . . . . . . . . . . . . . . . 31

3.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4 Uncertainty 3 344.1 Marginal Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.1.1 Computer Example . . . . . . . . . . . . . . . . . . . . . 35

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4.2 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . 364.2.1 Theatre Example . . . . . . . . . . . . . . . . . . . . . . 364.2.2 A More Complicated Example . . . . . . . . . . . . . . . 40

4.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.3.1 Marginal Analysis Problem . . . . . . . . . . . . . . . . 464.3.2 Sensitivity Problem 1 . . . . . . . . . . . . . . . . . . . . 464.3.3 Sensitivity Problem 2 . . . . . . . . . . . . . . . . . . . . 46

5 Decision Trees 1 475.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.2 Theatre Problem in Tree Form . . . . . . . . . . . . . . . . . . . 49

5.2.1 Without Cost Gates . . . . . . . . . . . . . . . . . . . . . 515.2.2 With Cost Gates . . . . . . . . . . . . . . . . . . . . . . 55

5.3 The Expected Value of Perfect Information . . . . . . . . . . . . 605.4 Sequential Decision Making . . . . . . . . . . . . . . . . . . . . 62

5.4.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 625.4.2 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 635.4.3 Finding the EVPI . . . . . . . . . . . . . . . . . . . . . . 66

5.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

6 Decision Trees 2 726.1 Case: New Detergent Marketing Campaign . . . . . . . . . . . . 72

6.1.1 Problem Description . . . . . . . . . . . . . . . . . . . . 726.1.2 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . 756.1.3 Solution and Recommendation . . . . . . . . . . . . . . . 81

6.2 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

7 Decision Trees 3 887.1 Airfare Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

7.1.1 Problem Description . . . . . . . . . . . . . . . . . . . . 887.1.2 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . 887.1.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 907.1.4 The EVPI . . . . . . . . . . . . . . . . . . . . . . . . . . 92

7.2 New Detergent Case: Some Extensions . . . . . . . . . . . . . . 947.2.1 The EVPI . . . . . . . . . . . . . . . . . . . . . . . . . . 947.2.2 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . 95

7.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

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8 Imperfect Information 1 1018.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1018.2 Example – Seismic Testing . . . . . . . . . . . . . . . . . . . . . 101

8.2.1 Problem Description . . . . . . . . . . . . . . . . . . . . 1018.2.2 Problem Formulation . . . . . . . . . . . . . . . . . . . . 1018.2.3 Bayesian Revision . . . . . . . . . . . . . . . . . . . . . 1048.2.4 Solution and Recommendation . . . . . . . . . . . . . . . 1168.2.5 Commentary . . . . . . . . . . . . . . . . . . . . . . . . 118

8.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

9 Imperfect Information 2 1219.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1219.2 Example – Wood Finishers . . . . . . . . . . . . . . . . . . . . . 121

9.2.1 Problem Description . . . . . . . . . . . . . . . . . . . . 1219.2.2 Part A . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1229.2.3 Part B . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1249.2.4 Part C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

9.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

10 Imperfect Information 3 13710.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

10.1.1 Binomial Probability Distribution . . . . . . . . . . . . . 13710.1.2 Destructive vs. Non-Destructive Testing . . . . . . . . . . 138

10.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13810.2.1 Problem Description . . . . . . . . . . . . . . . . . . . . 13810.2.2 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 13910.2.3 Recommendation . . . . . . . . . . . . . . . . . . . . . . 152

10.3 Commentary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15210.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

11 Linear Optimization 1 15411.1 Example – Cement Problem . . . . . . . . . . . . . . . . . . . . 15411.2 Making a Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

11.2.1 Verbal, Algebraic, and Spreadsheet Models . . . . . . . . 15511.2.2 Definition of the Variables . . . . . . . . . . . . . . . . . 15511.2.3 The Objective Function . . . . . . . . . . . . . . . . . . . 15611.2.4 The Constraints . . . . . . . . . . . . . . . . . . . . . . . 15611.2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 158

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11.3 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . 15911.4 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

11.4.1 General Form . . . . . . . . . . . . . . . . . . . . . . . . 16711.4.2 A Right-Hand Side Value of 0 . . . . . . . . . . . . . . . 16711.4.3 Comment . . . . . . . . . . . . . . . . . . . . . . . . . . 173

11.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17311.5.1 Garment Problem . . . . . . . . . . . . . . . . . . . . . . 17311.5.2 Baseball Bat Problem . . . . . . . . . . . . . . . . . . . . 17411.5.3 Quarry Problem . . . . . . . . . . . . . . . . . . . . . . . 174

12 Linear Optimization 2 17512.1 Example – Diet Problem . . . . . . . . . . . . . . . . . . . . . . 175

12.1.1 Problem Description . . . . . . . . . . . . . . . . . . . . 17512.1.2 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . 17512.1.3 Graphical Solution . . . . . . . . . . . . . . . . . . . . . 179

12.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18312.2.1 Problem Description . . . . . . . . . . . . . . . . . . . . 18312.2.2 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . 183

12.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19312.3.1 Office Rental . . . . . . . . . . . . . . . . . . . . . . . . 19312.3.2 Diet Problem . . . . . . . . . . . . . . . . . . . . . . . . 19312.3.3 Fruit Buying Problem . . . . . . . . . . . . . . . . . . . 193

13 Linear Optimization 3 19413.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19413.2 Slack and Surplus . . . . . . . . . . . . . . . . . . . . . . . . . . 19413.3 LINDO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

13.3.1 LINDO for Linear Optimization . . . . . . . . . . . . . . 19613.3.2 LINDO for Integer Optimization . . . . . . . . . . . . . . 199

13.4 Optimization using Spreadsheets . . . . . . . . . . . . . . . . . . 20213.5 Blending Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 205

13.5.1 Problem Description . . . . . . . . . . . . . . . . . . . . 20513.5.2 Formulation and Solution . . . . . . . . . . . . . . . . . 207

13.6 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21213.6.1 Problem Description . . . . . . . . . . . . . . . . . . . . 21213.6.2 Decision Variables . . . . . . . . . . . . . . . . . . . . . 21213.6.3 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . 21313.6.4 What Needs to be Done . . . . . . . . . . . . . . . . . . 213

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14 Linear Optimization 4 21414.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21414.2 New Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

14.2.1 Solution and Non-Solution Variables . . . . . . . . . . . 21514.2.2 Reduced Cost . . . . . . . . . . . . . . . . . . . . . . . . 21514.2.3 Dual Price and Shadow Price . . . . . . . . . . . . . . . . 21514.2.4 Allowable Range . . . . . . . . . . . . . . . . . . . . . . 216

14.3 Example 1: Maximization . . . . . . . . . . . . . . . . . . . . . 21814.3.1 Changes to the Objective Function Coefficients . . . . . . 22014.3.2 Changes to the Right-Hand-Side Values . . . . . . . . . . 221

14.4 Example 2: Minimization . . . . . . . . . . . . . . . . . . . . . . 22314.4.1 Changes to the Objective Function Coefficients . . . . . . 22514.4.2 Changes to the Right-Hand-Side Values . . . . . . . . . . 226

14.5 Example 3: Blending Model . . . . . . . . . . . . . . . . . . . . 22814.5.1 LINDO Sensitivity Output . . . . . . . . . . . . . . . . . 22814.5.2 Changes to the Objective Function Coefficients . . . . . . 23114.5.3 Changes to the Right-Hand Side Values . . . . . . . . . . 233

14.6 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23514.6.1 A Maximization Problem . . . . . . . . . . . . . . . . . . 23514.6.2 A Minimization Problem . . . . . . . . . . . . . . . . . . 235

15 Markov Chains 1 23715.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23715.2 Example: Brand Switching . . . . . . . . . . . . . . . . . . . . . 237

15.2.1 Description of the Situation . . . . . . . . . . . . . . . . 23715.2.2 State Transition (Probability) Diagram . . . . . . . . . . . 23815.2.3 State Transition (Probability) Matrix . . . . . . . . . . . . 24215.2.4 Finite Number of Transitions . . . . . . . . . . . . . . . . 243

15.3 Determining Steady-State Probabilities Analytically . . . . . . . . 24615.3.1 Three State Brand Switching Example . . . . . . . . . . . 24615.3.2 Two State Markov Chain . . . . . . . . . . . . . . . . . . 24915.3.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 24915.3.4 Solution by Computer . . . . . . . . . . . . . . . . . . . 250

15.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25215.4.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 25215.4.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 252

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16 Markov Chains 2 25316.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25316.2 Absorbing States Analysis . . . . . . . . . . . . . . . . . . . . . 25416.3 Solution by Computer . . . . . . . . . . . . . . . . . . . . . . . . 25816.4 An Application: Credit Card Analysis . . . . . . . . . . . . . . . 261

16.4.1 Situation Description . . . . . . . . . . . . . . . . . . . . 26116.4.2 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . 26216.4.3 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 262


17 Markov Chains 3 26717.1 Gambler’s Ruin . . . . . . . . . . . . . . . . . . . . . . . . . . . 26717.2 Educational Model . . . . . . . . . . . . . . . . . . . . . . . . . 271

17.2.1 Extension 1 . . . . . . . . . . . . . . . . . . . . . . . . . 27117.2.2 Extension 2 . . . . . . . . . . . . . . . . . . . . . . . . . 271

17.3 Repair Problem A . . . . . . . . . . . . . . . . . . . . . . . . . . 27217.3.1 Situation Description . . . . . . . . . . . . . . . . . . . . 27217.3.2 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . 272

17.4 Repair Problem B . . . . . . . . . . . . . . . . . . . . . . . . . . 27517.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

17.5.1 Educational Model . . . . . . . . . . . . . . . . . . . . . 27817.5.2 Repair Problem . . . . . . . . . . . . . . . . . . . . . . . 278

18 Utility Theory 1 27918.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27918.2 Obtaining the Data for a Utility Function . . . . . . . . . . . . . . 28018.3 Using the Graph with a Payoff Matrix . . . . . . . . . . . . . . . 28418.4 A More Complicated Graph . . . . . . . . . . . . . . . . . . . . 28718.5 Utility Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 292

18.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 29218.5.2 A Simple Decision . . . . . . . . . . . . . . . . . . . . . 293


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19 Utility Theory 2 29819.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29819.2 Three Useful Properties . . . . . . . . . . . . . . . . . . . . . . . 29819.3 Finding the EVPI . . . . . . . . . . . . . . . . . . . . . . . . . . 29919.4 Using Utility Functions with Decision Trees . . . . . . . . . . . . 302

19.4.1 The EVPI . . . . . . . . . . . . . . . . . . . . . . . . . . 31019.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311

19.5.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 31119.5.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 311

20 Game Theory 1 31220.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31220.2 Pure Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

20.2.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . 31320.2.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . 314

20.3 Mixed Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . 31620.3.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 31620.3.2 Dominance . . . . . . . . . . . . . . . . . . . . . . . . . 31620.3.3 DARS Algorithm . . . . . . . . . . . . . . . . . . . . . . 321

20.4 Prisoner’s Dilemma . . . . . . . . . . . . . . . . . . . . . . . . . 32320.4.1 Situation Description . . . . . . . . . . . . . . . . . . . . 32320.4.2 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 324

20.5 Obtaining the Payoffs . . . . . . . . . . . . . . . . . . . . . . . . 32420.6 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326

20.6.1 Mixed Strategy Game . . . . . . . . . . . . . . . . . . . 32620.6.2 Jack and Jill Problem . . . . . . . . . . . . . . . . . . . . 32620.6.3 Formulation Problem . . . . . . . . . . . . . . . . . . . . 326

21 Game Theory 2 32721.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . 327

21.1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . 32721.1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . 33521.1.3 General Methodology . . . . . . . . . . . . . . . . . . . 342

21.2 Formulation as a Linear Optimization Model . . . . . . . . . . . 34321.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347

21.3.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 34721.3.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 34721.3.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 347

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21.3.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 34821.3.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . 348

22 Moving Forward 349

23 Supplement – Extra Problems 35323.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354

23.1.1 Advanced Problems . . . . . . . . . . . . . . . . . . . . 35623.2 Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35723.3 Decision Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . 36823.4 Imperfect Information . . . . . . . . . . . . . . . . . . . . . . . . 37523.5 Linear Optimization – Introduction . . . . . . . . . . . . . . . . . 39423.6 Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . 40223.7 Utility Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41123.8 Game Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42123.9 Linear Optimization – Applications . . . . . . . . . . . . . . . . 428

c© April 28, 2010 D.M. Tulett 1

1 Introduction

Management Science involves the application of mathematics to managementproblems. As mentioned in the Preface, the mathematics that we need is cov-ered in Mathematics for Management Science. We will, for example, use suchthings as Bayesian revision and graphing for linear optimization that are coveredin that book. Also, the numerical calculations can often be left to a spreadsheet orother software tools. If there’s a difficulty with this subject, it’s probably not themathematics.

Instead, the difficulty is likely to be the building of the model which the math-ematics seeks to solve. The important thing is always going from a problem de-scription to a model for the problem. This is part of the paradigm of managementscience, which is the focus of this section.

1.1 The Paradigm of Management Science

A paradigm is a pattern for the general understanding of something. A commonparadigm for management science is to think of it being composed of four phases:

• problem definition

• model building

• solution

• implementation.

When the fourth phase has been done, it is appropriate to ask whether or not itaddressed the original problem. Because we are working in an academic context,we cannot observe the entire paradigm. The “problem” is not for us a real-worldobservation, but instead it a written description (a “word” or “story” problem).Also, we cannot implement the solutions. We are left with looking at the secondand third phases of the paradigm.

1.2 Example – Cellular Phone Plans

We use this example as a way to illustrate the paradigm.

2 c© April 28, 2010 D.M. Tulett

1.2.1 Problem Identification

Alison has decided to buy a cellular telephone, partly for safety in case of a break-down in an isolated area, but also because of the convenience that it will provide.She’s not concerned about the initial cost of the telephone itself, especially whensome cellular phone companies give the phones away in order to attract business.However, she is concerned about the monthly operating cost, especially since shewould use the cellphone almost entirely during working hours Monday to Friday.

Most plans operate on the user paying a fixed charge to obtain a set numberof minutes per month, with a per-charge premium if extra minutes are used. Forexample, one company offers a 200 minute calling plan for $25 per month with asurcharge of 25 cents per minute for all calls (local or long-distance) beyond the200 minutes, and long-distance charges are 35 cents per minute at any time. Forexample if a user of this plan made 160 minutes of local calls, and 90 minutes oflong-distance calls, the cost would be $25 plus $0.25(160 + 90− 200) + $0.35(90)for a total of $69.00. Other plans treat both local and long-distance (within thecountry) calls the same.

A summary of the plans available appears in the following table (the per-minute column refers to charges for calls not covered by the plan):

Plan Type Cost Minutes Per-MinutePay-as-you-go None None $0.35 all calls

$0.35 long-distanceBasic User Local $25 200 $0.25 all calls

$0.35 long-distanceRegular User Local $45 400 $0.20 all calls

$0.35 long-distanceExtended User Local $69 600 $0.20 all calls

$0.35 long-distanceBasic User Anywhere $39 200 $0.30 anywhereRegular User Anywhere $59 400 $0.25 anywhereHeavy User Anywhere $109 800 $0.25 anywhere

Alison plans to make 240 minutes of local calls and 60 minutes of long-distance calls each month. Which plan should she buy? Hers is a specific example,but without too much extra work we can solve a more general problem: for anygiven numbers of minutes of local and long-distance calls, which plan should bebought? By building a spreadsheet model, we can easily solve this problem.


1.2.2 Building the Model

Minimizing the total monthly cost should be her objective. The monthly cost isthe fixed cost of the plan, plus the cost of the calls (if any) which exceed theplan’s limit. We can quickly solve Alison’s problem by hand, but let’s look at thegeneral problem of a person who needs x minutes of local calls and y minutes oflong-distance calls each month. The cost of the plans (the seven alternatives) canbe written terms of x and y. The Pay-as-you-go plan is easy; its cost is

$0.35(x+ y)+$0.35y

For the Basic User Local plan, the fixed charge is of course $25, and the long-distance charge is $0.35y. The cost of extra calls is, however, a bit trickier. Ifx + y≤ 200, then there is no extra charge. However, if x + y > 200, then the usermust pay for an extra x + y−200 minutes. Hence the number of extra minutes iseither 0 or x + y− 200, whichever is greater. The number of extra minutes canbe represented by the expression max{0,x + y−200}, and the cost of these extraminutes is $0.25max{0,x + y−200}. Overall, the expression for the cost of thisplan is:

$25+$0.25max{0,x+ y−200}+$0.35y

Continuing to develop cost expressions for each alternative, we obtain:

Alternative Cost with x local and y long-distance minutes1 Pay-as-you-go $0.35(x+ y)+$0.35y2 Basic User Local $25+$0.25max{0,x+ y−200}+$0.35y3 Regular User Local $45+$0.20max{0,x+ y−400}+$0.35y4 Extended User Local $69+$0.20max{0,x+ y−600}+$0.35y5 Basic User Anywhere $39+$0.30max{0,x+ y−200}6 Regular User Anywhere $59+$0.25max{0,x+ y−400}7 Heavy User Anywhere $109+$0.25max{0,x+ y−800}

Every plan is of the form

a+bmax{0,x+ y− c}+dy

where some of parameters a, b, c, and d may be 0. These four parameters rep-resent, respectively: the fixed cost per month; the per-minute cost when over thelimit of the plan; the number of minutes for the plan; and the per-minute long-distance supplementary charge. Seeing patterns like this helps when making a


spreadsheet model, because then a formula can be entered once, and then becopied for every alternative. This is why, for example, the pay-as-you-go plan(which costs $0.35(x+ y)+$0.35y), would be thought of as

0+$0.35max{0,x+ y−0}+$0.35y

so that it fits the pattern, rather than being simplified to $0.35x+$0.70y.

1.2.3 Model Solution

What we have done so far is make an algebraic model of the problem. Becausethis problem is simple, we could solve it using a calculator with specific values ofx and y. However, it is useful to make a spreadsheet model to do the calculationsif we wish to calculate the cost of each plan for several values of x and y.1

We need to reserve two cells, one for x, and one for y. In one column every al-ternative is listed, their values for a, b, c, and d are put into the next four columns,and in the sixth column the monthly cost is calculated. For example, we could putthe values for the number of local and long-distance minutes into cells B1 and F1respectively. Allowing some room for column labels, the values of a, b, c, and dfor alternatives 1 to 7 could go into the range B8:E14. Doing this much we have:

A B C D E F1 Local Long-Distance23 Alternative Fixed Per-Minute Plan Long- Total4 Charge Charge Limit Distance Monthly5 Per over the (minutes) Per-Minute Cost6 Month Plan Limit78 1 $0.00 $0.35 0 $0.359 2 $25.00 $0.25 200 $0.3510 3 $45.00 $0.20 400 $0.3511 4 $69.00 $0.20 600 $0.3512 5 $39.00 $0.30 200 $0.0013 6 $59.00 $0.25 400 $0.0014 7 $109.00 $0.25 800 $0.00

1It is assumed that the reader has at least a basic familiarity with spreadsheets.


In columns B, C, and E, the data was entered as a number, but the cell formatwas set to currency with two decimal places, hence entering 25 produces $25.00.Cells B8, and the range E12:E14 are not left blank, but instead a 0 is entered(which the spreadsheet formats as $0.00). This is because the formulas to beentered in column F need to read a 0; a blank cell would produce an error message.

In the sixth column (F) we calculate the cost of the plans. In cell F8 we input

=B8+C8*MAX(0,$B$1+$F$1-D8)+E8*$F$1

This is then copied in the range F8:F14.Doing this we see that with the specific values of 240 in cell B1 and 60 in cell

F1, that alternatives 1 to 7 cost respectively $126, $71, $66, $90, $69, $59, and$109. The cheapest plan for Alison is the Regular User Anywhere Plan (alterna-tive 6), at a cost of $59 per month.

A B C D E F1 Local 240 Long-Distance 6023 Alternative Fixed Per-Minute Plan Long- Total4 Charge Charge Limit Distance Monthly5 Per over the (minutes) Per-Minute Cost6 Month Plan Limit78 1 $0.00 $0.35 0 $0.35 $126.009 2 $25.00 $0.25 200 $0.35 $71.0010 3 $45.00 $0.20 400 $0.35 $66.0011 4 $69.00 $0.20 600 $0.35 $90.0012 5 $39.00 $0.30 200 $0.00 $69.0013 6 $59.00 $0.25 400 $0.00 $59.0014 7 $109.00 $0.25 800 $0.00 $109.00

The developer of the model, whether using a calculator or a spreadsheet, mustmake the recommendation clear. The customer of the model (in this case, Alison)might not be familiar with spreadsheets, so the emphasis should be on giving therecommendation:


Recommendation

Alison should sign up for the Regular User Anywhere plan, at a cost of $59 per month.

Now suppose that Alison isn’t sure about her long-distance usage, and that itcould be anywhere from 40 to 80 minutes, rather than exactly 60. It’s easy to plugthe values 40 and 80 into the spreadsheet and see what happens. At 40 minutes,plans 2, 3 and 6 are tied at $59. At 80 minutes, plan 6 is still uniquely the best.Hence alternative 6 is best everywhere from 40 to 80 minutes of long-distanceuse, at a constant cost of $59. What we have done here is a primitive form of

sensitivity analysis , in which a parameter of the model is varied to examine theeffect (if any) on the recommended solution.

1.2.4 Implementation

Models only approximate reality. Sometimes, things can be left out because theydon’t affect the choice. For example, we have ignored taxes. Whatever the taxrate, be it 15% or something different, the cheapest alternative will still be thecheapest alternative. On the other hand, a model cannot capture every nuance,even if it might change the optimal choice. It may be that some plans have extrafeatures like call forwarding, but some plans do not. If we try to capture this inthe model, it will quickly become very big. For this reason, the recommendedsolution is only optimal for the model, and is not necessarily best at solving theoriginal problem. To complete the paradigm, we should go back to Alison to seeif she is happy with the recommended plan.

1.2.5 Commentary

The four phases of the management science paradigm are not totally distinct.When we had completed the algebraic model, we saw that it was useful to buildanother model, this one using a spreadsheet, so that we could solve it.

This model only involved cost, so we found the alternative with the minimumcost. In many management science examples, however, we seek the alternativewith the highest profit.


1.3 ExerciseTwo types of big boxes are about to be loaded onto a small cargo plane. A Type1 box has a volume of 2.9 cubic metres (m3), and a mass of 470 kilograms (kg),while a Type 2 box has a volume of 1.8 m3 and a mass of 530 kg. There are sixType 1 boxes and eight Type 2 boxes waiting to be loaded. There is only onecargo plane, and it has a volume capacity of 15 m3 and a mass capacity of 3600kg. Obviously, not all the boxes can be put onto the plane, therefore suppose thatthe objective is to maximize the value of the load. We will consider the followingthree situations: (i) both type of boxes are worth $400 each; (ii) a Type 1 box isworth $600, and a Type 2 box is worth $250; and (iii) a Type 1 box is worth $300,and a Type 2 box is worth $750.

Later in this course we shall see an efficient approach for solving this type ofproblem, but for now we use the following simple approach:

(a) Let x and y represent the number of Type 1 and Type 2 boxes respectivelywhich are put onto the plane. Where x and y are of course positive integers (includ-ing 0), determine all the feasible combinations (x,y), using a spreadsheet to helpwith the calculations. (To be feasible the total volume carried must be ≤ 15m3,and the total mass carried must be ≤ 3600 kg.)

(b) Consider a combination found in (a) which can be augmented by addingone box (of either type) with the capacities still not being exceeded. One exampleis (3,1), i.e. three Type 1 boxes, plus one Type 2 box, because (3+1,1) = (4,1),which is feasible, would be a better solution, as would the feasible solution (3,1+1) = (3,2). This combination (3,1) (and all others like it) is therefore triviallysub-optimal, because we would obtain more money by adding the extra box.

Therefore, we should narrow the search by looking only at the feasible com-binations which are so near the limit of either the mass or volume capacity thatputting one more box (of either type) onto the plane would make it unable to fly.Mathematically, these are the combinations for which (x,y) is feasible, but both(x+1,y) and (x,y+1) are not feasible. Find these combinations.

(c) Make a spreadsheet in which the alternatives are the combinations from(b), and which has two cells reserved for the value of each type of box. Usethe spreadsheet to determine, for each of the three financial scenarios, how manyboxes of each type are carried, and the value of the load.

Work on this on your own and come up with your own method. If you’re stuckafter 15 minutes or so, then look at the hints on the next page.


Hints

Obviously carrying no boxes is feasible, so this is a good starting point. Thissolution is represented as (x,y) = (0,0). We could then determine using trial-and-error if (0,1) is feasible, and if so, then see if (0,2) is feasible, and so on. A fasterway, however, is to start by fixing x = 0, and then find the largest value for y. Thereare three restrictions: we cannot exceed the volume available; we cannot exceedthe mass available; and y must be an integer. When x = 0 the volume available isof course the full 15 m3, and the mass available is 3600 kg. Each unit of y (eachType 2 box) takes up 1.8 m3 and 530 kg, therefore y is the largest integer such thatboth 1.8y ≤ 15 and 530y ≤ 3600. Hence y ≤ 8.333..., and y ≤ 6.792.... Hencethe most that y can be is 6. Therefore all combinations (x,y) = (0,0), (0,1), (0,2),(0,3), (0,4), (0,5), and (0,6) are feasible.

Now suppose that x = 1. This takes up 2.9 m3 and 470 kg, therefore the type 2boxes can use up to 15−2.9 = 12.1m3 and up to 3600−470 = 3130 kg. Based onthis, it can be seen that y can be at most 5. Keep repeating this for higher valuesof x until no more type 1 boxes can be carried, even if no type 2 boxes are carried.You should find a total of 26 feasible combinations. Using the rules of part (b),we see that the search can be limited to just five combinations.

A spreadsheet to do the calculations for part (a) could begin as follows:

A B C D E F1 x Remaining Remaining Max. y Max. y Overall2 Volume Mass (Volume) (Mass) Max. y34 0 =15-2.9*A45 16 27 38 49 510 611 7

Find the formula for each of the cells C4, D4, E4, and F4. For F4, you willneed the INT function (look at your spreadsheet’s Help menu). The range B4:F4 isthen copied to the rows below. There will be a row in which it and all subsequent


rows contain one or more negative numbers; this means that the correspondingvalue of x is infeasible.


2 Uncertainty 1

2.1 IntroductionThe simplest situation involving decision making under uncertainty has the fol-lowing attributes:

• There is one decision; the decision maker must choose one of several alter-natives.

• There is one event; one of several possible outcomes will occur.

• For each combination of alternative and outcome we can calculate the pay-off (which may be negative) to the decision maker.

The order is very important: the decision must precede the event . First analternative is chosen, and then an outcome occurs. Here are some common exam-ples:

1. At 8 a.m. you must decide whether or not to carry an umbrella; later thatday you find out whether or not it rains.

2. Before a hockey game, you decide whether or not to place a bet on theoutcome; and at the end of the game you find out which team has won.

2.2 Example2.2.1 Problem Description

A theatre company wishes to mount a play. A three night run is planned. Theyhave already spent or have committed to spend $2500 for such things as costumes,makeup, and so on. They are definitely going ahead with the play; the only de-cision they must make is where to hold it. Small, medium, and large theatres areavailable for rent which hold 100, 400, and 1200 people respectively. Three nightsrent at each theatre would cost $600, $1800, and $5500 respectively. They mustmake a commitment to one of these theatres several weeks before the run begins.

The demand for the play is uncertain until the run begins. Demand is heavilyinfluenced by the critics’ reviews. The critics will attend a dress rehearsal the nightbefore the first performance, and their opinions will be printed and broadcast inthe media the next morning. The theatre company has already decided to price all


the tickets at $12.50 each. From this, they must pay $2.50 per ticket sold in taxes,leaving them with a net revenue of $10 per ticket sold. They know from experiencethat demand for plays falls into four broad categories of interest: fringe; average;great; and heavy. The number of people who wish to see the play is typically250 for fringe, 800 for average, 2300 for great, and 4500 for heavy. These aredemand levels, not necessarily the number of tickets sold. For example, if a playsells every seat in a 250 seat theatre for three nights, and if another 50 people werewait-listed for tickets but could not obtain them, then 750 tickets were sold, butthe demand was for 800 tickets.

The demand is an event in which one of four outcomes will occur. To estimatethe probabilities of these four outcomes, the theatre company could look at thehistorical data for plays of this type with tickets sold in this price range. Supposethat of one hundred plays in the past, the interest attracted was twenty for fringe,seventy for average, nine for great, and one for heavy. We would then estimatethe chance of the next play attracting fringe interest as

P(fringe interest) =20100

= 0.20

Continuing in this manner we would estimate the probabilities for average, great,and heavy as 0.70, 0.09, and 0.01 respectively.

Using historical data to estimate probabilities ignores such factors as changingconsumer tastes and economic conditions, but we have to start somewhere. Usingthese numbers we will obtain one conclusion after solving the model, but anotherset of numbers will often lead to a different conclusion. We will examine some ofthese issues later in the course.

This model has been kept simple in that everything has been decided exceptone thing – which theatre to rent. This is the problem which we shall now solve.

2.2.2 Model Formulation

In all models with decision making under uncertainty, we must define the deci-sions, their alternatives, the events, and their outcomes. Some textbooks stress theuse of symbols for this purpose, however another approach is to use words only,and then define a shortcut word to use in place of each longer phrase.

For both approaches we have the following:

They must decide where to hold the play. The alternatives are to rent a smalltheatre with 100 seats, or rent a medium-sized theatre with 400 seats, or rent a


large theatre with 1200 seats. The event is the demand for tickets. The possibleoutcomes are as follows: there is fringe interest with demand for 250 tickets; thereis average interest with demand for 800 tickets; there is great interest with demandfor 2300 tickets; or there is heavy interest with demand for 4500 tickets.

Because it takes a lot of space to write all these words every time we wishto refer to them, we need a shortcut form. In the method of using symbols, thedecision is symbolized with the letter D, and the three alternatives have subscriptson the letter A, making them A1, A2, and A3. The event is symbolized with theletter E, and its four outcomes have subscripts on the letter O, making them O1,O2, O3, and O4.

The alternative and outcome symbols mean the following:

Alternative CostA1 rent a small theatre with 100 seats $600A2 rent a medium-sized theatre with 400 seats $1800A3 rent a large theatre with 1200 seats $5500

Outcome ProbabilityO1 there is fringe interest; the demand is for 250 tickets 0.20O2 there is average interest; the demand for 800 tickets 0.70O3 there is great interest; the demand is for 2300 tickets 0.09O4 there is heavy interest; the demand is for 4500 tickets 0.01

The other approach is to use one word (or a very short phrase) to mean theentire long phrase. Such words must be unique . For example, we cannot use“medium” to refer to both a medium-sized theatre and to average interest. Usingthis approach we could use the following words:

Alternatives Costsmall rent a small theatre with 100 seats $600

medium rent a medium-sized theatre with 400 seats $1800large rent a large theatre with 1200 seats $5500

Outcomes Probabilityfringe there is fringe interest; the demand is for 250 tickets 0.20

average there is average interest; the demand for 800 tickets 0.70great there is great interest; the demand is for 2300 tickets 0.09

heavy there is heavy interest; the demand is for 4500 tickets 0.01


Whichever method is used, the important thing is that the person making themodel must understand what the alternatives and outcomes are.

The only other pieces of information we need from the problem description isthat the net revenue after taxes is $10 per ticket sold, and that the play runs forthree nights. The other expenses such as costumes, makeup, and so on are whatare called sunk costs . A sunk cost is money which is either already spent or hasalready been committed, and is therefore irrelevant to the decision. Indeed, even ifthese fixed expenses (which total $2500) were not already committed, they wouldnot affect the decision in this example, because all alternatives would contain thesesame expenses.

2.2.3 Model Solution

There are three alternatives, and four outcomes, hence there are three times fourequals twelve situations which need to be evaluated. First we see what happens ifa small theatre is rented, and the play only attracts fringe interest. The 100-seatsmall theatre can hold 300 people over three nights, but only 250 people wantto see the play, so only 250 tickets are sold. The net revenue from the ticketsales is therefore $10 times 250 = $2500. We can now find what is often calledthe “profit”, but we define a new term payoff , which can mean profit, cost, orrevenue depending on the context. The payoff is found by subtracting the $600rent from the $2500 from the sales of tickets, i.e. $1900.

If a small theatre is rented, but the demand turns out to be average, then thereare more willing customers (800) than there are seats (300). The number of ticketsales is therefore just 300. For any situation, we can say that the number of ticketssold is the capacity of the theatre (over three nights), or the demand for tickets,whichever is less. The payoff is

$10(300)−$600 = $2400

We do not need to analyze in detail what happens if more potential customers(great or heavy) show up when only a small theatre has been rented; no moretickets can be sold, so the payoff will remain at $2400.

Now suppose that a medium-sized theatre is rented at a cost of $1800. Witha 400 seat capacity, a three-night run gives a maximum sales capacity of 1200tickets. There’s plenty of space with fringe or average demand, but the capacityof 1200 is reached with great or heavy demand. With fringe interest the payoff is:

$10(250)−$1800 = $700


With average interest the payoff is:

$10(800)−$1800 = $6200

With either great or heavy demand the payoff is

$10(1200)−$1800 = $10,200

If a large theatre with 1200 seats is rented for $5500, the three-night capacity is3600 people. This is sufficient for all but heavy demand. The number of ticketssold will equal the demand if interest is fringe, average, or great, and will equalthe total capacity (3600) if there is heavy demand. Hence we have:

Outcome fringe average great heavy3-Night Capacity 3600 3600 3600 3600

Demand 250 800 2300 4500# of Tickets Sold 250 800 2300 3600

Net Ticket Revenue $2500 $8000 $23,000 $36,000Rent $5500 $5500 $5500 $5500

Payoff −$3000 $2500 $17,500 $30,500

The preceding calculations do not need to be always explicitly written out aswe have done here. Often the calculations can be done on a calculator, with justthe final payoffs being written down. Or, as we soon shall see, we can use aspreadsheet to do the calculations. Of course, to do this by any means we mustunderstand how the final payoff is derived. In all twelve cases, the payoff is com-puted as:

payoff = $10 min{ three-night capacity, demand } − rent

All of this information can be conveniently summarized in what is called apayoff matrix (also called a payoff table). In doing this by hand, just one payoffmatrix is drawn . However, to help explain it, we draw it once with just the bor-

ders, then with the main body filled in, and then with the right-hand side filledin.

In the main body of the payoff matrix, each row represents an alternative, andeach column represents an outcome. Labels for the alternatives appear on the left-hand side, and labels for the outcomes appear on the top. The final row lists the


probabilities of the outcomes. The final column is reserved for the expected valueof each alternative – this will be explained shortly.

It is helpful if we put the theatre capacity (over three nights) and the cost ofthe rent next to the name of the alternative, and the demand as a number next tothe names for the four levels of demand. Doing this the payoff matrix begins as:

Demand for TicketsTheatre 3-Night Fringe Average Great Heavy ExpectedSize Capacity Rent 250 800 2300 4500 ValueSmall 300 $600Medium 1200 $1800Large 3600 $5500

Probability 0.20 0.70 0.09 0.01

Using the formula “=$10 min{ three-night capacity, demand } − rent”, eachpayoff is calculated and put into the table. If we are doing these calculations usinga calculator, we would look for shortcuts like noticing the repetition of the “2400”for the first alternative.

We of course have already done these calculations by hand, and hence we have(dropping the dollar signs):

Demand for TicketsTheatre 3-Night Fringe Average Great Heavy ExpectedSize Capacity Rent 250 800 2300 4500 ValueSmall 300 $600 1900 2400 2400 2400Medium 1200 $1800 700 6200 10,200 10,200Large 3600 $5500 −3000 2500 17,500 30,500

Probability 0.20 0.70 0.09 0.01

If we wish to use a spreadsheet, we will input the theatre size, and let the 3-night capacity be found as part of the formula, which is entered once and thenis copied. Besides doing the calculations, using a spreadsheet makes it easy tochange colours and/or fonts to highlight information. The real advantage, how-ever, is that it easily allows us to see what happens when some of the informationis changed.

In spreadsheet form we begin with:


A B C D E F G H1 Demand for Tickets2 Theatre # of Fringe Average Great Heavy Expected3 Size Seats Rent 250 800 2300 4500 Value4 Small 100 $6005 Medium 400 $18006 Large 1200 $55007 Probability 0.20 0.70 0.09 0.01

We want to make a formula in cell D4 which we can copy to the range D4:G6.The number of seats available is in cell B4, hence the 3-night capacity is 3*B4.The demand is in cell D3, and the cost of the rent is in cell C4. Remember that adollar sign creates an absolute rather than a relative cell address. Hence we mustuse a dollar sign to freeze the ‘B’ in ‘B4’, the ‘3’ in ‘D3’, and the ‘C’ in ‘C4’. Theformula to be placed in cell D4 is therefore:

=10*MIN(3*$B4,D$3)-$C4

With the numbers in the main body of the payoff matrix being calculated bythe spreadsheet (commas will not appear unless special formatting is used) wehave:

A B C D E F G H1 Demand for Tickets2 Theatre # of Fringe Average Great Heavy Expected3 Size Seats Rent 250 800 2300 4500 Value4 Small 100 $600 1900 2400 2400 24005 Medium 400 $1800 700 6200 10200 102006 Large 1200 $5500 −3000 2500 17500 305007 Probability 0.20 0.70 0.09 0.01

There is an expected value associated with each alternative. Recall from hav-ing studied random variables that in general, if there are n outcomes, and the prob-ability of outcome i is pi, and the payoff of outcome i is xi, then the expected valueis defined as:

E(X) =n

∑i=1

pixi (1)


The current example has four outcomes. The expected value associated with rent-ing a large theatre is

EV(large) = 0.20(−3000)+0.70(2500)+0.09(17,500)+0.01(30,500)= −600+1750+1575+305= 3030

What this figure means is that if the theatre company were to face the same situ-ation many times, and if they were to choose a large theatre each time, then overtime their profits/losses would average out to $3,030. The actual payoff on a par-ticular play will be either −$3000, or $2500, or $17,500, or $30,500. Hence theexpected value is none of the actual values; it is simply a long-term average value.

When some of the outcomes are the same, as occurs for the medium-sizedtheatre alternative, we can factor the numbers if we wish:

EV(medium) = 0.20(700)+0.70(6200)+(0.09+0.01)(10,200)= 140+4340+1020= 5500

The small theatre alternative is even easier:

EV(small) = 0.20(1900)+(0.70+0.09+0.01)(2400)= 380+1920= 2300

We have shown these calculations in detail because the material is new, but fromnow on we will simply calculate the numbers and write only the final answer.Filling in the numbers in the Expected Value column we have:

Demand for TicketsTheatre 3-Night Fringe Average Great Heavy ExpectedSize Capacity Rent 250 800 2300 4500 ValueSmall 300 $600 1900 2400 2400 2400 $2300Medium 1200 $1800 700 6200 10,200 10,200 $5500Large 3600 $5500 −3000 2500 17,500 30,500 $3030

Probability 0.20 0.70 0.09 0.01


Now let us see how to do this using a spreadsheet. In cell H4, we wish to writea formula which will find the “dot product” of the probabilities in D7:G7 with thepayoffs in D4:G4. One way to do this (ignoring absolute cell addresses for themoment) is:

=D7*D4+E7*E4+F7*F4+G7*G4

Because we only have four outcomes, we could do it this way. However, thisapproach would be very cumbersome if we had say twenty outcomes. Therefore,we will instead use the spreadsheet SUMPRODUCT function.

The SUMPRODUCT function finds the dot product of the numbers in range1with the numbers in range2, where both ranges are rows (or columns) of equalsize. The syntax is SUMPRODUCT(range1,range2). We must put absolutecell addresses on row 7 (the probabilities), hence the formula to be placed in cellH4 is:

=SUMPRODUCT(D$7:G$7,D4:G4)

This formula is copied into cells H5 and H6. If we format the range H4:H6 ascurrency, we will obtain:

A B C D E F G H1 Demand for Tickets2 Theatre # of Fringe Average Great Heavy Expected3 Size Seats Rent 250 800 2300 4500 Value4 Small 100 $600 1900 2400 2400 2400 $2300.005 Medium 400 $1800 700 6200 10200 10200 $5500.006 Large 1200 $5500 −3000 2500 17500 30500 $3030.007 Probability 0.20 0.70 0.09 0.01

The formatting of the numbers is a matter of individual preference. For example,any of 2300, $2300, or $2300.00 could be used.

On average, the best alternative is the one with the highest expected value. Inthe next section, we shall look at alternate decision criteria, but in the absenceof reason to the contrary the preferred criterion for decision making under uncer-tainty will be to choose the alternative with the highest expected value.


2.2.4 Recommendation

For the example at hand, the best alternative is clearly to rent a medium-sizedtheatre, with an expected payoff of $5500. As we said in the introductory section,the developer of the model must make the recommendation clear to the customerof the model. In this example, the customer is the theatre company. They mightnot be familiar with payoff matrices or spreadsheets, so we focus on giving therecommendation - the spreadsheet itself is just an appendix. For the sake of thiscourse, let’s say that the term “expected payoff” can be used; in real life moreexplanation would be required. Hence within this course we would write therecommendation as:

Recommendation

Rent a medium-sized theatre, with an expected payoff of $5500.

In giving a recommendation to the theatre company in real-life, somethingalong the following lines might be appropriate:

To: The Management Committee, Amateur Theatre GroupFrom: J. Blow, MS Consulting CompanySubject: Theatre Rental

Thankyou for this opportunity to assist your theatre company, which I amhappy to provide on a pro-bono basis. After studying the three alternatives, Iconclude that renting a medium-sized theatre would be best. Based on the as-sumptions which you provided, the profit (before deducting fixed expenses suchas costumes and makeup) will be either $700, $6200, or $10,200; if a situationlike this were to be repeated many times the profit would average out to $5500.After deducting the $2500 in fixed expenses the company will be left with a profit(loss) of ($1800), $3700, or $7700; if a situation like this were to be repeatedmany times the profit would average out to $3000. A spreadsheet which I used tomake the gross profit calculations appears as an appendix to this memo.

J. BlowAnalyst


A B C D E F G H1 Demand for Tickets2 Theatre # of Fringe Average Great Heavy Expected3 Size Seats Rent 250 800 2300 4500 Value4 Small 100 $600 1900 2400 2400 2400 $2300.005 Medium 400 $1800 700 6200 10200 10200 $5500.006 Large 1200 $5500 −3000 2500 17500 30500 $3030.007 Probability 0.20 0.70 0.09 0.01

2.3 ExerciseA computer retailer is about to order some computers from the manufacturer. Overthe next two months the retailer believes that there is a 20% chance of demand forten computers, a 30% chance of demand for eleven, a 20% chance of demandfor twelve, a 20% chance for thirteen, and finally a 10% chance that fourteencomputers will be demanded. On a per-unit basis, the wholesale cost is $1500,and the retail price is $1950. [Assume that any computers leftover after the twomonth period become worthless.]

(a) Solve this problem by hand by using a payoff matrix. Rather than develop aformula, it is easier to first find out what happens if 10 computers are ordered, andthen 10 are demanded. Then, calculate the payoff when 11 are ordered, and 11 aredemanded, and then continue to find the rest of the payoffs on the main diagonal.Next, write the payoffs in the top right-hand triangle (very easy). Finally, writethe payoffs in the bottom left-hand triangle. Begin this latter step by determiningwhat happens if 11 computers are ordered, but only 10 are demanded. By howmuch worse should this be than the (10,10) situation? By how much worse shouldthis be than the (11,11) situation? Both answers should lead to the same payoff forthe (11,10) case. Continue in this manner to find all the payoffs. Then, computethe expected values and determine the best policy.

(b) Solve this problem using a spreadsheet. This will require the developmentof a formula for the (10,10) situation, which is then copied for all situations. Cre-ate a formula for the order 10 alternative, and copy this formula to all alternatives.[Save the spreadsheet file for later use.]

Obviously the recommendation which follows from (a) or (b) must be thesame. State this recommendation clearly.


3 Uncertainty 2

3.1 Salvage ValueIn this section we consider an extension to the basic model of decision makingunder uncertainty when there is one decision and one event. First, we introducethe concept of a salvage value , which is the remaining value of something whichhas not sold at the regular price. It could also be called a clearance price . Itis often used when a company needs to clear inventory quickly; here are someexamples:

1. A newspaper has a regular price of 75 cents. The next morning, the left-overcopies are sold to a paper recycling operation for 5 cents each.

2. A winter coat is priced at $300. If it’s not sold by the end of March, it’spriced to clear at $160.

3. A hardcover book lists for $39.95. Some people buy it at this price, butwhen sales drop to nothing, the book is priced to clear at $9.99.

Sometimes items for sale pass through multiple price levels. For example, avideotape of a recent release may be priced as high as $34.99, but then the price isprogressively lowered to $19.99, then $12.99, and finally the product is priced toclear at $5.00. However, we will not make models with more than two price levels,for this only makes the problem complex. Also, unless stated to the contrary, wewill assume that all the inventory which remains after trying to sell the productat the regular price can in fact be sold at the salvage value. Another assumptionis that the existence of a clearance price does not affect the regular sales. Thesolution to the model depends on the assumptions made – if the assumptions areunrealistic, then so too will be the “solution”.

3.1.1 Theatre Example with Salvage Value

Suppose that fifteen minutes before showtime, the theatre company decides toprice all unsold seats at $2.40 each, netting a revenue of $2 each after taxes. Asign is placed outside the theatre announcing the price reduction, and hopefullybargain-hunters and passers-by who see the sign will pay the reduced price to seethe play. We will begin by investigating what happens if we make the followingassumptions:


1. All seats not sold at the regular price will sell-out at the reduced price.

2. The demand at the regular price is not affected by the existence of thecheaper tickets.

Because we have already solved the problem without the salvage revenue,all we need do is find what the salvage revenue will be in each of the twelvesituations (3 alternatives; 4 outcomes) and add it to the previously found payoffin that situation. Clearly, the sell-out situations are unchanged. These are: smalltheatre with average, great, or heavy demand; a medium-sized theatre with greator heavy demand; and a large theatre with heavy demand. For the non-selloutsituations, the salvage revenue is:

$2 × (three-night capacity − the demand for tickets at the regular price)

Using this formula we obtain:

Fringe Average Great Heavy250 800 2300 4500

Small 300 2(300−250) – – –= 100

Medium 1200 2(1200−250) 2(1200−800) – –= 1900 = 800

Large 3600 2(3600−250) 2(3600−800) 2(3600−2300) –= 6700 = 5600 = 2600

Before proceeding further we should question whether these results seem reason-able. The extreme situation is when a large theatre has been rented, but the playonly attracts fringe interest. According to the above model, 250 people pay theregular price, and then ten minutes before showtime 3350 people (spread overthree nights) arrive to fill the theatre. This is clearly not reasonable. First of all,not that many people would walk by the theatre to obtain tickets, especially a playwhich has been panned by the critics. A new assumption about demand is there-fore required. One possibility would be to limit demand to say 300 tickets (100per night) at the discount price. Doing this would limit the salvage revenue to amaximum of $2(300) = $600. With this assumption the table becomes:

Fringe Average Great Heavy250 800 2300 4500

Small 300 100 – – –Medium 1200 600 600 – –Large 3600 600 600 600 –


If we now believe that this table seems reasonable we can proceed to the nextstep, which is to add these payoffs to those obtained before. Doing this, and thenfinding the new expected values, we obtain:


Probability 0.20 0.70 0.09 0.01

As an aside, we note that there are two ways to find the new expected values.Using the alternative of renting a large theatre to illustrate, one way is to calculate:

0.20(−2400)+0.70(3100)+0.09(18,100)+0.01(30,500) = 3624

The other way is to note that the previous EV in this row was 3030. We added 600to each of the first three columns, therefore the new EV is:

3030+(0.20+0.70+0.09)600 = 3624

With the assumption that discount tickets sales are limited to 300, we see thatwhile each of the three EVs changes, the optimal alternative remains the same,i.e. rent a medium-sized theatre.

To do these calculations on a spreadsheet, we modify what we did earlier(page 16). The formula in cell D4 is currently:

=10*MIN(3*$B4,D$3)-$C4

The number of unsold seats is either 0 or 3*$B4-D$3, whichever is greater. Thisis represented as MAX(0,3*$B4-D$3). By our assumption that we cannot sellmore than 300 discount tickets, the number of discount tickets sold is either 300,or MAX(0,3*$B4-D$3), whichever is fewer. Hence the number of discounttickets sold is

MIN(300,MAX(0,3*$B4-D$3))

They net $2 for each ticket, and hence the salvage revenue is

2*MIN(300,MAX(0,3*$B4-D$3))


Adding this revenue the formula in cell D4 becomes:

=10*MIN(3*$B4,D$3)+2*MIN(300,MAX(0,3*$B4-D$3))-$C4

There may be more than one way to correctly write a formula. For example, ifthe demand equals or exceeds the number of seats, then the revenue is the numberof seats multiplied by $10, otherwise the revenue is the demand multiplied by $10,plus $2 for each seat not sold at the regular price up to a maximum demand of 300at the lower price. This logic is captured in the following IF statement for cell D4,from which the rent is subtracted:

=IF(D$3>=3*$B4,10*3*$B4,10*D$3+2*MIN(300,3*$B4-D$3))-$C4

This alternate formula is no shorter, but it may be easier to understand. Eitherformula is entered into cell D4 and is then copied into the range D4:G6. Doingthis we obtain:

A B C D E F G H1 Demand for Tickets2 Theatre # of Fringe Average Great Heavy Expected3 Size Seats Rent 250 800 2300 4500 Value4 Small 100 $600 2000 2400 2400 2400 $2320.005 Medium 400 $1800 1300 6800 10200 10200 $6040.006 Large 1200 $5500 −2400 3100 18100 30500 $3624.007 Probability 0.20 0.70 0.09 0.018 Salvage Model with up to 300 discount-priced last-minute tickets

Note that a few words have been added on the spreadsheet to make it clear thatwe are looking at variation of the basic model in which up to 300 discount-pricedlast-minute tickets may be sold.

When a model has been made using a spreadsheet, it is easy to see what hap-pens if one or more of the assumptions of the model is changed. For example,suppose that we believed that up to 1800 (rather than just 300) discount-pricedlast-minute tickets could be sold. All we need do is replace the “300” in the for-mulas in cell D4 with “1800”, and then copy this new formula to the range D4:G6.The modified formula in cell D4 is:

=10*MIN(3*$B4,D$3)+2*MIN(1800,MAX(0,3*$B4-D$3))-$C4


Of course, a better spreadsheet design would be to place this number in its owncell, and have the formula in D4 refer to the cell that contains the number. Withthe change from “300” to “1800” everywhere in D4:G6 we obtain:

A B C D E F G H1 Demand for Tickets2 Theatre # of Fringe Average Great Heavy Expected3 Size Seats Rent 250 800 2300 4500 Value4 Small 100 $600 2000 2400 2400 2400 $2320.005 Medium 400 $1800 2600 7000 10200 10200 $6440.006 Large 1200 $5500 600 6100 18100 20100 $6504.007 Probability 0.20 0.70 0.09 0.018 Salvage Model with up to 1800 discount-priced last-minute tickets

Based on this assumption, the recommendation for the theatre rental wouldchange from a medium-sized to a large theatre. A decision analysis model willalways be mathematically easy to solve, but whether or not we have solved the realproblem (where to hold the play) depends heavily on the assumptions on whichthe model is based.

3.2 Expected Value of Perfect InformationSuppose that in situations of decision making under uncertainty, it might be possi-ble to obtain perfect information about the uncertain event. For example, supposethat tomorrow’s weather will be either sunny, cloudy, or rainy. Perfect informationabout this event would imply that today’s forecast for tomorrow is certain to becorrect. Of course, a perfect weather forecast is not possible, but the hypotheticalconstruct of perfect information is useful because it establishes an upper boundfor the expected value of any information about the event. For example, if a per-son would pay $5.00 to hear a perfect weather forecast, then a real forecast can beworth no more than $5.00.

We are interested in determining the expected value of perfect information(EVPI). We now show how to calculate the EVPI, using the theatre problem asan example. In this case, the uncertainty is the level of demand. Having perfectinformation means that we are told the demand level before having to committo one of the three theatres. With perfect information we can choose the bestalternative with respect to the level of demand. For any level of demand, we areinterested in the highest payoff (i.e. the highest payoff in the column). We recall


the payoff matrix for the basic model (i.e. no salvage value), and on this wehighlight the best payoff in each column:


Probability 0.20 0.70 0.09 0.01

If we are told that the demand will be “fringe”, then we will choose a small theatrefor a payoff of $1900 (the highest payoff in the “fringe” column). If we are toldthat the demand will be “average”, then we will choose a medium-sized theatre fora payoff of $6200. If we are told that the demand will be “great” or “heavy”, thenwe will choose a large theatre, with a payoff of $17,500 for “great” and $30,500for “heavy”.

There are now two ways to complete the calculation of the EVPI.

3.2.1 Direct Calculation of the EVPI

The new information only has value if it would change the recommendation thatwe had before. Before receiving the perfect information, we would have recom-mended renting a medium-sized theatre. If the perfect information is that demandwill be “average”, then we will still make the same recommendation. However, inthe other three outcomes of demand, we will change the recommendation, therebyincreasing the payoff over what it would have been. If the perfect information isthat demand will be “fringe”, then we would change the recommendation frommedium to small, thereby increasing the payoff from 700 to 1900. If the perfectinformation is that demand will be “great”, then we would change the recom-mendation from medium to large, thereby increasing the payoff from 10,200 to17,500. If the perfect information is that demand will be “heavy”, then we wouldchange the recommendation from medium to large, thereby increasing the payofffrom 10,200 to 30,500. The probabilities of the perfect information being thatthese outcomes will occur are 0.20 for “fringe”, 0.09 for “great”, and 0.01 for“heavy”. Hence there is a 20% chance of increasing the payoff from 700 to 1900,a 9% chance of increasing the payoff from 10,200 to 17,500, and a 1% chance of


increasing the payoff from 10,200 to 30,500. The EVPI is therefore:

EVPI = 0.20(1900−700)+0+0.09(17,500−10,200)+0.01(30,500−10,200)= 0.20(1200)+0+0.09(7300)+0.01(20,300)= 240+0+657+203= 1100

The expected value of perfect information in the theatre example is $1100. [Note:We could have written the second term of the first line of the EVPI calculation as0.70(6200−6200), which of course is 0, or we could have omitted it altogether.]

3.2.2 Indirect Calculation of the EVPI

To indirectly calculate the EVPI, we first find the expected value with perfectinformation. To avoid confusion with the EVPI, the short form is EV with PI .The EV with PI is found by calculating the expected payoff based on the bestalternative for each outcome. This is done by calculating the expected payoffusing the highest payoff in each column.

EV with PI = 0.20(1900)+0.70(6200)+0.09(17,500)+0.01(30,500)= 380+4340+1575+305= 6600

The EV with PI is $6600. If we did not have the perfect information, we wouldhave chosen the medium-sized theatre alternative, which has an expected payoffof $5500. The EVPI is the expected amount of the profit increase from not havingperfect information to having it. The EVPI is therefore:

EVPI = EV with PI−EV without PI= 6600−5500= 1100

As before, the EVPI is $1100. You are expected to know how to calculate theEVPI using both of these methods.

For the theatre example, the $1100 establishes an upper bound to the value ofany information concerning the demand. If more information were available, themost that they would pay for it would be $1100. If the price were say $500, then itmight be worthwhile purchasing it; it would depend on how good the information


is. However, if the price were $2000, it would not be worth purchasing no matterhow good it is. While a small theatre company would not try to obtain moreinformation about the demand, a company with millions of dollars at risk probablywould.

3.3 Decision Criteria

Up to this point our sole decision criterion has been Expected Value. For a profitmaximization example, we would choose the alternative with the highest expectedvalue. For a cost minimization example (in which all the payoffs are costs) wewould choose the alternative with the lowest expected value. This will remainour preferred decision criterion, but there are other criteria as well, and they arereviewed here. They are illustrated using the theatre example:

Demand for TicketsTheatre 3-Night Fringe Average Great HeavySize Capacity Rent 250 800 2300 4500Small 300 $600 1900 2400 2400 2400Medium 1200 $1800 700 6200 10,200 10,200Large 3600 $5500 −3000 2500 17,500 30,500

Probability 0.20 0.70 0.09 0.01

3.3.1 Pessimism

If a small theatre is chosen, the payoff will be either $1900 or $2400. Hence, thepayoff will be at least $1900. If a medium-sized theatre is chosen, the payoff willbe either $700, or $6200, or $10,200, hence the payoff will be at least $700. Fromthe four outcomes in the “Large” alternative row, we see that the payoff will be atleast −$3000. Of these three minimum payoffs, $1900, $700, and −$3000, thehighest is the $1900 payoff. A pessimist would choose the alternative associatedwith this payoff, i.e. the small theatre.

A pessimist would recommend that the small theatre be rented.

For any maximization problem, the alternative associated with pessimism isthe one which contains the maximum of the row minimums. [Note: In this ex-ample, all the rows minimums were in the same column, but this will not be true


in general.] For any minimization problem, the alternative associated with pes-simism is the one which contains the minimum of the row maximums. Pessimismis an extreme form of risk-aversion which ignores all the information about prob-abilities. The subject of risk-aversion will be studied in more detail later in thecourse.

3.3.2 Optimism

An optimist seeks the maximum for each alternative, and then seeks the maxi-mum of the maximums. For the theatre example, the row maximums for Small,Medium, and Large are $2400, $10,200, and $30,500 respectively. The maximumof these three is $30,500, and the alternative associated with this payoff is to renta large theatre.

An optimist would recommend that the large theatre be rented.

Like pessimism, optimism ignore the information about probabilities. Whenoptimism is applied to a cost minimization problem, we find the minimum of therow minimums.

3.3.3 Hurwicz

The Hurwicz criterion is a mixture of the criteria of Pessimism and Optimism.Either a coefficient of Pessimism (CoP) or a coefficient of Optimism (CoO) (oneis the complement of the other) is chosen, and then (for maximization) a weightedaverage of the row minimums and maximums is found; the alternative with thehighest weighted average is then chosen.

For the purposes of this course the CoP or CoO will be an exogenously givennumber. For example, suppose we wish to solve the theatre problem with anexogenously given coefficient of pessimism of 0.85. Hence, the coefficient ofoptimism is 1−0.85 = 0.15, and we have:

Pess. Opt. HurwiczSmall 1900 2400 1975Medium 700 10,200 2125Large −3000 30,500 2025

0.85 0.15


The highest weighted average in the Hurwicz column is 2125; a medium-sizedtheatre is recommended.

Based on Hurwicz with CoP = 0.85, a medium-sized theatre is recommended.

For a cost minimization problem, the pessimism column is based on row max-imums, the optimism column is based on row minimums, and the chosen alterna-tive is based on the lowest number in the Hurwicz column.

3.3.4 Laplace

The Laplace and Expected Value criteria are similar, except that for the Laplaceequal probabilities are used. With n outcomes, the probability that any one ofthem occurs is 1/n. The ranking for the Laplace criterion is conveniently found bysumming, for every alternative, all n payoffs, and then dividing by n. We choosethe highest ranking for maximization, and the lowest ranking for minimization.

For the theatre example with its four outcomes we have:

Small (1900 + 3(2400))/4 = 2275Medium (700 + 6200 + 2(10,200))/4 = 6825Large (−3000 + 2500 + 17,500 + 30,500)/4 = 11,875

The highest of these is 11,875, which is associated with the large theatre.

Based on the Laplace criterion the large theatre is recommended.

The calculations for the four criteria of Pessimism, Optimism, Hurwicz, andLaplace can all be done on one payoff matrix. The best payoff for each criterion ishighlighted; from these payoffs the best alternative for each criterion can be seen.

Theatre Demand for TicketsSize Fringe Average Great Heavy Pess. Opt. Hurwicz LaplaceSmall 1900 2400 2400 2400 1900 2400 1975 2275Medium 700 6200 10,200 10,200 700 10,200 2125 6825Large −3000 2500 17,500 30,500 −3000 30,500 2025 11,875

0.85 0.15


3.3.5 The Regret Matrix

The regret matrix gives the cost of having not chosen, with hindsight, the bestalternative for a given outcome. For example, in the theatre problem if the demandturns out to be “average”, then the payoff is $2400 for small, $6200 for medium,and $2500 for large. With hindsight, renting a medium-sized theatre is best forthis particular outcome. In this case, there is no foregone profit. However, if thesmall theatre alternative were chosen, the foregone profit would be

$6200−$2400 = $3800

and if the large theatre alternative were chosen, the foregone profit would be

$6200−$2500 = $3700

The foregone profit is also called the opportunity loss . To find the opportunityloss for each situation from an already existing payoff matrix, we work with onecolumn at a time. In every column of the payoff matrix, we subtract each numberin the column from the largest number in that column. It is also possible to obtainthe regret matrix without first finding the payoff matrix, and indeed one reasonfor obtaining the regret matrix is that it is sometimes easier to calculate than thepayoff matrix. No matter how it is obtained, one property that the regret matrixwill always have is that there must be at least one zero in every column .

In a payoff matrix we found the expected value for every alternative; in aregret matrix we find the expected opportunity loss (EOL) for every alternative.It is found in an analogous manner to the EV – by finding the dot product of theprobability row with every payoff row. The objective is to minimize the expectedopportunity loss, so the best alternative is the one with the lowest number in theEOL column. The regret matrix along with EOL column for the theatre exampleis:

Regret Matrix Demand for TicketsTheatre 3-Night Fringe Average Great HeavySize Capacity Rent 250 800 2300 4500 EOLSmall 300 $600 0 3800 15,100 28,100 $4300Medium 1200 $1800 1200 0 7,300 20,300 $1100Large 3600 $5500 4900 3700 0 0 $3570

Probability 0.20 0.70 0.09 0.01

Therefore, the recommendation is:


Rent a medium-sized theatre, with an expected opportunity loss of $1100.

Some interesting comparisons can be made with the solution obtained by usinga payoff matrix:

1. We obtained the same alternative using the regret matrix as we did whenusing the payoff matrix. This is not a coincidence – whether we max-imize the expected value or minimize the expected opportunity loss, wealways obtain the same alternative . Hence minimizing EOL, unlike pes-

simism, optimism, Hurwicz, and Laplace, is not a new decision criterion,but instead is just a variation on the maximizing expected value approach.

2. The minimum EOL, which is $1100, equals the EVPI. Again, this is nota coincidence – it is always true. While this gives us a third method forfinding the EVPI, usually one does not not take this approach to finding it ifthe payoff matrix has already been found.

3. For every alternative, the sum of the expected value and the expected op-portunity loss is the same. Moreover, this sum is the EV with PI.

EV + EOL = EV with PISmall 2300 + 4300 = 6600Medium 5500 + 1100 = 6600Large 3030 + 3570 = 6600

This property is true for all examples.

3.4 Exercise(a) For the Exercise at the end of Uncertainty 1 (Exercise 2), determine the EVPIboth directly and indirectly.

For parts (b) to (f), the data is the same as in part (a), except that now anyleftover computers are sold for $1400 each.

(b) Re-solve Exercise 2 by hand, this time using the $1400 salvage value.

(c) Using a spreadsheet, re-solve Exercise 2, now using a $1400 salvage value.[To save work, begin by copying the spreadsheet you made before. Obviously, (b)and (c) should produce the same result.]


(d) Determine the EVPI both directly and indirectly.

(e) Which alternative would be picked using: (i) Pessimism; (ii) Optimism;(iii) Hurwicz with a coefficient of optimism of 0.8; (iv) Laplace?

(f) By hand, find the regret matrix, and determine the alternative with theminimum EOL. Try to find the regret matrix just from the data of the problem,without looking at the payoff matrix from part (b).


4 Uncertainty 3

4.1 Marginal AnalysisSome problems involving one decision and one event can be solved by a methodthat requires less work than is required for making a payoff matrix. This newmethod is called marginal analysis . It is applicable in problems such as the com-puter example (with or without salvage) in which there is a cost per unit ordered(w), a price at which items are sold (r) where the demand has any kind of discretedistribution (P(d)), and a price per unit at which all leftover items are sold (s).The marginal analysis method is not applicable for irregular problems such as thetheatre example.

Let x (an integer) represent the optimal order quantity. The parameters of themodel are:

Symbol Meaningr r etail prices s alvage pricew w holesale price

The context requires that the retail price be greater than the wholesale price, forotherwise the business could not exist. Also, the salvage value must be less thanthe wholesale price, for otherwise any amount of stock could be ordered at no riskto the retailer. Putting these observations into symbolic terms we have:

r > w > s

The distribution P(d) gives the probability that the demand at price r is for exactlyd units. We let F(d) represent the cumulative probability function, which is theprobability that d or fewer units are demanded. We can write F(d) in terms ofP(d) as follows:

F(0) = P(0)F(1) = P(0)+P(1)F(2) = P(0)+P(1)+P(2)F(3) = P(0)+P(1)+P(2)+P(3)

and so on. Also, for d ≥ 1, we can use the recursive formula:

F(d) = F(d−1)+P(d) (d ≥ 1)


The optimal order quantity is given by the marginal analysis formula . It is notproved here, because the proof is somewhat advanced for an introductory course.[You will not be tested on the proof – all that is required is that you know how touse the formula.] The value of x (the optimal order quantity) is chosen such that:

F(x−1) <r−wr− s

≤ F(x) (2)

Another way of saying this is that we want the smallest value of x such that:

r−wr− s

≤ F(x)

4.1.1 Computer Example

In the computer example, the retail price is $1950, the wholesale price is $1500,and the salvage value is $1400. Therefore,

r−wr− s

=1950−15001950−1400

=450550

≈ 0.818

The chance that the demand is for exactly 10, 11, 12, 13, and 14 units is 20%, 30%,20%, 20%, and 10% respectively. Hence P(10) = 0.2, P(11) = 0.3, P(12) = 0.2,P(13) = 0.2, and P(14) = 0.1. All P(d) from d = 0 to d = 9 inclusive are 0, henceF(d) = 0 from d = 0 to d = 9 inclusive. Hence

F(10) = F(9)+P(10)= 0+0.2= 0.2

We can make a table to find F(d), in which d goes from 10 to 14 inclusive, P(d)comes from the given probabilities, and F(d) is found recursively (for examplethe two numbers highlighted in blue are summed to find the number highlightedin red):

d 10 11 12 13 14P(d) 0.2 0.3 0.2 0.2 0.1F(d) 0.2 0.5 0.7 0.9 1.0


The critical value is about 0.818; the number just above this in the F(d) line is 0.9(highlighted in yellow), and this is in the d = 13 column. In terms of the formulawe have:

F(13−1) = 0.7 < 0.818≤ 0.9 = F(13)

The recommendation is to order 13 computers.

In this case, the critical fraction 0.818 is closer to 0.9 than it is to 0.7, but thisfact is irrelevant . Even if the critical fraction were say 0.703, we would order 13(and not 12) computers.

Note that while the formula gives us the best order quantity, it does not alsogive us the expected payoff associated with this quantity. If we want this too, wecan easily find it by solving for only the optimal row in the payoff matrix.

4.2 Sensitivity AnalysisThe subject of sensitivity analysis (also called what-if analysis ) is a recurringtheme throughout the field of management science. The whole point to buildinga model is that it is much cheaper than building what the model represents. Wecan play around with the model quite inexpensively, and one of the things that weshould do is see how sensitive it is to changes in the built-in assumptions of themodel.

The Greek symbol ∆ (pronounced delta) is often used to represent a change tosomething. We might use ∆p (read as “delta p”) to represent a change in probabil-ity, or ∆c to represent a change in cost; where the context is clear, we can simplyuse ∆. Usually, ∆ can be either positive or negative. To keeps things simple, weoften just vary one parameter at a time, but changing probabilities is an impor-tant exception. If one probability is increased, then at least one other probabilitymust be decreased (by the same absolute amount) so that the probabilities remainsummed to one. Also, in this situation, we must establish a domain for ∆p basedon the fact that no probability can go below 0 or above 1.

4.2.1 Theatre Example

Included in the original parameters are that the probability of fringe interest is 0.2,and the probability of average interest is 0.7. Suppose that we now wish to seewhat happens if we vary these two probabilities, with everything else remaining


constant. Suppose that the first is increased by ∆p, and the other is decreased by∆p. (Doing it the other way around would be fine; everything will work out in theend.) Hence we have:

p(fringe) = 0.2+∆pp(average) = 0.7−∆p

We must ensure that neither probability goes below 0. If we do this, we willautomatically ensure that neither probability goes above 1. The condition that0.2+∆p≥ 0 will be true provided that ∆p≥−0.2. The condition that 0.7−∆p≥0 will be true provided that ∆p≤ 0.7. Hence, the domain of ∆p is:

−0.2≤ ∆p≤ 0.7

[Note: In general if the two probabilities are a + ∆p and b−∆p, then we musthave −a≤ ∆p≤ b.]

When we first solved the problem we obtained:


Probability 0.20 0.70 0.09 0.01

Now, in the probability row, the 0.20 becomes 0.20 + ∆p, and the 0.70 becomes0.70−∆p, and we wish to determine the revised Expected Values.

Demand for TicketsTheatre 3-Night Fringe Average Great Heavy ExpectedSize Capacity Rent 250 800 2300 4500 ValueSmall 300 $600 1900 2400 2400 2400Medium 1200 $1800 700 6200 10,200 10,200Large 3600 $5500 −3000 2500 17,500 30,500

Probability 0.20 0.70 0.09 0.01+∆p −∆p


The long way to do this, using the “Small” alternative to illustrate, is to re-computethe entire dot product (“Small” row and the Probability row).

EV(small) = (0.2+∆p)1900+(0.7−∆p)2400+ .09(2400)+ .01(2400)= 0.2(1900)+1900∆p+0.7(2400)−2400∆p+ .09(2400)+ .01(2400)= 0.2(1900)+(0.7+0.09+0.01)2400+(1900−2400)∆p= 2300−500∆p

The short way to do this is to recognize that the “2300” has been computed already– all we need to do is include the terms involving “∆p”. [If ∆p = 0, we must obtainthe original result.] All we need are the columns which contain the ∆p’s and theoriginal Expected Values.

Expected ValueFringe Average Original ∆p Term

Small 1900 2400 2300Medium 700 6200 5500

Large −3000 2500 3030∆p −∆p

The short way is simply:

EV(small) = 2300+1900∆p+2400(−∆p)= 2300+1900∆p−2400∆p= 2300−500∆p

For the medium and large theatre alternatives we have:

EV(medium) = 5500+700∆p+6200(−∆p)= 5500−5500∆p

EV(large) = 3030+(−3000)∆p+2500(−∆p)= 3030−5500∆p

The completed table is:


Expected ValueFringe Average Original ∆p Term

Small 1900 2400 2300 −500∆pMedium 700 6200 5500 −5500∆p

Large −3000 2500 3030 −5500∆p∆p −∆p

Comparing Medium with Large, we see that for any value of ∆p,

5500−5500∆p > 3030−5500∆p

and hence Medium is better than Large. These lines are parallel (because of the−5500) and therefore they never intercept.

If we compare Small with Medium, we have EV(Small) = 2300−500∆p ver-sus EV(Medium) = 5500−5500∆p. We are indifferent between two alternativeswhen neither is preferred to the other. To find the point of indifference, we set thetwo payoffs equal to each other:

EV(Small) = EV(Medium)2300−500∆p = 5500−5500∆p

5000∆p = 3200∆p = 0.64

This value is within the domain −0.2≤ ∆p≤ 0.7. [Were it not so, there would beno point of indifference.]

We know that Medium is preferred at ∆p = 0 (the current situation), and wehave found that we would switch to Small at ∆p = 0.64. Since these are the onlyalternatives (because Large was eliminated), Medium must be best for any ∆p inthe domain < 0.64, there’s a tie at 0.64, and Small is best for all other values. Byletting both alternatives be considered “best” at the tie, we can state the regions ofpreference as:

−0.2≤ ∆p≤ 0.64 Medium0.64≤ ∆p≤ 0.70 Small

We can also show this information on a number line for ∆p (where −0.2 ≤∆p≤ 0.7), highlighting with colour the regions for the recommended theatre size.


−.2 0 .2 .4 .6 60.64

∆p

SmallMedium

∆p = 0.64 is a very large change for a probability. If we believe that the initialestimate of 0.2 couldn’t be off the true value by all that much, then we would bequite confident that our initial choice of Medium is correct.

The point of indifference can also be expressed in terms of the original proba-bilities. These are:

p(fringe) = 0.2+0.64= 0.84

p(average) = 0.7−0.64= 0.06

4.2.2 A More Complicated Example

The preceding example was fairly easy in that we were able to reduce it down totwo alternatives, and so we only had to find a single point of indifference. Usually,however, we cannot eliminate any alternative simply by inspection. When thishappens, a conceptually easy approach is to make a graph of EV versus ∆p. Doingit first this way gives us a shorter analytical method for this type of problem. Theexample presented here provides an illustration of these concepts.

Consider an example with four alternatives and three outcomes, for which webegin with all payoffs having been found, and the expected values having beencalculated:

O1 O2 O3 EVA1 7 5 4 5.1A2 5 5 6 5.3A3 4 6 3 4.7A4 6 4 6 5.0

Prob. .2 .5 .3


Hence the recommendation is to choose alternative A2, with an expected pay-off of 5.3. Now suppose that we wish to vary the probabilities for O2 and O3. Wewill let the probability of O2 be 0.5 + ∆p, and the probability of O3 be 0.3−∆p.The domain for ∆p is therefore:

−0.5≤ ∆p≤ 0.3

The new expected values are:

O2 O3 EVA1 5 4 5.1+∆pA2 5 6 5.3−∆pA3 6 3 4.7+3∆pA4 4 6 5.0−2∆p

∆p −∆p

Unlike the previous example, it is difficult to remove an alternative simply byinspection; one approach is to draw a graph.

Graphical Solution Each of the EV equations is a straight line. For each ofthese, all we need do is find the EV for any ∆p 6= 0, and this along with theoriginal EV (i.e. at ∆p = 0) gives the two distinct points needed to define the line.However, with a little bit of extra work we can obtain each line with a check onthe calculations. To do this we find the EV at the lower limit for ∆p, and thenfind the EV at the upper limit for ∆p. These two points are used to define the line,and the check on the calculations comes from making sure that the line passesthrough the original EV at ∆p = 0. In this example, the lower and upper limitsfor ∆p are at −0.5 and 0.3 respectively. For alternative 1, the EV ranges from5.1 + (−0.5) = 4.6 to 5.1 + 0.3 = 5.4. For alternative 2, the EV ranges from5.3− (−0.5) = 5.8 to 5.3− 0.3 = 5.0. For alternative 3, the EV ranges from4.7+3(−0.5) = 3.2 to 4.7+3(0.3) = 5.6. Finally, for alternative 4 the EV rangesfrom 5−2(−0.5) = 6.0 to 5−2(0.3) = 4.4. In summary we have:

EV at∆p =−0.5 ∆p = 0 ∆p = 0.3

A1 4.6 5.1 5.4A2 5.8 5.3 5.0A3 3.2 4.7 5.6A4 6.0 5.0 4.4


The horizontal axis goes from ∆p = −0.5 to ∆p = 0.3. Since the smallest EV is3.2, and the largest is 6.0, we can save vertical space by having the axis run onlybetween 3 and 6 (rather than starting at 0). It is helpful to draw this graph withthree vertical axes: one through the lower limit for ∆p; one through 0; and onethrough the upper limit for ∆p.

With the axes drawn, we proceed with drawing the four lines. The A1 linegoes from 4.6 on the left vertical axis to 5.4 on the right vertical axis. Next to thisline the A1 symbol is drawn. On the centre vertical axis, we see indeed that theline passes through the point (0,5.1). The other three lines are drawn with theirsymbols, each time verifying that the point on the centre vertical axis is where itshould be.

∆p

EV

−.4 −.2 0 .23.0

3.5

4.0

4.5

5.0

5.5

6.0

3.0

3.5

4.0

4.5

5.0

5.5

6.0

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A2

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A3

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A4


We want, of course, the line segments which maximize the expected value.These line segments have been highlighted on the graph. At ∆p = −0.5, A4 isbest. As we move to the right, the best alternative switches to A2, then A1, andthen A3. Hence we need to find where the following pairs of lines intercept: A4and A2; A2 and A1; and A1 and A3.

To find the value of ∆p at which the A4 and A2 lines intercept, we set the EVequations equal to each other:

EV(A4) = EV(A2)5.0−2∆p = 5.3−∆p

−∆p = 0.3∆p = −0.3

At this value for ∆p, EV(A4) = 5.0− 2(−0.3) = 5.6. [Also, EV(A2) = 5.3−(−0.3) = 5.6.] Hence the two lines intercept at (−0.3,5.6).

We then find the other two interception points:

EV(A2) = EV(A1)5.3−∆p = 5.1+∆p−2∆p = −0.2

∆p = 0.1

At this value for ∆p, EV(A2) = 5.3− (0.1) = 5.2.

EV(A1) = EV(A3)5.1+∆p = 4.7+3∆p−2∆p = −0.4

∆p = 0.2

At this value for ∆p, EV(A1) = 5.1+(0.2) = 5.3.There are three ways that we could report these values. As we did in the

previous example, we could draw a number line, highlighting the regions where aparticular alternative is best. Secondly, we could indicate this information on thegraph, which of course gives the absolute rather than just the relative ranking ofeach alternative. Thirdly, we could simply report the regions as follows:


Region for ∆p Best Alternative−0.5≤ ∆p≤−0.3 A4−0.3≤ ∆p≤ 0.1 A2

0.1≤ ∆p≤ 0.2 A10.2≤ ∆p≤ 0.3 A3

Sometimes, we do not need to know the best alternatives over the entire do-main of ∆p. Instead, we might only wish to determine the values for ∆p for whichthe current solution (i.e. at ∆p = 0) remains optimal. For this example, the currentsolution remains A2 provided that:

−0.3≤ ∆p≤ 0.1

Analytical Solution We can also solve such a problem quickly as follows. First,here are some general comments for any situation. Consider two alternatives, withexpected payoffs a+b∆, and c+d∆, where a > c and b 6= d (i.e. the two lines arenot parallel). (For simplicity, we are using ∆ rather than ∆p.) These alternativeshave the same expected payoff when:

c+d∆ = a+b∆

(d−b)∆ = a− c

∆ =a− cd−b

If the critical value (a−c)/(d−b) turns out to be outside of the domain of ∆, thenthe c + d∆ alternative is not best for any value of ∆. If, however, it is inside thedomain, then we must consider this alternative along with any others.

With many alternatives we find the critical value of ∆ for each one (wherethe comparison alternative is the one optimal at ∆ = 0); we seek the ones whosecritical values are immediately on either side of 0.

Now we solve the example from before, first going back to the table showingthe effects of ∆.

O2 O3 EVA1 5 4 5.1+∆

A2 5 6 5.3−∆

A3 6 3 4.7+3∆

A4 4 6 5.0−2∆

∆ −∆


At ∆ = 0, A2 is best. Hence we find where the A1, A3, and A4 lines meet the A2line:

EV(A2) = EV(A1)5.3−∆ = 5.1+∆

−2∆ = −0.2∆ = 0.1

EV(A2) = EV(A3)5.3−∆ = 4.7+3∆

−4∆ = −0.6∆ = 0.15

EV(A2) = EV(A4)5.3−∆ = 5.0−2∆

∆ = −0.3

Comparing the critical values 0.1, 0.15, and −0.3, the ones immediately oneither side of 0 are −0.3 (line A4) and 0.1 (line A1). Hence A2 remains optimalfrom −0.3 to 0.1. Below −0.3, A4 is best, and just above 0.1, A1 is best. Furtheron, the A1 and A3 lines will intercept:

EV(A1) = EV(A3)5.1+∆ = 4.7+3∆

−2∆ = −0.4∆ = 0.2

Over the entire domain of ∆ we have:

Region for ∆ Best Alternative−0.5≤ ∆≤−0.3 A4−0.3≤ ∆≤ 0.1 A2

0.1≤ ∆≤ 0.2 A10.2≤ ∆≤ 0.3 A3


4.3 Exercise4.3.1 Marginal Analysis Problem

A vendor has found that demand for newspapers can vary from 31 to 70 inclusivewith each number being equally likely. Newspapers are bought by the vendor at50 cents each, and are sold for 75 cents each. Any left-over copies at the end ofthe day are sold to a recycling operation at 5 cents per copy. By using the marginalanalysis formula, determine the number of copies that the vendor should order.

4.3.2 Sensitivity Problem 1

In this problem we use the data of the computer example with salvage value,except that we now allow the probabilities of demand for 10 and demand for 14computers to vary. Make a graph of Expected Value vs. ∆p for the five alternatives,and from this determine all the regions of ∆p where one of the alternatives is betterthan the others.

4.3.3 Sensitivity Problem 2

O1 O2 O3 EVA1 8 2 4A2 9 7 3A3 70 15 −30A4 −40 60 20

Prob. .3 .1 .6

(a) Find the best alternative, using expected value as the decision criterion.

Now suppose that the probability of O1 increases by ∆, the probability of O2increases by 3∆, the probability of O3 decreases by 4∆.

(b) What is the domain of ∆?

(c) Find, by the analytical method, the regions of ∆ where each alternative is best.


5 Decision Trees 1

5.1 Introduction

In the previous three sections, we examined simple situations in which there wasonly one decision, and each alternative was followed by the same event (same out-comes, with the same probabilities). Of course, problems in real-life are not thatsimple. Even when there is only one decision, the alternatives may be followedby different events. Also, there may be multiple decisions to be made. Either ofthese complications means that a payoff matrix cannot be used. Beginning withthis section, we show how to handle more complex examples. Such problems willbe analysed by first drawing what is called a decision tree . The drawing of thetree, along with the definitions of any symbols or abbreviated forms used on thetree, constitutes the formulation of the problem. The problem is then solved byperforming a rollback procedure on the tree. Finally, the recommendation shouldbe stated clearly.

The overall procedure involves three phases:

1. The tree is drawn from left to right.

2. The rollback procedure is performed from right to left.

3. The recommendation is given, both in writing and by highlighting on thetree, by going from left to right.

In this introduction we explain the graphical symbols used in the decision treemethod. Just as real trees have branches , so do decision trees. On a decision tree,the point at which two or more branches meet is called a node . All decision treeshave at least two kinds of nodes – decision nodes and event nodes . A decisionnode is drawn as a square, and an event node is drawn as a circle. Associatedwith each type of node is a corresponding branch emanating from the right sideof the node. A decision node is followed by an alternative branch , and an eventnode is followed by an outcome branch . An alternative branch is represented bya double line, and an outcome branch by a single line.

Some textbooks only use what we have described so far, but we find that it isuseful (whenever the decision criterion is expected value) to use three more sym-bols – a cost gate, a payoff node, and a null branch. A cost gate resembles a tollgate on a highway. It is drawn as two small squares (posts) joined by a straight


decision node

��event node

��@@�� payoff node

cost gate

alternative branch

outcome branch

null branch

Figure 1: Symbols for Drawing Decision Trees (from Left to Right)

line (the gate). The cost associated with the gate is written next to it; if it’s a rev-enue rather than a cost, then the number is placed in parentheses. A payoff nodeis represented by a crossed circle, which is used to keep track of payoffs whichoccur before the ending branches of the tree. At the ending branch of the tree(on the extreme right-hand side), the payoff at that point is simply written down– a payoff node is not used. A payoff node is followed by a null branch , whichis represented as a dashed line. Cost gates can appear on both alternative andnull branches, usually representing a cost on the former, and a revenue on thelatter. As an alternative to using cost gates (with the cost or revenue next to thegate), payoff nodes, and null branches, the maker of the decision tree could keeptrack of all costs and revenues and then subtract/add them to the appropriate finalpayoffs. Indeed, when dealing with utility functions (which we shall study later)that approach must be used . (This is why some textbooks avoid these three latter

symbols.) However, whenever expected value is the decision criterion, using thesethree extra symbols makes the solution easier to find.

These symbols with their meanings are shown in Figure 1. These seven (or justfour) symbols are all that are used when formulating a problem using a decisiontree. However, when solving the tree (from right to left), we will use an eighthsymbol, which is applied at every square to every branch except that of the bestalternative at that square. This symbol consists of two short parallel lines whichare drawn at a right angle to the alternative branch:

non-recommended alternative


In addition to doing this, we can highlight the alternatives which are recom-mended, should the decision maker possibly arrive at that square. We shall see inthe next section that some squares are not part of the optimal solution, hence thereis no point in highlighting the best alternative at such a square. This highlightingis part of the third phase, in which one proceeds from left to right.

recommended as part ofthe overall solution

As has been stated, a decision tree is drawn from left to right, and is then solved(“rolled-back”) from right to left. The tree is not drawn to scale with respect totime, but the relative position in time must be preserved. Hence if somethingappears to the right of something else, then the thing on the right must come after(or be at exactly the same time) as the thing on the left.

5.2 Theatre Problem in Tree Form

To illustrate the nature of this approach, we will begin by formulating the theatreproblem as a decision tree. [The problem description appears in Section 2.2.1.]This problem needs no payoff nodes; it can be done with or without cost gates onthe alternative branches. We will do it both ways, first without cost gates to showthe equivalence with the payoff matrix approach, and then with cost gates to showhow these are used. The basic shape of the tree is the same for both approaches,so we will start with that. It should be emphasized that in using this method onlyone tree needs to be drawn. However, to illustrate this methodology, several treesare shown for one problem so that the order in which the material is drawn ismade clear. We begin with a square on the left-hand side which represents thetheatre rental decision. Emanating from the right-hand side of the square are threealternative branches (double lines). Next to these branches is a word describingthe meaning of the alternative. These words are emphasized here in red, but ofcourse using colour is not required.


TheatreRental

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small

medium

@@@@@@@@@

@@@@@@@@@

large

At the end of each of the alternative branches, there is a circle for the demandfor tickets. Coming out of each circle there are four outcome branches (singlelines). Again, words are written to describe the meaning of each outcome. Thesewords have been emphasized in blue. Adding all this to the tree we obtain:


TheatreRental

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small

medium

@@@@@@@@@

@@@@@@@@@

large

Demand

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fringe 0.2

average 0.7

great 0.09heavy 0.01

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PPPP

PPP

fringe 0.2

average 0.7


��

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fringe 0.2

average 0.7


5.2.1 Without Cost Gates

When cost gates are not used, all costs are imbedded in the final payoffs. Thereare twelve final branches, and the payoffs which go to their right are in fact thetwelve numbers which we calculated earlier and placed in the main body of thepayoff matrix. Writing these numbers onto the tree we obtain:


TheatreRental

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small

medium

@@@@@@@@@

@@@@@@@@@

large

Demand

��

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hhhh

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PPPP

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fringe 0.2

average 0.7


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hhh

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average 0.7


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1900

2400

2400

2400

700

6200

10,200

10,200

−3000

2500

17,500

30,500

We have finished the left-to-right formulation of the model, and we now pro-ceed with the roll-back procedure, which proceeds from right to left.

At each circle, we compute the expected value. Just as we saw when we didit as a payoff matrix, the expected payoff at the circle which ends the “small”alternative branch is:

EV(small) = 0.20(1900)+(0.70+0.09+0.01)(2400)= 380+1920= 2300

Similarly,

EV(medium) = 0.20(700)+0.70(6200)+(0.09+0.01)(10,200)= 140+4340+1020= 5500


EV(large) = 0.20(−3000)+0.70(2500)+0.09(17,500)+0.01(30,500)= −600+1750+1575+305= 3030

Though not essential, it is helpful to differentiate numbers which are calcu-lated as part of the rollback procedure (right-to-left) from numbers which are partof the formulation (left-to-right). This can be done by highlighting the numberscalculated as part of the rollback. Putting these numbers onto the tree we obtain:

TheatreRental

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small

medium

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large

Demand

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average 0.7


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fringe 0.2

average 0.7


1900

2400

2400

2400

700

6200

10,200

10,200

−3000

2500

17,500

30,500

2300

5500

3030

Moving to the left, we come to the square. At a square the best (highest, forprofit maximization) payoff is chosen. Clearly, this is the $5500 associated withthe medium-sized theatre alternative, and this number goes next to the square. Thesub-optimal alternatives are marked with short double lines at right angles to thealternative branches. Putting these things on the tree we have:


TheatreRental

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small

medium

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large

Demand

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1900

2400

2400

2400

700

6200

10,200

10,200

−3000

2500

17,500

30,500

2300

5500

3030

5500

@@@@

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As with any problem, the recommendation should be stated clearly:


Highlighting the optimal alternative we have:


TheatreRental

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small

medium

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Demand

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1900

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2500

17,500

30,500

2300

5500

3030

5500

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5.2.2 With Cost Gates

In the theatre problem there is a cost associated with each alternative, which is therent for the theatre. Recall that this was $600 for the small theatre, $1800 for themedium-sized thatre, and $5500 for the large one. Though we are starting withthe tree having being drawn in this instance, normally one would begin to drawthe tree and put on the cost gates as the alternative branches are drawn. The treewith cost gates (but without the final payoffs) is:


TheatreRental

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small

medium

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large

Demand

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600

1800

5500

Now we must determine the final payoffs. For some problems, these payoffsare given exogenously in the problem description, but for this example we mustwork them out. These payoffs are all revenues from ticket sales. Recall that thesmall, medium, and large theatres can hold, over three nights, 300, 1200, and3600 people respectively. The demand levels for fringe, average, great, and heavyare 250, 800, 2300, and 4500 respectively. The tickets net $10 each. A smalltheatre obtains a revenue of $10(250) = $2500 with fringe demand, but otherwisethe theatre is filled for a revenue of $10(300) = $3000. A medium-sized theatrehas a revenue of $2500 for fringe demand, $10(800) = $8000 for average demand,but otherwise the theatre is filled for a revenue of $10(1200) = $12,000. A largetheatre has a revenue of $2500 for fringe demand, $8000 for average demand,$10(2300) = $23,000 for great demand, and is filled with heavy demand with arevenue of $10(3600) = $36,000. An advantage of using the cost-gate approachis that it is creates a fair amount of repetition in the final payoffs. Adding thesepayoffs to the tree we obtain:


TheatreRental

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small

medium

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large

Demand

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hhhh

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fringe 0.2

average 0.7


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average 0.7


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600

1800

5500

2500

3000

3000

3000

2500

8000

12,000

12,000

2500

8000

23,000

36,000

Now we calculate the expected value at each circle. Normally we would notwrite all the details out; we would simply calculate the numbers and then writethem on the tree. However, since this is the introductory section for this material,the full workings are shown:

EV(small) = 0.20(2500)+(0.70+0.09+0.01)(3000)= 500+2400= 2900

EV(medium) = 0.20(2500)+0.70(8000)+(0.09+0.01)(12,000)= 500+5600+1200= 7300

EV(large) = 0.20(2500)+0.70(8000)+0.09(23,000)+0.01(36,000)


= 500+5600+2070+360= 8530

Putting these numbers onto the tree we have:

TheatreRental

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small

medium

@@@@@@@@@

@@@@@@@@@

large

Demand

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hhhh

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average 0.7


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600

1800

5500

2500

3000

3000

3000

2500

8000

12,000

12,000

2500

8000

23,000

36,000

2900

7300

8530

We are now interested in finding the highest net payoff at the square, eachnet payoff being the expected value at the circle minus the cost at the gate. Thechoices are: small, 2900−600 = 2300; medium, 7300−1800 = 5500; and large,8530− 5500 = 3030. The best of these (as we saw before) is medium with anexpected payoff of 5500. We write the 5500 next to the square, and highlight thebest alternative to obtain:


TheatreRental

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small

medium

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large

Demand

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average 0.7


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average 0.7


@@

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600

1800

5500

2500

3000

3000

3000

2500

8000

12,000

12,000

2500

8000

23,000

36,000

2900

7300

8530

5500

@@@@

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Once again, it is emphasized that the recommendation should be stated clearly:



5.3 The Expected Value of Perfect InformationTo find the EVPI using a decision tree, the event must precede the decision. Thisis because the decision maker receives the perfect information (which is that aparticular outcome will occur) and then chooses the best alternative afterwards.Technically, the event is not the demand per se, but instead is the prediction aboutthe demand. However, because the prediction is perfect, it has the same outcomesand probabilities as the demand itself. The tree therefore begins with an eventnode, followed by the four outcomes.

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great 0.09heavy

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For each outcome, we can simplify the choices down to one or two reasonablealternatives. For example, if we are told that there will be fringe demand, it makesno sense to pay more rent for a medium-sized or large theatre, when a small onecan easily handle all the demand. At the other extreme, not even the large theatrecan handle heavy demand, so we wouldn’t even consider a small or medium-sized theatre in this situation. With the other outcomes, it’s not clear whetherwe should choose a theatre which is a size less than the demand (to save on the


rent), or whether we should rent a theatre which is a size bigger than the demand.Hence with average demand we could investigate both a small and a medium-sized theatre, and with great demand we could consider both a medium-sized anda large theatre. Adding the reasonable alternatives with their cost gates and finalpayoffs, we obtain:

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fring

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average 0.7

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((

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TheatreRental

small

small

medium

medium

large

large

2500

3000

8000

12,000

23,000

36,000

600

600

18001800

5500

5500

We now perform the rollback to obtain:


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fring

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average 0.7

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0.01

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TheatreRental

small

small

medium

medium

large

large

2500

3000

8000

12,000

23,000

36,000

600

600

18001800

5500

5500

1900

6200

17,500

30,500

6600

Calculating the EVPI by the indirect method, we obtain (as we did before):

EVPI = EV with PI−EV without PI= 6600−5500= 1100

The expected value of perfect information is $1100.

5.4 Sequential Decision Making5.4.1 Example

Bill operates a hardware store doing a reasonable amount of business for its size.There’s a possibility of a smelter being built nearby, which will boost the town’spopulation and his business if it goes ahead. Because of this, Bill wonders whether


or not he should expand the store. The company which would operate the smelterhas said that they will know one way or the other by late September, but that wouldbe too late to look for a contractor to get the work done before the onset of winter.At the present time, there’s about a 40% chance of the smelter going ahead.

Bill figures that relative to the profit that he would make anyway, the expan-sion would generate a profit margin of $5,000,000 (net present value, exclud-ing the capital costs of the expansion) if the smelter goes ahead, but only about$1,600,000 if it does not. A contractor has quoted him a firm construction cost of$2,900,000, provided that a contract is signed by July. If he does nothing beforeOctober, he could then make a deal for the expansion. If the smelter company hasthen said that they are indefinitely delaying the project, then the $2,900,000 priceis still available, but with a surcharge of $150,000 for winter work. On the otherhand, if the smelter company is going ahead with the project, then the construc-tion cost will jump to a total of $4,500,000, because everyone will be looking forconstruction work to be done.

5.4.2 Solution

Bill has two opportunities to expand his store. He could do it in July, when theconstruction cost would be lowest. Alternatively, he could do it in October, afterhe hears about a proposed smelter. The tree for this problem will have three parts:a decision about expanding now, an event concerning the smelter, and a decisionabout expanding in October.

We begin the tree by considering only the initial decision. He can either ex-pand now, or wait to hear about the smelter. Expanding now would cost $2,900,000,which we write on the tree as 2.9, making a note that all financial information isin millions of dollars.

The initial formulation of the tree is shown in Figure 2.No matter which alternative is chosen, we then hear some information about

the smelter. Either we find out that the smelter company is proceeding with itsconstruction, or they have decided to delay construction for an indefinite period.While this is a decision from the point of view of the smelter company, it is anevent from Bill’s perspective, because he has no control over it. Adding this eventwith its two outcomes, we obtain the partial formulation of the tree shown inFigure 3.

On the top part of the tree we place the two payoffs, which are the $5,000,000and $1,600,000 figures mentioned in the text of the case. On the bottom part ofthe tree, we draw the structure for the second decision. The construction costs are


Note: All financial informationis in millions of dollars.

PossibleEarlySigning

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sign in

July

@@@@@@@@@

@@@@@@@@@

do nothingbefore October

@@

2.9

Figure 2: Bill’s Hardware Store: Initial Part of the Formulation




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July

@@@@@@@@@

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��SmelterInfo

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proceeds 0.4

HHHHH

HHHH

delayed 0.6

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proceeds 0.4

HHHHHH

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delayed 0.6

@@

2.9

Figure 3: Bill’s Hardware Store: Partial Formulation of the Tree


different from what they were before, either because of a price increase causedby all the smelter activity, or because of the extra cost for construction during thewinter. There are four final payoffs; the two of these which are 0 are not mentionedexplicitly in the case. For these we must see that if Bill does not expand his store,then the profit margin relative to what he is doing now must be 0. The completeformulation of the tree is shown in Figure 4.

Proceeding from right to left, we rollback the tree, finding the highest netpayoff at each square, and the expected payoff at each circle. The rollbacked treeis shown in Figure 5.

The completed tree showing the recommended course of action using high-lighting is shown in Figure 6. The recommendation is to wait until October. If it’sannounced that the smelter will proceed, then Bill should expand his store. If it’sannounced that the smelter is indefinitely delayed, then Bill should do nothing.The ranking payoff is $200,000.

5.4.3 Finding the EVPI

We find the EV with PI by using a tree as shown in Figure 7. Note that becauseall the information comes at the outset, that if the expansion is to be done at all itwould be best to do it in July, when the construction cost would be lowest.

The EV with PI calculated on the tree is 0.84, i.e. $840,000. Since we wouldhave obtained $200,000 without the perfect information, the EVPI is:

EVPI = EV with PI−EV without PI= $840,000−$200,000= $640,000




��

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sign in

July

@@@@@@@@@

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proceeds 0.4

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1.6

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ContractinOctober

ContractinOctober

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5.0

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Figure 4: Bill’s Hardware Store: Complete Formulation of the Tree




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Figure 5: Bill’s Hardware Store: Rollbacked Tree




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Figure 6: Bill’s Hardware Store: Completed Tree



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[The perfectpredictionfor the]SmelterInfo

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5.0

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2.1

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Figure 7: Bill’s Hardware Store: Finding the EV with PI


5.5 ExerciseMake a decision tree for the situation described in the following case, and providea recommendation.

Case: Niagara Frontier Winery

Driving to her office in St. Catherines, Ontario, Betty Johnson, the productionmanager of Niagara Frontier Winery, heard a very disturbing weather forecast.“Environment Canada has issued a severe frost warning for the Niagara Regionfor later in the week”. This was only the 27th of September, and the grapes wouldnot be ready for harvest for another three weeks. Telephoning the weather officefor more detailed information, she was told that in three days time there would be a60% chance of a mild frost, and a 30% chance of a severe frost. A mild frost couldat least be contained by erecting heaters in the fields at a cost of $350,000. Usingheaters, the damage would be minimal; about 80% of the crop could still be madeinto high-quality wine, and a further 10% could be made into low-quality wine. Asevere frost, on the other hand, would destroy the crop entirely; even an attempt atusing heaters would be in vain. As long as she made up her mind by 1 p.m., therewas enough time to erect the heaters. Also, there was enough time to order thatall the crop be picked immediately, which would cost $400,000. It could eitherbe sold as grape juice or made into low-quality wine at a value of $1,200,000. Ifthe crop were picked in good condition in three weeks time, however, it would beworth about $3,000,000, but from this the $400,000 harvesting cost would have tobe paid.

There was one more complication. The research department had come up withsomething called “ice-wine”. If the heaters were not used and a mild frost wasexperienced, none of the crop could be made into high-quality wine, but perhapsthis could be made into ice-wine. This would cost $400,000 to pick the crop,and would have a 75% chance of success. A successful product would be worth$2,600,000, but a failure would be worth nothing. Alternatively, about 85% ofthe crop with mild frost damage could be sold as low-quality wine. Betty madeherself a pot of tea and then looked at her watch. The one o’clock deadline wasfast approaching.


6 Decision Trees 2

In the previous section, we saw most of the technical operations to handle decisiontrees. If there’s any difficulty using trees, it is the formulation – the rollbackprocedure is very straightforward. In this section we solve a fairly long case.Doing this case adds one more technical operation – the use of payoff nodes. Moreimportantly, though, solving this case illustrates the application of the decisiontree methodology to a somewhat complex situation.

6.1 Case: New Detergent Marketing Campaign

6.1.1 Problem Description

Elizabeth, John, and Susan work for a consumer products company. They comefrom widely different academic backgrounds. Elizabeth has a B.Sc. and M.Sc. inbiochemistry, and she has been on a research team which has come up with a newtype of detergent. John has a joint B.A. in English literature and art, and works onall advertising campaigns. Susan has a B.Comm., specializing in marketing, buthas taken a few management science electives as well.

They recently held a meeting to discuss what to do about the newly-developeddetergent. The meeting began with Elizabeth welcoming the others. “John andSusan, thanks for coming. The research team is very pleased with this new prod-uct. We tested it extensively in the laboratory, and found that there was virtuallyno fading of colours even after 100 washes. I hope that with your help we canbring this product to market.” “John and I have read the report,” Susan replied,“but it’s the part where it says that the cost will have to be 20% higher than evenfull-priced brands that has me worried. I fear that when it comes to the typicalshopper looking at the prices in the store, that a claim of technical excellence isnot going to amount to much.” “I’ve been thinking about that,” said John. “Wehave to make it clear right in the ad campaign that the consumer is paying morefor the detergent but saving much more than that in the long-term on the cost ofreplacing clothes. I grant you that the average shopper will be skeptical, but wehope that a least some segment of the market will understand the trade-off andtherefore buy our product.”

Susan knew that Elizabeth was excited about the new detergent because shehad helped develop it, and that John was looking forward to a challenge in writingthe ad copy. However, she also knew that only about one in ten new productseventually succeeded in the market place. Thinking that the others would want to


proceed, she had come up with some approximate numbers. “I’m assuming thatfor now at least, our market is Canada,” Susan said. “The United States is justnext door, with nearly ten times as many people, but we don’t have a distributionnetwork there, so the best that we could hope for in the States is a licensing agree-ment several years down the road, if everything works out here first. For now,we should see if this product will be profitable in the Canadian market alone.”Elizabeth and John nodded their heads, and Susan continued. “If we try for thewhole Canadian market, the start-up costs would be about $800,000. After thatwould come some revenue, whether the product turns out to be a success or not. Asuccess would bring in about $4,000,000, but a failure would provide only a tenthof that. If success or failure were 50/50, I’d proceed, but the chance of success isonly one in ten.”

Elizabeth wondered how accurate Susan’s figures were. Perhaps if the start-upcosts could be lowered, or the revenues raised, or the probability of success raised,the project would make sense. “Susan, your numbers are at best estimates. Withdifferent numbers this project could go ahead.” “Sure,” replied Susan, “and withdifferent numbers the project could be even less viable than it is now. I’m notsaying that this new detergent couldn’t do well for us, but maybe we should tryto test-market this product before launching it into the entire Canadian market.”John broke in when he heard this idea. “We did test-marketing when I was atmy former employer. Usually, if a product succeeded in the test-market, it didwell everywhere. There were exceptions, though. Chocolate-covered seaweeddid well when we tried it in Halifax, but bombed when we tried to go national –you couldn’t give it away in Toronto. On the other hand, we test-marketed a newquick-cook rice in Regina, and it didn’t do well, but when on a hunch we wentahead with a national campaign anyway, it suddenly became a success. It did bestin cities with large immigrant populations, and in hindsight we saw that Reginawasn’t a good test-market for that kind of product.”

“You’ve hit on a good point, John,” said Susan, “the test-market should ideallyreflect the country as a whole, but that’s not always easy to do. Since advertisingis expensive, we concentrate on small geographic areas away from high-pricedmedia buys in large cities. For example, Pickering [just east of Toronto] would bean expensive place to test-market, because we’d have to buy airtime on Torontostations and pay to reach eight million people in central southern Ontario, whenwe only want to reach the ones who live in Pickering. In Ontario, Peterborough isoften used as a test-market because we can buy air-time just in Peterborough at areasonable price. For the same reason, test-marketing in Alberta is often done inLethbridge, which is large enough to have its own media outlets, but doesn’t have


the high rates that are found in Edmonton and Calgary.”Elizabeth wondered aloud about some of John’s comments. “What does it

prove once we get the result from the test market? The detergent could be likethe chocolate-covered seaweed, or like the rice, rather than being a perfect pre-dictor for what should be done.” “You’re right, Elizabeth,” replied Susan, “test-marketing is not a perfect predictor, but it should give us a better idea of whatto do. If we believe that there’s one chance in ten of the product being a suc-cess in the country, then there should be more-or-less a 10% chance of success inany test-market. However, we know from past experience that people in BritishColumbia are most open to new products, and this figure generally declines as oneheads east. If we just test in one market, I’d say that there’s about a 12% chanceof success in Lethbridge, about 10% chance of success in Peterborough, and just8% in St. John’s. If we test in one of these places and it’s a failure, then the chanceof success in the rest of Canada certainly becomes less than 10% – I don’t knowhow much less, but it really doesn’t matter. This project is tenuous enough as itis, without having to deal with a negative test result. On the other hand, a successin a test market would be a good omen for the rest of the country. As John said,there’s no guarantee of success elsewhere, but I have to believe that on averagethe chance of success has increased from 10% to say 60%, though I think that thisfigure would range from 50% in Lethbridge to 70% in St. John’s. While the pur-pose of the test-marketing is to obtain information, there would be some revenuesas well, perhaps $30,000 for a success, but only a tenth of that for a failure.”

“We could test in two of these markets, or perhaps even all three of them,”Elizabeth suggested. “But if we test in two markets our advertising costs woulddouble, and if we test in three these costs will triple,” John said. “Testing inall three is probably not going to fly,” Susan said, “but a case could be made fortesting in two markets. I think that Lethbridge and St. John’s would give us a bettersense of the country as a whole than using either of these cities with Peterborough.That brings us to the question of how we should use these two markets. Shouldwe test simultaneously in both, or should we test sequentially, beginning with oneof the two cities, and then based on what we find there, possibly proceeding to theother?”

At this point Elizabeth jumped in. “Whatever we make in the lab, someoneelse can make too. I worry that if we do the test marketing, a competitor will buya litre of it, have it chemically analyzed, and then reverse-engineer it in their ownlabs. This would take some time, of course, but if we test in two markets, anddo it sequentially rather than simultaneously, we might just give them the timethat they need. After taking all the risk, we would then have to share the market


with someone else.” “Point taken,” said Susan. “If we test in both Lethbridge andSt. John’s, let’s agree that we will do the testing simultaneously. This gives usfour possibilities for the test results. If we fail in both places, we can forget aboutproceeding further with this product. Should we succeed in both, I’m almostcertain that we would have a winner on our hands; I’d put the probability at 0.99.If we fail in one, but succeed in the other, I would want to boost the advertisingexpenditures by $50,000, and based on that I’d put our chances of success in therest of the country at about 35%.”

From his experience, John had some figures on test-marketing. “Before wespend anything on advertising, we would have to spend about $15,000 to developan ad campaign. The three test markets aren’t much different in size. I’d say thatin each the cost to buy air time would be about $10,000.” Elizabeth wonderedif spending this money now would save some money later should they decide toundertake a national campaign. “Here’s a hypothetical one for you, John. Supposethat we test in Lethbridge and St. John’s, and both tests turn out to be a success,so we decide to go national. Having spent $15,000 plus two times $10,000 for atotal of $35,000, can we deduct this amount from the $800,000 cost of the nationalcampaign?” “I wouldn’t count on that,” John replied. “We would probably wantto modify the test-market advertising, so that will cost money. More importantly,when we buy national advertising, we obtain economies of scale by making onenation-wide media buy. It would be cheaper to do it that way than to buy airtime in every city individually except where we test-marketed. I’d say that nomatter what we do with test-marketing, the cost of the national campaign wouldbe $800,000. At the same time, this would be a new campaign as far as the test-market is concerned, so the national revenues wouldn’t be diminished.”

“It’s time to wrap this up for this morning,” Susan said. “Senior managementwill want to see a business plan, and the basis for this will be the recommendationwhich will come from making a decision tree of what we’ve been discussing. I’llwork on this later this morning, and we’ll meet again at 2 p.m. to discuss it.”

6.1.2 Formulation

We wish to develop and solve the decision tree to which Susan refers. It’s toocomplicated to think of all the decisions and events at once in a long case likethis. Instead, we should think about what must come first. It is more-or-less obvi-ous that the case presents us with at least four alternatives for the test-marketing:Lethbridge only; Peterborough only; St. John’s only; and testing simultaneouslyin both Lethbridge and St. John’s. The three persons seem to agree that other types


of multiple testing (sequential testing, or all three cities, or another pair of cities)should not be considered, and we will therefore leave these options out of thedecision tree. At the other extreme, going directly to a national campaign with-out doing any test-marketing was not clearly opposed by Elizabeth, so we mightwish to investigate this course of action. Also, we should consider doing nothingwhatsoever, which for many business situations may be best of all.

Based on the foregoing, we could begin with a square followed by six alterna-tive branches: one for each of the four testing alternatives; one for proceeding tonational marketing directly; and finally a do-nothing alternative. Doing it this waywould be correct, but things become clearer if we first have a test-marketing deci-sion which has just two alternatives: test market; and do not test market. The firstof these alternatives then requires a decision about the manner of the test market-ing. The second has a decision about the national campaign with two alternatives:proceed with the national campaign; or do nothing. Aside from the clarity pro-vided by this approach, it allows the $15,000 cost of preparing the test-market adcampaign to be by itself on the test market alternative branch, with the advertisingcosts being handled separately. If we proceed with the national campaign with notest marketing, then this alternative is followed by a result event, with its two out-comes: there is a 10% chance of making $4,000,000, and a 90% chance of making$400,000. Because of space limitations all financial figures will be written on thetree in thousands of dollars, hence for example $4,000,000 is written on the treesimply as 4000.

The manner of the test marketing could be one decision with four alternatives,but again it makes things conceptually easier if we have two decisions. First, wedecide whether we want one or two test markets. If one test market is chosen,then we must decide whether it will be in Lethbridge, Peterborough, or St. John’s,and if we want two test markets it is understood from the case that these will be inLethbridge and St. John’s, hence the event for the result on one of the test marketscomes next. In making the tree it turns out that we already have too much to puton one piece of letter-size paper. Hence, on this piece of paper we end with twonodes, one a decision node and one an event node, after alternative branches fortesting in one or two markets. Because the cost of test marketing is $10,000 permarket tested, we place $10,000 and $20,000 cost gates (written as 10 and 20) onthe test in one market and test in two markets alternative branches respectively.The beginning of the tree is shown in Figure 8.


Note: All financial informationis in thousands of dollars.

Preparationfor TestMarketing

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On Figure 8 two continuations are indicated. The first of these occurs at thedecision node for choosing between Lethbridge, Peterborough, and St. John’s asthe solitary test market. After each of these comes a similar structure, with onlysome of the numbers being different. First, there is a result event, with the testcampaign in every city being either a success or a failure. There is a payoff of$30,000 associated with a success, and a payoff of $3000 associated with a failure.Since it is implied in the case that a failure in a test market would immediately endthe venture, all we need do is put a ‘3’ (for $3000) at the end of every outcomebranch which represents failure. However, every ‘success’ outcome branch istreated differently. Because each of these is followed by more tree structure, wehandle the $30,000 in revenue by using a payoff node followed by a null branch.On the null branch we place a cost gate, with the figure placed in parenthesisindicating that we have a revenue rather than a cost. Hence a revenue of $30,000is indicated as:

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The second continuation comes after the alternative of testing in two markets.Since the places of these markets have been stated in the case as being Lethbridgeand St. John’s, we next have the result events for these two markets. Here is anexample where the order does not matter – it can either be the Lethbridge resultevent followed by the St. John’s result event, or vice versa. In real life, theseresults would be announced more-or-less simultaneously. This is why the payoffsare combined – for example, if we are successful in both markets, then $60,000in revenue (i.e. $30,000 from each place) is obtained. If one is a success, but theother is a failure, then $30,000 + $3000 = $33,000 is obtained. Finally, if both arefailures then the revenue is $6000 (i.e. $3000 from each place).

The rest of the tree is similar in structure to the first continuation, but we notethat the cost of a national campaign after one failing test market is now $800,000+ $50,000 = $850,000. The second continuation of the tree is shown in Figure 10.


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6.1.3 Solution and Recommendation

We perform the rollback beginning with the first continuation. Next, we performthe rollback for the second continuation. At the extreme left, we obtain the figure188.2224. The question arises as to how many decimal places we should report.Because the figures are in thousands of dollars, this figure represents $188,222.40,so at least we aren’t trying to report a fraction of a cent. Even so, some would ar-gue that it’s pretentious to report any figure closer than say the nearest ten dollars.My preference is to do things accurately, and then round the final answer, shouldthat be desirable. It turns out in this example that the final answer is unaffected bythis figure anyway.

The figures from the extreme left of the first and second continuations are thentransferred to the initial part of the tree. That part of the tree is then rolledback. Inwords, we can state the recommendation as:

Make the ads for a test market campaign, and run this campaignin Peterborough. If this turns out to be a success, then proceed withthe national campaign. If the test campaign turns out to be a failure,then abandon the project. The ranking payoff is $156,700.

The recommended courses of action have been highlighted on Figures 14 and15.


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6.2 ExerciseNewlab has come up with a new product in its research lab. The technical successis clear, but as with any new product the commercial success is risky. Becauseof this, they would sometimes test-market a product first, and then make a deci-sion about national marketing after the test-market results had come in; at othertimes they would proceed directly to national marketing. On some occasions, theywould abandon the product without even test-marketing it.

The test-marketing would cost about $120,000. If successful (probability 0.4)there would be revenues of $40,000; if unsuccessful the revenues would only be$10,000. Should the test market be successful, a followup national campaignat a cost of $500,000 would have a 70% chance of success with a revenue of$1,800,000, otherwise it would be a failure with a revenue of $150,000. Shouldthe test market be unsuccessful, a followup national campaign would have only a0.2 chance of success (with the same cost, and the same revenues for success andfailure).

A national campaign not preceded by a test campaign would have a 45%chance of success. It would cost $600,000, and would produce a revenue of$1,900,000 if successful, but only $175,000 otherwise.

(a) Draw and solve a decision tree for the situation (using payoff nodes whereappropriate), and state the recommendation for Newlab clearly.

(b) If the $600,000 figure in the last paragraph were changed to $800,000,what would be the revised recommendation?


7 Decision Trees 3In this section we begin with an example which contains only costs. We then lookat sensitivity analysis in the context of decision trees.

7.1 Airfare Problem


An office manager in St. John’s has been informed that a compulsory company-wide meeting might need to be held in Vancouver in fifteen days time. At thepresent time, there is about a 30% chance that the meeting will go ahead. Thereis about a 40% chance that in about five days from now they will know for surewhether or not the meeting will be held. If they still are not sure at that point, thenthere’s still, as there is now, only a 30% chance that the meeting will go ahead.There is a 100% chance that in ten days time they will know for sure about themeeting one way or the other. A full-fare economy return ticket, which would cost$3500, could be purchased as late as the day of the trip. Another option would beto buy a non-refundable ticket for $1300 which must be purchased at least sevendays before departure. Another choice is to buy a non-refundable seat-sale ticketfor $800, which would have to be purchased no later than tomorrow. Assumingthat a non-refundable ticket would be worthless should the meeting not go ahead,develop and solve a decision tree to analyze the manager’s problem.

7.1.2 Formulation

This example only mentions costs, not revenues, so if we put all the numbersonto the tree as we did in the previous section we will be rolling back negativenumbers. Instead of dealing with negative numbers, we could write costs on thetree as positive numbers, and rollback the tree as before, except that at each square,we would choose the alternative with the lowest cost . We will solve this problemusing the alternate approach; the final tree will have all financial information beingthe absolute value of what we would have had if we had not used this approach.

While there is a fifteen day continuum of time in this problem, only certainpoints in time are relevant. If we call today day 0, then there is the possibility ofmore information on day 5; if there’s no announcement on day 5, then there willbe an announcement on day 10. To attend a meeting on day 15, the manager mustfly across the country no later than day 14. Then there are the deadlines for the


purchase of the various classes of tickets: day 1 for the $800 ticket; day 7 for the$1300 ticket; and day 14 for the full-fare $3500 ticket. Hence our focus should beon days 1, 5, 7, 10, and 14.

While the $3500 ticket can be purchased at any time, there is no advantage topurchasing it early. If the ticket is bought early, then a few days extra interest ischarged, and more importantly, there would be the hassle of returning the ticketshould the trip become unnecessary. If the few days interest is not important, thenthe ticket could be bought after day 10. There is no sense in buying a $1300 tickettoday or tomorrow, because an $800 ticket is available during this time with thesame privileges. After tomorrow, the office manager might as well wait until atleast the end of day 5 to possibly obtain more information. Hence the $800 seat-sale ticket would be bought on either on days 0 or 1 (or not at all), the $1300 7day advance ticket would be bought on days 6 or 7 (or not at all), and the $3500full-fare ticket would be bought on days 11 to 14 inclusive (or not at all).

At the outset, the manager could buy an $800 ticket, or he could wait fivedays for more information. Hence the tree begins with two alternatives; thereare no further branches after the alternative branch to buy a seat-sale ticket for$800. Because this is a final branch, and because we are writing costs as positivenumbers, we do not need a cost gate – all we need to do is write 800 to the rightof the branch. The wait five days option, however, then has an event with twooutcomes: an announcement is made; or no announcement is made. The tree sofar is:

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After the outcome branch for the announcement being made there is an eventwith two outcomes: the meeting will go ahead, or it will not go ahead. It is


possible, but not advisable, to combine the two events into one event with threeoutcomes: the meeting will go ahead; it will not go ahead; and no announce-ment. Doing it this way would shorten the tree, but it would require computingsome joint probabilities – this confuses the formulation process with the solutionprocess. We will therefore write what is happening as two events.

After the “will go ahead” outcome branch, we could have a square with twoalternatives, one for buying a 7 day advance ticket, and one for buying a full-fareticket for $3500. However, it is obvious that the manager should buy a 7 dayadvance ticket for $1300, so we will only draw this alternative. After the “willnot go ahead” outcome branch, no action needs to be taken, so we simply write apayoff of 0 to the right of this branch.

If there’s no announcement after five days, then our choices are to either buya 7 day advance ticket or wait another five days. If the latter is chosen, then anevent occurs giving information about the meeting. After the “go ahead” branch,the manager must buy a full-fare ticket; after the “will not go ahead” branch, noaction is required.

The entire tree, printed in landscape form, appears in Figure 16. Because thefinancial information is all costs, a note has been placed on the figure to that effect.

7.1.3 Solution

To rollback the tree, we need to choose the lowest cost at each square. With thismodification, the rolled-back tree appears in Figure 17. The recommended courseof action can be followed on the tree, but the analyst should also make the rec-ommendation by clearing stating it in words. In trees with multiple decisions, weoften use the term ranking profit (or ranking cost in this example) to indicatethat the number being presented is a mixture of measures (best at a square, ex-pected value at a circle). This is reported along with the best course of action. Inwords, the recommendation is:


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RecommendationDo not buy the $800 seat-sale ticket, but instead wait to see if

there’s an announcement in five days time. If there’s an announce-ment that the meeting will go ahead, then buy a $1300 7 day advanceticket at that time. If there’s an announcement that the meeting is notgoing ahead, then do nothing. If there’s no announcement after fivedays, then wait for a further announcement. If the meeting is goingahead, then buy a $3500 full-fare ticket; otherwise, do nothing. Theranking cost is $786.

Although the $786 figure is the most important one on the tree, the otherrolled-back numbers are also important, because they give the ranking cost to beincurred for proceeding further down that path. For example, if an announcementis not made after five days, then the ranking cost increases from $786 to $1050.

7.1.4 The EVPI

In this example we need to find the expected cost with perfect information (ECwith PI). If at the outset we were to receive perfect information that the meetingwill be going ahead, then we would buy the seat-sale ticket for $800, otherwisewe would do nothing. The chance that we will be told that the meeting will begoing ahead is 30%, hence EC with PI = 0.3(800) + 0.7(0) = $240. The expectedcost without information is $786. We subtract to find the EVPI, in reverse orderbecause these are costs.2

EVPI = EC without PI − EC with PI= $786−$240= $546

2This is the same as saying that the EVPI is −$240− (−$786) = $546.


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7.2 New Detergent Case: Some Extensions

Here we look at some extensions to the case which was analyzed in the previoussection.

7.2.1 The EVPI

There are four pieces of uncertainty in the case: the result in each of the threetest markets; and the result of a national campaign. While it would be possibleto compute the EVPI based on knowing perfect information about any of thesefour things, or any combination of these four things, it is the uncertainty about thenational campaign which is of primary importance. If we know that the productwould be successful, then clearly we would spend $800,000 to make $4,000,000,for a net of $3,200,000. If we know that the product would be a failure, thenclearly we would not spend $800,000 to make only $400,000. There is a 10%chance of being told that a success will occur, and a 90% chance of being told thata failure will occur, hence the EV with PI is:

0.10($3,200,000)+0.90(0) = $320,000

Without perfect information, the ranking payoff is $156,700, hence the EVPIis:

EVPI = EV with PI−EV without PI= $320,000−$156,700= $163,300

Though we did not need to draw a tree the find the EV with PI, we can do soif we wish. We begin with the event, being the prediction about the success orfailure of the national campaign, followed by the decision about whether or not toproceed with the national campaign. Making this tree and performing the rollbackwe have:


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Hence the EV with PI is $320,000, and then subtracting the $156,700 we seethat the EVPI is $163,300.

7.2.2 Sensitivity Analysis

While we could examine the effect of changing one or more of the parameters byany amount, usually we are only interested in finding the point(s) at which therecommendation would change. Sometimes finding the region for which the samerecommendation hold also means that the ranking payoff is unchanged as well.At other times the ranking payoff will change as a parameter is changed. First, wewill look at changing costs, and then we shall look at changing probabilities.

Changing Costs The effect of changing the cost of making the ads is very easyto analyze. From Figure 14, we see that not making the ads leads to a payoff of0, and the payoff at the square on the right of the make ads alternative branchhas a ranking payoff of $171,700. Therefore, the ads can cost up to $171,700before making the other alternative better. At the other extreme, if the ads costnothing then the make ads alternative is of course still preferred. Since the adscurrently cost $15,000, we could say that the cost could be decreased by $15,000or increased by $156,700 without affecting the current recommendation.

The effect of changing the cost of the test market campaigns is a bit trickier.While only one cost is being changed, two costs are affected by this change. Sup-pose that the cost (in thousands of dollars) of a test market campaign is now 10 +


∆ (where ∆ is ≥ −10). The cost next to the alternative branch for testing in twomarkets is therefore 20+2∆. Therefore, the ranking payoff at the square labelled“Number of Test Markets” is either 181.70− (10 + ∆) or 188.2224− (20 + 2∆),whichever is higher. These expressions simplify to 171.70−∆ and 168.2224−2∆

respectively. We now need to find the value for ∆ which causes indifference be-tween the two alternatives. Setting the two expressions equal, we solve to find∆:

171.70−∆ = 168.2224−2∆

∆ = −3.4776

This figure is in thousands of dollars, hence if the cost per test market is less than$10,000−$3,477.60 = $6522.40, then the recommended solution is to test in twomarkets.

If ∆ >−3.4776, then one test market is preferred to two, but if ∆ becomes toohigh, then the company would be better off doing nothing. This critical value of∆ is where:

Ranking Payoff(make ads) = Ranking Payoff(do nothing)171.70−∆−15 = 0

−∆ = −156.70∆ = 156.70

Hence the recommendation remains unchanged provided that:

−3.4776≤ ∆≤ 156.70

The current cost of the test market campaign is $10,000. Hence, in absolute terms,the recommendation remains unchanged provided that the cost of the test marketcampaign remains between $6522.40 and $166,700.

Changing Probabilities To illustrate the effect of changing probabilities, con-sider the probabilities of success and failure in the national campaign, after twomarkets have been tested, and where one success and one failure has been ob-tained. These numbers are currently 0.35 and 0.65 for success and failure respec-tively. Now we will let the probability of success be 0.35+∆, and hence the prob-ability of failure is 0.65−∆. We must place the condition that −0.35≤ ∆≤ 0.65.These adjustments are shown on the appropriate outcome branches on the right-hand side of Figure 18.


These changes cascade through the tree, affecting most of the ranking pay-offs. At the two bottom circles on the right, we increase the payoffs by 4000∆ +400(−∆) = 3600∆. The 1660 figure does not need to be recomputed – this is theadvantage of dealing with changes to the current probabilities rather than lookingat absolute probabilities. At the squares immediately to the left, the payoffs willalso increase by 3600∆, provided that the “proceed” alternative remains betterthan the “abandon” alternative. This will be the case provided that:

810+3600∆≥ 0

This condition simplifies to ∆ ≥ −0.225. If ∆ goes below this figure, then theabandon alternative would be preferred to the proceed alternative, but this changewould not affect the overall recommendation, because this part of the tree is notpart of the current recommendation. Now let us suppose that ∆ ≥ −0.225, andsee what affect this has on the rest of the tree. The bottom two payoff nodes alsoincrease by 3600∆, and then the rollback increases the payoffs by 0.92(3600∆) =3312∆ after a success in Lethbridge, and by 0.08(3600∆) = 288∆ after a failurein Lethbridge. Finally, we obtain an increase of 0.12(3312∆) + 0.88(288∆) =650.88∆ at the circle on the extreme left, and this increase is transferred to theappropriate place on Figure 19.

For the current recommendation to remain unchanged, we must have

171.70 ≥ 188.2224+650.88∆−20171.70 ≥ 168.2224+650.88∆

3.4776 ≥ 650.88∆

0.00534... ≥ ∆

Hence the recommendation remains unchanged provided that ∆ ≤ 0.00534. Thisis not much, when the current probability of success nationally (after one success,and one failure) is 0.35, with all probabilities being reported to the nearest 5%. Allit would take is an increase to say 36%, and the recommendation would change totesting in two markets, and then proceeding if at least one of these turns out to bea success.


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7.3 ExerciseA farmer has been in the habit of always planting potatoes on his farm. In previousyears, the seeds for the potatoes were planted in the spring, and were ready toharvest in mid-July. After that, a second planting took place in late July, whichwas ready to harvest in early October.

This year, however, there is concern that a blight might destroy some or all ofthe potato crop. One thing he could do would be to plant a different crop suchas peas which would not be affected by the blight. The peas would have only asingle planting at a cost of $40,000. This planting would yield a crop in Octoberworth $70,000 if the weather turns out to be good, or $30,000 if the weather turnsout to be poor. There is a 60% chance that the weather will be good.

If, however, he decides to plant potatoes, he will have to worry about the blight(but the weather has little effect on the potato crop and can be ignored). The potatocrop would cost $60,000 to plant. There is a 10% chance of a severe blight, whichwould destroy the crop, and render any attempt at a second planting in late July notworth doing. A mild blight (20% chance) would partially destroy the crop, makingit worth only $35,000, while having no blight (70% chance) would produce a cropworth $80,000. After either a mild blight or no blight, a second planting couldbe undertaken, with the same costs and revenues as the first. The probability of asevere, mild, or no blight would be 15%, 30%, and 55% if the first planting had amild blight, but would be 0%, 5%, and 95% if the first planting had no blight.

NOTE: The crop planted in the Spring will be either peas or potatoes; doinga bit of both is not an option in this problem.

(a) Draw the tree, solve it using the rollback procedure, and state the recommen-dation and the ranking payoff. When drawing the tree, use payoff nodes forintermediate payoffs.

(b) Now suppose that the cost of planting potatoes is $60,000+∆c. For ∆c bothpositive and negative, find the limits for which the recommendation from part(a) does not change.

(c) Now suppose that the cost of planting potatoes is fixed at $60,000, but theprobability of a mild blight on the first planting is 0.2 + ∆p, with the proba-bility of a severe blight on the first planting being unchanged. For ∆p bothpositive and negative, find the limits for which the recommendation from part(a) does not change.




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8 Imperfect Information 1

8.1 Introduction

In this and the next two sections, we examine decision trees for which the use ofBayesian revision (covered extensively in Mathematics for Management Science)is needed in order to compute some of the probabilities. Starting with a problemdescription, we begin to develop the decision tree, except that not all of the proba-bilities can be written down immediately. We then perform a Bayesian revision tofind these probabilities, and then transfer these numbers to the decision tree. Thetree is then rolled-back to obtain a recommendation for the situation.

In Mathematics for Management Science, Bayesian revision is performed bymaking three tables. That method can be used here too, but we also demonstratea different method, that of using prior and posterior trees.

8.2 Example – Seismic Testing


An oil exploration company has identified a site under which there may be apocket of oil. The probability that oil exists at this location is 1%. It wouldcost $3,000,000 to drill for oil. If the oil exists, it would be worth $40,000,000. Aseismic test is available which would cost $40,000. The result of the test wouldbe one of the following: “positive”,“inconclusive”, or “negative”. If there reallyis oil present, then there is a 60% chance of a positive reading, a 30% chance ofan inconclusive reading, and a 10% chance of a negative reading. If there’s nooil at that location, then there’s a 0.04 probability of a positive reading, and a 0.2probability of an inconclusive reading.

We wish to develop a decision tree for this situation, and solve it to obtain arecommendation for the oil exploration company.

8.2.2 Problem Formulation

There are two decisions to be made in this situation. What we might call themajor decision is whether or not to spend $3,000,000 drilling for oil. The other

decision is whether or not to spend $40,000 to do the seismic test. The purpose ofthe seismic test is to obtain information which would help us with the major deci-sion. We will call the decision about the seismic test the information decision.


Not just in this situation, but in all problems of this type, the informationdecision must precede the major decision . Indeed, the information decision pre-

cedes everything else. This decision, with its two alternatives, is as follows:

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If the seismic is not done, then this becomes an easy problem. We must choosewhether or not to drill at a cost of $3,000,000, and if we drill we then have an oilevent with two outcomes: oil is present with probability 0.01; and oil is not presentwith probability 0.99. Adding these things to the tree we obtain:

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After the alternative to do the seismic test, comes the seismic event, with itsthree outcomes: positive; inconclusive; and negative. This is an example of acommon pattern in this type of problem – an alternative of the information deci-sion for which information is sought is followed by an information event, whichin turn is followed by the major decision. When we draw the outcome branchesfor this situation, we cannot immediately write the probabilities, for we do notknow what they are. We will find them later using Bayesian revision, and willthen transfer these numbers to the decision tree. Adding these outcome brancheswe obtain:

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At this point, a fair bit of repetition appears in the rest of the tree. After each


of the outcomes of the seismic result event, there is the decision about drilling.If the drilling is done, it is followed by the oil event. This of course is like whatwe have already drawn at the top of the tree, but there’s one important exception.The probabilities of oil and no oil are not 0.01 and 0.99 as they were before.Instead, these are now conditional probabilities, and they must be calculatedusing Bayesian revision. The decision tree with the probabilities absent on thebottom part of the tree is shown in Figure 20. To reduce the clutter on this partof the tree, the words “Drilling for Oil” and “Oil Event” only appear once ratherthan in all three places.

8.2.3 Bayesian Revision

Now, we must do the Bayesian revision. In Mathematics for Management Science,Bayesian revision was done using tables. If your memory of this topic is fuzzy,you might wish to go back and re-read this topic before proceeding further withthis discussion.

The event for which the marginal probabilities are known is that of the pres-ence of oil. These probabilities are 0.01 for the existence of oil at that location, and0.99 for the absence of oil. For the other event, the seismic testing, we have prob-abilities (given in the problem description) which are conditional on whether thereis or is not oil in the ground. These probabilities, and the two marginal probabili-ties, are given in the following table. Note that since one of the three outcomes ofthe seismic test must occur, we find P(negative/no oil) as 1−(0.04+0.20) = 0.76.

Seismic Eventpositive inconclusive negative

Oil oil 0.60 0.30 0.10 0.01Event no oil 0.04 0.20 0.76 0.99

Multiplying the conditional probabilities by the marginal probabilities of the oilevent we obtain the joint probabilities, and summing these gives the marginalprobabilities of the seismic event. The second table is:


Oil oil 0.0060 0.0030 0.0010 0.01Event no oil 0.0396 0.1980 0.7524 0.99P(seismic result) 0.0456 0.2010 0.7534


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Finally, dividing the joint probabilities by the marginal probabilities under-neath we obtain the posterior conditional probabilities. The third table using five-place decimals is:


Oil oil 0.13158 0.01493 0.00133Event no oil 0.86842 0.98507 0.99867

P(seismic) 0.0456 0.2010 0.7534

We read this as P(oil/positive) = 0.13158, P(no oil/positive) = 0.86842, and so on.These figures, even though there are five-place decimals, are approximations

of exact fractions. If we wish, we can use fractions instead. If this is done, it makessense to remove the decimals from the numerator and the denominator. However,it does not make sense to reduce the fraction to the lowest common denominator,as this only adds work. For example, instead of calculating the decimal quantity0.01493, we could have expressed 0.0030 divided by 0.2010 as the fraction 3

201 ,but we need not reduce this fraction to 1

67 . As unreduced fractions the third tableis:


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P(seismic) 0.0456 0.2010 0.7534

The concern about accuracy may seem to be misplaced when all the originalprobabilities in the first table are approximations anyway. However, some of theprobabilities will be multiplied by large numbers, specifically the $40,000,000figure. Here’s what happens depending on the level of accuracy when we approx-imate 60

456 using decimals. The decimal expansion is 0.13157947... When mul-tiplied by $40,000,000, we obtain (to the nearest cent) $5,263,157.90. If we ap-proximate the decimal we obtain (using rounding) 0.132 for three places, 0.1316for four places, and 0.13158 for five places. The values of these numbers times$40,000,000, and the differences between these values and the theoretical valueare:


Value Value × $40,000,000 Variation0.13157947... $5,263,157.90 –0.13158 $5,263,200.00 $42.100.1316 $5,264,000.00 $842.100.132 $5,280,000.00 $16,842.10

These variations are what would be present at the “Oil Event” node whichcomes after a “positive” seismic result. By the time everything is rolled back, theerror would be diminished, but it would still be considerable. For this reason, Istore all probabilities in my calculator’s memory, so that the nearly exact value isused, even if I only write five decimal places – doing it this way is equivalent tousing fractions. For student use, I would recommend that you either do it that way,or use five decimal places (rounded). At the very least, use four decimal places(rounded); using only three can cause substantial errors.

Before transferring these probabilities to the decision tree, we will look at theprior and posterior tree method of performing Bayesian revision. This methodtakes a little bit longer to do, but it’s conceptually easy because it mimics a subsetof the decision tree. We work with two probability trees, called the prior tree andthe posterior tree.

Both the prior and posterior trees contain two events. The prior tree (whichis done first) has the two events of the decision tree in the reverse order of howthey appear in the decision tree. The posterior tree (which is done after completingthe prior tree) has the two events in the reverse order of how they appear in theprior tree . Equivalently, the posterior tree has the two events in the same order

as they appear in the decision tree .

For this example, the two events of the decision tree are the seismic event andthe oil event, in that order. Since the prior tree contains these events in reverseorder, the prior tree consists of the oil event followed by the seismic event. Writ-ing the outcomes of the oil event with their marginal probabilities, and the threeoutcomes of the seismic event with their conditional probabilities, gives us thefollowing picture.


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On this tree we write the joint probabilities at each node and at the ends of thebranches. The node on the left begins with a probability of 1, meaning that it iscertain that something will occur. For any outcome branch on the tree, the jointprobability at the ending node (on the right) is the joint probability at the beginningnode (on the left) multiplied by the probability (be it marginal or conditional)on that outcome branch. For example, the joint probability at the top seismicevent node is 1 (the joint probability at the oil event node) multiplied by 0.01(the marginal probability along the “oil” outcome branch, which is simply 0.01).Similarly, the joint probability at the bottom seismic event node is 0.99. So far,everything is trivial.

The joint probability at the end of the top “positive” branch equals 0.01 (thejoint probability at the seismic event node) multiplied by 0.6 (the conditional prob-ability along the “positive” outcome branch), which is 0.006. Similarly, the jointprobabilities at the end of the top “inconclusive” and “negative” outcome branchesare 0.01(0.3) = 0.003 and 0.01(0.01) = 0.001 respectively. The joint probabil-ity at the end of the bottom “positive” branch equals 0.99 (the joint probability


at the seismic event node) multiplied by 0.04 (the conditional probability alongthe “positive” outcome branch), which is 0.0396. Similarly, the joint probabili-ties at the end of the bottom “inconclusive” and “negative” outcome branches are0.99(0.20) = 0.1980 and 0.99(0.76) = 0.7524 respectively.

Comparing this approach with the table method, it is seen that the prior tree issimply a visual way of displaying the information which appears in the first tableand in part of the second table. Adding the joint probabilities the completed priortree is:

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.006+ .003+ .001+ .0396+ .1980+ .7524 = 1.000√

This sum doesn’t have to be written out as it is here, but you should at least verifythat sum is 1 on your calculator. As stated earlier, the posterior tree containsthe same events, but in reverse order. For this problem, the posterior tree begins


with the seismic event, which is followed by the oil event. We begin drawingthe posterior tree by outlining its shape; the probabilities need to be computed bytransferring the final joint probabilities from the prior tree. The shape of the treeis:

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Throughout the development of the decision, prior, and posterior trees, wehave maintained consistency in the vertical ordering of the outcomes. We havealways placed “oil” above “no oil”, and “positive” above “inconclusive” whichin turn is above “negative”. This consistency will help us when transferring thejoint probabilities from the prior tree to the posterior tree, and when transferringmarginal and conditional probabilities from the posterior tree to the decision tree.

The final joint probabilities on the prior and posterior trees are the same, ex-cept that they are placed in a different order. The first (top) joint probability on theposterior tree is the joint probability of “positive” and “oil”. This is numericallythe same as the joint probability of “oil” and “positive”, which is found on theprior tree (at the top), and its value is 0.006.


The second (from the top) joint probability on the posterior tree is the jointprobability of “positive” and “no oil”. This is numerically the same as the jointprobability of “no oil” and “positive”, which is found on the prior tree (fourthfrom the top), and its value is 0.0396. Placing these values on the posterior treewe have:

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Going to the third place, we need the joint probability of “inconclusive” and“oil”, which from the prior tree is seen to be 0.003. Because of the consistentvertical labelling of the outcomes, a nice pattern emerges. The top three jointprobabilities on the prior tree become the joint probabilities at the top of each pairof outcomes on the posterior tree, and the bottom three joint probabilities on theprior tree become the joint probabilities at the bottom of each pair of outcomes onthe posterior tree.

Though the patterns differ from one Bayesian revision to the next, there willalways be a type of pattern to find whenever the outcomes have been labelledconsistently. If there is any doubt, remember that for each joint probability the


words on the outcomes of the prior and posterior trees must match up (in reverseorder).

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After transferring the ending joint probabilities from the prior tree, the nextstep is to compute the other joint probabilities on the posterior tree. This is donesimply by addition. At the top oil event node, we add 0.006 and 0.0396 to obtain0.0456, at the middle node we add 0.003 and 0.198 to obtain 0.201, and at thebottom the sum of 0.001 and 0.7524 is 0.7534. Each of these numbers is writtennext to its corresponding event node. Then, taking the numbers we have justcomputed, we sum them to obtain

0.0456+0.201+0.7534 = 1.0000

This number 1 is written next to the left-hand node. It is always true that we shouldobtain a 1 next to this node, so this acts as a check on our calculations. Had we not


obtained a 1, this would have indicated that an error had been made. The prob-ability on every outcome branch is obtained by dividing the ending (right-side)joint probability by the beginning (left-side) joint probability. For the outcomeson the left-side event, these are just the ending joint probabilities divided by 1,producing the same numbers. Doing this much the posterior tree becomes:

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At this point we have found all the information in the second table of the tablemethod of Bayesian revision. We now complete the Bayesian revision on theposterior tree, which provides the information found on the third table.

We continue the process of dividing joint probabilities, which provides theconditional probabilities. Starting with the top oil event (which comes after a“positive” seismic outcome), the conditional probability of oil is computed as

P(oil/positive) =0.006

0.0456≈ 0.13158

As stated earlier, we may wish to give the exact value by writing the conditionalprobability as an unreduced fraction, i.e. 60

456 . Writing all six conditional probabil-ities rounded to five decimal places the posterior tree becomes:


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While we have developed the prior and posterior trees slowly to illustrate theprocess, when doing this in practice all that is needed is a single sheet of paper onwhich both trees are written. This is shown in Figure 21.

This methodology for performing Bayesian revision is entirely optional – ifyou feel more comfortable using the table method, then by all means use the tablemethod. However, you should try the tree method of a couple of problems beforeyou make this decision. When you become used to both methods, there reallyisn’t much difference in time.

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The main advantage of the table approach is that it can be done on a spread-sheet (though this won’t help you on a test). The main advantage of the treemethod is that it ties in nicely with the decision tree for which the Bayesian re-vision is being performed. Once the posterior tree has been completed, it is veryeasy to see where to transfer the marginal and conditional probabilities onto thedecision tree.

To review, the steps involved in making the prior and posterior trees are:

1. Taking the events from the decision tree in reverse order, make the prior treeshowing its shape, putting labels on the outcomes, and write the marginaland conditional probabilities.

2. Using multiplication, find all the joint probabilities on the prior tree, andverify that they sum to 1.

3. Taking the events from the prior tree in reverse order, make the posteriortree showing its shape, and put labels on the outcomes.

4. Transfer the final (right side) joint probabilities from the prior tree to theappropriate places (i.e. matching pairs of outcomes) on the right side of theposterior tree.

5. Using addition, find the other joint probabilities, and verify that the initial(extreme left side) joint probability is 1.

6. For every outcome branch, find the probability on the branch by dividingthe joint probability at the end of the branch by the joint probability at thebeginning of the branch.

8.2.4 Solution and Recommendation

Now we can complete the formulation of the decision tree. Again, the consistencyin the vertical labelling of the outcomes makes the transfer of the marginal andconditional probabilities from the posterior tree to the decision tree very easy.With Figures 20 and 21 in hand, we can easily see what needs to be transferredwhere. The events are in the same order, the difference being that in the decisiontree there is a decision between the two events.

There are a total of three marginal probabilities, and six conditional probabili-ties, to be transferred from the posterior tree to the decision tree. (Alternatively, ifthe table method is used, we transfer these numbers from the third table.) Doingthis, we obtain the decision tree shown in Figure 22.


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This tree is then rolled-back to obtain a recommendation. The rolled-back treewith highlighted recommended alternatives is shown in Figure 23. The condi-tional probabilities are shown to five-place accuracy, and the rolled-back payoffsare shown to the nearest dollar, but in fact all this information was stored to theaccuracy of the calculator.

Recommendation

Do the seismic test. If the result is “positive”, then drill for oil; otherwise, do notdrill. The ranking payoff is $63,200.

8.2.5 Commentary

In problems of this type, it is often useful to find the EVPI at the outset. If thecost of obtaining information is higher than the EVPI, then we can eliminate thealternative to seek information. In this case, the EVPI is:

[0.01(40,000,000−3,000,000)+0.99(0)]−0 = 370,000

The cost of the seismic test, which is $40,000, is much less than $370,000. Hence,the seismic test cannot be trivially eliminated.

You may be wondering why we even considered the “drill” alternative after a“negative” seismic test. This is because the information is not perfect, so there isa chance that doing the non-obvious thing may be right.

Sensitivity analysis can be performed as usual, except on the marginal andconditional probabilities which come after the seismic test alternative. To considera change to the probabilities, we would have to go back to the prior tree, changingat least two numbers (one increasing by ∆, and another decreasing by ∆), andthen observing how these changes cascade through the posterior tree. This sort ofanalysis can quickly become quite complex.


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8.3 ExerciseA consumer products company has identified a potential product which wouldrequire $1,800,000 in start-up costs to launch. If it turns out to be a major success,there will be a $10,000,000 contribution to profit. A minor success would give aprofit contribution of $2,000,000, while a failure would have a profit contributionof only $500,000. The company is most worried about this third possibility, sincein this case the net profit would be $500,000 minus $1,800,000, i.e. a loss of$1,300,000. In the past, only one new product in twenty became a major success,while three-quarters of them became failures; there is no reason to suspect thatthis product would be any different from the rest.

Some of their competitors use an outside independent market research firm togive them advice about new products. The fee for the research firm is $50,000; inreturn, the consumer products company would be told that the proposed producteither “looks good” or “looks bad”. The research company had established a trackrecord which gave them confidence about saying the following:

- if a product would be a major success, they would say “looks good” withprobability 0.8;

- if a product would be a minor success, they would say “looks bad” withprobability 0.7;

- if a product would be a failure, they would say “looks bad” with probability0.9.

Develop a decision tree and solve it to obtain a recommendation for the con-sumer products company.



9.1 Introduction

In this section we look at sequential Bayesian revision. This is used when newinformation is used to revise the probabilities, and then more new informationarrives. This necessitates a second revision of the probabilities. The example isquite long, so it has been analyzed one paragraph at a time. Part A can be solvedsimply by using a payoff matrix. From this we can find the EVPI, which gives anupper bound to the expected value of any information. We see that the cost of thisinformation is less than the EVPI, so in Part B we proceed with making a decisiontree to analyze this situation. This leads to a decision tree and prior and posteriortrees which are very similar to those of the oil drilling example of the previoussection. Then in Part C we present the concept which is new to this section.

9.2 Example – Wood Finishers


Wood Finishers produces a line of executive-type office desks. A high qualitydesk nets a profit of $1000. A poorly-made desk, however, due to refunds andloss of customer goodwill, has a net loss of $6000. (High or low quality does notrefer to the visible part of the desk, which is always of high quality, but rather tothe ability to last years of use.) Ninety-six per cent of the production is of highquality. Adding a rework section to the assembly line would guarantee that eachdesk would be of high quality, but this would cost $400 for each desk reworked.

Suppose that the company can inspect each desk at a cost of $50 (per desk)before deciding whether or not to rework it. The results of the inspection at thisstation would be one of the following: “looks good,” “inconclusive,” or “looksbad.” If the inspected desk is of high quality, then there is a 70% chance that theinspection will indicate “looks good,” a 20% chance that the inspection will beinconclusive, and a 10% chance of a “looks bad” result. If the inspected desk isof poor quality then there is a 90% chance of a “looks bad” result, an 8% chancethat the inspection will be inconclusive, and a 2% chance of “looks good” result.

In addition to the inspection station mentioned above, Wood Finishers can adda second inspection station (which can only inspect a desk which was inspectedat the first station). The result of the inspection at the second station is reported asbeing either “pass” or “fail.” If the desk is of high quality there is a 95% chance


of a “pass.” If the desk is of poor quality there is a 97% chance of a “fail.” Thecost of this test would be $60 per desk inspected. For now, let us suppose that weonly need to consider adding the second station if the result of the first test was“inconclusive”.

9.2.2 Part A

The first paragraph of the problem description contains a decision (rework) andan event (quality):

Wood Finishers produces a line of executive-type office desks. Ahigh quality desk nets a profit of $1000. A poorly-made desk, how-ever, due to refunds and loss of customer goodwill, has a net loss of$6000. (High or low quality does not refer to the visible part of thedesk, which is always of high quality, but rather to the ability to lastyears of use.) Ninety-six per cent of the production is of high quality.Adding a rework section to the assembly line would guarantee thateach desk would be of high quality, but this would cost $400 for eachdesk reworked.

We can analyze this situation with a payoff matrix or a decision tree. Therework decision has two alternatives: do not rework; and rework. If the rework isnot done, then there is an event for which there are two possible outcomes: highquality; and low quality.

We do not know how many desks are being made, so we cannot find the abso-lute level of profit. Instead, we will work out the profit per desk.

Rework Quality EventDecision OutcomesAlternatives High Low EVDo not Rework 1000 −6000 720Rework 600 600 600

Prob. 0.96 0.04

Hence, we would choose to not rework, for an expected profit of $720 perdesk. The EV with PI is:

EV with PI = 0.96(1000)+0.04(600)= 960+24= 984


Hence the EVPI is $984 − $720 = $264 per desk.Although a payoff matrix is perfectly adequate for solving this part of the

problem, it is also possible to use a decision tree. Using a tree now helps whendrawing the tree for Part B (the second paragraph), because the large tree containsthree sub-trees which are similar to the one drawn here. Using a tree we obtain:

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9.2.3 Part B

The second paragraph adds an inspection decision and an inspection event:

Suppose that the company can inspect each desk at a cost of $50(per desk) before deciding whether or not to rework it. The results ofthe inspection at this station would be one of the following: “looksgood,” “inconclusive,” or “looks bad.” If the inspected desk is of highquality, then there is a 70% chance that the inspection will indicate“looks good,” a 20% chance that the inspection will be inconclusive,and a 10% chance of a “looks bad” result. If the inspected desk is ofpoor quality then there is a 90% chance of a “looks bad” result, an8% chance that the inspection will be inconclusive, and a 2% chanceof “looks good” result.

The $50 cost (per desk) of doing the inspection is much less than the EVPI,which is $264 (per desk). Hence we must proceed with the analysis to see if itwould be worthwhile to do the inspection.

The inspection decision must precede the inspection event, which in turn mustprecede the main (rework) decision. After the “no inspection” alternative, weare left with the situation which was analyzed in Part A. Therefore, we do notneed to redraw this section, but instead merely write the ranking payoff which wecalculated to be $720. For now, we cannot write the probabilities on the inspectionoutcomes, as these must be determined using Bayesian revision. This part of thetree is shown in Figure 24.

After every outcome node we have a sub-tree which resembles the tree madein Part A. Indeed, the only differences are the probabilities, which we need tocalculate using Bayesian revision. The tree for Part B without the probabilities isshown in Figure 25. We then draw the prior and posterior trees for the Bayesianrevision. The completed trees are shown in Figure 26. These probabilities aretransferred to the decision tree shown in Figure 27. Finally, the tree is rolled backto obtain a recommendation. The rolled-back tree is shown in Figure 28. Basedon this, the recommendation is:

Recommendation

Inspect every desk and rework it if and only if a “looks bad” result is obtained.The ranking payoff is $869.20 per desk.


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9.2.4 Part C

Solving Parts A and B required knowing only what we saw in the previous section.However, dealing with the third paragraph requires a bit of thought on how tobegin the prior tree:

In addition to the inspection station mentioned above, Wood Fin-ishers can add a second inspection station (which can only inspect adesk which was inspected at the first station). The result of the inspec-tion at the second station is reported as being either “pass” or “fail.”If the desk is of high quality there is a 95% chance of a “pass.” If thedesk is of poor quality there is a 97% chance of a “fail.” The cost ofthis test would be $60 per desk inspected. For now, let us supposethat we only need to consider adding the second station if the resultof the first test was “inconclusive”.

We will only modify the part of the tree which is affected by this paragraph.We begin with the “inconclusive” outcome branch. After this we can either do ornot do the second inspection. If we choose the “no 2nd inspection” alternative,then we are left with the situation which was analyzed in Part B. Therefore, we donot need to redraw this section, but instead merely write the ranking payoff whichwe calculated to be $885.25.

On the other hand, if we choose to do the second test, then we have an alterna-tive branch with a $60 cost gate, followed by the test event with its two outcomes,“pass”, and “fail”. For now, we cannot write the probabilities on the inspectionoutcomes, as these must be determined using Bayesian revision. This part of thetree is shown in Figure 29. After both outcome nodes we have a sub-tree whichresembles the tree made in Part A. The tree for Part C without the probabilities isshown in Figure 30.

The only tricky thing about the Bayesian revision is the determination of thebeginning probabilities. The prior tree begins with the “high” and “low” qual-ity outcomes, but the associated probabilities are not the 0.96 and 0.04 that wehad originally. Instead, we must use the probabilities which are conditional onthe first test result being “inconclusive”, because they come after the “incon-clusive” outcome. Hence we want P(high/inconclusive), which is 0.98361, andP(low/inconclusive), which is 0.01639.

We then draw the prior and posterior trees for the Bayesian revision. Thecompleted trees are shown in Figure 31. These probabilities are transferred to thedecision tree shown in Figure 32. Finally, the tree is rolled back, which is shown in


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Figure 33. Because the payoff after the “inconclusive” branch has increased, theoverall recommendation is changed. The increased payoff will cascade throughthe tree for Part B, increasing the payoff at the outset by:

0.1952(910.52−885.25) = 4.93

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Recommendation

Inspect every desk. If a “looks good” result is obtained, then do not rework it. Ifan “inconclusive” result is obtained, then do the second inspection, and rework itif and only if the result of the second inspection is “fail”. If the result of the firstinspection is “looks bad”, then rework it. The ranking payoff is $874.13 per desk.

9.3 ExerciseNow suppose that in the desk rework problem the second inspection could also beused after a “looks good” or a “looks bad” outcome from the first test. Determineif the second inspection would be used in either (or both) of these situations, andif so restate the recommendation and the revised ranking payoff at the outset ofthe tree.


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10.1 IntroductionIn this section there is only one Bayesian revision to be done, but to obtain theprobabilities for the prior tree we need to use the formula for the binomial prob-ability distribution. The binomial is discussed extensively in Mathematics forManagement Science. A very quick overview is provided here.

10.1.1 Binomial Probability Distribution

Consider an event with two outcomes, one called “success” which occurs withprobability p, and the other called “failure” which occurs with probability 1− p.If this event is repeated n times (where each event is identical and independentof the others), then the probability of obtaining (exactly) x successes (where x isan integer from 0 to n inclusive) is given by the following binomial probabilityformula:

P(x;n, p) =n!

x!(n− x)!px(1− p)n−x

This formula is accessed on a spreadsheet using the BINOMDIST function.The expected value of this distribution is simply pn.

Example An assembly line has been producing defects at a rate of 5%. A sam-ple of twenty units is taken from the line. What is the probability of finding threedefective units?

[Note: Unless indicated to the contrary, in all problems of this type we meanexactly x, not at least x.]

Solution Here n = 20 and x = 3. A defect rate of 5% means that a unitchosen from the line at random has probability p = 0.05 of being defective. Hence

P(3;20, .05) =20!

3!(20−3)!.053(1− .05)20−3

=(20)(19)(18)17!

(6)17!.053.9517

= (1140)(0.000125)(0.418120...)= 0.059582...


Shortcuts When x is at or near either 0 or n, the formula simplifies greatly.For example if x = 0, then P(0;n, p) = (1− p)n, and if x = n, P(n;n, p) = pn.If x = 1, then P(1;n, p) = np(1− p)n−1, and if x = n− 1, then P(n− 1;n, p) =npn−1(1− p). For 2≤ x≤ n−2, the original formula should be used.

10.1.2 Destructive vs. Non-Destructive Testing

When something is tested, the test may destroy the item. An example of this is acar operated by remote control which is driven at a research laboratory into a wallto test its ability to withstand a real collision. This is an example of destructivetesting.

An example of a non-destructive test is that of testing the voltage of a battery.The battery is not destroyed by the test, so provided that it worked during the testit can then be used in the intended manner. For our purposes, if after testing anacceptable product it can be sent to a customer, then the test is non-destructive. Ifafter testing the item cannot be sent to a customer, then the test is destructive.

10.2 Example


At the outset of each production day, a machine which has a capacity to make1000 units per day is cleaned and made ready for use. On ninety percent of theproduction days, the defect rate is only 1%. Defects are independent from oneunit to the next. The other ten percent of the time, the defect rate is 20%. Untilnow, they have simply produced all 1000 units and have hoped for the best. Usinga special reset process would guarantee that the defect rate will be only 1%, butit costs $600, and during the time it takes to reset the machine using this specialprocess, 30 units could have been produced. Each unit costs $10 to make, andsells for $14. If a unit is defective, the customer receives a refund, plus a $15 giftcertificate to compensate for the inconvenience.

The production manager has suggested that after the machine has been cleanedand made ready for use that they produce two units, and then stop the productionwhile these units are tested (destructively). The direct testing cost is two times $10for the destroyed units plus another $70 for a total of $90, but there is an indirectcost as well: in the time it takes to test the two units, they could have producedanother 18 units.


10.2.2 Solution

This problem contains a major decision (whether or not to reset the machine),and an information decision (whether or not to destructively test two units). Theproblem is complicated enough that we should start by ignoring the informationdecision. Once we have determined what to do when no testing is done (Part A),we can then consider whether or not two units should be tested (Part B).

Part A If the special setup is not used, then 1000 units are produced. The pro-duction cost would be 1000($10) = $10,000. There is a 0.9 probability that defectsare produced at a rate of 1%. This does not necessarily imply that exactly 10 ofthe 1000 are defective, but rather than any one unit chosen at random has a 1%chance of being defective. The number of defectives in a 1000 unit productionrun is governed by the binomial probability distribution, for which the expectednumber of defectives is pn = .01(1000) = 10. There is 0.1 probability that defectsare produced at a rate of 20%. There is a revenue of $14 for each unit, but for eachdefective unit the $14 is returned to the customer plus there’s a further loss $15for a total of $29 each. Hence the net total revenue when 1% of the 1000 units aredefective is:3

1000($14−0.01($14+$15)) = 1000($14−0.01($29))= 1000($14.00−$0.29)= 1000($13.71)= $13,710

When 20% of the units are defective the net total revenue becomes:

1000($14−0.20($29)) = 1000($14.00−5.80)= 1000($8.20)= $8,200

3An alternate approach is to think of 99% as the non-defective rate, with a gain of $14 for agood unit and a net loss of $15 for a defective unit. Hence the net revenue is:

1000(0.99($14)+0.01(−$15)) = 1000($13.86−$0.15)= 1000($13.71)= $13,710


If the special setup is used, then 1000−30 = 970 units are produced. The cost willbe $600 for the setup plus a production cost of 970($10) for a total of $10,300.The defect rate will be 1%, hence we could find the net total revenue from the 970units as:

970($14−0.01($29)) = $13,298.70

An alternate approach it to recognize that since 970 is 97% of 1000, producing970 units gives 97% of the net revenue associated with producing 1000 units:

(9701000

)$13,710 = 0.97($13,710)

= $13,298.70

The tree has a fair amount of information on it, so we will build it in stages.First, we show the decision with its two alternatives. To save space, we simplywrite “no” to mean that the special setup is not done. Associated with this alterna-tive is the number of units made, which is 1000. Similarly, “yes” means that thespecial setup is done, which causes the number of units made to be 30 fewer than1000, which is 970.

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Part B We now come to the information decision, that is to say we must choosebetween testing or not testing two units. Intuitively, we hope to find that neitherunit is defective, for then it is highly likely that the machine is producing defectsat a 1% rate rather than a 20% rate. On the other hand, if either (or both) of thetwo units is defective, then the defect rate is more likely to be 20% rather than1%, and we will probably want to use the special reset procedure. The analysispresented here quantifies these intuitive ideas.

To make and then test two units would cost 2($10) = $20 plus $70 for a totalof $90. This is less than the EVPI, so we proceed with further analysis.4

4The indirect cost of not being able to produce 18 units while the testing is in progress adds to


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The tree begins with the decision concerning the testing of the two units. Ifthe testing is not done, then we have the situation which was analyzed in Part A,which had a ranking payoff of $3159. If the testing is done, then we have thetesting event. Although both units are destroyed by the test, what we find outis whether or not the tested units were defective. There are three possibilities:neither unit is defective; one unit is defective (it does not matter which); or bothunits are defective. Hence we need three outcome branches. The probabilities onthe branches will be found using Bayesian revision. The tree so far is shown inFigure 34.

At the end of every outcome branch we have a tree similar to the one found

the $90, but we cannot quantify this without further analysis. Indeed, we need not find this figure,because to do so would require more work than is required to solve the original problem.


in Part A. Because two units of the day’s production have already been made (forthe test), and because during the time of the test eighteen units could have beenmade, the day’s production capacity has been reduced by 2+18 = 20 units by thetime we reach the major decision. So, if the special setup is not done, then wewould produce 1000−20 = 980 units; if the special setup is used then we wouldproduce 970−20 = 950 units during the rest of the day. These changes affect thecosts on the cost gates, and the final net revenues. These final payoffs are foundby pro-rating what the figure would be if 1000 units were produced and sold.Adding these sub-trees with all costs and revenues (but without the probabilities)we obtain the tree shown in Figure 35.

Putting the two events from the decision tree in reverse order, we create theprior tree, putting the outcomes of the defect rate event first, and then the outcomesof the testing event (the number of defectives found) second. In doing this, wecan easily write the marginal probabilities of the defect rate event. However, theconditional probabilities for the second event need to be calculated. Before doingthese calculations the prior tree is:

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Now we wish to compute the six conditional probabilities.

We need to determine the parameters n and p for the binomial probabilitydistribution. Two units are being tested, therefore n = 2. The number of defectsx can range from 0 to n, i.e. in this situation x can be 0 (neither is defective), 1(exactly one is defective), or 2 (both are defective).

To obtain p, consider first the top branch where the defect rate is 1%. If anyunit is pulled at random from the production line, there is a 1% (or 0.01) chancethat it is defective, and therefore a 99% that it is not defective. Hence p = 0.01,i.e. it is simply the defect rate. For example, the probability of finding neither unitdefective (x = 0) is (1− .01)2 = 0.9801. On the bottom branch, where the defectrate is 20%, we have the binomial probability distribution with n = 2 and p = 0.2.and x being 0, 1, or 2. All six numbers (two values for p times three values for x)are worked out in the following table.

Defect Rate (p) Number of Defects (x) Formula (n = 2) Probability0 (1− .01)2 0.9801

1% 1 2(.01)(1− .01) 0.01982 (.01)2 0.00010 (1− .2)2 0.64

20% 1 2(.2)(1− .2) 0.322 (.2)2 0.04

Putting these six conditional probabilities on the tree we have:


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At the outset of the day, two units should be made and be tested. If neither unitis defective, then continue to produce for the rest of the day, making 980 unitsfor sale. However, if either (or both) of the units is defective, then use the specialreset procedure, and then proceed with making 950 units for the rest of the day.The ranking profit is $3161.87.

10.3 CommentaryThough the situation here involved destructive testing, it’s not hard to change theanalysis to handle non-destructive testing. Since only non-defective units wouldbe sent on to a customer, the final payoffs would increase by $14 when one defec-tive is found (because this means that the other unit was fine), and would increaseby $28 if neither unit was found to be defective.

The number of units to be tested (should the test be done at all) was set in theproblem description to be two. Perhaps a higher ranking payoff could be obtainedif only one unit were tested, or if three (or possibly more) units were tested. Toanswer this question, we would have to begin with a decision about the numberof units to be tested, be it 0, 1, 2, 3, and so on. It may seem as though up to1000 units could be tested in theory (though this wouldn’t make any sense whenthe test is destructive). A practical limit can be established by determining theranking profit when 0, 1, 2 and so on units are tested. If the profit obtained fortesting n +1 units is higher than that for testing n of them, then keep going on ton + 2. However, if it’s lower, then a peak has been obtained and there’s no pointin considering testing one more unit.

Sometimes, the testing of multiple units can be done sequentially rather thansimultaneously. In that case, we have a situation like that of the previous section.


10.4 ExerciseEvery day a machine can produce 1000 computer chips. The machine has bothnormal and erratic days. On 4 days out of 5 (normal day), the defect rate is only2%, but on 1 day in 5 (erratic day) the defect rate is 16%. A good chip is worth$50, but a bad one has a net cost of $300. As an alternative to producing defectsas described above, they could use a special reset procedure. This procedure costs$14,000 and takes up 50 units worth of production time; using it guarantees thatthere will be no defects.

(a) What should the company do?

(b) What is the expected value of perfect information?

Now suppose that they can make three units at the outset of the day and thentest all three units simultaneously. The test is destructive. The test would cost$200, and takes up 47 units worth of production time. Assume that if the test isdone, and if two or three defective units are found, that they will reset the machine.

(c) Should they do the test, and if so what should they do if none or one defectiveunit is found? Note: You may omit any calculations of posterior probabilitieswhich are not needed for the decision tree.


11 Linear Optimization 1

In this the first of four lectures on the subject of linear optimization , we beginwith an example to illustrate this topic. We then consider variations which lead toa more general understanding of what linear optimization is.

11.1 Example – Cement ProblemA cement company makes two types of cement, which they market under reg-istered tradenames, but for our purposes we will simply call them Type 1 andType 2. Cement is sold by the Tonne (= 1000 kg), and production is measured inTonnes per Day, abbreviated as TPD. The company has contractual sales obliga-tions to produce at least 40 TPD of Type 1 cement, and at least 30 TPD of Type 2cement.

The physical capacity of the plant, which is governed by such things as con-veyor belt speed, storage size, and so on, is limited to 200 TPD. A new labouragreement has increased the length of breaks, and restricts and makes more costlythe use of overtime. The company therefore wishes to find its best productionplan using the new work rules with everyone working a 40 hour week. Work ismeasured in this company by the labour-hour, which is one person working forone hour. Each type of cement is made in three departments, labeled A, B, and C.To make each Tonne of Type 1 cement requires three labour-hours in DepartmentA, one and a half labour-hours in Department B, and four labour-hours in Depart-ment C. The amounts of work per Tonne of Type 2 cement are two, five, and sixlabour-hours in Departments A, B, and C respectively.

Based on the current authorized strength in each department, and factoringin allowances for breaks, absenteeism, and so on, Department A has 585 labour-hours available each day. Departments B and C are allowed to use up to 500 and900 labour-hours per day respectively. These are the most they can use for themaking of cement. If a department has some time leftover (i.e. if the time to makethe cement is less than the number of labour-hours available), then the workerswill be idle for a few minutes at the end of the day. The three departments requireworkers with very different training and skills, so the possibility of transferringemployees from one department to another is not something that is factored intothe planning process.

Taking the market price of each type of cement and from this subtracting allthe variable costs of making the cement leaves the company with a profit of $8per Tonne of Type 1 cement, and $10 per Tonne of Type 2 cement. There are also


fixed costs (taxes, security, and so on) which total $1400 per day. The companywants to know how much should be produced of each type of cement, so that theprofit is maximized.

11.2 Making a Model

11.2.1 Verbal, Algebraic, and Spreadsheet Models

Someone has already gone into the cement plant to obtain the relevant facts andfrom this research a verbal model has been made, which appears as the “ProblemDescription”. This model is complete in that the final sentence states the essenceof the problem, and gives the objective. Often, only the data is provided with ageneral question of the “what should the company do?” variety.

In order to solve the problem, we need to transform the verbal model into analgebraic model . Models with just two variables can be solved graphically, but

of course this is of limited practical use. Algebraic models can be solved by asoftware package designed for this purpose, up to a size limit set by the writersof the software. One such package is LINDO, to be described later. Anotheroption is to transform the algebraic model into a spreadsheet model . It can thenbe solved by Excel or a similar package, as we shall later see. Indeed, for a verysimple problem like the cement problem, one can bypass the algebraic model andgo directly to the spreadsheet model. However, this shortcut will not help us formore complex models, so we will not take this route.

11.2.2 Definition of the Variables

In beginning to make an algebraic model, we wish to determine the unknownswhich will be represented using variables. The emphasis here is to focus in on theunknowns which are at the heart of the problem, and to skip those things which caneasily be determined once the essential unknowns have been determined. In thisproblem, these unknowns come from the last sentence of the problem description:the TPD of Type 1 cement that should be made; and the TPD of Type 2 cement thatshould be made. Everything else, such as the total profit, or the idle time (if any) inone of the departments, can be determined if we know these two essential things.With just two unknowns we could label them x and y, but it is more common to usesubscripts, calling them x1 and x2. This way of labelling the unknowns is what isrequired when we consider realistically sized models, which can have thousandsof variables. Hence we have:


x1 = the number of TPD of Type 1 cement madex2 = the number of TPD of Type 2 cement made

It is very important that the definitions of the variables be made as clearly aspossible. For example, a shorthand such as “x1 = Type 1” is not acceptable.

11.2.3 The Objective Function

We now need to write an expression for the profit in terms of the variables. Look-ing at the Type 1 cement alone, one Tonne gives a contribution of $8 to the profit.Since we are producing x1 TPD, the daily profit from the production of Type 1cement is 8x1. Similarly, the daily profit from the production of Type 2 cement is10x2. Putting these together we have 8x1 + 10x2. The $1400 in daily fixed costsneeds to be subtracted from this expression, but most traditional software (beforespreadsheets) is not set up to handle this. Therefore we omit subtracting it fornow, but we can easily subtract it at the very end when everything else has beencalculated. We write the word maximize in front of the expression, because that isthe objective in this situation. The word maximize is often abbreviated to simplymax. What we call the objective function is:

maximize f (x1,x2) = 8x1 +10x2

In this document we call refer to the value of the objective function as OFV (forobjective function value). (A more traditional (but less intuitive) symbol is Z.)Again, anticipating that we will use software such as LINDO for these problems,we do not write the “ f (x1,x2) =” (or anything else such as OFV or Z), becausethe syntax of the software does not handle this notation. Hence we simplify theobjective function to:

maximize 8x1 +10x2

11.2.4 The Constraints

The objective function is subject to a set of constraints which represent, in thisexample, the minimum sales contract requirements, the limit on total production,and the limit on labour availability in each of the three departments. Also presentin this and in almost every linear optimization model are non-negativity restric-tions on the variables. Since we cannot produce a negative quantity of cement, werequire that x1 be greater than or equal to 0, and that x2 be greater than or equal to0. When writing the algebraic model, we will indicate this by writing x1 ≥ 0 and


x2 ≥ 0 at the end, or in short form simply x1,x2 ≥ 0. (Most software programsassume these restrictions and therefore they do not need to be explicitly entered.)By convention, this short form is only used for the non-negativity restrictions; itis not used for the other constraints.

The first three constraints are quite easy. Their sales contracts for 40 TPD ofType 1 cement and 30 TPD of Type 2 cement means that we must have x1 ≥ 40and x2 ≥ 30. Theoretically, these constraints make the non-negativity restrictionssuperfluous, but we keep them anyway. This is because the model might laterchange – should the sales constraints be removed, then the non-negativity restric-tions would become the new lower bounds on the variables. The third constraintthat the total production cannot exceed 200 TPD is represented by x1 + x2 ≤ 200.So far the constraint list is:

Type 1 Sales x1 ≥ 40Type 2 Sales x2 ≥ 30

Total Production x1 + x2 ≤ 200

Now we determine the three labour constraints, one for each department. The datafor these constraints is written both from a product perspective and a departmentalperspective. From the product perspective we have:

To make each Tonne of Type 1 cement requires three labour-hoursin Department A, one and a half labour-hours in Department B, andfour labour-hours in Department C. The amounts of work per Tonneof Type 2 cement are two, five, and six labour-hours in DepartmentsA, B, and C respectively.

From the departmental perspective we have:

Based on the current authorized strength in each department, andfactoring in allowances for breaks, absenteeism, and so on, Depart-ment A has 585 labour-hours available each day. Departments B andC could use up to 500 and 900 labour-hours per day respectively.

It may be helpful to put all this data into a table with two rows, one for each typeof cement, and three columns for the labour-hours to make one Tonne of Type 1,the labour-hours to make one Tonne of Type 2, and the number of labour-hoursavailable per day. Note that in this example, the data from the problem descriptiongo into the columns. (Be careful about this, in other problems some of the datamight go into the rows).


Labour-Hours per Tonne Labour-HoursDepartment of Type 1 Cement of Type 2 Cement Available each day

A 3 2 585B 1.5 5 500C 4 6 900

In each department, the labour-hours (LH) used cannot exceed the labour-hoursavailable. Let’s look at Department A in particular.

LH used ≤ LH availableLH to make Type 1+LH to make Type 2 ≤ 585

3x1 +2x2 ≤ 585

Once the pattern has been established, it becomes easy to write the labour con-straints for Departments B and C. For Department B we must have 1.5x1 +5x2 ≤500, and for Department C, we require that 4x1 + 6x2 ≤ 900. Once you have be-come used to problems like this, you may wish to write the constraints directlyfrom the problem description without doing the table as an intermediate step. Insummary the labour constraints are:

Dept. A Labour 3x1 + 2x2 ≤ 585Dept. B Labour 1.5x1 + 5x2 ≤ 500Dept. C Labour 4x1 + 6x2 ≤ 900

11.2.5 Summary

The algebraic model needs to be summarized in one place. This summary consistsof: the definition of the variables; the objective function; the words subject tofollowed by the constraints with their word descriptions; and the non-negativityrestrictions written in one line at the end. For questions of this type on a test orexamination in this course, just writing such a summary will suffice. Doing thiswe have:



maximize 8x1 + 10x2

subject to


Total Production x1 + x2 ≤ 200Dept. A Labour 3x1 + 2x2 ≤ 585Dept. B Labour 1.5x1 + 5x2 ≤ 500Dept. C Labour 4x1 + 6x2 ≤ 900

non-negativity x1 , x2 ≥ 0

11.3 Graphical Solution

Because the methodology of graphing linear optimization models is studied ex-tensively in Mathematics for Management Science, we shall omit most of thedetails in solving the Cement Problem. From the Total Production constraintx1 + x2 ≤ 200, we can see that a 200 by 200 grid is adequate for solving thisproblem. Recall that the convention is that the x1 variable is on the horizontalaxis, and the x2 variable is on the vertical axis. A picture of the grid, with worddescriptions on the axes, is shown in Figure 39. Though you will no doubt workwith lined paper, here we suppress the printing of the grid lines to make the plottedlines easier to see.

We now need to plot the boundaries of the six constraints, and to do this wemust find two points for each boundary line. Also, since all the constraints areinequalities, for each we must determine the direction of the arrow which indicatesthe inequality. Some of the constraints are easy. The first one, x1 ≥ 40, is simplya vertical line through x1 = 40, and the arrow points to the right (because theinequality makes the origin false). The second one, x2 ≥ 30, is a horizontal linethrough x2 = 30, with the arrow pointing upwards. The third constraint, which isx1 +x2 ≤ 200, passes through 200 on both axes. Since the origin is true, the arrowpoints towards the origin. The other three constraints require some calculations.

The Department A Labour constraint is 3x1 +2x2≤ 585. The boundary line ofthis constraint is given by the equation 3x1 +2x2 = 585. Setting x1 = 0, we obtainx2 = 292.5, which is off the grid. When this happens we try to find an interceptionpoint on either the right-hand side or top boundary of the grid. In this situation,we find the value of x1 where the line crosses the top boundary, at which x2 = 200.


0 50 100 150 200Tonnes per Day of Type 1 Cement

0

50

100

150

200

Tonn

espe

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Figure 39: Cement Problem – Axes


Hence we solve

3x1 +2(200) = 5853x1 +400 = 585

3x1 = 185x1 = 61.666...

Hence the line passes through the point (6123 ,200). Now setting x2 = 0, we obtain

x1 = 195, which is on the grid. Therefore the boundary of the Department ALabour constraint passes through the points (612

3 ,200) and (195,0).For Department B we require that 1.5x1 + 5x2 ≤ 500. Setting x1 = 0, we

obtain x2 = 100, which is fine. Setting x2 = 0 makes x1 = 333.333..., which is offthe grid. Therefore we set x1 = 200 (the right-hand side of the grid), and solve toobtain x2 = 40. For Department C we require that 4x1 +6x2≤ 900. Setting x1 = 0,we obtain x2 = 150, which is fine. Setting x2 = 0 makes x1 = 225, which is offthe grid. Therefore we set x1 = 200 and solve to obtain x2 = 16.666.... The originis true for all three labour constaints, so the arrow for each one points towards theorigin.

In summary, the points for the boundary lines of the constraints are as follows:

Constraint First Point Second PointType 1 Sales (40,0) verticalType 2 Sales (0,30) horizontal

Total Production (0,200) (200,0)Dept. A Labour (612

3 ,200) (195,0)Dept. B Labour (0,100) (200,40)Dept. C Labour (0,150) (200,162

3 )

A picture of this is shown in Figure 40. We must remember not to plot thepoints for the constraints backwards. For example, (0,100) lies 100 points abovethe origin, not 100 points to the right. The title of each constraint is written next toits boundary line. With these titles on the constraints, and with word descriptionson the axes, it makes the graph easy to understand.

We have not drawn arrows to explicitly indicate the two non-negativity restric-tions, but of course these restrictions are present nevertheless. Considering all theconstraints and the non-negativity restrictions, we find the feasible region. Thisregion, which is labelled and highlighted, is shown in Figure 41.

We now find a trial isovalue line, a line in which all points have the sameobjective function value. This being done, we then the find the optimal isovalue



0

50

100

150

200

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Type 1 Sales

Type 2Sales

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Type 2Sales Feasible Region

Figure 41: Cement Problem – Feasible Region


line, a line parallel with the trial isovalue line which passes through the optimalsolution. This procedure is discussed extensively in Mathematics for ManagementScience; a quick summary is provided here.

In general the objective function is of the form

max or min c1x1 + c2x2

Except when either c1 = 0 or c2 = 0 (which lead to vertical and horizontal isovaluelines respectively), we pick any value v (except 0) and solve

c1x1 + c2x2 = v

Using this equation we set each variable equal to 0 to obtain the intercepts on theaxes. These two points define the isovalue line, which is indicated by drawing adashed line between them.

For example, suppose we have 8x1 +10x2 in the objective function and wish totry v = 200, i.e. 8x1 +10x2 = 200. If x1 = 0, then 10x2 = 200, and hence x2 = 20.If x2 = 0, then 8x1 = 200 and hence x1 = 25. Therefore this particular isovalueline passes through (x1,x2) = (0,20) and (25,0), and we connect these two pointswith a dashed line.

However, while any non-zero value of v can be used, the special case where vis the product of c1 and c2 leads to an easy shortcut:

c1x1 + c2x2 = c1c2

If x1 = 0, then x2 = c1, and if x2 = 0, then x1 = c2. In other words, the shortcutis that the coefficient of x1 goes on the vertical axis, and the coefficient of x2 goeson the horizontal axis. Of course, the shortcut may produce points that are tooclose to the origin to be able to draw the connecting line, in which case we needto multiple each intercept by a number greater than 1. At the other extreme, theshortcut may produce intercepts which are off the page, in which case we need tomultiply each intercept by a number between 0 and 1.

For the example at hand, we seek to maximize 8x1 +10x2. Using the shortcutwe obtain a vertical intercept of 8 and a horizontal intercept of 10. However,this does not help us much here, because (0,8) and (10,0) are in the bottom left-hand corner, so it’s hard to draw the line between them. Hence we multiple eachof these intercepts by a number greater than 1. For example, multiplying eachintercept by 10 we obtain a vertical intercept of 80 and a horizontal intercept of100. These points (0,80) and (100,0) are what we would have obtained if we had


set 8x1 + 10x2 to a trial value of v = 800, and then solved for the intercepts. Ofcourse, there are an infinite number of trial isovalue lines, and any one of themwill suffice.

We now find a line parallel with the trial isovalue line, which just passesthrough the boundary5 of the feasible region such that the objective function valueis maximized. A convenient means of doing this is to use a rolling ruler, but a tri-angle moved along a straightedge will work too. This optimal isovalue line is alsodrawn on the graph (again, as a dashed line), and the optimal solution is identified.From the graph we can see that the binding constraints are the ones for (i) TotalProduction and (ii) Department C Labour, and that the optimal solution appearsto be at about x1 = 150 and x2 = 50. A picture of this is shown in Figure 42. Bytaking the boundaries of the two binding constraints, we can obtain the solutionexactly:

Total Production x1 + x2 = 200Dept. C Labour 4x1 + 6x2 = 900

6x1 + 6x2 = 12004x1 + 6x2 = 900

2x1 + 0x2 = 300

x1 = 150

By substituting x1 = 150 into x1 + x2 = 200, we obtain x2 = 50. The optimalmathematical solution is x∗1 = 150 and x∗2 = 50. The objective function value atthe point of optimality is

OFV∗ = 8x∗1 +10x∗2= 8(150)+10(50)= 1200+500= 1700

Going back to the original problem, the solution expressed in managerial termsis:

Recommendation

5Usually the optimal solution occurs at a corner of the feasible region, but when there is mul-tiple optimality an entire edge of the feasible region will be optimal. In any case, no part of theoptimal isovalue line will appear inside the feasible region.



0

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100

150

200

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Type

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emen

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Dept. A

Labour


Type 1 Sales

Type 2Sales Feasible Region

Trial

Optimal

OptimalSolution

Figure 42: Cement Problem – Optimal Solution


The cement plant should produce 150 Tonnes per day of Type 1 cement, and 50Tonnes per day of Type 2 cement, for a contribution to profit of $1700 per day.After deducting the $1400 daily fixed costs, the net profit is $300 per day.

11.4 Extensions

11.4.1 General Form

To be considered a linear optimization model, we must have a linear objectivefunction and linear constraints. By linear we mean that in each expression: (1)each variable appear on its own (we cannot have something like 7x1x2) ; and (2)every variable is multiplied by a number (which can be positive, zero, or negative)only (we cannot have something like 5

√x1). Thirdly, no uncertainty is permitted.

These three assumptions are common to both linear and integer optimization.If we require that these three assumptions hold, and that some or all of the

variables must be integer, then we have a variation called integer optimization.If, however, all the variables may take on fractional values (e.g. 7.2), then wehave a linear optimization model. In the literature, you will often see integeroptimization and linear optimization referred to as integer programming andlinear programming .

We can, however, make the model with minimization rather than maximiza-tion as the objective, and the constraints can be equalities as well as the moreusual ≤ and ≥ inequalities. The number on the right-hand side can be zero (con-sidered below). The non-negativity restrictions are almost always present, butthese can be removed when it is appropriate to do so. The number of variablesand the number of constraints is theoretically unlimited, but the software to solvethe model will come with limitations. There is no problem in practice solvingmodels with thousands of variables, and indeed solving models with millions ofvariables is sometimes done. With a few exceptions, requiring that some or all ofthe variables be integer can drastically cut down the size of problem which can behandled.

11.4.2 A Right-Hand Side Value of 0

It is possible for the number on the right-hand side of a constraint to be 0. Sucha constraint can arise, for example, if the amount of Type 1 cement productioncannot exceed two-thirds of the total amount produced. This is not necessarily


two-thirds of 200 TPD, because we do not know that in advance that this constraintwill be binding. Hence we have a new unknown which we will denote as x3.

x3 = the total production of cement in TPD

The variable x1 cannot exceed two-thirds of x3, which we write as

x1 ≤ 23x3

or equivalentlyx1− 2

3x3 ≤ 0

We need to add this proportion constraint to the existing model. Also, we needwrite the relationship between x3 and the other variables, which is:

x3 = x1 + x2

As a constraint with all variables on the left, this is:

x3− x1− x2 = 0

which can be re-arranged to:

x1 + x2− x3 = 0

With this third variable present the total production constraint can be written interms of it. Doing this and then adding the new third variable and the two newconstraints we obtain:

x1 = the number of TPD of Type 1 cement madex2 = the number of TPD of Type 2 cement madex3 = the total production of cement in TPD

maximize 8x1 + 10x2

subject to


Total Production x3 ≤ 200Dept. A Labour 3x1 + 2x2 ≤ 585Dept. B Labour 1.5x1 + 5x2 ≤ 500Dept. C Labour 4x1 + 6x2 ≤ 900

Proportion x1 − 23x3 ≤ 0

Balance x1 + x2 − x3 = 0

non-negativity x1 , x2 , x3 ≥ 0


Modelling in this manner is the best form in that no calculations are requiredfor any of the parameters. The advantages of doing no calculations are twofold:the original data are preserved; and there’s less likely to be a mistake. If a problemon a test asks you to formulate a model without doing any calculations, then theapproach used here is what is desired. If we wish to use a computer to solve thismodel, all we need do is make the minor calculation of converting the two-thirdsto decimal form, i.e. 0.66667.

However, if we wish to solve by using the graphical approach, we must revertto a model with just two variables. In general, a three-variable model cannot beconverted to a two-variable model, but we can do it here, because of the equalityconstraint.

x1 ≤ 23x3

x1 ≤ 23(x1 + x2)

3x1 ≤ 2x1 +2x2

x1−2x2 ≤ 0

In the second line, we used the fact that x3 = x1 + x2. In the third line, we cross-multiplied by 3, to avoid the repeating decimal.6

If we think of the original two-variable model as part (a), we will call part (b)the model which now includes the proportion constraint. This new two-variablemodel is:


6If the fraction had been something like 34 , we would have used the decimal equivalent 0.75;

there would be no need for cross-multiplication.


maximize 8x1 + 10x2

subject to



Part (b) Proportion x1 − 2x2 ≤ 0


If a problem on a test involves both formulation and graphing, and if no spe-cific mention has been made of doing the formulation with no calculations, thenyou may simply define two variables and proceed with making the model.

What’s new here is that we now must plot a constraint whose right-hand-sidevalue is 0. The boundary of any constraint with a 0 on the right-hand-side willpass through the origin. Since at the boundary x1 = 2x2, this particular constraintalso passes through, for example, the point (200,100). On any constraint whichhas a negative number of the left-hand side, and especially for one where theright-hand side value is 0, a great deal of care must be taken to make sure that thearrow is drawn in the correct direction. We must pick a point which is not on theline, such as (100,0). Substituting x1 = 100 and x2 = 0 into x1−2x2 ≤ 0 gives us100−2(0) = 100, and 100≤ 0 is false, therefore the arrow points away from thepoint (100,0).

Superimposing this constraint on the existing solution produces an altered fea-sible region; a part of the former feasible region has now become infeasible. InFigure 43, the new feasible region is shown in gold, the now infeasible part ofthe former feasible region is shown in light blue, and the old and new optimalsolutions are shown. The binding constraints now are the Department B Labourconstraint and the Proportion constraint, with the optimal solution occuring atx1 ≈ 125 and x2 ≈ 60. At the boundaries of these constraints we obtain the exact



0

50

100

150

200

Tonn

espe

rD

ayof

Type

2C

emen

t

Dept. A

Labour


Type 1 Sales

Type 2Sales

Proportion

Feasible Region (a)F.R. (b)

Trial

Optimal

(b)

(a)

OptimalSolution

Figure 43: Cement Problem – Altered Optimal Solution


solution:Dept. B Labour 1.5x1 + 5x2 = 500

Proportion x1 − 2x2 = 0

3x1 + 10x2 = 10005x1 − 10x2 = 0

8x1 + 0x2 = 1000

x1 = 125

By substituting x1 = 125 into x1− 2x2 = 0, we obtain x2 = 62.5. The optimalmathematical solution for the altered model is x∗1 = 125 and x∗2 = 62.5. The ob-jective function value at the point of optimality is

OFV∗ = 8x∗1 +10x∗2= 8(125)+10(62.5)= 1000+625= 1625

Going back to the original problem, the solution expressed in managerial termsis:


Recommendation

With the added requirement that the level of Type 1 production cannot exceed two-thirds of the total production, the cement plant should produce 125 Tonnes per dayof Type 1 cement, and 62.5 Tonnes per day of Type 2 cement, for a contributionto profit of $1625 per day. After deducting fixed costs of $1400 per day, the netdaily profit will be $225.

11.4.3 Comment

The initial model with its six constraints leads to a solution which creates a (dailycontribution to) profit of $1700. Then, after adding a seventh constraint, the profitfell to $1625. Whenever a constraint is added, the profit can at best stay thesame, and often it will fall. In general, adding another constraint (or making anexisting one more stringent) can at best keep the OFV the same, otherwise it willbe impaired. By impaired, we mean that the OFV will decrease if the objective ismaximization, and will increase if the objective is minimization. Note that whilethe profit went down, only one of the variables did. The Type 1 cement productiondecreased from 150 to 125 TPD, but the Type 2 cement production increased from50 to 62.5 TPD.

11.5 Exercise

Using just two variables, formulate a linear optimization model for the followingproblems. Solve each model graphically, clearly indicating the feasible region,and both the trial and optimal isovalue lines. For each model, use algebra todetermine the exact solution.

11.5.1 Garment Problem

When solving the following model, use a piece of graph paper with each axislabelled from 0 to 300, and draw all lines within the 300 by 300 grid.

A garment factory makes blouses and dresses. Each blouse gives a profit of$2, while each dress gives a profit of $3. They can sell at most 190 dresses. Eachgarment spends time on three machines as follows:


Minutes per Garment MinutesMachine Blouse Dress AvailableCutting 3 6 1413Sewing 6 2 1218Assembly 5 4 1317

The number of dresses must be at least 30% of the total number of garments made.

11.5.2 Baseball Bat Problem

A baseball bat company makes two models, the “slugger” and the “whacker”.Each slugger requires four minutes of lathework, and one minute of varnishing.Each whacker requires five minutes of lathework and 45 seconds of varnishing.Each day, the combined production cannot exceed 980 bats. The woodworkingshop operates 16 hours/day, with one room containing five lathes, and one var-nishing room. Each of the five lathes is available for 55 minutes each hour, and thevarnishing room is available for 50 minutes each hour. Each slugger contributes$5 to profit, and each whacker contributes $6.

The company wishes to determine how many sluggers and whackers shouldbe made each day.

11.5.3 Quarry Problem

Background Note: The density of an object is its mass divided by its volume.Hence mass is density times volume, and volume is mass divided by density.

Two types of rock are mined in a quarry. “Softrock” has an density of 5 Tonnesper cubic metre, and “hardrock” has a density of 8 Tonnes per cubic metre. Upto 600 Tonnes of softrock can be mined each hour, and independent of this, up to300 Tonnes of hardrock can be mined each hour. The mined rock is crushed andthen travels on a conveyor belt. (To avoid mixing the two types of rock they willcrush one type of rock, then switch over to the other type, and then keep switchingback and forth).

The conveyor belt can handle up to 110 cubic metres of rock per hour. Thecrusher can handle up to 1000 Tonnes per hour when crushing softrock, or up to400 Tonnes per hour when crushing hardrock. The company makes $10/Tonne forsoftrock and $14/Tonne for hardrock. The quarry operator wishes to know howmany Tonnes of each type of rock they should produce each hour.



There are two examples presented here. The first is a minimization example, forwhich we shall obtain the data from the WWW, and then solve graphically. Thesecond example looks at a situation in which the variables must take on integervalues.

12.1 Example – Diet Problem


This example is made to illustrate linear optimization. Don’t take it as nutritionaladvice. A real diet shouldn’t contain only these two items.

A twenty-two year old student lives on a diet of double hamburgers and orangejuice. To make a double hamburger (bun, two patties of beef, and condiments)costs about $1.25, and a serving (249 g) of unsweetened orange juice costs about$0.32. She wants to minimize her daily cost of buying these things, but she hasdecided to make sure that she obtains the recommended daily intake of all vita-mins and minerals. To keep the problem simple, she wants the protein, iron, andVitamin C to meet or exceed the recommended amounts for a woman of her age,and to restrict the amount of iron from hamburgers to be no more than 90% of hertotal iron intake.

12.1.2 Formulation

The problem as stated doesn’t contain enough information. We do not know, forexample, how many grams of protein are contained in a double hamburger, or howmany grams per day of protein that a 22 year old woman needs. For this informa-tion, we must look elsewhere. The web 7 can be used to obtain information aboutthe nutrients contained in food and drinks. We do searches on: (1) hamburger,and then look up a hamburger, regular, with double patty and condiments; and (2)juice, and then look up orange juice, canned, unsweetened, 1 cup (249 g) size, wefind the following:

7The electronic form of this document is linked to the URLs. If you are view-ing this in hard copy, the URLs are http://www.nal.usda.gov/fnic/cgi-bin/nut search.pl andhttp://www.nal.usda.gov/fnic/etext/000105.html

http://www.nal.usda.gov/fnic/cgi-bin/nut_search.pl


Nutrient Double Hamburger Orange Juice(per 215 g sandwich) (per 249 g serving)

Protein (g) 31.820 1.469Iron (mg) 5.547 1.096Vitamin C (mg) 1.075 85.656

Looking for dietary requirements, we go to the Table of 1989 RDA and 1997-1998 RDI8 (a pdf file), and look up Female 19-24 to find a daily need for 46 g ofProtein, 15 mg of Iron, and 60 mg of Vitamin C.

Now that we have all the data, we can begin the formulation of the model.Looking at the cost information, we see that we need to know the amounts con-sumed each day. We must therefore have at least the two following decision vari-ables:

x1 = the number of double hamburgers eaten each dayx2 = the number of servings of orange juice drunk each day

Her objective is to minimize the cost of her diet. Each double hamburger costs$1.25, and each serving of orange juice costs $0.32, hence the objective functionis:

minimize 1.25x1 +0.32x2

We begin with the first three constraints, one for each of three nutrients. Thepurpose of these three constraints is to ensure that the recommended daily intake(RDI) is met. For any constraint the units must match up on the left-hand andright-hand sides. The amount of protein consumed each day is:

total protein = protein from hamburgers + protein from orange juice= 31.820 grams/hamburger × x1 hamburgers +

1.469 grams/serving of orange juice × x2 servings of orange juice= 31.820x1 grams +1.469x2 grams

Her RDI is for 46 grams of protein. To ensure that she obtains at least this amountwe use a ≥ constraint:

31.820x1 grams +1.469x2 grams ≥ 46 grams

8The original source is the Food and Nutrition Board - National Academy of Sciences, 1998(University of California, Davis).

http://www.nal.usda.gov/fnic/etext/000105.html


With the sameness of the units on both sides, we can remove the word grams toobtain:

31.820x1 +1.469x2 ≥ 46

The iron and Vitamin C constraints are in milligrams rather than grams, but theidea is the same. We obtain units of milligrams on both sides of the inequality,and hence the word milligrams can be dropped from both sides. The constraintfor the iron requirement is:

5.547x1 +1.096x2 ≥ 15

The constraint for the Vitamin C requirement is:

1.075x1 +85.656x2 ≥ 60

Now we must restrict the iron from hamburgers to be no more than 90% of thetotal iron consumed. We can represent the total amount of iron consumed by athird variable:

x3 = the amount of iron consumed each day (in mg)

The daily intake of iron from hamburgers (in mg) is 5.547x1. Hence we musthave:

5.547x1 ≤ 0.9x3

5.547x1−0.9x3 ≤ 0

The total iron intake x3 is the amount from hamburgers, which is 5.547x1, plus theamount from orange juice, which is 1.096x2. Therefore, we must have:

x3 = 5.547x1 +1.096x2

which we can re-arrange as

5.547x1 +1.096x2− x3 = 0

Finally, we have the non-negativity restrictions. The completed model is:

x1 = the number of double hamburgers eaten each dayx2 = the number of servings of orange juice drunk each dayx3 = the amount of iron consumed each day (in mg)


minimize 1.25x1 + 0.32x2

subject to

Protein RDI 31.820x1 + 1.469x2 ≥ 46Iron RDI 5.547x1 + 1.096x2 ≥ 15

Vitamin C RDI 1.075x1 + 85.656x2 ≥ 60Iron Proportion 5.547x1 − 0.9x3 ≤ 0

Iron Balance 5.547x1 + 1.096x2 − x3 = 0


The model required no calculations. If we wish to solve this problem graph-ically, then we will have to imbed the final constraint into the Iron Proportionconstraint. After doing some calculations to revise the Iron Proportion constraint,the Iron Balance constraint and the third variable are removed.

5.547x1−0.9x3 ≤ 05.547x1−0.9(5.547x1 +1.096x2) ≤ 0

5.547x1−4.9923x1−0.9864x2 ≤ 00.5547x1−0.9864x2 ≤ 0

Making the last line above the revised Iron Proportion constraint the modelbecomes:

x1 = the number of double hamburgers eaten each dayx2 = the number of servings of orange juice drunk each day

minimize 1.25x1 + 0.32x2

subject to

Protein RDI 31.820x1 + 1.469x2 ≥ 46Iron RDI 5.547x1 + 1.096x2 ≥ 15

Vitamin C RDI 1.075x1 + 85.656x2 ≥ 60Iron Proportion 0.5547x1 − 0.9864x2 ≤ 0



12.1.3 Graphical Solution

To establish a reasonable scale for the graph, we can think of the context fromwhich the model came. Suppose that she eats three meals a day, each being a dou-ble hamburger and a serving of orange juice. Mathematically, this would implythat x1 = 3, and x2 = 3. By plugging these values into the four constraints, wecan see that this solution is feasible. Since we are trying to minimize the cost, thesolution must be less than 3 for one of the two variables, and we can hope that itwill be less than 3 for both of them. If the grid from (0,0) to (3,3) turns out to betoo small, we can always expand it later.

We try to find where the boundary of every constraint intercepts the axes.When this yields a point outside the grid, we use find the intercept on the right-hand side (x1 = 3) or top (x2 = 3) boundary instead. For example, the boundaryof the Protein RDI constraint is

31.820x1 +1.469x2 = 46

Setting x1 = 0 causes x2 to be off the 3 by 3 grid. Hence we set x2 = 3, and solve31.820x1 + 1.469(3) = 46, obtaining x1 ≈ 1.307. Setting x2 = 0 causes x1 to beabout 1.446, which is on the grid. Hence the two points for this constraint are(1.307,3) and (1.446,0). Doing this for every constraint we obtain:

Constraint First Point Second PointProtein RDI (1.307,3) (1.446,0)

Iron RDI (2.111,3) (2.704,0)Vitamin C RDI (0,0.7005) (3,0.6628)Iron Proportion (0,0) (3,1.687)

Because the origin is false for each of the first three constraints, all three ar-rows point away from the origin. The fourth constraint passes through the origin,so we test a point which is not on the constraint boundary, such as (0,2). Thispoint is true with respect to the inequality, so the arrow points toward this point,i.e. upwards and to the left. These four constraints, along with their arrows andword descriptions, are shown in Figure 44.

We now find and highlight the feasible region. In this example, the feasibleregion is of infinite size, but it is clipped by the boundaries of the grid. Plottingthe trial isovalue line is quite easy in this situation. The objective function is tominimize 1.25x1 + 0.32x2, so we try the shortcut of plotting 1.25 on the verticalaxis and 0.32 on the horizontal axis, and connect them with a dashed line. We thenmove a rolling ruler over to the feasible region, stopping at the corner where the


0 0.5 1.0 1.5 2.0 2.5 3.0Number of Double Hamburgers

0

0.5

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1.5

2.0

2.5

3.0

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ofSe

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gsof

Ora

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Juic

e

ProteinR

DI

IronR

DI

Vitamin C RDI

Iron Proportio

n

Figure 44: Diet Problem – Constraints


boundaries of the Iron RDI constraint and the Iron Proportion constraint intercept.This is shown in Figure 45.

We can see that the optimal solution lies at about 2.4 double hamburgers perday, and 1.4 servings of orange juice per day. To find the exact solution, wefind the interception point of the boundaries of the Iron RDI and Iron Proportionconstraints.

Iron RDI 5.547x1 + 1.096x2 = 15Iron Proportion 0.5547x1 − 0.9864x2 = 0

5.547x1 + 1.096x2 = 155.547x1 − 9.864x2 = 0

0x1 + 10.96x2 = 15

x2 ≈ 1.369

By substituting this value into either of the original constraints, we obtain x1 ≈2.438. The objective function value is about $1.25(2.438) + $0.32(1.369) ≈$3.48. The question now arises as to whether we should recommend values forthe variables which are not integer. To answer this question we need to considerthe context of the problem. The orange juice is not a problem, because if we want1.369 servings of 249 g each, all we have to do is make 2 servings of 249/1.369≈ 181.9 g each. The hamburgers are more of a problem, however, since it’s hardto cook 0.438 of a burger. However, for both the hamburgers and the orange juice,we can interpret the DRI for each nutrient as an average to be obtained over aperiod of time. For example, suppose that she eats two hamburgers and drinksone serving of orange juice on one day, and then eats three hamburgers and drinkstwo servings of orange juice on the next, and repeats this cycle. She would av-erage 2.5 double hamburgers and 1.5 servings of orange juice over time. Thiswould certainly meet the requirements of the DRI constraints (2.5 > 2.438, and1.5 > 1.369), and in the Iron Proportion constraint we have:

0.5547(2.5)−0.9864(1.5) =−0.09285 < 0

Hence (2.5,1.5) is a feasible solution. Its daily cost is

$1.25(2.5)+$0.32(1.5) = $3.605

which is about 12.5 cents higher than the theoretical optimal solution. We are nowready to make a recommendation.


0 0.5 1.0 1.5 2.0 2.5 3.0Number of Double Hamburgers

0

0.5

1.0

1.5

2.0

2.5

3.0

Num

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ProteinR

DI

IronR

DI

Vitamin C RDI

Iron Proportio

n

FeasibleRegion

Trial

Optim

al

OptimalSolution

Figure 45: Diet Problem – Optimal Solution


Recommendation

Based on a self-imposed diet of double hamburgers and orange juice, andconsidering only the four stated nutritional requirements, a near-optimal solutioncan be implemented by eating two hamburgers and drinking one serving of orangejuice on one day, and then eating three hamburgers and drinking two servings oforange juice on the next, and repeating this cycle. This gives a daily cost of$3.605.

12.2 Example 2

In the first lecture, we presented the following exercise:


Two types of big boxes are about to be loaded onto a small cargoplane. A Type 1 box has a volume of 2.9 cubic metres (m3), and amass of 470 kilograms (kg), while a Type 2 box has a volume of 1.8m3 and a mass of 530 kg. There are six Type 1 boxes and eight Type 2boxes waiting to be loaded. There is only one cargo plane, and it has avolume capacity of 15 m3 and a mass capacity of 3600 kg. Obviously,not all the boxes can be put onto the plane, therefore suppose that theobjective is to maximize the value of the load. We will consider thefollowing three situations: (i) both type of boxes are worth $400 each;(ii) a Type 1 box is worth $600, and a Type 2 box is worth $250; and(iii) a Type 1 box is worth $300, and a Type 2 box is worth $750.

Back then, we used an enumerative method to find all potential solutions, and thenevaluated each of these to find the optimal ones. Now, we will formulate and solvethis problem using integer optimization.

12.2.2 Formulation

We need to determine how many boxes of each type are carried on the plane, sowe define:

x1 = the number of Type 1 boxes carried on the planex2 = the number of Type 2 boxes carried on the plane


There are three cases of profit data. Each gives rise to a different objective func-tion:

(i) maximize 400x1 +400x2

(ii) maximize 600x1 +250x2

(iii) maximize 300x1 +750x2

There is a constraint for the volume capacity of the plane. By now it should beeasy to write this constraint:

Volume 2.9x1 +1.8x2 ≤ 15

Next, there is a constraint for the mass capacity of the plane:

Mass 470x1 +530x2 ≤ 3600

The plane cannot carry more boxes than are available to be carried, therefore wehave two more constraints:

Type 1 x1 ≤ 6

andType 2 x2 ≤ 8

In this example, we have not only the non-negativity restrictions, but also therequirement that both variables must be integer. The complete formulation istherefore:


One of:(i) maximize 400x1 + 400x2

(ii) maximize 600x1 + 250x2(iii) maximize 300x1 + 750x2

subject to

Volume 2.9x1 + 1.8x2 ≤ 15Mass 470x1 + 530x2 ≤ 3600

Type 1 x1 ≤ 6Type 2 x2 ≤ 8

non-negativity x1 , x2 ≥ 0integer x1 , x2


We begin as always by making a grid and plotting the boundaries of the con-straints. The last two constraints tell us that the optimal solution must be containedwithin a 6 by 8 grid, so we will make it this size. The boundary of the volumeconstraint is:

2.9x1 +1.8x2 = 15

Setting x1 = 0 makes x2 = 8.333..., which is just above the grid. Hence we setx2 = 8, and solve

2.9x1 +1.8(8) = 15

obtaining x1 ≈ 0.207. Setting x2 = 0 makes x1 ≈ 5.172, which is on the grid. Weapply the same approach to the mass constraint, to obtain the following table:

Constraint First Point Second PointVolume (0.207,8) (5.172,0)

Mass (0,6.792) (6,1.472)Type 1 (6,0) (6,8)Type 2 (0,8) (6,8)

All the arrows are easy; the origin is true for every constraint, so every arrowpoints toward the origin.

These four constraints, along with their arrows and word descriptions, areshown in Figure 46.

We can now fill-in with colour what would be the feasible region if the modelhad not contained integer variables. This is shown in Figure 47.

Because the variables must be integer, only those points in the coloured areawhich represent integer values for both variables are feasible.9 Finding all thesepoints, which we represent as dots, is fairly easy except when a point is very nearone of the constraint boundaries. In this example, the points (1,6), (2,5) are nearthe boundary of the mass constraint, and the point (4,2) is near the boundary ofthe volume constraint. We can test these contentious points by substituting thevalues into the appropriate constraint. For example, for the point (1,6):

470(1)+530(6) = 470+3180= 36506≤ 3600

9If x1 were integer, but x2 not integer, then we would have a set of feasible vertical lines. If x2were integer, but x1 not integer, then we would have a set of feasible horizontal lines.


0 1 2 3 4 5 6Number of Type 1 Boxes Carried

0

1

2

3

4

5

6

7

8N

umbe

rof

Type

2B

oxes

Car

ried

Volume

Mass

Type1

Type 2

Figure 46: Cargo Plane Problem – Constraints



0

1

2

3

4

5

6

7

8

Num

ber

ofTy

pe2

Box

esC

arri

ed

Volume

Mass

Type1

Type 2

Figure 47: Cargo Plane Problem – Non-Integer Region


Hence the point (1,6) is infeasible, and is therefore excluded from consideration.On the other hand, for the point (2,5) we obtain:

470(2)+530(5) = 940+2650= 3590≤ 3600

√

Therefore, the point (2,5) is feasible. Finally, for the point (4,2) we use the volumeconstraint:

2.9(4)+1.8(2) = 11.6+3.6= 15.26≤ 15

We see that the point (4,2) is infeasible, and it is therefore excluded. We are leftwith 26 feasible points, which are shown in Figure 48.

Beginning with the first of the three objective functions, we seek to maximize400x1 +400x2. The shortcut produces points which are off the graph, but dividingby 100 gives the points 4 on the vertical axis and 4 on the horizontal axis. Theseare connected to form the first of three trial lines. We then move the rolling ruler,stopping not at the corner of the volume and mass constraints (because this pointis infeasible), but instead at the integer solution (2,5). This is shown in Figure 49.

The optimal objective function value is:

$400(2)+$400(5) = $2800

If an integer solution had not been required, we would have obtained a solutionat the corner of the volume and mass constraints. By using linear algebra wewould have found x∗1 ≈ 2.12735, x∗2 ≈ 4.90593, and OFV∗ ≈ $2813.31. Since wedo require integer values we have instead x∗1 = 2, x∗2 = 5, and OFV∗ = $2800.00.By imposing the requirement that the variables be integer, we have impaired theobjective function value by $13.31. This will always be true – adding a require-ment that the variables must be integers will impair (i.e. lower for a maximizationmodel, higher for a minimization model) the objective function value.

Using the same diagram we draw the trial and optimal isovalue lines for sit-uations (ii) (in green) and (iii) (in blue). In order to obtain the intercepts on theaxes for the trial line used here for situation (iii), the objective function coef-ficients were divided by 200. All this is shown in Figure 50. We identify the



0

1

2

3

4

5

6

7

8

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ofTy

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Volume

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Figure 48: Cargo Plane Problem – Set of Feasible Points



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Trial (i)Optimal (i)

(i) (2,5)

Figure 49: Cargo Plane Problem – Optimal Solution for Part (i)


optimal solution for case (ii) as (5,0), i.e. five boxes of Type 1 only, the OFV is$600(5)+ $250(0) = $3000. The optimal solution for case (iii) is (0,6), i.e. sixboxes of Type 2 only, the OFV is $300(0)+ $750(6) = $4500. In summary wehave:

Recommendation

Situation Profit per Box Optimal Load TotalType 1 Type 2 Type 1 Type 2 Profit

(i) 400 400 2 5 $2800(ii) 600 250 5 0 $3000(iii) 300 750 0 6 $4500



0

1

2

3

4

5

6

7

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Trial (i)Optimal (i)

(i) (2,5)Trial (ii)

Optim

al (ii)

(ii) (5,0)

Trial (iii)

Optimal (iii)

(iii)(0,6)

Figure 50: Cargo Plane Problem – Optimal Solution for Parts (i), (ii), (iii)


12.3 Exercises12.3.1 Office Rental

A company needs to rent space for its office employees both in the suburbs anddowntown. Space is available in suburbia at a rate of $100 per square metre (perannum), while downtown space rents for $210 per square metre (per annum). Insuburbia, only 30% of the space is “executive” quality, while the rest is ordinaryquality. At the downtown location, 60% of the space is executive quality, while therest is ordinary quality. The company needs a total of at least 900 square metresof space, of which at least 420 square metres must be executive quality. No morethan three quarters of the entire space is to be at either location.

They wish to know how much space they should rent in each place. Formulateand solve by the graphical method.

12.3.2 Diet Problem

Suppose that a kilogram of beef contains 600 grams of protein, and 80 grams offat, but no Vitamin C. A litre of orange juice contains 6 grams of protein, no fat,and four times the required daily intake of Vitamin C. A person needs 54 gramsof protein per day, and the fat should be between 10 and 60 grams per day. Nomore than 95% of the protein consumed should come from beef. A kilogram ofbeef costs $6, while a litre of orange juice costs $2.

Based on these two foods alone, and only the stated requirements, we seek theminimum cost daily diet. Formulate and solve by the graphical method.

12.3.3 Fruit Buying Problem

Jennifer is making a large fruit salad for a party. She has everything she needsat home, except for pineapples and bananas. She needs 12 pineapples, and 31bananas. She goes to a nearby fruit stand, where she finds two vendors sellingbags of mixed fruit. Vendor 1 is selling bags containing two pineapples and tenbananas for $3 per bag. Vendor 2 is selling bags containing four pineapples andfive bananas for $4 per bag. She wants to know how many bags she should buyfrom each vendor to meet (or exceed) the requirements for the punch, but at theleast cost possible. Formulate and solve by the graphical method to determine thebest integer solution.



13.1 IntroductionThe two-dimensional world of the previous two lectures is useful for providingan understanding of what linear optimization is about, but it has very limited use-fulness for practical problems. Real-world applications may involve thousands oreven millions of decision variables. We won’t be doing anything that big, but wedo want to extend what we can do beyond just two variables. To do this requiresan algorithm, which is a structured sequential approach for solving a problem.For example, the roll-back procedure for decision trees that we saw earlier is atype of algorithm. There are several algorithms for linear optimization, but theone most commonly used is called the simplex algorithm. At one time, learningthe basics of how the simplex algorithm works was a core topic of the compulsorymanagement science course. When solving problems using the simplex algorithmin which some or all of the variables must be integer, we may fortuitously find asolution which is integer anyway. Usually, however, an even more complicatedalgorithm is needed when some or all of the variables must be integer. This algo-rithm is called the branch-and-bound algorithm. Again, the study of this subjectis not part of this course.

The simplex and branch-and-bound algorithms have been incorporated intosoftware for linear and integer optimization. There are purpose-built softwarepackages for optimization (such as LINDO, to be seen here), and spreadsheetscan solve optimization problems as well. We will look at both types of software.

We conclude with an example which illustrates a particular type of problemfor which linear optimization is applicable, that of blending gasoline. The size ofthis problem requires that it be solved using a computer.

13.2 Slack and SurplusFor a≤ constraint, the slack is defined as the right-hand side value minus the valueof the left-hand side at the point of optimality. For a ≥ constraint, the surplus isdefined as the value of the left-hand side at the point of optimality minus the right-hand side value. For a model for which the optimal solution has been computed,the slack or surplus can easily be found by hand. For example, the formulation ofthe cement model is:



maximize 8x1 + 10x2

subject to




We know that the optimal solution is x1 = 150 and x2 = 50. Let’s calculate thesurplus on the Type 1 Sales constraint, and the slack on the Department A Labourconstraint. The Type 1 Sales constraint is x1≥ 40 (or 1x1 +0x2≥ 40). The optimalvalues of x1 and x2 are 150 and 50 respectively, hence the value of the left-handside of the constraint is 1(150) + 0(50) = 150. Since the number on the right-handside is only 40, the left-hand side value is 150− 40 = 110 more than it needs tobe; the surplus on the Type 1 Sales Constraint is 110. The Department A Labourconstraint is 3x1 + 2x2 ≤ 585. At the optimal solution of (150,50) the left-handside value is:

3(150)+2(50) = 550

By subtracting 550 from 585, we obtain a slack of 35.In general, if the slack or surplus is 0, then the constraint is binding; if the

slack or surplus is greater than 0, then the constraint is non-binding. Knowing theslack or surplus gives information for the easiest kind of sensitivity analysis. Fora ≤ constraint, the right-hand side value can be decreased by up to the amount ofthe slack (or be increased indefinitely) without affecting the optimal solution. Fora ≥ constraint, the right-hand side value can be increased by up to the amount ofthe surplus (or be decreased indefinitely) without affecting the optimal solution.For example, suppose that the number of labour-hours in Department A weredecreased from 585 to 555, i.e. a decrease of 30. Since the slack is 35, this changeis within the range in which the optimal solution is not affected. All that wouldhappen is that the new slack in Department A would be 35−30 = 5.

The traditional wording of slack and surplus is maintained by LINDO, butExcel uses the single term slack for both senses.


13.3 LINDO

LINDO is a registered trademark of LINDO Systems Inc., and is both part of thename of the company, and the name of its first software program. LINDO was oneof the first programs of its type to address the educational as well as the commer-cial/industrial market. On their website at lindo.com they make available “trialversions” of LINDO and other software. To obtain the version that is appropriatefor this course, click on “Downloads”, then in the page with the title “Downloadtrial versions of our products” click on “Download Classic LINDO”, then openthe zip file. LINDO is also available as a subset of LINGO 11.0 for Windows.(LINDO API is a different kind of software.)

This free version is of course the least powerful version of the software, but itis nevertheless more than adequate for anything in this course. It is also possibleto buy at an educational price ($25 US as of April 28, 2010) a “Solver Suite” CDwhich contains a version of LINDO API, LINGO, and What’s Best!, a spreadsheetoptimizer.

13.3.1 LINDO for Linear Optimization

To illustrate the use of LINDO for a linear optimization model (no integer vari-ables), we will use the formulation of the cement problem. We are not requiredto enter the variable definitions, because they are not needed to solve the prob-lem mathematically. However, we may wish to enter them as comments in orderto make the LINDO printout easier to understand. A comment is made by firsttyping an exclamation mark. Anything on the same line which follows an excla-mation mark is ignored by LINDO. Comments could also be made to give thename of the model, the name of the person who made it, the date of its creation,or anything else that might make the file easier to understand when viewing it ata later point in time. Also, blank lines may be inserted at will to help improve theappearance of the file.

We do not enter the non-negativity restrictions, because they are always as-sumed to be present. Some adjustments have to be made because of the limita-tions of the keyboard. We cannot enter a subscripted variable, hence x1 and x2 areentered as x1 and x2. Also, since there are no ≤ or ≥ symbols on the keyboard,we enter <= and >= instead. For the names of the constraints, each is restricted toeight spaces, and must be followed by a right parenthesis – this restriction has re-sulted in making modifications to the names given in the algebraic model. Thereare several shortcuts available (not used here) such as just typing st instead of

http://www.lindo.com/


subject to. Further information is available from the LINDO Help menu.We open the LINDO software and in the initial window we type (put the names

of your group as the “Analyst” when submitting homework problems):

! Cement Plant Model Analyst: D.M. Tulett

! x1 = the number of TPD of Type 1 cement made! x2 = the number of TPD of Type 2 cement made

maximize 8x1 + 10x2

subject to

Sales 1) x1 >= 40Sales 2) x2 >= 30Tot.Prod) x1 + x2 <= 200Labour A) 3x1 + 2x2 <= 585Labour B) 1.5x1 + 5x2 <= 500Labour C) 4x1 + 6x2 <= 900

This file can be saved using a .ltx extension. We could call it cement.ltx, forexample. We then go to the Solve menu, and release the mouse button underSolve (the first line of the menu). This action causes LINDO to solve the model,and then LINDO asks DO RANGE(SENSITIVITY) ANALYSIS?. For now,click on No. This leaves a box called LINDO Solver Status, which should sayOptimal under Status. We close this box, and go to the Window menu, and releasethe mouse button under Reports Window. Doing this we obtain the following(whether the input be small or capital letters, the output is always in capitals):


LP OPTIMUM FOUND AT STEP 3

OBJECTIVE FUNCTION VALUE

1) 1700.000

VARIABLE VALUE REDUCED COSTX1 150.000000 0.000000X2 50.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICESSALES 1) 110.000000 0.000000SALES 2) 20.000000 0.000000

TOT.PROD) 0.000000 4.000000LABOUR A) 35.000000 0.000000LABOUR B) 25.000000 0.000000LABOUR C) 0.000000 1.000000

NO. ITERATIONS= 3

The first and last lines give us a measure of how much work the computer did tofind the optimal solution – this is of technical rather than managerial interest, andwe shall not use this information. The 1) which appears after Objective FunctionValue refers to the fact that LINDO numbers the objective function as row 1,and internally calls the first constraint row 2, and so on. Had we not named theconstraints, LINDO would have numbered them as rows 2) to 7). The columnsmarked Reduced Cost and Dual Prices will be explained in the next lecture.

Clearly, we see from the output that x∗1 = 150, x∗2 = 50, and OFV∗ = 1700. Itis up to the user of the software to translate this into the words needed to express arecommendation. The other information provided by this output is in the columnlabelled Slack or Surplus. For a non-binding constraint, the information providedin the Slack or Surplus column lets us answer a question about a proposed changeto the right-hand side without having to solve an altered model.

Other than the substitution of the word minimize for maximize, there is no


special procedure required to handle a minimization model. In models with manyvariables, it is likely that many of them will have optimal values of 0. To suppressthe printing of these variables, go to the Report menu and release the mouse onSolution. A dialog box will appear with a radio button to indicate that only non-zeros should be reported.

13.3.2 LINDO for Integer Optimization

There is one more step that needs to be done when entering the data for a modelwith integer variables, and there’s usually one more step to obtain the solution. Toillustrate, we will use the cargo plane example:


One of:(i) maximize 400x1 + 400x2

(ii) maximize 600x1 + 250x2(iii) maximize 300x1 + 750x2

subject to

Volume 2.9x1 + 1.8x2 ≤ 15Mass 470x1 + 530x2 ≤ 3600

Type 1 x1 ≤ 6Type 2 x2 ≤ 8

non-negativity x1 , x2 ≥ 0integer x1 , x2

LINDO has an integer command, but this is used to declare 0 and 1 to be the onlypossible variables. For an integer variable which can take on values 0, 1, 2, 3, andso on, we use the gin command. After the model has been entered up to the lastconstraint, we write gin x1 followed by gin x2. There is a shortcut in whichwe could enter gin 2 to declare the first two variables found to be integer. Thisshortcut would work well here, but care must be taken for a model in which somebut not all of the variables are required to be integer.

Using the first objective function variation, the model is entered into LINDOas:


! Cargo Plane Model Analyst: D.M. Tulett

! x1 = the number of Type 1 boxes carried on the plane! x2 = the number of Type 2 boxes carried on the plane

maximize 400x1 + 400x2

subject to

Volume) 2.9x1 + 1.8x2 <= 15Mass) 470x1 + 530x2 <= 3600Type 1) x1 <= 6Type 2) x2 <= 8endgin x1gin x2

Solving using the same procedure as before we obtain:

LP OPTIMUM FOUND AT STEP 2OBJECTIVE VALUE = 2813.31396

SET X1 TO <= 2 AT 1, BND= 2800. TWIN= 2600. 12

NEW INTEGER SOLUTION OF 2799.99976 AT BRANCH 1 PIVOT 12BOUND ON OPTIMUM: 2800.000DELETE X1 AT LEVEL 1ENUMERATION COMPLETE. BRANCHES= 1 PIVOTS= 12

LAST INTEGER SOLUTION IS THE BEST FOUNDRE-INSTALLING BEST SOLUTION...


1) 2800.000

VARIABLE VALUE REDUCED COSTX1 2.000000 -400.000000X2 5.000000 -400.000000


ROW SLACK OR SURPLUS DUAL PRICESVOLUME) 0.200000 0.000000

MASS) 10.000000 0.000000TYPE 1) 4.000000 0.000000TYPE 2) 3.000000 0.000000

NO. ITERATIONS= 12BRANCHES= 1 DETERM.= 1.000E 0

This example is very small, so this report is fine as it is. It begins with sometechnical information about how the solution was obtained, but then describes thesolution as it would for a problem without integer variables. However, in a largeexample the amount of technical material can be enormous, and what is worse theoptimal solution might not be at the end, but instead be buried somewhere in themiddle. The reason for this is that the software often cannot tell that it has theoptimal solution until more potential solutions have been examined. Because ofthis, it is useful to delete the complete report and ask for the “Solution” under theReports menu. Doing this gives:


1) 2800.000

VARIABLE VALUE REDUCED COSTX1 2.000000 -400.000000X2 5.000000 -400.000000

ROW SLACK OR SURPLUS DUAL PRICESVOLUME) 0.200000 0.000000

MASS) 10.000000 0.000000TYPE 1) 4.000000 0.000000TYPE 2) 3.000000 0.000000

NO. ITERATIONS= 12BRANCHES= 1 DETERM.= 1.000E 0

Note that the concept of a “binding constraint” does not apply when integervariables are present. In this example, no constraint has a slack of 0. Again, it


is up to the user of LINDO to take this mathematical solution (x∗1 = 2, x∗2 = 5,OFV∗ = 2800) and report it as we did before in managerial terms. The objectivefunction can easily be altered to solve parts (ii) and (iii).

13.4 Optimization using Spreadsheets

Spreadsheets (which are really a multi-purpose mathematical tool) can be used foroptimization. However, entering the model is much slower on a spreadsheet thanit is with a dedicated program such as LINDO. Also, for anything but very smallmodels, the size of the worksheet will be much larger than the size of the monitor,making data entry even more difficult. A spreadsheet may make sense when thedata is already in spreadsheet form. We present a brief overview here, but thenreturn to LINDO exclusively.

The user begins by entering three types of information. First of all, there arelabels. Secondly, there is the numerical information of the problem. Thirdly, thereare formulas.

Labels are used to help make the model understood to the user and other per-sons who may look at the spreadsheet. Any cell containing a label has no effecton the calculations. Some of these labels are obvious, such as “Tonnes per Day”and “Total Production”. However, there is also a column which gives the directionof the inequality of the constraints, be it <= to mean ≤, or >= to mean ≥, or = foran equality constraint. These may appear to be commands, but they are simplylabels.

To illustrate the second and third types of information, suppose that the objec-tive function or one of the constraints contained an expression such as 5x1 +4x2−6x3 +3x4. This expression can be thought of as the dot product of a vector of num-bers (5, 4,−6, 3) and a vector of variables (x1, x2, x3, x4). To use a spreadsheet wewould input the numbers in one row, and leave space for the variables in anotherrow. On a spreadsheet, the dot product of two rows is made using the SUMPROD-UCT function (discussed in Mathematics for Management Science).10 The set ofvariables is used by the objective function and by every constraint. Hence we havea row for the variables, a row for the coefficients (numbers) of the objective func-tion, and a row for the coefficients of each of the constraints. The value calculatedby the SUMPRODUCT function goes to the right of the left-hand side data, the

10The syntax can be found using the Excel Help menu. This function can also handle more thanjust a dot product. It should also be noted that the SUM function could do this calculation, butwould have to be defined as an array.


label indicating the direction of the inequality goes to the right of this, and finallyon the extreme right we enter the right-hand side value of the constraint.

We calculate the OFV by using the SUMPRODUCT function, and we alsouse this function to calculate the numerical value of the left-hand side of eachconstraint. These numerical values must obey the relationship of the constraint.To save work, we can enter the SUMPRODUCT function for the OFV usingabsolute labels for the range containing the variables, and then copy the formulato where it is used by the constraints.

As an example, we use the model which we developed for the cement com-pany. In the spreadsheet which follows, the cells which are reserved for the vari-ables are highlighted in yellow, and the cells for which the copy command shouldbe used are highlighted in blue.

The formula for cell A3 is =SUMPRODUCT(B4:C4,B$5:C$5)The formula for cell D8 is =SUMPRODUCT(B8:C8,B$5:C$5)

A B C D E F1 Cement Model2 OFV x1 x23 =SUMPR... Type 1 Type 24 maximize 8 105 Tonnes per Day67 Constraints RHS8 Type 1 Sales 1 0 =SUMPR... >= 409 Type 2 Sales 0 1 copy D8 >= 30

10 Total Production 1 1 copy D8 <= 20011 Dept. A Labour 3 2 copy D8 <= 58512 Dept. B Labour 1.5 5 copy D8 <= 50013 Dept. C Labour 4 6 copy D8 <= 900

In optimizing a model, we let Excel choose the values of the variables. To dothis, we need to use the spreadsheet Solver , which is used in Mathematics forManagement Science for nonlinear functions. The overview provided here shouldbe sufficient, but if needed a Solver tutorial is available from Frontline Systems,Inc. at solver.com.

If the solver has not already been installed, the installation in Excel 2007 isaccessed as follows:

• Click on the “Office Button” (top left of the screen).

http://www.solver.com


• A pop-up box appears. Click on “Excel Options” at the bottom of the box.

• A new box appears. On the left, click on “Add-Ins”.

• Under Add-ins, click on Solver add-in, and below click on “Go”.

• After a minute or so, the Solver should be ready to be used.

After entering the model, the spreadsheet’s solver is invoked (in Excel 2007 clickon Data, then Analysis, then Solver). The user specifies the following:

1. the cell which is to be optimized (called the target cell), which is the cellwhich will contain the OFV)

2. the objective (e.g. maximization)

3. a range of cells which the computer may vary, i.e. the range of cells reservedfor the values of the variables (called the changing cells), and

4. the constraints.

[OpenOffice uses the same terminology as Excel, but other brands of software usedifferent names. The cell to be optimized is called the solution cell in Quattro Pro,the optimal cell in Lotus 1-2-3, and the optimum cell in the What-If Solver add-infor Lotus 1-2-3. The cells which the computer may vary are called the variablecells in Quattro Pro, the adjustable cells in Lotus 1-2-3, and the what-if cells inWhat-If Solver.]

For every constraint we will compare the cell which contains the value of theleft hand side with the cell which contains the right hand side value, specifyingthe relationship (≤, =, or ≥) between these two cells. Constraints which are nextto one another of the same type (≤, =, or ≥) can be entered as a range rather thanspecifying each one separately. In the Solver window, under “Options”, the userneeds to click on “ Assume Linear Model ” and “ Assume Non-Negative ”. Doingthe former invokes the appropriate solution procedure for this type of problem; do-ing the latter enters the non-negativity restrictions. For technical reasons, it mightbe also useful to click on “ Use Automatic Scaling ”, though it’s not required forany of the models in this document.11

11A model is said to be poorly scaled when the coefficients of one row are very muchgreater than those of another, for example if one constraint is 2x1 + 5x2 ≤ 41 while another is450,000x1 + 195,000x2 ≤ 2,715,000. When the computer tries to solve a poorly scaled model,


For this example, the target cell is A3, the objective is maximization, and thechanging cells are in the range B5:C5. The constraints are entered as D8:D9 ≥F8:F9, and D10:D13 ≤ F10:F13. Optimizing the model we obtain (the optimalvalues of the variables are highlighted):

A B C D E F1 Cement Model2 OFV x1 x23 1700 Type 1 Type 24 maximize 0 8 105 Tonnes per Day 150 5067 Constraints RHS8 Type 1 Sales 1 0 150 >= 409 Type 2 Sales 0 1 50 >= 3010 Total Production 1 1 200 <= 20011 Dept. A Labour 3 2 550 <= 58512 Dept. B Labour 1.5 5 475 <= 50013 Dept. C Labour 4 6 900 <= 900

As one would expect, cell A3 contains the optimal OFV of 1700, and cells B5and C5 contain 150 and 50 respectively, which are the optimal Tonnes per Day oftype 1 and type 2 cement respectively.

The user can request an “Answer Report” which will give the value of thetarget cell (the OFV), the values of all the variables, and the slack or surplus(described simply as “slack”), if any, on each constraint.

For a model in which some or all of the variables must be integer, the userenters a range under “Constraints” and declares the range to be int.

13.5 Blending Problem13.5.1 Problem Description

A small gasoline blending operation called Blendex buys gasoline from four nearbyrefineries. These are then blended to make three commercial products: low-octane

it may experience numerical problems in finding the optimal solution. Automatic rescaling helpseliminate such problems. See D. Flystra, A. Lasdon, J. Watson, and A. Waren, “Design and Useof the Microsoft Excel Solver”, Interfaces, 28:5 September-October 1998, pp. 29-55.


gasoline for cars, high-octane gasoline for cars, and fuel for propeller-driven air-craft. The (minimum) octane requirements are 86 for low-octane, 92 for high-octane, and 105 for fuel for propeller-driven aircraft.

Though consumers buy gasoline by the litre, these companies quote prices andquantities by the cubic metre (equivalent to 1000 litres). Every week someonefrom Blendex contacts the four refineries to see what they have to sell. Theyare given: the octane rating of the gasoline available; the quantity available incubic metres; and the price per cubic metre. These quotes are valid for one hour;after that time, the refineries are free to sell to anyone. At about the same timeeach week, Blendex obtains the wholesale market prices for the three types ofgasolines. Based on all this information, Blendex can decide to buy none, some,or all of what each refinery has to sell.

Each week the Blendex operation can handle up to 500 cubic metres of gaso-line in total. There are three storage tanks, each with a capacity of 260 cubicmetres, so this becomes the upper limit for the production of any one type ofgasoline.

The octane rating of a blend of gasolines is approximately the weighted aver-age of the octane ratings of the inputs to the blend. For example, if 20 cubic metresof 80 octane gasoline is mixed with 30 cubic metres of 100 octane gasoline, theoctane rating of the 20 + 30 = 50 cubic metres is about:

20(80)+30(100)20+30

= 92

It’s Saturday afternoon and the Blendex planning office must decide what itwants to receive by truck on Monday morning. They have just received the fol-lowing information:

Refinery Octane Quantity (m3) Price per m3

A-One Petroleum 81 280 $350Better Grade Fuels 87 400 $400Clearly Superior 98 200 $450Deluxe Gasoline 115 130 $600

The current market prices per cubic metre are $570 for low-octane gasoline,$600 for high-octane gasoline, and $700 for propeller-driven aviation fuel. Theyhave about 45 minutes to come up with a plan, leaving a few minutes for Blendexto confirm what they are buying from the four refineries.


13.5.2 Formulation and Solution

We could start naming the variables for this problem as x1, x2, x3, and so on, butinstead we will name them so that it will be easy to recall what each variable namemeans.

Clearly Blendex must decide how much gasoline to buy from each of the foursources. Hence we let A, B, C, and D represent the number of cubic metresof gasoline purchased from A-One Petroleum, Better Grade Fuels, Clearly Su-perior, and Deluxe Gasoline respectively. They must also decide how much tomake of each product, so we let L, H, P represent the number of cubic metres oflow-octane, high-octane, and propeller-driven aviation fuel made respectively. Todetermine the octane ratings of the final products, we need to know how muchgasoline comes from each source. For example, we need to know how muchgasoline from A-One Petroleum is used to make low-octane gasoline. Since A isassociated with A-One Petroleum, and L with low-grade gasoline, we can denotethis unknown with two letters, AL. Keeping this pattern going we have 4(3) = 12variables with double-letter names, the first letter indication the input gasoline,and the second indicating the output gasoline. We can summarize this by sayingthat AL, AH, AP, BL, BH, BP, CL, CH, CP, DL, DH, and DP represent the numberof cubic metres of gasoline from source {A, B, C, or D} used to make output {L,H, or P}.

In a problem like this, where there are both revenues and costs, we subtractthe costs from the revenues to make a profit maximization model. The objectivefunction is:

maximize 570L + 600H + 700P − 350A − 400B − 450C − 600D

If we assume that no gasoline is wasted, the total of the amounts of gasolinefrom the four refineries used to make low-octane gasoline becomes the amount oflow-octane gasoline made. Hence

AL + BL + CL + DL = L

Subtracting L from both sides gives:

AL + BL + CL + DL − L = 0

Similar constraints are needed for high-octane and aviation fuel.A volume balance on the gasoline from each refinery is also needed. For

example, for the gasoline from the first refinery we must have:


AL + AH + AP − A = 0

Similar constraints are required for the other three input gasolines.We require four constraints for the amount available of each input (A ≤ 280,

and so on), and there are three constraints for the outputs individually (L≤ 260),12

and so on, and one more constraint for the total production (L + H + P ≤ 500).Finally, we need a constraint for each product to ensure that the minimum

octane rating is met. For example, the octane rating of the low-octane gasolinemust be at least 86. Hence we must have

81AL + 87BL + 98CL + 115DLAL + BL + CL + DL

≥ 86

The denominator is simply L, hence we have

81AL + 87BL + 98CL + 115DLL

≥ 86

The constraint as it stands is not linear, but we can make it so by multiplying bothsides by L (doing this also avoids a potential division by 0 problem). This gives:

81AL + 87BL + 98CL + 115DL≥ 86L

Finally, to put the constraint into the standard form in which all variables are onthe left-hand side, we subtract 86L from both sides to obtain:

81AL + 87BL + 98CL + 115DL−86L≥ 0

Similar constraints are made for high-octane gasoline and aviation fuel.To solve this model using LINDO we need the complete algebraic model.

However with what we have done so far we can develop the algebraic modeldirectly on LINDO. Doing this we obtain:

! Blending Model Analyst: D.M. Tulett

! A, B, C, and D represent the number of cubic metres

! of gasoline purchased from A-One Petroleum, Better

12As mentioned in Linear Optimization 1, the shortcut that we used for the non-negativity re-strictions is not used for the constraints; we write these three constraints using three lines.


! Grade Fuels, Clearly Superior, and Deluxe Gasoline

! respectively. L, H, P represent the number of cubic

! metres of low-octane, high-octane, and propeller-driven

! aviation fuel made respectively. AL, AH, AP, BL, BH,

! BP, CL, CH, CP, DL, DH, and DP represent the number

! of cubic metres of gasoline from source {A, B, C, or

! D} used to make output {L, H, or P}.

maximize 570L + 600H + 700P - 350A - 400B - 450C - 600D

subject to

! Balance on the Products

Bal. L) AL + BL + CL + DL - L = 0

Bal. H) AH + BH + CH + DH - H = 0

Bal. P) AP + BP + CP + DP - P = 0

! Balance on the Inputs

Bal. A) AL + AH + AP - A = 0

Bal. B) BL + BH + BP - B = 0

Bal. C) CL + CH + CP - C = 0

Bal. D) DL + DH + DP - D = 0


! Input Availability

Input A) A <= 280Input B) B <= 400Input C) C <= 200Input D) D <= 130

! Production Limitations

Output L) L <= 260Output H) H <= 260Output P) P <= 260

Total) L + H + P <= 500

! Octane Rating Constraints

Octane L) 81AL + 87BL + 98CL + 115DL - 86L >= 0

Octane H) 81AH + 87BH + 98CH + 115DH - 92H >= 0

Octane P) 81AP + 87BP + 98CP + 115DP - 105P >= 0

Solving we obtain (only the non-zero variables and the binding constraints arelisted):


1) 94288.23

VARIABLE VALUE REDUCED COSTL 260.000000 0.000000P 240.000000 0.000000A 192.352936 0.000000C 200.000000 0.000000


D 107.647057 0.000000AL 183.529419 0.000000CL 76.470589 0.000000AP 8.823529 0.000000CP 123.529411 0.000000DP 107.647057 0.000000

ROW SLACK OR SURPLUS DUAL PRICESBAL. L) 0.000000 245.588242BAL. H) 0.000000 245.588242BAL. P) 0.000000 245.588242BAL. A) 0.000000 350.000000BAL. B) 0.000000 400.000000BAL. C) 0.000000 475.000000BAL. D) 0.000000 600.000000INPUT C) 0.000000 25.000000OUTPUT L) 0.000000 9.705882

TOTAL) 0.000000 173.529419OCTANE L) 0.000000 -7.352941OCTANE H) 0.000000 -7.352941OCTANE P) 0.000000 -7.352941

NO. ITERATIONS= 9

Blendex’s immediate concern is to notify the suppliers of what they want tobuy. They send emails to A-One Petroleum, Clearly Superior, and Deluxe Gaso-line confirming the amounts as 192.353, 200.000, and 107.647 cubic metres re-spectively. Better Grade Fuels is sent an email thanking them for their bid, butstating that none will be purchased this week.

For their own production crew which must blend the gasolines, more informa-tion is required. The recommendation could be stated as follows:

Recommendation

Produce low-grade gasoline and aviation fuel by blending fuel from A-OnePetroleum, Clearly Superior, and Deluxe Gasoline as follows (all figures are incubic metres):


Source Low-Grade Aviation Fuel TotalA-One Petroleum 183.529 8.824 192.353Clearly Superior 76.471 123.529 200.000Deluxe Gasoline – 107.647 107.647Total 260.000 240.000 500.000

The contribution to profit based on this plan is $94,288.23.

13.6 Exercise13.6.1 Problem Description

A woodworking company buys lumber from which they make tables and chairs.They then outsource the painting of the tables and chairs, and sell the finishedproducts. They buy the lumber at a cost of $2 per board-metre. Each table re-quires 30 board-metres of lumber, while each chair requires 4 board-metres. Eachproduct spends time in three operations: cutting; polishing; and assembly. Thetimes in minutes per unit are:

Cutting Polishing AssemblyTable 45 12 15Chair 18 7 9

Each day, the shop is available for six hours of productive time. There are threecutting machines, one polisher, and one person to do the assembly.

The painting firm charges $54 per hour. When painting tables, they can paint6 tables per hour; when painting chairs, they can paint 20 chairs per hour.

The woodworking company sells its products to a wholesaler at $90 per tableand $28 per chair. The market requires that at least four chairs be made for everytable made.

They wish to use a linear optimization model to help determine what theyshould do.

13.6.2 Decision Variables

We define (all on a daily basis):T = the number of tables madeC = the number of chairs madeL = the amount of lumber purchased (board-metres)P = the number of hours of painting purchased


13.6.3 Discussion

We are modelling this problem using the four variables as defined above. Alter-natively, a problem like this could be modelled with just the two product variables,which would allow for a graphical solution. However, since we can use a com-puter to solve the model, this advantage is not all that important. We are betteroff modelling this problem with four variables, because this preserves the originaldata of the problem, so if later a price changes (e.g. if the painting cost goes from$54 to $56 per hour) we can more easily determine the effect of this change. Also,by using four variables we avoid a potential calculation error which could occurwhen calculating the objective function coefficients of the two-variable version.

13.6.4 What Needs to be Done

Finish this problem by completing the algebraic model, and then obtain the opti-mal solution by solving it using either LINDO or a spreadsheet. As always, reportthe solution so that a manager can understand it. The variable names given aboveare to be used for the algebraic model. If you decide to use a spreadsheet, wordsinstead of symbols (e.g. “Tables Made” instead of “T”) may be used.



14.1 Introduction

In the previous lecture we saw the terms reduced cost and dual price on theLINDO output, which now need to be explained. Also, there is more output avail-able, which is obtained by answering Yes when prompted with the question “DORANGE (SENSITIVITY) ANALYSIS?”. Earlier in this course we did sensitivityanalysis by hand for solutions obtained using a payoff matrix. Sensitivity analysisby hand in the context of linear optimization can be done using either graphicalsolutions or the final iteration of the simplex algorithm. Here in this course, how-ever, we shall simply use the LINDO output to give us the data we need to answersome managerial what-if questions. Everything that we are doing here assumesthat none of the variables is required to be integer . Except for problems whichare naturally integer (i.e. the variables turn out to be integer without forcing themto be integer), sensitivity analysis in the presence of integer variables is essentiallymeaningless.

In what follows, we are looking at predicting the effect of a single change tothe model at one time. Analyzing the effect of two or more concurrent changes isbeyond the scope of this course; however, we can find this effect on the computerby re-running the model with these multiple changes. Also, we would need tore-run the model if a proposed single change falls outside of the allowable range(defined below) and if we desire the exact change to the OFV. The explanationwhich follows is based on the usual situation in which there is a unique non-degenerate optimal solution.13

We could, of course, run an altered model every time a proposed change wasmade. For a small model, this could be done in seconds on a computer. However,for large models, the analysis provided here gives a fast way to evaluate one-at-a-time changes just by looking at the sensitivity output from the initial run.

13The computer output is still valid for non-usual cases in the sense that within the allowablerange the predicted change is correct. However, the reported allowable range might understate thetrue range.


14.2 New Concepts

14.2.1 Solution and Non-Solution Variables

When we run anything other than a very small model we are likely to find thatsome (and indeed many) of the variables have an optimal value of 0. Just as weconsider separately the binding and non-binding constraints, so it is useful to con-sider separately the variables which have optimal values greater than 0 from thosevariables whose optimal value is 0. The former we call solution variables (orvariables in the solution). The latter we call non-solution variables (or variablesnot in the solution).

14.2.2 Reduced Cost

The term reduced cost was originally developed for minimization models. Theidea is that a variable has a value of 0 because the cost is too high. To make thevalue greater than 0, i.e. to make the variable part of the solution, the objectivefunction coefficient of that variable needs to be reduced. This reduction in costmust be at least the amount given by the reduced cost. For example, if a variablehas an objective function coefficient of 5 and a reduced cost of 2, then the coef-ficient would have to fall below 3 in order for the variable to become part of thesolution. In a minimization model, both the Solver in Excel and LINDO will findthe same value for the reduced cost associated with each variable.

However, in a maximization model, the Solver in Excel and LINDO treat thereduced cost concept differently. The reported numbers are the same in magni-tude, but are opposite in sign. For example, suppose that a variable with an objec-tive function coefficient of 7 has an optimal value of 0. If increasing the coefficientto 10 would cause the variable to start increasing from 0, then the magnitude ofthe required change in the coefficient is 3. Excel would report the reduced cost tobe −3, but LINDO would report it as 3.

14.2.3 Dual Price and Shadow Price

In Mathematics for Management Science, when we studied how to optimize anon-linear function subject to an equality constraint, we saw the concept of aLagrangian multiplier. This gives us the rate of change of the OFV with respectto a small change to the right-hand side of the equality constraint. Dual and/orshadow prices provide something similar in the context of linear optimization


(though the sign may be reversed). Dual prices are used by LINDO; spreadsheetsuse shadow prices.

A dual or shadow price can only be non-zero for a binding constraint; a non-binding constraint always has a dual or shadow price of 0. Within an allowablerange, the dual price gives the improvement in the OFV per unit change in theright-hand side value of the constraint. Improvement means increase for a maxi-mization model, and decrease for a minimization model. The related term shadowprice gives the change in the OFV per unit change in the right-hand side value ofthe constraint (within the allowable range). The terms dual price and shadowprice mean the same thing for a maximization model; for a minimization modelthey are opposite in sign.

14.2.4 Allowable Range

The computer will find an allowable range for both objective function coefficientsand right-hand side values. The word allowable has nothing to do with grantingpermission. What it gives instead is a range in which changes to the model occurin a predictable manner. This allowable range could be given as absolute upperand lower limits, but LINDO’s approach is to report both an allowable increaseand an allowable decrease compared with the current value. Sometimes, theallowable increase or decrease may be infinite.

Changes to the Objective Function Coefficients For a non-solution variable ina maximization model, the allowable increase equals (theoretically) the reducedcost, and the allowable decrease is infinite. For a non-solution variable in a min-imization model, the allowable increase is infinite, and the allowable decreaseequals the reduced cost. For changes to the coefficient within the allowable range,there is no change to the values of any of the variables, and there is no change tothe OFV.

For a variable which is in the solution, there will be an allowable range forchanges to its objective function coefficient in which the values of the variables donot change. However, the change in the objective function coefficient will changethe OFV. Within the allowable range, the change in the OFV will equal the changeto the coefficient multiplied by the current value of the variable. There’s nothingprofound about this – if instead of selling 150 books at $20 per copy you sell 150books at $23 per copy, then you’ll make an extra $3(150) = $450. Of course, mostsellers cannot by themselves raise the price of anything. However, we are lookingat what happens to the OFV if in the market as a whole the price rises or falls.


Changes to the Right-Hand Side Values First, we consider the easier case ofa non-binding constraint. Within the allowable range, the values of the variableswill remain the same, and the OFV will remain the same. As was mentioned inthe previous lecture, the right-hand side value of a non-binding ≤ constraint canbe decreased by up to the amount of the slack, or be increased indefinitely. Theright-hand side value of a non-binding ≥ constraint can be increased by up to theamount of the surplus, or be decreased indefinitely.

For a binding constraint, a change to the right-hand side value will change thevalues of the variables, and therefore also the OFV. However, within the allowablerange, the effect on the OFV is a predictable one. Suppose that the change (in-crease or decrease) to the right-hand side value is an amount called ∆ rhs, where∆ rhs is in the allowable range. Since the allowable decrease is reported as apositive number, we must have:

− allowable decrease ≤ ∆ rhs ≤ allowable increase

On LINDO, the output reports the dual price. We denote the change in the OFVas ∆ OFV.For a maximization model:

∆ OFV = (dual price) (∆ rhs)

For a minimization model:

∆ OFV =− (dual price) (∆ rhs)

For a computer output which reports the shadow price rather than the dualprice, we have for both maximization and minimization:

∆ OFV = (shadow price) (∆ rhs)

This analysis alone will not help us predict what will happen to the variables; allit does is predict the change to the OFV. If we want the new value of the OFV itis simply:

new OFV = old OFV+(∆ OFV)

Though we cannot predict the new OFV if ∆ rhs falls outside the allowablerange, we can establish bounds on this number, provided that the model remainsfeasible. (If a constraint is made more stringent it could eliminate the feasible


region, thereby making the model infeasible.) For a maximization model, if ∆ rhs> allowable increase, then:

∆ OFV≥ (dual price)(allowable increase)

If the ∆ rhs <− allowable decrease, then:

∆ OFV≤ (dual price)(− allowable decrease)

For a minimization model, if ∆ rhs > allowable increase, then:

∆ OFV≥− (dual price)(allowable increase)

If the ∆ rhs <− allowable decrease, then:

∆ OFV≤− (dual price)(− allowable decrease)

These concepts are now illustrated using three examples. The first two arequite simple, while the third is the more complex blending model which we sawin the previous section.

14.3 Example 1: Maximization

A chemical laboratory can make three types of chemical powders. The variablesx1, x2, and x3 represent the number of kilograms per day of the three chemicals.The chemical company has made the following profit-maximization model:

maximize 32x1 + 25x2 + 18x3

subject to

conveyor) 3x1 + 5x2 + 7x3 <= 550shipping) 5x1 + 6x2 + 3x3 <= 800min.prod) 2x1 + 4x2 + 8x3 >= 360mixing) 8x1 + 9x2 + 4x3 <= 880

Running this model on LINDO, and asking for the sensitivity analysis, weobtain:




1) 3600.000

VARIABLE VALUE REDUCED COSTX1 90.000000 0.000000X2 0.000000 11.590909X3 40.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICESCONVEYOR) 0.000000 0.363636SHIPPING) 230.000000 0.000000MIN.PROD) 140.000000 0.000000

MIXING) 0.000000 3.863636

NO. ITERATIONS= 2

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGESVARIABLE CURRENT ALLOWABLE ALLOWABLE

COEF INCREASE DECREASEX1 32.000000 4.000000 11.860464X2 25.000000 11.590908 INFINITYX3 18.000000 56.666664 2.000000

RIGHTHAND SIDE RANGESROW CURRENT ALLOWABLE ALLOWABLE

RHS INCREASE DECREASECONVEYOR 550.000000 990.000000 110.000000SHIPPING 800.000000 INFINITY 230.000000MIN.PROD 360.000000 140.000000 INFINITY

MIXING 880.000000 389.230774 565.714294


We see that the solution is to produce 90 kg (all units are per day) of chem-ical 1, none of chemical 2, and 40 kg of chemical 3. The solution variables aretherefore x1 and x3. The profit obtained using this production plan is $3600. Theslack on the conveyor and mixing constraints is zero, hence these are the bindingconstraints.

14.3.1 Changes to the Objective Function Coefficients

We consider what happens to the OFV in each of the following situations:

1. The price of powder 1: (a) decreases by $10; (b) increases by $5.

2. The price of powder 2: (a) decreases by $18; (b) increases by $9; (c) in-creases by $15.

3. The price of powder 3: (a) increases by $30; (b) decreases by $7.

The amount of powder 1 made and sold is represented by variable x1, whichis a solution variable. The price per kg is the coefficient of this variable, which iscurrently $32 (don’t confuse this with the value of the variable itself, which is 90kg). From the OBJ COEFFICIENT RANGES section of the sensitivity report,we see that the allowable increase is 4, and the allowable decrease is 11.860464.In other words, we would obtain the same solution even if the coefficient were torise from 32 to 32 + 4 = 36, or if it were to fall to 32−11.860464 = 20.139536.Hence a decrease of $10 (which is ≤ 11.860464) would have no effect on thesolution; they would still make 90 kg per day of powder 1, and 40 kg per day ofpowder 3. However, the OFV would decrease by $10(90) = $900, i.e. it wouldfall from $3600 to $2700. We could also state this as ∆ OFV =−$900. A rise of$5 (> 4) is beyond the allowable increase, so we would obtain a new solution, andwe therefore cannot predict the new value of the OFV exactly. We would have tore-run the model on the computer replacing the 32 with 37, if we wanted to knowthe new value exactly. However, we can state that the new OFV will be at leastwhat it would be based on the allowable increase. An increase of 4 would causethe profit to increase by $4(90) = $360, hence an increase of 5 would cause anincrease of at least this much, i.e. ∆ OFV≥ $360.

Variable x2 is not in the solution; the current price of $25 per kg isn’t highenough to justify making any quantity of powder 2. Ordinary logic therefore tells


us that a price decrease is not going to change anything; a decrease in the price of$18 per kg does not change either the solution or the OFV. Note that the allowabledecrease is infinite. For a price increase, we cannot determine what will happenby logic – we need to look at the allowable increase from the sensitivity report.This figure is seen to be 11.590908. Hence an increase of $9 per kg is less than theallowable increase, and there would be no change to the solution. Furthermore,there would be no change to the OFV, because we are not making any powder 2.If however the price were to rise by $15 per kg, this would surpass the allowableincrease. The solution would change, and the OFV would increase, but neither ofthese things could be quantified without re-running the model.

The amount of powder 3 made and sold is represented by variable x3. Thecurrent coefficient of this solution variable is $18. We see from the printout thatthe allowable increase is 562

3 , and the allowable decrease is 2. In other words,we would obtain the same solution even if the coefficient were to fall from 18 to18− 2 = 16, or if it were to rise to 18 + 562

3 = 7423 . Hence an increase of $30

(which is ≤ 5623 ) would have no effect on the solution; they would still make 90

kg per day of powder 1, and 40 kg per day of powder 3. However, the OFV wouldincrease by $30(40) = $1200. A decrease of $7 (> 2) is beyond the allowabledecrease, so we would obtain a new solution. The new OFV will be at most whatit would be based on the allowable decrease. A decrease of 2 in the rhs wouldcause the profit to decrease by $2(40) = $80, hence a decrease of 7 would causea decrease of at least this much. We must be careful with the inequality here; themagnitude is at least $80. Hence if the change is say $80 or more downwards,then ∆ OFV≤−$80.

14.3.2 Changes to the Right-Hand-Side Values


1. The right-hand side value (rhs) of the conveyor constraint: (a) decreases by100; (b) decreases by 800; (c) increases by 550; (d) increases by 1200.

2. The rhs of the shipping constraint: (a) decreases by 100; (b) increases by200; (c) decreases by 300.

3. The rhs of the minimum production constraint: (a) increases by 150; (b)increases by 110.

4. The rhs of the mixing constraint: (a) decreases by 330; (b) decreases by600; (c) increases by 275.


Since the conveyor constraint is binding, any change to the rhs will affect thesolution. While the new solution is not easily found without re-running the model,the change to the OFV is easy to predict within the allowable range. We see fromthe sensitivity report that this constraint has an allowable increase of 990 and anallowable decrease of 110. Hence a decrease of 100 is within the allowable range.To see the effect on the OFV, we need the dual price for this constraint, whichis 0.363636. [On Excel for a maximization model we would look at the shadowprice.] The OFV will therefore change by:

∆ OFV = (dual price) (∆ rhs)= 0.363636(−100)= −36.36

(Note: we can say that the change is −36.36, or the decrease is 36.36.) If wewant the new OFV this is 3600−36.3636 = 3563.64. A decrease of 800 would bebeyond the allowable decrease of 110. The decrease in the OFV would thereforebe at least 0.363636(110) = $40, or we could write ∆ OFV≤−$40. An increase of550 would be allowable, and would cause the OFV to increase by 0.363636(550)= $200. An increase of 1200 would exceed the allowable increase of 990, so theOFV would increase by at least 0.363636(990) = $360.

The shipping and minimum production constraints are non-binding, so thesensitivity analysis is very easy. If the proposed change is within the allowablerange, then there is no change to the OFV. If the proposed change is beyond thisrange, then the OFV will be impaired, i.e. it would decline for a maximizationmodel. The rhs of the shipping constraint can be increased indefinitely or bedecreased by up to 230. Hence a decrease of 100 or an increase of 200 wouldnot affect the OFV. A decrease of 300 would cause the OFV to decrease, thoughwe cannot predict by how much. The minimum production constraint has anallowable increase of 140, and it can be decreased indefinitely. An increase of150 would cause the OFV to fall; an increase of 110 would leave it unchanged.

Finally, the mixing constraint is binding. It has an allowable increase of389.230774, an allowable decrease of 565.714294, and a dual price of 3.863636.Hence a decrease of 330 is within the allowable range and the OFV will fall by3.863636(330) = 1275.00. A decrease of 600 would be beyond the allowablerange; the OFV would fall by at least 3.863636(565.714294) = 2185.71. An in-crease of 275 would be within the allowable range, and the OFV would increaseby 3.863636(275) = 1062.50.


14.4 Example 2: MinimizationA company buys food products from some or all of five suppliers. These are mixedtogether. The mixture must meet minimum requirements for three nutrients, haveno more than a specific amount of fat, and then be packed into 14.4 kg bags. Thevariables have been defined as the amount of input of each of the five suppliers thatgoes into one bag of mixed product, and are denoted as x1 to x5. In this examplethe objective function coefficients are costs rather than revenues.

The company has made the following cost minimization model:

minimize 3.7x1 + 8.3x2 + 5.1x3 + 2.9x4 + 3.1x5

subject to

N 1) 3x1 + 4x2 + 6x3 + 5x4 + 2x5 >= 40.5N 2) 8x1 + 6x2 + 2x3 + 3x4 + 5x5 >= 81.0N 3) 4x1 + 5x2 + 8x3 + 7x4 + 3x5 >= 54.9Fat) 5x1 + 3x2 + 5x3 + 6x4 + 4x5 <= 64.8Mass) x1 + x2 + x3 + x4 + x5 = 14.4

Solving the model on LINDO we obtain:



1) 49.94000

VARIABLE VALUE REDUCED COSTX1 4.700000 0.000000X2 0.000000 1.288889X3 1.300000 0.000000X4 0.600000 0.000000X5 7.800000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES


N 1) 0.000000 -1.088889N 2) 0.000000 -0.311111N 3) 1.900000 0.000000FAT) 0.000000 1.422222MASS) 0.000000 -5.055555

NO. ITERATIONS= 7



COEF INCREASE DECREASEX1 3.700000 1.657143 1.120000X2 8.300000 INFINITY 1.288889X3 5.100000 0.828572 2.327273X4 2.900000 1.706667 0.773334X5 3.100000 1.866667 8.533333


RHS INCREASE DECREASEN 1 40.500000 1.800000 1.710000N 2 81.000000 3.600000 16.919998N 3 54.900002 1.900000 INFINITYFAT 64.800003 2.127273 0.720000MASS 14.400000 0.327273 1.017391

The optimal solution is for each bag of product to be composed of 4.7 kg fromsupplier 1, none from supplier 2, 1.3 kg from supplier 3, 0.6 kg from supplier4, and 7.8 kg from supplier 5. The cost of the optimal mixture is $49.94. Allconstraints except the one for Nutrient 3 are binding.




1. The cost [per kg] from supplier 1: (a) increases by $1.50; (b) decreases by$2.50.

2. The cost from supplier 2: (a) decreases by $1.28; (b) increases by $50.

3. The cost from supplier 3: (a) increases by 30 cents; (b) increases by 85cents; (c) decreases by $2.

4. The cost from supplier 4: (a) decreases by 50 cents; (b) decreases by $1.00;increases by $1.50.

5. The cost from supplier 5: (a) decreases by $5; (b) increases by $5.

Doing a sensitivity analysis on the objective function coefficients is no dif-ferent for minimization than it is for maximization. The company is currentlypaying $3.70 per kg to purchase 4.7 kg (per bag of finished product) from Sup-plier 1; there is an allowable increase of 1.657143 and an allowable decrease of1.12. An increase of $1.50 is within the allowable range, and the OFV would in-crease by $1.50(4.7) = $7.05. A decrease of $2.50 is outside the allowable range,so the OFV would fall by at least $1.12(4.7) = $5.264.

The coefficient of x2 can be increased indefinitely or be decreased by 1.288889.Hence a decrease of $1.28 or an increase of $50 are both within the allowablerange, and since x2 is not in the solution, there would be no change to the OFV.

The range for the coefficient of x3 is an increase of 0.828572 and a decrease of2.327273 from its current value of 5.1. Since x3 = 1.3, an increase of 30 cents (i.e.0.30) would increase the OFV by 0.30(1.3) = $0.39. An increase of 85 cents wouldbe beyond the range; the OFV would increase by at least 0.828572(1.3)≈ $1.077.A decrease by $2 would cause the OFV to fall by $2(1.3) = $2.60.

The current cost for purchases from Supplier 4 is $2.90 per kg; this has anallowable increase of $1.706667 and an allowable decrease of $0.773334. Hencea decrease of 50 cents is within the range, a decrease of a dollar would be outsidethe range, and an increase of $1.50 would be within the range. Since x4 = 0.6, a50 cent decrease would cause the OFV to fall by $0.50(0.6) = $0.30, a one dollardecrease would cause the OFV to fall by at least $0.773334(0.6) ≈ $0.464, and$1.50 increase would cause the OFV to rise by $1.50(0.6) = $0.90.


They pay $3.10 per kg from Supplier 5 and are currently ordering 7.8 kg perbag of final product. The allowable increase is $1.866667 and the allowable de-crease is $8.533333. Hence a decrease of $5 would be within the allowable rangebut an increase of $5 would be outside the range. A decrease of $5 would causethe OFV to fall by $5.00(7.8) = $39.00. An increase of $5 would cause the OFVto rise by at least $1.866667(7.8) = $14.56.

14.4.2 Changes to the Right-Hand-Side Values

In doing the sensitivity analysis for minimization using the LINDO output, wemust be careful when using the dual price, which gives the improvement per unitchange to the rhs. As stated earlier, improvement for a minimization model refersto a decline in the OFV (if the “improvement” is negative, then the OFV rises). IfExcel is used, then the shadow price gives the change to the OFV per unit changeto the right-hand side. For a minimization model the dual price and the shadowprice are opposite in sign. If a question is asking about what happens to the OFV,as opposed to asking about its improvement, it may be helpful to work with theshadow price instead. Going to the LINDO printout we find the dual prices, andby multiplying by −1 we obtain the shadow price. 14


1. The right-hand side value (rhs) of the Nutrient 1 constraint: (a) increases by1.5; (b) decreases by 1.6; (c) increases by 2.1.

2. The rhs of the Nutrient 2 constraint: (a) increases by 3.0; (b) decreases by15; (c) decreases by 20.

3. The rhs of the Nutrient 3 constraint: (a) increases by 1.5; (b) increases by2.5.

4. The rhs of the fat constraint: (a) decreases by 0.5; (b) decreases by 1.8; (c)increases by 0.34.

5. The rhs of the mass constraint: (a) decreases by 0.9 kg; (b) increases by 300g; (c) increases by 700 g.

14Note that this is not the same thing as finding the absolute value, because the dual price mayalready be positive, in which case the shadow price is negative.


The rhs of the Nutrient 1 constraint is currently 40.5. This binding constrainthas an allowable increase of 1.8 and an allowable decrease of 1.71. The dualprice as given by LINDO is −1.088889; the Excel sensitivity report would statethat the shadow price is 1.088889. An increase of 1.5 would cause the OFV toimprove by −1.088889(1.5) ≈ −$1.633. A negative improvement for a mini-mization model is an increase, i.e. the OFV would increase by about $1.633.Alternatively, we can use the shadow price directly: the change to the OFV wouldbe 1.088889(1.5) ≈ $1.633. A decrease of 1.6 is allowable, so the improvementto the OFV would be (−1.088889)(−1.6) ≈ $1.742. This improvement beingpositive for a minimization model means that the OFV would decline by thisamount, i.e. ∆ OFV = −$1.742. Notice that decreasing the rhs of this ≥ con-straint makes the restriction less stringent, and the cost decreases as a result. Adecrease of 2.1 is beyond the allowable range; the OFV would improve by at least(−1.08889)(−1.71) = $1.862, or ∆ OFV≤−1.862.

For the Nutrient 2 constraint, the allowable range is −16.919998 ≤ ∆ rhs ≤3.6, and the dual price is −$0.311111. Hence an increase of 3 would be allow-able, as would a decrease of 15, but a decrease of 20 would be beyond the allow-able range. An increase of 3 would cause an improvement of (−$0.311111)3 ≈−$0.933, i.e. the OFV would increase by $0.933. A decrease of 15 would cause animprovement of (−$0.311111)(−15) ≈ $4.667, i.e. the OFV would decrease by$4.667. A decrease of 20 would cause an improvement of at least (−$0.311111)(−16.919998) = $5.264, i.e. the OFV would decrease by at least $5.264.

The Nutrient 3 constraint is non-binding, which makes things easy. The al-lowable increase is 1.9, hence an increase of 1.5 would have no effect at all, whilean increase of 2.5 would cause there to be a new solution. Because such a changewould reduce the feasible region, it would cause the OFV to be impaired (i.e. risein this situation).

The fat constraint has an allowable increase of 2.127273, an allowable de-crease of 0.72, and a dual price of 1.422222. A decrease of 0.5 is therefore al-lowable, and would cause the OFV to improve by 1.422222(−0.5) ≈ −$0.711,i.e. ∆ OFV = $0.711. A decrease of 1.8 is beyond the allowable range; theOFV would improve by at least 1.422222(−0.72) = −$1.024, i.e. ∆ OFV ≥1.024. An increase of 0.34 is allowable, and would cause the OFV to improveby 1.422222(0.34)≈ $0.484, i.e. ∆ OFV =−$0.484.

Finally the mass constraint has an allowable increase of 0.327273, an allow-able decrease of 1.017391, and a dual price of−5.055555. The current rhs value is14.4, and the units are kg (kilograms). A decrease of 0.9 kg is therefore allowable,and the improvement to the OFV would be (−5.055555)(−0.9) = $4.55. In other


words, the OFV would fall by $4.55. Occasionally a conversion factor is requiredto analyze something; we re-state the 300 grams as 0.3 kg for consistency withthe way the constraint was written. An increase of 0.3 kg is allowable, and theimprovement to the OFV is (−5.055555)0.3 ≈ −$1.5174, i.e. ∆ OFV = $1.517.An increase of 700 g or 0.7 kg exceeds the allowable increase, the OFV wouldrise by at least 5.055555(0.327273) = $1.655.

14.5 Example 3: Blending Model14.5.1 LINDO Sensitivity Output

In the previous lecture the solution for the Blending Model reported only the vari-ables which have non-zero values. Now we will show the full report, because itis only the variables whose value is zero which can have positive reduced costs.Also, we will answer Yes when prompted with the “DO RANGE (SENSITIVITY)ANALYSIS?” question. Doing this we obtain:



1) 94288.23

VARIABLE VALUE REDUCED COSTL 260.000000 0.000000H 0.000000 4.411765P 240.000000 0.000000A 192.352936 0.000000B 0.000000 0.000000C 200.000000 0.000000D 107.647057 0.000000AL 192.352936 0.000000BL 0.000000 5.882353CL 58.823528 0.000000DL 8.823529 0.000000AH 0.000000 0.000000BH 0.000000 5.882353CH 0.000000 0.000000DH 0.000000 0.000000


AP 0.000000 0.000000BP 0.000000 5.882353CP 141.176468 0.000000DP 98.823532 0.000000

ROW SLACK OR SURPLUS DUAL PRICESBAL. L) 0.000000 245.588242BAL. H) 0.000000 245.588242BAL. P) 0.000000 245.588242BAL. A) 0.000000 350.000000BAL. B) 0.000000 400.000000BAL. C) 0.000000 475.000000BAL. D) 0.000000 600.000000

INPUT A) 87.647057 0.000000INPUT B) 400.000000 0.000000INPUT C) 0.000000 25.000000INPUT D) 22.352942 0.000000OUTPUT L) 0.000000 9.705882OUTPUT H) 260.000000 0.000000OUTPUT P) 20.000000 0.000000

TOTAL) 0.000000 173.529419OCTANE L) 0.000000 -7.352941OCTANE H) 0.000000 -7.352941OCTANE P) 0.000000 -7.352941

NO. ITERATIONS= 13



COEF INCREASE DECREASEL 570.000000 INFINITY 9.705882H 600.000000 4.411753 INFINITYP 700.000000 9.705882 4.411753A -350.000000 11.538434 7.142880B -400.000000 5.882371 INFINITYC -450.000000 INFINITY 25.000000


D -600.000000 17.368422 11.538433AL 0.000000 50.000000 0.000018BL 0.000000 5.882371 INFINITYCL 0.000000 0.000009 16.499998DL 0.000000 50.000000 0.000018AH 0.000000 0.000022 9.090940BH 0.000000 5.882371 INFINITYCH 0.000000 6.818161 0.000011DH 0.000000 0.000022 INFINITYAP 0.000000 0.000018 INFINITYBP 0.000000 5.882371 INFINITYCP 0.000000 16.499998 0.000009DP 0.000000 0.000018 10.714256


RHS INCREASE DECREASEBAL. L 0.000000 3.703704 9.382716BAL. H 0.000000 0.000000 0.000000BAL. P 0.000000 8.695652 2.608696BAL. A 0.000000 192.352936 87.647057BAL. B 0.000000 0.000000 400.000000BAL. C 0.000000 17.647058 44.705883BAL. D 0.000000 107.647057 22.352942

INPUT A 280.000000 INFINITY 87.647057INPUT B 400.000000 INFINITY 400.000000INPUT C 200.000000 17.647058 44.705883INPUT D 130.000000 INFINITY 22.352942OUTPUT L 260.000000 59.999996 20.000000OUTPUT H 260.000000 INFINITY 260.000000OUTPUT P 260.000000 INFINITY 20.000000

TOTAL 500.000000 20.000000 29.999998OCTANE L 0.000000 760.000000 300.000000OCTANE H 0.000000 0.000000 0.000000OCTANE P 0.000000 300.000000 1000.000000

We will now examine the effect of some changes to the current parameters.All of these are done one at a time, with everything else remaining as it is in the


current model.


For this problem, we restrict the analysis to those variables which appear in the ob-jective function, i.e. L, H, P, A, B, C, and D. We ignore the other twelve variables(AL, BL, etc.).

Non-Solution Variables In the current solution there is no production of high-octane gasoline for cars. This is seen on the solution report, where variable Hhas a value of 0 and a reduced cost of 4.411765. Also, if we look at the variableH row on the sensitivity report, we see that its current coefficient is 600, the al-lowable increase is 4.411765, and the allowable decrease is infinite. The latterpiece of information is intuitively obvious, for if we do not wish to make and sellhigh-octane gasoline when the price is $600 per cubic metre, then we certainlywouldn’t want to make it and sell it if the market price were to fall. If the pricerises by say $4 per cubic metre, this would not be enough to change the solutionbecause 4 < 4.411765. If the price were to rise by say $5 per cubic metre, thenthe solution would change, but we would have to re-run the model with the newprice to determine exactly what would happen.

Variable B, representing how much is purchased from Better Grade Fuels, isalso a non-solution variable. Curiously, variable B has a reported reduced costof 0. Normally, a reduced cost of 0 for a non-solution variable would indicatethat an alternative optimal solution exists, but that’s not the case here. We cantell that the price must change by more than 0 by looking at the variable B rowin the sensitivity report. There we see that the current coefficient is −400, theallowable increase is 5.882371, and the allowable decrease is infinite. Note that inthis example and for this variable, the price ($400) and the coefficient (−400) areopposite in sign. Hence, for example, if the price were to fall by say $5 per cubicmetre (to $395), the coefficient would rise by 5 (to −395). Such a change wouldnot affect the optimal solution. A fall in price to say $390 would however affectthe solution (a change of 10 > 5.882371); we would have to re-run the model todetermine exactly how.

The computer has done more work to come up with the sensitivity report inaddition to the solution report. In this case, we see that the sensitivity reportprovides better information, so it is wise to check this report, especially if thesolution report shows a reduced cost of 0.


Solution Variables For changes to the objective function coefficient of a solu-tion variable, we must turn to the sensitivity report. The variables left to examineare L, P, A, C, and D. Extracting just these rows we have:


COEF INCREASE DECREASEL 570.000000 INFINITY 9.705882P 700.000000 9.705882 4.411753A -350.000000 11.538434 7.142880C -450.000000 INFINITY 25.000000D -600.000000 17.368422 11.538433

From the variable L row, we see that an increase in price for low-octane gaso-line would not affect the variables. However, Blendex would make more money.They are currently making 260 cubic metres of low-grade gasoline. If the pricewere to rise by say $19 per cubic metre (from $570 to $589) the extra profit wouldbe $19(260) = $4940. The new OFV would be $94,288.13 + $4940 = $99,228.13.Conversely, a fall in the price of low-octane gasoline would cause the OFV to fall.If the price were to fall by $8 per cubic metre, we verify that 8 < 9.705882 (allow-able decrease), meaning that the OFV would fall by $8(260) = $2080. The effectof a fall in price to say $558 (a drop of 12) is beyond the allowable range. How-ever, we can say that the OFV would fall by at least $9.705882(260) = $2523.53.

Variables A, C, and D, because they are quantities which are bought rather thansold, have negative objective function coefficients. Hence if we are asked “Whatwould happen if the price from A-One Petroleum were to rise by $9 per cubicmetre?” (i.e. from $350 to $359), this is equivalent to the coefficient decreasingfrom −350 to −359. Because the coefficient is decreasing we must examine theallowable decrease, which is 7.142880. The change proposed in the question isgreater than this amount, so a new solution needs to be obtained by re-runningLINDO. However, since A = 192.352936, we can predict that the OFV will de-cline (because the cost of an input is rising) by at least $7.14288(192.352936) =$1373.95. As another example, suppose that the price of Clearly Superior were tofall by $30 per cubic metre. This would cause the coefficient (which is negative)to rise by 30, which is fine (the allowable increase is infinite). Since C = 300, theprofit for Blendex would rise by $30(200) = $6000.


14.5.3 Changes to the Right-Hand Side Values

It makes no sense to do a sensitivity analysis on the volume balance constraints orthe octane constraints. For these, the context requires that the right-hand side valuemust be 0. Because they are equality constraints, all of the volume constraints arebinding. The octane constraints are binding in this example.

Removing these volume balance and octane constraints leaves us with:

ROW SLACK OR SURPLUS DUAL PRICESINPUT A) 87.647057 0.000000INPUT B) 400.000000 0.000000INPUT C) 0.000000 25.000000INPUT D) 22.352942 0.000000OUTPUT L) 0.000000 9.705882OUTPUT H) 260.000000 0.000000OUTPUT P) 20.000000 0.000000

TOTAL) 0.000000 173.529419

The non-binding constraints (the ones for which the slack or surplus is greaterthan 0) are for the amounts available of inputs A, B, and D, and the production of Hand P. Besides the volume balance and octane constraints, the binding constraints(the ones for which the slack or surplus is 0) are the ones for the amount availableof input C, the production of L, and the total production.

Non-Binding Constraints All of the non-binding constraints are in this exam-ple ≤ constraints, so we are dealing with a slack on each one. The right-handside of any of these ≤ constraints can be increased indefinitely, or be decreasedby the amount of the slack, without affecting the current solution. For example,the supply of gasoline from A-One Petroleum could be increased indefinitely, orbe decreased by up to 87.647 cubic metres, without affecting the current solution.A change such as reducing the supply from Deluxe Gasoline by 30 cubic metresexceeds the slack, and hence a new solution would have to be found using LINDO.

Binding Constraints Any change to the binding constraints will affect the val-ues of the variables. However, within the allowable range such a change wouldhave a predictable effect on the OFV. The allowable ranges come from the sensi-tivity report:



RHS INCREASE DECREASEINPUT C 200.000000 17.647058 44.705883

OUTPUT L 260.000000 59.999996 20.000000TOTAL 500.000000 20.000000 29.999998

The total production is currently limited to 500 cubic metres, and all of this ca-pacity is being utilized. The dual price is 173.529419. The allowable increase is20. An increase in the right-hand side from 500 to 510 would increase the OFVby $173.5294(10) = $1735.29. The allowable decrease is 30. A decrease from500 to 475 would cause the OFV to fall by $173.5294(25) = $4338.24. An in-crease of 35 (> 20) would change the solution; the OFV would go up by at least$173.5294(20) = $3470.59.

Now let’s take a look at changes to the availability of gasoline from ClearlySuperior. This may seem a bit more complicated, because increasing the right-hand side isn’t free in that we have to buy more @ $450 per cubic metre in orderto obtain more. Actually, it’s no more difficult than the previous case. The dualprice on this availability constraint is $25. The allowance increase is 17.647058;the allowable decrease is 44.705883. Now suppose that we increase the right-hand side by say 14 units, from 200 to 214. The dual price of $25 means thatthe OFV will increase by $25(14) = $350.00. The $450 per cubic metre that hasto be paid to buy the extra gasoline has already been factored into the dual price.The dual price nets out everything. As the right-hand side is changed, the valuesof the variables change. Some go up, and some go down, and there are manycredits and debits which sum up to the dual price. Hence this situation is not morecomplicated. When evaluating the effect of a right-hand side change, see if it’swithin the allowable range, and if so the improvement to the OFV is simply thedual price multiplied by the change to the right-hand side.


14.6 Exercise14.6.1 A Maximization Problem

A garment factory can make skirts, blouses, and dresses. After deducting allvariable costs, the net revenue is $32 per skirt, $27 per blouse, and $40 per dress.There are three operations, each of which limits the amount of production: cutting,assembly, and finishing. In addition, each garment must be inspected. Since unionrules require that at least one inspector be on duty at all times, they will make aconstraint to keep at least one inspector busy. The model has been formulated as:

Let x1, x2, and x3 represent respectively the number of skirts, blouses, anddresses to be made each hour.

maximize 32x1 + 27x2 + 40x3

subject to

Cutting 5x1 + 4x2 + 2x3 ≤ 64Assembly 12x1 + 6x2 + 8x3 ≤ 160Finishing 7x1 + 5x2 + 8x3 ≤ 146

Inspection 6x1 + 4x2 + 3x3 ≥ 72


(a) Solve using a computer, and print this solution and the sensitivity report.(b) State the solution in words, and indicate which constraints are binding.(c) By using the information from the sensitivity report (NOT by re-running

the model each time), give the predicted change to the objective function value(and the reasoning behind your answer) for the following situations (taken one ata time). If the OFV cannot be predicted exactly, then give an answer such as “theOFV will increase by at least $90”.

(i) The price of each skirt rises by $5.00.(ii) There are three fewer units of assembly.(iii) The price of each dress falls from $40 to $27.(iv) The number of units of cutting increases by 10.(v) The number of units of finishing increases by 6.

14.6.2 A Minimization Problem

A company which makes chocolate bars needs to buy some exotic nuts: walnuts,chestnuts, and hazelnuts. They do not have to buy any of any one type, but they doneed to satisfy certain combinations of types, which has been modelled using three


constraints. Also, there is a capacity restriction. The model has been formulatedas:

Let x1, x2, and x3 represent respectively the number of kilograms of walnuts,chestnuts, and hazelnuts to be used each hour in the chocolate bar plant.

minimize 2x1 + 7x2 + 4x3

subject to

Combination 1 5x1 + 8x2 + 6x3 ≥ 230Combination 2 2x1 + x2 + 4x3 ≥ 145Combination 3 3x1 + 4x2 + 5x3 ≥ 196

Capacity 8x1 + 9x2 + 4x3 ≤ 252


(a) Solve using a computer, and print this solution and the sensitivity report.(b) State the solution in words, and indicate which constraints are binding.(c) By using the information from the sensitivity report (NOT by re-running

the model each time), give the predicted change to the objective function value(and the reasoning behind your answer) for the following situations (taken one ata time). If the OFV cannot be predicted exactly, then give an answer such as “theOFV will decrease by at least $50”.

(i) The price of hazelnuts rises by $1.20 per kg.(ii) The price of chestnuts falls by $2.70 per kg.(iii) An extra 100 units of capacity becomes available.(iv) The requirement for combination 1 falls by 25 units.(v) The requirement for combination 3 increases by 92 units.


15 Markov Chains 1

15.1 Introduction

Whenever we have a managerial problem which requires formulation as a mathe-matical model, we look for the type of model which is most appropriate. Decisiontrees are useful when we have decisions and events. Linear optimization is use-ful when we have a model which obeys the assumptions of this subject. Markovchains are a third approach, used for multi-period models which are governed byprobabilities which are constant over time.

Learning about Markov chains involves formulation and problem solutiontechniques which can be difficult. Many textbooks which contain a chapter onMarkov chains will discuss the topic of first passage time analysis, however thissubject is omitted in this course. To learn about Markov chains, a good knowledgeof matrix algebra as seen in Mathematics for Management Science is required.Matrix operations such as addition, subtraction, multiplication, and inversion willbe used extensively. If the knowledge of these topics is rusty, you should reviewthem.

Unlike most of the material in this course, this is not a chapter about decisionmaking. What is seen here can be imbedded into a decision making model, butthat is a subject for a subsequent optional course. We begin with an example forwhich modelling as a Markov chain is appropriate.

15.2 Example: Brand Switching

15.2.1 Description of the Situation

A supermarket has a points program based on the amount purchased. The pur-pose of having the program is to give them information about how each consumerchanges (or doesn’t change) brands from month to month. In particular, theyhave studied the three brands of detergent which they sell, here referred to simplyas Brands 1, 2, and 3. Of the consumers who bought Brand 1 last month, 47%stayed with this brand. Thirty-six percent switched to Brand 2, and seventeen per-cent switched to Brand 3. [A small number of shoppers who did not buy in bothmonths are not considered, meaning that the percentages will add up to 100%.]Of those who bought Brand 2 last month, 52% stayed with brand 2, 9% went toBrand 1, and 39% are purchasing Brand 3 this month. Among last month’s Brand3 shoppers, 61% stayed with Brand 3, while eighteen percent switched to Brand


1, and 21% switched to Brand 2. They have been tracking the percentages of con-sumers who stay or switch for several months, and have found that these numbersare very stable.

Assuming that this stability continues into the future, they would like to beable to predict the long-term market share for each brand of detergent.

15.2.2 State Transition (Probability) Diagram

A Markov chain consists of what are called states in which the probabilities oftransitions between the states obey certain properties. The central idea in Markov

chains is that history is irrelevant . The only thing that matters is the present,called the state of the system. A system is a Markov chain if the number of statesis finite, if the probabilities of transition between the states are known, and if theseprobabilities do not change over time.

For example, what matters in the brand switching situation described aboveis what brand the consumer is buying this month. This chain has three brands,and hence three states. We sometimes display a Markov chain on what is called astate transition probability diagram (or simply the state transition diagram ). We

begin to make this diagram by making a circle for each state, labeled as Brand 1,Brand 2, and Brand 3. This is shown in Figure 51.

From one month to the next, the consumer either stays with the same brand, orswitches to another brand. On the Markov chain, staying with the same brand is atransition from one state to itself; switching from one brand to another brand is atransition from one state to another. A transition from a state to itself is drawn asa loop with an arrowhead entering the state. A transition from one state to anotheris drawn as an line or curve with an arrowhead touching the new state. This isshown in Figure 52.

Next to each transition we place the probability of this transition, called thestate transition probability . When we are given, for example, a statement that

36% of those buying Brand 1 will switch to Brand 2 next month, this is the sameas saying that a Brand 1 consumer chosen at random has a 0.36 probability ofswitching to Brand 2 next month. There are nine (3× 3) pieces of data to beextracted from the situation description. Doing this, and then placing each pieceof data next to the arrow which represents the transition, we obtain the completedstate transition diagram. This is shown in Figure 53.

In this example, it is possible to buy any brand next month no matter what isbeing purchased this month. Because of this, all the arrows needed to be drawn.However, in many examples of Markov chains there are some impossible tran-


Brand 1 Brand 2

Brand 3

Figure 51: Brand Switching: States


Brand 1 Brand 2

Brand 3

Figure 52: Brand Switching: States with Transitions


Brand 1 Brand 2

Brand 3

0.47

0.36

0.170.09

0.52

0.39

0.18 0.21

0.61

Figure 53: Brand Switching: State Transition Diagram


sitions. When this happens, we do not draw an arrow for a transition with a 0probability. By omitting such arrows, the clutter on the diagram is greatly re-duced.

State transition diagrams are useful for providing a conceptual grasp of thesituation. However, they are not suited as a computational tool. For doing calcu-lations based on the state transition probabilities, we are better off displaying thedata in a matrix.

15.2.3 State Transition (Probability) Matrix

A Markov chain can be described by a square matrix in which the number of rows(or the number of columns) is the same as the number of states. This matrix iscalled the state transition probability matrix (or simply the state transition matrix ).For example, the brand switching situation which has three states can be describedby a 3 by 3 matrix. Unless indicated to the contrary, the rows and columns matchup with the states, i.e. row 1 and column 1 refer to state 1, and so on. By con-vention, the row is the current state and the column is the subsequent state, i.e.the number in row i and column j is the probability of going in one transitionfrom state i to state j. Because these numbers are probabilities, every number isbetween 0 and 1 inclusive, and the sum of each row must be 1.

Obtaining the probabilities either from the original situation description, orfrom the state transition diagram, the state transition matrix is:

From 1State 2

3

To State1 2 3 0.47 0.36 0.17

0.09 0.52 0.390.18 0.21 0.61

For emphasis, we have added the state numbers on the left and at the top, but

where these are clear they will in future be omitted. In this more compact formthe state transition probability matrix, which we denote as P, is simply:

P =

0.47 0.36 0.170.09 0.52 0.390.18 0.21 0.61


However, when using a spreadsheet it is useful to put labels to the left andon top of the space reserved for the transition probabilities. On a spreadsheet theinformation could be displayed as:

A B C D E F G H1 To State Brand2 1 2 3 Switching3 Matrix From 1 0.47 0.36 0.17 Example4 P State 2 0.09 0.52 0.395 3 0.18 0.21 0.616

15.2.4 Finite Number of Transitions

The matrix P gives the probabilities of going from one state to another in onetransition. We denote the number in row i and column j of this matrix as pi j,which represents the probability of going from state i to state j in one transition.

Now suppose that we want to know the probability of going from state i tostate j in two transitions. For example if i = 3 and j = 1, we are asking “If aconsumer buys Brand 3 this month, what is the probability that he or she will buyBrand 1 two months from now?”. Note that we are not precluding the purchase ofBrand 1 next month. (If we wanted to exclude this from happening, we would beinvolved with first passage time analysis, which as stated earlier is not part of thiscourse). There are three ways for this to happen:

1. The consumer switches to Brand 1 next month, and then stays with Brand 1for the subsequent month.

2. The consumer switches to Brand 2 next month, and then switches to Brand1 for the subsequent month.

3. The consumer stays with Brand 3 next month, and then switches to Brand 1for the subsequent month.

The probability which we seek is the sum of the joint probabilities of eachof the three possibilities. In finding each joint probability, we are using the factthat the transition probabilities do not vary from one transition to another (this isa property of a Markov chain). Hence, for example, the probability of switchingto Brand 1 next month, and then staying with Brand 1 for the subsequent month,is given by p3,1 p1,1.


In general, we denote the probability of going from state i to state j in k tran-sitions as p(k)

i j . Here we seek p(2)3,1:

p(2)3,1 = p3,1 p1,1 + p3,2 p2,1 + p3,3 p3,1

= 0.18(0.47)+0.21(0.09)+0.61(0.18)= 0.0846+0.0189+0.1098= 0.2133

To find p(2)3,1, we calculated the dot product of row 3 and column 1. In general to

find p(2)i j , we find the dot product of row i and column j. For example, suppose

we want the probability that a consumer who is currently buying Brand 2 will buyBrand 3 in two months time. This is:

p(2)2,3 = (row 2) · (column 3)

= (0.09 0.52 0.39)

0.170.390.61

= 0.09(0.17)+0.52(0.39)+0.39(0.61)= 0.0153+0.2028+0.2379= 0.4560

Since matrix multiplication involves finding the dot products of all rows andcolumns, we can find all the p(2)

i j ’s by calculating P times P. We will denote thisproduct as P2.

P2 =

0.47 0.36 0.170.09 0.52 0.390.18 0.21 0.61

0.47 0.36 0.170.09 0.52 0.390.18 0.21 0.61

=

0.2839 0.3921 0.32400.1593 0.3847 0.45600.2133 0.3021 0.4846

While finding P2 by hand is easy enough, doing this operation on a spreadsheet

is very fast and avoids a potential calculation error. With P in the range D3:F5,we can put P2 below it, say in the range D8:F10. To do this we drag the mouseover the range D8:F10, and in cell D8 we enter the appropriate formula:


=MMULT(D3:F5,D3:F5)

The commands Ctrl (keep it held down) Shift (keep it held down) and Enterthen produce:

A B C D E F G H1 To State Brand2 1 2 3 Switching3 Matrix From 1 0.47 0.36 0.17 Example4 P State 2 0.09 0.52 0.395 3 0.18 0.21 0.6167 1 2 38 Matrix 1 0.2839 0.3921 0.32409 P2 2 0.1593 0.3847 0.4560

10 3 0.2133 0.3021 0.484611

We can continue in this fashion for any power of P. In general, matrix multi-plication is not commutative, but in this context it is commutative. Hence we canfind P3 either as P2P or as P P2. There are three ways to find P4: (1) P3P; (2)P2P2; or (3) PP3. Note that if we wanted something like P9, the fast way to findit starting with P would be to first obtain P2, then square this to obtain P4, thensquare this to obtain P8, and then multiply this by P to obtain P9. Doing this on aspreadsheet we find:

P9 =

0.209147670 0.349827600 0.4410247310.209188246 0.349799030 0.4410127240.209183334 0.349835786 0.440980880

We see that in every column the probabilities are essentially independent of theinitial state. For example, whether the consumer is buying Brand 1, 2, or 3 thismonth, there is about a 0.209 probability of buying Brand 1 nine months fromnow.

After many transitions (k of them), the nine probabilities in the Pk matrixare reduced to only three distinct probabilities, one for each column. Rather thanfinding these numbers numerically by finding Pk for higher and higher values of k,


these numbers can be found analytically. Things will work out properly providedthat for some value k all the numbers in the matrix Pk are non-zero. There’s clearlyno problem in this example, since even P itself is entirely non-zero.

15.3 Determining Steady-State Probabilities Analytically

The analytical method of finding steady-state (also called long-term) probabilitiesfinds the exact solution in one use of the procedure. This is in contrast with thenumerical method of the previous section, which after many uses of the procedurefinds successive approximations. The analytical method is what we shall use fromnow on.

The implementation of the analytical method is different depending on whetherthe chain has just two states, or whether it has three states and we seek a solutionby hand, or whether it has many states and we seek a solution using a computer.We begin with the three-state brand switching example to illustrate.

15.3.1 Three State Brand Switching Example

The numbers which we seek are the steady-state probabilities of the system beingin state 1, state 2, or state 3. In the context of switching brands, these are theprobabilities of the consumer’s purchase being Brand 1, Brand 2, or Brand 3 manymonths into the future. In this document these numbers are represented by thesymbols x1, x2, and x3. Each of these is an unknown, for which we shall solvethree equations in three unknowns. One of these three equations is easy to obtain;these probabilities must sum to 1, so we can write:

x1 + x2 + x3 = 1

Now we need two more equations. Here’s how to obtain another one. Aftermany iterations, the probability of being in state 1 next month is the same as theprobability of being in state 1 this month; both are x1. For any matrix P, we canbe at state 1 next month in one of three ways:

1. We are in state 1 this month (probability x1), and we stay in state 1 (proba-bility p1,1).

2. We are in state 2 this month (probability x2), and we move to state 1 (prob-ability p2,1).


3. We are in state 3 this month (probability x3), and we move to state 1 (prob-ability p3,1).

Each of these possibilities has a joint probability, and the sum of these joint prob-abilities is:

x1 = p1,1x1 + p2,1x2 + p3,1x3

Doing a similar analysis for state 2 we obtain:

x2 = p1,2x1 + p2,2x2 + p3,2x3

We now have three equations in three unknowns:

x1 = p1,1x1 + p2,1x2 + p3,1x3

x2 = p1,2x1 + p2,2x2 + p3,2x3

x1 + x2 + x3 = 1

There is also an equation which could be written for state 3; this equationcould replace either of the state 1 or state 2 equations. The one equation whichcannot be replaced is the one for the sum being 1.

Now let’s determine the steady-state probabilities for the brand-switching ex-ample, for which the transition probability matrix is:

P =

0.47 0.36 0.170.09 0.52 0.390.18 0.21 0.61

Substituting the relevant pi j’s, the system of equations in numerical form is:

x1 = 0.47x1 +0.09x2 +0.18x3

x2 = 0.36x1 +0.52x2 +0.21x3

x1 + x2 + x3 = 1

(Note: the numbers in the first two rows come from the first two columns of P.)This system can easily be reduced to two equations in two unknowns by rewrit-

ing the third equation as x3 = 1− x1− x2, and then substituting this into the firsttwo equations. This gives:

x1 = 0.47x1 +0.09x2 +0.18(1− x1− x2)x2 = 0.36x1 +0.52x2 +0.21(1− x1− x2)


Simplifying the right-hand side we obtain:

x1 = 0.29x1−0.09x2 +0.18x2 = 0.15x1 +0.31x2 +0.21

Putting all variables on the left we obtain:

0.71x1 +0.09x2 = 0.18−0.15x1 +0.69x2 = 0.21

Multiplying the first row by 15, and the second row by 71, gives:

10.65x1 +1.35x2 = 2.70−10.65x1 +48.99x2 = 14.91

Adding we obtain:

0x1 +50.34x2 = 17.61

and therefore x2 = 0.34982... We then substitute this result into one of the twoequations in two unknowns to find x1:

0.71x1 +0.09(0.34982...) = 0.18

Solving, we obtain x1 = 0.20917.... Now we find x3:

0.20917...+0.34982...+ x3 = 1

Solving, we obtain x3 = 0.44100...

Hence the steady-state probabilities are: x1 = 0.20917..., x2 = 0.34982..., andx3 = 0.44100... In the context of the brand-switching example, this means that themarket shares of the three brands are about 20.9% for Brand 1, 35.0% for Brand

2, and 44.1% for Brand 3.The method shown here is useful for models with three states. However, when

there are just two states, there is a special formula which should be used instead.Conversely, when there are more than three states, you will probably want to use acomputer to do part of the calculations. We now show how to use these methods.


15.3.2 Two State Markov Chain

For a Markov chain with only two states, there is a formula for determining eachstate’s steady-state probability. It’s easy to derive these formulas. An arbitrarymatrix P is of the form

P =[

p1,1 p1,2p2,1 p2,2

]where each number is at least 0 and each row must sum to 1. At steady-state, wemust have:

x1 = p1,1x1 + p2,1x2.

Also, we must have:x1 + x2 = 1

Therefore, x2 = 1− x1, and substituting this into the other equation we obtain:

x1 = p1,1x1 + p2,1(1− x1)

and therefore(1− p1,1 + p2,1)x1 = p2,1.

Since each row sums to 1, 1− p1,1 = p1,2. Therefore

(p1,2 + p2,1)x1 = p2,1.

Therefore:x1 =

p2,1

p1,2 + p2,1(3)

and since x1 + x2 = 1 we obtain

x2 =p1,2

p1,2 + p2,1(4)

15.3.3 Example

P =[

0.66 0.340.41 0.59

]Using equations 3 and 4 we obtain

x1 =0.41

0.41+0.34≈ 0.5467

andx2 =

0.340.41+0.34

≈ 0.4533


15.3.4 Solution by Computer

Although we can automate things right from the outset, it is easier to understandwhat is happening if we first do things by hand, and then switch to using thecomputer to finish the problem. For example, suppose that we wish to determinethe steady-state probabilities of the following system:

P =

0.31 0.25 0.38 0.060.08 0.44 0.32 0.160.24 0.17 0.57 0.020.04 0.09 0.36 0.51

We make four equations in four unknowns. Three of them come in the samemanner as before, in which each steady-state probability in written in terms of thepi j’s and the other steady-state probabilities (by going down the columns of P)(we call these the x = xP equations). The fourth equation comes from the fact thatall the steady-state probabilities must sum to 1. Doing this we obtain:

x1 = 0.31x1 +0.08x2 +0.24x3 +0.04x4

x2 = 0.25x1 +0.44x2 +0.17x3 +0.09x4

x3 = 0.38x1 +0.32x2 +0.57x3 +0.36x4

x1 + x2 + x3 + x4 = 1

Putting all the variables on the left gives:

0.69x1−0.08x2−0.24x3−0.04x4 = 0−0.25x1 +0.56x2−0.17x3−0.09x4 = 0−0.38x1−0.32x2 +0.43x3−0.36x4 = 0

x1 + x2 + x3 + x4 = 1

This is a set of equations of the form

Ax = b

where

A =

0.69 −0.08 −0.24 −0.04−0.25 0.56 −0.17 −0.09−0.38 −0.32 0.43 −0.36

1.00 1.00 1.00 1.00


x =

x1x2x3x4

and

b =

0001

We wish to determine x = A−1b. On a spreadsheet we can input the matrix A,invert it (using the MINVERSE function), and then multiply (using the MMULTfunction) this result by b to obtain the column vector x. Indeed, for any matrix P,the column vector b will always be a string of 0’s ending with a “1”. Therefore,the product of A−1 and b is simply the same as the final column of A−1. Doingthis we obtain x1 = 0.190812702, x2 = 0.240611623, x3 = 0.448344037, and x4 =0.120231638.

The process can be expedited by first finding the transpose of P. Secondly, thetransposed matrix is subtracted from an identity matrix of the same size. Thirdly,the final row of this subtracted matrix is overwritten with 1’s. Fourthly, we findthe inverse of the overwritten matrix, and the final column of the inverse givesthe steady-state probabilities. You might wish to try this method on the secondproblem of this lecture’s exercise.


15.4 Exercise15.4.1 Problem 1

Solve this problem by hand (i.e. use a calculator rather than a spreadsheet).

Three companies are fighting for market share. For each company, the propor-tion of the customers who switch to another company (or who stay with the samecompany) each month is given by the following matrix.

P =

.2 .7 .1.5 .2 .3.6 .1 .3

(a) If a consumer is currently purchasing from company 2, what is the proba-

bility that he or she will purchase from company 1 in three months’ time?

(b) Analytically determine the long-term percentage market share for eachcompany.

15.4.2 Problem 2

A Markov chain has the following state transition probability matrix:

P =

0.0 1.0 0.0 0.00.0 0.3 0.7 0.00.0 0.0 0.2 0.80.2 0.2 0.2 0.4

(a) Draw the state transition diagram (omit the impossible transitions).

Use a spreadsheet to solve parts (b) and (c) of this problem.

(b) If the system is currently in state 2, what is the probability that it will be instate 4 after five transitions?

(c) Determine by matrix inversion the percentage of time that will be spent in eachstate after a large number of transitions.


16 Markov Chains 2

16.1 IntroductionIn this lecture we examine what happens when a Markov chain has some stateswhich, once entered, cannot be left. Such states are called absorbing or trappingstates. In a state transition diagram, an absorbing state has a loop to itself withprobability 1, with no other arrow leaving for any other state. In a state transitionmatrix, an absorbing state has a ‘1’ on the main diagonal, and a ‘0’ everywhereelse in that row. Another way of saying this is that state i is an absorbing state ifand only if pii = 1. In a Markov chain in which some of the states are absorbing,the non-absorbing states are often called transient states.

In matrix P1, state 1 is absorbing and state 2 is transient. Clearly, such asystem will eventually end up in state 1.

P1 =[

1 00.2 0.8

]We note that a matrix such as the one shown in P2 (which has no 1’s on the

main diagonal) contains no absorbing states:

P2 =

0 0 11 0 00.4 0.5 0.1

In matrix P3, states 1 and 2 are absorbing and state 3 is transient.

P3 =

1 0 00 1 00.4 0.5 0.1

To obtain the probability of eventual absorption into state 1 (x1), all we need

do is solve a single equation. If the system starts in state 3, there is a 0.4 chance ofbeing absorbed into state 1 after one transition. There is a 0.1 chance of returningto state 3, at which point there is probability x1 of eventually ending up in state 1.Therefore:

x1 = 0.4+0.1x1

0.9x1 = 0.4

x1 =49≈ 0.4444


The probability of eventual absorption into state 2 (x2) is

x2 = 1− x1

= 1− 49

=59≈ 0.5556

This approach is simple because of the small size of the system, but what we needis a general procedure for an arbitrary number of absorbing states, and an arbitrarynumber of transient states. This is presented in the next section. After that, wepresent an application for which this type of analysis is needed.

16.2 Absorbing States AnalysisIn dealing with Markov chains which contain absorbing states, it makes thingseasier if the absorbing states appear together first, and the transient states appeartogether afterwards. We let m represent the number of absorbing states, and n thenumber of transient states. Hence the Markov chain has m + n states. Here is anexample with three absorbing states and two transient states, the absorbing onesappearing first.

P =

1 0 0 0 00 1 0 0 00 0 1 0 00.12 0.34 0.38 0.09 0.070.25 0.41 0.16 0.05 0.13

This matrix can be partitioned into four sub-matrices, splitting the absorbing

states from the transient states both horizontally and vertically. Showing this withfour colours we have:

P =

1 0 0 0 00 1 0 0 00 0 1 0 00.12 0.34 0.38 0.09 0.070.25 0.41 0.16 0.05 0.13

We see that the 5 by 5 matrix was partitioned as follows: the area in red is a 3 by3 identity matrix; the area in grey is a 3 by 2 rectangle in which every number is


0; the area in blue is a 2 by 3 rectangle which gives the probabilities of switchingfrom the transient states to the absorbing states; and the area in yellow is a 2 by 2square which gives the probabilities of switching between the transient states.

In general P can be partitioned into: an identity matrix of order m (upper-left);a rectangle with m rows and n columns consisting only of 0’s (upper-right); arectangle with n rows and m columns (lower-left); and a square matrix of order n(lower-right). Each of these is a sub-matrix denoted as: Im; 0m×n; Rn×m; and Qnrespectively.

When there are more absorbing states than transient states (m > n), the parti-tion looks like:

P =

Im 0m×n

Rn×m Qn

When there are fewer absorbing states than transient states (m < n), the parti-

tion looks like:

P =

Im 0m×n

Rn×m Qn


When the number of absorbing states equals the number of transient states(m = n), the partition results in four square sub-matrices of the same size:

P =

Im 0m×n

Rn×m Qn

It turns out that steady-state behaviour can be determined just by using sub-

matrix Rn×m and sub-matrix Qn. What we are seeking is a matrix which hasthe same size and shape as Rn×m. This matrix gives the probability of eventualabsorption into absorbing state j given that the system begins in transient statei, where j = 1, . . . ,m and i = m + 1, . . . ,m + n. The formula to find this n by mmatrix, here called Fn×m, is

Fn×m =(

In − Qn

)−1× Rn×m (5)

The use of colour in Equation 5 is for emphasis. The identity matrix in this equa-tion is In , a matrix which has the same size as Qn . It is not Im , the sub-matrixof P which is kitty-corner to Qn.

Returning to the numerical example, the sub-matrices of P which we need are:

Q2 =[

0.09 0.070.05 0.13

]and

R2×3 =[

0.12 0.34 0.380.25 0.41 0.16

]To find F, we first perform the subtraction within the parentheses:

I2−Q2 =[

1 00 1

]−[

0.09 0.070.05 0.13

]=

[0.91 −0.07−0.05 0.87

]


Since this is a 2 by 2 matrix, it can be inverted by hand, as we shall now do.For our purposes, any matrix which is 3 by 3 or larger will be inverted using aspreadsheet.

Recall how to find the inverse of a 2 by 2 matrix A (this is seen in Chapter 3of Mathematics for Management Science):

A =[

a1,1 a1,2a2,1 a2,2

]The determinant of this matrix is

detA = a1,1a2,2−a2,1a1,2

If det A = 0, then the inverse of A, written as A−1, does not exist. If det A 6= 0,then

A−1 =(

1detA

)[a2,2 −a1,2−a2,1 a1,1

]The matrix that we wish to invert is:

I2−Q2 =[

0.91 −0.07−0.05 0.87

]We begin by calculating the determinant.

det(I2−Q2) = 0.91(0.87)− (−0.05)(−0.07)= 0.7917−0.0035= 0.7882

Since the determinant is not 0, the inverse is defined, and is given by:

(I2−Q2)−1 =

(1

0.7882

)[ 0.87 0.070.05 0.91

]≈

[1.10378 0.088810.06344 1.15453

]This matrix has a special meaning. It gives the expected number of transitionsspent in each transient state before absorption. If the system starts in the first tran-sient state (state 4 in P), then the system will spend about 1.104 transitions in state4 and 0.089 transitions in state 5 for a total of 1.193 transitions before being ab-sorbed somewhere. If the system starts in the second transient state (state 5 in P),


then the system will spend about 0.063 transitions in state 4 and 1.155 transitionsin state 5 for a total of 1.218 transitions before being absorbed somewhere.

Finally, we multiply this matrix and R:

Fn×m =(

In − Qn

)−1× Rn×m

=[

1.10378 0.088810.06344 1.15453

]×[

0.12 0.34 0.380.25 0.41 0.16

]≈

[0.1547 0.4117 0.43360.2962 0.4949 0.2088

]All the numbers in Fn×m are probabilities, hence each number must be between0 and 1 inclusive, and each row must (theoretically) sum to 1. By inspection wesee that the first attribute is true; by summing each row we obtain 1.000 for thefirst (state 4) and 0.9999 for the second (state 5). The latter sum is fine; there issimply some rounding error when the probabilities were reported to the nearestfourth digit. The number in row i (state m+ i) and column j gives the probabilityof the system eventually being absorbed into state j. For example, starting in state5, there is about a 49.5% chance of being absorbed into state 2.

Alternate Approach If the inverted matrix is not needed in its own right, butis simply an intermediate calculation in determining Fn×m, then an alternate ap-proach exists which is numerically easier to do by hand. Using this approach,we postpone the division by the determinant until the end. Doing it this way weobtain:

Fn×m =(

In − Qn

)−1× Rn×m

=(

10.7882

)[ 0.87 0.070.05 0.91

]×[

0.12 0.34 0.380.25 0.41 0.16

]=

(1

0.7882

)[ 0.1219 0.3245 0.34180.2335 0.3901 0.1646

]≈

[0.1547 0.4117 0.43360.2962 0.4949 0.2088

]

16.3 Solution by ComputerWhenever the matrix inversion is 3 by 3 or larger, we will use a computer to dothe calculations. An example to illustrate this is given by the following six stateMarkov chain.


P =

1 0 0 0 0 00 1 0 0 0 00.13 0.24 0.32 0.28 0.02 0.010.07 0.52 0.21 0.08 0.03 0.090.06 0.32 0.09 0.15 0.26 0.120.04 0.07 0.18 0.64 0.04 0.03

We can put this onto a spreadsheet, using four colours to highlight the parti-

tioned matrix P.

A B C D E F G H12 1 0 0 0 0 03 0 1 0 0 0 04 Matrix 0.13 0.24 0.32 0.28 0.02 0.015 P 0.07 0.52 0.21 0.08 0.03 0.096 0.06 0.32 0.09 0.15 0.26 0.127 0.04 0.07 0.18 0.64 0.04 0.03

Since Q (yellow) is 4 by 4, we need a 4 by 4 identity matrix. This is:

A B C D E F G H89 1 0 0 0

10 Matrix 0 1 0 011 I4 0 0 1 012 0 0 0 1

We now find I4− Q. To put this into the range C14:F17, we input =C9-E4into cell C14, and then copy this formula into the range C14:F17. Doing this weobtain:

A B C D E F G H1314 0.68 −0.28 −0.02 −0.0115 Matrix −0.21 0.92 −0.03 −0.0916 I4− Q −0.09 −0.15 0.74 −0.1217 −0.18 −0.64 −0.04 0.97


Now we find (I4− Q)−1. To write the label, we can use Equation Editor inWord and then paste this into Excel. Alternatively, we can write the label sim-ply as (I4-Q)ˆ{-1} or something similar. To put the inverse into the rangeC19:F22, we first drag the mouse from cell C19 to cell F22 which blackens therange. Secondly we enter =MINVERSE(C14:F17) into cell C19. Thirdly wepress Ctrl (keeping it held down) Shift (keeping it held down) and Enter, whichproduces:

A B C D E F1819 1.680313349 0.579065465 0.073219698 0.08010864920 Matrix 0.458271813 1.334249642 0.073918296 0.13766534221 (I4− Q)−1 0.399523068 0.504437482 1.394678348 0.22345980022 0.630650810 1.008586654 0.119870504 1.145838936

Finally, we are ready to calculate:

Fn×m =(

In − Qn

)−1× Rn×m

F is a 4 by 2 matrix (it’s always the same size as R, the blue sub-matrix of P).

Hence we reserve a 4 by 2 range for this matrix, say C24:D27.(

In − Qn

)−1is

in the range C19:F22, and Rn×m is in the range C4:D7, so in cell C24 we enterthe formula =MMULT(C19:F22,C4:D7). We then press Ctrl (keeping it helddown) Shift (keeping it held down) and Enter, which produces:

A B C D E F G H2324 0.266572846 0.73342715425 Matrix 0.162914522 0.83708547826 F 0.179867716 0.82013228427 0.205611459 0.794388541

The spreadsheet output produces far more places after the decimal point thanwe need (and far more than we can justify, given that the initial data is not precise).To present the data to a manager, we might report it in a table, with the spreadsheetoutput being an appendix to the report. In a generic form the table could resemble:


Probability ofAbsorption into

Beginning in State 1 State 2State 3 0.267 0.733State 4 0.163 0.837State 5 0.180 0.820State 6 0.206 0.794

However, in reporting the analysis for a specific example, the labels for thestates should be replaced by word descriptions in order to make the report easy tounderstand.

16.4 An Application: Credit Card AnalysisOften once an absorbing state is entered, there are no further transitions, howeverit is still appropriate to model the situation as if there were perpetual transitionsall returning to the same state. Here follows an example of this.

16.4.1 Situation Description

A credit card company classifies each account into one of four categories:

1. The account is fully paid.

2. The account last had credit granted within the past one to thirty days.

3. The account last had credit granted within the past thirty-one to sixty dayswith either no payment or insufficient payment to pay off the account.

4. The account has been written off as bad debt.

Of the accounts currently in the 1 – 30 day category, 20% are paid in full, 70%have debits and credits which cause the account to be in the same category nextmonth, and on the other accounts nothing is paid. Of the accounts currently in the31 – 60 day category, 5% are paid in full, enough is paid on 30% of the accountsto upgrade the status to 1 – 30 day old debt, 58% of the accounts remain in thesame category next month, and on the rest nothing is paid. The company expectsthese patterns to continue into the future.

The company wishes to know how long on average their money is lent outbefore it is either repaid or is written off. The company currently has $108,000,000


lent out, of which $96,000,000 is in the 1 – 30 day category. The company wishesto determine how much money they can expect to recover.

16.4.2 Formulation

Each of the four categories is a state of a Markov chain. Once a dollar lent outhas been returned, or has become bad debt, it doesn’t keep making transitions, butfor the sake of doing the Markovian analysis, we treat the fully paid and bad debtsituations as absorbing states. The other two states, with debt in either the 1 – 30or 31 – 60 day categories, are transient.

From what we saw in the previous section, the analysis is made easier if weput the absorbing states together, and then put the transient states together. Thefour states are therefore:

1. Fully Paid (absorbing)

2. Bad Debt (absorbing)

3. 1 – 30 Days (transient)

4. 31 – 60 Days (transient)

We are now almost ready to make the state transition diagram. The per-centages become the probabilities. Where it states that “on the other accountsnothing is paid”, the probability is found by subtracting everything else from 1to obtain 1− 0.2− 0.7 = 0.1. In this case, the system moves from the 1 – 30Days state to the 31 – 60 Days state. Where it states that “on the rest nothingis paid”, the probability is found by subtracting everything else from 1 to obtain1− 0.05− 0.30− 0.58 = 0.07. In this case, the system moves from the 31 – 60Days state to the Bad Debt state.

In making the state transition diagram, the words which describe the state aregiven prominence over the state number. For emphasis, the absorbing states are inred, and the transient states are in yellow. The state transition diagram is shownin Figure 54.

16.4.3 Analysis

While the diagram makes things easy to understand, we need the state transitionprobability matrix from which the calculations will be made. This matrix is:


Fully Paid(1)

Bad Debt(2)

1 – 30Days(3)

31 – 60Days(4)

1

0.2

0.1

0.05 0.07

0.3

1

0.7 0.58

Figure 54: Credit Card State Transition Diagram


P =

1 0 0 00 1 0 00.2 0 0.7 0.10.05 0.07 0.3 0.58

We now wish to determine:

Fn×m =(

In − Qn

)−1× Rn×m

We begin with the subtraction:

I2−Q2 =[

1 00 1

]−[

0.7 0.10.3 0.58

]=

[0.3 −0.10−0.3 0.42

]Now we find the inverse:

det(I2−Q2) = 0.3(0.42)− (−0.3)(−0.1)= 0.126−0.03= 0.096

Since the determinant is not 0, the inverse is defined, and is given by:

(I2−Q2)−1 =

(1

0.096

)[ 0.42 0.10.3 0.3

]≈

[4.375 1.041673.125 3.125

]If an account is currently in the 1 – 30 Day state, the money is expected to

spend 4.375 months in this state, and 1.04167 months in the 31 – 60 Day state,for a total of 5.41667 months before being absorbed into either the Fully Paid orBad Debt states. If an account is currently in the 31 – 60 Day state, the money isexpected to spend 3.125 months in each transient state for a total of 6.25 monthsbefore being absorbed into either the Fully Paid or Bad Debt states.

Finally, we multiply this matrix and R:

Fn×m =(

In − Qn

)−1× Rn×m

=[

4.375 1.041673.125 3.125

]×[

0.2 00.05 0.07

]≈

[0.92708 0.072920.78125 0.21875

]


The interpretation of these numbers is that for each dollar in the 1 – 30 Day state,we expect that 92.708 cents will be recovered, and 7.292 cents will end up as baddebt. For each dollar in the 31 – 60 Day state, we expect that only 78.125 centswill be recovered, and 21.875 cents will end up as bad debt. As one would expect,a dollar in the 31 – 60 Day state has less chance of being recovered than a dollarin the 1 – 30 Day state.

Hence the data could be reported as:

Percentage which isCurrent Status Recovered Not Recovered

1 – 30 Days 92.7 7.331 – 60 Days 78.1 21.9

With $108,000,000 currently on loan, of which $96,000,000 is in the 1 – 30Day state, there is $12,000,000 in the 31 – 60 Day state. They should expect torecover:

0.92708($96,000,000)+0.78125($12,000,000)≈ $98,375,000

Based on the $108,000,000 in credit outstanding, this is a recovery rate of about91.1% overall.


A chocolatier makes white, light, medium, and dark chocolate. Once a customerbuys either white or dark chocolate, he or she continues to buy it forever. Of thosewho are currently eating light chocolate, 20% will switch to white, 50% will staywith light, and 30% will switch to medium next time. Of those who are currentlyeating medium chocolate, 15% will switch to light, 60% will stay with medium,and 25% will switch to dark next time.

(a) Define the states and draw the state transition diagram.

(b) Write the state transition matrix, and doing the matrix inversion by hand,determine the expected number of transitions before absorption depending onwhether the consumer is currently buying light or medium chocolate.

(c) What are the probabilities of absorption into white or dark chocolate con-ditional on whether the consumer is currently buying light or medium chocolate?


(d) At the outset, light chocolate has 30% of the market, and medium choco-late has the rest. Eventually, what will the market shares be for white and darkchocolate? (Hint: Draw a probability tree.)

16.5.2 Problem 2

A credit-card company categorizes its open accounts as being “current”, one-month overdue, two-months overdue, and three-months overdue. They find that70% of their current accounts are paid in full. However, only 40% of the one-month old accounts are paid in full; 20% of the two-month old accounts are paidin full, and only 10% of the three-month old accounts are paid in full. Of the ac-counts which are not paid in full, half stay in the same category, and the other halfare not paid at all. For example, for the one-month accounts, 0.5(1−40%) = 30%of them go into the two-month category in the following month. Accounts whichhave not been paid for four months are written off as bad debt, and the account isclosed.

(a) Define the states and draw the state transition diagram.

(b) Using a spreadsheet to perform the necessary matrix inversion, determinethe percentage of the debt which is eventually repaid as a function of the age ofthe account.

(c) The company currently has $66 million lent out in current accounts, $35million in one-month accounts, $7 million in two-month accounts, and $2 mil-lion in three-month accounts. What is the expected amount of money that willeventually be paid in full?


17 Markov Chains 3

We have seen all the technical operations about Markov chains suitable for a intro-ductory course in management science. However, we need to spend some moretime on model formulation. In this lecture we present a few more formulationexamples in increasing order of complexity.

17.1 Gambler’s Ruin

A gambler walks into a casino with one of the following amounts of cash: $20;$40; $60; or $80. Every time he plays a game, he bets $20. If he wins, he receives$40 for a net gain of $20. If he loses, he receives nothing for a net loss of $20.Every time he plays the probability of winning is p, and the probability of losingis 1− p. He will leave the casino if he runs out of cash, and also leave the casinoif his cash reaches the $100 level.

This is a Markov chain in which the states represent his level of cash. Havingno cash or $100 in cash are absorbing states (numbered 1 and 2), and having$20, $40, $60, or $80 are the transient states (numbered 3, 4, 5, and 6). Notethat winning and losing are not states – winning and losing are transitions, andare represented by the arrows between the states. The state transition diagram isshown in Figure 55.

The state transition matrix is:

P =

1 0 0 0 0 00 1 0 0 0 01− p 0 0 p 0 00 0 1− p 0 p 00 0 0 1− p 0 p0 p 0 0 1− p 0

In putting this onto a spreadsheet, one cell is reserved for the value of p. For

example, suppose that this cell is A6 (with a label p in cell A5), and P is placedin the range C2:H7. All cells are 0 except: C2 and D3, which are 1; F4, G5, H6,and D7 in which we place the formula =A6; and C4, E5, F6, and G7, in which weplace the formula =1-A6. These formulas can be seen by pressing Cntl (keepingit held down) and the ˜ (tilde) key.


Leaveswith $0

(1)

Leaveswith $100

(2)

$20(3)

$40(4)

$60(5)

$80(6)

1 1

1− p

1− p p

p

1− p

p

p

1− p

Figure 55: Gambler’s Ruin State Transition Diagram


A B C D E F G H12 Gambler’s 1 0 0 0 0 03 Ruin 0 1 0 0 0 04 =1-A6 0 0 =A6 0 05 p 0 0 =1-A6 0 =A6 06 0 0 0 =1-A6 0 =A67 0 =A6 0 0 =1-A6 0

If we put a number into cell A6 and then press Cntl (keeping it held down) andthe ˜ key, the numbers in the matrix will appear. For example, with 0.47 in cellA6, we obtain:

A B C D E F G H12 Gambler’s 1 0 0 0 0 03 Ruin 0 1 0 0 0 04 0.53 0 0 0.47 0 05 p 0 0 0.53 0 0.47 06 0.47 0 0 0 0.53 0 0.477 0 0.47 0 0 0.53 0

By the method seen in the last lecture, we can create I4, then find I4− Q,then (I4−Q)−1, and finally F. We will present only the final solution here. Forexample, when p = 0.47 we obtain:

Based on Probability ofp = 0.47 Leaving with

Beginning $0 $100with $20 0.845 0.155with $40 0.670 0.330with $60 0.473 0.527with $80 0.251 0.749

For example, for someone entering the casino with $40 the probability of even-tually winning a net of $60 to obtain a total of $100 is given by row 2 (state 4) andcolumn 2 (state 2) of the table (the matrix F); this number is 0.33.


The expected net loss from the gambler’s perspective is the amount the gam-bler has initially minus the expected winnings. For example, if the gambler enterswith $40 the expected loss based on p = 0.47 is:

$40− (0.67($0)+0.33($100)) = $7.00

(Without rounding the probabilities from the F matrix the expected loss works outto $7.01.) Doing this calculation on the spreadsheet for all levels of initial cashwe obtain: $4.50 for $20; $7.01 for $40; $7.30 for $60; and $5.30 for $80.

The preceding has been based on the arbitrarily chosen value of p = 0.47. Thevalue of p will be set by the casino. In order for them to make money to pay theirexpenses and taxes, they will set p to be less than 0.5. They don’t have to makeit much less for them to collect the lion’s share of the winnings. Here is whathappens as a function of p if someone begins with $60.

Beginning with $60Probability

p of leaving Expectedwith $100 Loss

0.50 0.600 none0.45 0.478 $12.20 (20.3%)0.40 0.360 $23.98 (40.0%)0.35 0.256 $34.37 (57.3%)0.30 0.172 $42.83 (71.4%)

Things are even more bleak for someone who starts out with only $40.

Beginning with $40Probability

p of leaving Expectedwith $100 Loss

0.50 0.400 none0.45 0.286 $11.41 (28.5%)0.40 0.190 $21.04 (52.6%)0.35 0.116 $28.39 (71.0%)0.30 0.065 $33.48 (83.7%)

Here we see for example that a 30% chance of winning on any one play (whichdoesn’t sound too bad) translates into only a 6.5% chance of winning overall. Thisis why the game is called gambler’s ruin. The Markov chain “drifts” towards theabsorbing state which represents leaving with no money left.


17.2 Educational ModelThe first part of the Exercise for this lecture is based on the content of this section.

Suppose that a college offers a course over two years, with an exam at the endof each year written on a pass/fail basis. If a student fails either year, he or she isrequired to withdraw (RTW). If the student passes Year 1, he or she then entersYear 2. If the student also passes the exam at the end of Year 2, then the studentgraduates.

We could model this situation as a Markov chain, in which there are two ab-sorbing states and two transient states.

State 1 - Grad. (the student has graduated from the program)State 2 - RTW (the student has been required to withdraw)State 3 - Year 1 (the student is currently enrolled in Year 1)State 4 - Year 2 (the student is currently enrolled in Year 2)The pass rate is 80% for Year 1, and 75% for Year 2. Clearly, the overall

success rate is 80%(75%) = 60%. Eventually, we expect that 60% of the studentswill end up in the Grad. state, and 40% will end up in the RTW state. Notethat passing or failing each year’s program are not states; they are the transitionsbetween the states.

17.2.1 Extension 1

Now we consider a more realistic (and more complicated) model. The pass ratesare the same as before, but students are now allowed to fail once. If the studentfails the same year twice, or fails Year 2 having previously failed Year 1, then heor she is required to withdraw.

The new Markov chain is not simply the old one with a loop from State 3 (Year1) to itself. This would only be correct if the year could be repeated any numberof times. Instead, we need two “Year 1”-type states: one for first-timers; and onefor repeaters. The same applies for Year 2. The drawing of the state transitiondiagram is left as part of the Exercise.

17.2.2 Extension 2

Things are as they were in Extension 1, but now the pass rates depend on thehistory. The Year 1 pass rate is still 80% for first-timers, but is 90% for repeaters.The Year 2 pass rate is still 75% for those who have never failed before, but is60% for those who repeated Year 1 and are now doing Year 2 for the first time,and is 85% for those who are doing Year 2 for the second time.


The assumption of a Markov chain is that history is irrelevant, so when it isrelevant, we have to try altering the chain. We can create another state to handlethe 60% vs. 85% pass rates. This leaves us with three “Year 2”-type states. Again,the drawing of the diagram is left as part of the Exercise.

17.3 Repair Problem A


A machine contains three identical components. Under normal operation, twoof the components are working, and the third is in working order but idle. Atthe end of every hour there is probability p that each operating component willcease to operate. This probability does not depend on how long the componenthas been working. When one component fails, the idle component then startsoperating, and a repairer begins to fix the non-functioning component. Thoughthe machine is under repair, the machine is still considered to be “up”, becausehaving two working components is sufficient. The repairs always take an hour, andare always successful. In the rare situation where two operating components failsimultaneously, the machine is then “down”. The repairer takes an hour to repairone of the two broken components, and then repairs the other one. (There’s noroom for two repairers to work simultaneously.) We wish to model this situationas a Markov chain.

17.3.2 Formulation

In this situation if we try to make the model at the component level it will havenine states, because there are three components, and each can be working, idle,or under repair. This would lead to a model which would be overly complex.Instead, because each component is identical, we will make the model at the levelof the machine. It can be seen that the machine must be at all times in one ofthree situations. To see this we begin with the state of normal operation (State 1),in which two components are working and the third is idle. If neither workingcomponent fails, then the system returns to this state. If one fails but the otherdoes not, then the machine is “up” but under repair (State 2), i.e. two componentsworking and the third is under repair. If the system is in state 1 and both workingcomponents fail, then the machine is “down” and under repair (State 3), with onebeing repaired, one waiting for repair, and one component idle. If the systemis in state 2, and if neither working component fails, then because the repair is


completed at the next transition the system will move into State 1. If the system isin state 2 and just as one component is repaired either (but not both) of the workingcomponents fails, then the system stays in state 2. If the system is in state 2 andboth components fail, the component being repaired becomes an idle component,one of the other two components is being repaired, and the third one is waiting forrepair, i.e. the system has moved to state 3. Finally, if the system is in state 3, itmust move to state 2 when the state under repair becomes operational. We havetracked down all the possibilities, so we are left with a three-state Markov chain.

With three coloured boxes for the components, using green to mean working,blue to mean idle, yellow to mean under repair, and red to mean waiting for repair,the states are (where the boxes can be in any order):

State 1 – Operating Operating Idle

State 2 – Operating Operating Being Repaired

State 3 – Being Repaired Waiting for Repair Idle

In this situation, the breakdowns of the components, and moment that thecomponent is switched on after being repaired, are transitions between the states.The probabilities of transition are all either 0 (if impossible), 1 (if required), or afunction of p. In states 1 and 2, the joint probabilities are given by the binomialformula with n = 2: p2 for two simultaneous failures; (1− p)2 for two simultane-ous non-failures; and 2p(1− p) for one failing but the other not failing. The statetransition diagram is shown in Figure 56.

The equivalent state transition matrix is:

P =

(1− p)2 2p(1− p) p2

(1− p)2 2p(1− p) p2

0 1 0

With a given numerical value for p, we can create a numerical diagram or

matrix. For example, when p = 0.03, the state transition probability matrix is:

P =

0.9409 0.0582 0.00090.9409 0.0582 0.00090 1 0

By using the method seen two lectures ago, we could find the steady-state

probabilities x1, x2, and x3.


NormalOperation

(1)

Up& Repair

(2)

Down& Repair

(3)

(1− p)2

2p(1− p)p 2

(1− p)2

2p(1− p)

p21

Figure 56: Repair Problem A


17.4 Repair Problem BNow we consider a more complicated repair problem. Things are as they werein problem A, but now the probability of successful repair in an hour is q (ratherthan 1). This probability does not depend on how many hours the repairer hasbeen working on the failed component.

We clearly need the three states that we defined earlier. What we must do issee what other states might be needed now. In state 1, nothing is being repaired,so the probabilities exiting from state 1 do not depend on q. In state 2, if therepairer fixes the component this hour, then the system moves to state 1 if neitherworking component fails, stays in state 2 if exactly one working component fails,and moves to state 3 if both working components fail. In state 2, if the repairerfails to fix the component this hour, then in the next hour the system will be instate 2 if neither operating component fails, and will be in state 3 if exactly oneoperating component fails. If both operating components fail, then there will bethree bad components. For this, we need a new state. State 4 has one componentbeing repaired, and two components waiting for repair. Fixing a component instate 3 (probability q) moves the system to state 2 (as before), but failing to fixit (probability 1−q) keeps the system in state 3. Fixing a component in the newstate 4 (probability q) moves the system to state 3, but failing to fix it (probability1−q) keeps the system in state 4. Hence we only need one more state, and usingthe previously defined colours we have:

State 1 – Operating Operating IdleState 2 – Operating Operating Being RepairedState 3 – Being Repaired Waiting for Repair IdleState 4 – Being Repaired Waiting for Repair Waiting for Repair

We need to determine the expressions (which depend on p and q) for the statetransition probabilities, which are shown on the state transition diagram in Fig-ure 57. For exiting from states 1, 3, and 4, these are easy to find. From state 1, wehave the same thing as in Repair Problem A: p1,1 = (1− p)2; p1,2 = 2p(1− p);and p1,3 = p2 (the system cannot go from 1 to 4). From state 3, we have: p3,2 = q;and p3,3 = (1−q). From state 4, we have: p4,3 = q; and p4,4 = (1−q).

The probabilities coming out of State 2 are much more difficult to determine.To go from 2 to 1, we want the joint probability of two working components notfailing and the other component being successfully repaired. This is:

p2,1 = (1− p)× (1− p)×q


NormalOperation

(1)

Up& Repair

(2)

Down& Repair

(3)

All Down& Repair

(4)

(1− p)2

2p(1− p)

(1− p)2q

2p(1−

p)(1−q)

+p2 q

p2q

p2(1−q)

q

2p(1− p)q+(1− p)2(1−q)

1−q 1−q

Figure 57: Repair Problem B


= (1− p)2q

There are three ways of going from State 2 to itself. There are two ways in whichone working component fails and the other does not combined with the third com-ponent being successfully repaired, and the other way is if neither working com-ponent fails combined with the third component not being successfully repaired.This is:

p2,2 = p(1− p)q+(1− p)pq+(1− p)(1− p)(1−q)

= 2p(1− p)q+(1− p)2(1−q)

There are three ways of going from State 2 to State 3. There are two ways inwhich one working component fails and the other does not combined with thethird component not being successfully repaired, and the other way is if bothworking components fail combined with the third component being successfullyrepaired. This is:

p2,3 = p(1− p)(1−q)+(1− p)p(1−q)+ ppq

= 2p(1− p)(1−q)+ p2q

Finally, there is one way to go from State 2 to State 4. This happens when bothworking components fail and the third component is not successfully repaired.This is:

p2,4 = p2(1−q)

The state transition matrix as a function of p and q is:

P =

(1− p)2 2p(1− p) p2 0(1− p)2q 2p(1− p)q+(1− p)2(1−q) 2p(1− p)(1−q)+ p2q p2(1−q)0 q 1−q 00 0 q 1−q

For example, if p = 0.03 and q = 0.8, then:

P =

0.9409 0.0582 0.0009 00.75272 0.23474 0.01236 0.000180 0.8 0.2 00 0 0.8 0.2


17.5 Exercise17.5.1 Educational Model

Draw the state transition diagrams for Extensions 1 and 2 as described for thefollowing problem.

Suppose that a college offers a course over two years, with an exam at the endof each year written on a pass/fail basis. If a student fails either year, he or she isrequired to withdraw (RTW). If the student passes Year 1, he or she then entersYear 2. If the student also passes the exam at the end of Year 2, then the studentgraduates. The pass rate is 80% for Year 1, and 75% for Year 2.

Extension 1 The pass rates are the same as before, but students are now allowedto fail once. If the student fails the same year twice, or fails Year 2 having previ-ously failed Year 1, then he or she is required to withdraw.

Extension 2 Things are as they were in Extension 1, but now the pass ratesdepend on the history. The Year 1 pass rate is still 80% for first-timers, but is 90%for repeaters. The Year 2 pass rate is still 75% for those who have never failedbefore, but is 60% for those who repeated Year 1 and are now doing Year 2 for thefirst time, and is 85% for those who are doing Year 2 for the second time.

17.5.2 Repair Problem

There are two transmission lines from a generating station to a nearby city. Nor-mally, both are operating (“up”). On any day on which line A is operating, thereis probability p that it will go down at the end of the day. On any day on whichline B is operating, there is probability q that it will go down at the end of the day.It takes the repair crew a day to repair a broken line. Only one line can be repairedat a time; if both are down, they repair line A first.

Model this situation as a Markov chain. (Hint: there are four states.)


18 Utility Theory 1

18.1 IntroductionIn all the material which we saw earlier on payoff matrices and decision trees,our principal decision criterion was that of expected value. At each point of un-certainty the payoffs were weighted by the probabilities to calculate an expectedpayoff at that point. This decision criterion works best when a decision is re-peated many times. However, this criterion may not be the best in a non-repetitivesituation in which a great deal of money is involved.

Actual human behaviour often tends to shy away from risky situations infavour of a more certain situation, even when the expected payoff of the former ishigher. For example, consider someone who is a contestant on the popular televi-sion show Who Wants To Be A Millionaire? Suppose that this person has won (ifhe or she walks away now) $64,000. If the next question is answered correctly, thecontestant will have $125,000, otherwise he or she will be left with just $32,000.The 50/50 lifeline has been used to narrow the correct answer down to two pos-sibilities, but each of these two is equally likely in the contestant’s eyes. Whatshould the contestant do: guess; or walk away? The expected value criterion (alsocalled expected monetary value or EMV) says that the contestant should guess.The expected value associated with guessing is

0.5($32,000)+0.5($125,000) = $78,500

This is greater than the value of not guessing (i.e. walking away) which is $64,000.While some contestants would guess in this situation, others would not, preferringthe $64,000 for sure. What we want to do in this lecture is to capture this effectin the analysis of competing alternatives. Instead of looking for the alternativewith the highest expected payoff, we are now looking for the alternative with thehighest expected utility . Where U(x) means the utility of the amount x, thecontestant who chooses to walk away is saying that

U($64,000) > 0.5U($32,000)+0.5U($125,000)

i.e. the utility of the sure thing is greater than the expected utility of the gamble.We are now looking for the best alternative(s) for a particular individual. An-

other person may choose to do something quite different. For the individual weneed to derive his or her utility function by obtaining a set of data points. Theseare most easily represented by a graph, but can also be given as a mathematical


function. All the work required to make any payoff matrix or decision tree mustbe done as before, but now every payoff must be converted to the utility of thatpayoff.

18.2 Obtaining the Data for a Utility Function

The data for a particular individual’s utility function is obtained by asking him orher a series of simple questions. For example, the interviewer could start out bysaying:

Suppose that you are given a choice between having $5000 forsure, or being able to play a game with an equal chance of winningnothing or winning $10,000. Would you (i) prefer the sure thing; (ii)prefer the gamble; or (iii) be indifferent between the two alternatives?

Obviously, using expected value as the decision criterion leads to indifference,because the expected value of the gamble (which is 0.5($0) + 0.5($10,000) =$5000) equals the value of the sure thing. However, the answer to the interviewer’squestion has no right or wrong answer. If in this situation the decision maker saysthat he or she prefers the sure thing, then that person is said to be risk aversefor this situation; this is the usual response. Other possibilities are to prefer thegamble ( risk preferring ) or to be indifferent ( risk neutral ). If the answer is (i)(the sure thing), then the interviewer could ask the following as the next question:

We have the same situation as before, but now the sure amountis only $3000. Would you (i) prefer the sure thing; (ii) prefer thegamble; or (iii) be indifferent between the two alternatives?

If the person still prefers the sure thing, the question can be repeated, using say$1000. If the person now prefers the gamble (as most people would), the amountof the sure thing can be increased, to say $4000. In this case the third questionwould be:

Now suppose the sure amount is $4000. Would you (i) prefer thesure thing; (ii) prefer the gamble; or (iii) be indifferent between thetwo alternatives?


This is repeated going up and down as necessary until we hit the point ofindifference. Perhaps at $4000 the decision-maker prefers the sure thing, at $3500still prefers the sure thing, then at $3400 prefers the gamble, then at $3450 prefersthe sure thing, and then finally at $3440 he or she is indifferent between the twoalternatives. In other words,

U($3440) = 0.5U($0)+0.5U($10,000)

This critical amount (in this case, $3440) is called the certainty equivalent of thegamble. In general the certainty equivalent (CE) of a gamble is the amount suchthat the utility of the certainty equivalent equals the expected utility (EU) of thegamble. As an equation this is:

U(CE) = EU(gamble)

It is useful to establish a scale for a graph; we can begin the scale by arbitrarilypicking two values of the utility function (just as the freezing point and boilingpoint of water were arbitrarily chosen). For example we can define:

U($10,000) = 100U($0) = 0

Based on these values we can easily find the utility of $3440:

U($3440) = 0.5U($0)+0.5U($10,000)= 0.5(0)+0.5(100)= 50

We want all subsequent questions to have only one unknown, so that we may solvefor it. We can do this by now asking a question based on the utility of $3440,because this is a known point. Also, rather than going up and down until the pointof indifference is obtained, we could simply ask for the point of indifference. Thequestion posed to the decision-maker could be:

What amount of money would make you indifferent between hav-ing this amount for sure, or being able to play a game with an equalchance of winning $3440 or winning $10,000?

The expected value of the gamble is:

0.5($3440)+0.5($10,000) = $6720


A risk-averse person will be indifferent at an amount less than this figure. For ex-ample, the point of indifference may be at $6420. Using the fact that U($3440) =50 (which we just calculated) we obtain:

U($6420) = 0.5U($3440)+0.5U($10,000)= 0.5(50)+0.5(100)= 75

Another question that could be asked is:

What amount of money would make you indifferent between hav-ing this amount for sure, or being able to play a game with an equalchance of winning nothing or winning $3440?

The expected value of the gamble is $1720; for a risk-averse decision-makerthe certainty equivalent might be only $1180. We then calculate:

U($1180) = 0.5U($0)+0.5U($3440)= 0.5(0)+0.5(50)= 25

So far we have found the following data points:

x $0 $1180 $3440 $6420 $10,000U(x) 0 25 50 75 100

We can plot U(x) as a function of x for these five points. This is shown inFigure 58.

What we want to do now is to draw a curve through these points. There is awhole area of mathematics which is devoted to estimating a function from a setof data points, but we’re not going to use anything that sophisticated. For ourpurposes, it will suffice to draw a curve by freehand through the known points.Because each person will draw the curve slightly differently, the numerical valueswill differ slightly from one person to the next.

This decision maker has made this curve already possessing a wealth level.Often the level of a person’s wealth will affect his or her attitude to risk. Forexample, suppose that a student with little money chooses a certainty equivalent


$0 $2500 $5000 $7500 $10,000x - Amount above Current Wealth

0

25

50

75

100

U(x)

�

�

�

�

�

Figure 58: Data Points for the Utility Function


of only $3000 for a 50/50 gamble of $10,000 or nothing. A few days later, thestudent wins $100,000 in a lottery. Now he or she can afford to take a risk, so nowthe certainty equivalent of the same gamble may well be closer to the expectedvalue, which is $5000. Because of this, to use a utility function on a decisionproblem it is important that the person’s wealth level does not change between thetime that the function was obtained and the function would be used.

A possible drawing of the curve is shown in Figure 59. In the caption it is em-phasized that this is for dollar amounts beyond the person’s wealth level. Indeed,this is how we shall treat all examples, unless indicated to the contrary.

18.3 Using the Graph with a Payoff Matrix

Suppose that the decision maker for whom the graph has been made faces thefollowing situation:

DemandLow Medium High

Keep Existing Capacity $2000 $4000 $8000Expand Capacity $0 $4800 $9400

Prob. 0.30 0.45 0.25

If we work out the expected value for each alternative we obtain $4400 for “KeepExisting Capacity” and $4510 for “Expand Capacity”. Hence a risk-neutral personwould choose to expand capacity for a ranking payoff of $4510.

Now, let us see what our risk-averse decision maker would do. One of sixdollar figures in the table has a known utility: U($0) = 0. For the other five, wemust use the graph. For example, to find the utility of $8000, we start at $8000 onthe horizontal axis, go up to the curve, and then go across to the vertical axis. If wedo this carefully we see that the utility of $8000 is about 86 or 87. This is shownin Figure 60. It’s hard to estimate a number any closer than this (magnificationshows that the utility is about 86.5). Using either 86 or 87 would be fine for ourpurposes.

We go through the same procedure for all the dollar amounts in the abovetable. When this is complete, we create the following utility matrix, and for eachalternative, we compute the expected utilities.



0

25

50

75

100

U(x)

�

�

�

�

�

Figure 59: The Utility Function Shown as a Smooth Curve



0

25

50

75

100

U(x)

�

�

�

�

�

x=

$800

0

U(x)≈ 87

Figure 60: Finding the Utility of $8000


Demand ExpectedLow Medium High Utility

Keep Existing Capacity 35 55 87 57.0Expand Capacity 0 62 96 51.9

Prob. 0.30 0.45 0.25

The best alternative is to “Keep Existing Capacity”. In this example, the opti-mal alternative is different than the one recommended when using expected valueas the decision criterion. This alternative has an expected utility of 57. By draw-ing a horizontal line through U(x) = 57 until we reach the curve as shown inFigure 61, and then drawing a vertical down from this point, we see that the cer-tainty equivalent is at about $4200. This figure is less than the expected value of$4510, because this risk-averse decision maker is prepared to forego some of theprofit to avoid the risk.

Recommendation

Using expected value as the decision criterion, we would recommend expand-ing the capacity for a ranking payoff of $4510. However, using the utility functionwe would recommend for this particular individual that he or she should keep thecurrent capacity. For this person, this alternative has a certainty equivalent ofabout $4200.

18.4 A More Complicated GraphThe questions which the interviewer poses to the decision maker in order to obtainthe points for the graph need not be as straightforward as they are in the exampleabove. For one thing, the probabilities do not have to be equal. Also, we can createutilities outside of the initial scale, meaning that they can be negative (just liketemperature). Suppose that an interview has produced the following information:

1. A gamble with a 90% chance of winning nothing and a 10% chance ofwinning $100,000 has a certainty equivalent (CE) of $25,000.

2. A gamble with an equal probability of winning either $50,000 or $100,000has a CE of $60,000.

3. A gamble with a two-thirds chance of winning nothing and a one-thirdchance of winning $50,000 has a CE of $25,000.



0

25

50

75

100

U(x)

�

�

�

�

�

U(x) = 57x≈

$420

0

Figure 61: Finding the Certainty Equivalent of U(x) = 57


4. A gamble with a 0.4 probability of losing $50,000 and a 0.6 probability ofwinning $50,000 has a CE of 0.

5. A gamble with a 0.45 probability of losing $70,000 and a 0.55 probabilityof winning nothing has a CE of −$50,000.

6. A gamble with a 2/7 probability of winning $60,000 and a 5/7 probabilityof winning $100,000 has a CE of $80,000.

7. A gamble with a 5/16 probability of losing $70,000 and a 11/16 probabilityof losing $30,000 has a CE of −$50,000.

For convenience we will express all financial amounts in thousands of dol-lars; for example $100,000 is written as 100. Arbitrarily setting U(0) = 0 andU(100) = 100, we work through each of the seven equivalencies to determine therest of the data points.

From the first we obtain:

U(25) = 0.9U(0)+0.1U(100)= 0.9(0)+0.1(100)= 10

From the second statement we obtain:

U(60) = 0.5U(50)+0.5U(100)

Since both U(50) and U(60) are unknowns, we will have to return to this equationlater.

From the third statement we obtain a single unknown on the right:

U(25) = 23U(0)+ 1

3U(50)

10 = 23(0)+ 1

3U(50)

10 = 13U(50)

From this we obtain U(50) = 30. Knowing this, we can now go back to the secondstatement to obtain U(60):

U(60) = 0.5U(50)+0.5U(100)= 0.5(30)+0.5(100)= 65


If we recognize the unknown before writing the equation, we can start with theunknown on the left. From the fourth equivalency we obtain:

0.4U(−50)+0.6U(50) = U(0)0.4U(−50)+0.6(30) = 0

0.4U(−50) = −18U(−50) = −45

The fifth equation gives us U(−70):

0.45U(−70)+0.55U(0) = U(−50)0.45U(−70)+0.55(0) = −45

0.45U(−70) = −45U(−70) = −100

Next we find U(80):

U(80) = 27U(60)+ 5

7U(100)

= 27(65)+ 5

7(100)= 90

Finally we find the utility of a loss of $30,000:

516U(−70)+ 11

16U(−30) = U(−50)516(−100)+ 11

16U(−30) = −45−500+11U(−30) = −720

11U(−30) = −220U(−30) = −20

Where x is in thousands of dollars we have obtained the following utility table:

x −70 −50 −30 0 25 50 60 80 100U(x) −100 −45 −20 0 10 30 65 90 100

To plot these points we need the negative as well as the positive regions for boththe horizontal and vertical axes. Plotting these nine points, and drawing a smoothcurve through them, gives us the graph shown in Figure 62.


−100 −50 0 50 100x in Thousands of Dollars

−100

−50

50

100

U(x)

�

�

�

�

�

�

�

�

�

Figure 62: Graph of the Utility Function for the Second Example


18.5 Utility Functions18.5.1 Introduction

It is possible to take the data points from a set of questions about certainty equiv-alents and from these create a mathematical function. We will not make suchtransformations ourselves, but we will use exogenously given utility functions.For now, let us assume that the functions are based on changes to the decisionmaker’s current wealth level.

A Square Root Function An example of such a function is

U(x) =√

5x+1000

Since we must have 5x+1000≥ 0, the domain of this function is x≥−200. Theutility of an amount in the domain, for example, x = $680, is found as:

U(680) =√

5(680)+1000

=√

4400= 66.3324958...

For this function, an inverse exists in analytical form. This is useful for when wehave found the expected utility of an alternative, and we wish to find its certaintyequivalent. For example, suppose that we have found EU = 97.56.

97.56 =√

5x+10009517.9536 = 5x+10008517.9536 = 5x1703.59... = x

Hence we have CE = $1703.59.

An Exponential Function A modified exponential function is one of the form

U(x) = a−be−cx

where x is the amount of money in dollars, and a, b, and c are given constants. Forour purposes b > 0 and c > 0. This type of function has special properties whichwe shall examine in the next lecture.


For example, where a = 7, b = 3, and c = 0.001, the function is:

U(x) = 7−3e−0.001x

[On some calculators, exponents may be handled as 2nd function ln , or some-thing similar.] At x = $5000, we obtain:

U(5000) = 7−3e−50001000

= 7−3e−5

= 7−3(0.00673795)= 7−0.02021384= 6.97978616

This type of function has an analytical inverse function, so if we have an ex-pected utility of say 4.9, and if we want the certainty equivalent, we proceed asfollows:

4.9 = 7−3e−.001x

3e−.001x = 2.1e−.001x = 0.7

ln(e−.001x) = ln(0.7)−.001x = −0.356675

x = 356.675

The certainty equivalent is $356.675.

18.5.2 A Simple Decision

Here is a simple decision problem in payoff matrix format:

O1 O2A1 90 55A2 110 30

Prob. 0.6 0.4

We wish to find the recommended alternative and the certainty equivalent forthe following functions:


(i) U(x) = 3−2e−x

200

and(ii) U(x) =

√2x+100

We begin (i) first finding U(90):

U(90) = 3−2e−90200

= 3−2e−0.45

= 3−2(0.637628)= 3−1.275256= 1.724744

Similarly, we find U(55), U(110), and U(30), and put these numbers into a utilitymatrix, and work out the expected utilities on the right.

O1 O2 EUA1 1.724744 1.480856 1.627189A2 1.846100 1.278584 1.619094

Prob. 0.6 0.4

The higher EU is in the A1 row, so we recommend the use of alternative 1. Thecertainty equivalent is the number CE such that U(CE) = 1.627189. Hence wesolve:

3−2e−x

200 = 1.627189−2e−

x200 = −1.372811

e−x

200 = 0.686406− x

200= −0.376286

x = 75.257...

The certainty equivalent is about $75.26.

Now we solve situation (ii) where U(x) =√

2x+100, beginning with x = $90.

U(90) =√

2(90)+100

=√

280= 16.733201


Similarly, we find U(55), U(110), and U(30), and put these numbers into a utilitymatrix, and work out the expected utilities on the right.

O1 O2 EUA1 16.733201 14.491377 15.836471A2 17.888544 12.649111 15.792771

Prob. 0.6 0.4

As with (i), we recommend alternative 1. The EU of this optimal alternative is15.836471. We now solve for the certainty equivalent.

U(x) = 15.836471√2x+100 = 15.8364712x+100 = 250.793814

2x = 150.793814x = 75.3969...

The certainty equivalent is about $75.40.



A golf course is considering whether to expand from 9 holes to 18 holes this com-ing season, or whether to wait until next year. The payoff for each alternativedepends on the weather, which may be rainy, cloudy, or sunny. The payoffs asso-ciated with each alternative and outcome are (in thousands of dollars) as follows:

WeatherAlternative Rainy Cloudy Sunny

Expand now −100 600 800Wait a year 100 400 600Probability 0.35 0.40 0.25

The company has decided that a $100,000 loss will have a utility of zero and a$1,000,000 profit a utility of 100. Furthermore, they are indifferent in the follow-ing four situations.

(i) $700,000 for certain; $1,000,000 profit (prob. 0.85) or a $100,000 loss(prob. 0.15).

(ii) $700,000 for certain; $1,000,000 profit (prob. 0.25) or $600,000 profit(prob. 0.75).

(iii) $600,000 for certain; $700,000 profit (prob. 1011 ) or $100,000 profit (prob.

111 ).

(iv) $400,000 for certain; $600,000 profit (prob. 23 ) or $100,000 profit (prob.

13 ).

(a) What is the recommended alternative using the EMV decision rule?

(b) Construct the utility curve.

(c) What strategy should the golf course adopt using the expected utility deci-sion rule?

(d) What is the certainty equivalent of the optimal strategy?


18.6.2 Problem 2

An investor is considering whether to expand her business by adding a new smallstore, or a new large one (or do nothing). The payoff for each alternative dependson the demand, which may be low, medium, or high. The payoffs associated witheach alternative and outcome are (in thousands of dollars) as follows:

DemandAlternative Low Medium HighDo nothing 0 0 0Small store −50 100 300Large store −200 180 400Probability 0.2 0.5 0.3

(a) If she is risk-neutral, what should she do?

(b) Suppose now that her utility function is

U(x) = 1− e−x

1000

where x is in thousands of dollars. Find the recommended alternative and thecertainty equivalent.

(c) Suppose now that her utility function is

U(x) =√

x100

where x is her level of wealth in thousands of dollars. Just prior to making thedecision, her wealth level is $200,000. Find the recommended alternative, and thecertainty equivalent (exclude the current wealth at the end).


19 Utility Theory 2

19.1 IntroductionWe begin this lecture by explaining three properties which may be useful for autility function to have; only the exponential utility function possesses all threeproperties. Then, we show how to compute the EVPI when a non-linear utilityfunction is used. We shall see that this is now quite complicated, except whenthe function is exponential. Thirdly, we illustrate the use of utility functions on adecision tree, which requires that instead of using cost gates and/or payoff nodes,that all payoffs and costs be consolidated on the extreme right-hand branches ofthe tree.

19.2 Three Useful PropertiesHere are three properties that may be useful for a decision maker’s utility function:

1. More is preferred to less.

2. The decision maker is risk-averse.

3. If the same amount is added to all payoffs, the relative ranking of the alter-natives does not change.

It seems reasonable that all utility functions would possess the first prop-erty, i.e. the higher the payoff, the higher the utility. The second property (risk-aversion) is empirically useful, but of course is not a requirement of all utilityfunctions. We can test a differentiable function for these two properties by takingthe first and second derivatives. The first property requires that the first derivativebe positive everywhere; the second property requires that the second derivative benegative everywhere. Based on these derivative tests, it can be shown that bothsquare-root type functions and exponential functions obey these two properties.

The third property is certainly true when the decision criterion is expectedvalue. Indeed, this was the basis for being allowed to ignore costs or payoffswhich are common to all alternatives. However, it turns out that the only type ofnon-linear utility function for which this is true is the family of exponential utilityfunctions. Several advantages arise because of this third property:

1. The current wealth level of the decision maker is irrelevant.


2. The expected value of perfect information (EVPI) is easier to compute thanit is for other non-linear utility functions.

3. Any cost or payoff which is common to all cells of a payoff matrix, orcommon to all ending branches of a decision tree, can be ignored, if desired.

19.3 Finding the EVPIIn this section we show how to calculate the EVPI. The different methods willhelp illustrate the third property described above. As an example we will use thepayoff matrix seen in the previous lecture:

O1 O2A1 90 55A2 110 30

Prob. 0.6 0.4

First, let us recall how to find the EVPI when the decision criterion is expectedvalue. We compute the EV for each alternative, then find the EV with PI, andfinally subtract the EV of the optimal alternative from the EV with PI.

O1 O2 EVA1 90 55 76A2 110 30 78

Prob. 0.6 0.4

Hence a risk-neutral person would choose alternative 2 with a ranking payoffof $78. The EV with PI is 0.6($110)+0.4($55) = $88, and the EVPI is therefore$88−$78 = $10.

With an exponential utility function the approach is more complicated butsimilar. From the previous lecture we had the utility function:

U(x) = 3−2e−x

200

Using this formula we found:

O1 O2 EUA1 1.724744 1.480856 1.627189A2 1.846100 1.278584 1.619094

Prob. 0.6 0.4


We found the CE of 1.627189 to be about $75.26. To find the EVPI using theexponential, we must find the CE of the EU with PI, and from this subtract the CEof the optimal alternative.

EU with PI = 0.6(1.846100)+0.4(1.480856)= 1.7000024

To find the CE of this we solve:

3−2e−x

200 = 1.7000024−2e−

x200 = −1.2999976

e−x

200 = 0.6499988− x

200= −0.4307848

x = 86.15695...

Hence the CE of the EU with PI is about $86.16. Hence the EVPI is about:

EVPI = CE(EU with PI)−CE(optimal alternative)= $86.16−$75.26= $10.90

The method for finding the EVPI for non-linear utility functions in general(i.e. other than the exponential) is more complex. As before, we will use the utilityfunction U(x) =

√2x+100 to illustrate the method. The utility matrix which we

found in the previous lecture is not used to find the EVPI, except for the EU of theoptimal alternative. Instead, we imagine that perfect information is available for afee, and we let the symbol f represent the amount of the fee. The dollar amountsfrom the original matrix are:

O1 O2A1 90 55A2 110 30

Prob. 0.6 0.4

After paying f to obtain perfect information, we are told that the outcomewill be O1 with probability 0.6, and we are told that the outcome will be O2 with


probability 0.4. If O1 will occur, we want alternative 2, with a gross payoff of$110. Because we must pay f to hear the information, the net dollar payoff is$110− f . If O2 will occur, we want alternative 1, for a net payoff of $55− f . Insummary, after paying f to hear the perfect information, there is probability 0.6 ofobtaining a net payoff of $110− f , and probability 0.4 of obtaining a net payoffof $55− f .

Now we bring in the utility function to find the utility of each net payoff:

U(110− f ) =√

2(110− f )+100

=√

320−2 f

U(55− f ) =√

2(55− f )+100

=√

210−2 f

Therefore, the expected utility with having paid f to obtain perfect information is:

0.6√

320−2 f +0.4√

210−2 f

We recall the following utility matrix from the preceding lecture:

O1 O2 EUA1 16.733201 14.491377 15.836471A2 17.888544 12.649111 15.792771

Prob. 0.6 0.4

Hence the EU obtained without perfect information is 15.836471. The EVPI isthe amount f such that the decision maker is indifferent between obtaining or notobtaining perfect information. In other words, we seek the value of f such that:

0.6√

320−2 f +0.4√

210−2 f = 15.836471

We are left with one equation in one unknown. An equation like this one was seenin Chapter 1 of Mathematics for Management Science. It is possible to obtain ananalytical solution for this example by squaring both sides, simplifying, and thensquaring both sides again, and then simplifying again. Such a solution would takea great deal of work. Instead of doing it this way, we will obtain a numericalsolution using Goal Seek on Excel. Leaving cell A1 reserved for the value of f ,we put the following into cell B1:

=0.6*SQRT(320-2*A1)+0.4*SQRT(210-2*A1)


We then use Goal Seek, setting cell B1 to the value of 15.836471, by changingcell A1. The number which Excel puts into cell A1 is 11.0877. This is the criticalvalue of f which causes indifference between obtaining or not obtaining perfectinformation, and this is the EVPI. To the nearest cent, the EVPI is $11.09.

19.4 Using Utility Functions with Decision TreesWhen a decision tree is made for a decision-maker with a non-utility function,several changes must be made to what we saw earlier. First, all costs must beimbedded at the end (on the right); there are no cost gates on the tree. Secondly,all payoffs must be imbedded at the end; there are no payoff nodes. This meansthat at each ending branch we must work out the net payoff, and then we findthe utility of each of these final payoffs. The rollback proceeds as usual, exceptthat we are rolling back the expected utility rather than the expected value. Whenthe rollback is complete, it is useful to find the certainty equivalent of the optimalexpected utility.

We will illustrate this procedure using a problem which we saw much earlierin the course in Lecture 8. The problem description is repeated here:

An oil exploration company has identified a site under which theremay be a pocket of oil. The probability that oil exists at this locationis 1%. It would cost $3,000,000 to drill for oil. If the oil exists, itwould be worth $40,000,000. A seismic test is available which wouldcost $40,000. The result of the test would be one of the following:“positive”,“inconclusive”, or “negative”. If there really is oil present,then there is a 60% chance of a positive reading, a 30% chance ofan inconclusive reading, and a 10% chance of a negative reading. Ifthere’s no oil at that location, then there’s a 0.04 probability of a pos-itive reading, and a 0.2 probability of an inconclusive reading.

As before, we wish to develop a decision tree for this situation, and solve it toobtain a recommendation for the oil exploration company, but now we will usethe utility function:

U(x) =√

2x+7

where x is in millions of dollars.We already know what the shape of the tree is. All that is different is that we no

longer have cost gates as we did before. The tree without any of the revised prob-abilities or any of the financial information is shown in Figure 63. The Bayesian


revision is unchanged; it appears in Figure 21. Putting the revised probabilities onthe tree gives us Figure 64.

The dollar amount to be placed at the end of every final branch reflects allpayoffs and costs on the path between the first decision node and the final branch.For example, at the top of the tree after the “no seismic” alternative we have the“drill” alternative followed by the “oil” outcome. This gives us:

−$3,000,000+$40,000,000 = $37,000,000

After “no oil”, we have the cost of the drilling with no offsetting revenue, hencethe payoff is − $3,000,000. Completing this portion of the tree, the path con-taining the “do not drill” alternative has neither payoffs nor costs, and so the netpayoff is $0.

The $40,000 cost of the seismic test must be imbedded at the end, hence after“seismic”, “drill” and “oil” the net payoff is:

−$40,000−$3,000,000+$40,000,000 = $36,960,000

Similarly the other two figures are also reduced by $40,000: after “seismic”,“drill”, and “no oil” the net payoff is −$3,040,000; after “seismic” and “do notdrill” the net payoff is −$40,000. These three figures appear in three places.Because we will be putting utilities onto the tree, we will put a dollar sign in frontof all financial figures for emphasis.

The tree with all the financial information added to the right of every finalbranch is shown in Figure 65.

The dollar amounts must be converted to the utility of each of these amounts.Since the utility function is based on x being in millions of dollars, for a figure of$37,000,000 we input 37 into the formula. This gives:

U(37) =√

2(37)+7

=√

81= 9

For −$3,000,000, we input −3 into the formula:

U(−3) =√

2(−3)+7

=√

1= 1


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U(0) =√

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7≈ 2.64575

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√7); the certainty equivalent is $0.

Recommendation

The investor should do nothing (i.e. do not use the seismic, and do not drill). Thiscourse of action has a certainty equivalent of $0.


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19.4.1 The EVPI

As with any problem using Bayesian revision, it is possible to find the EVPI, andthen compare this with the cost of doing the testing. If the EVPI < the cost oftesting, we need not evaluate the testing alternative.

In this example perfect information costing f (in millions of dollars) wouldlead us to drill if there is oil (prob. 0.01) which gives a utility of

√2(37− f )+7,

or lead us not to drill if there is no oil (prob. 0.99) with a utility of√

2(0− f )+7.The current EU is 2.64575, hence we wish to solve:

0.01√

2(37− f )+7+0.99√

2(0− f )+7 = 2.64575

which simplifies to:

0.01√

81−2 f +0.99√

7−2 f = 2.64575

We put the following formula into cell B1:

=0.01*SQRT(81-2*A1)+0.99*SQRT(7-2*A1)

We then use Goal Seek with cell A1 reserved for f , and B1 being set equal to2.64575. We find the solution to be f = 0.167934. Therefore, since f is in millionsof dollars, the EVPI is $167,934. Since the cost of doing the seismic ($40,000)is less than the EVPI, we cannot exclude the possibility that the seismic is worthdoing. Hence we needed to work out the solution on the entire tree.



This is a repetition of the second problem from the previous lecture, except thatnow the problem is to determine the EVPI for each of the three situations. Youwill need some of the answers from the previous exercise.

An investor is considering whether to expand her business by adding a newsmall store, or a new large one (or do nothing). The payoff for each alternative de-pends on the demand, which may be low, medium, or high. The payoffs associatedwith each alternative and outcome are, (in thousands of dollars), as follows:

DemandAlternative Low Medium HighDo nothing 0 0 0Small store −50 100 300Large store −200 180 400Probability 0.2 0.5 0.3

(a) She is risk-neutral.

(b) Her utility function, where x is in thousands dollars, is:

U(x) = 1− e−x

1000

(c) Her utility function is

U(x) =√

x100

where x is her level of wealth in thousands of dollars. Just prior to making the de-cision, her wealth level is $200,000. Use Goal Seek to find the EVPI numerically.

19.5.2 Problem 2

Reformulate the Niagara Winery problem (first seen in Section 5.5), putting allcosts and payoffs onto the ends of the final branches. Where x is in millions ofdollars, Betty now uses the utility function:

U(x) = 1− e−x2

Solve to obtain a recommendation for Betty, and calculate the certainty equivalent.


20 Game Theory 1

20.1 Introduction

In the material which we saw on making decisions in the presence of uncertainty,there was only one decision maker. Now in this material on game theory , welook at two decision makers in competition with each other. This competition istreated like a game, and we refer to each decision maker as a player . We willgenerically refer to these two persons as Player A and Player B.

In order to keep the complexity manageable, we restrict the competition to twoplayers, rather than more than two. Also, we must make some assumptions abouttheir behaviour:

1. Both players know each other’s possible moves, and the payoff for eachplayer for each combination of moves.

2. The players make their moves simultaneously (i.e. one player cannot waitto see what the other player does).

3. Each player tries to do what is best for himself or herself (i.e. there is nocooperation with the other player).

4. For any combination of moves, the two players share the same total netpayoff; this is called a constant-sum game . A special case of this is thezero-sum game , in which the gain to one player is equal in absolute value

to the loss of the other player.

At the end of this lecture we will relax the last assumption. Doing this allowsus to investigate what is called the prisoner’s dilemma . Both the terms zero-sumgame and prisoner’s dilemma are widely used in financial and economic publica-tions.

Unless specifically indicated to the contrary, we speak of the payoffs to aplayer, a positive number meaning that the player receives something and a neg-ative number meaning that the player loses that amount. For a zero-sum game,only one payoff table needs to be written. The table is written such that the playeron the left (we will call him/her Player A) obtains the payoffs shown. The playeron top (Player B) obtains the negative of the payoffs shown.


20.2 Pure StrategySome games are simple in that they lead to each player always choosing the samealternative. When this happens each player is said to be following a pure strategy .Here are two examples.

20.2.1 Example 1

In this simple example Player A has two alternatives labelled as A1 and A2, andPlayer B has two alternatives labelled as B1 and B2. Suppose that are given thepayoffs shown in the following table:

Player BB1 B2

Player A1 3 7A A2 2 5

For example, if Player A plays A1, and if Player B plays B2, then Player Awill receive $7 from Player B, i.e. a gain of $7 for A, and a loss of $7 for B. We donot need to use (or even see) the table from Player B’s perspective, but it is simpleto find, being the negative of the transpose of the original table.

Player AA1 A2

Player B1 −3 −2B B2 −7 −5

Going back to the original table, we now determine what each side should do.In this example, Player A will always do better with A1 rather than A2, no matterwhat Player B does:

A1 3 7≥ ≥

A2 2 5

Therefore, Player A chooses to play A1. To look at things from B’s perspective,we must remember that the payoffs are to Player A, so Player B wants low num-bers. Player B looks at the table and sees:

B1 B23 ≤ 72 ≤ 5


Player B chooses the column with the lower numbers, i.e. Player B chooses B1.Because each player has found an alternative which he or she prefers, we

say that this game has a stable solution (also called a saddle-point solution ).When a game has a stable solution, each player will pursue a pure strategy. Herewe have a stable solution with Player A choosing A1 and Player B choosingB1. The payoff located at the interception of these two alternatives is called thevalue of the game . In this example, this value is 3. It is the payoff to the player

on the left, and represents a loss of 3 for the player on top.We summarize this game as follows: There is a stable solution, with Player A

choosing A1 and Player B choosing B1, and the value of the game is $3. Right onthe table we can highlight the preferred alternatives and the value of the game:

Player BB1 B2

Player A1 3 7A A2 2 5

There is a simple condition to check to see if a game has a stable solution. Sta-bility is obtained if and only if the maximum of the row minimums (“maximin”)equals the minimum of the column maximums (“minimax”). In this example:

max{min{3,7},min{2,5}} = max{3,2}= 3

andmin{max{3,2},max{7,5}} = min{3,7}

= 3

Both maximin = 3 and minimax = 3, so maximin = minimax, and we have there-fore a stable solution.

In general, one could start with a game theory table by seeing if maximin= minimax. When this condition is true, maximin or minimax is the value of thegame, and the optimal solution is to play the corresponding alternatives. However,since this condition rarely holds, the methods to be shown later in this module andthe next are of more practical benefit.

20.2.2 Example 2

The number of alternatives for each player need not be equal. Here is an example:


Player BB1 B2 B3 B4

A1 7 −6 8 −5Player A A2 6 −9 −1 −7

A3 10 −3 9 2

Though all work is done in practice on one table, here we repeat the table severaltimes to show each step clearly. We begin with finding maximin. In every row, wefind the minimum in that row, and write this number down on the right. Then wefind the maximum of the row minimums. Doing this much we have:

Player B rowB1 B2 B3 B4 minimums

A1 7 −6 8 −5 −6Player A A2 6 −9 −1 −7 −9

A3 10 −3 9 2 −3 max

Then in every column, we find the maximum in that column, and write thisnumber down on the bottom. Then we find the minimum of the column maxi-mums. This yields:


A1 7 −6 8 −5 −6Player A A2 6 −9 −1 −7 −9

A3 10 −3 9 2 −3 maxcolumn 10 −3 9 2

maximums min

Hence we see that both maximin and minimax are−3, so we have a pure strat-egy: Player A chooses A3; and Player B chooses B2. Highlighting the solutionon the table we have:


A1 7 −6 8 −5 −6Player A A2 6 −9 −1 −7 −9

A3 10 −3 9 2 −3 maxcolumn 10 −3 9 2

maximums min


20.3 Mixed StrategyUnfortunately, very few games are played with a pure strategy, but instead mustbe played with a mixed strategy in which the players will vary the chosen alter-native. The next example will illustrate that maximin and minimax are not equal,and when this happens, a mixed strategy must be played by each player.

20.3.1 Example

Player BB1 B2 B3

A1 4 6 5Player A A2 7 2 1

A3 3 2 4A4 5 3 −2

We begin to solve the problem by checking for the existence of a pure strategy:

Player B rowB1 B2 B3 minimums

A1 4 6 5 4 maxPlayer A A2 7 2 1 1

A3 3 2 4 2A4 5 3 −2 −2

column 7 6 5maximums min

Unfortunately, we see that the maximum of the row minimums, which is 4, isnot equal to the minimum of the column maximums, which is 5. Therefore, thegame does not have a pure strategy. We must therefore do more work to find thesolution. There is a procedure using linear optimization which we shall see at theend of the next lecture for solving two-player zero-sum games. Rather than coverthis right away, we shall see how some mixed-strategy solutions can be foundquite easily.

20.3.2 Dominance

If one alternative is always better than another for one of the players no matterwhat his or her opponent does, then the better alternative is said to dominate over


the other. The dominated (worse) alternative can be removed from further consid-eration, while the dominant (better) alternative stays. We can observe dominancewhen comparing two rows (the higher payoff row dominates), or when compar-ing two columns (the lower payoff column dominates). We eliminate dominatedalternatives, hence dominance eliminates lower payoff rows and higher payoffcolumns. When we say “higher payoff row”, we mean that in each column thehigher payoff row has a payoff which is greater than or equal to the payoff in theother row, and for at least one column the relationship is strictly greater than. Forexample, a row such as 2 7 3 would dominate a row such as 1 5 3. When we say“lower payoff column”, we mean that in each row the lower payoff column hasa payoff which is less than or equal to the payoff in the other column, and for atleast one row the relationship is strictly less than.

We start looking for dominance among either the rows or the columns. Oncea row or column is dominated, it is removed from further consideration. We willshow this removal by darkening the removed row or column, and put a numberbeside it to indicate the order of the removal. At the outset we have for thisexample:

Player BB1 B2 B3

A1 4 6 5Player A A2 7 2 1

A3 3 2 4A4 5 3 −2

All the work is done on one table, but we will show each step to help makethe method clear. If we compare A1 with A2, we see that neither dominates theother; 7≥ 4, but 2 6≥ 6, so this comparison ends without even looking at the 1 andthe 5. However, in comparing A1 and A3, we see that:

A1 4 6 5≥ ≥ ≥

A3 3 2 4

Hence A1 dominates A3, and the A3 row is therefore removed from further con-sideration. We put a ‘1’ followed by A1 next to the A3 row to indicate that this isthe first row or column to be removed, and the justification for doing so is that itis dominated by A1:


Player BB1 B2 B3

A1 4 6 5Player A A2 7 2 1

A3 3 2 4 1 A1A4 5 3 −2

Comparing A1 with A4, we see that neither dominates the other. Also, A2neither dominates nor is dominated by A4. Having compared all pairs of rows,we turn our attention to the columns.

Comparing B1 and B2, we find no dominance. Also, comparing B1 and B3there is no dominance. Now we compare B2 and B3:

B2 B36 ≥ 52 ≥ 1– – –3 ≥ −2

Note that what lies underneath the A3 row is immaterial, because the A3 row hasbeen removed. Player B will choose to remove the higher payoff column (sincethese are payoffs to Player A). Hence column 3 dominates over column 2, andcolumn 2 is removed.

Player BB1 B2 B3

A1 4 6 5Player A A2 7 2 1

A3 3 2 4 1 A1A4 5 3 −2

2B3

Having checked all pairs of columns, we now return to the rows. We check thepairs A1 and A2, then A1 and A4, and then A2 and A4. For the last of these wefind that with B2 having been removed, A2 dominates A4:

A2 7 – 1≥ – ≥

A4 5 – −2


We remove A4 leaving us with:

Player BB1 B2 B3

A1 4 6 5Player A A2 7 2 1

A3 3 2 4 1 A1A4 5 3 −2 3 A2

2B3

There is no further dominance. Using dominance we have reduced a 4 by 3 tabledown to a 2 by 2:

B1 B3A1 4 5A2 7 1

In a mixed strategy, each player will choose his or her alternatives by using proba-bilities, where it is understood that there is no probability of choosing a dominatedalternative. We let p be the probability that Player A chooses A1. Since Player Amust choose either A1 or A2, the probability that Player A chooses A2 is 1− p.If Player B chooses B1, the expected payoff is 4p+7(1− p). If Player B choosesB3, the expected payoff is 5p+1(1− p). Similarly, q and 1−q are the probabili-ties that Player B chooses alternatives B1 and B3 respectively. If Player A choosesA1, the expected payoff is 4q + 5(1− q). If Player A chooses A2, the expectedpayoff is 7q+1(1−q). The four expected payoffs are equal and are referred to asthe Expected Value (E.V.) of the game. Hence we have:

B1 B3 Prob. E.V.A1 4 5 p 4q+5(1−q)A2 7 1 1− p 7q+1(1−q)

Prob. q 1−qE.V. 4p+7(1− p) 5p+1(1− p)

We need to determine p, q, and the Expected Value of the game.Player A chooses p so that the expected payoff does not depend on what Player

B does. This is accomplished by setting the two expressions for the ExpectedValue equal to one another, and solving to obtain p.


4p+7(1− p) = 5p+1(1− p)6(1− p) = p

6−6p = p6 = 7p

p =67≈ 0.8571

Hence the expected value of the game is:

4p+7(1− p) = 4(67)+7(1− 6

7)

=247

+77

=317

≈ 4.429

Player B chooses q such that Player A has no advantage to choosing one or theother, which is accomplished by setting the two expected payoffs equal to eachother and solving for q.

4q+5(1−q) = 7q+1(1−q)4(1−q) = 3q

4−4q = 3q4 = 7q

q =47≈ 0.5714

The expected value of the game has already been found, but we might want to findit again as an arithmetic check. Any one of the four expressions for the expectedpayoff can be used:

p with B1 chosen 4p+7(1− p)p with B3 chosen 5p+1(1− p)q with A1 chosen 4q+5(1−q)q with A2 chosen 7q+1(1−q)

For example, using the third of these we obtain:

4q+5(1−q) = 4(47)+5(1− 4

7)


=167

+157

=317

≈ 4.429

In expressing the solution, fractions or decimals can be used for the probabilities,but usually the expected value of the game is expressed in decimal form. The nextsection gives an easier approach for finding these things starting from a 2 by 2table; the recommendation is presented at that point.

20.3.3 DARS Algorithm

After using dominance to reduce the table to a 2 by 2, we can combine all theabove operations into one easy method, called the DARS Algorithm. The fourletters stand for:

D Difference (always an absolute value)A AddR RatioS Switch

The procedure is all done on one table, but we show it here broken into severalsteps. First we add space to the right and on the bottom of the table for the opera-tions D, R, and S.

B1 B3 D R SA1 4 5A2 7 1DRS

The first thing to do is to find the absolute value of four differences. In theA1 row, |4− 5| = | − 1| = 1 goes into the D column. In the A2 row, we have|7−1|= 6. Next, in the B1 column, we put |4−7|= 3 into the D row, and in theB3 column, we put |5−1|= 4 into the D row. This gives:


B1 B3 D R SA1 4 5 1A2 7 1 6D 3 4RS

Next, we add (this is the A of DARS) the numbers in the D column (1+6 = 7)and the D row (3+4 = 7). No matter what the original four numbers may be, thetwo sums must be the same. The sum is placed where the D column and the Drow intercept.

B1 B3 D R SA1 4 5 1A2 7 1 6D 3 4 7RS

The ratio (R of DARS) refers to the fraction obtained by dividing the numberin the neighbouring D column or row by the sum we just obtained. For example,in the A1 row, the ratio is 1 divided by 7. i.e. 1

7 . Filling in the four ratios weobtain:

B1 B3 D R SA1 4 5 1 1

7A2 7 1 6 6

7D 3 4 7R 3

747

S

Finally, the ratios are switched: the R column is switched to obtain the Scolumn; and the R row is switched to obtain the S row. This gives:

B1 B3 D R SA1 4 5 1 1

767

A2 7 1 6 67

17

D 3 4 7R 3

747

S 47

37


Now in each row and column we have the desired probabilities, for example PlayerA will play A1 with probability 6

7 , and so on. We now find the expected value; asbefore, there are four ways to do this. We can obtain the dot product of any oneof: the B1 column and the S column; the B3 column and the S column; the A1row and the S row; and the A2 row and the S row. The latter two operations arelike what we did earlier with payoff matrices, so they may be more intuitive. Forexample, doing the fourth option yields:

7(47)+1(3

7) = 4+ 37

≈ 4.429

Recommendation

Player A should play A1 with probability 67 , and A2 with probability 1

7 . Player Bshould play B1 with probability 4

7 , and B3 with probability 37 . The expected value

of the game is 4.429.

20.4 Prisoner’s DilemmaHere we present a situation which is not a zero-sum (or constant-sum) game. It isimportant because it has implications for the study of collusion in business.


A armed robbery has taken place. The police have caught two men named Murphyand Smith with the goods in their possession. Getting a conviction against bothof them for possession of stolen property would be easy, but there’s not enoughevidence at the scene of the crime to obtain a conviction for the armed robberyas well. However, if either of them confesses, the confession along with othercorroborating evidence from the crime scene would be enough to convict the othersuspect, should he not confess himself.

The police have put each suspect into a separate interview room. To one sus-pect they say: “Murphy, you know that we got you on the possession charge. Thatalone will get you two years in jail. Smith is down the hall – if he confesses to thearmed robbery, you’re going away for 15 years. Why let him do that to you? Tellus you did it, and here’s what we’ll do. If he doesn’t confess, we’ll use your tes-timony to convict him, and for co-operating, we’ll not only let you off the armedrobbery charge but we’ll knock a year off the possession sentence. If he confesses,


then we don’t need your testimony, but for being cooperative, we will give youjust five years rather than fifteen.”

Down the hall a police officer is speaking to Smith. He gives the same offerusing exactly the same words, except that the surnames have been switched.

20.4.2 Solution

Let us assume that both men did indeed commit the armed robbery, so there is nopossibility of a false confession. The payoffs are not money but the number ofyears in jail. To consider them as payoffs rather than costs all these numbers arenegative. The choices for each player are i) do not confess; and ii) confess. Thegame payoff table is:

SmithDo Not Confess Confess

Murphy Do Not Confess −2 −15Confess −1 −5

We see that confessing dominates not confessing, so Murphy will confess. Smithhas the same table:

MurphyDo Not Confess Confess

Smith Do Not Confess −2 −15Confess −1 −5

Therefore, Smith will also confess. The problem from their perspective (notsociety’s) is that the closed communication prevents them coming up with a so-lution which is better for them. By confessing, each receives five years, but ifneither man confesses, each receives only two years. This is the dilemma whicheach man faces: to do what looks best for himself assuming the worse from theother man; or do what is best for both of them, which requires that the other mandoes likewise.

20.5 Obtaining the Payoffs

We now return to constant-sum games, and we wish to obtain the numbers in thetable. Here is an example:


Jack and Jill are playing a game in which each player secretly putstwo, three, four, or five marbles into his or her hand, and then theysimultaneously open their hands. If the total number of marbles isodd, then Jack gets all the marbles in their hands. If the total numberof marbles is even, then Jill gets all the marbles in their hands.

In this problem, the payoff is the net number of marbles gained or lost. Forexample, if Jack puts three marbles into his hand, and if Jill puts two marbles intoher hand, then the sum of five is odd, so Jack obtains the five marbles from theirhands, but three of these came from his pocket in the first place, so there is a netgain of two marbles, i.e. Jack has won the two marbles in Jill’s hand. As anotherexample, suppose that Jack plays three and Jill plays three, then the sum of sixis even, so Jill gains three marbles from Jack. In the payoff table from Jack’sperspective, Jack has lost three marbles to Jill, so the payoff is −3. Working outall sixteen combinations we obtain:

Jill’s # of Marbles2 3 4 5

Jack’s 2 −2 3 −2 5# 3 2 −3 4 −3of 4 −4 3 −4 5Marbles 5 2 −5 4 −5

The solution of this problem is left as part of the exercise.


20.6 Exercise20.6.1 Mixed Strategy Game

Players X and Y are playing a zero-sum game. The payoffs to player X are:

Y1 Y2 Y3 Y4X1 4 2 0 −1X2 6 0 6 5X3 3 1 7 6X4 12 −1 3 4

Determine the optimal strategy for each player, and find the expected value ofthe game.

20.6.2 Jack and Jill Problem

Jack and Jill are playing a game in which each player secretly puts two, three,four, or five marbles into his or her hand, and then they simultaneously open theirhands. If the total number of marbles is odd, then Jack gets all the marbles in theirhands. If the total number of marbles is even, then Jill gets all the marbles in theirhands.

The payoff table is seen to be:

Jill’s # of Marbles2 3 4 5

Jack’s 2 −2 3 −2 5# 3 2 −3 4 −3of 4 −4 3 −4 5Marbles 5 2 −5 4 −5

Determine how Jack and Jill should play this game, and find the expectedvalue of the game.

20.6.3 Formulation Problem

Players A and B are playing a game in which A shows either one or two fingers,and B simultaneously shows none, one, or two fingers. If the sum of the numberof fingers is even, then A wins the sum in dollars from B. If the sum is odd, thenB wins the sum in dollars from A. Write the payoff matrix from A’s perspective.(Do not solve)


21 Game Theory 2In the previous lecture we saw that a 2 by 2 mixed-strategy game could be solvedby hand using the DARS algorithm. Examples which were larger than a 2 by 2were solvable provided that through the use of dominance the problem could bereduced to a 2 by 2. In this lecture we shall see that a solution by hand is possibleas long as one of the players has (or can be reduced to) only two alternatives.Starting with any size matrix, we need to reduce it using dominance to either a 2by 3, a 2 by 4, and so on, or else a 3 by 2, a 4 by 2 and so on. Then, by using agraphical procedure, we further reduce the problem to a 2 by 2, after which it canbe solved using the DARS algorithm.

Some problems are not reducible to either two rows or two columns. For suchexamples the hand procedure will not work, nevertheless we can still obtain asolution. To do this we model the problem as a linear optimization model. Thistechnique allows us to solve any zero-sum (or constant-sum) game.

21.1 Graphical SolutionWe present two examples of zero-sum games. In the first, the table is first reducedusing dominance to two rows before proceeding with the graphical solution. Inthe second, the table is first reduced to two columns.

21.1.1 Example 1

B1 B2 B3 B4 B5A1 −3 3 6 2.5 2A2 4 −3 −4 −1 −2A3 5 −2.5 −4 −2 −1A4 −2 3.5 7 4 3

It is easily seen that this example has no pure strategy. Using dominance weobtain:

B1 B2 B3 B4 B5A1 −3 3 6 2.5 2 1 A4A2 4 −3 −4 −1 −2 3 A3A3 5 −2.5 −4 −2 −1A4 −2 3.5 7 4 3

2B2


With the dominated rows and column removed, the problem reduces to:

B1 B2 B3 B5A3 5 −2.5 −4 −1A4 −2 3.5 7 3

Whenever a table is reduced to two rows with three or more columns, the graphis based on the probability for the player on the left (in this example, Player A).The probability that Player A will choose A3 is p, and the probability that PlayerA will choose A4 is 1− p.

B1 B2 B3 B5 Prob.A3 5 −2.5 −4 −1 pA4 −2 3.5 7 3 1− p

In the reduced table, the payoffs range from −4 to 7. To solve the problem wemake a graph for which the horizontal axis is the probability p (0 to 1), and thevertical axis is the expected value of the game (−4 to 7). The axes are shown inFigure 68.

The first column is that of alternative B1. If Player B plays B1, then theexpected value of the game (denoted as V ) is V = 5p +(−2)(1− p). This is anequation of a line. The easiest way to plot it is to evaluate the equation at p = 0and at p = 1, plot these two points, and then connect them with a straight line. Atp = 0 we obtain V = −2, and at p = 1 we obtain V = 5. There’s really no workto do: the top number (5) goes on the right, and the bottom number (−2) goes onthe left. Plotting and then connecting these two points (0,−2) and (1,5) we obtainthe line shown in Figure 69. The label B1 is placed next to the line for futurereference.

We now plot the lines for alternatives B2, B3, and B5. In each case the topnumber goes on the right, and the bottom number goes on the left. Expressedanother way, the vertical intercept on the left comes from the bottom number, andthe vertical intercept on the right comes from the top number. We also write thealternative name as a label next to the corresponding line. Doing this for all threealternatives we obtain the graph shown in Figure 70.

Both Player A and Player B have the same information, so both are able todraw this graph. Player B wants a low expected value, because the expected valueis from Player A’s perspective. For any value of p (chosen by Player A), Player Bwants to be on the bottom of the lines which represent Player B’s alternatives. Inthis example this is represented by three line segments involving alternatives B1,


.1 .2 .3 .4 .5 .6 .7p

.8 .9

−1

−2

−3

−4

0

1

2

3

4

5

6

7

−1

−2

−3

−4

0

1

2

3

4

5

6

7

ExpectedValue ofthe Game

Figure 68: Example 1 – Axes


.1 .2 .3 .4 .5 .6 .7p

.8 .9

−1

−2

−3

−4

0

1

2

3

4

5

6

7

−1

−2

−3

−4

0

1

2

3

4

5

6

7


B1

Figure 69: Example 1 – Alternative B1


.1 .2 .3 .4 .5 .6 .7p

.8 .9

−1

−2

−3

−4

0

1

2

3

4

5

6

7

−1

−2

−3

−4

0

1

2

3

4

5

6

7


B1

B2

B3

B5

Figure 70: Example 1 – Alternatives B1, B2, B3, and B5


B2, and B3 (B5 is not best for any value of p). These line segments are highlightedin Figure 71.

Player A is aware of how Player B perceives the situation. Looking at thehighlighted line segments, Player A wants to be at the top, because this point hasthe highest expected value. This point is located where alternatives B1 and B2intercept. By inspection on the graph, we can see that optimal solution to thisgame occurs at p ≈ 0.42. The expected value of the game is approximately 1.This approximate solution is shown in Figure 72.

The graph has allowed us to eliminate two more alternatives because of localdominance. We are left with:

B1 B2A3 5 −2.5A4 −2 3.5

We have an approximate solution for p and V already, but because the gamehas been reduced to a 2 by 2, we can find these things as well as q exactly. Usingthe DARS algorithm we obtain:

B1 B2 D R SA3 5 −2.5 7.5 7.5

13 = 1526

1126

A4 −2 3.5 5.5 5.513 = 11

261526

D 7 6 13R 7

136

13S 6

137

13

We see that p is 1126 ≈ 0.423.

One way (of four) to calculate the expected value of the game is:

V = 613(5)+ 7

13(−2.5)

=3013− 17.5

13

=12.513

=2526

≈ 0.962


.1 .2 .3 .4 .5 .6 .7p

.8 .9

−1

−2

−3

−4

0

1

2

3

4

5

6

7

−1

−2

−3

−4

0

1

2

3

4

5

6

7


B1

B2

B3

B5

Figure 71: Example 1 – Player B’s Preferences Highlighted


.1 .2 .3 .4 .5 .6 .7p

.8 .9

−1

−2

−3

−4

0

1

2

3

4

5

6

7

−1

−2

−3

−4

0

1

2

3

4

5

6

7


B1

B2

B3

B5

�EV ≈ 1

p≈ 0.42

Figure 72: Example 1 – Approximate Optimal Solution for Player A


Recommendation

Player A should play A3 with probability 1126 , and A4 with probability 15

26 . PlayerB should play B1 with probability 6

13 , and B2 with probability 713 . The expected

value of the game is 0.962.

21.1.2 Example 2

B1 B2 B3 B4 B5A1 12 3 2 9 4A2 18 −5 −4 16 −7A3 17 −1 −2 16 3A4 0 14 12 −2 13A5 3 17 14 −6 15

The elimination of the rows and columns using dominance is a bit tricky forthis example, so we will take this slowly. First we do the row comparisons:A1,A2; A1,A3; A1,A4; A1,A5; A2,A3; A2,A4; A2,A5; A3,A4; A3,A5; andA4,A5. No dominance is found, so we move on to the columns: B1,B2; B1,B3,and finally we find dominance when comparing B1 and B4.

B1 B412 ≥ 918 ≥ 1617 ≥ 160 ≥ −23 ≥ −6

B1 (the higher payoff column) is eliminated, producing:

B1 B2 B3 B4 B5A1 12 3 2 9 4A2 18 −5 −4 16 −7A3 17 −1 −2 16 3A4 0 14 12 −2 13A5 3 17 14 −6 15

1B4


We then continue comparing pairs of columns: B2,B3; B2,B4; B2,B5; B3,B4;B3,B5; and B4,B5. Finding no dominance in any of these, we go back to the rows:A1,A2; A1,A3; A1,A4; A1,A5; and A2,A3. We see that A3 dominates A2 (theeliminated B1 column is ignored):

A2 – −5 −4 16 −7– ≤ ≤ ≤ ≤

A3 – −1 −2 16 3

Removing row A2 we obtain:

B1 B2 B3 B4 B5A1 12 3 2 9 4A2 18 −5 −4 16 −7 2 A3A3 17 −1 −2 16 3A4 0 14 12 −2 13A5 3 17 14 −6 15

1B4

There’s no further row dominance, and so we return to the columns. It can beseen that both columns B2 and B5 are dominated by B3. Removing B2 and B5gives us:

B1 B2 B3 B4 B5A1 12 3 2 9 4A2 18 −5 −4 16 −7 2 A3A3 17 −1 −2 16 3A4 0 14 12 −2 13A5 3 17 14 −6 15

1 3 4B4 B3 B3

There is no further dominance using this method. We have reduced the tabledown to two columns:

B3 B4A1 2 9A3 −2 16A4 12 −2A5 14 −6


In the reduced table, the payoffs range from −6 to 16. We let q be the proba-bility that Player B chooses B3, and 1−q is therefore the probability that PlayerB chooses B4.

B3 B4A1 2 9A3 −2 16A4 12 −2A5 14 −6

Prob. q 1−q

To solve the problem we make a graph for which the horizontal axis is theprobability q (0 to 1), and the vertical axis is the expected value of the game (−6to 16). The axes are shown in Figure 73.

If Player A chooses A1 the expected payoff is

V = 2q+9(1−q)

When q = 0, the expected value is 2(0)+9(1−0) = 9; when q = 1, the expectedvalue is 2(1)+9(1−1) = 2. We draw therefore a line from (0,9) to (1,2) and labelthis line A1. The shortcut to this process is this: the number on the left side of thetable goes on the right side of the graph, and the number on the right side of thetable goes on the left side of the graph.

This procedure is repeated for alternatives A3, A4, and A5. The four lines areshown in Figure 74.

Both Player A and Player B have the same information, so both are able todraw this graph. Player A wants a high expected value. For any value of q (cho-sen by Player B), Player A wants to be on the top of the lines which representPlayer A’s alternatives. In this example this is represented by three line segmentsinvolving alternatives A3, A4, and A5 (A1 is not best for any value of q). Theseline segments are highlighted in Figure 75.

Player B is aware of how Player A perceives the situation. Looking at thehighlighted line segments, Player B wants to be at the bottom, because this pointhas the lowest expected value (Player B must pay this amount to Player A). Thispoint is located where alternatives A3 and A4 intercept. By inspection on thegraph, we can see that optimal solution to this game occurs at q ≈ 0.56. Theexpected value of the game is approximately 5.9. This approximate solution isshown in Figure 76.

The graph has allowed us to eliminate two more alternatives because of localdominance. We are left with:


.1 .2 .3 .4 .5 .6 .7q

.8 .9

−2

−4

−6

0

2

4

6

8

10

12

14

16

−2

−4

−6

0

2

4

6

8

10

12

14

16


Figure 73: Example 2 – Axes


.1 .2 .3 .4 .5 .6 .7q

.8 .9

−2

−4

−6

0

2

4

6

8

10

12

14

16

−2

−4

−6

0

2

4

6

8

10

12

14

16


A1

A3

A4A

5

Figure 74: Example 2 – Alternatives A1, A3, A4, and A5


.1 .2 .3 .4 .5 .6 .7q

.8 .9

−2

−4

−6

0

2

4

6

8

10

12

14

16

−2

−4

−6

0

2

4

6

8

10

12

14

16


A1

A3

A4

A5

Figure 75: Example 2 – Player A’s Preferences Highlighted


.1 .2 .3 .4 .5 .6 .7q

.8 .9

−2

−4

−6

0

2

4

6

8

10

12

14

16

−2

−4

−6

0

2

4

6

8

10

12

14

16


A1

A3

A4A

5

�EV ≈ 5.9

q≈ 0.56

Figure 76: Example 2 – Approximate Optimal Solution for Player B


B3 B4A3 −2 16A4 12 −2

We have an approximate solution for q and V already, but because the gamehas been reduced to a 2 by 2, we can find these things as well as p exactly. Usingthe DARS algorithm we obtain:

B3 B4 D R SA3 −2 16 18 18

32 = 0.5625 0.4375A4 12 −2 14 14

32 = 0.4375 0.5625D 14 18 32R 0.4375 0.5625S 0.5625 0.4375

We see that q is 0.5625 (from the B3 column). [It’s just a coincidence that weobtained the same number in the A4 row.]

One way (of four) to calculate the expected value of the game is:

V = 12(0.5625)+(−2)0.4375= 6.75−0.875= 5.875

Recommendation

Player A should play A3 with probability 0.4375, and A4 with probability 0.5625.Player B should play B3 with probability 0.5625, and B4 with probability 0.4375.The expected value of the game is 5.875.

21.1.3 General Methodology

To use the graphical solution methodology, we must reduce by dominance to tworows with many columns, or two columns with many rows. In the two-rowsituation, we seek the apex point at top of the bottom set of line segments. In thetwo-column situation, we seek the apex point at the bottom of the top set of line

segments. The apex point defines the choices for the other player, and the DARSalgorithm can then be applied.


21.2 Formulation as a Linear Optimization ModelA zero-sum game of any size can be solved by linear optimization. Indeed, thereis no need to try to first reduce the size of the model using dominance. Here is anexample:

B1 B2 B3 B4A1 8 7 −5 1A2 6 9 4 −6A3 −4 −8 7 5

This table cannot be reduced, so we must use linear optimization. The param-eters of the model are the numbers in the payoff table, i.e. 8, 7 and so on. Thevariables are the probabilities and the expected value of the game. We define V =expected value of the game. The probabilities for Player A choosing A1, A2, andA3 are p1, p2, and p3 respectively, such that p1 + p2 + p3 = 1. The probabilitiesfor Player B choosing B1, B2, B3, and B4 are q1, q2, q3, and q4 respectively, suchthat q1 +q2 +q3 +q4 = 1.

B1 B2 B3 B4 Prob.A1 8 7 −5 1 p1A2 6 9 4 −6 p2A3 −4 −8 7 5 p3

Prob. q1 q2 q3 q4

Player A wants V to be as high as possible; the objective is simply to maximizeV .

If Player B chooses B1, then the expected value of the game will be:

V = 8p1 +6p2 +(−4)p3

Player B will certainly not exceed this amount on the right, but V could be lessthan this amount, because Player B does not have to choose alternative B1. Thealternative B1 constraint is that:

V ≤ 8p1 +6p2−4p3

Subtracting the right hand side from the left we obtain:

V −8p1−6p2 +4p3 ≤ 0


Multiplying by −1 we obtain:

alternative B1 −V + 8p1 + 6p2 − 4p3 ≥ 0

Going down the B2, B3, and B4 columns we obtain three more similar constraints:

alternative B2 −V + 7p1 + 9p2 − 8p3 ≥ 0alternative B3 −V − 5p1 + 4p2 + 7p3 ≥ 0alternative B4 −V + p1 − 6p2 + 5p3 ≥ 0

The probabilities must sum to 1:

prob. sum to 1 p1 + p2 + p3 = 1

Each probability must be greater than 0; these are the non-negativity restrictions.The expected value V , however, is unrestricted in sign. In writing the completeformulation, we will stress this fact. From Player A’s perspective, the model is:

maximize V

subject to

alternative B1 −V + 8p1 + 6p2 − 4p3 ≥ 0alternative B2 −V + 7p1 + 9p2 − 8p3 ≥ 0alternative B3 −V − 5p1 + 4p2 + 7p3 ≥ 0alternative B4 −V + p1 − 6p2 + 5p3 ≥ 0prob. sum to 1 p1 + p2 + p3 = 1

non-negativity p1 , p2 , p3 ≥ 0

V is unrestricted

We will call this Model A. Another related model is obtained by looking at thingsfrom the perspective of Player B, who wishes to minimize V . If Player A choosesA1, then the expected value of the game will be:

V = 8q1 +7q2 +(−5)q3 +q4

Player A will certainly not accept less than this amount on the right, but V couldbe more than this amount, because Player A does not have to choose alternativeA1. The alternative A1 constraint is that:

V ≥ 8q1 +7q2−5q3 +q4


Subtracting the right hand side from the left, and then multiplying by −1 weobtain:

alternative A1 −V + 8q1 + 7q2 − 5q3 + q4 ≤ 0

We also have similar constraints for alternatives A2 and A3, and we requirethat the probabilities sum to 1. As before, V is unrestricted. The model fromPlayer B’s perspective (called Model B) is:

minimize V

subject to

alternative A1 −V + 8q1 + 7q2 − 5q3 + q4 ≤ 0alternative A2 −V + 6q1 + 9q2 + 4q3 − 6q4 ≤ 0alternative A3 −V − 4q1 − 8q2 + 7q3 + 5q4 ≤ 0prob. sum to 1 q1 + q2 + q3 + q4 = 1

non-negativity q1 , q2 , q3 , q4 ≥ 0

V is unrestricted

The only thing new as far as linear optimization is concerned is the existenceof the unrestricted variable V in both models. On Excel, this means that we cannotuse the radio button to declare all variables to be non-negative. Instead, we mustdeclare each of the probability variable cells to be non-negative, by adding a rangeas a set of ≥ constraints with a right-hand side value of 0. For LINDO, we enterthe model as usual, and then after the end statement we add the command free V.For example, the first model is entered as

max Vstalt. B1) -V + 8p1 + 6p2 - 4p3 >= 0alt. B2) -V + 7p1 + 9p2 - 8p3 >= 0alt. B3) -V + -5p1 + 4p2 + 7p3 >= 0alt. B4) -V + p1 -6p2 + 5p3 >= 0sum to 1) p1 + p2 + p3 = 1endfree V

We can now use LINDO to solve both models. As one would expect, the samevalue of V is obtained for both. We also see that the dual prices (DP) from onemodel give us (except for the sign) the values of the variables for the other model:


Model A Model BV 1.367021 V 1.367021p1 0.425532 DP A1 0.425532p2 0.175532 DP A2 0.175532p3 0.398936 DP A3 0.398936

DP B1 0 q1 0DP B2 −0.319149 q2 0.319149DP B3 −0.257979 q3 0.257979DP B4 −0.422872 q4 0.422872

Therefore, all we need to do is formulate and solve either model. The valuesof the probability variables for the other model are simply the absolute values ofthe dual prices of the first model.

In a theoretical sense, game theory is an important link between probabilitybased payoff matrices and the deterministic field of linear optimization. Also, thetwo models form what are called the primal and the dual, which provide alternateways to solve what is really the same problem. Another thing of note is that if thelinear optimization model is applied to a situation where there are just two rowsor columns, we can create a graphical solution which is like the one seen earlier,in which the highlighted line segments are in fact edges of the feasible region.



For problem 3 from the previous Exercise, determine the optimal strategy for eachplayer and the value of the game.

21.3.2 Problem 2

Players A and B are playing a zero-sum game. The payoffs to player A are:

B1 B2 B3A1 5 0 1A2 1 −1 2A3 4 2 1A4 7 4 0A5 6 −1 1

(a) Eliminate any dominated alternatives.

(b) Plot the remaining alternatives on a graph.

(c) Use the analytic technique to determine the optimal strategy for each player,and state the value of the game.

21.3.3 Problem 3

Players X and Y are playing a zero-sum game. The payoffs to player X are:

Y1 Y2 Y3 Y4 Y5X1 5 −3 −2 7 2X2 −1 8 6.5 −2.5 7X3 −2 6 5 −4 9X4 −2 7 6 −3 5

(a) Eliminate any dominated alternatives.

(b) Plot the remaining alternatives on a graph.

(c) Use the analytic technique to determine the optimal strategy for each player,and state the value of the game.


21.3.4 Problem 4

Solve the following zero-sum game by linear optimization.

B1 B2 B3A1 2 6 −9A2 3 5 −1A3 −5 −10 1A4 0 5 −5A5 −2 −6 0

21.3.5 Problem 5

There is no dominance here. This is a somewhat challenging problem because ofthe tie in row A3.

B1 B2A1 2 14A2 10 −4A3 7 7

(a) Solve by linear optimization.

(b) Solve by hand using the graphical technique. How is this solution moregeneral than the one found in part (a)?


22 Moving Forward

We have come to the end of the content to be covered on the final examinationfor this course. Obviously there is more to learn for those who are interested. Tosearch further, it needs to be pointed out that the term management science is justone of many for this field. Other terms are operations (or operational) research,decision analysis/science, optimization and others. Using the terms operationsresearch and the management sciences, a very comprehensive site is located athttp://www.informs.org/Resources/. This website, maintained by Prof. MichaelTrick, has links to all sorts of things: courses from around the world; free com-puter programs; references to journals; and so on. Also included are referencesto many professional societies from around the world. The following two may beof particular interest. In Canada, the professional organization is called the Cana-dian Operational Research Society (or CORS). In the United States, there is theInstitute for Operations Research and the Management Sciences (or INFORMS).

There are many more topics in the field of management science than theones we covered, and of course what we did cover can be studied in greaterdepth. For example, two mainstream topics not covered here are simulationand queueing theory . Simulation is a methodology which allows the model tocontain many real-world characteristics. Traditionally, a major drawback of us-ing simulation was that it was very time consuming, but as the speed of computerscontinuously increases this is becoming less of an issue. However, simulation doesnot guarantee optimality, so it tends to be used only when there are no alternativeprocedures. Queueing theory studies the behaviour of queues (waiting lines) thatform in banks, at telephone call centres, and so on. Other topics include the studyof algorithms, non-linear optimization, and the formulation of more difficult lin-ear and integer optimization models. If learning about more advanced topics is ofinterest to you, then ask your professor what other courses are available to you.

Prof. Trick’s site has a resources section which links to a long list of MS/ORbooks. Here are some books which I have found to be useful (alphabetical bysurname):

Bell, Peter. Management Science/Operations Research: A Strategic Perspective,South-Western Publishing, 1999.Bell’s book emphasizes the use of cases, one of the few books in this field ever todo so. Bell is at the Ivey School at the University of Western Ontario; many of thecases are set in Canada.

Denardo, Eric V. The Science of Decision Making: A Problem-Based Approach

http://www.informs.org/Resources/

http://www.cors.ca

http://www.informs.org/

http://www.wiley-vch.de/publish/en/books/bySubjectIE00/ISBN0-471-31827-2


Using Excel, Wiley, 2002.This book covers the material seen in this e-book (and more), but at a deeper level.

Hillier and Lieberman, Introduction to Operations Research, 9th edition, McGraw-Hill, 2010.This book deals with all aspects of operations research.

Rardin, Ronald L., Optimization in Operations Research, Prentice-Hall, 1998.Rardin’s book deals with deterministic problems (no probabilities). There arethree of fourteen chapters which deal solely with formulation, but the emphasis ison the algorithms for solving these type of models.

Schrage, Linus E., Optimization Modeling with LINDO, 5th edition, DuxburyPress, 1997.Schrage’s book gives an extensive overview of the main types of applications forlinear and integer optimization, using LINDO to provide solutions.

Winston, Wayne L., Operations Research – Applications and Algorithms, 4th edi-tion, Duxbury Press, 2004.Winston’s book gives a comprehensive coverage of operations research.

This lecture is short because being the last one it is principally used for review,for questions, or for covering material that may have been missed earlier. Theexercises from the first twenty-one lectures will prepare you well for the finalexamination. Best of luck with it!



http://highered.mcgraw-hill.com/sites/0073376299/information_center_view0/

http://www.brookscole.com/cgi-wadsworth/course_products_wp.pl?fid=M2b&product_isbn_issn=0534380581&discipline_number=17

Indexabsorbing state, 253algebraic model, 155alternative branch, 47

binding constraint, 195BINOMDIST function, 137branch-and-bound algorithm, 194

certainty equivalent, 281cumulative probability function, 34

decision node, 47decision tree, 47dual price, 216

EMV, 279EOL, 31event node, 47Excel R©, iexpected opportunity loss, 31expected utility, 279expected value, 16expected value of perfect information,

25

game theory, 312

indifference, 39integer optimization, 167

LINDO R©, ilinear optimization, 154

marginal analysis, 34Markov chains, 237model

algebraic, 155

spreadsheet, 155

null branch, 48

OpenOffice.org 3 R©, i, iiopportunity loss, 31outcome branch, 47

payoff matrix, 14payoff node, 48posterior tree, 101, 107prior tree, 101, 107pure strategy, 313

queueing theory, 349

ranking profit/cost, 90reduced cost, 215regret matrix, 31risk

averse, 280neutral, 280preferring, 280

rollback procedure, 47

salvage value, 21sensitivity analysis, 36, 214shadow price, 216simplex algorithm, 194simulation, 349slack, 194solution variables, 215Solver, 203spreadsheet

functionsBINOMDIST, 137

model, 4

351


SUMPRODUCT, 18spreadsheet model, 155stable solution, 314state, 238state

absorbing, 253transient, 253trapping, 253

state transition diagram, 238state transition matrix, 242SUMPRODUCT, 18sunk cost, 13surplus, 194

transient state, 253transition, 238trapping state, 253tree

posterior, 107prior, 107

utility, 279


23 Supplement – Extra ProblemsThe problems which follow come from a collection prepared not only by me butby Dr. A.R. Redlack. They are reproduced here with his permission. The followthe order of the lectures, with each section covering more than one lecture, exceptfor the final section on Linear Optimization – Applications, which can be thoughtof as an Appendix to the earlier Linear Optimization – Introduction section. Thereis a Solutions Manual available for all these problems.


23.1 Introduction1. Suppose that minimizing your average annual cost is your only criterion for

deciding how often you should buy a new car. You are given the follow-ing information (in dollars), based on maintaining the car so that the car isalways safe to drive:

Year Resale Value Maintenance Cost(end of year) (during the year)

1 14,000 1002 10,000 2003 7,000 4004 5,000 6005 3,500 1,0006 2,500 1,5007 2,000 2,0008 1,500 2,5009 1,000 3,000

10 500 3,00011 0 3,000

A new car costs $20,000. Ignoring inflation, taxes, and so on, how oftenshould a new car be replaced?

2. A knight sees four veiled women. He knows that one is the Queen, but hedoes not know which one. The Queen always tells the truth, but the otherthree are rogues who always lie. The first woman says that she is the Queen.The second woman says that the fourth is the Queen. The third woman saysthat she is the Queen. The fourth woman says that the first is the Queen.Which one is the Queen?

3. Mr and Mrs Brown have four children, none of whom was adopted. A pub-lic official at city hall who has never met the Brown family enquires by tele-phone about the ages of the children. “Your first clue,” said Mr Brown (notwanting to be too cooperative), “is that the product of their ages (expressedas integers) is 36.” The official started to write down some possibilities andsaid, “Only one potential answer is biologically impossible: What’s the sec-ond clue?” Mr Brown continued, “the sum of their ages is equal to number


of councillors on city council,” which of course the official knew. “I still donot have enough information,” the official replied, “what is the third clue?”“The difference in ages between our oldest and our youngest child is an oddnumber,” replied Mr Brown. What are the ages of the Brown children?

4. A ‘3’ appears in both cells A1 and B1 of a spreadsheet.15 Someone enters=A1*B1 in cell C1, and 11 appears. Why?

5. On an island with a population of a million people a murder has occurred.All ports and airports were immediately sealed by the police so the murderermust be on the island. A violent struggle at the murder scene shows that theattacker was a left-handed male. (Left-handed males comprise 5% of thepopulation). Most important of all, in addition to the blood of the victim,some blood was found which comes from a very rare blood type found inonly one person in 10,000. An expert has said that there is a 99% chancethat this blood belongs to the murderer.

A criminal conviction requires that there be at least a 95% chance that theaccused actually committed the crime. The police have arrested Ralph, aman with a conviction record for theft, who does not have an alibi, who isleft-handed, and who has the rare blood type found at the scene of the crime.His lawyer explains to the police that he cannot be found guilty based on thecurrent evidence, and Ralph is released. What was the lawyer’s argument?

15We use the syntax of Microsoft Excel everywhere in this book.


23.1.1 Advanced Problems

1. The Panther Company uses 1,000 thermocouples in its production process.The thermocouples have a maximum life of five months, but often fail ear-lier. Past experience has shown that 10% will fail in the first month, 20%in the second month, 20% in the third month, 30% in the fourth month andthe remaining 20% in the fifth month. Each failed thermocouple must bereplaced by the end of the month in which it fails. If all the thermocouplesare replaced at the same time, then the cost is only $14 per thermocouple. Ifonly the failed thermocouples are replaced, the cost is $25 per thermocouplereplaced.

(a) What are the alternatives which must be evaluated?

(b) What is the cost associated with each alternative? Remember that thereplacement thermocouples are also subject to failure.

(c) What replacement policy would you recommend to the company?

2. On a television game show a contestant is told that behind one of threedoors there is a new car to be won; behind each of the other two doors thereis nothing of value. The contestant is invited to guess which door leads tothe car. After the guess, the host (who knows where the car is), opens oneof the other doors to reveal that the car is not behind it. The host then asksthe contestant if he or she wishes to change his or her guess. For example,if the contestant initially chooses door 2, and the host reveals that the car isnot behind door 1, should the contestant switch his or her guess to door 3?There are three possible answers to this question:

(i) the contestant should stick with his or her initial guess

(ii) it doesn’t matter whether the contestant switches or not

(iii) the contestant should switch.

Which is right?


23.2 Uncertainty

1. A business has four alternatives. After an alternative has been chosen, anevent occurs which has four mutually exclusive outcomes. The conditionalpayoffs are given in the following matrix.

O1 O2 O3 O4A1 8 6 3 12A2 14 −3 10 7A3 9 3 5 13A4 4 5 7 6

Prob. 0.3 0.4 0.2 0.1

In parts (a) to (d), determine which alternative and ranking payoff would bechosen according to the given decision criterion (the probabilities are to beignored). Parts (e) to (h) use the probabilities.

(a) Criterion of Optimism

(b) Criterion of Pessimism

(c) The Hurwicz criterion if the coefficient of optimism is 0.7.

(d) Laplace criterion

(e) What is the preferred alternative using maximum expected valueas the decision criterion?

(f) Determine the EVPI by the direct method.

(g) Find the regret matrix.

(h) Show that the optimal alternative using minimum EOL is thesame as the one found in part (e), and that the minimum EOLitself is the same as the EVPI found in part (f).

2. A business must choose between one of five alternatives. After the alter-native has been chosen, an event occurs. This event has three mutuallyexclusive outcomes. The conditional payoffs are given in the following ma-trix.


O1 O2 O3A1 32 31 34A2 45 −8 51A3 50 30 40A4 12 35 68A5 3 12 70

Prob. 0.45 0.02 0.53

In parts (a) to (d), determine which alternative and ranking payoff would bechosen according to the given decision criterion (the probabilities are to beignored). Parts (e) to (h) use the probabilities.





(e) What is the preferred alternative using maximum expected valueas the decision criterion?

(f) Determine the EVPI by the direct method.

(g) Find the regret matrix.

(h) Show that the optimal alternative using minimum EOL is thesame as the one found in part (e), and that the minimum EOLitself is the same as the EVPI found in part (f).

3. A computer contains several integrated circuit boards. Over the life of thecomputer there is a 55% chance that no boards will fail, a 25% chance thatone board will fail, a 15% chance that two boards will fail, and a 5% chancethat three boards will fail.

The vendor of the computer offers the purchaser a special deal on replace-ment boards – they can be purchased (in any quantity) for $200 each, pro-vided that they are purchased with the computer. The regular price (whichis available anytime) is $500 per board. The purchaser wishes to minimizethe expected cost of the replacement boards.


(a) Write a payoff matrix for the problem, where the payoffs repre-sent costs rather than profits.

(b) What is the optimal solution and what is its expected cost?

(c) Find the expected cost with perfect information, and from thisdeduce the EVPI.

(d) Set up the opportunity loss matrix for the problem, rememberingthat the original payoffs are already costs.

(e) Show that the minimum EOL = EVPI.

4. A friend is a frequent traveller between two airports. One, located in thetown of Gander, has little problem with adverse weather. The other, locatedin the city of St. John’s, is sometimes closed because very thick fog. As faras cost is concerned, there is little difference between travelling by bus orby plane. However, there is a significant difference in the travel time. Thebus trip takes 4 hours, whereas the plane trip depends on weather. If theweather is fine, the trip only takes 45 minutes. If St. John’s is fogged in,then one has to wait an average of 20 hours for it to clear. According tohistorical weather records for this time of year, St. John’s is clear 8 timesout of 10. Suppose that reservations are required in advance for either thebus or the plane; it is not possible to reserve both.

(a) What would you recommend to your friend?

(b) What is the probability of foggy weather that would imply indif-ference between the two alternatives?For those times that St. John’s is fogged in, suppose now there isa 20% chance that the weather will clear early and the wait willonly be 6 hours, a 70% chance that the wait will still be 20 hours,and a 10% chance that the fog will continue for an additional 10hours.

(c) Would this affect your recommendation to your friend?

5. The Fres Froot Company operates in a region of good soil and temperateclimate in which there are many fruit farms and vineyards. The companyowns a number of roadside fruit stands which sell to tourists in this region.Fres Froot purchases ripe peaches for 60 cents per kilogram (kg) and sellsthem for $1.00 per kg. Peaches which are not sold at the end of the day are


given to local farms to use as animal food. Demand for the peaches duringthe last year (120 day tourist season) was 40 kg on 40 days, 50 kg on 30days, 60 kg on 30 days and 70 kg on 20 days.

Solve parts (a) to (d) using the payoff matrix approach.

(a) What ordering policy would you recommend to Fres Froot?

(b) What is the EVPI?The Fres Froot Company has been contacted by a company whichbuys low quality fruit and ships it to another region of the countryin which the soil is thin and the weather often miserable. Insteadof giving it away for animal food, they will receive a salvagevalue of 10 cents per kg.

(c) Will this affect your recommendation to the Fres Froot Com-pany?

(d) At what salvage value(s) would you change from one recommen-dation to Fres Froot to a different one?

(e) Solve part (c) using marginal analysis.

6. A popcorn stand is faced with the decision of how many boxes of popcornto prepare before a hockey game at a local stadium. Each box cost $0.40and sells for $1.00. Past records indicate that 15,000 boxes are enoughto prevent any shortage, and until now, this has been the number preparedbefore each game. Unsold popcorn is disposed of in industrial garbage bagswhich cost $0.15 per bag and hold 50 boxes of popcorn. The following datasummarizes the sales history:

Demand Probability5,000 0.108,000 0.20

12,000 0.3015,000 0.40

What would you recommend to the popcorn stand?

7. Greenleaf Groceries buys fresh vegetables at wholesale for $5 per crate. Acrate which they sell at retail on the same day brings a $1.50 profit contribu-tion. A crate which is not sold on the same day is sold later as animal food


for $1 per crate. During the last year the demand for vegetables behaved asfollows:

Demand Number(in crates) of Days

10 12011 9012 7513 15

(a) What is the expected daily demand?

(b) How many crates should the store order if Greenleaf wants tomaximize expected daily profit from selling vegetables?

(c) Repeat the above using marginal analysis.

8. The owner of a store which sells mainly compact disks must decide whichdisks and the number of each to order for the Christmas sales season. A newrecording of Mozart’s Laudate Dominum is sweeping the world. Orders forthe disk must be placed with the distributor in lots of 100. If she orders100 disks, the cost to her would be $17 per disk; 200 disks would cost $15per disk, and 300 or more in lots of 100 would cost $13 per disk. UntilChristmas Day the retail selling price will be $20 per disk; any left overafter Christmas will be sold to a discount house for $5 per disk. The ownerbelieves that at the regular price the possible demands are 50, 100, 150, 200,250, 300, or 350 disks, with probabilities 0.05, 0.1, 0.2, 0.3, 0.2, 0.1, and0.05 respectively. She must place her entire order now. Assume that shewill suffer no loss of goodwill if she happens to be out of stock.

(a) What would you recommend to the owner?

(b) Determine the expected value of perfect information.

(c) Suppose that the $5 to be received for each leftover disk is nego-tiable. Over what range for this value would the solution foundin part (a) be valid?

9. A couple of enterprising business professors are faced with a dilemma.They have spent a lot of effort preparing a course manual and would liketo give themselves a small reward. They can produce and sell the manual


themselves or have the campus bookstore sell it (in which case they receivenothing). There are 250 students registered in the course, but used manualsare available from previous years. The cost to print the manual is $21, andthe sale price is $25. Any unsold manuals have no value, but if not enoughmanuals are ordered initially, then they have to be specially printed at a costof $26 per manual. The professors have estimated the possible demands fornew manuals to be (150, 175, 200, 225, 250) with probabilities (0.2, 0.3,0.3, 0.1, 0.1).

In addition, if they sell the manual themselves, then 10 manuals would havebe given free of charge to their fellow instructors and to the tutorial assis-tants of the course. If the bookstore sells the manual, then the free copiescome out of the departmental budget instead.

(a) What is the mathematical model for this problem?

(b) What would you recommend to the professors?

(c) What is the EVPI for this situation?

(d) What is the standard deviation of the payoffs for your recommen-dation?

10. The Wax n’ Wave Sporting Goods store must place their order for skis inAugust. The skis cost $75.00 a pair and sell for $125.00 a pair. Up to 50pairs of skis not sold during the winter can be sold during the big springsale for $50.00 a pair (any other unsold skis are worthless). The store mustorder skis in lots of 40 pairs. Based on the long-run forecast of the winterweather, the estimates of the demand for skis are as follows:

Demand Probability50 0.1075 0.50

100 0.30125 0.10

(a) What is the mathematical model for this problem?

(b) What would you recommend to the store?

(c) What is the EVPI for this situation?


(d) What is the standard deviation of the payoffs for your recommen-dation?

(e) Suppose that the above probabilities are changed to

P(demand = 75) = 0.5+∆pP(demand = 100) = 0.3−∆p

At what value of ∆p would the store be indifferent between or-dering 80 or 120 skis?

11. You stayed up late on Sunday night watching Alien and have overslept onMonday morning. It now looks as though you may be late for work. Youhave three options open to you:

– You could ride your bicycle and arrive 20 minutes late.

– You could drive your car. This would involve a cost for gasoline of $0.50each way and a $3.00 cost for parking. If the traffic is light you will arriveon time, but if it is medium or heavy you will arrive 15 minutes late.

– You could take the bus, which costs $1.00 per trip. If the traffic is light youwill arrive 10 minutes late; if it is medium, 20 minutes late, and if heavy, 30minutes late.

For this time of year, 30% of the time the traffic is light and 50% of the timeit is medium. It will cost you 30 cents for each minute you are late for work.If you take your bicycle or your car to work then you will return home bythe same; if you take the bus to work then you will walk home.

(a) What should you do?

(b) If you are unable to accurately estimate the probabilities of thetraffic levels, what would you do? Justify your answer.

12. A department store is considering carrying a line of baby grand pianos.They can be ordered from the manufacturer at cost of $6000 for the first oneand $4800 for each subsequent one. (e.g. three pianos would cost $15,600).The selling price until Christmas will be set at $8000. Any left over afterChristmas will be sold to a liquidator for $4000 each. Demand is estimatedas being between 2 and 5 inclusive with probabilities 0.15 for 2, 0.20 for 3,0.40 for 4, and 0.25 for 5.


(a) Using expected value as the decision criterion, use a payoff ma-trix to determine how many pianos should be ordered. What isthe expected profit?

(b) What is the EVPI?

(c) Suppose that the probabilities of demand being 2 or 3 are fixed,but the other probabilities can vary. At what values for the prob-abilities (if any) would the optimal alternative change?

13. The Surf ’n Sun Sports Shop must place their orders for snorkels in January.The snorkels sell for $15 each, and cost $10 each. Up to ten snorkels thatare not sold in the summer can be sold at $12 each in the Fall clearancesale (any unsold snorkels are worthless). The store must order snorkels bythe dozen. Based on the long-range forecast of summer weather, the storeestimates the demand for snorkels to be:

Demand Probability30 0.1040 0.5050 0.3060 0.10

(a) Using expected value as the decision criterion, use a payoff ma-trix to determine how many snorkels the store should order. Whatis the expected profit?


(c) Suppose that the demands for forty or fifty snorkels can vary,but the other two probabilities are fixed. At what values for theprobabilities (if any) would you be indifferent between orderingforty-eight and sixty snorkels?

14. Demand for a particular newspaper at Mrs. Reid’s News Depot can rangeanywhere from 31 to 49 papers per day, according to the following triangu-lar probability distribution:

P(D = d) = 0 if d ≤ 30

P(D = d) =d−30

100if 30≤ d ≤ 40


P(D = d) =50−d

100if 40≤ d ≤ 50

P(D = d) = 0 if d ≥ 50

where d is an integer, and D is a random variable which takes on the valuesof the number of newspapers demanded.

Mrs. Reid sells the papers for $0.80 each. Her buying price depends on thequantity ordered:

Quantity Buying Price(per day) (per paper)20 – 44 $0.60

45 or more $0.54

Papers that are not sold by the end of the day are sold to a recycling firm for5 cents each .

She wishes to know how many papers she should order so that her expectedprofit is maximized.

(a) Create a spreadsheet model for this problem.

(b) Obtain a graph of the expected profit as a function of the numberof papers ordered.

(c) What is the smallest value for her selling price for which she willorder 45 newspapers?

15. A business has four alternatives. After an alternative has been chosen, anevent occurs which has four mutually exclusive outcomes. The conditionalpayoffs are given in the following matrix.

O1 O2 O3 O4A1 7 4 5 8A2 13 9 3 5A3 11 8 6 3A4 7 15 −4 11

Prob. 0.1 0.3 0.4 0.2

In parts (a) to (e), determine which alternative and ranking payoff would bechosen according to the given decision criterion (the probabilities are to beignored). Parts (f), (g) and (h) use the probabilities.






(e) What is the preferred alternative using the regret matrix (ignoringthe probabilities and using pessimism)?

(f) What is the preferred alternative using the regret matrix (usingthe probabilities)?

(g) What is the preferred alternative using maximum expected valueas the decision criterion?

(h) Determine the EVPI using the payoff matrix.

16. A computer store is considering carrying a line of new computers. The man-ufacturer of the new computers is trying to encourage sales and is willing togive quantity discounts according to the following table:

Number PriceOrdered per Item

1 30002 27003 27004 24005 2400

The selling price will be set at $4000. Any left over will be repurchased bythe supplier for $2000 each. Demand is estimated as being between 2 and 5inclusive with probabilities 0.25 for 2, 0.30 for 3, 0.30 for 4, and 0.15 for 5.Any computers which are demanded but not in stock are ordered individu-ally from the supplier with an additional $50 charge for speedy delivery.

(a) Using expected value as the decision criterion, use a payoff ma-trix to determine how many computers should be ordered. Whatis the expected profit?



(c) Suppose that the probabilities of demand being 2 or 3 are fixed,but the other probabilities can vary. At what values for the prob-abilities (if any) would the optimal alternative change?

Uncertainty (Advanced Problems)

1. Assuming that cumulative density function F(D≤ d), the retail selling pricer, the wholesale buying price w, and the salvage value s are given, wherer ≥ w≥ s, prove the marginal analysis formula given in the text.

2. The situation is as in problem 14 of the previous section (with no salvagevalue), but now the probability density function is

P(K = k) =18!

k!(18− k)!qk(1−q)18−k

where K takes on the values of the number of newspapers demanded inexcess of 31, ranging from 0 to 18 inclusive.

(a) Determine the optimal number of papers to order when q = 0.3.

(b) What is the smallest value of q for which she will order 45 newspapers?

(c) Obtain a graph of the number of papers ordered as a function of q.


23.3 Decision Trees1. It is now October. An offshore oil development has just been announced,

and an entrepreneur is now considering the development of a housing estate.A piece of land is available for $2,000,000. Alternatively, she can purchasean option on the land for $32,000. If she buys the option she would thenhave the right to purchase the land on or before May 1 of next year for$2,150,000. The option cannot be re-sold and if it is not exercised on orbefore May 1 it will become worthless on May 2.

She is worried about the proposed oil development, fearing that there is a30% chance that it will not go ahead. By early April of next year the statusof the oil development will be known for sure. If the oil development goesahead, then she believes that the proposed piece of land will increase invalue to $2,400,000, otherwise the value will fall to $1,500,000.

(a) Given that she wishes to maximize her expected profit what shouldshe do?

(b) What is the value of the probability of the oil development notgoing ahead that would make her indifferent between the twoalternatives?

(c) What is the expected value of perfect information (EVPI)?

2. Nova Beauty Products has learned that an unexpected side-effect of one ofits products is that it seems to work as an excellent insect repellant. A ma-jor company in the repellant industry has offered to buy the rights (as theypertain to the repellant) of this product for $50,000, after which they willtry to isolate and market the relevant ingredient which has the repellant ef-fect. Nova is also considering trying to develop the repellant property itself.Nova estimates that the development would cost $80,000 and has only a70% chance of being successful. If the product is successfully developed,then Nova believes there is 20% chance of selling the rights for $230,000,otherwise they will sell them for $140,000. If the development is unsuccess-ful, then they believe they will still be able to sell the rights for $30,000.

(a) What would you recommend to Nova’s management?Suppose now that if the development turns out to be successful,that Nova will consider marketing the product itself. The market-ing department estimates that there are three levels of demand.


The marginal revenues associated with each level of demand areestimated to be $100,000, $180,000 or $280,000 with probabili-ties 0.30, 0.40 and 0.30.

(b) Would this affect your recommendation to Nova’s management?

3. Meni Idias Inc. must decide whether or not to pursue the development ofa new product for which its recent research has indicated possibilities. Itwill cost $200,000 to complete the research and there is a 75% chance thatthe research will be successful. The cost for setting up for production is$250,000. If the demand turns out to be high, medium, or low the operatingpayoffs will be $800,000, $500,000 and $100,000 respectively.

A competitor also has some knowledge of the product and if Meni Idias issuccessful in development, then there is a 20% chance that this competitorwill also be successful. If the competitor has also developed the product, theprobabilities of high, medium, and low demand are 0.1, 0.3 and 0.6, whereasthey are 0.5, 0.3, and 0.2 if the competitor does not develop the product.Unfortunately, information about the competition is not available before thepoint in time at which the $250,000 production setup cost is incurred.

(a) What would you recommend that Meni Idias do?Suppose now that Meni Idias can defer the production decisionuntil after it knows whether or not the competition has success-fully developed a similar product.

(b) Would this affect your recommendation to Meni Idias?

4. ABC Construction Co. has just learned that a tender for a significant projecthas just been issued. The cost of preparing a proposal for the tender isestimated to be $75,000. If they try to develop a proposal, there is a 5%chance that the company will find that it is unable to meet the requirementsof the tender. Assume that if this happens such a discovery will have beenmade after $30,000 of the $75,000 has been spent.

When submitting the bid, the company can either submit a low end bid or ahigh end bid. The low end bid will have an 80% chance of successfully ob-taining the contract whereas the high end bid will only have a 40% chanceof obtaining the contract. If accepted, a low end bid would give an incre-mental profit of $200,000, while a high end bid would give an incrementalprofit of $400,000. If the contract is obtained, there is a 10% chance that


unforeseen problems will add an unexpected $100,000 to the cost of theproject.

(a) What would you recommend to the management of ABC Con-struction Co.?

(b) What would be the EVPI for the unforeseen problems?

5. The Plastic Production Company needs to expand its production capacityfor the next year. The marketing department has determined that the com-pany will need 5,000, 10,000 or 15,000 cases of additional capacity withprobabilities of (0.3, 0.5, 0.2) respectively. The demand level will be knownafter two months.

The company is considering two options to meet the situation. First of all,they can use overtime at a cost of $3.00 per case in addition to the regularcost of $10.00 per case. However, if the required demand turns out to be15,000 units, then there will be a 50% chance that there will be labour unrestif overtime continues for more than six months, which would add another$2.00 per case to the production cost. The labour unrest, should it occur,will begin to be felt at the outset of the sixth month, but there would be noextra cost because of the unrest in this month.

Secondly, they can operate an additional shift. This will entail a fixed costof $15,000 for startup expenses and a $1.00 per case shift premium. How-ever, if the extra demand turns out to be only 5,000 cases, then they will berequired to layoff the shift at a cost of $5,000 and convert to overtime. Ifdesired, the additional shift could be started anytime instead of being startedimmediately.

(a) What would you recommend to the management of Plastic Pro-duction given that they must meet the required demand?

(b) Suppose a marketing research company offers to perform a sur-vey which will exactly determine the additional demand. Shouldthe company be willing to pay $1,000 for this information?

6. A machine shop has received an order for 2,000 units to be made on oneof its automated machines. This is a multistation operation and once it isstarted it runs without interruption. The machine shop makes $2 profit oneach part of acceptable quality. Each unit which is classified as “defective”


needs rework at a cost of $3.50 before it is considered “acceptable.” Thefull order must be acceptable.

Historical data indicates that if the machine is used without any specialpreparation, it produces either 1% defectives (this happens 50% of the time),2% defectives (30% of the time), 3% defectives (12% of the time) or 5%defectives the rest of the time.

A trial run can be performed at a cost of $30 to determine the defect rate.With a minor adjustment that costs $42, the defect rate above 2% can bereduced to 2%; with a major adjustment that costs $100, the defect rate canbe reduced to 1%.

What adjustment policy should the company have, given that they maketheir judgement based on expected monetary payoff?

7. An oil company is considering expanding one of its refineries. One phase ofthe construction requires the welding of a large number of seams. The initialwork will be done using ordinary welders assisted by apprentices. Once theinitial work has been completed, the company will have the option of usinga group of master welders to check and, if necessary, re-work the welds ofthe first group. This would cost an additional $20 per seam checked, plusanother $100 per seam re-worked. If any seams are checked, then all arechecked. The re-work, however, is applicable only to the seams which arechecked and found to be defective. (The master welders will always find adefective seam, and will never re-work a seam which is not defective.)

If no rework is done, then based on past experience, 2% of the time 1%of the seams will be defective, 50% of the time 5% of the seams will bedefective, 30% of the time 10% of the seams will be defective, and 18%of the time 20% of the seams will be defective. If the master welders havechecked and re-worked the defective seams then all seams are guaranteedto be non-defective.

If a defective seam is not discovered until after the construction work hasbeen completed, then there will be a cost of $1,000 for each defective seam.

(a) What would you recommend to the oil company?(b) If the percentage defective amongst all the seams could be reli-

ably estimated after only one tenth of the welds had been checked(and, if necessary, re-worked) by the master welders, would thisaffect the recommendation given in part (a)?


8. In the selection process for new products, one of the criteria which is oftenused is called the project value index (PVI). This is the ratio of the expectedrevenue of the project to the expected cost of the project.

The product must go through two stages (a) technical development and (b)commercial development. Let r be the revenue if the product is successful;if it is a failure, then assume that there is no revenue. Defining symbols forthe costs and probabilities which you think are relevant, set up a decisiontree and derive a formula for the PVI.

9. Smuth Tauker is the chief counsel for Peek-a-Boo Magazine. The com-pany is presently facing two related lawsuits for invasion of privacy. One isscheduled to be heard on July 2 and the other on November 5. The prepara-tion cost for either lawsuit is $15,000, but if both are done at the same time,then the combined cost will be only $25,000. Either suit can be settled outof court for $80,000, which will avoid the associated preparation cost forthat suit.

If a suit is won, then no further costs are incurred for that suit. If the first suitis lost, then the company will be required to pay $150,000. If the secondlawsuit is lost, then the court may decree only a minor penalty of $50,000or a major penalty of $200,000.

The probabilities of the outcomes of the second suit depend on the outcomeof the first suit as follows:

First Second SuitSuit Win Minor Major

Settled 0.30 0.30 0.40Won 0.60 0.30 0.10Lost 0.20 0.40 0.40

What would you recommend to Mr. Tauker, and what is the expected costof the recommendation to the company?

10. The government has indicated that it is planning to open up a new collegein either town A or town B. The decision will be made in one year’s time.

A large real estate developer is planning to develop a new subdivision in oneof these towns and has located land in both communities which is suitable.Land is being purchased by speculators and the developer needs to make his


decision on the land soon. He can take a one-year option on both locationsfor $100,000 or purchase one of them outright now. (The option would givehim the right to buy either property one year from now at the current price.)He does not want to commit the resources to buying both pieces of land.The value of the land is estimated to be as follows:

Town Present Future Value if TownValue Chosen Not Chosen

A 500,000 800,000 300,000B 1,000,000 1,400,000 600,000

The government has initiated an independent commission to recommendone of the sites. The report is due in eleven months (about a month beforethe expiry of the option) and although the recommendation is usually taken,other political factors sometimes cause the recommendation to be rejected.

The developer has hired a real estate consultant at a cost of $20,000 to helpwith his decision. The consultant reports that at the present time, town Ahas a slight edge over town B with a probability of being recommendedby the commission of 60% versus 40% for town B. In addition, based onthe results of previous commissions, the commission recommendation wasfollowed 80% of the time. The government will make its decision just priorto the expiration of the option.

As with most government initiatives, there is a chance that the governmentwill have to divert the funds elsewhere and the college will be postponedindefinitely. After the government has either accepted or rejected the rec-ommended site, and after the option will have expired, there is a 15% chancethat neither town will be awarded the new college.

What should the developer do?

11. A ballpoint pen manufacturer produces pens with an ink designed to flowfreely in sub-zero temperatures. The manufacturer produces the pens inbatches of 1000. Depending on factors in the manufacturing process, some-times the entire batch produced has pens with ink that have turned solid.This happens about ten percent of the time. If a batch has solid-ink pens,then the entire batch is worthless. Otherwise, the pens sell for $3.00 each.


A test can be performed on the raw materials before a batch is produced. Ifperformed at this time, the test would cost $50.00. If the test result is “neg-ative,” this guarantees that there will be no problems with the entire batch(the pens will not have solid ink). If the test result is “positive,” the solid-ink problem will affect the entire batch to be produced. In this situation,there are two alternatives. A chemical can be added to the raw materials ata cost of $2000 which will completely correct the problem. Alternatively,the ink can be heated at a negligible cost, which will partially correct theproblem. Pens made with heated ink will work at normal temperatures, butnot at sub-zero temperatures. Such pens sell for $0.79 each.

The test can also be done when the batch is half completed. It would costonly $20.00 if performed at this time. If the test result is positive, the pensalready produced are not salvageable, but either the chemical or the heatingprocess can be applied to the remaining half of the batch. The effects wouldbe the same as before but adding the chemical will cost only $1250.

Use a decision tree to determine what the manufacturer should do.

12. A machine shop has received an order for 4,000 units to be made on oneof its automated machines. Once started the machine runs without interrup-tion. The shop makes a profit of $5 on each part of acceptable quality. Eachunit which is classified as “defective” needs rework at a cost of $9 before itis considered to be “acceptable.” The full order must be acceptable.

Historical data indicates that if the machine is used without any specialpreparation, it produces either 1% defectives (this happens 30% of the time),2% defectives (45% of the time), or 5% defectives the rest of the time.

A trial run can be performed at a cost of $200 to determine the defect rate.With a minor adjustment that costs $100, the defect rate above 2% can bereduced to 2%; with a major adjustment that costs $500, the defect rate canbe reduced to 1%.

Use a decision tree to find the optimal adjustment policy.


23.4 Imperfect Information1. Major Motors produces a line of passenger cars. Ninety-seven per cent of

the production is of high quality, requiring only $500 in warranty work, butthe rest are “lemons.” If a customer receives a lemon it will cost MajorMotors $10,000 to replace it with a high quality car (this includes the sub-sequent $500 to be spent on the new car). Adding a re-work section to theassembly line would guarantee that each car would be of high quality, butthis would cost $400 per car.

(a) Should the company re-work the cars?

(b) What is the EVPI?Suppose that the company can inspect each car at a cost of $50(per car) before deciding whether or not to re-work it. The re-sults of the inspection would be one of the following: “looksbad,” “inconclusive,” or “looks good.” If the inspected car is ofhigh quality, then there is an 80% chance that the inspection willindicate “looks good,” and a 20% chance that the test will be in-conclusive. If the inspected car is a lemon then there is a 75%chance of a “looks bad” result, otherwise the inspection will beinconclusive.

(c) What should the company do now?

2. A publishing company is considering whether or not to publish a manuscriptfrom an unknown author. To develop a manuscript into a book would cost$50,000. In the absence of an external review of the manuscript, past expe-rience suggests that when the author is unknown there is only a 5% chancethat the book would be a major success, and a 15% chance that the bookwould be a minor success. If the book is published and if it turns out to bea major success, then it will generate $500,000 in revenue for the company,while a minor success will generate $100,000. If the book turns out to be afailure then it will generate no revenue.

(a) Should the company develop the manuscript into a book?

(b) What is the EVPI?The company can send the book to a reviewer who will either“recommend” or “not recommend.” If the book would be a ma-jor success, then there is a 90% chance that the reviewer will


recommend that the manuscript be developed as a book. If thebook would be a minor success, then the probability that he willrecommend is 60%. If the book would be a failure, then there isan 80% chance that he will not recommend.

(c) If the reviewer demands a fee for his services, what is the mostthat the publishing company would be willing to pay?

3. J.R. Blossom is a prolific author of murder mysteries. She holds the copy-right on her works, and she is now considering whether she should sell theexclusive film rights to her novel Murder on Wall Street to a movie com-pany or a TV network. If she signs with the TV network, then she wouldreceive a fixed payment, regardless of the ratings. The movie company, onthe other hand, is offering a royalty based on box office performance. Thepayoffs can be summarized as follows:

Box OfficeAlternative Small Medium LargeMovie Co. 100,000 1,000,000 3,000,000

TV Network 1,100,000 1,100,000 1,100,000Prob. 0.3 0.5 0.2

(a) Based on the above information, to whom should Ms Blossomsell the rights?

(b) What is the EVPI?She also knows of a market research firm specializing in movieproductions. For $25,000, the firm will give either a “looksfavourable” or “looks unfavourable” recommendation. The trackrecord of the firm can be summarized as follows:

Actual “Looks “LooksOutcome Favourable” Unfavourable”

Small 0.20 0.80Medium 0.60 0.40

Large 0.80 0.20

(c) What would you recommend that Ms Blossom do?

(d) What is the EVII?


4. Due to recent mild weather, a ski resort operator must prepare his slopesfor the weekend crowd using artificial snow. He estimates that it will costhim about $1,000 per slope. From past experience, he estimates that onlythree or four slopes will be needed at this time. If the turnout is “light,” thenthe weekend receipts will be $10,000. If the turnout is “moderate,” thenthree slopes will give receipts of about $13,000 whereas four slopes willgive $14,000. If the turnout is “heavy,” then three slopes will give receiptsof $15,000 whereas four slopes will give receipts of $20,000. At presenthe estimates the probabilities of having “light,” “moderate” and “heavy”turnouts to be 0.5, 0.4 and 0.1 respectively.

(a) With only the above information, what would you recommend tothe operator?

(b) What is the EVPI?A small marketing research firm will carry out a quick surveyof sports shops to determine recent activity. They will reportthe level of activity as either “low” or “high.” Past experiencehas shown that when the turnout was “light,” the firm reported“low” 85% of the time, when the turnout was “moderate” thefirm reported “high” 60% of the time, and when the turnout was“heavy” the firm reported “high” 90% of the time. The cost ofthe survey is $200.

(c) What would you recommend to the operator?

(d) What is the EVII?

5. A coin is lying on the table with a head facing up. It is either two-headedor a normal coin. You are asked to guess the type of coin. If you guesscorrectly you win $2.00, but you lose $1.00 if you guess incorrectly. Youhave the opportunity of having an impartial observer flip the coin and tellyou if it came up heads or tails. The flip, if performed, would cost you$0.20.

(a) What should you do?

(b) What is the value of the flip?

(c) Suppose you can elect to buy either one or two flips at a cost of$0.20 per flip before any flipping takes place? What should youdo?


6. Consider the situation as in the previous problem, but now you know thatthe coin has been picked at random from a pile of five coins with two ofthem being two-headed. The coin has been placed with its head (or one ofits two heads) face up. Solve as per parts (a), (b), and (c) above.

7. A Management Science student feels that things are not going very well.He thinks that he has only a 30% chance of passing the course. The actualmark is not too important if he manages to obtain the pass. A special tutorwill cost $100 and will increase his chance of passing to 90%. He valuespassing the course to be worth $500.

Some enterprising Management Science graduates have developed a pretestwhich helps predict how well students will do on the final exam. (Assumethat the final mark in the course is based only on the final exam). Theresult of the pretest is either “pass” or “fail.” For the students who do notsubsequently use a tutor, the pre-test has been found to be reliable 18 timesof the last 20 times it was given. For example, if 20 students who do notsubsequently use a tutor are destined to fail the course, then 18 of themwould fail the pre-test.

Assume that if a student who takes the pre-test subsequently uses a tutor,then the chance of passing the final exam is still 90%, regardless of whetherthe student passed or failed the pre-test. It costs $10 to take the pretest.

What would you recommend that this student do?

8. Management at Toys, Inc., must decide whether the company should pro-duce a new toy. They feel that there is a 60% chance that the toy will besuccessful. If the product is successful, then the firm will earn $600,000.Otherwise an unsuccessful toy will involve a loss of $400,000.

Two market survey firms have submitted bids for interviewing toy storeowners. R Inc., is known to conduct highly accurate surveys at high cost.Its fee is $40,000 per report containing data on the owners’ impression ofthe product. Its past studies show the following results:

- When a product has been successful, 80% of the interviewed owners hadbeen favourably impressed;

- When a product has been unsuccessful, 75% of the interviewed ownershad been unimpressed.


I Inc. has a lower fee but provides less accurate information. It wants$20,000 for its report on the owners’ impression of the product. Its paststudies show the following results:

- When a product has been successful, 60% of the interviewed owners hada positive impression and 25% were neutral;

- When a product has been unsuccessful, 30% of the interviewed ownershad a neutral impression and 50% were negative.

(a) What would you recommend to the management of Toys, Inc.assuming that at most one survey would be undertaken? What isthe EVPI?

(b) What are the efficiencies of the two surveys?

9. You would like to pass Management Science, but you are not exactly sureif you will!

You have the opportunity of doing an Exam Preparation Session before writ-ing the exam. You estimate the chance of passing the course without anyhelp to be 60%. However, the uncertainty as to what you are missing willgive rise to mental anguish costs of about $7. The exam preparation sessionwill leave you with a feeling of confidence, mild confusion or despair. Frompast sessions the following results were observed:

Actual Feeling after sessionResult Confident Confusion DespairPass 0.60 0.30 0.10Fail 0.20 0.30 0.50

It has assessed that the cost associated with failing is $30 (need for alcoholto calm your shattered nerves and ego) and the bonus of passing is $25.

How much would you be willing to pay to take the exam prep session?

10. Janet Descartes owns 1000 shares of International Automated Machines, acompany whose stock symbol is IAM and whose slogan is “I think, there-fore IAM.” Her mortgage is up for renewal next week and she wishes to sellthese shares so that she can pay down her mortgage. IAM’s earnings are


due out in three days time. She must decide whether to sell these shares justprior to or just after the earnings announcement.

The current price of the stock is $15 per share. Assume that this will remainconstant until the earnings announcement. Assume that after the announce-ment the price will be either $10, $13, $15, $17, or $20 per share withprobabilities of 0.1, 0.2, 0.3, 0.2 and 0.2 respectively.

(a) What would you recommend to Ms Descartes and what wouldbe the expected payoff?

(b) What is the expected value of perfect information?

(c) Ms Descartes can obtain some information on IAM. For a fee,a financial research company will investigate IAM and will pro-vide one of the following three predictions:

(1) earnings will be worse than expected(2) earnings will be as expected(3) earnings will be better than expected.

Based on their past predictive ability, Ms Descartes will use thefollowing conditional probabilities:

Price Earnings Forecastper Share “worse” “as expected” “better”

$10 0.90 0.09 0.01$13 0.70 0.25 0.05$15 0.30 0.40 0.30$17 0.10 0.15 0.75$20 0.02 0.05 0.93

What is the highest fee that Ms Descartes would be willing topay for such information?

11. A farmer who lives not too far from a large city has been approached by acircus and a rock concert promoter with offers to rent land. Normal profitfrom the land would be $5,000. The circus, held under a tent, will pay thefarmer $10,000 to rent the land. The rock concert promoter will pay $5,000+ 10% of the ticket sales. Ticket sales depend on the weather as follows:$100,000 if sunny, $70,000 if cloudy, and $20,000 if rainy. From past ex-perience the farmer estimates that the probabilities are 0.60, 0.30, 0.10 for


sun, cloud and rain respectively. The circus and rock concert promoter areresponsible for all expenses associated with each respective event.

(a) What would be your recommendation to the farmer?

(b) What is the EVPI?To better predict the weather next summer, the national metero-logical service will provide the farmer with a long range forecast.There is a cost of $200.00 to receive the forecast this far in ad-vance. The weather forecasters have provided the farmer withthe following summary of their previous “track record” for pre-dicting weather this far in advance.

Actual PredictedSun Cloud Rain

Sun 0.8 0.1 0.1Cloud 0.3 0.6 0.1Rain 0.1 0.2 0.7

(c) What is your recommendation for the farmer now?

(d) What is the standard deviation of the payoff for your recommen-dation?

(e) What is the EVII?

For problems 12 to 24 inclusive, the solutions first give the recommendedalternative in the absence of the testing information, secondly they give theEVPI, and thirdly they provide the overall recommendation considering thetesting information; only the latter is provided in the “Answers” section atthe back of the book.

12. An oil company owns a plot of land under which there might be a pocketof oil. The chance that a hole drilled on this land will find oil is about 1%.Such a hole would cost $700,000, and if oil is found assume that it has avalue of $40,000,000. At a cost of $50,000, a seismic test could be per-formed. The results of a seismic would be either “positive,” “inconclusive,”or “negative.” In the area in which the land is located there is 70% chance


of a positive reading if there really is oil present, a 25% chance of an incon-clusive reading, and a 5% chance of a negative reading. If there really isn’tany oil, then there is a 10% chance of a positive reading, a 35% chance ofan inconclusive reading, and a 55% chance of a negative reading.

What should the oil company do?

13. A trawler is fishing in international waters. A deep-water net from thistrawler becomes tangled in a sunken sailing ship which lies on the oceanfloor. Noting the exact latitude and longitude, the captain does some re-search when he arrives back in port. He discovers that eight sailing shipshave sunk in that vicinity over the past several centuries. One of these eightships was a Spanish galleon laden with gold worth about $10,000,000 at to-day’s prices. In the absence of any other information he believes that thereis one chance in eight (0.125) that he has found the Spanish galleon. Heassumes that the other seven ships contain nothing of value.

The depth of the water precludes scuba diving. However, he could hirea submarine to search the ship and, if applicable, retrieve the gold. Thiswould cost $800,000 if only searching were performed. There would be anadditional charge of $1,200,000 to retrieve the gold if the ship turns out tobe the galleon.

The captain learns from a scientist friend of a procedure which can be usedto do underwater testing of the presence of gold. This test would involvedropping a cable with an electronic probe from the side of the trawler. Giventhat some monitoring equipment needs to be on the trawler, the captainwould need to make a special trip to the location. Adding up the costsinvolved, the captain figures that the test will cost about $50,000.

The test is not perfect. It will give one of three readings, “strong,” “medium,”or “weak.” When the probe is near gold there is a 50% chance of a “strong”reading, a 40% chance of a “medium” reading, and a 10% chance of a“weak” reading. When the probe is not near gold there is a 5% chanceof a “strong” reading, a 25% chance of a “medium” reading, and a 70%chance of a “weak” reading.

What is your recommendation to the captain?

14. Nu Products Inc. has recently developed a new product and is in the processof trying to develop a marketing strategy. They have come up with twomarketing strategies – Plan A and Plan B. Unfortunately, the success of


the strategies depends on the market receptiveness. Management estimatesthat there is a 60% chance that the market would be receptive to Plan B,otherwise the market would be receptive to Plan A.

If the market turns out to be receptive to Plan A, then the payoff would be$500,000 if Plan A had been used, but the payoff would be only $300,000if Plan B had been used. Similarly, if the market turns out to be receptive toPlan B, then the payoff would be $450,000 if Plan B had been used, but thepayoff would be only $200,000 otherwise.

There is a small survey available which will recommend using either PlanA or Plan B. The survey is 80% reliable. In other words, if Plan A willbe preferred by the market, there is an 80% chance that the survey willrecommend that Plan A be used; if Plan B will be preferred by the market,there is an 80% chance that the survey will recommend that Plan B be used.The cost of the survey is $10,000.

What would you recommend to Nu Products?

15. A thoughtful young man has decided that it would be advantageous to hiscareer path if he obtained an M.B.A. degree. The universities he is consid-ering are: OWU which concentrates on the case approach, and CBU whichconcentrates on the theoretical approach. A government study has shownthat in two years time (i.e. upon graduation), there is a 60% chance thatemployers will prefer a graduate from a case program. If this occurs, agraduate of OWU will receive a starting salary of $38,000 whereas a grad-uate of CBU will receive only $35,500. However, if employers prefer agraduate from a theory program, then a graduate of OWU will receive only$34,000 but a graduate of CBU will receive $39,000. We will assume thatthe salaries in later years are independent of the starting salary.

Before making his final decision, a friend informs him that there is a gen-uine futurist who has a reliability of 95% on his predictions. However, thefuturist charges $1,000 per prediction and gives only one prediction per cus-tomer.

Assuming that the costs of obtaining the degree and other non-monetaryfactors are the same at each of the universities, what should the young mando?

16. General Lemmor has run into some unexpected resistance of unknown strength.If he launches an all out assault on the position with his existing forces, then


he is sure that he will wipe out the position but the casualties might be ex-cessive. If the position is well fortified, then he will sustain about 2,000casualties whereas if it is lightly defended he will sustain only 50 casual-ties. He thinks that there is about a 30% chance that it is well fortified. If hewaits for reinforcements, then an all out assault will only sustain about 100casualties if it is well fortified and no casualties if it is lightly defended. Un-fortunately time is of the essence and waiting for reinforcements will giverise to an estimated 1,000 casualties later.

General Lemmor has the option of launching a probing attack. The lossesincurred during this attack will indicate to him how well the position isfortified. If it is well fortified he will sustain 100 casualties with probability0.9, but only 10 casualties otherwise. If is lightly defended then he willsustain 100 casualties with probability 0.05 but only 10 casualties otherwise.

What advice would you give General Lemmor?

17. A company is in a position to invest in a new venture which has only twooutcomes - a gain of $25,000 with a probability of 0.60 or a loss of $15,000with a probability of 0.40. The company has the option of consulting aninvestment firm which will either recommend or not recommend investing.The past experience of the investment firm is as follows:

Firm recommendsInvestment to invest not investSuccessful 0.70 0.30

Failure 0.35 0.65

The investment firm has two plans - (i) an upfront charge of $2,000 or (ii)no upfront charge but instead a fee of $500 if the company does not invest,or a fee of $3,000 if the company does invest.

(a) What would you recommend to the company?

(b) What is the expected fee which would be paid if the second planwere to be used?

18. At the outset of each working day, an assembly line is set up to producecolour picture tubes. The daily production capacity is 50 units. Normally,the line will produce defectives according to a Bernoulli process with p =


0.02 (this happens about 85% of the time). On the other days, however,defectives are produced according to a Bernoulli process with p = 0.30. Aninspection station can test up to two picture tubes, at a cost of $55 per tubetested. The test is not destructive (hence any tubes which are tested andare found to be acceptable will be sent to a customer). Acceptable tubesgenerate a revenue of $120 each. If a customer receives a defective tube, norevenue is received – instead there is a cost of $40 whenever this happens.A resetting procedure is available which will guarantee that no defectiveswill be produced. Resetting the assembly line costs $500 and reduces thatday’s capacity by three units.

Hint: Use a payoff node whenever one or two of the tested tubes are foundto be acceptable, since there is a revenue of $120 for each of these regardlessof what else happens.

Determine the optimal number of tubes to test, if any, and the total expectedcontribution to profit associated with such a policy.

19. The Sultan of Samarkan has decided to give you a bonus for the tremendousjob you did using management science techniques to solve his problems. Hehas come up with the following proposal.

In front of you are two identical urns - one is filled with 1000 gold coinsand the other with 600 gold coins and 400 lead coins. If you correctly tellthe Sultan the contents of one of the urns you will receive both urns. If youguess incorrectly, then you will receive only 100 gold coins as a consolationprize.

If you wish, you may have one of the Sultan’s attendants take one coinfrom one of the urns and show it to you. However, this will subtract 500gold coins from the total number of 1600 should you guess correctly, orwill subtract 20 gold coins from the consolation prize should you guessincorrectly.

What is your optimal course of action?

20. Researchers at Chembioco have just discovered a vaccine which retards theAIDS virus. They are now in a race against time to produce the vaccinein large quantities. It is known that either process X or process Y will besuccessful but not both. The probability that X will be successful is 30%.It will take two years to develop and try process X; it will take one year


to develop and try process Y. Due to a restriction on the amount of labspace available, the two processes cannot be developed simultaneously. Forexample, if X is tried and is found to be unsuccessful, an additional yearmust be spent to develop process Y (even though you know that it will besuccessful).

It is possible to perform an experiment which will give an indication of thecorrect process. It takes three months to perform and predicts X 80% of thetime that X will be successful and predicts Y 90% of the time that Y will besuccessful. Because of the shortage of lab space, the experiment cannot beperformed simultaneously with either of the processes.

What would your recommend to Chembioco?

21. First Rate Inc. is having some minor production problems. The productionquality for one of the components of the final product is quite variable fromday to day. The recent experience has been that on 40% of the days thereis a 1% defect rate, on 50% of the days there is a 4% defect rate, and onthe remaining days there is a 10% defect rate. The daily production hasaveraged 1000 units and each defective unit sent back costs the company$100 in rework and other charges. Each of the components can be reworkedbefore putting them into the final product at a cost of $4 per unit. Thisrework will guarantee that the component is reliable.

The company has at its disposal a testing procedure which determines thereliability of the component. Unfortunately it destroys the tested unit(s) fora net cost to the company of $50 per unit tested.

What would you recommend to First Rate to improve the quality of its prod-uct?

22. An electronics manufacturer is having varying production quality for theintegrated circuits which it produces. Sixty-seven integrated circuits areneeded per day. The plant has a capacity to produce seventy per day. On“normal” days, the number of defective integrated circuits produced hasa binomial distribution with a percentage defectives of 4%. Sometimes,however, the machine which produces the IC’s has an “erratic” day and thepercentage defectives increases to 30%. “Erratic” days occur about 12% ofthe time.

If a defective IC is soldered to any other part, it results in a $8 charge torework. If desired, the IC’s can be reworked before soldering, guaranteeing


that they will not be defective at a cost of $2 per IC.

There is a destructive test procedure available which will indicate whetheror not the IC being tested is defective. To produce an IC in excess of therequired 67 would cost $1.50 per IC, and the test would cost $1 per unittested.

What should the manufacturer do?

23. A microcomputer chip production line produces large batches of chips eachday. In 80% of the batches there are no defects, and the remaining 20%have 30% defectives. A good batch received by a customer is worth $5,000to the company, but a bad batch costs $2,000.

If you wish, you may non-destructively test one or two chips at a cost of$100 per chip tested which will clearly identify the chip as being defectiveor not. At any time the batch may be destroyed at the plant, at a cost of$500.

What would you recommend to the company?

24. A company is considering introducing a new product and has two prod-ucts under consideration. Product 1 will cost $10,000,000 to develop andhas possible demand levels of high, medium and low with net payoffs (ex-cluding development costs) of $15,000,000, $5,000,000 and −$5,000,000respectively. The probabilities associated with the product are 0.5, 0.4 and0.1 for each demand level respectively. Product 2 will cost $5,000,000 todevelop and has possible demand levels of good and poor with probabilities0.6 and 0.4 respectively and with payoffs of $7,000,000 and −$2,000,000.

If the market response is unfavourable, then the company can improve theproduct further. Product 1 will cost $10,000,000 and product 2 will cost$2,000,000. For product 1 the effect of this improvement will be to changethe probabilities of high and medium demand to be 0.3 and 0.7 respectively,whereas for product 2 it will change the uncertainty of good and poor de-mands to be equally likely.

A marketing research company has a survey available which will provideinformation on the market response. In the past for those times that a prod-uct made a net profit, the survey predicted a “go” result 80% of the time andwhen a loss was incurred, the survey predicted a “no go” result 70% of thetime. The cost of the survey is $15,000.


What would you recommend to the management? What are the expectedvalues of imperfect information?

Problems Involving Sequential Bayesian RevisionThe rest of the problems in this section require that two revisions be donein series. Some of the technical aspects of this are covered in the appendixto the chapter.

25. A company is considering implementing a new quality control system. Oneof its machines normally produces only 2% defectives, but occasionally(10% of the time) it produces 8% defectives.

A test can be performed which will give a reading of “good”, “inconclusive”or “poor”. If the setup is normal, it will give a “good” reading 60% of thetime and a “poor” reading 10% of the time. If the setup is abnormal, it willgive an “inconclusive” reading 50% of the time and a “poor” reading 40%of the time.

(a) What is the probability that the first test will give an “inconclu-sive” reading?

(b) If both the first and second readings are “inconclusive”, what arethe probabilities of normal and abnormal setups?

26. A coin lying on a table is either a two-headed coin or a fair coin. The coinhas a head facing up. You are asked to guess what kind of coin it is. If youguess correctly, you win $2. If you guess incorrectly, you lose $1.

You have the opportunity of having an impartial observer flip the coin upto three times and tell you after each flip if it came up heads or tails. Eachflip, if performed, would cost $0.20. You will defer the decision about sub-sequent flips until you know what happens on the current flip.

(a) If a flip comes up tails, what can you say about the coin?

(b) Using the result from part (a), draw a decision tree and determineyour optimal action.

27. ABC Corp has developed but not produced a new product and must decidewhether or not to market it. The possible demands are characterized as high,medium, and low with marginal profits respectively of $500,000, $100,000


and −$300,000. Management assesses the probability of high demand tobe 20% and medium demand to be 40%.

Two surveys, A and B, are available. Survey A costs only $10,000 and willgive a recommendation of “go” or “no go.” Based on past experience, a“go” recommendation was given 70% of the times when the demand washigh, 50% of the times when the demand was medium and 30% of the timeswhen the demand was low. A “no go” recommendation was given 30% ofthe times when the demand was high, 50% of the times when the demandwas medium and 70% of the times when the demand was low. Survey Bcosts $50,000 and gives predictions on the actual demand. Past experienceis given in the following table.

Predicted Actual demandDemand High Medium LowHigh 0.90 0.30 0.10Medium 0.10 0.50 0.40Low 0.00 0.20 0.50

Survey B can be used by itself or following survey A. Survey A will not bedone if survey B has already been done.

What would you recommend?

28. A Type I urn contains four red balls and one blue ball. A Type II urn con-tains one red ball and three blue balls. You are invited to play a game inwhich there is one urn which is either Type I or Type II. The object of thegame is to correctly identify the type of urn. It costs $5.75 to play the game.The prize for correctly identifying the type of urn is $10.00.

If you wish, you may draw one or two balls from the urn. If a second ball isdrawn, it is only after you have seen the first ball. It costs $2.00 to draw thefirst ball. If after seeing this ball you then decide to draw another, then thecost to draw it is $0.40.

(a) Suppose that you have decided to draw a ball, and after observingthis ball you decide to draw another. Without doing any calcula-tions, what can you conclude about the urn if (i) both balls turnout to be red? (ii) both balls turn out to be blue?


(b) Using the results from part (a), draw the decision tree, and deter-mine how, if at all, the game should be played.

29. Pleasure Craft Inc. produces a line of power boats. Ninety-five per cent ofthe production is of high quality, requiring only $2000 in warranty work,but the rest are “lemons.” If a customer receives a lemon it will cost Plea-sure Craft $40,000 to replace it with a high quality boat (this includes thesubsequent $2000 to be spent on the new boat). Adding a re-work section tothe assembly line would guarantee that each boat would be of high quality,but this would cost $1500 per boat.

(a) Should the company re-work the boats?

(b) What is the EVPI?Suppose that the company can inspect each boat at a cost of $200(per boat) before deciding whether or not to re-work it. The re-sults of the inspection at this station would be one of the fol-lowing: “looks bad,” “inconclusive,” or “looks good.” If the in-spected boat is of high quality, then there is an 80% chance thatthe inspection will indicate “looks good,” a 15% chance that thetest will be inconclusive, and a 5% chance of a “looks bad” re-sult. If the inspected boat is a lemon then there is a 79% chanceof a “looks bad” result, a 20% chance that the inspection will beinconclusive, and a 1% chance of “looks good” result.

(c) What should the company do now?In addition to the inspection station mentioned above, PleasureCraft can add a second inspection station (which can only inspecta boat which was inspected at the first station). The result of theinspection at the second station is reported as being either “pass”or “fail.” If the boat is of high quality there is a 90% chance of a“pass.” If the boat is a lemon there is a 98% chance of a “fail.”

(d) Draw and solve the new tree.

Imperfect Information (Advanced)

1. Life at Chembioco (see the preceding section) is not as simple as it could be.A third process which will take 18 months to develop is also available whichchanges the probability of Y being the correct process to 50% while leaving


the probability of X unchanged. The laboratory test results are incorporatedin the following table. If needed, the test can be repeated.

Experiment Actual Successful ProcessPredicts X Y Z

X 0.70 0.40 0.30Y 0.20 0.50 0.30Z 0.10 0.10 0.40

If one process has been shown to be unsuccessful, then the remaining prob-abilities must be re-scaled to 1.

What would you recommend to Chembioco in this case?

Note: This is a major problem with a significant number of sub-trees.

2. A company has the drilling rights for a promising tract of land. Initial indi-cations are that there is a 60% chance that the tract has standard oil bearingformations. If the formation is present, there is a 15% chance of no oil, a35% chance of small reserves and a 50% chance of large reserves. Other-wise, there is an 80% chance of no oil and a 20% chance of small reserves.Small reserves are worth 10 million dollars to the company and large re-serves are worth 100 million dollars.

A more detailed seismic survey can be done at a cost of $500,000.

The seismic result will be either “positive” or “negative.” If the formationexists, then there is an 80% chance that the survey will give a true positiveresult and a 15% chance it will give a false positive (the formation does notexist but the test is positive).

The land can be sold at any time. If there is no survey, the sale price is$500,000. If the survey is positive, then the land is worth $1,000,000; oth-erwise it is worth only $100,000.

A full scale drilling program will cost the company 5 million dollars andwill completely determine the nature of the reserves. If there are no re-serves, then the land will have a value of only $10,000.

What would you recommend that the company do?

3. Video Shack is a retail chain specializing in electronic equipment. It iscurrently having some problems with one of its high volume items. These


items are ordered from the supplier in lots of 100 units and give a profitcontribution of $10 per unit. Past experience indicates that the possiblepercentage defectives in a lot are 10%, 20% and 30% with probabilities of0.5, 0.3 and 0.2 respectively.

Video Shack could non-destructively test some or all of the units. A max-imum of three units could be tested sequentially (test one and see whetherit’s defective or acceptable, then test another, and so on); however, at anytime they may choose to test all remaining units. (For example, after testingtwo units, they could choose to do no more testing, or test the third unit, ortest all remaining 98 units.) No matter how the testing is done, the testingcosts $4 per unit tested. If Video Shack inspects a unit and finds it to be de-fective, then it will be replaced by the supplier at no cost. The replacementunit will be known to be non-defective. However, when a defective unit issent to a customer, it is replaced by Video Shack at a cost of $20.

What would you recommend to Video Shack?

4. This problem was first seen on page 370.

The Plastic Production Company needs to expand its production capacityfor the next year. The marketing department has determined that the com-pany will need 5,000, 10,000, or 15,000 cases of increased capacity withprobabilities of 0.3, 0.5, and 0.2 respectively.

The company is considering two options to meet the situation. First of all,they can use overtime at a cost of $3.00 per case in addition to the regu-lar cost of $10.00 per case. However, if the required demand turns out tobe 15,000 units, then there will be a 50% chance that there will be labourunrest after six months which will add another $2.00 per case. Secondly,they could operate a second shift. This will entail a fixed cost of $15,000for startup expenses and a $1.00 per case shift premium. However, if thedemand turns out to be only 5,000 cases, then they will be required to layoffthe shift at a cost of $5,000 and use overtime. If desired, the company couldstart the second shift in six months time instead of starting it immediately.

A survey which costs $2,000 is available which has had the following his-tory:

– it gave a positive response 80% of the time when the demand subsequentlyhad a large increase


– it gave a positive response 50% of the time when the demand subsequentlyhad a moderate increase

– it gave a positive response 30% of the time when the demand subsequentlyhad a small increase.

In addition, the company could monitor the labour situation at a cost of$500. This would have a 95% reliability in predicting labour unrest.

Find the optimal course of action for the company.

5. A game is played with fifteen cards – the four Kings, and eleven other cards.If you wish to play you give $1 to the dealer.

The cards are shuffled and dealt into three piles (A, B, and C) of five cardseach. The object of the game is to correctly guess how many Kings thereare in Pile C.

Pile A is given to you, while piles B and C are placed face down. Afterlooking at your cards, you can either fold, or continue playing by giving $2to the dealer. After paying the $2, you may guess the number of Kings inpile C. A correct guess wins a $8 prize, an incorrect guess wins nothing.Only one guess is permitted at any time in the game, of course. Beforeguessing you may sequentially draw one or more cards from pile B, at acost of $1.50 per card. You may draw all five cards if you wish.

What is the optimal procedure for playing this game?

6. Solve the preceding problem where the drawing of the cards must be madesimultaneously.


23.5 Linear Optimization – Introduction

1. A company produces two products. Each product must pass through twomachines. Each unit of the first product requires 10 minutes on machineA, and 6 minutes on machine B. Each unit of the second product requires8 minutes on machine A, and 15 minutes on machine B. On a daily basis,machine A is available for 7 hours, while machine B is available for 7.5hours.

Each unit of product 1 contributes $45 to profit, while each unit of product2 contributes $30. At least 20 units of product 1 and at least 15 units ofproduct 2 must be produced each day.

(a) Formulate a linear model which will maximize the daily contribution toprofit.

(b) Solve graphically to obtain a recommendation.

2. A student has just won $100,000 in a lottery, and she has decided to investsome or all of it in the stock market. She has identified 10 blue chip stocks;stocks 1 to 4 are utilities, 5, 6, and 7 are industrials, and 8, 9, and 10 arehigh-tech companies.

Let Xi represent the amount of money (in dollars) invested in security (stock)i, where i = 1, . . . ,10. Write the following constraints mathematically, usingsummation notation where appropriate:

(a) Her total investment must not exceed $100,000.

(b) She will invest at least $2,000 in each security.

(c) She will invest no more than $20,000 in any security.

(d) She will invest at least $21,000 in utilities, at least $18,000 in industrials,and at least $15,000 in high-tech companies.

(e) She will invest no more than $60,000 in utilities, no more than $50,000in industrials, and nor more than $40,000 in high-tech.

(f) The amount invested in utilities must be at least half as much as theamount invested in industrials and high-tech combined.

(g) The amount invested in industrials must not be over $20,000 more thanthe amount invested in high-tech.


3. A cat owner buys cat food from two sources and blends them to obtain aminimum cost nutritious diet. The two products have the following charac-teristics:

Product Cost Protein Carbohydratesper kg. (% by weight) (% by weight)

1 $0.60 20 252 $1.20 50 30

Suppose that a cat needs 80 grams of protein per day, and 75 grams ofcarbohydrates per day. The owner wishes to determine how much to buy ofeach product. Assume that the cat will eat everything that she is given.

(a) Formulate a model for this problem.

(b) Solve this model by using the graphical method.

(c) The company which produces product 2 decides to raise its price to$1.80 per kilogram. What will the owner do now?

4. Given the following model

min 2X1 + 5X2subject to

(1) 6X1 + 3X2 ≥ 24(2) 3X1 + 5X2 ≥ 30(3) 4X1 − 3X2 ≥ 12(4) 2X1 − 1X2 ≥ 0

X1 , X2 ≥ 0

(a) Solve using the graphical procedure.

(b) For each constraint, state whether or not it is binding, and state whichconstraints (if any) are redundant.

5. (a) Show the following model graphically, labelling all constraints, drawingan isovalue line, and indicating the feasible region:



(1) X1 + 3X2 ≤ 12(2) 3X1 + X2 ≥ 13(3) X1 − X2 ≤ 3

X1 , X2 ≥ 0

(b) Solve the model graphically.

(c) Find the value of the objective function at each feasible corner point.

6. (a) Show and solve the following model graphically, labelling all constraints,drawing an isovalue line, and indicating the feasible region:


(1) 2X1 + 5X2 ≥ 20(2) 14X1 + 7X2 ≥ 35(3) 4X1 + 6X2 ≤ 60(4) X2 ≤ 8

X1 , X2 ≥ 0

(b) Find the value of the objective function at each feasible corner point.

(c) What happens to the feasible region and the optimal solution if the con-straint

3X1 +2X2 = 15

is added to the model?

(d) Not including the constraint given in part (c), what happens if the con-straint

3X1−2X2 = 15

is added to the model?

7. A prepared food shop produces regular and deluxe frozen pizzas. Each reg-ular pizza uses 750 grams of dough mix and 200 grams of toppings, while


a deluxe pizza uses 600 grams of dough mix and 500 grams of toppings. Anew shipment of dough and toppings will arrive next week; until that timethey have 75 kilograms of dough and 40.4 kilograms of toppings. They esti-mate that at most they could sell 70 deluxe and 80 regular pizzas during thenext week. Given the above, they wish to maximize the total contributionto profit, which is based on $0.50 for each regular pizza sold, and $0.75 foreach deluxe pizza sold.

(a) Give the linear optimization model formulation.


8. A brewery has two brands of beer, Dark Horse and Wholesome Light. Eachis composed of yeast, hops, barley, and water, but in different proportions.Each week, the brewery receives the following quantities:

Yeast 51 kgs.Hops 96 kgs.Barley 243 kgs.Water Unlimited

The two products require the following amounts of inputs per litre of output:

Dark Horse Wholesome LightYeast 5 grams 3 gramsHops 8 grams 6 gramsBarley 18 grams 15 gramsWater 1.1 litre 1.1 litre

(Some of the water is lost due to evaporation, but this does not concern us.)

Each litre of Dark Horse gives a profit contribution of $0.24, while eachlitre of Wholesome Light gives a profit contribution of $0.30. The brewerywishes to know how much of each brand should be produced.

(a) Formulate the brewery’s problem as a linear optimization model.


Dark Horse consumers are incensed at the solution found in part (b). Themanager therefore decrees that neither brand can represent more than 80%of the total production.

(c) Add the new constraint(s), and re-solve the problem.


9. A problem has been formulated as:

max 7X1 + 2.8X2subject to

(1) 2X1 + X2 ≤ 6(2) X1 + 2X2 ≥ 4(3) 2X1 + 3X2 ≤ 12(4) 5X1 + 2X2 ≥ 10

X1 , X2 ≥ 0

(a) Solve this problem by the graphical method, labelling each constraintand showing the feasible region and an isovalue line. Use algebra to de-termine the exact optimal solution for X1, X2, and the objective functionvalue.

(b) Suppose now that 5X1−2X2 = 0 is added as a fifth constraint. Draw thisconstraint on the same graph, and show the new optimal solution. Again,use algebra to find the optimal solution exactly.

10. A problem has been formulated as:

max 2X1 + X2subject to

(1) 4X1 + 2X2 ≥ 24(2) 3X1 + 2X2 ≤ 27(3) 3X1 + 4X2 ≤ 40(4) 3X1 − 4X2 ≤ 0

X1 , X2 ≥ 0

(a) Solve this problem by the graphical method, labelling each constraintand showing the feasible region and an isovalue line. Use algebra to de-termine the exact optimal solution for X1, X2, and the objective functionvalue.

(b) Suppose now that the objective function is changed to

min 6X1−2X2.

Show the new optimal solution. Again, use algebra to find the optimal so-lution exactly.


11. Solve the following model graphically. Label all the constraints, indicatethe feasible region, the isovalue line, and the optimal solution. Use algebrato find the exact optimal solution.

max 3X1 + 2X2subject to

(1) 2X1 + 5X2 ≤ 10(2) 4X1 + 3X2 = 12(3) X1 + 4X2 ≥ 4

X1 , X2 ≥ 0

12. (a) Solve the following model graphically. Label all the constraints, indicatethe feasible region, an isovalue line, and the optimal solution.

max 3X1 + 4X2subject to

(1) X1 ≤ 8(2) X2 ≤ 10(3) X1 + 2X2 ≤ 22(4) 3X1 + 5X2 ≥ 15

X1 , X2 ≥ 0

(b) What is the solution to this model if the objective is to minimize ratherthan to maximize?

The next three problems are more difficult. The reader may wish to com-plete them after studying some of the formulation examples of the Applica-tions chapter.

13. A small distillery in the highlands of Scotland custom blends a barley whiskywith an ordinary grain whisky. They have received an order of blendedwhisky, in an amount not less than 2000 litres, nor more than 3000 litres,but otherwise at the distillery’s discretion. The blend must be at least 40%(by volume) barley whisky, and must be no more than two parts barley toone part grain. They can use up to 1250 litres of each type of input whiskyper week. They desire to complete the order within one week. The blend


is sold (before tax) for £2.25 per litre. The fixed costs of operating the dis-tillery are £2800 per week. In addition, there is a cost of £0.75 per litre ofbarley whisky, and a cost of £0.45 per litre of ordinary grain whisky.

(a) Formulate a profit maximizing linear optimization model for the dis-tillery.

(b) Solve the model graphically to give a recommendation to the distillery.

14. An electrical utility has had its nuclear energy program frozen by the gov-ernment, and has therefore decided to activate a mothballed conventionalthermal plant. New pollution laws enacted since the plant last operatedmean that a new operating plan will be required.

The plant produces electrical energy by burning coal which heats water toproduce steam. The steam drives large turbines which generate electricalenergy. The coal arrives by rail from two mines. One mine produces highquality anthracite coal, while the other mine produces lower quality bitu-minous coal. The railcars dump each type of coal into separate bins. Fromeach bin, there is a conveyor belt which takes the coal to a grinding mill.From the mill the coal falls by gravity into separate bins for each type ofcoal. From these bins coal is fed into the combustion chamber.

There are three relevant characteristics of each type of coal with respect tothis particular plant. There is the thermal value, which is the amount ofuseful energy measured in megajoules (MJ) per tonne, secondly the amountof sulphur oxides created in the flue gases, measured in parts per million(PPM), and thirdly the particulate emissions measured in kilograms (kg) ofemissions per tonne of coal. They are as follows:

Thermal Value Sulphur Oxides Particulate(MJ/tonne) (PPM) Emissions (kg/tonne)

Anthracite 24,000 1800 0.4Bituminous 20,000 3800 1.0

Government regulations limit sulphur oxide emissions to 2600 PPM, andparticulate emissions to 12.8 kg/hour. The grinding mill can handle up to 18tonnes per hour of anthracite coal, or up to 24 tonnes per hour of bituminouscoal. The conveyor belt can handle up to 20 tonnes of coal (regardless oftype) per hour. The storage bins are large enough so as to not constrain theoperation of the plant.


(a) Formulate a linear model for this plant which seeks to maximize theenergy output per hour. Hint: The grinding mill is represented by a singleconstraint, which can handle any combination of the two types of coal.

(b) Graph the constraints and show an isovalue line.

(c) What is the maximum plant capacity?

15. A student has formulated the following model.

min 2X1 +5X2−3X3subject to

(1) 7X1 +3X2 ≤ 40+2X2−2X3(2) 2X1 +4X2 +X3 ≤ −20(3) 9X1+5X2−2X3

X1+X3+1 ≥ 2(4) 3X1 +X2 +2X3 = 28+X3(5) 10X1 +5X2

1 −4X1X2−6X3 ≥ 3X1X3 +8X2

X1 ≥ 0 X2 ≤ 0 X3 ≥ 0

While this model is non-linear in its current form, it can be converted to alinear model. Help the student out by rewriting the model in the appropri-ate format. Hint: Begin by replacing X2 with a variable which cannot benegative.


23.6 Markov Chains1. A university’s computer is always in one of two states: “up” (operating) or

“down” (under repair). Transitions from state to state occur every hour andare subject to the following Markovian process:

Hour Hour t +1t up down

up 0.9 0.1down 0.8 0.2

(a) If the computer is currently “up,” what is the probability that itwill be “up” four hours from now?

(b) What is the expected first passage time from “up” to “down”?

(c) What is the expected first passage time from “down” to “up”?

2. The manager at a fish plant has noticed that there seems to be a relationshipbetween one day’s intake of fish to be processed and the next’s. Eightypercent of the time when today’s catch is good, tomorrow’s is also good.If today’s catch is bad, then there is a sixty percent chance that tomorrow’scatch is also bad.

(a) If today’s catch is good, what is the probability that the catch willbe bad eight days from now?

(b) In the long run what percentage of the time is the catch good?

3. The NF Tea Company currently has one brand which is in competition withall other brands of tea. Brand loyalty for NF Tea is 90% per week. Only 2%of the consumers of all other brands switch to NF Tea each week.

(a) What is NF Tea’s long run market share at this point in time?

(b) If a consumer is currently buying a non-NF brand, what is theprobability that her next purchase of NF Tea will occur i) in twoweek’s time? ii) in three week’s time?NF Tea is considering the introduction of a new brand (calledbrand B), which will take consumers away from other brands,including their existing brand (brand A). They estimate that each


week 3% of those buying non-NF brands will switch to NF brandB, and 5% of those buying brand A will switch to brand B. Ofthose who try brand B, 70% stick with it, while 20% will switchto non-NF brands, and 10% will switch to NF brand A. There isa 7% chance that a brand A buyer will switch to a non-NF brand,but only a 1% chance that a non-NF buyer will switch to brandA.

(c) What is the overall long-term market share of the NF Tea Com-pany now?

(d) If a consumer is currently buying a non-NF brand, what is theprobability that her next purchase of NF Tea Brand B will occuri) in two week’s time? ii) in three week’s time?

4. Consumers in a particular region of the country make purchases of gasolineevery week and brand loyalty (or lack of it) is described by the followingMarkovian process:

Week Week t +1t ESSOH URVING SEASHELL

ESSOH 0.8 0.1 0.1URVING 0.2 0.7 0.1

SEASHELL 0.2 0.3 0.5

(a) In the long-run, what will the market shares of the three compa-nies be?

(b) How long, on average, does a consumer take to switch fromURVING to ESSOH?

5. An investor has purchases 1000 shares of a speculative stock for $25 pershare. If the price rises to $28 then s/he will sell and if it falls to $23 s/he willalso sell. From past experience with this stock, there is a 20% chance thatthe stock price will increase by $1 and a 15% chance that it will decreaseby $1 from one day to the next. Otherwise, the stock price will remain thesame.

What is the expected gain for this speculation?


6. Four companies are fighting for market share. For each company, the pro-portion of the customers who switch to another company (or who stay withthe same company) each month is given by the following matrix.

P =

.2 .3 .4 .1.5 .2 .1 .2.1 .2 .3 .4.2 .3 .2 .3

(a) By using matrix inversion within a spreadsheet model, determine

the long-term percentage market share for each company.

(b) There is a feature of this matrix which causes the result found in(a). What do you think it is?

7. U-Drive Car Agency maintains four rental facilities at locations W, X, Yand Z. A car can be returned to any of the four facilities. Historical datahave shown that for every 150 cars picked up at each of the locations, thefollowing return patterns have been observed:

Picked Returned toUp From W X Y Z

W 74 30 26 20X 30 86 20 14Y 15 20 90 25Z 35 15 30 70

(a) If this return pattern were to continue, what would be the long-run distribution of these cars, assuming no redistribution of carsby management?

(b) At which location should a new service or repair facility be lo-cated?

(c) In six months time, there will be a major convention at locationY. The demand during that week is estimated to be (100, 150,250, 100). What redistribution of cars will be necessary to meetthis demand?


8. Region X, with a workforce population (includes those working and thosewho are seeking work) of 500,000, is adjacent to Region Y, with a work-force population of 1,000,000. Each year 10% of X’s workforce populationmoves to Y and 4% of Y’s moves to X.

If the current unemployment rate (percentage of those in the workforce whoare seeking work) in X is 8%, what will it be in the long run assuming nochange in the number of jobs in region X?

9. A examination body for a profession administers two examinations to eachcandidate. The result of each examination is either “pass,” “marginal,” or“fail.” If the candidate passes the first exam, then he writes the second atthe next sitting of the examinations, which is usually six months later. If hepasses the second exam then he is admitted to the profession. If the studentobtains a “marginal” on either exam, then he must re-sit the same exam atthe next sitting of the examinations. If the student fails the second exam,he must repeat the first exam at the next sitting. If the student fails the firstexam he is no longer a candidate and he must therefore pursue a differentcareer.

Suppose that on the first exam 25% of the students pass, and 40% aremarginal. On the second exam 30% pass and 50% are marginal.

Find the probabilities of eventual success or failure, conditional on whetherthe candidate is about to write the first or the second exam.

10. An Undergraduate Studies Committee is evaluating the performance of stu-dents in its program. Each term, each student will either pass (and is there-fore promoted to the next term), fail (must repeat the same term), or with-draw. A student is said to have withdrawn if s/he changes programs or isrequired to permanently leave the program due to poor performance.

The committee has found that the pass and withdrawal rates for each termare as follows:

Term Pass WithdrawalRate Rate

I 80% 10%II 85% 8%III 85% 6%IV 90% 4%


Given that 200 students enter Term I, what is the expected number of stu-dents who will eventually be promoted to Term V?

11. On January 1 of each year, bank accounts are classified as “active” or “in-active.” An active account is one in which there was at least one depositor withdrawal during the previous year, otherwise the account is inactive.During each calendar year, about 5% of all active accounts are closed, and2% of all active accounts become inactive.

In early January a registered letter is sent to the last known address of eachinactive account holder, advising the customer to make the account active,or to close the account. Furthermore, the letter states that any money notspoken for by the end of the year will be turned over to the government.In response to the registered letters, about 40% of the accounts are madeactive, and about 45% of the account holders, who are happy to learn aboutforgotten money, close their accounts. About 10% of the registered lettersresult in a visit to the bank by a relative or friend stating that the accountholder has died or has disappeared. Such accounts are handed over to thebank’s legal department. The other 5% of the time no response is received,and such accounts are turned over to the government.

What percentage of active accounts end up eventually being transferred tothe government?

12. Renteels, Inc. is in the car rental business. Its policy is to use only new orone-year old cars. Older cars are sold to the public, at the end of the secondyear. During the year, customers are sometimes involved in accidents andwreck the car completely. These wrecks are sold to junkyard dealers only.Past statistics indicate that 20% of the new cars and 10% of the one-year oldcars meet this fate. Also, some of the cars may develop a major componentfailure and would then be taken out of service and returned to the manufac-turer. 10% of the new cars and 5% of the one-year old cars fall under thiscategory.

At the year-end, under a special arrangement with a local dealer, Renteelscan obtain a new car by trading in a one-year old car plus $2,000, or they canobtain a one-year old car for a two-year old car plus $1000. These tradeshappen 20% and 10% of the time respectively.

Renteels presently has 60 new cars and 40 one-year old cars and wishes toknow what percentage of these cars (or their trades) will eventually be sold


to the public.

13. DEF Co. has the following current distribution of accounts in their accountspayable ledger.

Age Number Percentage(Months) Paid

1 200 60%2 150 40%3 75 30%4 20 20%

The percentage paid column refers to the percentage of the accounts whichwill be paid by the end of the next month. Accounts which are older thanfour months are written off as bad debts.

(a) What is the expected number of the existing accounts which willbe written off as bad debts?

(b) A collection agency has offered the following deal. As soonas DEF makes a sale, it will turn the bill over to the collectionagency, who will pay 95% of the amount of the bill. Should DEFaccept the deal?

14. Fine Wines Winery annually offers a selection of fine wines to a numberof preferred customers during the Christmas season. Some of the winesin their collection are not considered to have aged sufficiently and are notoffered for sale. There are 3500 in this category. If a wine is not sufficientlyaged, then there is a 20% chance that it will become available and a 10%chance that it will spoil and have to be discarded. Wines available for salehave a 40% chance of sale this season, and a 20% chance of spoiling.

If the current inventory is 10,000 bottles, how many will end up being sold?

15. A credit card company classifies its accounts into one of five categories:fully paid, bad debt, 0-30 days old, 31-60 days old, and 61-90 days old.Each month, there is a 20% chance that nothing will be paid on an account.There is a 25% chance that just enough is paid to keep the account in thesame category. If the account is in the 31-60 day category, then there is a20% chance that the account will be paid in full. If the account is in the


61-90 day category, then there is a 5% chance that it will be paid in full, anda 10% chance that it will move to the 0-30 day category.

Draw the transition diagram, and write the transition probability matrix.What is the long term probability of being fully paid, conditional on theinitial classification of debt?

16. Many businesses fail because they are under capitalized. This problem ex-amines how the probability of avoiding bankruptcy could be related to theinitial level of capitalization.

A new business starts off with n units of capital, where n is an integer be-tween 0 and 20 inclusive. Each month, whenever the company begins themonth with between 0 and 20 units of capital inclusive, there is probabilityp that the capital will increase during the month by one unit, probability qthat it will decrease during the month by one unit, and probability 1− p−qthat it will stay the same. Suppose that when a company’s capital reaches21 units it is in the clear and there is no further risk of bankruptcy. When acompany’s capital falls to −1 units it declares bankruptcy.

(a) Model this situation as a Markov chain, making a spreadsheetmodel in which the actual values for p and q are only enteredonce.

(b) Now suppose that p = 0.2, and q = 0.5. Obtain a graph of theprobability that a company will avoid bankruptcy, conditional onthe initial level of capitalization (n). How long, on average, willa company possess between 0 and 20 units of capital inclusive ifit starts out with 0 units of capital? . . . with 20 units of capital?

(c) Repeat (b) using p = q = 0.3.

(d) Repeat (b) using p = 0.5, and q = 0.4.

17. Professional baseball players begin their careers in the relatively poorly paidminor leagues. However, a good year with a minor league team can give riseto a well-paying major league position. Alternatively, a player may realisethat he is not cut out for professional baseball and will therefore resign ofhis own accord. Some other players may be released by the team againsttheir will, while the rest will return to play minor league baseball in thesubsequent season. Suppose that from season to season a minor league


player has a 20% chance of advancing to the major leagues, a 10% chanceof retiring, and a 15% chance of being released.

The career possibilities for major league players are similar, except that theymay be demoted to the minor leagues or they may be promoted to coach ormanager. Suppose that from season to season a major league player has a12% chance of being demoted, a 5% chance or retiring, a 4% chance ofbeing released, and a 2% chance of promotion to coach or manager.

(a) In a particular year, 80 new players are signed up by minor leagueteams. Eventually, what is the expected number of these playerswho will retire? who will be released? who will be promoted tocoach or manager?

(b) Suppose that we also wished to know the expected number ofthese players who will reach the major leagues, even if it is foronly one season. Show how to re-formulate the model so thatthis question can be answered.

(c) If a player reaches the major leagues, what is the expected num-ber of consecutive years that he will remain in the major leagues?

18. The Undergraduate studies committee of a business school is evaluating itsscholastic standards. It has found that the pass rate in its seven terms to beas follows:

Term Pass RateI 70%II 70%III 75%IV 85%V 90%VI 95%VII 99%

In addition, it has been found that about 10% of the students in terms I to IIIvoluntarily withdraw from the program to pursue other degrees or careers.About 5% of the students in terms I to IV are required to permanently leavethe program due to poor performance. All other students are required totemporarily leave the program for the duration of two terms, after whichthey re-enter the program to repeat the failed term.


(a) What percentage of students entering term I will finally gradu-ate?

(b) What percentage of students entering term I will be required topermanently leave the program?

19. A tiny island nation has a parliament of just three seats. There are two politi-cal parties, denoted as P and Q. All three seats are contested by both partieswhenever there is a general election. If one of the three constituencies isrepresented by a member from Party P, then Party P will retain the seat withprobability p. If a constituency is represented by a member from Party Q,then Party Q will retain the seat with probability q.

(a) We wish to model the parliament using a single Markov chain.Define the states using the minimum number possible.

(b) Derive the transition probability matrix as a function of p and q.

(c) If we were to use the matrix found in part (b), it would be possibleto derive, as a function of p and q, the probability that Party Pwill control all three seats, or that they will form a government(at least two seats), and so on. However, this would require alot of algebraic work. What is a simpler way to answer suchquestions?


23.7 Utility Theory1. A businessman’s utility function is scaled at U(−$100,000) = 0, and U($1,000,000)

= 100. He is indifferent between the alternatives in the following situations:

Situation 1. a 50-50 gamble of−$100,000 or $1,000,000, or $200,000for certain;

Situation 2. a 50-50 gamble of −$100,000 or $200,000, or nothingfor certain;

Situation 3. a 60-40 gamble of $200,000 or $1,000,000, or $400,000for certain.

(a) Draw the utility function, using a smooth curve between the knownpoints.

(b) Which would he prefer, a gamble with an equal chance of obtain-ing −$50,000 or $300,000, or $100,000 for certain?

2. A farmer has determined that he has three options at the present time. Thepayoff for each option depends on the weather as follows:

WeatherOption Bad Average Good

Plant 20 Hectares −10 15 25Plant 40 Hectares −50 25 50Plant 80 Hectares −100 40 90

Probability 0.2 0.3 0.5

where the payoffs are in 1,000’s of dollars.

Using the Certainty Equivalence approach, a management scientist was ableto determine the following utility indices:

Payoff Utility−100 0.00−40 0.075

25 0.2755 0.5275 0.82

100 1.00


(a) Construct the utility curve for the range−$100,000 to $100,000.

(b) What would you recommend to the farmer?

(c) What is the certainty equivalent of the recommended alternative?

3. The Fresh Milk Dairy is considering enlarging their plant now or waitinguntil next year. The payoff for each alternative depends on the demand fortheir product which may be low, moderate or high. The payoffs associatedwith each alternative and outcome are, (in 1,000’s of dollars), as follows:

DemandAlternative Low Moderate High

Enlarge now −10 70 90Wait 5 50 70

Probability 0.3 0.4 0.3

The company has decided that a $10,000 loss will have a utility of zero anda $100,000 profit a utility of 100. Furthermore, they are indifferent in thefollowing three situations.

Situation1. (i) $80,000 for certain and (ii) a lottery: $100,000 profit(prob. 0.9); $10,000 loss (prob. 0.1).

Situation2. (i) $80,000 for certain and (ii) a lottery: $100,000 profit(prob. 2

3 ); $60,000 profit (prob. 13 ).

Situation3. (i) $60,000 for certain and (ii) a lottery: $80,000 profit(prob. 0.75); $10,000 profit (prob. 0.25).

(a) Find the recommended alternative using the EMV decision rule.

(b) Construct the utility curve.

(c) What strategy should the dairy adopt using the expected utilitydecision rule?

(d) What is the certainty equivalent of the optimal strategy?

4. Here we consider five investors who have the utility functions given in parts(a) to (e). Each person has evaluated a risky venture at a given number ofutiles. In each situation, determine the certainty equivalent for that individ-ual.


(a) U(x) = xx+6000 , 0.63 utiles

(b) U(x) = x4x+10000 , 0.85 utiles

(c) U(x) = 1− e−x

1000 , 0.57 utiles

(d) U(x) = 1− e−3x2000 , 0.45 utiles

(e) U(x) = 3ln( x+15000), 0.52 utiles

5. Assume a friend of yours asks your advice on the following decision. Yourfriend can invest in investment 1 which will guarantee a payoff of $50 orin investment 2 which has possible payoffs of $100, $40, $0 with equalprobability.

(a) Which investment would you recommend if your friend wantedto maximize his expected payoff?

(b) You have determined that he is indifferent between (i) equal chancesof A or B, and (ii) a certain equivalent, in the following four sit-uations:

Situation A B Certain1. 100 0 402. 40 0 253. 100 40 504. 100 50 62.5

Sketch the utility function.

(c) Which investment would you recommend if your friend wantedto maximize the expected utility of the investment?

6. A manager whose utility function is described in the following table (inunits of 1,000 dollars), must choose between contract A and contract B.Contract A promises a loss of $45,000 with probability of 0.4 and a profitof $80,000 with a probability of 0.6. Contract B consists of two subcon-tracts each of which promises a loss of $22,500 with probability 0.4 and aprofit of $40,000 with probability 0.6. The outcomes of the subcontracts areindependent.


Dollar Utility−200 0.000−100 0.375−50 0.525−20 0.650

0 0.70024 0.75060 0.825

100 0.875200 0.925280 0.925400 1.000

(a) Which contract would you recommend to the manager?

(b) What is the certainty equivalent of the chosen contract?

7. The following set of utilities has been determined for a manager:

Payoff Utility−200 0.00−120 0.40−50 0.575

0 0.6550 0.72570 0.80

100 0.85150 0.90400 1.00

(a) Construct his/her utility curve.

(b) The manager is given the opportunity to buy a contract whichwill either give him/her $200,000 (prob. 0.55) or cost him/her$100,000 (prob. 0.45). What is the contract worth to the man-ager?

(c) If a second identical contract were made available, what shouldthe manager be willing to pay for it?


8. Storage Inc. is considering the purchase of a theft insurance policy for alocal warehouse. There are three outcomes which can occur in any year: notheft (95% chance), a “minor” theft of $100,000 (4% chance), or a “major”theft of $1,000,000. If the insurance were purchased, the premium wouldbe $20,000. The insurance will fully compensate any loss due to theft.

After considering the risk and other related factors, the company executivesbelieve that the utility rating is the appropriate measure to use in evaluat-ing alternatives. The company president has assigned a utility of 100 to$1,000,000 and a utility of 0 to $0. In order to determine the utility func-tion, the president has determined that he is indifferent between each of thetwo alternatives presented in the following four situations:

Situation 1. (i) do nothing and (ii) a lottery: a $1,000,000 payoff withprobability 0.8 and a loss of $1,000,000 with a probability of 0.2.

Situation 2. (i) a loss of $250,000 for certain and (ii) a lottery: a$1,000,000 loss with probability 0.1 and no loss with probability0.9.

Situation 3. (i) a loss of $20,000 for certain and (ii) a lottery: a loss of$250,000 with probability 0.1 and no loss with probability 0.9.

Situation 4. (i) a loss of $20,000 for certain and (ii) a lottery: a lossof $100,000 with probability of 0.25 and no loss with probability0.75.

(a) Construct and graph the utility function for the company presi-dent.

(b) Given that the company president wishes to maximize his ex-pected utility, what would you recommend?

(c) What is the maximum amount that the president is willing topay as an insurance premium in order to avoid the risk of theft?[Assume that his utility function is linear in the interval from−$100,000 to −$20,000.]

9. Susan Prudent is a businesswoman who carefully considers the risks of anynew business venture. She uses the utility function U(x) =

√x

100,000 , where

x is her level of wealth in dollars. Her current net worth is $100,000. She isconsidering an investment of $50,000 in a venture which will have revenues


of 0, $80,000, or $200,000 with probabilities of 0.2, 0.3, and 0.5 respec-tively.

(a) Should she undertake this venture?

(b) What is the EVPI (for her)?

10. In the above problem suppose that the $50,000 investment is not fixed butis subject to negotiation. What is the maximum amount that she will pay toinvest in this venture? [Hint: At the “maximum amount” she will be indif-ferent between the two alternatives. The expression that you will obtain forthe maximum amount can be solved either by trial and error or by graphingthe expression.]

11. Derive your own utility curve for the range −10,000 to 100,000 dollars.

12. Derive your own utility curve for final grades in a course. Assume that youare faced with the following dilemma. Your final exam is tomorrow andyou have only time to study one of two important topics - A and B. Frompast results you know that 40% of the time A is heavily weighted, 35% ofthe time B is heavily weighted. Otherwise they are equally weighted. Ifyou study for A, then you will receive 85%, 55% and 70% respectively foreach situation. If you study B, then you will receive 50%, 95% and 65%respectively for each situation.

(a) Which alternative do you prefer?

(b) Evaluate these situations using your utility curve.

(c) Are the results to (a) and (b) different? If they are different, ex-plain why they are different.

13. A student is confronted with the problem of deciding which of two coursesto take. She has decided to take either a statistics course or a managementcourse. For the statistics course, she estimates the probabilities of eachgrade as 0.20, 0.30, 0.50 and 0.00 for A, B, C and D respectively. Forthe management course the respective probabilities are 0.10, 0.50, 0.30 and0.10. Furthermore she has determined that she is indifferent between thealternatives in the following two situations:


Situation 1. Alternative 1. a C for certainAlternative 2. a lottery with a probability of 0.20 of an A and aprobability of 0.80 of a D.

Situation 2. Alternative 1. a B for certainAlternative 2. a lottery with a probability of 0.6 of an A and aprobability of 0.40 of a C.

Which course should she take?

14. An individual’s utility function for dollar values greater than −100,000 isgiven by

U(x) = 100− 10,000100+ x

where x is in units of $1,000.

(a) What is the utility of −$20,000?

(b) If the utility of an uncertain situation is 40, what is its certaintyequivalent?

15. An individual whose present wealth level is $10,000 (but who is able toborrow more) has the following utility function

��

�

��

��

Wealth in $100,000

Utility

100


This individual has two investment alternatives which have the followingpayoffs:

Outcome (in $1,000’s)Investment O1 O2 O3 O4A1 20 −30 80 −10A2 30 90 −40 30Probability 0.1 0.3 0.4 0.2

(a) If each alternative costs $10,000, which alternative should bechosen?

(b) If each alternative costs $20,000, which alternative should bechosen?


Utility Theory (Advanced)

1. A man with utility function U(x) = 1− e−x100 has $200 dollars of which he

may gamble all, some, or none on a lottery on which he will either gain theamount wagered with probability 0.6 or he will lose the amount wageredwith probability 0.4.

(a) Over what dollar range is gambling better than not gambling?

(b) What gamble will give him the greatest expected utility?

(c) Consider the general case of this problem where he has y dollarsand the probability of winning the lottery is p. Re-solve parts (a)and (b).

2. An investor has decided that for any amount x, he is indifferent betweenreceiving either x for sure or a lottery which has a 50% chance of receiving$100 less than x and a 50% chance of receiving $200 more than x. If theutility of x = 0 is 0, and if the utility of x = 300 is 1, give the function U(x).

Hint: this sort of function is independent of wealth.

3. A major car manufacturer is thinking of redesigning the regular and pre-mium versions of one of its models. The cost of the redesign will be$2,000,000 for the regular model and $3,000,000 for the premium model.If the redesigns are successful (independent probabilities of 0.5 and 0.6 re-spectively), then the regular model would result in an increase in profits of$10,000,000, and the premium model would result in an increase in profitsof $15,000,000. If the designs are not successful, then the company wouldlose $5,000,000 and $8,000,000 respectively.

The company’s management has determined that it is indifferent in the fol-lowing situations:

Situation 1. $3,000,000 profit for sure or a 30% chance of a 10 millionloss and a 70% chance of a 20 million profit.

Situation 2. $10,000,000 loss for sure or a 20% chance of a 50 millionloss and an 80% chance of a 3 million profit.

Situation 3. $6,000,000 profit for sure or a 15% chance of a 10 millionloss and a 85% chance of a 20 million profit.


Situation 4. $3,000,000 profit for sure or a 20% chance of a 20 millionloss and an 80% chance of a 20 million profit.

Situation 5. $10,000,000 loss for sure or a 75% chance of a 30 millionprofit and a 25% chance of a 20 million loss.

What would you recommend that the company do?

4. A well established fact is the benefit one derives from wearing a seatbeltwhen in an accident. On average most drivers will have at least two or moreaccidents of varying degrees of severity. We classify the outcomes of anaccident as no injury, serious injury, and death. The probabilities of theseoutcomes can be described as follows:

no injury serious injury deathusing seatbelts 0.95 0.04 0.01no seatbelt 0.60 0.30 0.10

(a) How would you approach this situation using utility theory?

(b) Evaluate the behaviour of people who use seatbelts versus thosewho do not.

5. Consider the following set of situations16:

Situation 1. Alternative A1 - $1,000,000 for certain.Alternative A2 - $5,000,000 with probability 0.1, $1,000,000 withprobability 0.89 and $0 with probability 0.01.

Situation 2. Alternative A3 - $5,000,000 with probability 0.1, $0 oth-erwise.Alternative A4 - $1,000,000 with probability 0.11, $0 otherwise.

(a) Comparing A1 with A2, which do you prefer?

(b) Comparing A3 with A4, which do you prefer?

(c) Evaluate the four alternatives using utility theory. Is your answerto (a) consistent with your answer to (b)?

16This problem is usually named after M. Allais who discussed it in the paper “Le Comporte-ment de l’Homme Rationnel devant le Risque: Critique des postulats et Axiomes de l’EcoleAmericaine,” Econometrica, vol. 21, 1953.


23.8 Game Theory1. The Fast Track and Quick Trip Bus Companies serve the same route be-

tween two major cities. The total number of passengers travelling by bushas been fairly stable over the recent years. To obtain a larger market share,both companies are thinking of improving their service by either loweringfares or serving soft drinks en route. The payoffs associated with each op-tion are as follows:

Fast Track Quick TripDo Nothing Lower Fare Soft Drinks

Do Nothing 0 −40 −100Lower Fare 45 0 −10Soft Drink 90 0 0

(a) Eliminate all dominated alternatives. What strategies would you recom-mend for each company?

(b) Use the Criterion of Pessimism to find the optimal solution. Is the saddlepoint the same as in part (a)?

2. A union is just about to negotiate a contract with company management.They have identified four bargaining strategies for themselves and they be-lieve that the company will choose from four strategies as well. They haveestimated the payoffs in millions of dollars, for their members as a wholeover the life of the contract, as a function of the union’s and management’sstrategies as follows:

Union Company StrategiesStrategies M1 M2 M3 M4

U1 22 17 21 7U2 27 15 10 12U3 42 5 14 7U4 −8 7 13 2

From the management perspective the above payoffs represent costs in-curred by the company.

What strategy should each side choose?


3. Two firms, A and B, are each considering three advertising plans. The pay-offs are in terms of market share, and the game is zero-sum. The marketshare payoffs to firm A are:

Company Company B StrategiesA No Plan Plan

Strategies Advertising I IINo Advertising 0 −25 −20Plan I 5 −5 −15Plan II −5 −10 15

What should be the advertising strategy of each firm?

4. An environmental group has a $4,000,000 fund from which to allocate toadvertising campaigns on two environmental issues: acid rain and watertreatment. To combat these campaigns, an industry group has put together a$5,000,000 war chest with which to fight the environmental group. Supposethat each side will allocate its money in $1,000,000 increments. For exam-ple, the environmental group might allocate $3,000,000 to fighting againstacid rain and $1,000,000 to fighting for water treatment.

Suppose that the probability of winning an issue (in terms of governmentalaction) is proportional to the amount of money spent by each side. For ex-ample, if the environmental group spends $3,000,000 fighting against acidrain, and the industry spends $2,000,000 denying acid rain effects, then theprobability that the environmental group wins the issue is

3,000,0003,000,000+2,000,000

= 0.6

If neither side spends anything on a particular issue, then there is a prob-ability of 0.5 of winning for each side. The objective for each side is tomaximize the expected number of issues won. The advertising money isinconsequential for both parties.

What should each side do?

5. Due to the relative unpopularity of big business polluters, the effectivenessof their advertising is only 80% of the effectiveness of the environmental


group’s advertising. How will this affect the recommendations obtained inthe previous question?

6. A union and a company have made their “final” contract offers. The unionhas demanded an increase of $2.25 per hour, while the company has offeredan increase of $1.00 per hour. An arbitrator has been called in.

The arbitrator has asked each side to submit a new proposal. Each side musteither re-offer its existing “final” offer, or submit a new offer between $1.00per hour and $2.25 per hour inclusive in 25 cent increments (for example,the union could now ask for $1.75 per hour, but not $1.81 per hour).

The arbitrator will make the final wage increase equal to the offer whichhas moved the most from its current position. (For example, if the companyincreases its offer by 25 cents, but the union decreases its demand by 50cents, then the increase will be $1.75 per hour). In the case of ties (or ifneither side moves), then the arbitrator will award an increase of $1.625 perhour.

Draw the game table from the union’s perspective. What should each sidedo? What is the value of the game?

7. Formulate the children’s game of “Scissors, Paper, Rock” as a zero-sumgame. Recall that each player chooses scissors, paper or rock. Scissors cutpaper, paper entraps rock, and rock breaks scissors. If both players choosethe same item, then no-one wins. What is the optimal strategy for eachplayer?

8. Three hotels have been built to take advantage of a bay with a beautifulsandy beach. The hotel situated at the western end of the beach has onaverage 1000 guests, the hotel at the centre of the beach has 500 guests,and the eastern hotel has 1500 guests. Two ice cream vendors service thebeach. If one of the vendors is closer than the other vendor to a particularhotel, then the closer vendor will obtain 75% of the customers at that hotel,whereas if they are equidistant then each vendor will receive 50% of thecustomers at that hotel.

(a) Construct the payoff matrix in terms of market share.

(b) Find the optimal strategy for each vendor.

9. Alison and Wendy enjoy gambling. They play a game in which each oppo-nent secretly chooses a coin from one of her piles of British coins. Alison


has three piles of 2p, 5p, and 10p coins; Wendy has three piles of 1p, 5p,and 10p coins. If the value of the sum of the values of the two coins is eventhen Alison wins Wendy’s coin; otherwise Wendy wins Alison’s coin. Allwinnings go into two separate piles, one for each player. (The players donot choose from these piles).

(a) Set up the payoff table for Alison.

(b) Which strategies can be removed using dominance?

(c) What are the optimal strategies for each player, and what is the expectedvalue of the game?

10. The following table gives the payoffs to Player A for a zero-sum game.

B1 B2 B3 B4A1 −1 6 5 0A2 3 −2 −2 4A3 0 2 3 3A4 0 5 4 1

(a) eliminate the dominated alternatives.

(b) use a graph to find the alternative which is dominated locally.

(c) solve the game and find the expected payoff to Player A.

11. Three pubs are located in the downtown area of a large city. The relativelocations of the three pubs, and the expected number of patrons per week ateach is given in the following picture.


Westland

7,500��

Northland

12,500

@@@@@@@@@@@@@@

Eastland

20,000500 metres

300 metres 400 metres

Two competitive hot dog vendors named Jack and Jill are planning to ser-vice the area. At the outset of each day, each person will park his or her hotdog stand outside one of the three pubs. You are given the following data:

(1) if Jack is closer than Jill to a particular pub then he will receive 80% ofthe business from that pub

(2) if Jill is closer than Jack to a particular pub then she will receive 70% ofthe business from that pub

(3) if each person locates at the same pub then Jack will receive 55% of thebusiness from each pub.

Assuming that the patrons of each pub have similar preferences for hot dogs,what would you recommend to Jack? to Jill?

12. The Prisoner’s Dilemma I. Two men are arrested for armed robbery. Thepolice, convinced that both are guilty, but lacking sufficient evidence toconvict either, put the following proposal to the men, and then separatethem.

(1) If one man confesses but the other does not, the first will go free and theother will receive the maximum sentence of 10 years.

(2) If both confess, then they will receive lighter sentences of 5 years each.

(3) If neither confesses, then they will be imprisoned for jaywalking, va-grancy, resisting arrest for a total of 5 years for each man.


Draw up a payoff matrix for one of the prisoners and explain why the policeare convinced that both men will confess (providing both are guilty).

13. The Prisoner’s Dilemma II. Two men are arrested for armed robbery. Thepolice, convinced that both are guilty, but lacking sufficient evidence toconvict either, put the following proposal to the men, and then separatethem.

(1) If one man confesses but the other does not, the first will go free and theother will receive the maximum sentence of 10 years.

(2) If both confess, then they will receive lighter sentences of 5 years each.

(3) If neither confesses, then both men will go free.

(a) Why is Prisoner’s Dilemma II significantly different from Prisoner’sDilemma I?

(b) Draw up a payoff matrix for each of the prisoners and explain why thepolice are convinced that both men will confess (providing both are guilty).

14. Two friends are playing a simple game called “Five Pennies”. The rulesof the game are that each player chooses from 1 to 5 pennies (hidden fromthe other player). If the difference between the choices is 2 or fewer, thenthe player with the lower choice pays the difference to the player with thehigher choice, whereas if the difference is 3 or more, then the player withthe higher choice pays the difference to the player with the lower choice.

(a) Set up the appropriate payoff matrix for this problem.

(b) Is there a pure strategy available for either player?

15. Two groups (A and B) are in a conflict situation and have a set of alternativesavailable to them which are given in the following table:

B1 B2 B3 B4A1 12 4 8 15A2 10 5 6 10A3 11 6 6 12A4 6 10 12 8A5 5 13 15 9

What strategy would you recommend to each group?


Game Theory (Advanced)

1. Two players, Alf and Welf, decide to play a version of a “four-finger” gamein which they simultaneously show either 1, 2 or 4 fingers. If the sum ofthe fingers shown is even, Alf wins that amount (in dollars) from Welf;otherwise Welf wins the sum from Alf.

What is the optimal strategy for each player and the value of the game?

2. A “morra” game involves the showing of fingers (cf. previous problem)but also requires that each player guess the number of fingers which theother player has chosen. Hence each player must choose how many fingersto show, and how many to guess. If neither player guesses correctly or ifboth players guess correctly, then the game is a draw. Otherwise, the singleplayer who guesses correctly wins the sum (in dollars) of the total numberof fingers shown.

Solve the situation where each player must show either one or two fingers.


23.9 Linear Optimization – Applications

Formulate the following problems as linear optimization models. Where appropri-ate, you may wish to number the commodities and then use subscripted variables.

1. An oil refinery has three types of inputs, with the following prices and char-acteristics:

Input Price % Sulphur Thermal Value# per litre (by mass) (kilojoules/litre)1 $0.42 2.2 15,0002 $0.76 0.4 20,0003 $0.60 1.0 17,000

The inputs are blended to produce two outputs, with the following outputsand promised specifications:

Output Price Maximum % Minimum Thermal Value# per litre Sulphur (by mass) (kilojoules/litre)1 $0.63 1.2 16,0002 $0.91 0.7 18,000

The refinery has a capacity of 1,000,000 litres/month overall. Subject to theoverall capacity, up to 500,000 litres of any input, or 650,000 litres of anyoutput can be handled.

We can assume that all three inputs have identical densities, thereby en-abling the sulphur percentages to be treated as if they were by volume. Wecan also assume that there are no losses in the blending process, and thatthe characteristics of the outputs are a weighted average (by volume) of thecharacteristics of the inputs.

2. A large unionized restaurant is planning its workforce schedule. The re-quirements for employees over the seven day work week are:


Minimum NumberDay of EmployeesSunday 110Monday 81Tuesday 85Wednesday 118Thursday 124Friday 112Saturday 120

The collective bargaining agreement states that all employees are to workfive consecutive days per week with two consecutive days off (Saturday andSunday are consecutive). Such a schedule might mean that some employeesshow up for work (for which they are paid) but they are not required (forexample, the schedule might assign 119 employees on Wednesday). Therestaurant manager wishes to minimize such wastage. (Ignore the fact thatthe number of employees must be integer.)

3. A farmer owns 500 hectares of land in an arid region. The state governmentgives him up to 1,000,000 cubic metres of water for irrigation each year.In addition, he may purchase up to an additional 300,000 cubic metres ofwater per annum at a cost of $0.04 per cubic metre.

He grows corn, peas, and onions. The net revenue per hectare of each com-modity (excluding the cost of purchased water, if any) and the water re-quirement in cubic metres per hectare are:

Commodity Revenue Water Requirementper Hectare (cubic metres per hectare)

Corn $200 3500Peas $400 6700Onions $300 2000

He wishes to diversify his crop in case one commodity suffers an unantic-ipated fall in price. Therefore, no commodity may occupy more than 50%of the total area planted, nor may any commodity occupy less than 10% ofthe total area planted.

4. Columbian, Peruvian, and Nigerian coffee beans can be purchased for $1.20,$1.00, and $0.90 per kilogram respectively. From these sources a company


makes a “regular” and a “premium” blend of coffee, which sell for $1.30and $1.60 per kilogram respectively. The regular blend contains at least10% (by mass) Columbian beans, and at least 20% Peruvian beans. Thepremium blend contains at least 50% Columbian beans, and no more than15% Nigerian beans. The maximum market demand is for 200,000 kilo-grams of regular coffee and for 130,000 kilograms of premium coffee.

5. City bus drivers work two three and a half-hour shifts per day. In somecases, the two shifts are consecutive (effectively one seven-hour shift), butusually they are not. Because of the inconvenience of breaking up the day,those who work non-consecutive shifts are paid a $5 per day bonus. All busdrivers earn a base rate of $10 per hour. The bus company has the followingdaily requirements:

Minimum NumberTime of Day of Drivers Needed5:30 a.m. to 8:59 a.m. 1509:00 a.m. to 12:29 p.m. 8012:30 p.m. to 3:59 p.m. 904:00 p.m. to 7:29 p.m. 1607:30 p.m. to 10:59 p.m. 70

Subject to meeting all its requirements for drivers, the bus company wishesto minimize its daily labour cost (regular and bonus).

6. Wines from the Alsace, Burgundy, and Marseille regions of France are soldat wholesale for 20, 15, and 10 FF (French francs) per litre respectively. AParisian company buys these wines and blends them into two types, “do-mestic”, and “foreign”, which they sell for 25FF and 35FF per litre respec-tively. The domestic wine contains (by volume), no more than 12% Alsaceand at least 50% Burgundy wine. The foreign wine must be at least 70% Al-sace and no more than 5% Marseille. The supply of Alsace, Burgundy, andMarseille wines is limited to 15,000, 18,000, and 21,000 litres respectively.

7. Food service employees work two four-hour shifts per day. In some cases,the two shifts are consecutive (effectively one eight-hour shift), but usuallythey are not. However, the management has stipulated that no one will beasked to work from 7 a.m. to 10:59 a.m. and then again at night from 11p.m. to 2:59 a.m.


All food service workers earn a base rate of $5 per hour. In addition, there isan evening bonus of $1.20 per hour paid for each hour worked from 7 p.m.to 2:59 a.m. The restaurant has the following daily requirements:

Minimum NumberTime of Day of Workers Needed7 a.m. to 10:59 a.m. 10011:00 a.m. to 2:59 p.m. 2303:00 p.m. to 6:59 p.m. 1107:00 p.m. to 10:59 p.m. 18011:00 p.m. to 2:59 a.m. 80

Subject to meeting all its requirements for workers, the restaurant wishes tominimize its daily labour cost (regular and bonus).

8. A company has firm orders for the following quantities over the next sixmonths:

Month 1 2 3 4 5 6Demand 200 300 700 500 100 400

To change the production level from one month to the next costs $2 perunit increased or $5 per unit decreased. To hold a unit in inventory for onemonth costs $4. No shortages are permitted.

The company starts off (think of this as month 0) with 50 units in inventory.There must be no inventory left over at the end of month six. The previousmonth’s (month 0) production level was 240 units. There is no restrictionon the level of production in month six. The company wishes to minimizethe sum of production level change costs and inventory holding costs overthe six month horizon.

9. A company which produces a single product has definite orders for thisproduct over the next four quarters as follows:

Quarter 1 2 3 4Demand 350 680 275 590


The company ended the previous year with an inventory of 200 units, andthe final quarter production level was 400 units. The company wishes toend this year with an inventory of at least 250 units.

It costs $3.50 per unit to increase the production level from one quarter tothe next, and $6.00 per unit to decrease it. The cost of holding inventoryfrom one quarter to the next is $4.80 per unit per quarter. No shortages arepermitted. The company wishes to minimize the sum of production levelchange costs and inventory costs over the four quarter planning horizon.

10. A furniture manufacturing company has three plants which need 474, 620,and 870 tonnes of lumber per week. There are three lumber mills, eachof which can supply any of the three plants. The price of lumber is $240per tonne, at each lumber mill. Shipping charges, which vary according towhere the lumber is sent, are additional.

Shipping Charges in Dollars per TonneLumber Furniture Plant

Mill 1 2 31 2.20 5.50 4.102 3.20 4.40 3.003 1.50 3.60 3.50

Because of commitments to other customers, lumber mills 1 and 2 are lim-ited to 700 tonnes per week (each) to the furniture company, while mill 3can supply any demand. Mills 1 and 3 can only ship 350 tonnes each to anyone furniture plant; there is no such restriction on mill 2.

Hint: Use doubly-subscripted variables.

11. A military hospital in an isolated area needs to make 25 kilograms of avaccine. The vaccine consists of four ingredients, each of which is costlyand in short supply:

Ingredient Amount on Hand Cost(kgs.) ($/kg.)

1 22 282 18 353 20 524 24 26


In order for the vaccine to be effective, the following restrictions must bemet.

(i) Ingredient 1 must be between 45% and 60% of the total mixture (bymass).

(ii) Ingredients 2 and 3 must each comprise at least 10% of the mixture, buttogether must not exceed 25%.

(iii) Ingredient 4 must not be more than 50% of the total.

(a) Formulate a cost minimization model.

(b) Suppose that because of heavy casualties that as much vaccine as pos-sible is desired, and that money is of no concern. What is the objectivefunction now?

12. Precision Printers estimates that it will need 300 tonnes, 400 tonnes and450 tonnes of paper for the next three months. At present it has 50 tonneson hand and considers a “safety stock” of 50 tonnes at the end of the thirdmonth to be desirable. The firm’s storage area will not hold more than 500tonnes.

During the first month the price of paper is $6,000 per tonne and rises to$7,500 in the second and third month due to an anticipated strike at theregular supplier. As a result of humidity and other storage factors, 10% ofthe paper inventory on hand at the end of a month must be discarded.

Formulate the appropriate model for the management of Precision Printers.

13. We-Make-Um-Well has received a lucrative contract to manufacture twotypes of industrial components. The contract calls for certain quantitieseach year for the next four years. There is no prospect of production afterthat. The contracted quantities are:

DemandYear Type A Type B

1 4,000 6,0002 15,000 20,0003 10,000 8,0004 5,000 4,000


Each type of component needs a special machine at a particular stage of themanufacturing. On an annual basis, one machine will handle 3000 Type Acomponents or 2000 Type B components or any proportional combinationof them.

These machines can be either purchased or leased. To buy a machine costs$100,000 in year 1 and will increase by $5,000 each following year. Eachyear of its life the machine loses $10,000 of its value, except in the first yearwhen it loses $15,000. At the end of the four year planning horizon, ma-chines which have purchased can be sold at their salvage value. Machinescan be rented for either a 1-year or 2-year term at rental charges starting inyear 1 at $24,000 and $20,000 per year, respectively. Each year the rentalcharge will increase by $2,000, except that machines already on hire stay attheir original charge until the rental period expires. All expenditures are as-sumed to occur at the beginning of each year. The firm wishes to minimizethe present value of the net cost associated with these machines where thecompany’s discount rate is 20%. Inventory may be carried over from oneyear to the next at a cost of $2 per unit.

Formulate an appropriate model for We-Make-Um-Well’s management.

14. Due to recent market forecasts, Buns Bakery estimates that it will need anadditional 100 donut machines for its various facilities for the next twoyears. After two years a more efficient model is due to be available andall machines will be replaced at this time. Thus any machines on hand atthe end of two years will be sold at their depreciated value. Each machinehas a two year life and is depreciated linearly to a salvage value of $1,000.Machines can be purchased for $5,000 or leased for two years at $2,800 peryear payable at the beginning of each year. To break a lease would cost$300 per machine.

Buns has $150,000 in uncommitted funds that can be used to lease and/orbuy machines at the outset of the first year. In addition, Buns can obtaina loan at the outset of the first year for up to $300,000 at 13% interest peryear. The terms of the loan require Buns to repay the amount borrowed plusinterest at the end of the first year.

Each of the 100 machines will contribute $3,500 to profits each year, there-fore $350,000 will become available at the end of each year for use in thefollowing year. Unused cash earns 10% per year.


Formulate the appropriate model for the management of Buns Bakery. (Hint:Maximize the ending cash position.)

15. The owners of a fast food restaurant must make several staffing decisionsover the next several months. For January, February, March, April, May,and June, they require 1670, 1550, 1830, 1980, 1770, and 2040 labour hoursrespectively. This work is performed by employees who have completed aone month training program.

At the outset of the year, there are 12 trained employees available and will-ing for work. Each trained employee works up to 160 hours per month. Atthe end of each month, approximately 10% of the trained employees whoworked that month will be fired, and another 15% will resign for personalreasons. Each employee makes $960 per month (or $6 per hour).

Because of local labour conditions, it is unrealistic to expect that more thanseven trainees can be hired in any month. Each trainee is paid only $760per month, but costs another $440 in non-salary training expenses. The non-salary training cost will rise to $690 per trainee in April. Furthermore, theirinexperience means that the restaurant receives only a negligible amountof useful productive work from the trainees. However, the trainees are allkeeners, so firings and resignations are virtually non-existent.

Upon completion of their training, the restaurant can hire them as regularemployees immediately, or it can put them on retainer at $75 per personper month. Anyone on retainer can be re-called at the restaurant’s whim atany time. Such re-called employees do not require any further training.If the restaurant does not pay the $75 retainer fee, the right to recall islost. Because of this, every trainee is either hired as a regular employeethe following month or is put on retainer.

Formulate a linear optimization model to determine a minimum cost staffingschedule for the six month period.

Hint: What are the decisions to be made at the beginning of each month?

16. The state of Kaybeck produces three major types of outputs for the exportmarket: electrical energy, aluminum ingots, and high-tech products. Eachcommodity can be sold on the world market in units worth $50, $90, and$200 respectively.

Each unit of electrical energy produced requires 0.1 units of aluminum and0.05 units of high-tech products. To produce 1 unit of aluminum requires 1.2


units of electrical energy, $15 worth of imported bauxite ore, and 0.04 unitsof high-tech products. Each unit of high-tech product is produced using0.7 units of energy, 0.2 units of aluminum, and $120 worth of importedmaterials.

The state of Kaybeck has a capacity to produce up to 500,000 units of energyand 350,000 units of aluminum. There is no production capacity limit forhigh-tech products, but it is unlikely that more than 100,000 units could beexported.

(a) Assume for the moment that energy, aluminum, and high-tech productscannot be imported. Formulate a model which will maximize the dollarvalue of the trading surplus.

(b) Suppose now that there are no import restrictions. Make the appropriatemodifications to the model given in part (a).

(c) For the model given in part (b), give an argument as to why some someof the variables will be at their upper limits. From this deduce the values ofall the variables.

17. A paper mill produces rolls of paper in standard widths of 90 cm and 200cm. All paper produced has a thickness of 0.1 mm, and each roll has alength of 1000 metres. The customers all desire rolls of this thickness andthis length, but not necessarily either of the two standard widths. The millcurrently has the following non-standard width orders:

Width (cm) Number95 2580 3146 2321 68

The paper which is leftover on the cut rolls is re-cycled. Formulate a modelwhich will minimize the amount of paper which needs to be re-cycled.

18. J. W. Aut Co. has three major products: X, Y and Z. These are producedusing two parallel production lines. The single-product capacities of theselines over the planning horizon are 1000, 1400 and 3000 on line one and2500, 1600 and 1000 on line two for X, Y and Z respectively. If more thanone product is produced on either line, then the capacity of that line must


be determined in an appropriate manner. The lines are such that you mayswitch from one product to another without any setup cost.

There are two components (A and B) used in making these products whichare in relatively short supply. There are only 2500 units of component A,and 3000 units of component B. Products X, Y and Z need one, one, andtwo units of component A, and two, one, and one units of component Brespectively.

In addition, the production of Z must not be less than the combined pro-duction of X and Y. Marketing requires that there be at least twice as manyunits of product Y as of product X. The profit contributions are 100, 150and 300 for products X, Y and Z respectively.

(a) Formulate this problem.

(b) Use a computer to help give a recommendation to the management of J.W. Aut Co.

19. A student needs to prepare for three mid-term exams, each of which is ofequal importance to him. Without doing any preparation, he believes thathis marks will be 20, 25, and 30 in courses 1, 2, and 3 respectively. He hastwo methods of preparation: reading his texts and problem solving. He has20 hours available for studying. Based on his past experience, the studentestimates that one hour of study will improve his marks according to thefollowing table:

If devoted toCourse 1 Course 2 Course 3

Reading 5 marks 7 marks 6 marksProblems 9 marks 4 marks 8 marks

The student wishes obtain at least 50 on each exam and of course cannotobtain more than 100 in any exam. In addition, neither method of prepara-tion (nor any combination of the two) will improve his mark by more than60 marks in any course.

(a) Formulate this problem, omitting any constraints which are trivially re-dundant.

(b) Use a computer to help decide what the student should do.


20. The Three Hills Gas Company supplies its distributors using a fleet of gaso-line tankers. A new region has been acquired which will require expansionof the existing fleet. The company has $1,600,000 available to finance thecapital costs of the expansion. Three models of gasoline tanker are avail-able.

Truck Capacity Purchase MonthlyModel (litres) Cost Operating

(in $) Costs (in $)Super 20,000 150,000 800Regular 10,000 100,000 500Econoline 5,000 75,000 200

Depreciation has been included in the monthly operating cost. The com-pany estimates that the monthly demand for the new region will be a totalof two million litres of gasoline. Due to the size and speed differences ofthe trucks, the different truck models vary in terms of the number of deliv-eries or round trips possible per month. The maximum number of trips areestimated at twenty per month for the Super, twenty-five per month for theregular and thirty per month for the Econoline. Based on driver availabilityand maintenance capability, the firm does not want to add more than twentynew vehicles to the fleet. In addition, the company would like to make sureit purchases at least four of the new Econoline tankers to use on short run,low demand routes. Finally, the company does not want more than halfof the new models to be Supers. An added complication is that there is a$150 per month per vehicle fleet charge for insurance which has not beenincluded in the operating costs.

(a) Formulate a model for the management of Three Hills.

(b) Use a computer to help give a recommendation.

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