Quantitative methods notes

1. Quantitative Methods for the Management Sciences 45-760 Course Notes Instructors: Gerard Cornuejols Michael Trick Graduate School of Industrial Administration Carnegie Mellon University, Pittsburgh, PA 15213 USA Fall 1998

2. Contents 0.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 4 0.2 Grading Components : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5 0.3 Schedule : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6 0.4 Mathematical Methods in Business : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6 0.4.1 History : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6 0.4.2 Mathematical Modeling Today : : : : : : : : : : : : : : : : : : : : : : : : : : 7 1 Basic Linear Algebra 9 1.1 Linear Equations : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9 1.2 Operations on Vectors and Matrices : : : : : : : : : : : : : : : : : : : : : : : : : : : 12 1.3 Linear Combinations : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 15 1.4 Inverse of a Square Matrix : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 20 1.5 Determinants : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 23 1.6 Positive De

3. nite Matrices : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 23 2 Unconstrained Optimization: Functions of One Variable 25 2.1 Derivatives : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 25 2.2 Maximum and Minimum : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 26 2.3 Binary Search : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 30 2.4 Convexity : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 33 3 Unconstrained Optimization: Functions of Several Variables 37 3.1 Gradient : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 37 3.2 Maximum and Minimum : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 39 3.3 Global Optima : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 41 4 Constrained Optimization 43 4.1 Equality Constraints (Lagrangians) : : : : : : : : : : : : : : : : : : : : : : : : : : : : 43 4.1.1 Geometric Interpretation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 45 4.1.2 Economic Interpretation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 45 4.2 Equality and Inequality Constraints : : : : : : : : : : : : : : : : : : : : : : : : : : : 49 4.3 Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 53 5 Modeling with Linear Programming 57 5.1 Introductory Example : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 57 5.1.1 Modeling : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 57 5.1.2 Solving the Model : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 58 5.1.3 Using Solver : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 61 1

4. 2 CONTENTS 5.2 Second Example: Market Allocations : : : : : : : : : : : : : : : : : : : : : : : : : : : 62 5.2.1 Modeling : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 63 5.2.2 Solving the Model : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 63 5.2.3 Using Solver : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 64 5.3 Common Features : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 64 5.4 Linear Programming Models : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 65 5.4.1 Diet Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 65 5.4.2 Workforce Planning : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 66 5.4.3 Financial Portfolio : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 67 5.4.4 Production over Time : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 69 5.4.5 Investing over Time : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 70 5.4.6 Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 72 6 Case Study 79 7 The Simplex Method 87 7.1 Linear Programs in Standard Form : : : : : : : : : : : : : : : : : : : : : : : : : : : : 87 7.2 Solution of Linear Programs by the Simplex Method : : : : : : : : : : : : : : : : : : 88 7.3 Alternate Optimal Solutions, Degeneracy, Unboudedness, Infeasibility : : : : : : : : 92 8 Sensitivity Analysis for Linear Programming 97 8.1 Tableau Sensitivity Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 97 8.1.1 Cost Changes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 97 8.1.2 Right Hand Side Changes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 99 8.1.3 New Variable : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 100 8.2 Solver Output : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 104 8.2.1 Tucker Automobiles : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 104 8.2.2 Carla's Maps : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 106 9 Decision Theory 113 9.1 Decisions Under Risk : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114 9.2 Decision Trees : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 115 10 Game Theory 125 10.1 Two-Person Games : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 125 10.1.1 Pure Strategies : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 125 10.1.2 Mixed Strategies : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 129 10.2 n-Person Games : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 132 11 Network Optimization 135 11.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 135 11.2 Terminology : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 135 11.3 Examples : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 136 11.3.1 Shortest Paths : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 136 11.3.2 Maximum Flow : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 137 11.3.3 Transportation Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 138 11.3.4 Assignment Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 141 11.4 Unifying Model: Minimum Cost Network Flows : : : : : : : : : : : : : : : : : : : : : 142

5. CONTENTS 3 11.5 Generalized Networks : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 144 11.6 Conclusions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 145 12 Data Envelopment Analysis 147 12.1 Numerical Example : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 147 12.2 Graphical Example : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 148 12.3 Using Linear Programming : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 149 12.4 Applications : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 155 12.5 Strengths and Limitations of DEA : : : : : : : : : : : : : : : : : : : : : : : : : : : : 155 12.6 References : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 156

6. 4 CONTENTS Course Syllabus 0.1 Introduction Objectives of the Course: to provide a formal quantitative approach to problem solving; to give an intuition for managerial situations where a quantitative approach is appropriate; to introduce some widely used quantitative models; to introduce software for solving such models. To accomplish these objectives we provide: background in linear algebra and nonlinear optimization; description of the linear programming model, decision theory, etc.; explanation of the mathematical ideas behind these widely used models; hands-on experience with software such as SOLVER; examples of business applications; a case study that shows how models are used in practice. Instructors Gerard Cornuejols, GSIA 232A, phone 268-2284, fax 268-7357 email: gc0v@andrew.cmu.edu secretary: Barbara Carlson 268-1342, GSIA 232. Michael Trick, GSIA 257C, phone 268-3697, fax 268-7057 email: trick@andrew.cmu.edu Required Material for the Course The course notes are self contained. The software for this course is installed in your notebooks. Instruction on using it will be provided in class.

7. 0.2. GRADING COMPONENTS 5 0.2 Grading Components The course grade will be based on homework, a case study, and two exams. Homework The only way to learn mathematical material is by working problems. Homework is due at the beginning of class on the due date (see schedule). No late homework will be accepted. You are encouraged to work exercises that are not assigned. Some can be found in the course notes. More can be found in the textbook by Winston, Operations Research: Applications and Algorithms, Duxbury Press (1994). Copies are on reserve in the Hunt library. You should do the homework without consulting other students. If you need help, you may consult the instructor or a teaching assistant. For nonassigned exercises, it is a good idea to work them out with other students. Every attempt will be made to return graded homeworks promptly. Solutions to the homework will be distributed in class after the due date. The following rules apply to all homework: { Use only 11 by 8.5 in. paper. { Start each problem on a new page. { Assemble the homework in proper order and staple it together. { PRINT your name on the top right hand corner. Case Study The case study has two parts. The

8. rst part will require formulating a linear program. The second part requires solving the linear program and analyzing the results for managerial purposes. Exams There will be one midterm and one

9. nal exam. The precise dates of the exams will be announced by the GSIA administration. The questions in the exams will be similar to the homework problems and the problems discussed in class both in form and content. The lectures may occasionally range over peripheral topics, but the content of the exercises is a reliable guide to which topics can appear on the exam. All exams will be open-notes. Except for provable unforeseen circumstances, makeup exams will not be given. A student is required to inform the instructor as soon as possible in such a case. In case of an unavoidable con ict, it is the student's responsibility to schedule a makeup exam before the exam date after discussing with the instructor. Grading Weights Midterm exam = 25%, homework = 15%, case study = 25%,

10. nal exam = 35%.

11. 6 CONTENTS 0.3 Schedule It is advisable to read through the material in the course notes prior to when it is discussed in class. Date Topic Chapter Work Due Aug 25 Linear Algebra Chapter 1 Aug 27 Functions of One Variable Chapter 2 Sept 1 Functions of Several Variables Chapter 3 HW 1 due Sept 3 Optimization with Constraints Chapter 4 Sept 8 LP modeling / Solver Chapter 5 HW 2 due Sept 10 LP modeling Chapter 5 Sept ?? MIDTERM EXAM TBA Sept 15 The Simplex Method Chapter 7 HW 3 due Sept 17 Sensitivity Analysis Chapter 8 Sept 22 Case Study Part I Chapter 6 Case Part I due Sept 24 Decision Theory Chapter 9 Sept 29 Game Theory Chapter 10 HW 4 due Oct 1 Network Models Chapter 11 Oct 6 Data Envelopment Analysis Chapter 12 Case Part II due Oct 8 Problem Session HW 5 due Oct ?? FINAL EXAM TBA 0.4 Mathematical Methods in Business 0.4.1 History Mathematical methods have long played an important role in management and economics but have become increasingly important in the last few decades. Business mathematics, such as the computation of compound interest, appeared in ancient Mesopotamia but became prevalent in the mercantile economy of the Renaissance. Mathematics later contributed to the engineering advances that made the Industrial Revolution possible. Mathematics crept into economics during the 18th and 19th centuries and became

12. rmly established in the 20th. It is only in the last few decades that management per se has received the sort of rigorous study that permits the application of mathematics. In the early part of this century scienti

13. c management, called Taylorism after its founder F. Taylor, was fashionable. This movement made some worthwhile contributions, such as time-and-motion studies, but it was coupled with dreadful labor relations practices. The symbol of the movement was the eciency expert policing the shop oor with stopwatch and clipboard. After Taylorism waned, interest shifted more to making ecient use of labor and other factors than making employees work harder. Shortly before the Second World War a team of scientists solved the operational problems of coordinating Britain's radar stations and thereby created operational research, called operations research in the U.S. During the war this practice of putting scientists and mathematicians to work on operational problems was surprisingly successful in both the British and American military, partly because members of the professional class who had not previously dirtied their hands had their

14. rst opportunity to apply their training to operational

15. 0.4. MATHEMATICAL METHODS IN BUSINESS 7 problems. Their success attracted attention, and operations research spread to industry during the

16. fties. After the war, G. Dantzig and others developed linear programming at about the same time that computers became available to solve linear programming models. Linear programming proved a remarkably useful tool for the ecient allocation of resources, and this gave a great boost to operations research. In the early postwar years J. von Neumann and O. Morgenstern invented game theory (closely related to linear programming), and H. W. Kuhn and A. W. Tucker broke ground in nonlinear programming. W. E. Deming and others developed statistical techniques for quality control, and the statistical methods

17. rst designed for psychometrics in the early 20th century became the mathematical basis for econometrics. Meanwhile business schools began teaching many of these techniques along with microeconomics and other quantitative

18. elds. As a result of all this, mathematical modeling has played an increasingly important role in management and economics. In fact one can make a case that overreliance on a mathematical approach has sometimes led to neglect of other approaches. After its

19. rst few years, operations research focused almost exclu- sively on mathematical modeling. The reigning orthodoxy in much of economics is the neoclas- sical paradigm, which is heavily mathematical. There is little communication between people in quantitative

20. elds and those in such soft

21. elds as organizational science and general management. In fact both approaches appear to be in a crisis. Economics and operations research have achieved much, but neither can adequately understand phenomena that involve human beings. The soft sciences take on the dicult, ill-structured, human-oriented problems, but it is unclear even what counts as a satisfactory theory in these areas. One can hope that this double crisis will lead in the coming years to a paradigm shift that will supplant the hard and soft approaches with an integrated science with the rigor of one and the breadth of the other. There are faint signs of such a shift, but they are not suciently crystallized to show up in a business school curriculum. Despite the limitations of the quantitative methods discussed in this course, in many speci

22. c areas they are quite useful. In fact the potential of most mathematical techniques, particularly those of operations research, is only beginning to be realized in practice. Aordable personal computers and user-friendly software are motivating managers to learn about mathematical models and use them. Also the line between managerial and technical sta is blurring, so that managers now run models on their notebook computers that systems analysts once ran on mainframes. This again leads to more widespread use. 0.4.2 Mathematical Modeling Today The applications of mathematical methods in management and economics today are so manifold that it is dicult to

23. nd a single person who is aware of their full scope. The following list can only be incomplete. Economists use linear and nonlinear programming, the theory of variational inequalities, optimal control theory, dynamic programming, game theory, probability choice models, utility theory, regression and factor analysis and other techniques to study equilibrium, optimal investment, competition, consumer behavior, and a host of other phenomena. People in operations management use statistical sampling and estimation theory, linear and integer programming, network programming, dynamic programming and optimal control theory, queuing theory, simulation, arti

24. cial intelligence techniques, and combinatorial optimization methods to solve problems in quality control, allocation of resources, logistics, project

25. 8 CONTENTS scheduling, labor and machine scheduling, job shop scheduling and assembly line balancing, and facility layout and location. The introduction of exible manufacturing systems, robots and other automated devices has posed a whole new array of unsolved mathematical problems. People in

26. nance use linear, nonlinear and integer programming, optimal control theory and dynamic programming, Markov decision theory, regression and time series to determine optimal resource allocation, multiperiod investments, capital budgeting, and investment and loan portfolio design, and to try to forecast market behavior. People in marketing use regression and factor analysis, time series, game theory, Markov decision theory, location theory, mathematical programming, probability choice models and utility theory to study consumer preferences, determine optimal location in product space, allocate advertising resources, design distribution systems, forecast market behavior, and study competitive strategy. People in information systems and decision support systems use arti

27. cial intelligence techniques, propositional and quanti

28. ed logic, Bayesian methods, probabilistic logic, data struc- tures and other computer science techniques, mathematical programming, and statistical decision theory to design expert and other knowledge-based systems, develop ecient inference and retrieval methods, and evaluate the economic and organizational eects of information systems. The peculiar nature of mathematics unfortunately raises two obstacles to learning it. One is that students often believe that mathematics can be learned simply by studying a book. Mathematics is learned by working through problems. A second obstacle is that students often believe that they can learn to work problems by studying a book. A book can get one started, but learning to work problems is like learning to play basketball or play the piano|it requires practice, practice, practice. Mathematical skills are very athletic in this sense. (The phrase quant jock is appropriate.) That is why this course assigns a lot of homework problems. Acknowledgements: We would like to thank John Hooker, Bill Hrusa, Rick Green and Anuj Mehrotra for their inputs.

29. Chapter 1 Basic Linear Algebra Linear equations form the basis of linear programming. If you have a good understanding of the Gauss-Jordan method for solving linear equations, then you can also understand the solution of linear programs. In addition, this chapter introduces matrix notation and concepts that will be used in Chapters 3 and 4 (optimizing functions of several variables). 1.1 Linear Equations The Gauss-Jordan elimination procedure is a systematic method for solving systems of linear equations. It works one variable at a time, eliminating it in all rows but one, and then moves on to the next variable. We illustrate the procedure on three examples. Example 1.1.1 (Solving linear equations) x1 + 2x2 + x3 = 4 2x1 x2 + 3x3 = 3 x1 + x2 x3 = 3 In the

30. rst step of the procedure, we use the

31. rst equation to eliminate x1 from the other two. Speci

32. cally, in order to eliminate x1 from the second equation, we multiply the

33. rst equation by 2 and subtract the result from the second equation. Similarly, to eliminate x1 from the third equation, we subtract the

34. rst equation from the third. Such steps are called elementary row operations. We keep the

35. rst equation and the modi

36. ed second and third equations. The resulting equations are x1 + 2x2 + x3 = 4 5x2 + x3 = 5 x2 2x3 = 1 Note that only one equation was used to eliminate x1 in all the others. This guarantees that the new system of equations has exactly the same solution(s) as the original one. In the second step of the procedure, we divide the second equation by -5 to make the coecient of x2 equal to 1. Then, we use this equation to eliminate x2 from equations 1 and 3. This yields the following new system of equations. x1 + 7 5x3 = 2 x2 1 5x3 = 1 11 5 x3 = 0 9

37. 10 CHAPTER 1. BASIC LINEAR ALGEBRA Once again, only one equation was used to eliminate x2 in all the others and that guarantees that the new system has the same solution(s) as the original one. Finally, in the last step of the procedure, we use equation 3 to eliminate x3 in equations 1 and 2. x1 = 2 x2 = 1 x3 = 0 So, there is a unique solution. Note that, throughout the procedure, we were careful to keep three equations that have the same solution(s) as the original three equations. Why is it useful? Because, linear systems of equations do not always have a unique solution and it is important to identify such situations. Another example: x1 + 2x2 + x3 = 4 x1 + x2 + 2x3 = 1 2x1 + 3x2 + 3x3 = 2 First we eliminate x1 from equations 2 and 3. x1 + 2x2 + x3 = 4 x2 + x3 = 3 x2 + x3 = 6 Then we eliminate x2 from equations 1 and 3. x1 + 3x3 = 2 x2 x3 = 3 0 = 3 Equation 3 shows that the linear system has no solution. A third example: x1 + 2x2 + x3 = 4 x1 + x2 + 2x3 = 1 2x1 + 3x2 + 3x3 = 5 Doing the same as above, we end up with x1 + 3x3 = 2 x2 x3 = 3 0 = 0 Now equation 3 is an obvious equality. It can be discarded to obtain x1 = 2 3x3 x2 = 3 + x3 The situation where we can express some of the variables (here x1 and x2) in terms of the remaining variables (here x3) is important. These variables are said to be basic and nonbasic respectively. Any choice of the nonbasic variable x3 yields a solution of the linear system. Therefore the system has in

38. nitely many solutions.

39. 1.1. LINEAR EQUATIONS 11 It is generally true that a system of m linear equations in n variables has either: (a) no solution, (b) a unique solution, (c) in

40. nitely many solutions. The Gauss-Jordan elimination procedure solves the system of linear equations using two elementary row operations: modify some equation by multiplying it by a nonzero scalar (a scalar is an actual real number, such as 1 2 or -2; it cannot be one of the variables in the problem), modify some equation by adding to it a scalar multiple of another equation. The resulting system of m linear equations has the same solution(s) as the original system. If an equation 0 = 0 is produced, it is discarded and the procedure is continued. If an equation 0 = a is produced where a is a nonzero scalar, the procedure is stopped: in this case, the system has no solution. At each step of the procedure, a new variable is made basic: it has coecient 1 in one of the equations and 0 in all the others. The procedure stops when each equation has a basic variable associated with it. Say p equations remain (remember that some of the original m equations may have been discarded). When n = p, the system has a unique solution. When n p, then p variables are basic and the remaining n p are nonbasic. In this case, the system has in

41. nitely many solutions. Exercise 1 Solve the following systems of linear equations using the Gauss-Jordan elimination procedure and state whether case (a), (b) or (c) holds. (1) 3x1 4x3 = 2 x1 + x2 + x3 = 4 2x2 + x3 = 3 (2) 2x1 + 2x2 x3 = 1 4x1 + x3 = 2 x1 x2 + x3 = 2 (3) x1 x2 + x3 = 1 2x1 + 2x2 2x3 = 2 x1 + x2 x3 = 1

42. 12 CHAPTER 1. BASIC LINEAR ALGEBRA Exercise 2 Indicate whether the following linear system of equations has 0, 1 or in

43. nitely many solutions. x1 + 2x2 + 4x3 + x4 + 3x5 = 2 2x1 + x2 + x3 + 3x4 + x5 = 1 3x2 + 7x3 x4 + 5x5 = 6 1.2 Operations on Vectors and Matrices It is useful to formalize the operations on vectors and matrices that form the basis of linear algebra. For our purpose, the most useful de

44. nitions are the following. A matrix is a rectangular array of numbers written in the form A =0B@ a11 a12 : : : a1n ... ... ... am1 am2 : : : amn 1CA : The matrix A has dimensions mn if it has m rows and n columns. When m = 1, the matrix is called a row vector; when n = 1, the matrix is called a column vector. A vector can be represented either by a row vector or a column vector. Equality of two matrices of same dimensions: Let A =0B@ a11 : : : a1n ... ... am1 : : : amn 1CA and B =0B@ b11 : : : b1n ... ... bm1 : : : bmn . 1CA Then A = B means that aij = bij for all i; j. Multiplication of a matrix A by a scalar k: kA =0B@ ka11 ka12 : : : ka1n ... ... ... kam1 kam2 : : : kamn 1CA : Addition of two matrices of same dimensions: Let A =0B@ a11 : : : a1n ... ... am1 : : : amn 1CA and B =0B@ b11 : : : b1n ... ... bm1 : : : bmn . 1CA Then A + B =0B@ a11 + b11 : : : a1n + b1n ... ... am1 + bm1 : : : amn + bmn 1CA : Note that A + B is not de

45. ned when A and B have dierent dimensions. Exercise 3 Compute 3 2 3 1 0 5 4 ! 2 0 4 1 1 6 6 !.

46. 1.2. OPERATIONS ON VECTORS AND MATRICES 13 Multiplication of a matrix of dimensions m n by a matrix of dimensions n p: Let A = 0BBBBBBB@ a11 : : : a1n ... ... ai1 : : : ain ... ... am1 : : : amn 1CCCCCCCA and B =0B@ b11 : : : b1j : : : b1n ... ... ... bm1 : : : bmj : : : bmn . 1CA Then AB is a matrix of dimensions m p computed as follows. AB =0B@ a11b11 + : : :+ a1nbn1 : : : a11b1p + : : :+ a1nbnp ... ai1b1j + : : :+ ainbnj ... am1b11 + : : :+ amnbn1 : : : am1b1p + : : :amnbnp 1CA : As an example, let us multiply the matrices A = 2 3 4 5 1 0 !and B =0B@ 11 12 14 15 17 18 1CA : The result is AB = 2 11 + 3 14 + 4 17 2 12 + 3 15 + 4 18 5 11 + 1 14 + 0 17 5 12 + 1 15 + 0 18 ! =0B@ 132 141 69 75 1CA : Note that AB is de

47. ned only when the number of columns of A equals the number of rows of B. An important remark: even when both AB and BA are de

48. ned, the results are usually dierent. A property of matrix multiplication is the following: (AB)C = A(BC): That is, if you have three matrices A, B, C to multiply and the product is legal (the number of columns of A equals the number of rows of B and the number of columns of B equals the number of rows of C), then you have two possibilities: you can

49. rst compute AB and multiply the result by C, or you can

50. rst compute BC and multiply A by the result. Exercise 4 Consider the following matrices x = 1 1 1 y = 3 2 1 z =0B@ 1 1 1 1CA : When possible, compute the following quantities: (a) xz (b) zx (c) yzx (d) xzy (e) (x + y)z

51. 14 CHAPTER 1. BASIC LINEAR ALGEBRA (f) (xz) + (yz). Remark: A system of linear equations can be written conveniently using matrix notation. Namely, a11x1 + : : : + a1nxn = b1 : : : : : : : : : am1x1 + : : : + amnxn = bm can be written as 0B@ a11 : : : a1n ... ... am1 : : : amn 1CA 0B@ x1 ... xn 1CA =0B@ b1 ... bm 1CA or as 0B@ a11 ... am1 1CA x1 + : : :+0B@ a1n ... amn 1CA xn =0B@ b1 ... bm 1CA : So a matrix equation Ax = b where A is a given m n matrix, b is a given m-column vector and x is an unknown n-column vector, is a linear system of m equations in n variables. Similarly, a vector equation a1x1+: : :+anxn = b where a1; : : :; an; b are given m-column vectors and x1; : : :; xn are n unknown real numbers, is also a system of m equations in n variables. Exercise 5 (a) Solve the matrix equation 0B@ 1 2 1 2 1 3 1 1 1 1CA 0B@ x1 x2 x3 1CA =0B@ 4 3 3 1CA : (b) Solve the vector equation 0B@ 1 2 1 1CA x1 +0B@ 2 1 1 1CA x2 +0B@ 1 3 1 1CA x3 =0B@ 4 3 3 1CA : [Hint: Use Example 1.1.1] The following standard de

52. nitions will be useful: A square matrix is a matrix with same number of rows as columns. The identity matrix I is a square matrix with 1's on the main diagonal and 0's elsewhere, i.e. I =0BBB@ 1 0 : : : 0 0 1 : : : 0 : : : : : : : : : : : : 0 0 : : : 1 . 1CCCA The zero matrix contains only 0's. In particular, the zero vector has all components equal to 0. A nonzero vector is one that contains at least one nonzero component.

53. 1.3. LINEAR COMBINATIONS 15 The transpose of matrix A =0B@ a11 a12 : : : a1n ... ... ... am1 am2 : : : amn 1CA is the matrix AT =0BBBB@ a11 : : : am1 a12 : : : am2 ... ... a1n : : : amn 1CCCCA . A square matrix A is symmetric if its elements satisfy aij = aji for all i and j. So a symmetric matrix satis

54. es AT = A. 1.3 Linear Combinations Suppose that the vector (1; 3) represents the contents of a Regular can of mixed nuts (1 lb cashews and 3 lb peanuts) while (1; 1) represents a Deluxe can (1 lb cashews and 1 lb peanuts). Can you obtain a mixture of 2 lb cashews and 3 lb peanuts by combining the two mixtures in appropriate amounts? The answer is to use x1 cans of Regular and x2 cans of Deluxe in order to satisfy the equality x1 1 3!+ x2 1 1! = 2 3!: This is none other than a system of two linear equations: x1 +x2= 2 3x1 +x2= 3 The solution of these equations (obtained by the Gauss-Jordan procedure) is x1 = 1=2, x2 = 3=2. So the desired combination is to mix 1/2 can of Regular nuts with 3/2 cans of Deluxe nuts. Thus if some recipe calls for the mixture (2; 3), you can substitute 1/2 can of Regular mix and 3/2 can of Deluxe mix. Suppose now that you want to obtain 1 lb cashews and no peanuts. This poses the equations, x1 1 3!+ x2 1 1! = 1 0!: The solution is (x1; x2) = (1=2; 3=2). Thus you can obtain a pound of pure cashews by buying 3/2 cans of Deluxe mix and removing enough nuts to form 1/2 can Regular mix, which can be sold. In this case it is physically possible to use a negative amount of some ingredient, but in other cases it may be impossible, as when one is mixing paint. A vector of the form x1 1 3!+ x2 1 1! is called a linear combination of the vectors 1 3! and 1 1!. In particular we just saw that the vector 2 3! is a linear combination of the vectors 1 3! and 1 1!. And so is 1 0!. The question arises: can one obtain any mixture whatever by taking the appropriate combina- b2 ! a linear combination of 1 tion of Regular and Deluxe cans? Is any vector b1 3! and 1 1!? To answer the question, solve the equations in general. You want a mixture of b1 lb cashews and b2 lb peanuts and set x1 1 3!+ x2 1 1! = b1 b2 !;

55. 16 CHAPTER 1. BASIC LINEAR ALGEBRA or equivalently, x1 +x2 = b1 3x1 +x2 = b2: These equations have a unique solution, no matter what are the values of b1 and b2, namely, x1 = b2 b1 2 ; x2 = 3b1 b2 2 : No matter what vector (b1; b2) you want, you can get it as a linear combination of (1; 3) and (1; 1). The vectors (1; 3) and (1; 1) are said to be linearly independent. Not all pairs of vectors can yield an arbitrary mixture (b1; b2). For instance, no linear combination of (1; 1) and (2; 2) yields (2,3). In other words, the equations x1 1 1!+ x2 2 2! = 2 3! have no solution. The reason is that (2; 2) is already a multiple of (1; 1), that is (2; 2) is a linear combination of (1; 1). The vectors (1; 1) and (2; 2) are said to be linearly dependent. If (2,2) represents, for instance, a large Deluxe can of nuts and (1,1) a small one, it is clear that you cannot obtain any mixture you want by combining large and small Deluxe cans. In fact, once you have small cans, the large cans contribute nothing at all to the mixtures you can generate, since you can always substitute two small cans for a large one. A more interesting example is (1,2,0), (1,0,1) and (2,2,1). The third vector is clearly a linear combination of the other two (it equals their sum). Altogether, these three vectors are said to be linearly dependent. For instance, these three vectors might represent mixtures of nuts as follows: Brand A B C cashews 1 1 2 peanuts 2 0 2 almonds 0 1 1 Then once you have brands A and B, brand C adds nothing whatever to the variety of mixtures you can concoct. This is because you can already obtain a can of brand C from brands A and B anyway. In other words, if a recipe calls for brand C, you can always substitute a mixture of 1 can brand A and 1 can brand B. Suppose you want to obtain the mixture (1; 2; 1) by combining Brands A, B, and C. The equations are x1 + x2 + 2x3 = 1 2x1 + 2x3 = 2 x2 + x3 = 1 If you try to solve these equations, you will

56. nd that there is no solution. So the vector (1; 2; 1) is not a linear combination of (1; 2; 0), (1; 0; 1) and (2; 2; 1). The concepts of linear combination, linear dependence and linear independence introduced in the above examples can be de

57. ned more formally, for any number of n-component vectors. This is done as follows.

58. 1.3. LINEAR COMBINATIONS 17 A vector b having n components is a linear combination of the k vectors v1; : : :; vk, each having n components, if it is possible to

59. nd k real numbers x1; : : :; xk satisfying: x1v1 + : : :xkvk = b (1.1) To

60. nd the numbers x1; : : :; xk, view (1.1) as a system of linear equations and solve by the Gauss-Jordan method. A set of vectors (all having n components) is linearly dependent if at least one vector is a linear combination of the others. Otherwise they are linearly independent. Given n linearly independent vectors v1; : : :; vn, each having n components, any desired vector b with n components can be obtained as a linear combination of them: x1v1 + : : :xnvn = b (1.2) The desired weights x1; : : :; xn are computed by solving (1.2) with the Gauss-Jordan method: there is a unique solution whenever the vectors v1; : : :; vn are linearly independent. Exercise 6 A can of Brand A mixed nuts has 1 lb cashews, 1 lb almonds, 2 lb peanuts. Brand B has 1 lb cashews and 3 lb peanuts. Brand C has 1 lb almonds and 2 lb peanuts. Show how much of each brand to buy/sell so as to obtain a mixture containing 5 lb of each type of nut. Exercise 7 Determine whether the vector0B@ 1 2 2 1CA is a linear combination of (a)0B@ 1 0 2 1CA ,0B@ 2 0 1 1CA (b)0B@ 2 1 1 ,0B@ 1CA 1 2 1 ,0B@ 1CA 1 1 2 . 1CA In linear programming, there are typically many more variables than equations. So, this case warrants looking at another example. Example 1.3.1 (Basic Solutions.) A craft shop makes deluxe and regular belts. Each deluxe belt requires a strip of leather and 2 hours of labor. Each regular belt requires a leather strip and 1 hour of labor. 40 leather strips and 60 hours of labor are available. How many belts of either kind can be made? This is really a mixing problem. Rather than peanuts and cashews, the mixture will contain leather and labor. The items to be mixed are four activities: manufacturing a deluxe belt, manufacturing a regular belt, leaving a leftover leather strip in inventory, and leaving an hour of labor

61. 18 CHAPTER 1. BASIC LINEAR ALGEBRA idle (or for other work). Just as each Regular can of mixed nuts contributes 1 pound of cashews and 3 pounds of peanuts to the mixture, each regular belt will consume 1 leather strip and 2 hours of labor. The aim is to combine the four activities in the right proportion so that 40 strips and 60 hours are accounted for: 2!+ x2 1 x1 1 1!+ s1 1 0!+ s2 0 1! = 40 60!: So, x1 = number of deluxe belts made x2 = number of regular belts made s1 = number of leather strips left over s2 = number of labor hours left over In tableau form, the equations are: x1 x2 s1 s2 RHS 1 1 1 0 40 2 1 0 1 60 (1.3) Since there are only two equations, you can only solve for two variables. Let's solve for x1 and x2, using Gauss-Jordan elimination. After the

62. rst iteration you get, x1 x2 s1 s2 RHS 1 1 1 0 40 0 1 2 1 20 (1.4) A second iteration yields the solution, x1 x2 s1 s2 RHS 1 0 1 1 20 0 1 2 1 20 (1.5) This tableau represents the equations x1 s1 +s2= 20 x2 +2s1 s2= 20 or, x1 =20+ s1 s2 x2 =202s1 +s2 (1.6) You can't say how many deluxe belts x1 and regular belts x2 the plant will make until you specify how much leather s1 and labor s2 will be left over. But (1.6) is a formula for computing how many belts of either type must be made, for any given s1 and s2. So the equations have many solutions (in

63. nitely many). For example, if you want to have nothing left over (s1 = s2 = 0), you will make 20 of each. If you want to have 5 strips and 5 hours left over, you will make x1 = 20 deluxe and x2 = 15 regular belts. The variables x1; x2 you solved for are called basic variables. The other variables are nonbasic. You have control of the nonbasic variables. Once you give them values, the values of the basic

64. 1.3. LINEAR COMBINATIONS 19 variables follow. A solution in which you make all the nonbasic variables zero is called a basic solution. Can you have basic variables other than x1; x2? Sure. Any pair of variables can be basic, provided the corresponding columns in (1.3) are linearly independent (otherwise, you can't solve for the basic variables). Equations (1.3), for instance, are already solved in (1.3) for basic variables s1; s2. Here the two basic activities are having leftover leather and having leftover labor. The basic solution is (s1; s2) = (40; 60). This means that if you decide to produce no belts (x1 = x2 = 0), you must have 40 leftover leather strips and 60 leftover labor hours. The intermediate step (1.4) solves the equations with basic variables x1 and s2. Here the basic solution is unrealizable. If you decide to participate only in the basic activities (making deluxe belts and having leftover labor), you must make 40 belts and have 20 leftover hours (i.e., use 20 more than you have), which you can't do within your current resources. Exercise 8 Consider the system of equations x1 x2 x3 x4 RHS 1 1 2 4 100 3 1 1 2 200 where the

65. rst four columns on the left represent a Regular mixture of nuts, a Deluxe mixture, a small can of Premium mixture, and a large can of Premium. a) Solve the system with x1 and x2 basic. b) You want 100 cans of mixed nuts, each of which contains 1 lb cashews and 2 lb peanuts (i.e., you want the right-hand side of the above equation). How can you get them by blending 10 small cans of Premium with proper amounts of the Regular and Deluxe blends? Hint. Set x3 = 10 and x4 = 0 in the expression for the solution values of x1, x2 found in (a). c) How can you obtain a small can of Premium mix by combining (and decombining) the Regular and Deluxe blends? d) How can you obtain one can of Regular mix by combining large and small cans of Premium? If you cannot do it, why not? Hint. It has to do with linear dependence. Exercise 9 A can of paint A has 1 quart red, 1 quart yellow. A can of paint B has 1 quart red, 1 quart blue. A can of paint C has 1 quart yellow, 1 quart blue. a) How much of each paint must be mixed to obtain a mixture of 1 quart red, 1 quart yellow and 1 quart blue? b) How much of each paint must be mixed to obtain one quart of pure red? What do you conclude about the physical feasibility of such a mixture? Exercise 10 An eletronics plant wants to make stereos and CB's. Each stereo requires 1 power supply and 3 speakers. Each CB requires 1 power supply and 1 speaker. 100 power supplies and 200 speakers are in stock. How many stereos and CB's can it make if it wants to use all the power supplies and all but 10 of the speakers? Hint. Use the following equations. x1 x2 s1 s2 RHS 1 1 1 0 100 3 1 0 1 200

66. 20 CHAPTER 1. BASIC LINEAR ALGEBRA Exercise 11 A construction foreman needs cement mix containing 8 cubic yards (yd3) cement, 12 yd3 sand and 16 yd3 water. On the site are several mixer trucks containing mix A (1 yd3 cement, 3 yd3 sand, 3 yd3 water), several containing mix B (2 yd3 cement, 2 yd3 sand, 3 yd3 water) and several containing mix C (2 yd3 cement, 2 yd3 sand, 5 yd3 water). How many truckloads of each mix should the foreman combine to obtain the desired blend? 1.4 Inverse of a Square Matrix If A and B are square matrices such that AB = I (the identity matrix), then B is called the inverse of A and is denoted by A1. A square matrix A has either no inverse or a unique inverse A1. In the

67. rst case, it is said to be singular and in the second case nonsingular. Interestingly, linear independence of vectors plays a role here: a matrix is singular if its columns form a set of linearly dependent vectors; and it is nonsingular if its columns are linearly independent. Another property is the following: if B is the inverse of A, then A is the inverse of B. Exercise 12 (a) Compute the matrix product0B@ 1 3 2 0 5 4 2 3 3 1CA 0B@ 3 3 2 8 7 4 10 9 5 1CA : (b) What is the inverse of0B@ 3 3 2 8 7 4 10 9 5 ? 1CA (c) Show that the matrix A =0B@ 0 0 0 0 1 1 1 0 1 1CA is singular. [Hint: Assume that the inverse of A is B =0B@ b11 b12 b13 b21 b22 b23 b31 b32 b33 1CA and perform the matrix product AB. Then show that no choice of bij can make this product equal to the identity matrix I =0B@ 1 0 0 0 1 0 0 0 1 .] 1CA An important property of nonsingular square matrices is the following. Consider the system of linear equations 0BBBB@ a11 : : : a1n a21 : : : a2n ... ... an1 : : : ann 1CCCCA 0B@ x1 ... xn 1CA = 0BBBB@ b1 b2 ... bn 1CCCCA ; simply written as Ax = b. When A is a square nonsingular matrix, this linear system has a unique solution, which can be obtained as follows. Multiply the matrix equation Ax = b by A1 on the left: A1Ax = A1b: This yields Ix = A1b and so, the unique solution to the system of linear equations is x = A1b:

68. 1.4. INVERSE OF A SQUARE MATRIX 21 Exercise 13 Solve the system 3x1 3x2 + 2x3 = 1 8x1 + 7x2 4x3 = 1 10x1 + 9x2 5x3 = 1 using the result of Exercise 12(b). Finding the Inverse of a Square Matrix Given A =0B@ a11 : : : a1n ... ... an1 : : : ann , we must

69. nd B =0B@ 1CA b11 : : : b1n ... ... bn1 : : : bnn 1CA such that AB = I (the identity matrix). Therefore, the

70. rst column of B must satisfy A0B@ b11 ... bn1 1CA =0BBBB@ 1 0 ... 0 1CCCCA (this vector is the 1st column of I). Similarly, for the other columns of B. For example, the jth column of B satis

71. es A0 B@ bij ... bnj 1CA = 0BBBBBBBBBBB@ 0... 0 1 0... 0 1CCCCCCCCCCCA (the jth column of I). So in order to get the inverse of an n n matrix, we must solve n linear systems. However, the same steps of the Gauss-Jordan elimination procedure are needed for all of these systems. So we solve them all at once, using the matrix form. Example: Find the inverse of A = 3 2 4 3 !. We need to solve the following matrix equation 3 2 4 3 !B = 1 0 0 1 ! We divide the

72. rst row by 3 to introduce a 1 in the top left corner. 4 3 !B = 1 1 2 3 3 0 0 1 ! Then we add four times the

73. rst row to the second row to introduce a 0 in the

74. rst column. 1 2 1 3 !B = 3 0 1 3 0 4 3 1 ! Multiply the second row by 3. 1 2 !1 3 B = 0 1 3 0 4 3 !

75. 22 CHAPTER 1. BASIC LINEAR ALGEBRA Add 2 3 the second row to the

76. rst. (All this is classical Gauss-Jordan elimination.) 1 0 0 1 !B = 3 2 4 3 ! As IB = B, we get B = 3 2 4 3 !: It is important to note that, in addition to the two elementary row operations introduced earlier in the context of the Gauss-Jordan elimination procedure, a third elementary row operation may sometimes be needed here, namely permuting two rows. Example: Find the inverse of A =0B@ 0 1 0 2 1 0 0 0 1 . 1CA 0B@ 0 1 0 2 1 0 0 0 1 1CA B =0B@ 1 0 0 0 1 0 0 0 1 1CA Because the top left entry of A is 0, we need to permute rows 1 and 2

77. rst. 0B@ 2 1 0 0 1 0 0 0 1 1CA B =0B@ 0 1 0 1 0 0 0 0 1 1CA Now we divide the

78. rst row by 2. 0B@ 1 1 2 0 0 1 0 0 0 1 1CA B =0B@ 0 1 2 0 1 0 0 0 0 1 1CA Next we add 1 2 the second row to the

79. rst. 0B@ 1 0 0 0 1 0 0 0 1 1CA B =0B@ 1 2 1 2 0 1 0 0 0 0 1 1CA and we are done, since the matrix in front of B is the identity. Exercise 14 Find the inverse of the matrix A =0B@ 1 4 2 0 1 0 1 4 2 1CA . Exercise 15 Find the inverse of the matrix A =0B@ 0 1 2 4 0 5 3 0 4 1CA .

80. 1.5. DETERMINANTS 23 1.5 Determinants To each square matrix, we associate a number, called its determinant, de

81. ned as follows: If A = (a11), then det(A) = a11, If A = a11 a12 a21 a22 !, then det(A) = a11a22 + a12a21. For a square matrix A of dimensions n n, the determinant can be obtained as follows. First, de

82. ne A1j as the matrix of dimensions (n 1) (n 1) obtained from A by deleting row 1 and column j. Then det(A) = a11det(A11) a12det(A12) + a13 det(A13) : : :a1ndet(A1n): Note that, in this formula, the signs alternate between + and . For example, if A =0B@ a11 a12 a13 a21 a22 a23 a31 a32 a33 , then 1CA det(A) = a11(a22a33 a23a32) a12(a21a33 a23a31) + a13(a21a32 a22a31): Determinants have several interesting properties. For example, the following statements are equivalent for a square matrix A: det(A) = 0, A has no inverse, i.e. A is singular, the columns of A form a set of linearly dependent vectors, the rows of A form a set of linearly dependent vectors. For our purpose, however, determinants will be needed mainly in our discussion of classical optimization, in conjunction with the material from the following section. Exercise 16 Compute the determinant of A =0B@ 4 2 3 1 0 1 0 1 2 1CA . 1.6 Positive De

83. nite Matrices When we study functions of several variables (see Chapter 3!), we will need the following matrix notions. A square matrix A is positive de

84. nite if xTAx 0 for all nonzero column vectors x. It is negative de

85. nite if xTAx 0 for all nonzero x. It is positive semide

86. nite if xTAx 0 and negative semide

87. nite if xTAx 0 for all x. These de

88. nitions are hard to check directly and you might as well forget them for all practical purposes. More useful in practice are the following properties, which hold when the matrix A is symmetric (that will be the case of interest to us), and which are easier to check. The ith principal minor of A is the matrix Ai formed by the

89. rst i rows and columns of A. So, the

90. rst principal minor of A is the matrix A1 = (a11), the second principal minor is the matrix aaA= 11 12 2 !, and so on. a21 a22

91. 24 CHAPTER 1. BASIC LINEAR ALGEBRA The matrix A is positive de

92. nite if all its principal minors A1, A2; : : :An have strictly positive determinants. If these determinants are nonzero and alternate in signs, starting with det(A1) 0, then the matrix A is negative de

93. nite. If the determinants are all nonnegative, then the matrix is positive semide

94. nite, If the determinant alternate in signs, starting with det(A1) 0, then the matrix is negative semide

95. nite. To

96. x ideas, consider a 2 2 symmetic matrix A = a11 a12 a12 a22 !. It is positive de

97. nite if: (i) det(A1) = a11 0 (ii) det(A2) = a11a22 a12a12 0 and negative de

98. nite if: (i) det(A1) = a11 0 (ii) det(A2) = a11a22 a12a12 0. It is positive semide

99. nite if: (i) det(A1) = a11 0 (ii) det(A2) = a11a22 a12a12 0 and negative semide

100. nite if: (i) det(A1) = a11 0 (ii) det(A2) = a11a22 a12a12 0. Exercise 17 Check whether the following matrices are positive de

101. nite, negative de

102. nite, positive semide

103. nite, negative semide

104. nite or none of the above. (a) A = 2 1 1 4 ! (b) A = 2 4 4 8 ! (c) A = 2 2 2 4 ! 4 3 !. (d) A = 2 4

105. Chapter 2 Unconstrained Optimization: Functions of One Variable This chapter reviews basic results about functions of one variable, including the notions of derivative, extremum and convexity. It also describes a numerical method for

106. nding x such that f(x) = 0, known as binary search. 2.1 Derivatives It is important to know the following dierentiation formulas: Function Derivative f(x) + g(x) f0(x) + g0(x) af(x) af0(x) f(x)g(x) f0(x)g(x)+ f(x)g0(x) f(x)=g(x) [g(x)f0(x) f(x)g0(x)]=g2(x) f(g(x)) f0(g(x))g0(x) xa axa1 ex ex ln x (x 0) where ln x = loge x 1=x Exercise 18 Compute the derivatives of the following functions: (a) x3 3x + 1 (b) ln(1 + x2) (c) ln(1 + x)2 (d) (3 2x)ex2 25

107. 26 CHAPTER 2. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF ONE VARIABLE For a function f of one variable x, recall that the derivative f0(x) is equal to the slope of a tangent line at point x. So, if the function has a positive derivative at point x, then the function is increasing, and if it has a negative derivative, it is decreasing. Since the function and its tangent line are close around point x, the following formula can be used when is small. f(x+) f(x)+f0(x) Example 2.1.1 Suppose that the demand for gasoline at price x (in dollars) is ae0:2x gallons, where a is a constant. Suppose the price of gas rises by 10 cents. By how much does demand fall? Let f(x) = ae0:2x denote the demand for gas at price x. The rate of change is given by the derivative f0(x) = 0:2f(x): Since = 0:10, we get f(x + 0:10) f(x) + 0:10(0:2f(x)) = 0:98f(x): So demand drops by 2%. The factor relating change in demand to change in price is known as price elasticity of demand in economics (You will learn more about this in 45-749 Managerial Economics and in marketing courses such as 45-720 Marketing Management and 45-834 Pricing). Here f0(x) = 0:2f(x), so price elasticity of demand is -0.2. Exercise 19 Suppose that, if x dollars are spent on advertising for a product during a given year, f(x) = k(1 ecx) customers will purchase the product (k 0 and c 0 are two constants). (a) As x grows large, the number of customers purchasing the product approaches a limit. Find this limit. Can you give an interpretation for the constant k? (b) The sales response from a dollar of advertising is f(x + 1) f(x). Using the formula for small changes, show that the sales response from a dollar of advertising is proportional to the number of potential customers who are not purchasing the product at present. 2.2 Maximum and Minimum Let f be a function of one variable de

108. ned for all x in some domain D. A global maximum of f is a point x0 in D such that f(x) f(x0) for all x in D. For a constant 0, the neighborhood N(x0) of a point x0 is the set of all points x such that x0 x x0 + . A point x0 is a local maximum of f if there exists 0 such that f(x) f(x0) for all x in N(x0) where f(x) is de

109. ned. Similarly one can de

110. ne local and global minima. In Figure 2.1, the function f is de

111. ned for x in domain [a; e]. Points b and d are local maxima and b is a global maximum, whereas a, c, and e are local minima and e is a global minimum. Finding extrema Extrema, whether they are local or global, can occur in three places:

112. 2.2. MAXIMUM AND MINIMUM 27 a b c d e x f(x) Figure 2.1: local maxima and minima 1. at the boundary of the domain, 2. at a point without a derivative, or 3. at a point x0 with f0(x0) = 0. The last case is particularly important. So we discuss it further. Let f be dierentiable in a neighborhood of x0. If x0 is a local extremum of f, then f0(x0) = 0. Conversely, if f0(x0) = 0, three possibilities may arise: x0 is a local maximum, x0 is a local minimum, or neither!!! To decide between these three possibilities, one may use the second derivative. Let f be twice dierentiable in a neighborhood of x0. If f0(x0) = 0 and f00(x0) 0 then x0 is a local minimum. If f0(x0) = 0 and f00(x0) 0 then x0 is a local maximum. If f0(x0) = 0 and f00(x0) = 0 then x0 may or may not be a local extremum. Figure 2.2 illustrates these three possibilities. Example 2.2.1 Suppose an oil cartel can price gasoline as it wishes. How will it maximize revenue, given that demand for gasoline at price x is f(x) = ae0:2x The revenue at price x is g(x) = xf(x): We compute the derivative of g and set it to 0. g0(x) = f(x) + x(0:2f(x)) = (1 0:2x)f(x)

113. 28 CHAPTER 2. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF ONE VARIABLE f(x) x Figure 2.2: f00(x0) 0, f00(x0) 0 and one of the possibilities with f00(x0) = 0 Since f(x) 0 for all x, setting g0(x) = 0 implies 1 0:2x = 0. So x = 5. This is the only possible local optimum. To determine whether it is a maximum or a minimum, we compute the second derivative. g00(x) = 2(0:2f(x))+ x(0:2)2f(x) = 0:2(2 0:2x)f(x): Putting in x = 5 shows g00(x) 0, so this is a maximum: the oil cartel maximizes its revenue by pricing gasoline at $5 per gallon. Example 2.2.2 Economic Order Quantity We want to keep paper towels on hand at all times. The cost of stockpiling towels is $ h per case per week, including the cost of space, the cost of tying up capital in inventory, etc. The time and paperwork involved in ordering towels costs $ K per order (this is in addition to the cost of the towels themselves, which is not relevant here because we are supposing that there are no price discount for large orders.) Towels are used at a rate of d cases per week. What quantity Q should we order at a time, to minimize cost? We must

114. rst calculate the holding cost (cost of stockpiling). If each order period begins with Q cases and ends with zero cases, and if usage is more or less constant, then the average stock level is Q=2. This means that the average holding cost is hQ=2. Since each order of Q cases lasts Q=d weeks, the average ordering cost per week is K Q=d = Kd Q : Thus the average total cost per week is hQ 2 + Kd Q : To

115. nd Q that minimizes cost, we set to zero the derivative with respect to Q. h 2 Kd Q2 = 0: This implies that the optimal order quantity is Q = s2Kd h :

116. 2.2. MAXIMUM AND MINIMUM 29 This is the classical economic order quantity (EOQ) model for inventory management. (You will learn when the EOQ model is appropriate and when to use other inventory models in 45-765). Exercise 20 Find the local maxima and minima, if any, of the following functions. (a) x3 3x + 1 (b) x3 3x2 + 3x 1 (c) (3 2x)ex2 Exercise 21 We want to design a cone-shaped paper drinking cup of volume V that requires a minimum of paper to make. That is, we want to minimize its surface area rpr2 + h2, where r is the radius of the open end and h the height. Since V = 3 r2h, we can solve for h to get h = 3V=(r2). Thus the area in terms of r is A(r) = r r2 + 9 V 2 2r4!1=2 = 2r4 + 9 V 2 r2 !1=2 : a) Compute the derivative A0(r). b) Find the radius r at which A(r) is minimized (the solution you get is in fact a minimum). c) What is the ratio h=r when r has the optimal value? Exercise 22 An air courier company operates a single depot in a large city and wants to decen- tralize by establishing several depots, each covering the same delivery radius r. The total area to be covered is A, and the area covered by each depot is r2. Thus if n depots are installed, we must have A = nr2, so that n = A=(r2). Expenses and revenues break down as follows. Trucking expenses. The number of trucks required at each depot is r3, so that the total number of trucks required is nr3 = (A=r2)r3 = Ar. Each truck costs $K per day, so that total expense for trucks is $KAr. Overhead. Each depot incurs an overhead (

117. xed) expense of $F, for a total overhead of nF = FA=(r2). Revenue from customers. Since the total area is

118. xed, the total revenue from customers is independent of the delivery radius. Since the total revenue is

119. xed, we can simply minimize costs: minKAr + FA r2 : Your problem. Find the delivery radius r that minimizes cost. Your solution should indicate that the optimal delivery radius is proportional to the cube root (the 1/3 power) of the

120. xed cost. Thus if the

121. xed cost increases eightfold (and other costs were constant), the optimal delivery radius would only double. This is fortunate, since it means that the rent and other

122. xed costs must rise substantially (relative to other costs) before they justify relocating depots.

123. 30 CHAPTER 2. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF ONE VARIABLE 2.3 Binary Search Binary search is a very simple idea for solving numerically f(x) = 0. It requires: (i) the function f to be continuous, (ii) values (a; b) such that f(a) 0 and f(b) 0. Then a desired value x0 such that f(x0) = 0 can be found between the values a and b. The method consists in computing f(c) for c = a+b 2 . If f(c) 0, the procedure is repeated with the values (a; b) replaced by the values (c; b). If f(c) 0, the procedure is repeated with the values (a; b) replaced by (a; c). If f(c) = 0, the procedure terminates as c is the value we are looking for. This method is easy to program and converges rapidly (20 iterations will give you x0 with 5 signi

124. cant

125. gures). Example 2.3.1 Binary search can be used to compute the internal rate of return of an investment. The internal rate of return of an investment is the interest rate r that makes the present value of the cash ows from the investment equal to the cost of the investment. Mathematically, r is the interest rate that satis

126. es the equation F1 1 + r + F2 (1 + r)2 + F3 (1 + r)3 + : : :+ FN (1 + r)N C = 0 where Ft = cash ow in year t N = number of years C = cost of the investment For a bond, the internal rate of return r is known as the yield to maturity or simply yield. Let us consider the case of a noncallable bond, i.e. a bond that the issuer cannot retire prior to its stated maturity date. For such a bond, the cash ows consist of periodic interest payments to the maturity date and the par value paid at maturity. As an example, consider a 4-year noncallable bond with a 10% coupon rate paid annually and a par value of $1000. Such a bond has the following cash ows: t Years from now Ft 1 $ 100 2 100 3 100 4 1100 Suppose this bond is now selling for $900. Compute the yield of this bond.

127. 2.3. BINARY SEARCH 31 The yield r of the bond is given by the equation 100 1 + r + 100 (1 + r)2 + 100 (1 + r)3 + 1100 (1 + r)4 900 = 0 Let us denote by f(r) the left-hand-side of this equation. We

128. nd r such that f(r) = 0 using binary search. We start by

129. nding values (a; b) such that f(a) 0 and f(b) 0. In this case, we expect r to be between 0 and 1. Since f(0) = 500 and f(1) = 743:75, we have our starting values. Next, we let c = 0:5 (the midpoint) and calculate f(c). Since f(0:5) = 541:975, we replace our range with a = 0 and b = 0:5 and repeat. When we continue, we get the following table of values: Iter. a c b f(a) f(c) f(b) 1 0 0.5 1 500 -541.975 -743.75 2 0 0.25 0.5 500 -254.24 -541.975 3 0 0.125 0.25 500 24.85902 -254.24 4 0.125 0.1875 0.25 24.85902 -131.989 -254.24 5 0.125 0.15625 0.1875 24.85902 -58.5833 -131.989 6 0.125 0.140625 0.15625 24.85902 -18.2181 -58.5833 7 0.125 0.132813 0.140625 24.85902 2.967767 -18.2181 8 0.132813 0.136719 0.140625 2.967767 -7.71156 -18.2181 9 0.132813 0.134766 0.136719 2.967767 -2.39372 -7.71156 10 0.132813 0.133789 0.134766 2.967767 0.281543 -2.39372 11 0.133789 0.134277 0.134766 0.281543 -1.05745 -2.39372 12 0.133789 0.134033 0.134277 0.281543 -0.3883 -1.05745 So the yield of the bond is r = 13.4%. Of course, this routine sort of calculation is perfectly set up for calculation by computer. In particular, we can use Microsoft's Excel (or any spreadsheet program) to do these calculations. On the course home page is a spreadsheet that implements binary search for this problem. We will show this spreadsheet in class. Cells B3:G3 down to B20:G20 contain the body of the table. You can generate the same table by following these steps: In cell B3, type 0 [comment: this is initial value of a.] In cell D3, type 1 [comment: this is the initial value of b.] In cell C3, type =(B3+D3)/2 [comment: this is the middle point c.] In cell E3, type =100/(1+B3) +100/(1+B3)^2 +100/(1+B3)^3 +1100/(1+B3)^4 -900 [comment: this is f(a).] Use the Copy Command to copy cell E3 into cells F3 and G3 [comment: cell F3 contains the same function as cell E3, except that B3 is replaced by C3. So cell F3 contains f(c). Similarly, G3 contains f(b).] In cell B4, type =IF(F3 0,C3,B3) [comment: If F3 0, cell B4 contains C3 and otherwise, cell B4 contains B3.] In cell D4, type =IF(F3 0,D3,C3) Use the Copy Command to copy cell B4 down column B Similarly, copy C3 down column C, D4 down column D, E3 down column E, F3 down column F, and G3 down column G. Now you should have the full table! Read down column C to

130. nd the value of r that makes f(r) = 0.

131. 32 CHAPTER 2. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF ONE VARIABLE Excel also has an addin called Solver that can search for any particular value for a function. We will also review this system in class (we will use Solver extensively in this course). Exercise 23 (a) Show the cash ows for the three bonds below, each of which pays interest annually: Bond Coupon Rate % Par Value Years to Maturity Price X 14 $ 1000 5 $ 900 Y 16 2000 7 1900 Z 17 1000 4 950 (b) Calculate the yield to maturity for the three bonds. Golden Section Search Golden section search is similar in spirit to binary search. It is used to compute the maximum of a function f(x) de

132. ned on an interval [a; b], when the method presented in Section 2.2 is not applicable for one reason or another. It assumes that (i) f is continuous (ii) f has a unique local maximum in the interval [a; b]. The golden search method consists in computing f(c) and f(d) for a d c b. If f(c) f(d), the procedure is repeated with the interval (a; b) replaced by (d; b). If f(c) f(d), the procedure is repeated with the interval (a; b) replaced by (a; c). Note: The name golden section comes from a certain choice of c and d that yields fast conver- pgence, namely c = a + r(b a) and d = b + r(a b), where r = 51 = :618034 : : :. This is the 2 golden ratio, already known to the ancient greeks. Example 2.3.2 Find the maximum of the function x5 10x2 + 2x in the interval [0; 1]. In this case, we begin with a = 0 and b = 1. Using golden section search, that gives d = 0:382 and c = 0:618. The function values are f(a) = 0, f(d) = 0:687, f(c) = 2:493, and f(b) = 7. Since f(c) f(d), our new range is a = 0, b = :618. Recalculating from the new range gives d = :236, c = :382 (note that our current c was our previous d: it is this reuse of calculated values that gives golden section search its speed). We repeat this process to get the following table:

133. 2.4. CONVEXITY 33 Iter. a d c b f(a) f(d) f(c) f(b) 1 0 0.382 0.618 1 0 -0.6869 -2.4934 -7 2 0 0.2361 0.382 0.618 0 -0.0844 -0.6869 -2.4934 3 0 0.1459 0.2361 0.382 0 0.079 -0.0844 -0.6869 4 0 0.0902 0.1459 0.2361 0 0.099 0.079 -0.0844 5 0 0.0557 0.0902 0.1459 0 0.0804 0.099 0.079 6 0.0557 0.0902 0.1115 0.1459 0.0804 0.099 0.0987 0.079 7 0.0557 0.077 0.0902 0.1115 0.0804 0.0947 0.099 0.0987 8 0.077 0.0902 0.0983 0.1115 0.0947 0.099 0.1 0.0987 9 0.0902 0.0983 0.1033 0.1115 0.099 0.1 0.0999 0.0987 10 0.0902 0.0952 0.0983 0.1033 0.099 0.0998 0.1 0.0999 11 0.0952 0.0983 0.1002 0.1033 0.0998 0.1 0.1 0.0999 12 0.0983 0.1002 0.1014 0.1033 0.1 0.1 0.1 0.0999 13 0.0983 0.0995 0.1002 0.1014 0.1 0.1 0.1 0.1 14 0.0995 0.1002 0.1007 0.1014 0.1 0.1 0.1 0.1 15 0.0995 0.0999 0.1002 0.1007 0.1 0.1 0.1 0.1 16 0.0995 0.0998 0.0999 0.1002 0.1 0.1 0.1 0.1 17 0.0998 0.0999 0.1 0.1002 0.1 0.1 0.1 0.1 18 0.0999 0.1 0.1001 0.1002 0.1 0.1 0.1 0.1 19 0.0999 0.1 0.1 0.1001 0.1 0.1 0.1 0.1 20 0.0999 0.1 0.1 0.1 0.1 0.1 0.1 0.1 21 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 Again we can use Excel to implement golden section search or use Solver to maximize this function directly. 2.4 Convexity Finding global extrema and checking that we have actually found one is harder than

134. nding local extrema, in general. There is one nice case: that of convex and concave functions. A convex function is one where the line segment connecting two points (x; f(x)) and (y; f(y)) lies above the function. Mathematically, a function f is convex if, for all x; y and all 0 1, f(x + (1 )y) f(x) + (1 )f(y): See Figure 2.3. The function f is concave if f is convex. There is an easy way to check for convexity when f is twice dierentiable: the function f is convex on some domain [a; b] if (and only if) f00(x) 0 for all x in the domain. Similarly, f is concave on some domain [a; b] if (and only if) f00(x) 0 for all x in the domain. If f(x) is convex, then any local minimum is also a global minimum. If f(x) is concave, then any local maximum is also a global maximum. Exercise 24 Find whether the following functions are concave, convex or neither. (a) x4 4x3 + 6x2 + 3x + 1 (b) ex2 Exercise 25 Find a local extremum of the function f(x) = xex. Indicate whether it is a local maximum, a local minimum or neither. Is it a global optimum on the domain [-2,2]?

135. 34 CHAPTER 2. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF ONE VARIABLE x function y x f(x) x y x f(x) convex function concave Figure 2.3: Convex function and concave function

136. 2.4. CONVEXITY 35 Exercise 26 Consider the linear demand function D(P) = K aP, where P( K=a) is the price and K and a are positive constants. Thus demand for a free good is K, and demand drops linearly until the price reaches K=a, at which point demand is zero. a) Find the price P that maximizes revenue PD(P) by setting the derivative of revenue equal to zero. b) Use the second derivative to show that you found a local maximum. c) Use the second derivative to show that the revenue function is concave. It follows that your local maximum is a global maximum. Exercise 27 Oil is to be shipped in barrels having an undetermined height h and radius r. Each barrel must have a

137. xed volume V. We want to pick r and h so as to minimize the surface area of each barrel, given by A = 2r2 + 2rh, and thereby minimize the cost of making the barrels. Since V = r2h, we can solve for h in terms of r to get h = V=r2, and we can substitute this in the formula for A to get a formula for area solely in terms of r, namely, A(r) = 2r2 + 2V=r: a) Compute A0(r), A00(r) and show that the function A is convex on the set of positive reals. b) Find the radius at which A(r) has its global minimum. Explain how you know the minimum is global. c) What is the ratio h=r when r has the optimal value you just derived?

138. 36 CHAPTER 2. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF ONE VARIABLE

139. Chapter 3 Unconstrained Optimization: Functions of Several Variables Many of the concepts for functions of one variable can be extended to functions of several variables. For example, the gradient extends the notion of derivative. In this chapter, we review the notion of gradient, the formula for small changes, how to

140. nd extrema and the notion of convexity. 3.1 Gradient Given a function f of n variables x1; x2; : : :; xn, we de

141. ne the partial derivative relative to variable xi, written as @f @xi , to be the derivative of f with respect to xi treating all variables except xi as constant. Example 3.1.1 Compute the partial derivatives of f(x1; x2) = (x1 2)2 + 2(x2 1)2. The answer is: @f @x1 (x1; x2) = 2(x1 2); @f @x2 (x1; x2) = 4(x2 1). Let x denote the vector (x1; x2; : : :; xn). With this notation, f(x) = f(x1; x2; : : :; xn), @f @xi (x) = @f @xi (x1; x2; : : :; xn), etc. The gradient of f at x, written rf(x), is the vector 0BBBBB@ @f @x1 (x) @f @x2 (x) ... @f @xn (x) . The 1CCCCCA gradient vector rf(x) gives the direction of steepest ascent of the function f at point x. The gradient acts like the derivative in that small changes around a given point x can be estimated using the gradient. f(x +) f(x)+rf(x) where = (1; : : :;n) denotes the vector of changes. Example 3.1.2 If f(x1; x2) = x2 1 3x1x2 + x2 2, then f(1; 1) = 1. What about f(1:01; 1:01)? @x1 (x1; x2) = 2x1 3x2 and @f In this case, x = (1; 1) and = (0:01; 0:01). Since @f @x2 (x1; x2) = 3x1 + 2x2, we get 37

142. 38CHAPTER 3. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF SEVERAL VARIABLES rf(1; 1) = 1 1 ! . So f(1:01; 1:01) = f((1; 1)+(0:01; 0:01)) f(1; 1)+(0:01; 0:01)rf(1; 1) = 1+(0:01; 0:01) 1 1 ! = 1:02. Example 3.1.3 Suppose that we want to put away a

143. shing pole in a closet having dimensions 3 by 5 by 1 feet. If the ends of the pole are placed at opposite corners, there is room for a pole of length, f(x1; x2; x3) = f(3; 5; 1) = px2 1 + x2 2 + x2 3 = 5:9ft: It turns out that the actual dimensions of the closet are 3 + 1, 5+2 and 1 + 3 feet, where 1, 2 and 3 are small correction terms. What is the change in pole length, taking into account these corrections? By the formula for small changes, the change in pole length is f(3+1;5+2;1+3) f(3; 5; 1) (1;2;3)rf(3; 5; 1): So, we need to compute the partial derivatives of f. For i = 1; 2; 3 @ @xi f(x1; x2; x3) = xi px2 1 + x2 2 + x2 3 : Now we get (1;2;3)rf(3; 5; 1) = (1;2;3)0 B@ 0:51 0:85 0:17 1CA = 0:511 + 0:852 + 0:173: Exercise 28 Consider the function f(x1; x2) = x1 ln x2. (a) Compute the gradient of f. (b) Give the value of the function f and give its gradient at the point (3; 1). (c) Use the formula for small changes to obtain an approximate value of the function at the point (2:99; 1:05). Exercise 29 Consider a conical drinking cup with height h and radius r at the open end. The volume of the cup is V (r; h) = 3 r2h. a) Suppose the cone is now 5 cm high with radius 2 cm. Compute its volume. b) Compute the partial derivatives @V=@r and @V=@h at the current height and radius. c) By about what fraction (i.e., percentage) would the volume change if the cone were lengthened 10%? (Use the partial derivatives.) d) If the radius were increased 5%?

144. 3.2. MAXIMUM AND MINIMUM 39 e) If both were done simultaneously? Hessian matrix Second partials @2f @xi@xj (x) are obtained from f(x) by taking the derivative relative to xi (this yields the

145. rst partial @f @xi (x) ) and then by taking the derivative of @f @xi (x) relative to xj. So we can compute @2f @x1@x1 (x), @2f @x1@x2 (x) and so on. These values are arranged into the Hessian matrix H(x) = 0BBBBB@ @2f @x1@x1 (x) @2f @x1@x2 (x) @2f @x1@xn (x) @2f @x2@x1 (x) @2f @x2@x2 (x) @2f @x2@xn (x) ... ... . . . ... @2f @xn@x1 (x) @2f @xn@x2 (x) @2f @xn@xn (x) 1CCCCCA The Hessian matrix is a symmetric matrix, that is @2f @xi@xj (x) = @2f @xj@xi (x). Example 3.1.1 (continued): Find the Hessian matrix of f(x1; x2) = (x1 2)2 + 2(x2 1)2: The answer is H(x) = 2 0 0 4 !: 3.2 Maximum and Minimum Optima can occur in three places: 1. at the boundary of the domain, 2. at a nondierentiable point, or 3. at a point x with rf(x) = 0. We will identify the

146. rst type of point with Kuhn{Tucker conditions (see next chapter). The second type is found only by ad hoc methods. The third type of point can be found by solving the gradient equations. In the remainder of this chapter, we discuss the important case where rf(x) = 0. To identify if a point x with zero gradient is a local maximum or local minimum, check the Hessian. If H(x) is positive de

147. nite then x is a local minimum. If H(x) is negative de

148. nite, then x is a local maximum. Remember (Section 1.6) that these properties can be checked by computing the determinants of the principal minors. Example 3.2.1 Find the local extrema of f(x1; x2) = x3 1 + x3 2 3x1x2.

149. 40CHAPTER 3. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF SEVERAL VARIABLES This function is everywhere dierentiable, so extrema can only occur at points x such that rf(x) = 0. 2 3x1 ! rf(x) = 3x2 1 3x2 3x2 This equals 0 i (x1; x2) = (0; 0) or (1; 1). The Hessian is H(x) = 6x1 3 3 6x2 ! So, H(0; 0) = 0 3 3 0 ! Let H1 denote the

150. rst principal minor of H(0; 0) and let H2 denote its second principal minor (see Section 1.6). Then det(H1) = 0 and det(H2) = 9. Therefore H(0; 0) is neither positive nor negative de

151. nite. H(1; 1) = 6 3 3 6 ! Its

152. rst principal minor has det(H1) = 6 0 and its second principal minor has det(H2) = 369 = 25 0. Therefore H(1; 1) is positive de

153. nite, which implies that (1; 1) is a local minimum. Example 3.2.2 Jane and Jim invested $20,000 in the design and development of a new product. They can manufacture it for $2 per unit. For the next step, they hired marketing consultants XYZ. In a nutshell, XYZ's conclusions are the following: if Jane and Jim spend $a on advertizing and sell the product at price p (per unit), they will sell 2; 000+ 4pa 20p units. Using this

154. gure, express the pro

155. t that Jane and Jim will make as a function of a and p. What price and level of advertising will maximize their pro

156. ts? The revenue from sales is (2; 000+ 4pa 20p)p. The production costs are (2; 000+ 4pa 20p)2, the development cost is $20,000 and the cost of advertizing is a. Therefore, Jane and Jim's pro

157. t is f(p; a) = (2; 000+ 4pa 20p)(p 2) a 20; 000 To

158. nd the maximum pro

159. t, we compute the partial derivatives of f and set them to 0: @f @p (p; a) = 2; 040+ 4pa 40p = 0: @f @a(p; a) = 2(p 2)=pa 1 = 0: Solving this system of two equations yields p = 63:25 a = 15; 006:25 We verify that this is a maximum by computing the Hessian.

160. 3.3. GLOBAL OPTIMA 41 H(x) = 40 2=pa 2=pa (p 2)=apa ! det(H1) = 40 0 and det(H2) = 40(p2)=apa4=a 0 at the point p = 63:25, a = 15; 006:25. So, indeed, this solution maximizes pro

161. t. 1 + (x1 + x2)2 + (x1 + x3)2. Example 3.2.3 Find the local extrema of f(x1; x2; x3) = x2 @f (x) = 2x1 + 2(x1 + x2) + 2(x1 + x3) @x1 @f (x) = 2(x1 + x2) @x2 @f @x3 (x) = 2(x1 + x3) Setting these partial derivatives to 0 yields the unique solution x1 = x2 = x3 = 0. The Hessian matrix is H(0; 0; 0) =0B@ 6 2 2 2 2 0 2 0 2 1CA The determinants of the principal minors are det(H1) = 6 0, det(H2) = 12 4 = 8 0 and det(H3) = 24 8 8 = 8 0. So H(0; 0; 0) is positive de

162. nite and the solution x1 = x2 = x3 = 0 is a minimum. Exercise 30 Find maxima or minima of the following functions when possible. (a) f(x1; x2; x3) = x2 1 3x2 2 10x2 3 + 4x1 + 24x2 + 20x3 (b) f(x1; x2; x3) = x1x2 + x2x3 + x3x1 2x1 2x2 2x3 Exercise 31 Consider the function of three variables given by f(x1; x2; x3) = x2 1 x1 x1x2 + x2 2 x2 + x4 3 4x3: (a) Compute the gradient rf(x1; x2; x3). (b) Compute the Hessian matrix H(x1; x2; x3). (c) Use the gradient to

163. nd a local extremum of f. Hint: if x3 3 = 1, then x3 = 1. (d) Compute the three principal minors of the Hessian matrix and use them to identify this extremum as a local minimum or a local maximum. 3.3 Global Optima Finding global maxima and minima is harder. There is one case that is of interest. We say that a domain is convex if every line drawn between two points in the domain lies within the domain. We say that a function f is convex if the line connecting any two points lies above the function. That is, for all x; y in the domain and 0 1, we have f(x+(1)y) f(x) + (1)f(y), as before (see Chapter 2).

164. 42CHAPTER 3. UNCONSTRAINED OPTIMIZATION: FUNCTIONS OF SEVERAL VARIABLES If a function is convex on a convex domain, then any local minimum is a global minimum. If a function is concave on a convex domain, then any local maximum is a global maximum. To check that a function is convex on a domain, check that its Hessian matrix H(x) is positive semide

165. nite for every point x in the domain. To check that a function is concave, check that its Hessian is negative semide

166. nite for every point in the domain. Example 3.3.1 Show that the function f(x1; x2; x3) = x4 1 + (x1 + x2)2 + (x1 +x3)2 is convex over 3. @f @x1 (x) = 4x3 1 + 2(x1 + x2) + 2(x1 + x3) @f @x2 (x) = 2(x1 + x2) @f @x3 (x) = 2(x1 + x3) H(x1; x2; x3) =0B@ 12x2 1 + 4 2 2 2 2 0 2 0 2 1CA The determinants of the principal minors are det(H1) = 12x2 1 +4 0, det(H2) = 12x2 1 0 and det(H3) = 48x2 1 0. So H(x1; x2; x3) is positive semide

167. nite for all (x1; x2; x3) in 3. This implies that f is convex over 3. Exercise 32 For each of the following, determine whether the function is convex, concave, or neither over 2. (a) f(x) = x1x2 x2 1 x2 2 (b) f(x) = 10x1 + 20x2 (c) f(x) = x4 1 + x1x2 (d) f(x) = x2 1 x1x2 2x2 2 Exercise 33 Let the following function be de

168. ned for all points (x; y) in the plane. f(x; y) = 2xy x4 x2 y2: (a) Write the gradient of the function f. (b) Write the Hessian matrix of f. (c) Is the function f convex, concave or neither? (d) Use the gradient to

169. nd a local extremum of f. (e) Identify this extremum as a minimum, a maximum or neither.

170. Chapter 4 Constrained Optimization 4.1 Equality Constraints (Lagrangians) Suppose we have a problem: Maximize 5 (x1 2)2 2(x2 1)2 subject to x1 + 4x2 = 3 If we ignore the constraint, we get the solution x1 = 2; x2 = 1, which is too large for the constraint. Let us penalize ourselves for making the constraint too big. We end up with a function L(x1; x2; ) = 5 (x1 2)2 2(x2 1)2 + (3 x1 4x2) This function is called the Lagrangian of the problem. The main idea is to adjust so that we use exactly the right amount of the resource. = 0 leads to (2; 1). = 1 leads to (3=2; 0) which uses too little of the resource. = 2=3 gives (5=3; 1=3) and the constraint is satis

171. ed exactly. We now explore this idea more formally. Given a nonlinear program (P) with equality constraints: Minimize (or maximize) f(x) subject to g1(x) = b1 g2(x) = b2 ... gm(x) = bm a solution can be found using the Lagrangian: L(x; ) = f(x) + m Xi=1 i(bi gi(x)) m(Note: this can also be written f(x) Pi=1 i(gi(x) bi)). Each i gives the price associated with constraint i. 43

172. 44 CHAPTER 4. CONSTRAINED OPTIMIZATION The reason L is of interest is the following: Assume x = (x1; x2; : : :; xn) maximizes or minimizes f(x) subject to the constraints gi(x) = bi, for i = 1; 2; : : :;m. Then either (i) the vectors rg1(x);rg2(x); : : :;rgm(x) are linearly dependent, or (ii) there exists a vector = (1; 2; : : :; m) such that rL(x; ) = 0. I.e. @L @x1 (x; ) = @L @x2 (x; ) = = @L @xn (x; ) = 0 and @L @1 (x; ) = = @L @m (x; ) = 0 Of course, Case (i) above cannot occur when there is only one constraint. The following example shows how it might occur. Example 4.1.1 Minimize x1 + x2 + x2 3 subject to x1 = 1 x2 1 + x2 2 = 1: It is easy to check directly that the minimum is acheived at (x1; x2; x3) = (1; 0; 0). The associated Lagrangian is L(x1; x2; x3; 1; 2) = x1 + x2 + x2 3 + 1(1 x1) + 2(1 x2 1 x2 2): Observe that @L @x2 (1; 0; 0; 1; 2) = 1 for all 1; 2; and consequently @L @x2 does not vanish at the optimal solution. The reason for this is the following. Let g1(x1; x2; x3) = x1 and g2(x1; x2; x3) = x2 1 + x2 2 denote the left hand sides of the constraints. Then rg1(1; 0; 0) = (1; 0; 0) and rg2(1; 0; 0) = (2; 0; 0) are linearly dependent vectors. So Case (i) occurs here! Nevertheless, Case (i) will not concern us in this course. When solving optimization problems with equality constraints, we will only look for solutions x that satisfy Case (ii). Note that the equation @L @i (x; ) = 0 is nothing more than bi gi(x) = 0 or gi(x) = bi: In other words, taking the partials with respect to does nothing more than return the original constraints.

173. 4.1. EQUALITY CONSTRAINTS (LAGRANGIANS) 45 Once we have found candidate solutions x, it is not always easy to

174. gure out whether they correspond to a minimum, a maximum or neither. The following situation is one when we can conclude. If f(x) is concave and all of the gi(x) are linear, then any feasible x with a corresponding making rL(x; ) = 0 maximizes f(x) subject to the constraints. Similarly, if f(x) is convex and each gi(x) is linear, then any x with a making rL(x; ) = 0 minimizes f(x) subject to the constraints. Example 4.1.2 Minimize 2x2 1 + x2 2 subject to x1 + x2 = 1 1 + x2 L(x1; x2; ) = 2x2 2 + 1(1 x1 x2) @L @x1 (x1 ; x2 ; ) = 4x1 1 = 0 @L @x2 (x1 ; x2 ; ) = 2x2 1 = 0 @L @ (x1 ; x2 ; ) = 1 x1 x2 = 0 Now, the

175. rst two equations imply 2x1 = x2. Substituting into the

176. nal equation gives the solution x1 = 1=3, x2 = 2=3 and = 4=3, with function value 2/3. Since f(x1; x2) is convex (its Hessian matrix H(x1; x2) = 4 0 0 2 ! is positive de

177. nite) and g(x1; x2) = x1 + x2 is a linear function, the above solution minimizes f(x1; x2) subject to the constraint. 4.1.1 Geometric Interpretation There is a geometric interpretation of the conditions an optimal solution must satisfy. If we graph Example 4.1.2, we get a picture like that in Figure 4.1. Now, examine the gradients of f and g at the optimum point. They must point in the same direction, though they may have dierent lengths. This implies: rf(x) = rg(x) which, along with the feasibility of x, is exactly the condition rL(x; ) = 0 of Case (ii). 4.1.2 Economic Interpretation The values i have an important economic interpretation: If the right hand side bi of Constraint i is increased by , then the optimum objective value increases by approximately i . In particular, consider the problem Maximize p(x) subject to g(x) = b,

178. 46 CHAPTER 4. CONSTRAINED OPTIMIZATION 2/3 0 f = lambda g 1/3 1 1 f(x)=2/3 g(x)=1 x* Figure 4.1: Geometric interpretation where p(x) is a pro

179. t to maximize and b is a limited amount of resource. Then, the optimum Lagrange multiplier is the marginal value of the resource. Equivalently, if b were increased by , pro

180. t would increase by . This is an important result to remember. It will be used repeatedly in your Managerial Economics course. Similarly, if Minimize c(x) subject to d(x) = b, represents the minimum cost c(x) of meeting some demand b, the optimum Lagrange multiplier is the marginal cost of meeting the demand. In Example 4.1.2 Minimize 2x2 1 + x2 2 subject to x1 + x2 = 1, if we change the right hand side from 1 to 1:05 (i.e. = 0:05), then the optimum objective function value goes from 2 3 to roughly 2 3 + 4 3 (0:05) = 2:2 3 :

181. 4.1. EQUALITY CONSTRAINTS (LAGRANGIANS) 47 If instead the right hand side became 0:98, our estimate of the optimum objective function value would be 2 3 + 4 3 (0:02) = 1:92 3 Example 4.1.3 Suppose we have a re

182. nery that must ship

183. nished goods to some storage tanks. Suppose further that there are two pipelines, A and B, to do the shipping. The cost of shipping x units on A is ax2; the cost of shipping y units on B is by2, where a 0 and b 0 are given. How can we ship Q units while minimizing cost? What happens to the cost if Q increases by r%? Minimize ax2 + by2 Subject to x + y = Q L(x; y; ) = ax2 + by2 + (Q x y) @L @x (x; y; ) = 2ax = 0 @L @y (x; y; ) = 2by = 0 @L @ (x; y; ) = Q x y = 0 The

184. rst two constraints give x = ba y, which leads to x = bQ a + b ; y = aQ a + b ; = 2abQ a + b a+b . The Hessian matrix H(x1; x2) = 2a 0 and cost of abQ2 0 2b ! is positive de

185. nite since a 0 and b 0. So this solution minimizes cost, given a; b;Q. If Q increases by r%, then the RHS of the constraint increases by = rQ and the minimum cost increases by = 2abrQ2 a+b . That is, the minimum cost increases by 2r%. Example 4.1.4 How should one divide his/her savings between three mutual funds with expected returns 10%, 10% and 15% repectively, so as to minimize risk while achieving an expected return of 12%. We measure risk as the variance of the return on the investment (you will learn more about measuring risk in 45-733): when a fraction x of the savings is invested in Fund 1, y in Fund 2 and z in Fund 3, where x + y + z = 1, the variance of the return has been calculated to be 400x2 + 800y2 + 200xy + 1600z2 + 400yz: So your problem is min 400x2 + 800y2 + 200xy + 1600z2 + 400yz s:t: x + y + 1:5z = 1:2 x + y + z = 1 Using the Lagrangian method, the following optimal solution was obtained x = 0:5 y = 0:1 z = 0:4 1 = 1800 2 = 1380

186. 48 CHAPTER 4. CONSTRAINED OPTIMIZATION where 1 is the Lagrange multiplier associated with the

187. rst constraint and 2 with the second constraint. The corresponding objective function value (i.e. the variance on the return) is 390. If an expected return of 12.5% was desired (instead of 12%), what would be (approximately) the correcponding variance of the return? Since = 0:05, the variance would increase by 1 = 0:05 1800 = 90: So the answer is 390+90=480. Exercise 34 Record'm Records needs to produce 100 gold records at one or more of its three studios. The cost of producing x records at studio 1 is 10x; the cost of producing y records at studio 2 is 2y2; the cost of producing z records at studio 3 is z2 + 8z. (a) Formulate the nonlinear program of producing the 100 records at minimum cost. (b) What is the Lagrangian associated with your formulation in (a)? (c) Solve this Lagrangian. What is the optimal production plan? (d) What is the marginal cost of producing one extra gold record? (e) Union regulations require that exactly 60 hours of work be done at studios 2 and 3 combined. Each gold record requires 4 hours at studio 2 and 2 hours at studio 3. Formulate the problem of

188. nding the optimal production plan, give the Lagrangian, and give the set of equations that must be solved to

189. nd the optimal production plan. It is not necessary to actually solve the equations. Exercise 35 (a) Solve the problem max 2x + y subject to 4x2 + y2 = 8 (b) Estimate the change in the optimal objective function value when the right hand side increases by 5%, i.e. the right hand side increases from 8 to 8.4. Exercise 36 (a) Solve the following constrained optimization problem using the method of Lagrange multipliers. max ln x+ 2lny + 3lnz subject to x + y + z = 60 (b) Estimate the change in the optimal objective function value if the right hand side of the constraint increases from 60 to 65.

190. 4.2. EQUALITY AND INEQUALITY CONSTRAINTS 49 4.2 Equality and Inequality Constraints How do we handle both equality and inequality constraints in (P)? Let (P) be: Maximize f(x) Subject to g1(x) = b1 ... gm(x) = bm h1(x) d1 ... hp(x) dp If you have a program with constraints, convert it into by multiplying by 1. Also convert a minimization to a maximization. The Lagrangian is L(x; ; ) = f(x) + m Xi=1 i(bi gi(x)) + p Xj=1 j(dj hj (x)) The fundamental result is the following: Assume x = (x1; x2; : : :; xn) maximizes f(x) subject to the constraints gi(x) = bi, for i = 1; 2; : : :;m and hj (x) dj , for j = 1; 2; : : :; p. Then either (i) the vectors rg1(x); : : :;rgm(x), rh1(x); : : :;rhp(x) are linearly dependent, or (ii) there exists vectors = (1; : : :; m) and = (1; : : :; p ) such that rf(x) m Xi=1 i rgi(x) p Xj=1 j rhj(x) = 0 j (hj(x) dj) = 0 (Complementarity) j 0 In this course, we will not concern ourselves with Case (i). We will only look for candidate solutions x for which we can

191. nd and satisfying the equations in Case (ii) above. In general, to solve these equations, you begin with complementarity and note that either j must be zero or hj(x) dj = 0. Based on the various possibilities, you come up with one or more candidate solutions. If there is an optimal solution, then one of your candidates will be it. The above conditions are called the Kuhn{Tucker (or Karush{Kuhn{Tucker) conditions. Why do they make sense? For x optimal, some of the inequalities will be tight and some not. Those not tight can be ignored (and will have corresponding price j = 0). Those that are tight can be treated as equalities which leads to the previous Lagrangian stu. So

192. 50 CHAPTER 4. CONSTRAINED OPTIMIZATION j (hj(x) dj) = 0 (Complementarity) forces either the price j to be 0 or the constraint to be tight. Example 4.2.1 Maximize x3 3x Subject to x 2 The Lagrangian is L = x3 3x + (2 x) So we need 3x2 3 = 0 x 2 (2 x) = 0 0 Typically, at this point we must break the analysis into cases depending on the complementarity conditions. If = 0 then 3x2 3 = 0 so x = 1 or x = 1. Both are feasible. f(1) = 2, f(1) = 2. If x = 2 then = 9 which again is feasible. Since f(2) = 2, we have two solutions: x = 1 and x = 2. Example 4.2.2 Minimize (x 2)2 + 2(y 1)2 Subject to x+ 4y 3 x y First we convert to standard form, to get Maximize (x 2)2 2(y 1)2 Subject to x+ 4y 3 x + y 0 L(x; y; 1; 2) = (x 2)2 2(y 1)2 + 1(3 x 4y) + 2(0 + x y) which gives optimality conditions 2(x 2) 1 + 2 = 0 4(y 1) 41 2 = 0 1(3 x 4y) = 0

193. 4.2. EQUALITY AND INEQUALITY CONSTRAINTS 51 2(x y) = 0 x+ 4y 3 x + y 0 1; 2 0 Since there are two complementarity conditions, there are four cases to check: 1 = 0; 2 = 0: gives x = 2, y = 1 which is not feasible. 1 = 0; x y = 0: gives x = 4=3; y = 4=3; 2 = 4=3 which is not feasible. 2 = 0; 3 x 4y = 0 gives x = 5=3; y = 1=3; 1 = 2=3 which is O.K. 3x4y = 0; xy = 0 gives x = 3=5; y = 3=5; 1 = 22=25; 2 = 48=25 which is not feasible. Since it is clear that there is an optimal solution, x = 5=3; y = 1=3 is it! Economic Interpretation The economic interpretation is essentially the same as the equality case. If the right hand side of a constraint is changed by a small amount , then the optimal objective changes by , where is the optimal Lagrange multiplier corresponding to that constraint. Note that if the constraint is not tight then the objective does not change (since then = 0). Handling Nonnegativity A special type of constraint is nonnegativity. If you have a constraint xk 0, you can write this as xk 0 and use the above result. This constraint would get a Lagrange multiplier of its own, and would be treated just like every other constraint. An alternative is to treat nonnegativity implicitly. If xk must be nonnegative: 1. Change the equality associated with its partial to a less than or equal to zero: @f(x) @xk Xi i @gi(x) @xk Xj j @hj(x) @xk 0 2. Add a new complementarity constraint: 0@ @f(x) @xk Xi i @gi(x) @xk Xj j @hj(x) @xk 1A xk = 0 3. Don't forget that xk 0 for x to be feasible.

194. 52 CHAPTER 4. CONSTRAINED OPTIMIZATION Suciency of conditions The Karush{Kuhn{Tucker conditions give us candidate optimal solutions x. When are these conditions sucient for optimality? That is, given x with and satisfying the KKT conditions, when can we be certain that it is an optimal solution? The most general condition available is: 1. f(x) is concave, and 2. the feasible region forms a convex region. While it is straightforward to determine if the objective is concave by computing its Hessian matrix, it is not so easy to tell if the feasible region is convex. A useful condition is as follows: The feasible region is convex if all of the gi(x) are linear and all of the hj(x) are convex. If this condition is satis

195. ed, then any point that satis

196. es the KKT conditions gives a point that maximizes f(x) subject to the constraints. Example 4.2.3 Suppose we can buy a chemical for $10 per ounce. There are only 17.25 oz available. We can transform this chemical into two products: A and B. Transforming to A costs $3 per oz, while transforming to B costs $5 per oz. If x1 oz of A are produced, the price we command for A is $30 x1; if x2 oz of B are produced, the price we get for B is $50 x2. How much chemical should we buy, and what should we transform it to? There are many ways to model this. Let's let x3 be the amount of chemical we purchase. Here is one model: Maximize x1(30 x1) + x2(50 2x2) 3x1 5x2 10x3 Subject to x1 + x2 x3 0 x3 17:25 The KKT conditions are the above feasibility constraints along with: 30 2x1 3 1 = 0 50 4x2 5 1 = 0 10 + 1 2 = 0 1(x1 x2 + x3) = 0 2(17:25 x3) = 0 1; 2 0 There are four cases to check: 1 = 0; 2 = 0. This gives us 10 = 0 in the third constraint, so is infeasible. 1 = 0; x3 = 17:25. This gives 2 = 10 so is infeasible. x1x2+x3 = 0, 2 = 0. This gives 1 = 10; x1 = 8:5; x2 = 8:75; x3 = 17:25, which is feasible. Since the objective is concave and the constraints are linear, this must be an optimal solution. So there is no point in going through the last case (x1 x2 +x3 = 0, x3 = 17:25). We are done with x1 = 8:5; x2 = 8:75; and x3 = 17:25. What is the value of being able to purchase 1 more unit of chemical?

197. 4.3. EXERCISES 53 This question is equivalent to increasing the right hand side of the constraint x3 17:25 by 1 unit. Since the corresponding lagrangian value is 0, there is no value in increasing the right hand side. Review of Optimality Conditions. The following reviews what we have learned so far: Single Variable (Unconstrained) Solve f0(x) = 0 to get candidate x. If f00(x) 0 then x is a local min. f00(x) 0 then x is a local max. If f(x) is convex then a local min is a global min. f(x) is concave then a local max is a global max. Multiple Variable (Unconstrained) Solve rf(x) = 0 to get candidate x. If H(x) is positive de

198. nite then x is a local min. H(x) is negative de

199. nite x is a local max. If f(x) is convex then a local min is a global min. f(x) is concave then a local max is a global max. Multiple Variable (Equality constrained) Form Lagrangian L(x; ) = f(x) +Pi i(bi gi(x)) Solve rL(x; ) = 0 to get candidate x (and ). Best x is optimum if optimum exists. Multiple Variable (Equality and Inequality constrained) Put into standard form (maximize and constraints) Form Lagrangian L(x; ) = f(x) +Pi i(bi gi(x)) +Pj j (dj hj(x)) Solve rf(x) Pi irgi(x) Pj jrhj (x) = 0 gi(x) = bi hj (x) dj j (dj hj (x)) = 0 j 0 to get candidates x (and , ). Best x is optimum if optimum exists. Any x is optimum if f(x) concave, gi(x) convex, hj(x) linear. 4.3 Exercises Exercise 37 Solve the following constrained optimization problem using the method of Lagrange multipliers. max 2 lnx1 + 3lnx2 + 3lnx3 s.t. x1 + 2x2 + 2x3 = 10

200. 54 CHAPTER 4. CONSTRAINED OPTIMIZATION Exercise 38 Find the two points on the ellipse given by x2 1+4x2 2 = 4 that are at minimum distance of the point (1; 0). Formulate the problem as a minimization problem and solve it by solving the Lagrangian equations. [Hint: To minimize the distance d between two points, one can also minimize d2. The formula for the distance between points (x1; x2) and (y1; y2) is d2 = (x1y1)2+(x2y2)2.] Exercise 39 Solve using Lagrange multipliers. 1 + x2 a) min x2 2 + x2 3 subject to x1 + x2 + x3 = b, where b is given. b) maxpx1 + px2 + px3 subject to x1 + x2 + x3 = b, where b is given. c) min c1x2 1 +c2x2 2 +c3x2 3 subject to x1 +x2 +x3 = b, where c1 0, c2 0, c3 0 and b are given. 1 + x2 d) min x2 2 + x2 3 subject to x1 + x2 = b1 and x2 + x3 = b2, where b1 and b2 are given. Exercise 40 Let a, b and c be given positive scalars. What is the change in the optimum value of the following constrained optimization problem when the right hand side of the constraint is increased by 5%, i.e. a is changed to a + 5 100a. max by x4 s.t. x2 + cy = a Give your answer in terms of a, b and c. Exercise 41 You want to invest in two mutual funds so as to maximize your expected earnings while limiting the variance of your earnings to a given

201. gure s2. The expected yield rates of Mutual Funds 1 and 2 are r1 and r2 respectively, and the variance of earnings for the portfolio (x1; x2) is 2x2 1 + x1x2 + 2x2 2. Thus the problem is max r1x1 + r2x2 s.t. 2x2 1 + x1x2 + 2x2 2 = s2 (a) Use the method of Lagrange multipliers to compute the optimal investments x1 and x2 in Mutual Funds 1 and 2 respectively. Your expressions for x1 and x2 should not contain the Lagrange multiplier . (b) Suppose both mutual funds have the same yield r. How much should you invest in each? Exercise 42 You want to minimize the surface area of a cone-shaped drinking cup having

202. xed volume V0. Solve the problem as a constrained optimization problem. To simplify the algebra, minimize the square of the area. The area is rpr2 + h2. The problem is, min 2r4 + 2r2h2 s.t. 1 3 r2h = V0: Solve the problem using Lagrange multipliers. Hint. You can assume that r6= 0 in the optimal solution.

203. 4.3. EXERCISES 55 Exercise 43 A company manufactures two types of products: a standard product, say A, and a more sophisticated product B. If management charges a price of pA for one unit of product A and a price of pB for one unit of product B, the company can sell qA units of A and qB units of B, where qA = 400 2pA + pB; qB = 200 + pA pB: Manufacturing one unit of product A requires 2 hours of labor and one unit of raw material. For one unit of B, 3 hours of labor and 2 units of raw material are needed. At present, 1000 hours of labor and 200 units of raw material are available. Substituting the expressions for qA and qB, the problem of maximizing the company's total revenue can be formulated as: max 400pA + 200pB 2p2 A p2 B + 2pApB s.t. pA pB 400 pB 600 (a) Use the Khun-Tucker conditions to

204. nd the company's optimal pricing policy. (b) What is the maximum the company would be willing to pay for { another hour of labor, { another unit of raw material?

205. 56 CHAPTER 4. CONSTRAINED OPTIMIZATION

206. Chapter 5 Modeling with Linear Programming 5.1 Introductory Example SilComputers makes quarterly decisions about their product mix. While their full product line includes hundreds of products, we will consider a simpler problem with just two products: notebook computers and desktop computers. SilComputers would like to know how many of each product to produce in order to maximize pro

207. t for the quarter. There are a number of limits on what SilComputers can produce. The major constraints are as follows: 1. Each computer (either notebook or desktop) requires a Processing Chip. Due to tightness in the market, our supplier has allocated 10,000 such chips to us. 2. Each computer requires memory. Memory comes in 16MB chip sets. A notebook computer has 16MB memory installed (so needs 1 chip set) while a desktop computer has 32MB (so requires 2 chip sets). We received a great deal on chip sets, so have a stock of 15,000 chip sets to use over the next quarter. 3. Each computer requires assembly time. Due to tight tolerances, a notebook computer takes more time to assemble: 4 minutes versus 3 minutes for a desktop. There are 25,000 minutes of assembly time available in the next quarter. Given current market conditions, material cost, and our production system, each notebook computer produced generates $750 pro

208. t, and each desktop produces $1000 pro

209. t. There are many questions SilComputer might ask. The most obvious are such things as How many of each type computer should SilComputer produce in the next quarter? What is the maximum pro

210. t SilComputer can make? Less obvious, but perhaps of more managerial interest are How much should SilComputer be willing to pay for an extra memory chip set? What is the eect of losing 1,000 minutes of assembly time due to an unexpected machine failure? How much pro

211. t would we need to make on a 32MB notebook computer to justify its production? Linear programming gives us a mechanism for answering all of these questions quickly and easily. There are three steps in applying linear programming: modeling, solving, and interpreting. 5.1.1 Modeling We begin by modeling this problem. Modeling a problem using linear programming involves writing it in the language of linear programming. There are rules about what you can and cannot do within 57

212. 58 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING linear programming. These rules are in place to make certain that the remaining steps of the process (solving and interpreting) can be successful. Key to a linear program are the decision variables, objective, and constraints. Decision Variables. The decision variables represent (unknown) decisions to be made. This is in contrast to problem data, which are values that are either given or can be simply calculated from what is given. For this problem, the decision variables are the number of notebooks to produce and the number of desktops to produce. We will represent these unknown values by x1 and x2 respectively. To make the numbers more manageable, we will let x1 be the number of 1000 notebooks produced (so x1 = 5 means a decision to produce 5000 notebooks) and x2 be the number of 1000 desktops. Note that a value like the quarterly pro

213. t is not (in this model) a decision variable: it is an outcome of decisions x1 and x2. Objective. Every linear program has an objective. This objective is to be either minimized or maximized. This objective has to be linear in the decision variables, which means it must be the sum of constants times decision variables. 3x1 10x2 is a linear function. x1x2 is not a linear function. In this case, our objective is to maximize the function 750x1 +1000x2 (what units is this in?). Constraints. Every linear program also has constraints limiting feasible decisions. Here we have four types of constraints: Processing Chips, Memory Sets, Assembly, and Nonnegativity. In order to satisfy the limit on the number of chips available, it is necessary that x1 + x2 10. If this were not the case (say x1 = x2 = 6), the decisions would not be implementable (12,000 chips would be required, though we only have 10,000). Linear programming cannot handle arbitrary restrictions: once again, the restrictions have to be linear. This means that a linear function of the decision variables must be related to a constant, where related can mean less than or equal to, greater than or equal to, or equal to. So 3x1 2x2 10 is a linear constraint, as is x1 + x3 = 6. x1x2 10 is not a linear constraint, nor is x1 + 3x2 3. Our constraint for Processing Chips x1 + x2 10 is a linear constraint. The constraint for memory chip sets is x1 + 2x2 15, a linear constraint. Our constraint on assembly can be written 4x1 + 3x2 25, again a linear constraint. Finally, we do not want to consider decisions like x1 = 5, where production is negative. We add the linear constraints x1 0, x2 0 to enforce nonnegativity of production. Final Model. This gives us the complete model of this problem: Maximize 750x1 + 1000x2 Subject to x1 + x2 10 x1 + 2x2 15 4x1 + 3x2 25 x1 0 x2 0 Formulating a problem as a linear program means going through the above process to clearly de

214. ne the decision variables, objective, and constraints. 5.1.2 Solving the Model Models are useful in their own right: they allow for a formal de

215. nition of a problem and can be useful in thinking about a problem. Linear programming models are particularly useful, however, because it is easy to have a computer solve for the optimal decision values.

216. 5.1. INTRODUCTORY EXAMPLE 59 For a model with only two variables, it is possible to solve the problem without a computer by drawing the feasible region and determining how the objective is optimized on that region. We go through that process here. The purpose of this exercise is to give you intuition and understanding of linear programming models and their solution. In any real application, you would use a computer to solve even two variable problems (we outline how to use Excel's Solver routine to

217. nd solutions in the next section). We can represent a model with two variables by labeling the axes of a graph with each of the variables. The entire graph then represents possible decisions. Constraints are represented by lines on the graph, with the feasible region lying on one side of the line. The following

218. gure illustrates this with the constraint x1 + x2 10. 2 4 6 8 10 8 10 6 4 2 x1 x2 x1 + x2 10 Figure 5.1: Single Constraint We can continue this process and add in all of the constraints. Since every constraint must be satis

219. ed, the resulting feasible region is the intersection of the feasible region for each constraint. This is shown in the following

220. gure. Note that just graphing the model gives us information we did not have before. It seems that the Chip constraint (x1 + x2 10) plays little role in this model. This constraint is dominated by other constraints. Now, how can we

221. nd the optimal solution? We can include the objective function on this diagram by drawing iso-pro

222. t lines: lines along which the pro

223. t is the same. Since our goal is to maximize the pro

224. t, we can push the isopro

225. t line out until moving it any further would result in no feasible point (see diagram, z represents pro

226. t). Clearly the optimal pro

227. t occurs at point X. What are the variable values at point X. Note that X is the intersection of the constraints:

228. 60 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING 2 4 6 8 10 10 8 6 4 2 0 x1 4x1 + 3x2 25 x2 x1 + x2 10 x1 + 2x2 15 Figure 5.2: All Constraints z=6000 2 4 6 8 10 10 8 6 4 2 0 X x1 4x1 + 3x2 25 x2 x1 + x2 10 x1 + 2x2 15 z=3000 Figure 5.3: Finding Optimal Pro

229. t

230. 5.1. INTRODUCTORY EXAMPLE 61 x1 + 2x2 = 15 4x1 + 3x2 = 25 We discussed in the previous section how to solve such equations: the solution here is x1 = 1 and x2 = 7. The optimal decision is to produce 1,000 notebooks and 7,000 desktops, for a pro

231. t of $7,750,000. 5.1.3 Using Solver Rather than the somewhat tedious and error-prone graphical method (which is limited to linear programs with 2 variables), special computer programs can be used to

232. nd solutions to linear programming models. The most widespread program is undoubted Solver, included in all recent versions of the Excel spreadsheet program. Solver, while not a state of the art code (which can cost upwards of $15,000 per copy) is a reasonably robust, easy-to-use tool for linear programming. Solver uses standard spreadsheets together with an interface to de

233. ne variables, objective, and constraints to de

234. ne a linear program. It is dicult to describe in words how to create a Solver spreadsheet, so we will do one or two in class. Here is a brief outline and some hints and shortcuts: We will start with a spreadsheet that has all of the data entered in some reasonably neat way. We will create the model in a separate part of the spreadsheet. We will have one cell for each variable. Solver will eventually put the optimal values in each cell. We will have a single cell to represent the objective. We will enter a formula that represents the objective. This formula must be a linear formula, so it must be of the form: cell1*cell1'+cell2*cell2'+: : : , where cell1, cell2 and so on contain constant values and cell1', cell2' and so are the variable cells. Helpful Hint: Excel has a function sumproduct() that is designed for linear programs. sumproduct(a1..a10,b1..b10) is identical to a1*b1+a2*b2+a3*b3+: : :+a10*b10. This function will save much time and aggravation. All that is needed is that the length of the

235. rst range is the same as the length of the second range (so one can be horizontal and the other vertical). Helpful Hint: It is possible to assign names to cells and ranges (under the Insert-Name menu). Rather than use a1..a10 as the variables, you can name that range var (for example) and then use var wherever a1..a10 would have been used. We then have a cell to represent the left hand side of each constraint (again a linear function) and another cell to represent the right hand side (a constant). We then select Solver under the Tools menu. This gives a form to

236. ll out to de

237. ne the linear program. In the Set Cell box, select the objective cell. Choose Maximize or Minimize. In the By Changing Cells, put in the range containing the variable cells.

238. 62 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING We next add the constraints. Press the Add... button to add constraints. The dialog box has three parts for the left hand side, the type of constraint, and the right hand side. Put the cell references for a constraint in the form, choose the right type, and press Add. Continue until all constraints are added. On the

239. nal constraint, press OK. We need to explicitly include nonnegativity constraints. Helpful Hint: It is possible to include ranges of constraints, as long as they all have the same type. c1..e1 = c3..e3 means c1 = c3, d1 = d3, e1 = e3. a1..a10 = 0 means each individual cell must be greater than or equal to 0. Push the options button and toggle the Assume Linear Model in the resulting dialog box. This tells Excel to call a linear rather than a nonlinear pgrogramming routine so as to solve the problem more eciently. This also gives you sensitivity ranges, which are not available for nonlinear models. Push the Solve button. In the resulting dialog box, select Answer and Sensitivity. This will put the answer and sensitivity analysis in two new sheets. Ask Excel to Keep Solver values, and your worksheet will be updated so that the optimal values are in the variable cells. 5.2 Second Example: Market Allocations MegaMarketing is planning a concentrated one week advertising campaign for their new CutsEv- erything SuperKnife. The ads have been designed and produced and now they wish to determine how much money to spend in each advertising outlet. In reality, they have hundreds of possible outlets to choose from. We will illustrate their problem with two outlets: Prime-time TV, and newsmagazines. The problem of optimally spending advertising dollars can be formulated in many ways. For instance, given a

240. xed budget, the goal might be to maximize the number of target customers reached (a target customer is a customer with a reasonable chance of purchasing the product). An alternative approach, which we adopt here, is to de

241. ne targets for reaching each market segment and to minimize the money spent to reach those targets. For this product, the target segments are Teenage Boys, Auent Women (ages 40-49), and Retired Men. Each minute of primetime TV and page of newsmagazine advertisement reaches the following number of people (in millions): Outlet Boys Women Men Cost TV 5 1 3 600 Mag 2 6 3 500 Target 24 18 24 Again, MegaMarketing is interested in straightforward answers like how many units of each outlet to purchase to meet the segment goals. They are also interested in such questions as How much will it cost to reach an extra million retired men?, One radio spot reaches 1 million boys, 1 million women, and 1 million men: how much are we willing to pay for such a spot?, and similar questions.

242. 5.2. SECOND EXAMPLE: MARKET ALLOCATIONS 63 5.2.1 Modeling The process of modeling using linear programming is fairly consistent: de

243. ne the decision variables, objective, and constraints. Decision Variables. In this case, the decision variables are the number of units of each outlet to purchase. Let x1 be the number of TV spots, and x2 the number of magazine pages. Objective. Our objective is to minimize the amount we spend: 600x1 + 500x2. Constraints. We have one constraint for each market segment: we must be certain that we reach sucient people of each segment. For boys, this corresponds to the constraint: 5x1 + 2x2 24 Similar constraints for the other segments gives us the full formulation as: Minimize 600x1 + 500x2 Subject to 5x1 + 2x2 24 x1 + 6x2 18 3x1 + 3x2 24 x1 0 x2 0 5.2.2 Solving the Model Since this model has only two variables, we can solve the problem in two ways: graphically and using SOLVER (with SOLVER being much easier). We start by solving this graphically. The

244. rst step is to graph the feasible region, as given in the following

245. gure: 10 8 6 4 2 x1 + 6x2 18 0 2 4 6 8 10 12 x2 x1 3x1 + 3x2 24 5x1 + 2x2 24 Figure 5.4: Marketing example

246. 64 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING The next step is to put some iso-cost lines on the diagram: lines representing points that have the same cost. We mark the optimal point with an X. 10 8 6 4 2 X z = 6000000 x1 + 6x2 18 0 2 4 6 8 10 12 x1 3x1 + 3x2 24 5x1 + 2x2 24 x2 Figure 5.5: Marketing example with Isocost Line X is the intersection of the constraints: 5x1 + 2x2 = 24 3x1 + 3x2 = 24 The optimal solution is x1 = 2:67 and x2 = 5:33 5.2.3 Using Solver The SOLVER model for this problem is on the course web pages. 5.3 Common Features Hidden in our models of these problems are a number of assumptions. The usefulness of a model is directly related to how close reality matches up with these assumptions. The

247. rst assumption is due to the linear form of our functions. Since the objective is linear, the contribution to the objective of any decision variable is proportional to the value of the decision variable. Producing twice as much of a product produces twice as much pro

248. t; buying twice as many pages of ads costs twice as much. This is the Proportionality Assumption. Furthermore, the contribution of a variable to the objective is independent of the values of the other variables. One notebook computer is worth $750, independent of how many desktop computers we produce. This is the Additivity Assumption.

249. 5.4. LINEAR PROGRAMMING MODELS 65 Similarly, since each constraint is linear, the contribution of each variable to the left hand side of each constraint is proportional to the value of the variable and independent of the values of any other variable. These assumptions are quite restrictive. We will see, however, that clever modeling can handle situations that may appear to violate these assumptions. The next assumption is the Divisibility Assumption: it is possible to take any fraction of any variable. Rethinking the Marketing example, what does it mean to purchase 2.67 television ads? It may be that the divisibility assumption is violated in this example. Or, it may be that the units are such that 2.67 ads actually corresponds to 2666.7 minutes of ads, in which case we can round o our solution to 2667 minutes with little doubt that we are getting an optimal or nearly optimal solution. Similarly, a fractional production quantity may be worisome if we are producing a small number of battleships or be innocuous if we are producting millions of paperclips. If the Divisibility Assumption is important and does not hold, then a technique called integer programming rather than linear programming is required. This technique takes orders of magnitude more time to

250. nd solutions but may be necessary to create realistic solutions. You will learn more about this in 45-761. The

251. nal assumption is the Certainty Assumption: linear programming allows for no uncertainty about the numbers. An ad will reach the given number of people; the number of assembly hours we give will certainly be available. It is very rare that a problem will meet all of the assumptions exactly. That does not negate the usefulness of a model. A model can still give useful managerial insight even if reality diers slightly from the rigorous requirements of the model. For instance, the knowledge that our chip inventory is more than sucient holds in our

252. rst model even if the proportionality assumptions are not satis

253. ed completely. 5.4 Linear Programming Models Linear programming models are found in almost every

254. eld of business (and beyond!). The next sections go through a number of problems, showing how to model them by the appropriate choice of decision variables, objective, and constraints. In all cases, we will describe the problem and give a model. Solver worksheets for each of these is available on the web page. 5.4.1 Diet Problem Problem De

255. nition What is the perfect diet? An ideal diet would meet or exceed basic nutritional requirements, be inexpensive, have variety, and be pleasing to the palate. How can we

256. nd such a diet? Suppose the only foods in the world are as follows: Energy Protein Calcuim Price Limit Food Serving Size (kcal) (g) (mg) (cents/serving) (servings/day) Oatmeal 28g 110 4 2 3 4 Chicken 100g 205 32 12 24 3 Eggs 2 Large 160 13 54 13 2 Whole Milk 237cc 160 8 285 9 8 Cherry Pie 170g 420 4 22 20 2 PorkBeans 260g 260 14 80 19 2

257. 66 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING After consulting with nutitionists, we decree that a satisfactory diet has at least 2000 kcal of energy, 55 g of protein, and 800 mg of calcium (vitamins and iron are supplied by pills). While some of us would be happy to subsist on 10 servings of pork and beans, we have decided to impose variety by having a limit on the number of servings/day for each of our six foods. What is the least cost satisfactory diet? Problem Modeling First we decide on decision variables. Let us label the foods 1, 2, : : : 6, and let xi represent the number of servings of food i in the diet. Our objective is to minimize cost, which can be written 3x1 + 24x2 + 13x3 + 9x4 + 20x5 + 19x6 We have constraints for energy, protein, calcuim, and for each serving/day limit. This gives the formulation: Minimize 3x1 + 24x2 + 13x3 + 9x4 + 20x5 + 19x6 Subject to 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800 x1 4 x2 3 x3 2 x4 8 x5 2 x6 2 xi 0 (for all i) Discussion The creation of optimal diets was one of the

258. rst uses of linear programming. Some diculties with linear programming include diculties in formulating palatability requirements, and issues of divisibility (no one wants to eat half a green bean). These linear programming models do give an idea on how much these palatability requirements are costing. 5.4.2 Workforce Planning Problem De

259. nition. Consider a restaurant that is open seven days a week. Based on past experience, the number of workers needed on a particular day is given as follows: Day Mon Tue Wed Thu Fri Sat Sun Number 14 13 15 16 19 18 11 Every worker works

260. ve consecutive days, and then takes two days o, repeating this pattern inde

261. nitely. How can we minimize the number of workers that sta the restaurant?

262. 5.4. LINEAR PROGRAMMING MODELS 67 Model. A natural (and wrong!)

263. rst attempt at this problem is to let xi be the number of people working on day i. Note that such a variable de

264. nition does not match up with what we need to

265. nd. It does us no good to know that 15 people work Monday, 13 people Tuesday, and so on because it does not tell us how many workers are needed. Some workers will work both Monday and Tuesday, some only one day, some neither of those days. Instead, let the days be numbers 1 through 7 and let xi be the number of workers who begin their

266. ve day shift on day i. Our objective is clearly: x1 + x2 + x3 + x4 + x5 + x6 + x7 Consider the constraint for Monday's stang level of 14. Who works on Mondays? Clearly those who start their shift on Monday (x1). Those who start on Tuesday (x2) do not work on Monday, nor do those who start on Wednesday (x3). Those who start on Thursday (x4) do work on Monday, as do those who start on Friday, Saturday, and Sunday. This gives the constraint: x1 + x4 + x5 + x6 + x7 14 Similar arguments give a total formulation of: Minimize Pi xi Subject to x1 + x4 + x5 + x6 + x7 14 x1 + x2 + x5 + x6 + x7 13 x1 + x2 + x3 + x6 + x7 15 x1 + x2 + x3 + x4 + x7 16 x1 + x2 + x3 + x4 + x5 19 x2 + x3 + x4 + x5 + x6 18 x3 + x4 + x5 + x6 + x7 11 xi 0 (for all i) Discussion. Workforce modeling is a well developed area. Note that our model has only one type of shift, but the model is easily extended to other types of shifts, with diering shift costs. 5.4.3 Financial Portfolio Problem De

267. nition. In your

268. nance courses, you will learn a number of techniques for creating optimal portfolios. The optimality of a portfolio depends heavily on the model used for de

269. ning risk and other aspects of

270. nancial instruments. Here is a particularly simple model that is amenable to linear programming techniques. Consider a mortgage team with $100,000,000 to

271. nance various investments. There are

272. ve categories of loans, each with an associated return and risk (1-10, 1 best):

273. 68 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING Loan/investment Return (%) Risk First Mortgages 9 3 Second Mortgages 12 6 Personal Loans 15 8 Commercial Loans 8 2 Government Securities 6 1 Any uninvested money goes into a savings account with no risk and 3% return. The goal for the mortgage team is to allocate the money to the categories so as to: (a) Maximize the average return per dollar (b) Have an average risk of no more than 5 (all averages and fractions taken over the invested money (not over the saving account)). (c) Invest at least 20% in commercial loans (d) The amount in second mortgages and personal loans combined should be no higher than the amount in

274. rst mortgages. Model Let the investments be numbered 1: : :5, and let xi be the amount invested in investment i. Let xs be the amount in the savings account. The objective is to maximize 9x1 + 12x2 + 15x3 + 8x4 + 6x5 + 3xs subject to x1 + x2 + x3 + x4 + x5 + xs = 100; 000; 000: Now, let's look at the average risk. Since we want to take the average over only the invested amount, a direct translation of this constraint is 3x1 + 6x2 + 8x3 + 2x4 + x5 x1 + x2 + x3 + x4 + x5 5 This constraint is not linear, but we can cross multiply, and simplify to get the equivalent linear constraint: 2x1 + x2 + 3x3 3x4 4x5 0 Similarly we need x4 0:2(x1 + x2 + x3 + x4 + x5) or 0:2x1 0:2x2 0:2x3 + 0:8x4 0:2x5 0 The

275. nal constraint is x2 + x3 x1 0 Together with nonnegativity, this gives the entire formulation.

276. 5.4. LINEAR PROGRAMMING MODELS 69 Discussion Optimal portfolios do not just happen: they must be calculated, and there is a constant interplay between models and solvability. Linear programming models provide great modeling power with a great limit: the handling of risk must be done in a linear fashion (like our Risk factors here). Other models you will see in

277. nance will look at the co-variance of returns between investments, a fundamentally nonlinear eect. This can give rise to nonlinear models like those that try to minimize variance subject to return requirements. It is very dicult to embed idiosyncratic constraints (like (c) and (d) here) in such models. 5.4.4 Production over Time Problem Description Many problems are multiperiod, where a series of decisions have to be made over time. For instance, in our very

278. rst SilComputer example, production decisions for computers have to be made with the future in mind. In this example, it would be easy to work with both desktop and notebook computers but let's work just with notebook computers. SilComputer needs to meet the demand of its largest corporate and educational customers for notebook computers over the next four quarters (before its current model becomes obsolete). SilComputer currently has 5,000 notebook computers in inventory. Expected demand over the next four quarters for its notebook is 7,000; 15,000; 10,000; and 8,000. SilComputer has sucient capacity and material to produce up to 10,000 computers in each quarter at a cost of $2000 per notebook. By using overtime, up to an additional 2,500 computers can be produced at a cost of $2200 each. Computers produced in a quarter can be used either to meet that quarter's demand, or be held in inventory for use later. Each computer in inventory is charged $100 to re ect carying costs. How should SilComputer meet its demand for notebooks at minimum cost? Modeling. The decisions seem to be how many of each computer to produce in each period at regular time, how many to produce at overtime, and how much inventory to carry in each period. Let's denote our time periods t = 1; 2; 3; 4. Let xt be the number of notebooks produced in period t at regular time and yt be the number of notebooks produced in period t at overtime. Finally, let it be the inventory at the end of period t. Look at the

279. rst quarter: how are these variables restricted and related? Clearly we need x1 10000 and y1 2500. Now, anything that starts as inventory or is produced in the period must either be used to meet demand or ends up as inventory at the end of period 1. This means: 5000 + x1 + y1 = 7000 + i1 For period 2, in addition to the upper bounds, we get the constraint i1 + x2 + y2 = 15000 + i2 For period 3, we get i2 + x3 + y3 = 10000 + i3

280. 70 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING and for period 4 (assuming no inventory at the end): i3 + x4 + y4 = 8000 Our objective charges 2000 for each x, 2200 for each y, and 100 for each i: 2000x1 + 2000x2 + 2000x3 + 2000x4 + 2200y1 + 2200y2 + 2200y3 + 2200y4 + 100i1 + 100i2 + 100i3 Discussion. The constraints tracking where production goes are known as conservation constraints, and model the requirement that production never disappears. Whenever you have multiple types of variables in a formulation, be certain to think about whether there are necessary constraints that link them. 5.4.5 Investing over Time We are going to manage an investment portfolio over a 6-year time horizon. We begin with $1000, and at various times we can invest in one or more of the following: Savings account X, annual yield 5%. Security Y, 2-year maturity, total yield 12% if bought now, 11% thereafter. Security Z, 3-year maturity, total yield 18%. Security W, 4-year maturity, total yield 24%. To keep things simple we will assume that each security can be bought in any denomination. (This assumption can be relaxed if one uses integer or dynamic programming .) We can make savings deposits or withdrawals anytime. We can buy Security Y any year but year 3. We can buy Security Z anytime after the

281. rst year. Security W, now available, is a one-time opportunity. We let xt be the amount of money invested in the savings account X at the beginning of year t, and similarly for yt, zt, and wt. We will put any money not tied up in securities into the savings account. The situation is summed up in the

282. gure. The problem is really a kind of inventory problem. In a given year, the amount of money carried forward from the previous year (in savings), plus the yield from securities that mature that year, equals the amount invested in new securities plus the amount of money left over for next year. There is one twist: inventory grows while in storage. (In real inventory problems, stock often decreases over time due to spoilage, etc., and this can be re ected in negative interest rates.) Let v be the

283. nal yield when all securities are cashed in at the end of the sixth year. Then if the objective is to maximize

284. nal yield, the LP is, max v s.t. x1 + y1 + w1 = 1000 x2 + y2 + z2 = 1:05x1 x3 + z3 = 1:05x2 + 1:12y1 x4 + y4 + z4 = 1:05x3 + 1:11y2 x5 + y5 = 1:05x4+ 1:18z2+ 1:24w1 x6 = 1:05x5 + 1:11y4 + 1:18z3 v = 1:05x6 + 1:11y5 + 1:18z4; xt; yt; zt; wt 0

285. 5.4. LINEAR PROGRAMMING MODELS 71 1 2 3 4 5 6 7 x1 x2 x3 x4 x5 x6 5% y1 y5 12% 11% y2 y4 11% 11% 18% z2 18% z3 18% z4 w1 24% Figure 5.6: Possible investments over a 6-year horizon

286. 72 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING The

287. nal assets come to $1387.6, an average annual yield of 5.61% 5.4.6 Problems Modeling is an art, and it takes practice. The following examples show the variety of problems that can be attacked by linear programming, and give you the opportunity to try your hand at some problems. Exercise 44 A plant produces two types of refrigerators, A and B. There are two production lines, one dedicated to producing refrigerators of Type A, the other to producing refrigerators of Type B. The capacity of the production line for A is 60 units per day, the capacity of the production line for B is 50 units per day. A requires 20 minutes of labor whereas B requires 40 minutes of labor. Presently, there is a maximum of 40 hours of labor per day which can be assigned to either production line. Pro

288. t contributions are $20 per refrigerator of Type A produced and $30 per Type B produced. What should the daily production be? Solve graphically and by Solver. Exercise 45 Albert, Bill, Charles, David, and Edward have gotten into a bind. After a series of

289. nancial transactions, they have ended up each owing some of the others huge amounts of money. In fact, near as the lawyers can make out, the debts are as follows Debtor Creditor Amount ($millions) A E 10 A C 3 B A 5 C B 6 C D 4 D A 4 E C 7 E D 3 The question is, who is bankrupt? We will say that a person i is bankrupt if there is no possible transfer of funds among the people such that i completely pays o his obligations, and the transfer of funds satis

290. es the following condition: for every two persons j and k, the amount paid by person j to person k is no greater than the debt of j to k. For instance, Albert is bankrupt since he owes 10, and is only owed 9. Formulate the problem of determining whether Bill is bankrupt as a linear program. Then modify your formulation to determine if each of the others is bankrupt. This example may look contrived, but it is inspired by a solution to the debts involved in a crash of Kuwait's al-Mankh stock market. Exercise 46 Due to an unexpected glut of orders, Blaster Steel has decided to hire temporary workers for a

291. ve day period to clear out the orders. Each temporary worker can work either a two day shift or a three day shift for this period (shifts must be consecutive days). At least 10 workers are needed on days 1, 3, 5, and at least 15 workers are needed on days 2 and 4. A worker on a two day shift gets paid $125/day, while those on a three day shift gets paid $100/day. (a) Formulate the problem of hiring temporary workers to minimize cost while meeting the demand for workers. (b) Due to a limited number of training personnel, no more than 10 workers can start their shift on any day. Update your formulation in (a) to take this into account. (c) Union regulations require that at least half of all money spent on workers go to those who work three day shifts. Update your formulation in (a) to handle this requirement.

292. 5.4. LINEAR PROGRAMMING MODELS 73 (d) There are four people who are willing to work a shift consisting of days 1, 2, and 5, for a payment of $110/day. Update your formulation in (a) to handle this possibility. Exercise 47 You have $1000 to invest in Securities 1 and 2; uninvested funds are deposited in a savings account with annual yield of 5%. You will sell the securities when they reach maturity. Security 1 matures after 2 years with a total yield of 12%. Security 2 matures after 3 years with a total yield of 19%. Your planning horizon is 7 years, and you want to maximize total assets at the end of the seventh year. Suppose either security may be purchased in arbitrarily small denominations. a) Let xit be the amount invested in Security i at the beginning of year t; the savings account can be considered Security 0. Write the appropriate LP model. b) Solve the problem by computer and indicate your optimal portfolio in each year 1,...,7. Exercise 48 Chemco produces two chemicals: A and B. These chemicals are produced via two manufacturing processes. Process 1 requires 2 hours of labor and 1 lb. of raw material to produce 2 oz. of A and 1 oz. of B. Process 2 requires 3 hours of labor and 2 lb. of raw material to produce 3 oz. of A and 2 oz. of B. Sixty hours of labor and 40 lb. of raw material are available. Chemical A sells for $16 per oz. and B sells for $14 per oz. Formulate a linear program that maximizes Chemco's revenue. Hint: De

293. ne the amounts of Process 1 and Process 2 used, as your decision variables. Exercise 49 Gotham City National Bank is open Monday-Friday from 9 am to 5 pm. From past experience, the bank knows that it needs the following number of tellers. Time Period Tellers Required 9 10 4 10 11 3 11 noon 4 noon 1 6 1 2 5 2 3 6 3 4 8 4 5 8 The bank hires two types of tellers. Full-time tellers work 9-5

294. ve days a week, except for 1 hour o for lunch, either between noon and 1 pm or between 1 pm and 2 pm. Full-time tellers are paid (including fringe bene

295. ts) $25/hour (this includes payment for lunch hour). The bank may also hire up to 3 part-time tellers. Each part-time teller must work exactly 4 consecutive hours each day. A part-time teller is paid $20/hour and receives no fringe bene

296. ts. Formulate a linear program to meet the teller requirements at minimum cost. Exercise 50 Sales forecasts for the next four months are (in thousand of units): October 10 November 16 December 10 January 12 September's production was set at 12,000 units. Varying production rate incurs some cost: production can be increased from one month to the next at a cost of $2 per unit and it can be

297. 74 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING decreased at a cost of $0.50 per unit. In addition, inventory left at the end of a month can be stored at a cost of $1 per unit per month. Given current demand, there will be no inventory at the end of September. No inventory is desired at the end of January. Formulate a linear program that minimizes the total cost (varying production rate + inventory costs) of meeting the above demand. Exercise 51 An oil company blends gasoline from three ingredients: butane, heavy naphta and catalytic reformate. The characteristics of the ingredients as well as minimum requirements for regular gasoline are given below: Catalytic Heavy Butane Reformate Naphta Gasoline Octane 120 100 74 89 Vapor Pressure 60 2:5 4:0 11 Volatility 105 3 12 17 The cost (per gallon) of butane is $0.58, it is $1.55 for catalytic reformate and $0.85 for heavy naphta. How many gallons of the three ingredients should be blended in order to produce 12,000 gallons of gasoline at minimum cost? Exercise 52 An electric utility has six power plants on the drawing board. The anticipated useful life of these plants is 30 years for the coal-

298. red plants (plants 1, 2 and 3) and 40 years for the fuel-

299. red plants (plants 4, 5 and 6). Plants 1, 2 and 4 will be on line in year 5. Plants 3 and 5 in year 15. Plant 6 in year 25. The cost of installing generating capacity at Plant i, discounted to the present, is ci per megawatt, for i = 1; 2; : : :; 6. Projected power demand in year t is Dt megawatts, for t = 10; 20; 30; 40: Due to environmental regulations, the fraction of generating capacity at coal-

300. red plants in year t, relative to total generating capacity in year t, can be at most rt for t = 10; 20; 30; 40. The generating capacity of each plant has yet to be determined. Write a linear program that assigns capacities to the six plants so as to minimize total present cost of installing generating capacity, while meeting demand and satisfying the environmemtal constraints in years t = 10; 20; 30; 40: Exercise 53 (Wagner) You must decide how many tons x1 of pure steel and how many tons x2 of scrap metal to use in manufacturing an alloy casting. Pure steel costs $300 per ton and scrap $600 per ton (larger because the impurities must be skimmed o). The customer wants at least 5 tons but will accept a larger order, and material loss in melting and casting is negligible. You have 4 tons of steel and 7 tons of scrap to work with, and the weight ratio of scrap to pure cannot exceed 7/8 in the alloy. You are allotted 18 hours melting and casting time in the mill; pure steel requires 3 hrs per ton and scrap 2 hrs per ton. a) Write a linear programming model for the problem assuming that the objective is to minimize the total steel and scrap costs. b) Graph the problem obtained in (a). c) Solve the problem in (a) using Solver and indicate the optimal values of x1, x2. What is the weight ratio of scrap to pure steel? Exercise 54 (Wagner) An airline must decide how many new ight attendants to hire and train over the next six months. The sta requirements in person- ight-hours are respectively 8000, 9000, 7000, 10,000, 9000, and 11,000 in the months January through June. A new ight attendant is

301. 5.4. LINEAR PROGRAMMING MODELS 75 trained for one month before being assigned to a regular ight and therefore must be hired a month before he or she is needed. Each trainee requires 100 hours of supervision by experienced ight attendants during the training month, so that 100 fewer hours are available for ight service by regular ight attendants. Each experienced ight attendant can work up to 150 hours a month, and the airline has 60 regular ight attendants available at the beginning of January. If the maximum time available from experienced ight attendants exceeds a month's ying and training requirements, they work fewer hours, and none are laid o. At the end of each month, 10% of the experienced ight attendants quit their jobs. An experienced ight attendant costs the company $1700 and a trainee $900 a month in salary and bene

302. ts. a) Formulate the problem as a linear programming model. Let xt be the number of ight attendants that begin training in month t. Hint. Let yt be the number of experienced ight attendants available in month t; also let x0 = 0, y1 = 60. The problem is really an inventory problem, with two kinds of stock: trainees and experienced employees. The holding costs are the salaries, and the demands are the number of ight hours needed. Spoilage is attrition of experienced sta. b) Solve the model with Solver and indicate the solution value for the xt's and yt's. Exercise 55 (Wagner) A lumber company operates a sawmill that converts timber to lumber or plywood. A marketable mix of 1000 board feet of lumber products requires 1000 board feet of spruce and 4000 board feet of Douglas

303. r. Producing 1000 square feet of plywood requires 2000 board feet of spruce and 4000 board feet of

304. r. The company's timberland yields 32,000 board feet of spruce and 72,000 board feet of

305. r each season. Sales commitments require that at least 5000 board feet of lumber and 12,000 board feet of plywood be produced during the season. The pro

306. t contributions are $45 per 1000 board feet of lumber and $60 per 1000 square feet of plywood. a) Express the problem as a linear programming model. b) Graph the problem and indicate the optimal solution on the graph. Exercise 56 (Wagner) An electronics

307. rm manufactures radio models A, B and C, which have pro

308. t contributions of 16, 30 and 50 respectively. Minimum production requirements are 20, 120 and 60 for the three models, respectively. A dozen units of Model A requires 3 hours for manufacturing of components, 4 for assembling, and 1 for packaging. The corresponding

309. gures for a dozen units of Model B are 3.5, 5, and 1.5, and for Model C are 5, 8, and 3. During the forthcoming week the company has available 120 plant-hours for manufacturing components, 160 for assembly, and 48 for packaging. Formulate this production planning problem as a linear programming problem. Exercise 57 (Wagner) Process 1 in an oil re

310. nery takes a unit feed of 1 bbl (barrel) of crude oil A and 3 bbl of crude B to make 50 gallons (gal) of gasoline X and 20 gal of gasoline Y. From a unit feed of 4 bbl crude A and 2 bbl crude B, process 2 makes 30 gal of X and 80 gal of Y. Let xi be the number of units of feed type i; e.g., x1 = 1 indicates that process 1 uses 1 bbl crude A and 3 bbl crude B. There are 120 bbl of crude A available and 180 bbl of crude B. Sales commitments call for at least 2800 gal of gasoline X and 2200 gal of Y. The unit pro

311. ts of process 1 and 2 are p1 and p2, respectively. Formulate an LP model.

312. 76 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING Exercise 58 (Wagner) An air cargo

313. rm has 8 aircraft of type 1, 15 of type 2 and 11 of type 3 available for today's ights. A type 1 craft can carry 45 tons, type 2, 7 tons and type 3, 5 tons. 20 tons of cargo are to be own to city A and 28 tons to city B. Each plane makes at most one ight a day. The costs of ying a plane from the terminal to each city are as follows. Type 1 Type 2 Type 3 City A 23 15 1.4 City B 58 20 3.8 Let xi be the number of type i planes sent to A and yi the number to B. Formulate an LP for this routing problem. Exercise 59 (Wagner) A manufacturing

314. rm produces widgets and distributes them to

315. ve wholesalers at a

316. xed delivered price of $2.50 per unit. Sales forecasts indicate that monthly deliveries will be 2700, 2700, 9000, 4500 and 3600 widgets to wholesalers 1-5 respectively. The monthly production capacities are 4500, 9000 and 11,250 at plants 1, 2 and 3, respectively. The direct costs of producing each widget are $2 at plant 1, $1 at plant 2 and $1.80 at plant 3. The transport cost of shipping a widget from a plant to a wholesaler is given below. Wholesaler 1 2 3 4 5 Plant 1 .05 .07 .11 .15 .16 Plant 2 .08 .06 .10 .12 .15 Plant 3 .10 .09 .09 .10 .16 Formulate an LP model for this production and distribution problem. Exercise 60 (Wagner) Ft. Loudoun and Watts Bar are two large hydroelectric dams, the former upstream of the latter. The level of Watts Bar Lake must be kept within limits for recreational purposes, and the problem is to plan releases from Ft. Loudoun to do so. In reality this problem is solved simultaneously for numerous dams covering an entire watershed, but we focus on a single reservoir. There are also sophisticated models for predicting runo into the reservoirs, but we will suppose that runo is negligible. Thus any water entering Watts Bar Lake must be released through Ft. Loudoun Dam. The planning period is 20 months. During month t, let xt be the average water level of Watts Bar Lake before augmentation by water from Ft. Loudoun; x1 = 25. Let yt be the number of feet added to the average level in month t from Ft. Loudoun. Lt and Ut are the lower and upper bounds on the lake level in month t (more restrictive in summer). To model seepage, evaporation and hydroelectric release through Watts Bar Dam we suppose that Watts Bar Lake begins month t+1 at a level equal to .75 times the average level of the previous month (including the augmentation from Ft. Loudoun). The cost of water from Ft. Loudoun Lake is ct for every foot added to the level of Watts Bar. Formulate the appropriate LP model. (In reality the model is a huge nonlinear program.) Exercise 61 Red Dwarf Toasters needs to produce 1000 of their new Talking Toaster. There are three ways this toaster can be produced: manually, semi-automatically, and robotically. Man- ual assembly requires 1 minute of skilled labor, 40 minutes of unskilled labor, and 3 minutes of assembly room time. The corresponding values for semiautomatic assembly are 4, 30, and 2; while those for robotic assembly are 8, 20, and 4. There are 4500 minutes of skilled labor, 36,000 minutes of unskilled labor, and 2700 minutes of assembly room time available for this product. The

317. 5.4. LINEAR PROGRAMMING MODELS 77 total cost for producing manually is $7/toaster; semiautomatically is $8/toaster; and robotically is $8.50/toaster. (a) Formulate the problem of producing 1000 toasters at minimum cost meeting the resource requirements. Clearly de

318. ne your variables, objective and constraints. (b) Our union contract states that the amount of skilled labor time used is at least 10% of the total labor (unskilled plus skilled) time used. Update your formulation in (a) to handle this requirement. (c) Any unused assembly oor time can be rented out at a pro

319. t of $0.50/minute. Update your formulation to include this possibility. Answers to Exercise 61: (a) Let x1 be the number of toasters produced manually, x2 be the number produced semiautomatically, and x3 be the number produced robotically. The objective is to Minimize 7x1 + 8x2 + 8:5x3. The constraints are: x1 + x2 + x3 = 1000 (produce enough toasters) x1 + 4x2 + 8x3 4500 (skilled labor used less than or equal to amount available). 40x1 + 30x2 + 20x3 36000 (unskilled labor constraint) 3x1 + 2x2 + 4x3 2700 (assembly time constraint) x1 0; x2 0; x3 0 (nonnegativity of production) (b) Add a constraint x1 + 4x2 + 8x3 :1(41x1 + 34x2 + 28x3) (c) Add a variable sa to represent the assembly time slack. Add +0:5sa to the objective. Change the assembly time constraint to 3x1 + 2x2 + 4x3 + sa = 2700 (assembly time constraint) sa 0

320. 78 CHAPTER 5. MODELING WITH LINEAR PROGRAMMING

321. Chapter 6 Case Study The Global Oil Company You may complete the project individually or in groups. In the latter case, which is encouraged, members of the group submit a joint report and all receive the same grade. The ideal size of a group is three. Groups of up to four are allowed. There should be no collaboration among groups and/or students working individually. The Global Oil Company is an international producer, re

322. ner, transporter and distributor of oil, gasoline and petrochemicals. Global Oil is a holding company with subsidiary operating companies that are wholly or partially owned. A major problem for Global Oil is to coordinate the actions of these various subsidiaries into an overall corporate plan, while at the same time maintaining a reasonable amount of operating autonomy for the subsidiary companies. To deal with this dilemma, the logistics department at Global Oil Headquarters develops an annual corporate-wide plan which details the pattern of shipments among the various subsidiaries. The plan is not rigid but provides general guidelines and the plan is revised periodically to re ect changing conditions. Within the framework of this plan, the operating companies can make their own decisions and plans. This corporate-wide plan is presently done on a trial and error basis. There are two problems with this approach. First, the management of the subsidiaries often complains that the plan does not re ect properly the operating conditions under which the subsidiary operates. The plan sometimes calls for operations or distribution plans that are impossible to accomplish. And secondly, corporate management is concerned that the plan does not optimize for the total company. The technique of linear programming seems a possible approach to aid in the annual planning process, that will be able to answer at least in part, the two objections above. In addition the building of such a model will make it possible to make changes in plans quickly when the need arises. Before embarking on the development of a world-wide model, Global Oil asks you to build a model of the Far Eastern operations for the coming year. Far Eastern Operations The details of the 1998 planning model for the Far Eastern Operations are now described. There are two sources of crude oil, Saudi Arabia and Borneo. The Saudi crude is relatively heavier (24 API), and the Far Eastern sector could obtain as much as 60,000 barrels per day at a 79

323. 80 CHAPTER 6. CASE STUDY cost of $18.50 per barrel during 1998. A second source of crude is from the Brunei

324. elds in Borneo. This is a light crude oil (36 API). Under the terms of an agreement with the Netherlands Petroleum Company in Borneo, a

325. xed quantity of 40,000 b/d of Brunei crude, at a cost of $19.90 per barrel is to be supplied during 1998. There are two subsidiaries that have re

326. ning operations. The

327. rst is in Australia, operating a re

328. nery in Sydney with a capacity of 50,000 b/d throughput. The company markets its products throughout Australia, as well as having a surplus of re

329. ned products available for shipment to other subsidiaries. The second subsidiary is in Japan, which operates a 30,000 b/d capacity re

330. nery. Marketing operations are conducted in Japan, and excess production is available for shipment to other Far Eastern subsidiaries. In addition, there are two marketing subsidiaries without re

331. ning capacity of their own. One of these is in New Zealand and the other is in the Philippines. Their needs can be supplied by shipments from Australia, Japan, or the Global Oil subsidiary in the United States. The latter is not a regular part of the Far Eastern Operations, but may be used as a source of re

332. ned products. Finally, the company has a eet of tankers that move the crude oil and re

333. ned products among the subsidiaries. Re

334. nery Operations The operation of a re

335. nery is a complex process. The characteristics of the crudes available, the desired output, the speci

336. c technology of the re

337. nery, etc., make it dicult to use a simple model to describe the process. In fact, management at both Australia and Japan have complex linear programming models involving approximately 300 variables and 100 constraints for making detailed decisions on a daily or weekly basis. For annual planning purposes the re

338. nery model is greatly simpli

339. ed. The two crudes (Saudi and Brunei) are input. Two general products are output { (a) gasoline products and (b) other products such as distillate, fuel oil, etc. In addition, although the re

340. nery has processing exibility that permits a wide range of yields, for planning purposes it was decided to include only the values at highest and lowest conversion rates (process intensity). Each re

341. nery could use any combination of the two extreme intensities. These yields are shown in Table 6.1. The incremental costs of operating the re

342. nery depend somewhat upon the type of crude and process intensity. These costs are shown in Table 6.1. Also shown are the incremental transportation costs from either Borneo or Saudi Arabia. Marketing Operations Marketing is conducted in two home areas (Australia and Japan) as well as in the Philippines and New Zealand. Demand for gasoline and distillate in all areas has been estimated for 1998. Area 1998 Demand (thous. of b/d) Gasoline Distillate Australia 9.0 21.0 Japan 3.0 12.0 Philippines 5.0 8.0 New Zealand 5.4 8.7 TOTAL 22.4 49.7

343. 81 Table 6.1: Re

344. nery Operations Australia Japan Capacity (b/d of input) 50,000 30,000 Saudi Crude Transportation Cost ($/b) 0.65 0.70 High Process Intensity ($/b) 1.19 1.26 Yield of Gasoline 0.31 0.30 Yield of Distillate 0.61 0.62 Low Process Intensity ($/b) 0.89 0.88 Yield of Gasoline 0.19 0.18 Yield of Distillate 0.73 0.74 Brunei Crude Transportation Cost ($/b) 0.15 0.25 High Process Intensity ($/b) 0.93 0.91 Yield of Gasoline 0.36 0.35 Yield of Distillate 0.58 0.59 Low Process Intensity ($/b) 0.61 0.55 Yield of Gasoline 0.26 0.25 Yield of Distillate 0.69 0.70

345. 82 CHAPTER 6. CASE STUDY Variable costs of supplying gasoline or distillate to New Zealand and the Philippines are: Variable costs of shipment of gasoline/distillate in $/b From: To: New Zealand Philippines Australia .20 .30 Japan .25 .40 Tanker Operations Tankers are used to bring crude from Saudi Arabia and Borneo to Australia and Japan and to transport re

346. ned products from Australia and Japan to the Philippines and New Zealand. The variable costs of these operations are included above. However, there is a limited capacity of tankers available. The eet has a capacity of 6.5 equivalent (standard sized) tankers. The amount of capacity needed to deliver one barrel from one destination to another depends upon the distance traveled, port time, and other factors. The table below lists the fraction of one standard sized tanker needed to deliver 1,000 b/d over the indicated routes. Tanker Usage Factors (Fraction of Standard Sized Tanker Needed to Deliver 1,000 b/d) Between and Australia Japan Saudi Arabia .12 .11 Borneo .05 .05 Philippines .02 .01 New Zealand .01 .06 It is also possible to charter independent tankers. The rate for this is $5,400 per day for a standard sized tanker. United States Supply United States operations on the West Coast expect a surplus of 12,000 b/d of distillate during 1998. The cost of distillate at the loading port of Los Angeles is $20.70 per barrel. There is no excess gasoline capacity. The estimated variable shipping costs and tanker requirements of distillate shipments from the United States are: Variable costs Tanker of shipments requirements New Zealand 1.40 .18 Philippines 1.10 .15

347. 83 Questions Part I: Formulate a linear program which can be used to generate a comprehensive plan for the whole Far Eastern operations. Clearly de

348. ne every variable used in your formulation. Each group should be prepared to make a presentation of their linear programming model in class, on September 22. No written report is due on this date. Part II: Written report due on October 6. Start your report with an executive summary (at most two pages) containing the most important results and recommendations. Attach supporting material in Appendices. 1. Solve your linear program using SOLVER. How many barrels of crude should Global Oil purchase from Saudi Arabia for its Far Eastern operations? How much crude should be re

349. ned in Australia? How much in Japan? Provide tables showing the quantities of gasoline and distillate shipped from each of the two re

350. neries and from the U.S. to each of the four market areas. 2. Use sensitivity analysis to answer the following questions. { What is the marginal value of increasing supply from Brunei

351. elds? Can this marginal value be used to estimate the total savings for Far Eastern operations if 41,000 b/d are supplied from Brunei

352. elds instead of 40,000? Explain. { What is the marginal value of increasing the tanker eet? Can this marginal value be used if we want to increase the size to 7 tankers (from the current size of 6.5)? Explain. { What is the additional cost to Far Eastern operations if demand for gasoline in the Philippines increases to 5,200 b/d? What is the minimum price of gasoline in the Philippines that would make it pro

353. table for Global Oil to consider such an increase in distribution? { By how much should the production costs be reduced at the re

354. nery in Japan when operating at high process intensity in order to make it cost eective to use Saudi crude? { Global Oil is planning a three day shutdown of its Australian re

355. nery for maintenance purposes, in the coming year. It has storage facilities and at least two weeks of inventories, so a re

356. nery shutdown for a few days will not disrupt distribution. What would be the cost of a planned shutdown of the re

357. nery in Australia for three days per year? Same question for the re

358. nery in Japan. { Currently, it is not economical to ship US distillate to the Philippines. What is the cost of US distillate at which Global Oil should consider starting such shipments? 3. Several opportunities present themselves to the Global Oil company (see the attached memos). Consider combinations of these options and prepare a recommendation. Doc- ument your report.

359. 84 CHAPTER 6. CASE STUDY Memo to: Global Oil Headquarters From: Australian Aliate Re: Supplements to Annual Plan Since submitting data for annual planning purposes, two additional opportunities have arisen. We would like to include these in the plans. A. Bid on Gasoline Contract with Australian Government The government of Australia will submit to bid a contract for 1.5 thousand b/d of gasoline. We expect we could win this bid at a price of $26.40 per barrel. Estimated costs per barrel as follows: Variable Costs (crude, re

360. ning, transportation) $24.90 Allocated Overhead .80 Total $25.70 At these costs, the contract would have a pro

361. t contribution of $0.70 per barrel. We would like permission to bid on the contract. We hope that the linear programming wizards who are working for your logistics department will not contradict us! B. Expansion of Australian Re

362. nery For the past two years, the Australian re

363. nery has been operating at full capacity. We request authorization for capital expenditures to increase the re

364. nery capacity to 55 thousand b/d. There are several reasons for the need for this expansion: 1. Australia can supply the current requirements in New Zealand and the Philippines more cheaply than can Japan. 2. The proposed bid on Australian government gasoline contract (above). 3. We understand the New Zealand aliate is considering increasing its requirements by 4.5 thousand b/d. (See below.) The cost of this expansion is 4.0 million dollars. To recover this investment1, we need an annual savings of $702,000. 1This assumes a cost of capital rate of 20%. Depreciation tax eects are included. With these considerations, the $702,000 savings per year is equivalent to the $4 million investment

365. 85 Memo to: Global Oil Headquarters From: New Zealand Aliate Re: Supplement to Annual Plan Negotiations have been begun with the NOZO Oil Company in New Zealand. This company is a distributor, with sales of 1.6 thousand b/d of gasoline and 3.2 thousand b/d of distillate. If negotiations are successfully completed, these requirements would be added to current requirements for New Zealand, making total requirements of: Gasoline: 7.0 thousand b/d Distillate: 11.9 thousand b/d The anticipated revenue (after subtracting variable marketing costs) for this acquisition are $30.20 per barrel for gasoline and $24.60 per barrel for distillate. The purchase cost of NOZO oil is expected to be about $21.0 million. On an annual basis, this would require $3.5 million per year incremental pro

366. t to justify the purchase. Memo to: Global Oil Headquarters From: Tanker Aliate Re: Supplement to Annual Plan We have been made aware that expansions in requirements are being considered in New Zealand and Australia. We are currently operating the tanker eet at capacity. Additional requirements will increase the transport requirements both for crude and re

367. ned products. This will necessitate spot chartering unless additional tanker capacity is added. We can lease on a long time basis additional tankers at a rate of $4.8 thousand dollars per day per 1 unit tanker equivalent. We propose a lease of 0.5 equivalent tanker units giving us a total capacity of 7.0 equivalent units. The cost of this would be $2.4 thousand per day or $876,000 per year. If you prefer other arrangements, we are willing to discuss them. Memo to: Global Oil Headquarters From: Borneo Oce Re: Supplement to Annual Plan We have just been oered the opportunity to increase our contract with our supplier of Brunei crude. They are willing to supply us an additional 5 thousand b/d at a cost of $20.65 per barrel. Should we accept the oer? If not, is there a counter oer that we should propose?

368. 86 CHAPTER 6. CASE STUDY

369. Chapter 7 The Simplex Method In this chapter, you will learn how to solve linear programs. This will give you insights into what SOLVER and other commercial linear programming software packages actually do. Such an understanding can be useful in several ways. For example, you will be able to identify when a problem has alternate optimal solutions (SOLVER never tells you this: it always give you only one optimal solution). You will also learn about degeneracy in linear programming and how this could lead to a very large number of iterations when trying to solve the problem. 7.1 Linear Programs in Standard Form Before we start discussing the simplex method, we point out that every linear program can be converted into standard form Max c1x1 + c2x2 + : : :+ cnxn subject to a11x1 + a12x2 + : : :+ a1nxn = b1 : : : : : : : : : am1x1 + am2x2 + : : :+ amnxn = bm x1 0; : : :xn 0 where the objective is maximized, the constraints are equalities and the variables are all nonnegative. This is done as follows: If the problem is min z, convert it to max z. If a constraint is ai1x1+ai2x2+: : :+ainxn bi, convert it into an equality constraint by adding a nonnegative slack variable si. The resulting constraint is ai1x1+ai2x2+: : :+ainxn+si = bi, where si 0. If a constraint is ai1x1 + ai2x2 + : : : + ainxn bi, convert it into an equality constraint by subtracting a nonnegative surplus variable si. The resulting constraint is ai1x1+ai2x2+: : :+ ainxn si = bi, where si 0. If some variable xj is unrestricted in sign, replace it everywhere in the formulation by x0j x0j0, where x0j 0 and x0j0 0. 87

370. 88 CHAPTER 7. THE SIMPLEX METHOD Example 7.1.1 Transform the following linear program into standard form. Min 2x1 +3x2 x1 3x2 +2x3 3 x1 +2x2 2 x1 urs; x2 0; x3 0 Let us

371. rst turn the objective into a max and the constraints into equalities. Max 2x1 3x2 x1 3x2 +2x3 +s1 = 3 x1 +2x2 s2 = 2 x1 urs; x2 0; x3 0 s1 0 s2 0 The last step is to convert the unrestricted variable x1 into two nonnegative variables: x1 = x01 x010. Max 2x01 2x010 3x2 x01 x010 3x2 +2x3 +s1 = 3 x01 +x010 +2x2 s2 = 2 x01 0; x010 0; x2 0; x3 0 s1 0 s2 0 7.2 Solution of Linear Programs by the Simplex Method For simplicity, in this course we solve by hand only the case where the constraints are of the form and the right-hand-sides are nonnegative. We will explain the steps of the simplex method while we progress through an example. Example 7.2.1 Solve the linear program max x1 +x2 2x1 +x2 4 x1 +2x2 3 x1 0; x2 0 First, we convert the problem into standard form by adding slack variables x3 0 and x4 0. max x1 +x2 2x1 +x2 +x3 = 4 x1 +2x2 +x4 = 3 x1 0; x2 0 x3 0; x4 0 Let z denote the objective function value. Here, z = x1 + x2 or, equivalently, z x1 x2 = 0: Putting this equation together with the constraints, we get the following system of linear equations. z x1 x2 = 0 Row 0 2x1 +x2 +x3 = 4 Row 1 x1 +2x2 +x4 = 3 Row 2 (7.1)

372. 7.2. SOLUTION OF LINEAR PROGRAMS BY THE SIMPLEX METHOD 89 Our goal is to maximize z, while satisfying these equations and, in addition, x1 0, x2 0, x3 0, x4 0. Note that the equations are already in the form that we expect at the last step of the Gauss- Jordan procedure. Namely, the equations are solved in terms of the nonbasic variables x1, x2. The variables (other than the special variable z) which appear in only one equation are the basic variables. Here the basic variables are x3 and x4. A basic solution is obtained from the system of equations by setting the nonbasic variables to zero. Here this yields x1 = x2 = 0 x3 = 4 x4 = 3 z = 0: Is this an optimal solution or can we increase z? (Our goal) By looking at Row 0 above, we see that we can increase z by increasing x1 or x2. This is because these variables have a negative coecient in Row 0. If all coecients in Row 0 had been nonnegative, we could have concluded that the current basic solution is optimum, since there would be no way to increase z (remember that all variables xi must remain 0). We have just discovered the

373. rst rule of the simplex method. Rule 1 If all variables have a nonnegative coecient in Row 0, the current basic solution is optimal. Otherwise, pick a variable xj with a negative coecient in Row 0. The variable chosen by Rule 1 is called the entering variable. Here let us choose, say, x1 as our entering variable. It really does not matter which variable we choose as long as it has a negative coecient in Row 0. The idea is to pivot in order to make the nonbasic variable x1 become a basic variable. In the process, some basic variable will become nonbasic (the leaving variable). This change of basis is done using the Gauss-Jordan procedure. What is needed next is to choose the pivot element. It will be found using Rule 2 of the simplex method. In order to better understand the rationale behing this second rule, let us try both possible pivots and see why only one is acceptable. First, try the pivot element in Row 1. z x1 x2 = 0 Row 0 2x1 +x2 +x3 = 4 Row 1 x1 +2x2 +x4 = 3 Row 2 This yields z 1 2x2 +1 3x3 = 2 Row 0 x1 +1 x+1 x= 2 Row 1 22 23 3 x1 2 x3 +x4 = 1 Row 2 22with basic solution x2 = x3 = 0 x1 = 2 x4 = 1 z = 2: Now, try the pivot element in Row 2. z x1 x2 = 0 Row 0 2x1 +x2 +x3 = 4 Row 1 x1 +2x2 +x4 = 3 Row 2 This yields

374. 90 CHAPTER 7. THE SIMPLEX METHOD z +x2 +x4 = 3 Row 0 3x2 +x3 2x4 = 2 Row 1 x1 +2x2 +x4 = 3 Row 2 with basic solution x2 = x4 = 0 x1 = 3 x3 = 2 z = 3: Which pivot should we choose? The

375. rst one, of course, since the second yields an infeasible basic solution! Indeed, remember that we must keep all variables 0. With the second pivot, we get x3 = 2 which is infeasible. How could we have known this ahead of time, before actually performing the pivots? The answer is, by comparing the ratios Right Hand Side Entering Variable Coecient in Rows 1 and 2 of (7.1). Here we get 4 2 in Row 1 and 3 1 in Row 2. If you pivot in a row with minimum ratio, you will end up with a feasible basic solution (i.e. you will not introduce negative entries in the Right Hand Side), whereas if you pivot in a row with a ratio which is not minimum you will always end up with an infeasible basic solution. Just simple algebra! A negative pivot element would not be good either, for the same reason. We have just discovered the second rule of the simplex method. Rule 2 For each Row i, i 1, where there is a strictly positive entering variable coecient, compute the ratio of the Right Hand Side to the entering variable coecient. Choose the pivot row as being the one with MINIMUM ratio. Once you have idendi

376. ed the pivot element by Rule 2, you perform a Gauss-Jordan pivot. This gives you a new basic solution. Is it an optimal solution? This question is addressed by Rule 1, so we have closed the loop. The simplex method iterates between Rules 1, 2 and pivoting until Rule 1 guarantees that the current basic solution is optimal. That's all there is to the simplex method. After our

377. rst pivot, we obtained the following system of equations. z 1 x2 +1 x3 = 2 Row 0 23x1 +1 x+1 x= 2 Row 1 22 23 3 x1 2 x3 +x4 = 1 Row 2 22with basic solution x2 = x3 = 0 x1 = 2 x4 = 1 z = 2: Is this solution optimal? No, Rule 1 tells us to choose x2 as entering variable. Where should we pivot? Rule 2 tells us to pivot in Row 2, since the ratios are 2 1=2 = 4 for Row 1, and 1 3=2 = 2 3 for Row 2, and the minimum occurs in Row 2. So we pivot on 3 2x2 in the above system of equations. This yields z +1 3x3 +1 3x4 = 7 3 Row 0 x1 +2 3x3 1 3x4 = 5 3 Row 1 x2 1 3x3 +2 3x4 = 2 3 Row 2 with basic solution x3 = x4 = 0 x1 = 5 3 x2 = 2 3 z = 7 3: Now Rule 1 tells us that this basic solution is optimal, since there are no more negative entries in Row 0. All the above computations can be represented very compactly in tableau form.

378. 7.2. SOLUTION OF LINEAR PROGRAMS BY THE SIMPLEX METHOD 91 z x1 x2 x3 x4 RHS Basic solution 1 1 1 0 0 0 basic x3 = 4 x4 = 3 0 2 1 1 0 4 nonbasic x1 = x2 = 0 0 1 2 0 1 3 z = 0 1 0 1 1 0 2 basic x1 = 2 x4 = 1 2 2 0 1 1 2 1 2 0 2 nonbasic x2 = x3 = 0 0 0 3 2 1 2 1 1 z = 2 1 0 0 1 3 1 3 7 3 basic x1 = 5 3 x2 = 2 3 0 1 0 2 3 1 3 5 3 nonbasic x3 = x4 = 0 0 0 1 1 3 2 3 2 3 z = 7 3 Since the above example has only two variables, it is interesting to interpret the steps of the simplex method graphically. See Figure 7.1. The simplex method starts in the corner point (x1 = 0; x2 = 0) with z = 0. Then it discovers that z can increase by increasing, say, x1. Since we keep x2 = 0, this means we move along the x1 axis. How far can we go? Only until we hit a constraint: if we went any further, the solution would become infeasible. That's exactly what Rule 2 of the simplex method does: the minimum ratio rule identi

379. es the

380. rst constraint that will be encountered. And when the constraint is reached, its slack x3 becomes zero. So, after the

381. rst pivot, we are at the point (x1 = 2; x2 = 0). Rule 1 discovers that z can be increased by increasing x2 while keeping x3 = 0. This means that we move along the boundary of the feasible region 2x1 + x2 = 4 until we reach another constraint! After pivoting, we reach the optimal point (x1 = 5 3; x2 = 2 3 ). z=7/3 x 2 x z=0 z=2 1 Figure 7.1: Graphical Interpretation

382. 92 CHAPTER 7. THE SIMPLEX METHOD Exercise 62 Solve the following linear program by the simplex method. max 4x1 +x2 x3 x1 +3x3 6 3x1 +x2 +3x3 9 x1 0; x2 0 x3 0 7.3 Alternate Optimal Solutions, Degeneracy, Unboudedness, In- feasibility Alternate Optimal Solutions Let us solve a small variation of the earlier example, with the same constraints but a slightly dierent objective: max x1 +1 2x2 2x1 +x2 4 x1 +2x2 3 x1 0; x2 0 As before, we add slacks x3 and x4, and we solve by the simplex method, using tableau repre- sentation. z x1 x2 x3 x4 RHS Basic solution 1 1 1 2 0 0 0 basic x3 = 4 x4 = 3 0 2 1 1 0 4 nonbasic x1 = x2 = 0 0 1 2 0 1 3 z = 0 1 0 0 1 2 0 2 basic x1 = 2 x4 = 1 0 1 1 2 1 2 0 2 nonbasic x2 = x3 = 0 0 0 3 2 1 2 1 1 z = 2 Now Rule 1 shows that this is an optimal solution. Interestingly, the coecient of the nonbasic variable x2 in Row 0 happens to be equal to 0. Going back to the rationale that allowed us to derive Rule 1, we observe that, if we increase x2 (from its current value of 0), this will not eect the value of z. Increasing x2 produces changes in the other variables, of course, through the equations in Rows 1 and 2. In fact, we can use Rule 2 and pivot to get a dierent basic solution with the same objective value z = 2. z x1 x2 x3 x4 RHS Basic solution 1 0 0 1 2 0 2 basic x1 = 5 3 x2 = 2 3 0 1 0 2 3 1 3 5 3 nonbasic x3 = x4 = 0 0 0 1 1 3 2 3 2 3 z = 2 Note that the coecient of the nonbasic variable x4 in Row 0 is equal to 0. Using x4 as entering variable and pivoting, we would recover the previous solution!

383. 7.3. ALTERNATE OPTIMAL SOLUTIONS, DEGENERACY, UNBOUDEDNESS, INFEASIBILITY93 Degeneracy Example 7.3.1 max 2x1 +x2 3x1 +x2 6 x1 x2 2 x2 3 x1 0; x2 0 Let us solve this problem using the {by now familiar{ simplex method. In the initial tableau, we can choose x1 as the entering variable (Rule 1) and Row 2 as the pivot row (the minimum ratio in Rule 2 is a tie, and ties are broken arbitrarily). We pivot and this yields the second tableau below. z x1 x2 x3 x4 x5 RHS Basic solution 1 2 1 0 0 0 0 basic x3 = 6 x4 = 2 x5 = 3 0 3 1 1 0 0 6 nonbasic x1 = x2 = 0 0 1 1 0 1 0 2 z = 0 0 0 1 0 0 1 3 1 0 3 0 2 0 4 basic x1 = 2 x3 = 0 x5 = 3 0 0 4 1 3 0 0 nonbasic x2 = x4 = 0 0 1 1 0 1 0 2 z = 4 0 0 1 0 0 1 3 Note that this basic solution has a basic variable (namely x3) which is equal to zero. When this occurs, we say that the basic solution is degenerate. Should this be of concern? Let us continue the steps of the simplex method. Rule 1 indicates that x2 is the entering variable. Now let us apply Rule 2. The ratios to consider are 0 4 in Row 1 and 3 1 in Row 3. The minimum ratio occurs in Row 1, so let us perform the corresponding pivot. z x1 x2 x3 x4 x5 RHS Basic solution 1 0 0 3 4 1 4 0 4 basic x1 = 2 x2 = 0 x5 = 3 0 0 1 1 4 3 4 0 0 nonbasic x3 = x4 = 0 0 1 0 1 4 1 4 0 2 z = 4 0 0 0 1 4 3 4 1 3 We get exactly the same solution! The only dierence is that we have interchanged the names of a nonbasic variable with that of a degenerate basic variable (x2 and x3). Rule 1 tells us the solution is not optimal, so let us continue the steps of the simplex method. Variable x4 is the entering variable and the last row wins the minimum ratio test. After pivoting, we get the tableau: z x1 x2 x3 x4 x5 RHS Basic solution 1 0 0 2 3 0 1 3 5 basic x1 = 1 x2 = 3 x4 = 4 0 0 1 0 0 1 3 nonbasic x3 = x5 = 0 0 1 0 1 3 0 1 3 1 z = 5 0 0 0 1 3 1 4 3 4 By Rule 1, this is the optimal solution. So, after all, degeneracy did not prevent the simplex method to

384. nd the optimal solution in this example. It just slowed things down a little. Unfor- tunately, on other examples, degeneracy may lead to cycling, i.e. a sequence of pivots that goes

385. 94 CHAPTER 7. THE SIMPLEX METHOD through the same tableaus and repeats itself inde

386. nitely. In theory, cycling can be avoided by choosing the entering variable with smallest index in Rule 1, among all those with a negative coef-

387. cient in Row 0, and by breaking ties in the minimum ratio test by choosing the leaving variable with smallest index (this is known as Bland's rule). This rule, although it guaranties that cycling will never occur, turns out to be somewhat inecient. Actually, in commercial codes, no eort is made to avoid cycling. This may come as a surprise, since degeneracy is a frequent occurence. But there are two reasons for this: Although degeneracy is frequent, cycling is extremely rare. The precision of computer arithmetic takes care of cycling by itself: round o errors accumu- late and eventually gets the method out of cycling. Our example of degeneracy is a 2-variable problem, so you might want to draw the constraint set in the plane and interpret degeneracy graphically. Unbounded Optimum Example 7.3.2 max 2x1 +x2 x1 +x2 1 x1 2x2 2 x1 0; x2 0 Solving by the simplex method, we get: z x1 x2 x3 x4 RHS Basic solution 1 2 1 0 0 0 basic x3 = 1 x4 = 2 0 1 1 1 0 1 nonbasic x1 = x2 = 0 0 1 2 0 1 2 z = 0 1 0 5 0 2 4 basic x1 = 2 x3 = 3 0 0 1 1 1 3 nonbasic x2 = x4 = 0 0 1 2 0 1 2 z = 4 At this stage, Rule 1 chooses x2 as the entering variable, but there is no ratio to compute, since there is no positive entry in the column of x2. As we start increasing x2, the value of z increases (from Row 0) and the values of the basic variables increase as well (from Rows 1 and 2). There is nothing to stop them going o to in

388. nity. So the problem is unbounded. Interpret the steps of the simplex method graphically for this example. Infeasible Linear Programs It is easy to construct constraints that have no solution. The simplex method is able to identify such cases. We do not discuss it here. The interested reader is referred to Winston, for example. Properties of Linear Programs There are three possible outcomes for a linear program: it is infeasible, it has an unbounded optimum or it has an optimal solution.

389. 7.3. ALTERNATE OPTIMAL SOLUTIONS, DEGENERACY, UNBOUDEDNESS, INFEASIBILITY95 If there is an optimal solution, there is a basic optimal solution. Remember that the number of basic variables in a basic solution is equal to the number of constraints of the problem, say m. So, even if the total number of variables, say n, is greater than m, at most m of these variables can have a positive value in an optimal basic solution. Exercise 63 The following tableaus were obtained in the course of solving linear programs with 2 nonnegative variables x1 and x2 and 2 inequality constraints (the objective function z is maximized). Slack variables s1 and s2 were added. In each case, indicate whether the linear program (i) is unbounded (ii) has a unique optimum solution (iii) has an alternate optimum solution (iv) is degenerate (in this case, indicate whether any of the above holds). (a) z x1 x2 s1 s2 RHS 1 0 3 2 0 20 0 1 2 1 0 4 0 0 1 0 1 2 (b) z x1 x2 s1 s2 RHS 1 0 1 0 2 20 0 0 0 1 2 5 0 1 2 0 3 6 (c) z x1 x2 s1 s2 RHS 1 2 0 0 1 8 0 3 1 0 2 4 0 2 0 1 1 0 (d) z x1 x2 s1 s2 RHS 1 0 0 2 0 5 0 0 1 1 1 4 0 1 1 1 0 4 Exercise 64 Suppose the following tableau was obtained in the course of solving a linear program with nonnegative variables x1; x2; x3 and two inequalities. The objective function is maximized and slack variables s1 and s2 were added. z x1 x2 x3 s1 s2 RHS 1 0 a b 0 4 82 0 0 2 2 1 3 c 0 1 1 3 0 5 3 Give conditions on a, b and c that are required for the following statements to be true: (i) The current basic solution is a feasible basic solution. Assume that the condition found in (i) holds in the rest of the exercise. (ii) The current basic solution is optimal.

390. 96 CHAPTER 7. THE SIMPLEX METHOD (iii) The linear program is unbounded (for this question, assume that b 0). (iv) The currrent basic solution is optimal and there are alternate optimal solutions (for this question, assume a 0). Exercise 65 A plant can manufacture

391. ve products P1, P2, P3, P4 and P5. The plant consists of two work areas: the job shop area A1 and the assembly area A2. The time required to process one unit of product Pj in work area Ai is pij (in hours), for i = 1; 2 and j = 1; : : :; 5. The weekly capacity of work area Ai is Ci (in hours). The company can sell all it produces of product Pj at a pro

392. t of sj , for i = 1; : : :; 5. The plant manager thought of writing a linear program to maximize pro

393. ts, but never actually did for the following reason: From past experience, he observed that the plant operates best when at most two products are manufactured at a time. He believes that if he uses linear programming, the optimal solution will consist of producing all

394. ve products. Do you agree with him? Explain, based on your knowledge of linear programming. Exercise 66 Consider the linear program Maximize 5x1 + 3x2 + x3 Subject to x1 + x2 + x3 6 5x1 + 3x2 + 6x3 15 x1; x2; x3 0 and an associated tableau z x1 x2 x3 s1 s2 RHS 1 0 0 5 0 1 15 0 0 .4 -0.2 1 -.2 3 0 1 .6 1.2 0 .2 3 (a) What basic solution does this tableau represent? Is this solution optimal? Why or why not? (b) Does this tableau represent a unique optimum. If not,

395. nd an alternative optimal solution. Answer: (a) The solution is x1 = 3, x2; x3 = 0, objective 15. It is optimal since cost row is all at least 0. (b) It is not unique (since x2 has reduced cost 0 but is not basic). Alternative found by pivoting in x2 (question could have asked for details on such a pivot) for solution x2 = 5, x1; x3 = 0, objective 15.

396. Chapter 8 Sensitivity Analysis for Linear Programming Finding the optimal solution to a linear programming model is important, but it is not the only information available. There is a tremendous amount of sensitivity information, or information about what happens when data values are changed. Recall that in order to formulate a problem as a linear program, we had to invoke a certainty assumption: we had to know what value the data took on, and we made decisions based on that data. Often this assumption is somewhat dubious: the data might be unknown, or guessed at, or otherwise inaccurate. How can we determine the eect on the optimal decisions if the values change? Clearly some numbers in the data are more important than others. Can we

397. nd the important numbers? Can we determine the eect of misestimation? Linear programming oers extensive capability for addressing these questions. We begin by showing how data changes show up in the optimal table. We then give two examples of how to interpret Solver's extensive output. 8.1 Tableau Sensitivity Analysis Suppose we solve a linear program by hand ending up with an optimal table (or tableau to use the technical term). We know what an optimal tableau looks like: it has all non{negative values in Row 0 (which we will often refer to as the cost row), all non{negative right{hand{side values, and a basis (identity matrix) embedded. To determine the eect of a change in the data, we will try to determine how that change eected the

398. nal tableau, and try to reform the

399. nal tableau accordingly. 8.1.1 Cost Changes The

400. rst change we will consider is changing a cost value by in the original problem. We are given the original problem and an optimal tableau. If we had done exactly the same calculations beginning with the modi

401. ed problem, we would have had the same

402. nal tableau except that the corresponding cost entry would be lower (this is because we never do anything except add or subtract scalar multiples of Rows 1 through m to other rows; we never add or subtract Row 0 to other rows). For example, take the problem 97

403. 98 CHAPTER 8. SENSITIVITY ANALYSIS FOR LINEAR PROGRAMMING Max 3x+2y Subject to x+y = 4 2x+y = 6 x,y = 0 The optimal tableau to this problem (after adding s1 and s2 as slacks to place in standard form) is: z x y s1 s2 RHS 1 0 0 1 1 10 0 0 1 2 1 2 0 1 0 1 1 2 Suppose the cost for x is changed to 3+ in the original formulation, from its previous value 3. After doing the same operations as before, that is the same pivots, we would end up with the tableau: z x y s1 s2 RHS 1 0 1 1 10 0 0 1 2 1 2 0 1 0 1 1 2 Now this is not the optimal tableau: it does not have a correct basis (look at the column of x). But we can make it correct in form while keeping the same basic variables by adding times the last row to the cost row. This gives the tableau: z x y s1 s2 RHS 1 0 0 1 1+ 10 + 2 0 0 1 2 1 2 0 1 0 1 1 2 Note that this tableau has the same basic variables and the same variable values (except for z) that our previous solution had. Does this represent an optimal solution? It does only if the cost row is all non{negative. This is true only if 1 0 1 + 0 which holds for 1 1. For any in that range, our previous basis (and variable values) is optimal. The objective changes to 10 + 2. In the previous example, we changed the cost of a basic variable. Let's go through another example. This example will show what happens when the cost of a nonbasic variable changes. Max 3x+2y + 2.5w Subject to x+y +2w = 4 2x+y +2w = 6 x,y,w = 0

404. 8.1. TABLEAU SENSITIVITY ANALYSIS 99 Here, the optimal tableau is : z x y w s1 s2 RHS 1 0 0 1:5 1 1 10 0 0 1 2 2 1 2 0 1 0 0 1 1 2 Now suppose we change the cost on w from 2.5 to 2:5+ in the formulation. Doing the same calculations as before will result in the tableau: z x y w s1 s2 RHS 1 0 0 1:5 1 1 10 0 0 1 2 2 1 2 0 1 0 0 1 1 2 In this case, we already have a valid tableau. This will represent an optimal solution if 1:5 0, so 1:5. As long as the objective coecient of w is no more than 2.5+1.5=4 in the original formulation, our solution of x = 2; y = 2 will remain optimal. The value in the cost row in the simplex tableau is called the reduced cost. It is zero for a basic variable and, in an optimal tableau, it is non{negative for all other variables (for a maximization problem). Summary: Changing objective function values in the original formulation will result in a changed cost row in the

405. nal tableau. It might be necessary to add a multiple of a row to the cost row to keep the form of the basis. The resulting analysis depends only on keeping the cost row non{negative. 8.1.2 Right Hand Side Changes For these types of changes, we concentrate on maximization problems with all constraints. Other cases are handled similarly. Take the following problem: Max 4x + 5y Subject to (8.1) 2x + 3y = 12 x + y = 5 x; y = 0 The optimal tableau, after adding slacks s1 and s2 is z x y s1 s2 RHS 1 0 0 1 2 22 0 0 1 1 2 2 0 1 0 1 3 3 Now suppose instead of 12 units in the

406. rst constraint, we only had 11. This is equivalent to forcing s1 to take on value 1. Writing the constraints in the optimal tableau long{hand, we get z + s1 + 2s2 = 22

407. 100 CHAPTER 8. SENSITIVITY ANALYSIS FOR LINEAR PROGRAMMING y + s1 2s2 = 2 x s1 + 3s2 = 3 If we force s1 to 1 and keep s2 at zero (as a nonbasic variable should be), the new solution would be z = 21, y = 1, x = 4. Since all variables are nonnegative, this is the optimal solution. In general, changing the amount of the right{hand{side from 12 to 12+ in the

408. rst constraint changes the tableau to: z x y s1 s2 RHS 1 0 0 1 2 22 + 0 0 1 1 2 2+ 0 1 0 1 3 3 This represents an optimal tableau as long as the righthand side is all non{negative. In other words, we need between 2 and 3 in order for the basis not to change. For any in that range, the optimal objective will be 22 + . For example, with equals 2, the new objective is 24 with y = 4 and x = 1. Similarly, if we change the right{hand{side of the second constraint from 5 to 5 + in the original formulation, we get an objective of 22 + 2 in the

409. nal tableau, as long as 1 1. Perhaps the most important concept in sensitivity analysis is the shadow price i of a constraint: If the RHS of Constraint i changes by in the original formulation, the optimal objective value changes by i . The shadow price i can be found in the optimal tableau. It is the reduced cost of the slack variable si. So it is found in the cost row (Row 0) in the column corresponding the slack for Constraint i. In this case, 1 = 1 (found in Row 0 in the column of s1) and 2 = 2 (found in Row 0 in the column of s2). The value i is really the marginal value of the resource associated with Constraint i. For example, the optimal objective value (currently 22) would increase by 2 if we could increase the RHS of the second constraint by = 1. In other words, the marginal value of that resource is 2, i.e. we are willing to pay up to 2 to increase the right hand side of the second constraint by 1 unit. You may have noticed the similarity of interpretation between shadow prices in linear programming and Lagrange multipliers in constrained optimization. Is this just a coincidence? Of course not. This parallel should not be too surprising since, after all, linear programming is a special case of constrained optimization. To derive this equivalence (between shadow prices and optimal Lagrange multipiers), one could write the KKT conditions for the linear program...but we will skip this in this course! In summary, changing the right{hand{side of a constraint is identical to setting the corresponding slack variable to some value. This gives us the shadow price (which equals the reduced cost for the corresponding slack) and the ranges. 8.1.3 New Variable The shadow prices can be used to determine the eect of a new variable (like a new product in a production linear program). Suppose that, in formulation (8.1), a new variable w has coecient 4 in the

410. rst constraint and 3 in the second. What objective coecient must it have to be considered for adding to the basis? If we look at making w positive, then this is equivalent to decreasing the right hand side of the

411. rst constraint by 4w and the right hand side of the second constraint by 3w in the original formulation. We obtain the same eect by making s1 = 4w and s2 = 3w. The overall eect of this is to decrease the objective by 1(4w)+ 2(3w) = 1(4w)+2(3w) = 10w. The objective value must

412. 8.1. TABLEAU SENSITIVITY ANALYSIS 101 be sucient to oset this, so the objective coecient must be more than 10 (exactly 10 would lead to an alternative optimal solution with no change in objective). Example 8.1.1 A factory can produce four products denoted by P1, P2, P3 and P4. Each product must be processed in each of two workshops. The processing times (in hours per unit produced) are given in the following table. P1 P2 P3 P4 Workshop 1 3 4 8 6 Workshop 2 6 2 5 8 400 hours of labour are available in each workshop. The pro

413. t margins are 4, 6, 10, and 9 dollars per unit of P1, P2, P3 and P4 produced, respectively. Everything that is produced can be sold. Thus, maximizing pro

414. ts, the following linear program can be used. MAX 4 X1 + 6 X2 + 10 X3 + 9 X4 SUBJECT TO 3 X1 + 4 X2 + 8 X3 + 6 X4 = 400 Row 1 6 X1 + 2 X2 + 5 X3 + 8 X4 = 400 Row 2 X1, X2, X3, X4 = 0 Introducing slack variables s1 and s2 in Rows 1 and 2, respectively, and applying the simplex method, we get the

415. nal tableau: z x1 x2 x3 x4 s1 s2 RHS 1 0:5 0 2 0 1:5 0 600 0 0:75 1 2 1:5 0:25 0 100 0 4:5 0 1 5 0:5 1 200 (a) How many units of P1, P2, P3 and P4 should be produced in order to maximize pro

416. ts? (b) Assume that 20 units of P3 have been produced by mistake. What is the resulting decrease in pro

417. t? (c) In what range can the pro

418. t margin per unit of P1 vary without changing the optimal basis? (d) In what range can the pro

419. t margin per unit of P2 vary without changing the optimal basis? (e) What is the marginal value of increasing the production capacity of Workshop 1? (f) In what range can the capacity of Workshop 1 vary without changing the optimal basis? (g) Management is considering the production of a new product P5 that would require 2 hours in Workshop 1 and ten hours in Workshop 2. What is the minimum pro

420. t margin needed on this new product to make it worth producing? Answers: (a) From the

421. nal tableau, we read that x2 = 100 is basic and x1 = x3 = x4 = 0 are nonbasic. So 100 units of P2 should be produced and none of P1, P3 and P4. The resuting pro

422. t is $ 600 and that is the maximum possible, given the constraints.

423. 102 CHAPTER 8. SENSITIVITY ANALYSIS FOR LINEAR PROGRAMMING (b) The reduced cost for x3 is 2 (found in Row 0 of the

424. nal tableau). Thus, the eect on pro

425. t of producing x3 units of P3 is 2x3. If 20 units of P3 have been produced by mistake, then the pro

426. t will be 2 20 = $40 lower than the maximum stated in (a). (c) Let 4 + be the pro

427. t margin on P1. The reduced cost remains nonnegative in the

428. nal tableau if 0:5 0. That is 0:5. Therefore, as long as the pro

429. t margin on P1 is less than 4:5, the optimal basis remains unchanged. (d) Let 6 + be the pro

430. t margin on P2. Since x2 is basic, we need to restore a correct basis. This is done by adding times Row 1 to Row 0. This eects the reduced costs of the nonbasic variables, namely x1, x3, x4 and s1. All these reduced costs must be nonnegative. This implies: 0:5+ 0:75 0 2+ 2 0 0+ 1:5 0 1:5+ 0:25 0. Combining all these inequalities, we get 0. So, as long as the pro

431. t margin on P2 is 6 or greater, the optimal basis remains unchanged. (e) The marginal value of increasing capacity in Workshop 1 is 1 = 1:5. (f) Let 400 + be the capacity of Workshop 1. The resulting RHS in the

432. nal tableau will be: 100 + 0:25 in Row 1, and 200 0:5 in Row 2. The optimal basis remains unchanged as long as these two quantities are nonnegative. Namely, 400 400. So, the optimal basis remains unchanged as long as the capacity of Workshop 1 is in the range 0 to 800. (g) The eect on the optimum pro

433. t of producing x5 units of P5 would be 1(2x5)+2(10x5) = 1:5(2x5) + 0(10x5) = 3x5. If the pro

434. t margin on P5 is sucient to oset this, then P5 should be produced. That is, we should produce P5 if its pro

435. t margin is at least 3. Exercise 67 A paper mill converts pulpwood to low, medium and high grade newsprint. The pulpwood requirements for each newsprint, availability of each pulpwood, and selling price (per ton) are shown below: Low Medium High Available grade grade grade (tons) Virginia pine 2 2 1 180 White pine 1 2 3 120 Loblolly pine 1 1 2 160 Price $900 $1000 $1200 The associated linear program is MAX 900 X1 + 1000 X2 + 1200 X3 SUBJECT TO 2 X1 + 2 X2 + X3 + S1 = 180 X1 + 2 X2 + 3 X3 + S2 = 120 X1 + X2 + 2 X3 + S3 = 160 X1, X2, X3, S1, S2, S3 = 0 with the optimal tableau

436. 8.1. TABLEAU SENSITIVITY ANALYSIS 103 z x1 x2 x3 s1 s2 s3 RHS 1 0 200 0 300 300 0 90; 000 0 1 0:8 0 0:6 0:2 0 84 0 0 0:4 1 0:2 0:4 0 12 0 0 0:6 0 0:2 0:6 1 52 (a) In what range can the price of low grade paper vary without changing the optimal basis? (b) What is the new optimal solution if the price of low grade paper changes to $800? (c) At what price should medium grade paper be sold to make it pro

437. table to produce? (d) In what range can the availability of Virginia pine vary without changing the optimal basis? (e) If 10 additional tons of Virginia pine are obtained, by how much will the optimal pro

438. t increase? (f) If the pulpwood resources are increased as in (e), what is the new optimal solution (i.e. the optimal production levels of low, medium and high grade newsprint)? (g) What would the plant manager be willing to pay for an additional ton of Loblolly pine? Exercise 68 Snacks 'R Us is deciding what to produce in the upcoming month. Due to past purchases, they have 500 lb of walnuts, 1000 lb of peanuts, and 500 lb of chocolate. They currently sell three mixes: Trail Mix, which consists of 1 lb of each material; Nutty Crunch, which has two pounds of peanuts and 1 lb of walnut; and Choc-o-plenty, which has 2 pounds of chocolate and 1 lb of peanuts. They can sell an unlimited amount of these mixes, except they can sell no more than 100 units of Choc-o-plenty. The income on the three mixes is $2, $3, and $4, respectively. The problem of maximizing total income subject to these constraints is Maximize 2x1 + 3x2 + 4x3 Subject to x1 + x2 500 x1 + 2x2 + x3 1000 x1 + 2x3 500 x3 100 xi 0 for all i The optimal tableau for this problem is (after adding slacks for the four constraints): z x1 x2 x3 s1 s2 s3 s4 RHS 1 0 0 0 1 1 0 3 1800 0 1 0 0 2 1 0 1 100 0 0 1 0 1 1 0 1 400 0 0 0 0 2 1 1 3 200 0 0 0 1 0 0 0 1 100 Using the above tableau, determine the following: (a) What is the solution represented by this tableau (both production quantities and total income)? How do we know it is the optimal tableau? Production quantities:

439. 104 CHAPTER 8. SENSITIVITY ANALYSIS FOR LINEAR PROGRAMMING Total income: Optimal? Reason: (b) Suppose the income for Trail Mix (x1) is only an estimate. For what range of values would the basis given by the above tableau still be optimal? What would be the solution (both production quantities and total income), if the income was only $2.75? Range: Lower: Upper: Solution for $2.75: Total Income Production: (c) How much should Snacks 'R Us willing to spend to procure an extra pound of peanuts? How about a pound of walnuts? How about an extra pound of chocolate? Peanuts: Walnuts: Chocolate: (d) For what range of peanut{stock values is the basis given by the above tableau still optimal? What would be the solution (both production quantities and total objective) if there were only 900 lbs of peanuts available? Range: Lower: Upper: Solution for 900 lbs: Total Income Production: (e) A new product, Extra-Walnutty Trail Booster, consists of 1 lb of each of peanuts and chocolate, and 2 lb of walnuts. What income do we need to make on this product in order to consider producing it? Income: Reason: 8.2 Solver Output A large amount (but not all) of the previous analysis is available from the output of Solver. To access the information, simply ask for the sensitivity report after optimizing. Rather than simply give rules for reading the report, here are two reports, each with a set of questions that can be answered from the output. 8.2.1 Tucker Automobiles Tucker Inc. needs to produce 1000 Tucker automobiles. The company has four production plants. Due to diering workforces, technological advances, and so on, the plants dier in the cost of producing each car. They also use a dierent amount of labor and raw material at each. This is summarized in the following table: Plant Cost ('000) Labor Material 1 15 2 3 2 10 3 4 3 9 4 5 4 7 5 6 The labor contract signed requires at least 400 cars to be produced at plant 3; there are 3300 hours of labor and 4000 units of material that can be allocated to the four plants. This leads to the following formulation: MIN 15 X1 + 10 X2 + 9 X3 + 7 X4 SUBJECT TO 2 X1 + 3 X2 + 4 X3 + 5 X4 = 3300

440. 8.2. SOLVER OUTPUT 105 3 X1 + 4 X2 + 5 X3 + 6 X4 = 4000 X3 = 400 X1 + X2 + X3 + X4 = 1000 X1, X2, X3, X4 = 0 Attached is the Solver output for this problem. Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$11 X1 400 0 15 1E+30 3.5 $B$12 X2 200 0 10 2 1E+30 $B$13 X3 400 0 9 1E+30 4 $B$14 X4 0 7 7 1E+30 7 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$18 Labor 3000 0 3300 1E+30 300 $B$19 Material 4000 -5 4000 300 200 $B$20 Plant3Min 400 4 400 100 400 $B$21 Total 1000 30 1000 66.6667 100 Here are some questions to be answered: 1. What are the current production quantities? What is the current cost of production? 2. How much will it cost to produce one more vehicle? How much will we save by producing one less? 3. How would our solution change if it cost only $8,000 to produce at plant 2? For what ranges of costs is our solution (except for the objective value) valid for plant 2? 4. How much are we willing to pay for a labor hour? 5. How much is our union contract costing us? What would be the value of reducing the 400 car limit down to 200 cars? To 0 cars? What would be the cost of increasing it by 100 cars? by 200 cars? 6. How much is our raw material worth (to get one more unit)? How many units are we willing to buy at that price? What will happen if we want more?

441. 106 CHAPTER 8. SENSITIVITY ANALYSIS FOR LINEAR PROGRAMMING 7. A new plant is being designed that will use only one unit of workers and 4 units of raw material. What is the maximum cost it can have in order for us to consider using it? 8. By how much can the costs at plant 1 increase before we would not produce there? 8.2.2 Carla's Maps Carla Lee, a current MBA student, decides to spend her summer designing and marketing bicycling maps of Western Pennsylvania. She has designed 4 maps, corresponding to four quadrants around Pittsburgh. The maps dier in size, colors used, and complexity of the topographical relief (the maps are actually 3-dimensional, showing hills and valleys). She has retained a printer to produce the maps. Each map must be printed, cut, and folded. The time (in minutes) to do this for the four types of maps is: Print Cut Fold A 1 2 3 B 2 4 2 C 3 1 5 D 3 3 3 Avail 15000 20000 20000 The printer has a limited amount of time in his schedule, as noted in the table. The pro

442. t per map, based on the projected selling price minus printers cost and other variable cost, comes out to approximately $1 for A and B and $2 for C and D. In order to have a suciently nice display, at least 1000 of each type must be produced. This gives the formulation: MAX A + B + 2 C + 2 D SUBJECT TO A + 2 B + 3 C + 3 D = 15000 2 A + 4 B + C + 3 D = 20000 3 A + 2 B + 5 C + 3 D = 20000 A = 1000 B = 1000 C = 1000 D = 1000 Attached is the Solver output. Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$11 XA 1500 0 1 1 0.333333 $B$12 XB 1000 0 1 0.333333 1E+30 $B$13 XC 1000 0 2 0.333333 1E+30 $B$14 XD 2833.333 0 2 1 0.5

443. 8.2. SOLVER OUTPUT 107 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$17 Print 15000 0.5 15000 100 366.6667 $B$18 Cut 16500 0 20000 1E+30 3500 $B$19 Fold 20000 0.166667 20000 7000 1000 $B$20 MinA 1500 0 1000 500 1E+30 $B$21 MinB 1000 -0.333333 1000 1750 1000 $B$22 MinC 1000 -0.333333 1000 500 1000 $B$23 MinD 2833.333 0 1000 1833.333 1E+30 Here are some questions to answer: 1. What are the production quantities and projected pro

444. t? 2. How much is Carla willing to pay for extra printing time? cutting time? folding time? For each, how many extra hours are we willing to buy at that price? 3. Suppose we reduced the 1000 limit on one item to 900. Which map should be decreased, and how much more would Carla make? 4. A

445. fth map is being thought about. It would take 2 minutes to print, 2 minutes to cut, and 3 minutes to fold. What is the least amount of pro

446. t necessary in order to consider producing this map? What is the eect of requiring 1000 of these also? 5. The marketing analysis on D is still incomplete, though it is known that the pro

447. t of $2 per item is within $.25 of the correct value. It will cost $500 to complete the analysis. Should Carla continue with the analysis? Exercise 69 Recall the Red Dwarf toaster example (Exercise 61). From Answer report: Final Cost: 7833.33 Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$1 Manual 633.3333333 0 7 0.5 1E+30 $B$2 SemiAut 333.3333333 0 8 1E+30 0.25 $B$3 Robotic 33.33333333 0 8.5 0.5 2.5 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease

448. 108 CHAPTER 8. SENSITIVITY ANALYSIS FOR LINEAR PROGRAMMING $B$5 Total 1000 10.83333333 1000 170 100 $B$6 Skilled 2233.333333 0 4500 1E+30 2266.666667 $B$7 Unskill 36000 -0.083333333 36000 1000 6800 $B$8 Assembly 2700 -0.166666667 2700 500 100 (a) What is the optimal allocation of production? What is the average cost/toaster of production. (b) By how much can the cost of robots increase before we will change that production plan. (c) How much is Red Dwarf willing to pay for more assembly room time? How many units is Red Dwarf willing to purchase at that price? (d) How much will we save if we decide to produce only 950 toasters? (e) A new production process is available that uses only 2 minutes of skilled labor, 10 minutes of unskilled labor, and an undetermined amount of assembly oor time. Its production cost is determined to be $10. What is the maximum assembly oor time that the process can take before it is deemed too expensive to use? Answer: (a) 633.3 should be produced manually, 333.3 should be produced semiautomatically, and 33.3 produced robotically, for an average cost of $7.383/toaster. (b) It can increase by $0.50. (c) Value is $0.16/minute, willing to purchase 500 at that price. (d) Objective will go down by 50(10.833). (e) Cost of $10 versus marginal cost of $10.833, leave 0.83. Unskilled labor costs $0.0833/unit. Therefore, if the new process takes any time at all, it will be deemed too expensive. Exercise 70 Kennytrail Amusement park is trying to divide its new 50 acre park into one of three categories: rides, food, and shops. Each acre used for rides generates $150/hour pro

449. t; each acre used for food generates $200/hour pro

450. t. Shops generate $300/hour pro

451. t. There are a number of restrictions on how the space can be divided. 1. Only 10 acres of land is suitable for shops. 2. Zoning regulations require at least 1000 trees in the park. A food acre has 30 trees; a ride acre has 20 trees; while a shop acre has no trees. 3. No more than 200 people can work in the park. It takes 3 people to work an acre of rides, 6 to work an acre of food, and 5 to work an acre of shops. The resulting linear program and Solver output is attached: MAX 150 RIDE + 200 FOOD + 300 SHOP SUBJECT TO RIDE + FOOD + SHOP = 50 SHOP = 10 20 RIDE + 30 FOOD = 1000 3 RIDE + 6 FOOD + 5 SHOP = 200 -------------------------------------------------------------------

452. 8.2. SOLVER OUTPUT 109 Answer report: Target Cell (Max) Cell Name Original Value Final Value $B$5 Cost 0 9062.5 Adjustable Cells Cell Name Original Value Final Value $B$1 Ride 0 31.25 $B$2 Food 0 12.5 $B$3 Shop 0 6.25 Constraints Cell Name Cell Value Formula Status Slack $B$6 Land 50 $B$6=$D$6 Binding 0 $B$7 ShopLim 6.25 $B$7=$D$7 Not Binding 3.75 $B$8 Trees 1000 $B$8=$D$8 Binding 0 $B$9 Workers 200 $B$9=$D$9 Binding 0 ----------------------------------------------------------------------------- Sensitivity report: Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$1 Ride 31.25 0 150 83.33333333 76.6666667 $B$2 Food 12.5 0 200 115 125 $B$3 Shop 6.25 0 300 1E+30 116.666667 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$6 Land 50 143.75 50 10 16.66666667 $B$7 ShopLim 6.25 0 10 1E+30 3.75 $B$8 Trees 1000 -4.375 1000 166.66667 100 $B$9 Workers 200 31.25 200 30 50 For each of the following changes, either

453. nd the answer or state that the information is not available from the Solver output. In the latter case, state why not. (a) What is the optimal allocation of the space? What is the pro

454. t/hour of the park? Optimal Allocation: Pro

455. t/hour: (b) Suppose Food only made a pro

456. t of $180/hour. What would be the optimal allocation of the park, and what would be the pro

457. t/hour of the park? Optimal Allocation: Pro

458. t/hour:

459. 110 CHAPTER 8. SENSITIVITY ANALYSIS FOR LINEAR PROGRAMMING (c) City Council wants to increase our tree requirement to 1020. How much will that cost us (in $/hour). What if they increased the tree requirement to 1200? Increase to 1020: Increase to 1200: (d) A construction

460. rm is willing to convert 5 acres of land to make it suitable for shops. How much should Kennytrail be willing to pay for this conversion (in $/hour). Maximum payment: (e) Kennytrail is considering putting in a waterslide. Each acre of waterslide can have 2 trees and requires 4 workers. What pro

461. t/hour will the waterslide have to generate for them to consider adding it? Minimum Pro

462. t: Reason: (f) An adjacent parcel of land has become available. It is

463. ve acres in size. The owner wants to share in our pro

464. ts. How much $/hour is Kennytrail willing to pay? Maximum payment: Exercise 71 Attached is the sensitivity report for the Diet Problem (see Section 5.4.1 in Chap- ter 5). Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$14 oatmeal 4 -3.1875 3 3.1875 1E+30 $B$15 chicken 0 12.46875 24 1E+30 12.46875 $B$16 eggs 0 4 13 1E+30 4 $B$17 milk 4.5 0 9 2.6923 1.38095 $B$18 pie 2 -3.625 20 3.625 1E+30 $B$19 porkbeans 0 4.375 19 1E+30 4.375 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$23 Energy 2000 0.05625 2000 560 100 $B$24 Protein 60 0 55 5 1E+30 $B$25 Calcium 1334.5 0 800 534.5 1E+30 Answer each of the following questions independently of the others. 1. What does the optimal diet consist of? 2. If the cost of oatmeal doubled to 6 cents/serving, should it be removed from the diet? 3. If the cost of chicken went down to half its current price, should it be added to the diet? 4. At what price would eggs start entering the diet? 5. In what range can the price of milk vary (rounding to the nearest tenth of a cent) while the current diet still remaining optimal?

465. 8.2. SOLVER OUTPUT 111 6. During midterms, you need a daily diet with energy content increased from 2000 kcal to 2200 kcal. What is the resulting additional cost? 7. Your doctor recommends that you increase the calcium requirement in your diet from 800 mg to 1200 mg. What is the eect on total cost? 8. Potatoes cost 12 cents/serving and have energy content of 300 kcal per serving, but no protein nor calcium content. Should they be part of the diet? Exercise 72 Attached is the sensitivity report for the Workforce Planning Problem (see Sec- tion 5.4.2 in Chapter 5). Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$14 Shift1 4 0 1 0.5 1 $B$15 Shift2 7 0 1 0 0.333333 $B$16 Shift3 1 0 1 0.5 0 $B$17 Shift4 4 0 1 0.5 0 $B$18 Shift5 3 0 1 0 0.333333 $B$19 Shift6 3 0 1 0.5 1 $B$20 Shift7 0 0.333333 1 1E+30 0.333333 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$24 Monday 14 0.333333 14 1.5 6 $B$25 Tuesday 17 0 13 4 1E+30 $B$26 Wednesday 15 0.333333 15 6 3 $B$27 Thursday 16 0 16 3 4 $B$28 Friday 19 0.333333 19 4.5 3 $B$29 Saturday 18 0.333333 18 1.5 6 $B$30 Sunday 11 0 11 4 1 Answer each of the following questions independently of the others. 1. What is the current total number of workers needed to sta the restaurant? 2. Due to a special oer, demand on thurdays increases. As a result, 18 workers are needed instead of 16. What is the eect on the total number of workers needed to sta the restaurant? 3. Assume that demand on mondays decreases: 11 workers are needed instead of 14. What is the eect on the total number of workers needed to sta the restaurant? 4. Currently, 15workers are needed on wednesdays. In what range can this number vary without changing the optimal basis?

466. 112 CHAPTER 8. SENSITIVITY ANALYSIS FOR LINEAR PROGRAMMING 5. Currently, every worker in the restaurant is paid $1000 per month. So the objective function in the formulation can be viewed as total wage expenses (in thousand dollars). Workers have complained that Shift 4 is the least desirable shift. Management is considering increasing the wages of workers on Shift 4 to $1100. Would this change the optimal solution? What would be the eect on total wage expenses? 6. Shift 1, on the other hand, is very desirable (sundays o while on duty fridays and saturdays, which are the best days for tips). Management is considering reducing the wages of workers on Shift 1 to $ 900 per month. Would this change the optimal solution? What would be the eect on total wage expenses? 7. Management is considering introducing a new shift with the days o on tuesdays and sundays. Because these days are not consecutive, the wages will be $ 1200 per month. Will this increase or reduce the total wage expenses?

467. Chapter 9 Decision Theory Decision theory treats decisions against nature. This refers to a situation where the result (return) from a decision depends on action of another player (nature). For example, if the decision is to carry an umbrella or not, the return (get wet or not) depends on what action nature takes. It is important to note that, in this model, the returns accrue only to the decision maker. Nature does not care what the outcome is. This condition distinguishes decision theory from game theory. In game theory, both players have an interest in the outcome. The fundamental piece of data for decision theory problems is a payo table: decision state of nature 1 2 : : : m d1 r11 r12 : : : r1m d2 r21 r22 : : : r2m : : : : : : : : : : : : : : : dn rn1 rn2 : : : rnm The entries rij are the payos for each possible combination of decision and state of nature. The decision process is the following. The decision maker selects one of the possible decisions d1; d2; : : :; dn. Say di. After this decision is made, a state of nature occurs. Say state j. The return received by the decision maker is rij . The question faced by the decision maker is: which decision to select? The decision will depend on the decision maker's belief concerning what nature will do, that is, which state of nature will occur. If we believe state j will occur, we select the decision di associated with the largest number rij in column j of the payo table. Dierent assumptions about nature's behavior lead to dierent procedures for selecting the best decision. If we know which state of nature will occur, we simply select the decision that yields the largest return for the known state of nature. In practice, there may be in

468. nitely many possible decisions. If these possible decisions are represented by a vector d and the return by the real-valued function r(d), the decision problem can then be formulated as max r(d) subject to feasibility constraints on d: 113

469. 114 CHAPTER 9. DECISION THEORY The linear programming model studied earlier in this course

470. ts in this category. In this case, r(d) is a linear function and the feasibility constraints are linear as well. Many other deterministic models of operations research also fall in this category. In the remainder of this chapter, we consider decisions under risk. 9.1 Decisions Under Risk We make the assumption that there is more than one state of nature and that the decision maker knows the probability with which each state of nature will occur. Let pj be the probability that state j will occur. If the decision maker makes decision di, her expected return ERi is ERi = ri1p1 + ri2p2 + : : : + rimpm: She will make the decision di that maximizes ERi, namely ERi = maximum over all i of ERi: Example 9.1.1 Let us consider the example of the newsboy problem: a newsboy buys papers from the delivery truck at the beginning of the day. During the day, he sells papers. Leftover papers at the end of the day are worthless. Assume that each paper costs 15 cents and sells for 50 cents and that the following probability distribution is known. p0 = Prob f demand = 0 g = 2=10 p1 = Prob f demand = 1 g = 4=10 p2 = Prob f demand = 2 g = 3=10 p3 = Prob f demand = 3 g = 1=10 How many papers should the newsboy buy from the delivery truck? To solve this exercise, we

471. rst construct the payo table. Here rij is the reward acheived when i papers are bought and a demand j occurs. decision state of nature 0 1 2 3 0 0 0 0 0 1 15 35 35 35 2 30 20 70 70 3 45 5 50 105 Next, we compute the expected returns for each possible decision: ER0 = 0(2=10)+ 0(4=10) + 0(3=10)+ 0(1=10) = 0 ER1 = 15(2=10)+ 35(4=10)+ 35(3=10)+ 35(1=10) = 25 ER2 = 40(2=10)+ 10(4=10)+ 60(3=10)+ 60(1=10) = 30 ER3 = 45(2=10)+ 5(4=10)+ 50(3=10)+ 105(1=10) = 18:50

472. 9.2. DECISION TREES 115 The maximum occurs when the newsboy buys 2 papers from the delivery truck. His expected return is then 30 cents. The fact that the newsboy must make his buying decision before demand is realized has a considerable impact on his revenues. If he could

473. rst see the demand being realized each day and then buy the corresponding number of newspapers for that day, his expected return would increase by an amount known as the expected value of perfect information. Millions of dollars are spent every year on market research projects, geological tests etc, to determine what state of nature will occur in a wide variety of applications. The expected value of perfect information indicates the expected gain from any such endeavor and thus places an upper bound on the amount that should be spent in gathering information. Let us compute the expected value of perfect information EVPI for the above newsboy example. If demand were known before the buying decision is made, the newsboy's expected return would be 0(2=10)+ 35(4=10)+ 70(3=10)+ 105(1=10) = 45:5 So, EV PI = 45:5 30 = 15:5 It should be pointed out that the criterion of maximizing expected return can sometimes produce unacceptable results. This is because it ignores downside risk. Most people are risk averse, which means they would feel that the loss of x dollars is more painful than the bene

474. t obtained from the gain of the same amount. Decision theory deals with this problem by introducing a function that measures the attractiveness of money. This function is called the utility function. You are referred to 45-749 (managerial economics) for more on this notion. Instead of working with a payo table containing the dollar amounts rij, one would instead work with a payo table containing the utilities, say uij . The optimal decision di is that which maximizes the expected utility EUi = ui1p1 + ui2p2 + : : :+ uimpm over all i. 9.2 Decision Trees Example 9.2.1 Company ABC has developed a new line of products. Top management is at- tempting to decide on the appropriate marketing and production strategy. Three strategies are being considered, which we will simply refer to as A (aggressive), B (basic) and C (cautious). The market conditions under study are denoted by S (strong) or W (weak). Management's best estimate of the net pro

475. ts (in millions of dollars) in each case are given in the following payo table. decision state of nature S W A 30 8 B 20 7 C 5 15 Management's best estimates of the probabilities of a strong or a weak market are 0.45 and 0.55 respectively. Which strategy should be chosen?

476. 116 CHAPTER 9. DECISION THEORY Using the approach introduced earlier, we can compute the expected return for each decision and select the best one, just as we did for the newsboy problem. ERA = 30(0:45) 8(0:55) = 9:10 ERB = 20(0:45)+ 7(0:55) = 12:85 ERC = 5(0:45)+ 15(0:55) = 10:50 The optimal decision is to select B. A convenient way to represent this problem is through the use of decision trees, as in Figure 9.1. A square node will represent a point at which a decision must been made, and each line leading from a square will represent a possible decision. A circular node will represent situations where the outcome is uncertain, and each line leading from a circle will represent a possible outcome. A B C S W 30 7 5 S W S W P(S)=.45 P(W)=.55 P(S)=.45 P(W)=.55 P(S)=.45 P(W)=.55 -8 20 15 Figure 9.1: Decision Tree for the ABC Company Using a decision tree to

477. nd the optimal decision is called solving the tree. To solve a decision tree, one works backwards. This is called folding back the tree. First, the terminal branches are folded back by calculating an expected value for each terminal node. See Figure 9.2. Management now faces the simple problem of choosing the alternative that yields the highest expected terminal value. So, a decision tree provides another, more graphic, way of viewing the same problem. Exactly the same information is utilized, and the same calculations are made.

478. 9.2. DECISION TREES 117 A B C ER A = 9.10 = 10.50 ER B ER C = 12.85 Figure 9.2: Reduced Decision Tree for the ABC Company Sensitivity Analysis The expected return of strategy A is ERA = 30P(S) 8P(W) or, equivalently, ERA = 30P(S) 8(1 P(W)) = 8 + 38P(S): Thus, this expected return is a linear function of the probability that market conditions will be strong. Similarly ERB = 20P(S) + 7(1 P(S)) = 7 + 13P(S) ERC = 5P(S) + 15(1 P(S)) = 15 10P(S) We can plot these three linear functions on the same set of axes (see Figure 9.3). This diagram shows that Company ABC should select the basic strategy (strategy B) as long as the probability of a strong market demand is between P(S) = 0:348 and P(S) = 0:6. This is reassuring, since the optimal decision in this case is not very sensitive to an accurate estimation of P(S). However, if P(S) falls below 0.348, it becomes optimal to choose the cautious strategy C, whereas if P(S) is above 0.6, the agressive strategy A becomes optimal. Sequential Decisions Example 9.2.2 Although the basic strategy B is appealing, ABC's management has the option of asking the marketing research group to perform a market research study. Within a month, this group can report on whether the study was encouraging (E) or discouraging (D). In the past, such studies have tended to be in the right direction: When market ended up being strong, such studies were encouraging 60% of the time and they were discouraging 40% of the time. Whereas, when market ended up being weak, these studies were discouraging 70% of the time and encouraging 30% of the time. Such a study would cost $500,000. Should management request the market research study or not?

479. 118 CHAPTER 9. DECISION THEORY ER A 20 15 10 5 0 -5 -10 ERC ERB 0.348 0.45 0.6 P(S)=0 P(S)=1 Figure 9.3: Expected Return as a function of P(S) Let us construct the decision tree for this sequential decision problem. See Figure 9.4. It is important to note that the tree is created in the chronological order in which information becomes available. Here, the sequence of events is Test decision Test result (if any) Make decision Market condition. The leftmost node correspond to the decision to test or not to test. Moving along the Test branch, the next node to the right is circular, since it corresponds to an uncertain event. There are two possible results. Either the test is encouraging (E), or it is discouraging (D). The probabilities of these two outcomes are P(E) and P(D) respectively. How does one compute these probabilities? We need some fundamental results about probabilities. Refer to 45-733 for additional material. The information we are given is conditional. Given S, the probability of E is 60% and the probability of D is 40%. Similarly, we are told that, given W, the probability of E is 30% and the probability of D is 70%. We denote these conditional probabilities as follows P(EjS) = 0:6 P(EjW) = 0:3 P(DjS) = 0:4 P(DjW) = 0:7 In addition, we know P(S) = 0:45 and P(W) = 0:55. This is all the information we need to compute P(E) and P(D). Indeed, for events S1; S2; : : :; Sn that partition the space of possible

480. 9.2. DECISION TREES 119 Test No Test E D P(E) P(D) A B C A B C A 29.5 -8.5 19.5 6.5 4.5 14.5 15 B C S W S W S W S W S W S W S W S W S W P(S|E) P(W|E) P(S|E) P(W|E) P(S|E) P(W|E) P(S|D) P(W|D) P(S|D) P(W|D) P(S|D) P(W|D) P(S) P(W) P(S) P(W) P(S) P(W) 29.5 -8.5 19.5 6.5 4.5 14.5 30 -8 20 7 5 Figure 9.4: Test versus No-Test Decision Tree

481. 120 CHAPTER 9. DECISION THEORY outcomes and an event T, one has P(T) = P(TjS1)P(S1) + P(TjS2)P(S2) + : : :+ P(TjSn)P(Sn): Here, this gives P(E) = P(EjS)P(S)+ P(EjW)P(W) = (0:6)(0:45)+ (0:3)(0:55) = 0:435 and P(D) = P(DjS)P(S) + P(DjW)P(W) = (0:4)(0:45)+ (0:7)(0:55) = 0:565 As we continue to move to the right of the decision tree, the next nodes are square, corresponding to the three marketing and production strategies. Still further to the right are circular nodes corresponding to the uncertain market conditions: either weak or strong. The probability of these two events is now conditional on the outcome of earlier uncertain events, namely the result of the market reseach study, when such a study was performed. This means that we need to compute the following conditional probabilities: P(SjE); P(WjE); P(SjD) and P(WjD). These quantities are computed using the formula P(RjT) = P(TjR)P(R) P(T) ; which is valid for any two events R and T. Here, we get P(SjE) = P(EjS)P(S) P(E) = (0:6)(0:45) 0:435 = 0:621 Similarly, P(SjE) = 0:379 P(SjD) = 0:318 P(WjD) = 0:682 Now, we are ready to solve the decision tree. As earlier, this is done by folding back. See Figures 9.5, 9.6 and 9.7. You fold back a circular node by calculating the expected returns. You fold back a square node by selecting the decision that yields the highest expected return. The expected return when the market research study is performed is 12.96 million dollars, which is greater than the expected return when no study is performed. So the study should be undertaken. As a

482. nal note, let us compare the expected value of the study (denoted by EV SI, which stands for expected value of sample information) to the expected value of perfect information EV PI. EVSI is computed without incorporating the cost of the study. So EV SI = (12:96 + 0:5) 12:85 = 0:61 whereas EV PI = (30)(0:45)+ (15)(0:55) 12:85 = 21:75 12:85 = 8:90 We see that the market research study is not very eective. If it were, the value of EV SI would be much closer to EV PI. Yet, its value is greater than its cost, so it is worth performing.

483. 9.2. DECISION TREES 121 Test No Test E D 8.29 P(E) P(D) A B C A B C A B C 15.10 14.57 3.58 10.63 11.32 9.10 12.85 10.50 Figure 9.5: Solving the Tree

484. 122 CHAPTER 9. DECISION THEORY Test No Test 12.85 E D P(E) P(D) 15.10 11.32 Figure 9.6: Solving the Tree

485. 9.2. DECISION TREES 123 Test 12.85 No Test 12.96 Figure 9.7: Solving the Tree Example 9.2.3 An art dealer has a client who will buy the masterpiece Rain Delay for $50,000. The dealer can buy the painting now for $40,000 (making a pro

486. t of $10,000). Alternatively, he can wait one day, when the price will go down to $30,000. The dealer can also wait another day when the price will be $25,000. If the dealer does not buy by that day, then the painting will no longer be available. On each day, there is a 2/3 chance that the painting will be sold elsewhere and will no longer be available. (a) Draw a decision tree representing the dealers decision making process. (b) Solve the tree. What is the dealers expected pro

487. t? When should he buy the painting? (c) What is the Expected Value of Perfect Information (value the dealer would place on knowing when the item will be sold)? Answer: (a) 10 20 Buy Buy Buy 0 Don’t 1/3 1/3 Avail Avail Don’t Don’t Sold Sold 2/3 2/3 25 0 Figure 9.8: Solution (b) The value is 10, for an expected pro

488. t of $10,000. He should buy the painting immediately.

489. 124 CHAPTER 9. DECISION THEORY (c) With probability 2/3, the painting will be sold on the

490. rst day, so should be bought immediately. With probability 1/3(2/3) it will be sold on the second day, so should be bought after one day. Finally, with probability 1/3(1/3) it will not be sold on the

491. rst two days, so should be bought after two days. The value of this is 2/3(10)+1/3(2/3)20+1/3(1/3)25 = 13.89. The EVPI is therefore $3,889. Exercise 73 The Scrub Professional Cleaning Service receives preliminary sales contracts from two sources: its own agent and building managers. Historically, 3 8 of the contracts have come from the Scrub agent and 5 8 from building managers. Unfortunately, not all preliminary contracts result in actual sales contracts. Actually, only 1 2 of those preliminary contracts received from building managers result in a sale, whereas 3 4 of those received from the Scrub agent result in a sale. The net return to Scrub from a sale is $6400. The cost of processing and following up on a preliminary contract that does not result in a sale is $320. What is the expected return associated with a preliminary sales contract? Exercise 74 Walter's Dog and Pony show is scheduled to appear in Cedar Rapids on July 4. The pro

492. ts obtained are heavily dependent on the weather. In particular, if the weather is rainy, the show loses $28,000 and if sunny, the show makes a pro

493. t of $12,000. (We assume that all days are either rainy or sunny.) Walter can decide to cancel the show, but if he does, he forfeits a $1,000 deposit he put down when he accepted the date. The historical record shows that on July 4, it has rained 1 4 of the time for the last 100 years. (a) What decision should Walter make to maximize his expected net dollar return? (b) What is the expected value of perfect information? Walter has the option to purchase a forecast from Victor's Weather Wonder. Victor's accuracy varies. On those occasions when it has rained, Victor has been correct (i.e. predicted rain) 90% of the time. On the other hand, when it has been sunny, he has been right (i.e. he predicted sun) only 80% of the time. (c) If Walter had the forecast, what strategy should he follow to maximize his expected net dollar return? (d) How much should Walter be willing to pay to have the forecast? Exercise 75 Wildcat Oil is considering spending $100,000 to drill at a particular spot. The result of such a drilling is either a DryWell (of no value), a Wet Well (providing $150,000 in revenues) or a Gusher (providing $250,000 in revenues). The probabilities for these three possibilities are .5, .3 and .2 respectively. (a) Draw a decision tree for the problem of deciding whether to drill or not. (b) Solve the decision tree assuming the goal is to maximize the expected net revenue. Should the company drill? The following two problems should be done independently. (c) SureFire Consultants is able to determine for certain the type of well before drilling. They oer to tell Wildcat the type for $50,000. Should Wildcat accept their oer? (d) Close-enough Consultants oer to use their specialized seismic hammer. This hammer returns either encouraging or discouraging results. In the past, when applied to a Gusher, the hammer always returned encouraging results. When applied to a Wet Well, it was encouraging 75% of the time and discouraging 25% of the time. When applied to a Dry Well, it was encouraging one-third of the time and discouraging two-thirds of the time. What is the maximum Wildcat should pay Close-enough Consultants for use of their hammer?

494. Chapter 10 Game Theory In previous chapters, we have considered situations where a single decision maker chooses an optimal decision without reference to its eect on other decision makers. In many business situations, however, the outcome for a given

495. rm depends not just on what strategy it chooses, but also on what strategies its competitors choose. Game theory is an attempt to formalize such competitive situations. We present some of the main ideas in the context of Two-Person games and then discuss brie y n-Person Games. 10.1 Two-Person Games Each player has several strategies. If the

496. rst player chooses Strategy i while the second player chooses Strategy j, then Player 1 gains aij while Player 2 gains bij. This outcome is represented by (aij; bij. 2-person games where the players' interests are completely opposed are called zero-sum or constant-sum games: one player's gain is the other player's loss. Games where the players' interests are not completely opposed are called variable-sum games. Such games arise in business on an everyday basis, and solving them is not an easy task. Certain 2-person games admit pure strategies whereas others require mixed strategies. A pure strategy is one where, each time the players play the game, they choose the same strategy. A mixed strategy is one where the players introduce a random element in their choice of a strategy, thus leaving the opponent guessing. 10.1.1 Pure Strategies We start with a constant-sum game: for every possible outcome of the game, the utility u1 of Player 1 plus the utility u2 of Player 2, adds to a constant. For example, if two

497. rms are competing for market shares, then u1 + u2 = 100%. Example 10.1.1 (Battle of the Networks) Two television networks are battling for viewer shares. Viewer share is important because, the higher it is, the more money the network can make from selling advertising time during that program. Consider the following situation: the networks make their programming decisions independently and simultaneously. Each network can show either sports or a sitcom. Network 1 has a programming advantage in sitcoms and Network 2 has one in sports: If both networks show sitcoms, then Network 1 gets a 56% viewer share. If both networks show sports, then Network 2 gets a 54% viewer share. If Network 1 shows a sitcom and Network 2 shows sports, then Network 1 gets a 51% viewer share and Network 2 gets 49%. Finally, if Network 1 shows sports and Network 2 shows a sitcom, then each gets a 50% viewer share. 125

498. 126 CHAPTER 10. GAME THEORY The possible outcomes are best represented in a table. Network 2 Sitcom Sports Sitcom (56%; 44%) (51%; 49%) Network 1 Sports (50%; 50%) (46%; 54%) We can solve Battle of the Networks as follows: from Network 1's point of view, it is better to show a sitcom, whether Network 2 shows a sitcom or sports. The strategy Show a Sitcom is said to dominate the strategy Show Sports for Network 1. So Network 1 will show a sitcom. Similarly, Network 2 is better o showing sports whether Network 1 shows a sitcom or sports. In other words, Network 2 also has a dominating strategy. So Network 2 shows sports. The resulting outcome, namely 51% viewer share to Network 1 and 49% to Network 2, is an equilibrium, since neither of the two players in this game can unilaterally improve their outcome. If Network 1 were to switch from sitcom to sports, its viewer share would drop 5%, from 51% to 46%. If Network 2 were to switch from sports to sitcom, its viewer share would also drop, from 49% to 44%. Each network is getting the best viewer share it can, given the competition it is up against. In a general 2-person game, Strategy i for Player 1 dominates Strategy k for Player 1 if aij akj for every j. Similarly, Strategy j for Player 2 dominates Strategy ` for Player 2 if bij bi` for every i. In a general 2-person game, Strategy i for Player 1 and Strategy j for Player 2 is an equilibrium if the corresponding outcome (aij; bij) has the following property: aij is the largest element akj in column j and bij is the largest element bi` in row i. When such a pair i; j exists, a pure strategy for each player provides an optimal solution to the game. When no such a pair exists, the players must play mixed strategies to maximize their gains (see Section 10.1.2). There is an easy way to relate constant-sum games and zero-sum games. Let u1 and u2 be the payos of players 1 and 2, respectively, in a constant-sum game: u1 + u2 = k. Now consider a new set of payos of the form v1 = u1 u2 v2 = u2 u1 Clearly, v1 + v2 = 0, so we now have a zero-sum game. Using the relation u1 + u2 = k, we can rewrite the payos v1 and v2 as: v1 = 2u1 k v2 = 2u2 k These are positive, linear transformations of utility, and so, according to the expected utility theorem (see 45-749), they have no eect on decisions. For the Battle of the Networks example, the new table becomes

499. 10.1. TWO-PERSON GAMES 127 Network 2 Sitcom Sports Sitcom (12%;12%) (2%;2%) Network 1 Sports (0%; 0%) (8%; 8%) It represents each network's advantage in viewer share over the other network. Whether you reason in terms of total viewer shares or in terms of advantage in viewer shares, the solution is the same. Example 10.1.2 Competitive Advantage In a technologically advanced economy like that of the United states,

500. rms constantly encounter the following situation. A new technological advance becomes available. If one

501. rm adopts the new technology, it gains an advantage over its competitors, a competitive advantage. If all

502. rms adopt the new technology, then the advantage vanishes. This is represented in the next table, where a measures the size of the competitive advantage. Firm 2 Adopt Stay put Adopt (0; 0) (a;a) Firm 1 Stay put (a; a) (0; 0) Each

503. rm has two strategies, either Stay put, or Adopt the new technology. Firm 1 has an incentive to adopt the new technology: in the event Firm 2 stays put, then Firm 1 gets the competitive advantage a, and in the event Firm 2 adopts the new technology, then Firm 1 erases its competitive disadvantage a. So, whatever Firm 2's decision is, Firm 1 is better o adopting the new technology. This is Firm 1's dominant strategy. Of course, the situation for Firm 2 is identical. So the equilibrium of Competitive Advantage is for both

504. rms to adopt the new technology. As a result, both

505. rms get a payo of 0. This may seem like a paradox, since the payos would have been the same if both

506. rms had stayed put. But, in Competitive Advantage, neither

507. rm can aord to stay put. The

508. rms in this game are driven to adopt any technology that comes along. Take, for example, the hospital industry. Magnetic Resonance Imaging (MRI) is a new technology that enhances conventional X rays. It allows doctors to see body damage in ways that were not previously possible. Once MRI became available, any hospital that installed an MRI unit gained a competitive advantage over other hospitals in the area. From a public policy standpoint, it may not make much sense for every hospital to have its own MRI unit. These units are expensive to buy and to operate. They can eat up millions of dollars. Often, one MRI unit could handle the trac of several hospitals. But an individual hospital would be at a competitive disadvantage if it did not have its own MRI. As long as hospitals play Competitive Advantage, they are going to adopt every new technology that comes along.

509. 128 CHAPTER 10. GAME THEORY The two-person games we have encountered so far have had unique pure strategy equilibria. However, a two-person zero-sum game may have multiple equilibria. For example, consider the game: Player 2 A B C A (0; 0) (1;1) (0; 0) Player 1 B (1; 1) (0; 0) (1; 1) C (0; 0) (1;1) (0; 0) Each player can play indierently strategy A or C and so there are four pure strategy equilibria, corresponding to the four corners of the above table. Note that these equilibria all have the same payo. This is not a coincidence. It can be shown that this is always the case. Every equilibrium of a 2-person zero-sum game has the same payo. For pure strategy equilibria, there is a simple proof: Suppose two equilibria had payos (a;a) and (b;b) where a6= b. If these two solutions lied on the same row or column, we would get an immediate contradiction to the de

510. nition of an equilibrium. Let (a;a) lie in row i and column k and let (b;b) lie in row j and column `. The subtable corresponding to these two rows and columns looks as follows Player 2 k ` i (a;a) (c;c) Player 1 j (d;d) (b;b) Since (a;a) is an equilibrium, we must have a d (from Player 1) and a c (from Player 2). Similarly, since (b;b) is an equilibrium, we must have b c and b d. Putting these inequalities together, we get a d b c a This implies that a = b, which completes the proof. Example 10.1.3 Cigarette Advertising on Television On January 1, 1971, cigarette advertizing was banned on american television. It came as something of a surprise to the industry, and a welcome one at that, that pro

511. ts rose by $91 million in 1971. Why did this happen? Consider the strategic interaction in the industry before and after the agreement went into eect. Although there were four large tobacco companies involved, let us consider the strategic interaction between only two of them, for simplicity. In 1970, the payo matrix was:

512. 10.1. TWO-PERSON GAMES 129 Company 2 No TV ads TV Ads No TV Ads (50; 50) (20; 60) Company 1 TV Ads (60; 20) (27; 27) It is clear from this table that advertising on television is a powerful marketing tool. If Com- pany 1 switches from not advertising to advertising on television, its pro

513. ts go up 20%, everything else being equal, when Company 2 does not advertise on TV, and they go up 35% when Company 2 advertises on TV; the same is true if the roles are reversed. In other words, advertising on television is a dominant strategy for each of the companies. Hence the equilibrium is when both companies advertise on television, with a payo of $27 million. We say that an outcome (aij; bij) is ecient if there is no other outcome (akl; bkl) that pays both players at least as much, and one or both players strictly more. Namely, the pair i; j is ecient if there is no pair k; l satisfying akl aij and bkl bij and akl+bkl aij+bij. Obviously, the equilibrium in the cigarette advertizing game is not ecient. There is one ecient outcome, however, in this game: that is when neither company advertises on TV. Then both companies enjoy a payo of $50 million. The ban of cigarette advertizing on television in 1971 only left the strategy No TV Ads and forced the industry into the ecient outcome! In large part, this is why pro

514. ts went up in 1971. It often happens in variable-sum games that the solution is not ecient. One of the biggest dierences between constant-sum and variable-sum games is that solutions to the former are always ecient, whereas solutions to the latter rarely are. 10.1.2 Mixed Strategies Example 10.1.4 Market Niche Two

515. rms are competing for a single market niche. If one

516. rm occupies the market niche, it gets a return of 100. If both

517. rms occupy the market niche, each loses 50. If a

518. rm stays out of the market, it breaks even. The payo table is: Firm B Enter Stay out Enter (50;50) (100; 0) Firm A Stay out (0; 100) (0; 0) This game has two pure strategy equilibria, namely one of the two

519. rms enters the market niche and the other stays out. But, unlike the games we have encountered thus far, neither player has a dominant strategy. When a player has no dominant strategy, she should consider playing a mixed strategy. In a mixed strategy, each of the various pure strategies is played with some probability, say p1 for Strategy 1, p2 for Strategy 2, etc with p1 + p2 + : : : = 1. What would be the best

520. 130 CHAPTER 10. GAME THEORY mixed strategies for Firms A and B? Denote by p1 the probability that Firm A enters the market niche. Therefore p2 = 1p1 is the probability that Firm A stays out. Similarly, Firm B enters the niche with probability q1 and stays out with probability q2 = 1 q1. The key insight to a mixed strategy equilibrium is the following. Every pure strategy that is played as part of a mixed strategy equilibrium has the same expected value. If one pure strategy is expected to pay less than another, then it should not be played at all. The pure strategies that are not excluded should be expected to pay the same. We now apply this principle. The expected value of the Enter strategy for Firm A, when Firm B plays its mixed strategy, is EVEnter = 50q1 + 100q2: The expected value of the Stay out strategy for Firm A is EVStay out = 0. Setting EVEnter = EVStay out we get 50q1 + 100q2 = 0: Using q1 + q2 = 1, we obtain q1 = 2 3 q2 = 1 3 : Similarly, p1 = 2 3 p2 = 1 3 : As you can see, the payos of this mixed strategy equilibrium, namely (0; 0), are inecient. One of these

521. rms could make a lot of money by entering the market niche, if it was sure that the other would not enter the same niche. This assurance is precisely what is missing. Each

522. rm has exactly the same right to enter the market niche. The only way for both

523. rms to exercise this right is to play the inecient, but symmetrical, mixed strategy equilibrium. In many industrial markets, there is only room for a few

524. rms { a situation known as natural oligopoly. Chance plays a major role in the identity of the

525. rms that ultimately enter such markets. If too many

526. rms enter, there are losses all around and eventually some

527. rms must exit. From the mixed strategy equilibrium, we can actually predict how often two

528. rms enter a market niche when there is only room for one: with the above data, the probability of entry by either

529. rm is 2/3, so the probability that both

530. rms enter is (2=3)2 = 4=9. That is a little over 44% of the time! This is the source of the ineciency. The ecient solution has total payo of 100, but is not symmetrical. The fair solution pays each player the same but is inecient. These two principles, eciency and fairness, cannot be reconciled in a game like Market Niche. Once

531. rms recognize this, they can try to

532. nd mecanisms to reach the ecient solution. For example, they may consider side payments. Or

533. rms might simply attempt to scare o competitors by announcing their intention of moving into the market niche before they actually do so. An important case where mixed strategies at equilibrium are always ecient is the case of constant-sum games. Interestingly, the optimal mixed strategies can be computed using linear programming, one linear program for each of the two players. We illustrate these ideas on an example. Example 10.1.5 Stone, Paper, Scissors Each of two players simultaneously utters one of the three words stone, paper, or scissors. If both players utter the same word, the game is a draw. Otherwise, one player pays $1 to the other according to the following: Scissors defeats (cuts) paper, paper defeats (covers) stone, and stone defeats (breaks) scissors. Find the optimal strategies for this game.

534. 10.1. TWO-PERSON GAMES 131 The payo matrix is as follows. Player 2 Stone Paper Scissors Stone (0; 0) (1; 1) (1;1) Player 1 Paper (1;1) (0; 0) (1; 1) Scissors (1; 1) (1;1) (0; 0) Suppose Player 1 chooses the mixed strategy (x1; x2; x3), where x1 = Probability that Player 1 chooses Stone x2 = Probability that Player 1 chooses Paper x3 = Probability that Player 1 chooses Scissors The expected payo of Player 1 is x2 x3 when Player 2 plays Stone, x3 x1 when Player 2 plays Paper and x1x2 when Player 2 plays Scissors. The best Player 2 can do is to achieve the minimum of these three values, namely v = min(x2x3; x3x1; x1x2). Player 1, on the other hand, should try to make the quantity v as large as possible. So, Player 1 should use probabilities x1, x2 and x3 that maximize v = min(x2 x3; x3 x1; x1 x2). This can be written as the following linear program. max v v x2 x3 v x3 x1 v x1 x2 x1 + x2 + x3 = 1 x1; x2; x3 0: Similarly, we obtain Player 2's optimal mixed strategy (y1; y2; y3), where y1 = Probability that Player 2 chooses Stone y2 = Probability that Player 2 chooses Paper y3 = Probability that Player 2 chooses Scissors by solving the linear program max z z y2 y3 z y3 y1 z y1 y2 y1 + y2 + y3 = 1 y1; y2; y3 0:

535. 132 CHAPTER 10. GAME THEORY For the Stone, Paper, Scissors game, the optimal solutions are x1 = x2 = x3 = 1=3 and y1 = y2 = y3 = 1=3 with v = z = 0. In general, for any 2-person zero-sum game, let v and w be the optima of the linear programs for Players 1 and 2 respectively. Because of the special form of these linear programs, it can be shown that z = v. So, the payo at equilibrium is (v;v). The fact that, for each player, the equilibrium solutions are the optimal solutions of a linear has another consequence: All equilibrium solutions have the same payo. 10.2 n-Person Games When there are more than two players, fundamental dierences arise. For example, the theorem about 2-person zero-sum games which says that all equilibrium solutions have the same payo cannot be generalized to three players. Once there are more than two players, a zero-sum game can have equilibrium solutions that do not have the same payos. Here is an example with three players: each player has two strategies, say A and B. Whatever Players 1 and 2 play, the payo vector is (1; 1 2 ; 1 2 ) when Player 3 chooses Strategy A and it is (1 2 ; 1; 1 2 ) when Player 3 chooses Strategy B. We will call the player that receives the payo of 1 the winner and the players that receive a payo of 1 2 , the loosers. In this game, although Player 3 can never be the winner, he determines which of Players 1 or 2 will be the winner. When this situation occurs, we say that Player 3 is a spoiler. For example, take the 1992 presidential election: 1=Bush, 2=Clinton, 3=Perot. Perot could not win, but he could play the spoiler in this campaign. In a two-way race, noone can play a spoiler; in a three-way race, there is room for a spoiler. Example 10.2.1 Competitive Advantage with Three Firms As before, each

536. rm has the choice between staying put or adopting the new technology. If no

537. rm adopts the new technology, then there is no competitive advantage and the payo vector is (0; 0; 0). If exactly one

538. rm adopts the new technology, then that

539. rm gets the competitive advantage a, while each

540. rm at a competitive disadvantage looses a 2 . Thus if only Firm 1 adopts the new technology, then the payo vector is (a; a 2 ; a 2 ). You can think of this as Firm 1 taking market share from both Firm 2 and Firm 3. If exactly two

541. rms adopt the new technology, then these two

542. rms split the competitive advantage, each gaining a 2 , and the

543. rm at a disadvantage looses a. Finally, if all

544. rms adopt the new technology, there is no competitive advantage and the payo vector is (0; 0; 0). We can represent the payos in a table as follows. Firm 3 Adopt Firm 2 Adopt Stay put Adopt (0; 0; 0) ( a 2 ;a; a 2 ) Firm 1 Stay put (a; a 2; a 2) (a 2 ; a 2 ; a) Firm 3 Stay put Firm 2 Adopt Stay put Adopt ( a 2; a 2;a) (a; a 2 ; a 2 ) Firm 1 Stay put (a 2 ; a; a 2 ) (0; 0; 0) Each

545. rm has a dominant strategy, which is to adopt the new technology. So the unique equilibrium occurs when all three

546. rms play the pure strategy: Adopt the new technology. This is

547. 10.2. N-PERSON GAMES 133 presicely what happened with two

548. rms. No

549. rm can be left behind in the race to adopt the new technology. This is just as true for n players as it is for two or three. So, here, the insight from the 2-person game version of Competitive Advantage generalizes nicely to n-person games. An important aspect of n-person games is that subsets of players can form coalitions. A useful concept in this case is the core of the game. Another important aspect of game theory is that of fairness: the Shapley value provides an answer. Game theory also considers games with a sequential structure, games with imperfect information, market games (you learned about Cournot and Bertrand competition in 45-749). To learn more about these and other aspects of game theory, you are referred to the excellent book Games for Business and Economics by Roy Gardner (1995), which was used as a basis for this chapter. Exercise 76 Here is a game that arises in resource economics. (a) There are two

550. shing spots, one good and the other excellent. The good spot has 6

551. sh whereas the excellent spot has 10

552. sh. Two

553. shermen know how good each of the spots is. Each

554. sherman must choose one of the two spots to go

555. shing. If they end up going to the same spot, they divide the catch equally. If they go to dierent spots, the

556. sherman at the good spot gets 6

557. sh, the one at the excellent spot gets 10. Find three equilibria for this game. (b) In such

558. shing games, con icts often break over the

559. shing grounds. How might an assymetrical equilibrium, such as the one provided by the 200-mile

560. shing limit among nations, lead to a more ecient outcome? Exercise 77 Find each player's optimal strategy and the value of the two-person zero-sum game with the following payo table. Player 2 S1 S2 S3 S4 T1 (4;4) (5;5) (1;1) (4;4) Player 1 T2 (2;2) (1;1) (6;6) (3;3) T3 (1;1) (0; 0) (0; 0) (2;2)

561. 134 CHAPTER 10. GAME THEORY

562. Chapter 11 Network Optimization 11.1 Introduction Network optimization is a special type of linear programming model. Network models have three main advantages over linear programming: 1. They can be solved very quickly. Problems whose linear program would have 1000 rows and 30,000 columns can be solved in a matter of seconds. This allows network models to be used in many applications (such as real-time decision making) for which linear programming would be inappropriate. 2. They have naturally integer solutions. By recognizing that a problem can be formulated as a network program, it is possible to solve special types of integer programs without resorting to the ineective and time consuming integer programming algorithms. 3. They are intuitive. Network models provide a language for talking about problems that is much more intuitive than the variables, objective, and constraints language of linear and integer programming. Of course these advantages come with a drawback: network models cannot formulate the wide range of models that linear and integer programs can. However, they occur often enough that they form an important tool for real decision making. 11.2 Terminology A network or graph consists of points, and lines connecting pairs of points. The points are called nodes or vertices. The lines are called arcs. The arcs may have a direction on them, in which case they are called directed arcs. If an arc has no direction, it is often called an edge. If all the arcs in a network are directed, the network is a directed network. If all the arcs are undirected, the network is an undirected network. Two nodes may be connected by a series of arcs. A path is a sequence of distinct arcs (no nodes repeated) connecting the nodes. A directed path from node i to node j is a sequence of arcs, each of whose direction (if any) is towards j. An undirected path may have directed arcs pointed in either direction. A path that begins and ends at the same node is a cycle and may be either directed or undirected. A network is connected if there exists an undirected path between any pair of nodes. A connected network without any cycle is called a tree, mainly because it looks like one. 135

563. 136 CHAPTER 11. NETWORK OPTIMIZATION 11.3 Examples There are many examples of using network ows in practice. Here are a few: 11.3.1 Shortest Paths Consider a phone network. At any given time, a message may take a certain amount of time to traverse each line (due to congestion eects, switching delays, and so on). This time can vary greatly minute by minute and telecommunication companies spend a lot of time and money tracking these delays and communicating these delays throughout the system. Assuming a centralized switcher knows these delays, there remains the problem of routing a call so as to minimize the delays. So, in

564. gure 11.1, what is the least delay path from LA to Boston? How can we

565. nd that path quickly? 7 1 2 1 3 2 8 3 5 7 LA BOS Figure 11.1: Phone Network This is an example of a particular type of network model, called the shortest path problem. In such a problem, you have a network with costs on the edges and two special nodes: a start node and a

566. nish node. The goal is to

567. nd a path from the start node to the

568. nish node whose total weight is minimized. Here is another problem that might not appear to be a shortest path problem, but really is: At a small but growing airport, the local airline company is purchasing a new tractor for a tractor-trailer train to bring luggage to and from the airplanes. A new mechanized luggage system will be installed in 3 years, so the tractor will not be needed after that. However, because it will receive heavy use, and maintenance costs are high, it may still be more economical to replace the tractor after 1 or 2 years. The following table gives the total net discounted cost associated with purchasing a tractor in year i and trading it in in year j (where year 0 is now): j 1 2 3 0 8 18 31 i 1 | 10 21 2 | | 12 The problem is to determine at what times the tractor should be replaced (if ever) to minimize the total costs for tractors. How can this be formulated as a shortest path problem?

569. 11.3. EXAMPLES 137 11.3.2 Maximum Flow Another type of model again has a number on each arc, but now the number corresponds to a capacity. This limits the ow on the arc to be no more than that value. For instance, in a distribution system, the capacity might be the limit on the amount of material (in tons, say) that can go over a particular distribution channel. We would then be concerned with the capcity of the network: how much can be sent from a source node to the destination node? Using the same network as above, treating the numbers as capacities, how much can be sent from LA to Boston? 7 1 2 1 3 2 8 3 5 7 LA BOS Figure 11.2: Distribution Network Associated with the maximum ow is a bottleneck: a set of arcs whose total capacity equals the maximum ow, and whose removal leaves no path from source to destination in the network. It is actually a nontrivial result to show that the maximum ow equals the size of the minimum bottleneck. It is also an interesting task to

570. nd the bottleneck in the above example. Maximum ow models occur in applications where cost is not an issue, and the goal is to maximize the number of items handled (in a broad sense). Here is a similar problem that can be formulated as a maximum ow problem (this was part of a conversation with a Chicago consulting

571. rm on how to improve their internal operations): Over the next three months, there are four projects to be completed. Project A requires 8 person-months of work, and must be done by the end of the second month. Project B requires 3 person-months and must be done by end of the

572. rst month. Project C can start at the beginning of the second month, and requires 7 person-months of work, no more than 2 of which are in the second month. Project D requires 6 person-months, no more than 3 of which occur in any month. There are 9 consultants available in the

573. rst and second month, and 6 in the third. Is it possible to meet the project deadlines? Surprisingly, this is a maximum ow problem. One nice (and surprising) aspect of the result is that this shows that no consultant need split a month's work between jobs. Exercise 78 The Smith's, Jones', Johnson's, and Brown's are attending a picnic. There are four members of each family. Four cars are available to transport them: two cars that can hold four people, and two that can hold three. No more than 2 members of any family can be in any car. The goal is to maximize the number of people who can attend the picnic. Formulate this as a maximum ow problem. Give the nodes and arcs of the network and state which is the starting node and which is the destination. For each arc, give its capacity.

574. 138 CHAPTER 11. NETWORK OPTIMIZATION 2 4 4 4 4 4 4 3 3 2 2 2 2 start end Figure 11.3: Solution of the exercise. 11.3.3 Transportation Problem Consider the following snow removal problem: there are a number of districts in a city. After a snowfall, the snow in each area must be moved out of the district into a convenient location. In Montreal (from where this example is taken), these locations are large grates (leading to the sewer system), a couple large pits, and a couple entry points to the river. Each of these destinations has a capacity. The goal is to minimize the distance traveled to handle all of the snow. This problem is an example of a transportation problem. In such a problem, there are a set of nodes called sources, and a set of nodes called destinations. All arcs go from a source to a destination. There is a per{unit cost on each arc. Each source has a supply of material, and each destination has a demand. We assume that the total supply equals the total demand (possibly adding a fake source or destination as needed). For the snow removal problem, the network might look like that in

575. gure 11.4. AREAS 1 1000 2000 2500 1500 3000 2000 2000 2 3 10 9 7 5 3 2 2 4 7 Figure 11.4: Snow Transportation Network Transportation problems are often used in, surprise, transportation planning. For instance,

576. 11.3. EXAMPLES 139 in an application where goods are at a warehouse, one problem might be to assign customers to a warehouse so as to meet their demands. In such a case, the warehouses are the sources, the customers are the destinations, and the costs represent the per{unit transportation costs. Here is another example of this: One of the main products of PT Company is canned peas. The peas are prepared at three canneries (near Bellingham, Washington; Eugene, Oregon; and Albert Lea, Minnesota) and are then shipped by truck to four distributing warehouses in Sacremento, California; Salt Lake City, Utah; Rapid City, South Dakota; and Albuquerque, New Mexico. Because shipping costs are a major expense, management has begun a study to reduce them. For the upcoming season, an estimate has been made of what the output will be from each cannery, and how much each warehouse will require to satisfy its customers. The shipping costs from each cannery to each warehouse has also been determined. This is summarized in the next table. Shipping cost Warehouse per truckload 1 2 3 4 Output 1 464 513 654 867 75 Cannery 2 352 416 690 791 125 3 995 682 388 685 100 Requirement 80 65 70 85 You should

577. nd it an easy exercise to model this as a linear program. If we let xij be the number of truckloads shipped from cannery i to warehouse j, the problem is to minimize 464x11 + 513x12 + 654x13 + 867x14 + 352x21 + : : :+ 685x34 subject to x11 + x12 + x13 + x14 = 75 x21 + x22 + x23 + x24 = 125 x31 + x32 + x33 + x34 = 100 x11 + x21 + x31 = 80 x12 + x22 + x32 = 65 x13 + x23 + x33 = 70 x14 + x24 + x34 = 85 xij 0 for all i and j. This is an example of the transportation model. As has been pointed out, this problem has a lot of nice structure. All the coecients are 1 and every variable appears in exactly two constraints. It is this structure that lets the simplex algorithm be specialized into an extremely ecient algorithm. What de

578. nes a transportation model? In general, the transportation model is concerned with distributing (literally or

579. guratively) a commodity from a group of supply centers, called sources to a group of receiving centers, called destinations to minimize total cost. In general, source i has a supply of si units, and destination j has a demand for dj units. The cost of distributing items from a source to a destination is proportional to the number of units. This data can be conveniently represented in a table like that for the sample problem. We will generally assume that the total supply equals the total demand. If this is not true for a particular problem, dummy sources or destinations can be added to make it true. The text

580. 140 CHAPTER 11. NETWORK OPTIMIZATION refers to such a problem as a balanced transportation problem. These dummy centers may have zero distribution costs, or costs may be assigned to represent unmet supply or demand. For example, suppose that cannery 3 makes only 75 truckloads. The total supply is now 25 units too little. A dummy supply node can be added with supply 25 to balance the problem, and the cost from the dummy to each warehouse can be added to represent the cost of not meeting the warehouse's demand. The transportation problem has a couple of nice properties: Feasibility. As long as the supply equals demand, there exists a feasible solution to the problem. Integrality. If the supplies and demands are integer, every basic solution (including optimal ones) have integer values. Therefore, it is not necessary to resort to integer programming to

581. nd integer solutions. Linear programming suces. Note that this does not mean that each destination will be supplied by exactly one source. In the pea shipping example, a basic solution might be to ship 20 truckloads from cannery 1 to warehouse 2 and the remaining 55 to warehouse 4, 80 from cannery 2 to warehouse 1 and 45 to warehouse 2 and,

582. nally, 70 truckloads from cannery 3 to warehouse 3 and 30 to warehouse 4. Even though the linear programming formulation of the pea shipping example has seven constraints other than nonnegativity, a basic solution has only six basic variables! This is because the constraints are linearly dependent: the sum of the

583. rst three is identical to the sum of the last four. As a consequence, the feasible region de

584. ned by the constraints would remain the same if we only kept six of them. In general, a basic solution to the transportation model will have a number of basic variables equal to the number of sources plus the number of destinations minus one. Exercise 79 Formulate the following production problem as a transportation model. The demands for a given item are 150, 250, 200 units for the next three months. The demand may be satis

585. ed by excess production in an earlier month held in stock for later consumption, production in the current month, excess production in a later month backordered for preceding months. The variable production cost per unit in any month is $6.00. A unit produced for later consumption will incur a storage cost at the rate of $1 per unit per month. On the other hand, backordered items incur a penalty cost of $3.00 per unit per month. The production capacity in each of the next three months is 200 units. The objective is to devise a minimum cost production plan. Exercise 80 Avertz RentaCar needs to redeploy its automobiles to correct imbalances in the system. Currently Avertz has too many cars in New York (with 10 cars excess) and Chicago (12 cars excess). Pittsburgh would like up to 6 cars, Los Angeles up to 14 cars, and Miami up to 7 cars (note that more cars are demanded than are available). The cost of transporting a car from one city to another is given by: Pittsburgh Los Angeles Miami New York 50 250 100 Chicago 25 200 125 (a) Formulate this problem as a transportation problem. Clearly give the nodes and arcs. For each node, give the supply or demand at the node. For each arc, give the cost. Assume that unmet demand at a city has no cost but that no city can end up with excess cars.

586. 11.3. EXAMPLES 141 (b) It turns out that unmet demand costs $50/car in Pittsburgh, $75/car in LA, and $100/car in Miami. Update your transportation formulation in (a) to account for this change. 11.3.4 Assignment Problem A special case of the transportation problem is the assignment problem which occurs when each supply is 1 and each demand is 1. In this case, the integrality implies that every supplier will be assigned one destination and every destination will have one supplier. The costs give the charge for assigning a supplier and destination to each other. Example 11.3.1 A company has three new machines of dierent types. There are four dierent plants that could receive the machines. Each plant can only receive one machine, and each machine can only be given to one plant. The expected pro

587. t that can be made by each plant if assigned a machine is as follows: Plant 1 2 3 4 1 13 16 12 11 Machine 2 15 0 13 20 3 5 7 10 6 This is a transportation problem with all supplies and demands equal to 1, so it is an assignment problem. Note that a balanced problem must have the same number of supplies and demands, so we must add a dummy machine (corresponding to receiving no machine) and assign a zero cost for assigning the dummy machine to a plant. Exercise 81 Albert, Bob, Carl, and David are stranded on a desert island with Elaine, Francine, Gert, and Holly. The compatibility measures in the next table indicate the happiness each couple would experience if they spent all their time together. If a couple spends only a partial amount of time together, then the happiness is proportional to the fraction of time spent. So if Albert and Elaine spend half their time together, they earn happiness of 7/2. E F G H A 7 5 8 2 B 7 8 9 4 C 3 5 7 9 D 5 5 6 7 (a) Let xij be the fraction of time that man i spends with woman j. Formulate a linear program that maximizes the total happiness of the island (assume that no man spends time with any other man, no woman spends time with any other woman, and no one spends time alone). (b) Explain why exactly four xij will be 1 and the rest will be 0 in an optimal solution to this linear program. This result is called the Marriage Theorem for obvious reasons.

588. 142 CHAPTER 11. NETWORK OPTIMIZATION (c) Do you think that this theorem will hold if we allow men to spend time with men and women to spend time with women? 11.4 Unifying Model: Minimum Cost Network Flows All of the above models are special types of network ow problems: they each have a specialized algorithm that can

589. nd solutions hundreds of times faster than plain linear programming. They can all also be seen as examples of a much broader model, the minimum cost network ow model. This model represents the broadest class of problem that can be solved much faster than linear programming while still retaining such nice properties as integrality of solution and appeal of concept. Like the maximum ow problem, it considers ows in networks with capacities. Like the shortest path problem, it considers a cost for ow through an arc. Like the transportation problem, it allows multiple sources and destinations. In fact, all of these problems can be seen as special cases of the minimum cost ow problem. Consider a directed network with n nodes. The decision variables are xij , the ow through arc (i; j). The given information includes: cij : cost per unit of ow from i to j (may be negative), uij : capacity (or upper bound) on ow from i to j, bi: net ow generated at i. This last value has a sign convention: bi 0 if i is a supply node, bi 0 if i is a demand node, bi = 0 if i is a transshipment node. The objective is to minimize the total cost of sending the supply through the network to satisfy the demand. Note that for this model, it is not necessary that every arc exists. We will use the convention that summations are only taken over arcs that exist. The linear programming formulation for this problem is: Minimize PiPj cijxij Subject to Pj xij Pj xji = bi for all nodes i, 0 xij uij for all arcs (i; j). Again, we will assume that the network is balanced, so Pi bi = 0, since dummies can be added as needed. We also still have a nice integrality property. If all the bi and uij are integral, then the resulting solution to the linear program is also integral. Minimum cost network ows are solved by a variation of the simplex algorithm and can be solved more than 100 times faster than equivalently sized linear programs. From a modeling point of view, it is most important to know the sort of things that can and cannot be modeled in a single network ow problem: Can do

590. 11.4. UNIFYING MODEL: MINIMUM COST NETWORK FLOWS 143 1. Lower bounds on arcs. If a variable xij has a lower bound of lij , upper bound of uij , and cost of cij, change the problem as follows: Replace the upper bound with uij lij , Replace the supply at i with bi lij , Replace the supply at j with bi + lij , Now you have a minimum cost ow problem. Add cij lij to the objective after solving and lij to the ow on arc (i; j) to obtain a solution of the original problem. 2. Upper bounds on ow through a node. Replace the node i with nodes i0 and i00. Create an arc from i0 to i00 with the appropriate capacity, and cost 0. Replace every arc (j; i) with one from j to i0 and every arc (i; j) with one from i00 to j. Lower bounds can also be handled this way. 3. Convex, piecewise linear costs on arc ows (for minimization). This is handled by introducing multiple arcs between the nodes, one for each portion of the piecewise linear function. The convexity will assure that costs are handled correctly in an optimal solution. Can't do 1. Fixed cost to use a node. 2. Fixed cost to use an arc. 3. Proportionality of ow. That is, if one unit enters node i, then you insist that .5 units go to node j and .5 to node k. 4. Gains and losses of ow along arcs, as in power distribution. Note that although these cannot be done in a single network, it may be possible to use the solutions to multiple networks to give you an answer. For instance, if there is only one arc with a

591. xed cost, you can solve both with and without the arc to see if it is advantageous to pay the

592. xed cost. Exercise 82 Here is an example of a problem that doesn't look like a network ow problem, but it really is: A company must meet the following demands for cash at the beginning of each of the next six months: Month 1 2 3 4 5 6 Needs $200 $100 $50 $80 $160 $140 At the beginning of month 1, the company has $150 in cash and $200 worth of bond 1, $100 worth of bond 2 and $400 worth of bond 3. Of course, the company will have to sell some bonds to meet demands, but a penalty will be charged for any bonds sold before the end of month 6. The penalties for selling $1 worth of each bond are shown in the table below. Month of sale 1 2 3 4 5 6 1 $0.07 $0.06 $0.06 $0.04 $0.03 $0.03 Bond 2 $0.17 $0.17 $0.17 $0.11 $0 $0 3 $0.33 $0.33 $0.33 $0.33 $0.33 $0

593. 144 CHAPTER 11. NETWORK OPTIMIZATION (a) Assuming that all bills must be paid on time, formulate a balanced transportation problem that can be used to minimize the cost of meeting the cash demands for the next six months. (b) Assume that payment of bills can be made after they are due, but a penalty of $0.02 per month is assessed for each dollar of cash demands that is postponed for one month. Assuming all bills must be paid by the end of month 6, develop a transshipment model that can be used to minimize the cost of paying the next six months' bills. [ Hint: Transshipment points are needed, in the form Ct = cash available at beginning of month t after bonds for month t have been sold, but before month t demand is met. Shipments into Ct occur from bond sales and Ct1. Shipments out of Ct occur to Ct+1 and demands for months 1; 2; : : :; t.] Exercise 83 Oil R' Us has oil

594. elds in San Diego and Los Angeles. The San Diego

595. eld can produce up to 500,000 barrels per day (bpd); Los Angeles can produce up to 400,000 bpd. Oil is sent to a re

596. nery, either in Dallas or Houston. It costs $700 to re

597. ne 100,000 bpd in Dallas and $900 to re

598. ne 100,000 bpd in Houston. The re

599. ned oil is then shipped to either Chicago or New York. Chicago requires exactly 400,000 bpd and New York requires exactly 300,000 bpd. The shipping costs (per 100,000 bpd) are given as follows: Dallas Houston New York Chicago LA $300 $110 | | San Diego $420 $100 | | Dallas | | $450 $550 Houston | | $470 $530 (a) Formulate this problem as a minimum cost ow problem. Clearly give the network. Give the cost and capacity on each arc and the net supply requirement on each node. (b) Suppose Dallas could process no more than 300,000 bpd. How would you modify your formulation? Exercise 84 A company produces a single product at two plants, A and B. A customer demands 12 units of the product this month, and 20 units of the product next month. Plant A can produce 7 items per month. It costs $12 to produce each item at A this month and $18 to produce at A next month. Plant B can produce 14 items per month. It costs $14 to produce each item at B this month and $20 to produce each item at B next month. Production from this month can be held in inventory to satisfy next month's demand at a cost of $3/unit. (a) Formulate the problem of determining the least cost way of meeting demand as a balanced transportation problem. Give the costs and node supplies, and describe the interpretation of each node and arc. (b) Suppose no more than six items can be held in inventory between this month and next month. Formulate the resulting problem as a minimum cost ow problem. Draw the network, give the interpretation for each node and arc, and give the supply/demand at each node and the cost and capacity for each arc. 11.5 Generalized Networks The

600. nal model I would like to familiarize you with is called the generalized network model. In this model, there may be gains or loses as ow goes along an arc. Each arc has a multiplier to represent

601. 11.6. CONCLUSIONS 145 these gains and loses. For instance, if 3 units of ow enter an arc with a multiplier of .5, then 1.5 unit of ow exit the arc. Such a model can still be represented as a linear program, and there are specialized techniques that can solve such models much faster than linear programming (but a bit slower than regular network ow problems). The optimal solution need not be integer however. The standard example of a generalized network is in power generation: as electricty moves over wires, there is some unavoidable loss along the way. This loss is represented as a mutliplier. Here is another example on how generalized networks might be used: Consider the world's currency market. Given two currencies, say the Yen and the USDollar, there is an exchange rate between them (currently about 110 Yen to the Dollar). It is axiomatic of a arbitrage-free market that there is no method of converting, say, a Dollar to Yen then to Deutsch Marks, to Francs, then Pounds, and to Dollars so that you end up with more than a dollar. How would you recognize when there is an arbitrage possibility? For this model, we use a node to represent each currency. An arc between currency i and j gives the exchange rate from i to j. We begin with a single dollar and wish to send ow through the network and maximize the amount we generate back at the USDollar node. This can be drawn as in

602. gure 11.5. .013493 USD Yen Mark Franc 5.0852 111.52 21.933 .045593 .1966 .008966 3.3823 Max flow out .6659 Figure 11.5: Financial Exchange Network These are actual trades made on November 10, 1996. It is not obvious, but the Dollar-Yen- Mark-Dollar conversion actually makes $0.002. How would you formulate a linear program to recognize this? 11.6 Conclusions One class can only provide a bare introduction to this area. The points we would like you to take away are: 1) Networks are an important subclass of linear programs that are intuitive, easy to solve, and have nice integrality properties. 2) Problems that might not look like networks might be networks. 3) Networks provide a useful way to think about problems even if there are additional constraints or variables that preclude use of networks for modeling the whole problem.

603. 146 CHAPTER 11. NETWORK OPTIMIZATION In practice, you would generally save using the fastest network codes until the

604. nal implemen- tation phase. Until then, linear programming codes will tend to be suciently fast to prove the concepts.

605. Chapter 12 Data Envelopment Analysis Data Envelopment Analysis (DEA) is an increasingly popular management tool. This write-up is an introduction to Data Envelopment Analysis (DEA) for people unfamiliar with the technique. For a more in-depth discussion of DEA, the interested reader is referred to Seiford and Thrall [1990] or the seminal work by Charnes, Cooper, and Rhodes [1978]. DEA is commonly used to evaluate the eciency of a number of producers. A typical statistical approach is characterized as a central tendency approach and it evaluates producers relative to an average producer In contrast, DEA compares each producer with only the best producers. By the way, in the DEA literature, a producer is usually referred to as a decision making unit or DMU. DEA is not always the right tool for a problem but is appropriate in certain cases. (See Strengths and Limitations of DEA.) In DEA, there are a number of producers. The production process for each producer is to take a set of inputs and produce a set of outputs. Each producer has a varying level of inputs and gives a varying level of outputs. For instance, consider a set of banks. Each bank has a certain number of tellers, a certain square footage of space, and a certain number of managers (the inputs). There are a number of measures of the output of a bank, including number of checks cashed, number of loan applications processed, and so on (the outputs). DEA attempts to determine which of the banks are most ecient, and to point out speci

606. c ineciencies of the other banks. A fundamental assumption behind this method is that if a given producer, A, is capable of producing Y(A) units of output with X(A) inputs, then other producers should also be able to do the same if they were to operate eciently. Similarly, if producer B is capable of producing Y(B) units of output with X(B) inputs, then other producers should also be capable of the same production schedule. Producers A, B, and others can then be combined to form a composite producer with composite inputs and composite outputs. Since this composite producer does not necessarily exist, it is typically called a virtual producer. The heart of the analysis lies in

607. nding the best virtual producer for each real producer. If the virtual producer is better than the original producer by either making more output with the same input or making the same output with less input then the original producer is inecient. The subtleties of DEA are introduced in the various ways that producers A and B can be scaled up or down and combined. 12.1 Numerical Example To illustrate how DEA works, let's take an example of three banks. Each bank has exactly 10 tellers (the only input), and we measure a bank based on two outputs: Checks cashed and Loan 147

608. 148 CHAPTER 12. DATA ENVELOPMENT ANALYSIS applications. The data for these banks is as follows: Bank A: 10 tellers, 1000 checks, 20 loan applications Bank B: 10 tellers, 400 checks, 50 loan applications Bank C: 10 tellers, 200 checks, 150 loan applications Now, the key to DEA is to determine whether we can create a virtual bank that is better than one or more of the real banks. Any such dominated bank will be an inecient bank. Consider trying to create a virtual bank that is better than Bank A. Such a bank would use no more inputs than A (10 tellers), and produce at least as much output (1000 checks and 20 loans). Clearly, no combination of banks B and C can possibly do that. Bank A is therefore deemed to be ecient. Bank C is in the same situation. However, consider bank B. If we take half of Bank A and combine it with half of Bank C, then we create a bank that processes 600 checks and 85 loan applications with just 10 tellers. This dominates B (we would much rather have the virtual bank we created than bank B). Bank B is therefore inecient. Another way to see this is that we can scale down the inputs to B (the tellers) and still have at least as much output. If we assume (and we do), that inputs are linearly scalable, then we estimate that we can get by with 6.3 tellers. We do that by taking .34 times bank A plus .29 times bank B. The result uses 6.3 tellers and produces at least as much as bank B does. We say that bank B's eciency rating is .63. Banks A and C have an eciency rating of 1. 12.2 Graphical Example The single input two-output or two input-one output problems are easy to analyze graphically. The previous numerical example is now solved graphically. (An assumption of constant returns to scale is made and explained in detail later.) The analysis of the eciency for bank B looks like the following: 150 50 20 O 200 400 1000 Checks Loans C B A V

609. 12.3. USING LINEAR PROGRAMMING 149 If it is assumed that convex combinations of banks are allowed, then the line segment connecting banks A and C shows the possibilities of virtual outputs that can be formed from these two banks. Similar segments can be drawn between A and B along with B and C. Since the segment AC lies beyond the segments AB and BC, this means that a convex combination of A and C will create the most outputs for a given set of inputs. This line is called the eciency frontier. The eciency frontier de

610. nes the maximum combinations of outputs that can be produced for a given set of inputs. Since bank B lies below the eciency frontier, it is inecient. Its eciency can be determined by comparing it to a virtual bank formed from bank A and bank C. The virtual player, called V, is approximately 54% of bank A and 46% of bank C. (This can be determined by an application of the lever law. Pull out a ruler and measure the lengths of AV, CV, and AC. The percentage of bank C is then AV/AC and the percentage of bank A is CV/AC.) The eciency of bank B is then calculated by

611. nding the fraction of inputs that bank V would need to produce as many outputs as bank B. This is easily calculated by looking at the line from the origin, O, to V. The eciency of player B is OB/OV which is approximately 63%. This

612. gure also shows that banks A and C are ecient since they lie on the eciency frontier. In other words, any virtual bank formed for analyzing banks A and C will lie on banks A and C respectively. Therefore since the eciency is calculated as the ratio of OA/OV or OA/OV, banks A and C will have eciency scores equal to 1.0. The graphical method is useful in this simple two dimensional example but gets much harder in higher dimensions. The normal method of evaluating the eciency of bank B is by using an linear programming formulation of DEA. Since this problem uses a constant input value of 10 for all of the banks, it avoids the com- plications caused by allowing dierent returns to scale. Returns to scale refers to increasing or decreasing eciency based on size. For example, a manufacturer can achieve certain economies of scale by producing a thousand circuit boards at a time rather than one at a time - it might be only 100 times as hard as producing one at a time. This is an example of increasing returns to scale (IRS.) On the other hand, the manufacturer might

613. nd it more than a trillion times as dicult to produce a trillion circuit boards at a time though because of storage problems and limits on the worldwide copper supply. This range of production illustrates decreasing returns to scale (DRS.) Combining the two extreme ranges would necessitate variable returns to scale (VRS.) Constant Returns to Scale (CRS) means that the producers are able to linearly scale the inputs and outputs without increasing or decreasing eciency. This is a signi

614. cant assumption. The assumption of CRS may be valid over limited ranges but its use must be justi

615. ed. As an aside, CRS tends to lower the eciency scores while VRS tends to raise eciency scores. 12.3 Using Linear Programming Data Envelopment Analysis, is a linear programming procedure for a frontier analysis of inputs and outputs. DEA assigns a score of 1 to a unit only when comparisons with other relevant units do not provide evidence of ineciency in the use of any input or output. DEA assigns an eciency score less than one to (relatively) inecient units. A score less than one means that a linear combination of other units from the sample could produce the same vector of outputs using a smaller vector of inputs. The score re ects the radial distance from the estimated production frontier to the DMU under consideration.

616. 150 CHAPTER 12. DATA ENVELOPMENT ANALYSIS There are a number of equivalent formulations for DEA. The most direct formulation of the exposition I gave above is as follows: Let Xi be the vector of inputs into DMU i. Let Yi be the corresponding vector of outputs. Let X0 be the inputs into a DMU for which we want to determine its eciency and Y0 be the outputs. So the X's and the Y 's are the data. The measure of eciency for DMU0 is given by the following linear program: Min s:t: PiXi X0 PiYi Y0 0 where i is the weight given to DMU i in its eorts to dominate DMU 0 and is the eciency of DMU 0. So the 's and are the variables. Since DMU 0 appears on the left hand side of the equations as well, the optimal cannot possibly be more than 1. When we solve this linear program, we get a number of things: 1. The eciency of DMU 0 (), with = 1 meaning that the unit is ecient. 2. The unit's comparables (those DMU with nonzero ). 3. The goal inputs (the dierence between X0 and PiXi) 4. Alternatively, we can keep inputs

617. xed and get goal outputs ( 1 Pi Yi) DEA assumes that the inputs and outputs have been correctly identi

618. ed. Usually, as the number of inputs and outputs increase, more DMUs tend to get an eciency rating of 1 as they become too specialized to be evaluated with respect to other units. On the other hand, if there are too few inputs and outputs, more DMUs tend to be comparable. In any study, it is important to focus on correctly specifying inputs and outputs. Example 12.3.1 Consider analyzing the eciencies of 3 DMUs where 2 inputs and 3 outputs are used. The data is as follows: DMU Inputs Outputs 1 5 14 9 4 16 2 8 15 5 7 10 3 7 12 4 9 13 The linear programs for evaluating the 3 DMUs are given by: LP for evaluating DMU 1: min THETA st 5L1+8L2+7L3 - 5THETA = 0 14L1+15L2+12L3 - 14THETA = 0 9L1+5L2+4L3 = 9 4L1+7L2+9L3 = 4 16L1+10L2+13L3 = 16 L1, L2, L3 = 0

619. 12.3. USING LINEAR PROGRAMMING 151 LP for evaluating DMU 2: min THETA st 5L1+8L2+7L3 - 8THETA = 0 14L1+15L2+12L3 - 15THETA = 0 9L1+5L2+4L3 = 5 4L1+7L2+9L3 = 7 16L1+10L2+13L3 = 10 L1, L2, L3 = 0 LP for evaluating DMU 3: min THETA st 5L1+8L2+7L3 - 7THETA = 0 14L1+15L2+12L3 - 12THETA = 0 9L1+5L2+4L3 = 4 4L1+7L2+9L3 = 9 16L1+10L2+13L3 = 13 L1, L2, L3 = 0 The solution to each of these is as follows: DMU 1. Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficien Increase Decrease $B$10 theta 1 0 1 1E+30 1 $B$11 L1 1 0 0 0.92857142 0.619047619 $B$12 L2 0 0.24285714 0 1E+30 0.242857143 $B$13 L3 0 0 0 0.36710963 0.412698413 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$16 IN1 -0.103473 0 0 1E+30 0 $B$17 IN2 -0.289724 -0.07142857 0 0 1E+30 $B$18 OUT1 9 0.085714286 9 0 0 $B$19 OUT2 4 0.057142857 4 0 0 $B$20 OUT3 16 0 16 0 1E+30

620. 152 CHAPTER 12. DATA ENVELOPMENT ANALYSIS DMU 2. Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficien Increase Decrease $B$10 theta 0.773333333 0 1 1E+30 1 $B$11 L1 0.261538462 0 0 0.866666667 0.577777778 $B$12 L2 0 0.22666667 0 1E+30 0.226666667 $B$13 L3 0.661538462 0 0 0.342635659 0.385185185 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$16 IN1 -0.24820512 0 0 1E+30 0.248205128 $B$17 IN2 -0.452651 -0.0666667 0 0.46538461 1E+30 $B$18 OUT1 5 0.08 5 10.75 0.655826558 $B$19 OUT2 7 0.0533333 7 1.05676855 3.41509434 $B$20 OUT3 12.78461538 0 10 2.78461538 1E+30 DMU 3. Adjustable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficien Increase Decrease $B$10 theta 1 0 1 1E+30 1 $B$11 L1 0 0 0 1.08333333 0.722222222 $B$12 L2 0 0.283333333 0 1E+30 0.283333333 $B$13 L3 1 0 0 0.42829457 0.481481481 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$16 IN1 -0.559375 0 0 1E+30 0 $B$17 IN2 -0.741096 -0.08333333 0 0 1E+30 $B$18 OUT1 4 0.1 4 16.25 0 $B$19 OUT2 9 0.066666667 9 0 0 $B$20 OUT3 13 0 13 0 1E+30 Note that DMUs 1 and 3 are overall ecient and DMU 2 is inecient with an eciency rating of 0.773333. Hence the ecient levels of inputs and outputs for DMU 2 are given by: Ecient levels of Inputs: 0:261538 5 14 #+ 0:661538 7 12 # = 5:938 11:6 #

621. 12.3. USING LINEAR PROGRAMMING 153 Ecient levels of Outputs: 0:261538 2 64 9 4 16 375 + 0:661538 2 64 4 9 13 375 =264 5 7 12:785 375 Note that the outputs are at least as much as the outputs currently produced by DMU 2 and inputs are at most as big as the 0.773333 times the inputs of DMU 2. This can be used in two dierent ways: The inecient DMU should target to cut down inputs to equal at most the ecient levels. Alternatively, an equivalent statement can be made by

622. nding a set of ecient levels of inputs and outputs by dividing the levels obtained by the eciency of DMU 2. This focus can then be used to set targets primarily for outputs rather than reduction of inputs. Alternate Formulation There is another, probably more common formulation, that provides the same information. We can think of DEA as providing a price on each of the inputs and a value for each of the outputs. The eciency of a DMU is simply the ratio of the inputs to the outputs, and is constrained to be no more than 1. The prices and values have nothing to do with real prices and values: they are an arti

623. cial construct. The goal is to

624. nd a set of prices and values that puts the target DMU in the best possible light. The goal, then is to Max uT Y0 vTX0 s:t: uT Yj vTXj 1; j = 0; :::; n; uT 0; vT 0: Here, the variables are the u's and the v's. They are vectors of prices and values respectively. This fractional program can be equivalently stated as the following linear programming problem (where Y and X are matrices with columns Yj and Xj respectively). Max uT Y0 s:t: vTX0 = 1; uT Y vTX 0; uT 0; vT 0: We denote this linear program by (D). Let us compare it with the one introduced earlier, which we denote by (P): Min s:t: PiXi X0 PiYi Y0 0:

625. 154 CHAPTER 12. DATA ENVELOPMENT ANALYSIS To

626. x ideas, let us write out explicitely these two formulations for DMU 2, say, in our example. Formulation (P) for DMU 2: min THETA st -5 L1 - 8 L2 - 7 L3 + 8 THETA = 0 -14L1 -15 L2 -12 L3 + 15THETA = 0 9 L1 + 5 L2 + 4 L3 = 5 4 L1 + 7 L2 + 9 L3 = 7 16L1 +10 L2 +13 L3 = 10 L1=0, L2=0, L3=0 Formulation (D) for DMU 2: max 5 U1 + 7 U2 + 10 U3 st - 5 V1 - 14V2 + 9 U1 + 4 U2 + 16 U3 = 0 - 8 V1 - 15V2 + 5 U1 + 7 U2 + 12 U3 = 0 - 7 V1 - 12V2 + 4 U1 + 9 U2 + 13 U3 = 0 8 V1 + 15V2 = 1 V1=0, V2=0, U1=0, U2=0, U3=0 Formulations (P) and (D) are dual linear programs! These two formulations actually give the same information. You can read the solution to one from the shadow prices of the other. We will not discuss linear programming duality in this course. You can learn about it in some of the OR electives. Exercise 85 Consider the following baseball players: Name At Bat Hits HomeRuns Martin 135 41 6 Polcovich 82 25 1 Johnson 187 40 4 (You need know nothing about baseball for this question). In order to determine the eciency of each of the players, At Bats is de

627. ned as an input while Hits and Home Runs are outputs. Consider the following linear program and its solution: MIN THETA SUBJECT TO - 187 THETA + 135 L1 + 82 L2 + 187 L3 = 0 41 L1 + 25 L2 + 40 L3 = 40 6 L1 + L2 + 4 L3 = 4 L1, L2, L3 = 0

628. 12.4. APPLICATIONS 155 Adjustable Cells Final Reduced Objective Cell Name Value Cost Coefficient $B$10 THETA 0.703135 0 1 $B$11 L1 0.550459 0 0 $B$12 L2 0.697248 0 0 $B$13 L3 0 0.296865 0 Constraints Final Shadow Constraint Cell Name Value Price R.H. Side $B$16 AT BATS 0 -0.005348 0 $B$17 HITS 0 0.017515 40 $B$18 HOME RUNS 0 0.000638 4 (a) For which player is this a DEA analysis? Is this player ecient? What is the eciency rating of this player? Give the virtual producer that proves that eciency rating (you should give the At bats, Hits, and Home Runs for this virtual producer). (b) Formulate the linear program for Jonhson using the alternate formulation and solve using Solver. Compare the Final Value and Shadow Price columns from your Solver output with the solution given above. 12.4 Applications The simple bank example described earlier may not convey the full view on the usefulness of DEA. It is most useful when a comparison is sought against best practices where the analyst doesn't want the frequency of poorly run operations to aect the analysis. DEA has been applied in many situations such as: health care (hospitals, doctors), education (schools, universities), banks, manufacturing, benchmarking, management evaluation, fast food restaurants, and retail stores. The analyzed data sets vary in size. Some analysts work on problems with as few as 15 or 20 DMUs while others are tackling problems with over 10,000 DMUs. 12.5 Strengths and Limitations of DEA As the earlier list of applications suggests, DEA can be a powerful tool when used wisely. A few of the characteristics that make it powerful are: DEA can handle multiple input and multiple output models. It doesn't require an assumption of a functional form relating inputs to outputs. DMUs are directly compared against a peer or combination of peers. Inputs and outputs can have very dierent units. For example, X1 could be in units of lives saved and X2 could be in units of dollars without requiring an a priori tradeo between the two.

629. 156 CHAPTER 12. DATA ENVELOPMENT ANALYSIS The same characteristics that make DEA a powerful tool can also create problems. An analyst should keep these limitations in mind when choosing whether or not to use DEA. Since DEA is an extreme point technique, noise (even symmetrical noise with zero mean) such as measurement error can cause signi

630. cant problems. DEA is good at estimating relative eciency of a DMU but it converges very slowly to absolute eciency. In other words, it can tell you how well you are doing compared to your peers but not compared to a theoretical maximum. Since DEA is a nonparametric technique, statistical hypothesis tests are dicult and are the focus of ongoing research. Since a standard formulation of DEA creates a separate linear program for each DMU, large problems can be computationally intensive. 12.6 References DEA has become a popular subject since it was

631. rst described in 1978. There have been hundreds of papers and technical reports published along with a few books. Technical articles about DEA have been published in a wide variety of places making it hard to

632. nd a good starting point. Here are a few suggestions as to starting points in the literature. 1. Charnes, A., W.W. Cooper, and E. Rhodes. Measuring the eciency of decision making units. European Journal of Operations Research (1978): 429-44. 2. Banker, R.D., A. Charnes, and W.W. Cooper. Some models for estimating technical and scale ineciencies in data envelopment analysis. Management Science 30 (1984): 1078-92. 3. Dyson, R.G. and E. Thanassoulis. Reducing weight exibility in data envelopment analysis. Journal of the Operational Research Society 39 (1988): 563-76. 4. Seiford, L.M. and R.M. Thrall. Recent developments in DEA: the mathematical programming approach to frontier analysis. Journal of Econometrics 46 (1990): 7-38. 5. Ali, A.I., W.D. Cook, and L.M. Seiford. Strict vs. weak ordinal relations for multipliers in data envelopment analysis. Management Science 37 (1991): 733-8. 6. Andersen, P. and N.C. Petersen. A procedure for ranking ecient units in data envelopment analysis. Management Science 39 (1993): 1261-4. 7. Banker, R.D. Maximum likelihood, consistency and data envelopment analysis: a statistical foundation. Management Science 39 (1993): 1265-73. The

633. rst paper was the original paper describing DEA and results in the abbreviation CCR for the basic constant returns-to-scale model. The Seiford and Thrall paper is a good overview of the literature. The other papers all introduce important new concepts. This list of references is certainly incomplete. A good source covering the

634. eld of productivity analysis is The Measurement of Productive Eciency edited by Fried, Lovell, and Schmidt, 1993, from Oxford University Press. There is also

635. 12.6. REFERENCES 157 a recent book from Kluwer Publishers, Data Envelopment Analysis: Theory, Methodology, and Applications by Charnes, Cooper, Lewin, and Seiford. To stay more current on the topics, some of the most important DEA articles appear in Man- agement Science, The Journal of Productivity Analysis, The Journal of the Operational Research Society, and The European Journal of Operational Research. The latter just published a special issue, Productivity Analysis: Parametric and Non-Parametric Approaches edited by Lewin and Lovell which has several important papers.

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