Slides by John Loucks St. Edward’s University

Chapter 5 Advanced Linear Programming Applications Data Envelopment Analysis Revenue Management Portfolio Models and Asset Allocation Game Theory

Data Envelopment Analysis Data envelopment analysis (DEA) is an LP application used to determine the relative operating efficiency of units with the same goals and objectives. DEA creates a fictitious composite unit made up of an optimal weighted average (W1, W2,…) of existing units. An individual unit, k, can be compared by determining E, the fraction of unit k’s input resources required by the optimal composite unit. If E < 1, unit k is less efficient than the composite unit and be deemed relatively inefficient. If E = 1, there is no evidence that unit k is inefficient, but one cannot conclude that k is absolutely efficient.

Data Envelopment Analysis The DEA Model MIN E s.t. Weighted outputs > Unit k’s output (for each measured output) Weighted inputs < E [Unit k’s input] (for each measured input) Sum of weights = 1 E, weights > 0

Data Envelopment Analysis The Langley County School District is trying to determine the relative efficiency of its three high schools. In particular, it wants to evaluate Roosevelt High. The district is evaluating performances on SAT scores, the number of seniors finishing high school, and the number of students who enter college as a function of the number of teachers teaching senior classes, the prorated budget for senior instruction, and the number of students in the senior class.

Data Envelopment Analysis Input Roosevelt Lincoln Washington Senior Faculty 37 25 23 Budget ($100,000's) 6.4 5.0 4.7 Senior Enrollments 850 700 600

Data Envelopment Analysis Output Roosevelt Lincoln Washington Average SAT Score 800 830 900 High School Graduates 450 500 400 College Admissions 140 250 370

Data Envelopment Analysis Define the Decision Variables E = Fraction of Roosevelt's input resources required by the composite high school w1 = Weight applied to Roosevelt's input/output resources by the composite high school w2 = Weight applied to Lincoln’s input/output resources by the composite high school w3 = Weight applied to Washington's input/output resources by the composite high school

Data Envelopment Analysis Define the Objective Function Minimize the fraction of Roosevelt High School's input resources required by the composite high school: MIN E

Data Envelopment Analysis Define the Constraints Sum of the Weights is 1: (1) w1 + w2 + w3 = 1 Output Constraints: Since w1 = 1 is possible, each output of the composite school must be at least as great as that of Roosevelt: (2) 800w1 + 830w2 + 900w3 > 800 (SAT Scores) (3) 450w1 + 500w2 + 400w3 > 450 (Graduates) (4) 140w1 + 250w2 + 370w3 > 140 (College Admissions)

Data Envelopment Analysis Define the Constraints (continued) Input Constraints: The input resources available to the composite school is a fractional multiple, E, of the resources available to Roosevelt. Since the composite high school cannot use more input than that available to it, the input constraints are: (5) 37w1 + 25w2 + 23w3 < 37E (Faculty) (6) 6.4w1 + 5.0w2 + 4.7w3 < 6.4E (Budget) (7) 850w1 + 700w2 + 600w3 < 850E (Seniors) Nonnegativity of variables: E, w1, w2, w3 > 0

Data Envelopment Analysis Computer Solution OBJECTIVE FUNCTION VALUE = 0.765 VARIABLE VALUE REDUCED COSTS E 0.765 0.000 W1 0.000 0.235 W2 0.500 0.000 W3 0.500 0.000

Data Envelopment Analysis Computer Solution (continued) CONSTRAINT SLACK/SURPLUS DUAL VALUES 1 0.000 -0.235 2 65.000 0.000 3 0.000 -0.001 4 170.000 0.000 5 4.294 0.000 6 0.044 0.000 7 0.000 0.001

Data Envelopment Analysis Conclusion The output shows that the composite school is made up of equal weights of Lincoln and Washington. Roosevelt is 76.5% efficient compared to this composite school when measured by college admissions (because of the 0 slack on this constraint (#4)). It is less than 76.5% efficient when using measures of SAT scores and high school graduates (there is positive slack in constraints 2 and 3.)

Revenue Management Another LP application is revenue management. Revenue management involves managing the short-term demand for a fixed perishable inventory in order to maximize revenue potential. The methodology was first used to determine how many airline seats to sell at an early-reservation discount fare and many to sell at a full fare. Application areas now include hotels, apartment rentals, car rentals, cruise lines, and golf courses.

Revenue Management LeapFrog Airways provides passenger service for Indianapolis, Baltimore, Memphis, Austin, and Tampa. LeapFrog has two WB828 airplanes, one based in Indianapolis and the other in Baltimore. Each morning the Indianapolis based plane flies to Austin with a stopover in Memphis. The Baltimore based plane flies to Tampa with a stopover in Memphis. Both planes have a coach section with a 120-seat capacity.

Revenue Management LeapFrog uses two fare classes: a discount fare D class and a full fare F class. Leapfrog’s products, each referred to as an origin destination itinerary fare (ODIF), are listed on the next slide with their fares and forecasted demand. LeapFrog wants to determine how many seats it should allocate to each ODIF.

Revenue Management Fare Class D F ODIF Code IMD IAD ITD IMF IAF ITF BMD BAD BTD BMF BAF BTF MAD MTD MAF MTF ODIF 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Origin Indianapolis Baltimore Memphis Destination Memphis Austin Tampa Tampa Austin Fare 175 275 285 395 425 475 185 315 290 385 525 490 190 180 310 295 Demand 44 25 40 15 10 8 26 50 42 12 16 9 58 48 14 11

Revenue Management Define the Decision Variables There are 16 variables, one for each ODIF: IMD = number of seats allocated to Indianapolis-Memphis- Discount class IAD = number of seats allocated to Indianapolis-Austin- Discount class ITD = number of seats allocated to Indianapolis-Tampa- Discount class IMF = number of seats allocated to Indianapolis-Memphis- Full Fare class IAF = number of seats allocated to Indianapolis-Austin-Full Fare class

Revenue Management Define the Decision Variables (continued) ITF = number of seats allocated to Indianapolis-Tampa- Full Fare class BMD = number of seats allocated to Baltimore-Memphis- Discount class BAD = number of seats allocated to Baltimore-Austin- BTD = number of seats allocated to Baltimore-Tampa- BMF = number of seats allocated to Baltimore-Memphis- BAF = number of seats allocated to Baltimore-Austin-

Revenue Management Define the Decision Variables (continued) BTF = number of seats allocated to Baltimore-Tampa- Full Fare class MAD = number of seats allocated to Memphis-Austin- Discount class MTD = number of seats allocated to Memphis-Tampa- MAF = number of seats allocated to Memphis-Austin- MTF = number of seats allocated to Memphis-Tampa-

Revenue Management Define the Objective Function Maximize total revenue: Max (fare per seat for each ODIF) x (number of seats allocated to the ODIF) Max 175IMD + 275IAD + 285ITD + 395IMF + 425IAF + 475ITF + 185BMD + 315BAD + 290BTD + 385BMF + 525BAF + 490BTF + 190MAD + 180MTD + 310MAF + 295MTF

Revenue Management Define the Constraints Indianapolis-Memphis leg There are 4 capacity constraints, one for each flight leg: Indianapolis-Memphis leg (1)   IMD + IAD + ITD + IMF + IAF + ITF < 120 Baltimore-Memphis leg (2)    BMD + BAD + BTD + BMF + BAF + BTF < 120 Memphis-Austin leg (3)    IAD + IAF + BAD + BAF + MAD + MAF < 120 Memphis-Tampa leg (4)    ITD + ITF + BTD + BTF + MTD + MTF < 120

Revenue Management Define the Constraints (continued) There are 16 demand constraints, one for each ODIF: (5) IMD < 44 (11) BMD < 26 (17) MAD < 5 (6) IAD < 25 (12) BAD < 50 (18) MTD < 48 (7) ITD < 40 (13) BTD < 42 (19) MAF < 14 (8) IMF < 15 (14) BMF < 12 (20) MTF < 11 (9) IAF < 10 (15) BAF < 16 (10) ITF < 8 (16) BTF < 9

Revenue Management Computer Solution Objective Function Value = 94735.000 Variable Value Reduced Cost IMD 44.000 0.000 IAD 3.000 0.000 ITD 40.000 0.000 IMF 15.000 0.000 IAF 10.000 0.000 ITF 8.000 0.000 BMD 26.000 0.000 BAD 50.000 0.000

Revenue Management Computer Solution (continued) Variable Value Reduced Cost BTD 7.000 0.000 BMF 12.000 0.000 BAF 16.000 0.000 BTF 9.000 0.000 MAD 27.000 0.000 MTD 45.000 0.000 MAF 14.000 0.000 MTF 11.000 0.000

Portfolio Models and Asset Management Asset allocation involves determining how to allocate investment funds across a variety of asset classes such as stocks, bonds, mutual funds, real estate. Portfolio models are used to determine percentage of funds that should be made in each asset class. The goal is to create a portfolio that provides the best balance between risk and return.

Portfolio Model John Sweeney is an investment advisor who is attempting to construct an "optimal portfolio" for a client who has $400,000 cash to invest. There are ten different investments, falling into four broad categories that John and his client have identified as potential candidate for this portfolio. The investments and their important characteristics are listed in the table on the next slide. Note that Unidyde Corp. under Equities and Unidyde Corp. under Debt are two separate investments, whereas First General REIT is a single investment that is considered both an equities and a real estate investment.

Portfolio Model Exp. Annual After Tax Liquidity Risk Category Investment Return Factor Factor Equities Unidyde Corp. 15.0% 100 60 (Stocks) CC’s Restaurants 17.0% 100 70 First General REIT 17.5% 100 75 Debt Metropolis Electric 11.8% 95 20 (Bonds) Unidyde Corp. 12.2% 92 30 Lewisville Transit 12.0% 79 22 Real Estate Realty Partners 22.0% 0 50 First General REIT ( --- See above --- ) Money T-Bill Account 9.6% 80 0 Money Mkt. Fund 10.5% 100 10 Saver's Certificate 12.6% 0 0

Portfolio Model Formulate a linear programming problem to accomplish John's objective as an investment advisor which is to construct a portfolio that maximizes his client's total expected after-tax return over the next year, subject to the limitations placed upon him by the client for the portfolio. (Limitations listed on next two slides.)

Portfolio Model Portfolio Limitations 1. The weighted average liquidity factor for the portfolio must to be at least 65. 2. The weighted average risk factor for the portfolio must be no greater than 55. 3. No more than $60,000 is to be invested in Unidyde stocks or bonds. 4. No more than 40% of the investment can be in any one category except the money category. 5. No more than 20% of the total investment can be in any one investment except the money market fund. continued

Portfolio Model Portfolio Limitations (continued) 6. At least $1,000 must be invested in the Money Market fund. 7. The maximum investment in Saver's Certificates is $15,000. 8. The minimum investment desired for debt is $90,000. 9. At least $10,000 must be placed in a T-Bill account.

Portfolio Model Define the Decision Variables X1 = $ amount invested in Unidyde Corp. (Equities) X2 = $ amount invested in CC’s Restaurants X3 = $ amount invested in First General REIT X4 = $ amount invested in Metropolis Electric X5 = $ amount invested in Unidyde Corp. (Debt) X6 = $ amount invested in Lewisville Transit X7 = $ amount invested in Realty Partners X8 = $ amount invested in T-Bill Account X9 = $ amount invested in Money Mkt. Fund X10 = $ amount invested in Saver's Certificate

Portfolio Model Define the Objective Function Maximize the total expected after-tax return over the next year: Max .15X1 + .17X2 + .175X3 + .118X4 + .122X5 + .12X6 + .22X7 + .096X8 + .105X9 + .126X10

Portfolio Model Define the Constraints Total funds invested must not exceed $400,000: (1) X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 = 400,000 Weighted average liquidity factor must to be at least 65: 100X1 + 100X2 + 100X3 + 95X4 + 92X5 + 79X6 + 80X8 + 100X9 > 65(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10) Weighted average risk factor must be no greater than 55: 60X1 + 70X2 + 75X3 + 20X4 + 30X5 + 22X6 + 50X7 + 10X9 < 55(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10) No more than $60,000 to be invested in Unidyde Corp: X1 + X5 < 60,000

Portfolio Model Define the Constraints (continued) No more than 40% of the $400,000 investment can be in any one category except the money category: (5) X1 + X2 + X3 < 160,000 (6) X4 + X5 + X6 < 160,000 X3 + X7 < 160,000 No more than 20% of the $400,000 investment can be in any one investment except the money market fund: (8) X2 < 80,000 (12) X7 < 80,000 (9) X3 < 80,000 (13) X8 < 80,000 (10) X4 < 80,000 (14) X10 < 80,000 (11) X6 < 80,000

Portfolio Model Define the Constraints (continued) At least $1,000 must be invested in the Money Market fund: (15) X9 > 1,000 The maximum investment in Saver's Certificates is $15,000: (16) X10 < 15,000 The minimum investment the Debt category is $90,000: (17) X4 + X5 + X6 > 90,000 At least $10,000 must be placed in a T-Bill account: (18) X8 > 10,000 Non-negativity of variables: Xj > 0 j = 1, . . . , 10

Portfolio Model Solution Summary Total Expected After-Tax Return = $64,355 X1 = $0 invested in Unidyde Corp. (Equities) X2 = $80,000 invested in CC’s Restaurants X3 = $80,000 invested in First General REIT X4 = $0 invested in Metropolis Electric X5 = $60,000 invested in Unidyde Corp. (Debt) X6 = $74,000 invested in Lewisville Transit X7 = $80,000 invested in Realty Partners X8 = $10,000 invested in T-Bill Account X9 = $1,000 invested in Money Mkt. Fund X10 = $15,000 invested in Saver's Certificate

Introduction to Game Theory In decision analysis, a single decision maker seeks to select an optimal alternative. In game theory, there are two or more decision makers, called players, who compete as adversaries against each other. It is assumed that each player has the same information and will select the strategy that provides the best possible outcome from his point of view. Each player selects a strategy independently without knowing in advance the strategy of the other player(s). continue

Introduction to Game Theory The combination of the competing strategies provides the value of the game to the players. Examples of competing players are teams, armies, companies, political candidates, and contract bidders.

Two-Person Zero-Sum Game Two-person means there are two competing players in the game. Zero-sum means the gain (or loss) for one player is equal to the corresponding loss (or gain) for the other player. The gain and loss balance out so that there is a zero-sum for the game. What one player wins, the other player loses.

Two-Person Zero-Sum Game Example Competing for Vehicle Sales Suppose that there are only two vehicle dealer-ships in a small city. Each dealership is considering three strategies that are designed to take sales of new vehicles from the other dealership over a four-month period. The strategies, assumed to be the same for both dealerships, are on the next slide.

Two-Person Zero-Sum Game Example Strategy Choices Strategy 1: Offer a cash rebate on a new vehicle. Strategy 2: Offer free optional equipment on a new vehicle. Strategy 3: Offer a 0% loan

Two-Person Zero-Sum Game Example Payoff Table: Number of Vehicle Sales Gained Per Week by Dealership A (or Lost Per Week by Dealership B) Dealership B Cash Rebate b1 Free Options b2 0% Loan b3 Dealership A Cash Rebate a1 Free Options a2 0% Loan a3 2 2 1 -3 3 -1 3 -2 0

Two-Person Zero-Sum Game Step 1: Identify the minimum payoff for each row (for Player A). Step 2: For Player A, select the strategy that provides the maximum of the row minimums (called the maximin).

Two-Person Zero-Sum Game Example Identifying Maximin and Best Strategy Dealership B Cash Rebate b1 Free Options b2 0% Loan b3 Row Minimum Dealership A Cash Rebate a1 Free Options a2 0% Loan a3 2 2 1 1 -3 -2 -3 3 -1 3 -2 0 Best Strategy For Player A Maximin Payoff

Two-Person Zero-Sum Game Step 3: Identify the maximum payoff for each column (for Player B). Step 4: For Player B, select the strategy that provides the minimum of the column maximums (called the minimax).

Two-Person Zero-Sum Game Example Identifying Minimax and Best Strategy Dealership B Best Strategy For Player B Cash Rebate b1 Free Options b2 0% Loan b3 Dealership A Cash Rebate a1 Free Options a2 0% Loan a3 2 2 1 -3 3 -1 Minimax Payoff 3 -2 0 Column Maximum 3 3 1

Pure Strategy Whenever an optimal pure strategy exists: the maximum of the row minimums equals the minimum of the column maximums (Player A’s maximin equals Player B’s minimax) the game is said to have a saddle point (the intersection of the optimal strategies) the value of the saddle point is the value of the game neither player can improve his/her outcome by changing strategies even if he/she learns in advance the opponent’s strategy

Pure Strategy Example Saddle Point and Value of the Game Dealership B game is 1 Cash Rebate b1 Free Options b2 0% Loan b3 Row Minimum Dealership A Cash Rebate a1 Free Options a2 0% Loan a3 2 2 1 1 -3 -2 -3 3 -1 3 -2 0 Column Maximum 3 3 1 Saddle Point

Pure Strategy Example Pure Strategy Summary Player A should choose Strategy a1 (offer a cash rebate). Player A can expect a gain of at least 1 vehicle sale per week. Player B should choose Strategy b3 (offer a 0% loan). Player B can expect a loss of no more than 1 vehicle sale per week.

Mixed Strategy If the maximin value for Player A does not equal the minimax value for Player B, then a pure strategy is not optimal for the game. In this case, a mixed strategy is best. With a mixed strategy, each player employs more than one strategy. Each player should use one strategy some of the time and other strategies the rest of the time. The optimal solution is the relative frequencies with which each player should use his possible strategies.

Mixed Strategy Example Consider the following two-person zero-sum game. The maximin does not equal the minimax. There is not an optimal pure strategy. Player B Row Minimum b1 b2 Player A Maximin a1 a2 4 5 4 8 11 5 Column Maximum 11 8 Minimax

Mixed Strategy Example p = the probability Player A selects strategy a1 (1 - p) = the probability Player A selects strategy a2 If Player B selects b1: EV = 4p + 11(1 – p) If Player B selects b2: EV = 8p + 5(1 – p)

Mixed Strategy Example To solve for the optimal probabilities for Player A we set the two expected values equal and solve for the value of p. 4p + 11(1 – p) = 8p + 5(1 – p) 4p + 11 – 11p = 8p + 5 – 5p 11 – 7p = 5 + 3p Hence, (1 - p) = .4 -10p = -6 p = .6 Player A should select: Strategy a1 with a .6 probability and Strategy a2 with a .4 probability.

Mixed Strategy Example q = the probability Player B selects strategy b1 (1 - q) = the probability Player B selects strategy b2 If Player A selects a1: EV = 4q + 8(1 – q) If Player A selects a2: EV = 11q + 5(1 – q)

Mixed Strategy Example Expected gain per game for Player A Value of the Game For Player A: EV = 4p + 11(1 – p) = 4(.6) + 11(.4) = 6.8 For Player B: Expected loss per game for Player B EV = 4q + 8(1 – q) = 4(.3) + 8(.7) = 6.8

Dominated Strategies Example Suppose that the payoff table for a two-person zero- sum game is the following. Here there is no optimal pure strategy. Player B Row Minimum b1 b2 b3 Player A Maximin a1 a2 a3 6 5 -2 -2 -3 1 0 3 3 4 -3 Column Maximum 6 5 3 Minimax

Dominated Strategies Example If a game larger than 2 x 2 has a mixed strategy, we first look for dominated strategies in order to reduce the size of the game. Player B b1 b2 b3 Player A a1 a2 a3 6 5 -2 1 0 3 3 4 -3 Player A’s Strategy a3 is dominated by Strategy a1, so Strategy a3 can be eliminated.

Dominated Strategies Example We continue to look for dominated strategies in order to reduce the size of the game. Player B b1 b2 b3 Player A a1 a2 6 5 -2 1 0 3 Player B’s Strategy b2 is dominated by Strategy b1, so Strategy b2 can be eliminated.

Dominated Strategies Example The 3 x 3 game has been reduced to a 2 x 2. It is now possible to solve algebraically for the optimal mixed-strategy probabilities. Player B b1 b3 Player A a1 a2 6 -2 1 3

Two-Person Zero-Sum Game Example #2 Competing for Vehicle Sales Let us continue with the two-dealership game presented earlier, but with a change to one payoff. If both Dealership A and Dealership B choose to offer a 0% loan, the payoff to Dealership A is now an increase of 3 vehicle Sales per week. (The revised payoff table appears on the next slide.)

Two-Person Zero-Sum Game Example #2 Payoff Table: Number of Vehicle Sales Gained Per Week by Dealership A (or Lost Per Week by Dealership B) Dealership B Cash Rebate b1 Free Options b2 0% Loan b3 Dealership A Cash Rebate a1 Free Options a2 0% Loan a3 2 2 1 -3 3 -1 3 -2 3

Two-Person Zero-Sum Game Example #2 The maximin (1) does not equal the minimax (3), so a pure strategy solution does not exist for this problem. The optimal solution is for both dealerships to adopt a mixed strategy. There are no dominated strategies, so the problem cannot be reduced to a 2x2 and solved algebraically. However, the game can be formulated and solved as a linear program.

Two-Person Zero-Sum Game Example #2 Let us first consider the game from the point of view of Dealership A. Dealership A will select one of its three strategies based on the following probabilities: PA1 = the probability that Dealership A selects strategy a1 PA2 = the probability that Dealership A selects strategy a2 PA3 = the probability that Dealership A selects strategy a3

Two-Person Zero-Sum Game Example #2 Weighting each payoff by its probability and summing provides the expected value of the increase in vehicle sales per week for Dealership A. Dealership B Strategy Expected Gain for Dealership A b1 EG(b1) = 2PA1 – 3PA2 + 3PA3 b2 EG(b2) = 2PA1 + 3PA2 – 2PA3 b3 EG(b3) = 1PA1 – 1PA2 + 3PA3

Two-Person Zero-Sum Game Example #2 Define GAINA to be the optimal expected gain in vehicle sales for Dealership A, which we want to maximize. Thus, the individual expected gains, EG(b1), EG(b2) and EG(b3) must all be greater than or equal to GAINA. For example, 2PA1 – 3PA2 + 3PA3 > GAINA Also, the sum of Dealership A’s mixed strategy probabilities must equal 1. This results in the LP formulation on the next slide …..

Two-Person Zero-Sum Game Example #2 Dealership A’s Linear Programming Formulation Max GAINA s.t. 2PA1 – 3PA2 + 3PA3 – GAINA > 0 (Strategy b1) 2PA1 + 3PA2 – 2PA3 – GAINA > 0 (Strategy b2) 1PA1 – 1PA2 + 0PA3 – GAINA > 0 (Strategy b3) PA1 + PA2 + PA3 = 1 (Prob’s sum to 1) PA1, PA2, PA3, GAINA > 0 (Non-negativity)

Two-Person Zero-Sum Game Example #2 Computer Solution: Dealership A OBJECTIVE FUNCTION VALUE = 1.333 VARIABLE VALUE REDUCED COSTS PA1 0.833 0.000 PA2 0.000 1.000 PA3 0.167 0.000 GAINA 1.333 0.000

Two-Person Zero-Sum Game Example #2 Computer Solution: Dealership A CONSTRAINT SLACK/SURPLUS DUAL VALUES 1 0.833 0.000 2 0.000 -0.333 3 0.000 -0.667 4 0.000 1.333

Two-Person Zero-Sum Game Example #2 Dealership A’s Optimal Mixed Strategy Offer a cash rebate (a1) with a probability of 0.833 Do not offer free optional equipment (a2) Offer a 0% loan (a3) with a probability of 0.167 The expected value of this mixed strategy is a gain of 1.333 vehicle sales per week for Dealership A.

Two-Person Zero-Sum Game Example #2 Let us now consider the game from the point of view of Dealership B. Dealership B will select one of its three strategies based on the following probabilities: PB1 = the probability that Dealership B selects strategy b1 PB2 = the probability that Dealership B selects strategy b2 PB3 = the probability that Dealership B selects strategy b3

Two-Person Zero-Sum Game Example #2 Weighting each payoff by its probability and summing provides the expected value of the decrease in vehicle sales per week for Dealership B. Dealership A Strategy Expected Loss for Dealership B a1 EL(a1) = 2PB1 + 2PB2 + 1PB3 a2 EL(a2) = -3PB1 + 3PB2 – 1PB3 a3 EL(a3) = 3PB1 – 2PB2 + 3PB3

Two-Person Zero-Sum Game Example #2 Define LOSSB to be the optimal expected loss in vehicle sales for Dealership B, which we want to minimize. Thus, the individual expected losses, EL(a1), EL(a2) and EL(a3) must all be less than or equal to LOSSB. For example, 2PA1 + 2PA2 + 1PA3 < LOSSB Also, the sum of Dealership B’s mixed strategy probabilities must equal 1. This results in the LP formulation on the next slide …..

Two-Person Zero-Sum Game Example #2 Dealership B’s Linear Programming Formulation Min LOSSB s.t. 2PB1 + 2PB2 + 1PB3 – LOSSB < 0 (Strategy a1) -3PB1 + 3PB2 – 1PB3 – LOSSB < 0 (Strategy a2) 3PB1 – 2PB2 + 3PB3 – LOSSB < 0 (Strategy a3) PB1 + PB2 + PB3 = 1 (Prob’s sum to 1) PB1, PB2, PB3, LOSSB > 0 (Non-negativity)

Two-Person Zero-Sum Game Example #2 Computer Solution: Dealership B OBJECTIVE FUNCTION VALUE = 1.333 VARIABLE VALUE REDUCED COSTS PB1 0.000 0.833 PB2 0.333 0.000 PB3 0.667 0.000 LOSSB 1.333 0.000

Two-Person Zero-Sum Game Example #2 Computer Solution: Dealership B CONSTRAINT SLACK/SURPLUS DUAL VALUES 1 0.000 0.833 2 1.000 0.000 3 0.000 0.167 4 0.000 -1.333

Two-Person Zero-Sum Game Example #2 Dealership B’s Optimal Mixed Strategy Do not offer a cash rebate (b1) Offer free optional equipment (b2) with a probability of 0.333 Offer a 0% loan (b3) with a probability of 0.667 The expected payoff of this mixed strategy is a loss of 1.333 vehicle sales per week for Dealership B. Note that expected loss for Dealership B is the same as the expected gain for Dealership A. (There is a zero- sum for the expected payoffs.)

Other Game Theory Models Two-Person, Constant-Sum Games (The sum of the payoffs is a constant other than zero.) Variable-Sum Games (The sum of the payoffs is variable.) n-Person Games (A game involves more than two players.) Cooperative Games (Players are allowed pre-play communications.) Infinite-Strategies Games (An infinite number of strategies are available for the players.)

End of Chapter 5