Explanation Given Information: A payoff matrix, [ 3 − 6 − 5 4 ] . Formula used: Value of the game is: v = 1 z 1 + z 2 + ⋯ z n The optimal strategy for C is: [ c 1 c 2 ⋮ c n ] Where, c 1 = v z 1 , c 2 = v z 2 , … c n = v z n . Calculation: Consider the payoff matrix, [ 3 − 6 − 5 4 ] . As only two entries are positive, add 7 to each entry to make all the entries positive. Thus, [ 3 − 6 − 5 4 ] = [ 3 + 7 − 6 + 7 − 5 + 7 4 + 7 ] = [ 10 1 2 11 ] So, the given payoff matrix, [ 3 − 6 − 5 4 ] is transformed to [ 10 1 2 11 ] . These two payoff matrices have the same optimal strategies. But the value of the game in new matrix is 7 more than that of the given matrix. Let z 1 , z 2 be chosen so as to maximize z 1 + z 2 subject to the constraints, { 10 z 1 + 1 z 2 ≤ 1 2 z 1 + 11 z 2 ≤ 1 z 1 ≥ 0 , z 2 ≥ 0 Hence, the linear programming problem corresponding to the provided payoff matrix is, Maximize z 1 + z 2 Subjects to the constraints { 10 z 1 + 1 z 2 ≤ 1 2 z 1 + 11 z 2 ≤ 1 z 1 ≥ 0 , z 2 ≥ 0 Now, solve the above problem as below: Introduce the slack variables t and u and write the problem in terms of a system of linear equations. { 10 z 1 + 1 z 2 + t = 1 2 z 1 + 11 z 2 + u = 1 z 1 ≥ 0 , z 2 ≥ 0 , t ≥ 0 , u ≥ 0 The objective function z 1 + z 2 gives the equation, M = z 1 + z 2 . So, the corresponding system of linear equation with slack variables is, { 10 z 1 + 1 z 2 + t = 1 2 z 1 + 11 z 2 + u = 1 − 1 z 1 − 1 z 2 + M = 0 Where, M is as large as possible. Thus, the initial simplex table corresponding to above system is: z 1 z 2 t u M t 10 1 1 0 0 1 u 2 11 0 1 0 1 M − 1 − 1 0 0 1 0 As the left part of the bottom row has negative entries − 1 , the above table does not give the maximum value of the objective function. So, create a new simplex table by pivoting. Since, − 1 is the most negative entry in the last row, the pivot column can be chosen arbitrarily. Let first column becomes the pivot column. Now, to determine the pivot element, compute ratios as below: z 1 z 2 t u M Ratios t 10 1 1 0 0 1 1 10 u 2 11 0 1 0 1 1 2 M − 1 − 1 0 0 1 0 Since, the smallest non-negative ratio is 1 10 , so the pivot element is 10

Finite Mathematics & Its Applications (12th Edition)

12th Edition

ISBN: 9780134437767

Author: Larry J. Goldstein, David I. Schneider, Martha J. Siegel, Steven Hair

Publisher: PEARSON

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Chapter 9.3, Problem 7E

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To calculate: The value of the game and optimal strategy for R and C for the payoff matrix, [3−6−54].

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Chapter 9.3, Problem 6E

Chapter 9.3, Problem 8E

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Chapter 9 Solutions

Finite Mathematics & Its Applications (12th Edition)

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The value of the game and optimal strategy for R and C for the payoff matrix, [ 3 − 6 − 5 4 ] .

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To calculate: The value of the game and optimal strategy for R and C for the payoff matrix, [3−6−54].

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Chapter 9 Solutions