Explanation Given Information: The provided row matrix is R = [ 0 0.5 0 0.5 ] and column matrix is C = [ 0.5 0.5 0 0 ] T . Formula used: The expected payoff e of the game is product of the matrices in the order RPC , where R is the row matrix, P is the payoff Matrix and C is the column matrix. For matrices A with dimension m × n and B with dimension k × l , the product AB is defined if the number of columns in A equal number of columns in B , n = k For matrices A with dimension m × n and B with dimension n × k , the product AB is the matrix of dimension m × k and i j t h entry of AB is the sum of product of corresponding entries of row i of A and column j of B . Calculation: Consider the payoff matrix P = [ 2 0 − 1 2 − 1 0 0 − 2 − 2 0 0 1 3 1 − 1 1 ] , row matrix R = [ 0 0.5 0 0.5 ] and column matrix C = [ 0.5 0.5 0 0 ] T . Take the matrix C = [ 0.5 0.5 0 0 ] T . This contains 1 row and 4 columns along with transpose sign. Transpose interchange the rows with columns and vice versa. Thus, matrix C is rewritten as, C = [ 0.5 0.5 0 0 ] T = [ 0.5 0.5 0 0 ] Recall that the expected payoff E of the game is product of the matrices in the order RPC , where R is the row matrix, P is the payoff Matrix and C is the column matrix. Consider the product, e = R P C = [ 0 0.5 0 0.5 ] [ 2 0 − 1 2 − 1 0 0 − 2 − 2 0 0 1 3 1 − 1 1 ] [ 0.5 0.5 0 0 ] Recall that for matrices A with dimension m × n and B with dimension k × l , the product AB is defined if the number of columns in A equal number of columns in B , n = k . For matrix R , Since, number of rows in [ 0 0.5 0 0.5 ] is 4 and number of columns is 1, Substitute 4 for m and 1 for n in m × n . Thus, dimension of [ 0 0.5 0 0.5 ] is 1 × 4 . For matrix P , Since, number of rows in [ 2 0 − 1 2 − 1 0 0 − 2 − 2 0 0 1 3 1 − 1 1 ] is 4 and number of columns is 4 . Substitute 4 for m and 4 for n in m × n . Thus, dimension of [ 2 0 − 1 2 − 1 0 0 − 2 − 2 0 0 1 3 1 − 1 1 ] is 4 × 4

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ISBN: 9781337280426

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 4.4, Problem 4E

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To calculate: The expected payoff of the game with the payoff matrix P=[20−12−100−2−200131−11] and mixed strategies supplied where R=[00.500.5] and C=[0.50.500]T.

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Chapter 4.4, Problem 3E

Chapter 4.4, Problem 5E

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Chapter 4 Solutions

Finite Mathematics

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The expected payoff of the game with the payoff matrix P = [ 2 0 − 1 2 − 1 0 0 − 2 − 2 0 0 1 3 1 − 1 1 ] and mixed strategies supplied where R = [ 0 0.5 0 0.5 ] and C = [ 0.5 0.5 0 0 ] T .

Solution Summary: The author explains how to calculate the expected payoff of the game with matrices in the order RPC.

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To calculate: The expected payoff of the game with the payoff matrix P=[20−12−100−2−200131−11] and mixed strategies supplied where R=[00.500.5] and C=[0.50.500]T.

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Chapter 4 Solutions