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solutions_hw5

# solutions_hw5 - Solutions to homework 5 3.7#1 Consider the...

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Solutions to homework 5 3.7#1 Consider the game with matrix A = - 2 2 - 1 1 1 1 3 0 1 . (a) a 23 = 1 is a saddle point, since it is simultaneously the minimum value of the second row and the maximum of the third column. (b) Since det A = - 2 + 6 + 0 - 2 - 0 - ( - 3) = 5 6 = 0, A is invertible. The inverse A can be computed in different ways. For example, by performing Gauss elimination on the matrix and mimicking the same operations on the identity matrix: - 2 2 - 1 1 0 0 1 1 1 0 1 0 3 0 1 0 0 1 swap ----→ R 1 ,R 2 1 1 1 0 1 0 - 2 2 - 1 1 0 0 3 0 1 0 0 1 R 0 2 = R 2 +2 R 1 --------→ R 0 3 =3 R 1 - R 3 1 1 1 0 1 0 0 4 1 1 2 0 0 3 2 0 3 - 1 R 0 2 = R 2 / 4 ------→ 1 1 1 0 1 0 0 1 1 / 4 1 / 4 1 / 2 0 0 3 2 0 3 - 1 R 0 1 = R 1 - R 2 --------→ R 0 3 = R 3 - 3 R 2 1 0 3 / 4 1 / 4 1 / 2 0 0 1 1 / 4 1 / 4 1 / 2 0 0 0 5 / 4 - 3 / 4 3 / 2 - 1 R 0 1 = R 1 - 3 5 R 3 ---------------→ R 0 2 = R 2 - 1 5 R 3 ,R 0 3 = 4 5 R 3 1 0 0 1 / 5 - 2 / 5 3 / 5 0 1 0 2 / 5 1 / 5 1 / 5 0 0 1 - 3 / 5 6 / 5 - 4 / 5 . Hence, A - 1 = 1 5 1 - 2 3 2 1 1 - 3 6 - 4 . (c) Consider, q = ( 1 6 , 1 2 , 1 3 ) T . This is an optimal strategy (see (2) on page II – 17 for the definition), since, no matter which row Player I choses the payoff is no more than the value V = 1: Aq = 1 3 1 5 6 1 1 1 . Caution, not all strategies of Player II with positive values will work! For example, q = ( 1 3 , 1 3 , 1 3 ) T does not. If Player II were to play this strategy then Player I could increase his winning to 4 / 3 by playing the last strategy with probability 1. (d) The equation (16) does only give an optimal strategy for Player II if the assumption that Player I has an optimal strategy giving positive weight to each of the rows.” holds (first paragraph of section 3, page II –27). Since this assumption is false in our case we cannot use equation (16) to compute the optimal strategy of Player II. It can be seen by a short computation that Player I has only 1 optimal strategy, giving positive weight to only one of the rows. Since the only solution to p T A (1 , 1 , 1) is p T = (0 , 1 , 0) .

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