cp - OR 4350 Game Theory Prof. Bland March 1, 2012 A...

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OR 4350 Game Theory March 1, 2012 Prof. Bland A Complementary Pivot Algorithm to Compute a Nash Equilibrium in Bimatrix Games an abbreviated version of notes by Prof MJ Todd Consider the bimatrix game ( A,B ) with A and B m × n matrices. Then (¯ p, ¯ q ) P × Q is a Nash equilibrium iff pA ¯ q ¯ pA ¯ q for all p P, and ¯ pBq ¯ pB ¯ q for all q Q, or equivalently, ( i ) A i ¯ q ¯ pA ¯ q for all i, ( ii ) ¯ pB j ¯ pB ¯ q for all j. Note that, if we add a constant to every entry of A to get A 0 , and a constant to every entry of B to get B 0 , then (¯ p, ¯ q ) is a Nash equilibrium of ( ) iff it is one for ( A 0 ,B 0 ). This is because for every p P , q Q , pA 0 q is more than pAq by the same constant used to form A 0 from A ; and similarly for B 0 and B . So we’ll assume that all entries of A and B are nonnegative, with a positive entry in each column of A and in each row of B . This will en- sure that ¯ pA ¯ q and ¯ pB ¯ q will be positive for any Nash equilibrium: ˜ p = (1 /m, 1 /m,. .., 1 /m ) T guarantees a positive payoff for I for any q , and hence for ¯ q , so a best response like ¯ p guar- antees at least as much, and similarly for B . Example: We take A = " 5 1 1 1 1 2 # , B = " 0 1 3 2 1 0 # . (We could eliminate player II’s second column, since it is strongly dominated by q = (3 / 5 , 0 , 2 / 5) T , but we’ll leave it in to better illustrate the algorithm.) The pair (¯ p = (2 / 5 , 3 / 5) , ¯ q = (1 / 5 , 0 , 4 / 5) T ) is a Nash equilibrium and gives payoffs 9 / 5 and 6 / 5 to players I and II respectively. We’ll find this using our complementary pivoting algorithm below. We want to find a solution to (i) and (ii) above. Unfortunately, while the left-hand sides are linear in ¯ p and ¯ q , the right-hand sides aren’t. Consider instead: There is some α > 0, β > 0 such that ( i 0 ) A i ¯ q α for all i , with equality if ¯ p i > 0 , ( ii 0 ) ¯ pB j β for all j , with equality if ¯ q j > 0 . 1
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Then (i)–(ii) hold iff (i’)–(ii’) do, with α and β necessarily equal to ¯ pA ¯ q and ¯ pB ¯ q respectively. Since we don’t know α and β , but we do know they are positive, divide both sides of ( i 0 ) by α and ( ii 0 ) by β . Replace ¯ q/α by y and ¯ p/β by x T and add slack vectors to get u + Ay = e, u 0 , y 0 , B T x + v = f, x 0 , v 0 , and (1) u i x i = 0 for all i, v j y j = 0 for all j. (2) (Here e and f denote column vectors of all ones; e
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This note was uploaded on 03/18/2012 for the course ORIE 4350 taught by Professor Shmoys during the Spring '08 term at Cornell University (Engineering School).

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cp - OR 4350 Game Theory Prof. Bland March 1, 2012 A...

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