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Unformatted text preview: ORIE 435 2006 A Complementary Pivot Algorithm to Compute a Nash Equilibrium in Bimatrix Games Consider the bimatrix game ( A, B ) with A and B m × n matrices. Then (¯ p, ¯ q ) ∈ P × Q is a Nash equilibrium iff p T A ¯ q ≤ ¯ p T A ¯ q for all p ∈ P, and ¯ p T Bq ≤ ¯ p T B ¯ q for all q ∈ Q, or equivalently, ( i ) A i ¯ q ≤ ¯ p T A ¯ q for all i, ( ii ) ¯ p T B j ≤ ¯ p T B ¯ q for all j. Note that, if we add a constant to every entry of A to get A , and a constant to every entry of B to get B , then (¯ p, ¯ q ) is a Nash equilibrium of ( A, B ) iff it is one for ( A , B ). This is because for every p ∈ P , q ∈ Q , p T A q is the same constant used to form A from A more than p T Aq ; and similarly for B 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 ensure that ¯ p T A ¯ q and ¯ p T B ¯ 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 guarantees at least as much, and similarly for B . Example: We take A = " 5 1 1 1 1 2 # , B = " 1 3 2 1 # . We could eliminate player II’s second column, since it is strongly dominated by q = (3 / 5 , , 2 / 5) T , but we’ll leave it in to better illustrate the algorithm. If we did eliminate this column, we would have a 2 × 2 game which we could solve by plotting response curves. You can check that the result is the Nash equilibrium (¯ p = (2 / 5 , 3 / 5) T , ¯ q = (1 / 5 , , 4 / 5) T ). This gives payoffs 9 / 5 and 6 / 5 to players I and II respectively. We’ll find this using our complementary pivoting algorithm below. 1 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 ) A i ¯ q ≤ α for all i , with equality if ¯ p i > , ( ii ) ¯ p T B j ≤ β for all j , with equality if ¯ q j > . Then (i)–(ii) hold iff (i’)–(ii’) do, with α and β necessarily equal to ¯ p T A ¯ q and ¯ p T B ¯ q respectively. We now use the same trick we used in zero-sum games: we divide by α and β (they’re positive!), replace ¯ q/α by y and ¯ p/β by x and add slack vectors to get u + Ay = e, u ≥ , y ≥ , B T x + v = f, x ≥ , v ≥ , and (1) u i x i = 0 for all i, v j y j = 0 for all j. (2) (The last relations can be written more shortly as u T x = v T y = 0.) This looks very like the complementary slackness conditions for optimal- ity in linear programming, but with two different matrices A and B , and there is no objective function in sight! Note that we have written the slack variables u first in the ( u, y...
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This note was uploaded on 02/06/2011 for the course ORIE 4350 at Cornell.