ORIE
Section 12 Notes

# Section 12 Notes - (p q NE of bimatrix game(A,B u Ay = e...

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(p, q) NE of bimatrix game (A,B) (x,y, u, v) solve u + Ay = e, u>=0, y>=0 B T x + v = f, x>=0, v>=0 u i x i = 0 for all i, v j y j = 0 for all j, And (x,y,u,v) not equal to (0,0, e, f)

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(p, q) NE of bimatrix game (A,B) (x,y, u, v) solve u + Ay = e, u>=0, y>=0 B T x + v = f, x>=0, v>=0 u i x i = 0 for all i, v j y j = 0 for all j, And (x,y,u,v) not equal to (0,0, e, f) Pivoting algorithm: 1. Start with (x,y, u, v) = (0,0,e, f) 2. Relax one of the complementarity conditions 3. While we don’t have a complementary solution 1. Find an entering variable (increase from 0 to a positive value) 2. Find a leaving variable (goes to zero)
(BoS): Find an NE of the following bimatrix game A\B t1 t2 t3 s1 1 3 0 0 2 -1 s2 0 0 3 1 0 0

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3 0 -1 0 1 0 A’ = 1 0 2 0 3 0 B’ =
3 0 -1 0 1 0 A’ = 1 0 2 0 3 0 B’ = 4 1 0 1 2 1 A = 1 0 2 0 3 0 B =

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We want to find a solution (other than x=0, y=0, u=e, v=f ) to: Ay + u = e, y>=0, u >= 0, B T x + v = f, x>=0, v>=0, u i x i = 0 for all i, v j y j = 0 for all j,
We want to find a solution (other than x=0, y=0, u=e, v=f ) to: Ay + u = e, y>=0, u >= 0, B T x + v = f, x>=0, v>=0, u i x i = 0 for all i, v j y j = 0 for all j, i.e., 4y1 + y2 + u1 = 1 y1 + 2y2 + y3 + u2 = 1 x1+ v1 = 1 3x2 + v2 = 1 2x1 +v3 = 1 y1, y2, y3, u1, u2 >=0, v1, v2, v3, x1, x2 >=0, u1 x1 =0, u2 x2=0 v1 y1 = 0, v2 y2 = 0, v3 y3 =0

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4y1 + y2 + u1 = 1 y1 + 2y2 + y3 + u2 = 1 x1+ v1 = 1 3x2 + v2 = 1 2x1 +v3 = 1 u1 x1 =0, u2 x2=0 v1 y1 = 0, v2 y2 = 0, v3 y3 =0 = = 0 0 1 1 1 1 1 0 0 0 x v u y Initial bfs: (Note: Complementary soln, but not NE)
4y1 + y2 + u1 = 1 y1 + 2y2 + y3 + u2 = 1 x1+ v1 = 1 3x2 + v2 = 1 2x1 +v3 = 1 u1 x1 =0, u2 x2=0 v1 y1 = 0, v2 y2 = 0, v3 y3 =0 = = 0 0 1 1 1 1 1 0 0 0 x v u y Initial bfs: (Note: Complementary soln, but not NE) u1 = 1 – 4y1 –y2 u2 = 1 – y1 -2y2 –y3 v1 = 1 –x1 v2 = 1 -3x2 v3 = 1 -2x1 u1 x1 =0, u2 x2=0 v1 y1 = 0, v2 y2 = 0, v3 y3 =0

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= = 0 0 1 1 1 1 1 0 0 0 x v u y Let’s relax the complementarity condition v1 y1 =0 Now, we can increase y1 By how much can we increase it? Iteration 1
= = 0 0 1 1 1 1 1 0 0 0 x v u y Let’s relax the complementarity condition v1 y1 =0 Now, we can increase y1 By how much can we increase it?

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• Fall '06
• TODD,M.
• ASCII, Complementarity, Complementarity theory, complementary solution, bimatrix

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