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Unformatted text preview: COMPLEMENTARY SLACKNESS I We will state this theorem for the normal form: Let x 1 , x 2 , , x n be the variables and s 1 , s 2 , , s m be the slacks variables of the max problem. Let y 1 , y 2 , , y m be the variables and e 1 , e 2 , , e n be the excess variables of the min problem. then the theorem states that THEOREM Let x T = ( x 1 , x 2 , , x n ) be a feasible solution of the primal and y T = ( y 1 , y 2 , , y m ) be a feasible solution of the dual, then they are optimal solution for the respective problem if and only if y i s i = 0 for all i = 1 , , m and x j e j = 0 for all j = 1 , , n . Saigal (U of M) IOE 310 1 / 25 THE PRIMAL: max z = 60 x 1 +30 x 2 +20 x 3 slack s 1 : 8 x 1 +6 x 2 + x 3 48 slack s 2 : 4 x 1 +2 x 2 +1 . 5 x 3 20 slack s 3 : 2 x 1 +1 . 5 x 2 +0 . 5 x 3 8 x 1 x 2 x 3 Its DUAL is min w = 48 y 1 +20 y 2 +8 y 3 slack e 1 : 8 y 1 +4 y 2 +2 y 3 60 slack e 2 : 6 y 1 +2 y 2 +1 . 5 y 3 30 slack e 3 : y 1 +1 . 5 y 2 +0 . 5 y 3 20 y 1 y 2 y 3 Saigal (U of M) IOE 310 2 / 25 USE OF COMPLEMENTARY SLACKNESS I With slack variables the primal problem is: max 60 x 1 + 30 x 2 + 20 x 3 = z 8 x 1 + 6 x 2 + x 3 + s 1 = 48 4 x 1 + 2 x 2 + 1 . 5 x 3 + s 2 = 20 2 x 1 + 1 . 5 x 2 + 0 . 5 x 3 + s 3 = 8 We computed the optimal solution and found the following breakup of the optimal primal bfs: BV = { s 1 , x 3 , x 1 } NBV = { x 2 , s 2 , s 3 } Saigal (U of M) IOE 310 3 / 25 USE OF COMPLEMENTARY SLACKNESS I Show that z = 280 , x 1 = 2 , x 2 = 0 , x 3 = 8 is an optimal solution of the primal. Compute the slack variables. Set up the dual system complementary to this solution. Solve the dual system. Check if feasible for dual constraints. Argue that if dual is feasible then it must be optimal....
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This note was uploaded on 04/02/2008 for the course IOE 310 taught by Professor Saigal during the Spring '08 term at University of Michigan.
 Spring '08
 Saigal

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