primal-dual-ex

primal-dual-ex - Example(using the dual to solve the...

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Example (using the dual to solve the primal) (P) Min z = 2x 1 + 3x 2 + 5x 3 + 2x 4 + 3x 5 s.t. x 1 + x 2 + 2x 3 + x 4 + 3x 5 4 y1 2x 1 2x 2 + 3x 3 + x 4 + x 5 3 y2 x i 0 (D) Max w = 4y 1 + 3y 2 s.t. y 1 + 2y 2 2 x1 y 1 2y 2 3 x2 2y 1 + 3y 2 5 x3 y 1 + y 2 2 x4 3y 1 + y 2 3 x5 y i 0 Graphically, the optimal dual solution is y* 1 = 5 4 , y* 2 = 5 3 , and w* = ( 4 x 5 4 ) + ( 3 x 5 3 ) = 5 = z* . Complementary slackness implies that x 2 *( y 1 * 2y 2 * 2 ) = 0 , x 3 * ( 2y 1 * + 3y 2 * 5 ) = 0 , x 4 * ( y 1 * + y 2 * 2 ) = 0 , x* 2 = 0, x* 3 = 0, and x* 4 = 0 (since dual constraints (2), (3) and (4) are not binding). y1 y2 1 2 2 5 3 1 3 5 3 -1 2 3 4 ( 4, 3 ) 2 c

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In addition, y* 1 ( x* 1 + x* 2 + 2x* 3 + x* 4 + 3x* 5 4 ) = 0 , y* 2 ( 2x* 1 - 2x* 2 + 3x* 3 + x* 4 + x* 5 3 ) = 0 , x* 1 + 3x* 5 = 4 , 2x* 1 + x* 5 = 3 , x* 1 = 1 and x* 5 = 1. Then, the optimal Solution is x* 1 = 1, x* 2 = 0, x* 3 = 0, x* 4 = 0, x* 5 = 1, and Z* = 5 . In tabular form Max w = 4y 1 + 3y 2 s.t. y 1 + 2y 2 + S 1 = 2 x1 y 1 2y 2 + S 2 = 3 x2 2y 1 + 3y 2 + S 3 = 5 x3 y 1 + y 2 + S 4 = 2 x4 3y 1 + y 2 + S 5 = 3 x5 y i 0 y 1 y 2 S 1 S 2 S 3 S 4 S 5
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primal-dual-ex - Example(using the dual to solve the...

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