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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online