{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW6_sol

# HW6_sol - IEOR 262A Mathematical Programming I Fall 2007...

This preview shows pages 1–2. Sign up to view the full content.

IEOR 262A Mathematical Programming I Fall 2007 Homework 6 Solution Exercise 4.2 Solution The dual is max p 0 b s.t. p 0 A c 0 p 0 Convert it into its equivalent minimization problem: max ( - b 0 ) p s.t. ( - A 0 ) p - c p 0 This problem is identical to the primal iff A 0 = - A and b = - c . For example, let A = 0 - 1 1 0 , b = 1 2 and c = - 1 - 2 . Exercise 4.4 Solution The dual is max p 0 c s.t. Ap c p 0 by the symmetry of A . If Ax * = c , then x * is both primal and dual feasible. Moreover, it has the same primal and dual cost c 0 x * . Therefore, it is an optimal solution of both problems. Exercise 4.8 Solution (a) By the optimality of x * under c , we have c 0 x - x * ) 0; by the optimality of ˜ ( x ) under ˜ c , we have ˜ c 0 ( x * - ˜ x ) 0. Add the two inequality together, we get ( c 0 - ˜ c 0 )(˜ x - x * ) 0 or c - c ) 0 x - x * ) 0 . (b) By strong duality, c 0 x * = ( p * ) 0 ˜ x . By weak duality, since p * is feasible under ˜ b , we have ( p * ) 0 ˜ b c 0 ˜ x . ( p * ) 0 ˜ b - ( p * ) 0 b c 0 ˜ x - c 0 x * or ( p * ) 0 ( ˜ b - b ) c 0 x - x * ) .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}