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

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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
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This note was uploaded on 09/16/2010 for the course IEOR 262A taught by Professor Don'trknow during the Fall '09 term at University of California, Berkeley.

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

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