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Unformatted text preview: ORIE 3300/5300 ASSIGNMENT 9 SOLUTION Fall 2008
Problem 1 (a) see ﬁle hw9.mod for AMPL input. AMPL returns that x∗ = [x1 , x2 , x3 , x4 , x5 ] = [−1, 0, 0, 2, 0] is an optimal solution to the original problem and the optimal value is −27. (b) Let y = [y1 , y2 , y3 ] be the dual variables, the dual problem (D) of (P) is min 6y1 3y1 − y1 2y1 y1 + − − + + + 10y2 y2 2y2 3y2 4y2 2y2 − + + − − − 7y3 3y3 2y3 y3 2y3 3y3 = ≥ ≥ ≥ ≥ 11 2 −3 −8 −12 s.t. − 2y1 y1 free, y2 ≥ 0, y3 free (c) Change the objective to ”bty” in AMPL and solve again. We have y1 = −1, y2 = 0, y3 = 3 optimal to (D) with objective value −27. Call this solution y ∗ . (d) First, check feasibilities. Plug in x∗ and y ∗ to all constraints of (P) and (D) respectively, we get that they are both feasible. Then, since cT x∗ = bT y ∗ = −27, then by certiﬁcate of optimality, x∗ and y ∗ are optimal to (P) and (D). Therefore, x∗ is optimal to the original problem. 1 ...
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This homework help was uploaded on 02/20/2009 for the course ORIE 3300 taught by Professor Todd during the Fall '08 term at Cornell University (Engineering School).
- Fall '08