E the negative gradient lies in the cone spanned by

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Unformatted text preview: ) = −c − Cx∗ = 37 16 , and the gradients of the active constraints are given by (−1, 1) and (1, 0) , so −1 1 + 53 , − f (x∗ ) = 16 1 0 i.e., the negative gradient lies in the cone spanned by the gradients of the active constraints. 3 (5.2) Note that the optimality conditions are Ax∗ ≤ b, − f (x∗ ) = A u, u ≥ 0, u (Ax∗ − b) = 0, For x∗ = (1, 1, 1) we have Ax∗ = (3, 5) = b, so the first (primal feasibility) and last (complementary slackness) conditions above are satisfied. Therefore, we need only check the dual feasibility condition. Here, we have − f (x∗ ) = −c − Cx∗ = −c − (1, 3, 4) , so dual feasibility becomes 1 u1 + 2u2 , u ≥ 0. −c − 3 = u1 4 u1 + 3u2 When c = (−4, −4, −8) , this becomes 3 u1 + 2u2 1 = u1 , 4 u1 + 3u2 which is satisfied with u = (1, 1) , and u ≥ 0 so x∗ is indeed the optimal solution. When c = (0, −4, −2) , this becomes −1 u1 + 2u2 1 = u1 , −2 u1 + 3u2 which is only satisfied with u = (1, −1) . However, since u ≥ 0, this implies that x∗ is not optimal for this problem. When c = (1, 1, 1) , t...
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This document was uploaded on 04/10/2014.

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