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Unformatted text preview: 5 1 0 0 4 3 0 0 b = [3 2 6 6]' x = A\b; x = 1.0909 0.5455 -0.1074 0.3388 2(c): Yes, there is a feasible point other than a vertex where the objective function takes on an even lower value. It's computed by: 2 0 4 A= 0 2 3 4 3 0 b=[3 2 6]'; x=A\b x = 1.0200 0.6400 0.2400 This point satisfies KKT conditions 1 and 5 and the Lagrange multiplier (0.24) is positive which also satisfies KKT condition 4. It also passes KKT condition 3 and KKT condition 2 does not apply here....
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- Spring '11
- nihao
- Mathematical optimization, lagrange multipliers, Joseph Louis Lagrange, Karush–Kuhn–Tucker conditions
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