HW6ans - 5 1 0 0 4 3 0 0 b = [3 2 6 6]' x = A\b; x = 1.0909...

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HW6 Problem 1 1(a):  See figure 1 (b): Vertices: (0,0), (0,2), (6/5,0), (12/11, 6/11). f(0,0) = 0 f(0,2) = -4 f(6/5,0) = -3.6 f(12/11, 6/11) = -4.363  * * objective function has the min. value at this vertex. 1(c): At vertex (12/11, 6/11) the Lagrange multipliers (lambdas) are 0.0909 and 0.6364 respectively.  At vertex (0,0) the Lagrange multipliers are negative, hence violated KKT condition 4. 
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Problem 2 2(a): At vertex (12/11, 6/11) the objective function takes on its lowest value. The constraints that are active  are 5x1 + x2 <= 6 and 4x1 + 3x2 <=6. 2(b): At the chosen vertex from part a, the Lagrange multipliers are in violation of KKT condition 4. The  Lagrange multipliers are -0.107 and 0.338 respectively. They are computed by:           2     0     5     4   A =   0     2     1     3 
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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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This note was uploaded on 02/06/2012 for the course CS 5555 taught by Professor Nihao during the Spring '11 term at Cosumnes River College.

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HW6ans - 5 1 0 0 4 3 0 0 b = [3 2 6 6]' x = A\b; x = 1.0909...

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