HW8_Sol - 2 = g(t ≤ t ≤ 1 Then g(t = 14 58t = 0 for t =...

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E160 Operations Research I Spring 2009 Homework #8 Solution Chapter 12 Section 12.12 1. x 0 = [1/2 1/2] T f(x,y) = [4 - 4x - 2y, 6 - 2x -4y] f(.5,.5) = [1 3] Find d 0 by solving max z = d1 + 3d2 st d1 + 2d2 2 d1,d2 0 Optimal solution is d 0 = [0 1]T Choose x 1 = [.5 .5] T + t 0 [-.5 .5] T = [.5 -.5t 0 .5 + .5t 0 ] where t 0 solves max f(.5 - .5t, .5 + .5t) = 3.5 + t - t 2 /2 = g(t). 0 t 1 Then g'(t) = 1 - t = 0 for t = 1. Since g"(t)<0, t 0 = 1 and x 1 = [0 1] T . Here z = f(0, 1) = 4. f(x 1 ) = [2 2]. We find d 1 by solving max z = 2d1 + 2d2 st d1 + 2d2 2 d1,d2 0 Optimal solution is d 1 = [2 0] T . Now x 2 = [0 1] T + t 1 [2 -1] T = [2t 1 1 - t 1 ] where t 1 solves max f(2t, 1 - t) = 4 + 2t - 6t 2 = h(t). Then h'(t) = 2 - 12t = 0 0 t 1 for t = 1/6. Since h"(t)<0, t 1 = 1/6 and x 2 = [1/3 5/6]. At this point z = 4.17. 2. x 0 = [1 0] T f(x,y) = [3y - 2x 3x - 2y] f(1, 0) = [-2 3] Find d 0 by solving max z = -2d1 + 3d2 st 3d1 + d2 4 d1,d2 0 Optimal solution is d 0 = [0 4]T Choose x 1 = [1 0] T + t 0 [-1 4] T = [1 - t 0 4t 0 ] where t 0 solves
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max f(1 - t, 4t) = -1 + 14t - 29t
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Unformatted text preview: 2 = g(t) ≤ t ≤ 1 Then g'(t) = 14 - 58t = 0 for t = 7/29. Since g"(t)<0, t = 7/29 and x 1 = [22/29, 28/29] T . Here z = f(22/29, 28/29) = .69. ∇ f(x 1 ) = [40/29 10/29]. We find d 1 by solving max z = 40d1/29 + 10d2/29 st 3d1 + d2 ≤ 4 d1,d2 ≥ Optimal solution is d 1 = [4/3 0] T . Now x 2 = [22/29 28/29] T + t 1 [50/87 -28/29] T = [22/29 + 50t/87 28/29 - 28t/29] where t 1 solves max f(22/29 + 50t/87, 28/29 - 28t/29) = .69+.46t - 2.927t 2 = h(t). ≤ t ≤ 1 Then h'(t) = .46 - 5.854t = 0 for t = .0786. Since g"(t)<0, t 1 = .0786 and x 2 = [.804 .890] T For this point z = .708. Review Problem 13. We want to show that q maximizes g(p), where g(p) = q[1 - (1-p) 2 ] + (1 - q)(1 - p 2 ). Then g'(p) = 2q(1-p) + (1-q)(-2p) = 0 for q = p. Since g''(p) = -2q - 2(1-q)<0, we know that choosing p = q will maximize g(p), as desired....
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This note was uploaded on 09/06/2009 for the course IEOR 160 taught by Professor Hochbaum during the Spring '07 term at Berkeley.

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HW8_Sol - 2 = g(t ≤ t ≤ 1 Then g(t = 14 58t = 0 for t =...

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