HW9_Sol - E160 Operations Research I Fall 2008 Homework#9...

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E160 Operations Research I Fall 2008 Homework #9 Solution Chapter 12 Section 12.12 1. By the method from the book: 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. By the method during discussion section: Iteration 1: Step 1: max z = d 1 + 3d 2 st -1 ≤d 1 1 -1 ≤d 2 1 => d0=(1,1). Step 2: x1=x0+t0*d0
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Find largest tbar such that 1/2+ tbar+2(1/2+tbar) ≤2 1/2+ tbar≥0 =>tbar=1/6 Max 4(1/2+t0)+6(1/2+t0)-2(1/2+t0)^2-2(1/2+t0)(1/2+t0)- 2(1/2+t0)^2= 5 + 10t0-6(1/2+t0)^2 s.t. 0≤t0≤1/6 =>t0=1/6 =>x1=(2/3, 2/3)
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