HW8Sol - Fall 2009 IEOR 160 Industrial Engineering...

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Fall 2009 IEOR 160 October 25, 2009 Page 1 of 2 HOMEWORK 8 SOLUTIONS Section 12.12 1. x 0 = [1 / 2 , 1 / 2] f ( x,y ) = [4 - 4 x - 2 y, 6 - 2 x - 4 y ]. Then f ( . 5 ,. 5) = [1 , 3]. Find d 0 by solving max z = d 1 + 3 d 2 subject to d 1 + 2 d 2 2 d 1 ,d 2 0 Optimal solution is d 0 = [0 , 1]. Choose x 1 = [ . 5 ,. 5] + t 0 [ - . 5 ,. 5] = [ . 5 - . 5 t 0 ,. 5 + . 5 t 0 ] where t 0 solves max f ( . 5 - . 5 t,. 5 + . 5 t ) = 3 . 5 + t - t 2 / 2 = g ( t ) subject to 0 t 1. Then g 0 ( t ) = 1 - t = 0 for t = 1. Since g 00 ( t ) < 0, t 0 = 1 and x 1 = [0 , 1]. Here z = f (0 , 1) = 4. f ( x 1 ) = [2 , 2]. We find d 1 by solving max z = 2 d 1 + 2 d 2 subject to d 1 + 2 d 2 2 d 1 ,d 2 0 Optimal solution is d 1 = [2 , 0]. Now x 2 = [0 , 1] + t 1 [2 , - 1] = [2 t 1 , 1 - t 1 ] where t 1 solves max f (2 t, 1 - t ) = 4 + 2 t - 6 t 2 = h ( t ) subject to 0 t 1. Then
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This note was uploaded on 02/17/2010 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at Berkeley.

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HW8Sol - Fall 2009 IEOR 160 Industrial Engineering...

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