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HW9_soln

# HW9_soln - 2< 40 20 = 60 Thus for b 2< 60 the current...

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IEOR 162 Spring 2009 Homework 9 6.1.1 Typical isoprofit line is 3x 1 + c 2 x 2 = z. This has slope -3/c 2 . If slope of isoprofit line is <-2, then Point C is optimal. Thus if -3/c 2 < -2 or c 2 < 1.5 the current basis is no longer optimal. Also if the slope of the isoprofit line is >-1 Point A will be optimal. Thus if -3/c 2 > -1 or c 2 > 3 the current basis is no longer optimal. Thus for 1.5 c 2 3 the current basis remains optimal. For c 2 = 2.5, x 1 = 20, x 2 = 60, but z = 3(20) + 2.5(60) = \$210. 6.1.2 Currently Number of Available Carpentry Hours = b 2 = 80. If we reduce the number of available carpentry hours we see that when the carpentry constraint moves past the point (40,20) the carpentry and finishing hours constraints will be binding at a point where x 1 > 40. In this situation, b
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Unformatted text preview: 2 < 40 + 20 = 60. Thus for b 2 < 60 the current basis is no longer optimal. If we increase the number of available carpentry hours we see that when the carpentry constraint moves past (0,100) the carpentry and finishing hours constraints will both be binding at a point where x 1 < 0. In this situation b 2 > 100.Thus if b 2 > 100 the current basis is no longer optimal. Thus the current basis remains optimal for 60 ≤ b 2 ≤ 100. If 60 ≤ b 2 ≤ 100, the number of soldiers and trains produced will change. 6.5.3 min w = 5y 1 + 7y 2 + 6y 3 + 4y 4 s.t. y 1 + 2y 2 + y 4 ≥ 4 y 1 + y 2 + 2y 3 = -1 y 3 + y 4 = 2 y 1 , y 2 ≥ 0; y 3 ≤ 0; y 4 u.r.s. 6.5.4 max z = 6x 1 + 8x 2 s.t. x 1 + x 2 ≤ 4 2x 1- x 2 ≤ 2 2x 2 = -1 x 1 ≤ 0; x 2 u.r.s....
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