HW_Solution_Chapter 05_EMSE102-202

HW_Solution_Chapter 05_EMSE102-202 - SECTION 5.2 2a...

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1 SOLUTIONS TO OR CHAPTER 5 PROBLEMS SECTION 5.1 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 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.

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Unformatted text preview: SECTION 5.2 2a. Decision variables remain the same. New z-Value = Old z-value+10(88) = \$33,420 2b. Relevant Shadow Price is -\$20. Current basis remains optimal if demand is decreased by up to 3 cars, so Dual Price may be used to compute new z-value. New Profit = \$32,540 + (-2) (-20) = \$32,580. 4a. Reduced Cost for X12 is 1, so if cost of producing Product 1 at Plant 1 is less than or equal to 6 - 1 = \$5, we would produce product 2 at Plant 1. 4b. Shadow Price is \$2 and AD = 1000 so new z-value = 128,000- (-1000)2 = \$130,000. 4c. AI = \$1, so current basis remains optimal. Decision variables remain the same and new z- value = 128,000 + 4000(1) = \$132,000. 2 SECTION 5.3 2. Cannot answer this question because current basis is no longer optimal if one more machine is available....
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HW_Solution_Chapter 05_EMSE102-202 - SECTION 5.2 2a...

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