# Ch05 - 1 SOLUTIONS TO MP CHAPTER 5 PROBLEMS SECTION 5.1 1...

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SOLUTIONS TO MP CHAPTER 5 PROBLEMS SECTION 5.1 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. 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. 3. If b 3 , the demand for soldiers, is increased then the current basis remains feasible and therefore optimal. If, however, b 3 <20 then the point where the finishing and carpentry constraints are binding is no longer feasible (it has s 3 <0). Thus for b 3 >20 the current basis remains optimal. For b 3 >20 the point where the carpentry and finishing constraints are still binding remains at (20, 60) so producing 20 soldiers and 60 trains remains optimal. 4a. If isocost line is flatter than HIM constraint, point C is optimal. If isocost line is steeper than HIW, point B is optimal. Thus current basis remains optimal if -7/2 -c 1 /100 -1/6 or 50/3 c 1 350. 4b. Current basis remains optimal if -7/2 -50/c 2 -1/6 or 100/7 c 2 300. 1

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4c. If we decrease HIW requirement optimal solution moves towards D = (0, 2). We lose feasibility if HIW requirement is 2(2) = 4. Increase HIW requirement and optimal solution moves towards C = (12, 0). We lose feasibility when HIW requirement = 7(12) = 84. Thus current basis remains optimal for 4,000,000 HIW Requirement 84,000,000. If HIW = 28 + , the optimal solution is where 7x 1 + 2x 2 = 28 + and 2x 1 + 12x 2 = 24. This yields x 1 = 3.6 + .15 and x 2 = 1.4 -.025 . 4d. If we decrease HIM requirement optimal solution moves towards A = (4, 0). We lose feasibility if HIM requirement is 2(4) = 8. Increase HIM requirement and optimal solution moves towards B = (0, 14). We lose feasibility when HIM requirement = 14(12) = 168. Thus current basis remains optimal for 8,000,000 HIW Requirement 168,000,000. If HIM = 28 + , the optimal solution is where 7x 1 + 2x 2
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## This homework help was uploaded on 04/02/2008 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at Berkeley.

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Ch05 - 1 SOLUTIONS TO MP CHAPTER 5 PROBLEMS SECTION 5.1 1...

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