Linear programming applications in marketing finance

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Linear Programming Applications in Marketing, Finance and Operations Management FP 2714.28571 3.57692 SM 10000.00000 0.00000 SP 0.00000 0.00000 TM 0.00000 1.15385 TP 5000.00000 0.00000 Constraint Slack/Surplus Dual Value 1 0.00000 -2.69231 2 3171.42857 0.00000 3 7714.28571 0.00000 4 0.00000 47.42308 5 0.00000 15.00000 6 0.00000 7.50000 18. a. Let x 1 = number of Super Tankers purchased x 2 = number of Regular Line Tankers purchased x 3 = number of Econo-Tankers purchased Min 550 x 1 + 425 x 2 + 350 x 3 s.t. 6700 x 1 + 55000 x 2 + 4600 x 3 600,000 Budget 15(5000) x 1 + 20(2500) x 2 + 25(1000) x 3 550,000 or 75000 x 1 + 50000 x 2 + 25000 x 3 550,000 Meet Demand x 1 + x 2 + x 3 15 Max. Total Vehicles x 3 3 Min. Econo-Tankers x 1 1/2( x 1 + x 2 + x 3 ) or 1/2 x 1 - 1/2 x 2 - 1/2 x 3 0 No more than 50% Super Tankers x 1 , x 2 , x 3 0 Solution: 5 Super Tankers, 2 Regular Tankers, 3 Econo-Tankers Total Cost: \$583,000 Monthly Operating Cost: \$4,650 b. The last two constraints in the formulation above must be deleted and the problem resolved. The optimal solution calls for 7 1/3 Super Tankers at an annual operating cost of \$4033. However, since a partial Super Tanker can't be purchased we must round up to find a feasible solution of 8 Super Tankers with a monthly operating cost of \$4,400. Actually this is an integer programming problem, since partial tankers can't be purchased. We were fortunate in part (a) that the optimal solution turned out integer. 4 - 17
Chapter 4 The true optimal integer solution to part (b) is x 1 = 6 and x 2 = 2 with a monthly operating cost of \$4150. This is 6 Super Tankers and 2 Regular Line Tankers. 19. a. Let x 11 = amount of men's model in month 1 x 21 = amount of women's model in month 1 x 12 = amount of men's model in month 2 x 22 = amount of women's model in month 2 s 11 = inventory of men's model at end of month 1 s 21 = inventory of women's model at end of month 1 s 12 = inventory of men's model at end of month 2 s 22 = inventory of women's model at end of month The model formulation for part (a) is given. Min 120 x 11 + 90 x 21 + 120 x 12 + 90 x 22 + 2.4 s 11 + 1.8 s 21 + 2.4 s 12 + 1.8 s 22 s.t. 20 + x 11 - s 11 = 150 or x 11 - s 11 = 130 Satisfy Demand  30 + x 21 - s 21 = 125 or x 21 - s 21 = 95 Satisfy Demand  s 11 + x 12 - s 12 = 200 Satisfy Demand  s 21 + x 22 - s 22 = 150 Satisfy Demand  s 12 25 Ending Inventory  s 22 25 Ending Inventory  Labor Hours: Men’s = 2.0 + 1.5 = 3.5 Women’s = 1.6 + 1.0 = 2.6 3.5 x 11 + 2.6 x 21 900 Labor Smoothingfor  3.5 x 11 + 2.6 x 21 1100 Month 1  3.5 x 11 + 2.6 x 21 - 3.5 x 12 - 2.6 x 22 100 Labor Smoothingfor  -3.5 x 11 - 2.6 x 21 + 3.5 x 12 + 2.6 x 22 100 Month 2  x 11 , x 12 , x 21 , x 22 , s 11 , s 12 , s 21 , s 22 0 The optimal solution is to produce 193 of the men's model in month 1, 162 of the men's model in month 2, 95 units of the women's model in month 1, and 175 of the women's model in month 2. Total Cost = \$67,156
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