We223 - The optimal solution is then X13 = 30 X21 = 20 X24 = 10 X32 = 5 X34 = 25 Minimum total cost = $345 Note that store 2 receives 20 units less

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Chapter 5-Suggested Problems and Solution # 14 a. This is a transportation problem. Note that demand exceeds supply by 20 units. You need to draw three supply nodes (the warehouses) and four demand nodes (the stores, along with the arcs, costs on each arc, and supply and demand numbers corresponding to each node. b. MIN 5 X11 + 4 X12 + 6 X13 + 5 X14 +3 X21 + 6 X22 + 4 X23 + 4 X24 +4 X31 + 3 X32 + 3 X33 + 2 X34 ST -X11 - X12 - X13 - X14 = -30 -X21 - X22 - X23 - X24 = -30 -X31 - X32 - X33 - X34 = -30 +X11 + X21 + X31 + XD1 +20 +X12 + X22 + X32 + XD2 +25 +X13 + X23+ X33 + XD3 +30 +X14 + X24+ X34 + XD4 +35 X ij 0 c. See file: Prb5_23.xls The optimal solution is: X12 = 25, X14 = 5, X21 = 20, X23 = 10, X34 = 30, XD3 = 20. Minimum total cost = $285. Note that store 3 receives 20 units less than demanded. d. Assign arbitrarily large costs (such as $999) to the arcs representing these flows.
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Unformatted text preview: The optimal solution is then: X13 = 30, X21 = 20, X24 = 10, X32 = 5, X34 = 25. Minimum total cost = $345. Note that store 2 receives 20 units less than demanded. # 26 Notes: 0-1 = keep current equipment to use during the coming year 0-2 = trade-in current equipment immediately and use new equipment during the coming year b. See file: Prb5_13.xls The solution is: X01=X13= X37=1 with a minimum total cost of $67,825. c. The problem could be made more realistic by considering the tax savings associated with depreciation on the equipment. Also, the current problem does not consider the time value of money (i.e., we might attempt to minimize the net present value of the cash flows). Both of these considerations could be accommodated easily by altering the objective function coefficients. # 31 #34...
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This note was uploaded on 05/29/2010 for the course BKASD 546 taught by Professor Amsd;lkamsd during the Spring '98 term at 카이스트, 한국과학기술원.

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We223 - The optimal solution is then X13 = 30 X21 = 20 X24 = 10 X32 = 5 X34 = 25 Minimum total cost = $345 Note that store 2 receives 20 units less

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