CH13RP - 1 MP CHAPTER 13 REVIEW PROBLEMS 1. ft(i) = xt(i) =...

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MP CHAPTER 13 REVIEW PROBLEMS 1. f t (i) = length of shortest path from node i to node 10. x t (i) = point to which we should travel from node i to attain f t (i). f 3 (6) = 3, x 3 (6) = 10 f 3 (7) = 4, x 3 (7) = 10 f 3 (8) = 2, x 3 (8) = 10 f 3 (9) = 3, x 3 (9) = 10 3 + f 3 (6) = 6* x 2 (3) = 6 or 7 f 2 (3) = min 2 + f 3 (7) = 6* 4 + f 3 (6) = 7 f 2 (4) = min 2 + f 3 (8) = 4* x 2 (4) = 8 5 + f 3 (9) = 8 2 + f 3 (7) = 6 f 2 (5) = min 2 + f 3 (8) = 4* x 2 (5) = 8 3 + f 2 (3) = 9 f 1 (1) = min 2 + f 2 (4) = 6* x 1 (1) = 4 Shortest path from node 1 to node 10 is 1-4-8-10. 4 + f 2 (3) = 10 f 1 (2) = min 2 + f 2 (4) = 6 1 + f 2 (5) = 5* x 2 (2) = 5 Shortest path from node 2 to node 10 is 2-5-8-10. 2a. Let f t (i) = minimum cost (including only setup and holding costs) incurred during months t, . .. 4 given that i units are on hand at the beginning of month t. Note that by the end of the problem 5 units will be produced (producing < 5 will result in some unmet demand while producing >5 will incur unnecessary production costs). This information limits the possible states during each period. Since exactly 5 units will be produced, it is unnecessary to know the per unit production cost( this assumes that the per unit production cost is the same during each period). Let x t (i) = a production level during month t attaining f t (i). Let c(x) = 4 for x>0 and c(0) = 0. Then f 4 (0) = f 4 (1) = $4, x 4 (0) = 2 x 4 (1) = 1 1
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f 4 (2) =$0, x 4 (2) = 0. Then f 3 (i) = min {c(x) + 2(i + x - 2) + f 4 (i + x - 2)} x where x must satisfy i + x - 2 0 and i + x 4(the latter condition ensures that no units will be leftover at the end of month 4) 4 + f 4 (0) = 8* (x 3 = 2) f 3 (0) = min 6 + f 4 (1) = 10 (x 3 = 3) 8 + f 4 (2) = 8* (x 3 = 4) x 3 (0) = 2 or 4 4 + f 4 (0) = 8* (x 3 = 1) x 3 (1) = 1 or 3 f 3 (1) = min 6 + f 4 (1) = 10 (x 3 = 2) 8 + f 4 (2) = 8* (x 3 = 3) 0 + f 4 (0) = 4* (x 3 = 0) f 3 (2) = min 6 + f 4 (1) = 10 (x 3 = 1) x 3 (2) = 0 8 + f 4 (2) = 8 (x 3 = 2) 2 + f 4 (1) = 6* (x 3 = 0) f 3 (3) = min 8 + f 4 (2) = 8 (x 3 = 1) x 3 (3) = 0 f 3 (4) = 4 + f 4 (2) = 4 (x 3 = 0) x 3 (4) = 0 f 2 (x) = min {c(x) + 2(i + x - 1) + f 3 (i + x - 1)} x where i + x - 1 0 and i + x 5. Thus 4 + f 3 (0) = 12* (x 2 = 1) f 2 (0) = min 6 + f 3 (1) = 14 (x 2 = 2) 8 + f 3 (2) = 12* (x 2 = 3) 10 + f 3 (3) = 16 (x 2 = 4) x 2 (0) = 1 or 3 12 + f 3 (4) = 16 (x 2 = 5) 0 + f 3 (0) = 8* (x 2 = 0) f 2 (1) = min 6 + f 3 (1) = 14 (x 2 = 1) 8 + f 3 (2) = 12 (x 2 = 2) x 2 (1) = 0 10 + f 3 (3) = 16 (x 2 = 3) 12 + f 3 (4) = 16 (x 2 = 4) 2 + f 3 (1) = 10* (x 2 = 0) f 2 (2) = min 8 + f 3 (2) = 12 (x 2 = 1) 10 + f 3 (3) = 16 (x 2 = 2) 12 + f 3 (4) = 16 (x 2 = 3) x 2 (2) = 0 4 + f 3 (2) = 8* (x 2 = 0) x 2 (3) = 0 f 2 (3) = min 10 + f 3 (3) = 16 (x 2 = 1) 12 + f 3 (4) = 16 (x 2 = 2) 6 + f 3 (3) = 12* (x 2 = 0) x 2 (4) = 0 2
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f 2 (4) = min 12 + f 3 (4) = 16 (x 2 = 1) f 2 (5) = 8 + f 3 (4) = 12* x 2 (5) = 0 Finally f 1 (1) = min {c(x) + f 2 (1 + x - 1)} x where 1 + x - 1 0 and 1 + x - 1
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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 University of California, Berkeley.

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CH13RP - 1 MP CHAPTER 13 REVIEW PROBLEMS 1. ft(i) = xt(i) =...

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