# Set5-1 - Shortest Paths 2: Application to lot-sizing...

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Shortest Paths 2: Application to lot-sizing

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2 Application 19.19. Dynamic Lot Sizing K periods of demand for a product. The demand is d j in period j. Assume that d j > 0 for j = 1 to K. Cost of producing p j units in period j: a j + b j p j h j : unit cost of carrying inventory from period j Question: what is the minimum cost way of meeting demand?
3 Application 19.19. Dynamic Lot Sizing 1 2 3 4 K-1 K 0 D -d 1 -d 2 -d 3 -d 4 -d K-1 -d K Flow on arc (0, j): amount produced in period j Flow on arc (j, j+1): amount carried in inventory from period j Lemma: There is production in period j or there is inventory carried over from period j-1, but not both.

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4 Lemma: There is production in period j or there is inventory carried over from period j-1, but not both. Suppose now that there is inventory from period j-1 and production in period j. Let period i be the last period in which there was production prior to period j, e.g., j = 7 and i = 4. Claim : There is inventory stored in periods i, i+1, …, j-1 4 5 6 7 K-1 K 0 D -d 4 -d 5 -d 6 -d 7 -d K-1 -d K
5 Thus there is a cycle C with positive flow. C = 0-4-5-6-7-0. Let x 07 be the flow in (0,7). The cost of sending units of flow around C is linear (ignoring the fixed charge for production). Let Q = b 4 + h 4 + h 5 + h 6 – b 7 . If Q < 0, then the solution can be improved by sending a unit of flow around C. If Q > 0, then the solution can be improved by decreasing flow in C by a little. If Q = 0, then the solution can be improved by increasing flow around C by x 07 units (and thus eliminating the fixed cost a 7 ). This contradiction establishes the lemma. 4 5 6 7 K-1 K 0 D -d 4 -d 5 -d 6 -d 7 -d K-1 -d K

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6 i i+1 i+2 j-1 j 0 -d i -d i+1 -d i+2 -d j-1 -d j Let c ij be the (total) cost of this flow. c ij = a i + b i (d i + d i+1 + … + d j-1 ) + h i (d i+1 + d i+2 + … + d j-1 ) + h i+1 (d i+2 + d i+3 + … + d j-1 ) + … h j-2 (d j-1 ) Corollary . Production in period i satisfies demands exactly in periods i, i+1, …, j-1 for some j. Consider 2 consecutive production periods i and j. Then production in period i must meet demands in i+1 to j-1.
7 1 2 3 4 K K+1 1 6 8 11 K+1 Interpretation: produce in periods 1, 6, 8 and 11. Each path from 1 to K+1 gives a production and inventory schedule. The cost of the path is the cost of the schedule. Let c ij be the cost of producing in period i to meet demands in periods i, i+1, …, j-1 (including cost of inventory). Create a graph on nodes 1 to K+1, where the cost of (i,j) is c ij . Conclusion : The minimum cost path from node 1 to node K+1 gives the minimum cost lot-sizing solution.

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8 Complexity of Dijkstra’s Algorithm O(m+nC): Dial’s implementation O(mlogn): Binary heap implementation O(m+nlogn): Fibonacci heap implementation O(m+nlog(nC)): Radix heap implementation
The Label Correcting Algorithm

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10 Overview of the Lecture A generic algorithm for solving shortest path problems negative costs permitted but no negative cost cycle (at least for now) The use of reduced costs All pair shortest path problem INPUT G = (N, A) with costs c Node 1 is the source node There is no negative cost cycle We will relax that assumption later
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## This note was uploaded on 04/15/2010 for the course INDUSTRIAL ie513 taught by Professor Zeynephuygur during the Spring '10 term at Bilkent University.

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Set5-1 - Shortest Paths 2: Application to lot-sizing...

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