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

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Shortest Paths 2: Application to lot-sizing
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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?
Background image of page 2
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.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 4
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
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.
Background image of page 6
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.
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 8
The Label Correcting Algorithm
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 62

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

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online