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Lecture 6 Knapsack Problem Quadratic Assignment Problem

Lecture 6 Knapsack Problem Quadratic Assignment Problem. Outline. knapsack problem quadratic assignment problem. Knapsack Problem. Knapsack Problem. very useful sub-problems books Knapsack Problems by Hans Kellerer, Ulrich Pferschy, and David Pisinger (2004)

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Lecture 6 Knapsack Problem Quadratic Assignment Problem

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  1. Lecture 6 Knapsack ProblemQuadratic Assignment Problem 1

  2. Outline knapsack problem quadratic assignment problem 2

  3. Knapsack Problem 3

  4. Knapsack Problem very useful sub-problems books Knapsack Problems by Hans Kellerer, Ulrich Pferschy, and David Pisinger (2004) Knapsack Problems: Algorithms and Computer Implementations by Silvano Martello and Paolo Toth (1990) 4

  5. Problem Statement n types of products weight of type i: ai value of type i: pi capacity of truck: b kg question: selection of products to maximize the total value in a truck special tpye: 0-1 knapsack problem: i  {0, 1} i = number of type-i products put in the truck 5

  6. Common Knapsack Problems project selection problems capital budgeting allocation problems inventory stocking problems … 6

  7. Multi-Dimensional Knapsack Problem n types of products weight of type i: ai volume of type i: vi value of type i: pi capacity of truck b kg c m3 question: selection of products to maximize the total value in a truck i = number of type-i products put in the truck 7

  8. Cutting-Stock Problem knapsack problem: a sub-problem in solving cutting-stock problem by column generation 7-meter steel pipes of a given diameter order 150 pieces of 1.5-meter segments 250 pieces of 2-meter segments 200 pieces of 4-meter segments objective: minimize the amount of trim off 8

  9. Formulation of the Cutting-Stock Problem decisions: how to cut the pipes three types of segments type-1 segment: 1.5 m type-2 segment: 2 m type-3 segment: 4 m aij = the number of j type segments produced by the ith cut pattern the ith cut pattern: (ai1, ai2, ai3)T with trim loss ti e.g., a1 = (a11, a12, a13)T = (0, 0, 1) and t1 = 3 9

  10. Formulation of the Cutting-Stock Problem totally 15 cutting patterns 10

  11. Formulation of the Cutting-Stock Problem xi = the # of steel pipes cut in the ith pattern 11

  12. Solution of the Cutting-Stock Problem for Simplex Method, the reduced cost of the ith variable (i.e., the ith cutting pattern) can be shown to relate to a Knapsack Problem max 1a1 + 2a2 + 3a3, s.t. 1.5a1 + 2a2 + 4a3 7, where j= value of the jth dual variable of the current basic feasible solution 12

  13. Solution of the Cutting-Stock Problem Formulate a LP Solve the LP to get a BFS Resolve the LP to get a new BFS Get dual variables jof the BFS Solve a knapsack problem to get a new, valuable cutting pattern any new cutting pattern? Yes Add a cutting pattern (i.e., new column) to the LP No Stop. Current BFS is optimum 13

  14. Solution of the Knapsack Problem dynamic program: more appropriate branch and bound: not as appropriate 14

  15. Quadratic Assignment Problem 15

  16. Door Assignment in a Distribution Center a DC with two-inbound and two-outbound doors cross-docking operations to handle goods of two suppliers and two retailers objective: to minimize the total goods-distance Amount of Goods from Suppliers to Retailers Inbound Doors Outbound Doors Door a Door A Distances Among the Doors Door b Door B 16

  17. Door Assignment in a Distribution Center goods-distance for supplier 1  Door A, supplier 2  Door B, retailer 1  Door a, retailer 2  Door b B b b a a B A A Distances Among the Doors 7 2 total goods-distance = (2)(7) + (3)(4) + (2)(6) Amount of Goods from Suppliers to Retailers 3 Inbound Doors Outbound Doors 4 3 6 2 Door a Door A formulate an optimization to find the door assignment that minimizes the total goods-distance Door b Door B 17

  18. Formulation of the Door Assignment Problem Distances Among the Doors Cost Coefficients cijkl, for xij = ykl = 1 Amount of Goods from Suppliers to Retailers 18

  19. Formulation of the Door Assignment Problem 19

  20. Quadratic Assignment Problem four sets S1, T1, S2, T2, S1 and T1 of m items, Sn and Tnof n items two groups of assignments the pairing of items in S1 and T1 the pairing of items in S2 and T2 cost of assigning item i S1, j T1, k S2, l T2 = cijkl 20

  21. Quadratic Assignment Problem 21

  22. Another Quadratic Assignment Problem two sets S and T, each of n items the pairing of items in S and T as in an assignment problem cost of pairing i S with j T and k S with l T: cijkl 22

  23. Another Quadratic Assignment Problem 23

  24. Example three factories a, b, c three cities A, B, C Quantities Shipped Among the Factories Distance Between the Cities 24

  25. Distance Between the Cities Example Quantities Shipped Among the Factories 25

  26. Example 26

  27. Linearization of the Quadratic Assignment Problem non-linear objective function with terms such as ijkl to linearize the non-linear term ijkl let ijkl = ij kl need to ensure that ijkl= 1 ij kl= 1 27

  28. Linearization of the Quadratic Assignment Problem ijkl= 1 ij kl= 1 ijkl= 1 ij = 1 and kl= 1 two parts ijkl= 1 ij = 1 and kl= 1 ij = 1 and kl= 1 ijkl= 1 tricks ijkl  ij and ijkl   kl ij + kl 1 + ijkl 28

  29. Linearization of the Quadratic Assignment Problem drawback of the method addition of one variable and three constraints for one cross-product n(n1)/2 cross products for nij’s any method to add less variables or less constraints 29

  30. Linearization of the Quadratic Assignment Problem three cross-products 212+31314 of four variables 1, 2, 3, 4 the previous method: 3 new variables and 9 new constraints another method with 1 new variable and 3 new constraints 30

  31. Linearization of the Quadratic Assignment Problem let w = 212+31314 = 1(22+334) hope to have: 1 = 1  w = 22+334, and 1 = 0  w= 0 possible values of w  {-1, 0, 1, 2, 3, 4, 5} 1 = 0  w = 0: w  51 how to model 1 = 1  w = 22+334? 31

  32. Linearization of the Quadratic Assignment Problem to model 1 = 1  w = 22+334 w = 22+334  w  22+334 and w  22+334 1 = 1  w  22+334 and w  22+334 when 1 = 1: two valid constraints w  22+334 and w  22+334 when 1 = 0: two redundant constraints w  22+334+551 and w  22+334+515 32

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