1. Lecture 6 Knapsack Problem Quadratic Assignment Problem1

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
Cut Patterns x1 x2 x3 x4 x5 x6 x7 x8 x9
x10
x11
x12
x13
x14
x15
1.5-meter segment 1 2 4 3
2-meter segment
4-meter segment
trim loss (meter)
1.5
3.5 5
2.5
5.5 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
Get dual variables j of the BFS
Resolve the LP to get a new BFS
Solve a knapsack problem to get a new, valuable cutting pattern
any new
cutting pattern?
Add a cutting pattern (i.e., new column) to the LP
Yes
Stop. Current BFS is optimum No 13

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

15. Quadratic Assignment Problem15

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
Door A
Door B
Door a
Door b
Inbound
Doors
Outbound
Amount of Goods from Suppliers to Retailers
Retailer 1
Retailer 2
Supplier 1 7
Supplier 2 4 6
Distances Among the Doors
Door a
Door b
Door A 2 3
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
Door A
Door B
Door a
Door b
Inbound
Doors
Outbound
Distances Among the Doors
Door a
Door b
Door A 2 3
Door B 3 2 A B a b
total goods-distance
= (2)(7) + (3)(4) + (2)(6)
Amount of Goods from Suppliers to Retailers
Retailer 1
Retailer 2
Supplier 1 7
Supplier 2 4 6
formulate an optimization to find the door assignment that minimizes the total goods-distance 4 6 7 A B a b 17

18. Formulation of the Door Assignment Problem
Distances Among the Doors
Door a
Door b
Door A 2 3
Door B
Cost Coefficients cijkl, for xij = ykl = 1
y1a
y1b
y2a
y2b
x1A 14 21
x1B
x2A 8 12 18
x2B
Amount of Goods from Suppliers to Retailers
Retailer 1
Retailer 2
Supplier 1 7
Supplier 2 4 6 18

19. Formulation of the Door Assignment Problem
y1a
y1b
y2a
y2b
x1A 14 21
x1B
x2A 8 12 18
x2B 19

20. Quadratic Assignment Problem
four sets S1, T1, S2, T2, S1 and T1 of m items, Sn and Tn of 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 Problem21

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 Problem23

24. Quantities Shipped Among the Factories Distance Between the Cities
Example
Quantities Shipped Among the Factories
three factories a, b, c
three cities A, B, C a b c 
1 0 15 18
Distance Between the Cities A B C  14 8 11 24

25. Distance Between the Cities
Quantities Shipped
Among the Factories A B C  14 8 11
Example a b c 
1 0 15 18
aA
aB
aC
bA
bB
bC
cA
cB
cC 
140 80
210
120
110
165
252
144
198 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