1. EMIS 8373: Integer Programming
Column Generation
updated 18 April 2005

2. Example 1: Adapted from “Optimal Placement of ADD/DROP Multiplexers: Heuristic and Exact Algorithms” by Alain Sutter, Francis Vanderbeck, Laurence Wolsey Operations Research, Vol. 46 (5), pp , 1998

3. Formulation Sets Parameters N is the set of nodes in the demand graph
E  N  N is the set of edges in the demand graph
M is a set of candidate SONET rings to may be used to construct the network
Parameters
de is weight of edge e  E
b is the ring capacity
aej = 1 if demand e can be routed on ring j
cj is cost of ring j (the number of ADMs required)

4. Formulation Continued
Decision Variables
zj = 1 if ring j is select, and 0, otherwise.
BIP Formulation:
Constraint set (1) ensures that each demand is assigned to exactly one of the candidate rings.
Views the problem as edge partitioning

5. Example Demand Graph and Candidate Set with b = 183
{(1, 2)} c1 = 2
{(1, 3)} c2 = 2
{(1, 4)} c3 = 2
{(2, 3)} c4 = 2
{(2, 5)} c5 = 2
{(3, 4)} c6 = 2
{(3, 5)} c7= 2
{(4, 5)} c8 = 2
{(1, 2), (1, 3), (2, 3)} c9 = 3
{(1, 2), (1, 3), (1, 4), (2, 3), (3, 5), (4, 5)} c10 = 5 1 2 3 2 3 4 3 8 2 4 5 2
Optimal Solution:
z5 = z6 = z10 = 1
cost = 9 ADMs

6. Comments
The best solution for the given demand graph uses two rings and eight ADMs:
{(1, 2), (2, 3), (2, 5), (3, 5)}
{(1, 3), (1, 4), (3, 4), (4, 5)}
The edge-partitioning formulation cannot find this solution unless these candidate rings are given as part of the input.
Let BIP(M) be the edge-partitioning formulation for a given candidate set M.
The exact formulation is BIP(M*) where M* is the set of all feasible candidate rings.
For a given demand graph |M*| = O(2|E|)

7. Matrix Representation of BIP(M)
BIP(M*) could have up to 28 –1 = 255 columns depending on b.

8. LP Relaxation of BIP(M)
Dual Problem

9. Additional Notation
For a given basic feasible solution (BFS) for LP(M)
Let B denote the basis matrix
Let Aj denote column j of the constraint matrix A
Let cB denote the vector of objective coefficients of the basic variables
Let  denote the vector of corresponding dual variables
Recall that the reduced cost for variable zj is given by the formula cj- cB B-1 Aj = cj-  Aj.
- B is optimal if all variables have a non-negative reduced cost.

10. Reduced Cost Example 1 Suppose z1,z2, …, z8 are the basic variables.
The current BFS uses 16 ADMs.
Bringing ring 9 into the basis reduces this to 16 – 3 = 13

11. Reduced Cost Example 1 Suppose z1,z2, …, z8 are the basic variables
Each demand assigned to its own ring
Using ring 9 to cover three demands saves 3 ADMs.
The current BFS uses 16 ADMs.
Bringing ring 9 into the basis reduces this to 16 – 3 = 13

12. Reduced Cost Example 2 Suppose z1,z2, …, z8 are the basic variables.
The current BFS uses 16 ADMs.
Bringing ring 10 into the basis reduces this to 16 – 5 = 11

13. A Column-Generation Heuristic
Solve restricted LP master problem LP(M) and let B be the optimal basis matrix
LP(M*) is referred to as the linear programming master problem.
Look for a ring (column) j that is not in M, but would have a negative reduced cost if it were added to M
This is referred as as solving the pricing problem.
If j is found then add j to M and goto step 1.
Solve BIP(M)
At this point an optimal solution to LP(M) is also an optimal solution to LP(M*).

14. Feasible Rings with Negative Reduced Costs
Reduced cost for the new ring:
We want to find a ring with
negative reduced cost

15. Generating Feasible Rings
Let yij = 1 if the new ring contains edge (i,j), and 0, otherwise
Let xi = 1 if the new ring requires an ADM at node i
For the x’s and y’s to represent a feasible ring we need

16. Ring Generation BIP
This problem is NP-hard, but in practical terms is much easier to solve than BIP(M*).
If the optimal value of the objective function is zero then the optimal basis for LP(M) is optimal for L(M*).

17. Column Generation Flow Chart
Solve LP(M)
Add column to M
Solve the Pricing Problem
New column found ?
Yes No
Restore Integrality Constraints and Solve BIP(M)

18. Restricted LP Master Problem (LP(M)) for Iteration 12 z 1 + 3 4 5 6 7 8 9 s . t = ( d e a b w )

19. Optimal Solution for LP(M): Iteration 1

20. Pricing Problem for Iteration 1
Optimal Solution:
Objective function value = -7
Add ring 11 = {(1,3), (1, 4), (2, 3), (2, 5), (3, 5), (4,5)}

21. Restricted LP Master Problem (LP(M)) for Iteration 2z 1 + 3 4 5 6 7 8 9 s . t = ( d e a b w )

22. Optimal Solution for LP(M): Iteration 2

23. Pricing Problem for Iteration 2
Optimal Solution:
Objective function value = -7
Add ring 12 = {(1,2), (1, 4), (2, 3), (2, 5), (3, 5), (4,5)}

24. Restricted LP Master Problem (LP(M)) for Iteration 32 z 1 + 3 4 5 6 7 8 9 s . t x = ( d e a b w )

25. Optimal Solution for LP(M): Iteration 3

26. Pricing Problem for Iteration 3
Optimal Solution:
Objective function value = -5
Add ring 13 = {(1, 3), (2, 5), (3, 4), (3, 5), (4,5)}

27. Restricted LP Master Problem (LP(M)) for Iteration 42 z 1 + 3 4 5 6 7 8 9 s . t x = ( d e a b w )

28. Optimal Solution for LP(M): Iteration 4

29. Comments
Repeat until optimal solution to the pricing problem has objective function value zero.
In many cases, the pricing problem is NP-hard.
There is no guarantee that the column-generation procedure will generate all of the columns that are selected in the optimal solution to BIP(M*)
A Branch-and-Price procedure does column generation at each node of the branch-and-bound tree
The extra constraints added by branching usually complicate the pricing the problem.