1. EMIS 8373: Integer Programming
Branch-and-Bound: Combinatorial Relaxations
Revised 11 April 2005

2. Example 1: Assignment Relaxation for TSP3 7 10 6 5

3. Assignment Problem 1 1 -1 3 1’ 3 1 2 2’ -1 5 7 5 6 1 3 3’ -1 7 7 5 610 7 1 4 4’ -1

4. Mapping Tour 1234 to an Assignment
assignment cost =
= 21.
cost of tour  2  3  4 1 = 21. 1 1 -1 1’ 3 1 2 2’ -1 6 1 3 3’ -1 7 5 1 4 4’ -1

5. Optimal Assignment 1 1 -1 3 1’ 3 1 2 2’ -1 1 3 3’ -1 7 4 7 1 4’ -1

6. Assignment Relaxation of TSP
Every TSP tour solution corresponds to a feasible assignment with the same objective function value:
Tour 1  2  3  4 1  x12 = x23 = x34 = x41 = 1
Tour 1  3  2  4 1  x13 = x32 = x24 = x41 = 1
Some assignments do not map back to tours
Assignment x12 = x21 = x34 = x41 = 1 gives sub-tours 1  2  1 and 3  4  3
Since every TSP tour is a feasible assignment and the cost of the tour = cost of the assignment, the assignment problem is a relaxation of TSP.

7. Branch-and-Bound with Assignment Relaxation
Solve assignment problems to get lower bound.
ZU(S) = 
ZL(S) = 20 S
Visit city 2 first.
S13
S14
S12
Visit city 4 first.
Visit city 3 first.
Branch on the next city in the tour.

8. Assignment Problem for S12-1 1’ 3 1 2 2’ -1 5 5 6 1 3 3’ -1 7 7 5 6 10 1 4 7 4’ -1

9. Optimal Assignment for S12-1 1’ 3 1 2 2’ -1 6 1 3 3’ -1 7 5 10 1 4 4’ -1

10. Branch-and-Bound with Assignment Relaxation
Solve assignment problems to get lower bound.
ZU(S) = 21
ZL(S) = 20 S
S13
S14
S12
ZU(S) = 21
ZL(S) = 21

11. Assignment Problem for S13-1 3 1’ 1 2 2’ -1 7 5 6 1 3 3’ -1 7 7 5 6 4 7 1 4’ -1

12. Optimal assignment for S13-1 3 1’ 1 2 2’ -1 7 1 3 3’ -1 7 6 1 4 4’ -1

13. Branch-and-Bound with Assignment Relaxation
Solve assignment problems to get lower bound.
ZU(S) = 21
ZL(S) = 20 S
S13
S14
S12
ZU(S12) = 21
ZU(S13) = 23
ZL(S12) = 21
ZL(S13) = 23
1234
1342

14. Assignment Problem for S14-1 3 1’ 1 2 2’ -1 5 5 6 1 3 3’ -1 7 7 6 10 4 7 1 4’ -1

15. Optimal assignment for S14-1 3 1’ 1 2 2’ -1 5 1 3 3’ -1 10 4 7 1 4’ -1

16. Branch-and-Bound with Assignment Relaxation
Solve assignment problems to get lower bound.
ZU(S) = 21
ZL(S) = 21 S
S13
S14
S12
ZU(S12) = 21
ZU(S13) = 23
ZU(S13) = 25
ZL(S12) = 21
ZL(S13) = 23
ZL(S13) = 25
1234
1342
1432
1234 is an optimal tour.

17. Example 2: Symmetric TSP
tij Table
Graph Representation 9 2 3 10 5 2 6 1 4 1 3 6 5 4 2

18. The 1-Tree Relaxation Given an graph G=(V, E)
Let E1 be the set of edges adjacent to node 1.
Let E2 = E \ E1 be the set of edges not adjacent 1.
Let H =(V\{1}, E2) be the subgraph induced by E2.
A 1-tree is subgraph T=(V,F) of G where F consists of a spanning tree of H and two edges from E1.

19. 1-Trees G 9
T1 is TSP tour of G
T2 is not a TSP tour of G
All tours of G are 1-trees
Not all 1-trees are tours
1-tree is a relaxation of TSP 2 3 10 5 2 6 1 4 1 3 6 5 4 2 9 2 3 2 3 10 10 2 6 T1 6 T2 1 1 6 5 4 5 4 2 2

20. Finding a Minimum-Cost 1-Tree9 2 10 1 3 5 4 6
1. Find MST in H=(V\{1}, E\E1)
2. Add shortest two edges in E1 2 3 2 5 1
Cost = 14 1 4 5 4 2

21. Branch-and-Bound Tree
ZU(S) = 
ZL(S) = 0 S
S{{1,3}}
S{{1,4}}
S{{2,5}}
S{{3,5}}
S{{4,5}}
Branching rule: Let S{F} be the set of solutions that do not use edges in the set F.

22. 1-Trees for S{{1,3}} and S{{1,4}}2 3 2 3 5 2 1 5 1 1 4 1 6 5 4 6 5 4 2 2
Cost = 18
Cost = 16

23. Optimal 1-Tree for S{{2,5}}9
1. Find MST in H=(V\{1}, E\E1) 2 3 10 5
2. Add shortest two edges in E1 2 6 1 4 3 6 5 4 2 2 3 2 3 5
Cost = 16 1 4 5 4 2

24. Optimal 1-Tree for S{{3,5}}9
1. Find MST in H=(V\{1}, E\E1) 2 3 10
2. Add shortest two edges in E1 2 6 1 4 1 3 6 5 4 2 2 3 2 1 6
Cost = 15 1 4 5 4 2

25. Optimal 1-Tree for S{{4,5}}9
1. Find MST in H=(V\{1}, E\E1) 2 3 10 5
2. Add shortest two edges in E1 2 6 1 4 1 3 6 5 4 2 3 2 3 5 1
Cost = 15 1 4 5 4

26. Branch-and-Bound Tree
ZU(S) = 
ZL(S) = 0 15 S
S{{1,3}}
S{{1,4}}
S{{2,5}}
S{{3,5}}
S{{4,5}} 15
ZL(S{{1,3}}) = 18
ZL(S{{2,5}}) = 16
ZL(S{{4,5}}) = 15
ZL(S{{1,4}}) = 16
ZU(S{{4,5}}) = 15
ZL(S{{3,5}}) = 15
Optimal TSP tour found at node S{{4,5}}.

27. Example 3: Symmetric TSP
tij Table
Graph Representation 9 2 3 10 6 2 5 1 4 1 3 6 5 4 2

28. Finding a Minimum-Cost 1-Tree9 2 10 1 3 5 4 6
1. Find MST in (V\{1}, E\E1)
2. Add shortest two edges in E1 2 3 2 1 5
Cost = 14 1 4 5 4 2

29. Branch-and-Bound Tree
ZU(S) = 
ZL(S) = 14 S
S{{1,3}}
S{{1,4}}
S{{2,5}}
S{{3,4}}
S{{4,5}}
Branching rule: Let S{F} be the set of solutions that do not use edges in the set F.

30. 1-Trees for S{{1,3}} and S{{1,4}}2 3 2 3 2 1 1 1 4 5 1 5 6 5 4 6 5 4 2 2
Cost = 18
Cost = 16

31. 1-Tree for S{{2,5}} 9
Let F = E \ (E1  {2,5})
Find MST in (V\{1}, F) 2 3 10 6 2
3. Add shortest two edges in E1 5 1 4 3 6 5 4 2 2 3 2 5
Cost = 16 1 4 3 5 4 2

32. 1-Tree for S{{3,4}} 9
Let F = E \ (E1  {3,5})
Find MST in (V\{1}, F) 2 3 10 6 2
3. Add shortest two edges in E1 1 4 1 3 6 5 4 2 2 3 2 6
Cost = 15 4 1 1 5 4 2

33. 1-Tree for S{{4,5}} 9
Let F = E \ (E1  {4,5})
Find MST in (V\{1}, F) 2 3 10 6 2
3. Add shortest two edges in E1 5 1 4 1 3 6 5 4 2 3 2 5
Cost = 15 4 1 1 3 5 4

34. Branch-and-Bound Tree
ZU(S) = 
ZL(S) = 14 S 15
S{{1,3}}
S{{1,4}}
S{{2,5}}
S{{3,4}}
S{{4,5}}
ZL(S{{1,3}}) = 18
ZL(S{{2,5}}) = 16
ZL(S{{4,5}}) = 15
ZL(S{{1,4}}) = 16
ZL(S{{3,4}}) = 15
We must continue branching since we don’t have an upper (primal) bound.
Try a depth-first strategy.