EMIS 8373: Integer Programming

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Presentation transcript:

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

Example 1: Assignment Relaxation for TSP 3 7 10 6 5

Assignment Problem 1 1 -1 3 1’ 3 1 2 2’ -1 5 7 5 6 1 3 3’ -1 7 7 5 6 10 7 1 4 4’ -1

Mapping Tour 1234 to an Assignment assignment cost = 3 + 6 +7 + 5 = 21. cost of tour 1  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

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

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.

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.

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

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

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

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

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

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

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

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

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.

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

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.

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

Finding a Minimum-Cost 1-Tree 9 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

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.

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

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

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

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

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}}.

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

Finding a Minimum-Cost 1-Tree 9 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

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.

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

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

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

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

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.