1 / 24

Lecture 19 Discrete Optimization Methods

Lecture 19 Discrete Optimization Methods. Branch-And-Bound Algorithm Also called Divide-And-Conquer 12.16 A linear inequality is a valid inequality for an ILP if it holds for all integer solutions. Valid inequalities are also called cutting planes. Example #1.

shelley
Download Presentation

Lecture 19 Discrete Optimization Methods

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 19 Discrete Optimization Methods • Branch-And-Bound Algorithm • Also called Divide-And-Conquer • 12.16 A linear inequality is a valid inequality for an ILP if it holds for all integer solutions. Valid inequalities are also called cutting planes.

  2. Example #1 There are only 4 feasible integer points.

  3. Example #1 Continued Area Cut Off There are only 4 feasible integer points. Valid Cut

  4. Perfect CutsCould Solve With LP Solver New Reduced Feasible Region All Extreme Points Are Integer

  5. Example #2 • Minimize v = 7x1 + 12x2 + 5x3 + 14x4 • Subject To • 300x1 + 600x2 + 500x3 + 1600x4> 700 (1) • x1, x2, x3, x4 binary • LP Relaxation • Minimize v = 7x1 + 12x2 + 5x3 + 14x4 • Subject To • 300x1 + 600x2 + 500x3 + 1600x4> 700 (1) • 0 < x1, x2, x3, x4< 1

  6. Example #2 Continued • LP Relaxation Optimal Solution Is • (x1, x2, x3, x4) = (0, 0, 0, 0.44) • Objective Value = v0 = 6.12 • Therefore, a lower bound on the ILP is 6.12. • Divide-And-Conquer Strategy • Branch on x4 by creating two new problems • One problem has x4 = 1 and the other has x4 = 0

  7. Example #2 Continued • Problem 1 • Minimize v = 7x1 + 12x2 + 5x3 + 14x4 • Subject To • 300x1 + 600x2 + 500x3 + 1600x4> 700 (1) • x1, x2, x3, x4 binary and x4 = 1 • Problem 2 • Minimize v = 7x1 + 12x2 + 5x3 + 14x4 • Subject To • 300x1 + 600x2 + 500x3 + 1600x4> 700 (1) • x1, x2, x3, x4 binary and x4 = 0

  8. Divide-And-Conquer • We could not solve the original problem, so we divided the problem into two smaller problems. • How are they smaller? • If we could solve Problem 1 and Problem 2, then we could take the best solution as the solution to the original problem. • Problems 1 and 2 are still hard because they are ILPs. However, we can solve the LP relaxations of Problems 1 and 2. These are relatively easy problems.

  9. Branch-And-Bound Tree 0 v0 =6.12 (0,0,0,0.44) x4 = 0 x4 = 1 2 1 v1 = 14, (0,0,0,1) Upper Bound Feasible for ILP Incumbent = Best Found So Far = (0,0,01) v2 = 9, (0,0.33,1,0)

  10. Current Status Problem 1 (0,0,0,1) Incumbent Upper Bound 14 Gap = (14-9)/14 = 36% Problem 2 (0,0.33,1,0) Lower Bound 9

  11. Branch On x2 • From Problem 2 and the solution to its LP relaxation, we create two new problems. • Problem 3 has x2 = 1 and Problem 4 has x2 = 0

  12. Problem 3 and 4 • Problem 3 • Minimize v = 7x1 + 12x2 + 5x3 + 14x4 • Subject To • 300x1 + 600x2 + 500x3 + 1600x4> 700 (1) • x1, x2, x3, x4 binary and x4 = 0, x2 = 1 • Problem 4 • Minimize v = 7x1 + 12x2 + 5x3 + 14x4 • Subject To • 300x1 + 600x2 + 500x3 + 1600x4> 700 (1) • x1, x2, x3, x4 binary and x4 = 0, x2 = 0

  13. The Strategy • If we could solve Problems 3 and 4, then the optimum for the original problem is the best solution from Problems 1, 3, and 4. • Why are Problems 3 and 4 still hard? • However, we can solve the LP relaxations of problems 3 and 4 relatively easy. • Why?

  14. The Branch-And-Bound Tree 0 x4 = 0 x4 = 1 2 1 x2 = 0 x2 = 1 4 3 v3 = 13, (0,1,0.2,0) v4 = 9.67, (0.67,0,1,0)

  15. Current Status Problem 1 (0,0,0,1) Incumbent 14 Gap = (14-9.67)/14 = 31% 13 Problem 3 (0,1,0.2,0) 9.67 Problem 4 (0.67,0,1,0) Lower Bound

  16. Branch On Problem 3 • Problem 5: Original plus x4 = 0, x2 = 1, x3 = 1 • Problem 6: Original plus x4 = 0, x2 = 1, x3 = 0

  17. Branch-And-Bound TreeIncumbent = (0,0,0,1) With v1 = 14 0 x4 = 0 2 1 x2 = 1 3 4 x3 = 1 x3 = 0 5 6 v5 = 17, (0,1,1,0) Fathom v6 = 14.33, (0.33,1,0,0) Fathom

  18. Current Status Problem 1 (0,0,0,1) Incumbent 14 Gap = (14-9.67)/14 = 31% 9.67 Problem 4 (0.67,0,1,0) Lower Bound

  19. Branch On Problem 4 • Problem 7: original plus x4 = 0, x2 = 0, x1 = 1 • Problem 8: original plus x4 = 0, x2 = 0, x1 = 0

  20. Branch-And-Bound TreeIncumbent = (0,0,0,1) With v1 = 14 0 x4 = 0 2 1 x2 = 0 3 4 x1 = 0 x1 = 1 5 6 7 8 infeasible Why? v7 = 11, (1,0,0.8,0)

  21. Current Status Problem 1 (0,0,0,1) Incumbent 14 Gap = (14-11)/14 = 21% 11 Problem 7 (1,0,0.8,0) Lower Bound

  22. Branch On Problem 7 • Problem 9: original plus x4 = 0, x2 = 0, x1 = 1, x3 = 1 • Problem 10: original plus x4 = 0, x2 = 0, x1 = 0, x3 = 0 • Both problem are easy!!!!!!!!!!! • Why?

  23. Branch-And-Bound TreeIncumbent = (0,0,0,1) With v1 = 14 0 x4 = 0 2 1 x2 = 0 3 4 x1 = 0 x1 = 1 5 6 7 8 x3 = 1 x3 = 0 New Incumbent 9 10 v9 = 12, (1,0,1,0) infeasible

  24. Current Status Problem 9 (1,0,1,0) Optimal Solution 12 We had to solve 10 LPs to solve an ILP with 4 binary variables! Complete enumeration = 24 = 16.

More Related