# lec20 - 15.053 Thursday May 3 z Cutting plane techniques...

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1 15.053 Thursday, May 3 z Cutting plane techniques for getting improved bounds

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2 Quick Review of Branch and Bound maximize 16x 1 + 22x 2 + 12x 3 + 8x 4 +11x 5 + 19x 6 subject to 5x 1 + 7x 2 + 4x 3 + 3x 4 +4x 5 + 6x 6 14 x j binary for j = 1 to 6 Budget: 14 IHTFP points. Prize P o i n t s 5 7 4 3 6 Utility 16 22 12 8 11 19 1 2 3 4 5 6 iPod server brass rat 6.041 15.053 gift certificate AS you will see, it is a very quick review.
18 19 4 8 9 10 11 13 15 17 3 Branch and Bound 44 43 - 44 44 44 z i = 43 1 2 3 x 1 = 0 x 1 = 1 44 44 44 x 2 = 0 42 5 x 2 = 1 6 x 2 = 0 7 x 2 = 1 4 x 3 = 0 x 3 = 1 x 3 = 0 x 3 = 1 12 x 3 = 0 x 3 = 1 43 43 43 44 9 10 11 13 8 14 16 44 - 19 38 18 - 17 - 15 Explore subproblems one at a time. Fathom node j if z LP (j) z i . The only thing I wanted to focus on was that we fathom a node j of the branch and bound tree whenever z LP (j) z i .

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On Improving B&B Through Better Bounds Explore subproblems one at a time. Fathom node j if z LP (j) z i . Way to speed up things: fathom nodes sooner! Strategy 1: Improve z i . This means seeking out better solutions either at the beginning of B&B or during it. Strategy 2: Improve z LP (j). This means coming up with better bounds. What is a better bound? How does one get it? The branch and bound tree is huge. With 100 binary variables, it is of size 2 100 . To give you a sense for how big that is, the universe has only existed for less than 2 60 seconds. The only chance for branch and bound to work with 100 binary variables is to fathom nodes that are not far from the root node (that is node 1) of the tree. Improving the quality of an incumbent is very important. But here, we will focus on improving the bound. 4
On Bounds from Linear Programming Better Bounds means bounds that are closer to the true optimum integer value. For maximization problems we are computing upper bounds. But we want the upper bound to be as low as possible. 44 < z LP (1) < 45. We cannot fathom node 1, even if we start with the optimum solution as the incumbent. NEXT: an IP formulation s.t. z LP (1) < 44. We could fathom node 1, if z i = 43. We will next show that we can easily improve the bound for the IHTFP game show problem. Moreover, the branch and bound algorithm can possibly terminate after looking only at node 1 of the tree. 5

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6 Budget: 14 IHTFP points. Prize P o i n t s 5 7 4 3 6 1 2 3 4 5 6 Claim: Nooz can select at most two of prizes 1, 2, 3 and 6. (Check it out). Therefore: we can add the constraint x 1 + x 2 + x 3 + x 6 2 to the original IP and it is still a correct IP model. It has the same set of integer solutions as before. iPod server brass rat 6.041 15.053 gift certificate We note that the number of points needed for prizes 1, 2, 3 and 6 are such that at most two of the prizes can be selected. Any three of the prizes would use up at least 15 IHTFP points. So, we can add the constraint x 1 + x 2 + x 3 + x 6 2. This new constraint does not eliminate any feasible integer solutions, and so we can add it to the integer program.
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## This note was uploaded on 01/17/2012 for the course MGMT 15.053 taught by Professor Jamesorli during the Spring '07 term at MIT.

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lec20 - 15.053 Thursday May 3 z Cutting plane techniques...

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