s02lec15

s02lec15 - 15.053 Overview of Techniques for Solving...

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1 15.053 Tuesday, April 9 z Branch and Bound Handouts: Lecture Notes 2 Overview of Techniques for Solving Integer Programs z Enumeration Techniques Complete Enumeration list all “solutions” and choose the best Branch and Bound Implicitly search all solutions, but cleverly eliminate the vast majority before they are even searched Implicit Enumeration Branch and Bound applied to binary variables z Cutting Plane Techniques Use LP to solve integer programs by adding constraints to eliminate the fractional solutions. 3 Investment 1 2 3 4 5 6 Cash Required (1000s) $5 $7 $4 $3 $4 $6 NPV added (1000s) $16 $22 $12 $8 $11 $19 Capital Budgeting Example Investment budget = $14,000 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 4 Complete Enumeration z Systematically considers all possible values of the decision variables. If there are n binary variables, there are 2 n different ways. z Usual idea: iteratively break the problem in two. At the first iteration, we consider separately the case that x 1 = 0 and x 1 = 1. 5 An Enumeration Tree x 1 = 0 x 1 = 1 x 2 = 0 x 2 = 1 x 2 = 0 x 2 = 1 x 3 = 0 x 3 = 1 x 3 = 0 x 3 = 1 x 3 = 0 x 3 = 1 x 3 = 0 x 3 = 1 Original problem 6 On complete enumeration z Suppose that we could evaluate 1 billion solutions per second. z Let n = number of binary variables z Solutions times n = 30, 1 second n = 40, 17 minutes n = 50 11.6 days n = 60 31 years
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7 On complete enumeration z Suppose that we could evaluate 1 trillion solutions per second, and instantaneously eliminate 99.9999999% of all solutions as not worth considering z Let n = number of binary variables z Solutions times n = 70, 1 second n = 80, 17 minutes n = 90 11.6 days n = 100 31 years 8 Branch and Bound The essential idea: search the enumeration tree, but at each node 1. Solve the linear program at the node 2. Eliminate the subtree (fathom it) if 1. The solution is integer (there is no need to go further) or 2. The best solution in the subtree cannot be as good as the best available solution (the incumbent) or 3. There is no feasible solution 9 Branch and Bound 1 44 3/7 Solution at node 1: x 1 =1 x 2 = 3/7 x 3 = x 4 = x 5 = 0 x 6 =1 z = 44 3/7 Node 1 is the original LP Relaxation 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 0 x j 1 for j = 1 to 6 The IP cannot have value higher than 44 3/7. 10 Branch and Bound 1 2 x 1 = 0 44 3/7 44 Solution at node 2: x 1 = 0 x 2 = 1 x 3 = 1/4 x 4 = x 5 = 0 x 6 = 1 z = 44 Node 2 is the original LP Relaxation plus the constraint x 1 = 0. 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 0 x j 1 for j = 1 to 6, x 1 = 0 Branch and Bound 1 2 x 1 = 0 44 3/7 44 Node 3 is the original LP Relaxation plus the constraint x 1 = 1.
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This note was uploaded on 12/20/2011 for the course BUS 15.053 taught by Professor Prof.jamesorlin during the Spring '05 term at MIT.

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s02lec15 - 15.053 Overview of Techniques for Solving...

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