s02lec16b

s02lec16b - 15.053 Example: Fire company location....

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1 15.053 Thursday, April 11 z Some more applications of integer programming z Cutting plane techniques for getting improved bounds Handouts: Lecture Notes 2 Example: Fire company location. z Consider locating fire companies in different districts. z Objective: place fire companies so that each district either has a fire company in it, or one that is adjacent, and so as to minimize cost. 3 Example for the Fire Station Problem 123 4 56 7 89 11 10 12 14 15 13 16 Let x j = 1 if a fire station is placed in district j. x j = 0 otherwise. Let c j = cost of putting a fire station is district j. With partners, formulate the fire station problem. 4 Set covering problem z set S= {1, …, m} of items to be covered districts that need to have a fire station or be next to a district with one z n subsets of S For each possible location j for a fire station, the subset is district j plus the list of districts that are adjacent to district j. a ij = 1 if district i is adjacent to district j or if i = j. jj j cx Minimize 1 for each i ij j j ax is binary for each j j x subject to 5 Covering constraints are common z Fleet routing for airlines: Assigning airplanes to flight legs. Each flight must be included z Assigning crews to airplanes Each plane must be assigned a crew z Warehouse location Each retailer needs to be served by some warehouse 6 Independent Set (set packing) 4 7 11 10 12 14 15 13 16 What is the maximum number of districts no two of which have a common border? With your partner, formulate the independent set problem. List all constraints containing x 10 .
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7 Set Packing constraints arise often in manufacturing and logistics z Cutting shapes out of sheet metal z Manufacturing many items at once which share resources (how much work can be done in parallel?) 8 Review from Last Lecture 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 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 9 Branch and Bound 1 2 3 x 1 = 0 x 1 = 1 44 44 44 4 x 2 = 0 42 5 x 2 = 1 6 x 2 = 0 7 x 2 = 1 44 44 44 The incumbent solution has value 43 x 3 = 0 x 3 = 1 x 3 = 0 x 3 = 1 12 x 3 = 0 13 x 3 = 1 43 43 43 43 44 Inf 8 9 10 11 Inf Inf 14 15 16 17 44 44 18 19 Inf 38 10 On Bounds from Linear Programming We found an incumbent with a value of 43. The best LP solution value is between 44 and 45. Is there a way that we could have directly established an upper bound between 43 and 44? Perhaps we could form a “better linear program.” The closer the LP is to the Integer Program, the better.
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s02lec16b - 15.053 Example: Fire company location....

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