Lecture20 - Integer Programming IE418 Lecture 20 Dr Ted...

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Integer Programming IE418 Lecture 20 Dr. Ted Ralphs
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IE418 Lecture 20 1 Reading for This Lecture Nemhauser and Wolsey Sections II.2.2 Wolsey Chapter 9
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IE418 Lecture 20 2 Valid Inequalities for the Knapsack Problem Consider the set S = { x B n | n j =1 a j x k b } where a Z n + and b Z + are positive integers. This is the feasible set for a 0-1 knapsack problem. Let C N be such that j C a j > b (called a dependent set ). Then the inequality X j C x j ≤ | C | - 1 is a valid inequality for S . These inequalities are known as cover inequalities .
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IE418 Lecture 20 3 Minimal Dependent Sets and Extended Cover Inequalities Consider again a knapsack set S . A dependent set is minimal if all of its subsets are independent. The extension of a minimal dependent set C is E ( C ) = C ∪ { k N \ C | a k a j for all j C } . Proposition 1. If C is a minimal dependent set, then X j E ( C ) x j ≤ | C | - 1 is a valid inequality for S . Under certain conditions, the extended cover inequalities are facet- defining for conv( S ) .
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IE418 Lecture 20 4 Facet-defining Cover Inequalities Proposition 2. If C is a minimal dependent set for S and ( C 1 , C 2 ) is any partition of C with C 1 ± = , then j C 1 x j ≤ | C 1 | - 1 is facet-defining for conv( S ( C 1 , C 2 )) , where S ( C 1 , C 2 ) = S ∩ { x B n | x j = 0 for j N \ C, x j = 1 for j C 2 } Hence, beginning with any minimal dependent set, we can use lifting to derive a variety of facet-defining inequalities.
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IE418 Lecture 20 5 Using Valid Inequalities for a Relaxation Why do we care about strong inequalities for the knapsack problem? Every IP has a family of relaxations that are knapsack problems.
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Lecture20 - Integer Programming IE418 Lecture 20 Dr Ted...

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