Lecture 20110408

# Lecture 20110408 - IMSE2008 Operational Research Techniques...

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IMSE2008 Operational Research Techniques ecture 04/08 Lecture 04/08 Integer Programming Miao Song Dept of Industrial & Manufacturing Systems Engineering

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genda Agenda Linear Programming Duality Rules to Create Dual Weak Duality and Strong Duality omplementary Slackness Complementary Slackness Integer Programming Introduction to Integer Programming teger Programming Formulation Integer Programming Formulation How to Solve IP 4/8/2011 2
omplementary Slackness Complementary Slackness et x* denote a primal feasible solution Let x* denote a primal feasible solution Let y* denote a dual feasible solution Complementary slackness conditions For all i, if the i-th constraint is an inequality constraint that is not tight, then p* i = 0 If the x* j 0, then the j-th constraint of the dual problem is tight, and thus the reduced cost of x* j is 0 Theorem: If x* and p* are optimal primal and dual solutions, then they satisfy the complementary slackness conditions. Theorem: If x* is feasible for the primal, and if p* is feasible for the dual, and if the pair (x*,p*) satisfy complementary slackness conditions, then x* is optimal for the primal problem and p* is optimal for the dual problem 4/8/2011 3

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Complementary Slackness: An Alternative Representation min 5x 1 + 4x 2 + 2x 3 Primal Dual max 12p 1 + 8p 2 + 20p 3 s.t. 4x 1 + 2x 2 12 3x 1 + 4x 2 8 x x 2x 20 s.t. 4p 1 + 3p 2 + 4p 3 5 2p 1 + 4p 2 + p 3 4 p 2 4x 1 + x 2 + 2x 3 x 1 ,x 2 ,x 3 0 2p 3 p 1 ,p 2 ,p 3 0 = 3 = 1/4 x 1 3 x 2 = 0 x 3 = 4 p 1 1/4 p 2 = 0 p 3 = 1 x 1 ×(4p 1 + 3p 2 + 4p 3 –5)±=±0 x 2 ×(2p 1 + 4p 2 + 1p 3 –4)±=±0 p 1 ×(4x 1 + 2x 2 – 12) = 0 p 2 ×(3x 1 + 4x 2 –8) ±=±0 x x 0) = 0 4/8/2011 4 x 3 × (2p 3 – 2) = 0 p 3 × (4x 1 + x 2 + 2x 3 – 20) = 0
ual Simplex Dual Simplex implex: from feasibility to optimality Simplex: from feasibility to optimality BFS in each iteration, i.e., x is nonnegative but reduced cost is allowed to be negative Terminate when optimality condition (nonnegative reduced cost) or unboundedness condition is satisfied Dual Simplex: from optimality to feasibility Dual BFS in each iteration, i.e., reduced cost is nonnegative but x is allowed to be negative Terminate when feasibility condition (x nonnegative) or unboundedness condition of the dual is satisfied Why do we need dual simplex?

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Lecture 20110408 - IMSE2008 Operational Research Techniques...

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