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Lecture 20110401

Lecture 20110401 - IMSE2008 Operational Research Techniques...

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IMSE2008 Operational Research Techniques Lecture 04/01 Duality Miao Song D t f I d t i l & M f t i Dept of Industrial & Manufacturing Systems Engineering
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Agenda D lit Duality Rules for creating a dual LP Weak duality and strong duality Complementary slackness 4/1/2011 2
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Primal and Dual Primal LP min z = –52x 1 – 30x 2 – 20x 3 s.t. 2x 1 + 4x 2 + 5x 3 + x 4 = 100 x + x + x + x = 30 max 52x 1 + 30x 2 + 20x 3 s.t. 2x 1 + 4x 2 + 5x 3 100 x + x + x 30 x 1 + x 2 + x 3 + x 5 10x 1 + 5x 2 + 2x 3 + x 6 = 204 x 1 ,x 2 ,x 3 ,x 4 ,x 5 ,x 6 0 x 1 + x 2 + x 3 10x 1 + 5x 2 + 2x 3 204 x 1 ,x 2 ,x 3 0 max 100p 1 + 30p 2 + 204p 3 s t 2p + p + 10p 52 Dual LP min 100p 1 + 30p 2 + 204p 3 s t 2p + p + 10p 52 s.t. 2p 1 + p 2 + 10p 3 –52 4p 1 + p 2 + 5p 3 –30 5p 1 + p 2 + 2p 3 –20 s.t. 2p 1 + p 2 + 10p 3 4p 1 + p 2 + 5p 3 30 5p 1 + p 2 + 2p 3 20 p 2p 20 p 1 0 p 2 0 p 2p p 1 ,p 2 ,p 3 0 4/1/2011 3 p 3 0
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More on Primal and Dual Primal LP x is feasible to the primal min z = –52x 1 – 30x 2 – 20x 3 s.t. 2x 1 + 4x 2 + 5x 3 + x 4 = 100 x + x + x + x = 30 p is feasible to the dual Which of these two values is larger? x 1 + x 2 + x 3 + x 5 10x 1 + 5x 2 + 2x 3 + x 6 = 204 x 1 ,x 2 ,x 3 ,x 4 ,x 5 ,x 6 0 –52x 1 – 30x 2 – 20x 3 100p 1 + 30p 2 + 204p 3 Wh t i th ti l max 100p 1 + 30p 2 + 204p 3 s t 2p + p + p 52 Dual LP What is the optimal solution to the dual? Shadow Prices! s.t. 2p 1 + p 2 + p 3 –52 4p 1 + p 2 + 5p 3 –30 5p 1 + p 2 + 2p 3 –20 What do we know about the optimal values of the primal and dual? p 2p 20 p 1 0 p 2 0 Opt value of primal Opt value of dual Opt value of primal = O t l f d l 4/1/2011 4 p 3 0 Opt value of dual
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Rules for Creating a Dual LP Primal LP (min) Dual LP (max) Objective Coefficients RHS Coefficients RHS Coefficients Objective Coefficients Column Coefficients Row Coefficients Row Coefficients Column Coefficients Primal LP (min) Dual LP (max) i-th Constraint b i i-th Variable 0 i th C t i t b i th V i bl 0 i-th Constraint i i-th Variable i-th Constraint = b i i-th Variable free j-th Variable 0 j-th Constraint c j j th Variable j th Constraint j-th Variable 0 j-th Constraint c j j-th Variable free j-th Constraint = c j 4/1/2011 5
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Example min 2x 1 – 3x 2 + 5x 3 max 10p 1 + 7p 2 + 15p 3 s.t. 2x 1 + 4x 2 + x 3 10 x 1 + x 2 + x 3 7 3 + 5 + 2 15 s.t. 2p 1 + p 2 + 3p 3 2 4p 1 + p 2 + 5p 3 –3 + + 2 5 p 1 p 2 3x 1 + 5x 2 + 2x 3 = 15 x 1 0, x 2 0, x 3 free p 1 + p 2 + 2p 3 = 5 p 1 0, p 2 0, p 3 free p 3 Primal LP (min) Dual LP (max) i-th Constraint b i i-th Variable 0 i th C t i t b i th V i bl 0 i-th Constraint i i-th Variable i-th Constraint = b i i-th Variable free j-th Variable 0
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