Lec8 - IE521 Advanced Optimization Lecture 8 Dr Zeliha Akca December 2011 1 23 Hw3 and Midterm 1 Grades Hw3 is posted Midterm 1 was graded Max

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IE521 Advanced Optimization Lecture 8 Dr. Zeliha Akc ¸a December 2011 1 / 23
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Hw3 and Midterm 1 Grades I Hw3 is posted. I Midterm 1 was graded. I Max grade = 86 I Average = 44.3 2 / 23
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Recall Previous Lectures I Example: For the following LP, x 1 and x 2 are basic variables in the optimal tableau. Construct the optimal tableau. min - 3 x 1 - x 2 2 x 1 - x 2 2 - x 1 + x 2 4 x 1 , x 2 0 3 / 23
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Recall Lecture 7: Motivation of Duality I Allow the constraint to be violated for a price . Under a specific value of p , the presence or absence of the constraint does not affect the optimal cost I These prices can be found by solving a new linear program called dual problem . I We have seen the procedure to construct a dual problem from a Primal Problem: min c > x s.t. Ax = b x 0 I We relax constraints, form function g ( p ) , and construct and LP to maximize the value of g ( p ) , the dual problem: max p > b s.t. p > A c > 4 / 23
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Recall: How to Come Up with the Dual Problem I Consider the Primal Problem: min c > x s.t. Ax = b x 0 I Relaxing the constraints for a price p : min c > x + p > ( b - Ax ) s.t. x 0 I Define a function of p , g ( p ) which is the optimal cost to the new problem. g ( p ) = min x 0 ± c > x + p > ( b - Ax ) ² I g ( p ) is a lower bound for the optimal solution of primal problem. I To find the best lower bound maximize g ( p ) . I Dual problem: Maximize g ( p ) . 5 / 23
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Recall: Constructing the Dual Problem I We have, g ( p ) = min x 0 ± c > x + p > ( b - Ax ) ² = p > b + min x 0 ( c > - p > A ) x I Note that min x 0 ( c > - p > A ) x = ( 0 , if c > - p > A 0 > , -∞ , otherwise , I We only need to consider the first case. I Therefore, maximization of g ( p ) becomes the following dual problem: max p > b s.t. p > A c > 6 / 23
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Example min - x 1 - x 2 x 1 + 8 3 x 2 + x 3 = 4 x 1 + x 2 + x 4 = 2 x 1 , x 2 , x 3 , x 4 0 Construct function g ( p ) . Maximize g ( p ) and construct dual problem. 7 / 23
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From the Primal to the Dual We can dualize general LPs as follows PRIMAL minimize maximize DUAL b i 0 constraints b i 0 variables = b i free 0 c j variables 0 c j constraints free = c j 8 / 23
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From the Primal to the Dual min c > x s.t. a > i x b i i M 1 a > i x b i i M 2 a > i x = b i i M 3 x j 0 j N 1 x j 0 j N 2 x j free j N 3 max p > b s.t. p i 0 i M 1 p i 0 i M 2 p i free i M 3 p > A j c j j N 1 p > A j c j i N 2 p > A j = c j i N 3 9 / 23
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This note was uploaded on 01/09/2012 for the course IE 521 taught by Professor Zelihaakça during the Fall '11 term at Fatih Üniversitesi.

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Lec8 - IE521 Advanced Optimization Lecture 8 Dr Zeliha Akca December 2011 1 23 Hw3 and Midterm 1 Grades Hw3 is posted Midterm 1 was graded Max

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