Handout13 - ENGRI 1101 Engineering Applications of OR Fall...

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Fall ’08 Handout 13 Lectures on Duality Theory 1 Motivation of a dual problem Consider the following LP: min 6 x 1 +4 x 2 2 x 3 s . t . 4 x 1 +2 x 2 + x 3 5 x 1 + x 2 3 x 2 + x 3 4 x 1 , x 2 , x 3 0 Without thinking too much about it, we can show that the value of this LP is at least 5. Why? We have, for any feasible solution ( x 1 , x 2 , x 3 ) : 6 x 1 + 4 x 2 + 2 x 3 4 x 1 + 2 x 2 + x 3 5 Hence, value of the LP is at least 5. We can do even better than this: 6 x 1 + 4 x 2 + 2 x 3 (4 x 1 + 2 x 2 + x 3 ) + 2( x 1 + x 2 ) 5 + 2 · 3 = 11 , so the value of the LP is at least 11. Can we do even better? 6 x 1 + 4 x 2 + 2 x 3 (4 x 1 + 2 x 2 + x 3 ) + ( x 1 + x 2 ) + ( x 2 + x 3 ) 5 + 3 + 4 = 12 , so the value of the LP is at least 12. How can we generalize and describe this process, and figure out what is the best lower bound on the value of the LP we can derive this way? Answer — use LP! Variables: let y i = number of “copies” of constraint i we will use Constraints: The number of “copies” of x 1 we create should be at most 6, so 4 y 1 + y 2 6 The number of “copies” of x 2 we create should be at most 4, so 2 y 1 + y 2 + + y 3 4 The number of “copies” of x 3 we create should be at most 2, so y 1 + y 3 2 Other constraints: y 1 , y 2 , y 3 0 What is the objective function? max 5 y 1 + 3 y 2 + 4 y 3 The resulting LP is called the dual LP. The original LP is called the primal LP. Given any feasible solution y 1 , y 2 , y 3 to the dual LP, and any feasible solution x 1 , x 2 , x 3 to the primal LP, how do their values relate to each other? 1
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This note was uploaded on 11/12/2009 for the course ENGRD 2700 taught by Professor Staff during the Spring '05 term at Cornell.

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Handout13 - ENGRI 1101 Engineering Applications of OR Fall...

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