6_lp_duality

# 6_lp_duality - Duality Consider the following LP z = min 3...

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Unformatted text preview: Duality Consider the following LP: z = min 3 x 1 + 2 x 2 s . t . x 1 + 3 x 2 ≥ 4 2 x 1 + x 2 ≥ 1 x 1- x 2 ≥ 1 x 1 ,x 2 ≥ . Suppose ( x 1 ,x 2 ) (say, e.g. (2,1)) is a feasible solution. Then we know that z ≤ 3 x 1 + 2 x 2 (= 8), i.e. we get an upper bound. How can we get a lower bound (or a certificate of optimality)? Note that any feasible solution should satisfy 1 ≤ 2 x 1 + x 2 ≤ 3 x 1 + 2 x 2 . Therefore 1 is a lower bound on the optimum value. Let us take a weighted combination of the constraints with weights ( 2 3 , 1 2 , 1), to get the constraint: 8 3 x 1 + 3 2 x 2 ≥ 3 1 3 . Any feasible solution has to satisfy the above con- straint (why?), therefore z ≥ 3 1 3 . 1 Duality (contd.) In general, choose non-negative weights ( y 1 ,y 2 ,y 3 ) and combine the three inequalities to get ( y 1 + 2 y 2 + y 3 ) x 1 + (3 y 1 + y 2- y 3 ) x 2 ≥ 4 y 1 + y 2 + y 3 . Note that the weights have to be non-negative to ensure the inequality does not reverse. For 4 y 1 + y 2 + y 3 to be an lower bound, the weights must satisfy y 1 + 2 y 2 + y 3 ≤ 3 3 y 1 + y 2- y 3 ≤ 2 y 1 , y 2 , y 3 ≥ To obtain the greatest lower bound, we should chose weights that maximizes 4 y 1 + y 2 + y 3 , i.e. solve the following LP: max 4 y 1 + y 2 + y 3 s . t . y 1 + 2 y 2 + y 3 ≤ 3 3 y 1 + y 2- y 3 ≤...
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## This note was uploaded on 02/04/2012 for the course ECON 6673 taught by Professor Ahmed during the Spring '11 term at Georgia Tech.

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6_lp_duality - Duality Consider the following LP z = min 3...

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