mthsc810-lecture13-1x2

mthsc810-lecture13-1x2 - MthSc 810: Mathematical...

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Unformatted text preview: MthSc 810: Mathematical Programming Lecture 13 Pietro Belotti Dept. of Mathematical Sciences Clemson University October 13, 2011 Reading for today: Section 4.5 Reading for Oct. 20: Section 4.6-4.7 Midterm 2: November 1, 2011, 2pm Dual feasibility Consider a primal LP and its dual: max c ⊤ x min u ⊤ b s . t . A x ≤ b s . t . u ⊤ A ≥ c ⊤ x ≥ u ≥ Add slack variables to the primal, surplus 1 variables to the dual. max c ⊤ x min u ⊤ b s . t . A x + x ′ = b s . t . u ⊤ A − u ′ = c ⊤ x , x ′ ≥ u , u ′ ≥ Note: primal ≡ − min {− c ⊤ x : A x + x ′ = b , ( x , x ′ ) ≥ } . 1 Called “surplus” because we subtract from the left-hand side. Example max 3 x 1 + 2 x 2 + x 3 min 4 u 1 + 6 u 2 s . t . x 1 + x 2 + x 3 ≤ 4 s . t . u 1 + 2 u 2 ≥ 3 2 x 1 + x 2 + 3 x 3 ≤ 6 u 1 + u 2 ≥ 2 x 1 , x 2 , x 3 ≥ u 1 + 3 u 2 ≥ 1 u 1 , u 2 ≥ become max 3 x 1 + 2 x 2 + x 3 ( ≡ − min − 3 x 1 − 2 x 2 − x 3 ) s . t . x 1 + x 2 + x 3 + x 4 = 4 2 x 1 + x 2 + 3 x 3 + x 5 = 6 x 1 , x 2 , x 3 , x 4 , x 5 ≥ min 4 u 1 + 6 u 2 s . t . u 1 + 2 u 2 − u 3 = 3 u 1 + u 2 − u 4 = 2 u 1 + 3 u 2 − u 5 = 1 u 1 , u 2 , u 3 , u 4 , u 5 ≥ Primal and dual dictionaries Make the primal a min problem. Then for B = { 4 , 5 } , x 4 = 4 − x 1 − x 2 − x 3 x 5 = 6 − 2 x 1 − x 2 − 3 x 3 z = − 3 x 1 − 2 x 2 − x 3 . Feasible 2 solution ( 4 , 6 ≥ 0) but not optimal ( − 3 , − 2 , − 1 < 0). In the dual, consider the basis ( u 3 , u 4 , u 5 ) and the dictionary u 3 = − 3 + u 1 + 2 u 2 u 4 = − 2 + u 1 + u 2 u 5 = − 1 + u 1 + 3 u 2 w = 4 u 1 6 u 2 . Infeasible solution ( − 3 , − 2 , − 1 < 0). If it wasn’t, it would be optimal (the reduced costs are 4 , 6 ≥ 0)....
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This note was uploaded on 03/14/2012 for the course MTHSC 810 taught by Professor Staff during the Fall '08 term at Clemson.

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mthsc810-lecture13-1x2 - MthSc 810: Mathematical...

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