mthsc810-lecture13-2x2

mthsc810-lecture13-2x2 - Dual feasibility MthSc 810...

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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 0 u 0 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 0 u , u 0 Note: primal ≡ − min {− c x : A x + x = b , ( x , x ) 0 } . 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 0 u 1 + 3 u 2 1 u 1 , u 2 0 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 0 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 0 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). 2

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mthsc810-lecture13-2x2 - Dual feasibility MthSc 810...

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