3 2 y 3 c 2 a 1 3 y 1 a 2 3 y 2 a 3 3 y 3 c 3 Lizhi Wang lzwangiastateedu IE

3 2 y 3 c 2 a 1 3 y 1 a 2 3 y 2 a 3 3 y 3 c 3 lizhi

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September 10, 2012 19 / 22
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Primal-dual table It’s important to memorize the primal-dual table. Here are some tips: max min Constraints variables The two sides are not symmetric: max or min makes a difference Define “natural” direction for constraints. For a max problem, it’s natural to have a bound from above, so is defined as the natural direction for constraints. Similarly, is the natural direction of constraints for a min problem. Constraints in the natural direction correspond to 0 dual variables; non-natural constraints correspond to 0 dual variables; equality constraints correspond to free dual variables; and vice versa. Lizhi Wang ([email protected]) IE 534 Linear Programming September 10, 2012 20 / 22
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Example 3 Find the dual min x 1 + 2 x 2 + 3 x 3 s . t . x 1 - 3 x 2 = 5 2 x 1 - x 2 + 3 x 3 6 x 3 4 x 1 0 x 2 0 x 3 free . max 5 y 1 + 6 y 2 + 4 y 3 s . t . y 1 free y 2 0 y 3 0 y 1 + 2 y 2 1 - 3 y 1 - y 2 2 3 y 2 + y 3 = 3 . Lizhi Wang ([email protected]) IE 534 Linear Programming September 10, 2012 21 / 22
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Example 4 Check the optimality of ( x * 1 = 5 , x * 2 = 0 , x * 3 = - 4 / 3) and ( y * 1 = - 1 , y * 2 = 1 , y * 3 = 0) . 1 x * is primal feasible 2 y * is dual feasible 3 primal and dual solutions are complementary: 0 = ( x * 1 - 3 x * 2 - 5) y * 1 free 0 (2 x * 1 - x * 2 + 3 x * 3 - 6) y * 2 0 0 (4 - x * 3 ) - y * 3 0 0 x * 1 (1 - y * 1 - 2 y * 2 ) 0 0 ≤ - x * 2 ( - 3 y * 1 - y * 2 - 2) 0 free x * 3 (3 y * 2 + y * 3 - 3) = 0 Lizhi Wang ([email protected]) IE 534 Linear Programming September 10, 2012 22 / 22
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