# lp2 - 1 Math 1b Practical March 9 2009 Intoduction to...

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1 Math 1b Practical, March 9, 2009 Intoduction to Linear Programming. Duality Recall the ‘primal’ and ‘dual’ canonical forms for LP problems: (PRIMAL) CANONICAL: maximize cx subject to A x b and x 0 . DUAL CANONICAL: minimize yb subject to y A c and y 0 . (1) Here A is an m by n real matrix; x and b will be column vectors of lengths n and m , respectively; c and y will be row vectors of lengths n and m , respectively. (We do not mean to imply that m n even though it may appear that way below.) We think of A , b ,and c as ‘given’ and x and y as ‘unknowns’. A b x c y The idea of duality in linear programming is motivated by trying to bound an ob- jective function. Consider the following example of a canonical form LP problem: Find the maximum value M of 4 x +6 y +5 z +3 w subject to x +2 y z + w 7 , 2 x y + z w 20 , x, y, z, w 0 . To ±nd lower bounds for the optimal value of the objective function, we can play around by evaluating it on feasible vectors, assuming that we can guess some. For example, (1 , 1 , 1 , 1) is feasible, and we get M 18. We can do better than that: (1 , 2 , 0 , 2) is feasible and we get M 22. To get an upper bound, we can add the ±rst inequality to twice the second to get 5 x +8 y z w 47; then 4 x y z w 5 x y z w 47 . So M 47. We can do better: add the ±rst inequality to twice the second to get 4 x +7 y z +4 w 34; then 4 x y z w 4 x y z w 34 . So M 34. We get an upper bound 7 s +20 t of this type when we add s times the ±rst inequality to t times the second inequality where ( s, t ) is feasible for the dual canonical problem. Think about this. Anyway, we know at this point that 22 M 34. (The true answer is M = 28, by the way.)

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2 Theorem 1. If x and y satisfy the inequalities in (1) ,then cx yb . If there exist vectors x and y satisfying the inequalities in (1) , then there exist vectors x 0 and y 0 so that cx 0 = y 0 b .
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lp2 - 1 Math 1b Practical March 9 2009 Intoduction to...

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