Algorithms_Part12

# Algorithms_Part12 - S Dasgupta C.H Papadimitriou and U.V...

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Unformatted text preview: S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 221 Let’s investigate the issue by describing what we expect of these three multipliers, call them y 1 , y 2 , y 3 . Multiplier Inequality y 1 x 1 ≤ 200 y 2 x 2 ≤ 300 y 3 x 1 + x 2 ≤ 400 To start with, these y i ’s must be nonnegative, for otherwise they are unqualified to multiply inequalities (multiplying an inequality by a negative number would flip the ≤ to ≥ ). After the multiplication and addition steps, we get the bound: ( y 1 + y 3 ) x 1 + ( y 2 + y 3 ) x 2 ≤ 200 y 1 + 300 y 2 + 400 y 3 . We want the left-hand side to look like our objective function x 1 + 6 x 2 so that the right-hand side is an upper bound on the optimum solution. For this we need y 1 + y 3 to be 1 and y 2 + y 3 to be 6 . Come to think of it, it would be fine if y 1 + y 3 were larger than 1 —the resulting certificate would be all the more convincing. Thus, we get an upper bound x 1 + 6 x 2 ≤ 200 y 1 + 300 y 2 + 400 y 3 if    y 1 , y 2 , y 3 ≥ y 1 + y 3 ≥ 1 y 2 + y 3 ≥ 6    . We can easily find y ’s that satisfy the inequalities on the right by simply making them large enough, for example ( y 1 , y 2 , y 3 ) = (5 , 3 , 6) . But these particular multipliers would tell us that the optimum solution of the LP is at most 200 · 5 + 300 · 3 + 400 · 6 = 4300 , a bound that is far too loose to be of interest. What we want is a bound that is as tight as possible, so we should minimize 200 y 1 + 300 y 2 + 400 y 3 subject to the preceding inequalities. And this is a new linear program ! Therefore, finding the set of multipliers that gives the best upper bound on our original LP is tantamount to solving a new LP: min 200 y 1 + 300 y 2 + 400 y 3 y 1 + y 3 ≥ 1 y 2 + y 3 ≥ 6 y 1 , y 2 , y 3 ≥ By design, any feasible value of this dual LP is an upper bound on the original primal LP. So if we somehow find a pair of primal and dual feasible values that are equal, then they must both be optimal. Here is just such a pair: Primal : ( x 1 , x 2 ) = (100 , 300); Dual : ( y 1 , y 2 , y 3 ) = (0 , 5 , 1) . They both have value 1900 , and therefore they certify each other’s optimality (Figure 7.9). Amazingly, this is not just a lucky example, but a general phenomenon. To start with, the preceding construction—creating a multiplier for each primal constraint; writing a constraint 222 Algorithms Figure 7.9 By design, dual feasible values ≥ primal feasible values. The duality theorem tells us that moreover their optima coincide. Primal Primal feasible This duality gap is zero opt Dual feasible Objective value opt Dual Figure 7.10 A generic primal LP in matrix-vector form, and its dual. Primal LP: max c T x Ax ≤ b x ≥ Dual LP: min y T b y T A ≥ c T y ≥ in the dual for every variable of the primal, in which the sum is required to be above the objective coefficient of the corresponding primal variable; and optimizing the sum of the mul- tipliers weighted by the primal right-hand sides—can be carried out for any LP, as shown in...
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Algorithms_Part12 - S Dasgupta C.H Papadimitriou and U.V...

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