MIT6_079F09_hw05 - braceleftBigg 6.079/6.975, Fall...

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Unformatted text preview: braceleftBigg 6.079/6.975, Fall 2009–10 S. Boyd & P. Parrilo Homework 5 additional problems 1. Heuristic suboptimal solution for Boolean LP. This exercise builds on exercises 4.15 and 5.13 in Convex Optimization , which involve the Boolean LP minimize c T x subject to Ax precedesequal b x i ∈ { , 1 } , i = 1 , . . . , n, with optimal value p ⋆ . Let x rlx be a solution of the LP relaxation minimize c T x subject to Ax precedesequal b 0 precedesequal x precedesequal 1 , so L = c T x rlx is a lower bound on p ⋆ . The relaxed solution x rlx can also be used to guess a Boolean point ˆ x , by rounding its entries, based on a threshold t ∈ [0 , 1]: rlx 1 x i ≥ t x ˆ i = 0 otherwise , for i = 1 , . . ., n . Evidently ˆ x is Boolean ( i.e. , has entries in { , 1 } ). If it is feasible for the Boolean LP, i.e. , if Ax ˆ precedesequal b , then it can be considered a guess at a good, if not optimal, point for the Boolean LP. Its objective value, U = c T x ˆ, is an upper bound on p ⋆ . If U and L are close, then ˆ x is nearly optimal; specifically, ˆ x cannot be more than ( U − L )-suboptimal for the Boolean LP. This rounding need not work; indeed, it can happen that for all threshold values, ˆ x is infeasible. But for some problem instances, it can work well. Of course, there are many variations on this simple scheme for (possibly) constructing a feasible, good point from x rlx . Finally, we get to the problem. Generate problem data using rand(’state’,0); n=100; m=300; A=rand(m,n); b=A*ones(n,1)/2; c=-rand(n,1); You can think of x i as a job we either accept or decline, and − c i as the (positive) revenue we generate if we accept job i . We can think of Ax precedesequal b as a set of limits on 1 summationdisplay...
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This note was uploaded on 01/31/2012 for the course ECE 202 taught by Professor Boyde during the Fall '06 term at Goldsmiths.

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MIT6_079F09_hw05 - braceleftBigg 6.079/6.975, Fall...

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