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Unformatted text preview: CME 305: Discrete Mathematics and Algorithms Instructor: Professor Amin Saberi ([email protected]) February 18, 2010 Lecture 13: Approximation Algorithms II In this lecture, we continue our discussion on approximation algorithms. In particular, we focus on techniques to define an appropriate relaxed problem that can be solved in polynomial time and then to convert the solution to an appropriate integral one. 1 Minimum Makespan Revisited Recall the minimum makespan problem. We have set J of jobs to schedule on set M of machines and we want to minimize the makespan, the maximum load on a machine. In the previous lecture, we assumed that each job has the same processing time on all machines. Here we consider a more general version of the problem in which job j can have different processing time on different machines. Let p ij be the time it takes machine i to process job j . We want to minimize the makespan T = max i ∑ j x ij p ij where the variable x ij ∈ { , 1 } indicates whether job j is assigned to machine i . We can write this as an integer program with constraints ensuring that each job is assigned to exactly one machine and that the load of each machine does not exceed T , the makespan. minimize T subject to X i ∈ M x ij = 1 ∀ j ∈ J X j ∈ J x ij p ij ≤ T ∀ i ∈ M x ij ∈ { , 1 } First, one may try to relax variable x ij and let x ij ∈ [0 , 1], however, the solution of the corresponding LP may be too far from the solution of the IP. Thus solution of LP does not serve as a tight lower bound for OPT. For instance, if we have only 1 job, m machines, and p i 1 = m , i ∈ M , then OPT= m , however, solution of LP is 1 by assigning x i 1 = 1 m . The maximum ratio of the optimal IP and LP solutions is called the integrality gap . We need to define the relaxed problem in such a way that the integrality gap is small. The idea is to ensure that if p ij > T we assign x ij = 0. We do this by defining a series of feasibility LPs with a makespan parameter T as follows. X i : p ij ≤ T x ij = 1 ∀ j ∈ J X j ∈ J x ij p ij ≤ T ∀ i ∈ M x ij ≥ ∀ i ∈ M j ∈ J Using binary search we can obtain the smallest value of T , T * , for which the above feasibility linear program (FLP) has a solution. It is easy to show that in such a solution we assign x ij = 0 if p ij > T * . 2 CME 305: Discrete Mathematics and Algorithms  Lecture 13 Claim 1 FLP assigns at most n + m (fractional) jobs to the machines. Proof: Recall that a solution x is a vertex of the polyhedron formed by the constraints. Note that there are  x  + n + m total constraints, where  x  of them are of the from x ij ≥ 0. At a vertex,  x  linearly independent constraints must be satisfied; at least  x  ( m + n ) of them are of the form x ij = 0, therefore the number of nonzero x ij ’s is at most m + n ....
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This document was uploaded on 06/17/2010.
 Winter '09

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