pset4 solution

# pset4 solution - EECS 310 Fall 2011 Instructor Nicole...

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Unformatted text preview: EECS 310, Fall 2011 Instructor: Nicole Immorlica Problem Set #4 Due: October 20, 2011 1. (20 points) You have n jobs J = { 1 ,...,n } that must be scheduled on m machines M = { 1 ,...,m } . Each job i can only be processed by a subset M i ⊆ M of the machines. Furthermore, each machine can process only a total number of k i jobs. Derive a condition similar to that in Hall’s Theorem to determine whether all the jobs can be scheduled. Use Hall’s Theorem to prove that your condition is necessary and sufficient. Solution: The similar condition will the following ∀ S ⊆ J, | S | ≤ X i ∈ N ( S ) k i Where N ( S ) is the neighbors of set S (which in this case, will only be machines). In words, this is stating that for any set of jobs, the total capacity of the machines that can process those jobs must be at least as big as the set of jobs. In graph theoretic terms, we can interpret k i as the maximum degree that machine i can have in any matching. First we prove this is necessary. That is, if there is a valid scheduling, then this condition must hold. Consider any set of jobs S . Since we have a valid scheduling, each job in S must be assigned to a machine. Let N ( S ) be the set of all machines that jobs in S are assigned to. Let d ( m i ) be the degree of machine i when the edges are limited to only those edges which are incident upon vertices in S. Then the total degree of machines in this subgraph is exactly | S | . And since this a a valid matching, the degree of each individual machine must be at most k i , its max capacity. So we get the following | S | = X d ( m i ) ≤ X k i So we get that if there is a valid scheduling, this condition must hold. Thus it is necessary. Now we prove that if this condition holds, then we can find a valid scheduling. We do this by transforming the graph in a similar manner, apply Hall’s theorem, and then argue this would give as a valid scheduling. Transform the graph in the following manner: For each machine m i , add k i- 1 more copies of m i to this graph (so there are now a total k i copies of m i ). Add edges such that any edge that was incident upon the original m i is now incident upon all m i . Claim: This graph satisfies the conditions for Hall’s theorem Proof: Take any set of nodes S . Since our modified condition holds, we get the following | S | ≤ X i ∈ N ( S ) k i In our modified graph, there are exactly k i nodes for each machine in the old graph. Therefore the neighbor set size is exactly ∑ i ∈ N ( S ) k i . Thus Hall’s theorem holds in this new graph for any S. Since Hall’s theorem holds, we know there exists a matching of jobs to machines. Now any job j that was assigned to a copy of machine i in the modified graph is scheduled on m i in the old graph. Since we had a matching, every job can be scheduled in the manner. And since we had only added k i copies for each machine m i , then each machine will have at most k i jobs scheduled on it. So this implies that we have a valid scheduling. Thus this condition is sufficient.implies that we have a valid scheduling....
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pset4 solution - EECS 310 Fall 2011 Instructor Nicole...

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