030707 - Lecture 13: 03/07/2007 Recall: Simulated Annealing...

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Unformatted text preview: Lecture 13: 03/07/2007 Recall: Simulated Annealing Given S, : S->R, find x S minimizing (x) ((i)). SA solution : for each of a bunch of T-values, run metropolis at each T using some Q(I,j) (symmetric) for proposals acceptance prob ( , )= < ( ) -(- ) > ( ) i j 1 if j i e j i kT if j i At end of run, use best I to seed another run at a lower T-value, etc. One hopes that , as T->0, end up with populations of near minima for . More precisely: have a sequence {T n }; T n->0, n->. This is called an annealing schedule . For each n, run metropolis at temp T n . This is same as a given Markov chain on S; get a population of points in S. Let S * =set of all i S maximizing . Let Tn =unique stationary distribution of T n-chain. Let Tn (S * )=amount of probability concentrated on S * at end of T n-mn (or is mn up???) = * ( ) j S Tn j Can prove: [ ( *)]= limn Tn S 1 n-> as temp->0 Let X k,Tn = state at time k of T n chain. The last equation says: , *= limn limk ProbXk Tn S 1 Inside of = Tn (S * ) A stronger convergence result (due to B. Hajek and J. Tstsiklis (at MIT)):A stronger convergence result (due to B....
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030707 - Lecture 13: 03/07/2007 Recall: Simulated Annealing...

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