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Comparison-Based Complexityof Multi-Objective Optimization Paper by O. Teytaud Presented by M. Schoenauer TAO, Inria, Lri, UMR CNRS 8623, Université Paris-Sud
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Evolutionary multi-objectiveoptimization Generate a population Evaluate fitness values Select the « best » (various rules are possible here) Generate offspring Go back to the Evaluation step.
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Model of fitness functions ?No assumption ? Unrealistically pessimistic results Unreadable lower boundsLets do as in: ==> quadratic convex objective functions
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Complexity upper boundsEach objective function = a sphereBelow just a short overview of algorithms; ==> the real algorithms are a bit more trickyFor finding the whole Pareto front : Optimize each objective separately The PF is the convex hullFor finding a single point : Optimize any single objective
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Finding one point of the Pareto Setd objective functionsIn dimension NOne point at distance at most e of the PFcost=O( (N+1-d) log (1/e) )Proof : M log(1/e) in monoobjective optimization where M is the codimension of the set of optima (Gelly Teytaud, 2006)
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Finding the whole Pareto Setd objective functionsIn dimension NOne point at distance at most e of the PF for the Hausdorff metriccost=O( (Nd) log (1/e) )Proof : d times the monoobjective case.
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OutlineMultiobjective optimizationComplexity upper boundsComplexity lower bounds Proof technique (monoobjective) The MO case
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We want to know how many iterations we need for reaching precision in an evolutionary algorithm.Key observation: (most) evolutionary algorithms are comparison-basedLets consider (for simplicity) a deterministic selection-based non-elitist algorithmFirst idea: how many different branches we have in a run ? We select points among Therefore, at most K = ! / ( ! ( - )!) different branchesSecond idea: how many different answers should we able to give ? Use packing numbers: at least N() different possible answersConclusion: the number n of iterations should verify Kn ≥ N ( )Frédéric Lemoine MIG 11/07/2008 10
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We want to know how many iterations we need for reaching precision in an evolutionary algorithm.Key observation: (most) evolutionary algorithms are comparison-basedLets consider (for simplicity) a deterministic selection-based non-elitist algorithmFirst idea: how many different branches we have in a run ? We select points among Therefore, at most K = ! / ( ! ( - )!) different branchesSecond idea: how many different answers should we able to give ? Use packing numbers: at least N() different possible answersConclusion: the number n of iterations should verify Kn ≥ N ( )Frédéric Lemoine MIG 11/07/2008 11
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We want to know how many iterations we need for reaching precision in an evolutionary algorithm.Key observation: (most) evolutionary algorithms are comparison-basedLets consider (for simplicity) a deterministic selection-based non-elitist algorithmFirst idea: how many different branches we have in a run ? We select points among Therefore, at most K = ! / ( ! ( - )!) different branchesSecond idea: how many different answers should we able to give ? Use packing numbers: at least N() different possible answersConclusion: the number n of iterations should verify Kn ≥ N ( )Frédéric Lemoine MIG 11/07/2008 12
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We want to know how many iterations we need for reaching precision in an evolutionary algorithm.Key observation: (most) evolutionary algorithms are comparison-basedLets consider (for simplicity) a deterministic selection-based non-elitist algorithmFirst idea: how many different branches we have in a run ? We select points among Therefore, at most K = ! / ( ! ( - )!) different branchesSecond idea: how many different answers should we able to give ? Use packing numbers: at least N() different possible answersConclusion: the number n of iterations should verify Kn ≥ N ( )Frédéric Lemoine MIG 11/07/2008 13
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We want to know how many iterations we need for reaching precision in an evolutionary algorithm.Key observation: (most) evolutionary algorithms are comparison-basedLets consider (for simplicity) a deterministic selection-based non-elitist algorithmFirst idea: how many different branches we have in a run ? We select points among Therefore, at most K = ! / ( ! ( - )!) different branchesSecond idea: how many different answers should we able to give ? Use packing numbers: at least N() different possible answersConclusion: the number n of iterations should verify Kn ≥ N ( )Frédéric Lemoine MIG 11/07/2008 14
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We want to know how many iterations we need for reaching precision in an evolutionary algorithm.Key observation: (most) evolutionary algorithms are comparison-basedLets consider (for simplicity) a deterministic selection-based non-elitist algorithmFirst idea: how many different branches we have in a run ? We select points among Therefore, at most K = ! / ( ! ( - )!) different branchesSecond idea: how many different answers should we able to give ? Use packing numbers: at least N() different possible answersConclusion: the number n of iterations should verify Kn ≥ N ( )Frédéric Lemoine MIG 11/07/2008 15
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We want to know how many iterations we need for reaching precision in an evolutionary algorithm.Key observation: (most) evolutionary algorithms are comparison-basedLets consider (for simplicity) a deterministic selection-based non-elitist algorithmFirst idea: how many different branches we have in a run ? We select points among Therefore, at most K = ! / ( ! ( - )!) different branchesSecond idea: how many different answers should we able to give ? Use packing numbers: at least N() different possible answersConclusion: the number n of iterations should verify Kn ≥ N ( )Frédéric Lemoine MIG 11/07/2008 16
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We want to know how many iterations we need for reaching precision in an evolutionary algorithm.Key observation: (most) evolutionary algorithms are comparison-basedLets consider (for simplicity) a deterministic selection-based non-elitist algorithmFirst idea: how many different branches we have in a run ? We select points among Therefore, at most K = ! / ( ! ( - )!) different branchesSecond idea: how many different answers should we able to give ? Use packing numbers: at least N() different possible answersConclusion: the number n of iterations should verify Kn ≥ N ( )Frédéric Lemoine MIG 11/07/2008 17
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OutlineMultiobjective optimizationComplexity upper boundsComplexity lower bounds Proof technique (monoobjective) The MO case
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How to apply this in MOO ?Covering numbers can be computed also for Hausdorff distance ==>Plus a little bit of boring mathsLeads to bounds as expected Nd log(1/e) for the whole Pareto set (Hausdorff) (N+1-d) log(1/e) for pointwise convergence (distance to one point of the Pareto set) N=dimension, d=nb of objectives
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Results in multiobjective casesThe proof method is not new (Fournier & Teytaud, Algorithmica 2010)Its application to MOO is new : Tight bounds But no result in case of use of surrogate models (as for corresponding results in the monoobjective case) ; in fact, the problem becomes unrealistically easy with surrogate models...
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Sorry for not being here ==> really impossibleDiscussion ==> all email questions welcomeTight bounds thanks to a realistic modelCombining previous papers Complexity bounds Relevant modelMaybe an extension : using VC-dimensionThis paper did it(single objective)
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