26 MIN-MAX MULTI-OBJECTIVE OPTIMIZATION 85 26 Min-max Multi-Objective Optimization Select the best candidate solution below [A-F], based on min-max optimization. This exercise presents some of the considerations you might encounter in the purchase of a large machine, such as an engine. The objectives are to minimize weight and cost, while maximizing the other attributes. A high score in reputation means the company and product have a good reputation, and a high score in specification means that the product fits the application very well. Hence, candidate D has a terrific machine on paper for the application, but with a rather poor reputation relative to the others. Do you agree with what the min-max calculation finds? A B C D E F Weight (N) 1260 1190 1470 1540 952 1358 Cost (dollars) 24700 23920 28860 33800 24700 27300 Reputation (nom=1) 1.1 0.7 1.2 0.5 0.8 0.9 Warranty (years) 5 6 8 5 3 6 Efficiency (pct.) 0.51 0.52 0.38 0.36 0.45 0.41 Specifications (nom=1) 1 1.1 1.2 1.4 0.85 1.1 The maximum normalized deviation from peak performance, for each of the six candidates is [0.72, 0.71, 0.88, 1.0, 1.0, 0.69], and so Candidate F is the winner. See the attached code. You might not agree with this result. Perhaps we can weight the deviations to distinguish between those properties that are important vs. those that are not? The purist will say that the less important performance measures shouldn’t even make it onto the chart! %------------------------------------------------------------------------- % minMax optimization % clear all; % give properties for each candidate weight = [.9 .85 1.05 1.1 .68 .97] * 1400 ; cost = [.95 .92 1.11 1.3 .95 1.05] * 26000 ; reputation = [1.1 .7 1.2 .5 .8 .9] * 1 ; warranty =[5 6 8 5 3 6] ; efficiency = [.91 .92 .78 .76 .85 .81]-.4; spec = [1 1.1 1.2 1.4 .85 1.1] ; % write this data to a file for latex f = fopen(’minMax.dat’,’w’); fprintf(f,’Weight (N)’);fprintf(f,’ & %g’,weight) ; fprintf(f,’\\\\ \n’);26 MIN-MAX MULTI-OBJECTIVE OPTIMIZATION 86 fprintf(f,’Cost (dollars)’);fprintf(f,’ & %g’,cost) ; fprintf(f,’\\\\ \n’); fprintf(f,’Reputation (nom=1)’);fprintf(f,’ & %g’,reputation) ; fprintf(f,’\\\\ \n’); fprintf(f,’Warranty (years)’);fprintf(f,’ & %g’,warranty) ; fprintf(f,’\\\\ \n’); fprintf(f,’Efficiency (pct.)’);fprintf(f,’ & %g’,efficiency) ; fprintf(f,’\\\\ \n’); fprintf(f,’Specifications (nom=1)’);fprintf(f,’ & %g’,spec) ; fclose(f); % make matrix: each candidate gets a column, each attribute is a row. perf = [-weight ; -cost ; reputation; warranty ; efficiency ; spec] meanPerf = sum(perf’)’ ; maxPerf = max(perf’)’ ; % the maximum for each attribute minPerf = min(perf’)’ ; % minimum for each attribute rangePerf = maxPerf-minPerf ; % range for each attribute devPerf = maxPerf*ones(1,size(perf,2)) - perf ; % deviation from peak, % all cands. and attributes normDevPerf = devPerf ./ (rangePerf*ones(1,size(perf,2))) % normalize deviations with attribute ranges maxNormDevPerf = max(normDevPerf) % get the max deviation for the % candidates [junk,ind] = sort(maxNormDevPerf) ; disp(sprintf(’The best min-max candidate is Candidate %d’, ind(1))); %-------------------------------------------------------------------------MIT OpenCourseWarehttp://ocw.mit.edu 2.017J Design of Electromechanical Robotic Systems Fall 2009 For information about citing these materials or our Terms of Use, visit:
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