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資源簡介

通過自己的整合,可以用來求解多元函數最大值,保證可以運行,希望對你們有用

資源截圖

代碼片段和文件信息

clc
clear
%遺傳算法參數
maxgen=30;
???size=40;
pcross=0.6;
pmutation=0.01;
lenchrom=[1?1?1?1?1];
bound=[0?99;0?99;?0?99;?0?99;0?99];
for?b=1:1
%個體初始化
indi=struct(‘fitness‘zeros(1size(b))‘chrom‘[]);
%avgfitness=[];%種群的平均適應度
bestfitness=[];%種群的最佳適應度
bestchrom=[];%適應度最好的染色體
for?i=1:size(b)
????indi.chrom(i:)=Code(lenchrombound);
????x=indi.chrom(i:);
????indi.fitness(i)=fun(x);
end
[bestfitnessbestindex]=max(indi.fitness);
bestchrom=indi.chrom(bestindex:);
avgfitness=sum(indi.fitness)/size(b);
%記錄每一代中最好的適應度和平均適應度
for?i=1:maxgen
????%選擇
????indi=select(indisize(b));
????avgfitness=sum(indi.fitness)/size(b);
????%交叉
????indi.chrom=Cross(pcrosslenchromindi.chromsize(b)bound);
????%變異
????indi.chrom=Mutation(pmutationlenchromindi.chromsize(b)[i?maxgen]bound);
???%每進化10代,以所得值為初始值進行非線性尋優
???if?mod(i10)==0
???????indi.chrom=nonlinear(indi.chromsize(b));
???end
???for?j=1:size(b)
???????x=indi.chrom(j:);
???????indi.fitness(j)=fun(x);
???end
???%找到最優染色體急他們在種群中的位置
???[newbestfitnessnewbestindex]=max(indi.fitness);
???%代替上一次進化中最好的染色體
???if?bestfitness???????bestfitness=newbestfitness;
???????bestchrom=indi.chrom(newbestindex:);
???end
???%avgfitness=sum(indi.fitness)/size;
???trace(bi)=bestfitness?;
???
end


end
x=1:maxgen;
plot(xtrace(1:))
hold?on?
?title(‘遺傳算法求解最大值‘);
?%legend(‘種群為20‘‘種群為30‘)
?xlabel(‘迭代次數‘)
?ylabel(‘最大值‘)
?grid
????

????
????




?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????1317??2017-10-10?08:47??Cross.m
?????文件????????1232??2017-10-10?08:41??Mutation.m
?????文件????????1600??2017-10-01?10:25??changeyichuan1.m
?????文件?????????331??2017-09-28?08:34??nonlinear.m
?????文件?????????465??2017-09-27?21:17??select.m
?????文件?????????405??2017-10-10?08:39??Code.m

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