📄 蚁群算法matlab源码 - 我的文章 - 阳光的味道.htm
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<P class=MsoNormal style="MARGIN: 0cm 0cm 0pt"><SPAN style="FONT-SIZE: 9pt; COLOR: black; FONT-FAMILY: 宋体">该程序试图对具有<SPAN lang=EN-US>31</SPAN>个城市的<SPAN lang=EN-US>VRP</SPAN>进行求解,已知的最优解为<SPAN lang=EN-US>784.1</SPAN>,我用该程序只能优化到<SPAN lang=EN-US>810</SPAN>左右,应该是陷入局部最优,但我不知问题出在什么地方。请用过蚁群算法的高手指教。<SPAN lang=EN-US><?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /><o:p></o:p></SPAN></SPAN></P>
<P class=MsoNormal style="MARGIN: 0cm 72pt 0pt; LINE-HEIGHT: 13pt; mso-margin-top-alt: auto; mso-margin-bottom-alt: auto"><SPAN lang=EN-US style="FONT-SIZE: 9pt; COLOR: black; FONT-FAMILY: 宋体">% the procedure of ant colony algorithm for VRP<BR>%<BR>% % % % % % % % % % %<BR><BR>%initialize the parameters of ant colony algorithms<BR>load data.txt;<BR>d=data(:,2:3);<BR>g=data(:,4);<o:p></o:p></SPAN></P>
<P class=MsoNormal style="MARGIN: 0cm 0cm 0pt"><SPAN lang=EN-US style="FONT-SIZE: 9pt; COLOR: black; FONT-FAMILY: 宋体">m=31; % </SPAN><SPAN style="FONT-SIZE: 9pt; COLOR: black; FONT-FAMILY: 宋体">蚂蚁数<SPAN lang=EN-US><BR>alpha=1;<BR>belta=4;% </SPAN>决定<SPAN lang=EN-US>tao</SPAN>和<SPAN lang=EN-US>miu</SPAN>重要性的参数<SPAN lang=EN-US><BR>lmda=0;<BR>rou=0.9;</SPAN> <SPAN lang=EN-US>%</SPAN>衰减系数<SPAN lang=EN-US><BR>q0=0.95;<BR>% </SPAN>概率<SPAN lang=EN-US><BR>tao0=1/(31*841.04);%</SPAN>初始信息素<SPAN lang=EN-US><BR>Q=1;%</SPAN> 蚂蚁循环一周所释放的信息素<SPAN lang=EN-US><BR>defined_phrm=15.0; % initial pheromone level value <BR>QV=100; % </SPAN>车辆容量<SPAN lang=EN-US><BR>vehicle_best=round(sum(g)/QV)+1;</SPAN> <SPAN lang=EN-US>%</SPAN>所完成任务所需的最少车数<SPAN lang=EN-US><BR>V=40;<BR>% </SPAN>计算两点的距离<SPAN lang=EN-US> <BR>for i=1:32;<BR> for j=1:32;<BR> dist(i,j)=sqrt((d(i,1)-d(j,1))^2+(d(i,2)-d(j,2))^2);<BR> end;<BR>end;<BR>%</SPAN>给<SPAN lang=EN-US>tao miu</SPAN>赋初值<SPAN lang=EN-US><BR>for i=1:32;<BR> for j=1:32;<BR> if i~=j;<BR> %s(i,j)=dist(i,1)+dist(1,j)-dist(i,j);<BR> tao(i,j)=defined_phrm;<BR> miu(i,j)=1/dist(i,j); <BR> end; <BR> end;<BR>end;<BR><BR>for k=1:32;<BR> for k=1:32;<BR> deltao(i,j)=0;<BR> end;<BR>end; <BR>best_cost=10000; <BR>for n_gen=1:50;<BR> print_head(n_gen); <BR> for i=1:m;<BR> %best_solution=[];<BR> print_head2(i);<BR> sumload=0;<BR> cur_pos(i)=1;<BR> rn=randperm(32);<BR> n=1;<BR> nn=1;<BR> part_sol(nn)=1;<BR> %cost(n_gen,i)=0.0;<BR> n_sol=0; % </SPAN>由蚂蚁产生的路径数量<SPAN lang=EN-US><BR> M_vehicle=500;<BR> t=0; %</SPAN>最佳路径数组的元素数为<SPAN lang=EN-US>0<BR> <BR> while sumload<=QV;<BR> <BR> for k=1:length(rn);<BR> if sumload+g(rn(k))<=QV;<BR> gama(cur_pos(i),rn(k))=(sumload+g(rn(k)))/QV;<BR> A(n)=rn(k);<BR> n=n+1;<BR> end;<BR> end;<BR> fid=fopen('out_customer.txt','a+');<BR> fprintf(fid,'%s %i\t','the current position is:',cur_pos(i)); <BR> fprintf(fid,'\n%s','the possible customer set is:')<BR> fprintf(fid,'\t%i\n',A);<BR> fprintf(fid,'------------------------------\n');<BR> fclose(fid);<BR> <SPAN style="mso-tab-count: 1"> </SPAN>p=compute_prob(A,cur_pos(i),tao,miu,alpha,belta,gama,lmda,i);<BR> maxp=1e-8;<BR> na=length(A);<BR> for j=1:na;<BR> if p(j)>maxp<BR> maxp=p(j);<BR> index_max=j;<BR> end;<BR> end;<BR> <BR> old_pos=cur_pos(i);<BR> if rand(1)<q0<BR> cur_pos(i)=A(index_max);<BR> else <BR> krnd=randperm(na);<BR> cur_pos(i)=A(krnd(1)); <BR> bbb=[old_pos cur_pos(i)];<BR> ccc=[1 1];<BR> if bbb==ccc;<BR> cur_pos(i)=A(krnd(2)); <BR> end;<BR> end;<BR> <BR> tao(old_pos,cur_pos(i))=taolocalupdate(tao(old_pos,cur_pos(i)),rou,tao0);%</SPAN>对所经弧进行局部更新<SPAN lang=EN-US><BR> <BR> sumload=sumload+g(cur_pos(i));<BR><BR> nn=nn+1;<BR> part_sol(nn)=cur_pos(i);<BR> temp_load=sumload;<BR> <BR> if cur_pos(i)~=1;<BR> rn=setdiff(rn,cur_pos(i));<BR> n=1;<BR> A=[];<BR> end;<BR> <BR> if cur_pos(i)==1; % </SPAN>如果当前点为车场<SPAN lang=EN-US>,</SPAN>将当前路径中的已访问用户去掉后<SPAN lang=EN-US>,</SPAN>开始产生新路径<SPAN lang=EN-US><BR> if setdiff(part_sol,1)~=[];<BR> n_sol=n_sol+1; % </SPAN>表示产生的路径数<SPAN lang=EN-US>,n_sol=1,2,3,..5,6...,</SPAN>超过<SPAN lang=EN-US>5</SPAN>条对其费用加上车辆的派遣费用<SPAN lang=EN-US><BR> fid=fopen('out_solution.txt','a+');<BR> fprintf(fid,'%s%i%s','NO.',n_sol,'</SPAN>条路径是<SPAN lang=EN-US>:');<BR> fprintf(fid,'%i ',part_sol);<BR> fprintf(fid,'\n');<BR> fprintf(fid,'%s','</SPAN>当前的用户需求量是<SPAN lang=EN-US>:');<BR> fprintf(fid,'%i\n',temp_load);<BR> fprintf(fid,'------------------------------\n');<BR> fclose(fid); <BR> <BR> % </SPAN>对所得路径进行路径内<SPAN lang=EN-US>3-opt</SPAN>优化<SPAN lang=EN-US><BR> final_sol=exchange(part_sol);<BR> <BR> for nt=1:length(final_sol); % </SPAN>将所有产生的路径传给一个数组<SPAN lang=EN-US><BR> temp(t+nt)=final_sol(nt);<BR> end;<BR> t=t+length(final_sol)-1;<BR> <BR> sumload=0;<BR> final_sol=setdiff(final_sol,1);<BR> rn=setdiff(rn,final_sol);<BR> part_sol=[];<BR> final_sol=[];<BR> nn=1;<BR> part_sol(nn)=cur_pos(i);<BR> A=[];<BR> n=1;<BR> <BR> end; <BR> end;<BR> <BR> if setdiff(rn,1)==[];% </SPAN>产生最后一条终点不为<SPAN lang=EN-US>1</SPAN>的路径<SPAN lang=EN-US><BR> n_sol=n_sol+1;<BR> nl=length(part_sol);<BR> part_sol(nl+1)=1;%</SPAN>将路径的最后<SPAN lang=EN-US>1</SPAN>位补<SPAN lang=EN-US>1<BR> <BR> % </SPAN>对所得路径进行路径内<SPAN lang=EN-US>3-opt</SPAN>优化<SPAN lang=EN-US><BR> final_sol=exchange(part_sol); <BR> <BR> for nt=1:length(final_sol); % </SPAN>将所有产生的路径传给一个数组<SPAN lang=EN-US><BR> temp(t+nt)=final_sol(nt); <BR> end;<BR> <BR> cost(n_gen,i)=cost_sol(temp,dist)+M_vehicle*(n_sol-vehicle_best); %</SPAN>计算由蚂蚁<SPAN lang=EN-US>i</SPAN>产生的路径总长度<SPAN lang=EN-US><BR> <BR> for ki=1:length(temp)-1;<BR> deltao(temp(ki),temp(ki+1))=deltao(temp(ki),temp(ki+1))+Q/cost(n_gen,i);<BR> end; <BR> <BR> if cost(n_gen,i)<best_cost;<BR> best_cost=cost(n_gen,i);<BR> old_cost=best_cost;<BR> best_gen=n_gen; % </SPAN>产生最小费用的代数<SPAN lang=EN-US><BR> best_ant=i; %</SPAN>产生最小费用的蚂蚁<SPAN lang=EN-US><BR> best_solution=temp;<BR> end;<BR> <BR> if i==m;</SPAN> <SPAN lang=EN-US>%</SPAN>如果所有蚂蚁均完成一次循环,,则用最佳费用所对应的路径对弧进行整体更新<SPAN lang=EN-US><BR> for ii=1:32;<BR> for jj=1:32;<BR> tao(ii,jj)=(1-rou)*tao(ii,jj);<BR> end;<BR> end;<BR> <BR> for kk=1:length(best_solution)-1;<BR> tao(best_solution(kk),best_solution(kk+1))=tao(best_solution(kk),best_solution(kk+1))+deltao(best_solution(kk),best_solution(kk+1));<BR> end; <BR> end; <BR> <BR> fid=fopen('out_solution.txt','a+');<BR> fprintf(fid,'%s%i%s','NO.',n_sol,'</SPAN>路径是<SPAN lang=EN-US>:');<BR> fprintf(fid,'%i ',part_sol);<BR> fprintf(fid,'\n');<BR> fprintf(fid,'%s %i\n','</SPAN>当前的用户需求量是<SPAN lang=EN-US>:',temp_load);<BR> fprintf(fid,'%s %f\n','</SPAN>总费用是<SPAN lang=EN-US>:',cost(n_gen,i));<BR> fprintf(fid,'------------------------------\n');<BR> fprintf(fid,'%s\n','</SPAN>最终路径是<SPAN lang=EN-US>:');<BR> fprintf(fid,'%i-',temp);<BR> fprintf(fid,'\n');<BR> fclose(fid); <BR> temp=[];<BR> break;<BR> end;<BR> end;<BR> <BR> end;<BR>end;<o:p></o:p></SPAN></SPAN></P>
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