蚁群算法求解TSP问题的源程序及简要说明

2025-03-04 16:25:58
推荐回答(5个)
回答1:

该程序试图对具有31个城市的VRP进行求解,已知的最优解为784.1,我用该程序只能优化到810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。
蚁群算法的matlab源码,同时请指出为何不能优化到已知的最好解

%
%
% the procedure of ant colony algorithm for VRP
%
% % % % % % % % % % %

%initialize the parameters of ant colony algorithms
load data.txt;
d=data(:,2:3);
g=data(:,4);

m=31; % 蚂蚁数
alpha=1;
belta=4;% 决定tao和miu重要性的参数
lmda=0;
rou=0.9; %衰减系数
q0=0.95;
% 概率
tao0=1/(31*841.04);%初始信息素
Q=1;% 蚂蚁循环一周所释放的信息素
defined_phrm=15.0; % initial pheromone level value
QV=100; % 车辆容量
vehicle_best=round(sum(g)/QV)+1; %所完成任务所需的最少车数
V=40;

% 计算两点的距离
for i=1:32;
for j=1:32;
dist(i,j)=sqrt((d(i,1)-d(j,1))^2+(d(i,2)-d(j,2))^2);
end;
end;

%给tao miu赋初值
for i=1:32;
for j=1:32;
if i~=j;
%s(i,j)=dist(i,1)+dist(1,j)-dist(i,j);
tao(i,j)=defined_phrm;
miu(i,j)=1/dist(i,j);
end;
end;
end;

for k=1:32;
for k=1:32;
deltao(i,j)=0;
end;
end;

best_cost=10000;
for n_gen=1:50;

print_head(n_gen);

for i=1:m;
%best_solution=[];
print_head2(i);
sumload=0;
cur_pos(i)=1;
rn=randperm(32);
n=1;
nn=1;
part_sol(nn)=1;
%cost(n_gen,i)=0.0;
n_sol=0; % 由蚂蚁产生的路径数量
M_vehicle=500;
t=0; %最佳路径数组的元素数为0

while sumload<=QV;

for k=1:length(rn);
if sumload+g(rn(k))<=QV;
gama(cur_pos(i),rn(k))=(sumload+g(rn(k)))/QV;
A(n)=rn(k);
n=n+1;
end;
end;

fid=fopen('out_customer.txt','a+');
fprintf(fid,'%s %i\t','the current position is:',cur_pos(i));
fprintf(fid,'\n%s','the possible customer set is:')
fprintf(fid,'\t%i\n',A);
fprintf(fid,'------------------------------\n');
fclose(fid);

p=compute_prob(A,cur_pos(i),tao,miu,alpha,belta,gama,lmda,i);
maxp=1e-8;
na=length(A);
for j=1:na;
if p(j)>maxp
maxp=p(j);
index_max=j;
end;
end;

old_pos=cur_pos(i);
if rand(1) cur_pos(i)=A(index_max);
else
krnd=randperm(na);
cur_pos(i)=A(krnd(1));
bbb=[old_pos cur_pos(i)];
ccc=[1 1];
if bbb==ccc;
cur_pos(i)=A(krnd(2));
end;
end;

tao(old_pos,cur_pos(i))=taolocalupdate(tao(old_pos,cur_pos(i)),rou,tao0);%对所经弧进行局部更新

sumload=sumload+g(cur_pos(i));

nn=nn+1;
part_sol(nn)=cur_pos(i);
temp_load=sumload;

if cur_pos(i)~=1;
rn=setdiff(rn,cur_pos(i));
n=1;
A=[];
end;

if cur_pos(i)==1; % 如果当前点为车场,将当前路径中的已访问用户去掉后,开始产生新路径
if setdiff(part_sol,1)~=[];
n_sol=n_sol+1; % 表示产生的路径数,n_sol=1,2,3,..5,6...,超过5条对其费用加上车辆的派遣费用
fid=fopen('out_solution.txt','a+');
fprintf(fid,'%s%i%s','NO.',n_sol,'条路径是:');
fprintf(fid,'%i ',part_sol);
fprintf(fid,'\n');
fprintf(fid,'%s','当前的用户需求量是:');
fprintf(fid,'%i\n',temp_load);
fprintf(fid,'------------------------------\n');
fclose(fid);

% 对所得路径进行路径内3-opt优化
final_sol=exchange(part_sol);

for nt=1:length(final_sol); % 将所有产生的路径传给一个数组
temp(t+nt)=final_sol(nt);
end;
t=t+length(final_sol)-1;

sumload=0;
final_sol=setdiff(final_sol,1);
rn=setdiff(rn,final_sol);
part_sol=[];
final_sol=[];
nn=1;
part_sol(nn)=cur_pos(i);
A=[];
n=1;

end;
end;

if setdiff(rn,1)==[];% 产生最后一条终点不为1的路径
n_sol=n_sol+1;
nl=length(part_sol);
part_sol(nl+1)=1;%将路径的最后1位补1

% 对所得路径进行路径内3-opt优化
final_sol=exchange(part_sol);

for nt=1:length(final_sol); % 将所有产生的路径传给一个数组
temp(t+nt)=final_sol(nt);
end;

cost(n_gen,i)=cost_sol(temp,dist)+M_vehicle*(n_sol-vehicle_best); %计算由蚂蚁i产生的路径总长度

for ki=1:length(temp)-1;
deltao(temp(ki),temp(ki+1))=deltao(temp(ki),temp(ki+1))+Q/cost(n_gen,i);
end;

if cost(n_gen,i) best_cost=cost(n_gen,i);
old_cost=best_cost;
best_gen=n_gen; % 产生最小费用的代数
best_ant=i; %产生最小费用的蚂蚁
best_solution=temp;
end;

if i==m; %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
for ii=1:32;
for jj=1:32;
tao(ii,jj)=(1-rou)*tao(ii,jj);
end;
end;

for kk=1:length(best_solution)-1;
tao(best_solution(kk),best_solution(kk+1))=tao(best_solution(kk),best_solution(kk+1))+deltao(best_solution(kk),best_solution(kk+1));
end;
end;

fid=fopen('out_solution.txt','a+');
fprintf(fid,'%s%i%s','NO.',n_sol,'路径是:');
fprintf(fid,'%i ',part_sol);
fprintf(fid,'\n');
fprintf(fid,'%s %i\n','当前的用户需求量是:',temp_load);
fprintf(fid,'%s %f\n','总费用是:',cost(n_gen,i));
fprintf(fid,'------------------------------\n');
fprintf(fid,'%s\n','最终路径是:');
fprintf(fid,'%i-',temp);
fprintf(fid,'\n');
fclose(fid);
temp=[];
break;
end;
end;

end;
end;
我现在也在研究它,希望能共同进步.建义可以看一下段海滨的关于蚁群算法的书.讲的不错,李士勇的也可以,还有一本我在图书馆见过,记不得名字了.

回答2:

简单蚁群算法求解TSP的源程序(我帮你找的)

蚁群算法是新兴的仿生算法,最初是由意大利学者Dorigo M于1991年首次提出,由于具有较强的鲁棒性,优良的分布式计算机制和易于与其它方法结合等优点,成为人工智能领域的一个研究热点。本程序是实现简单的蚁群算法,TSP问题取的是att48,可从http://www.iwr.uni-heidelberg.de/groups/comopt/software/TSPLIB95获取,程序运行时间可能会比较长,在我的这台CPU 1.6G+内存256M的机器上运行时间大概是13分钟左右。我用的语言是MATLAB 7.1。此程序仅供学习所用,如有问题请反馈。谢谢。(注:程序没有计算最后一个城市回来起点城市的距离)

function [y,val]=QACS
tic
load att48 att48;
MAXIT=300; % 最大循环次数
NC=48; % 城市个数
tao=ones(48,48);% 初始时刻各边上的信息最为1
rho=0.2; % 挥发系数
alpha=1;
beta=2;
Q=100;
mant=20; % 蚂蚁数量
iter=0; % 记录迭代次数
for i=1:NC % 计算各城市间的距离
for j=1:NC
distance(i,j)=sqrt((att48(i,2)-att48(j,2))^2+(att48(i,3)-att48(j,3))^2);
end
end
bestroute=zeros(1,48); % 用来记录最优路径
routelength=inf; % 用来记录当前找到的最优路径长度
% for i=1:mant % 确定各蚂蚁初始的位置
% end
for ite=1:MAXIT
for ka=1:mant %考查第K只蚂蚁
deltatao=zeros(48,48); % 第K只蚂蚁移动前各边上的信息增量为零
[routek,lengthk]=travel(distance,tao,alpha,beta);
if lengthk routelength=lengthk;
bestroute=routek;
end
for i=1:NC-1 % 第K只蚂蚁在路径上释放的信息量
deltatao(routek(i),routek(i+1))=deltatao(routek(i),routek(i+1))+Q/lengthk;
end
deltatao(routek(48),1)=deltatao(routek(48),1)+Q/lengthk;
end
for i=1:NC-1
for j=i+1:NC
if deltatao(i,j)==0
deltatao(i,j)=deltatao(j,i);
end
end
end
tao=(1-rho).*tao+deltatao;
end
y=bestroute;
val=routelength;
toc

function [y,val]=travel(distance,tao,alpha,beta) % 某只蚂蚁找到的某条路径
[m,n]=size(distance);
p=fix(m*rand)+1;
val=0; % 初始路径长度设为 0
tabuk=[p]; % 假设该蚂蚁都是从第 p 个城市出发的
for i=1:m-1
np=tabuk(length(tabuk)); % 蚂蚁当前所在的城市号
p_sum=0;
for j=1:m
if isin(j,tabuk)
continue;
else
ada=1/distance(np,j);
p_sum=p_sum+tao(np,j)^alpha*ada^beta;
end
end
cp=zeros(1,m); % 转移概率
for j=1:m
if isin(j,tabuk)
continue;
else
ada=1/distance(np,j);
cp(j)=tao(np,j)^alpha*ada^beta/p_sum;
end
end
NextCity=pchoice(cp);
tabuk=[tabuk,NextCity];
val=val+distance(np,NextCity);
end
y=tabuk;

function y=isin(x,A) % 判断数 x 是否在向量 A 中,如在返回 1 ,否则返回 0
y=0;
for i=1:length(A)
if A(i)==x
y=1;
break;
end
end

function y=pchoice(A)
a=rand;
tempA=zeros(1,length(A)+1);
for i=1:length(A)
tempA(i+1)=tempA(i)+A(i);
end
for i=2:length(tempA)
if a<=tempA(i)
y=i-1;
break;
end
end

回答3:

长见识了.多多学习啊.
关于 算法 不好发上来 提取密码 与 QQ 515521882 联系.
http://pickup.mofile.com/0410386614837552

回答4:

蚁群算法属于现代控制理论中的热门啊
真想能够搞明白
呵呵

回答5:

哇,我学习一下,没能力帮你,很抱歉!

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