急求matlab车辆调度遗传算法代码，需求车辆行驶最优路径。

function [path,lmin]=ga(data,d) %data为点集,d为距离矩阵，即赋权图

tic

%======================

sj0=data%开环最短路线

%=================================

% sj0=[datadata(1,:)]%闭环最短路线

%=========================

x=sj0(:,1)y=sj0(:,2)

N=length(x)

%=========================

% d(N,:)=d(1,:)%闭环最短路线

% d(:,N)=d(:,1)%距离矩阵d

%======================

L=N %sj0的长度

w=800dai=1000

%通过改良圈算法选取优良父代A

for k=1:w

c=randperm(L-2)

c1=[1,c+1,L]

flag=1

while flag>0

flag=0

for m=1:L-3

for n=m+2:L-1

if d(c1(m),c1(n))+d(c1(m+1),c1(n+1))<d(c1(m),c1(m+1))+d(c1(n),c1(n+1))

flag=1

c1(m+1:n)=c1(n:-1:m+1)

<a href="https://www.baidu.com/s?wd=end&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">end</a>

<a href="https://www.baidu.com/s?wd=end&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">end</a>

<a href="https://www.baidu.com/s?wd=end&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">end</a>

end

J(k,c1)=1:L

end

J=J/L

J(:,1)=0J(:,L)=1

rand('state',sum(clock))

%遗传算法实现过程

A=J

for k=1:dai %产生0～1 间随机数列进行编码

B=A

c=randperm(w)

%交配产生子代B

for i=1:2:w

F=2+floor(100*rand(1))

temp=B(c(i),F:L)

B(c(i),F:L)=B(c(i+1),F:L)

B(c(i+1),F:L)=temp

end

%变异产生子代C

by=find(rand(1,w)<0.1)

if length(by)==0

by=floor(w*rand(1))+1

end

C=A(by,:)

L3=length(by)

for j=1:L3

<a href="https://www.baidu.com/s?wd=bw&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">bw</a>=floor(1+fix(rand(1,3)*N)) %产生1-N的3个随机数

<a href="https://www.baidu.com/s?wd=bw&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">bw</a>=sort(<a href="https://www.baidu.com/s?wd=bw&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">bw</a>)

C(j,:)=C(j,[1:bw(1)-1,bw(2)+1:bw(3),bw(1):bw(2),bw(3)+1:L])

end

G=[ABC]

<a href="https://www.baidu.com/s?wd=TL&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">TL</a>=size(G,1)

%在父代和子代中选择优良品种作为新的父代

[<a href="https://www.baidu.com/s?wd=dd&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">dd</a>,IX]=sort(G,2)

temp=[]

temp(1:<a href="https://www.baidu.com/s?wd=TL&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">TL</a>)=0

for j=1:<a href="https://www.baidu.com/s?wd=TL&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">TL</a>

for i=1:L-1

temp(j)=temp(j)+d(IX(j,i),IX(j,i+1))

end

end

[DZ,IZ]=sort(temp)

A=G(IZ(1:w),:)

end

path=IX(IZ(1),:)

% for i=1:length(path)

% path(i)=path(i)-1

% end

% path=path(2:end-1)

lmin=0l=0

for j=1:(length(path)-1)

<a href="https://www.baidu.com/s?wd=t1&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">t1</a>=path(j)t2=path(j+1)

l=d(<a href="https://www.baidu.com/s?wd=t1&tn=44039180_cpr&fenlei=mv6quAkxTZn0IZRqIHckPjm4nH00T1Y3PvF-rjf1rj6LryD3Pjm30ZwV5Hcvrjm3rH6sPfKWUMw85HfYnjn4nH6sgvPsT6KdThsqpZwYTjCEQLGCpyw9Uz4Bmy-bIi4WUvYETgN-TLwGUv3EnW0dnHmvrHcYnWTYnjn4rHb3Ps" target="_blank" class="baidu-highlight">t1</a>,t2)

lmin=lmin+l

end

xx=sj0(path,1)yy=sj0(path,2)

plot(xx,yy,'r-o')

axis equal

toc

现有两个工厂A1,A2，产量都是10万个，工厂有三个仓库T1,T2,T3，如何用matlab求解向四个顾客配送产品，以达到总运费最少的运输方案？

这个运输问题可以按下列思路来考虑：

1、A1工厂向T1，T2，T3仓库配送的最少运费，即其关系为

min 4x1+2y1+5z1

2、A2工厂向T1，T2，T3仓库配送的最少运费，即其关系为

min 3x2+y2+2z2

3、T1仓库向B1、B2、B3、B4顾客配送的最少运费，即其关系为

min 5*B1T1+7*B2T1+10*B3T1+20*B4T1

4、T2仓库向B1、B2、B3、B4顾客配送的最少运费，即其关系为

min 9*B1T2+6*B2T2+7*B3T2+15*B4T2

5、T3仓库向B1、B2、B3、B4顾客配送的最少运费，即其关系为

min 20*B1T3+6*B2T3+7*B3T3+4*B4T3

6、A1,A2工厂分别向T1,T2,T3仓库输送产品大于等于10万个

7、T1,T2,T3仓库分别向B1、B2、B3、B4顾客配送的量必须满足3，5，4，5万个

8、T1,T2,T3仓库分别向B1、B2、B3、B4顾客配送的量同时满足A1,A2工厂分别向T1,T2,T3仓库配送的量

9、从上述分析，该运输问题转变为最小化问题。所以，其解决的途径可以用ga函数、fmincon函数等等来完成。

10、解决方法：①创建目标函数，②创建约束条件函数，③确定变量的上下限值，④确定变量的初值（ga函数不考虑），⑤使用fmincon函数求解

11、按上述思路和方法，编程后运行，可得到如下部分结果

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