【TWVRP】基于matlab节约算法求解带时间窗的电动车路径规划问题【含Matlab源码 1169期】
作者：互联网
一、简介
基本思想
行时通过这一条弧。
迭代步骤
二、源代码
function [outcome1,outcome2,outcome3]=cw(Numberoffacilities,assignofpoint,ttimeu,timewindow,distMatrix,quantity,Qofcar,p1)
H=1;
outcome3=zeros(1,Numberoffacilities);%与设施点的伪编号是一一对应的
%%%%%%%找到实际分配的设施点，因为之前的chrom中为1的点，可能%%%%%%%%%%%%%
facilitypop=zeros(1:Numberoffacilities);
facilitypop(assignofpoint)=1;
trueSelectefacilities=find(facilitypop(1,:)==1); %被选择的设施编号
trueselectNumberoffacilities=size(trueSelectefacilities,2);
outcome1=cell(1,trueselectNumberoffacilities);
outcome2=cell(1,trueselectNumberoffacilities);
for i=1:trueselectNumberoffacilities%此时的设施为i
a=trueSelectefacilities(i);%设施的伪编号
pointofsubroute=find(assignofpoint==a);%属于该设施点的所有需求点的伪编号
Numberofpointsofsubroute=size(pointofsubroute,2);%该设施点的需求点数量
judge=zeros(1,Numberofpointsofsubroute); %判断需求点的位置情况
chrom1=zeros(1,Numberofpointsofsubroute);%最后输入的是伪编号
chrom2=sort(pointofsubroute); %需求点伪编号从小到大排序
originalchrom2=chrom2;
%%对关键节点先分配一个车辆给它？？？？？？？？？？？？？？？？？？？？？对不对？
%for j=1:Numberofpointsofsubroute
%if timewindow(chrom2(j),2)==timewindow(chrom2(j),3)
%chrom1(j)=H;
%H=H+1;
%end
%end
%%%%%%%%计算到达时间%%%%%%%%%%%%%%%%%%%%
arrivetime=zeros(1,Numberofpointsofsubroute);%需求点的到达时间,与当前的originalchrom2位置一一对应，与分配给该设施点的需求点一一对应
for j=1:Numberofpointsofsubroute
if ttimeu(chrom2(j)+Numberoffacilities,a)<timewindow(chrom2(j),2)
arrivetime(j)=timewindow(chrom2(j),2);
else
arrivetime(j)=ttimeu(chrom2(j)+Numberoffacilities,a);
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%路径节约值列表%%%%%%%%%%%%%%%%%%%%%%
savingnumber=zeros(Numberofpointsofsubroute);%与分配给该设施的需求点的伪编号一一对应
for j=1:Numberofpointsofsubroute
for z=1:Numberofpointsofsubroute
if j~=z
savingnumber(j,z)=distMatrix(chrom2(j)+Numberoffacilities,a)+distMatrix(chrom2(z)+Numberoffacilities,a)-distMatrix(chrom2(j)+Numberoffacilities,chrom2(z)+Numberoffacilities);
end
end
end
[a b]=max(savingnumber);%a输入值，b中对应的数字为行，对应的列数为列
[c d]=max(a);%c输入具体的当前节约值，d为列数
e=b(d);%e为行
%%%在节约值列表中e行d列是当前最大的节约量
E=chrom2(e);%此时的E为节约量最大的需求点伪编号之一
D=chrom2(d);%此时的D为节约量最大的需求点伪编号之1
done=1;
while(done<2)
if judge(e)==0&&judge(d)==0
PD=0;
a=quantity(E,2)+quantity(D,2);
aa=quantity(E,3)+quantity(D,3);
pp1=normcdf(Qofcar,a,sqrt(aa));
a2=find(chrom2==D);
b2=find(chrom2==E);
if pp1>p1
EFj=arrivetime(e)+ttimeu(E+Numberoffacilities,D+Numberoffacilities)-arrivetime(d);
if EFj<0
aheadtime=arrivetime(d)-timewindow(D,2);
if aheadtime>=-EFj
PD=1;
end
elseif EFj==0
PD=1;
elseif EFj>0
delaytime=timewindow(D,3)-arrivetime(d);
if delaytime>=EFj
PD=1;
end
end
if PD==1
chrom1(e)=H;
chrom1(d)=H;%H为车辆编号
judge(e)=1;%与设施直接相连的起点为1
judge(d)=2;%与设施直接相连的终点为2
H=H+1;
arrivetime(d)=arrivetime(d)+EFj;
if b2>a2
chrom2(a2)=E;
chrom2(b2)=D;
end
end
end
savingnumber(e,d)=0;
elseif judge(e)==0&&judge(d)==1;
a2=find(chrom2==D);
b2=find(chrom2==E);
a=find(chrom1==chrom1(a2));
b=sum(quantity(chrom2(a),2))+quantity(E,2);
bb=sum(quantity(chrom2(a),3))+quantity(E,3);
pp1=normcdf(Qofcar,b,sqrt(bb));
c=size(a,2);
if pp1>p1
EFj=arrivetime(e)+ttimeu(E+Numberoffacilities,D+Numberoffacilities)-arrivetime(d);
if EFj<0
a1=10000;
for j=1:c
b1=find(originalchrom2==chrom2(a(j)));
Aheadtime=arrivetime(b1)-timewindow(chrom2(a(j)),2);
if Aheadtime<a1
aheadtime=Aheadtime;
a1=Aheadtime;
end
end
if aheadtime>=-EFj
PD=1;
end
elseif EFj==0
PD=1;
elseif EFj>0
a1=10000;
for j=1:c
b1=find(originalchrom2==chrom2(a(j)));
Delaytime=timewindow(chrom2(a(j)),3)-arrivetime(b1);
if Delaytime<a1
delaytime=Delaytime;
a1=Delaytime;
end
end
if delaytime>=EFj
PD=1;
end
end
if PD==1
chrom1(b2)=chrom1(a2);
a1=find(chrom1==chrom1(a2));
CHROM2=chrom2;
chrom2(a1(1))=CHROM2(b2);
k=2;
while(k<=c+1)
chrom2(a1(k))=CHROM2(a(k-1));
k=k+1;
end
judge(d)=10;%10为内点
judge(e)=1;
end
end
savingnumber(e,d)=0;
elseif judge(e)==0&&judge(d)==2
a2=find(chrom2==D);
b2=find(chrom2==E);
a=find(chrom1==chrom1(a2));
b=sum(quantity(chrom2(a),2))+quantity(E,2);
bb=sum(quantity(chrom2(a),3))+quantity(E,3);
pp1=normcdf(Qofcar,b,sqrt(bb));
c=size(a,2);
if pp1>p1
EFj=arrivetime(d)+ttimeu(E+Numberoffacilities,D+Numberoffacilities)-arrivetime(e);
if EFj<0
aheadtimee=arrivetime(e)-timewindow(E,2);
aheadtimed=arrivetime(d)-timewindow(D,2);
if aheadtimee<aheadtimed
aheadtime=aheadtimee;
else
aheadtime=aheadtimed;
end
if aheadtime>=-EFj
PD=1;
end
elseif EFj==0
PD=1;
elseif EFj>0
delaytimee=timewindow(E,3)-arrivetime(e);
delaytimed=timewindow(D,3)-arrivetime(d);
if delaytimee<delaytimed
delaytime=delaytimee;
else
delaytime=delaytimed;
end
if delaytime>=EFj
PD=1;
end
end
if PD==1
chrom1(b2)=chrom1(a2);
a1=find(chrom1==chrom1(a2));
CHROM2=chrom2;
k=1;
while(k<=c)
chrom2(a1(k))=CHROM2(a(k));
k=k+1;
end
chrom2(a1(k))=E;
judge(d)=10;
judge(e)=2;
end
end
savingnumber(e,d)=0;
elseif judge(e)==1&&judge(d)==0
a2=find(chrom2==D);
b2=find(chrom2==E);
a=find(chrom1==chrom1(b2));
b=sum(quantity(chrom2(a),2))+quantity(D,2);
bb=sum(quantity(chrom2(a),3))+quantity(D,3);
pp1=normcdf(Qofcar,b,sqrt(bb));
c=size(a,2);
if pp1>p1
EFj=arrivetime(e)+ttimeu(E+Numberoffacilities,D+Numberoffacilities)-arrivetime(d);
if EFj<0
a1=10000;
for j=1:c
b1=find(originalchrom2==chrom2(a(j)));
Aheadtime=arrivetime(b1)-timewindow(chrom2(a(j)),2);
if Aheadtime<a1
aheadtime=Aheadtime;
a1=Aheadtime;
end
end
if aheadtime>=-EFj
PD=1;
end
elseif EFj==0
PD=1;
elseif EFj>0
a1=10000;
for j=1:c
b1=find(originalchrom2==chrom2(a(j)));
Delaytime=timewindow(chrom2(a(j)),3)-arrivetime(b1);
if Delaytime<a1
delaytime=Delaytime;
a1=Delaytime;
end
end
三、运行结果
四、备注
版本：2014a
标签：end,arrivetime,1169,TWVRP,EFj,源码,PD,chrom1,chrom2 来源： https://blog.csdn.net/TIQCmatlab/article/details/119279887