数学建模社区-数学中国

标题: 一段lingo代码（急！） [打印本页]

作者: starbinbin 时间: 2009-8-16 08:44
标题: 一段lingo代码（急！）

model:
!n个货车8收点运输问题;
sets:
  place/place1..place10/:time1,time2,timexie;
car/car1..car3/:time;

jiu/1..9/;
er/1..2/;

lc(car,jiu):lucheng,ct;
plan(place,place,car):a;
distance(place,place):dd,dt,which;
endsets

!这里是数据;
data:
s=1;
time1=-10 1 4 1 4 3.5 2 5 1.5 1;
time2=7 4 6 2 7 5 5 8 4 10 ;
timexie=0 1 2 1 3 2 2.5 3 0.8 0;

  dd=
0       40       60       75       90       90       100       135       80       0
40       0       65       40       100       50       75       110       100       40
60       65       0       75       100       100       75       75       75       60
75       40       75       0       100       50       90       90       125       75
90       100       100       100       0       100       75       75       100       90
90       50       100       50       100       0       70       90       75       90
100       75       75       90       75       70       0       70       100       100
135       110       75       90       75       90       70       0       100       135
80       100       75       125       100       75       100       100       0       80
0       40       60       75       90       90       100       135       80       0 ;
dt=    0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000       0
0.8000       0 1.3000 0.8000 2.0000 1.0000 1.5000 2.2000 2.0000 0.8000
1.2000 1.3000       0 1.5000 2.0000 2.0000 1.5000 1.5000 1.5000 1.2000
1.5000 0.8000 1.5000       0 2.0000 1.0000 1.8000 1.8000 2.5000 1.5000
1.8000 2.0000 2.0000 2.0000       0 2.0000 1.5000 1.5000 2.0000 1.8000
1.8000 1.0000 2.0000 1.0000 2.0000       0 1.4000 1.8000 1.5000 1.8000
2.0000 1.5000 1.5000 1.8000 1.5000 1.4000       0 1.4000 2.0000 2.0000
2.7000 2.2000 1.5000 1.8000 1.5000 1.8000 1.4000       0 2.0000 2.7000
1.6000 2.0000 1.5000 2.5000 2.0000 1.5000 2.0000 2.0000       0 1.6000
      0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000       0;
enddata

!目标函数;
  min=@sum(lc(I,J): lucheng(I,J));

!整数约束;
@for(plan

bin(a));
@for(car(I)

free(time(I)) ;

!从零点开始最终回来;
@for(car(I):
  a(1,1,I)=1;a(10,10,I)=1);

!每个任务只能有一个;
@sum(car(I):
@sum(place(K):
   @sum(place(J): a(J,K,I))))=1;

!每个车的任务最多有四个;
@for(car(I):
  @sum(plan(J,K,I):a(J,k,I))<=6 );

!前一辆车任务大于等于后一辆车;
@for(car(I)|I#lt#3:
  @sum(plan:a(J,K,I))>=@sum(plan:a(J,K,I+1) );

!求which  例子若a（1，：）=[1 0 0 1 0 1 0 0 0 1];
   ! 则  which（1，：）=[1 1 1 4 4 6 6 6 6 10] ;
@for(car(I):
  @for(place(K)|@sum(place(J):a(K,J,I)) #eq#1:
   @for(place(J)|a(K,J,I) #eq#1:
      which(I,K)=J ););
@for(place(K)|@sum(place(J):a(K,J,I)) #ne#1:
which(I,K)=which(I,K-1));

!时间限制;
@for(car(J):
time(J)>=-10;time(J)<=7);

@for(jiu(J):
ct(I,J)=@if(which(I,J)#ne#which(I,J+1),dt(which(I,J),which(I,J+1))+timexie(which(I,J)),0));

@for(jiu(J):
time(I)+ct(I,J)>=time1(J+1);time(I)+ct(I,J)<=time2(J+1));

!求每辆车行驶的距离;
@for(jiu(J):
lucheng(I,J)=@if(which(I,J)#ne#which(I,J+1),dd(which(I,J),which(I,J+1)),0) );
);
end

bin(a));
@for(car(I)

本人新手，这段代码实在是看不懂，请问有没有哪位大侠能够指教一下！跪谢！
model:
!n个货车8收点运输问题;
sets:
  place/place1..place10/:time1,time2,timexie;
car/car1..car3/:time;
jiu/1..9/;
er/1..2/;
lc(car,jiu):lucheng,ct;
plan(place,place,car):a;
distance(place,place):dd,dt,which;
endsets

!这里是数据;
data:
s=1;
time1=-10 1 4 1 4 3.5 2 5 1.5 1;
time2=7 4 6 2 7 5 5 8 4 10 ;
timexie=0 1 2 1 3 2 2.5 3 0.8 0;

  dd=
0 40 60 75 90 90 100 135 80 0
40 0 65 40 100 50 75 110 100 40
60 65 0 75 100 100 75 75 75 60
75 40 75 0 100 50 90 90 125 75
90 100 100 100 0 100 75 75 100 90
90 50 100 50 100 0 70 90 75 90
100 75 75 90 75 70 0 70 100 100
135 110 75 90 75 90 70 0 100 135
80 100 75 125 100 75 100 100 0 80
0 40 60 75 90 90 100 135 80 0 ;
dt=    0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000       0
0.8000       0 1.3000 0.8000 2.0000 1.0000 1.5000 2.2000 2.0000 0.8000
1.2000 1.3000       0 1.5000 2.0000 2.0000 1.5000 1.5000 1.5000 1.2000
1.5000 0.8000 1.5000       0 2.0000 1.0000 1.8000 1.8000 2.5000 1.5000
1.8000 2.0000 2.0000 2.0000       0 2.0000 1.5000 1.5000 2.0000 1.8000
1.8000 1.0000 2.0000 1.0000 2.0000       0 1.4000 1.8000 1.5000 1.8000
2.0000 1.5000 1.5000 1.8000 1.5000 1.4000       0 1.4000 2.0000 2.0000
2.7000 2.2000 1.5000 1.8000 1.5000 1.8000 1.4000       0 2.0000 2.7000
1.6000 2.0000 1.5000 2.5000 2.0000 1.5000 2.0000 2.0000       0 1.6000
      0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000       0;
enddata

!目标函数;
  [email=min=@sum(lc(I,J]min=@sum(lc(I,J[/email]): lucheng(I,J));
!整数约束;
@for(plan

bin(a));
@for(car(I)

free(time(I)) ;
!从零点开始最终回来;
@for(car(I):
  a(1,1,I)=1;a(10,10,I)=1);
!每个任务只能有一个;
@sum(car(I):
@sum(place(K):
   @sum(place(J): a(J,K,I))))=1;

!每个车的任务最多有四个;
@for(car(I):
  @sum(plan(J,K,I):a(J,k,I))<=6 );
!前一辆车任务大于等于后一辆车;
@for(car(I)|I#lt#3:
  @sum(plan:a(J,K,I))>=@sum(plan:a(J,K,I+1) );
!求which  例子若a（1，：）=[1 0 0 1 0 1 0 0 0 1];
   ! 则  which（1，：）=[1 1 1 4 4 6 6 6 6 10] ;
@for(car(I):
  @for(place(K)|@sum(place(J):a(K,J,I)) #eq#1:
   @for(place(J)|a(K,J,I) #eq#1:
      which(I,K)=J ););
@for(place(K)|@sum(place(J):a(K,J,I)) #ne#1:
which(I,K)=which(I,K-1));
!时间限制;
@for(car(J):
time(J)>=-10;time(J)<=7);
@for(jiu(J):
ct(I,J)=@if(which(I,J)#ne#which(I,J+1),dt(which(I,J),which(I,J+1))+timexie(which(I,J)),0));
@for(jiu(J):
time(I)+ct(I,J)>=time1(J+1);time(I)+ct(I,J)<=time2(J+1));
!求每辆车行驶的距离;
@for(jiu(J):
lucheng(I,J)=@if(which(I,J)#ne#which(I,J+1),dd(which(I,J),which(I,J+1)),0) );
);
end

作者: starbinbin 时间: 2009-8-16 08:45
好像发了很多遍。。。。
对了，表情那里的符号原本是@

作者: wxj880516 时间: 2009-8-16 08:52
你想问什么哦？

作者: starbinbin 时间: 2009-8-16 08:56
这段代码怎么报错啊？

欢迎光临数学建模社区-数学中国 (http://www.madio.net/)