Question

请求高手帮我看看这个lingo程序错在哪里？ 急急急急！！！

store/1..12/:x,q,m,n;!m为备用容量上限,n为容量上限，因为有的仓储可能没有被选取;
sall/1..13/:p;
link(sall,store):cost,volume;
endsets
data:
q=5600 8000 9500 4000 3500 8500 6500 3500 4000 9500 9000 3800;
n=300 300 200 270 280 350 260 220 280 300 350 230;
p=125 90 80 100 105 100 90 70 30 130 95 125 100;
!用足够大的数10000000来代替无穷大;
cost=0.880000 1.111111 1.062500 0.650000 0.904762 1.050000 1.222222 1.285714 2.000000 0.615385 0.736842 0.680000 0.980000
0.960000 1.311111 1.187500 0.780000 1.000000 1.200000 1.500000 1.785714 2.833333 0.692308 0.947368 0.624000 1.220000
1.080000 1.611111 1.312500 0.980000 1.047619 2.000000 1.111111 1.928571 3.000000 0.730769 1.105263 0.760000 1.080000
1.000000 1.222222 1.125000 1.200000 0.904762 1.800000 1.166667 2.071429 3.333333 10000000 10000000 0.880000 1.200000
0.760000 2.077778 1.250000 10000000 10000000 10000000 1.000000 2.642857 5.166667 10000000 10000000 1.008000 0.950000
0.904000 2.166667 1.600000 10000000 10000000 10000000 0.944444 2.285714 4.333333 0.923077 1.263158 1.040000 1.000000
0.704000 1.166667 1.187500 0.880000 1.047619 0.900000 1.388889 2.928571 5.266667 0.753846 1.400000 0.624000 1.700000
0.944000 1.355556 1.625000 0.990000 1.000000 1.250000 0.977778 1.857143 4.266667 1.046154 1.052632 0.760000 1.350000
1.624000 2.000000 1.750000 1.250000 0.761905 0.900000 1.277778 2.571429 4.400000 1.076923 1.315789 0.864000 1.300000
1.120000 1.777778 1.937500 1.450000 1.238095 0.850000 2.222222 2.500000 6.000000 0.923077 10000000 10000000 1.250000
1.056000 2.000000 1.850000 1.300000 10000000 10000000 10000000 1.971429 5.333333 0.769231 10000000 10000000 1.400000
1.040000 2.055556 2.075000 1.000000 10000000 10000000 10000000 2.000000 5.833333 0.884615 1.473684 1.240000 1.700000 ;
enddata
@for(store:m(j)=@if(x(j)#eq#0,0,n(j)));!让没被选取的仓储的容量为0;
[obj] min=@sum(link:volume*cost)+@sum(store:q(j)*x(j));!目标函数;
!容量限制;
@for(store(j):@sum(sall:volume(i,j))<=m(j));
!需求限制;
@for(sall:@sum(store:volume(i,j))=p(i));
!0-1变量;
@for(store:@bin(x));

Answer 1

经过修改之后，已经没有语法错误了，但是由于我的软件是盗版，所以算不了这么多变量，你自己试试吧。
存在的问题主要有：刚开始的时候要有“set:”，后面的i，j该加的加上就好了，没有啥大问题。如果不加就会出现语法错误。
我做的如下：
sets:
store/1..12/:x,q,m,n;!m为备用容量上限,n为容量上限，因为有的仓储可能没有被选取;
sall/1..13/:p;
link(sall,store):cost,volume;
endsets
data:
q=5600 8000 9500 4000 3500 8500 6500 3500 4000 9500 9000 3800;
n=300 300 200 270 280 350 260 220 280 300 350 230;
p=125 90 80 100 105 100 90 70 30 130 95 125 100;
!用足够大的数10000000来代替无穷大;
cost=0.880000 1.111111 1.062500 0.650000 0.904762 1.050000 1.222222 1.285714 2.000000 0.615385 0.736842 0.680000 0.980000
0.960000 1.311111 1.187500 0.780000 1.000000 1.200000 1.500000 1.785714 2.833333 0.692308 0.947368 0.624000 1.220000
1.080000 1.611111 1.312500 0.980000 1.047619 2.000000 1.111111 1.928571 3.000000 0.730769 1.105263 0.760000 1.080000
1.000000 1.222222 1.125000 1.200000 0.904762 1.800000 1.166667 2.071429 3.333333 10000000 10000000 0.880000 1.200000
0.760000 2.077778 1.250000 10000000 10000000 10000000 1.000000 2.642857 5.166667 10000000 10000000 1.008000 0.950000
0.904000 2.166667 1.600000 10000000 10000000 10000000 0.944444 2.285714 4.333333 0.923077 1.263158 1.040000 1.000000
0.704000 1.166667 1.187500 0.880000 1.047619 0.900000 1.388889 2.928571 5.266667 0.753846 1.400000 0.624000 1.700000
0.944000 1.355556 1.625000 0.990000 1.000000 1.250000 0.977778 1.857143 4.266667 1.046154 1.052632 0.760000 1.350000
1.624000 2.000000 1.750000 1.250000 0.761905 0.900000 1.277778 2.571429 4.400000 1.076923 1.315789 0.864000 1.300000
1.120000 1.777778 1.937500 1.450000 1.238095 0.850000 2.222222 2.500000 6.000000 0.923077 10000000 10000000 1.250000
1.056000 2.000000 1.850000 1.300000 10000000 10000000 10000000 1.971429 5.333333 0.769231 10000000 10000000 1.400000
1.040000 2.055556 2.075000 1.000000 10000000 10000000 10000000 2.000000 5.833333 0.884615 1.473684 1.240000 1.700000 ;
enddata
[obj] min=@sum(link(i,j):volume(i,j)*cost(i,j))+@sum(store(j):q(j)*x(j));!目标函数;
@for(store(j):m(j)=@if(x(j)#eq#0,0,n(j)));!让没被选取的仓储的容量为0;
!容量限制;
@for(store(j):@sum(sall(i):volume(i,j))<=m(j));
!需求限制;
@for(sall(i):@sum(store(j):volume(i,j))=p(i));
!0-1变量;
@for(store(j):@bin(x(j)));