Question

lingo两次求和

model
sets:
p1/1..4/:b;
p2/1..4/;
link(p2,p1):x,v;
endsets
data:
v=2800 4500 6000 7300
2800 4500 6000 7300
2800 4500 6000 7300
2800 4500 6000 7300 ;
b=15,10,20,12;
enddata
min=@sum(link:x*v);
@sum(p1(i):(@sum p2(j):x(i,j)))<=b(i);
end
哪里错了？？？
目标函数：∑∑ v(i,j)x（i,j）
bi=(1,2,3,4)

∑ i （∑ j x(i,j)<b(i)）
怎么解决？

Answer 1

看看是不是这样。。。

model:
sets:
p1/1..4/:b;
p2/1..4/;
link(p2,p1):x,v;
endsets
data:
v=2800 4500 6000 7300
2800 4500 6000 7300
2800 4500 6000 7300
2800 4500 6000 7300 ;
b=15,10,20,12;
enddata
min=@sum(link:x*v);
@for(p1(i):@sum(p2(j):x(i,j))<=b(i));
end

追问

model:
sets:
p1/1..4/:b;
p2/1..4/;
p3/1..4/:k;
link(p1,p2):x,v;
endsets
data:
v=2800 4500 6000 7300
2800 4500 6000 7300
2800 4500 6000 7300
2800 4500 6000 7300 ;
b=15,10,20,12;
enddata
min=@sum(p1(i):@sum( p2(j):x(i,j)*v(i,j)));
@for(p1(i):@sum (p3(k):@sum (p2(j):x(x+1-k,j)))>=b(i));
@for(link:@gin(x));
end
三次累加呢。这个呢···

追答

数据是二维的何来三次累加。。。

追问

∑（i=1到4） ( ∑(k=1到i) ∑(k到4)x(i+1-k,j)>=b(i) )
多了一个K怎么解？

追答

model:
sets:
p1/1..4/:b;
p2/1..4/;
p3/1..4/:k;
link(p1,p2):x,v;
endsets
data:
v=2800 4500 6000 7300
2800 4500 6000 7300
2800 4500 6000 7300
2800 4500 6000 7300 ;
b=15,10,20,12;
enddata
min=@sum(p1(i):@sum( p2(j):x(i,j)*v(i,j)));
@for(p1(i):@sum (p3(k)|k#le#i:@sum (p2(j):x(i+1-k,j)))>=b(i));
@for(link:@gin(x));
end

追问

如何让 j=(k到4）??

追答

@for(p1(i):@sum (p3(k)|k#le#i:@sum (p2(j)|j#ge#k:x(i+1-k,j)))>=b(i));