求懂lingo的大佬​帮忙简洁一下程序,让​代码更少点。

SETS:
M/1..5/:;
N/1..4/:;
link1(N,M):x,y;
ENDSETS
DATA:
!价格;
y=10.8,10.95,11.1,11.25,0,100,11.1,11.25,11.4,0,100,100,11,11.15,0,100,100,100,11.3,0;
ENDDATA
MIN = @SUM(link1(i,j):x(i,j)*y(i,j));
!交货量;
x(1,1)+x(2,1)+x(3,1)+x(4,1)=10;
x(1,2)+x(2,2)+x(3,2)+x(4,2)=15;
x(1,3)+x(2,3)+x(3,3)+x(4,3)=25;
x(1,4)+x(2,4)+x(3,4)+x(4,4)=20;
!生产能力;
x(1,1)+x(1,2)+x(1,3)+x(1,4)+x(1,5)<=35;
x(2,1)+x(2,2)+x(2,3)+x(2,4)+x(2,5)<=35;
x(3,1)+x(3,2)+x(3,3)+x(3,4)+x(3,5)<=30;
x(4,1)+x(4,2)+x(4,3)+x(4,4)+x(4,5)<=10;
@GIN(x(1,1));
@GIN(x(1,2));
@GIN(x(1,3));
@GIN(x(1,4));
@GIN(x(1,5));
@GIN(x(2,1));
@GIN(x(2,2));
@GIN(x(2,3));
@GIN(x(2,4));
@GIN(x(2,5));
@GIN(x(3,1));
@GIN(x(3,2));
@GIN(x(3,3));
@GIN(x(3,4));
@GIN(x(3,5));
@GIN(x(4,1));
@GIN(x(4,2));
@GIN(x(4,3));
@GIN(x(4,4));
@GIN(x(4,5));

 채택된 답변

knwgkb
knwgkb 2023년 5월 21일

0 개 추천

参考下:
SETS:
M/1..5/:;
N/1..4/:;
link1(N,M):x,y;
link2(N):v,p;
ENDSETS
DATA:
!价格;
y=10.8,10.95,11.1,11.25,0,100,11.1,11.25,11.4,0,100,100,11,11.15,0,100,100,100,11.3,0;
v=10,15,25,20;
p=35,35,20,10;
ENDDATA
MIN = @SUM(link1(i,j):x(i,j)*y(i,j));
!交货量;
@for(link2(i):@sum(link2(j):x(j,i))=v(i));
!生产能力;
@for(link2(i):@sum(link1(i,j):x(i,j))<p(i));
@for(link1(i,j):@GIN(x(i,j)));

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2023년 5월 21일

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2023년 5월 21일

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