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lingo for函数 错误

答案:2  悬赏:80  手机版
解决时间 2021-12-22 11:26
  • 提问者网友:温旧梦泪无声
  • 2021-12-21 20:42
model:
title location problem;
sets:
demand/1..15/:a,b,d;
supply/1..5/:x,y,e;
link(demand,supply):c;
endsets
data:
a=1.25,8.75,0.5,5.75,3,7.25,3.24,8.75,1.75,7.75,2.25,4.25,6.25,5,6.25;
b=1.25,0.75,4.75,5,6.5,7.75,4.75,4.25,7.75,7.25,2.25,5.25,4.50,3.0,5.75;
d=3,5,4,7,6,11,8,10,6,4,6,3,4,3,7;
e=15,20,15,18,22;
enddata
init:
x=1.5,8.25,1.75,7.0,4.25;
y=1.5,1.75,7.5,7.0,5.0;
endinit
[OBJ] min=@sum(link(i,j):c(i,j)*((x(j)-a(i))^2+(y(j)-b(i))^2)^(1/2);
@FOR(demand(i):[DEMAND_CON] @sum(demand(j):c(i,j)) =d(i););
@for(supply(i):[SUPPLY_CON] @sum((j):c(j,i))<=e(i););
@for(supply:@free(X);@free(Y););
END
最佳答案
  • 五星知识达人网友:woshuo
  • 2021-12-21 21:19
model:
title location problem;
sets:
demand/1..15/:a,b,d;
supply/1..5/:x,y,e;
link(demand,supply):c;
endsets
data:
a=1.25,8.75,0.5,5.75,3,7.25,3.24,8.75,1.75,7.75,2.25,4.25,6.25,5,6.25;
b=1.25,0.75,4.75,5,6.5,7.75,4.75,4.25,7.75,7.25,2.25,5.25,4.50,3.0,5.75;
d=3,5,4,7,6,11,8,10,6,4,6,3,4,3,7;
e=15,20,15,18,22;
enddata
init:
x=1.5,8.25,1.75,7.0,4.25;
y=1.5,1.75,7.5,7.0,5.0;
endinit
[OBJ] min=@sum(link(i,j):c(i,j)*((x(j)-a(i))^2+(y(j)-b(i))^2)^(1/2));
@FOR(demand(i):[DEMAND_CON] @sum(supply(j):c(i,j)) =d(i););
@for(supply(j):[SUPPLY_CON] @sum(demand(i):c(i,j))<=e(j););
@for(supply:@free(X);@free(Y););
END
全部回答
  • 1楼网友:天凉才是好个秋
  • 2021-12-21 22:31
你再取一个变量z y=x*z; z>=1 @gin(z); 直接发自己写的函数来看 这样写不对,你自己找个教程好好看一下,要不就百度hi找我,一直改问题不嫌麻烦 你的jk是什么 当然不行
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