[英]linear programming in Lingo
I want to do a linear programming with Lingo, I have the solution but I want to improve the code. 我想用Lingo进行线性编程,我有解决方案,但我想改善代码。
Here is what I want to do: 这是我想做的:
SETS:
SEMANA/ 1..12/: D, X, I, Y, Z, R, n;
ENDSETS
X(1)>=D(1);
X(2)+I(1)>=D(2);
X(3)+I(2)>=D(3);
X(4)+I(3)>=D(4);
X(5)+I(4)>=D(5);
X(6)+I(5)>=D(6);
X(7)+I(6)>=D(7);
X(8)+I(7)>=D(8);
X(9)+I(8)>=D(9);
X(10)+I(9)>=D(10);
X(11)+I(10)>=D(11);
X(12)+I(11)>=D(12);
I have tried this option but there is a mistake that says: Subscript out of range on attribute I. 我已经尝试过该选项,但是有一个错误提示:下标超出属性I的范围。
@FOR (SEMANA(j):
X(j)+ I(j-1)>= D(j)) ;
I(j-1)
is out of range so I can´t solve the problem. I(j-1)
超出范围,所以我无法解决问题。
Thank you 谢谢
Notice that your code will try to get into I(0) when j=1. 请注意,当j = 1时,您的代码将尝试进入I(0)。 I(0) does not defined and that's why it's out of range problem.
I(0)没有定义,这就是为什么它超出范围的问题。
You need a index filter 您需要索引过滤器
@FOR (SEMANA(j) | j#GT#1:
X(j)+ I(j-1)>= D(j)) ;
This would be equivalent to having I(0)=0, but only implicitly. 这将等效于I(0)= 0,但仅是隐式的。 If you have I(0)>0, eg you have positive initial inventory, then you will need to extend your index set to include '0' and put an additional constraint I(0)=INITIAL_VALUE
如果您的I(0)> 0,例如初始库存为正,那么您将需要扩展索引集以使其包括“ 0”,并附加一个约束I(0)= INITIAL_VALUE
Consequently, your @for loop will have to be 因此,您的@for循环必须为
@FOR (SEMANA(j) | j#GT#0:
X(j)+ I(j-1)>= D(j)) ;
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