Re: [Help-glpk] Implementing conditional amount constraints using binary variables

[Xypron]

Hello Yaron,

param M := 1000;
var a, >=0, <= M;
var b, >=0, <= M;
var x, binary;

minimize opt: a - .3 * b;
s.t. c0: a <= x * M;
s.t. c1: a - b <= ( 1 - x ) * M;
s.t. c2: b - a <= ( 1 - x ) * M;

solve;
display a, b, x;
end;

Best regards

Xypron

Yaron Kretchmer wrote:
> Both a and b can be bounded by a  large M.
> So let me reformulate my problem:
>
> 0 <= a <= M
>
> if a > 0 : a=b
>
> if a=0 : 0<=b<=M
>
> Thanks
> Kretch
>
>