You are saying that this formulation will give the right results.
#######
param number := 4;
set NUMBER := {1..number};
var x{n in NUMBER}, integer;
s.t. c1: - x[1] + x[2] + x[3]+ 10 * x[4], <= 20;
s.t. c2: x[1] - 3 * x[2] + x[3], <= 30;
s.t. c3: x[2] - 3.5 * x[4], == 0;
s.t. x1: 0 <= x[1] <= 40;
s.t. x4: 2 <= x[4] <= 3;
maximize obj: x[1] + 2 * x[2] + 3 * x[3] + x[4];
#####
X2 and X3 needs to Float / Real, based on the results of the Original
Cplex problem.
This is the answer of the CPLEX problem.
######
No. Column name Activity Lower bound Upper bound
------ ------------ ------------- ------------- -------------
1 x1 40 0 40
2 x2 10.5 0
3 x3 19.5 0
4 x4 * 3 2 3
######
Still the question remains the same,
How you exactly translate this variable
var x{n in NUMBER}?
Noli
On 11/7/12, Raniere Gaia Silva <r.gaia...@gmail.com> wrote:
> Maybe it answer your question:
>
> ###
>
> param number := 4;
>
> set NUMBER := {1..number};
>
> var x{n in NUMBER};
>
> s.t. c1: - x[1] + x[2] + x[3]+ 10 * x[4], <= 20;
> s.t. c2: x[1] - 3 * x[2] + x[3], <= 30;
> s.t. c3: x[2] - 3.5 * x[4], == 0;
> s.t. x1: 0 <= x[1] <= 40;
> s.t. x4: 2 <= x[4] <= 3;
>
> maximize obj: x[1] + 2 * x[2] + 3 * x[3] + x[4];
>
> end;
>
> ###
>
> Once your problem is a MIP, for the integer variables you must use:
>
> ###
>
> var x{n in NUMBER}, integer;
>
> ###
>
> And for binary variables:
>
> ###
>
> var x{n in NUMBER}, binary;
>
> ###
>
> Raniere
>
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