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## Re: [Help-glpk] ILP problem , unexpected solution

 From: Jeffrey Kantor Subject: Re: [Help-glpk] ILP problem , unexpected solution Date: Tue, 29 Dec 2009 07:35:32 -0500

On Tue, Dec 29, 2009 at 6:56 AM, satish raman wrote:
Hi all,
I am trying to solve a simple integer linear program
with the following constraints

x1 + x2 +x3 +x4 = 1
10.00x1 + (-1.00)x2 < 5.00
1.00x2 and +(-2.00)x3  < 5.00
2.00x3 + (-0.50)x4  < 5.00
I have no objective function to optimise.

I am expecting output as x1 = 0 x2 = 1 x3 = 0 and x4 = 0

But when i solved using glpk 4.23 , i always get the output x1 = 0 x2 = 0 x3 = 0 and x4 = 0

All zeros appears to be a feasible solution to your problem. If you need a different solution, you'll need to give glpk a few more hints.

My code is as follows. Can somebody pls help me pointing out the problem.

int main()
{
glp_prob *lp;
int ia[1+1000], ja[1+1000];
double ar[1+1000];
int  x1, x2, x3,x4;
;

outfile = fopen("lpout.out","w");
float cur1, cur2, cur3, cur4;
float bound_p, bound_q, bound_r;

glp_iocp      parm;
glp_smcp simplex_parm;
int ret;

/*10.00x1 + (-1.00)x2 < 5.00
1.00x2 and +(-2.00)x3  < 5.00
2.00x3 + (-0.50)x4  < 5.00
*/
lp = glp_create_prob();
glp_set_prob_name(lp, "sample");
glp_set_obj_dir(lp, GLP_MAX);

bound_p = 5 ;
bound_q =  5;
bound_r = 5 ;

fprintf(outfile, "bound_p = %.2f bound_q %.2f bound_r =  %.2f \n ", bound_p, bound_q, bound_r);

glp_set_row_name(lp, 1, "p");
glp_set_row_bnds(lp, 1, GLP_UP, -DBL_MAX, bound_p);
glp_set_row_name(lp, 2, "q");
glp_set_row_bnds(lp, 2, GLP_UP, -DBL_MAX, bound_q);
glp_set_row_name(lp, 3, "r");
glp_set_row_bnds(lp, 3, GLP_UP, -DBL_MAX, bound_r);

glp_set_row_name(lp, 4, "o");
glp_set_row_bnds(lp, 4, GLP_FX, 1, 1);

glp_set_col_name(lp, 1, "x1");
glp_set_col_bnds(lp, 1, GLP_DB, 0, 1);
glp_set_col_kind(lp, 1, GLP_IV);

glp_set_col_name(lp, 2, "x2");
glp_set_col_bnds(lp, 2, GLP_DB, 0, 1);
glp_set_col_kind(lp, 2, GLP_IV);

glp_set_col_name(lp, 3, "x3");
glp_set_col_bnds(lp, 3, GLP_DB, 0, 1);
glp_set_col_kind(lp, 3, GLP_IV);

glp_set_col_name(lp, 4, "x4");
glp_set_col_bnds(lp, 4, GLP_DB, 0, 1);
glp_set_col_kind(lp, 4, GLP_IV);

ia = 2, ja = 1, ar = 10.0;
ia = 2, ja = 2, ar = -1.0;
ia = 2, ja = 3, ar = 0.0;
ia = 2, ja = 4, ar = 0.0;

ia = 3, ja = 1, ar = 0.0;
ia = 3, ja = 2, ar = 1.0;
ia = 3, ja = 3, ar = -2.0;
ia = 3, ja = 4, ar = 0.0;

ia = 4, ja = 1, ar = 0.0;
ia = 4, ja = 2, ar = 0.0;
ia = 4, ja = 3, ar = 2.0;
ia = 4, ja = 4, ar = -0.5;

fprintf(outfile, "x1 + x2 +x3 +x4 = 1 \n");
/*x1 + x2+ x3 + x4  = o */
ia = 1, ja = 1, ar = 1;
ia = 1, ja = 2, ar = 1;
ia = 1, ja = 3, ar = 1;
ia = 1, ja = 4, ar = 1;

glp_init_smcp(&simplex_parm);
simplex_parm.presolve = GLP_OFF;

if(glp_simplex(lp, &simplex_parm) != 0)
{
printf("failure of simplex\n");
exit(-1);
}

x1 = glp_get_col_prim(lp, 1);
x2 = glp_get_col_prim(lp, 2);
x3 = glp_get_col_prim(lp, 3);
x4 = glp_get_col_prim(lp, 4);

fprintf(outfile,"\n x1 = %d; x2 = %d; x3 = %d x4 = %d \n", x1, x2, x3, x4);

glp_delete_prob(lp);

return 0;
}
The terminal output is
"      0:   objval =   0.000000000e+00   infeas =   1.000000000e+00 (0)
1:   objval =   0.000000000e+00   infeas =   0.000000000e+00 (0)
OPTIMAL SOLUTION FOUND"

Thanks,
satish

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