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
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);
glp_add_rows(lp, 4);
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_add_cols(lp, 4);
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[1] = 2, ja[1] = 1, ar[1] = 10.0;
ia[2] = 2, ja[2] = 2, ar[2] = -1.0;
ia[3] = 2, ja[3] = 3, ar[3] = 0.0;
ia[4] = 2, ja[4] = 4, ar[4] = 0.0;
ia[5] = 3, ja[5] = 1, ar[5] = 0.0;
ia[6] = 3, ja[6] = 2, ar[6] = 1.0;
ia[7] = 3, ja[7] = 3, ar[7] = -2.0;
ia[8] = 3, ja[8] = 4, ar[8] = 0.0;
ia[9] = 4, ja[9] = 1, ar[9] = 0.0;
ia[10] = 4, ja[10] = 2, ar[10] = 0.0;
ia[11] = 4, ja[11] = 3, ar[11] = 2.0;
ia[12] = 4, ja[12] = 4, ar[12] = -0.5;
fprintf(outfile, "x1 + x2 +x3 +x4 = 1 \n");
/*x1 + x2+ x3 + x4 = o */
ia[13] = 1, ja[13] = 1, ar[13] = 1;
ia[14] = 1, ja[14] = 2, ar[14] = 1;
ia[15] = 1, ja[15] = 3, ar[15] = 1;
ia[16] = 1, ja[16] = 4, ar[16] = 1;
glp_load_matrix(lp, 16, ia, ja, ar);
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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