[Help-glpk] ILP problem , unexpected solution

[satish raman]

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