Complete Program: mipex1.c

The example derives from lpex8.c. Here are the differences between that linear program and this mixed integer program:

We do not get dual variables this way. If we want dual variables, we must do the following:

  /*------------------------------------------------------------------------*/
  /*  File: examples/src/mipex1.c                                           */
  /*  Version 8.1                                                           */
  /*------------------------------------------------------------------------*/
  /*  Copyright (C) 1997-2002 by ILOG.                                      */
  /*  All Rights Reserved.                                                  */
  /*  Permission is expressly granted to use this example in the            */
  /*  course of developing applications that use ILOG products.             */
  /*------------------------------------------------------------------------*/
  
  /* mipex1.c - Entering and optimizing a MIP problem */
  
  /* Bring in the CPLEX function declarations and the C library
     header file stdio.h with the include of cplex.h. */
  
  #include <ilcplex/cplex.h>
  
  /* Bring in the declarations for the string functions */
  
  #include <string.h>
  #include <stdlib.h>
  
  /* Include declaration for function at end of program */
  
  static int
     setproblemdata (char **probname_p, int *numcols_p, int *numrows_p,
                     int *objsen_p, double **obj_p, double **rhs_p,
                     char **sense_p, int **matbeg_p, int **matcnt_p,
                     int **matind_p, double **matval_p,
                     double **lb_p, double **ub_p, char **ctype_p);
  
  static void
     free_and_null (char **ptr);
  
  
  /* The problem we are optimizing will have 2 rows, 3 columns
     and 6 nonzeros.  */
  
  #define NUMROWS    3
  #define NUMCOLS    4
  #define NUMNZ      9
  
  
  int
  main (void)
  {
  /* Declare pointers for the variables and arrays that will contain
     the data which define the LP problem.  The setproblemdata() routine
     allocates space for the problem data.  */
  
     char     *probname = NULL;
     int      numcols;
     int      numrows;
     int      objsen;
     double   *obj = NULL;
     double   *rhs = NULL;
     char     *sense = NULL;
     int      *matbeg = NULL;
     int      *matcnt = NULL;
     int      *matind = NULL;
     double   *matval = NULL;
     double   *lb = NULL;
     double   *ub = NULL;
     char     *ctype = NULL;
  
     /* Declare and allocate space for the variables and arrays where we will
        store the optimization results including the status, objective value,
        variable values, and row slacks. */
  
     int      solstat;
     double   objval;
     double   x[NUMCOLS];
     double   slack[NUMROWS];
  
  
     CPXENVptr     env = NULL;
     CPXLPptr      lp = NULL;
     int           status;
     int           i, j;
     int           cur_numrows, cur_numcols;
  
     /* Initialize the CPLEX environment */
  
     env = CPXopenCPLEX (&status);
  
     /* If an error occurs, the status value indicates the reason for
        failure.  A call to CPXgeterrorstring will produce the text of
        the error message.  Note that CPXopenCPLEX produces no output,
        so the only way to see the cause of the error is to use
        CPXgeterrorstring.  For other CPLEX routines, the errors will
        be seen if the CPX_PARAM_SCRIND indicator is set to CPX_ON.  */
  
     if ( env == NULL ) {
        char  errmsg[1024];
        fprintf (stderr, "Could not open CPLEX environment.\n");
        CPXgeterrorstring (env, status, errmsg);
        fprintf (stderr, "%s", errmsg);
        goto TERMINATE;
     }
  
     /* Turn on output to the screen */
  
     status = CPXsetintparam (env, CPX_PARAM_SCRIND, CPX_ON);
     if ( status ) {
        fprintf (stderr,
                 "Failure to turn on screen indicator, error %d.\n", status);
        goto TERMINATE;
     }
  
     /* Fill in the data for the problem.  */
  
     status = setproblemdata (&probname, &numcols, &numrows, &objsen, &obj,
                              &rhs, &sense, &matbeg, &matcnt, &matind, &matval,
                              &lb, &ub, &ctype);
     if ( status ) {
        fprintf (stderr, "Failed to build problem data arrays.\n");
        goto TERMINATE;
     }
  
     /* Create the problem. */
  
     lp = CPXcreateprob (env, &status, probname);
  
     /* A returned pointer of NULL may mean that not enough memory
        was available or there was some other problem.  In the case of
        failure, an error message will have been written to the error
        channel from inside CPLEX.  In this example, the setting of
        the parameter CPX_PARAM_SCRIND causes the error message to
        appear on stdout.  */
  
     if ( lp == NULL ) {
        fprintf (stderr, "Failed to create LP.\n");
        goto TERMINATE;
     }
  
     /* Now copy the problem data into the lp */
  
     status = CPXcopylp (env, lp, numcols, numrows, objsen, obj, rhs,
                         sense, matbeg, matcnt, matind, matval,
                         lb, ub, NULL);
  
     if ( status ) {
        fprintf (stderr, "Failed to copy problem data.\n");
        goto TERMINATE;
     }
  
     /* Now copy the ctype array */
  
     status = CPXcopyctype (env, lp, ctype);
     if ( status ) {
        fprintf (stderr, "Failed to copy ctype\n");
        goto TERMINATE;
     }
  
  
     /* Optimize the problem and obtain solution. */
  
     status = CPXmipopt (env, lp);
     if ( status ) {
        fprintf (stderr, "Failed to optimize MIP.\n");
        goto TERMINATE;
     }
  
     solstat = CPXgetstat (env, lp);
  
     /* Write the output to the screen. */
  
     printf ("\nSolution status = %d\n", solstat);
  
     status = CPXgetmipobjval (env, lp, &objval);
     if ( status ) {
        fprintf (stderr,"No MIP objective value available.  Exiting...\n");
        goto TERMINATE;
     }
  
     printf ("Solution value  = %f\n\n", objval);
  
     /* The size of the problem should be obtained by asking CPLEX what
        the actual size is, rather than using what was passed to CPXcopylp.
        cur_numrows and cur_numcols store the current number of rows and
        columns, respectively.  */
  
     cur_numrows = CPXgetnumrows (env, lp);
     cur_numcols = CPXgetnumcols (env, lp);
  
     status = CPXgetmipx (env, lp, x, 0, cur_numcols-1);
     if ( status ) {
        fprintf (stderr, "Failed to get optimal integer x.\n");
        goto TERMINATE;
     }
  
     status = CPXgetmipslack (env, lp, slack, 0, cur_numrows-1);
     if ( status ) {
        fprintf (stderr, "Failed to get optimal slack values.\n");
        goto TERMINATE;
     }
  
     for (i = 0; i < cur_numrows; i++) {
        printf ("Row %d:  Slack = %10f\n", i, slack[i]);
     }
  
     for (j = 0; j < cur_numcols; j++) {
        printf ("Column %d:  Value = %10f\n", j, x[j]);
     }
  
     /* Finally, write a copy of the problem to a file. */
  
     status = CPXwriteprob (env, lp, "mipex1.lp", NULL);
     if ( status ) {
        fprintf (stderr, "Failed to write LP to disk.\n");
        goto TERMINATE;
     }
  
  
  TERMINATE:
  
     /* Free up the problem as allocated by CPXcreateprob, if necessary */
  
     if ( lp != NULL ) {
        status = CPXfreeprob (env, &lp);
        if ( status ) {
           fprintf (stderr, "CPXfreeprob failed, error code %d.\n", status);
        }
     }
  
     /* Free up the CPLEX environment, if necessary */
  
     if ( env != NULL ) {
        status = CPXcloseCPLEX (&env);
  
        /* Note that CPXcloseCPLEX produces no output,
           so the only way to see the cause of the error is to use
           CPXgeterrorstring.  For other CPLEX routines, the errors will
           be seen if the CPX_PARAM_SCRIND indicator is set to CPX_ON. */
  
        if ( status ) {
           char  errmsg[1024];
           fprintf (stderr, "Could not close CPLEX environment.\n");
           CPXgeterrorstring (env, status, errmsg);
           fprintf (stderr, "%s", errmsg);
        }
     }
  
     /* Free up the problem data arrays, if necessary. */
  
     free_and_null ((char **) &probname);
     free_and_null ((char **) &obj);
     free_and_null ((char **) &rhs);
     free_and_null ((char **) &sense);
     free_and_null ((char **) &matbeg);
     free_and_null ((char **) &matcnt);
     free_and_null ((char **) &matind);
     free_and_null ((char **) &matval);
     free_and_null ((char **) &lb);
     free_and_null ((char **) &ub);
     free_and_null ((char **) &ctype);
  
     return (status);
  
  }  /* END main */
  
  
  /* This function fills in the data structures for the mixed integer program:
  
        Maximize
         obj: x1 + 2 x2 + 3 x3 + x4
        Subject To
         c1: - x1 + x2 + x3 + 10x4  <= 20
         c2: x1 - 3 x2 + x3         <= 30
         c3:       x2       - 3.5x4  = 0
        Bounds
         0 <= x1 <= 40
         2 <= x4 <= 3
        Integers
          x4
        End
   */
  
  
  static int
  setproblemdata (char **probname_p, int *numcols_p, int *numrows_p,
                  int *objsen_p, double **obj_p, double **rhs_p,
                  char **sense_p, int **matbeg_p, int **matcnt_p,
                  int **matind_p, double **matval_p,
                  double **lb_p, double **ub_p, char **ctype_p)
  {
     char     *zprobname = NULL;     /* Problem name <= 16 characters */
     double   *zobj = NULL;
     double   *zrhs = NULL;
     char     *zsense = NULL;
     int      *zmatbeg = NULL;
     int      *zmatcnt = NULL;
     int      *zmatind = NULL;
     double   *zmatval = NULL;
     double   *zlb = NULL;
     double   *zub = NULL;
     char     *zctype = NULL;
     int      status = 0;
  
     zprobname = (char *) malloc (16 * sizeof(char));
     zobj      = (double *) malloc (NUMCOLS * sizeof(double));
     zrhs      = (double *) malloc (NUMROWS * sizeof(double));
     zsense    = (char *) malloc (NUMROWS * sizeof(char));
     zmatbeg   = (int *) malloc (NUMCOLS * sizeof(int));
     zmatcnt   = (int *) malloc (NUMCOLS * sizeof(int));
     zmatind   = (int *) malloc (NUMNZ * sizeof(int));
     zmatval   = (double *) malloc (NUMNZ * sizeof(double));
     zlb       = (double *) malloc (NUMCOLS * sizeof(double));
     zub       = (double *) malloc (NUMCOLS * sizeof(double));
     zctype    = (char *) malloc (NUMCOLS * sizeof(char));
  
     if ( zprobname == NULL || zobj    == NULL ||
          zrhs      == NULL || zsense  == NULL ||
          zmatbeg   == NULL || zmatcnt == NULL ||
          zmatind   == NULL || zmatval == NULL ||
          zlb       == NULL || zub     == NULL ||
          zctype    == NULL                       )  {
        status = 1;
        goto TERMINATE;
     }
  
     strcpy (zprobname, "example");
  
     /* The code is formatted to make a visual correspondence
        between the mathematical linear program and the specific data
        items.   */
  
       zobj[0]  = 1.0;   zobj[1]   = 2.0;  zobj[2]    = 3.0;    zobj[3] = 1.0;
  
     zmatbeg[0] = 0;    zmatbeg[1] = 2;    zmatbeg[2] = 5;   zmatbeg[3] = 7;
     zmatcnt[0] = 2;    zmatcnt[1] = 3;    zmatcnt[2] = 2;   zmatcnt[3] = 2;
  
     zmatind[0] = 0;    zmatind[2] = 0;    zmatind[5] = 0;   zmatind[7] = 0;
     zmatval[0] = -1.0; zmatval[2] = 1.0;  zmatval[5] = 1.0; zmatval[7] = 10.0;
  
     zmatind[1] = 1;    zmatind[3] = 1;    zmatind[6] = 1;
     zmatval[1] = 1.0;  zmatval[3] = -3.0; zmatval[6] = 1.0;
  
                        zmatind[4] = 2;                      zmatind[8] = 2;
                        zmatval[4] = 1.0;                    zmatval[8] = -3.5;
  
     zlb[0] = 0.0;   zlb[1] = 0.0;      zlb[2] = 0.0;     zlb[3] = 2.0;
     zub[0] = 40.0;  zub[1] = CPX_INFBOUND; zub[2] = CPX_INFBOUND; zub[3] = 3.0;
  
      zctype[0] = 'C';    zctype[1] = 'C';   zctype[2] = 'C';   zctype[3] = 'I';
  
    /* The right-hand-side values don't fit nicely on a line above.  So put
       them here.  */
  
     zsense[0] = 'L';
     zrhs[0]   = 20.0;
  
     zsense[1] = 'L';
     zrhs[1]   = 30.0;
  
     zsense[2] = 'E';
     zrhs[2]   = 0.0;
  
  TERMINATE:
  
     if ( status ) {
        free_and_null ((char **) &zprobname);
        free_and_null ((char **) &zobj);
        free_and_null ((char **) &zrhs);
        free_and_null ((char **) &zsense);
        free_and_null ((char **) &zmatbeg);
        free_and_null ((char **) &zmatcnt);
        free_and_null ((char **) &zmatind);
        free_and_null ((char **) &zmatval);
        free_and_null ((char **) &zlb);
        free_and_null ((char **) &zub);
        free_and_null ((char **) &zctype);
     }
     else {
        *numcols_p   = NUMCOLS;
        *numrows_p   = NUMROWS;
        *objsen_p    = CPX_MAX;   /* The problem is maximization */
  
        *probname_p  = zprobname;
        *obj_p       = zobj;
        *rhs_p       = zrhs;
        *sense_p     = zsense;
        *matbeg_p    = zmatbeg;
        *matcnt_p    = zmatcnt;
        *matind_p    = zmatind;
        *matval_p    = zmatval;
        *lb_p        = zlb;
        *ub_p        = zub;
        *ctype_p     = zctype;
     }
     return (status);
  
  }  /* END setproblemdata */
  
  
  
  /* This simple routine frees up the pointer *ptr, and sets *ptr to NULL */
  
  static void
  free_and_null (char **ptr)
  {
     if ( *ptr != NULL ) {
        free (*ptr);
        *ptr = NULL;
     }
  } /* END free_and_null */
  
  


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