/********************************************************
  Xpress-BCL C Example Problems
  =============================

  file folioheur.c
  ````````````````
  Modeling a small MIP problem
  to perform portfolio optimization.
   -- Heuristic solution --

  (c) 2008-2024 Fair Isaac Corporation
      author: S.Heipcke, Dec. 2003, rev. Mar. 2011
********************************************************/

#include <stdio.h>
#include "xprb.h"
#include "xprs.h"

#define MAXNUM 4                  /* Max. number of different assets */

#define NSHARES 10                /* Number of shares */
#define NRISK 5                   /* Number of high-risk shares */
#define NNA 4                     /* Number of North-American shares */

void solveHeur(void);

double RET[] = {5,17,26,12,8,9,7,6,31,21};  /* Estimated return in investment */
int RISK[] = {1,2,3,8,9};         /* High-risk values among shares */
int NA[] = {0,1,2,3};             /* Shares issued in N.-America */

XPRBprob prob;
XPRBvar frac[NSHARES];            /* Fraction of capital used per share */
XPRBvar buy[NSHARES];             /* 1 if asset is in portfolio, 0 otherwise */

int main(int argc, char **argv)
{
 int s;
 XPRBctr Risk,Na,Return,Cap,Num;

 prob = XPRBnewprob("FolioMIPHeur");  /* Initialize a new problem in BCL */

/* Create the decision variables (including upper bounds for `frac') */
 for(s=0;s<NSHARES;s++) frac[s]=XPRBnewvar(prob, XPRB_PL,="frac" buy[s]=XPRBnewvar(prob, XPRB_BV,="buy" Objective: total return Return=XPRBnewctr(prob, for(s=0;s<NSHARES;s++) Set the objective function Limit percentage of high-risk values Risk=XPRBnewctr(prob, Minimum amount North-American Na=XPRBnewctr(prob, Spend all capital Cap=XPRBnewctr(prob, number assets Num=XPRBnewctr(prob, Linking variables XPRBnewprec(prob,="Link" Solve problem heuristically XPRBmipoptimize(prob,="" Solution printing if(XPRBgetmipstat(prob)==XPRB_MIP_SOLUTION XPRBgetmipstat(prob)==XPRB_MIP_OPTIMAL) solution: Total return: : else solution is void XPRBbasis int double Disable automatic cuts Switch presolve off Get feasibility tolerance LP-problem basis=XPRBsavebasis(prob); Save current Fix for which at or a relatively large value fsol[s]=XPRBgetsol(frac[s]);> 0,2-TOL)  XPRBsetlb(buy[s], 1);
 }

 XPRBmipoptimize(prob, "c");       /* Continue solving the MIP-problem */
 ifgsol=0;
 if(XPRBgetmipstat(prob)==XPRB_MIP_SOLUTION ||
    XPRBgetmipstat(prob)==XPRB_MIP_OPTIMAL)
 {                                 /* If an integer feas. solution was found */
  ifgsol=1;
  solval=XPRBgetobjval(prob);      /* Get the value of the best solution */
  printf("Heuristic solution: Total return: %g\n", XPRBgetobjval(prob));
  for(s=0;s<NSHARES;s++) : Re-initialize the global search Reset variables to their original bounds for(s=0;s<NSHARES;s++)> 0,2-TOL))
  {
   XPRBsetlb(buy[s], 0);
   XPRBsetub(buy[s], 1);
  }

 XPRBloadbasis(basis);      /* Load the saved basis: bound changes are
                               immediately passed on from BCL to the
                               Optimizer if the problem has not been modified
                               in any other way, so that there is no need to
                               reload the matrix */
 XPRBdelbasis(basis);       /* No need to store the saved basis any longer */
 if(ifgsol==1)
  XPRSsetdblcontrol(XPRBgetXPRSprob(prob), XPRS_MIPABSCUTOFF, solval+TOL);
                            /* Set the cutoff to the best known solution */
}