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Portfolio - Quadratic Programming with discrete variables

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Source Files
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Data Files

xbportf.cxx

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

file xbportf.cxx


(c) 2008-2023 Fair Isaac Corporation
author: S.Heipcke, Jan. 2000, rev. Mar. 2011
********************************************************/

/* In this model, a choice has to be made which values are taken   *
* into a portfolio in order to minimize the total cost. The costs *
* for some values are interrelated, introducing a quadratic part  *
* to the objective function. Upper bounds are given on the total  *
* number of values and the share of each value that may be taken. */

#include <iostream>
#include <cstring>
#include <cstdio>
#include "xprb_cpp.h"
#include "xprs.h"

using namespace std;
using namespace ::dashoptimization;

#define NVal 30                       /* Total number of values */
#define LIMIT 20                      /* Maximum number to be chosen */

#define QFILE XPRBDATAPATH "/portf/pfqcost.dat" /* Quadratic cost coeff.s */
#define BFILE XPRBDATAPATH "/portf/pfubds.dat"  /* Upper bds. on percentages */
#define CFILE XPRBDATAPATH "/portf/pflcost.dat" /* Linear cost coefficients */

/**** DATA ****/
double Cost[NVal];                    /* Coeff. of lin. part of the obj. */
double QCost[NVal][NVal];             /* Coeff. of quad. part of the obj. */
double UBnd[NVal];                    /* Upper bound values */

/***********************************************************************/

void modFolio(XPRBprob &p)
{
XPRBexpr le, qobj;
XPRBvar x[NVal];                 /* Amount of a value taken into
the portfolio */
XPRBvar y[NVal];                 /* 1 if value i is chosen, else 0 */
int i,j;

/**** VARIABLES ****/
for(i=0;i<NVal;i++)
{
x[i] = p.newVar(XPRBnewname("x_%d",i+1), XPRB_PL, 0, UBnd[i]);
y[i] = p.newVar(XPRBnewname("y_%d",i+1), XPRB_BV);
}

/****OBJECTIVE****/
for(i=0;i<NVal;i++)              /* Define objective: total cost */
{
qobj += Cost[i]*x[i];
qobj += QCost[i][i]*sqr(x[i]);
for(j=i+1;j<NVal;j++)
qobj += QCost[i][j]*x[i]*x[j];
}
p.setObj(qobj);                  /* Set objective function */

/**** CONSTRAINTS ****/
/* Amounts of values chosen must add up to 100% */
for(i=0;i<NVal;i++) le += x[i];
p.newCtr("C1", le == 100);

for(i=0;i<NVal;i++)              /* Upper limits */
p.newCtr("UL", x[i] <= UBnd[i]*y[i]);

le = 0;                          /* Limit on total number of values */
for(i=0;i<NVal;i++) le += y[i];
p.newCtr("Card", le <= LIMIT);

/****SOLVING + OUTPUT****/
// p.print();			  ~~COMMENT24~~
p.exportProb(XPRB_MPS,"Portf");  /* Output the matrix in MPS format */
p.exportProb(XPRB_LP,"Portf");   /* Output the matrix in LP format */

p.setSense(XPRB_MINIM);      	  /* Choose the sense of the optimization */
p.lpOptimize("");                /* Solve the QP-problem, use 'mipOptimize' to
solve MIQP */

cout << "Objective function value: " << p.getObjVal() << endl;
for(i=0;i<NVal;i++)
cout << x[i].getName() << ": " << x[i].getSol() << ", ";
cout << endl;
}

/***********************************************************************/

/**** Read data from files ****/
{
int i,j;
double value;
FILE *datafile;

memset(QCost,0,NVal*NVal*sizeof(double));      /* Initialize Q to 0 */
datafile=fopen(QFILE,"r");
while (XPRBreadlinecb(XPRB_FGETS, datafile, 200, "i,i,g", &i, &j,
&value) == 3)
QCost[i-1][j-1]=value;
fclose(datafile);

/* Read the linear cost data file */
datafile=fopen(CFILE,"r");
while (XPRBreadlinecb(XPRB_FGETS, datafile, 200, "i,g", &i, &value) == 2)
Cost[i-1]=value;
fclose(datafile);

/* Read the bounds data file */
datafile=fopen(BFILE,"r");
while (XPRBreadlinecb(XPRB_FGETS, datafile, 200, "i,g", &i, &value) == 2)
UBnd[i-1]=value;
fclose(datafile);
}

/***********************************************************************/

int main(int argc, char **argv)
{
XPRBprob p("Portfolio");             /* Initialize a new problem in BCL */
readData();                          /* Data input from file */
modFolio(p);                         /* Formulate and solve the problem */

return 0;
}