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Burglar - Use of index sets, formulating logical constraints

Description
Several versions of a simple knapsack problem:
  • xbburg: standard formlation
  • xbburgi: shows how to index an array of variables by an index set
  • xbburgl: adds several indicator constraints to state logical conditions
Further explanation of this example: Quick reference guide 'MIP formulations and linearizations', Section 4 Indicator constraints


Source Files





xbburg.cxx

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

  file xbburg.cxx
  ```````````````
  Burglar problem, binary variable formulation.

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

#include <iostream>
#include "xprb_cpp.h"

using namespace std;
using namespace ::dashoptimization;

#define NItems 8                  /* Number of items */

/****DATA****/
/* Item:             1   2   3   4   5   6   7   8 */
double VALUE[]  =  {15,100, 90, 60, 40, 15, 10,  1};    /* Value of items */
double WEIGHT[] =  { 2, 20, 20, 30, 40, 30, 60, 10};    /* Weight of items */
double WTMAX    = 102;            /* Max weight allowed for haul */

int main(int argc, char **argv)
{
 XPRBvar x[NItems];  
 XPRBexpr lobj, kn;  
 int i;
 XPRBprob p("Burglar");           /* Initialize a new problem in BCL */
 
/****VARIABLES****/
                                  /* 1 if we take item i; 0 otherwise */
 for(i=0;i<NItems;i++)  x[i]=p.newVar("x",XPRB_BV);
 
/****OBJECTIVE****/
 for(i=0;i<NItems;i++)  lobj += VALUE[i]*x[i]; 
 p.setObj(p.newCtr("OBJ",lobj));  /* Set objective: maximize total value */ 

/****CONSTRAINTS****/
 for(i=0;i<NItems;i++)  kn += WEIGHT[i]*x[i];  
 p.newCtr("WtMax", kn <= WTMAX);  /* Weight restriction */

/****SOLVING + OUTPUT****/
 p.setSense(XPRB_MAXIM);          /* Choose the sense of the optimization */
 p.mipOptimize("");               /* Solve the MIP-problem */
 cout << "Objective: " << p.getObjVal() << endl;  /* Get objective value */

 for(i=0;i<NItems;i++)            /* Print out the solution */
  cout << x[i].getName() << ":" << x[i].getSol() << " ";  
 cout << endl;

 return 0;
} 


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