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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
By clicking on a file name, a preview is opened at the bottom of this page.
xbburg.cxx[download]
xbburgi.cxx[download]
xbburgl.cxx[download]





xbburgi.cxx

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

  file xbburgi.cxx
  ````````````````
  Burglar problem.
  Binary variable formulation with index sets.

  (c) 2008-2024 Fair Isaac Corporation
      author: S.Heipcke, Jan. 2000, rev. Mar. 2011
********************************************************/


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

using namespace std;
using namespace ::dashoptimization;

/****DATA****/
/* Item:           ca  ne  va  pi  tv  vi  ch  br */
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 */

char *ITEMNAMES[] = {"camera", "necklace", "vase", "picture", "tv", "video", 
            "chest", "brick"};
            
int NItems;                      /* Number of items */

int main(int argc, char **argv)
{
 XPRBvar *x;     
 XPRBindexSet ITEMS;             /* Set of items */
 int i;
 XPRBexpr lobj, kn;  
 XPRBprob p("Burglari");         /* Initialize a new problem in BCL */
 
/****INDICES****/
 ITEMS=p.newIndexSet("Items",8);   /* Create the index set */
 for(i=0;i<8;i++)  ITEMS+=ITEMNAMES[i];
 
 NItems=ITEMS.getSize();         /* Get the size of the index set */
 
/****VARIABLES****/
 x = new XPRBvar[NItems];
 for(i=0;i<NItems;i++)
  x[i] = p.newVar(XPRBnewname("x_%s",ITEMS[i]), XPRB_BV);    
                                 /* 1 if we take item i; 0 otherwise */
 
/****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 chosen items */
  if(x[i].getSol()>0) 
   cout << ITEMS[i] << ": " << x[i].getSol() << endl;  

 delete [] x;
 
 return 0;
} 


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