/******************************************************** Xpress-BCL C++ Example Problems =============================== file xbburgl.cxx ```````````````` Burglar problem. Binary variable formulation with index sets. -- Formulating logical conditions with indicator constraints -- (c) 2009-2024 Fair Isaac Corporation author: S.Heipcke, June 2009, 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;  
 XPRBctr Log3a,Log3b;
 XPRBprob p("BurglarL");         /* 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]), if we take item otherwise for(i=0;i<NItems;i++) lobj +=VALUE[i]*x[i]; Set objective: maximize total value kn <=WTMAX); Weight restriction Logic constraint: Either and="picture" or="tv" not both Values within each pair are the same x[ITEMS["vase"]]== x[ITEMS["tv"]]== Choose exactly one Log3a=p.newCtr("Log3a", Log3b=p.newCtr("Log3b",>= 2);

/* Turn the 2 constraints into indicator constraints */
 Log3a.setIndicator(1, x[ITEMS["vase"]]);
                                   // x["vase"]=1 -> x["tv"]+x["video"]=0
 Log3b.setIndicator(-1, x[ITEMS["vase"]]);
                                   // x["vase"]=0 -> x["tv"]+x["video"]=2

/* Alternative MIP formulation (instead of Log3a and Log3b) */
// p.newCtr("Log3", x[ITEMS["tv"]] = 1 - x[ITEMS["vase"]]); 


/****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>0) 
   cout << ITEMS[i] << ": " << x[i].getSol() << endl;  

 delete [] x;
 
 return 0;
}