/******************************************************** BCL Example Problems ==================== file xbburgl.c `````````````` Burglar problem. Binary variable formulation with index sets. -- Formulating logical conditions with indicator constraints -- (c) 2009-2023 Fair Isaac Corporation author: S.Heipcke, June 2009, rev. Mar. 2011 ********************************************************/ #include <stdio.h>
#include <stdlib.h>
#include "xprb.h"

/****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;
 XPRBidxset ITEMS;                     /* Set of items */
 int i;
 XPRBctr ctr,cobj,Log3a,Log3b;
 XPRBprob prob;

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

/****INDICES****/
 ITEMS=XPRBnewidxset(prob,"Items",8);   /* Create the index set */
 for(i=0;i<8;i++) XPRBaddidxel(ITEMS, ITEMNAMES[i]);

 NItems=XPRBgetidxsetsize(ITEMS);       /* Get the size of the index set */

/****VARIABLES****/
 x = (XPRBvar *)malloc(NItems*sizeof(XPRBvar));
 for(i=0;i<NItems;i++) x[i]=XPRBnewvar(prob,XPRB_BV, if we take item otherwise cobj=XPRBnewctr(prob,"OBJ",XPRB_N); Maximize the total value for(i=0;i<NItems;i++) Select objective function ctr=XPRBnewctr(prob,"WtMax", Weight restriction Logic constraint: Either and="picture" or="tv" not both Values within each pair are same x["vase"]== x["tv"]== Choose exactly one Log3a=XPRBnewctr(prob, <=0 Log3b=XPRBnewctr(prob,>= 2 */
 XPRBaddterm(Log3b, x[XPRBgetidxel(ITEMS,"tv")], 1);
 XPRBaddterm(Log3b, x[XPRBgetidxel(ITEMS,"video")], 1);
 XPRBaddterm(Log3b, NULL, 2);

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

/* Alternative MIP formulation (instead of Log3a and Log3b) */
/*
 ctr = XPRBnewctr(prob, "Log3", XPRB_E);      // x["tv"] = 1 - x["vase"]
 XPRBaddterm(ctr, x[XPRBgetidxel(ITEMS,"tv")], 1);
 XPRBaddterm(ctr, x[XPRBgetidxel(ITEMS,"vase")], 1);
 XPRBaddterm(ctr, NULL, 1);
*/

/****SOLVING + OUTPUT****/
 XPRBsetsense(prob,XPRB_MAXIM);         /* Choose the sense of optimization */
 XPRBmipoptimize(prob,"");              /* Solve the MIP-problem */
 printf("Objective: %g\n",XPRBgetobjval(prob));   /* Get objective value */

 for(i=0;i<NItems;i++)
  if(XPRBgetsol(x[i])>0)
   printf("%s : %g\n", XPRBgetidxelname(ITEMS, i), XPRBgetsol(x[i]));
                                        /* Print out the chosen items */
 return 0;
}