Burglar - Use of index sets, formulating logical constraints

• 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

/********************************************************
  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 
#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

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

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

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


#include 
#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;i0) 
   cout << ITEMS[i] << ": " << x[i].getSol() << endl;  

 delete [] x;
 
 return 0;
} 

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

  file xbburgl.cxx
  ````````````````
  Burglar problem.
  Binary variable formulation with index sets.
  -- Formulating logical conditions 
     with indicator constraints --

  (c) 2009 Fair Isaac Corporation
      author: S.Heipcke, June 2009, rev. Mar. 2011
********************************************************/

#include 
#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= 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;i0) 
   cout << ITEMS[i] << ": " << x[i].getSol() << endl;  

 delete [] x;
 
 return 0;
}