/********************************************************
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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/********************************************************
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 <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;
}
|
/********************************************************
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 <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]), 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 */
/** Logic constraint:
** Either take "vase" and "picture" or "tv" and "video" (but not both pairs). */
/* Values within each pair are the same */
p.newCtr("Log1", x[ITEMS["vase"]] == x[ITEMS["picture"]]);
p.newCtr("Log2", x[ITEMS["tv"]] == x[ITEMS["video"]]);
/* Choose exactly one pair */
Log3a = p.newCtr("Log3a", x[ITEMS["tv"]]+x[ITEMS["video"]] <= 0);
Log3b = p.newCtr("Log3b", x[ITEMS["tv"]]+x[ITEMS["video"]] >= 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 */
if(x[i].getSol()>0)
cout << ITEMS[i] << ": " << x[i].getSol() << endl;
delete [] x;
return 0;
}
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