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Els - An economic lot-sizing problem solved by cut-and-branch and branch-and-cut heuristics

Description
The version 'xbels' of this example shows how to implement cut-and-branch (= cut generation at the root node of the MIP search) and 'xbelsc' implements a branch-and-cut (= cut generation at the MIP search tree nodes) algorithm using the cut manager.

Source Files
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xbels.cxx

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

file xbels.cxx

Economic lot sizing, ELS, problem, solved by adding
(l,S)-inequalities) in several rounds looping over
the root node.

ELS considers production planning over a horizon
of T periods. In period t, t=1,...,T, there is a
given demand DEMAND[t] that must be satisfied by
production prod[t] in period t and by inventory
carried over from previous periods. There is a
set-up up cost SETUPCOST[t] associated with
production in period t. The unit production cost
in period t is PRODCOST[t]. There is no inventory
or stock-holding cost.

(c) 2008 Fair Isaac Corporation
author: S.Heipcke, 2001, rev. Mar. 2011
********************************************************/

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

using namespace std;
using namespace ::dashoptimization;

#define EPS    1e-6

#define T 6                             /* Number of time periods */

/****DATA****/
int DEMAND[]    = { 1, 3, 5, 3, 4, 2};  /* Demand per period */
int SETUPCOST[] = {17,16,11, 6, 9, 6};  /* Setup cost per period */
int PRODCOST[]  = { 5, 3, 2, 1, 3, 1};  /* Production cost per period */
int D[T][T];                            /* Total demand in periods t1 - t2 */

XPRBvar prod[T];                        /* Production in period t */
XPRBvar setup[T];                       /* Setup in period t */

/***********************************************************************/

void modEls(XPRBprob &p)
{
int s,t,k;
XPRBexpr cobj,le;

for(s=0;s<T;s++)
for(t=0;t<T;t++)
for(k=s;k<=t;k++)
D[s][t] += DEMAND[k];

/****VARIABLES****/
for(t=0;t<T;t++)
{
prod[t]=p.newVar(XPRBnewname("prod%d",t+1));
setup[t]=p.newVar(XPRBnewname("setup%d",t+1), XPRB_BV);
}

/****OBJECTIVE****/
for(t=0;t<T;t++)                       /* Minimize total cost */
cobj += SETUPCOST[t]*setup[t] + PRODCOST[t]*prod[t];
p.setObj(cobj);

/****CONSTRAINTS****/
/* Production in period t must not exceed the total demand for the
remaining periods; if there is production during t then there
is a setup in t */
for(t=0;t<T;t++)
p.newCtr("Production", prod[t] <= D[t][T-1]*setup[t]);

/* Production in periods 0 to t must satisfy the total demand
during this period of time */
for(t=0;t<T;t++)
{
le=0;
for(s=0;s<=t;s++) le += prod[s];
p.newCtr("Demand", le >= D[0][t]);
}

}

/**************************************************************************/
/*  Cut generation loop at the top node:                                  */
/*    solve the LP and save the basis                                     */
/*    get the solution values                                             */
/*    identify and set up violated constraints                            */
/**************************************************************************/
void solveEls(XPRBprob &p)
{
double objval;               /* Objective value */
int t,l;
int starttime;
int ncut, npass, npcut;      /* Counters for cuts and passes */
double solprod[T], solsetup[T];   /* Solution values for var.s prod & setup */
double ds;
XPRBbasis basis;
XPRBexpr le;

starttime=XPRB::getTime();
/* Disable automatic cuts - we use our own */
/* Switch presolve off */
ncut = npass = 0;

do
{
npass++;
npcut = 0;
p.lpOptimize("p");          /* Solve the LP */
basis = p.saveBasis();      /* Save the current basis */
objval = p.getObjVal();     /* Get the objective value */

/* Get the solution values: */
for(t=0;t<T;t++)
{
solprod[t]=prod[t].getSol();
solsetup[t]=setup[t].getSol();
}

/* Search for violated constraints: */
for(l=0;l<T;l++)
{
for (ds=0.0, t=0; t<=l; t++)
{
if(solprod[t] < D[t][l]*solsetup[t] + EPS)  ds += solprod[t];
else  ds += D[t][l]*solsetup[t];
}

/* Add the violated inequality: the minimum of the actual production
prod[t] and the maximum potential production D[t][l]*setup[t]
in periods 0 to l must at least equal the total demand in periods
0 to l.
sum(t=1:l) min(prod[t], D[t][l]*setup[t]) >= D[0][l]
*/
if(ds < D[0][l] - EPS)
{
le=0;
for(t=0;t<=l;t++)
{
if (solprod[t] < D[t][l]*solsetup[t] + EPS)
le += prod[t];
else
le += D[t][l]*setup[t];
}
p.newCtr(XPRBnewname("cut%d",ncut+1), le >= D[0][l]);
ncut++;
npcut++;
}
}

cout << "Pass " << npass << " (" << (XPRB::getTime()-starttime)/1000.0;
cout << " sec), objective value " << objval << ", cuts added: " << npcut;
cout << " (total " << ncut << ")" << endl;

if(npcut==0)
cout << "Optimal integer solution found:" << endl;
else
{
basis.reset();               /* No need to keep the basis any longer */
}
} while(npcut>0);

/* Print out the solution: */
for(t=0;t<T;t++)
{
cout << "Period " << t+1 << ": prod " << prod[t].getSol() << " (demand: ";
cout << DEMAND[t] << ", cost: " << PRODCOST[t] << "), setup ";
cout << setup[t].getSol() << " (cost: " << SETUPCOST[t] << endl;
}
}

/***********************************************************************/

int main(int argc, char **argv)
{
XPRBprob p("Els");             /* Initialize a new problem in BCL */
modEls(p);                     /* Model the problem */
solveEls(p);                   /* Solve the problem */

return 0;
}