/********************************************************
  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) 2008-2024 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                            */
/*    load the modified problem and load the saved basis                  */
/**************************************************************************/
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();
 XPRSsetintcontrol(p.getXPRSprob(), XPRS_CUTSTRATEGY, 0); 
                              /* Disable automatic cuts - we use our own */
 XPRSsetintcontrol(p.getXPRSprob(), XPRS_PRESOLVE, 0);
 XPRSsetintcontrol(p.getXPRSprob(), XPRS_MIPPRESOLVE, 0);
 XPRSsetintcontrol(p.getXPRSprob(), XPRS_PREPROBING, 0);
                              /* 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
  { 
   p.loadMat();                 /* Reload the problem */
   p.loadBasis(basis);          /* Load the saved basis */
   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;
}