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Cutstk - Column generation for a cutting stock problem

Description
This example features iteratively adding new variables, basis in/output and working with subproblems. The column generation algorithm is implemented as a loop over the root node of the MIP problem.

xbcutstkcpp.zip[download all files]

Source Files





xbcutstk.cxx

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

  file xbcutstk.cxx
  `````````````````
  Cutting stock problem, solved by column (= cutting 
  pattern) generation heuristic looping over the
  root node.

  (c) 2008 Fair Isaac Corporation
      author: S.Heipcke, 2001, rev. Mar. 2014
********************************************************/

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

using namespace std;
using namespace ::dashoptimization;

#define NWIDTHS 5
#define MAXWIDTH 94

#define EPS 1e-6
#define MAXCOL 10

/****DATA****/
double WIDTH[] = {17, 21, 22.5,  24, 29.5};  /* Possible widths */
int DEMAND[] =  {150, 96,   48, 108,  227};  /* Demand per width */
int PATTERNS[NWIDTHS][NWIDTHS];              /* (Basic) cutting patterns */

XPRBvar pat[NWIDTHS+MAXCOL];               /* Rolls per pattern */
XPRBctr dem[NWIDTHS];                      /* Demand constraints */
XPRBctr cobj;                              /* Objective function */

XPRBprob p("CutStock");                    /* Initialize a new problem in BCL */

double knapsack(int N, double *c, double *a, double R, int *d, int *xbest);

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

void modCutStock()
{
 int i,j;
 XPRBexpr le;

 for(j=0;j<NWIDTHS;j++) PATTERNS[j][j]=(int)floor(MAXWIDTH/WIDTH[j]);
 
 /****VARIABLES****/
 for(j=0;j<NWIDTHS;j++)
  pat[j]=p.newVar(XPRBnewname("pat_%d",j+1), XPRB_UI, 0,
       (int)ceil((double)DEMAND[j]/PATTERNS[j][j]));
       
/****OBJECTIVE****/
 for(j=0;j<NWIDTHS;j++) le += pat[j];    /* Minimize total number of rolls */
 cobj = p.newCtr("OBJ", le);
 p.setObj(cobj);

/****CONSTRAINTS****/
 for(i=0;i<NWIDTHS;i++)                  /* Satisfy the demand per width */
 { 
  le = 0;
  for(j=0;j<NWIDTHS;j++)
   le += PATTERNS[i][j] * pat[j];
  dem[i] = p.newCtr("Demand", le >= DEMAND[i]);
 }
}

/**************************************************************************/
/*  Column generation loop at the top node:                               */
/*    solve the LP and save the basis                                     */
/*    get the solution values                                             */
/*    generate new column(s) (=cutting pattern)                           */
/*    load the modified problem and load the saved basis                  */
/**************************************************************************/
void solveCutStock()
{
 double objval;                  /* Objective value */
 int i,j;
 int starttime;
 int npatt, npass;               /* Counters for columns and passes */
 double solpat[NWIDTHS+MAXCOL];  /* Solution values for variables pat */
 double dualdem[NWIDTHS];        /* Dual values of demand constraints */
 XPRBbasis basis;
 double dw,z;
 int x[NWIDTHS];

 starttime=XPRB::getTime();
 npatt = NWIDTHS;

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

      /* Get the solution values: */
  for(j=0;j<npatt;j++)  solpat[j]=pat[j].getSol();
  for(i=0;i<NWIDTHS;i++)  dualdem[i]=dem[i].getDual();
  
      /* Solve integer knapsack problem  z = min{cx : ax<=r, x in Z^n}
         with r=MAXWIDTH, n=NWIDTHS */
  z = knapsack(NWIDTHS, dualdem, WIDTH, (double)MAXWIDTH, DEMAND, x);
  cout << "(" << (XPRB::getTime()-starttime)/1000.0 << " sec) Pass " << npass+1 << ": ";
   
  if(z < 1+EPS)
  {
   cout << "no profitable column found." << endl << endl;
   basis.reset();                  /* No need to keep the basis any longer */
   break;
  }
  else 
  {
      /* Print the new pattern: */
   cout << "new pattern found with marginal cost " << z-1 << endl << "   ";
   cout << "Widths distribution: ";
   dw=0;
   for(i=0;i<NWIDTHS;i++)  
   {
    cout << WIDTH[i] << ":" << x[i] << "  ";
    dw += WIDTH[i]*x[i];
   } 
   cout << "Total width: " << dw << endl;

      /* Create a new variable for this pattern: */
   pat[npatt]=p.newVar(XPRBnewname("pat_%d",npatt+1),XPRB_UI);

   cobj += pat[npatt];             /* Add new var. to the objective */
   dw=0;
   for(i=0; i<NWIDTHS; i++)        /* Add new var. to demand constraints*/
    if(x[i] > EPS)
    {
     dem[i] += x[i]*pat[npatt];
     if((int)ceil((double)DEMAND[i]/x[i]) > dw)
      dw = (int)ceil((double)DEMAND[i]/x[i]);
    }
   pat[npatt].setUB(dw);           /* Change the upper bound on the new var.*/  

   npatt++;

   p.loadMat();                    /* Reload the problem */
   p.loadBasis(basis);             /* Load the saved basis */
   basis.reset();                  /* No need to keep the basis any longer */
  }
 }

 p.mipOptimize("");                /* Solve the MIP */
  
 cout << "(" << (XPRB::getTime()-starttime)/1000.0 << " sec) Optimal solution: " << p.getObjVal() << " rolls, " << npatt << " patterns" << endl << "   "; 
 cout << "Rolls per pattern: ";
 for(i=0;i<npatt;i++) cout << pat[i].getSol() << ", ";
 cout << endl;   
}

/**************************************************************************/
/* Integer Knapsack Algorithm for solving the integer knapsack problem    */
/*    z = max{cx : ax <= R, x <= d, x in Z^N}                             */
/*                                                                        */
/* Input data:                                                            */
/*   N:        Number of item types                                       */
/*   c[i]:     Unit profit of item type i, i=1..n                         */
/*   a[i]:     Unit resource use of item type i, i=1..n                   */
/*   R:        Total resource available                                   */
/*   d[i]:     Demand for item type i, i=1..n                             */
/* Return values:                                                         */
/*   xbest[i]: Number of items of type i used in optimal solution         */
/*   zbest:    Value of optimal solution                                  */
/**************************************************************************/
double knapsack(int N, double *c, double *a, double R, int *d, int *xbest)
{
 int j;
 double zbest = 0.0;
 XPRBvar *x;
 XPRBexpr klobj, knap;
 XPRBprob pk("Knapsack");

/*
 cout << "Solving z = max{cx : ax <= b; x in Z^n}\n   c   =";
 for (j = 0; j < N; j++) cout << " " << c[j] << ",";
 cout << "\n   a   =";
 for (j = 0; j < N; j++) cout << " " << a[j] << ",";
 cout << "\n   c/a =";
 for (j = 0; j < N; j++) cout << " " << c[j]/a[j] << ",";
 cout << "\n   b   = " << R << endl;
*/
 
 x = new XPRBvar[N];
 if(x==NULL) cout << "Allocating memory for variables failed." << endl;
 for(j=0;j<N;j++) x[j]=pk.newVar("x", XPRB_UI, 0, d[j]);

 for(j=0;j<N;j++) klobj += c[j]*x[j];
 pk.setObj(pk.newCtr("OBJ",klobj));  
 for(j=0;j<N;j++) knap += a[j]*x[j];
 pk.newCtr("KnapXPRBctr", knap <= R);
 pk.setSense(XPRB_MAXIM);
 pk.mipOptimize("");

 zbest = pk.getObjVal();
 for(j=0;j<N;j++) xbest[j]=(int)floor(x[j].getSol() + 0.5);
/*
 cout << "   z = " << zbest << endl << "   x =";
 for(j=0; j<N; j++) cout << " " << xbest[j] << ",";
 cout << endl;
*/

 delete [] x;
 
 return (zbest);
}

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

int main(int argc, char **argv)
{
 modCutStock();                    /* Model the problem */
 solveCutStock();                  /* Solve the problem */
  
 return 0;
} 


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