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

  file xbrecurs.cxx
  `````````````````
  Recursion solving a non-linear financial planning problem
  The problem is to solve
  	net(t) = Payments(t)  - interest(t)
  	balance(t) = balance(t-1) - net(t)
  	interest(t) = (92/365) * balance(t) * interest_rate
  where
        balance(0) = 0
        balance[T] = 0
  for interest_rate

  (c) 2008-2023 Fair Isaac Corporation
      author: S.Heipcke, 2001, rev. Mar. 2011
********************************************************/

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

using namespace std;
using namespace ::dashoptimization;

#define T 6

/****DATA****/
double X = 0.00;                /* An INITIAL GUESS as to interest rate x */
double B[] = /* {796.35, 589.8918, 398.1351, 201.5451, 0.0, 0.0}; */
             {1,1,1,1,1,1};
                                /* An INITIAL GUESS as to balances b(t) */
double P[] = {-1000, 0, 0, 0, 0, 0};                 /* Payments */
double R[] = {206.6, 206.6, 206.6, 206.6, 206.6, 0}; /*  "       */
double V[] = {-2.95, 0, 0, 0, 0, 0};                 /*  "       */

XPRBvar b[T];                   /* Balance */
XPRBvar x;                      /* Interest rate */
XPRBvar dx;                     /* Change to x */
XPRBctr interest[T], ctrd;
 
/***********************************************************************/

void modFinNLP(XPRBprob &p)
{
 XPRBvar i[T];                  /* Interest */
 XPRBvar n[T];                  /* Net */
 XPRBvar epl[T], emn[T];        /* + and - deviations */
 XPRBexpr cobj, le;
 int t;
 
/****VARIABLES****/
 for(t=0;t<T;t++) i[t]=p.newVar("i"); n[t]=p.newVar("n", b[t]=p.newVar("b", epl[t]=p.newVar("epl"); emn[t]=p.newVar("emn"); x=p.newVar("x"); dx=p.newVar("dx", for(t=0;t<T;t++) Objective: get feasible cobj +=epl[t]+emn[t]; Select objective function Choose the sense of optimization net=payments interest Money balance across periods>0) p.newCtr("bal", b[t] == b[t-1]-n[t]);
  else  p.newCtr("bal", b[t] == -n[t]);
 }

 for(t=1;t<T;t++) i(t)=(92/365)*( approx. interest[t]=p.newCtr("int", emn[t]== ctrd=p.newCtr("def", x== X Recursion loop until variation of converges to save the current basis and solutions for variables set balance estimates value interest rate estimate reload problem saved solve LP calculate void XPRBbasis double int ct=0; Switch automatic cut generation off Solve LP-problem>0)
 {
  ct++;
  basis=p.saveBasis();           /* Save the current basis */

                                 /* Get the solution values for b and x */
  for(t=1;t<T;t++) BC[t-1]=b[t-1].getSol();
  XC=x.getSol();
  cout << "Loop " << ct << ": " << x.getName() << ":" << x.getSol();
  cout << "  (variation: " << variation << ")" << endl;
                   
  for(t=1;t<T;t++)               /* Change coefficients in interest[t] */
  {   
   interest[t].setTerm(dx, BC[t-1]);
   B[t-1]=BC[t-1];
   interest[t].setTerm(b[t-1], XC);
  }
  ctrd.setTerm(XC);              /* Change constant term of ctrd */ 
  X=XC;

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

  p.lpOptimize("");              /* Solve the LP-problem */
  variation=fabs(x.getSol()-oldval);
 }

 cout << "Objective: " << p.getObjVal() << endl; /* Get objective value */
 cout << "Interest rate: " << x.getSol()*100 << "%" << endl << "Balances:  ";
 for(t=0;t<T;t++)                /* Print out the solution values */
  cout << b[t].getName() << ":" << b[t].getSol() << " ";
 cout << endl;
} 

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

int main(int argc, char **argv)
{
 XPRBprob p("Fin_nlp");          /* Initialize a new problem in BCL */
 modFinNLP(p);                   /* Model the problem */
 solveFinNLP(p);                 /* Solve the problem */
  
 return 0;
}