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

  file foliomip1.cpp
  ``````````````````
  Modeling a small MIP problem 
  to perform portfolio optimization.
   -- Limiting the total number of assets --

  (c) 2008 Fair Isaac Corporation
      author: S.Heipcke, Aug. 2003, rev. Mar. 2011
********************************************************/

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

using namespace std;
using namespace ::dashoptimization;

#define MAXNUM 4                   // Max. number of different assets

#define NSHARES 10                 // Number of shares
#define NRISK 5                    // Number of high-risk shares
#define NNA 4                      // Number of North-American shares

double RET[] = {5,17,26,12,8,9,7,6,31,21};  // Estimated return in investment
int RISK[] = {1,2,3,8,9};          // High-risk values among shares
int NA[] = {0,1,2,3};              // Shares issued in N.-America

int main(int argc, char **argv)
{
 int s;
 XPRBprob p("FolioMIP1");          // Initialize a new problem in BCL
 XPRBexpr Risk,Na,Return,Cap,Num;
 XPRBvar frac[NSHARES];            // Fraction of capital used per share
 XPRBvar buy[NSHARES];             // 1 if asset is in portfolio, 0 otherwise

// Create the decision variables (including upper bounds for `frac')
 for(s=0;s<NSHARES;s++) 
 {
  frac[s] = p.newVar("frac", XPRB_PL, 0, 0.3);
  buy[s] = p.newVar("buy", XPRB_BV);
 } 
 
// Objective: total return
 for(s=0;s<NSHARES;s++) Return += RET[s]*frac[s]; 
 p.setObj(Return);                // Set the objective function

// Limit the percentage of high-risk values
 for(s=0;s<NRISK;s++) Risk += frac[RISK[s]]; 
 p.newCtr(Risk <= 1.0/3);

// Minimum amount of North-American values
 for(s=0;s<NNA;s++) Na += frac[NA[s]]; 
 p.newCtr(Na >= 0.5);

// Spend all the capital
 for(s=0;s<NSHARES;s++) Cap += frac[s]; 
 p.newCtr(Cap == 1);

// Limit the total number of assets
 for(s=0;s<NSHARES;s++) Num += buy[s];
 p.newCtr(Num <= MAXNUM);

// Linking the variables
 for(s=0;s<NSHARES;s++) p.newCtr(frac[s] <= buy[s]);

// Solve the problem
 p.setSense(XPRB_MAXIM);
 p.mipOptimize("");


 char *MIPSTATUS[] = {"not loaded", "not optimized", "LP optimized",  
		      "unfinished (no solution)", 
		      "unfinished (solution found)", "infeasible", "optimal",
                      "unbounded"};
 
 cout << "Problem status: " << MIPSTATUS[p.getMIPStat()] << endl;

 
// Solution printing
 cout << "Total return: " << p.getObjVal() << endl;
 for(s=0;s<NSHARES;s++) 
  cout << s << ": " << frac[s].getSol()*100 << "% (" << buy[s].getSol()
       << ")" << endl;  

 return 0;
}