How to get Balance for 8.5?

[Blueshark1]

Hello, I currently have a program to give me a line for the best fit of the y(Balance) graph. But I am also trying to find how much the Balance is on week 8.5. Below is my code along with the data table for the weeks and their balances.
Week Balance
1 490
2 720
3 970
4 1265
5 1210
6 1521
7 1165
8 1365
9 1850
10 2050
11 1350
12 1550
13 2350
14 2450
15 2650
16 1550
17 1665
18 1800
19 2465
20 2165
21 2900
22 3600

/* This program computes computes a linear model for a list of weekly balances. */
#include <iostream>		//Required for cin, cout
#include <fstream>		//Required for ifstream
#include <string>		//Required for string
using namespace std;
int
main ()
{				// Declare and initialize objects.
  int count (0), k;
  double Week, Balance, first, last, sumWeek (0), sumBalance (0),
    sumWeek2 (0), sumWeekBalance (0), denominator, m, b;
  string filename;
  ifstream TimeBal;		// read a file and use for getting data in
  ofstream report;		// create and write to a file
// first we will create a data file

  cout << "Enter the name of the data file you want";	// Prompt user for name of input file.
  cin >> filename;

  report.open (filename.c_str ());
  cout << " enter the week along with its corresponding balance." << endl;
  cin >> Week >> Balance;	// get the values from keypad
  report << Week << " " << Balance << "  " << endl;	// send values to data file

  for (k = 1; k <= 21; k = k + 1)
    {				//How many times we want it to ask
      cout << " enter the week along with its corresponding balance." << endl;
      cin >> Week >> Balance;
      report << Week << " " << Balance << "  " << endl;
    }
  cout << "Enter the name of the data file you want to analyse";	// Prompt user for name of input file.
  cin >> filename;
  report.close ();		// close created data file

  TimeBal.open (filename.c_str ());	// Open file and check if it exists.
  if (TimeBal.fail ())
    {
      cerr << "Error opening input file\n";
      exit (1);
    }
  report.open ("MonaesReport");	//Open Report File
  TimeBal >> Week >> Balance;	// While not at the end of the file, read and accumulate information
  while (!TimeBal.eof ())
    {
      ++count;
      if (count == 1)
	first = Week;
      sumWeek += Week;
      sumBalance += Balance;
      sumWeek2 += Week * Week;
      sumWeekBalance += Week * Balance;
      TimeBal >> Week >> Balance;
    }
  last = Week;
  denominator = sumWeek * sumWeek - count * sumWeek2;	// Compute slope and y-intercept.
  m = (sumWeek * sumBalance - count * sumWeekBalance) / denominator;	//Slope
  b = (sumWeek * sumWeekBalance - sumWeek2 * sumBalance) / denominator;	//Y-intercept
// Set format flags
  cout.setf (ios::fixed);
  cout.precision (2);
  report.setf (ios::fixed);
  report.setf (ios::showpoint);
  report.precision (2);
// Print summary information.
  cout << "The Change in Balances from week: \n";
  cout << first << " to week " << last << endl << endl;
  cout << "Linear model: \n";
  cout << " The added or subtracted funds (slope) is:\n " << m <<
    "and the accumulated  amount each week (y-intercept) is: " << b << endl;
  cout << "The algebraic linear equation is " << "y" << "=" << m <<
    "*" << "x" << "+" << b << endl;
  cout << "The balance for week 8.5 is:";
//Report
  report << "The Change in Balances: \n";
  report << first << " to " << last << endl << endl;
  report << "Linear model: \n";
  report << " added or subtracted funds " << m <<
    " accumulated  amount each week " << b << endl;
  report << "The algebraic linear equation is " << "y" << "=" << m <<
    "*" << "x" << "+" << b << endl;
  report << "The balance for week 8.5 is:";

// Close file and exit program.
  report.close ();
  TimeBal.close ();
  return 0;
}				// end of main  

[againtry]

https://www.mathsisfun.com/data/least-squares-regression.html is my guide. Using your data, it might be a bit clearer as follows:

#include <iostream>
#include <fstream>
#include <string>

using namespace std;

int main ()
{
    string filename("weekly_balance.txt"); // HARD CODE FOR TESTING
    ifstream TimeBal{filename};
    
    double y{0}; // BALANCE
    double x{0}; // WEEK
    
    double sum_x{0}, sum_y{0};
    double sum_x2{0}, sum_y2{0};
    double sum_xy{0};
    
    int N{0};
    if (TimeBal.is_open())
    {
        while ( TimeBal >> x >> y ) // eof() IS NOT A GOOD CHOICE
        {
            N++;
            cout << x << ' ' << y << '\n';
            
            sum_x += x; sum_y += y;
            sum_x2 += (x*x); sum_y2 += (y*y);
            sum_xy += (x*y);
            
        }
        TimeBal.close();
    }
    else
    {
        cout << "Unable to open file";
        exit(1);
    }
    
    cout << "No. of records: " << N << '\n';
    
    double m = (N * sum_xy - sum_x*sum_y)/(N*sum_x2 - sum_x * sum_x); // SLOPE <--
    double b = (sum_y - m*sum_x)/N; // INTERCEPT
    
    cout
    << "Linear best fit is y = " << m << "x + " << b << '\n'
    << " where y is the weekly balance and x is the week number\n"
    << "\n\n"
    << "Balance at 8.5 weeks: " << m * 8.5 + b << '\n';
    
    return 0;
}

1 490
2 720
3 970
4 1265

... blah blah
20 2165
21 2900
22 3600
No. of records: 22
Linear best fit is y = 96.476x + 667.844
 where y is the weekly balance and x is the week number


Balance at 8.5 weeks: 1487.89
Program ended with exit code: 0

Then proceed to outputting to the report file. If you use any of the above make sure you check it.

[lastchance]

Are you sure that you don't want to use interpolation rather than regression? That data is quite a long way away from a linear relationship.

[lastchance]

#include <iostream>
#include <fstream>
#include <vector>
#include <cmath>
using namespace std;

//======================================================================

void regression( const vector<double> &x, const vector<double> &y, double &m, double &c, double &rho )
{
   int N = x.size();
   double Sx = 0, Sy = 0, Sxx = 0, Syy = 0, Sxy = 0;
   for ( int i = 0; i < N; i++ )
   {
      Sx  += x[i];
      Sy  += y[i];
      Sxx += x[i] * x[i];
      Syy += y[i] * y[i];
      Sxy += x[i] * y[i];
   }
   double varxx = Sxx - Sx * Sx / N, varyy = Syy - Sy * Sy / N, varxy = Sxy - Sx * Sy / N;
   m = varxy / varxx;
   c = ( Sy - m * Sx ) / N;
   rho = varxy / sqrt( varxx * varyy );
}

//======================================================================

double interpolation( const vector<double> &x, const vector<double> &y, double xval )    // x sorted and increasing
{
   int i = 0, N = x.size();
   if ( xval >= x[N-1] )
      i = N - 2;
   else if ( xval > x[0] )
      while( x[i+1] < xval ) i++;

   return y[i] + ( xval - x[i] ) * ( y[i+1] - y[i] ) / ( x[i+1] - x[i] );
}

//======================================================================

int main()
{
   vector<double> x, y;
   ifstream in( "input.txt" );
   for ( double xpt, ypt; in >> xpt >> ypt; )
   {
      x.push_back( xpt );
      y.push_back( ypt );
   }

   double m, c, rho;
   regression( x, y, m, c, rho );
   cout << "Linear regression line y = mx + c\n"
        << "   m = " << m << '\n'
        << "   c = " << c << '\n'
        << "Correlation coefficient rho = " << rho << "\n\n";

   double test = 8.5;
   cout << "At test value " << test << '\n'
        << "   linear regression gives y = " << m * test + c << '\n'
        << "   interpolation     gives y = " << interpolation( x, y, test ) << '\n';
}

Linear regression line y = mx + c
   m = 96.476
   c = 667.844
Correlation coefficient rho = 0.837577

At test value 8.5
   linear regression gives y = 1487.89
   interpolation     gives y = 1607.5