Need Help with a Table for an investment calculator

[Dave Man]

my calculator only outputs the total value, need it to show the total after each year. Do not worry about the other two cases just need help with the first calculator


#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <cmath>
#include <string>
#include <iomanip>
#include <locale>
using namespace std;
const double E = 2.718;
void displayMenu();

void investmentcalculator();
void mortgageloancalculator();
void autoloancalculator();
int main()
{
void displayMenu();
int choice; // Menu choice should be 1, 2, 3 or 4
// Display the menu of choices

displayMenu();
// Get the user's choice
cin >> choice;
// Tell the user what he or she picked
switch (choice) {

case 1:
investmentcalculator();

break;
case 2:
mortgageloancalculator();

break;
case 3:
autoloancalculator();
break;
case 4:
break;
default:
cout << "Error, invalid selection, please choose one of the following" << endl;
displayMenu();
break;
}

system("pause");
return 0;
}

void displayMenu()
{
cout << "Choose a one of the calculators. \n\n";
cout << "1 Investment Calculator \n" << "2 Mortagage Loan Calculator \n" << "3 Auto loan Calculator \n" << "4 quit \n";
}
void investmentcalculator()
{
double total, principle, rate, years;// using double as we might have decimal values
int counter, choice;
string fname, lname;
counter = 0;

cout << "What is your first name and last name? " << endl;
cin >> fname >> lname;
cout << "How much would you like to invest? " << endl;
cin >> principle;
cout << "How much is your expected anual rate of return? " << endl;
cin >> rate;
cout << "How many years would you like to invest for?" << endl;
cin >> years;
total = (principle * (pow(E, (rate * years))));
cout << "The amount invested $ " << principle << " at an intrest rate of " << rate << " percent for " << years << " year(s)." << endl;
cout << fname + " " + lname << " will have $" << (principle * (pow(E, (rate * years)))) << " in" << years << " years" << endl;
cout << "do you want to see your annual yeilds? (1 = Yes/2 = No)" << endl;
cin >> choice;
switch (choice)
{
case 1:


do
{
cout.setf(ios::fixed);
cout.precision(2);
cout << counter << setw(21) << "$ " << total << endl;
} while (counter++ >= years);
break;

case 2:
{cout << "Ok, thank you. this program will now exit.";
system("pause");
break;

}
default:
{
cout << "Invalid choice, choose yes or no";
}
system("Pause");

}
}

[Chervil]

First, you need to fix the existing calculations.

You can delete the line const double E = 2.718;, it isn't needed. As an aside, if you ever did need to find E raised to some power, you would use the exp() function.

The interest calculation:

    total = (principle * (pow(E, (rate * years))));

as I said, you don't need E. What you do need, is the interest rate (which is a percentage) converted to a decimal. Then raise 1 + interest rate to the power years.
It would look like this:

    total = principal * pow( 1.0 + rate/100.0, years ) ;

Example, if the interest rate is 3% the calculation uses 1 + 3/100 or 1.03 .

For the individual years, you need a loop, a for-loop is easiest, something like

    for (int y = 1; y <= years; y++)

and then in the calculation, use y instead of years:

    total = principal * pow(1.0 + rate/100.0, y ) ;