Enter Principle | |
Enter Time | |
Enter Rate | |
This calculator uses the compound interest formula to find principal plus interest. It uses this same formula to solve for principal, rate or time given the other known values. You can also use this formula to set up a compound interest calculator in Excel®1.
A = P(1 + r/n)nt
In the formula
The basic compound interest formula A = P(1 + r/n)nt can be used to find any of the other variables. The tables below show the compound interest formula rewritten so the unknown variable is isolated on the left side of the equation.
Say you have an investment account that increased from $30,000 to $33,000 over 30 months. If your local bank offers a savings account with daily compounding (365 times per year), what annual interest rate do you need to get to match the rate of return in your investment account?
In the calculator above select "Calculate Rate (R)". The calculator will use the equations: r = n((A/P)1/nt - 1) and R = r*100.
Enter:
Showing the work with the formula r = n((A/P)1/nt - 1):
\[ r = 365 \left(\left(\frac{33,000}{30,000}\right)^\frac{1}{365\times 2.5} - 1 \right) \] \[ r = 365 (1.1^\frac{1}{912.5} - 1) \] \[ r = 365 (1.1^{0.00109589} - 1) \] \[ r = 365 (1.00010445 - 1) \] \[ r = 365 (0.00010445) \] \[ r = 0.03812605 \] \[ R = r \times 100 = 0.03812605 \times 100 = 3.813\% \]Your Answer: R = 3.813% per year
So you'd need to put $30,000 into a savings account that pays a rate of 3.813% per year and compounds interest daily in order to get the same return as the investment account.
A common definition of the constant e is that:
\[ e = \lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^m \]With continuous compounding, the number of times compounding occurs per period approaches infinity or n → ∞. Then using our original equation to solve for A as n → ∞ we want to solve:
\[ A = P{(1+\frac{r}{n})}^{nt} \] \[ A = P \left( \lim_{n\rightarrow\infty} \left(1 + \frac{r}{n}\right)^{nt} \right) \]This equation looks a little like the equation for e. To make it look more similar so we can do a substitution we introduce a variable m such that m = n/r then we also have n = mr. Note that as n approaches infinity so does m.
Replacing n in our equation with mr and cancelling r in the numerator of r/n we get:
\[ A = P \left( \lim_{m\rightarrow\infty} \left(1 + \frac{1}{m}\right)^{mrt} \right) \]Rearranging the exponents we can write:
\[ A = P \left( \lim_{m\rightarrow\infty} \left(1 + \frac{1}{m}\right)^{m} \right)^{rt} \]Substituting in e from our definition above:
\[ A = P(e)^{rt} \]And finally you have your continuous compounding formula.
\[ A = Pe^{rt} \]