I don't know what value to use for continuously (n) in this formula. EPR=[{1+(r/n)}^n]-1-->EPR=[{1+(.09/n)}^n]-1. Do you know what does n equal to?

A bank offers 9.00% on savings accounts. What is the effective annual rate if interest is compounded continuously?

1 answer

In the formula you provided, \( EPR = \left(1 + \frac{r}{n}\right)^n - 1 \), the variable \( n \) represents the number of compounding periods per year. For example:

  • If interest is compounded annually, \( n = 1 \).
  • If compounded semi-annually, \( n = 2 \).
  • If compounded quarterly, \( n = 4 \).
  • If compounded monthly, \( n = 12 \).
  • If compounded daily, \( n = 365 \).

However, when interest is compounded continuously, \( n \) approaches infinity, and we can use a different formula to find the effective annual rate (EAR):

\[ EPR = e^r - 1 \]

where \( e \) is the base of the natural logarithm (approximately equal to 2.71828), and \( r \) is the nominal interest rate (as a decimal).

In your case, with an interest rate of 9.00% (or 0.09 as a decimal), we can calculate the effective annual rate (EAR) using the continuous compounding formula:

\[ EPR = e^{0.09} - 1 \]

Let's do the calculation:

  1. Calculate \( e^{0.09} \): \[ e^{0.09} \approx 1.09417 \]

  2. Now subtract 1: \[ EPR \approx 1.09417 - 1 \approx 0.09417 \]

  3. Convert to percentage: \[ EPR \approx 0.09417 \times 100 \approx 9.42% \]

So, the effective annual rate (EAR) when the interest is compounded continuously at a rate of 9.00% is approximately 9.42%.