2
$\begingroup$

The compound interest formula $A=P\left(1+\frac{r}{n}\right)^{nt}$ is usually used in examples where you are given a nominal annual rate and then calculate the accrued amount, where $\frac{r}{n}$ is the "monthly" rate.

e.g. nominal interest rate $= 5\% \,\,\text{p.a}, P = \$100 $ and this is compounded monthly.

So we say $ A= 100\left(1+\frac{0.05}{12}\right)^{12}=\$105.116...$

We have in fact earned $\approx5.116\%$ interest. This is what is called the "effective interest rate".

So which is the "correct" annual rate and what has $\frac{r}{n}$ got to do with $r$? Why do we say $\frac{r}{n}$ is the monthly rate when the "true" monthly rate would be in this example $\sqrt[12]{1.05}-1$, giving us an annual interest of $5\%$. I guess my question is really what logic would lead you to divide $\boldsymbol{r}$ by the number of compounding periods? I can't seem to understand the reasoning behind that. As in the example, the effective rate seems unrelated or somehow arbitrary compared to the nominal rate.

The naive answer would be if you have $5\%$ interest per year, you will get $\frac{1}{12}$ of this rate every month, but given that you know this will be compounded, why would you do this?

Apologies if this has been asked before, this is a question that has occasionally bothered me and I have never found an answer.

$\endgroup$
3
  • $\begingroup$ The amount is being compounded every month, i.e. being multiplied by a certain factor of the form $(1+r’)$. Now since the interest rate for the whole year is $r$, the interest rate must be ‘uniformly distributed’ into the months giving that the monthly interest rate is $\frac{r}{12}$. $\endgroup$
    – Tavish
    Apr 11, 2021 at 11:49
  • 2
    $\begingroup$ @Tavish why must the annual rate be uniformly distributed? And why is $\frac{r}{n}$ used instead of for example $\sqrt[n]{1+r}-1$ which would give the same accrued amount at the end of the year? $\endgroup$
    – Joseph
    Apr 11, 2021 at 18:48
  • 1
    $\begingroup$ Note that $\sqrt[n]{1 + r} - 1 = \frac{r}{n} + O(r^2)$ so $\frac{r}{n}$ is the linear approximation of the effective per-period rate, close for small values of $r$. $\endgroup$
    – heiner
    Apr 13, 2021 at 19:19

1 Answer 1

Reset to default
2
$\begingroup$

The answer has as much to do with terminology as anything else. When we say that the nominal annual interest rate is $5\%$, that means something very specific:

If interest is accrued at a nominal annual rate of $5\%$, compounded $n$ times per year, then on each compounding period the interest rate is $5\%/n$.

Another way of interpreting the nominal rate is that if there were no compounding, i.e. $n = 1$, then the nominal annual rate would tell us the actual (or effective) annual growth rate.

Quoting a nominal rate gives a convenient, if sometimes misleading, way of talking about the interest rate in a way that is independent of the number of compounding periods. Meaning that whether we are discussing a $5\%$ nominal rate compounded quarterly, daily, continuously, etc., the nominal rate stays the same.

In some sense, then, the nominal rate is more practical to work with, because you need less information to understand it and make informed decisions. If, for example, you are comparing two loans with nominal annual rates of $5\%$ and $6\%$, respectively, then you clearly want to go with the first, because no amount of compounding is going to raise the effective rate of the first above $6\%$. In fact, the highest effective rate a $5\%$ nominal rate can have is if it were compounded continuously, in which case the effective rate is $$e^{0.05} - 1 \approx 0.0513 = 5.13\%.$$

The comparison would be more nuanced if, for example, you were comparing a nominal interest rate of $5\%$ compounded daily with a nominal interested rate of $5.05\%$ compounded semiannually. In the first case, the effective rate is $$\left(1+\frac{0.05}{365}\right)^{365} - 1 \approx 0.0513 = 5.13\%,$$ whereas the effective rate of the second is $$\left(1 + \frac{0.0505}{2}\right)^{2} - 1 \approx 0.0511 \approx 5.11\%.$$ Here the compounding makes a (slight, but appreciable) difference, and you would actually choose the account with the higher nominal rate because it has a lower effective rate.

So, it's important to understand both types of interest rates, as well as when to pay close attention to detail.

$\endgroup$
1
  • 1
    $\begingroup$ This not only explains the usefulness of a nominal rate, it also explains why I might care about the limit of $\left(1+\frac{r}{n}\right)^{nt}$. Superb. $\endgroup$
    – Joseph
    Apr 12, 2021 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.