Financial Mathematics Terms: Effective Versus Nominal Rates of Interest

One of the most confusing topics for many people when starting to learn about financial mathematics is the various ways of measuring compound interest.

Terms like “effective interest rate”, “nominal interest rate”, “continuous interest rate”, and “real interest rate” show up in various articles and textbooks. Exploring the distinctions and relationships between effective and nominal interest rates is the main purpose of this article. Continuous and real interest rates will be discussed in the future.

As with most topics, it is best to start with examples. You will want to have your calculator ready to check the calculations that follow.

Savings Account Example

A common situation for thinking about earning interest is in savings accounts. The current low interest rate environment of the 2000’s makes for less interesting examples because the distinction between effective and nominal rates is not as extreme, so let’s go back in time to the early 1980’s.

Back in about 1983, I had a savings account that paid me an 8% annual rate of interest. Yes, you read that right! That was a basic savings account, not even a certificate of deposit.

Of course, inflation was rampant in the early 1980’s, so banks had to offer high interest rates in order to entice depositors (supply and demand do affect interest rates). Interest rates on loans, such as home mortgages, were also very high — so the banks could certainly use their depositors’ money to make more money on loans, in spite of paying the depositors a lot of interest.

If you had such a savings account, you would probably quote the interest rate just as I did — as an 8% rate of interest. But what does this mean?

Does it mean that a deposit of $100 would grow by 8% to $108 after one year? Most people would probably guess that this is true. However, it is (usually) false.

Usually, if you are quoted an interest rate by a bank for a savings account, it will be a nominal interest rate — meaning “in name only“. Because of “compounding”, you will actually receive more than 8% for a given year. Good news, right?!?!

It is certainly good news if you have a savings accounts, but on the flip side, it can be bad news if you have a loan and are not paying it back, such as what happens for many people with credit cards — more on that later.

When banks compound your interest, it means they give you interest not only on your original deposit, but also on any interest you have already earned. Banks typically pay interest every quarter of a year (every three months).

Suppose the bank does this for your $100 deposit. How much interest should you get after one quarter? For the 8% nominal annual interest rate to be compounded quarterly, we divide by 4 (since there are 4 quarters in a year) to get 2%. This 2% is an effective interest rate over every quarter.

This means that your $100 will indeed grow by 2% to $102 over the first quarter. Congratulations! You earned $2 of interest on a relatively small deposit. In the year 2019, your deposit might have to be $10,000 to earn $2 per quarter because of the low interest rates on savings accounts.

This process then continues. You earn 2% on your new balance every quarter. After the next quarter, your interest earned will be $\$102\times 2\%=\$102\times 0.02=\$2.04$ and your balance will be $\$102+\$2.04=\$104.04$ — also note that $\$102\times 1.02=\$104.04.$ After the third quarter, your balance will be $\$104.04\times 1.02=\$106.1208\approx \$106.12$ (you earn $2.08 in interest during the third quarter). And after the end of one year, your balance will be

$\$106.1208\times 1.02=\$108.243216\approx \$108.24$

(you earn $2.12 in interest during the fourth quarter).

Since your final balance after 1 year is $108.24, which is 8.24% more than $100, the 8% savings account with quarterly compounding has an equivalent effective annual interest rate of 8.24% (a.k.a. an “annual equivalent rate“). This is the the actual percentage that your money grows by each year, and it is independent of the initial deposit, though rounding can affect the answer.

In general, after $n$ quarters have gone by, your balance will be $100\times (1.02)^{n}$ , which is called an exponential function of $n$ . In the long run, this can lead to quite a lot of growth — “exponential growth”, in fact. For example, after 50 years (200 quarters) your balance would be $100\times (1.02)^{200}\approx \$5248.49,$ which is 5148.49% growth over that amount of time (and no, that’s not a mistake — that should really be a 1 in that last percentage and not a 2).

Paid-Back Home Loan Example

As emphasized above, compound interest is good for you personally when depositing money in savings. On the other hand, it is bad for you personally when you take out a loan and then you get behind on your payments.

Let’s first consider what happens in the “good” case where a loan is being paid back, though this is an extreme example from the early 1980’s again.

According to the website https://www.macrotrends.net/2604/30-year-fixed-mortgage-rate-chart, the 30-year fixed rate mortgage annual interest rate peaked in the United States at 18.44% in October of 1981. This interest rate is a nominal interest rate compounded monthly — twelve times per year! The effective monthly rate would then be $\frac{18.44\%}{12} \approx 1.536666667\%.$

If you took out a mortgage for $100,000 on that date, your interest paid for the first month’s payment would have been $\$100000\times 0.01536666667\approx \$1536.67!$ That’s a lot of interest on a relatively small (by today’s standards) mortgage!

It turns out, for reasons that I won’t go into, that your total principal and interest (P & I) payment at the end of every month for 30 years would be $1543.04. Almost your entire first payment would be interest!!! Your balance would go down by only $1543.04 – $1536.67 = $6.37 to $99,993.63 after one month!!! This type of extreme disparity between principal paid and interest paid will happen early in payback periods for loans with high interest rates, such as 18.44%, and long terms (payback periods), such as 30 years.

For the second month, the interest owed would have been $\$99993.63\times 0.01536666667\approx \$1536.57$ and the new balance would go down by $1543.04 – $1536.57 = $6.47 to $99,987.16 after two months.

If you are good at using formulas on spreadsheets, you can determine the amount of interest paid in the first year on this loan to be $18,433.197, or about 18.4332% of the original loan amount.

You effectively paid 18.4332% interest on the loan during the first year, which is less than the nominal rate of 18.44%. This is because you are in the “good” case of paying the loan back.

If you continued paying your monthly payments, you would pay $18,416.48 of interest in the second year, or about 18.4165% of the original loan amount and about 18.4318% of the balance of $99,916.76 at the end of the first year. These are both less than 18.44%, again because you are paying the loan back.

Overall, however, you are still paying a ton of interest over the life of the loan, even relative to the size of the loan itself, because the interest rate is so high and the term is so long. The total interest paid will be $455,493.16 on a loan of $100,000. The total amount paid is $555,493.16, which is the same as $1543.04\times 360,$ up to rounding error (the “true” monthly payment, including fractions of a cent, is actually $1543.036551, and $\$1543.036551\times 360\approx \$555,\hspace{-.03in}493.16.$ )

Unpaid Credit Card Example

If you use a credit card and don’t pay it back, it’s definitely bad for you. Not only do they charge high interest rates, but if you don’t make the minimum payment, there will also be penalties incurred. Furthermore, if you continue to use the credit card, your balance will go up even faster (though they will probably “freeze” your card to stop you from using it if you are a couple months behind).

If there are no penalties for non-payment, then the situation is just the reverse of the savings account example from above, though with higher interest rates.

Let’s say you use a credit card to pay for a $1,000 television. Let us also assume that the interest rate is a 24% nominal annual rate, compounded monthly, and that you make no other charges and that there is no minimum payment and no late fees. Then, your effective monthly interest rate will be $\frac{24\%}{12}=2\%$ and your balance after $n$ months will be $\$1000\times (1.02)^{n}$ . After 1 year, your balance will be $\$1000\times (1.02)^{12}\approx \$1268.24$ , a 26.824% increase and therefore a 26.824% effective annual interest rate.

It gets worse over time. For example, your balance after 5 years (60 months) will be $\$1000\times (1.02)^{60}\approx \$3281.03.$

More realistically, the company will charge late fees, making it even worse for you. If their monthly late fee is $30, this will need to get added on to the balance every month. Your balance after one month would be $\$1000\times 1.02+\$30=\$1050.$ Your balance after two months would be

$\$1050\times 1.02+\$30=(\$1000\times 1.02+\$30)\times 1.02+\$30$ $=\$1000\times (1.02)^{2}+\$30\times 1.02+\$30=\$1101.$

Your balance after three months would be

$\$1000\times (1.02)^{3}+\$30\times (1.02)^{2}+\$30\times 1.02+\$30=\$1153.02.$

Following this pattern, in general, your balance after $n$ months would be

$\$1000\times (1.02)^{n}+\$30\times (1.02)^{n-1}+\$30\times (1.02)^{n-2}+\cdots+30\times 1.02+\$30.$

If $n=12$ (one year), this balance will be $1670.60, for an effective annual interest rate of 67.06% — this is actually dependent on the initial loan because the $30 fee does not depend on the amount borrowed. Any way you slice it, this is bad for you.

Advanced: Formulas and Graphical Relationships Between Effective and Nominal Rates

A common notation for a nominal annual interest rate, compounded $m$ times per year, where $m$ is a positive integer, is $i^{(m)}$ . Note that this is not the same as raising $i$ to the mth power and it is also not the mth derivative of $i$ .

Let $i$ be the corresponding effective annual interest rate.

Suppose $A$ represents an initial deposit and suppose no other deposits or withdrawals are made during the year. Using the logic described in the examples above, these quantities are related by the equation

$A\left(1+\frac{i^{(m)}}{m}\right)^{m}=A(1+i).$

The quantity $A$ can be cancelled from both sides of this equation, so it simplifies to

$\left(1+\frac{i^{(m)}}{m}\right)^{m}=1+i.$

We can solve this equation for $i$ as a function of $i^{(m)}$ to get

$i=f_{m}(i^{(m)})=\left(1+\frac{i^{(m)}}{m}\right)^{m}-1.$

We can also solve for $i^{(m)}$ as a function of $i$ to get, for a fixed value of $m$ , the inverse function

$i^{(m)}=f_{m}^{-1}(i)=m\left((1+i)^{1/m}-1\right)=m(1+i)^{1/m}-m.$

The graphs shown below are both for $i=f_{m}(i^{(m)})$ when $m=4$ . The one on the left is a close up view for “typical” ranges $0\leq i^{(m)}\leq 0.2=20\%$ , while the one on the right shows a wider view for “unrealistic” ranges $0\leq i^{(m)}\leq 1.5=150\%.$ The $45^{\circ}$ diagonal line is also shown so that the graph of interest (pun-alert) can be compared to the graph of $i=i^{(m)},$ which has a constant slope of 1 and goes through the origin.

Two views of the graph of $i=f_{m}(i^{(m)})$ when $m=4$ . Notice the graph is increasing and concave up. Also notice that, when $i^{(m)}$ is close to zero, the graph is very close to, but slightly above, the line $i=i^{(m)}$ of slope 1 going through the origin. The graphs diverge from each other as $i^{(m)}$ continues to increase.

Notice that the graph of $i=f_{m}(i^{(m)})$ is increasing and concave up for $m=4$ . Also notice that, when $i^{(m)}$ is close to zero, the graph is very close to, but slightly above, the line $i=i^{(m)}$ of slope 1 going through the origin. The graphs diverge from each other as $i^{(m)}$ continues to increase. All this is true for other positive integer values of $m$ as well.

The real-life upshot of this is that, for a given $i^{(m)}>0,$ the equivalent effective annual interest rate $i$ satisfies both $i>i^{(m)}$ and $i\approx i^{(m)}$ when $i^{(m)}\approx 0.$ Furthermore, the difference $i-i^{(m)}$ grows larger as $i^{(m)}$ grows.

We can also view the graph of the inverse function, though it really tells us the same information, even though we have switched the roles of independent and dependent variables (and switched the labeling of the axes).

Two views of the graph of $i^{(m)}=f_{m}^{-1}(i)$ when $m=4$ . Notice the graph is increasing and concave down. Also notice that, when $i$ is close to zero, the graph is very close to, but slightly below, the line $i^{(m)}=i$ of slope 1 going through the origin. The graphs diverge from each other as $i$ continues to increase.

Two views of the graph of $i^{(m)}=f_{m}^{-1}(i)$ when $m=4$ . Notice the graph is increasing and concave down. Also notice that, when $i$ is close to zero, the graph is very close to, but slightly below, the line $i^{(m)}=i$ of slope 1 going through the origin. The graphs diverge from each other as $i$ continues to increase.

Finally, we can inquire as to the effect of the value of $m$ , the number of compounding periods per year. This is best done with an animation like the one below.

We see the graph of $i=f_{m}(i^{(m)})$ move further away from the graph of $i=i^{(m)}$ as $m$ increases from $m=1$ to $m=24$ . However, after about $m=12$ , the effect is not really visible anymore.

In this animation we see the graph of $i=f_{m}(i^{(m)})$ move further away from the graph of $i=i^{(m)}$ as $m$ increases from $m=1$ to $m=24.$ However, after about $m=12,$ the effect is not really visible anymore.

The real-life implication of this animation is that the effect of compounding gets bigger the more interest is compounded every year, but the “marginal” effect of increasing $m$ becomes negligible when $m$ is sufficiently large.

For those who are interested, the Wolfram Mathematica code for this animation is shown below.