8.2: Compound Interest

Learning Objectives

• Understand the difference between simple interest and compound interest.
• Be able to calculate the future value of an investment given compounding information (including compounding continuously).
• Understand and be able to calculate the Annual Percentage Yield (APY) for an investment.

With simple interest, we were assuming that we pocketed the interest when we received it. In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on that interest in future years. This reinvestment of interest is called compounding.

Suppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow?

The 3% interest is an annual percentage rate (APR) – the total interest to be paid during the year. Since interest is being paid monthly, each month, we will earn \(\frac{3 \%}{12}=0.25 \%\) per month.

In the first month,

\(P_{0}=\$ 1000\)

\(r=0.0025(0.25 \%)\)

\(I=\$ 1000(0.0025)=\$ 2.50\)

\(A=\$ 1000+\$ 2.50=\$ 1002.50\)

In the first month, we will earn $2.50 in interest, raising our account balance to $1002.50.

In the second month,

\(P_{0}=\$ 1002.50\)

\(I=\$ 1002.50(0.0025)=\$ 2.51\) (rounded)

\(A=\$ 1000+\$ 2.50=\$ 1002.50\)

Notice that in the second month we earned more interest than we did in the first month. This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the $2.50 of interest we earned the first month. This is the key advantage that compounding of interest gives us.

Calculating out a few more months:

\(\begin{array}{|l|l|l|l|}
\hline \textbf { Month } & \textbf { Starting balance } & \textbf { Interest earned } & \textbf { Ending Balance } \\
\hline 1 & 1000.00 & 2.50 & 1002.50 \\
\hline 2 & 1002.50 & 2.51 & 1005.01 \\
\hline 3 & 1005.01 & 2.51 & 1007.52 \\
\hline 4 & 1007.52 & 2.52 & 1010.04 \\
\hline 5 & 1010.04 & 2.53 & 1012.57 \\
\hline 6 & 1012.57 & 2.53 & 1015.10 \\
\hline 7 & 1015.10 & 2.54 & 1017.64 \\
\hline 8 & 1017.64 & 2.54 & 1020.18 \\
\hline 9 & 1020.18 & 2.55 & 1022.73 \\
\hline 10 & 1022.73 & 2.56 & 1025.29 \\
\hline 11 & 1025.29 & 2.56 & 1027.85 \\
\hline 12 & 1027.85 & 2.57 & 1030.42 \\
\hline
\end{array}\)

To find an equation to represent this, if \(P_{m}\) represents the amount of money after \(m\) months, then we could write the recursive equation:

\(P_{0}=\$ 1000\)

\(P_{m}=(1+0.0025) P_{m-1}\)

You probably recognize this as the recursive form of exponential growth. If not, we could go through the steps to build an explicit equation for the growth:

\(P_{0}=\$ 1000\)

\(P_{1}=1.0025 P_{0}=1.0025(1000)\)

\(P_{2}=1.0025 P_{1}=1.0025(1.0025(1000))=1.0025^{2}(1000)\)

\(P_{3}=1.0025 P_{2}=1.0025\left(1.0025^{2}(1000)\right)=1.0025^{3}(1000)\)

\(P_{4}=1.0025 P_{3}=1.0025\left(1.0025^{3}(1000)\right)=1.0025^{4}(1000)\)

Observing a pattern, we could conclude

\(P_{m}=(1.0025)^{m}(\$ 1000)\)

Notice that the $1000 in the equation was \(P_0\), the starting amount. We found 1.0025 by adding one to the growth rate divided by 12, since we were compounding 12 times per year.

Generalizing our result, we could write

\(P_{m}=P_{0}\left(1+\frac{r}{k}\right)^{m}\)

In this formula:

\(m\) is the number of compounding periods (months in our example)

\(r\) is the annual interest rate

\(k\) is the number of compounds per year.

While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If \(N\) is the number of years, then \(m = N k\). Making this change gives us the standard formula for compound interest.

Compound Interest

\(P_{t}=P_{0}\left(1+\frac{r}{k}\right)^{k t}\)

\(P_t\) is the balance in the account after \(t\) years.

\(P_0\) is the starting balance of the account (also called initial deposit, or principal)

\(r\) is the annual interest rate in decimal form

\(k\) is the number of compounding periods in one year.

If the compounding is done annually (once a year), \(k = 1\).

If the compounding is done quarterly, \(k = 4\).

If the compounding is done monthly, \(k = 12\).

If the compounding is done daily, \(k = 365\).

The most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest.

Example 1

A certificate of deposit (CD) is a savings instrument that many banks offer. It usually gives a higher interest rate, but you cannot access your investment for a specified length of time. Suppose you deposit $3000 in a CD paying 6% interest, compounded monthly. How much will you have in the account after 20 years?

Solution

In this example,

\(\begin{array} {ll} P_{0}=\$ 3000 & \text{the initial deposit} \\ r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{12 months in 1 year} \\ t = 20 & \text{since we’re looking for how much we’ll have after 20 years} \end{array}\)

So \(P_{20}=3000\left(1+\frac{0.06}{12}\right)^{12 \times 20}=\$ 9930.61\) (round your answer to the nearest penny)

Let us compare the amount of money earned from compounding against the amount you would earn from simple interest

\(\begin{array}{|l|r|r|}
\hline \text { Years } & \begin{array}{l}
\text { Simple Interest } \\
\text { (\$15 per month) }
\end{array} & \begin{array}{l}
6 \% \text { compounded } \\
\text { monthly }=0.5 \% \\
\text { each month. }
\end{array} \\
\hline 5 & \$ 3900 & \$ 4046.55 \\
\hline 10 & \$ 4800 & \$ 5458.19 \\
\hline 15 & \$ 5700 & \$ 7362.28 \\
\hline 20 & \$ 6600 & \$ 9930.61 \\
\hline 25 & \$ 7500 & \$ 13394.91 \\
\hline 30 & \$ 8400 & \$ 18067.73 \\
\hline 35 & \$ 9300 & \$ 24370.65 \\
\hline
\end{array}\)

As you can see, over a long period of time, compounding makes a large difference in the account balance. You may recognize this as the difference between linear growth and exponential growth.

 

 

 

Using your calculator

Evaluating exponents on the calculator

When we need to calculate something like \(5^3\) it is easy enough to just multiply \(5 \cdot 5 \cdot 5=125\). But when we need to calculate something like \(1.005^{240}\), it would be very tedious to calculate this by multiplying 1.005 by itself 240 times! So to make things easier, we can harness the power of our scientific calculators.

Most scientific calculators have a button for exponents. It is typically either labeled like:

\([\wedge ]\), \([y^x]\), or \([x^y]\)

To evaluate \(1.005^{240}\) we'd type 1.005 \([^]\) 240, or 1.005 \([y^x]\) 240. Try it out - you should get something around 3.3102044758.

Example 2

You know that you will need $40,000 for your child’s education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?

Solution

We’re looking for \(P_0\).

\(\begin{array} {ll} r = 0.04 & 4\% \\ k = 4 & \text{4 quarters in 1 year} \\ t = 18 & \text{Since we know the balance in 18 years} \\ P_{18} = \$40,000 & \text{The amount we have in 18 years} \end{array}\)

In this case, we’re going to have to set up the equation, and solve for \(P_0\).

\(40000=P_{0}\left(1+\frac{0.04}{4}\right)^{4 \times 18}\)

\(40000=P_{0}(2.0471)\)

\(P_{0}=\frac{40000}{2.0471}=\$ 19539.84\)

So you would need to deposit $19,539.84 now to have $40,000 in 18 years.

Rounding

It is important to be very careful about rounding when calculating things with exponents. In general, you want to keep as many decimals during calculations as you can. Be sure to keep at least 3 significant digits (numbers after any leading zeros). Rounding 0.00012345 to 0.000123 will usually give you a “close enough” answer, but keeping more digits is always better.

Example 3

To see why not over-rounding is so important, suppose you were investing $1000 at 5% interest compounded monthly for 30 years.

Solution

\(\begin{array} {ll} P_0 = \$1000 & \text{the initial deposit} \\ r = 0.05 & 5\% \\ k = 12 & \text{12 months in 1 year} \\ t = 30 & \text{since we’re looking for the amount after 30 years} \end{array}\)

If we first compute \(\frac{r}{k}\), we find \(\frac{0.05}{12} = 0.00416666666667\)

Here is the effect of rounding this to different values:

\(\begin{array}{|l|l|l|}
\hline r / k \text { rounded to: } & \text { Gives } \boldsymbol{P}_{30} \text { to be: } & \text { Error } \\
\hline 0.004 & \$ 4208.59 & \$ 259.15 \\
\hline 0.0042 & \$ 4521.45 & \$ 53.71 \\
\hline 0.00417 & \$ 4473.09 & \$ 5.35 \\
\hline 0.004167 & \$ 4468.28 & \$ 0.54 \\
\hline 0.0041667 & \$ 4467.80 & \$ 0.06 \\
\hline \text { no rounding } & \$ 4467.74 & \\
\hline
\end{array}\)

If you’re working in a bank, of course you wouldn’t round at all. For our purposes, the answer we got by rounding to 0.00417, three significant digits, is close enough - $5 off of $4500 isn’t too bad. Certainly keeping that fourth decimal place wouldn’t have hurt.

Using your calculator

In many cases, you can avoid rounding completely by how you enter things in your calculator. For example, in the example above, we needed to calculate

\(P_{30}=1000\left(1+\frac{0.05}{12}\right)^{12 \times 30}\)

We can quickly calculate \(12 \times 30=360\), giving \(P_{30}=1000\left(1+\frac{0.05}{12}\right)^{360}\).

Now we can use the calculator.

\(\begin{array}{|c|c|}
\hline \textbf { Type this } & \textbf { Calculator shows } \\
\hline 0.05 [\div] 12 [=] & 0.00416666666667 \\
\hline [+] 11 [=] & 1.00416666666667 \\
\hline [\mathrm{y}^{\mathrm{x}}] 360 [=] & 4.46774431400613 \\
\hline [\times] 1000 [=] & 4467.74431400613 \\
\hline \hline
\end{array}\)

Using your calculator continued

The previous steps were assuming you have a “one operation at a time” calculator; a more advanced calculator will often allow you to type in the entire expression to be evaluated. If you have a calculator like this, you will probably just need to enter:

1000 \([\times]\) ( 1 \([+]\) 0.05 \([\div]\) 12 ) \([y^x]\) 360 \([=]\).

Solving for time

Often we are interested in how long it will take to accumulate money.

Note: This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in Chapter 5.

Example 4

If you invest $2000 at 6% compounded monthly, how long will it take the account to double in value?

Solution

This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,

\(\begin{array}{ll} P_0 = \$2000 & \text{the initial deposit} \\ r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{12 months in 1 year} \end{array}\)

So our general equation is \(P_{t}=2000\left(1+\frac{0.06}{12}\right)^{12 \times t}\). We also know that we want our ending amount to be double of \(\$ 2000,\) which is \(\$ 4000,\) so we're looking for \(t\) so that \(P_{t}=4000 .\) To solve this, we set our equation for \(P_{t}\) equal to \$4000.

\(\begin{array}{ll} 4000=2000\left(1+\frac{0.06}{12}\right)^{12 \times t} & \text{Divide both sides by 2000} \\ 2=(1.005)^{12 t} & \text{To solve for the exponent, take the log of both sides} \\ \log (2)=\log \left((1.005)^{12 t}\right) & \text{Use the exponent property of logs on the right side} \\ \log (2)=12 t \log (1.005) & \text{Now we can divide both sides by } 12\log{1.005} \\ \frac{\log (2)}{12 \log (1.005)}=t & \text{Approximating this to a decimal} \\ t=11.581 & \end{array}\)

It will take about 11.581 years for the account to double in value. Note that your answer may come out slightly differently if you had evaluated the logs to decimals and rounded during your calculations, but your answer should be close. For example if you rounded log(2) to 0.301 and log(1.005) to 0.00217, then your final answer would have been about 11.577 years.