3.1: Simple and Compound Interest
We have to work with money every day. Balancing your checkbook or calculating your monthly expenditures on espresso requires only arithmetic, but we need more complex mathematics when we start saving, plan for retirement, or need a loan.
Simple Interest
Discussing interest starts with the principal, or amount your account starts with. This could be a starting investment or the starting amount of a loan. Interest, in its most simple form, is calculated as a percent of the principal. For example, if you borrowed $100 from a friend and agree to repay it with 5% interest, then the amount of interest you would pay would just be 5% of 100: $100(0.05) = $5. The total amount you would repay would be $105, the original principal plus the interest.
Simple One-time Interest
\(I=P_{0} r \)
\(A=P_{0}+I=P_{0}+P_{0} r=P_{0}(1+r) \)
\(I\) is the interest
\(A\) is the end amount: principal plus interest
\(P_{0}\) is the principle (starting amount)
\(r\) is the interest rate (in decimal form. Example: 5%=0.05)
Example \(\PageIndex{1}\)
A Friend asks to borrow $300 and agrees to repay it in 30 days with 3% interest. How much interest will you earn?
Solution
\(P_0\)=$300 The principal
\(r\)=0.03 3% rate
\(I\)=$300(0.03)=9 You will earn $9 interest
One-time simple interest is only common for extremely short-term loans. For longer term loans, it is common for interest to be paid on a daily, monthly, quarterly, or annual basis. In that case, interest would be earned regularly. For example, bonds are essentially a loan made to the bond issuer (a company or government) by you, the bond holder. In return for the loan, the issuer agrees to pay interest, often annually. Bonds have a maturity date, at which time the issuer pays back the original bond value.
Example \(\PageIndex{2}\)
Suppose your city is building a new park, and issues bonds to raise the money to build it. You obtain a $1,000 bond that pays 5% interest annually that matures in 5 years. How much interest will you earn?
Solution
Each year, you would earn 5% interest: $1000(0.05) = $50 in interest. So over the course of five years, you would earn a total of $250 in interest. When the bond matures, you would receive back the $1,000 you originally paid, leaving you with a total of $1,250.
We can generalize this idea of simple interest over time.
Simple Interest over Time
\(I=P_{0} r t\)
\(A=P_{0}+I=P_{0}+P_{0} r t=P_{0}(1+r t) \)
\(I\) is the interest
\(A\) is the end amount: principle plus interest
\(P_{0}\) is the principal (starting amount)
\(r\) is the interest rate in decimal form
\(t\) is time
The units of measurement (year, months, etc.) for the time should match the time period for the interest rate.
APR - Annual Percentage Rate
Interest rates are usually given as an annual percentage rate (APR), the total interest that will be paid in the year. If the interest is paid in smaller time increments, the APR will be divided up.
For example, a 6% APR paid monthly would be divided into twelve 0.5% payments.
A 4% annual rate paid quarterly would be divided into four 1% payments.
Example \(\PageIndex{3}\)
Treasury Notes (T-notes) are bonds issued by the federal government to cover its expenses. Suppose you obtain a $1000 T-note with a 4% annual rate, paid semi-annually, with a maturity in 4 years. How much interest will you earn?
Solution
Since interest is being paid semi-annually (twice a year), the 4% interest will be divided into two 2% payments.
\(P_0\)= $1000 The principal
\(r\)=0.02 2% rate per half-year
\(t\)=8 4 years= 8 half-years
\(I\)=$1000(.02)(8) = $160 You will earn $160 interest total over the four years.
Try it Now 1
A loan company charges $30 interest for a one month loan of $500. Find the annual interest rate they are charging.
Compound Interest
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). (Note that this does NOT mean you get 3% every month, as explained below. That would be too good to be true.) Since interest is being paid monthly, each month, we will earn \(3\% \div 12 = 0.25\%\) per month.
In the first month,
\(P_0\)=$1000
\(r\)=0.0025 (0.25%)
\(I\)=$1000 (0.0025)=$2.50 (Interest for the first month)
\(A\)=$1000+$250=$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)
\(A\) = $1002.50+$2.51=$1005.01
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:
|
Month |
Starting balance |
Interest earned |
Ending Balance |
|---|---|---|---|
|
1 |
1000.00 |
2.50 |
1002.50 |
|
2 |
1002.50 |
2.51 |
1005.01 |
|
3 |
1005.01 |
2.51 |
1007.52 |
|
4 |
1007.52 |
2.52 |
1010.04 |
|
5 |
1010.04 |
2.53 |
1012.57 |
|
6 |
1012.57 |
2.53 |
1015.10 |
|
7 |
1015.10 |
2.54 |
1017.64 |
|
8 |
1017.64 |
2.54 |
1020.18 |
|
9 |
1020.18 |
2.55 |
1022.73 |
|
10 |
1022.73 |
2.56 |
1025.29 |
|
11 |
1025.29 |
2.56 |
1027.85 |
|
12 |
1027.85 |
2.57 |
1030.42 |
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
\(P_{m} = (1+0.0025) P_{m-1}\). Note that adding 0.25% to \(P\) amounts to calculating 100.25% of \(P\).
This is a type of exponential growth to be studied in a later chapter. Let’s 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
\[P_{m}=P_{0}\left(1+\dfrac{r}{k}\right)^{m} \nonumber \]
In this formula:
\(m\) is the number of compounding periods (months in our example)
\(r\) is the annual percentage rate (APR)
\(k\) 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 \(N k\) is the total number of compounding periods. Making this change gives us the standard formula for compound interest.
Compound Interest
\[P_{N}=P_{0}\left(1+\dfrac{r}{k}\right)^{N k}\nonumber \]
\(P_{N}\) is the balance in the account after \(N\) years.
\(P_{0}\) is the starting balance of the account (also called initial deposit, or principal)
\(r\) is the annual interest rate (APR) 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 \(\PageIndex{4}\)
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,
\(P_0\)=$3000 the initial deposit
\(r\)=0.06 6% annual rate
\(k\)=12 12 months in 1 year
\(N\)=20 since we’re looking for how much we’ll have after 20 years
So, \(P_{20}=3000\left(1+\dfrac{0.06}{12}\right)^{20 \times 12}=\$ 9930.61 \) (round your answer to the nearest penny)
|
Years |
Simple Interest ($15 per month) |
6% compounded monthly = 0.5% each month. |
|---|---|---|
|
5 |
$3900 |
$4046.55 |
|
10 |
$4800 |
$5458.19 |
|
15 |
$5700 |
$7362.28 |
|
20 |
$6600 |
$9930.61 |
|
25 |
$7500 |
$13394.91 |
|
30 |
$8400 |
$18067.73 |
|
35 |
$9300 |
$24370.65 |
Let us compare the amount of money earned from compounding against the amount you would earn from simple interest
As you can see, over a long period of time, compounding makes a large difference in the account balance. You may want to keep this in mind as a main difference between linear growth and exponential growth. We will see this in Chapter 6 again.
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 \(1.005^{\wedge} 240\), or \(1.005 y^{x} 240\). Try it out- you should get something around 3.3102044758
Example \(\PageIndex{5}\)
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
In this example,
We’re looking for \(P_0\)
\(r\)=0.04 4%
\(k\)=4 4 quarters in 1 year
\(N\)=18 Since we know the balance in 18 years
\(P_{18}\)=$40,000 The amount we have in 18 years
In this case, we’re going to have to set up the equation, and solve for \(P_0\).
\[\begin{array}{l}
40000=P_{0}\left(1+\dfrac{0.04}{4}\right)^{18 \times 4} \\
40000=P_{0}(2.0471) \\
P_{0}=\dfrac{40000}{2.0471}=\$ 19539.84
\end{array} \nonumber \]
So, you would need to deposit $19,539.84 now to have $40,000 in 18 years.
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
\[P_{30}=1000\left(1+\dfrac{0.05}{12}\right)^{12 \times 30} \nonumber \]
We can quickly calculate \(12 \times 30=360\), giving \(P_{30}=1000\left(1+\dfrac{0.05}{12}\right)^{360}\)
Now we can use the calculator.
| Type this | Calculator shows |
|---|---|
| \(0.05 \div 12=\) | 0.00416666666667 |
| +1 = | 1.00416666666667 |
| \(y^{x} 360=\) | 4.46774431400613 |
| \(\times 1000 = \) | 4467.74431400613 |
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:
\[1000 \times(1+0.05 \div 12) y^{x} 360= \nonumber \]
Reference
- References (5)
Contributors and Attributions
-
Saburo Matsumoto
CC-BY-4.0