4.2: Simple and Compound Interest
- Page ID
- 130968
In this section, we'll study how to describe interest accrual using both simple and compound interest, and relate these concepts to our study of exponential functions.
- Use the simple interest formulas to compute and interpret quantities in simple interest scenarios
- Describe how compound interest arises from repeated simple interest calculations
- Use the compound interest formula to compute and interpret quantities in compound interest scenarios
Simple Interest: Ideas and Formulas
The essential idea behind interest is: it costs money to borrow money. There are numerous historical reasons why this is true. You've studied inflation already, so you know that a particular amount of money loses value over time. So, in general, if someone loans someone else money, they expect to get back the amount loaned plus some additional amount — at least enough to make up for inflation, if not more. This additional amount of money owed, beyond the initial amount of the loan itself, is called interest. Roughly, interest is the fee paid to borrow a certain amount of money.
Interest can be calculated in a variety of ways. You might agree on a fixed extra amount, but what's more common is to describe the interest as percent of the original amount borrowed. As an example: let's say you loan \($3000\) to a friend who is trying to start a company, and you ask for \(12\%\) interest on your loan. That means that you would expect to be paid back the original \($3000\) plus an additional \(12\%\) of \(\$3000\) at the end of the time period you mutually specify. (This time period is often called the "term" of the loan.) Using the Basic Percent Equation, we can calculate that \(12\%\) of \(\$3000\) is \(\$360\). Therefore, you would expect to get paid back a total of \($3360\), \(\$3000\) of which will be repaying your initial investment, and \(\$360\) of which is a fee that you are charging your friend for borrowing your money.
The author acknowledges the many current and historical problems caused by usurious interest rate practices; this book will not cover the very interesting and important debate about whether current interest practices make sense and whether or not interest should exist at all. The fact is that if you want to borrow or loan money through an institutional method (a bank, payday loan company, credit card, or similar structure), there is going to be an interest calculation involved. So we will endeavor to understand some things about these calculations.
This leads to our first definition, which is really an explanation of two formulas:
A loan accrues simple interest, if the total interest is the product of the initial loaned amount known as the principal, a fixed percent known as an interest rate, and the length of the loan known as the term. The balance is the total amount in the account at the end of the term, whose length is a time \(t\) in years. We use the formulas:
\[I = Prt \quad \text{ and } \quad B = P + I\]
where \(I\) is interest, \(P\) is principal, \(t\) is time in years, \(r\) is the interest rate (as a decimal), and \(B\) is the balance to describe simple interest calculations.
Notice that we can actually combine these two formulas in the following way by substituting \(I = Prt\) into the second formula and then factoring out a P: \[B = P + Prt = P(1+rt)\]
This will give us a quick way to find the balance of a loan -- that is, the total amount that is owed -- if we know the rate, initial amount, and the length of the loan. Let's see an example of how these formulas are used.
Erin takes out a \(\$1,200\) loan to buy a new laptop. The loan has \(4\%\) interest, and a term of \(24\) months. How much interest will she pay on this loan? How much will she pay in total?
Solution
First of all: if someone knows where to find a loan like this, please contact the author immediately! Unfortunately this is not a realistic situation due to the low interest rate and the simplicity of its structure; rather, it is meant to illustrate the use of the formula using relatively nice numbers. We will see more realistic situations later on.
We need to look through the problem and identify which numbers correspond to which variables in the simple interest formula. Remember that those variables are: \[\begin{align*} B &= \text{Balance} \\ I&= \text{Interest} \\ P &= \text{Principal} \\ r &= \text{rate} \\ t &= \text{time in years}\\ \end{align*}\]
The balance is actually not yet known to us: it's what we're trying to find. This will usually be the case. However, we are told the initial amount of the loan, which is \(\$1200\). The initial value of a loan is called the principal, so we'll use \(P = 1200\). The interest rate is \(4\%\), which we'll convert to \(r = .04\). In general, the interest rate will be the only percent value in the question. Finally, we're told that the term of the loan is \(24\) months. We need to convert this to years, and since \(24\) months is \(2\) years, we will use \(t = 2\). We also don't know the value of \(I\); we will calculate it from the given information.
Revisiting our variables, we then have: \[\begin{align*} B &= \text{Balance} = \; ? \\ I&= \text{Interest} = \; ? \\ P &= \text{Principal} = 1200 \\ r &= \text{rate} = .04\\ t &= \text{time in years} = 2\\ \end{align*}\]
The first question asks: how much interest will she pay on this loan? In order to calculate that, we'll use the formula: \[I = Prt\]
Substituting in our values \(P = 1200, r = .04, t = 2\), we get: \[I = 1200 \times .04 \times 2 = 96\]
Therefore, she will pay \(\$96\) in interest on this loan.
To find the total amount paid, we can use the formula: \[B = P + I\]
Now that we know that \(I = 96\), and we previously knew that \(P = 1200\), we can calculate: \[B = 1200 + 96 = 1296\]
Therefore, the total amount that she pays back is \(\$1296\).
Now, note that if we'd want to calculate the total amount paid back directly, we could use the more direct formula: \[B = P(1+rt)\] and calculate: \[B = 1200(1+.04(2)) = 1200 (1.08) = 1296\] which gives the same answer that we had before.
In general, it's not necessary to calculate the balance twice — the point was to illustrate the utility of the \(B = P(1+rt)\) formula. This will become important in the next section. Before then, though, we illustrate how to find a different value in the simple interest formula.
A bank lends Michael \(\$20,000\) so he can buy a car. Michael signs a contract saying that he'll pay back \(\$26,000\) at the end of \(4\) years. Assuming this is a simple interest loan, what is its interest rate?
Solution
We'll set up our variables as we did before. We have different information than we did in the previous example. In this case, we know the initial loan amount is \(\$20,000\), so \(P = 20000\). Also, we know the final amount paid back — the balance — is \(\$26000.\). Therefore, we have \(B = 26000\). We're also given that the length of the loan is \(t = 4\) years. Looking at our variables, we have: \[\begin{align*} B &= \text{Balance} = 26000 \\ I&= \text{Interest} = \; ? \\ P &= \text{Principal} = 20000 \\ r &= \text{rate} = \; ?\\ t &= \text{time in years} = 4\\ \end{align*}\]
Our goal is to find \(r\), but in order to do that, we need to find \(I\). Otherwise, it will be difficult to find \(r\) using the formulas we have. So, our first step will be use the formula \(B = P+ I\) to solve for \(I\). We have:\[B = P+I\]
Substituting in \(P = 20000\) and \(B = 26000\), we can calculate: \[26000 = 20000 + I\]
Now, to find \(I\), we will subtract \(20000\) from both sides. We get: \[6000 = I\] which means that \(\$6000\) was paid in interest. From there, we can use the formula \(I = Prt\) to find \(r\). Substituting \(I = 6000, P = 20000\), and \(t = 4\), we get: \[6000 = 20000 \times r \times 4\] Now to calculate \(r\), we need to know that the three terms multiplied on the right can have their orders switched. (This is a mathematical property known as commutativity.) In any case, we can rewrite the equation above as \[6000 = 20000 \times 4 \times r\] which simplifies to \[6000 = 80000r\]
Now, we can use Division undoes Multiplication to get \[r = \frac{6000}{80000} = .075\]
This is the interest rate, expressed as a decimal. So as a percent, the interest rate is \(7.5\%\).
In general, it is possible to manipulate the equations \(I = Prt\) and \(B = P + I\) to find any of the variables involved, given enough information.
However, it is much more useful in most cases to use the simplified formula \(B = P(1+rt)\) because it leads to the formula for a much more common type of interest: compound interest.
Compound Interest
In the previous section, we studied simple interest. While simple interest is relatively straightforward to compute, it is not the type of interest that is typically used in most actual loans. Instead, most loan accounts use some form of compound interest.
A loan accrues compound interest if the balance is computed as a repeated simple interest calculation on a periodic schedule. The balance on a compound interest loan can be expressed as a function of the time of the loan. This function is given by: \[B(t) = P\left(1+\frac{r}{n}\right)^{nt} \]
where \(B(t)\) is the balance at time \(t\), \(P\) is the principal, \(r\) is the interest rate, \(n\) is the number of times per year the interest is calculated, and \(t\) is the time in years. This may seem complicated, but we will see soon how it makes sense.
Take a moment to compare the equations for simple vs compound interest. We said that, for simple interest, \[B = P (1+rt)\] and for compound interest, \[ B(t) = P\left(1+\frac{r}{n}\right)^{nt} \]
If you stare at these for a few minutes, you will likely see some similarities. On the left side, there is the balance. (In the second equation, we happened to write the balance as a function of time, but it's still standing for the same thing.) On the right side, there is the principal \(P\), multiplied by a factor that involves the interest rate. However, in the compound interest equation, the variable \(t\) is in the exponent. For this reason, compound interest is an exponential function.
In the previous definition, we are familiar with all of the variables besides \(n\) from the simple interest formulas. The idea behind \(n\) is that it counts the number of times per year the interest is calculated. Since the interest rate is annual, we take that rate, \(r\), and divide it by the number of times per year the interest is calculated. This evenly distributes the percent interest calculation throughout the year. However, since the interest is being calculated on a higher and higher balance each time, the amount of interest continues to grow over time.
In general, you can determine what \(n\) is by looking for a keyword that indicates the compounding schedule. Here is a table that shows the most common values of \(n\) and their corresponding keywords. Remember that \(n\) is the number of times per year interest is calculated, so these values are obtained by simply counting the number of a given period in a year.
| Keyword | \(n\) |
| Annually | 1 |
| Monthly | 12 |
| Weekly | 52 |
| Daily | 365 |
Let's see an example to understand.
Leanne would like to purchase an iPad Pro using her credit card. Her credit card has \(24\%\) interest rate compounded monthly. Assuming that she does not make any payments on the purchase, how much will she owe after \(2\) years? Compare this with a simple interest rate for the same rate and time period.
Solution
In order to solve this question, we must note the word "compounded" in the question. That tells us that this is a compound interest question. That tells us to use the formula: \[B(t) = P\left(1+\frac{r}{n}\right)^{nt} \]
where \(B(t)\) is the balance at time \(t\), \(P\) is the principal of the loan, \(r\) is the rate, and \(n\) is the number of times per year that the interest is compounded. We are trying to find the balance when the time \(t = 2\) years. The initial loan amount, \(P\), is \(\$649\). So we'll plug in \(P = 649\). The rate is given by \(r= .24\), which is computed by converting the rate of \(24\%\) to a decimal.
The only missing variable is \(n\). That can be found by looking for one of the keywords and finding the associated \(n\) value. We see in this question that the interest is compounded monthly. Therefore, we will choose \(n = 12\).
We've shown that: \[ \begin{align*} B &= \text{Balance} = \; ? \\ P &= \text{Principal} = 649 \\ n &= \text{monthly} = 12 \\ r &= \text{rate} = .24\\ t &= \text{time in years} = 2\\ \end{align*}\]
From here, we can use the formula: \(B(t) = P (1+\frac{r}{n})^{nt}\) with the values given above. That is, we have: \[B(2) = 649\left(1+ \frac{.24}{12}\right)^{12 \times 2}\]
Let's make a couple simplifications on the right side: \[B(2) = 649\left(1+ \frac{.24}{12}\right)^{12 \times 2} = 649(1.02)^{24}\]
Take a moment to note that this has the same format as the exponential functions we saw in the previous section. We will use the same techniques to calculate it, and we find that: \[B(2) = 649(1.02)^{24} = 1043.86\]
That is, after \(2\) years, Leanne will owe \(\$1043.86\).
To contrast the situation above with simple interest, note that if the same loan were applied with simple interest instead of compound interest, we'd use \(r = .24, t = 2\), and \(P = 649\). That would result in a balance of: \[B(t) = P(1+rt) = 649(1+(.24)(2)) = 960.52\]
That is, the same loan with a simple interest structure would be \(\$960.52\), as compared to the \(\$1043.86\) price tag of compound interest. The reason is that compound interest was calculated multiple times over the course of the loan; even though it was a smaller percent interest each time, the exponential growth caused it to grow larger over time.
Exercises
- You take out a simple interest loan from a bank for \(\$35,000\). The interest rate is \(7\%\) and you plan to pay the loan back \(3\) years later. How much interest will you pay, and how much will you pay in total?
- An entrepreneur plans to open a coffee shop in Monmouth. A bank finances his equipment by granting him a \(\$65,000\) loan. The loan has \(1.7\%\) annual simple interest. He eventually pays the bank \(\$90,000\). How long did it take him to pay back the loan? Give your answer as a number of years rounded to one decimal place.
- Mr. and Mrs. Gonzales want to save money to help their son Phil go to college. When Phil is born, they deposit \(\$9,000\) into a savings account with \(7.3\%\) annual interest that compounds daily. How much will be in the account when Phil turns \(18\)?