Opening the book…
- Compute simple interest.
- Understand and compute future value.
- Compute simple interest loans with partial payments.
- Understand and compute present value.
- Understand and compute future value.
- Compute present value.
- Compute and interpret effective annual yield.
There is truth in the phrase “You need to have money to make money.” In essence, if you have money to lend, you can lend it at a cost to a borrower and make money on that transaction.
When money is borrowed, the person borrowing the money (the borrower) typically has to pay the person or entity that lent the money (the lender) more than the amount borrowed. This extra money is the interest that is to be paid. Interest is sometimes referred to as the cost of borrowing, the cost of the loan, or the finance charge.
This idea also applies when someone deposits money into a bank account or invests in another form of investment. That person is essentially lending the money to the bank or company. The money earned by the depositor is also referred to as interest. The interest is typically based on the amount borrowed, or the principal.
The pairing of borrower and lender can take various forms. The borrower may be a consumer using a credit card or taking out a loan from a bank, the lender. Companies also borrow from lending banks. Someone who invests in a company’s stock is the lender in this case; the company is essentially the borrower.
In this section, we examine the basic building block of interest paid on loans and borrowed credit, as well as the returns on investments such as bank accounts and simple interest.
Simple Interest and Future Value
Let’s get some terminology understood. The interest to be paid by a borrower is often expressed as an annual percentage rate, which is the percentage of the principal paid as interest each year the money is borrowed. This means that the more that is borrowed, the more that must be paid back. Sometimes, the interest to be repaid is simple interest, meaning it is calculated only on the amount borrowed.
The length of time until the loan must be repaid is the loan term. The date when the loan must be paid off is the due date. The day that the loan is issued is the origination date. We’ll put this terminology to use in the following examples. Note that in this section, we will use letters, called variables, to represent the different parts of the formulas we’ll be using. This will help keep our formulas and calculations manageable.
Calculating simple interest is similar to the percent calculations we made in Understanding Percent and Discounts, Markups, and Sales Tax, but it must be multiplied by the loan term (in years when using an annual percentage rate).
The simple interest, \(I\), to be paid on a loan with annual interest rate \(r\), with principal \(P\), is found using
\[I=P\times r\times t\nonumber\]where the decimal form of the interest rate, \(r\), is used and time, \(t\), is expressed in years.
The amount that is loaned is called the present value, \(PV\), or just \(P\).
The future value, , of an investment that yields simple interest is
\[\begin{align*}
FV&=P+I \\
&Or\\
FV&=P(1+rt)\\
\end{align*} \]
Also, the interest that is earned by investing \(P\) can be calculated as
\(P\) is the principal (amount invested at the start), \(r\) is the annual interest rate in decimal form, and \(t\) is the length of time the money is invested (in years).
\[I = FV- P \nonumber\]
Note: The future value of the amount is also denoted by just \(A\).
The Money can also be invested for a certain period, earning simple interest during that time. The terms and formulas we use are the same as before.
The investor wants to know how much the investment will be worth after interest is added. This is called the future value.
As we see above, future value is the amount of money you end up with after the principal (the amount originally invested) and all the interest earned during that time are added together.
Example \(\PageIndex{1}\): Simple Interest on Loans with Whole Number Year Terms
- Calculate the simple interest to be paid on a loan with the given principal, annual percentage rate, and number of years. Then, calculate the loan payoff amount. (Round your answers to the nearest cent.): Principal P = \($4,000,\) annual interest rate r = \(5.5\%\), and number of years t = \(4.\)
- Riley runs an auto repair shop and needs to purchase a new brake lathe, which costs \($11,995.\) She takes out a two-year, simple interest loan at an annual interest rate of \(14.9\%.\) How much interest will she pay, and how much will she repay on the loan? (Round your answers to the nearest cent)
- Answer 1
-
Substitute the principal P = \($4,000,\) the decimal form of the annual interest rate r = \(0.055,\) and number of years \(t=4\) into the formula for simple interest, and calculate. \[I=4,000\times 0.055\times4=880\nonumber\]The simple interest, or cost of the loan, to be paid on the loan is \($880,\) which is \(I\). The loan payoff amount, or the total to be repaid, is \[A=P+I=$4,000+880=$4,880\nonumber\] So \($4,880.00\) is the loan payoff amount.
- Answer 2
-
Determine the variables or parts of the formula. The principal P = \($11,995.\) The interest rate Riley pays is \(14.9 \%,\) or \(r=0.149\) in decimal form. The length of the loan is two years, so \(t=2\). We are first asked to find \(I\), the interest Riley will pay. Substitute the known variables into the formula for simple interest \(I=P\times r\times t\) and solve for \(I.\)
This tells us that the simple interest, or cost to borrow, to be paid on the loan is \($3,574.51.\)
Use the formula \(A=P+I\) to determine the total amount Riley will repay.
\[A=P+I=$11,995+$3,574.51=$15,569.51\nonumber\]
The total to be repaid is \($15,569.51\).
Simple Interest Loans with Other Lengths of Terms
In the previous example and the Your Turn exercise, the loans were repaid in a single payment after an integer number of years. However, there are also loans lasting a length of time not equal to an integer number of years (like \(1\), \(2\), or \(3\) years or more), but in a number of months (like \(4\) months, \(18\) months, and so on). What model would apply to these situations?
When the loan is repaid after a term that is not a year but rather a number of months, days, or weeks, \(t\) must be expressed in years. Here are some examples.
- For a \(5\)-month loan, the time in years is \(t=\frac{5}{12}\).
- For a \(18\)-month loan, the time in years is \(t=\frac{18}{12}\).
- For a \(100\)-day term, the term in years is \(t=\frac{100}{365}\).
- For a \(30\)-week loan, the time in years is \(t=\frac{30}{52}\).
Do not forget to convert time to years before you put it in the formula.
Example \(\PageIndex{2}\): Loan to Purchase Equipment
- Abeje needs a loan to purchase equipment for the gym she is going to open. She visits the bank and secures a \(4\)-month loan of \($20,000.\) Her annual percentage rate is \(6.75\%.\) How much interest will Abeje pay, and what is her loan payoff amount?
- David plans to move his family from Raleigh, North Carolina, to Tempe, Arizona. His company will reimburse (pay after the move) David for the move. David does research and determines that movers will cost \($5,600\) to move his family’s belongings to Tempe. He takes out a simple interest, \(45\)-day loan at \(11.75\%\) interest to pay this cost. How much interest will be paid on this \(45\)-day loan, and what is David’s loan payoff amount? (Round your answers to the nearest cent)
- Answer 1
-
Abeje’s loan is for \($20,000,\) so her principal is \(P=$20,000.\) The interest rate Abeje will pay is \(6.75\%,\) or \(r=0.0675\) in decimal form. The length of the loan is \(4\) months, so \(t=\frac{4}{12}\) . Substituting these in the formula for simple interest, we find her interest to be\[I=20,000\times 0.0675\times\frac{4}{12}=450\nonumber \] The simple interest, or cost to borrow, to be paid on the loan is \($450.00.\) The payoff is \($20,000 + $ 450 = $ 20,450\).
- Answer 2
-
The principal for the loan is the moving cost, or \(P= $5,600.\) The annual interest rate that David will pay is \(11.75\%,\) which in decimal is \(r=0.1175.\) The length of time for the loan is \(45\) days, so \(t=\frac{45}{365}\) year. Substituting these values into the formula and calculating, we find that the interest to be paid is
\[I=5600\times 0.1175\times\frac{45}{365}=81.12\nonumber\]
\($81.12\) is the interest for \(45\) days.
For payoff
\[A=P+I=$5,600+$81.12=$5,681.12\nonumber\]
The payoff for the loan is \($5,681.12.\)
It seems reasonable to use \(365\) as the number of days in the year since there are \(365\) days in most years. However, sometimes, banks have used (and continue to use) \(360\) as the number of days in a year. They may also treat all months as if they have \(30\) days. These differences lead to (sometimes small) differences in how much interest is paid. Since the number of days is in the denominator, a smaller denominator (\(360\)) will result in larger numbers (interest); therefore, \(365\) is used for the denominator. See this page from ACRE(opens in new window) for a comparison.
Example \(\PageIndex{3}\): Simple Interest on a Investment
In the following, determine how much interest was earned on the investment and the future value of the investment, if the investment yields simple interest. Round your answer to the nearest cent.
- Principal is $\(1,000,\) annual interest rate is \(2.01\%,\) and time is \(5\) years.
- Principal is $\(10,000,\) annual interest rate is \(1.25\%,\) and time is \(18\) months.
- Principal is $\(7,000,\) annual interest rate is \(3.26\%,\) and time is \(100\) days.
- Answer 1
-
The principal is \(P=$1,000,\) the annual interest rate, in decimal form, is \(r=0.0201,\) and the term is \(5\) years, or \(t=5.\) \[FV=P(1+rt)=1000(1+0.0201\times 5)=1100.50 \nonumber\]The interest earned on the investment is\[I = FV- P = 1100.50-1000=100.50\nonumber \]The future value of the investment at the end of \(5\) years is $\(1,100.50,\) where \($100.50\) comes from interest and \($1000\) comes from principal..
- Answer 2
-
The principal is \(P=$10,000,\) the annual interest rate, in decimal form, is \(0.0125,\) and the term is \(18\) months. Since the term is in months, we have to write the months in terms of years. For \(18\) months, we have used \(\frac{18}{12}\) as t.\begin{align*} FV&=P(1+rt)\\
&=10,000\left(1+0.0125\times \frac{18}{12}\right)\\
&=10,187.50 \\
\end{align*}The interest earned on the investment is \[I = FV- P = 10,187.50-10,000=187.50\nonumber\]The future value of the investment at the end of \(18\) months is \($10,187.50,\) where \($187.50\) comes from interest and \($10,000\) comes from principal..
- Answer 3
-
The principal is \(P=$7,000,\) the annual interest rate, in decimal form, is \(r=0.0326,\) and the term is \(100\) days. Since the term is in days, we have to write the time in years, or \(t = \frac{100}{365}.\)
\[\begin{align*}
FV&=P(1+rt)\\
&=7,000\left(1+0.0326\times \frac{100}{365}\right)\\
&=7,062.52\\
\end{align*} \]The interest earned on the investment is
\[I = FV- P = $7,062.52-$7,000=$62.52\nonumber\]
The future value of the investment at the end of \(100\) days is \($7,062.52,\) where \($62.52\) comes from interest and \($7,000\) comes from principal.
Understand and Compute Present Value for Simple Interest Investments
When finding the future value of an investment, we know how much is deposited, but we have no idea how much that money will be worth in the future. If we set a future goal, it would be useful to know how much to deposit now for the account to reach it. The amount that needs to be deposited now to reach a goal in the future is called the present value, PV, or simply P, which is a notation used for a loan.
The present value, \(PV\) of money deposited at an annual, simple interest rate of \(r\) (in decimal form) for time \(t\) (in years) with a specified future value of \(FV\) is calculated with the formula
\[\text{PV}=\frac{FV}{(1+rt)}\nonumber\]
Note: Present value, in this calculation, is always rounded UP. Otherwise, future value may fall short of the target future value. Sometimes, present value may be denoted by just \(P.\)
Understanding what this tells you is important. When you find the present value, that is how much you need to invest now to reach the goal, under the conditions (time and rate) at which the money will be invested.
Example \(\PageIndex{4}\): Find Present Value
Compute the present value of the investment described. Interpret the result.
Beatriz will invest some money in a CD that yields \(3.99\%\) simple interest when invested for \(30\) years. How much must Beatriz invest so that after those \(30\) years, her CD is worth \($300,000?\)
- Answer
-
Beatriz needs to know how much to deposit now so that her CD is worth \($300,000\) after \(30\) years. This means she needs to know the present value of that \($300,000.\) The time is \(30\) years and the annual simple interest rate, in decimal form, is \(r=0.0399.\) Using that information and the formula for present value, we calculate the present value of that \($300,000.\)
\[\begin{align*}
PV&=\frac{FV}{(1+rt)}\\
&= \frac{300,000}{(1+0.0399\times30)}\\
&=\frac{300,000}{2.197}\\
&=136,459.8407\\
\end{align*} \]Rounding up, Beatriz needs to invest \($136,549.85\) so that she has \($300,000\) in \(30\) years.
Determining Simple Interest Rate and Time
Using the same formula, we can also calculate the simple interest rate \(r\) and the time \(t\).
\[\begin{align*}
\text{Interest rate:}~r&=\frac{FV-P}{P\times t}\\[6pt]
\text{Time:}~t&=\frac{FV-P}{P\times r}\
\end{align*} \]
Example \(\PageIndex{5}\): Find \(r\) and \(t\)
- A simple interest loan for \($20,000\) is taken at a \(12.6\%\) annual rate when the loan's future value is $\(25,000.\)
- A simple interest loan for \($1500\) is taken, and you pay it back to double in \(4\) years. What simple interest rate do you pay?
- Answer 1
-
Here we have, \(PV=$20,000,\) \(FV=$25,000,\) \(r=0.126,\) and \(t\) = Unknown
Use the formula for \(t\)
\[\begin{align*}
\text{Time:}~t&=\frac{FV-P}{P\times r}\\
&=\frac{25,000-20,000}{20,000\times 0.126}\\
&=\frac{5,000}{2520}\\
&=1.98\\
\end{align*} \]So A simple interest loan for \($20,000\) will be \($25,000\) in almost \(2\) years with a \(12.6\%\) interest rate.
- Answer 2
-
Here we have, \(PV=$1500,\) \(FV=$3000,\) \(t = 4,\) and \(r\) = Unknown
Use the formula for \(r\)
\[\begin{align*}
\text{Time:}~r&=\frac{FV-P}{P\times t}\\
&=\frac{3,000-1,500}{1,500\times 4}\\
&=\frac{1,500}{6000}\\
&=0.25\\
\end{align*} \]So \($1500\) loan will be doubled in \(4\) year with \(25\%\) simple interest rate.
CAREFUL: When solving problems in the Simple Interest section, make sure to read each question carefully. Look for the word “simple” so you know to use the simple interest formula. Always ensure you understand which formula is required to solve the problem correctly.
Compound Interest
For a very long time in certain parts of the world, interest was not charged due to religious dictates. Once this restriction was relaxed, interest-bearing loans became possible. Initially, such loans had short terms, so only simple interest was applied. However, when loans began to stretch out for years, it was natural to add interest at the end of each year to the loan principal. After another year, the interest was calculated on the initial principal plus the interest from year \(1\), or the interest earned. Each year, more interest was added to the money owed, and that interest continued to earn interest.
Since the amount in the account grows each year, more money earns interest, increasing the account faster. This growth follows a geometric series (Geometric Sequences). It is this feature that gives compound interest its power. This module covers the mathematics of compound interest.
As we saw in Simple Interest, an account that pays simple interest only pays based on the original principal and the term of the loan. Accounts offering compound interest pay interest at regular intervals. After each interval, the interest is added to the original principal. Later, interest is calculated on the original principal plus the interest previously added.
After each period, the interest on the account is calculated and then added to the account balance. Then, after the next period, interest is computed based on the original principal AND the interest added in previous periods. The period of time between two interest payments is called the compounding period.
Compound interest is interest calculated on both the principal (the original amount of money) and any interest that has already been earned. In other words, interest is added to the balance, allowing you to earn interest on previously earned interest.
With compound interest, your money grows faster because the interest earned is added to the balance. In each subsequent period, interest is calculated on the new, larger balance.
The comic in the following figure illustrates the concept of compound interest and helps you understand how it works.
\[A=P\left(1+\frac{r}{n}\right)^{nt}\nonumber\]
\(A\): The balance (future value) in the account after \(t\) years.
\(P\): The starting balance of the account (also called initial deposit, or principal)
\(r\): The annual interest rate in decimal form.
\(n\): The number of compounding periods in one year.
Additionally, we can calculate the compound interest earned by subtracting the principal from the future value.
\[I=A-P\nonumber\]
Note: Sometimes, future value \(A\) can also be represented as \(FV\), and \(P\) can also be represented as \(PV\).
In finance and mathematics, a compounding period is the time interval between interest calculations.
When you invest money or take out a loan, interest is usually not just added once at the end—it’s added many times along the way. Each time interest is added (or "compounded"), that's a compounding period. We use \(n\) to denote the compounding period.
Example \(\PageIndex{6}\): Computing Future Value for Compound Interest
In the following, compute the future value of the investment with the given conditions.
Cody invests \($18,500,\) in an account that earns \(6.25\%,\) compounded quarterly, for \(17\) years. Determine the value of Cody’s investment after \(17\) years. How much interest is included in Cody's investment? (Round your final answer to the nearest cent.)
- Answer
-
The principal is \(P=$18,500,\) interest rate, in decimal form, \(r=0.0625,\) compounded quarterly so \(n=4,\) and for \(t=17\) years. Substituting these values into the formula, we
\begin{align*}
The future value of the investment is \($53093.55.\)
A&=P\left(1+\frac{r}{n}\right)^{nt}\\
&=18,500\left(1+\frac{0.0625}{4}\right)^{4\times 17}\\
&=18,500\left(1+0.015625\right)^{68}\\
&=18,500\left(1.015625\right)^{68}\\
&=18,500\times2.86992152\\
&=53,093.54812\\
\end{align*}Interest earned
\[I= A-P=\$53,093.55-$18,500=\$34,593.55\]
You can also find future value using the following TVM (Time Value of Money) Calculator. But you should learn how to do it manually using a formula. For example, input the following in the TVM solver.
\(6.25\)
\(18,500\)
CAREFUL: When solving problems in the Compound Interest section, make sure to read each question carefully. Look for the word “compounded,” so you know to use the compound interest formula. Always be sure you understand which formula is required to solve the problem correctly.
If you are using the TVM Solver:
-
Do not use commas and dollar signs in the input boxes.
-
Do not convert the interest rate to a decimal.
-
If your answer comes out negative, enter the final answer as a positive value.
-
Choose the value of \(n\) compounding periods based on the words monthly, quarterly, etc. (See Figure \(\PageIndex{6}\))
-
Remember that \(N=n×t\).
A Certificate of Deposit (CD) is a type of savings account offered by banks and credit unions that holds a fixed amount of money for a fixed period of time (such as \(6\) months, \(1\) year, or \(5\) years).
In exchange for keeping your money locked away for the agreed term, the financial institution pays you a fixed interest rate—typically higher than what standard savings accounts offer.
Almost all major traditional banks, online banks, credit unions, and brokerage firms offer Certificates of Deposit (CDs). See the figure below for a list of some financial institutions that offer CDs.
Example \(\PageIndex{7}\): Computing Future Value for Compound Interest
In the following, compute the future value of the investment with the given conditions.
You invest \($5,000,\) into a CD at Allay Bank, which gives you a fixed annual interest rate of \(3.80\%,\) compounded monthly, for \(5\) years. How much will you have in the account in \(5\) years? How much interest is included in the future value? (Round your final answer to the nearest cent)
- Answer
-
The principal is \(P=\$5,000,\) interest rate, in decimal form, \(0.038,\) compounded monthly so =\(12,\) and for =\(5\) years. Substituting these values into the formula, we find
\begin{align*}
A&=P\left(1+\frac{r}{n}\right)^{nt}\\
&=5000\left(1+\frac{0.038}{12}\right)^{12\times 5}\\
&=5000\left(1+0.00316666\right)^{60}\\
&=5000\left(1.00316666\right)^{60}\\
&=5000\times1.20888615\\
&=6044.4307\\
\end{align*}
The future value of the investment is \($6,044.43 \).Interest earned
\[I= A-P=\$6,044.43-$5,000=\$1,044.43\]
You can also find future value using the following TVM (Time Value of Money) Calculator. But you should learn how to do it manually using a formula. Input the following in TVM solver.
\(3.80\)
\(5,000\)
| When you do not use a Formula | When you use a Formula |
|
|
Let us compare the amount of money earned from compounding against the amount you would earn from simple interest
\[\begin{array}{|l|r|r|}
\hline \text { Years } & \begin{array}{l}
\text { Simple Interest } \\
\text { (\$15 per month) }
\end{array} & \begin{array}{l}
6 \% \text { compounded } \\
\text { monthly }=0.5 \% \\
\text { each month. }
\end{array} \\
\hline 5 & \$ 3900 & \$ 4046.55 \\
\hline 10 & \$ 4800 & \$ 5458.19 \\
\hline 15 & \$ 5700 & \$ 7362.28 \\
\hline 20 & \$ 6600 & \$ 9930.61 \\
\hline 25 & \$ 7500 & \$ 13394.91 \\
\hline 30 & \$ 8400 & \$ 18067.73 \\
\hline 35 & \$ 9300 & \$ 24370.65 \\
\hline
\end{array}\nonumber\]
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 see figure \(\PageIndex{7}\).
Understand and Compute Present Value from Compound Interest
When investing, there is often a goal to reach, such as “after \(20\) years, I’d like the account to be worth \($100,000.\)” The question to be answered in this case is “How much money must be invested now to reach the goal?” As with simple interest, this is referred to as the present value.
The money invested in an account bearing an annual interest rate of \(r\) (in decimal form), compounded \(n\) times per year for \(t\) years, is called the present value, \(PV\) (or just \(P\)), of the account (or of the money) and found using the formula
\begin{aligned}
P =\frac{A}{\left(1+\frac{r}{n}\right)^{nt}}
\end{aligned}
where \(A\) is the value of the account at the investment’s end. Always round this value UP to the nearest cent unless mentioned otherwise.
Example \(\PageIndex{8}\): Computing Present Value
Pilar plans for retirement early, believing she will need \(1,500,000\) to live comfortably after age \(67\). How much will she need to deposit at age \(23\) in an account bearing \(6.35\%\) annual interest compounded monthly? How much of \($1,500,000\) comes from interest?
- Answer
-
She’s \(23\) now and will leave the money in the account until the age of \(67\). She leaves the money for \(67-23=44\) years, making \(t=44.\) \(r=6.35\%=0.0635\), and \(n=12 \) because it is compounded monthly. Using this information, substitute into the formula for present value given above.
\[\begin{align*}
P &=\frac{A}{\left(1+\frac{r}{n}\right)^{nt}}\\
&=\frac{1,150,000}{\left(1+\frac{0.0635}{12}\right)^{12\times44}}\\
&=\frac{1,150,000}{\left(1.005291667\right)^{528}}\\
&=70,872.5737\\
\end{align*} \]Rounding up to the nearest cent, we get \(P=$70,872.58.\)
So, for this account to reach \($1,150,000\) after \(44\) years, \($70,872.58\) needs to be invested at the given interest rate, compounded monthly. Interest earned during that period is: \[I = A-P=\$1,150,000 -$70,872.58=\$1,079,127.42\nonumber\]Also, you can find the present value using the TVM solver
Compute and Interpret Effective Annual Yield
As we’ve seen, quarterly compounding pays interest \(4\) times a year or every months; monthly compounding pays \(12\) times a year; daily compounding pays interest every day, and so on. Effective annual yield allows direct comparisons between simple interest and compound interest by converting compound interest to its equivalent simple interest rate. We can even directly compare different compound interest situations. This provides information to identify the best investment from a yield perspective.
Using a formula, we can express compound interest in terms of simple interest. The effective annual yield formula stems from the compound interest formula and is based on an investment of \($1\) for one year.
\[\begin{align*}
Y&=\left(1+\frac{r}{n}\right)^n-1\\
\end{align*} \]
where \(Y\) = effective annual yield, \(r\) = interest rate in decimal form, and \(n\)= number of times the interest is compounded in a year.
\(Y\) is interpreted as the equivalent annual simple interest rate.
Note: Effective annual yield is also called Annual Percentage Yield (APY).
Example \(\PageIndex{9}\): Determine and Interpret Effective Annual Yield for 6% Compounded Quarterly
Suppose you have an investment paying a rate of \(6\%\) compounded quarterly. Determine and interpret the effective annual yield of the investment.
- Answer
-
Here, \(n=4\) (quarterly) and \(r=0.06\) (decimal form). Substituting into the formula, we find that the effective annual yield is
\[\begin{align*}
Y&=\left(1+\frac{r}{n}\right)^n-1\\
&=\left(1+\frac{0.06}{4}\right)^n-1\\
&=(1.015)^4-1\\
&=0.06136
\end{align*} \]Therefore, a rate of \(6\%\) compounded quarterly is equivalent to a simple interest rate of \(6.14\%.\)
This page titled 4.3: Simple and Compound Interest was last modified on Mon, 19 Oct 2026 21:45:04 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Cheban Acharya via source content that was edited to the style and standards of the LibreTexts platform.
The book could not be loaded just now.