6.1: Simple and Compound Interest
 Page ID
 67128
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 Onetime Interest
\[I=P r\]
\[A=P+I=P+P r=P(1+r)\]
where
 \(I\) is the interest
 \(A\) is the end amount: principal plus interest
 \(P\) is the principal (starting amount)
 \(r\) is the interest rate (in decimal form. Example: \(5\% = 0.05\))
Example 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
\(\begin{array}{ll} P=\$ 300 & \text{the principal } \\ r=0.03 & 3 \%\text{ rate} \\
I=\$ 300(0.03)=\$ 9. & \text{You will earn }\$ 9 \text{ interest.}\end{array}\)
Onetime simple interest is only common for extremely shortterm 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 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
\(I=P r t\)
\(A=P+I=P+P r t=P(1+r t)\)
where
 \(I\) is the interest
 \(A\) is the end amount: principal plus interest
 \(P\) is the principal (starting amount)
 \(r\) is the interest rate in decimal form
 \(t\) is time
The units of measurement (years, 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 3: Treasury Notes
Treasury Notes (Tnotes) are bonds issued by the federal government to cover its expenses. Suppose you obtain a $1,000 Tnote with a 4% annual rate, paid semiannually, with a maturity in 4 years. How much interest will you earn?
Solution
Since interest is being paid semiannually (twice a year), the 4% interest will be divided into two 2% payments.
\(\begin{array}{ll} P=\$ 1000 & \text{the principal } \\ r=0.02 & 2 \%\text{ rate} \\ t = 8 & \text{4 years = 8 halfyears} \\
I=\$ 1000(0.02)(8)=\$ 160. & \text{You will earn }\$ 160 \text{ interest total over the four years.}\end{array}\)
Excercies 1
A loan company charges $30 interest for a one month loan of $500. Find the annual interest rate they are charging.
 Answer

\(I=\$ 30\) of interest
\(P=\$ 500\) principal
\(r=\) unknown
\(t=1\) month
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. We looked at this situation earlier, in the chapter on exponential growth.
Compound Interest
\(A=P\left(1+\frac{r}{k}\right)^{kt}\)
\(A\) is the balance in the account after t years.
\(P\) 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 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,
\(\begin{array} {ll} P=\$ 3000 & \text{the initial deposit} \\ r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{12 months in 1 year} \\ t = 20 & \text{since we’re looking for how much we’ll have after 20 years} \end{array}\)
So \(A=3000\left(1+\frac{0.06}{12}\right)^{20 \times 12}=\$ 9930.61\) (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
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 
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.
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 would type 1.005 \([^]\) 240, or 1.005 \([y^x]\)^{ } 240. Try it out  you should get something around 3.3102044758.
Example 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
We’re looking for \(P\).
\(\begin{array} {ll} r = 0.04 & 4\% \\ k = 4 & \text{4 quarters in 1 year} \\ t = 18 & \text{Since we know the balance in 18 years} \\ A = \$40,000 & \text{The amount we have in 18 years} \end{array}\)
In this case, we’re going to have to set up the equation, and solve for \(P\).
\[\begin{align*} 40000 &=P\left(1+\frac{0.04}{4}\right)^{18 \times 4} \\ 40000 &=P(2.0471) \\ P &=\frac{40000}{2.0471}=\$ 19539.84\end{align*}\]
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 6
To see why not overrounding is so important, suppose you were investing $1000 at 5% interest compounded monthly for 30 years.
Solution
\(\begin{array} {ll} P = \$1000 & \text{the initial deposit} \\ r = 0.05 & 5\% \\ k = 12 & \text{12 months in 1 year} \\ t = 30 & \text{since we’re looking for the amount after 30 years} \end{array}\)
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:
r/k rounded to: 
Gives A_{ }to be: 
Error 

0.004 
$4208.59 
$259.15 
0.0042 
$4521.45 
$53.71 
0.00417 
$4473.09 
$5.35 
0.004167 
$4468.28 
$0.54 
0.0041667 
$4467.80 
$0.06 
no rounding 
$4467.74 

If you are 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 would not 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
\(A=1000\left(1+\frac{0.05}{12}\right)^{12 \times 30}\)
We can quickly calculate \(12 \times 30=360\), giving \(A=1000\left(1+\frac{0.05}{12}\right)^{360}\).
Now we can use the calculator.
\(\begin{array}{cc}
\hline \textbf { Type this } & \textbf { Calculator shows } \\
\hline 0.05 [\div] 12 [=] & 0.00416666666667 \\
\hline [+] 11 [=] & 1.00416666666667 \\
\hline [\mathrm{y}^{\mathrm{x}}] 360 [=] & 4.46774431400613 \\
\hline [\times] 1000 [=] & 4467.74431400613 \\
\hline \hline
\end{array}\)
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 ) \([\text{y}^\text{x}]\) 360 \([=]\).
Exercise \(\PageIndex{2}\)
If $70,000 are invested at 7% compounded monthly for 25 years, find the end balance.
 Answer

\[A = 70,000\left(1 + \frac{0.07}{12} \right)^{12(25)} = 400,779.27 \nonumber\]
Because of compounding throughout the year, with compound interest the actual increase in a year is more than the annual percentage rate. If $1,000 were invested at 10%, the table below shows the value after 1 year at different compounding frequencies:
Frequency 
Value after 1 year 

Annually 
$1100 
Semiannually 
$1102.50 
Quarterly 
$1103.81 
Monthly 
$1104.71 
Daily 
$1105.16 
If we were to compute the actual percentage increase for the daily compounding, there was an increase of $105.16 from an original amount of $1,000, for a percentage increase of \(\frac{105.16}{1000} = 0.10516= 10.516\% \) increase. This quantity is called the annual percentage yield (APY).
Notice that given any starting amount, the amount after 1 year would be
\[A = P\left(1 + \frac{r}{k} \right)^k\nonumber\].
To find the total change, we would subtract the original amount, then to find the percentage change we would divide that by the original amount:
\[\frac{P\left(1 + \frac{r}{k} \right)^k  P}{P} = \left(1 + \frac{r}{k} \right)^k  1.\nonumber \]
Definition: Annual Percentage Yield
The annual percentage yield is the actual percent a quantity increases in one year. It can be calculated as
\[ APY = \left(1 + \frac{r}{k} \right)^k  1\nonumber\]
Notice this is equivalent to finding the value of $1 after 1 year, and subtracting the original dollar.
Example \(\PageIndex{7}\)
Bank \(A\) offers an account paying 1.2% compounded quarterly. Bank \(B\) offers an account paying 1.1% compounded monthly. Which is offering a better rate?
Solution
We can compare these rates using the annual percentage yield – the actual percent increase in a year.
Bank \(A\): \(APY = \left(1 + \frac{0.012}{4} \right)^4  1 = 0.012054 = 1.2054\% \)
Bank \(B\): \(APY = \left(1 + \frac{0.011}{12} \right)^{12}  1 = 0.011056 = 1.1056\% \)
Bank \(B\)’s monthly compounding is not enough to catch up with Bank \(A\)’s better APR. Bank \(A\) offers a better rate.
Example \(\PageIndex{8}\)
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,
\(P = \$2000\) the initial deposit
\(r = 0.06\) 6% annual rate
\(k = 12\) 12 months in 1 year
So our general equation is \(A = 2000\left( 1 + \frac{0.06}{12} \right)^{12t}\). 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 \(A = 4000\). To solve this, we set our equation for \(A\) equal to 4000.
\(4000 = 2000\left(1 + \frac{0.06}{12} \right)^{12t}\) Divide both sides by 2000
\(2 = \left(1.005 \right)^{12t}\) To solve for the exponent, take the log of both sides
\(\log \left( 2 \right) = \log \left( \left(1.005 \right)^{12t} \right)\) Use the exponent property of logs on the right side
\(\log \left( 2 \right) = 12t\log \left( {1.005} \right)\) Now we can divide both sides by \(12\log(1.005)\)
\(\frac{\log \left( 2 \right)}{12\log \left(1.005 \right)} = t\) Approximating this to a decimal
\(t = 11.581\)
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.
Important Topics of this Section
APR
Simple interest
Compound interest
Compounding frequency
APY
Evaluating on a calculator