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# Simple and Compound Interest

We have to work with money every day. While balancing your checkbook or calculating your monthly expenditures on espresso requires only arithmetic, when we start saving, planning for retirement, or need a loan, we need more mathematics.

### Learning Objectives

The leaning objectives for this section include:
• Calculate one-time simple interest, and simple interest over time
• Determine APY given an interest scenario
• Calculate compound interest

## 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

[latex-display]\begin{align}&I={{P}_{0}}r\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}r={{P}_{0}}(1+r)\\\end{align}[/latex-display]
• I is the interest
• A is the end amount: principal plus interest
• \begin{align}{{P}_{0}}\\\end{align} is the principal (starting amount)
• r is the interest rate (in decimal form. Example: 5% = 0.05)

Answer: 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. Further explanation about solving this example can be seen here. https://youtu.be/rNOEYPCnGwg We can generalize this idea of simple interest over time. ### Simple Interest over Time [latex-display]\begin{align}&I={{P}_{0}}rt\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}rt={{P}_{0}}(1+rt)\\\end{align}[/latex-display] • I is the interest • A is the end amount: principal plus interest • \begin{align}{{P}_{0}}\\\end{align} 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. [latex-display]6\div{12}=0.5[/latex-display] A 4% annual rate paid quarterly would be divided into four 1% payments. [latex-display]4\div{4}=1[/latex-display] ### Example Treasury Notes (T-notes) are bonds issued by the federal government to cover its expenses. Suppose you obtain a1,000 T-note with a 4% annual rate, paid semi-annually, with a maturity in 4 years. How much interest will you earn?

Answer: Since interest is being paid semi-annually (twice a year), the 4% interest will be divided into two 2% payments.

 \begin{align}{{P}_{0}}\\\end{align} = $1000 the principal r = 0.02 2% rate per half-year t = 8 4 years = 8 half-years I =$1000(0.02)(8) = $160. You will earn$160 interest total over the four years.

This video explains the solution. https://youtu.be/IfVn20go7-Y

## 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) – 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, • P0 =$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, • P0 =$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 interest gives us. Calculating out a few more months gives the following:  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 We want to simplify the process for calculating compounding, because creating a table like the one above is time consuming. Luckily, math is good at giving you ways to take shortcuts. To find an equation to represent this, if Pm represents the amount of money after m months, then we could write the recursive equation: P0 =$1000 Pm = (1+0.0025)Pm-1 You probably recognize this as the recursive form of exponential growth. If not, we go through the steps to build an explicit equation for the growth in the next example.

### Example

Build an explicit equation for the growth of $1000 deposited in a bank account offering 3% interest, compounded monthly. Answer: • P0 =$1000
• 1 = 1.00250 = 1.0025 (1000)
• 2 = 1.00251 = 1.0025 (1.0025 (1000)) = 1.0025 2(1000)
• 3 = 1.00252 = 1.0025 (1.00252(1000)) = 1.00253(1000)
• 4 = 1.00253 = 1.0025 (1.00253(1000)) = 1.00254(1000)
Observing a pattern, we could conclude
• Pm = (1.0025)m($1000) Notice that the$1000 in the equation was P0, 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 [latex-display]{{P}_{m}}={{P}_{0}}{{\left(1+\frac{r}{k}\right)}^{m}}[/latex-display]   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.

View this video for a walkthrough of the concept of compound interest. https://youtu.be/xuQTFmP9nNg
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

[latex-display]P_{N}=P_{0}\left(1+\frac{r}{k}\right)^{Nk}[/latex-display]
• PN is the balance in the account after N years.
• P0 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.
In the next example, we show how to use the compound interest formula to find the balance on a certificate of deposit after 20 years.

### Example

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? Answer: In this example,  P0 =$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 ${{P}_{20}}=3000{{\left(1+\frac{0.06}{12}\right)}^{20\times12}}=\9930.61$ (round your answer to the nearest penny)

A video walkthrough of this example problem is available below. https://youtu.be/8NazxAjhpJw
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. ### Try It Now ### 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 To see why not over-rounding is so important, suppose you were investing$1000 at 5% interest compounded monthly for 30 years.
 P0 = $1000 the initial deposit r = 0.05 5% k = 12 12 months in 1 year N = 30 since we’re looking for the amount after 30 years If we first compute r/k, we find 0.05/12 = 0.00416666666667 Here is the effect of rounding this to different values:  r/k rounded to: Gives P­30­ 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’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. View the following for a demonstration of this example. https://youtu.be/VhhYtaMN6mo

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 ${{P}_{30}}=1000{{\left(1+\frac{0.05}{12}\right)}^{12\times30}}$ We can quickly calculate 12×30 = 360, giving ${{P}_{30}}=1000{{\left(1+\frac{0.05}{12}\right)}^{360}}$. Now we can use the calculator.
 Type this Calculator shows 0.05 ÷ 12 = . 0.00416666666667 + 1 = . 1.00416666666667 yx 360 = . 4.46774431400613 × 1000 = . 4467.74431400613

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 ×  ( 1 + 0.05 ÷ 12 ) yx 360 =

## Solving For Time

Note: This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.
Often we are interested in how long it will take to accumulate money or how long we’d need to extend a loan to bring payments down to a reasonable level.

### Examples

If you invest $2000 at 6% compounded monthly, how long will it take the account to double in value? Answer: This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,  P0 =$2000 the initial deposit r = 0.06 6% annual rate k = 12 12 months in 1 year
So our general equation is ${{P}_{N}}=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}$. We also know that we want our ending amount to be double of $2000, which is$4000, so we’re looking for N so that PN = 4000. To solve this, we set our equation for PN equal to 4000.

$4000=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}$              Divide both sides by 2000

$2={{\left(1.005\right)}^{12N}}$                                   To solve for the exponent, take the log of both sides

$\log\left(2\right)=\log\left({{\left(1.005\right)}^{12N}}\right)$             Use the exponent property of logs on the right side

$\log\left(2\right)=12N\log\left(1.005\right)$                     Now we can divide both sides by 12log(1.005)

$\frac{\log\left(2\right)}{12\log\left(1.005\right)}=N$                              Approximating this to a decimal

N = 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.

Get additional guidance for this example in the following: https://youtu.be/zHRTxtFiyxc

• Introduction and Learning Objectives. Provided by: Lumen Learning License: CC BY: Attribution.

### CC licensed content, Shared previously

• interest. Authored by: NY. Located at: https://www.picserver.org/i/interest.html. License: CC BY: Attribution.
• Math in Society. Authored by: David Lippman. Located at: http://www.opentextbookstore.com/mathinsociety/. License: CC BY-SA: Attribution-ShareAlike.
• money-grow-interest-save-invest-1604921. Authored by: TheDigitalWay. License: CC0: No Rights Reserved.
• One time simple interest. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
• Simple interest over time. Authored by: OCLPhase2's channel. License: CC BY: Attribution.
• Simple interest T-note example. Authored by: OCLPhase2's channel. License: CC BY: Attribution.