Compound Interest Formula

If you put $\,\$1000\,$ in a bank that offers $\,5\%\,$ simple annual interest,
then your interest is added in only after one year has passed.
If you withdraw your savings one day before the year is over,
you'll have only your original $\,\$1000\,$ (drat).
However, if you wait the extra day (making one full year),
then you'll have $\,\$1000 + (5\%)(\$1000) = \$1050\,$.

Wouldn't it be better to add in $\,\frac{1}{12}\,$ of your annual interest after the first month,
giving a slightly greater amount to earn interest for the next month?
Or, why not add in $\,\frac{1}{365}\,$ of your annual interest after the first day,
giving a slightly greater amount to earn interest for the next day?

When you add in interest at regular intervals, this is called compound interest.
With compound interest, you are earning interest on your interest, not only on your original principal!
The Compound Interest Formula is a non-recursive formula that is convenient for working with compound interest situations,
and is the subject of this section.

Before deriving the compound interest formula, let's go back to that $\,\$1000\,$ with $\,5\%\,$ interest,
and see how much benefit is gained from:

compounding monthly (adding in interest each month);
there are $\,12\,$ months in one year, so the monthly interest rate is $\,\frac{1}{12}(0.05)=\frac{0.05}{12}$
compounding weekly (adding in interest each week);
there are $\,52\,$ weeks in one year, so the weekly interest rate is $\,\frac{1}{52}(0.05)=\frac{0.05}{52}$
compounding daily (adding in interest each day);
there are $\,365\,$ days in one year, so the daily interest rate is $\,\frac{1}{365}(0.05)=\frac{0.05}{365}$

EFFECTS OF COMPOUNDING
$\,\$1000\,$ initial deposit, $\,5\%\,$ interest rate
(all units are dollars, rounded to the nearest cent)

time	simple annual interest	compounding monthly	compounding weekly	compounding daily
$1\,$ day	$1000$	$1000$	$1000$	$1000+\frac{0.05}{365}(1000)=1000.14$
$2\,$ days	$1000$	$1000$	$1000$	$1000.14+\frac{0.05}{365}(1000.14)=1000.28$
$1\,$ week	$1000$	$1000$	$1000+\frac{0.05}{52}(1000)=1000.96$	$1000.96$
$2\,$ weeks	$1000$	$1000$	$1000.96+\frac{0.05}{52}(1000.96)=1001.92$	$1001.92$
$1\,$ month	$1000$	$1000+\frac{0.05}{12}(1000)=1004.17$	$1004.17$	$1004.18$
$2\,$ months	$1000$	$1004.17+\frac{0.05}{12}(1004.17)=1008.35$	$1008.36$	$1008.37$
$6\,$ months	$1000$	$1025.26$	$1025.30$	$1025.31$
$1\,$ year	$1000 + 0.05(1000) = 1050.00$	$1051.16$	$1051.25$	$1051.27$
$2\,$ years	$1050 + 0.05(1050) = 1102.50$	$1104.94$	$1105.12$	$1105.16$
$10\,$ years	$1628.89$	$1647.01$	$1648.33$	$1648.66$

The added savings from earning interest on interest is perhaps not quite as much as you'd hope.
For example, in one year you'd earn an additional $\,\$1051.27 - \$1050 = \$1.27\,$ over simple annual interest, by adding in interest daily.
In ten years, you'd earn an additional $\,\$1648.66 - \$1628.89 = \$19.77\,$ over simple annual interest, by adding in interest daily.
As the length of time and the amount of money invested increase, though, the savings do go up—and every little bit helps!

DERIVATION OF THE COMPOUND INTEREST FORMULA

The compound interest formula results from using variables to represent a general investing situation,
writing down several computations, and seeing a pattern emerge.

In a nutshell, you're going to invest $\,P\,$ dollars at annual interest rate $\,r\,$,
add in interest $\,n\,$ times per year, and see how much you have after $\,t\,$ years.

Here are the details:

You are investing $\,P\,$ dollars.
You will start the clock at the moment you make this initial investment;
i.e., let $\,t=0\,$ correspond to the initial investment time.
Let $\,r\,$ denote the (simple) annual interest rate for this investment.
Express $\,r\,$ as a decimal. For example, $\,5\%\,$ corresponds to $\,r=0.05\,$.
Assume that interest is added in at regular intervals, $\,n\,$ times per year.
For monthly compounding, $\,n=12\,$.
For weekly compounding, $\,n=52\,$.
For daily compounding, $\,n=365\,$.
Thus, $\,n\,$ represents the number of compounding periods per year.
Let $\,A\,$ represent the total amount of money (principal plus interest) after $\,t\,$ years.

The table below shows the accumulations after various numbers of compounding periods:

after this time...	you'll have this much money...	NOTE:
$1\,$ compounding period	$P+\frac{r}{n}\cdot P=P(1+\frac{r}{n})=P{(1+\frac{r}{n})}^1$	factor out $\,P\,$
$2\,$ compounding periods	$P(1+\frac{r}{n})+\frac{r}{n}\cdot P(1+\frac{r}{n})=P(1+\frac{r}{n})(1+\frac{r}{n}) =P{(1+\frac{r}{n})}^2$	factor out $\,P(1+\frac{r}{n})$
$3\,$ compounding periods	$P{(1+\frac{r}{n})}^2+\frac{r}{n}\cdot P{(1+\frac{r}{n})}^2 =P{(1+\frac{r}{n})}^2(1+\frac{r}{n}) =P{(1+\frac{r}{n})}^3$	factor out $\,P{(1+\frac{r}{n})}^2$
...	...	...
$n \text{ compounding periods} = 1 \text{ year}$	$P{(1+\frac{r}{n})}^n$	notice the emerging pattern
$2n \text{ compounding periods} = 2 \text{ years}$	$P{(1+\frac{r}{n})}^{2n}$
$tn \text{ compounding periods} = t \text{ years}$	$A=P{(1+\frac{r}{n})}^{tn}=P{(1+\frac{r}{n})}^{nt}$	the compound interest formula!

Thus, we have:

THE COMPOUND INTEREST FORMULA

Suppose you invest $\,P\,$ dollars at (simple) annual interest rate $\,r\,$,
and add in interest $\,n\,$ times per year (that is, there are $\,n\,$ compounding periods per years).
The amount, $\,A\,$ (principal plus interest), that you have after $\,t\,$ years is given by
the compound interest formula:

$$A=P{(1+\frac{r}{n})}^{nt}$$

USING THE COMPOUND INTEREST FORMULA

Question:
Suppose $\,\$2500\,$ is invested at $\,3\%\,$ annual interest, compounded daily, for seven years.
How much money will you have?

Solution:
It's a good idea to do a crude estimate first to get a ‘ballpark’ figure.
This catches a lot of calculator mistakes.

If there were no compounding at all, then in one year the interest would be

$(3\%)(\$2500) = \$75\,$.
Do this for $\,7\,$ years: $7(\$75) = \$525\,$
Put together the principal plus the interest: $\,\$2500 + \$525 = \$3025\,$
You'll do better than this by adding in interest at regular intervals,
so you know you'll have more than $\,\$3025\,$.

Now, use the compound interest formula:

$P=2500\,$, $\,r=0.03\,$, $\,n=365\,$, $\,t=7\,$

$\displaystyle A=2500{\left(1+\frac{0.03}{365}\right)}^{(365\cdot 7)}=\ \$3084.17$

Compare with $\,\$3025\,$; believable!

If you want, you can do the computation up at WolframAlpha (cut-and-paste):

2500(1 + 0.03/365)^(365*7)

(Notice that the exponent computation must be put inside parentheses, to get the correct order of operations.)

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Graphs of Functions

(MAX is 21; there are 21 different problem types.)