Section 3.4 Compound Growth

Supplemental Videos.

The main topics of this section are also presented in the following videos:

Subsection 3.4.1 Compounding Once a Year

If the interest on a savings account is compounded annually, the amount of money in the account grows exponentially.

Consider a principal of $100 invested at 5% interest compounded annually. At the end of $1$ year, the amount is

\begin{align*} \text{Amount} \amp = \text{Principal} + \text{Interest}\\ A \amp = P + Pr\\ \amp = 100 + 100(0.05) = 105. \end{align*}

It will be more useful to write the formula for the amount after $1$ year in factored form.

\begin{align*} A \amp = P + Pr \amp\amp \text{Factor out P.}\\ \amp = P(1 + r) \end{align*}

With this version of the formula, the calculation for the amount at the end of $1$ year looks like this:

\begin{align*} A \amp = P(1 + r )\\ \amp = 100(1 + 0.05)\\ \amp = 100(1.05) = \alert{105}. \end{align*}

The amount, $105, becomes the new principal for the second year. To find the amount at the end of the second year, we apply the formula again, with $P = 105\text{.}$

\begin{align*} A \amp = P(1 + r )\\ \amp = 105(1 + 0.05)\\ \amp = \alert{105}(1.05) = 110.25. \end{align*}

Observe that to find the amount at the end of each year, we multiply the principal by a factor of $1 + r = 1.05\text{.}$ Thus, we can express the amount at the end of the second year as

\begin{align*} A \amp = [100(1.05)](1.05)\\ \amp = 100(1.05)^2 \end{align*}

and at the end of the third year as

\begin{align*} A \amp = \left[100(1.05)^2\right](1.05)\\ \amp = 100(1.05)^3. \end{align*}

At the end of each year, we multiply the old balance by another factor of $1.05$ to get the new amount. We organize our results into Table 3.23, where $A(t)$ represents the amount of money in the account after $t$ years. For this example, a formula for the amount after $t$ years is

\begin{equation*} A(t) = 100(1.05)^t. \end{equation*}

$t$	$P(1 + r)^t$	$A(t)$
$0$	$100$	$100$
$1$	$100(1.05)$	$105$
$2$	$100(1.05)^2$	$110.25$
$3$	$100(1.05)^3$	$115.76$

Table 3.23.

In general, for an initial investment of $P$ dollars at an interest rate $r$ compounded annually, we have the following formula for the amount accumulated after $t$ years.

Annual Interest.

The amount, $A(t)\text{,}$ accumulated (principal plus interest) in an account bearing interest compounded annually is

\begin{equation*} A(t) = P(1 + r)^t \end{equation*}

where

\begin{align*} \amp P \amp\amp \text{is the principal invested,}\\ \amp r \amp\amp \text{is the effective annual interest rate (as a decimal),}\\ \amp t \amp\amp \text{is the time period, in years,}\\ \amp b=1+r \amp\amp \text{is the growth factor.} \end{align*}

This function describes exponential growth with an initial value of $P$ and a growth factor of $b = 1 + r\text{.}$ The interest rate $r\text{,}$ which indicates the percent increase in the account each year, corresponds to a growth factor of $1 + r\text{.}$ The notion of percent increase is often used to describe the growth factor for quantities that grow exponentially.

Subsection 3.4.2 Compounding $n$-Times a Year

If we have a principal of $P$ and an interest rate $r\text{,}$ then by compounding this interest rate $n$-times a year we mean the following:

Rather than apply the interest rate $r$ one time over the course of a year, we instead apply an interest of $\dfrac{r}{n}$ a total of $n$ times over the course of the year.

Compound Interest.

The amount $A(t)$ accumulated (principal plus interest) in an account bearing interest compounded $n$-times a year is

\begin{equation*} A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt} \end{equation*}

where

\begin{align*} \amp P \amp\amp \text{is the principal invested,}\\ \amp r \amp\amp \text{is the nominal interest rate,}\\ \amp n \amp\amp \text{is the number of times the interest is compounded,}\\ \amp t \amp\amp \text{is the time period, in years.} \end{align*}

Caution 3.24.

The growth factor of

\begin{equation*} A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt} \end{equation*}

is not $1+\dfrac{r}{n}\text{.}$ Recall that when a function is written in the form $P(b)^t $ the growth factor is $b\text{.}$ So, we need to rewrite our function in this form. We do this as follows:

\begin{equation*} A(t)=P\left(1 + \dfrac{r}{n}\right)^{nt} = P\left[\left(1+\dfrac{r}{n}\right)^n\right]^t. \end{equation*}

If we set

\begin{equation*} b=\left(1+\dfrac{r}{n}\right)^n, \end{equation*}

then we have rewritten $A(t)$ in the form $P(b)^t\text{,}$ and thus the growth factor is

\begin{equation*} \left(1+\dfrac{r}{n}\right)^n. \end{equation*}

If you have a sharp eye, then you’ve realized that the Annual Interest formula can come from the Compound Inteterest formula! Since annual means once a year, we let $n=1$ in the Compound Interest formula to get:

\begin{equation*} A(t) = P\left(1 + \dfrac{r}{1}\right)^{1t} = P\left(1 + r\right)^{t}. \end{equation*}

So, the nominal interest rate is the same as the annual interest rate if your $n$ is equal to $1\text{.}$ In general, these are NOT the same. You will be asked to translate between nominal and annual. The NOTE after Example 3.25 shows how to go from nominal to annual.

Example 3.25.

During a period of rapid inflation, prices of a pound of butter rose by $24$% compounded twice a year. At the beginning of the inflationary period, a pound of butter cost $$2\text{.}$

Write a function that gives the price of a pound of butter $t$ years after inflation began.
How much did a pound of butter cost after $3$ years? After $15$ months?

Solution.

Looking at our compound interest formula

\begin{equation*} A(t) = P\left(1 + \dfrac{r}{n}\right)^{nt}, \end{equation*}

we know that we are compounding twice a year, so $n=2\text{,}$ and the nominal rate is $24$%, so $r=.24\text{.}$ The problem also tells us the original price of a pound of butter was $$2\text{,}$ so $P=2\text{.}$ Using this information, we get the formula

\begin{equation*} A(t) = 2\left(1 + \dfrac{.24}{2}\right)^{2t}= 2(1.12)^{2t}. \end{equation*}
To find the price of butter at any time after inflation began, we evaluate the function at the appropriate value of $t\text{.}$

\begin{align*} P(\alert{3}) \amp = 2(1.12)^{2(\alert{3})}\\ \amp = 2(1.12)^6 \approx 3.95 \end{align*}

After $3$ years, the price was $$3.95\text{.}$ Fifteen months is $1.25$ years, so we evaluate $P(1.25)\text{.}$

\begin{align*} P(\alert{1.25}) \amp = 2(1.12)^{2(\alert{1.25})}\\ \amp = 2(1.12)^{2.5} \approx 2.66 \end{align*}

After $15$ months, the price of butter was $$2.66\text{.}$

NOTE: In Example 3.25, we can rewrite the formula for $P(t)$ as follows:

\begin{align*} P(t)\amp = 2(1.12)^{2t}\\ \amp = 2\left[(1.12)^2\right]^t = 2(1.2544)^t. \end{align*}

Thus, the annual growth factor for the price of butter is $1.2544\text{,}$ and the annual percent growth rate, or effective annual rate, is $25.44$%.

Example 3.26.

In $1998\text{,}$ the average annual cost of attending a public college was $\$10,069\text{,}$ and costs were climbing by $6$% per year.

Write a formula for $C(t)\text{,}$ the cost of one year of college $t$ years after $1998\text{.}$
Complete the table and sketch a graph of $C(t)\text{.}$

$t$ $0$ $5$ $10$ $15$ $20$ $25$

$C(t)$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$
If the growth rate remained steady, how much did a year of college cost in $2005\text{?}$
If the growth rate continues to remain steady, how much will a year of college cost in $2020\text{?}$

Solution.

$\displaystyle C(t) = 10,069(1.06)^t$
$t$ $0$ $5$ $10$ $15$ $20$ $25$

$C(t)$ $10,069$ $13,475$ $18,032$ $24,131$ $32,293$ $43,215$

exponential growth
Since $2005$ is $7$ years after $1998\text{,}$

\begin{equation*} C(7)=10,069(1.06)^7\approx15,140 \end{equation*}

S o in $2005\text{,}$ a year of college will cost about $\$15,140$ per year
Since $2020$ is $22$years after $1998\text{,}$

\begin{equation*} C(22)=10,069(1.06)^22\approx36,284 \end{equation*}

S o in $2020\text{,}$ a year of college will cost about $\$36284$ per year

Exercises 3.4.3 Exercises

1. Compound Growth Over Time.

Suppose an investment grows by 7 percent per year for a period of 8 years. By what total percentage does it increase over this period?

percent

Hint.

Using the general form of an exponential equation, we have $I(t) = a (b)^t\text{,}$ where $b$ is our growth factor and $a$ is the initial value.

To find the percentage of increase over the period, consider the following formula:

\begin{equation*} I(8)=a(b)^{8}. \end{equation*}

You have enough information to fill in $a$ and $b\text{.}$ The increase over the period will be given by $b^{8}-1\text{.}$ Make sure you convert to a percentage.

2. Effective Annual Rate.

Suppose an investment decreases by a total of 26 percent over a period of 11 years. At what effective annual percent rate is it decreasing? Hint: You can figure this out even without knowing the initial value.

percent per year

Hint.

As indicated by the hint, we don’t need the initial value of the investment to solve this problem. Note that we can write two equations.

One comes from the general exponential function equation: $I(11) = a (1+r)^{11}\text{.}$

The other comes from the given information in the problem that the investment has decreased 26 percent in 11 years, that is, the investment is at 74 percent of its original value: $I(11) = 0.74(a)\text{.}$

We can now set these two equations equal to each other and solve for $r\text{.}$ That is, we have $a (1+r)^{11} = (0.74)a$ which we can solve for $r\text{.}$

Note that the question asks for the effective annual percent rate that the investment is decreasing, so you need to convert the value you find to a percentage.

3. Compound Growth: Annually vs. Daily.

Suppose a bank pays interest at the nominal rate of 6 percent per year. Find the balance after one year of an account whose initial balance is 1,000 dollars if interest is compounded:

a) Annually

dollars

b) Daily

dollars

Hint.

a) Recalling the formula for annual interest, we have $A(t) = P(1+r)^t\text{,}$ where $P$ is the principal amount invested and $r$ is the nominal interest rate.

b) To find the investment when compounding daily, we need to use the compound growth formula: $A(t) = P(1+\frac{r}{n})^{nt}\text{,}$ where $n$ is the number of times we compound per year.

In both parts, we want to know how much money is in the account after 1 year, so you should evaluate the equations at $t=1\text{.}$

4. Compound Growth at Various Intervals.

Suppose a bank pays interest at the nominal rate of 17 percent per year. Find the effective annual rate if interest is compounded:

a) Annually

percent

b) Monthly

percent

c) Weekly

percent

d) Daily

percent

e) Hourly

percent

Hint.

For each of parts a) - e), we will use the compound interest formula $A(t) = P (1 + \frac{r}{n})^{nt}\text{.}$

First we calculate the growth factor using $(1+\frac{r}{n})^n$ since the initial balance doesn’t affect the rate, and we are only looking at one year (i.e., $t = 1$) in each part. Note that this gives the growth factor.

We can get the effective annual rate as a percent by subtracting 1 from the growth factor and multiplying by 100.

For example, if interest is compounded biannually, we would have the effective annual rate give by

\begin{equation*} \left(1+\frac{0.17}{2}\right)^2-1. \end{equation*}

This value should be converted into a percent before entering into answer box.

5. Savings Account.

Jordan’s parents would like to invest some money to help pay his future college tuition. Their bank will pay interest on their investment at a nominal rate of 3 percent per year.

a) Assuming interest is compounded weekly, and Jordan’s parents deposit 2000 dollars, how much money will be in the account after 15 years?

dollars

b) If their goal is to have 15000 dollars in the account in 15 years, what is the minimum amount of money Jordan’s parents should deposit now, assuming that interest is compounded monthly?

dollars

c) Answer the same question in (b) if interest is compounded daily.

dollars

d) Did you find that they need to deposit more money if interest is compounded monthly or daily?

Monthly, because the growth factor is smaller
Monthly, because the growth factor is larger
Daily, because the growth factor is smaller
Daily, because we compound more frequently

Hint.

a) We wish to find the amount of money in the account with the formula $A(t) = P (1 + \frac{r}{n})^{nt}\text{,}$ where $P$ is the principal amount invested, $r$ is the nominal interest rate, $n$ is the number of times compounded each year, and $t$ is the number of years money is invested.

b) We wish to find the amount of money initially invested by using the formula $A(t) = P (1 + \frac{r}{n})^{nt}\text{.}$ Try to plug in all the information you know. In this case you should have an equation where the only unknown is $P\text{.}$ You will then solve the equation for this value.

c) You will use the same approach as in part b), with a different value of $n\text{.}$

d) Compare your initial deposit values in parts b) and c).

6. Nominal Interest Rate.

Suppose Thea invests 200 dollars, and 23 years later has 1200 dollars. If interest on her investment was compounded weekly, what was the nominal interest rate of the account?

Hint.

When compounding more frequently than annually, the nominal rate is not the same as the effective growth rate. In the formula for compound interest, we identify the nominal rate with the $r$ in $A(t) = P \left(1 + \frac{r}{n}\right)^{nt}\text{,}$ so we need to write an equation representing Thea’s situation, and then solve it for $r\text{.}$ Try to plug in all the information you are given to an equation where $r$ is the only unknown.

You will need to solve the equation for $r$ to get your answer.

7. Compound Growth.

Suppose the balance of a bank account, $B(t)\text{,}$ $t$ years after the initial deposit, is given by the equation $B(t)=4400 (1.049)^{t}$

a) What was the initial deposit into the account?

b) What is the effective annual interest rate for this account?

percent

c) If the bank compounds interest quarterly, what is the nominal annual interest rate for this account?

percent

Hint.

a) The initial deposit is the amount in the account at $t = 0\text{.}$ This is also the principal amount invested.

b) The effective annual rate is the rate at which the account grows per year. This is also called the annual percent growth rate sometimes. We find this by subtracting 1 from the growth factor. Remember to convert this into a percentage before entering it into the answer box.

c) The compound interest formula is given by $A(t)=P(1+\frac{r}{n})^{nt}.$ Plug in the information you know and set this equal to the given formula. You are trying to solve for $r\text{,}$ which is the nominal interest rate. Remember to convert $r$ into a percentage before entering it into the answer box.

Prev Top Next

\(t\)	\(P(1 + r)^t\)	\(A(t)\)
\(0\)	\(100\)	\(100\)
\(1\)	\(100(1.05)\)	\(105\)
\(2\)	\(100(1.05)^2\)	\(110.25\)
\(3\)	\(100(1.05)^3\)	\(115.76\)

\(t\)	\(0\)	\(5\)	\(10\)	\(15\)	\(20\)	\(25\)
\(C(t)\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)

\(t\)	\(0\)	\(5\)	\(10\)	\(15\)	\(20\)	\(25\)
\(C(t)\)	\(10,069\)	\(13,475\)	\(18,032\)	\(24,131\)	\(32,293\)	\(43,215\)

Section 3.4 Compound Growth

Supplemental Videos.

Subsection 3.4.1 Compounding Once a Year

Annual Interest.

Subsection 3.4.2 Compounding \(n\)-Times a Year

Compound Interest.

Caution 3.24.

Example 3.25.

Example 3.26.

Exercises 3.4.3 Exercises

1. Compound Growth Over Time.

2. Effective Annual Rate.

3. Compound Growth: Annually vs. Daily.

4. Compound Growth at Various Intervals.

5. Savings Account.

6. Nominal Interest Rate.

7. Compound Growth.