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Business Mathematics: Applications Using Algebra and Calculus

Alfred Dahma

Section 5.3 Interest Models

Subsection 5.3.1 Simple Interest

Exponential functions are commonly used to model the growth and decay of money. But, let's first revisit a topic introduced in Section 1.3 for comparison. In particular, recall Example 1.3.10.
Simple Interest is interest computed only on the principal amount invested or borrowed. For this reason, when money grows (or declines) according to a simple interest model, it does so linearly.
Let
  • \(P\) = principal dollar amount invested or borrowed
  • \(r\) = yearly interest rate (as a decimal)
  • \(t\) = number of years
Then \(I(t) = Prt\) is the amount of (simple) interest earned (or owed) in \(t\) years. Notice that this is a linear model with vertical intercept \((0,0)\) and slope \(m=Pr\text{.}\) You should be aware that \(t\) may represent a unit of time other than years, such as days, weeks, months, etc. So, pay attention to the given context for the appropriate units. However, unless otherwise mentioned, we will assume \(t\) to be in years.
Maggie borrowed $4,000 at 3% simple interest for 9 months. How much interest will she pay? How much will Maggie owe in 9 months? Solution.
We use the simple interest formula with
  • \(P = 4,000\) and
  • \(\displaystyle r= .03\)
so that our model for this problem is
\begin{equation*} I(t) = 4000(.03)t = 120t. \end{equation*}
In \(t = \frac{9}{12} = \frac{3}{4}\) years,
\begin{equation*} I\left (\frac{3}{4} \right) = 120\left( \frac{3}{4} \right) = 90. \end{equation*}
So, Maggie will pay $90 in simple interest. In 9 months she will owe her principal and interest, which is
\begin{equation*} P + I = $4000 + $90 = $4,090. \end{equation*}
The total amount owed in Example 5.3.1 is referred to as a future value. In other words, $4,090 is the future value, \(F\text{,}\) of $4,000 invested at 3% simple interest for 9 nine months. In general, the future value, \(F\text{,}\) of a principal amount \(P\) dollars in \(t\) years at a simple interest rate \(r\) (as a decimal) is
\begin{equation*} F = P + I = P + Prt = P(1+rt). \end{equation*}
Notice that we can solve this expression above for the principal amount (or present value) if we know the future value \(F\text{:}\)
\begin{equation*} P = \frac{F}{1+rt} \end{equation*}

Remark 5.3.2. Summary of Simple Interest Formulas.

  1. Future Value: \(F = P(1+rt)\)
  2. Interest: \(I = F - P = Prt\)
  3. Present Value: \(P = \frac{F}{1+rt}\)
Notice that in this context \(P\) is the principal amount invested or borrowed, and the present value. The future value \(F\) is sometimes referred to as the maturity value.
Randy borrows $73,000 at 6% simple interest per year for 90 days. Find the maturity value. Solution.
We use the future value formula above with
  • \(\displaystyle P = 73,000\)
  • \(r= .06\) and
  • \(t= \frac{90}{365}\text{.}\)
Substituting these values
\begin{align*} F \amp = P(1+rt) \\ F \amp = 73,000 \left(1+.06 \cdot \dfrac{90}{365}\right) \\ F \amp = 74,080 \end{align*}
Randy will owe a total of $74,080 in 90 days. That is, $74,080 is the future value of $73,000 at 6% simple interest for 90 days.
Larry owes $6,500 to Margaret. The loan is payable in one year at 5\% simple interest. Margaret needs cash immediately, so she decides to sell the loan to a bank 4 months before it is due. The bank wants an 8% return. How much should the bank pay? Solution.
We first need to determine the amount the bank will get when the loan is paid off. That is, we need the future value of the $6,500 Margaret loaned to Larry. This is what the bank will receive at the end of the year. We use Item 1.
\begin{align*} F \amp = P(1+rt) \\ F \amp = 6,500 \left(1+.05 \cdot 1\right) \: (\textrm{the loan is payable in} \: t=1 \: \textrm{year})\\ F \amp = 6,825 \end{align*}
$6,825 is the amount the bank will get when the year is up in 4 months. To determine what they should pay in order to get an 8% return, we need the present value, \(P\text{,}\) of $6,890 at 8% simple interest for four months. To calculate this, we use Item 3.
\begin{align*} P \amp = \frac{F}{1+rt} \\ \\ P \amp = \frac{6,825}{1+.08\left(\frac{4}{12}\right)} \\ \\ P \amp = 6,647.73 \end{align*}
The bank should pay $6,647.73 to get an 8% return on the loan they purchase from Margaret.
Treasury Bills, or T-bills, often involve simple interest calculations. A T-bill is a simple discount loan, meaning the interest is deducted in advance of the loan, but the full value of the loan must be repaid. T-bills are short term discount loans by investors, backed by the U.S. Treasury Department.
An investor bought a 10 month, $400 T-bill that sold at a simple discount rate of 3%. Find the interest earned on the T-bill, the price of the T-bill, and the simple interest rate the Treasury paid. Solution.
Using Item 2 with \(t=\frac{10}{12}\text{,}\) the interest is calculated as \(400(.03)(10/12) = 10\text{.}\) So the interest the investor earned on the T-bill is $10. Since the interest is deducted in advance of the loan, we calculate
\begin{equation*} 400 - 10 = 390. \end{equation*}
The price of the T-bill is $390. To find the interest rate will solve Item 1 for \(r\text{,}\) using \(F=400\) and \(P=390\text{:}\)
\begin{align*} F \amp = P(1+rt) \\ \\ 400 \amp = 390 \left(1+\frac{10}{12}r\right) \\ \\ 1+\frac{10}{12}r \amp = \frac{400}{390} \\ \\ r \amp = \frac{12}{10}(\frac{400}{390} - 1)\\ \\ r \amp \approx .030769 \end{align*}
Rounding to three decimal places, the simple interest rate paid by the Treasury was 3.077%.

Subsection 5.3.2 Compound Interest

Compound interest is interest on both the initial principal and the accumulated interest from previous periods. Because of this, compound interest is calculated using exponential models. Under the same interest rate, compound interest will make an amount of money grow faster than simple interest, which is calculated only on the principal amount (recall that simple interest is linear).
Let
  • \(P\) = principal dollar amount invested or borrowed
  • \(r\) = yearly or annual interest rate (as a decimal)
  • \(n\) = number of compounding periods per year
  • \(t\) = number of years
Consider principal amount $\(P\) accumulating interest for \(n\) compounding periods per year at an annual interest rate of \(r\text{.}\) Note that in this case, the interest per period is \(\frac{r}{n}\text{.}\) What will this amount grow to by the end of the year under compound interest?
  • Amount at end of first period:
    \begin{equation*} P + \frac{r}{n}P = P\left(1+\frac{r}{n}\right) \end{equation*}
    This amount now collects interest during the second period. That is, this is the "new" principal amount for periond 2.
  • Amount at end of second period:
    \begin{equation*} P\left(1+\frac{r}{n}\right) + \frac{r}{n}\left(P\left(1+\frac{r}{n}\right) \right) = P\left(1+\frac{r}{n}\right)^2 \end{equation*}
    This amount now collects interest during the third period. That is, this is the "new" principal amount for periond 3.
  • Amount at end of third period:
    \begin{equation*} P\left(1+\frac{r}{n}\right)^2 + \frac{r}{n}\left(P\left(1+\frac{r}{n}\right) \right)^2 = P\left(1+\frac{r}{n}\right)^3 \end{equation*}
    This amount now collects interest during the fourth period. That is, this is the "new" principal amount for periond 4.
This process continues \(n-3\) more times. By the end of the year, this amount will have grown to
\begin{equation*} P\left(1+\frac{r}{n}\right)^n. \end{equation*}
Using the notation described above, let \(A(t)\) be the model that gives the dollar amount after \(t\) years under compound interest. Then
\begin{equation*} A(t) = P\left(1+ \dfrac{r}{n}\right)^{nt}. \end{equation*}
\(A(t)\) is the future value of the principal \(P\) in \(t\) years. So, we will sometimes write \(F = A(t)\) and call \(P\) the present value. Just like with simple interest, we can write the above model in present and future value form.

Remark 5.3.6. Summary of Compund Interest Formulas.

For principal amount $\(P\) invested or borrowed at an annual interest rate of \(r\) with \(n\) compounding periods per year, in \(t\) years we have
  1. Future Value = \(A(t) = F = P\left(1+\dfrac{r}{n}\right)^{nt}\)
  2. Interest earned = \(I = F - P\)
  3. Present Value = \(P = \dfrac{A(t)}{\left(1+\dfrac{r}{n}\right)^{nt}} = \dfrac{F}{\left(1+\dfrac{r}{n}\right)^{nt}}\)
Sam is planning to open a bar and grill in Boston, MA in 54 months. He invests $28,500 in account that pays 3.5% interest compounded quarterly. How much money will Sam have by the time he is ready to open his bar? How much interest will he have earned? Solution.
We us Item 1 with \(P = 28,500\text{,}\) \(r = .035\text{,}\) and \(n=4\text{.}\) The compound interest model is then
\begin{equation*} A(t) = 28,500\left(1+\frac{.035}{4}\right)^{4t}. \end{equation*}
Since 54 months corresponds to \(t = \frac{54}{12} = 4.5\) years, we evaluate
\begin{equation*} A(4.5) = 28,500\left(1+\frac{.035}{4}\right)^{4(4.5)} \approx 33,338.70 \end{equation*}
When Sam opens his bar in 54 months, his investment will have grown to $33,338.70. Hence, Sam will have earned $33,338.70 - $28,500 = $4,838.70 in interest.
Coach Pantusso needs $2,500 in 18 months for equipment for his little league baseball teams. He finds an account offering 5.75% interest compounded monthly. How much should Coach invest so that he has enough money to purchase the equipment? How much interest did he earn? Solution.
To determine now much Coach should invest, we use Item 3 given a future value in \(t=\frac{18}{12}\) years of \(F = A(\frac{18}{12}) = 2,500\) with interest \(r = .0575\) and \(n = 12\) compounding periods per year.
\begin{equation*} P = \dfrac{2,500}{\left(1+\dfrac{.0575}{12}\right)^{12\left(\frac{18}{12}\right)}} \approx 2,293.88. \end{equation*}
Coach needs to invest $2,293.88 so that he has enough money to purchase the equipment. In this case he will earn $2,500 - $2,293.88 = $206.12 interest.
The average annual inflation rate on Echo Island for the years 2020-2022 was 1.95%. How much did an item that sold for \$1,000 early 2023 cost 3 years prior? Assuming the same rate of inflation, how long will it take this price ($1,000) to double? Solution.
We use Item 3 (with present value corresponding to early 2020) given a future value in \(t=3\) years (the year 2023) of \(F = A(3) = 1,000\) with interest \(r = .0195\) and \(n = 1\) (for inflation, the compounding is 1 time a year).
\begin{equation*} P = \dfrac{1000}{\left(1+ \frac{.0195}{1}\right)^{1 \cdot 3}} \approx 943.71 \end{equation*}
An item that sold for $1,000 at the beginning of 2023 cost $943.71 at the start of 2020.
We now want to determine the doubling time under this inflation rate. To do this, we solve Item 1 for \(t\text{,}\) using \(P=1,000\) and \(F = 2 \cdot 1000\)
\begin{align*} 2 \cdot 1000 \amp = 1000 \left(1+\dfrac{.0195}{1}\right)^{1 \cdot t} \\ \\ (1.0195)^t \amp = 2 \quad (\textrm{apply the natural log to both sides}) \\ \\ t\ln(1.0195) \amp = \ln(2) \quad (\textrm{recall} \knowl{./knowl/fact-log-tool.html}{\text{Fact 5.1.12}})\\ \\ t \amp = \dfrac{\ln(2)}{\ln(1.0195)} \\ \\ t \amp \approx 35.89 \end{align*}
Rounding up, it will take about 36 years for this price to double. Observe that this doubling time is independent of cost of the item. An item that cost $1 would also take about 36 years until the price doubled.
Carla and Eddie managed to save $11,559.23 for their wedding. They invested $10,000 at an annual interest rate of 7.25% compounded weekly. For how long did they invest thier money? How many compounding periods were there in this time. Solution.
We solve Item 1 for \(t\text{,}\) using \(P=10,000\) and \(F = 11,559.23\text{,}\) \(r = .0725\text{,}\) and \(n=52\text{.}\)
\begin{align*} 11,559.23 \amp = 10,000 \left(1+\dfrac{.0725}{52}\right)^{52 \cdot t} \\ \\ \left(1+\dfrac{.0725}{52}\right)^{52 \cdot t} \amp = \dfrac{11,559.23}{10,000} \quad (\textrm{apply the natural log to both sides}) \\ \\ 52t\ln\left(1+\dfrac{.0725}{52}\right) \amp = \ln\left( \dfrac{11,559.23}{10,000}\right) \quad (\textrm{recall} \knowl{./knowl/fact-log-tool.html}{\text{Fact 5.1.12}})\\ \\ t \amp = \dfrac{\ln\left( \dfrac{11,559.23}{10,000}\right)}{52 \ln\left(1+\dfrac{.0725}{52}\right)} \\ \\ t \amp = 2 \end{align*}
Carla and Eddie invested their money for 2 years with a total of \(2 \cdot 52 = 104 \) compounding periods.

Subsection 5.3.3 Continuous Compounding

Let's look at the effect of increasing the number of yearly compounding periods on an investment. For simplicity, suppose $1 is invested at an annual interest rate of 100% for \(t=1\) year, compounded \(n\) times a year. For our calculations we use Item 1 with \(P=r=1\text{.}\) The future value of $1 at \(t=1\) year is then
\begin{equation*} A(1) = F = \left(1 + \dfrac{1}{n}\right)^n \end{equation*}
Let's look at how $1 grows as we increase \(n\text{.}\) We round to the nearest cent:
  • \(n=1\) (yearly): \(\quad F = (1+1)^1 = $2\)
  • \(n=2\) (semiannual): \(\quad F = \left(1+\dfrac{1}{2}\right)^2 \approx $2.25\)
  • \(n=4\) (quarterly): \(\quad F = \left(1+\dfrac{1}{4}\right)^4 \approx $2.44\)
  • \(n=12\) (monthly): \(\quad F = \left(1+\dfrac{1}{12}\right)^{12} \approx $2.61\)
  • \(n=52\) (weekly): \(\quad F = \left(1+\dfrac{1}{52}\right)^{52} \approx $2.69\)
  • \(n=365\) (daily): \(\quad F = \left(1+\dfrac{1}{52}\right)^{365} \approx $2.71\)
We can continue to increase the number of compounding periods by letting \(n \to \infty\text{.}\) Recalling Definition 5.1.4, we have
\begin{equation*} \lim_{n \to \infty} \left( 1 + \dfrac{1}{n}\right)^n \approx $2.72 \end{equation*}
We obtain similar results using any interest rate. Indeed, for any value \(r\) one can show
\begin{equation*} \lim_{n \to \infty} \left( 1 + \dfrac{r}{n}\right)^n = e^r \end{equation*}
Using the properties of limits, we can continue to extend the above result. For the compound interest model \(A(t) = P\left(1 + \dfrac{r}{n} \right)^{nt}\text{,}\) we obtain the following by letting the number of compounding perionds increase without bound:
\begin{align*} \lim_{n \to \infty}A(t) \amp =\lim_{n \to \infty} \left( P \left(1 + \dfrac{r}{n} \right)^{nt} \right) \\ \\ \amp = P \left[ \lim_{n \to \infty} \left( 1 + \dfrac{r}{n} \right)^{nt} \right]\\ \\ \amp = P \left[ \lim_{n \to \infty} \left( 1 + \dfrac{r}{n} \right)^{n} \right]^t\\ \\ \amp = P \left[ e^r \right]^t\\ \\ \amp = Pe^{rt} \end{align*}
The continuous compounding model is the result of letting the number of yearly compounding periods tend to infinity in the compound interest model. As we observed above, if we let the number of yearly compounding periods, \(n\text{,}\) tend to infinity in the model given in Item 1, then we obtain the continuous compounding model
\begin{equation*} A(t) = Pe^{rt}, \end{equation*}
where as before, \(P\) is the principal amount invested or borrowed, \(r\) is the annual interest rate as a decimal, and \(t\) is time in years. As with our other interest models, continous compounding has a future and present value form. The present value form is found by rewriting the above model in terms of \(P\text{.}\)

Remark 5.3.11. Continuous Compounding Formulas.

For principal amount $\(P\) invested or borrowed at an annual interest rate of \(r\) with \(n\) compounding periods per year, in \(t\) years we have
  1. Future Value = \(A(t) = F = Pe^{rt}\)
  2. Interest earned = \(I = F - P\)
  3. Present Value = \(P = A(t)e^{-rt} = Fe^{-rt}\)
Norm, Cliff, and Woody are saving money to travel cross county on a snipe hunt. They decide to invest their money with Howe's Investment Firm, which is the only finanacial establishment they could only find that offered accounts with continuous compounding.
  1. How much will Norm, Cliff, and Woody have if they invest $750 in an account with interest rate 3.86% compounded continuously for 15 months? How much did they earn in interest? Solution.
    Using Item 1 with the given information, the continous compouding model is
    \begin{equation*} A(t) = 750e^{.0386t}. \end{equation*}
    Evaluating at \(t=\dfrac{15}{12}\) years gives
    \begin{equation*} A\left( \dfrac{15}{12}\right) = 750e^{.0386 \cdot \frac{15}{12}} \approx 787.07. \end{equation*}
    In 15 months, Norm, Cliff, and Woody will have $787.07. This means they earned $787.07 - $750 = $37.07 in interest.
  2. If the desired amount in 15 months is $977, how much should Norm, Cliff, and Woody invest? If they let this investment grow for an additional 6 months, how much interest would they earn. Solution.
    Using Item 3 with \(F = 977\text{,}\) \(r=.0386\) and \(t=\dfrac{15}{12}\text{,}\)
    \begin{equation*} P = 977e^{-.0386 \cdot \frac{15}{12}} \approx 930.98. \end{equation*}
    If Norm, Cliff, and Woody want $977 in 15 months, they should invest $930.98 in the account.
    In this case their their continuous compounding model is
    \begin{equation*} A(t) = 930.98e^{.0386t}. \end{equation*}
    If they let their investment grow for an additional 6 months, then evaluating at \(t=\dfrac{21}{12}\) gives
    \begin{equation*} A\left( \dfrac{21}{12}\right) = 930.98e^{.0386 \cdot \frac{21}{12}} \approx 996.04. \end{equation*}
    In this case, Norm, Cliff, and Woody would earn $996.04 - $930.98 = $65.06 in interest.
Jaco wants to buy a new bass guitar which costs $1,799.99 plus an additional 8.7396% in tax and surcharge. He invested $1,835 in an account with continuous compounding and was able to use this money to purchase his bass 19 months after his deposit. What was Jaco's interest rate? Solution.
First, including tax and surcharge, Jaco's bass cost $1,799.99(1+.087396) = $1,957.30 (rounded to the nearest cent). To determine the account's interest rate, we can use Item 1 with \(F= 1,957.30\text{,}\) \(P=1,835\text{,}\) and \(t=\dfrac{19}{12}\) to solve for the interest rate, \(r\text{:}\)
\begin{align*} 1,835e^{\frac{19}{12}r} \amp = 1,957.30\\ \\ e^{\frac{19}{12}r} \amp = \dfrac{1,957.30}{1,835} \: (\textrm{take the natural log of both sides})\\ \\ \dfrac{19}{12}r \amp = \ln\left( \dfrac{1,957.30}{1,835} \right) \: (\textrm{recall} \knowl{./knowl/obs-apply-log.html}{\text{Item 5}}; \: \textrm{here,} \: x=\frac{19}{12}r)\\ \\ r \amp = \dfrac{12}{19}\ln\left( \dfrac{1,957.30}{1,835} \right) \\ \\ r \amp \approx .04075 \end{align*}
Jaco's account had an interest rate of 4.075%.
Suppose $P is invested.
  1. If the money is earning 6.21% compounded continuously, how long will it take the investment to increase by 50%. Solution.
    We use Item 1 with \(F = 1.5P\) and \(r=.0621\) to solve for \(t\text{.}\)
    \begin{align*} Pe^{.0621t} \amp = 1.5P \\ \\ e^{.0621t} \amp = 1.5 \: (\textrm{take the natural log of both sides}) \\ \\ .0621t \amp = \ln(1.5) \: (\textrm{recall} \knowl{./knowl/obs-apply-log.html}{\text{Item 5}}; \: \textrm{here,} \: x=.0621t)\\ \\ t \amp = \dfrac{\ln(1.5)}{.0621}\\ \\ t \amp \approx 6.5 \end{align*}
    It will take approximately 6.5 years (or about 6 years and 6 months) for the investment to increase by 50%.
  2. Under the same conditions, how long will it take the investment to double? Solution.
    We use Item 1 with \(F = 2P\) and \(r=.0621\) to solve for \(t\text{.}\)
    \begin{align*} Pe^{.0621t} \amp = 2P \\ \\ e^{.0621t} \amp = 2 \: (\textrm{take the natural log of both sides}) \\ \\ .0621t \amp = \ln(2) \: (\textrm{recall} \knowl{./knowl/obs-apply-log.html}{\text{Item 5}}; \: \textrm{here,} \: x=.0621t)\\ \\ t \amp = \dfrac{\ln(2)}{.0621}\\ \\ t \amp \approx 11.16 \end{align*}
    It will take approximately 11.16 years (or about 11 years and 2 months) for the investment to double.
  3. If the value of the investment decreases (or decays) continuously at rate 5.53%, how long will it take for the investment to decrease to half its value? Solution.
    We use Item 1 with \(F = .5 P\) to solve for \(t\text{.}\) Because the value of the investemet is decaying, however, we must use \(r=-.0553\text{.}\)
    \begin{align*} Pe^{-.0553t} \amp = .5P \\ \\ e^{-.0553t} \amp = .5 \: (\textrm{take the natural log of both sides}) \\ \\ -.0553t \amp = \ln(.5) \: (\textrm{recall} \knowl{./knowl/obs-apply-log.html}{\text{Item 5}}; \: \textrm{here,} \: x=-.0553t)\\ \\ t \amp = \dfrac{\ln(.5)}{-.0553}\\ \\ t \amp \approx 12.5 \end{align*}
    It will take approximately 12.5 years (or about 12 years and 6 months) for the investment to decay to half its value.

Subsection 5.3.4 APR and APY

APR (the annual percentage rate), also called the nominal rate, represents the stated yearly interest rate. While we will also use this term for investments, it is often used as the rate charged for borrowing money, including fees. APR does not take into consideration the number of compounding periods per year. For our purposes, it would be like the rate \(r\) used in all our interest models above.
APY (the annual percentage yield), also called the effective rate, describes the interest rate under the effect of compounding periods. Because APY includes a calculation of how compounded interest impacts the rate over the course of one year, the APY will usually be larger than the APR. Also, the greater the number of yearly compounding periods, the bigger the difference between the APR and APY.
Suppose we are given an APR (annual interest rate) of \(r\) compounded \(n\) times a year. Then in \(t\) years $P will grow according to the compound interest model
\begin{equation*} A(t) = P\left(1+\dfrac{r}{n} \right)^{nt}. \end{equation*}
The associated APY is the (actual) growth rate of this exponential model. Rewriting the above model,
\begin{equation*} A(t) = P\left[\left(1+\dfrac{r}{n} \right)^{n}\right]^t \end{equation*}
and we see that \(A(t)\) is a base \(b=\left(1+\dfrac{r}{n} \right)^{n}\) exponential growth model. Recall Remark 5.2.1. The (actual) growth rate, or APY, associated with \(A(t)\) is \(b-1\text{,}\) or
\begin{equation*} \left(1+\dfrac{r}{n} \right)^{n} - 1. \end{equation*}
This value is the associated APY and for a fixed APR, will incease as \(n\) increases.

Definition 5.3.15.

With an APR (or annual interest rate) of \(r\) compounded \(n\) times a year, the APY (annual percentage yield, or effective rate), denoted \(r_{E}\text{,}\) is
\begin{equation*} \textrm{APY} \: = r_{E} = \left(1 + \dfrac{r}{n} \right)^{n} - 1. \end{equation*}
In the case of continuous compounding, the APY is
\begin{equation*} \textrm{APY} \: = r_{E} = e^r - 1. \end{equation*}
For example, suppose $\(P\) is invested at an annual rate (APR) of 2.5%. The APY is the (actual) growth rate of this investment, which increase with the compounding periods:
  • Quarlterly compounding:
    \begin{equation*} \textrm{APY} \: = r_{E} =\left(1 + \dfrac{.025}{4} \right)^{4} - 1 \approx .025235 \end{equation*}
  • Montly compounding:
    \begin{equation*} \textrm{APY} \: = r_{E} =\left(1 + \dfrac{.025}{12} \right)^{12} - 1 \approx .025288 \end{equation*}
  • Daily compounding:
    \begin{equation*} \textrm{APY} \: = r_{E} =\left(1 + \dfrac{.025}{365} \right)^{365} - 1 \approx .025314 \end{equation*}
  • Continuous compounding:
    \begin{equation*} \textrm{APY} \: = r_{E} =e^{.025} - 1 \approx .025315 \end{equation*}
Because the APY gives the (actual) growth rate of an investment, it can be used to determine which account provides a better investment option.
Chambers and Crane Investment Bank is offering Account 1 with an APR of 3.245% compounded semiannually and Account 2 with an APR 3.223% compounded weekly. Which is the better investment option. Solution.
While Account 1 has a higher APR, Account 2 offers more compounding periods. To compare, we find the APY (or (actual) growth rate) for each account.
  • Account 1
    \begin{equation*} \textrm{APY} \: = r_{E} =\left(1 + \dfrac{.03245}{2} \right)^{2} - 1 \approx .03271 \end{equation*}
  • Account 2
    \begin{equation*} \textrm{APY} \: = r_{E} =\left(1 + \dfrac{.03223}{52} \right)^{52} - 1 \approx .03274 \end{equation*}
Since the APY for Account 2 (3.274%) is highter than the APY for Account 1 (3.271%), Account 2 is the better investment option. In other words, Account 2 is the better investment option since it offers a higher growth rate.
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