022 Exam 2 Sample B, Problem 4

Set up the equation to solve. You only need to plug in the numbers-not solve for the particular values!

What is the present value of $3000, paid 8 years from now, in an investment that pays 6% interest,

(a) compounded quarterly?

(b) compounded continuously?

Foundations:
The primary purpose of this problem is to demonstrate that you understand the difference between continuous compounding and compounding on an interval of time, as well as the concept of present and future value. When we compound on an interval, say monthly, the value in the account only changes at the end of each interval. In other words, there is no interest accrued for a week or a day. As a result, we use the formula
$A\,=\,P\left(1+{\frac {r}{n}}\right)^{nt},$
where $A$ is the value of the account, $P$ is the principal (original amount invested), $r$ is the annual rate and $n$ is the number of compoundings per year. The value of $n$ is $365$ for compounding daily, $52$ for compounding weekly, and $12$ for compounding monthly. As a result, the exponent $nt$, where $t$ is the time in years, is the number of compounding periods where we actually earn interest. Similarly, $r/n$ is the rate per compounding period (the annual rate divided by the number of compoundings per year).
For example, if we compound monthly for $7$ years at a $6\%$ rate, we would compound $nt\,=\,12\cdot 7\,=\,84$ times, once per month, at a rate of $0.06/12\,=\,0.005$ per monthly period. Notice that we always use the decimal version for interest rates when using these equations.
On the other hand, interest compounded continuously earns rate in just that way - continuously. I can have any value I want for time $t$, and the total amount in the account will change with each and every second. Therefore, we express the account value as
$A\,=\,Pe^{rt}.$
Notice that in both formulas, the value at the initial time $t\,=\,0$ is just our initial investment $P$.
However, a present value/future value problem requires you to think a little differently. When we assess these, we are usually given the eventual payout (future value, or $A$ in the equations above), and then are asked to find the present value, which would be $P$ the equations given. As a result, we can rewrite the equations to solve for $P$. The equations then become
$P\,=\,A\left(1+{\frac {r}{n}}\right)^{-nt},$
for the non-continuous case, and
$P\,=\,Ae^{-rt}$
for the continuous case.

(a):
We are given all the pieces required. We begin with $\$3000$ of principal, and compound quarterly, or $4$ times per year. Using the formula in 'Foundations', the equation for the present value is
$P\,=\,A\left(1+{\frac {r}{n}}\right)^{-nt}\,=\,3000\left(1+{\frac {0.06}{4}}\right)^{-4\cdot 8}.$

(b):
Again, we need to apply the formula from foundations to find the present value is
$P\,=\,Ae^{-rt}\,=\,3000e^{-0.06\cdot 8}.$

Final Answer:
(a) $P\,=\,3000(1.015)^{-32}.$
(b) $P\,=\,3000e^{-0.48}.$

Return to Sample Exam