Present Value Examples

Example 1

Present value

How much money should be deposited in a bank paying a yearly interest rate of \(6\%\) compounded monthly so that after 3 years, the accumulated amount will be $20,000?

Step 1:   Recall appropriate present value formula.

Notice that this is a present value problem since we’re given the accumulated amount and we’re asked to find the principal. And since interest is compounded monthly, we’ll use the present value formula for discrete compounding of interest.

\[P = A\left(1 + \frac{r}{m}\right)^{-mt}\]

Step 2:   Plug in the given values.

\(A = 20000\), \(r = 0.06\), \(m=12\), and \(t=3\).

\[\begin{align*} P &=20000\left(1+\frac{0.06}{12}\right)^{-(12)(3)}\\ \\ &=20000(1.005)^{-36} \approx \$16,712.90 \end{align*}\]

Therefore, $16,712.90 invested at 6% interest compounded monthly will be worth $20,000 in 3 years.

Observation

In the previous example, instead of having to recall the present value formula, we could have started with the accumulated formula, and then solved for \(P\).

Step 1:   Recall appropriate accumulated amount formula.

Start with the formula for accumulated amount for discrete compounding of interest.

\[A = P\left( 1 + \frac{r}{m}\right)^{mt}\]

Step 2:   Plug in the given values.

\(A = 20000\), \(r = 0.06\), \(m=12\), and \(t=3\).

\[\begin{align*} 20000 &= P\left(1+\frac{0.06}{12}\right)^{(12)(3)}\\ &=P(1.005)^{36} \end{align*}\]

Step 3:   Solve for \(P\).

\[\begin{align*} P = \frac{20000}{1.005^{36}} \approx \$16,712.90 \end{align*}\]

Example 2

Present value

Parents wish to establish a trust fund for their child’s education. If they need $170,000 in 7 years, how much should they set aside now if the money is invested at \(20\%\) compounded continuously?

Step 1:   Recall appropriate present value formula.

Notice that this is a present value problem since we’re given the accumulated amount and we’re asked to find the principal. And since interest is compounded continuously, we’ll use the present value formula for continuous compounding of interest.

\[P = Ae^{-rt}\]

Step 2:   Plug in the given values.

\(A = 170000\), \(r = 0.2\), and \(t=7\).

\[\begin{align*} P &=170,000e^{-(0.2)(7)}\\ \\ &=170,000e^{-1.4} \approx {\$41,921.48} \end{align*}\]

Therefore, $41,921.48 invested at 20% interest compounded continuously will be worth $170,000 in 7 years.

Obervation

In the previous example, instead of having to recall the present value formula, we could have started with the accumulated formula, and then solved for \(P\).

Step 1:   Recall appropriate accumulated amount formula.

Start with the formula for accumulated amount for continuous compounding of interest.

\[A = Pe^{rt}\]

Step 2:   Plug in the given values.

\(A = 170000\), \(r = 0.2\), and \(t=7\).

\[\begin{align*} 170000 &= Pe^{(0.2)(7)}\\ &= Pe^{1.4} \end{align*}\]

Step 3:   Solve for \(P\).

\[\begin{align*} P &= \frac{170000}{e^{1.4}} \approx \$41,921.48 \end{align*}\]