In this example,

We’re looking for P_0: how much money needs to be in the account at the beginning.

Putting this into the equation:

[latex]\begin{align} &{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-20(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\ &{{P}_{0}}=\frac{1000\times\left(1-{{\left(1.005\right)}^{-240}}\right)}{\left(0.005\right)}\\ &{{P}_{0}}\approx 139580.7717 \\ \end{align}[/latex]

You will need to have $139,580.78 in your account when you retire.

Notice two things:

First, even though there was a 1 in the next place, we still rounded up. The formula calculated the minimum we would need to start with in the account in order for the account to last for 20 years. Since that minimum was above $139,580.77, this amount would be too small, so we need to round up.

Second, you withdrew a total of $240,000 ($1000 a month for 240 months). The difference between what you pulled out and what you started with is the interest earned. In this case it is $240,000 – $139,580.78 = $100,419.22 in interest.

In this example, we’re looking for d.

In this case, we’re going to have to set up the equation, and solve for d.

[latex]\begin{align} &500,000=\frac{d\left(1-{{\left(1+\frac{0.08}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.08}{12}\right)}\\ &500,000\left(\frac{0.08}{12}\right)=d\left(1-{{\left(1+\frac{0.08}{12}\right)}^{-360}}\right)\\ &\frac{500,000\left(\frac{0.08}{12}\right)}{1-\left(1+\frac{0.08}{12}\right)^{-360}}=d\\ &d\approx\$3668.82 \\\end{align}[/latex]

You would be able to withdraw $3,668.82 each month for 30 years.