Sir Subhotosh Khan said:
Funner problem is to do the problem in the reverse order : i.e. first year $5/month, second year 10/month, and so on - how long would it take to deplete $16,090.80 at a rate 12%/year (I wish) compounded monthly.
From Dr. Howard W. Eves’ Return to Mathematical Circles (ca 1988)
143º The equality of null sets.
The man in the wilderness asked of me
How many strawberries grew in the sea.
I answered him as I thought good,
As many as red herring grow in the wood.
-Anonymous
how long would it take to deplete $16,090.80 at a rate 12%/year (I wish) compounded monthly?
I’d say the probability of my winning the powerball lottery (near zero as it is) is better than the probability of this fund getting depleted at the stated rate of withdrawal. Accordingly, the present value of such a problem is represented by
\(\displaystyle \begin{gathered} A = W \cdot \frac{{\left( {1 + i} \right) \cdot s_{\left. {\overline {\, n \,}}\! \right| i} - n}}{i} \cdot \left( {1 + i} \right)^{ - n} \hfill \\ \Leftrightarrow \hfill \\ \end{gathered}\)
\(\displaystyle \begin{gathered} A = \frac{W}{i}\left[ {\left( {1 + i} \right) \cdot a_{\left. {\overline {\, n \,}}\! \right| i} - n\left( {1 + i} \right)^{ - n} } \right] \hfill \\ \Leftrightarrow \hfill \\ A \approx \$ 4,236.430938... \hfill \\ \end{gathered}\)
where
\(\displaystyle \begin{gathered} W = \left( {5 \cdot s_{\left. {\overline {\, {12} \,}}\! \right| .01} } \right) \hfill \\ n = 40 \hfill \\ i = \left( {1.01} \right)^{12} - 1 \hfill \\ \end{gathered}\)
One could almost see that it would take exactly 40 years to deplete a present value of $4,236.43. Applying the so-called retrospective method for determining the fund balance on a certain date, we have:
\(\displaystyle A\left( {1 + i} \right)^{40} - W \cdot \frac{{\left( {1 + i} \right) \cdot s_{\left. {\overline {\, {40} \,}}\! \right| i} - 40}}{i} = 0\)
On the other hand, if the present value to be depleted is $16,090.80, it can be shown that unless the world suddenly ends for whatever reason, the fund value will never be depleted since, among other factors, the growth rate is greater than the depletion or withdrawal rate. As an example, at the end of 40 years, the fund balance (using Sir Denis’ results) would be
\(\displaystyle \left( {{\text{\$16,090}}{\text{.788674169358}}....} \right)\left( {1 + i} \right)^{40} - W \cdot \frac{{\left( {1 + i} \right) \cdot s_{\left. {\overline {\, {40} \,}}\! \right| i} - 40}}{i} \approx \$ {\text{1,406,492}}{\text{.57798946}}...\)
Sir Subhotosh Khan’s problem is therefore a type of a growing (arithmetic) perpetuity.
Incidentally, to determine the fund balance at the end of a month that is between years, say 2 3/12 years, with
\(\displaystyle A_1 = {\text{\$16,090}}{\text{.788674169358}}....\)
and
\(\displaystyle P = A_1 \left( {1 + i} \right)^2 - W \cdot \frac{{\left( {1 + i} \right) \cdot s_{\left. {\overline {\, 2 \,}}\! \right| i} - 2}}{i}\)
the fund balance would be
\(\displaystyle P\left( {1.01} \right)^3 - 15 \cdot s_{\left. {\overline {\, 3 \,}}\! \right| .01} \approx \$ {\text{20,800}}{\text{.3731827106}}...\)