Chapter 3, Problem 54P

To determine

Calculate the time period.

Explanation of Solution

The present value (P) is $20,000. The first withdraw (D) is $5,000 that has taken place now. The second year’s withdrawal (D2) is $4,500 and decreases the withdrawal (d) by $500. Thereafter, interest rate (i) is 8%.

Time period (n) can be calculated as follows:

$\begin{array}{c}\text{P}=\text{W}+\text{W}2\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)-\text{d}\times \frac{1}{\text{i}}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(1+\text{i}\right)}^{\text{n}}}\right)\\ 20,000=5,000+4,500\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}\right)-\text{500}\times \frac{1}{\text{0}\text{.08}}\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}\right)\\ 20,000-5,000=4,500\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}\right)-\text{500}\times 12.5\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}\right)\\ 15,000=4,500\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}\right)-\text{500}\times 12.5\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(1+\text{0}\text{.08}\right)}^{\text{n}}}\right)\end{array}$

Substitute the time period as 5 years in the above calculation.

$\begin{array}{l}15,000=4,500\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{5}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{5}}}\right)-\text{500}\times 12.5\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{5}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{5}}}-\frac{\text{5}}{{\left(1+\text{0}\text{.08}\right)}^{\text{5}}}\right)\\ 15,000=4,500\left(\frac{1.4693-1}{\text{0}\text{.08}\left(1.4693\right)}\right)-\text{500}\times 12.5\left(\frac{1.4693-1}{\text{0}\text{.08}\left(1.4693\right)}-\frac{\text{5}}{1.4693}\right)\\ 15,000=4,500\left(\frac{0.4693}{0.1175}\right)-\text{500}\times 12.5\left(\frac{0.4693}{0.1175}-3.403\right)\\ 15,000=4,500\left(3.994\right)-\text{500}\times 12.5\left(3.994-3.403\right)\\ 15,000=17,973-3,693.75\\ 15,000>14,279.25\end{array}$

The calculated value is less than $15,000. Thus, it increases the time period to 5.6.

$\begin{array}{c}15,000=4,5000\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{5}\text{.6}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{5}\text{.6}}}\right)-\text{500}\times 12.5\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{5}\text{.6}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{5}\text{.6}}}-\frac{\text{5}\text{.6}}{{\left(1+\text{0}\text{.08}\right)}^{\text{5}\text{.6}}}\right)\\ 15,000=4,500\left(\frac{1.5388-1}{\text{0}\text{.08}\left(1.5388\right)}\right)-\text{500}\times 12.5\left(\frac{1.5388-1}{\text{0}\text{.08}\left(1.5388\right)}-\frac{\text{5}\text{.6}}{1.5388}\right)\\ 15,000=4,500\left(\frac{0.5388}{0.1231}\right)-\text{500}\times 12.5\left(\frac{0.5388}{0.1231}-3.6392\right)\\ 15,000=4,500\left(4.3769\right)-\text{500}\times 12.5\left(4.3769-3.6392\right)\\ 15,000=19,696.05-\text{500}\times 12.5\left(0.7377\right)\\ 15,000=19,696.05-4,610.625\\ 15,000\approx 15,085.43\end{array}$

Since the calculated value is nearly equal to $15,000, it is confirmed that the time period is 5.6 years.