Question 5.3: A savings account earns interest at the rate of 6% per year,......
A savings account earns interest at the rate of 6% per year, compounded continuously. How much money must initially be placed in the account to provide for twenty end-of-year withdrawals, if the first withdrawal is $1000 and each subsequent withdrawal increases by $200?
The solution may be formulated after (2.11):
P = A_0 × (P/A, 6%, 15) + G × (A/G, 6%, 15) × (P/A, 6%, 15) (2.11)
P = $1000[P/A, 6%, 20] + $200[A/G, 6%, 20] [P/A, 6%, 20]
From (5.7) and (5.8),
P/A = \frac{1 – e^{-m}}{e^r – 1} (5.7) \\\\ A/G = \frac{1}{e^r – 1} – \frac{n}{e^m – 1} (5.8) \\\\ [P/A, 6\%, 20] = \frac{1 – e^{(0.06)(20)}}{e^{0.06} – 1} = 11.3009 \\\\ [A/G, 6\%, 20] = \frac{1}{e^{0.06} – 1} – \frac{20}{e^{(0.06)(20)} – 1} = 7.5514(these values can also be obtained from Appendix C); hence,
P = $1000(11.3009) + $200(7.5514)(11.3009) = $28 368.42
If interest is compounded continuously but payments are made p times a year, formulas (5.2) through (5.8) remain valid with r replaced by r/p and with n replaced by np. [These substitutions do not, of course, alter the forms of (5.2) and (5.3).
F/P = e^m (5.2) \\\\ P/F = e^{-m} (5.3)F/A={\frac{\mathrm{e}^{\mathrm{m}}-1}{\mathrm{e}^{\mathrm{r}}-1}} (5.4)
A/F={\frac{e^{r}-1}{e^{m}-1}} (5.5)
A/P={\frac{\mathrm{e}^{r}-1}{1-e^{-m}}} (5.6)