1. Problem statement: We have an ordinary annuity with a 7.92% annual interest rate compounded monthly. (A) A person deposits $100 monthly for 30 years and then makes equal monthly withdrawals for 15 years, reducing the balance to zero. We want to find the monthly withdrawals and the total interest earned over 45 years. 2. Given data: - Annual interest rate, $r_{annual} = 7.92\% = 0.0792$ - Monthly interest rate, $r = \frac{0.0792}{12} = 0.0066$ - Deposit period, $n_1 = 30 \times 12 = 360$ months - Withdrawal period, $n_2 = 15 \times 12 = 180$ months - Monthly deposit, $P = 100$ 3. Calculate the future value (FV) of deposits after 30 years: $$ FV = P \times \frac{(1+r)^{n_1} - 1}{r} = 100 \times \frac{(1 + 0.0066)^{360} - 1}{0.0066} $$ First, calculate $(1 + 0.0066)^{360}$: $$ (1.0066)^{360} \approx e^{360 \times \ln(1.0066)} = e^{360 \times 0.006578} = e^{2.368} \approx 10.68 $$ Then, $$ FV = 100 \times \frac{10.68 - 1}{0.0066} = 100 \times \frac{9.68}{0.0066} = 100 \times 1466.7 = 146670 $$ 4. Let the monthly withdrawal be $W$. The balance will be depleted after withdrawals for 180 months. The present value of withdrawals at the beginning of withdrawals equals the future value at the end of deposits: $$ FV = W \times \frac{1 - (1+r)^{-n_2}}{r} $$ Solving for $W$: $$ W = FV \times \frac{r}{1 - (1+r)^{-n_2}} = 146670 \times \frac{0.0066}{1 - (1.0066)^{-180}} $$ Calculate denominator: $$ (1.0066)^{-180} = \frac{1}{(1.0066)^{180}} \approx \frac{1}{e^{180 \times 0.006578}} = \frac{1}{e^{1.184}} = \frac{1}{3.267} = 0.306 $$ So, $$ 1 - 0.306 = 0.694 $$ Then, $$ W = 146670 \times \frac{0.0066}{0.694} = 146670 \times 0.00951 = 1394.45 $$ 5. Total amount deposited over 30 years: $$ Total\ deposits = 100 \times 360 = 36000 $$ 6. Total amount withdrawn over 15 years: $$ Total\ withdrawals = W \times n_2 = 1394.45 \times 180 = 251001 $$ 7. Total interest earned is total withdrawals minus total deposits: $$ Interest = 251001 - 36000 = 215001 $$ Final answers: - Monthly withdrawals: $\boxed{1394.45}$ - Interest earned over 45 years: $\boxed{215001}$ --- (B) If monthly withdrawals for 15 years are $1500$, find monthly deposit for 30 years. 8. Given withdrawals $W = 1500$, find necessary $P$ for deposits. Use formula from step 4: $$ FV = W \times \frac{1 - (1+r)^{-n_2}}{r} = 1500 \times \frac{0.694}{0.0066} = 1500 \times 105.15 = 157725 $$ 9. Now solve for $P$ from future value formula: $$ P = \frac{FV \times r}{(1+r)^{n_1} - 1} = \frac{157725 \times 0.0066}{10.68 - 1} = \frac{1040.69}{9.68} = 107.56 $$ Final answer: - Monthly deposit: $\boxed{107.56}$