Let’s break down the problem step by step: ### **Step 1: Find the amount accumulated in Hardaway's deferred compensation account.** **Given:** - Annual deposit (\(PMT\)) = $8,000 (at end of each year) - Number of years (\(n\)) = 30 - Interest rate (\(r\)) = 7% This is a **Future Value of an Ordinary Annuity** problem: \[ FV = PMT \times \frac{(1 + r)^n - 1}{r} \] \[ FV = 8,000 \times \frac{(1.07)^{30} - 1}{0.07} \] Now, calculate \((1.07)^{30}\): - \((1.07)^{30} \approx 7.612255\) Plug in the values: \[ FV = 8,000 \times \frac{7.612255 - 1}{0.07} = 8,000 \times \frac{6.612255}{0.07} = 8,000 \times 94.4608 = 755,686.40 \] **Rounded to the nearest whole number:** \[ \boxed{755,686} \] --- ### **Step 2: How much can Hardaway withdraw at the beginning of each year for 16 years?** Now, he wants to withdraw an equal amount at the **beginning** of each year for 16 years. This is a **Present Value of an Annuity Due** (since withdrawals are at the beginning). We set up: - \(PV = 755,686\) (the amount accumulated) - \(n = 16\) - \(r = 7\%\) - Find \(PMT\) (payment per year, at beginning of each year) \[ PVAD = PMT \times \left[\frac{1 - (1 + r)^{-n}}{r}\right] \times (1 + r) \] Solve for \(PMT\): \[ PMT = \frac{PVAD}{\left[\frac{1 - (1 + r)^{-n}}{r}\right] \times (1 + r)} \] First, calculate the bracketed term: - \((1 + r)^{-n} = (1.07)^{-16} \approx 0.3677\) So, \[ \frac{1 - 0.3677}{0.07} = \frac{0.6323}{0.07} = 9.033 \] Now, multiply by \((1 + r) = 1.07\): \[ 9.033 \times 1.07 = 9.665 \] Now, \(PMT = \frac{755,686}{9.665} = 78,218\) **Rounded to nearest whole number:** \[ \boxed{78,218} \] --- ### **Step 3: For how many years can Hardaway withdraw $85,000 at the beginning of each year?** Now, \(PMT = 85,000\) (withdrawals at **beginning** of each year), \(PVAD = 755,686\), \(r = 7\%\), solve for \(n\): \[ PVAD = PMT \times \left[\frac{1 - (1 + r)^{-n}}{r}\right] \times (1 + r) \] \[ \frac{PVAD}{PMT \times (1 + r)} = \frac{1 - (1 + r)^{-n}}{r} \] \[ \frac{755,686}{85,000 \times 1.07} = \frac{1 - (1.07)^{-n}}{0.07} \] \[ \frac{755,686}{90,950} = 8.313 \] So, \[ 8.313 = \frac{1 - (1.07)^{-n}}{0.07} \] \[ 0.5819 = 1 - (1.07)^{-n} \] \[ (1.07)^{-n} = 1 - 0.5819 = 0.4181 \] Take natural log of both sides: \[ -n \ln(1.07) = \ln(0.4181) \] \[ -n \times 0.06766 = -0.8715 \] \[ n = \frac{0.8715}{0.06766} = 12.89 \] **Rounded to nearest whole number:** \[ \boxed{13} \] --- ## **Summary Table** | Requirement | Calculation | Final Answer | |-------------|-------------|--------------| | 1. Accumulated value after 30 years | FV of ordinary annuity | **$755,686** | | 2. Annual withdrawal for 16 years (beginning of year) | PV of annuity due | **$78,218** | | 3. Number of years for $85,000 withdrawals (beginning of year) | PV of annuity due | **13 years** | --- ## **Explanation** - **Step 1:** Use the future value formula for an ordinary annuity to find how much is accumulated after 30 years. - **Step 2:** Use the present value formula for an annuity due to determine the annual withdrawal for 16 years. - **Step 3:** Rearranged the annuity due formula to solve for the number of years withdrawals can be made at $85,000 per year. **All answers are rounded to the nearest whole number.**