Question

A retired person has a sum \(S(t)\) invested so as to draw interest at an annual rate \(r\) compounded continuously. Withdrawals for living expenses are made at a rate of \(k\) dollars/year; assume that the withdrawals are made continuously. $$ \begin{array}{l}{\text { (a) If the initial value of the investment is } S_{0} \text { , determine } S(t) \text { at any time. }} \\ {\text { (b) Assuming that } S_{0} \text { and } r \text { are fixed, determine the withdrawal rate } k_{0} \text { at which } S(t) \text { will }} \\ {\text { remain constant. }} \\ {\text { (c) If } k \text { exceeds the value } k_{0} \text { found in part (b), then } S(t) \text { will decrease and ultimately }} \\ {\text { become zero. Find the time } T \text { at which } S(t)=0 \text { . }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Determine } T \text { if } r=8 \% \text { and } k=2 k_{0} \text { . }} \\ {\text { (e) Suppose that a person retiring with capital } S_{0} \text { wishes to withdraw funds at an annual }} \\ {\text { rate } k \text { for not more than } T \text { years. Determine the maximum possible rate of withdrawal. }} \\ {\text { (f) How large an initial investment is required to permit an annual withdrawal of } \$ 12,000} \\\ {\text { for } 20 \text { years, assuming an interest rate of } 8 \% ?}\end{array} $$

Short answer

Question: Determine the maximum possible withdrawal rate k for not more than T years. Answer: To determine the maximum possible withdrawal rate k for not more than T years, use the formula derived in part (e): $$k_{\text{max}} = rS_0 \frac{1 - e^{-rT}}{1 - e^{-rT}}$$

Step-by-step solution

(a) Find S(t) with initial value S0

: To begin, let's consider the initial value of investment \(S_0\). The differential equation representing the net rate of change of \(S(t)\) with respect to time is given by: $$\frac{dS}{dt} = r \cdot S(t) - k$$ Here, r is the annual interest rate compounded continuously, and k is the withdrawal rate (both constants). We can solve this first-order linear differential equation to find the function \(S(t)\). First, find the integrating factor \(e^{\int rp\,dt}= e^{rt}\). Multiply both sides of the equation by the integrating factor to make the left side an exact derivative: $$e^{rt} \cdot \frac{dS}{dt} + re^{rt} \cdot S(t) = k \cdot e^{rt}$$ This can be simplified to: $$\frac{d}{dt} \Bigl( S(t) \cdot e^{rt} \Bigr) = k \cdot e^{rt}$$ Now, integrate both sides with respect to t: $$S(t) \cdot e^{rt} = k \int e^{rt} dt + C = ke^{rt} \cdot \frac{1}{r} + C$$ Finally, divide both sides by \(e^{rt}\) and apply the initial investment amount \(S_0\): $$S(t) = S_0 + \frac{k}{r}(1 - e^{-rt})$$

(b) Find the withdrawal rate k0 to keep S(t) constant

: To keep \(S(t)\) constant, the derivative with respect to time must be equal to zero: $$\frac{dS}{dt} = rS(t) - k = 0$$ Now, set \(k_0\) as the withdrawal rate that keeps \(S(t)\) constant and insert \(S_0\): $$k_0 = rS_0$$

(c) Find the time T when S(t) = 0 if k > k0

: If the withdrawal rate k is greater than \(k_0\), then \(S(t)\) will eventually decrease to zero. Use the equation for \(S(t)\) from part (a) and set S(t) equal to 0: $$0 = S_0 + \frac{k}{r}(1 - e^{-rt})$$ Now rearrange the equation to solve for T: $$T = \frac{-\ln{\frac{k - rS_0}{k}}}{r}$$

(d) Determine T with r = 8% and k = 2 * k0

: Plug in the specified rates for r and k: $$T = \frac{-\ln \frac{(2rS_0) - rS_0}{2rS_0}}{0.08} = \frac{-\ln \frac{1}{2}}{0.08}$$ Thus, the time when the investment becomes zero: $$T = \frac{\ln 2}{0.08} \approx 8.66 \text{ years}$$

(e) Determine the maximum possible rate of withdrawal k for not more than T years

: From the equation in part (c), we can rewrite the equation for k in terms of T: $$k_{\text{max}} = rS_0 \frac{1 - e^{-rT}}{1 - e^{-rT}}$$

(f) Initial investment required for annual withdrawal of $12,000 for 20 years at 8% interest

: Using the equation from part (e), set k = \(12,000\) and T = 20 years, and r = 0.08, then solve for \(S_0\): $$12000 = 0.08S_0 \frac{1 - e^{-0.08(20)}}{1 - e^{-0.08(20)}}$$ Therefore, the required initial investment is: $$S_0 = \frac{12,000}{0.08} \approx \$ 150,000$$