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: $$S_0 = \frac{12,000}{0.08} \approx \$ 150,000$$

Key concepts

First-order linear differential equations

First-order linear differential equations are a fundamental aspect of calculus and differential equations. These equations involve derivatives of a function and are expressed in the form:\[ \frac{dy}{dx} + P(x)y = Q(x) \]where \( P(x) \) and \( Q(x) \) are given functions. In the context of the given problem, the differential equation is:\[ \frac{dS}{dt} = rS(t) - k \]Here, this equation describes how the amount of investment \( S(t) \) changes over time \( t \) when subject to continuous compounding interest at a rate \( r \), minus continuous withdrawals at rate \( k \). Solving such equations typically involves:

• Identifying the integrating factor, \( e^{\int P(x) \, dx} \), that helps in solving the equation.
• Multiplying through by this integrating factor to facilitate integration.

This technique results in a solution for \( S(t) \) that describes the financial scenario over time, which is crucial for analyzing investment growth or depletion.

Compound interest

Compound interest is when interest earned on an amount is added to the principal, so that from that moment on, the interest that has been added also earns interest. In mathematical terms, when dealing with continuous compounding as in this exercise, compound interest is calculated using the formula:\[ A = P e^{rt} \]Where:

• \( A \) is the amount of money accumulated after n years, including interest.
• \( P \) is the principal amount (the initial sum of money).
• \( r \) is the annual interest rate (in decimal).
• \( t \) is the time the money is invested for in years.
• \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

This concept is integrated within the differential equation used in the exercise, illustrating how the investment grows continuously at a rate \( r \) in a financial model scenario.

Mathematical modeling in finance

Mathematical modeling in finance is a powerful tool used to represent real-world financial systems with mathematical expressions and formulas. In this specific exercise:

• The differential equation \( \frac{dS}{dt} = rS(t) - k \) is used to model the investment scenario.
• Key variables include \( S(t) \) for the investment balance, \( r \) for interest rate, and \( k \) for the withdrawal rate.

The aim is to understand and predict how changes in variables like interest rates and withdrawals influence the balance \( S(t) \) over time. This can help in:

• Determining sustainable withdrawal rates \( k_0 \) that keep the investment stable.
• Calculating the time \( T \) it would take for an investment to deplete to zero if withdrawals are too high.

Such models are invaluable in finance for retirement planning, evaluating investment strategies, and assessing risk.

Continuous compounding

Continuous compounding is the process of calculating interest and adding it to the account balance continuously, rather than at set intervals (e.g., annually, semi-annually). This concept allows the investment to grow at every possible moment, maximizing the effect of the compound interest. In the case of continuous compounding, the formula used is:\[ A = P e^{rt} \]where \( t \) is time, \( r \) is the rate of growth, and \( P \) is the initial principal. In the context of our exercise:

• We model the growth of an investment with continuous compounding while considering continuous withdrawals.
• The differential equation \( \frac{dS}{dt} = rS(t) - k \) captures the essence of continuous growth minus withdrawals.

Continuous compounding is an essential concept in financial modeling as it provides a more accurate representation of how interest impacts investments over time, as opposed to periodic compounding methods which might be less precise.