Question

A young person with no initial capital invests \(k\) dollars per year at an annual rate of return \(r\). Assume that investments are made continuously and the return is compounded continuously. $$ \begin{array}{l}{\text { (a) Determine the sum } S(t) \text { accumulated at any time } t} \\ {\text { (b) If } r=7.5 \% \text { , determine } k \text { so that } \$ 1 \text { million will be available for retirement in } 40 \text { years. }} \\ {\text { (c) If } k=\$ 2000 / \text { year, determine the return rate } r \text { that must be obtained to have } \$ 1 \text { million }} \\ {\text { available in } 40 \text { years. }}\end{array} $$

Short answer

Answer: Approximately $823.20 per year.

Step-by-step solution

Understanding continuous compounding formula

The general formula for continuous compounding with a constant rate of return is: $$S(t) = P \cdot e^{r \cdot t}$$ where S(t) is the accumulated amount at time t, P is the initial capital, r is the rate of return, and t is the time in years.

(a) Step 2: Determine S(t) in terms of k and r

Since investments are made continuously, we need to use integration. Let's represent the continuous investment flow with the function f(t) = k. We can compute the total sum, S(t), as follows: $$S(t) = \int_0^t k e^{r(u-t)} du$$

(b) Step 3: Calculate the value of k for r = 7.5% and t = 40 years

We want to have 1 million dollars by the end of 40 years. We set S(40) = 1000000 and r = 0.075. We can now solve for k: $$1000000 = \int_0^{40} k e^{0.075(u-40)} du$$, $$k = \frac{1000000}{\int_0^{40} e^{0.075(u-40)} du} $$ Calculating this integral using the fundamental theorem of calculus, we get: $$k = \frac{1000000}{\left[\frac{1}{0.075} e^{0.075(u-40)}\right]_0^{40}} = 823.20$$ Thus, the person would need to invest approximately $823.20 per year.

(c) Step 4: Calculate the value of r for k = $2000/year and t = 40 years

We want to have 1 million dollars at the end of 40 years. We set S(40) = 1000000 and k = 2000. We can now solve for r: $$1000000 = \int_0^{40} 2000 e^{r(u-40)} du$$, $$r = \frac{\ln{\left(\frac{1000000}{2000}\cdot\int_0^{40} e^{-(u-40)} du\right)}}{(40-0)}$$ Calculating this integral using the fundamental theorem of calculus, we get: $$r = \frac{\ln{\left(\frac{1000000}{2000}\cdot\left[-e^{-(u-40)}\right]_0^{40}\right)}}{40} = 0.0566$$ Thus, a return rate of approximately 5.66% is required to have $1 million available in 40 years.