How to find the required interest rate?? (Help isolating for r)

Question

I have $300,000$ (A) that I want to last for $10$ years (Y). Every year I will be spending $55,000$ (C) all withdrawn at the end of the year. What constant interest rate would I need every year to make this work? If possible could you also include the equation to find the interest rate?

The equation I'm using is: A = C x (1 - (1 / (1 +r)^Y )) / r

Answer 1

the net present value of the future cash flows at the required rate equals the initial balance. $\sum_\limits{i=1}^{10} \frac {55,000}{(1+y)^i} = 300,000\\ 55,000\frac {(1-(1+y)^{-10})}{y} = 300,000\\ 55,000(1+y)^{10} - 1) = 300,000y(1+y)^{10}\\ 300(1+y)^{11} -355(1+y)^{10} + 55,000 = 0$

I don't know of a way to solve that analytically.

But there is only one real root where $y> 0$ and that is $y = 0.1287$

$12.87\%$

Answer 2

The balance $A_t$ at time $t$, where $A_0 = A$, is equal to:

$$A_t = A_{t-1} \cdot (1+r) - C$$

So we have:

$$\begin{align} A_1 &= A \cdot (1+r) - C \\ A_2 &= A \cdot (1+r)^2 - C\cdot (1+r) - C\\ A_3 &= A \cdot (1+r)^3 - C\cdot (1+r)^2 - C\cdot (1+r) - C\\ \vdots \\ A_{t} &= A\cdot (1+r)^t - C\sum_{k=0}^{t-1}(1+r)^k = A \cdot (1+r)^t - C \cdot \frac{(1 + r)^t - 1}{r} \\ A_{t} &= \frac{(r + 1)^t (A\cdot r - C) + C}{r} \end{align}$$

So if we want $A_{10} = 0$, i.e. have just enough money to last, we can solve for $r$ numerically and get

$$r = 0.1287$$

Here is a sample Python script that uses a binary search to hone in on the value for $r$:

a, c, t = 300000, 55000, 10
lo = 0
hi = 1

while hi - lo > 0.000000001:
    r = (lo + hi) / 2.0
    ending_bal = ((r + 1)**t * (a * r - c) + c) / r

    if ending_bal < 0:
        lo = r
    else:        
        hi = r 

print(round(r, 4)) #outputs 0.1287