Last time we discussed a retirement calculation in terms of the Savings Ratio. Today, we’ll complete the thought and derive a remaining aspect of the simplest retirement calculation.
Recommend first reading The Savings Ratio.
Setup
Last time, we had N=25/R years (or N=300/R months) until retirement at savings ratioR=\frac{p}{1-p} and savings rate p (starting from net worth of zero). The major limitation of this simple division is that the R values aren’t compounding pre-retirement. Let’s fix that now.
Compounding
Way, way back in high school, we learned the differences among simple interest, compounding interest (over various compounding periods), and continuously compounding interest. Let’s see how much the compounding interval actually matters at 4%.
\begin{darray}{rcll}(1 + 0.04)^n &=& 1.04^n&\textrm{(annual compounding)}\\\\ \left(1+\frac{0.04}{12}\right)^{12\cdot n} &=& 1.040742^n&\textrm{(monthly compounding)} \\\\e^{0.04\cdot n} &=& 1.040811^n&\textrm{(continous compounding)}\end{darray}As you can see, it doesn’t really matter. 4% vs 4.074% vs 4.08% is really small potatoes. While it will pack a punch over a 40-year time horizon, today we’re going to squint enough that they’re essentially the same. In fact, one of the concessions we’re going to make for the sake of producing the simplest retirement calculation on the internet is to substitute 5% annual compounding for 4% continuous compounding (4.08% is certainly less than 5% and hence, more conservative).
Finally, smaller compounding periods receive smaller deposits more often. Since R is a unitless multiple of expenses (per compounding period), this fact has been transparent so far.
Discrete Interest on Savings
What happens when we give the R values interest (compounded according to the timeframe chosen)? The math gets harder fast, but not too hard. Before, we solved for N in the following to get years until retirement (replace 25 with 300 for the months version).
25 = R\cdot N = R\left(\sum_{k=0}^{N-1}1\right)=\underbrace{R\left(\int_0^N1~dt\right)}_\textrm{foreshadowing}Now, we add the interest to R with the appropriate compounding period and again solve for N to get years (replace 25 with 300 and r with \frac{r}{12} for months).
\begin{darray}{rcl}25 &=&R\sum_{k=0}^{N-1}(1+r)^k\\\\&=& R\frac{(1+r)^N-1}{(1+r)-1} = R\frac{(1+r)^N-1}{r}\end{darray}In his now widely cited Shockingly Simple Math Behind Early Retirement, MMM uses some conservative assumptions. Specifically, he assumes the standard r_1=0.04 withdrawal rate post-retirement but r_0=0.05 real rate of return pre-retirement. We’re going to simplify and use the same r=r_0=r_1 for both sides. The left-hand side can be written in those terms as well because 25 = r^{-1} (for r=0.04) is the inverse of the withdrawal rate.
25 = \frac{1}{r} = R\frac{(1+r)^N-1}{r}Recalling that R=\frac{p}{1-p}, 0<p<1 and cancelling r, we can simplify.
\begin{darray}{rcl}(1+r)^N&=&1+\frac{1-p}{p} = \frac{1}{p}\\\\N&=&\frac{-\ln{p}}{\ln(1+r)}\end{darray}Continuous Interest on Savings
It turns out that when we use continuously compounding interest, the equation gets simpler still. For the calculus-aware, as our compounding periods shrink, so do our deposits and in the limit, R\cdot\Delta t \to R\cdot dt.
\begin{darray}{rcl}25 = \frac{1}{r} = R\int_0^N e^{r\cdot t}dt &=&\left.R\cdot \frac{e^{r\cdot t}}{r}\right|_{t=0}^N\\\\&=&R\left(\frac{e^{r\cdot N}-1}{r}\right)\end{darray}Solving for N this time leads to something even better.
\begin{darray}{rcl}N&=&-\frac{1}{r}\ln(p),~~0<p<1\\\\&=&-25\ln(p)\textrm{~years} = -300\ln(p)\textrm{~months, when }r=0.04\\\\&=&-20\ln(p)\textrm{~years} = -240\ln(p)\textrm{~months, when }r=0.05\end{darray}Examples
We had a couple of examples in the savings ratio article. For savings rate p_1\approx0.10, the time till retirement is -25\ln(0.1)=57 years. In reality, another 12.4\% of their paycheck is going to social security (only half of that appears on their pay stub), so their savings rate is higher and standard retirement will happen on time.
With a high savings rate of p_2=0.75, retirement comes much sooner in only -25\ln(0.75)=7 years! Finally for a more reasonable p_3=0.5, one can retire in -25\ln(0.5)=17 years.
Summary
There we have it. The easiest retirement calculator on the web. Easy enough to discuss on the run even.
\begin{darray}{rcl}\textrm{years} &=& -25\ln(p),~~\textrm{for savings rate}~0<p\leq1\\\\\textrm{months} &=& -300\ln(p)\end{darray}The best part is that we don’t even have to choose between the discrete and continuous versions. A fun fact from Calculus class is that \ln(1+r)\approx r for small values of r. So they’re essentially the same equation… which is a pleasant result since the effective interest rates are also essentially the same.
One tiny caveat casually mentioned in the introduction is that this estimate is from a beginning point of zero (invested) net worth. Thankfully, this gives us something to overthink another day when we introduce the notion of Effective Savings Rate.