The Idea Shop: Some Interesting Mathematics of the Rule of 72

Some Interesting Mathematics of the Rule of 72

Most people first encounter the “rule of 72” in a basic finance course. The idea is simple: divide 72 by your growth rate times 100, and the result is the number of years it takes to double the investment. However, there's some interesting math going on behind this rule that most people never learn.

In this post I’ll explain the hidden math of the rule of 72, show you how it’s derived algebraically, and point out some important limitations.

Some Background
The key idea behind the rule of 72 is that it’s designed for quick mental calculations. Seventy-two gives a decent ballpark approximation for doubling times. But more importantly, 72 is evenly divisible by lots of common growth rates -- 2, 3, 4, 6, 8, 9, 12, 18, and so on. That makes it ideal for back-of-the-envelope approximations.

However, 72 has some major drawbacks. It's inaccurate in many cases, sometimes giving answers that are off by years. Recognizing this, some textbooks offer more accurate (but harder to use) alternatives. Some recommend a “rule of 70,” while others suggest a “rule of 69.” While these are both harder to use, they're considerably more accurate in many cases. As we'll see in a minute, the rule of 69 is probably the most mathematically defensible of all, despite being pretty useless for mental calculation.

Deriving the Optimal Rule of Thumb
So where do these rules come from? In this section, I'll show you how they're derived algebraically from basic compound interest formulas.

When we say we're looking for a rule of thumb for doubling time, here's what we really mean. We want some constant M (like 72, 70, 69, etc.) that we can divide by our growth rate times 100 and get a good estimate of the time to double an investment. The first step is to write an expression for the value of the investment at time t, given some initial value a0. Then we set this expression equal to twice the initial value, or 2a0. That is

$a_{0} \left ( 1+\frac{r}{k} \right )^{kt}=2a_{0}$

The a0s cancel out on both sides of the above, leaving us with

$\left ( 1+\frac{r}{k} \right )^{kt}=2$

Taking the natural log of both sides and solving for t, we get

$ln \left ( 1+\frac{r}{k} \right )^{kt}=ln 2 \\ \\ \Rightarrow (kt)ln \left ( 1+\frac{r}{k} \right )=ln 2 \\ \\ \Rightarrow t=\frac{ln 2}{(k)ln\left ( 1 + \frac{r}{k} \right )}$

This expression gives the time t it takes to double the investment, given some r and k. What we want is a number M such that when we divide by 100r, it gives us t. That is, we want an M such that M/(100r) = t. Solving for M, we see that M = t(100r). So our rule of thumb M is simply given by 100r times our equation for t above, or

$M = (100r)t = \frac{(100r)ln 2}{(k)ln\left ( 1 + \frac{r}{k} \right )}$

As you can see, the optimal rule of thumb M is a function of both r and k. So for different growth rates and frequency of compounding, you’ll want different rules of thumb. The ubiquitous M=72 is really only a good rule of thumb for a very small number of cases.

The graph below is a plot of the function M(r,k) for r between 1 percent and 30 percent and k between 1 and 30. R and k are shown on the horizontal axes, and the optimal rule M is given on the vertical axis. The colored bands show areas where each rule is optimal. For example, the light green band near the center shows where the rule of 70 is optimal, and the flat red plane shows where the rule of 69 is best, and so on.

As you can see, the optimal rules are all over the map for small values of k. In the extreme case of k=1, the usual rules of 70 and 72 are only close for growth rates between 1 percent and 10 percent. Beyond that they quickly break down. When k=1, the optimal rule M grows linearly as r increases, so no rule holds up well in that range. Overall, 72 looks pretty lousy from this perspective -- it's the small aqua-blue band in the center-left. Seventy looks slightly better -- the green band corresponding to 70 covers much more area -- but it also only works over a pretty small range.

A Special Case: The Rule for Continuous Compounding
The rule that seems to hold up best is actually the awkward rule of 69, which corresponds to the broad, flat plane of red on the right-hand side of the graph. It works for a huge range of values as the function flattens out to the right and roughly approaches 69. Beyond k=15, the graph is essentially flat regardless of the growth rate r. This is no surprise, as the rule of 69 turns out to be a special case for continuously compounded growth rates.

With continuously compounded growth, k is very large. To see how this affects our optimal rule M, let's return to our first equation for the time to double our investment.

If growth is continuously compounded, we have the following equation for the value at time t

$a_{0}e^{rt}=2a_{0}$

Again, the a0s cancel out on both sides, giving us

$e^{rt}=2$

Solving for t we get

$ln (e^{rt})=ln2 \\ \\ \Rightarrow (rt)ln(e)=ln2 \\ \\ \Rightarrow t=\frac{ln2}{r}$

Since we're looking for an M such that M/(100r)=t, or M=(100r)t, we get the following optimal rule M

$M=\frac{(100r)ln2}{r}=(100)ln2=69.314718\hdots$

So with continuously compounded growth, the optimal rule becomes pretty simple. Just divide (100)ln 2 -- or roughly 69.3 -- by your growth rate times 100. In the graph, that’s the value our function M is approaching as it flattens out to the right. Put differently, the limit of M as k approaches infinity is (100)ln 2 or about 69.3.

We can easily see this result analytically by taking the limit of the function M as k grows to infinity. That is, we can find the following limit

$\lim_{k \to \infty}M(r,k)=\lim_{k \to \infty}\left ( \frac{(100r)ln2}{(k)ln\left ( 1+\frac{r}{k} \right )} \right )$

To work out this limit, we need to do some tricky algebra to the denominator. First, let's move the k from the left side up into the exponent on the log to its right. Then let's define a new number x such that x = k/r. This trick will let us collapse most of the denominator into the constant e, using the fact that e equals the limit of (1 + 1/x)^x as x grows to infinity.

Substituting in x = k/r, we get

$\lim_{k%20\to%20\infty}M(r,k)=\lim_{k%20\to%20\infty}\left%20(%20\frac{(100r)ln2}{ln\left%20(%201+\frac{1}{x}%20\right%20)^{xr}}%20\right%20) = \lim_{k%20\to%20\infty}\left%20(%20\frac{(100r)ln2}{(r)ln\left%20(1+\frac{1}{x}%20\right%20)^{x}}%20\right%20)$

Since x = k/r, letting k grow to infinity is equivalent to letting x grow to infinity. But we know that the term in the denominator (1+1/x)^x just becomes e as x grows to infinity. So we have

$\lim_{x \to \infty}\left ( \frac{(100r)ln2}{(r)ln\left ( 1+\frac{1}{x} \right )^{x}} \right )=\left ( \frac{(100r)ln2}{(r)ln(e)}\right )=(100)ln2=69.314718\hdots$

And that's the result we alluded to above. As k grows, the function M converges very quickly to 69.314... So that's the optimal rule for a huge number of values for r and k -- infinitely many. So in some long-run mathematical sense, the most defensible rule actually turns out to be the awkward and practically useless rule of 69.

To see how this argument works in a spreadsheet, here's an Excel file with the equations, data and graph from above.

Posted by Andrew on Tuesday May 26, 2009 | Feedback?

* * *