John Huber wrote a series of wonderful articles on the relationship between ROIC and the long term rate of return on compounding machines. It fleshes out a key idea that Charlie Munger once summarized as:

Over the long term, it’s hard for a stock to earn a much better return that the business which underlies it earns.

That is, in the long run ROIC = IRR.

If you like spreadsheets, you can plug in some numbers, and try out a few cases, as John does so skillfully on his webpage. It is really not too hard to convince yourself.

Different people like their truths delivered to them in different formats. Some like pictures, some like numbers, some like stories, and so on. I like math, and whenever possible, I prefer my fundamental truths delivered to me in the equation form.

Let’s try to derive the mathematical equation underlying Munger’s observation.

Suppose at time t = 0, you have a business with book value {B_0}, and cash earnings {E_0}. For simplicity, assume the company has no debt and ROIC = ROE = {\alpha}. Thus, {E_0/B_0 = \alpha}.

Let us further assume that the company shoves all its earnings into the business (reinvestment rate = 100%), and all that additional money can be deployed at the average ROE to generate more earnings. The earnings pile on and increase the book value in a steady fashion.

The growth rate in book value and earnings is equal to the ROE.

This means that in year {n}, the earnings are {E_n = E_0 (1 + \alpha)^n} and the book value is {B_n = B_0 (1 + \alpha)^n}.

Suppose we buy the business at a price {P_0} at time t = 0, and sell it for {P_n} after {n} years. What is our rate of return {r}?

The math is simple:

\displaystyle \begin{array}{rcl} P_n & = & P_0 (1 + r)^n \\ (P_n/E_n) E_0 (1 + \alpha)^n & = & (P_0/E_0) E_0 (1 + r)^n\\ \left(\dfrac{P_0/E_0}{P_n/E_n}\right) & = & \left(\dfrac{1+\alpha}{1+r}\right)^n\\ \left(\dfrac{\text{PE}_0}{\text{PE}_n}\right)^{1/n} & = & \left(\dfrac{1+\alpha}{1+r}\right)\\ \end{array}

Time for some elementary calculus. As {n \rightarrow \infty}, the LHS = 1. Any finite number raised to 0 is 1.

Thus, {1+\alpha = 1 + r}, which means, … DING DING DING DING

\displaystyle \boxed{r = \alpha}


Suppose I buy a security with ROE = 20% at an initial PE ratio of {\text{PE}_0} and sell it at a final PE ratio of {\text{PE}_n}, {n} years later.

What does my rate of return look like as a function of time {n}?


If the buy and sell PE ratio is the same, then my rate of return is {\alpha} regardless of how long I hold the stock. Clearly, if I can buy it at a lower PE and sell it at a higher PE multiple, then IRR > ROIC. Conversely, if you buy at a high multiple, you might have to hold on for nearly two decades to get a 15% rate of return.

You might think this is bad. But what is worse is is buying a security with poor ROE at a high price. You might wait 15 years, and barely break even. Warren Buffett said,“Time is the friend of a wonderful business and the enemy of a poor business”.


If you hold on to a high ROE business for a long time, you might not do great, but you will do fine, even if you pay up for quality. If you catch a poor business sufficiently cheap, you should probably flip it over as soon as the market gives you an opportunity.

The former are classic Warren Buffett high-quality, large moat companies.

The latter are Ben Graham cigar butts.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s