$$P = E (8.5 + 2G)$$
where P is the value or price, E are current (or normal) earnings and G is the growth rate.
In a footnote, Graham remarks:
He mainly used the formula to show how absurd some implied growth expectations are. In other words:Note that we do not suggest that this formula gives the "true value" of a growth stock, but only that it approximates the results of the more elaborate calculations in vogue.
$$G = \frac{P/E - 8.5}{2}$$
For example, a company with a P/E of 100 has an implied growth rate of about 46% which seems pretty excessive for long-term growth (over the long-term, company growth rates are more or less constrained by overall economic growth, say 2-4%.)
We can derive Graham's formula from a variation of the Gordon Growth Model.
Let's suppose that you have a company which next year will earn E in earnings, grows at a perpetual constant rate g and has a required rate of return (discount rate) R (with R > g). Then it can be shown (with a bit of calculus) that the present value of those earnings is given by:
$$P = \frac{E}{R-g}$$
Next, we'll take the Taylor expansion (with respect to g) of this and just look at the first two terms which will give us a linear approximation of the above formula.
$$P = E (\frac{1}{R}+\frac{1}{R^2}g)$$
To get the 8.5 in Graham's formula, we have to use R~11.75%:
$$P = E (8.5 + 72g)$$
And since Graham's growth rate G = 100g, we can substitute that in to obtain Graham's formula:
$$P = E (8.5 + 0.72G)$$
So it's not an exact match but it's pretty close. Considering that we eliminated a number of the terms from the Taylor expansion, some of which were non-negligible, it's reasonable to conclude that Graham's formula is a decent approximation to a future earnings discount model.
If you're having trouble reading the above post it's due to a latex script which seems to be working in my end. I'm curious if it works in comments as well so I'll give it a try.
ReplyDeleteHere's one way to derive the Gordon's Growth Model.
We assume continuous perpetual growth at rate g and discounted at rate r:
$$P = \int_0^\infty \! E\exp^{(r-g)} \, \mathrm{d}t.$$
$$P = \frac{E}{r-g}$$
Correction, it should read:
ReplyDelete$$P = \int_0^\infty \! E e^{(g-r)} \, \mathrm{d}t. $$
since we're applying $e^g$ growth to the earnings and discounted by a factor of $e^-r$.
Otherwise the Latex appears to be working.