# Logic behind Gordon Growth Model in a DCF analysis?

Sorry, I wanted to ask this on the finance/money forum, but they don't support LaTeX there.

Let's say we are valuing a company using the DCF methodology with a 5-year projection period.

We project free cash flows of $F_{1},\ldots,F_{5}$. Then if $w$ is the WACC of this company and $g$ is the perpetual growth rate from year 5 forward, the sum of the future cash flows discounted at $w$ is

$$V_{1}:=F_{1}(1+w)^{-1}+\ldots+F_{5}(1+w)^{-5}+F_{5}\sum_{t=6}^{\infty}\frac{(1+g)^{t-5}}{(1+w)^{t}}.$$

This formula for the Gordon Growth model replaces the infinite sum with the easily computed geometric series $$F_{5}\sum_{t=1}^{\infty}\frac{(1+g)^{t}}{(1+w)^{t}}=F_{5}\frac{1+g}{w-g},$$ and therefore (basically) DOUBLE COUNTS (!!) the cash flows $F_{1},\ldots,F_{5}$ to get \begin{align*} V_{2}&:=F_{1}(1+w)^{-1}+\ldots+F_{5}(1+w)^{-5}+F_{5}\sum_{t=1}^{\infty}\frac{(1+g)^{t}}{(1+w)^{t}}\\ &=F_{1}(1+w)^{-1}+\ldots+F_{5}(1+w)^{-5}+F_{5}\frac{1+g}{w-g}\\ &\gg V_{1}.\end{align*}

What am I missing here?

EDIT

Even if you could convince me of the legitimacy of $F_{5}(1+g)^{t-5}\mapsto F_{1}(1+g)^{t}$ in order to get a uniformly indexed sum (and hence a geometric series), i.e. $F_{5}$ equals the 6-fold growth of $F_{0}$ before we first start to sum it, I would still be very hard to convince of the legitimacy that we should also not truncate the series and re-index the sum at $t=1$.

Your last cash flow is not correctly expressed as you forgot the $(1+r)^{-5}$ when you reinjected.

A $t= 5$ (in 5 years), your PV of the remaining cash flows is: $F_5 \sum_{k=1}^\infty (\frac{1+g}{1+r})^k$. That is the formula for receiving a cash-flow $F_5$ growing at $1+g$, discounted at $(1+r)$ each year, receiving the first cash-flow in year 6.

Now discount that to the present, you need to multiply by $\frac{1}{(1+r)^5}$. If you think that $F_5 = (1+g) ^5 F_0$ and the same for the previous ones, you easily demonstrate that

$V_{1}:=F_{1}(1+r)^{-1}+\ldots+F_{5}(1+r)^{-5}+F_{5}\sum_{t=6}^{\infty}\frac{(1+g)^{t-5}}{(1+r)^{t}} = \sum_{t=1}^{5} F_0 (\frac{1+g}{1+r})^t + F_0 (\frac{1+g}{1+r})^5\frac{(1+g)}{(r-g)} = F_0 \frac{1+g}{r-g}$

• Your question doesn't resolve the problem. First, you are assuming my first comment from my edit: you are assuming in the second "equality" that $F_{i}=F_{1}(1+g)^{i}$ for $1<i\leq 5$. If we accept this approximation, then my $V_{1}$ becomes your $V_{1}$ and I am satisfied. However, see investopedia.com/university/dcf/dcf4.asp as an example of what every DCF tutorial does: they add $V_{1}$ and the cash flows $F_{1},\ldots,F_{5}$ (after discounting), so that those cash flows are then double-counted. – Sargera Jul 16 '14 at 17:48
• Actually, even worse, they discount the terminal value by the WACC, even though the terminal value was computed from individually discounted cash flows! So they double-discount it appears... – Sargera Jul 16 '14 at 17:52
• Do you agree with the terminal value ? The trick is that they transform the terminal value "as" a cash flow in year 5. Then, like any cash flow you discount it to today. There is absolutely no relationship between the cashflow $F_1$ etc. and the exit value. – Matt B. Jul 16 '14 at 19:37

What you're missing in your interpretation is what the sixth payment actually is. It is not the fifth cash flow or even the value of the fifth cash flow. It is the value of an annuity at t5 whose first payment occurs in one years/time periods time and is F5*(1+g). Or the last known dividend grown by one years worth of the growth rate. It is important to remember that the value of a annuity is calculated based on the assumption that the first payment occurs at t+1 and not t.