Valuation of growing streams of cash flows

A look at geometric and arithmetic annuities with a soupçon of stock valuation

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A few weeks back, I derived the monthly mortgage payment formula. The mathematical star of the show was a geometric series of 360 terms, which can be written (surprisingly) as a compact, closed formula. The monthly payment amount is calculated by insisting that the present value of the mortgage payments—discounted by the monthly interest rate—is equal to the loan amount. Things simplify nicely because the payment amount is constant. However, there are many cases where you might have a variable stream of cash flows. Valuation is the process of determining the fair value of an asset. To value a generic stream of cash flows (CF) occurring at fixed time intervals, we find its present value based on a prescribed discount rate r:

\(PV=\displaystyle\sum_{i=1}^N\frac{CF_i}{(1+r)^i}\)

If each CF_i is the same (as with a mortgage), you can factor that term out and directly apply the formula for a geometric series to the sum. If the cash flows change over time, then all bets are off. There are, however, a few cases where you can have non-constant cash flow streams that still have formulas for the sums. That’s what I want to explore in this post.

Arithmetic annuities

Note: This section uses calculus. If you find that you’re miserable while reading this, please feel free to stop and move to geometric growth annuities. There are no dependencies between the two sections, and the math is gentler in the next section.

The first stream of cash flows that I’ll consider is an arithmetic annuity. This refers to a payment occurring at fixed intervals, which increases each period by a fixed amount d. In other words, if the first payment is P, then the second payment is P + d, the third is P + 2d, and so on. These annuities arise most frequently in leases, rental agreements, membership dues, and insurance payments. Let’s derive a formula for the present value of an arithmetic annuity of n payments with interest rate r and payment step d. I’ll assume that the first payment occurs at the end of the first period, although it wouldn’t be difficult to adjust the formula if it occurs at a different time. The sum we need to evaluate is:

\(\begin{aligned}PV&=\displaystyle\sum_{i=1}^{n}\frac{P+(i-1)d}{(1+r)^i}\\[3pt] &=\frac{P}{1+r}+\frac{P+d}{(1+r)^2}+\dots+\frac{P+(n-1)d}{(1+r)^{n}}\end{aligned}\)

I’ll tackle this sum by splitting it into two parts: one with the fixed payment, and one with the variable part. The fixed payment part is a pure geometric series, the formula for which you can find in my previous post on the topic.

\(\begin{aligned}PV&=\displaystyle\sum_{i=1}^{n}\frac{P+(i-1)d}{(1+r)^i}\\[3pt] &=\displaystyle\sum_{i=1}^{n}\frac{P}{(1+r)^i}+\displaystyle\sum_{i=1}^{n}\frac{(i-1)d}{(1+r)^i}\\[3pt]&\text{ (Split up the sum)}\\[3pt] &= P\cdot\frac{\frac{1}{(1+r)^{n+1}}-1}{\frac{1}{1+r}-1}+\displaystyle\sum_{i=1}^{n}\frac{(i-1)d}{(1+r)^i}\\[3pt]&\text{ (Geometric series formula)}\\[3pt] &= P\cdot\frac{1-\frac{1}{(1+r)^{n}}}{r}+d\cdot\displaystyle\sum_{i=1}^{n}\frac{(i-1)}{(1+r)^i}\\[3pt]&\text{ (Algebra)} \end{aligned}\)

Now we just have to figure out the second sum in the final equation. Let’s call that sum S and write out the terms (omitting d for now).

\(\begin{aligned} S&= \displaystyle\sum_{i=1}^n\frac{i-1}{(1+r)^i}\\[3pt] &=\frac{1}{(1+r)^2}+\frac{2}{(1+r)^3}+\dots+\frac{n-1}{(1+r)^n} \end{aligned}\)

This is where the magic happens. Notice that numerator and exponent are off by 1 in each term in the sum. Those of you who have experienced the thrill of calculus may be reminded of the power rule for taking derivatives. We can rewrite S as

\(\begin{aligned} S&= \displaystyle\sum_{i=1}^n\frac{i-1}{(1+r)^i}\\[3pt] & = \displaystyle\sum_{i=1}^{n-1}\frac{d}{dr}\left(-\frac{1}{(1+r)^i}\right)\\[3pt] &=-\frac{d}{dr}\left(\frac{1}{(1+r)}+\dots+\frac{1}{(1+r)^{n-1}}\right). \end{aligned}\)

The negative sign appears because 1/(1 + r)^n is the same as (1+r)^(-n). The power rule says that the derivative of that term is (-n)*(1+r)^(-n-1), which is -n/(1+r)^(n+1). Now, notice that the series that we’re differentiating is itself geometric, so we can apply the geometric series formula before taking the derivative.

\(\begin{aligned} S&= -\frac{d}{dr}\displaystyle\sum_{i=1}^{n-1}\frac{1}{(1+r)^i}\\[3pt] &=-\frac{d}{dr}\left(\frac{1-(1+r)^{-(n-1)}}{r} \right)\\[3pt] &= \left(\frac{1}{r^2}-\frac{n-1}{r(1+r)^n}-\frac{1}{r^2(1+r)^{n-1}} \right). \end{aligned}\)

The last equality is a result of the quotient rule. (This is turning into quite the trip down memory lane.) Now we just have to combine the formulae for the constant and increasing parts:

\(\begin{aligned}PV&=\displaystyle\sum_{i=1}^{n}\frac{P}{(1+r)^i}+\displaystyle\sum_{i=1}^{n}\frac{(i-1)d}{(1+r)^i}\\[3pt] &= P\cdot\frac{1-\frac{1}{(1+r)^{n}}}{r}\\[3pt] &+d\left(\frac{1}{r^2}-\frac{n-1}{r(1+r)^n}-\frac{1}{r^2(1+r)^{n-1}} \right). \end{aligned}\)

I never said it would be a nice formula. But it is a closed formula. Actuaries developed a special notation for formulas like this to make them easier to manipulate. Before moving to the next topic, I want to emphasize that sometimes you need to do a little work before applying this formula. For example, maybe your apartment lease has a fixed monthly payment throughout the year, but each year at renewal it goes up by $100/month. To find the present value of five years of rent payments, you would apply the formula to each calendar month separately and then sum the 12 resulting present values.

Cash flows with geometric growth

The other category of variable cash flows that we can handle are those with geometric growth. This means that there is a fixed growth rate g, so that each payment is (1 + g) times the previous one. As before, let’s assume that there are n payments, the first of which occurs at the end of the first period. With an initial payment of P, we therefore need to sum the following series:

\(\begin{aligned} PV&=\frac{P}{(1+r)}+\frac{P(1+g)}{(1+r)^2}+\dots+\frac{P(1+g)^{n-1}}{(1+r)^n}\\[3pt] &=\displaystyle\sum_{i=1}^n \frac{P(1+g)^{i-1}}{(1+r)^i}. \end{aligned}\)

The math is going to be a lot easier this time around (i.e., no calculus needed). What we’ll do is factor out one copy of (1 + r) from the denominator. The payoff of doing so is that the exponents in the numerator and denominator will then match. That allows us to define x = (1 + g)/(1 + r) and apply the geometric series formula directly:

\(\begin{aligned} PV&=\displaystyle\frac{1}{(1+r)}\sum_{i=1}^n \frac{P(1+g)^{i-1}}{(1+r)^{i-1}}\\[3pt] &= \frac{P}{(1+r)}\sum_{i=0}^{n-1} x^i\\[3pt] &= \frac{P}{(1+r)}\cdot\frac{1-x^n}{(1-x)}. \end{aligned}\)

The last thing to do is get rid of x, using the formula

\(\begin{aligned}1-x&=1-\frac{1+g}{1+r}\\[3pt] &= \frac{1+r}{1+r}-\frac{1+g}{1+r}\\[3pt] &= \frac{r-g}{1+r}. \end{aligned}\)

Now put it all together:

\(\begin{aligned} PV&=\frac{P}{(1+r)}\cdot\frac{1-x^n}{(1-x)}\\[3pt] &=\frac{P}{(1+r)}\frac{1-\left(\frac{1+g}{1+r}\right)^n}{\frac{r-g}{1+r}}\\[3pt] &= \frac{P-P\cdot \left(\frac{1+g}{1+r}\right)^n}{r-g}. \end{aligned}\)

(Note that this formula doesn’t apply if g = r. In that case, the sum is n*P/(1+r).) Typically, this formula is used when the growth rate is less than the discount rate. Because of the term (r - g) in the denominator, the present value can get big really quickly when the rates are close to each other.

Gordon growth model

Geometric annuities are very useful in the valuation of companies and stocks. While there are many methods of stock valuation, the most appealing to me (as a first principles guy) is the dividend discount model, which says that the value of a stock is given by the present value of all future dividend payments. It’s hard to argue with that premise, right? The problem is that you have to estimate what those dividends will be. One thing people do is assume that dividends will grow at a constant rate. This may be defensible if the company’s growth has stabilized. Concretely, we assume that the initial payment P is today’s dividend amount D. Analysts then decide on an appropriate discount rate r and dividend growth rate g.

Under the constant growth assumption, the present value calculation looks just like the geometric annuity sum above. The one difference is that I considered finite sums (i.e., n total cashflows). With stock valuation, you have to consider the dividends as a perpetuity, meaning that dividend payments go on forever. (Even if you expire, the stock will continue to pay dividends, which means it has value to someone.) If the idea of unbounded dividend growth seems unrealistic, remember that it is also competing with the time value of money: the further away the payments are, the less value they have today. There’s also inflation, so those sums won’t seem astronomical on the day that they arrive. This is why the assumption that g < r is critical. If that’s the case, then you can take the limit as n goes to infinity in the geometric series formula. Doing so yields the Gordon growth model (GGM) for stock valuation:

\(\text{Price}=\frac{D}{r-g}.\)

If you’re very confident in the selection of r and g, you can use this calculation to decide if a stock is priced appropriately. If you’re less confident, you can start with the market price and try to back out r or g to see the implied assumption that The Market is making.

You can also apply the Gordon growth model to cash flow growth, which is used in the discounted cash flow model for company valuation (e.g., investment banking). The paradigm there is to do detailed financial statement forecasts over some period—say, 10 years—and then calculate the terminal value based on GGM and an assumption about future cash flow growth. If you do that exercise a few times, you’ll see that the terminal value tends to be a huge chunk of the total firm value. The reason is that that the denominator (r - g) can be very tiny if r and g are close. You can insulate yourself from assumptions about g by extending your detailed forecasts further into the future. Since you have to discount the terminal value to the present, doing 15 years of financial statement forecasts instead of 10 decreases the size of the terminal value by a factor of (1 + r)^5. (Of course, forecasting financial statements comes with its own set of assumptions.)

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