Two-stage DDM: the “never seen” shape

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Fast growth for a few years, then Gordon forever: discount the explicit dividends with NPV, add a terminal value that sits at year n and uses D(n+1), and discount it n periods — not n + 1.

When growth is not constant

A young company grows fast, then settles. The Gordon model cannot take a 20 % growth rate — it is not sustainable and it may exceed $r$ — so the valuation is split in two: an explicit phase of individually forecast dividends, and a terminal value that is the Gordon price of everything after it.

$P_{0} = explicit dividends t = 1 \sum n \frac{D _{t}}{( 1 + r ) ^{t}} + terminal value at year n, discounted \frac{1}{( 1 + r ) ^{n}} \times \frac{D _{n + 1}}{r - g _{2}}$

Worked example — Lakehead Robotics

Lakehead just paid $1.00 * * ($ D_0$). Dividends will grow 20 % a year for three years, then 4 % forever; investors require 10 %**.

Step 1 — the explicit dividends. =1.00*1.2^1 = 1.20; =1.00*1.2^2 = 1.44; =1.00*1.2^3 = 1.728.

Step 2 — the terminal value. At the end of year 3 the share is a Gordon share: its next dividend, $D_{4}$ , is the first of the slow phase — =1.728*(1+0.04) = 1.797 — and its value at year 3 is $D_{4} / (r - g_{2})$ : =1.728*1.04/(0.10-0.04) = 29.952.

Step 3 — discount everything to today. The three dividends, with Excel's NPV (which starts at the end of period 1 — exactly right here): =NPV(0.10,1.20,1.44,1.728) = 3.58. The terminal value is a year-3 number, so it comes back three periods: =29.952/1.1^3 = 22.50.

$P_{0} = 3.58 + 22.50 = 26.08$

The terminal value is 86 % of the price — normal, and the reason its two rules matter.

The two rules of the terminal value

It sits at year $n$ , so discount it $n$ periods. Students see $D_{n + 1}$ in the numerator and discount by $n + 1$ : 29.952 ÷ 1.1⁴ = 20.46, price 24.04 — wrong. The Gordon formula already places the value one period before the dividend it uses; $D_{4} / (r - g)$ is a value at year 3.
It uses $D_{n + 1}$ , the first dividend of the constant-growth phase — $D_{3} \times (1 + g_{2})$ , grown at the new rate. Using $D_{3}$ itself, or growing at 20 % once more, are the neighbouring errors.

The single-stage trap

Run Lakehead through the plain Gordon model at 4 %: 1.00 × 1.04 ÷ 0.06 = 17.33. It throws away three years of 20 % growth and undervalues the share by a third. The paper's "never seen" question is this shape with different numbers; the method is always the same three steps.

Recompute

A five-year fast phase: dividends 1.20, 1.44, 1.728, 2.074, 2.488; TV at year 5 = 2.488 × 1.04 ÷ 0.06 = 43.131; =NPV(0.10,…) = 6.54; TV ÷ 1.1⁵ = 26.78; $P_{0}$ = 33.32. Where does the terminal value sit, cold: at year n — the last year of the fast phase — discounted n periods.

Three explicit dividends and a terminal value that stands at year 3 — the Gordon price of every dividend from year 4 onward. All four pieces come back to t = 0 at 10 %; the terminal value comes back three periods.