Implied Cost Of Equity With Multistage Dividends

Why Use a Multistage Dividend Model?

A company’s dividends may not follow one steady growth rate from year to year. A firm might temporarily retain more earnings, make a planned distribution, or move through a period of unusually rapid growth before settling into a sustainable pattern. In those cases, applying one constant-growth rate to every future dividend can misrepresent the forecast. A multistage dividend model discounts the dividends forecast individually and applies stable growth only after the explicit forecast period.

The model can also be used in reverse. If the current share price and dividend forecasts are known, the implied cost of equity is the required return that makes the present value of those forecast dividends equal to the share price. There is usually no convenient algebraic solution for that return, so it is found numerically.

Definition: Let $P_0$ be the share price at time 0, $D_t$ the dividend paid at the end of year $t$, $N$ the final year of the explicit dividend forecast, $g$ the annual stable dividend growth rate beginning after year $N$, and $k_e$ the annual required return on equity. The implied cost of equity is a value of $k_e$ for which the present value of the forecast dividends and the stable-growth value after year $N$ equals $P_0$.

Set Up the Valuation Equation

Assume dividends are paid annually at each year-end. Forecast $D_1,\ldots,D_N$ individually. From year $N+1$ onward, assume dividends grow at the constant annual rate $g$, so that $D_{N+1}=D_N(1+g)$. Under these assumptions, the value at time $N$ of the dividends from year $N+1$ onward is the stable-growth terminal value:

P_N=\frac{D_{N+1}}{k_e-g} =\frac{D_N(1+g)}{k_e-g}.

The share price at time 0 is the present value of the explicit dividends plus the present value of that terminal value:

P_0=\sum_{t=1}^{N}\frac{D_t}{(1+k_e)^t} +\frac{P_N}{(1+k_e)^N} =\sum_{t=1}^{N}\frac{D_t}{(1+k_e)^t} +\frac{D_N(1+g)}{(k_e-g)(1+k_e)^N}.

This form assumes $k_e\ne g$, $1+k_e\ne0$, and convergence of the stable-growth stream. Specifically, its geometric-series ratio must satisfy $\left|(1+g)/(1+k_e)\right|<1$. Under the usual financial assumptions $k_e>-1$ and $g>-1$, this condition reduces to $k_e>g$. In typical applications the rates are nonnegative, and $k_e>g$ is essential: if the required return does not exceed stable growth, the stable-growth terminal value is not finite under this model.

Timing rule: The terminal value is calculated at time $N$, using the first stable-growth dividend $D_{N+1}$. It is then discounted $N$ years to time 0. Do not discount it $N+1$ years, and do not use $D_N$ as its numerator without applying the growth factor.

Why the Terminal-Value Formula Holds

At time $N$, the first dividend in the stable period, $D_{N+1}$, is one year away. The next is $D_{N+1}(1+g)$, two years away, and so on. Discounting the stable stream to time $N$ gives a geometric series:

P_N= \frac{D_{N+1}}{1+k_e} +\frac{D_{N+1}(1+g)}{(1+k_e)^2} +\frac{D_{N+1}(1+g)^2}{(1+k_e)^3} +\cdots.

The first term is $a=D_{N+1}/(1+k_e)$. Each successive term is the previous term multiplied by $r=(1+g)/(1+k_e)$. When $|r|<1$, the infinite geometric series has sum $a/(1-r)$. Substitution and simplification give:

P_N= \frac{D_{N+1}/(1+k_e)} {1-(1+g)/(1+k_e)} = \frac{D_{N+1}}{k_e-g}.

This derivation also explains why the condition on the ratio matters: without convergence, the discounted stable-growth dividends do not sum to a finite terminal value.

Finding the Implied Return

For a trial value of $k_e$, calculate the right-hand side of the valuation equation and compare it with the observed $P_0$. A root finder changes $k_e$ until the difference is zero. Define that difference as $f(k_e)=\text{model value at }k_e-P_0$. A spreadsheet’s goal-seek or root-finding function can solve $f(k_e)=0$; a calculator can also use bisection or Newton’s method.

With positive dividends and $k_e>g$, increasing $k_e$ reduces the present value of each positive forecast dividend. It also reduces the terminal value at time $N$, because $D_{N+1}/(k_e-g)$ decreases, and reduces the present value of that terminal value. Thus the model value decreases as $k_e$ increases. If trial rates on either side of a range give model values above and below $P_0$, a root is bracketed. In this conventional setting the decreasing relationship gives a unique solution within that bracket.

Worked Example: Solving with Two Forecast Dividends

A hypothetical share sells for $P_0=\$45.45$. Forecast dividends are $D_1=\$2.00$ and $D_2=\$2.40$; starting in year 3, dividends are expected to grow at a stable annual rate of 5%. Estimate the implied cost of equity to two decimal places. Assume year-end annual dividends and that stable growth continues indefinitely.

Method. Use the multistage valuation equation and solve for the rate that equates its value to the market price. This setup avoids treating the year 1 and year 2 dividends as if they already grew at the stable rate.

The first stable dividend is $D_3=2.40(1.05)=\$2.52$. The terminal value at time 2 is therefore $P_2=2.52/(k_e-0.05)$, and the equation to solve is:

45.45=\frac{2.00}{1+k_e} +\frac{2.40}{(1+k_e)^2} +\frac{2.52}{(k_e-0.05)(1+k_e)^2}.

A numerical root search gives $k_e$ approximately 10.00%. Verify by substituting 10.00%:

\begin{aligned} D_3&=2.40(1.05)=2.52,\\ P_2&=\frac{2.52}{0.10-0.05}=50.40,\\ \text{PV}(D_1)&=\frac{2.00}{1.10}=1.8182,\\ \text{PV}(D_2)&=\frac{2.40}{1.10^2}=1.9835,\\ \text{PV}(P_2)&=\frac{50.40}{1.10^2}=41.6529,\\ \text{Model value}&=1.8181818+1.9834711+41.6528926 =45.4545455. \end{aligned}

The model value rounds to $\$45.45$ to the nearest cent, matching the market price. The implied cost of equity is 10.00% to two decimal places. The small difference between the unrounded model value and the stated price reflects rounding the price to cents.

Worked Example: Four Cash-Flow Components and a Numerical Root

A hypothetical company has a share price of $\$20.59$. Its forecast dividends are $D_1=\$1.00$, $D_2=\$1.40$, and $D_3=\$1.60$. After year 3, dividends are expected to grow at 4% annually. Estimate the implied cost of equity to two decimal places. Assume annual year-end dividends.

Method. First form the terminal value at year 3, then discount it along with all three forecast dividends. A numerical root search is required because $k_e$ appears in several powers and in the terminal-value denominator. This method avoids discounting the terminal value from the wrong date.

The first stable dividend is $D_4=1.60(1.04)=\$1.664$. The time-3 terminal value is $P_3=D_4/(k_e-0.04)$. The equation is:

20.59= \frac{1.00}{1+k_e} +\frac{1.40}{(1+k_e)^2} +\frac{1.60}{(1+k_e)^3} +\frac{1.664}{(k_e-0.04)(1+k_e)^3}.

Numerical solution gives $k_e$ approximately 11.00%. At 11.00%, the terminal value and discounted components are:

\begin{aligned} P_3&=\frac{1.664}{0.11-0.04}=23.7714,\\ \text{PV}(D_1)&=\frac{1.00}{1.11}=0.9009,\\ \text{PV}(D_2)&=\frac{1.40}{1.11^2}=1.1363,\\ \text{PV}(D_3)&=\frac{1.60}{1.11^3}=1.1699,\\ \text{PV}(P_3)&=\frac{23.7714286}{1.11^3}=17.3815,\\ \text{Model value}&=0.9009009+1.1362714+1.1699062+17.3814638\\ &=20.5885423. \end{aligned}

The model value rounds to $\$20.59$ to the nearest cent, so the implied cost of equity is 11.00% to two decimal places. The terminal value is discounted by $1.11^3$, not $1.11^4$, because it is measured at the end of year 3.

Worked Example: Checking the Transition Year

A hypothetical share sells for $\$30.79$. Forecast dividends are $D_1=\$1.25$ and $D_2=\$1.60$, and stable growth of 3% begins in year 3. Estimate the implied cost of equity to two decimal places. Assume year-end annual dividends.

Method. Locate the transition carefully: year 2 is the final explicit forecast year, so the terminal value is at time 2 and uses $D_3$. A numerical root search is used to match that value to the price.

At a trial rate of 8.00%, the first stable dividend is $D_3=1.60(1.03)=\$1.648$, and the terminal value at year 2 is:

P_2=\frac{1.648}{0.08-0.03}=32.96.

The corresponding present values are:

\begin{aligned} \text{PV}(D_1)&=\frac{1.25}{1.08}=1.1574,\\ \text{PV}(D_2)&=\frac{1.60}{1.08^2}=1.3717,\\ \text{PV}(P_2)&=\frac{32.96}{1.08^2}=28.2579,\\ \text{Model value}&=1.1574074+1.3717421+28.2578850=30.7870345. \end{aligned}

The model value rounds to $\$30.79$, so the numerical solution is 8.00% to two decimal places.

Common mistake and correction. Discounting the terminal value by $1.08^3$ would give $32.96/1.08^3$, incorrectly treating a value at year 2 as though it were at year 3. The correct present value is $32.96/1.08^2$, because the terminal value is dated at the end of the final explicit forecast year.

Interpreting the Estimate and Avoiding Traps

An implied cost of equity is conditional on the inputs: the share price, the explicit dividend forecasts, the stable growth rate, and the timing convention. It is not a directly observed return. A different stable growth assumption can change the terminal value substantially, so a reported estimate should be accompanied by its key assumptions.

Use $D_{N+1}$ in the terminal value. If the forecast ends at $D_N$, grow that dividend once to obtain the first dividend in the stable period.
Date the terminal value correctly. $P_N$ is a value at time $N$, so discount it $N$ periods to time 0.
Check the stable-growth condition. Under usual financial assumptions, $k_e$ must exceed $g$; otherwise the perpetual stream does not have a finite value under the model.
Substitute the answer back. Calculate each present value and compare their total with $P_0$. This catches numerical-root, timing, and rounding errors.
Do not overstate precision. A numerical answer cannot be more reliable than the price and forecast assumptions that produced it.

If all modeled dividends are positive and the rate is restricted to $k_e>g$, the valuation equation declines as the required return rises. A numerical solver should therefore be used within a plausible bracket that satisfies the stable-growth condition. If the model value does not cross the observed price over the relevant range, recheck the inputs and assumptions rather than reporting an unverified root. This technique is aligned with the SOA VEE Accounting and Finance syllabus.

Check Your Understanding

For each problem, assume annual dividends paid at year-end and stable growth continuing indefinitely where specified. Round implied returns to two decimal places.

A hypothetical share is priced at $\$26.70$. Forecast $D_1=\$1.10$ and $D_2=\$1.40$, followed by stable growth of 4% from year 3. Estimate the implied cost of equity using a numerical method. (Answer: 9.00%)
A share price is $\$35.80$. Forecast $D_1=\$2.00$ and $D_2=\$2.20$, then assume stable growth of 2% beginning in year 3. Set up the terminal value at year 2 and estimate the implied cost of equity. (Answer: 8.00%)
At the end of year 4, a firm’s last explicit forecast dividend is $D_4=\$2.80$. Stable growth begins in year 5 at 3%, and the required return is 8%. Calculate $D_5$ and the terminal value at year 4. (Answer: $D_5=\$2.884$; terminal value $=\$57.68$)
A firm has $D_1=\$1.50$, $D_2=\$1.80$, and stable growth of 3% beginning in year 3. At a trial required return of 8%, calculate the terminal value at year 2 and state the power by which it must be discounted to find its present value. (Answer: terminal value $=\$37.08$; discount by $(1.08)^2$)
For a stable-growth stream with $k_e=-2$ and $g=-3$, does $k_e>g$ alone ensure convergence? Apply the absolute-value condition for the geometric-series ratio. (Answer: No; $\left|(1+g)/(1+k_e)\right|=2$, so the series diverges.)

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