Sensitivity of Gordon Value to Growth and Return

Why Sensitivity Matters

A stock value from the Gordon growth model can change substantially when its inputs change. This is especially important when the required return is only slightly above the assumed dividend growth rate: the model divides by the difference between those rates, so a small adjustment to either input can noticeably change the denominator. Sensitivity analysis makes that effect measurable.

The calculations below distinguish a local sensitivity, measured by a derivative at a particular set of inputs, from an exact price change, found by recalculating the model after an input changes. They also make explicit whether the next dividend or the current dividend is being held fixed. That condition matters because changing growth can also change the forecast dividend.

Definition: In the constant-growth Gordon model, $P_0$ is the value at time 0, $D_1$ is the dividend expected at the end of year 1, $r$ is the annual required return, and $g$ is the annual constant dividend growth rate. Under the model’s assumptions, $$ P_0=\frac{D_1}{r-g}, \qquad r>g. $$ Rates must use the same time unit and be expressed consistently, such as annual decimal rates. This sensitivity analysis assumes the model’s dividends grow at the constant rate $g$ indefinitely and are discounted at the constant required return $r$. The formula and its derivatives do not give a finite conventional value when $r\leq g$.

Local Sensitivity to Growth and Required Return

First hold $D_1$ fixed and differentiate the value with respect to each rate. The partial derivative with respect to $g$ measures the approximate dollar change in $P_0$ for a small change in growth, with $r$ and $D_1$ unchanged. The derivative with respect to $r$ has the analogous interpretation, with $g$ and $D_1$ unchanged.

\frac{\partial P_0}{\partial g}=\frac{D_1}{(r-g)^2}, \qquad \frac{\partial P_0}{\partial r}=-\frac{D_1}{(r-g)^2}.

For a positive $D_1$, the growth derivative is positive: increasing $g$ while holding the other inputs fixed increases the model value. The return derivative is negative: increasing $r$ reduces the value. Both sensitivities grow in magnitude as $r-g$ narrows. The derivatives are measured in dollars per unit change in the rate. For example, a one-percentage-point change is $0.01$, not 1, so a local dollar estimate is the derivative multiplied by $0.01$.

A short derivation explains both signs and the increasing sensitivity. Write the model as $D_1(r-g)^{-1}$. Differentiating with respect to $g$ gives a positive derivative because differentiating $r-g$ contributes a factor of $-1$, which cancels the negative power’s sign. Differentiating with respect to $r$ gives a negative derivative:

\frac{\partial}{\partial g}\left[D_1(r-g)^{-1}\right] =D_1(r-g)^{-2}, \qquad \frac{\partial}{\partial r}\left[D_1(r-g)^{-1}\right] =-D_1(r-g)^{-2}.

The squared denominator shows why the sensitivity rises rapidly as the spread shrinks. For example, halving $r-g$, with $D_1$ unchanged, makes either derivative’s absolute value four times as large. This is a local statement: for a sizeable rate change, the derivative at the starting point may not accurately predict the full price change.

Useful exact comparison: If $D_1$ is fixed and the spread $r-g$ changes from $s$ to $s+\Delta s$, then the old and new values are $D_1/s$ and $D_1/(s+\Delta s)$. Thus, $$ \frac{P_{0,\mathrm{new}}}{P_{0,\mathrm{old}}} =\frac{s}{s+\Delta s}. $$ This exact ratio shows that the same absolute change in the spread has a larger proportional effect when the starting spread $s$ is small.

Worked Examples

Worked Example: Estimating and Recalculating a Growth Change

A hypothetical share is valued using an expected next-year dividend of $D_1=\$3.60$, a required return of $r=10\%$, and a constant growth rate of $g=4\%$. Estimate the effect of increasing growth to $5\%$ using the local derivative, then calculate the exact new value. Keep $D_1$ fixed in both calculations.

Method and starting value. Convert the rates to decimals. The starting spread is $0.10-0.04=0.06$, so $$ P_0=\frac{\$3.60}{0.06}=\$60.00. $$

Local estimate. At the starting inputs, the growth derivative is $$ \frac{\partial P_0}{\partial g} =\frac{\$3.60}{(0.06)^2} =\frac{\$3.60}{0.0036} =\$1{,}000 $$ per unit change in the growth rate. The increase in growth is $0.05-0.04=0.01$, so the local estimate of the price increase is $\$1{,}000(0.01)=\$10.00$. This estimates a new value of $\$70.00$.

Exact recalculation. At $g=5\%$, the spread is $0.10-0.05=0.05$. Therefore, $$ P_{0,\mathrm{new}}=\frac{\$3.60}{0.05}=\$72.00. $$ The exact increase is $\$72.00-\$60.00=\$12.00$, which is larger than the local estimate because the change is not infinitesimal and the denominator has narrowed.

Check and trap avoided. The exact value is higher than the derivative estimate, as expected for a growth increase under this convex model. Treating one percentage point as a change of 1 would produce an erroneous estimate of $\$1{,}000$, rather than $\$10.00$.

Worked Example: Comparing Changes When Rates Are Close

A hypothetical company has $D_1=\$2.40$, $r=9\%$, and $g=7\%$. Calculate the initial value and the effect of (a) increasing growth to $7.5\%$, or (b) increasing the required return to $9.5\%$. For each change, compare the exact repricing with the local derivative estimate. Hold $D_1$ fixed.

Method and initial value. The starting spread is $0.09-0.07=0.02$. Thus, $$ P_0=\frac{\$2.40}{0.02}=\$120.00. $$ At this point, the absolute value of either derivative is $$ \frac{\$2.40}{(0.02)^2} =\frac{\$2.40}{0.0004} =\$6{,}000 $$ per unit change in a rate.

(a) Growth increase. The rate change is $0.075-0.070=0.005$. The derivative predicts an increase of $\$6{,}000(0.005)=\$30.00$. The exact new spread is $0.09-0.075=0.015$, giving $$ P_{0,\mathrm{new}}=\frac{\$2.40}{0.015}=\$160.00. $$ The exact increase is $\$160.00-\$120.00=\$40.00$.

(b) Required-return increase. The rate change is $0.095-0.090=0.005$. The return derivative predicts a decrease of $\$6{,}000(0.005)=\$30.00$. The exact new spread is $0.095-0.070=0.025$, giving $$ P_{0,\mathrm{new}}=\frac{\$2.40}{0.025}=\$96.00. $$ The exact decrease is $\$120.00-\$96.00=\$24.00$.

Check and trap avoided. The initial spread is only two percentage points, so even a half-percentage-point input change has a substantial effect. The local estimates use the initial derivative and need not match exact changes. Using the starting derivative as if it were an exact repricing rule would give $\$150.00$ for part (a) or $\$90.00$ for part (b), neither of which is the exact value.

Worked Example: Distinguishing a Fixed Next Dividend from a Fixed Current Dividend

A hypothetical firm has just paid a current dividend $D_0=\$2.00$. Its required return is $r=10\%$, and its initial constant growth assumption is $g=4\%$. Find the model value, then find the value if growth rises to $5\%$, holding $D_0$, rather than $D_1$, fixed. Compare the change with the result that would follow from incorrectly holding $D_1$ fixed.

Method and initial value. When the current dividend is supplied, first calculate the next dividend using $D_1=D_0(1+g)$. Initially, $$ D_1=\$2.00(1.04)=\$2.08, \qquad P_0=\frac{\$2.08}{0.10-0.04} =\frac{\$2.08}{0.06} =\$34.6667. $$ The initial value is $\$34.67$ to the nearest cent.

Revalue with $D_0$ fixed. At $g=5\%$, the next dividend becomes $$ D_1=\$2.00(1.05)=\$2.10. $$ The new spread is $0.10-0.05=0.05$, so $$ P_{0,\mathrm{new}}=\frac{\$2.10}{0.05}=\$42.00. $$ The increase is $\$42.00-\$34.6667=\$7.3333$, or $\$7.33$ to the nearest cent.

Check and trap avoided. If $D_1=\$2.08$ were incorrectly held fixed, the new value would be $\$2.08/0.05=\$41.60$, understating the correct value by $\$0.40$. Changing $g$ changes the forecast $D_1$ when $D_0$ is fixed. The fixed-$D_1$ partial derivative alone does not include that dividend effect.

Interpreting the Results Carefully

The partial derivatives are useful for ranking and explaining input sensitivity, but they must be interpreted with their “holding other inputs fixed” condition. In the first two examples, $D_1$ was fixed by design. In the third, $D_0$ was fixed, so the forecast next dividend changed with $g$. More generally, if $D_0$ is fixed, substitute $D_1=D_0(1+g)$ before differentiating: $$ P_0=\frac{D_0(1+g)}{r-g}, \qquad \frac{\partial P_0}{\partial g} =\frac{D_0(1+r)}{(r-g)^2}. $$ This derivative includes both the effect of growth on the denominator and its effect on next year’s dividend. It is not the same question as changing $g$ while holding $D_1$ fixed.

For small simultaneous changes in $r$ and $g$, with $D_1$ fixed, the differential approximation is $$ \Delta P_0\approx \frac{D_1}{(r-g)^2}(\Delta g-\Delta r). $$ This is an approximation around the original inputs, not a substitute for recalculating the model when the changes are large. If growth and required return rise by the same small amount, the spread is unchanged, so the model value is unchanged under the fixed-$D_1$ condition. If $D_1$ also changes, that conclusion no longer follows.

A common exam trap is to report a derivative as though it were a percentage effect or an exact dollar effect. The derivative is in dollars per unit rate. Multiply it by the rate change expressed in decimals to get a local dollar estimate; then, when asked for the resulting model value, reprice using the new inputs. Another trap is to overlook the narrow-spread condition: when $r-g$ is small but positive, modest input uncertainty can dominate the result. If assumptions imply $r\leq g$, the standard constant-growth formula is not applicable as a finite valuation.

Key takeaway: With $D_1$ fixed, growth raises Gordon value and required return lowers it. Both effects become more pronounced as $r-g$ narrows. State what dividend is held fixed, use decimal rate changes, and distinguish a derivative-based local estimate from an exact recalculation.

Check Your Understanding

For each problem, state any intermediate rate changes in decimal form and distinguish local estimates from exact values where requested.

A hypothetical share has $D_1=\$2.40$, $r=11\%$, and $g=5\%$. Find its current value, its exact value if $g$ rises to $6\%$, and the local derivative estimate of the price increase at the original inputs. (Answer: Current value $\$40.00$; new value $\$48.00$; exact increase $\$8.00$; local estimate $\$6.67$.)
For $D_1=\$4.00$, $r=10\%$, and $g=8\%$, calculate the starting value and the exact value if $r$ increases to $10.5\%$, holding $D_1$ fixed. What is the local estimated decrease? (Answer: Starting value $\$200.00$; new value $\$160.00$; exact decrease $\$40.00$; local estimated decrease $\$50.00$.)
A share has $D_1=\$3.00$, $r=12\%$, and $g=4\%$. Find its value, then calculate the exact value and percentage increase if growth rises to $5\%$, with $D_1$ fixed. (Answer: Initial value $\$37.50$; new value $\$42.86$; increase $\$5.36$, or $14.2857\%$ before rounding.)
A hypothetical firm has $D_1=\$1.80$, $r=9\%$, and $g=5\%$. Calculate the value before and after growth falls to $4\%$, holding $D_1$ fixed, and find the percentage decrease. (Answer: Initial value $\$45.00$; new value $\$36.00$; decrease $\$9.00$, or $20\%$.)
A firm has current dividend $D_0=\$2.00$, required return $r=10\%$, and growth $g=4\%$. Revalue it after growth increases to $5\%$, holding $D_0$ fixed, and calculate the increase. (Answer: Initial value $\$34.67$; new value $\$42.00$; increase $\$7.33$, to the nearest cent.)

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