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I am doing an exercise (I don't know for how long now) that I just can't understand, worst, it's quite simple.

opportunity cost for equity holders = $10\%$

Stock C is expected to pay a dividend of $\$5$ next year. Thereafter, dividend growth is expected to be $20\%$ a year for five years (i.e., years 2 through 6) and zero thereafter. PV? The guy gave us the solution:

(1) PV growing annuity: $PV = 5 [ (1/ 0.10 − 0.20) − (1.20^5 /(0.10 − 0.20) × 1.105^5) ] = 27.25$

(2)PV perpetuity: $5\cdot 1.20^5/0.10 =.../1.105=77.25$

$PC = 27.25 + 77.25 = \$104.50$

How on earth does he find $27.25$? In my calculations $( 0.10 - 0.20)= - 0.1$ I'm always stuck with a negative number. I don't get it

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    $\begingroup$ The second term of the equation is (1.20^5 /(0.10 − 0.20) × 1.105^5). This is negative but you subtract it. This means it is actually added. $\endgroup$
    – Guenterino
    Commented Jul 23, 2021 at 19:13

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Note that the dividend amount changes every year from year $1$ to year $6$. It is constant from year $6$ to infinity. So, we'll take the sum from year $1$ to year $5$ first, then the sum from $6$ to infinity.

For years $1$-$5$ (denoted by $t$), the dividend can be given by $D_t = 5*1.2^{t-1}$. That covers the $\$5$ next year, plus the $20\%$ dividend growth for the four following years.

For $t=1$ to $t=5$, the present value of $D_t$ is $PV_t = \frac{D_t}{1.1^t}$. Putting these two together, we have that the present value of the growing annuity for $t=1$ to $t=5$ is:

$$ \sum_{t = 1}^5 {PV_t} = \sum_{t = 1}^5 {\frac{5*1.2^{t-1}}{1.1^t}} $$ $$ = \frac{5}{1.2} \sum_{t = 1}^5 {\frac{1.2^{t}}{1.1^t}} $$ $$ = \frac{5}{1.2} \sum_{t = 1}^5 \left(\frac{1.2}{1.1}\right)^t $$ $$ = \frac{5}{1.2} * \frac{1.2}{1.1}*\frac{1-\left(\frac{1.2}{1.1}\right)^5}{1-\left(\frac{1.2}{1.1}\right)} $$ $$ \approx \$27.25 . $$

Then, from $t = 6$ as $t\to \infty $, the dividend growth goes to $0$, meaning $D_t = 5*1.2^5$ (this includes the final year of growth in the dividend between $t = 5$ and $t = 6$). The PV is still $PV_t = \frac{D_t}{1.1^t}$, so we have

$$ \sum_{t = 6}^{\infty} {PV_t} = \sum_{t = 6}^{\infty} {\frac{5*1.2^5}{1.1^t}} $$ $$ = 5*1.2^5 * \sum_{t = 6}^{\infty} {\frac{1}{1.1^t}} $$ $$ = 5*\frac{1.2^5}{1.1^6} * \sum_{t = 0}^{\infty} {\frac{1}{1.1^t}} $$ $$ = 5*\frac{1.2^5}{1.1^6} * \frac{1}{1-\frac{1}{1.1}} $$ $$ \approx \$77.25 . $$


For $t = 1$ to $t = 5$, we can get to a similar formula to the one in the original question, but it's not exactly the same, which leads me to believe there's a typo in that formula. Starting from the last step above, we have:

$$ \sum_{t = 1}^5 {PV_t} = \frac{5}{1.2} * \frac{1.2}{1.1}*\frac{1-\left(\frac{1.2}{1.1}\right)^5}{1-\frac{1.2}{1.1}} $$

$$ = 5 * \frac{1-\left(\frac{1.2}{1.1}\right)^5}{1.1\left(1-\frac{1.2}{1.1}\right)} $$

$$ = 5 * \frac{1-\left(\frac{1.2}{1.1}\right)^5}{1.1-1.2} $$

$$ = 5 * \frac{1-\left(\frac{1.2}{1.1}\right)^5}{0.1-0.2} $$

$$ = 5 *\left[ \frac{1}{0.1-0.2}-\frac{\left(\frac{1.2}{1.1}\right)^5}{0.1-0.2} \right]$$

$$ = 5 *\left[ \frac{1}{0.1-0.2}-\frac{1.2^5}{(0.1-0.2)*1.1^5} \right] $$

$$ \approx \$27.25. $$

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  • $\begingroup$ Thank you! I would never get there with the formula that he used, actually I think that the problem is there he must have made a mistake $\endgroup$ Commented Jul 25, 2021 at 9:38
  • $\begingroup$ You're welcome! I think there's a typo in that first formula, you can get to something similar algebraically from my last step, but it's not exactly the same. I've edited the answer to show that. $\endgroup$
    – Amaan M
    Commented Jul 25, 2021 at 21:06

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