# WHY? Present Value of a growing annuity

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

• 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. Commented Jul 23, 2021 at 19:13

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:

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

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

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

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

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

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

$$\approx \27.25.$$

• 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 Commented Jul 25, 2021 at 9:38
• 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. Commented Jul 25, 2021 at 21:06