Assume that a stock is expected to pay dividends at the end of Year 1 and Year 2 of
$1.25 and $1.56, respectively. Dividends are expected to grow at a 5% rate
thereafter. Assuming that ke is 11%, the value of the stock is closest to:
A. $22.30.
B. $23.42.
C. $24.55.
Why is it like this?
“5. C ($1.25 / 1.11) + [1.56 / (0.11 − 0.05)] / 1.11 = $24.55 (LOS 48.h)”
And not (1,25/1,11)+(1,56/1,11^2)+(1,56/0,05^2)/0,11^2
I seem to not get how to really implement the formula.
I seem to not get how to really implement the formula.
and helpfully you didn’t include the formula you want to know how to implement.
Starting from scratch,
V=\sum_{n=1}^{\infty}\frac{D_{n}}{(1+k_{e})^{n}}
where D_{n} is expected dividend at year n and k_{e}=11\%
with D_{1}=1.25 and for n\ge 2, D_{n}=D_{2}(1+g)^{n-2} with g=5\%
Using the values of the dividends,
V=\frac{D_{1}}{1+k_{e}}
+\frac{D_{2}}{(1+k_{e})^{2}}\sum_{n=0}^{\infty}\left(\frac{1+g}{1+k_{e}}\right)^{n}
The infinite sum is a standard result:
If S=\sum_{n=0}^{\infty}x^{n} then Sx=\sum_{n=1}^{\infty}x^{n}=S-1 so S=\frac{1}{1-x}
So
V=\frac{D_{1}}{1+k_{e}}
+\frac{D_{2}}{(1+k_{e})^{2}}\frac{1}{1-\frac{1+g}{1+k_{e}}}
=\frac{D_{1}}{1+k_{e}}
+\frac{D_{2}}{(1+k_{e})^{2}}\frac{1+k_{e}}{k_{e}-g}
=\frac{D_{1}}{1+k_{e}}
+\frac{D_{2}}{(1+k_{e})(k_{e}-g)}
Put in the numbers for D_{1}, D_{2}, g and k_{e} and you get the official answer
The part I’ve bolded is the incorrect part. This part of the DDM is called the terminal value, at year 2, which is based on year 3’s dividend. You’re supposed to find the terminal value using the formula D2*(1+g) / (r - g), and then you discount everything back to today.
To actually answer his question, which boils down to why would you use
V = (contribution from year 1) + (contribution from year 2 onwards)
rather than
V = (contribution from year 1) + (contribution from year 2) +(contribution from year 3 onwards)
the answer is that you can use both, and they should both give the same numerical value, but you’ve made a boo-boo.
Because you haven’t given us the formula you were using, we don’t know if you’ve got the formula wrong or if the formula is correct but you put the numbers in wrong.
As to why you’d use one rather than the other, I would note that one has 2 terms and is simpler, the other has 3 terms and is more complicated. As Albert Einstein used to say, Make everything as simple as possible, but not simpler.
I’ve already given you the two term formula,
V=\frac{D_{1}}{1+k_{e}}+\frac{D_{2}}{(1+k_{e})(k_{e}-g)}
If you follow the same procedure but write
V=\frac{D_{1}}{1+k_{e}}+\frac{D_{2}}{(1+k_{e})^{2}}
+\frac{D_{2}(1+g)}{(1+k_{e})^{3}}\sum_{n=0}^{\infty}\left(\frac{1+g}{1+k_{e}}\right)^{n},
you get
V=\frac{D_{1}}{1+k_{e}}+\frac{D_{2}}{(1+k_{e})^{2}}
+\frac{D_{2}(1+g)}{(1+k_{e})^{3}}\frac{1+k_{e}}{k_{e}-g}
=\frac{D_{1}}{1+k_{e}}+\frac{D_{2}}{(1+k_{e})^{2}}
+\frac{D_{2}(1+g)}{(1+k_{e})^{2}(k_{e}-g)}
These two formulae should give the same numerical value:
The first one is V=\frac{1.25}{1.11}+\frac{1.56}{(1.11)(0.11-0.05)}=24.55, which is the official answer
The second one should give V=\frac{1.25}{1.11}+\frac{1.56}{1.11^{2}}
+\frac{1.56(1.05)}{1.11^{2}(0.11-0.05)}=24.55, the same
Okay, thank you very much.