I need some help tin questions D and E please. Notation: Let P n = Price at time
ID: 2745401 • Letter: I
Question
I need some help tin questions D and E please.
Notation: Let
Pn = Price at time n
Dn = Dividend at time n
Yn = Earnings in period n
r = retention ratio = (Yn– Dn) / Yn = 1 – Dn/ Yn = 1 - dividend payout ratio
En = Equity at the end of year n
k = discount rate
g = dividend growth rate = r x ROE
ROE = Yn / En-1 for all n>0.
We will further assume that k and ROE are constant, and that r and g are constant after the first dividend is paid.
A. Using the Discounted Dividend Model, calculate the price P0 if
D1 = 20, k = .15, g = r x ROE = .8 x .15 = .12, and Y1 = 100 per share
B. What, then, will P5 be if:
D6 = 20, k = .15, and g = r x ROE = .8 x .15 = .12?
C. If P5 = your result from part B, and assuming no dividends are paid until D6, what would be P0? P1? P2?
D. Again, assuming the facts from part B, what is the relationship between P2 and P1 (i.e., P2/P1)? Explain why this is the result.
E. If k = ROE, we can show that the price P0 doesn’t depend on r. To see this, let
g = r x ROE, and ROE = Yn / En-1, and
since r = (Yn – Dn) / Yn , then D1 = (1 – r) x Y1 and
P0
=
D1 / (k – g)
P0
=
[(1 – r) x Y1] / (k – g)
P0
=
[(1 – r) x Y1] / (k – g), but, since k = ROE = Y1 / E0
P0
=
[(1 – r) x Y1] / (ROE – r x ROE)
P0
=
[(1 – r) x Y1] / (Y1 / E0 – r x Y1 / E0)
P0
=
[(1 – r) x Y1] / (1 – r) x Y1 / E0), and cancelling (1 – r)
P0
=
Y1 / (Y1/E0) = Y1 x (E0 / Y1) = E0
P0
=
D1 / (k – g)
P0
=
[(1 – r) x Y1] / (k – g)
P0
=
[(1 – r) x Y1] / (k – g), but, since k = ROE = Y1 / E0
P0
=
[(1 – r) x Y1] / (ROE – r x ROE)
P0
=
[(1 – r) x Y1] / (Y1 / E0 – r x Y1 / E0)
P0
=
[(1 – r) x Y1] / (1 – r) x Y1 / E0), and cancelling (1 – r)
P0
=
Y1 / (Y1/E0) = Y1 x (E0 / Y1) = E0
Explanation / Answer
Requirement A:
Po = D1 (1+g) / (k-g)
= 20 / (0.15 – 0.12)
= 20 / 0.03
= $666.67
Po = $666.67
Requirement B:
P5 = D6 / (k-g)
= 20 / (0.15 – 0.12)
= 20 / 0.03
= $666.67
P5 = $666.67
Requirement C:
P5 = $666.67
Po = P5 / (1+0.15)5
= $666.67 * 0.4971767
= $331.45
P1 = Po (1+0.15)
= $331.45 * 1.15
= $381.1675
P2 = P1 (1+0.15)
= $381.1675 (1.15)
= $438.34
Requirement D:
P2 / P1 = 438.34 / 381.1675 = 1.15 = 1+0.15 = 1+K
The result is 1+k since P2 is the 1 year compounding value of P1 at the rate of k i.e., 15% and P1 is the 1 year discounting value of P2 at the rate of k i.e., 15%.