9.1 | The Constant-Growth-Rate Discounted Dividend Model, as described equation 9.5 on page 247, says that:xx

P0 = D1 / (k – g) | | A. | rearrange the terms to solve for:

i. g; and

ii. D1.

As an example, to solve for k, we would do the following:

1. Multiply both sides by (k – g) to get: P0 (k – g) = D1

2. Divide both sides by P0 by to get: (k – g) = D1 / P0

3. Add g to both sides: k = D1 / P0 + g (8 marks) | 9.2 | Notation: Let

Pn = Price at time nDn = Dividend at time nYn = 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 rateg = dividend growth rate = r x ROEROE = 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