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Holt Enterprises recently paid a dividend, D0, of $1.25. It expects to have nonconstant growth of 19% for 2 years followed by a constant rate of 10% thereafter. The firm's required return is 20%. How far away is the horizon date? The terminal, or horizon, date is the date when the growth rate becomes constant. This occurs at the end of Year 2. The terminal, or horizon, date is infinity since common stocks do not have a maturity date. The terminal, or horizon, date is Year 0 since the value of a common stock is the present value of all future expected dividends at time zero. The terminal, or horizon, date is the date when the growth rate becomes nonconstant. This occurs at time zero. The terminal, or horizon, date is the date when the growth rate becomes constant. This occurs at the beginning of Year 2. What is the firm's horizon, or continuing, value? Do not round intermediate calculations. Round your answer to the nearest cent. $ What is the firm's intrinsic value today, ? Do not round intermediate calculations. Round your answer to the nearest cent. $

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Answer:

The terminal, or horizon, date is infinity since common stocks do not have a maturity date.

the firm's intrinsic value today=15.99

Step-by-step explanation:

What is the firm's horizon, or continuing, value?

Price of the stock at t = 0 is the sum of the present values of all expected cash-flows.

[tex]P0=\frac{D1}{1+ke} +\frac{D2}{(1+ke)^2}+\frac{D3}{(ke-g)(1+ke)^2}[/tex]

Where D1 = D0(1+g1) = $1.25(1.19)=$1.4875

            D2 = D1(1+g2) = $1.25(1.19)(1.19)= $1.770125

            D3= D2(1+g3)=$1.25(1.19)(1.19)(1.1)= $1.9471375

              ke=0.2

Note, at the end of year 2, we can value the present value of all the remaining dividends from D3 being the next expected dividend till infinity by using the constant growth model where [tex]P2=\frac{D3}{ke-g}[/tex]. This value at the end of year 2 still has to be discounted 2 years back to get the PV at t=0.

Therefore [tex] PO =\frac{1.4875}{1+0.2} +\frac{1.770125}{(1+0.2)^2}+\frac{1.9471375}{(0.2-0.1)(1+0.2)^2}[/tex] = $15.99

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