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fm13 10 - this isn’t 10 as you might hope The 2-year...

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Answers and Solutions: 13 - 10 is at the end of year 2. P = \$18,646 X = \$20,000 t = 2. r RF = 0.06. σ 2 = 0.2025 Although the problem stated to assume the variance of the project’s rate of return was 0.2025, we’ll also calculate it using the direct method. First calculate the rates of return using the decision tree. To do this, calculate the present values of the two branches as of the exercise date, year 2, and the rates of return assuming the initial value of the investment was P = \$18,646. 0 1 2 3 4 40% Prob . | | | | | Good 43,388 = PV of two 25,000 cash flows Bad | | | | | 60% Prob. 8,678 = PV of two 5,000 cash flows Annual rate of return in good state = [43,388/18,646] 1/2 -1 = 52.54% Annual rate of return in bad state = [8,678/18,646] 1/2 – 1 = -31.78% The expected value of these two returns is 52.54(0.40) – 31.78(0.60) = 1.95% [Note:
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Unformatted text preview: this isn’t 10% as you might hope! The 2-year returns are 43,388/18,646 – 1 = 132.69% in the good state and 8,678/18,646 – 1 = -53.46% in the bad state. The expected value of these is 132.69% (0.40) - 53.46% (0.60) = 21.0%, which is exactly 10% compounded twice: 0.21 = (1.10) 2 – 1.] The variance is σ 2 = (0.5254 - 0.0195) 2 (0.40) + (-0.3178 – 0.0195) 2 (0.60) = 0.1706 To calculate the variance of the project’s returns using the indirect method, first calculate the standard deviation of the value at year 2. The value is either 43,388 (probability 40%) or 8,678 (probability 60%). Expected value at year 2: 43,388(0.4) + 8,678(0.6) = 22,562. Standard deviation = [(43,388-22,562) 2 (0.40) + (8,678 – 22,562) 2 (0.60)] 1/2 = 17,004....
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