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# fm13 24 - σ value CV = \$74.61 \$28.90 = 0.39 To find the...

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Mini Case: 13 - 24 The indirect approach: First, find the coefficient of variation for the value of the project at the time the option expires (year 3). We previously calculated the value of the project at the time the option expires, and we can use this to calculate the expected value and the standard deviation. Value At Expiration Year 3 High \$111.91 Average \$74.61 Low \$37.30 Expected Value =.3(\$111.91)+.4(\$74.61)+.3(\$37.3) = \$74.61. σ value = [.3(\$111.91-\$74.61) 2 + .4(\$74.61-\$74.61) 2 + .3(\$37.30-\$74.61) 2 ] 1/2 = \$28.90. Coefficient of Variation = CV = Expected Value /
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Unformatted text preview: σ value CV = \$74.61 / \$28.90 = 0.39. To find the variance of the project’s rate or return, we use the formula below: σ 2 = LN[CV 2 + 1]/T = LN[0.39 2 + 1]/3 = 4.7%. Now, we proceed to use the OPM: V = \$56.06[N(d 1 )] - \$75e-(0.06)(3) [N(d 2 )]. d 1 = 5 . ) 3 ( 5 . ) 047 . ( ) 3 )]( 0.047/2 06 . [( ) \$56.06/\$75 ln( + + = -0.1085. d 2 = d 1- (0.047) 0.5 (3) 0.5 = -.1085 - 0.3755 = -0.4840. N(d 1 ) = N(-0.1080) = 0.4568. N(d 2 ) = N(-0.4835) = 0.3142....
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