Fm13 11 - 2 = 0.3966(0.2025 0.5(2 0.5 =-0.2398 From Excel function NORMSDIST or approximated from the table in Appendix A N(d 1 = 0.6542 N(d 2 =

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Answers and Solutions: 13 - 11 Then calculate the coefficient of variation: CV = 17,004/22,562 = 0.7537. σ 2 = ln(CV 2 +1)/t = ln(0.7537 2 +1)/2 = 0.2249 so σ 2 = 0.2249 Notice that in this case the direct method and the indirect method give very similar results for σ . P = $18,646 X = $20,000 t = 2. r RF = 0.06. σ 2 = 0.2025 d 1 = ln[18.646/20] + [0.06 + .5(0.2025)](2) = 0.3966 (0.2025) 0.5 (2) 0.5 d
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Unformatted text preview: 2 = 0.3966 - (0.2025) 0.5 (2) 0.5 = -0.2398 From Excel function NORMSDIST, or approximated from the table in Appendix A: N(d 1 ) = 0.6542 N(d 2 ) = 0.4053 Using the Black-Scholes Option Pricing Model, you calculate the option’s value as: V = P[N(d 1 )] - t r RF Xe − [N(d 2 )] = $18.646(0.6542) - $20e (-0.06)(2) (0.4053) = $12.198 - $7.189 = $5.009 thousand....
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This note was uploaded on 07/13/2011 for the course FIN 4414 taught by Professor Staff during the Spring '08 term at University of Florida.

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