Consider an asset that trades at $100 today. Suppose that the European call and put options on this asset are available both with a strike price of $100. The options expire in 275 days, and the volatility is 45%. The continuously compounded risk-free rate is 3%. Determine the value of the European call and put options using the Black-Scholes-Merton model. Assume that the continuously compounded yield on the asset is 1,5% and there are 365 days in the year.
| As per Black Scholes Model | ||||||
| Value of call option = (S*e^(q*t))*N(d1)-N(d2)*K*r^(-r*t) | ||||||
| Where | ||||||
| S = Current price = | 100 | |||||
| t = time to expiry = | 0.753424 | |||||
| K = Strike price = | 100 | |||||
| r = Risk free rate = | 3.000% | |||||
| q = Dividend Yield = | 1.50% | |||||
| σ = Std dev = | 45% | |||||
| d1 = (ln(S/K)+(r-q+σ^2/2)*t)/(σ*t^(1/2) | ||||||
| d1 = (ln(100/100)+(0.03-0.015+0.45^2/2)*0.753424)/(0.45*0.753424^(1/2)) | ||||||
| d1 = 0.224233 | ||||||
| d2 = d1-σ*t^(1/2) | ||||||
| d2 =0.224233-0.45*0.753424^(1/2) | ||||||
| d2 = -0.166367 | ||||||
| N(d1) = Cumulative standard normal dist. of d1 | ||||||
| N(d1) =0.588712 | ||||||
| N(d2) = Cumulative standard normal dist. of d2 | ||||||
| N(d2) =0.433934 | ||||||
| Value of call= 100*e^(-0.015*0.753424)*0.588712-0.433934*100*e^(-0.03*0.753424) | ||||||
| Value of call= 15.79 | ||||||
| As per Black Scholes Model | ||||||
| Value of put option = N(-d2)*K*e^(-r*t)-(S*e^(q*t))*N(-d1) | ||||||
| Where | ||||||
| S = Current price = | 100 | |||||
| t = time to expiry = | 0.753424 | |||||
| K = Strike price = | 100 | |||||
| r = Risk free rate = | 3.000% | |||||
| q = Dividend Yield = | 1.50% | |||||
| σ = Std dev = | 45% | |||||
| d1 = (ln(S/K)+(r-q+σ^2/2)*t)/(σ*t^(1/2) | ||||||
| d1 = (ln(100/100)+(0.03-0.015+0.45^2/2)*0.753424)/(0.45*0.753424^(1/2)) | ||||||
| d1 = 0.224233 | ||||||
| d2 = d1-σ*t^(1/2) | ||||||
| d2 =0.224233-0.45*0.753424^(1/2) | ||||||
| d2 = -0.166367 | ||||||
| N(-d1) = Cumulative standard normal dist. of -d1 | ||||||
| N(-d1) =0.411288 | ||||||
| N(-d2) = Cumulative standard normal dist. of -d2 | ||||||
| N(-d2) =0.566066 | ||||||
| Value of put= 0.566066*100*e^(-0.03*0.753424)-100*e^(-0.015*0.753424)*0.411288 | ||||||
| Value of put= 14.67 | ||||||
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