The inputs to the Black-Scholes model are the current price of the underlying asset (S), the exercise or strike price of the option (K), the time to expiration of the option in fractions of a year (t), the variance of the underlying asset (cr2), and the continuously-compounded risk-free interest rate (r).

In this problem, the inputs are: S = $55 σ2 = 0.0625

K =$50 r = 0.10

t = 1

After identifying the inputs, solve for dl and d2:

dl = [In(S/K) + (r + 0.5*σ2)(t) ] / (σ2t)1/2

= [In(55/50) + {0.10 + ½(0.0625)}(1) ] / (0.0625*1)1/2

= 0.9062

d2 = dl- (σ2t) 1/2

= 9062 - (0.0625*1) 1/2

= 0.6562

Find N(dl) and N(d2), the area under the normal curve from negative infinity to dl and negative infinity to d2, respectively.

N(dl) = N(0.9062) = 0.8176

N(d2) = N(0.6562) = 0.7442

According to the Black-Scholes formula, the price of a European call option (C) on a non-dividend paying common stock is:

C = SN(dl) - Ke-rtN(d2)

= (55)(0.8176) - (50)e-(.10)(1) (0.7442)

= $11.30

The Black-Scholes Price of the call option is $11.30.

For Part 2 and 3,

Please refer to the Excel spreadsheet. The Black Scholes values can be computed very handily using EXCEL. For the mathematical formulas, just use log (x, 2.7182818) for the natural logarithmic function, use exp(x) for the exponential function and normsdist (x) for the standard normal distribution function.

It can be easily noted that for part 2 the answer is $8.24, and for part 3 the answer is $24.01. Hence it’s clear that the intrinsic value method is wrong and that option with stock price equaling the exercise price can be quite valuable.

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