458 Part IV: Additional Considerations 
This approximation is quite accurate for option pricing purposes, since one is not 
really interested in thousandths of a point where option prices are concerned. 
Example: Suppose that XYZ is trading at 45 and we are interested in evaluating the 
July 50 call, which has 60 days remaining until expiration. Furthermore, assume that 
the volatility of XYZ is 30% and that the risk-free interest rate is currently 10%. The 
theoretical value calculation is shown in detail, in order that those readers who wish 
to program the model will have something to compare their calculations against. 
page: 
Initially, determine t, d1, and d2, by referring to the formulae on the previous 
t = 60/365 = .16438 years 
d _ In (45/50) + (.1 + .3 x .3/2) x .16438 
1-
.3 X ✓.16438 
= -.10536 + (.145 X .16438) = __ 67025 .3 X .40544 
d2 = -.67025 - .3 ✓.16438 = -.67025 - (.3 x .40544) = -.79189 
Now calculate the cumulative normal distribution function for d1 and d2 by 
referring to the above formulae: 
dl = -.67025 
l 1 
y = l + (.2316419 I -.67025 I) = 1.15526 = ·86561 
z = .3989423e--(-.67025 X -.67025)/2 
= .3989423e-0·22462 = .31868 
There are too many calculations involved in the computation of the fifth-order 
polynomial to display them here. Only the result is given: 
X = .74865 
Since we are determining the cumulative normal distribution of a negative 
number, the distribution is determined by subtracting x from l. 
N(d1) = N(-.67O25) = l -x = l - .74865 = .25134 
In a similar manner, which requires computing new values for x, y, and z, 
N(d2) = N(-.79179) = 1- .78579 = .21421