## oBS_IX( ) Example - Equity Call Option

 Description Consider a European call option on a stock that has a current spot price of \$80.00 and a volatility of 30%. The call option matures on 1 March 2003 and has a market value of \$12.60. The risk-free interest rate (on an actual/365 basis) is 6.0%. What is the implied strike rate of this option as at 1 June 2002? Function Specification =oBS_IX(1, 12.60, "1/6/02", "1/3/03", 80, 0.30, 0.06) Solution As there is no closed form solution for implied strike prices, the Newton-Raphson iteration procedure is used to solve for X. When calculating implied strike prices, the Newton-Raphson iteration procedure uses the spot rate as the initial estimate of the strike price, i.e., x0 = \$80. The procedure will iterate using more and more precise estimates of the spot price until the difference between the option value derived from the spot price estimate and the given market option value is less than the desired accuracy level (see Newton-Raphson). In this example the desired accuracy level is 9 decimal places.   The continuous equivalent of the actual/365 risk-free interest rate is calculated as follows: Referring to the equations for d1 and d2 (see model definition), if S = X = 80, vol = 0.30, r = 0.0583, and T =0.7479 (273/365 days), d1 = 0.2977 and d2 = 0.0383. As iPC = 1 (call), N(d1) is 0.6170 and N(d2) is 0.5153 (see oCumNorm( ) function), the oBS( ) equation becomes: Since \$9.9002 is below the market value of the option, \$12.60, the strike rate of \$80 is too high. The oBS( ) value is therefore computed at a lower strike price, i.e., x1 < x0. Referring to the Newton-Raphson iteration procedure, x1 is determined as: Using the same parameter values as above with a new strike rate estimate of 74.5268, the oBS( ) equation returns \$12.8792. As this value is above the market value of the option the next strike trial is: This process continues until the convergence criteria is met, which for this example occurs on the 5th iteration at a strike rate of of 75.00.