## oBSw( ) Example 1

Description

Consider a stock that has 1 million outstanding shares, a current spot price of \$75, a volatility of 20% and pays no dividends. There are 100,000 warrants on issue each entitling the holder to one share in the stock. The warrants have a strike of \$80 and mature on 1 May 2003, while the risk-free interest rate is 6% (on an actual/365 basis). What is the value of each warrant as at 1 May 2002?

Function Specification

=oBSw("1/5/02", "1/5/03", 75, 80, 0.2, 0.06, 1000000, 100000, 1, A1:B1, 0)

It is assumed the cell references for the dividend schedule contain empty cells.

Solution

As the value of a warrant is a function of itself (see model definition) an iteration procedure is used to solve for W.

When calculating warrant values, the iteration procedure uses zero as the initial estimate of the warrant price. The procedure will iterate using more and more precise estimates of the warrant price until the outputted warrant value is within 15 decimal places of the inputted warrant value .

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 W = 0, S = 75, X = 80, r = 0.0583, vol = 0.2, and T = 1 (365/365 days), d1 = 0.0687 and d2 -0.1313.

As N(d1) is 0.5274 and N(d2) is 0.4477 (see oCumNorm( ) function), the Black Scholes warrant equation becomes:

Since the output warrant value (\$5.2364) is above the inputted warrant price (\$0.00), the valuation is re-run with W = \$5.2364. This gives a warrant price of 5.4908.

This process continues until the convergence criteria is met, which for this example occurs on the 12th iteration at a warrant price of \$5.5039.

Greeks

The following Greeks are computed using a discrete approximation of the partial derivative (see oBSw() Model Greeks):

0.518158

-0.027969

-4.809824

28.66088

33.35855