\documentclass[12pt]{article}

\input{../../../../etc/macros}
\input{../../../../etc/mzlatex_macros}
%\input{../../../../etc/pdf_macros}


\begin{document}

\myheader
\mytitle

\hr

\sectiontitle{Solution of the Black-Scholes Equation}

\hr
\section*{Rating} %one of 
% Everyone: contains no mathematics.
% Student: contains scenes of mild algebra or calculus that may require guidance.
Mathematically Mature: may contain mathematics beyond calculus with proofs.
% Mathematicians Only: prolonged scenes of intense rigor.

\hr
\visual{QuestionofDay}{../../../../CommonInformation/Lessons/question_mark.jpeg}
\section*{Question of the Day}
What is the solution method for the Cauchy-Euler type of ordinary
differential equation:
\[
    x^2 \frac{d^2 v}{dx^2} + a x \frac{dv}{dx} + b v = 0?
\]

\hr
\visual{Key Concepts}{../../../../CommonInformation/Lessons/keyconcepts.jpeg}
\section*{Key Concepts}

\begin{enumerate}
    \item
        We solve the Black-Scholes equation for the value of a European call
        option on a security by judicious changes of
        variables that reduce the equation to the heat equation.  The
        heat equation has a solution formula.  Using the
        solution formula with the changes of variables gives the
        solution to the Black-Scholes equation.
      \item Solving the Black-Scholes equation is an example of how to
        choose and execute changes of variables to solve a partial
        differential equation.
\end{enumerate}

\hr
\visual{Key Concepts}{../../../../CommonInformation/Lessons/keyconcepts.jpeg}
\section*{Vocabulary}
\begin{enumerate}
    \item
        A differential equation with auxiliary initial conditions and
        boundary conditions, that is an initial value problem, is said
        to be \defn{well-posed} if the solution exists, is unique, and
        small changes in the equation parameters, the initial conditions
        or the boundary conditions produce only small changes in the
        solutions.
\end{enumerate}

\hr
\visual{Mathematical Ideas}{../../../../CommonInformation/Lessons/mathematicalideas.jpeg}
\section*{Mathematical Ideas}

\subsection*{Analytic Solution of the Black-Scholes Equation }

This discussion is drawn from Section 4.2, pages 59--63; Section 4.3,
pages 66--69; Section 5.3, pages 75--76; and Section 5.4, pages 77--81 of
\emph{The Mathematics of Financial Derivatives:  A Student Introduction}
by P. Wilmott, S. Howison, J. Dewynne, Cambridge University Press,
Cambridge, 1995.  Some ideas are also taken from Chapter 11 of \emph{Stochastic
Calculus and Financial Applications} by J. Michael Steele, Springer, New
York, 2001.

We have to start somewhere, and to avoid the problem of deriving
everything back to calculus, we will assert that the \emph{initial value
problem for the heat equation on the real line is well-posed},
that is, the solution to the partial differential equation
$$
    \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2}
    \qquad -\infty < x < \infty, \quad \tau > 0
$$
with
$$
    u(x,0) = u_0(x)
$$
where the initial condition and the solution satisfy the following
technical requirements:
\begin{enumerate}
    \item
        $ u_0(x) $ has no more than a finite number of discontinuities
        of the jump kind,
    \item
        $ \lim_{|x| \to \infty} u_0(x) e^{-a x^2} = 0 $ for any $ a >0 $,
    \item
        $ \lim_{|x| \to \infty} u(x, \tau) e^{-a x^2} = 0 $ for any $ a
        >0 $
\end{enumerate}
exists for all time, is unique, and most importantly, can be represented
as
$$
    u(x, \tau) = \frac{1}{2\sqrt{\pi \tau}} \int_{-\infty}^{\infty} u_0(s)
    e^{-(x-s)^2/4\tau} \; ds
$$

\begin{remark}
This solution can derived in several different ways, I
think the easiest way is to use Fourier transforms.  The derivation of
this solution representation is standard in any course or book on
partial differential equations.
\end{remark}

\begin{remark}
Mathematically, the conditions above are unnecessarily
restrictive, and can be considerably weakened, however, they will be
more than sufficient for all practical situations we encounter in
mathematical finance.
\end{remark}

\begin{remark}
The use of $ \tau $ for the time variable (instead of the
more natural $ t $) is to avoid a conflict of notation in the several
changes of variables we will soon have to make.
\end{remark}

The Black-Scholes terminal value problem for the value $ V(S,t) $ of a
European call option on a security with price $ S $ at time $ t $ is
$$
    \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2
    V}{\partial S^2} + r S \frac{\partial V}{\partial S} - r V = 0
$$
with $ V(0,t) = 0 $, $ V(S,t) \sim S $ as $ S \to \infty $ and
$$
    V(S,T) = \max( S- K, 0).
$$
Note that this looks a little like the heat equation on the infinite
interval in that it has a first derivative of the unknown with respect
to time and the second derivative of the unknown with respect to the
other (space) variable, but notice:
\begin{enumerate}
    \item
        Each time the unknown is differentiated with respect to $ S $,
        it also multiplied by the independent variable $ S $, so the
        equation is not a constant coefficient equation.
    \item
        There is a first derivative of $ V $ with respect to $ S $ in
        the equation.
    \item
        There is a zero-th order term $ V $ in the equation.
    \item
        The sign on the second derivative is the opposite of the heat
        equation form, so the equation is of backward parabolic form.
    \item
        The data of the problem is given at the final time $ T $ instead
        of the initial time $ 0 $, consistent with the backward
        parabolic form of the equation.
    \item
        There is a \emph{boundary condition} $ V(0,t) = 0 $ specifying
        the value of the solution at one sensible boundary of the
        problem.  The boundary is sensible since security values must
        only be zero or positive.  This boundary condition says that any
        time the security value is 0, then the call value (with strike
        price $ K $) is also worth 0.
    \item
        There is another boundary condition $ V(S,t) \sim S $, as $ S
        \to \infty $, but although this is financially sensible, (it
        says that for very large security prices, the call value with
        strike price $ K $ is approximately $ S $) it is more in the
        nature of a technical condition, and we will ignore it without
        consequence.
\end{enumerate}

We eliminate each objection with a suitable
change of variables.  The plan is to change variables to reduce the
Black-Scholes terminal value problem to the heat equation, then to
use the known solution of the heat equation to represent the solution,
and finally change variables back.  This is a standard solution technique in
partial differential equations.  All the transformations are
standard, well-motivated, and well known.

First we take $ t = T - \frac{\tau}{ (1/2)\sigma^2} $ and $ S = K e^x
$, and we set
$$
    V(S,t) = K v(x,\tau).
$$
Remember, $ \sigma $ is the volatility, $ r $ is the interest rate on a
risk-free bond, and $ K $ is the strike price.  In the changes of
variables above, the choice for $ t $ reverses the sense of time,
changing the problem from backward parabolic to forward parabolic.  The
choice for $ S $ is a well-known transformation based on experience with
the Euler equidimensional equation in differential equations.  In
addition, the variables have been carefully scaled so as to make the
transformed equation expressed in dimensionless quantities.  All of
these techniques are standard and are covered in most courses and books
on partial differential equations and applied mathematics.

Some extremely wise advice adapted from page 186 of \emph{Stochastic
Calculus and Financial Applications} by J. Michael Steele, Springer,
New York, 2001 is appropriate here.
\begin{quotation}
    ``There is nothing particularly difficult about changing variables
    and transforming one equation to another, but there is an element of
    tedium and complexity that slows us down.  There is no universal
    remedy for this molasses effect, but the calculations do seem to go
    more quickly if one follows a well-defined plan.  If we know that $
    V(S,t) $ satisfies an equation (like the Black-Scholes equation) we
    are guaranteed that we can make good use of the equation in the
    derivation of the equation for a new function $ v(x, \tau) $ defined
    in terms of the old if we write the old $ V $ as a function of the
    new $ v $ and write the new $ \tau $ and $ x $ as functions of the
    old $ t $ and $ S $.  This order of things puts everything in the
    direct line of fire of the chain rule; the partial derivatives $ V_t
    $, $ V_S $ and $ V_{SS} $ are easy to compute and at the end, the
    original equation stands ready for immediate use.''
\end{quotation}

Following the advice, write
$$
    \tau = (1/2)\sigma^2( T - t)
$$
and
$$
    x = \log(S/K)
$$
so
$$
    \frac{\partial V}{\partial t} = K \frac{\partial v}{\partial \tau}
    \cdot \frac{ d\tau}{dt} = K \frac{\partial v}{\partial \tau} \cdot
    \frac{-\sigma^2}{2}
$$
and
$$
    \frac{\partial V}{\partial S} = K \frac{\partial v}{\partial x}
    \cdot \frac{dx}{dS} = K \frac{\partial v}{\partial x}\cdot\frac{1}{S}
$$
and

\begin{align*}
    \frac{\partial^2 V}{\partial S^2} &= \frac{\partial}{\partial S}
    \left( \frac{\partial V}{\partial S} \right) \\
    &= \frac{\partial}{\partial S} \left( K \frac{\partial v}{\partial
    x} \frac{1}{S} \right) \\
    &= K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} + K \frac{\partial}
    {\partial S} \left( \frac{\partial v}{\partial x} \right) \cdot
    \frac{1}{S} \\
    &= K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} + K \frac{\partial}
    {\partial x} \left( \frac{\partial v}{\partial x} \right) \cdot
    \frac{dx}{dS} \cdot \frac{1}{S} \\
    &= K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} + K \frac{\partial^2
    v}{\partial x^2} \cdot \frac{1}{S^2}
\end{align*}
and the terminal condition
$$
    V(S,T) = \max(S-K, 0) = \max(K e^x - K, 0)
$$
but $ V(S,T) = K v(x,0) $ so $ v(x,0) = \max( e^x - 1, 0) $.

Now substitute all of the derivatives into the Black-Scholes equation to
obtain:
$$
    K \frac{\partial v}{\partial \tau} \cdot \frac{-\sigma^2}{2} + \frac
    {\sigma^2}{2} S^2 \left( K \frac{\partial v}{\partial x} \cdot \frac
    {-1}{S^2} + K \frac{\partial^2 v}{\partial x^2} \cdot \frac{1}{S^2}
    \right) + r S \left( K \frac{\partial v}{\partial x} \cdot \frac{1}{S}
    \right) - r K v = 0
$$
Now begin the simplification:
\begin{enumerate}
    \item
        Isolate the common factor $ K $ and cancel.
    \item
        Transpose the $ \tau $-derivative to the other side, and divide
        through by $ \sigma^2/2 $
    \item
        Rename the remaining constant $ r/(\sigma^2/2) $ as $ k $.  $ k $
        measures the ratio between the risk-free interest rate and the
        volatility.
    \item
        Cancel the $ S^2 $ terms in the second derivative.
    \item
        Cancel the $ S $ terms in the first derivative.
    \item
        Gather up like order terms.
\end{enumerate}

What remains is the rescaled, constant coefficient equation:
$$
    \frac{\partial v}{\partial \tau} = \frac{\partial^2 v}{\partial x^2}
    + (k-1) \frac{\partial v}{\partial x} - k v.
$$
We have made considerable progress, because
\begin{enumerate}
    \item
        Now there is only one dimensionless parameter $ k $ measuring
        the risk-free interest rate as a multiple of the volatility and
        a rescaled time to expiry $ (1/2)
        \sigma^2 T $, not the original $ 4 $ dimensioned quantities
         $ K $, $ T $, $ \sigma^2 $ and $ r $.
    \item
        The equation is defined on the interval $ -\infty < x < \infty $,
        since this $ x $-interval defines $ 0 < S < \infty $ through the
        change of variables $ S = K e^x $.
    \item
        The equation is now constant coefficient.  
\end{enumerate}
In principle, we could now solve the equation directly.

Instead, we will change the dependent variable scale yet again, by
$$
    v = e^{\alpha x + \beta \tau} u(x,\tau).
$$
where $ \alpha $ and $ \beta $ are yet to be determined.  Using the
product rule:
$$
    v_\tau = \beta e^{\alpha x + \beta \tau} u(x,\tau) + e^{\alpha x +
    \beta \tau} u_\tau
$$
and
$$
    v_x = \alpha e^{\alpha x + \beta \tau} u(x,\tau) + e^{\alpha x +
    \beta \tau} u_x
$$
and
$$
    v_{xx} = \alpha^2 e^{\alpha x + \beta \tau} u(x,\tau) + 2\alpha e^{\alpha
    x + \beta \tau} u_x + e^{\alpha x + \beta \tau} u_{xx}.
$$
Put these into our constant coefficient partial differential equation,
cancel the common factor of $ e^{\alpha x + \beta \tau} $ throughout and
obtain:
$$
    \beta u + u_\tau = \alpha^2 u + 2\alpha u_x + u_{xx} + (k - 1) (\alpha
    u + u_x) - k u
$$
Gather like terms:
$$
    u_\tau = u_{xx} + [ 2\alpha + (k-1)] u_x + [ \alpha^2 + (k-1) \alpha
    - k - \beta] u.
$$
Choose $ \alpha = -(k-1)/2 $ so that the $ u_x $ coefficient is $ 0
$, and then choose $ \beta = \alpha^2 + (k-1) \alpha - k = -(k+1)^2/4 $
so the $ u $ coefficient is likewise $ 0 $.  With this choice, the
equation is reduced to
$$
    u_\tau = u_{xx}.
$$
We also need to transform the initial condition too.  This
transformation is
\begin{eqnarray*}
    u(x,0) &= e^{-(-(k-1)/2)x - (-(k+1)^2/4)0} v(x,0) \\
    &= e^{((k-1)/2)x} \max( e^x - 1, 0) \\
    &= \max( e^{((k+1)/2)x} - e^{((k-1)/2)x}, 0 ).
\end{eqnarray*}
For future reference, we notice that this function is strictly positive
when the argument $ x $ is strictly positive, that is $ u_0(x) > 0 $
when $ x > 0 $, otherwise, $ u_0(x) = 0 $ for $ x \le 0 $.

We are in the final stage since we are ready to apply the solution
representation formula:
$$
    u(x, \tau) = \frac{1}{2\sqrt{\pi \tau}} \int_{-\infty}^{\infty} u_0(s)
    e^{-(x-s)^2/4\tau} \; ds.
$$
However, first we want to make a change of variable in the integration,
by taking $ z = (s-x)/\sqrt{2\tau} $, (and thereby $ dz = (-1/\sqrt{2\tau})
\; dx $) so that the integration becomes:
$$
    u(x,\tau) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u_0(z
    \sqrt{2\tau} + x) e^{-z^2/2} \; dz.
$$
We may as well only integrate over the domain where $ u_0 > 0 $, that is
for $ z > -x/\sqrt{2\tau} $.  On that domain, $ u_0 = e^{((k+1)/2)(x + z\sqrt
{2\tau})} - e^{((k-1)/2)(x + z\sqrt{2\tau}) } $ so we are down to:
$$
    \frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty} e^{\frac{k+1}{2}
    (x + z\sqrt{2\tau})} e^{-z^2/2} \; dz - \frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt
    {2\tau}}^{\infty} e^{\frac{k-1}{2}(x + z\sqrt{2\tau})} e^{-z^2/2} \;
    dz
$$
Call the two integrals $ I_1 $ and $ I_2 $ respectively.

We will evaluate $ I_1 $ ( the one with the $ k+1 $ term) first.  This
is easy, completing the square in the exponent yields a standard,
tabulated integral.  The exponent is
\begin{eqnarray*}
    ((k+1)/2)(x + z\sqrt{2\tau}) - z^2/2 &= (-1/2)( z^2 -\sqrt{2\tau}(k+1)z)
    + ((k+1)/2) x \\
    & = & (-1/2)(z^2 - \sqrt{2\tau}(k+1)z + \tau(k+1)^2/2) + ((k+1)/2)x
    + \tau(k+1)^2/4 \\
    & = & (-1/2)(z - \sqrt{\tau/2}(k+1))^2 + (k+1)x/2 +\tau (k+1)^2/4
\end{eqnarray*}
Therefore
\[
    \frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty} e^{\frac{k+1}{2}
    (x + z\sqrt{2\tau})} e^{-z^2/2} \; dz = \frac{ e^{(k+1)x/2 +\tau (k+1)^2/4}}
    {\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^\infty e^{\frac{-1}{2} (z -
    \sqrt{\tau/2}(k+1))^2} \; dz
\]
Now, change variables again on the integral, choosing $ y = z - \sqrt{\tau/2}
(k+1) $ so $ dy = dz $, and all we need to change are the limits of
integration:
\[
    \frac{ e^{ (k+1)x/2 +\tau (k+1)^2/4} }{ \sqrt{2\pi}} \int_{-x/\sqrt{2\tau}
    - \sqrt{\tau/2}(k+1)}^\infty e^{(-1/2)y^2} \; dz
\]
The integral can be represented in terms of the cumulative distribution
function of a normal random variable, usually denoted $ \Phi $.  That
is,
\[
    \Phi(d) = (1/\sqrt{2\pi})\int_{-\infty}^{d} e^{-y^2/2} \; dy
\]
so
\[
    I_1 = e^{ (k+1)x/2 +\tau (k+1)^2/4} \Phi(d_1)
\]
where $ d_1 = x/\sqrt{2\tau} + \sqrt{\tau/2}(k+1) $ (note the use of the
symmetry of the integral!) The calculation of $ I_2 $ is identical,
except that $ (k+1) $ is replaced by $ (k-1) $ throughout.

The solution of the transformed heat equation initial value problem is
\[
    u(x,\tau) = e^{ (k+1)x/2 +\tau (k+1)^2/4} \Phi(d_1) - e^{ (k-1)x/2 +\tau
    (k-1)^2/4} \Phi(d_2)
\]
where $ d_1 = x/\sqrt{2\tau} + \sqrt{\tau/2}(k+1) $ and $ d_2 = x/\sqrt{2\tau}
+ \sqrt{\tau/2}(k-1). $

Now we must systematically unwind each of the changes of variables,
from $ u $.  First, $ v(x,\tau) = e^{(-1/2)(k-1)x - (1/4)(k+1)^2 \tau} u(x,\tau)
$.  Notice how many of the exponentials neatly combine and cancel!
Next put $ x = \log(S/K) $, $ \tau = (1/2)\sigma^2(T-t) $ and $ V(S,t) =
Kv(x,\tau) $.

The final solution is the Black-Scholes formula for the value of a
European call option at time $ T $ with strike price $ K $, if the
current time is $ t $ and the underlying security price is $ S $, the
risk-free interest rate is $ r $ and the volatility is $ \sigma $:
\[
    V(S,t) = S \Phi\left( \frac{\log(S/K) + ( r + \sigma^2/2)(T-t)}{\sigma
    \sqrt{T-t}} \right) - K e^{-r(T-t)} \Phi\left( \frac{\log(S/K) + ( r
    - \sigma^2/2)(T-t)}{\sigma \sqrt{T-t}} \right).
\]

Usually one doesn't see the solution as this full closed form
solution.  Most versions of the solution write intermediate
steps in small pieces, and then present the solution as an algorithm
putting the pieces together to obtain the final answer.

\subsection*{Graphical Solution of the Black-Scholes Equation}

Consider for purposes of graphical illustration the value of a call
option with strike price $ K = 100 $.  The risk-free interest rate per
year, continuously compounded is 12\%, so $ r = 0.12 $, the time to
expiration is $ T = 1 $ measured in years, and the standard deviation
per year on the return of the stock, or the volatility is \( \sigma =
0.10 \).  The value of the call option at maturity plotted over a range
of stock prices $ 70 \le S \le 130 $ surrounding the strike price is
illustrated below:

\begin{figure}
    \centering
    \includegraphics{calloption.jpg}
    \caption{Value of the call option at maturity
    \label{fig:calloption}}
\end{figure}

We use the Black-Scholes formula above to compute the value of the
option prior to expiration.  With the same parameters as above the value
of the call option is plotted over a range of stock prices $ 70 \le S
\le 130 $ at time remaining to expiration $ t = 1 $ (red), $ t=0.8 $, (orange),
$ t=0.6 $ (yellow), $ t=0.4 $ (green), $ t = 0.2 $ (blue) and at
expiration $ t = 0 $ (black).

\begin{figure}
    \centering
    \includegraphics[scale=0.50]{optionvalueS.jpg}
    \caption{Value of the call option at various times
    \label{fig:optionvalueS}}
\end{figure}

Notice a couple of trends in the value from this graph:
\begin{enumerate}
    \item
        As the stock price increases, for a fixed time the option value
        increases,
    \item
        As the time to expiration decreases, for a fixed stock value
        price the value of the option decreases to the value at
        expiration.
\end{enumerate}
Both of these trends were predicted from our intuitive analysis of
options.  The Black-Scholes option pricing formula makes the intuition precise.

We can also plot the solution of the Black-Scholes equation as a
function of security price and the time to expiration as value surface:

\begin{figure}
    \centering
    \includegraphics{valuesurface.jpg}
    \caption{Value surface from the Black-Scholes formula
    \label{fig:valuesurface}}
\end{figure}

\hr
\visual{Problems to Work}{../../../../CommonInformation/Lessons/solveproblems.jpeg}
\section*{Problems to Work for Understanding}
\begin{enumerate}
    \item
        Explicitly evaluate the integral $ I_2 $ in terms of the c.d.f. $
        \Phi $ and other elementary functions as was done for the
        integral $ I_1 $.

        %%\HCode{<a href="problem1.shtml">} Solution \HCode{</a>}

    \item
        What is the price of a European call option on a
        non-dividend-paying stock when the stock price is \$52, the
        strike price is \$50, the risk-free interest rate is 12\% per
        annum (compounded continuously), the volatility is 30\% per
        annum, and the time to maturity is 3 months?
        %% adapted from Hull, Problem 10.21, page 241.

        %%\HCode{<a href="problem2.shtml">} Solution \HCode{</a>}

    \item
        What is the price of a European call option on a non-dividend
        paying stock when the stock price is \$30, the exercise price is
        \$29, the risk-free interest rate is 5\%, the volatility is 25\%
        per annum, and the time to maturity is 4 months?
        % adapted from Hull, Problem 10.23 (a) page 241.

        %%\HCode{<a href="problem3.shtml">} Solution \HCode{</a>}

    \item
        Show that the Black-Scholes formula for the price of a call
        option tends to $ \max( S-K, 0) $ as $ t \to T $.

        %%\HCode{<a href="problem4.shtml">} Solution \HCode{</a>}

\end{enumerate}

\hr
\visual{Books}{../../../../CommonInformation/Lessons/books.jpeg}
\section*{Reading Suggestion:}
\begin{enumerate}
    \item
        Section 4.2, pages 59--63; Section 4.3, pages 66--69; Section 5.3,
        pages 75--76; and Section 5.4, pages 77--81 of \emph{The
        Mathematics of Financial Derivatives:  A Student Introduction }
        by P. Wilmott, S. Howison, J. Dewynne, Cambridge University
        Press, Cambridge, 1995.
    \item
        Chapter 11 of \emph{Stochastic Calculus and Financial
        Applications } by J. Michael Steele.  Springer, New York, 2001.
\end{enumerate}

\hr
\visual{Links}{../../../../CommonInformation/Lessons/chainlink.jpeg}
\section*{Outside Readings and Links:}
\begin{enumerate}
    \item
        \link{http://www.cs.cornell.edu/Info/Courses/Spring-98/CS522/content/lecture2.math.pdf}
        {Cornell University, Department of Computer Science, Prof. T.
        Coleman Rhodes and Prof.  R. Jarrow} Numerical Solution
        of Black-Scholes Equation, Submitted by Chun Fan, Nov. 12, 2002.
    \item
        \link{http://http://www.maths.monash.edu.au/mth3251/Lectures/NumericalSol.pdf}
        { Monash University, Department of Mathematical Science, Eric.
        W. Chu} This link gives some examples and maple commands, Submitted by Chun
        Fan, Nov.  12, 2002.
    \item
        \link{http://www.blobek.com/black-scholes.html}{
        An applet for calculating the option value} based on the
        Black-Scholes model.  Also contains tips on options, business
        news and literature on options.  Submitted by Yogesh Makkar,
        November 19, 2003.

    \item
        \link{www.xleverywhere.com/samples/bs/bs.htm}{ ExcelEverywhere},
        a commercial application for spreadsheets on the Web.  A
        sample spreadsheet based calculator for calculating the option
        values, based on Black-Scholes model.  Submitted by Yogesh
        Makkar, November 19,2003
\end{enumerate}

\hr

\mydisclaim
\myfooter

Last modified: \flastmod

\end{document}

