first time matlab user, trying to use fmincon for a simple question

Hi, I am trying to use fmincon to get the best result for an input variable. I think I am close. Student here!

my input variables are

S0 = 1433.32; K = 1430; r = 0.0011; T = 30/365; sigma = .15;

y = ??

the y is my variable I need to change, so that the result of this function will be 24.5.

x0 = 24.9;
x = fmincon(blackScholesCallPrice,x0,[],[])
function [cprice] = blackScholesCallPrice( K, T, S0, r, y, sigma )
numerator = log(S0./K) + (r-y+0.5*sigma.^2).*T;
denominator = sigma.*sqrt(T);
d1 = numerator./denominator;
d2 = d1 - denominator;
cprice = S0 *exp(-y*T).* normcdf(d1) - exp(-r.*T).*K.*normcdf(d2);
end

Answer 1

[Walter Roberson]

S0 = 1433.32; K = 1430; r = 0.0011; T = 30/365; sigma = .15;

x0 = 24.9;
x = fmincon(@(y)blackScholesCallPrice(K,T,S0,r,y,sigma), x0, [], [])

Initial point is a local minimum that satisfies the constraints.

Optimization completed because at the initial point, the objective function is non-decreasing 
in feasible directions to within the value of the optimality tolerance, and 
constraints are satisfied to within the value of the constraint tolerance.

x = 24.9000

syms Y
X = blackScholesCallPrice( K, T, S0, r, Y, sigma )

X = 

dX = diff(X,Y)

dX = 

bestX = vpasolve(dX == 0)

 
bestX =
 
Empty sym: 0-by-1
 

fplot(dX, [0 1])

limit(dX, Y, 0)

ans = 

vpa(ans)

ans = 

limit(dX, Y, inf)

ans = 

0

vpa(ans)

ans = 

0.0

function [cprice] = blackScholesCallPrice( K, T, S0, r, y, sigma )
numerator = log(S0./K) + (r-y+0.5*sigma.^2).*T;
denominator = sigma.*sqrt(T);
d1 = numerator./denominator;
d2 = d1 - denominator;
cprice = S0 *exp(-y*T).* normcdf(d1) - exp(-r.*T).*K.*normcdf(d2);
end

So the minimum of the function is at Y = infinity

[Matt J]

Brian O'Connell's reply moved here:

HI, Thanks.

I am way out of my depth here. The lecturer said the cprice should be the midpoint of 24.1 - 24.9, which is why I put x0 = 24.5.

I think y is supped to equal 0.3. he just wants us to familiarise ourselves with using fmincon. Apparently its better than excel solver, which is where I got the result 0.3 for Y.

regards

brian