fsolve not working, always say near zero?

my Black-Scholes Function:

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function C = bs ( Interest, Volatility, Stock, StrikePrice, TimeToMaturity )

d1 = (log(Stock ./ StrikePrice) + (Interest + (Volatility .^ 2) ./ 2) .* TimeToMaturity) ./ (Volatility .* sqrt(TimeToMaturity));

d2 = (log(Stock ./ StrikePrice) + (Interest - (Volatility .^ 2) ./ 2) .* TimeToMaturity) ./ (Volatility .* sqrt(TimeToMaturity));

C = normcdf(d1) .* Stock - normcdf(d2) .* StrikePrice .* exp(-Interest .* TimeToMaturity);

end

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My new function for fsolve

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function F = myfunc(C, Interest, Volatility, Stock, StrikePrice, TimeToMaturity)

F = C - bs(Interest, Volatility, Stock, StrikePrice, TimeToMaturity);

end

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Now I make C =261

volatility = fsolve(@(x) myfunc(261,0.05,x,1033.56, 775, 1/52), 1);

The answer should be around 1.12, but I only get these:

fsolve completed because the vector of function values is near zero as measured by the default value of the function tolerance, and the problem appears regular as measured by the gradient.

Equation solved. The sum of squared function values, r = 1.289886e-20, is less than sqrt(options.TolFun) = 1.000000e-03. The relative norm of the gradient of r, 9.923745e-10, is less than options.TolFun = 1.000000e-06.

Please Help!! Thank you very much!

Answer 1

[Amit]

type

volatility

This is the variable that you stored the solution. I just tested it. The volatility comes to be 1.1195 which is close to 1.12.

The 'Equation solved. The sum of squared function values, r = 1.289886e-20, is less than sqrt(options.TolFun) = 1.000000e-03. The relative norm of the gradient of r, 9.923745e-10, is less than options.TolFun = 1.000000e-06.' is telling you that a solution is found.

[Tiancong]

YES! Finally I got this right, thank you sooooooo much!

[Tiancong]

You are awesome !!! Thank you!