problem with fsolve optimization error message

Hi all,

I am trying to calculate the implied volatility of this problem:

>> T = 10;
>> CDS_bps = 337.5;
>> sigma_start = 1;
>> r = 0.03;
>> R_rel = 0.4
sigma_opt(sigma_start, CDS_bps, R_rel, T, r);

This is the function I wrote:

function [sigma,fval] = sigma_opt(sigma_start, CDS_bps, R_rel, T, r)

[sigma,fval] = fsolve(@nested_sigma, sigma_start);

% Nested function that computes the objective function 
function prices = nested_sigma(sigma)

    Lambda = CDS_bps/10000/(1-R_rel);
    Ts = [1:T];
    Default_probs = [0;1- exp(-Lambda .* Ts')];
    partial_Nominator = zeros(1, T+1);
    for i = 1:T
    partial_Nominator(i) = (1-R_rel)*exp(-r*i)*(Default_probs(i+1) - ...
        Default_probs(i));
    end    
    Nominator = sum(partial_Nominator,2);
prices(1) =  Nominator/(1-R_rel);
  d1 = (log(100/5)+(r+sigma^2/2*T))/(sigma*sqrt(T));
  d2 = d1 - sigma *sqrt(T);
prices(2) = exp(-r*T) * normcdf(-d2);

end
end

and this is the error I get:

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
default value of the function tolerance.

<stopping criteria details>


fsolve stopped because the sum of squared function values, r, has gradient with relative
norm 6.524979e-19; this is less than 1e-4*options.OptimalityTolerance = 1.000000e-10.
However, r = 1.377972e-01, exceeds sqrt(options.FunctionTolerance) = 1.000000e-03.

Optimization Metric                                Options
norm(grad r) =   6.52e-19         1e-4*OptimalityTolerance =   1e-10 (default)
r =   1.38e-01                     sqrt(FunctionTolerance) =  1.0e-03 (default)

I would be very happy if anyone could help me on this.

Thank you.

Answer 1

[Star Strider]

There is actually no error or any other problem. The fsolve function is a root-finder, meaning of course that it search for values whre the function values are zero. If it cannot find the roots, it returns that message.

[MW]

Thank you for answering.

I get the following result for sigma in excel: 0,779385279680472. How can I retrieve this result using my formula?

[Star Strider]

My pleasure.

Excel must be doing something different, probably using a different optimization algorithm. When I run your code, I get:

sigma =
   0.141848077754023

fval =
   0.371210469413452   0.000000000025851

This is not to say that the MATLAB result is wrong, it just means that fsolve may not be the correct solver. Note that most of tthe solvers require that your objective function return a scalar value, not the (1x2) vector ‘nested_sigma’ returns.

[Star Strider]

‘I am actually trying to find sigma where price 2 matches price 1. Can you tell me the name of the solver with which such a problem can be solved?’

Solving that is relatively straight forward, and fsolve is ideal in that situation, since you are searching for the ‘sigma’ value where ‘diff(price)’ is zero.

Change your fsolve call to:

[sigma,fval] = fsolve(@(x)diff(nested_sigma(x)), sigma_start);

producing:

sigma =
   0.778769515215809

fval =
    -1.883382338974116e-12

as desired, and reasonably close to the Excel result.

[MW]

Thank you very much! Have a nice week :)

[Star Strider]

As always, my pleasure!

You, too!

Answer 2

[John D'Errico]

You seem to misunderstand how things work.

You have provided ONE unknown to solve for.

However, there are TWO equations that you are trying to solve. For example, at the start point of sigma==1, we see TWO results.

prices
prices =
      0.37121      0.54361

Fsolve is a root finder. It tries to find a value of the unknown variable (here sigma) that reduces BOTH of those results to zero. Of course, no single number for sigma will exist that kills both of the prices off to zero, at the same time. But worse, consider your function. It never even gets to zero for any value of sigma.

I took your function, then varied sigma. So, in this next output, the first column is sigma, the second and third columns are the prices your function computes, as a function of sigma.

[siglist',pricelist]
ans =
          0.1      0.37121   1.8361e-21
          0.2      0.37121   2.9262e-06
          0.3      0.37121    0.0024545
          0.4      0.37121     0.029068
          0.5      0.37121     0.096828
          0.6      0.37121      0.19197
          0.7      0.37121       0.2944
          0.8      0.37121      0.39075
          0.9      0.37121      0.47452
            1      0.37121      0.54361
          1.1      0.37121      0.59835
          1.2      0.37121      0.64032
          1.3      0.37121      0.67155
          1.4      0.37121      0.69415
          1.5      0.37121      0.71009
          1.6      0.37121      0.72104
          1.7      0.37121      0.72838
          1.8      0.37121      0.73317
          1.9      0.37121      0.73623
            2      0.37121      0.73812
          2.1      0.37121      0.73927
          2.2      0.37121      0.73995
          2.3      0.37121      0.74034
          2.4      0.37121      0.74056
          2.5      0.37121      0.74069
          2.6      0.37121      0.74075
          2.7      0.37121      0.74078
          2.8      0.37121       0.7408
          2.9      0.37121      0.74081
            3      0.37121      0.74081

So your start value is 1. I varied sigma from 0.1 to 3. Note that the first price NEVER varies. The second price does not cross zero, although it might if I went lower.

You cannot use fsolve to solve a problem that has no solution.

Worse, you cannot use it to solve an overdetermined problem like this, because it will never satisfy both equatinos at once.

[MW]

I am actually trying to find sigma where price 2 matches price 1. Can you tell me the name of the solver with which such a problem can be solved?