Calibrating a non-linear array in MATLAB using 2 variables

Hello, Currently, I am trying to optimize the following equation:

function fcpred = bup(x)
load calset
c1 = x(1);
c2 = x(2);
fcpred = 100./(1+exp(((c1.*(-2*cprime))+(c2.*cprime.*ln100d))));
end

And claset contains obs, ln100d and cprime 44*1 arrays. Subjected to the contraint:

obs - fcpred = 0

and wish to minimize COV of the above. I have tried:

[x,fval] = fmincon(@bup,[0 0],[],[],[],[],[],[],@opt)

Where [opt]:

function [c, ceq]=opt(x)
load calset_LC
ceq = obs - fcpred
c = []
end

But to the error requesting that the supplied function should return a scalar value. Have I used the wrong function? or should I look to solve it as a system instead?

Answer 1

[Matt J]

I have the impression that like you are trying to solve

   100./(1+exp(((c1.*(-2*cprime))+(c2.*cprime.*ln100d))))  - obs = 0;

If so, and if c1 and c2 are unconstrained, you would probably want to use fsolve,

[x,fval] = fsolve(@(x) bup(x) - obs, x0 )

The initial guess x0 can probably be obtained by rearranging this as a linear equation

   log(100./obs-1) =  c1.*(-2*cprime) + c2.*cprime.*ln100d

and solving. Or, maybe the linear solution is already good enough for your purposes.

[Iliya Nemtsov]

Hi Matt,

Thank you for your suggestion but I am trying to eliminate the residual of the result, hence:

obs - fcpred = 0

c1 and c2 values are calibration factors and yes they are unconstrained.

But using the same logic I had is set up in Excel Solver and C1 and C2 should yield approximately .5236 and 0.1404 respectively.

This code is a stepping stone as I am going to utilize it in a Jack Knife Procedure (custom code) utilizing n-2 sets.

[Matt J]

Thank you for your suggestion but I am trying to eliminate the residual of the result, hence: obs - fcpred = 0

I don't see how you can do that. You have 44 residuals to drive to zero, but only 2 degrees of freedom (c1 and c2).