Problem with optimizing a function which has indirect variables (variables are a set of data that is generated with the help of another function)

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I desperately need to find a way to solve this for my work ,
i have a certain function which calculates the fit of my chosen parameters, i have to find the parameters which gives the min possible fit value
above is the fit function
R(Tn) and I(Tn) is the true data
Rn^(k) and In^(k) are the variables .
Rn^(k) and In^(k) are generated from another function which is as follows
function [Z, phase] = modelMT(resistivities, thicknesses,frequency)
Rn^(k) - real(Z);
In^(k) - imag(Z);
it takes set of resistivities and thicknesses , i need to choose these values for which i get the min of fit function .
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harshit vaddi
harshit vaddi 2015 年 6 月 16 日
編集済み: harshit vaddi 2015 年 6 月 16 日
Sorry about that K there means something different it is not kth power , k is just an indication of type of model that is being used , in depth k=2 means 2 layer model . you can just ignore K there and you can consider it as Rn(Tn) and In(Tn). R(Tn), I(Tn), Rn(Tn), In(Tn) here Tn -- nth period those are vectors which change with time period n= 1,2,3,4.....n
Samar Kenkre
Samar Kenkre 2015 年 6 月 16 日
I have a very similar scenario and posted my question about an hour ago. Will let u know if I get any hits.

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回答 (1 件)

Alan Weiss
Alan Weiss 2015 年 6 月 16 日
I am not completely sure, but it appears that you are trying to minimize a sum of squares. If so, then take a look at lsqnonlin in Optimization Toolbox.
It is often the case that such problems have multiple local solutions. If you want to try to find a global solution, then look at this documentation section using Global Optimization Toolbox.
Alan Weiss
MATLAB mathematical toolbox documentation
  1 件のコメント
harshit vaddi
harshit vaddi 2015 年 6 月 17 日
It is minimization of sum of squares but the variables there is a set of data or a vector which is generated by an another function, using the actual variables (i need to find these variables) so that the sum of squares becomes minimum.

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