Nonlinear least-squares data fit

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I am trying to make a data fit for the data attached to this post,Nu=f(Re,Theta,Beta).I use lsqnonlin(fun,x0) function for this purpose.I have created a script file for this fitting,but everytime I try to run the script,the program always shows error messages.So,what is the problem with this script.
clc
clear all
% Create an anonymous function that describes the expected relationship
% between X and Y
f=@(c,x) c(1).*(x(:,1).^c(2)).*(x(:,2).^c(3)).*(x(:,3).^c(4))./x(:,4)-1;
% data set
% Specify x variables from data file,Re,Theta and Beta columns.
x=xlsread('all data for fitting');
% Specify y variable from data file ,(Nu)column.
y=x(:,4);
% Specify a vector of starting conditions for the solvers
c0=[1;1;1;1];
% Perform a nonlinear regression
c=lsqnonlin(f,c0);

Accepted Answer

Star Strider
Star Strider on 14 Aug 2020
The objective function needs to be coded as:
ffcn = @(c) f(c,x) - y;
with the complete lsqnonlin call being:
f=@(c,x) c(1).*(x(:,1).^c(2)).*(x(:,2).^c(3)).*(x(:,3).^c(4))./x(:,4)-1;
ffcn = @(c) f(c,x) - y;
c0=[1;1;1;1];
C = lsqnonlin(ffcn, c0);
producing:
C =
1.0308e-01
1.3246e+00
1.9801e-06
-4.6017e-01
.
  13 Comments
Star Strider
Star Strider on 19 Aug 2020
I used this:
D = xlsread('all data for fitting.xlsx');
x = D;
y = x(:,4);
f=@(c,x) c(1).*(x(:,1).^c(2)).*(x(:,2).^c(3)).*(x(:,3).^c(4));
ffcn = @(c) (f(c,x) - y)./y;
ftns = @(c) norm(ffcn(c));
PopSz = 500;
Parms = 4;
opts = optimoptions('ga', 'PopulationSize',PopSz, 'InitialPopulationMatrix',randi(1E+4,PopSz,Parms)*1E-4, 'MaxGenerations',2E3, 'PlotFcn',@gaplotbestf, 'PlotInterval',1);
t0 = clock;
fprintf('\nStart Time: %4d-%02d-%02d %02d:%02d:%07.4f\n', t0)
[theta,fval,exitflag,output] = ga(ftns, Parms, [],[],[],[],-Inf(Parms,1),Inf(Parms,1),[],[],opts)
t1 = clock;
fprintf('\nStop Time: %4d-%02d-%02d %02d:%02d:%07.4f\n', t1)
GA_Time = etime(t1,t0)
QQQ1 = datetime([zeros(1,5) GA_Time], 'Format','HH:mm:ss.SSS')
fprintf('\nElapsed Time: %23.15E s ', GA_Time)
fprintf(1,'\tRate Constants:\n')
for k1 = 1:length(theta)
fprintf(1, '\t\tTheta(%d) = %12.5E\n', k1, theta(k1))
end
and when I ran that just now, got these parameter estimates:
theta =
2.8517e+000 431.7000e-003 99.6000e-003 -324.6437e-003
with a fitness value of:
fval =
5.5386e+000
.

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