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How to use Nlinfit for a function with two independent variables?

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Hi here is my data and code and I am trying to predict parameters for a function with two independent variables but Nlinfit is giving me error.
clc
clear %all; % Clear the workspace.
close all; % Close all figures.
format compact
%% Read in data
data =readmatrix('Flexible_BUC.xlsx');
x1=[7.50000000000000
7.50000000000000
7.70000000000000
5
5
5
5
5
5
5
5
5
7.50000000000000
7.50000000000000
7.50000000000000
7.50000000000000
7.50000000000000
6
6
6
6
6
6
3.75000000000000
3.75000000000000
3.75000000000000
3.75000000000000
3.75000000000000
8
8
8
8
8
8
7.50000000000000
7.50000000000000
7.50000000000000
8
8
8];
x2=[0.00153000000000000
0.00522000000000000
0.000189000000000000
3.73000000000000e-06
3.73000000000000e-06
1.17000000000000e-05
8.66000000000000e-05
1.17000000000000e-05
1.51000000000000e-05
2.99000000000000e-05
7.00000000000000e-05
7.92000000000000e-05
6.06000000000000e-05
0.000163000000000000
0.000656000000000000
0.000818000000000000
0.00129000000000000
9.78000000000000e-06
0.000161000000000000
0.000183000000000000
0.000204000000000000
0.000297000000000000
0.000343000000000000
0.00705000000000000
0.00850000000000000
0.0233000000000000
0.0250000000000000
0.0267000000000000
0.00125000000000000
0.00245000000000000
0.00391000000000000
0.00878000000000000
0.00111000000000000
0.00122000000000000
3.23000000000000e-05
4.26000000000000e-05
4.53000000000000e-05
4.13000000000000e-05
8.24000000000000e-05
8.33000000000000e-05];
yobs=[5.95833333300000
0.300000000000000
0.0625000000000000
0.111111111000000
0.213809289000000
0.140625000000000
0.651315789000000
0.351694915000000
12.0555555600000
0.626846311000000
0.555555556000000
0.136363636000000
6.38793103400000
0.233051458000000
0.540000000000000
0.0277777780000000
1.05555555600000
0.113636364000000
0.0933323590000000
0.352272727000000
3.20833333300000
0.897435897000000
1.17046404700000
1.41666666700000
1.79545454500000
1.15384615400000
1.85576923100000
10.8333333300000
0.848684211000000
6.18835443000000
0.767441860000000
0.527777778000000
1.54872306000000
0.337691494000000
1.08333333300000
1.87477002000000
1.81654734900000
1.97222222200000
0.550000000000000
0.340020401000000];
xm=[x1 x2];
%% Initial parameter guesses
C1=0.25;
C2=0.73;
beta0(1)=C1; %initial guess beta 1
beta0(2)=C2; %initial guess beta 2
p=length(beta0); %p = # parameters
%% define function to be used for inverse problem
fINV=@Project_funcINV;
%fnameINV=@forderexpINV;
[beta,resids,J,COVB,mse] = nlinfit(xm,yobs,fINV,beta0);
beta
%% Functions
function y=Project_funcINV(beta0,x1,x2)
c2s=@(x)-2.40874-39.748*(1+x).^-2.856;
y=100./(1+exp(-beta0(1).*c2s(x1)+(beta0(2).*c2s(x1).*log10(100.*x2))));
end
Someone can please help, I would appreciate it.

Risposta accettata

Star Strider
Star Strider il 23 Feb 2023
You are using the correct approach with:
xm=[x1 x2];
In the function, refer to ‘x1’ as ‘xm(:,1)’ and ‘x2’ as ‘xm(:,2)’ , passing ‘xm’ as the independent variable to ‘Project_funcINV’. I made those changes, and added a fitnlm call to display the statistics, and provided a plot of the data and the fit to it (as a line plot).
Try this —
x1=[7.50000000000000
7.50000000000000
7.70000000000000
5
5
5
5
5
5
5
5
5
7.50000000000000
7.50000000000000
7.50000000000000
7.50000000000000
7.50000000000000
6
6
6
6
6
6
3.75000000000000
3.75000000000000
3.75000000000000
3.75000000000000
3.75000000000000
8
8
8
8
8
8
7.50000000000000
7.50000000000000
7.50000000000000
8
8
8];
x2=[0.00153000000000000
0.00522000000000000
0.000189000000000000
3.73000000000000e-06
3.73000000000000e-06
1.17000000000000e-05
8.66000000000000e-05
1.17000000000000e-05
1.51000000000000e-05
2.99000000000000e-05
7.00000000000000e-05
7.92000000000000e-05
6.06000000000000e-05
0.000163000000000000
0.000656000000000000
0.000818000000000000
0.00129000000000000
9.78000000000000e-06
0.000161000000000000
0.000183000000000000
0.000204000000000000
0.000297000000000000
0.000343000000000000
0.00705000000000000
0.00850000000000000
0.0233000000000000
0.0250000000000000
0.0267000000000000
0.00125000000000000
0.00245000000000000
0.00391000000000000
0.00878000000000000
0.00111000000000000
0.00122000000000000
3.23000000000000e-05
4.26000000000000e-05
4.53000000000000e-05
4.13000000000000e-05
8.24000000000000e-05
8.33000000000000e-05];
yobs=[5.95833333300000
0.300000000000000
0.0625000000000000
0.111111111000000
0.213809289000000
0.140625000000000
0.651315789000000
0.351694915000000
12.0555555600000
0.626846311000000
0.555555556000000
0.136363636000000
6.38793103400000
0.233051458000000
0.540000000000000
0.0277777780000000
1.05555555600000
0.113636364000000
0.0933323590000000
0.352272727000000
3.20833333300000
0.897435897000000
1.17046404700000
1.41666666700000
1.79545454500000
1.15384615400000
1.85576923100000
10.8333333300000
0.848684211000000
6.18835443000000
0.767441860000000
0.527777778000000
1.54872306000000
0.337691494000000
1.08333333300000
1.87477002000000
1.81654734900000
1.97222222200000
0.550000000000000
0.340020401000000];
xm=[x1 x2];
%% Initial parameter guesses
C1=0.25;
C2=0.73;
beta0(1)=C1; %initial guess beta 1
beta0(2)=C2; %initial guess beta 2
p=length(beta0); %p = # parameters
%% define function to be used for inverse problem
fINV=@Project_funcINV;
%fnameINV=@forderexpINV;
[beta,resids,J,COVB,mse] = nlinfit(xm,yobs,fINV,beta0);
beta
beta = 1×2
1.3585 0.1749
mdl = fitnlm(xm,yobs,fINV,beta0) % ADDED
mdl =
Nonlinear regression model: y ~ Project_funcINV(b,X) Estimated Coefficients: Estimate SE tStat pValue ________ _______ ______ __________ b1 1.3585 0.13278 10.231 1.8021e-12 b2 0.17495 0.11064 1.5812 0.12212 Number of observations: 40, Error degrees of freedom: 38 Root Mean Squared Error: 2.81 R-Squared: -0.0164, Adjusted R-Squared -0.0431 F-statistic vs. zero model: 7.46, p-value = 0.00185
figure % ADDED
stem3(x1, x2, yobs, 'filled')
hold on
plot3(x1, x2, Project_funcINV(beta,xm), '-r')
hold off
%% Functions
function y=Project_funcINV(beta0,xm)
c2s=@(x)-2.40874-39.748*(1+x).^-2.856;
y=100./(1+exp(-beta0(1).*c2s(xm(:,1))+(beta0(2).*c2s(xm(:,1)).*log10(100.*xm(:,2)))));
end
The fit is reasonably good, although ‘beta(2)’ may not be significnatly different from zero.
.
  13 Commenti
Faizan Lali
Faizan Lali il 27 Feb 2023
How you parameter estimates are so high?
Which code did you use?
Walter Roberson
Walter Roberson il 27 Feb 2023
Alex uses a commercial program named 1stOpt that does some very nice optimization. Sometimes I am able to improve a little on his results, but not usually, and when I do manage then it is only after a couple of days of continuous computations.

Accedi per commentare.

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