nonlinear regression for multiple variable parameter

I'm trying to write code to perform a nonlinear regression with multiple variables, but i have no idea how to do it.

i need to use Keq and pressure like Xdata and define the x1,x2,x3 from the regression.

I hope I have explained myself.

Function value and YDATA sizes are not equal.

Error in prova3 (line 49)

x_fit = lsqcurvefit(T, x0, [pressure Keq], temperature);

clc
clear all
close all
%%
%read all .xlsx file inside the folder
path = dir('C:\Users\Andrea\Desktop\matlab\input_file\*.xlsx');

%cicle for obtain the sample data (P-T-Ni) content directly from the .xlsx file
for i=1:1
    table = readtable(strcat('C:\Users\Andrea\Desktop\matlab\input_file\',path(i).name));
    temperature(:,i) = table.temperature;
    pressure(:,i) = table.pressure;
    Ni_grt(:,i) = table.Ni_Grt;
    Ni_ol(:,i) = table.Ni_Ol;
    Cr_grt(:,i) = table.Wt_Cr;
    Ca_grt(:,i) = table.Wt_Ca;
end
%% calculation
%Keq 
Ni_grt = Ni_grt(~isnan(Ni_grt), :);
Keq= log(Ni_grt/2900);

%temperature from C to K
temperature = temperature + 273.15;

%olivine mean
Ni_ol = Ni_ol(~isnan(Ni_ol), :);
mean_Ni_ol = mean(Ni_ol);

%% ignore the NaN value
temperature = temperature(~isnan(temperature), :);
pressure = pressure(~isnan(pressure), :);
Ca_grt = Ca_grt(~isnan(Ca_grt), :);
Cr_grt = Cr_grt(~isnan(Cr_grt), :);
%% Calculate the parameters (ΔH, ΔV, ΔS)

% function of T
T = @(x,pressure,Keq) (x(1) + pressure * x(2)) ./ (x(3) - log(Keq));

% random start parameters (ΔH, ΔV, ΔS)
x0 = [1, 1, 1];

% nonlinear regression with lsqcurvefit
x_fit = lsqcurvefit(T, x0, [pressure Keq], temperature);

% Extract the fitted parameters
DeltaH = x_fit(1);
DeltaV = x_fit(2);
DeltaS = x_fit(3);

% the new value od temperature from Ni_grt of database
predicted_temperature = T(x_fit, pressure);

% print the result
fprintf('Parametri adattati:\n');
fprintf('ΔH = %.2f\n', DeltaH);
fprintf('ΔV = %.2f\n', DeltaV);
fprintf('ΔS = %.2f\n', DeltaS);
fprintf('Ni_ol = %.2f\n', mean_Ni_ol);

%% diff TNi-T
deltaT = predicted_temperature-temperature;
mean_deltaT = mean(deltaT);
fprintf('mean_detaT = %.2f\n', mean_deltaT);

Answer 1

[Star Strider]

Your code works. The only change required was to create the ‘pressure_Keq’ matrix as the independent variable and then change ‘T’ to refer to it correctly internally.

Try this —

clc
clear all
close all
%%
%read all .xlsx file inside the folder
% path = dir('C:\Users\Andrea\Desktop\matlab\input_file\*.xlsx');
path = dir('*.xlsx');

%cicle for obtain the sample data (P-T-Ni) content directly from the .xlsx file
for i=1:1
    % table = readtable(strcat('C:\Users\Andrea\Desktop\matlab\input_file\',path(i).name));
    table = readtable(path(i).name)
    temperature(:,i) = table.temperature;
    pressure(:,i) = table.pressure;
    Ni_grt(:,i) = table.Ni_Grt;
    Ni_ol(:,i) = table.Ni_Ol;
    Cr_grt(:,i) = table.Wt_Cr;
    Ca_grt(:,i) = table.Wt_Ca;
end

table = 171×6 table
    pressure    temperature    Ni_Grt    Wt_Cr     Wt_Ca     Ni_Ol
    ________    ___________    ______    ______    ______    _____

     61.627       1292.9       119.99    9.5358    6.7344     NaN 
     59.458         1280        91.07    4.7708    5.4524    3858 
     59.035       1269.2       199.38    11.087     6.926     NaN 
     57.332       1198.3       51.712    11.273    6.2995    3078 
      58.42       1266.4       84.844    5.8212    5.8972    3110 
     66.891       1265.2       74.004    2.0122    4.1251    3734 
     62.266         1276       114.78    4.7802    5.1415     NaN 
     63.981       1223.7       79.805    9.2159     6.561     NaN 
     66.462       1284.9       109.54    2.6306    4.3678    3650 
     68.121         1315        104.8    2.6699     4.372     NaN 
     60.293       1291.5       143.75    9.6895    6.3098     NaN 
     48.298       911.33        25.03    6.5673     6.485     NaN 
     39.735       889.06        138.1    7.4448    7.0952     NaN 
     29.302       761.53        19.58    5.1597    6.3017     NaN 
     48.895       894.94        28.51    5.4462     5.466     NaN 
     59.131       987.97        35.22    6.7152    6.3878     NaN 

%% calculation
%Keq 
Ni_grt = Ni_grt(~isnan(Ni_grt), :);
Keq= log(Ni_grt/2900);

%temperature from C to K
temperature = temperature + 273.15;

%olivine mean
Ni_ol = Ni_ol(~isnan(Ni_ol), :);
mean_Ni_ol = mean(Ni_ol);

%% ignore the NaN value
temperature = temperature(~isnan(temperature), :);
pressure = pressure(~isnan(pressure), :);
Ca_grt = Ca_grt(~isnan(Ca_grt), :);
Cr_grt = Cr_grt(~isnan(Cr_grt), :);
%% Calculate the parameters (ΔH, ΔV, ΔS)

% function of T
pressure_Keq = [pressure Keq];                              % Create The Independent Variable As A Matrix, Then Address Its Columns Appropriately In The Objective Function
T = @(x,pressure_Keq) (x(1) + pressure_Keq(:,1) * x(2)) ./ (x(3) - log(pressure_Keq(:,2)));

% random start parameters (ΔH, ΔV, ΔS)
x0 = [1, 1, 1];

% nonlinear regression with lsqcurvefit
x_fit = lsqcurvefit(T, x0, pressure_Keq, temperature);

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

% Extract the fitted parameters
DeltaH = x_fit(1);
DeltaV = x_fit(2);
DeltaS = x_fit(3);

% the new value od temperature from Ni_grt of database
predicted_temperature = T(x_fit, pressure_Keq);

% print the result
fprintf('Parametri adattati:\n');

Parametri adattati:

fprintf('ΔH = %.2f\n', DeltaH);

ΔH = -2319.97

fprintf('ΔV = %.2f\n', DeltaV);

ΔV = -9.90

fprintf('ΔS = %.2f\n', DeltaS);

ΔS = -0.67

fprintf('Ni_ol = %.2f\n', mean_Ni_ol);

Ni_ol = 2759.78

%% diff TNi-T
deltaT = predicted_temperature-temperature;
mean_deltaT = mean(deltaT);
fprintf('mean_detaT = %.2f\n', mean_deltaT);

mean_detaT = 0.34

.

[andrea rigotto]

thank you for reply, but the results are in complex numbers, how can I obtain the results in real numbers?

[Star Strider]

That may be because you are calculating the logarithm of a negative number.

I just noticed this now, that these two calculations appear to be redundant:

Keq = log(Ni_grt/2900);

that division is likely to result in values less than 1 (so the log of ‘Keq’ will be negative in those instances), and later you take the log of ‘Keq’ in ‘T’:

T = @(x,pressure_Keq) (x(1) + pressure_Keq(:,1) * x(2)) ./ (x(3) - log(pressure_Keq(:,2)));

Perhaps taking the log of ‘Keq’ in ‘T’ needs to be changed, so that you are only subtracting ‘Keq’ and not the log of it.

I do not know what you are calculating, so I am hesitant to make any other changes to your code.

.