How can i solve this problem' when i am running, this message appears: Failure in initial objective function evaluation. FMINUNC cannot continue.

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Khadija 2024-4-15，8:44

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2106941-how-can-i-solve-this-problem-when-i-am-running-this-message-appears-failure-in-initial-objective

评论： Khadija 2024-4-16，21:13

采纳的回答： Torsten

I'm writing a matlab program to try to fit a mathematical model to real data.

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Answer 1

Torsten 2024-4-15，9:58

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编辑：Torsten 2024-4-15，9:59

Call your function in which you define the sum of differences squared between real and modelled data for the initial values of your parameters and see what the problem is. Maybe the function returns something undefined or a vector of values instead of a single value.

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Khadija 2024-4-15，13:32

function dy=model_CVD(~,y,k)

delta=0.02 ;

deltaC=0.03 ;

h=1244;

xi = k(1);

sigma = k(2);

alpha= k(3);

gammaH= k(4);

gammaA= k(5);

deltaOC= k(6);

p1= k(7);

p2= k(8);

p3= k(9);

psi=k(10);

dy = zeros(5,1);

dy(1)= h-(delta+sigma+xi+gammaH)*y(1);

dy(2)=sigma*y(1) -(delta+alpha+gammaA)*y(2);

dy(3)=gammaH*y(1)+ psi*y(4) -(delta+deltaOC+p1*sigma+p3*xi)*y(3);

dy(4)=gammaA*y(2)+ p1*sigma*y(3) -(psi+delta+p2*alpha)*y(4);

dy(5)=xi*y(1)+alpha*y(2)+p2*alpha*y(4)+p3*xi*y(3) -(delta+deltaC)*y(5);

end

function error_in_data = moderCVD(k)

% Données spécifiques

specific_data = [

2014 52204 2000 36728 68;

2015 43000 2007 41899 94;

2016 30400 3124 51385 91;

2017 26300 1320 55583 97;

2018 20800 2153 59948 99;

2019 16613 2000 64288 99;

2020 13500 2523 65877 100;

2021 12700 2154 67446 10;

2022 10150 1473 68279 100

];

tforward = 2014:0.01:2022;

tmeasure = [1:100:901]';

% Utilisez les données spécifiques

tdata = specific_data(:, 1);

Hdata = specific_data(:, 2);

Adata=specific_data(:, 3);

THdata=specific_data(:, 4);

TAdata=specific_data(:, 5);

[T, Y] = ode23s(@(t,y)(model_CVD(t,y,k)),tforward,[ 52204.0 2000.0 36728.0 68.0 1000.0]);

Hq = Y(tmeasure(:),1);

Aq = Y(tmeasure(:),2);

THq = Y(tmeasure(:),3);

TAq = Y(tmeasure(:),4);% assignts the y-coordinates of ...

%the solution at

A=(Hq - Hdata).^2;

error_in_data =sum(A);

%%

end

clear all ; clf ; clc ;

% Données spécifiques

specific_data = [

2014 52204 2000 36728 68;

2015 43000 2007 41899 94;

2016 30400 3124 51385 91;

2017 26300 1320 55583 97;

2018 20800 2153 59948 99;

2019 16613 2000 64288 99;

2020 13500 2523 65877 100;

2021 12700 2154 67446 10;

2022 10150 1473 68279 100

];

tforward = 2014:0.01:2022;

tmeasure = [1:100:801]';

% Utilisez les données spécifiques

tdata = specific_data(:, 1);

Hdata = specific_data(:, 2);

Adata = specific_data(:, 3);

THdata = specific_data(:, 4);

TAdata = specific_data(:, 5);

% Paramètres initiaux

%deltaOC= 0.01;

p1= 0.5;

delta=0.016+0.0004 ;

deltaC=0.3 ;

h=1244;

sigma=0.06 ;

xi= 0.02;

alpha= 0.04;

epsilon= 0.03;%epsilon=xi*1.5

beta= 0.06;% beta=alpha*1.5

phi= 0.03;%phi=sigma*0.05

psi= 0.3;

gammaH=0.2 ;

gammaA=0.3 ;

delta=0.0016 ;

deltaA=0.0004 ;

deltaC=0.01 ;

deltaOC= 0.01;

p2=1.5;

p3=2;

k0 = [xi sigma alpha gammaH gammaA deltaOC p1 p2 p3 psi];

% Résolution de l'équation différentielle avec les paramètres initiaux

[T, Y] = ode23s(@(t, y) model_CVD(t, y, k0), tforward, [52204.0 2000.0 36728.0 68.0 1000.0]);

yintH = Y(tmeasure(:),1);

yintA = Y(tmeasure(:),2);

yintTH = Y(tmeasure(:),3);

yintTA = Y(tmeasure(:),4);

% Affichage des données spécifiques et de la solution de l'équation différentielle

% Display the results

figure(1)

%subplot(1,2,1);

plot(tdata, Hdata, 'r.');

hold on

plot(tdata, yintH, 'b-');

xlabel('time in days');

ylabel('Number of cases');

legend('Data', 'Model estimation');

axis([2014 2022 0 100000]);

% Minimization routine using Nelder & Mead Simplex algorithm (a derivative-free method)

% Assigns the new values of parameters to k and the error to fval

% Minimization routine using Nelder & Mead Simplex algorithm (a derivative-free method)

% Assigns the new values of parameters to k and the error to fval

[k,fval] = fminunc(@moderCVD,k0);

%print final values of fitted parametres

disp(k);

xi=k(1);

sigma=k(2);

alpha= k(3);

gammaH=k(4) ;

gammaA=k(5) ;

deltaOC= k(6);

p1= k(7);

p2= k(8);

p3= k(9);

psi= k(10);

%Draw the data with the final ODE

[T Y] = ode45(@(t,y)(model_CVD(t,y,k)),tforward,[52204.0 2000.0 36728.0 68.0 1000.0]);

yintH = Y(tmeasure(:),1);

yintA = Y(tmeasure(:),2);

yintTH = Y(tmeasure(:),3);

yintTA = Y(tmeasure(:),4);

residuals = Hdata - yintH;% a chercher comment calcumer les residus pour plusieurs variables

%subplot(1,2,2);

figure(2)

plot(tdata,Hdata,'r.');

hold on

plot(tdata,yintH,'g-');

xlabel('Time in days');

ylabel('Number of newly infected cases with fitted parameters');

axis([2014 2022 0 100000]);

%Graphe des résidus

%subplot(1,2,2);

figure(3)

plot(tdata, residuals, 'r+');

xlabel('Time in days');

ylabel('Residuals');

title('Residuals Plot');

axis([2014 2022 min(residuals) max(residuals)]);

% % Affichage de la légende

legend('Residuals');

%

% % Vous pouvez également ajouter une ligne horizontale à zéro pour référence

% hold on

% plot([3, 14], [0, 0], 'k--');

%

%Graphe des résidus

%subplot(1,2,2);

%figure(3)

%plot(tdata, residuals, 'g.');

%xlabel('Time in days');

%ylabel('Residuals');

%title('Residuals Plot');

%axis([2014 2022 min(residuals) max(residuals)]);

% Affichage de la légende

%legend('Residuals');

% Ajout de lignes verticales reliant chaque point de résidu à la ligne horizontale à zéro

%hold on

for i = 1:length(tdata)

%plot([tdata(i), tdata(i)], [0, residuals(i)], 'k--');

end

% Ajout de la ligne horizontale à zéro

%plot([2014, 2022], [0, 0], 'k--');

Torsten 2024-4-15，13:51

在 MATLAB Online 中打开

Most probably you mean something like this:

clear all ; clf ; clc ;

% Données spécifiques

specific_data = [

2014 52204 2000 36728 68;

2015 43000 2007 41899 94;

2016 30400 3124 51385 91;

2017 26300 1320 55583 97;

2018 20800 2153 59948 99;

2019 16613 2000 64288 99;

2020 13500 2523 65877 100;

2021 12700 2154 67446 10;

2022 10150 1473 68279 100

];

tforward = 2014:1:2022;

%tmeasure = [1:100:801]';

% Utilisez les données spécifiques

tdata = specific_data(:, 1);

Hdata = specific_data(:, 2);

Adata = specific_data(:, 3);

THdata = specific_data(:, 4);

TAdata = specific_data(:, 5);

% Paramètres initiaux

%deltaOC= 0.01;

p1= 0.5;

delta=0.016+0.0004 ;

deltaC=0.3 ;

h=1244;

sigma=0.06 ;

xi= 0.02;

alpha= 0.04;

epsilon= 0.03;%epsilon=xi*1.5

beta= 0.06;% beta=alpha*1.5

phi= 0.03;%phi=sigma*0.05

psi= 0.3;

gammaH=0.2 ;

gammaA=0.3 ;

delta=0.0016 ;

deltaA=0.0004 ;

deltaC=0.01 ;

deltaOC= 0.01;

p2=1.5;

p3=2;

k0 = [xi sigma alpha gammaH gammaA deltaOC p1 p2 p3 psi];

% Résolution de l'équation différentielle avec les paramètres initiaux

[T, Y] = ode23s(@(t, y) model_CVD(t, y, k0), tforward, [52204.0 2000.0 36728.0 68.0 1000.0]);

yintH = Y(:,1);

yintA = Y(:,2);

yintTH = Y(:,3);

yintTA = Y(:,4);

% Affichage des données spécifiques et de la solution de l'équation différentielle

% Display the results

figure(1)

%subplot(1,2,1);

plot(tdata, Hdata, 'r.');

hold on

plot(tdata, yintH, 'b-');

xlabel('time in days');

ylabel('Number of cases');

legend('Data', 'Model estimation');

axis([2014 2022 0 100000]);

% Minimization routine using Nelder & Mead Simplex algorithm (a derivative-free method)

% Assigns the new values of parameters to k and the error to fval

% Minimization routine using Nelder & Mead Simplex algorithm (a derivative-free method)

% Assigns the new values of parameters to k and the error to fval

[k,fval] = fminunc(@moderCVD,k0);

Local minimum possible. fminunc stopped because the size of the current step is less than the value of the step size tolerance.

%print final values of fitted parametres

disp(k);

0.0099 0.0499 0.0400 0.1899 0.3000 0.0100 0.5000 1.5000 2.0000 0.3000

xi=k(1);

sigma=k(2);

alpha= k(3);

gammaH=k(4) ;

gammaA=k(5) ;

deltaOC= k(6);

p1= k(7);

p2= k(8);

p3= k(9);

psi= k(10);

%Draw the data with the final ODE

[T Y] = ode45(@(t,y)(model_CVD(t,y,k)),tforward,[52204.0 2000.0 36728.0 68.0 1000.0]);

yintH = Y(:,1);

yintA = Y(:,2);

yintTH = Y(:,3);

yintTA = Y(:,4);

residuals = Hdata - yintH;% a chercher comment calcumer les residus pour plusieurs variables

%subplot(1,2,2);

figure(2)

plot(tdata,Hdata,'r.');

hold on

plot(tdata,yintH,'g-');

xlabel('Time in days');

ylabel('Number of newly infected cases with fitted parameters');

axis([2014 2022 0 100000]);

%Graphe des résidus

%subplot(1,2,2);

figure(3)

plot(tdata, residuals, 'r+');

xlabel('Time in days');

ylabel('Residuals');

title('Residuals Plot');

axis([2014 2022 min(residuals) max(residuals)]);

% % Affichage de la légende

legend('Residuals');

%

% % Vous pouvez également ajouter une ligne horizontale à zéro pour référence

% hold on

% plot([3, 14], [0, 0], 'k--');

%

%Graphe des résidus

%subplot(1,2,2);

%figure(3)

%plot(tdata, residuals, 'g.');

%xlabel('Time in days');

%ylabel('Residuals');

%title('Residuals Plot');

%axis([2014 2022 min(residuals) max(residuals)]);

% Affichage de la légende

%legend('Residuals');

% Ajout de lignes verticales reliant chaque point de résidu à la ligne horizontale à zéro

%hold on

for i = 1:length(tdata)

%plot([tdata(i), tdata(i)], [0, residuals(i)], 'k--');

end

% Ajout de la ligne horizontale à zéro

%plot([2014, 2022], [0, 0], 'k--');

function error_in_data = moderCVD(k)

% Données spécifiques

specific_data = [

2014 52204 2000 36728 68;

2015 43000 2007 41899 94;

2016 30400 3124 51385 91;

2017 26300 1320 55583 97;

2018 20800 2153 59948 99;

2019 16613 2000 64288 99;

2020 13500 2523 65877 100;

2021 12700 2154 67446 10;

2022 10150 1473 68279 100

];

tforward = 2014:1:2022;

%tmeasure = [1:100:901]';

% Utilisez les données spécifiques

tdata = specific_data(:, 1);

Hdata = specific_data(:, 2);

Adata=specific_data(:, 3);

THdata=specific_data(:, 4);

TAdata=specific_data(:, 5);

[T, Y] = ode23s(@(t,y)(model_CVD(t,y,k)),tforward,[ 52204.0 2000.0 36728.0 68.0 1000.0]);

Hq = Y(:,1);

Aq = Y(:,2);

THq = Y(:,3);

TAq = Y(:,4);% assignts the y-coordinates of ...

%the solution at

A=(Hq - Hdata).^2;

error_in_data =sum(A);

%%

end

function dy=model_CVD(~,y,k)

delta=0.02 ;

deltaC=0.03 ;

h=1244;

xi = k(1);

sigma = k(2);

alpha= k(3);

gammaH= k(4);

gammaA= k(5);

deltaOC= k(6);

p1= k(7);

p2= k(8);

p3= k(9);

psi=k(10);

dy = zeros(5,1);

dy(1)= h-(delta+sigma+xi+gammaH)*y(1);

dy(2)=sigma*y(1) -(delta+alpha+gammaA)*y(2);

dy(3)=gammaH*y(1)+ psi*y(4) -(delta+deltaOC+p1*sigma+p3*xi)*y(3);

dy(4)=gammaA*y(2)+ p1*sigma*y(3) -(psi+delta+p2*alpha)*y(4);

dy(5)=xi*y(1)+alpha*y(2)+p2*alpha*y(4)+p3*xi*y(3) -(delta+deltaC)*y(5);

end

Torsten 2024-4-16，12:19

编辑：Torsten 2024-4-16，12:21

Star Strider's code under

https://uk.mathworks.com/matlabcentral/answers/43439-monod-kinetics-and-curve-fitting

demonstrates how to use curve fitting for a system of ODEs.

But your code is ok - I doubt you can learn much from the link (except for "lsqcurvefit" as a different possible solver).

By the way: it would help if you used the google translator french -> english and post your comments in English:

https://translate.google.com/?hl=de

Khadija 2024-4-16，21:13

Thank you sir for your help, and I will take your comment into consideration.

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How can i solve this problem' when i am running, this message appears: Failure in initial objective function evaluation. FMINUNC cannot continue.

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