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# The issue of optimization (minimization) of the average relative error between experimental and calculated data

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roborrr
on 22 Aug 2024

hello

I want to share the difficulties that I faced. Can someone help

problem statement:

there is a 'x' column where the values of the independent variable are written and there is a 'y' column where the experimental values of the dependent variable are written.

approximation model is considered:

y_calculate=A*x^B+C,

and based on this model, an objective function is created, which is equal to the average value of the relative deviations between y and y_calculate:

error_function = mean(abs(y - y_calculate)) / y)=mean(abs(y - =A*x^B+C)) / y);

Our goal is to select parameters A,B,C in such a way that 'error_function' takes the value of the global minimum.

I calculated the optimal values of A, B, C and got:

A = 85.5880, B = -0.0460, C = 4.8824,

at which error function value for optimized parameters: 0.0285.

but I know in advance the specific values of A, B, C:

A = 1005.6335852931, B = -1.59745963925582, C = 73.54149744754400,

at which error function value for specific parameters: 0.002680472178434,

which is much better than with optimization

Below is the code with visualization, which confirms the above.

clear

close all

% Data

x = [7.3392, 14.6784, 22.0176, 29.3436, 36.6828, 44.0088, 51.3348, 58.674, 66, 73.3392, 80.6652, 88.0044, 95.3304, 102.6696, 109.9956, 117.3348, 124.6608, 132];

y = [115.1079, 87.7698, 80.5755, 78.1611, 76.5743, 75.7074, 74.9375, 74.9453, 74.59, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990];

% Initial guesses for parameters A, B, C

initial_guess = [1, 1, 1];

% Error function

error_function = @(params) mean(abs(y - (params(1) * x.^params(2) + params(3))) ./ y);

% Optimization of parameters

optimized_params = fminsearch(error_function, initial_guess);

% Results of optimization

A_optimized = optimized_params(1);

B_optimized = optimized_params(2);

C_optimized = optimized_params(3);

% Calculation of the fitted function for optimized parameters

y_calculate_optimized = A_optimized * x.^B_optimized + C_optimized;

% Calculate and display the error function value for optimized parameters

value_error_optimized = error_function(optimized_params);

fprintf('Optimized parameters:\nA = %.4f\nB = %.4f\nC = %.4f\n', A_optimized, B_optimized, C_optimized);

fprintf(' error function value for optimized parameters: %.4f\n', value_error_optimized);

% Other specific parameters A, B, C

A_specific = 1005.63358529310;

B_specific = -1.59745963925582;

C_specific = 73.541497447544;

% Calculation of the fitted function for specific parameters

y_calculate_specific = A_specific * x.^B_specific + C_specific;

% Calculate and display the error function value for specific parameters

value_error_specific = error_function([A_specific, B_specific, C_specific]);

fprintf('Specific parameters:\nA = %.10f\nB = %.14f\nC = %.14f\n', A_specific, B_specific, C_specific);

fprintf(' error function value for specific parameters: %.4f\n', value_error_specific);

% Visualization

figure;

plot(x, y, 'bo-', 'DisplayName', 'Experimental data');

hold on;

plot(x, y_calculate_optimized, 'r--', 'DisplayName', 'Fitted model (Optimized)');

plot(x, y_calculate_specific, 'g-.', 'DisplayName', 'Fitted model (Specific)');

xlabel('x');

ylabel('y');

legend('Location', 'best');

title('Approximation of experimental data');

grid on;

Obviously, my optimization code does not lead to a global minimum of the objective function, since there is a better approximation for specific values of A,B,C. Maybe this is caused by a random selection of the initial values of the parameters A=1, B=1, c=1 and therefore my code is stuck in a local minimum?

who can write a code that will select the A,B,C parameters so as to achieve the global minimum of the target function 'error_function', for any initial iteration data of the variables A,B,C. Thoughts for testing: the value of the target function 'error_function' should not be worse (that is, more) than 0.002680472178434, which is obtained with the specific value of A,B,C: A = 1005.6335852931, B = -1.59745963925582, C = 73.54149744754400

### Accepted Answer

Matt J
on 22 Aug 2024

Edited: Matt J
on 5 Sep 2024

If you download minL1lin from

then you can do,

% Data

x = [7.3392, 14.6784, 22.0176, 29.3436, 36.6828, 44.0088, 51.3348, 58.674, 66, 73.3392, 80.6652, 88.0044, 95.3304, 102.6696, 109.9956, 117.3348, 124.6608, 132];

y = [115.1079, 87.7698, 80.5755, 78.1611, 76.5743, 75.7074, 74.9375, 74.9453, 74.59, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990];

% Initial guess for parameter B

initial_guess = 1;

% Error function

error_function = @(B) objFun(B, x,y);

% Optimization of parameters

B = fminsearch(error_function, initial_guess);

[fval,ABC]=error_function(B);

[A,B,C]=deal(ABC{:})

fval

function [fval,p]=objFun(theta, x,y)

arguments

theta; x (:,1); y(:,1);

end

B=theta;

d=ones(size(y));

Q=[x.^B, d]./y;

[AC,fval]=minL1lin(Q,d,[],[],[],[],[],[],[],optimoptions('linprog','Display','off'));

fval=fval/numel(y);

A=AC(1); C=AC(2);

p={A, B, C};

end

##### 31 Comments

roborrr
on 23 Aug 2024

Edited: roborrr
on 23 Aug 2024

Matt J, thank you for your response.

I ran your code in my matlab and saw the following error:

"Error: Line: 26 Column: 2

Function definitions in a script must appear at the end of the file.

Move all statements after the "objFun" function definition to before the first local function definition."

As I found out the reason of this error is that my version of matlab R(2018b) does not support the following construction:

arguments

B; x (:,1); y(:,1);

end

this construction was introduced starting with version R(2019b)

please rework this part of the code so that it works in my version of matlab - R2018b.

roborrr
on 4 Sep 2024

Edited: roborrr
on 4 Sep 2024

Dear Matt J, I checked the code you suggested and got a stunning result for the model

y=A*x^B+C

However, I did not understand much about how it works.

My goal is to check other models on the same experimental data x,y and choose the best model from them. I tried to do it myself, following your example, but I did not succeed and gave up.

Here are the models that I want to check:

Model 1

y_calculate = eta_inf + (eta_0 - eta_inf) / (1 + (lambda * x)^m);

In this model, two options need to be considered:

1.1- all parameters need to be optimized - eta_0 , eta_inf , lambda , m

1.2 - we assume that m=1.5 is known and we optimize only the parameters - eta_0, eta_inf, lambda

Model 2

y_calculate = eta_inf + (eta_0 - eta_inf) / (1 + (lambda * x)^a)^((1 - n) / a);

in this model, as in model 1, two cases are considered:

2.1. - all parameters need to be optimized - eta_0 , eta_inf , lambda , n,a

2.2 - we assume that parameters eta_0 , eta_inf , lambda , n is known and we optimize only the parameter a.

Matt J
on 4 Sep 2024

Dear Matt J, I checked the code you suggested and got a stunning result for the model

Can I assume "stunning" means that it did what you want? If so, please Accept-click the answer to indicate that it resolved your question.

I tried to do it myself, following your example, but I did not succeed and gave up.

To see what went wrong, I would need to see what you tried.

all parameters need to be optimized - eta_inf , eta_inf , lambda , m

Here and elsewhere you have listed eta_inf twice, so I'm not sure what the model unknowns are.

roborrr
on 5 Sep 2024

Can I assume "stunning" means that it did what you want? If so, please Accept-click the answer to indicate that it resolved your question.

Yes, "stunning" does mean that you did the initial task perfectly, which I did not even expect, for which I am very grateful. I simply forgot to click the "Accept" button, for which I apologize.

Here and elsewhere you have listed eta_inf twice, so I'm not sure what the model unknowns are.

I made a mistake in the letter, where eta_inf is repeated twice, while in place of the first eta_inf, there should be eta_0. I immediately edited the letter, apparently you read the old version of the letter. Please read the new version of the previous letter.

To see what went wrong, I would need to see what you tried.

I didn't try anything because I couldn't figure out how it all worked. How do you solve nonlinear problems with 'minL1lin', which solves the linear problem C*x=d. I tried to figure it out, but I couldn't. Is it possible to write similar codes, as you did for the model y=A*x^B+C, for the models I presented in the previous letter?

Matt J
on 5 Sep 2024

Edited: Matt J
on 6 Sep 2024

Your objective function is, in general,

mean( abs( y_calculate./y -1 ) );

Originally, you had

y_calculate./y = A*x.^B./y + A./y

= [x.^B, d] * [A;C]

= Q * [A;C]

where we have defined the matrices,

d=ones(size(y));

Q=[x.^B, d]./y;

Now take one of your new models. It can be put in the same form:

y_calculate./y = A* 1./(1 + (lambda * x).^m)./y + C./y ;

= Q *[A;C]

but notice that Q has the new form,

Q=[ 1./(1 + (lambda * x).^m), d]./y;

So,

function [fval,p]=objFun(theta, x,y)

x=x(:);y=y(:); theta=theta(:)';

lambda=theta(1); m=theta(2);

d=ones(size(y));

Q=[ 1./(1 + (lambda * x).^m), d]./y;

[AC,fval]=minL1lin(Q,d,[],[],[],[],[],[],[],optimoptions('linprog','Display','off'));

fval=fval/numel(y);

A=AC(1); C=AC(2);

p={A, theta, C};

end

roborrr
on 6 Sep 2024

Following your tips I wrote a program for the model

y_calculate = A* (1 + (lambda * x).^m)+ C,

where it is necessary to calculate the best parameters

A,Lambda,m,C

or optimizing the objective functionЖ

mean( abs( y_calculate-y)/y) .

Here is the code that I wrote following your tips:

clear

% Data

x = [7.3392, 14.6784, 22.0176, 29.3436, 36.6828, 44.0088, 51.3348, 58.674, 66, 73.3392, 80.6652, 88.0044, 95.3304, 102.6696, 109.9956, 117.3348, 124.6608, 132];

y = [115.1079, 87.7698, 80.5755, 78.1611, 76.5743, 75.7074, 74.9375, 74.9453, 74.59, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990, 74.2990];

% Initial guess for parameter theta

initial_guess = 1;

% Error function

error_function = @(theta) objFun(theta, x,y);

% Optimization of parameters

theta = fminsearch(error_function, initial_guess);

[fval,AthetaC]=error_function(theta);

[A, theta,C]=deal(AthetaC{:})

fval

function [fval,p]=objFun(theta, x,y)

x=x(:);y=y(:); theta=theta(:)';

lambda=theta(1); m=theta(2);

d=ones(size(y));

Q=[ 1./(1 + (lambda * x).^m), d]./y;

[AC,fval]=minL1lin(Q,d,[],[],[],[],[],[],[],optimoptions('linprog','Display','off'));

fval=fval/numel(y);

A=AC(1); C=AC(2);

p={A, theta, C};

end

This code gives errors in my matlab. where am I making mistakes?

Torsten
on 6 Sep 2024

Matt J
on 6 Sep 2024

roborrr
on 11 Sep 2024 at 17:34

I corrected the code according to your tips and received:

clear

% Data

% Initial guess for parameter theta

initial_guess = [1,1];

% Error function

error_function = @(theta) objFun(theta, x,y);

% Optimization of parameters

theta = fminsearch(error_function, initial_guess);

[fval,AthetaC]=error_function(theta);

[A, theta,C]=deal(AthetaC{:})

fval

function [fval,p]=objFun(theta, x,y)

x=x(:);y=y(:); theta=theta(:)';

lambda=theta(1); m=theta(2);

d=ones(size(y));

Q=[ 1./(1 + (lambda * x).^m), d]./y;

[AC,fval]=minL1lin(Q,d,[],[],[],[],[],[],[],optimoptions('linprog','Display','off'));

fval=fval/numel(y);

A=AC(1); C=AC(2);

p={A, theta, C};

for the values of the variables x,y specified in the code, the result is fine.

but I encountered values of the variables x,y for which the code gives an error. for example, for the following values of x,y:

y = [21.58079135, 17.26809355, 15.3488609, 14.1074809, 13.81896005, 13.43865625, 13.49266905, 13.44, 13.44, 13.5349172, 13.4323965, 13.439328, 13.36635835, 13.365, 13.29, 13.29, 13.51, 13.365];

the code gives the following error:

Error using linprog (line 373)

A must be a real matrix.

Error in comment_3255504>minL1lin (line 147)

[xz,~, varargout{3:nargout-1}]=linprog(f,A,b,Aeq,beq,LB,UB,args{:});

Error in comment_3255504>objFun (line 23)

[AC,fval]=minL1lin(Q,d,[],[],[],[],[],[],[],optimoptions('linprog','Display','off'));

Error in comment_3255504>@(theta)objFun(theta,x,y) (line 8)

error_function = @(theta) objFun(theta, x,y);

Error in fminsearch (line 336)

fxr = funfcn(x,varargin{:});

how to make the code work correctly for all values of the x,y variables?

roborrr
on 11 Sep 2024 at 22:26

Use Q=[ 1./(1 + (lambda^2 * x).^m), d]./y;

great, this trick prevents complex numbers from appearing when negative lambda values are used. I modified your idea a bit and instead of 'lambda^2' I used 'abs(lambda)' and got the lambda value directly, without the need to square it.

Torsten
on 11 Sep 2024 at 22:52

I modified your idea a bit and instead of 'lambda^2' I used 'abs(lambda)' and got the lambda value directly, without the need to square it.

Usually, this is not a good idea since abs() is not differentiable, and most solvers for optimization need differentiability with respect to the parameters. But you are in luck: fminsearch does not belong to this class of solvers.

roborrr
on 12 Sep 2024 at 9:45

Usually, this is not a good idea since abs() is not differentiable, and most solvers for optimization need differentiability with respect to the parameters. But you are in luck: fminsearch does not belong to this class of solvers.

Thank you, I understand you. I want the code to be universal and not depend on luck. So I will remake the code exactly as you advised.

And one more question: is it possible to impose restrictions on the parameters, for example, for the parameter C impose such a restriction:

C>0 and Abs(C-min(y))/min(y)<0.1

roborrr
on 12 Sep 2024 at 13:09

In the call to minL1lin,

minL1lin(C,d,A,b,Aeq,beq,lb,ub,x0,options)

define the lower bound (lb) for C as 0.9*min(y) and the upper bound (ub) for C as 1.1*min(y).

I already call 'minL1lin' once in the code

[AC,fval]=minL1lin(Q,d,[],[],[],[],[],[],[],optimoptions('linprog','Display','off'));

Do I need to call 'minL1lin' again? Where in the code should I call it minL1lin(C,d,A,b,Aeq,beq,lb,ub,x0,options)? Should I define lb and ub before or after the call?

Matt J
on 12 Sep 2024 at 13:15

No you do not need to call minL1lin again. You must modify the call you have,

lb=[-inf,0.9,min(y)];

ub=[+inf, 1.1*min(y)];

[AC,fval]=minL1lin(Q,d,[],[],[],[],lb,ub,[],optimoptions('linprog','Display','off'));

Torsten
on 12 Sep 2024 at 13:26

Edited: Torsten
on 12 Sep 2024 at 13:38

function [fval,p]=objFun(theta, x,y)

x=x(:);y=y(:); theta=theta(:)';

lambda=theta(1); m=theta(2);

d=ones(size(y));

Q=[ 1./(1 + (lambda^2 * x).^m), d]./y;

%Q=[ 1./(1 + (lambda * x).^m), d]./y;

%Since you also want C>0, the below setting assumes min(y)>0.

lb = [-Inf,0.9*min(y)];

ub = [Inf,1.1*min(y)];

[AC,fval]=minL1lin(Q,d,[],[],[],[],lb,ub,[],optimoptions('linprog','Display','off'));

%[AC,fval]=minL1lin(Q,d,[],[],[],[],[],[],[],optimoptions('linprog','Display','off'));

fval=fval/numel(y);

A=AC(1); C=AC(2);

p={A, theta, C};

end

roborrr
on 12 Sep 2024 at 17:18

Edited: roborrr
on 12 Sep 2024 at 17:19

No you do not need to call minL1lin again. You must modify the call you have,

lb=[-inf,0.9,min(y)];

ub=[+inf, 1.1*min(y)];

[AC,fval]=minL1lin(Q,d,[],[],[],[],lb,ub,[],optimoptions('linprog','Display','off'));

Thank you for your answer. Please make changes to this design so that it also imposes restrictions on the rest of the parameters:

A>0, lambda0<lambda<lambda1, m0<m<m1

where the lambda0,lambda1,m0,m1 numbers will be preset in the code.

Torsten
on 12 Sep 2024 at 17:36

Edited: Torsten
on 12 Sep 2024 at 17:39

Please make changes to this design so that it also imposes restrictions on the rest of the parameters:

A>0, lambda0<lambda<lambda1, m0<m<m1

where the lambda0,lambda1,m0,m1 numbers will be preset in the code.

For A>0, set

lb=[0,0.9*min(y)];

ub=[+inf,1.1*min(y)];

For the other two, switch from "fminsearch" to an optimizer that allows to set bounds on the parameters (e.g. "fmincon"). This way, the artificial solution to replace lambda by lambda^2 can also be avoided.

roborrr
on 12 Sep 2024 at 19:47

For A>0, set

lb=[0,0.9*min(y)];

ub=[+inf,1.1*min(y)];

but, lb, ub we used to impose restrictions on parameter C, and where are the restrictions of parameter A? and can't restrictions of all parameters(A,lamda,m) be written into the following construction:

[AC,fval]=minL1lin(Q,d,[],[],[],[],lb,ub,[],optimoptions('linprog','Display','off'));

as Matt J did for parameter C?

Torsten
on 12 Sep 2024 at 20:00

Edited: Torsten
on 12 Sep 2024 at 20:06

but, lb, ub we used to impose restrictions on parameter C, and where are the restrictions of parameter A?

The restrictions on A are the 0 and the +inf in

lb=[0,0.9*min(y)];

ub=[+inf,1.1*min(y)];

A and C are solved for by minL1lin - so you have to set bounds for them in the call to minL1lin.

lambda and m are solved for by fminsearch - so you have to set bounds for them in the call to fminsearch. But since fminsearch does not have the option to impose bounds on the parameters, you have to use a different optimizer here:

theta = fminsearch(error_function, initial_guess);

I suggested "fmincon":

theta = fmincon(error_function, initial_guess, [],[],[],[],[lambda0,m0],[lambda1,m1]);

You should try to understand the code you are using.

roborrr
on 13 Sep 2024 at 3:47

Edited: roborrr
on 13 Sep 2024 at 3:48

A and C are solved for by minL1lin - so you have to set bounds for them in the call to minL1lin.

restrictions for parameter C (lb and ub) we wrote instead of five and six empty square brackets:

[AC,fval]=minL1lin(Q,d,[],[],[],[],lb,ub,[],optimoptions('linprog','Display','off'));

and where to write restrictions for parameter A? I don't know the syntax, how to write it.

Torsten
on 13 Sep 2024 at 12:57

Edited: Torsten
on 13 Sep 2024 at 13:02

The code requires "fminsearchbnd" from the file exchange:

clear

% Data

% Initial guess for parameter theta

initial_guess = [1,1];

% Error function

error_function = @(theta) objFun(theta, x,y);

% Optimization of parameters

%Bounds on lambda and m

lambda0 = 0;

lambda1 = inf;

m0 = -inf;

m1 = inf;

%Call the optimizer

theta = fminsearchbnd(error_function, initial_guess,[lambda0,m0],[lambda1,m1]);

%theta = fminsearch(error_function, initial_guess);

[fval,AthetaC]=error_function(theta);

[A, theta,C]=deal(AthetaC{:})

A = 131.5017

theta = 1×2

0.2036 1.9539

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

C = 73.9207

fval

fval = 0.0015

function [fval,p]=objFun(theta, x,y)

x=x(:);y=y(:); theta=theta(:)';

lambda=theta(1); m=theta(2);

d=ones(size(y));

Q=[ 1./(1 + (lambda * x).^m), d]./y;

%Bounds on A and C

A0 = 0;

A1 = inf;

C0 = 0.9*min(y);

C1 = 1.1*min(y);

lb=[A0,C0];

ub=[A1,C1];

%Call the optimizer

[AC,fval]=minL1lin(Q,d,[],[],[],[],lb,ub,[],optimoptions('linprog','Display','off'));

%[AC,fval]=minL1lin(Q,d,[],[],[],[],[],[],[],optimoptions('linprog','Display','off'));

fval=fval/numel(y);

A=AC(1); C=AC(2);

p={A, theta, C};

end

DGM
on 24 Sep 2024 at 17:27

roborrr
on 24 Sep 2024 at 20:12

Edited: roborrr
on 24 Sep 2024 at 22:06

TThanks to Torsen for perfecting the code, it was trully amazing. Also thanks to Matt J who wrote the initial version of the code, that was also excellent.

The main beauty of this code is that it eliminates the occurrence of local minima during optimization, that could arise when using built-in Matlab functions.

In this code, the dependent variable 'y' depends only on one variable 'x' (one-dimensional problem). Is it possible to write code for a two-dimensional problem, when the dependent variable depends nonlinearly on two independent variables at once, using the scheme you used for the one-dimensional problem, which excludes the occurrence of local minima? Here is an example of such a problem:

The experiments were carried out on the following unique values of the variable x_un length of 5 and y_un length of 10 :

x_un = [0.4, 0.5, 0.6, 0.7, 0.8];

y_un = [283.15, 288.15, 290.65, 293.15, 295.15, 298.15, 300.65, 303.15, 313.15, 323.15];

based on x_un and y_un, new variables x and y length of 50 (5*10) are created so that the pairs (x(i),y(i)) i=1,50 exhaust all combinations of pairs (x_un, y_un):

x = [0.4, 0.5, 0.6, 0.7, 0.8, 0.4, 0.5, 0.6, 0.7, 0.8, 0.4, 0.5, 0.6, 0.7, 0.8, ...

0.4, 0.5, 0.6, 0.7, 0.8, 0.4, 0.5, 0.6, 0.7, 0.8, 0.4, 0.5, 0.6, 0.7, 0.8, ...

0.4, 0.5, 0.6, 0.7, 0.8, 0.4, 0.5, 0.6, 0.7, 0.8, 0.4, 0.5, 0.6, 0.7, 0.8, ...

0.4, 0.5, 0.6, 0.7, 0.8];

y = [283.15, 283.15, 283.15, 283.15, 283.15, 288.15, 288.15, 288.15, 288.15, 288.15, ...

290.65, 290.65, 290.65, 290.65, 290.65, 293.15, 293.15, 293.15, 293.15, 293.15, ...

295.15, 295.15, 295.15, 295.15, 295.15, 298.15, 298.15, 298.15, 298.15, 298.15, ...

300.65, 300.65, 300.65, 300.65, 300.65, 303.15, 303.15, 303.15, 303.15, 303.15, ...

313.15, 313.15, 313.15, 313.15, 313.15, 323.15, 323.15, 323.15, 323.15, 323.15];

variable z is also created, the length of which is also 50, in such a way that experimental data are entered into z(i), which correspond to (x(i),y(i)) i=1,50:

z = [12.61323, 15.96843, 22.20326, 30.15335, 60.71139, 10.97453, 14.54062, 18.84638, 24.82916, ...

53.02121, 9.57412, 13.34543, 17.50987, 22.78405, 46.60942, 8.74127, 11.59837, 14.62923, ...

21.16607, 42.31916, 7.93542, 10.73998, 13.71756, 19.05432, 39.08479, 7.33475, 10.12368, ...

12.67264, 18.14767, 35.87183, 6.52910, 9.07151, 11.60863, 16.87576, 32.95969, 5.43683, ...

8.19517, 10.54588, 15.03453, 29.99945, 4.25370, 6.86437, 9.20690, 12.29640, 24.78083, ...

2.99082, 5.05884, 6.68976, 9.54918, 18.76493];

approximation model is considered:

z_calculate = A * exp(E/ (8.314 * y)) * (1 + k * x^n);

and based on this model, an objective function is created, which is equal to the average value of the relative deviations between z and z_calculate:

error_function = mean(abs(z - z_calculate)) /z)

Our goal is to select parameters A,E,k,n in such a way that 'error_function' takes the value of the global minimum.

if possible, set lower and upper limits for parameters:

lb_A = 10^(-3); ub_A = 100;

lb_E = 2000; ub_E = 50000;

lb_k = 0.1; ub_k = 10;

lb_n = 1; ub_n = 3;

roborrr
on 26 Sep 2024 at 7:29

and construct Q this time using x,y, and z.

in the linear part of the model z_calculate = A * exp(E/ (8.314 * y)) * (1 + k * x^n) there is only the coefficient A, and the free term C is missing. the nonlinear part of the model depends on two variables at once. how to construct Q? and how to replace those commands where C is present? I don't know the syntax, how to do this.

Matt J
on 26 Sep 2024 at 12:37

Edited: Matt J
on 26 Sep 2024 at 13:53

The role of minL1lin is to solve for the linear parameters, for a fixed and given set of values of the nonlinear parameters. So, if A is your only linear parameter, then Q should contain coefficients only for A. Bounds on the nonlinear parameters should be given to fminsearchbnd as before.

function [fval,p]=objFun(theta, x,y,z)

x=x(:);y=y(:);z=z(:); theta=theta(:)';

E=theta(1); k=theta(2); n=theta(3);

M=numel(z);

d=ones(M,1);

Q= exp(E./(8.314 .* y)) .* (1 + k .* x.^n) ./z;

lb_A=___;

ub_A=___;

%Solve the linear part of the model

[A,fval]=minL1lin(Q,d,[],[],[],[],lb_A,ub_A,[],optimoptions('linprog','Display','off'));

fval=fval/M;

p={A, theta};

end

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