I want to find constants in an equation that has got many parameters

I have an equation. But it has got more than 3 known variables. Because of that, I cannot use curve fitting toolbox. For example my equation is z=h*(a*x+b*y/x^2+c*u) I have got 1600 values of z, h, x, y and u values in excel. These are known variables. a, b, c are constants. And I want to find best a, b and c values. How can I do it? I am very new in Matlab. Please help me.

[Walter Roberson]

Effectively duplicates http://www.mathworks.com/matlabcentral/answers/301300-determining-constants-in-an-equation-that-have-more-than-3-variables

Answer 1

[Star Strider]

Something like this should work:

% % % % % z=h*(a*x+b*y/x^2+c*u) 
% % PARAMETER MAPPING: p(1) = a,  p(2) = b,  p(3) = c  
% % VARIABLE MAPPING: v(:,1) = z,  v(:,2) = h,  v(:,3) = x,  v(:,4) = y,  v(:,5) = u
z = @(p,v) v(:,2).*(p(1).*v(:,3) + p(2).*v(:,4) ./ v(:,3).^2 + p(3).*v(:,5));       % Objective Function
v = [z(:)  h(:)  x(:)  y(:)  u(:)];                                                 % Variable Array
P0 = [1; 2; 3];                                                                     % Choose Appropriate Initial Parameter Estimates
SSECF = @(p) sum((v(:,1) - z(p,v)).^2);                                             % Sum-Squared-Error Cost Function
[abc, SSE] = fminsearch(SSECF, P0);                                                 % Estimate Parameters
a = abc(1)
b = abc(2)
c = abc(3)

NOTE: I tested this on random data and it ran without error. I have obviously not been able to test it with your data.

[Can Ekici]

I was writing a new code for another equation with using your help. My equation is: z=y*(a*(1-exp((x)^b))+c*u In my code, there is an error like: "Maximum number of function evaluations has been exceeded - increase MaxFunEvals option. Current function value: Inf". I was setting these value with optimset but it does not work.

v = xlsread('I:\Yeni klasör\Tez 30 temmuz\Tez kilitlenme\Türkiye\Ankara veri\matlab model.xlsx')
z=v(:,1)
y=v(:,2)
x=v(:,3)
u=v(:,4)
syms a, syms b, syms c
p(1) = a,  p(2) = b,  p(3) = c
z = @(p,v) v(:,2).*(p(1).*(1-exp(v(:,3).^p(2)))+p(3).*v(:,4))
P0 = [1; 2; 3];                                                                     % Choose Appropriate Initial Parameter Estimates
SSECF = @(p) sum((v(:,1) - z(p,v)).^2);                                             % Sum-Squared-Error Cost Function
options = optimset('MaxFunEvals',1000000);
options = optimset('MaxIter',1000000);
[abc, SSE] = fminsearch(SSECF, P0, options);                                                 % Estimate Parameters
a = abc(1)
b = abc(2)
c = abc(3)

[Star Strider]

What you got was a Warning, not an error.

Please do not use the Symbolic Math Toolbox for this problem. It is not necessary.

So delete this line:

syms a, syms b, syms c

Use my code as I wrote it but with your ‘z’ function (that appears to have been coded correctly). The success of the convergence of any nonlinear parameter estimation routine in large part depends on the initial parameter estimates. You have an exponential term that was not in the code you originally posted, so I would experiment with different values for ‘P0’. For example, see if this initial parameter estimate set works:

P0 = [1; -1; 1];

If it doesn’t, keep experimenting. If you have the Global Optimization Toolbox, use the patternsearch function to find the best-fit parameters.

Also, please understand that ‘does not work’ gives us no useful information, especially in situations such as this that we cannot run your code. We need to know exactly what the problem is. If it is throwing an error, copy all the red error text from the Command Window and paste it to a Comment in your thread.

[Can Ekici]

It works, many thanks to you.

[Star Strider]

As always, my pleasure.

[Walter Roberson]

This is the same solution outline as I described in http://www.mathworks.com/matlabcentral/answers/301300-determining-constants-in-an-equation-that-have-more-than-3-variables .

My cautions from there apply: Your exponential term strongly implies you are going to get caught in local minima. The only difference is that here you are taking x to a power instead of z to a power. If you have negative x then you will have problems with your fitting.

[Riddle Riddle]

is it possible to use that method but for more than one equation? thank you

[Star Strider]

@Riddle Riddle — Possibly.

If you have a specific instance in mind, post it as a new Question.

[Walter Roberson]

No. Any sum of squares of residues fitting approach must fail when you have multiple equations.

When you have multiple equations then you seldom have the case where the minima of one is the minima of the other (which is a situation that reduces down to just doing either of the equations anyhow.)

Instead you typically run into the situation where you cannot minimize both simultaneously. Then to proceed you need information about the relative importance of minimizing each. And watch out because the scale of one might be larger than the scale of the other.

[Riddle Riddle]

@Star Strider I already made one in my profile. Would you be so kind to take look? Thank you.

@Walter Roberson Okay. I'll try to somehow estimate the coefficient per equation only.

[Nidhi]

@Star Strider Hi, how do I find the initial parameter estimates? Is there a particular way to estimate the values?

[Star Strider]

@NDC — That entirely depends on the model and the data. There is no general method.

When I have problems determining them, I use the ga (genetic algorithm) function that will generally find the best parameter set, because I have a fair amount of ecperience with it over the years. There are similar functions (such as GlobalSearch) that use a different approach to determining the best parameter set, and that may be preferable for a particular problem.

.

[Nidhi]

@Star Strider Can you please help me with this. I am a bit lost as to how I should determine the initial parameters as the value for the constants changes quite drastically with different values of initial parameter.

The code I have written is:

clc
v = xlsread('Creep data-Paeru.xlsx');
creep=v(1:43,2);
time=v(1:43,1);
creep = @(p,v) (6/p(1))*(1-(p(2)/(p(1)+p(2)))*exp(-p(1)*p(2).*v(1:43,1)/(p(3)*(p(1)+p(2)))));
p0= [1; 2; 3];                                                                     
SSECF = @(p) sum((v(1:43,2) - creep(p,v)).^2);                                             
[abc, SSE] = fminsearch(SSECF, p0);                                              
E1 = abc(1)
E2 = abc(2)
n1 = abc(3)

Here,I have randomly chosen the initial parameters but is there a more suitable way to determine them.

Thank you in advance!

[Star Strider]

If the estimated parameters provided a good fit to the data, then they worked. If they did not, use one of the techniques I previously suggested to estimate the best set of parameters.

[Nidhi]

Thank you! I will try doing that.

[Nidhi]

@Star Strider Is there any other way to solve such problems without having to estimate initial values?

[Star Strider]

No. Not for nonlinear problems. If you can transform it (for example logarithmically) and then linearise it, you can use linear methods to estimate the parameters, however that is not possible with the function posted. Also, don’t subscript ‘v’ inside the funciton. Pass it as whatever length it is suposed to be. It may also be encessary to use element-wise operations. I always default to that unless I know I need to use matrix operations.

[Walter Roberson]

Sometimes it is productive to take calculus approaches. Use symbolic variables for v and p and call SSECF() with the symbolic p to get a formula for SSECF.

At that point you would try to differentiate SSECF with respect to p(1) or p(2) or p(3) and then solve for the derivative being 0.

However... I ran those tests, and found that for this particular formula, MATLAB is not able to find a solution. WIth the number of exp() and additions involved, I would be surprised if there does turn out to be a closed form formula in this particular case.

Calculus is a useful tool in some cases, but not so much here.

.. Though I guess you could vpasolve() the derivative and hope that gave you something to work with.