Re: Help in parameter estimation

[Doug Stewart]

On Tue, Aug 30, 2016 at 10:45 AM, Bharath R <address@hidden> wrote:

Hi Nicholas,

Thanks for the mail. I am sorry that I have used mat earlier, now I changed the function as follows:

function F=f(x,p)

  A=xlsread('OGparameters.xlsx');

  x=[A(:,3),A(:,5),A(:,2),A(:,6),A(:,7)];

  p(1)=20;

  p(2)=1;

  p(3)=100;

  p(4)=1;

  p(5)=0.001;

  p(6)=1;

  p(7)=0.01;

  typeinfo(x)

  pin=[p(1);p(2);p(3);p(4);p(5);p(6);p(7)];

  m(1,1)=20;

  for i=1:length(x)

    m(i+1,1)=((p(1)-m(i,1))/(p(2)*p(3))+((p(4)*x(i,2)*(x(i,3)-m(i,1)))+(p(5)*x(i,4))+(p(6)*x(i,5)))/(p(3))+p(7))+m(i,1);

   endfor

   F=m(2:end);

endfunction

I am generating a matrix of [length(x), 1] as output from the function. My measured value is also a matrix of [length(x), 1]. 

What I would like to do now is to estimate parameters p.  I am using the following command to run the optimization program ( I have tried with nonlinear curve fit as well):

[L,p,cvg,iter]=leasqr(x,y,pin,f)

pin ,x and y are  initialized. The problem is when I run the optimization algorithm, it throws me the following error:

error: leasqr: subscript indices must be either positive integers less than 2^31 or logicals

error: called from

    leasqr at line 329 column 9

do you have some other variable named leasqr???

what does

which leasqr

show?

 

I look forward to your reply. 

Regards,

Bharath

On Tue, Aug 30, 2016 at 4:01 PM, Nicholas Jankowski <address@hidden> wrote:

On Tue, Aug 30, 2016 at 6:00 AM, Bharath R <address@hidden> wrote:

 

function m=f(x,p)

  m(1,:)=20;

  for i=1:length(x)

    m(i+1,:)=((p(1)-m(i,:))/(p(2)*p(3))+((p(4)*x(i,2)*(x(i,3)-m(i,:)))+(p(5)*x(i,4))+(p(6)*x(i,5)))/(p(3))+p(7))+m(i,:);

  endfor

  mat=m(2:end);

endfunction

is your intent to return the variable m or the variable mat from function f? you are currently returning m and never using mat. since mat is a different size than m, this could make a significant difference when you use the results of f later.

your declaration for m is odd. it asks to make all of the columns in row1 equal to 20.  since m has not yet been initialized, the number of rows is 1. so the output is a 1x1 array, or a scalar, and m = 20. then in the for loop, if the result of the m(i+1,:) equation was 6, you would be appending it as a new row to m, creating a 2x1 array [20 ; 6] . Since you went to the trouble of using the : operator, but have it doing nothing, I suspect this was not your intent?

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DAS