Re: Help in leasqr

[Bharath R]

 Dear Mr. Stewart,

Thank you very much for the clarification. I changed the maximum iterations to 200 and got the convergence. I checked that if the max iteration value isn't given, it takes 20 to be the default value. 

Thanks for trying my data.

Regards,

Bharath

On Fri, Aug 26, 2016 at 4:32 PM, Doug Stewart <address@hidden> wrote:

On Fri, Aug 26, 2016 at 6:17 AM, Bharath R <address@hidden> wrote:

Hello all, 

Thank you very much for such a helpful forum. I am using Octave for the past couple of days. I have been working to get the parameters that best fit my function. 

Function:   g=@(x,p) [(p(1).*p(2).*x(:,1)+(p(3).*x(:,2))+(p(4)*p(5)))./((p(1)*p(2))+p(3))]

where p indicates the parameters , x indicates the independent variables and y is measured variable.

[pin]=1.0000e-002

      1.0000e+003

      1.0000e+002

      1.0000e-002

      1.0000e+003

[x]=21.33700    0.80000

     21.28800    0.10000

     20.55900   -0.80000

     21.21800   -1.40000

     20.68600   -1.70000

[y] = 19.545

       19.300

       19.300

       19.137

       19.000

When I run this function ([f, p, cvg, iter]=leasqr(x,y,pin,g)) with leasqr function it gives me the following result:

 CONVERGENCE NOT ACHIEVED!

f =

   19.710

   19.578

   18.827

   19.326

   18.824

p =

  1.6953e-001

  1.6258e+003

  4.0087e+001

  7.8664e-002

  3.9346e+003

cvg = 0

iter =  20

I would like to know what does "f" in result give as output. Is it giving me the output of function "g" after implementing the newly calculated parameters or is it something else? If it is the output of g, then I feel it is relatively closer to the measured data (y). yet it is showing " CONVERGENCE NOT ACHIEVED!". 

Also , I would like to know how do I implement a differential equation in leasqr to calculate the parameters. I look forward to your answer. 

Thanks and regards,

Bharathy

I tried your data and see what you said.

I then set pin to the resultant p and ran it 2 more times and got it to converge whit this result

f =

   19.703

   19.577

   18.828

   19.339

   18.834

p =

   4.0188e-02

   5.8681e+03

   3.1513e+01

   3.2001e-02

   6.5753e+03

cvg = 1

iter =  1

So you are on the correct  path, just use the result of one run for a better starting point for the next run.

HTH

--

DAS