RE: fminunc and for loop

[Michael Teitelbaum]

Thanks for the quick responses Juan and Jordi.  I tried following both of your instructions, but ran into some difficulties with each:

Juanian Method:

I think I'm close, but don't have the syntax exactly right.  I tried:

function v = my_func(x,A)

> v = 0

> for i = 5:35

> v += ((((A(i,3) / 1000 + x(1) + A(i,4) / 1000 - ((A(i,3) / 1000 + x(1) + A(i,4) / 1000) ^ 2 - 4 * A(i,3) / 1000 * A(i,4) / 1000) ^ 0.5) / 2 * 1.4e-3 - (1.4e-3 - 8e-6) * (A(i-1,3) / 1000 + x(1) + A(i-1,4) / 1000 - ((A(i-1,3) / 1000 + x(1) + A(i-1,4) / 1000) ^ 2 - 4 * A(i-1,3) / 1000 * A(i-1,4) / 1000) ^ 0.5) / 2) * x(2)) * 1e6 - A(i,1)) ^ 2

> endfor

> endfunction

A = dlmread("/users/michael/desktop/MN4cp2Quin.CSV", ",")

fminunc(@(x)my_func(x,A),[0,0])

What I want as my output is the values of x(1) and x(2) that minimize the function, but what I got was around 100 lines of :

v = 0

v = 1515.1

v = 3311.6

v = 4734.4

and so on.

Jordian method:

I received the following error when I inputted your code:

error: `x' undefined near line 2 column 29

error: called from:

error:   midterm at line 2, column 5

error: evaluating argument list element number 1

error:   f at line 5, column 5

error:   /sw/share/octave/3.6.1/m/optimization/fminunc.m at line 149, column 6

Was your code meant to be put directly into octave, or was it kind of just a general idea of the type of thing I should try?  (Either way, thanks so much for the thought and effort you put into your response)  If it's the latter, I won't bother trying to pursue this because I don't know enough about programming to write something similar to what you sent me.

Thanks again guys,

Michael

> Date: Tue, 10 Jul 2012 19:34:59 +0200
> Subject: Re: fminunc and for loop
> From: address@hidden
> To: address@hidden
> CC: address@hidden
>
> On Tue, Jul 10, 2012 at 7:23 PM, Michael Teitelbaum
> <address@hidden> wrote:
> > Hi,
> > I am trying to perform the fminunc function, but I want to use a for loop
> > within my function, and I'm not sure exactly how to do that.
> > What I want is:
> >
> > fminunc(f,[0,0])
> > where f equals the sum of:
> >
> > f = @(x)((((A(i,3) / 1000 + x(1) + A(i,4) / 1000 - ((A(i,3) / 1000 + x(1) +
> > A(i,4) / 1000) ^ 2 - 4 * A(i,3) / 1000 * A(i,4) / 1000) ^ 0.5) / 2 * 1.4e-3
> > - (1.4e-3 - 8e-6) * (A(i-1,3) / 1000 + x(1) + A(i-1,4) / 1000 - ((A(i-1,3) /
> > 1000 + x(1) + A(i-1,4) / 1000) ^ 2 - 4 * A(i-1,3) / 1000 * A(i-1,4) / 1000)
> > ^ 0.5) / 2) * x(2)) * 1e6 - A(i,1)) ^ 2
> >
> > for i between 5 and 35.
> >
> > If it is not possible to use a for loop within fminunc, is there a good way
> > to get octave to write out that function for me instead of me having to
> > write it out manually?
> >
> > Any help would be greatly appreciated.
> > Thanks,
> > Michael
> >
> > _______________________________________________
> > Help-octave mailing list
> > address@hidden
> > https://mailman.cae.wisc.edu/listinfo/help-octave
> >
>
> You can create a function (not a anonymous one)
>
> function v = my_func(x,A)
> %% Do your magic here
> endfunction
>
> and then define A and call
> fminunc(@(x)my_func(x,A),[0,0])
>
> --
> M. Sc. Juan Pablo Carbajal
> -----
> PhD Student
> University of Zürich
> http://ailab.ifi.uzh.ch/carbajal/