Re: minimization problem, sqp

[Juan Pablo Carbajal]

On Mon, Feb 25, 2013 at 4:37 PM, Urs Hackstein
<address@hidden> wrote:
> Juan,
>
> I read the manual before posting my question (my attempt was the result of
> reading it) and I did it again, but it didn't help. I still get the error
> "`b' undefined near line 3 column 6" and don't know how to fix it. Perhaps
> the problem is that fun is not just a function of x, but also of b and the
> psj, but, as I said, I don't know how to fix it.
>
>
> 2013/2/25 Juan Pablo Carbajal <address@hidden>
>>
>> On Mon, Feb 25, 2013 at 3:32 PM, Urs Hackstein
>> <address@hidden> wrote:
>> >
>> >
>> > ---------- Forwarded message ----------
>> > From: Urs Hackstein <address@hidden>
>> > Date: 2013/2/25
>> > Subject: Re: minimization problem, sqp
>> > To: Juan Pablo Carbajal <address@hidden>
>> >
>> >
>> > Hi Juan,
>> >
>> > the problem is that my attempt does not work and I am not familiar
>> > enough
>> > with the sqp syntax to get the x(1),...,x(7) as output. Thus any
>> > suggestions
>> > for a correct program?
>> >
>> >
>> > Sincerely yours,
>> >
>> > Urs Hackstein
>> >
>> > 2013/2/25 Juan Pablo Carbajal <address@hidden>
>> >>
>> >> On Mon, Feb 25, 2013 at 2:30 PM, Urs Hackstein
>> >> <address@hidden> wrote:
>> >> > Dear list,
>> >> >
>> >> > I am relatively new to octave and dealing with the following
>> >> > minimization
>> >> > problem:
>> >> > Let b, ps1, ps2, ps3, ps4, ps5, ps6 and ps7 arbitrary complex
>> >> > numbers.
>> >> > My
>> >> > goal is to determine the real numbers x(1), x(2), x(3),
>> >> > x(4),x(5),x(6),x(7)
>> >> > with -1<=x(j)<=1, j=1,..,7, such that the function
>> >> >
>> >> >
>> >> >
>> >> > (b-x(1).*ps1-x(2).*ps2-x(3).*ps3-x(4).*ps4-x(5).*ps5-x(6).*ps6-x(7).*ps7).^2
>> >> > takes its minimal value.
>> >> >
>> >> > I made the following attempt using sqp:
>> >> >
>> >> > function erg=fun(x)
>> >> >
>> >> >
>> >> > erg=(b-x(1).*ps1-x(2).*ps2-x(3).*ps3-x(4).*ps4-x(5).*ps5-x(6).*ps6-x(7).*ps7).^2
>> >> > endfunction
>> >> >
>> >> > function minimi=minimize(b,ps1,ps2,ps3,ps4,ps5,ps6,ps7)
>> >> > global b;
>> >> >
>> >> > minimi=sqp([0;0;0;0;0;0;0],@fun,[-1;-1;-1;-1;-1;-1;-1],[1;1;1;1;1;1;1])
>> >> > endfunction
>> >> >
>> >> > Thanks a lot in advance!
>> >> >
>> >> > Sincerely yours,
>> >> >
>> >> > Urs Hackstein
>> >> >
>> >> > _______________________________________________
>> >> > Help-octave mailing list
>> >> > address@hidden
>> >> > https://mailman.cae.wisc.edu/listinfo/help-octave
>> >> >
>> >>
>> >>  Hi Urs,
>> >>
>> >> Great! and what is the problem?
>> >
>> >
>> >
>> >
>> > _______________________________________________
>> > Help-octave mailing list
>> > address@hidden
>> > https://mailman.cae.wisc.edu/listinfo/help-octave
>> >
>>
>> Urs,
>>
>> Ok, have you try reading the help of sqp?
>> > help sqp
>>
>> If you have questions after reading that we can help you, but do
>> please read the manual. also search the web, there are plenty of
>> examples using sqp
>
>

Hi Urs,

Yes, that may be a problem. Lets say you want of din the minimum of the function

f(x; a, b) = x*(a*x + b)

The function is parametrized by a and b, i.e. it represents a family
of functions. To optimize you have to specify which one you want to
work with.
So we create the parametrized function

function y = f (x, p)
  y = x .* (p(1)*x + p(2));
endfunction

Before we can pass it to sqp we have to define the values of the
parameters, one way of doing that is using a wrapper (I think is the
recommended way, though it may be less efficient than other ways, but
do not know for how much)

a = 1; b = -1;
fwrap = @(x) f (x,[a,b])

Then we can call sqp

 sqp (2*rand(1)-1, fwrap)

and I get

ans =  0.50000

as expected.

Hope this solves your problem