Re: Nonlinear Estimation with Optimize

[etienne]

  Hi Thomas,


I just switched from lists (a historic data structure in octave) to cells in
the minimization backend used by "minimize", which is nelder_mead_min(). I
send you a replacement in attachment.

  I also had to fix your function, because as you wrote it, it does not return
a scalar, but a vector and minimizing a vector is not what you want to do.

octave:103> function r = fit (a0, b, c), r = sum ((c -
[b.^2,b,ones(7,1)]*a0).^2); end

octave:111> x0 = [0,0,0]';
octave:112> t = [3.3598, 26.4377, 89.6760, 211.5766, 414.1391, 716.1501,
1128.1406 ]';
octave:113> s = 1000*(1:7)';
octave:114> [x,v,n] = minimize ("fit",{x0,s,t});
octave:115> x
x =

   3.9283e-05
  -1.3290e-01
   1.1586e+02

octave:116> v
v =  2046.5
octave:117> n
n =  435

  Check the result for correctness, because I have not.

  I also note that your values in t & s are very big and the order of
magnitude of x's elements are very different. You may want to normalize
that:

octave:118> s2 = (1:7)'
s2 =

   1
   2
   3
   4
   5
   6
   7

octave:119> t2 = [3.3598, 26.4377, 89.6760, 211.5766, 414.1391, 716.1501,
1128.1406 ]'/1000;
octave:123> [x2,v2,n2] = minimize ("fit",{x0,s2,t2});
octave:124> x2
x2 =

   0.039283
  -0.132897
   0.115857

octave:125> v2
v2 =  0.0020465
octave:127> n2
n2 =  233
octave:128> x2 ./ [1e6; 1000; 1]
ans =

   3.9283e-08
  -1.3290e-04
   1.1586e-01

octave:129> x
x =

   3.9283e-05
  -1.3290e-01
   1.1586e+02


   Ok, the two results look pretty much the same, normalization was not an 
issue.

  Hth,

  Etienne


PS: I could not check my fixes in octave-forge, I forgot my password. I should
do that soon enough.

On Sat, August 23, 2008 14:34, Thomas D. Dean wrote:
# Octave 3.0.2
#
# I installed optim-1.0.3.tar.gz, and changed is_list() to islist() to
# remove a warning message.  The files are:
#   optim-1.0.3/minimize.m
#   optim-1.0.3/d2_min.m
#   optim-1.0.3/nelder_mead_min.m
#
# I have a problem with argument syntax.  I think I need to use minimize()
# and pass a function that uses three arguments.  I have a simplified
# example.
#
# Here is what I want to do (wrong?):
#
# t = [3.3598, 26.4377, 89.6760, 211.5766, 414.1391, ...
#      716.1501, 1128.1406 ]';
# for i=1:7
#   s(i,1)=1000*i;
# end;
# ## I wanted to use s(i)=1000*i and then s=s'
# ## but, octave seemed to ignore that.
# x0=[0,0,0]';
# l={x0,s,t};  ## a list to pass to minimize
#
# function r = fit(l)
#   a0=nth(l,1); b=nth(l,2); c=nth(l,3);
#   r = c - [b.^2,b,ones(7,1)]*a0;
# end
#
# [x,v,n]=minimize("fit",l)  ## I want to use fit(x0,s,t)
#
# What am I doing wrong?
#
# tomdean
#
# _______________________________________________
# Help-octave mailing list
# address@hidden
# https://www-old.cae.wisc.edu/mailman/listinfo/help-octave
#


-- 
http://www.isr.ist.utl.pt/~etienne

nelder_mead_min.m
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