Re: multimin convergence problem

[Roland Roberts <roland at astrofoto dot org>]

>>>>> "Brian" == Brian Gough <bjg@network-theory.co.uk> writes:

    Brian> Are you doing least-squares fitting of the residuals
    Brian> between the two curves?  If so the nonlinear 'multifit'
    Brian> functions should handle that better.

Yes, actually I've tried that approach as well hoping to use the
covariance matrix as a crude sensitivity metric.  I get slightly
different answers compared to the multimin solution (at the 2% level,
which is larger than I expect, but I don't have a good quantitative
basis for that expectation yet).  Additionally, the covariance matrix
makes no sense to me at all.  The minimum is in the vicinity of
r=17.19, beta=2 and I get r=17.19+/-1909, beta=2+/-0.001.

    >> Is there some magic to picking a good value for STEP?
    >> Shouldn't GSL return an error rather than setting all my
    >> parameters to NaN?

    Brian> Yes, it should return an error not NaN.  There must be an
    Brian> unchecked division by zero that we need to catch in there
    Brian> somewhere.  Can send your program to the list.

Attached is both the multimin program.  Getting NaN is easy; just pick
a starting point away from the minimum.  For example, these two runs:

281 roland> make -k && ./barndoor -v -b 10 -c 5 -r 17 -E 600 --dt 60 -u 0 -w 0
gcc -DHAVE_CONFIG_H -I. -I. -I.   -DDEBUG  -g -O2 -c barndoor.c
gcc  -g -O2  -o barndoor  barndoor.o  -lgsl -lgslcblas -lm 
iter:   0 x =     17.00000000      2.00000000  f(x) =      38436.8  RMS =      59.1122  |g(x)| =      1512.46  g(x) = -350.614 -1471.26
iter:   1 x =     17.01159087      2.04863797  f(x) =      1821.51  RMS =      12.8682  |g(x)| =      68.4458  g(x) = -16.5877 -66.4053

Minimum found at:
iter:   2 x =     17.01491189      2.06257370  f(x) =  7.53164e-08  RMS =  8.27463e-05  |g(x)| =  1.66044e-09  g(x) = -4.07916e-10 -1.60956e-09

282 roland> make -k && ./barndoor -v -b 10 -c 5 -r 20 -E 600 --dt 60 -u 0 -w 0
iter:   0 x =     20.00000000      2.00000000  f(x) =  6.16155e+06  RMS =      748.425  |g(x)| =       204526  g(x) = -40530.4 -200470
iter:   1 x =             nan             nan  f(x) =          nan  RMS =          nan  |g(x)| =          nan  g(x) = nan nan

roland
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Roland B. Roberts, PhD                             RL Enterprises
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Attachment: barndoor.c
Description: Double-arm Barndoor Tracker Optimization Code