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From: Tim on 13 Apr 2010 11:22
I'm using lsqcurvefit with restrictions placed on the upper and lower bounds. When I wanted to compare the results to an unconstrained model I changed the restrictions on all parameters to -inf and inf. New regression parameters were generated but they were still within the original constraints.

Example...
GENERAL CODE STAYS SAME ACROSS RUNS
func = (b1.*x1.*x2.^b2.*x3.^b3);

options = optimset('Display', 'iter', 'FunValCheck', 'on', 'MaxFunEvals', 5000, 'MaxIter', 1000);

[betahat,...] = lsqcurvefit('func', beta0, xdata, ydata, LB, UB, options);

PARTS I'VE CHANGED BETWEEN RUNS

CONSTRAINED
LB = [-10,0,0];
UB = [50,5,5];
betahat returned [11.37, 1.83, 0.83]

UNCONSTRAINED
LB = [-inf,-inf,-inf];
UB = [inf,inf,inf];
betahat returned [10.11, 1.78, 0.82]

If the parameters changed between runs I would have expected at least one element from the new betahat from the unconstrained model to be outside the LB and UB of the constrained model.

Do I just need to put my inner engineer away and think of the two betahats as essentially the same? Any thoughts on why the slightly different answers? If I do accept them as the same can I return a level of significance?

Thanks,

Tim

From: Alan Weiss on 13 Apr 2010 13:40
That is an interesting result. You could tell from the original answer
CONSTRAINED
LB = [-10,0,0];
UB = [50,5,5];
betahat returned [11.37, 1.83, 0.83]
that the constraints don't matter. But the results differ. Hmm.

It seems to me that the minimum is probably pretty flat. In other words,
I am guessing that the resulting residuals were virtually the same for
your two runs. Why not try a less fussy objective function. Take the log
of everything:
lfunc = log(b1) + log(x1) + b2.*log(x2) + b3.*log(x3);
This transformation has the same minimum, but might be better-behaved.

Alan Weiss
MATLAB mathematical toolbox documentation

On 4/13/2010 11:22 AM, Tim wrote:
> I'm using lsqcurvefit with restrictions placed on the upper and lower
> bounds. When I wanted to compare the results to an unconstrained model I
> changed the restrictions on all parameters to -inf and inf. New
> regression parameters were generated but they were still within the
> original constraints.
>
> Example...
> GENERAL CODE STAYS SAME ACROSS RUNS
> func = (b1.*x1.*x2.^b2.*x3.^b3);
>
> options = optimset('Display', 'iter', 'FunValCheck', 'on',
> 'MaxFunEvals', 5000, 'MaxIter', 1000);
> [betahat,...] = lsqcurvefit('func', beta0, xdata, ydata, LB, UB, options);
> PARTS I'VE CHANGED BETWEEN RUNS
>
> CONSTRAINED
> LB = [-10,0,0];
> UB = [50,5,5];
> betahat returned [11.37, 1.83, 0.83]
>
> UNCONSTRAINED
> LB = [-inf,-inf,-inf];
> UB = [inf,inf,inf];
> betahat returned [10.11, 1.78, 0.82]
>
> If the parameters changed between runs I would have expected at least
> one element from the new betahat from the unconstrained model to be
> outside the LB and UB of the constrained model.
> Do I just need to put my inner engineer away and think of the two
> betahats as essentially the same? Any thoughts on why the slightly
> different answers? If I do accept them as the same can I return a level
> of significance?
>
> Thanks,
>
> Tim

From: Tim on 13 Apr 2010 16:04
Thanks for the quick response Alan. You are right, the residuals are very close in both cases. I also switched to a logarithmic expression and tried the same test. My betahats are much closer together but I still get the same behavior.

Tim

Alan Weiss <aweiss(a)mathworks.com> wrote in message <hq2aag$7e9$1(a)fred.mathworks.com>...
> That is an interesting result. You could tell from the original answer
> CONSTRAINED
> LB = [-10,0,0];
> UB = [50,5,5];
> betahat returned [11.37, 1.83, 0.83]
> that the constraints don't matter. But the results differ. Hmm.
>
> It seems to me that the minimum is probably pretty flat. In other words,
> I am guessing that the resulting residuals were virtually the same for
> your two runs. Why not try a less fussy objective function. Take the log
> of everything:
> lfunc = log(b1) + log(x1) + b2.*log(x2) + b3.*log(x3);
> This transformation has the same minimum, but might be better-behaved.
>
> Alan Weiss
> MATLAB mathematical toolbox documentation
>
> On 4/13/2010 11:22 AM, Tim wrote:
> > I'm using lsqcurvefit with restrictions placed on the upper and lower
> > bounds. When I wanted to compare the results to an unconstrained model I
> > changed the restrictions on all parameters to -inf and inf. New
> > regression parameters were generated but they were still within the
> > original constraints.
> >
> > Example...
> > GENERAL CODE STAYS SAME ACROSS RUNS
> > func = (b1.*x1.*x2.^b2.*x3.^b3);
> >
> > options = optimset('Display', 'iter', 'FunValCheck', 'on',
> > 'MaxFunEvals', 5000, 'MaxIter', 1000);
> > [betahat,...] = lsqcurvefit('func', beta0, xdata, ydata, LB, UB, options);
> > PARTS I'VE CHANGED BETWEEN RUNS
> >
> > CONSTRAINED
> > LB = [-10,0,0];
> > UB = [50,5,5];
> > betahat returned [11.37, 1.83, 0.83]
> >
> > UNCONSTRAINED
> > LB = [-inf,-inf,-inf];
> > UB = [inf,inf,inf];
> > betahat returned [10.11, 1.78, 0.82]
> >
> > If the parameters changed between runs I would have expected at least
> > one element from the new betahat from the unconstrained model to be
> > outside the LB and UB of the constrained model.
> > Do I just need to put my inner engineer away and think of the two
> > betahats as essentially the same? Any thoughts on why the slightly
> > different answers? If I do accept them as the same can I return a level
> > of significance?
> >
> > Thanks,
> >
> > Tim
From: Alan Weiss on 14 Apr 2010 08:34
Well, there is one more thing to try, though I am a bit reluctant to
suggest it. You could set TolFun to 1e-8 or 1e-10 and see if the results
get any closer (I am assuming that the reason lsqcurvefit stopped was
due to a the TolFun tolerance). The reason I am reluctant to suggest
this measure is people often overdo this by setting ludicrously small
tolerances. There is no sense in setting a tolerance smaller than eps,
and this is already too small for most purposes.

Alan Weiss
MATLAB mathematical toolbox documentation

On 4/13/2010 4:04 PM, Tim wrote:
> Thanks for the quick response Alan. You are right, the residuals are
> very close in both cases. I also switched to a logarithmic expression
> and tried the same test. My betahats are much closer together but I
> still get the same behavior.
>
> Tim
>
> Alan Weiss <aweiss(a)mathworks.com> wrote in message
> <hq2aag$7e9$1(a)fred.mathworks.com>...
>> That is an interesting result. You could tell from the original answer
>> CONSTRAINED
>> LB = [-10,0,0];
>> UB = [50,5,5];
>> betahat returned [11.37, 1.83, 0.83]
>> that the constraints don't matter. But the results differ. Hmm.
>>
>> It seems to me that the minimum is probably pretty flat. In other
>> words, I am guessing that the resulting residuals were virtually the
>> same for your two runs. Why not try a less fussy objective function.
>> Take the log of everything:
>> lfunc = log(b1) + log(x1) + b2.*log(x2) + b3.*log(x3);
>> This transformation has the same minimum, but might be better-behaved.
>>
>> Alan Weiss
>> MATLAB mathematical toolbox documentation
>>
>> On 4/13/2010 11:22 AM, Tim wrote:
>> > I'm using lsqcurvefit with restrictions placed on the upper and lower
>> > bounds. When I wanted to compare the results to an unconstrained
>> model I
>> > changed the restrictions on all parameters to -inf and inf. New
>> > regression parameters were generated but they were still within the
>> > original constraints.
>> >
>> > Example...
>> > GENERAL CODE STAYS SAME ACROSS RUNS
>> > func = (b1.*x1.*x2.^b2.*x3.^b3);
>> >
>> > options = optimset('Display', 'iter', 'FunValCheck', 'on',
>> > 'MaxFunEvals', 5000, 'MaxIter', 1000);
>> > [betahat,...] = lsqcurvefit('func', beta0, xdata, ydata, LB, UB,
>> options);
>> > PARTS I'VE CHANGED BETWEEN RUNS
>> >
>> > CONSTRAINED
>> > LB = [-10,0,0];
>> > UB = [50,5,5];
>> > betahat returned [11.37, 1.83, 0.83]
>> >
>> > UNCONSTRAINED
>> > LB = [-inf,-inf,-inf];
>> > UB = [inf,inf,inf];
>> > betahat returned [10.11, 1.78, 0.82]
>> >
>> > If the parameters changed between runs I would have expected at least
>> > one element from the new betahat from the unconstrained model to be
>> > outside the LB and UB of the constrained model.
>> > Do I just need to put my inner engineer away and think of the two
>> > betahats as essentially the same? Any thoughts on why the slightly
>> > different answers? If I do accept them as the same can I return a level
>> > of significance?
>> >
>> > Thanks,
>> >
>> > Tim

From: Tim on 14 Apr 2010 14:42
I set TolFun to 1e-8 and all of my betahats match through the first decimal place now. It seems for this particular case these values should not change but I'm willing to move on. For future reference do you have any guidance for what is a ludicrously small tolerance? Thanks again for your help.

Tim

Alan Weiss <aweiss(a)mathworks.com> wrote in message <hq4cos$c71$1(a)fred.mathworks.com>...
> Well, there is one more thing to try, though I am a bit reluctant to
> suggest it. You could set TolFun to 1e-8 or 1e-10 and see if the results
> get any closer (I am assuming that the reason lsqcurvefit stopped was
> due to a the TolFun tolerance). The reason I am reluctant to suggest
> this measure is people often overdo this by setting ludicrously small
> tolerances. There is no sense in setting a tolerance smaller than eps,
> and this is already too small for most purposes.
>
> Alan Weiss
> MATLAB mathematical toolbox documentation
>
> On 4/13/2010 4:04 PM, Tim wrote:
> > Thanks for the quick response Alan. You are right, the residuals are
> > very close in both cases. I also switched to a logarithmic expression
> > and tried the same test. My betahats are much closer together but I
> > still get the same behavior.
> >
> > Tim
> >
> > Alan Weiss <aweiss(a)mathworks.com> wrote in message
> > <hq2aag$7e9$1(a)fred.mathworks.com>...
> >> That is an interesting result. You could tell from the original answer
> >> CONSTRAINED
> >> LB = [-10,0,0];
> >> UB = [50,5,5];
> >> betahat returned [11.37, 1.83, 0.83]
> >> that the constraints don't matter. But the results differ. Hmm.
> >>
> >> It seems to me that the minimum is probably pretty flat. In other
> >> words, I am guessing that the resulting residuals were virtually the
> >> same for your two runs. Why not try a less fussy objective function.
> >> Take the log of everything:
> >> lfunc = log(b1) + log(x1) + b2.*log(x2) + b3.*log(x3);
> >> This transformation has the same minimum, but might be better-behaved.
> >>
> >> Alan Weiss
> >> MATLAB mathematical toolbox documentation
> >>
> >> On 4/13/2010 11:22 AM, Tim wrote:
> >> > I'm using lsqcurvefit with restrictions placed on the upper and lower
> >> > bounds. When I wanted to compare the results to an unconstrained
> >> model I
> >> > changed the restrictions on all parameters to -inf and inf. New
> >> > regression parameters were generated but they were still within the
> >> > original constraints.
> >> >
> >> > Example...
> >> > GENERAL CODE STAYS SAME ACROSS RUNS
> >> > func = (b1.*x1.*x2.^b2.*x3.^b3);
> >> >
> >> > options = optimset('Display', 'iter', 'FunValCheck', 'on',
> >> > 'MaxFunEvals', 5000, 'MaxIter', 1000);
> >> > [betahat,...] = lsqcurvefit('func', beta0, xdata, ydata, LB, UB,
> >> options);
> >> > PARTS I'VE CHANGED BETWEEN RUNS
> >> >
> >> > CONSTRAINED
> >> > LB = [-10,0,0];
> >> > UB = [50,5,5];
> >> > betahat returned [11.37, 1.83, 0.83]
> >> >
> >> > UNCONSTRAINED
> >> > LB = [-inf,-inf,-inf];
> >> > UB = [inf,inf,inf];
> >> > betahat returned [10.11, 1.78, 0.82]
> >> >
> >> > If the parameters changed between runs I would have expected at least
> >> > one element from the new betahat from the unconstrained model to be
> >> > outside the LB and UB of the constrained model.
> >> > Do I just need to put my inner engineer away and think of the two
> >> > betahats as essentially the same? Any thoughts on why the slightly
> >> > different answers? If I do accept them as the same can I return a level
> >> > of significance?
> >> >
> >> > Thanks,
> >> >
> >> > Tim
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