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MathGroup Archive 2006

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Re: variable metric method automatic gradient yields Indeterminate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70433] Re: [mg70397] variable metric method automatic gradient yields Indeterminate
  • From: "Chris Chiasson" <chris at chiasson.name>
  • Date: Mon, 16 Oct 2006 02:34:52 -0400 (EDT)
  • References: <200610150419.AAA12698@smc.vnet.net> <F474721B-1771-4C59-8CE9-7174C99CA71A@mimuw.edu.pl>

I've already implemented the limit approach, but I think your approach
is better. It would be able to handle problems where the indeterminate
expression arose from a different variable other than the one
corresponding to that position in the gradient.

On 10/15/06, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>
> On 15 Oct 2006, at 13:19, Chris Chiasson wrote:
>
> > A user-defined augmented lagrange multiplier method for NMinimize
> > drives a (user defined) variable metric method for FindMinimum. The
> > NMinimize routine ends up creating penalty functions that have
> > "discontinuous" first order derivatives due to the presence of
> > functions like Max. At the points of discontinuity in the derivative,
> > the Indeterminate result usually ends up multiplied by zero. The limit
> > of the derivative exists.
> >
> > for example, the function passed to FindMinimum is
> > func=Max[0,-X[1]]^2+Max[0,-1+X[1]+X[2]]^2+(-1+X[1])^2+(-1+X[2])^2
> >
> > its derivative with respect to X[1] is
> > 2*(-1+Max[0,-1+X[1]+X[2]]*Piecewise[{{1,X[1]+X[2]>1}},0]+
> > Max[0,-X[1]]*Piecewise[{{-1,X[1]<0},{0,X[1]>0}},Indeterminate]+X[1])
> >
> > Notice that when X[1] is zero, the Piecewise returns Indeterminate,
> > which is multiplied by zero from the nearby Max function. However,
> > 0*Indeterminate is still Indeterminate in Mathematica. When the
> > derivative is evaluated at X[1]=zero, the answer returned is
> > Indeterminate. This totally messes up the numerical routine.
> >
> > the limit of D[func,X[1]] as X[1]->0 is
> > Piecewise[{{-2,X[2]<1}},2*(-2 +X[2])]
> >
> > I am tempted to check the gradient vector for Indeterminate results,
> > look up the "corresponding" variable (heh, how do I really know which
> > one is responsible?), and take the limit as that variable approaches
> > the value I wanted to evaluate at. I don't know how well that will
> > work in practice.
> >
> > So, does anyone have any suggestions?
> > --
> > http://chris.chiasson.name/
> >
>
>
> It interesting to note that the following two methods give different
> looking answers:
>
>
> p = Block[{Indeterminate}, D[func, X[1]] /. X[1] -> 0]
>
>
> 2*Max[0, X[2] - 1]*Piecewise[{{1, X[2] > 1}}] - 2
>
>
> q = Limit[D[func, X[1]], X[1] -> 0]
>
>
> Piecewise[{{-2, X[2] < 1}}, 2*(X[2] - 2)]
>
> but are, of course, equivalent:
>
> Simplify[p == q]
> True
>
> (This is also meant as a suggestion of an alternative method, though
> I would not recommend its unthinking use in other situations where
> Indeterminate occurs).
>
> Andrzej Kozlowski
>
>
>
>
>


-- 
http://chris.chiasson.name/


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