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From: Greg Heath on 19 Nov 2009 08:18
On Nov 19, 4:14 am, "aurikel Radzali" <suriar...(a)gmail.com> wrote:
> Greg Heath <he...(a)alumni.brown.edu> wrote in message <f88c4240-7211-4544-9ca1-f5a656845...(a)f20g2000vbl.googlegroups.com
-----SNIP
> > > > size(P) = ?
> > > > minmax(P) = ?
> > > > size(T) = ?
> > > > minmax(T) = ?
>
> > Please answer the questions.
>
> ---->size(P) =3 24
> size(T) =1 24

MINMAX?

OK [0 1] from a recent reply.

-----SNIP

> > > > > SPREAD=5;
>
> > > > Why did you choose this value?
>
> > > ------------------> i just put some value, i don't know what
> > values should i use? (i thought the best spread is obtained
> > > through trial and error method)

Yes. That is your best bet.

Now that you realize that the input data is contained in a unit cube
it should be obvious the spread = 5 is inappropriate.

> > True, but why whistle in the dark when you can quickly estimate
> > a few charcteristic distances to help limit the search?
>
> ------------------>how do i estimate characteristic distances to help
limit the search? i even do not know what no should i start with
..
What can you tell about the data from N, minmax(P) and summary
statistics of dist(P,P') (exclude the diagonal)?
What if the data was uniformly distributed over the cube?

The input data is only 3-D. Therefore you can obtain 1,2,and
3-D scatter plots.

You can use cluster routines to visualize data groups,

etc

>
> > > > How does it compare to
>
> > > > 0.5*mean(median(dist(P,P')))?
>
> > Typical half-distance between data points. More useful
> > when the data points are either cluster centers from a previous
> > cluster analysis or data points selected as neurons.
>
> > > ---------> do you mean compare this and spread=5?
>
> > Yes.
>
> > > ------------->btw, 0.5*mean(median(dist(P,P')))=1.0815
>
> -----------> is this some kind of guideline to choose the spread?
the spread should be around that value, it is?less or more but not too
far
away?

It depends on the data. Other summary stats or even a histogram of
the distance matrix maybe enlightening w.r.t. what is a good value
for
spread.

Try to understand why the optimal value is good.


> > What is it for hidden neurons ?
>
> > > > > MN=24; % sample points
> > > > > DF=1;
>
> > > > > %net.layers{1}.transferFcn='radbas';
> > > > > %net.layers{2}.transferFcn='purelin';
>
> > > > > net=newrb(P,T,GOAL,SPREAD);
>
> > > > [net, tr] = newrb(P,T,GOAL,SPREAD);
>
> > > -----------------> what is the different between the those two?
> > >What is the purpose of putting [net,tr]?
>
> > tr is the training record
>
> > help newrb
> > doc newrb
>
> > > > > Y=sim(net,P);
> > > > > e=T-Y;
> > > > > msetrain=mse(e)
>
> > > > compare with tr.perf(end)
>
> > > --------------tr.perf(end)=0.0018
>
> > > > > --> this is the result:
> > > > > NEWRB, neurons = 0, MSE = 0.130203
>
> > > > If N ~ 24, This is not inconsistent with
> > > > mean(var(T)) = 0.1359.
>
> > > ___________> I'm really sorry, but i truly do not
> > understand what do you by If N ~ 24,
> > This is not inconsistent with mean(var(T)) = 0.1359
>
> > > ----------------------->how it suppose to be??
>
> > If neurons = 0. Then the the output is a constant
> > equal to the bias. To minimize SSE, the constant must
> > be mean(T). The resulting MSE is (N-1)*var(T)/N
>
> ---------->forgive me, i still don't understand.
> mean(T)=0.5389
> mean(var(T)) = 0.1359
> MSE i= (N-1)*var(T)/N= 0.1302 so?

It is a consistency check. It also warns you that there
must be a nonzero neuron solution with a much lower MSE.
>
> --->To minimize SSE, 'the constant' must be mean(T).
'the constant' refer to?

Yes

> > > > Where is the tabulation for neurons > 0?
>
> > > > > msetrain =
>
> > > > > 0.0018
>
> > > > For how many neurons?
> > > > -------------->Number of nodes in hidden layer=2.0000e+000

You should be able to tabulate No. of neurons and SSE from tr
and recreate the training plot.

> > > > > why the mse value is different?
>
> > > > Different number of neurons
> > > > What is tr.epoch(end)?
> > > > ---------> 2
> > > > >the mean(var(T)) is 0.1359
>
> > > > MSE(neurons = 0) = (N-1)*mean(var(T'))/N
> > > > -------------> owhh..thats why i get the MSE=0.1302
> > > > Hope this helps.
>
> > > > Greg
>
> > > ==============> Thanks, I still have some questions.
>
> > > based on your advice, %For newrb, it is desirable to have SSEgoal < SSE00/100
> > > SSE00=3.1249
> > > SSEm=SSE00/100 %produce 0.0312
>
> > > so i choose my goal 0.0005 % since SSEgoal < SSE00/100. is this right?
>
> > I would use 0.02.

0.0018 more than fits the bill.

Hope this helps.

GregGreg-
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