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Precisions obtained with network openloop NarX

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FRANCISCO
FRANCISCO 2013-3-3
Write data from network openloop with NarX:
N=2200
Ndof=1449
Neq=2200
Ntrneq=1540
Nw=91
I=8
O=1
R2=0.9911
R2a=0.9907
R2trn=0.9914
R2tst=0.9900
R2val=0.9905
MSE=8.48e-5
MSEa=8.84e-5
MSEtrn=9.2e-5
MSEtst=9.2e-5
MSEval=9.065e-5
If you know someone tell me if there is any sign of overfitting appreciate it because I think it achieves high precision and the graph I get a little offset from the original. Variables used are highly correlated. We also appreciate and would tell me ontener several steps ahead with real data and because I get to perform one iteration morning values but need to continue iterating my future values of inputs and that is not possible. thank you very much

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Greg Heath
Greg Heath 2013-3-4
net = narxnet(ID,FD,H);
What are ID, FD and H?
Typically, the sign of overfitting is very high R2trn with much lower R2trna, R2val and R2tst.
You're looking good.
Greg
  1 个评论
FRANCISCO
FRANCISCO 2013-3-4
ID = 1:2
FD = 1:2
H = I think more or less I move between 8-12. Is that I have made many tests and now I'm in college and I have my data here, but the number of H is handled in that interval.
As calculated R2trna? It is the only parameter that would fail me by check. All other conditions met.
Many thanks Greg

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