newnarx initialization and validation problem

조회 수: 3 (최근 30일)
Unnikrishnan PC
Unnikrishnan PC 2012년 10월 27일
Dear all,
I am working on the problem: y(k)=0.3*y(k-1)+0.6*y(k-2)+u(k)^3+0.3*u(k)^2-0.4*u(k) using narx. the code is listed below.
net = newnarx(u,yn,0,1:2,[15 10],{'tansig','tansig','purelin'},'trainscg');
net.trainParam.lr = 0.05; net.trainParam.lr_inc = 1.05;
net.trainParam.lr_dec = 0.7; net.trainParam.hide = 50;
net.trainParam.mc = 0.9; net.trainParam.epochs = s;
net.trainParam.goal = 1e-8; net.trainParam.time = 5*3600; [trainP,valP,testV,trainInd,valInd,testInd] = divideblock(u,0.6,0.2,0.2);
[trainT,valT,testT] = divideind(yn,trainInd,valInd,testInd);
net.divideFcn='divideblock';
net = init(net);
My training stops after 6 validation checks. Why? The trained net gives different outputs each time. Is there any mistake in the initialization?
Please help.

채택된 답변

Greg Heath
Greg Heath 2012년 10월 27일
See my previous post re your last program. Most of the comments are relevant for this post.
In particular, do not overwrite defaults unless you have a darned good reason.
Also, it is very seldom that a 2nd hidden layer is necessary.
Validation stopping was created to make sure the trained net is useful for nontraining data. Stopping after MSEval increases for 6 consecutive epochs is reasonable.
You will get different results every time because the initial weights are random.
For each different setting of delays and hidden nodes I usually loop over Ntrials = 10 designs.
You would be surprised how many times only 7 or 8 out of 10 are acceptable.
However, I do initialize the random number generator before the double loop over hidden nodes and random weight initializations.
Hope this helps.
Thank you for formally accepting my answer.
Greg

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Sequence and Numeric Feature Data Workflows에 대해 자세히 알아보기

태그

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by