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ANN

Malathy 님이 질문을 제출함. 24 Feb 2012
최근 활동 Walter Roberson 님이 댓글을 추가함. 20 Sep 2019 19:58
My input array is
35.0000 1.0000 1.2500 0.5000
65.0000 1.0000 1.2500 0.5000
35.0000 2.0000 1.2500 0.5000
65.0000 2.0000 1.2500 0.5000
35.0000 1.0000 1.7500 0.5000
65.0000 1.0000 1.7500 0.5000
35.0000 2.0000 1.7500 0.5000
65.0000 2.0000 1.7500 0.5000
35.0000 1.0000 1.2500 1.0000
65.0000 1.0000 1.2500 1.0000
35.0000 2.0000 1.2500 1.0000
65.0000 2.0000 1.2500 1.0000
35.0000 1.0000 1.7500 1.0000
65.0000 1.0000 1.7500 1.0000
35.0000 2.0000 1.7500 1.0000
65.0000 2.0000 1.7500 1.0000
20.0000 1.5000 1.5000 0.7500
80.0000 1.5000 1.5000 0.7500
50.0000 0.5000 1.5000 0.7500
50.0000 2.5000 1.5000 0.7500
50.0000 1.5000 1.0000 0.7500
50.0000 1.5000 2.0000 0.7500
50.0000 1.5000 1.5000 0.2500
50.0000 1.5000 1.5000 1.2500
50.0000 1.5000 1.5000 0.7500
50.0000 1.5000 1.5000 0.7500
50.0000 1.5000 1.5000 0.7500
50.0000 1.5000 1.5000 0.7500
50.0000 1.5000 1.5000 0.7500
50.0000 1.5000 1.5000 0.7500
My output array is
7.9700
15.2200
9.5300
12.6000
9.4400
13.1600
9.8100
11.6500
8.4700
14.0500
9.5000
11.7000
7.0900
11.1000
6.6900
10.4400
6.2900
14.8800
12.2500
10.3800
10.6500
9.5000
10.1600
10.6500
11.1300
11.5000
11.1000
10.9900
10.9800
11.0300
I used Neural Network tool(nntool). I clicked new in the network data manager and in the create network page I selected the Network type as Feed-forward backprop
Input data in a variable'c' as given above and selected it from the dropdown box and the target data 'd' same. other parameters like training function & Adaption learning function and Performance function as default values.
If I need only one hidden layer, Do I need to set the number of layers as 1?
Similarly I set the number of neurons as 3 for layer1 & the transfer function as LOGSIG and created the network by name network1. Now I opened the network and clicked the Train tab. In the training info I selected the Training Data Input as c and target as d and clicked Train. But I am not getting the ANN predicted output value corresponding to each input value.
Please help me at the earliest.

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Malathy 님의 답변 29 Feb 2012
 채택된 답변

we use the SIM function to see the predicted response of the ANN.

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husam eddin comments
me2

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Greg Heath 님의 답변 24 Feb 2012

If you plot output vs each of the inputs you will see that each plot consists of 3 vertical lines. The best model I can think of is a 2nd order polynomial that will intersect each line approximately in the middle.
Hope this helps.
Greg

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My question here is how to see the predicted response of ANN

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Greg Heath 님의 답변 4 Mar 2012

ANN : Malathy asked on 24 Feb 2012 at 7:49
My input array is
-----SNIP
My output array is
-----SNIP
I used Neural Network tool(nntool). I clicked new in the network data manager and in the create network page I selected the Network type as Feed-forward backprop
Input data in a variable'c' as given above and selected it from the dropdown box and the target data 'd' same. other parameters like training function & Adaption learning function and Performance function as default values.
If I need only one hidden layer, Do I need to set the number of layers as 1?
% NO. The number of weight layers is 2 (input-hidden and hidden- output)
Similarly I set the number of neurons as 3 for layer1 & the transfer function as LOGSIG
% TANSIG is preferred because inputs are transformed (via default MAPMINMAX) to be bipolar.
and created the network by name network1. Now I opened the network and clicked the Train tab. In the training info I selected the Training Data Input as c and target as d and clicked Train. But I am not getting the ANN predicted output value corresponding to each input value.
Please help me at the earliest.
% You'll have to quantify your performance.
% I ran 10 trials each for H = 0:4 (H=0 is just a linear regression but you need to use H = [] in newff). The summary statistics for an adjusted R^2 obtained from all of the data are tabulated below. If Ntrneq is the number of training equations and Nw is the number of weights to be estimated, the adjusted MSEtrn is obtained from SSEtrn/(Ntrneq-Nw)to reduce the optimistic bias caused by using the same data that was used in estimating the weights. Validation and Test error estimates are not usually adjusted even though the validation data is used in the design.
%Ignoring the default Ntrn/Nval/Ntst division 0.6/0.2/0.2, I just used all of the data to obtain MSEa = Neq*MSE/(Neq-Nw).
H = 0 1 2 3 4
minR2a = 0.7797 0.1428 0.1497 0.5269 0.4192
meanR2a = 0.8207 0.7342 0.7733 0.8689 0.8363
medianR2 = 0.8222 0.8199 0.8479 0.9062 0.9015
stdR2a = 0.0262 0.2267 0.2312 0.1461 0.1841
maxR2a = 0.8622 0.8599 0.9573 0.9969 0.9876
As you can see, for H = 3 and 4, the best of 10 designs yields R2a ~ 0.99.
Hope this helps.
Greg

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