MATLAB: How can i choose the parameters of the network

neural networkneural networks

Hi, I'm beginner in neural network,so to justify its using in my doctorate research, i want to do some comparatives with other tools to show the ability of this method.Now i want to create network, its input is x=[ -1 1 -1 1 -1 1 -1 1; -1 -1 1 1 -1 -1 1 1; -1 -1 -1 -1 1 1 1 1] and the ouptut for example y=[ 9 6 4 3 8 1 3 4] knowing that for example 9= a*(-1)+b*(-1)+c*(-1)*a1*(-1)*(-1)+b1*(-1)*(-1)+c1*(-1)*(-1)+d*(-1)*(-1)*(-1) the same way for others how can i choose the parameters of my network?

Best Answer

close all, clear all, clc
 x0 = [ -1 1 -1 1 -1 1 -1 1; -1 -1 1 1 -1 -1 1 1; -1 -1 -1 -1 1 1 1 1]
 x4 =[ 9 6 4 3 8 1 3 4]
 x = [ x0; x4]
 t = [ 23.72, 19.84, - 8.91 , 4.41 , -7.53, 1.41, 5.16, 0.66 ]
 [I N]  = size(x)  % [ 4 8 ]
 [O N ] = size(t)  % [ 1 8 ]
 Neq    = N*O      %   8
% Obviously, you have Neq = 8 equations.
 meant2 = mean(t,2)     %   4.845
 vart12 = var(t,1,2)    %  119.13 Biased
 vart02 = var(t,0,2)    % 136.15 Unbiased
% Therefore, if you try a constant model you get
 y00    = repmat(mean(t,2),1,N)  
 MSE00  = vart12
 MSE00a = vart02
% If you try a linear model solution (see my previous code) you get
 W0 = t/[x;ones(1,N)]    %3.3242  -2.9258  -3.9665  1.2714  -1.194
 Nw0  = numel(W0)        %5
 Ndof0 = Neq-Nw0         %3
 y0 = W0*[x;ones(1,N)]  
 e0 = t-y0 
 SSE0 = sse(e0)          %  536.7
 MSE0 = SSE0/Neq         %   67.1
 MSE0a =SSE0/Ndof0       %  178.9
 NMSE0 = MSE0/MSE00      %  0.563 
 NMSE0a = MSE0a/MSE00a   %  1.314          
 R20 = 1-NMSE0           %  0.437
 R20a = 1-NMSE0a         % -0.314
% R20 = 0.44 means that the linear model appears to account for 44% of the
% target variance and, therefore, is better than the constant model. However,
% R20a < 0 means that when the bias of testing with training data is taken
% into account, the linear model is probably worse and not much, if any,
% confidence should be put in using the R2 estimate for future data.
% In order to obtain a better coefficient model, higher order polynomial or
% neural network models can be tried. In general, however, this will result in
% decreased and negative (more unknowns than equations.) estimation degrees of
% freedom
% Apparently you have deduced a reduced term 3rd order polynomial model for
% the I/O relation. Was this done via underdetermined linear least squares and
% the coefficients for the 12 missing terms were negligible? If so, was
% regularization used?
% Neural Network Solutions:
% H = 0 hidden nodes corresponds to the linear classifier.
% H = 1 hidden node can be worse if not much better than H= 0
% H= 2 overfits the model Nw > Neq. Therefore R2a < 0. Nevertheless,
% It is interesting to see the results for Ntrials = 20 multiple designs for
% H=1 and H=2:
 % H=1 result =
 %          Trial      Nepochs     R2              R2a              
 %             1           11         0.242          -4.309
 %             2           10         0.014          -5.901
 %             3           10         0.038          -5.736
 %             4           11         0.260          -4.183
 %             5           11    => 0.837         -0.144     
 %             6             8          0.427         -3.010      
 %             7           10    => 0.837         -0.144    
 %             8           33    => 0.851         -0.047     
 %             9             7          0.178         -4.752       
 %            10          17    => 0.851         -0.075     
 %            11        458    => 0.867          0.061       
 %            12          20    => 0.851         -0.047       
 %            13        146         0.293         -3.948
 %            14          24    => 0.851         -0.047       
 %             15          84         0.293         -3.948
 %            16          11         0.290         -3.971
 %            17          16         0.293         -3.948
 %            18        255         0.298         -3.914           
 %            19            4         0.000          -5.998         
 %            20        317   => 0.866           0.061
 % 


 % maxresult =  
 %          20          458       0.866            0.061
% THERE ARE 7/20 R^2 RESULTS IN [ 0.837 0.867 ]
 % H=2 OVERFITTING (Nw > Neq) result =
 % 
 %            1           391      0.791 
 %             2          749      0.896
 %             3          456      0.896 
 %             4          247       ==>0.998  
 %             5          672      0.896 
 %             6              7      0.046 
 %             7          863      0.910      
 %             8            19      0.860      
 %             9            40       ==>0.990  
 %            10         669      0.913     
 %            11             7       ==>0.992   
 %            12           18      0.858       
 %            13           14       ==>0.993 
 %            14         526      0.910      
 %            15         625      0.579    
 %            16           16       ==>0.991   
 %            17       1000      0.911      
 %            18         952      0.896    
 %            19             9      0.036    
 %            20         132       ==>0.993      
 %  
 % maxresult =
 % 
 %            20        1000      0.998
 %
 % THERE ARE 6/20 R^2 RESULTS IN [ 0.990 0.998 ]
Hope this helps.
Thank you for formally accepting my answer
Greg
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