How can I solve this problem?

Why does the outcome of this code:
function Ks = nn_ks(v, f, t)
for i=1:27
v=[80;80;80;80;80;80;80;80;80;150;150;150;150;150;150;150;150;150;220;220;220;220;220;220;220;220;220;]
f=[0.05;0.05;0.05;0.1;0.1;0.1;0.15;0.15;0.15;0.05;0.05;0.05;0.1;0.1;0.1;0.15;0.15;0.15;0.05;0.05;0.05;0.1;0.1;0.1;0.15;0.15;0.15;]
t=[5;10;15;5;10;15;5;10;15;5;10;15;5;10;15;5;10;15;5;10;15;5;10;15;5;10;15;]
v(i,1)=(v(i,1)-80)./(220-80)
f(i,1)=(f(i,1)-0.05)./(0.15-0.05)
t(i,1)=(t(i,1)-5)./(15-5)
X=[v(i,1); f(i,1); t(i,1)]
W1 = [0.83953 1.416 4.3197;
0.70603 -4.1921 -0.30888;
-3.1168 2.7796 0.75788]
B1=[-3.1521;
-0.45802;
-2.9756]
W2 =[0.70956 1.1833 -0.059955]
B2 =[-0.98502];
Y1 = logsig(W1*X + B1*ones(1,size(X,2)));
Y2 = purelin(W2*Y1 + B2*ones(1,size(Y1,2)));
Ks(i,1) =Y2.'
end
is very different from the outcome (outputs) of NN Toolbox when the same weights, biases, learning functions and algorithms are used?

 Accepted Answer

% How can I solve this problem?
% Asked by Parwaz Ali about 6 hours ago
% Latest activity Commented on by Walter Roberson about 6 hours ago
% Why does the outcome of this code:
% function Ks = nn_ks(v, f, t)
% is very different from the outcome (outputs) of NN Toolbox when the same weights, biases, learning functions and algorithms are used?
Answer is below
Create column vectors w/o semicolons: columnvec = rowvec'
Avoid using the same variable name on both sides of an equation
Avoid unnecessary loops
close all, clear all, clc
V = [ 80 80 80 80 80 80 80 80 80 ...
150 150 150 150 150 150 150 150 150 ...
220 220 220 220 220 220 220 220 220 ];
F = [ 0.05 0.05 0.05 0.1 0.1 0.1 0.15 0.15 0.15 ...
0.05 0.05 0.05 0.1 0.1 0.1 0.15 0.15 0.15 ...
0.05 0.05 0.05 0.1 0.1 0.1 0.15 0.15 0.15 ] ;
T = [ 5 10 15 5 10 15 5 10 15 ...
5 10 15 5 10 15 5 10 15 ...
5 10 15 5 10 15 5 10 15 ] ;
X = [ V; F ; T ];
[ I N ] = size(X)
maxX = max(X')' * ones(1,N);
minX = min(X')' * ones(1,N);
x = ( X-minX)./(maxX-minX);
whos
W1 = [ 0.83953 1.416 4.3197;
0.70603 -4.1921 -0.30888;
-3.1168 2.7796 0.75788 ];
[ H I ] = size(W1)
b1 = [ -3.1521 -0.45802 -2.9756 ]'
B1 = b1*ones( 1,N);
W2 = [ 0.70956 1.1833 -0.059955 ];
[ O H ] = size(W2)
b2 = -0.98502 ;
h = logsig( W1*x + B1);
y = W2*h + b2;
No learning functions or algorithms were used here
MATLAB uses [-1,1] scaling, not [0,1]
Hope this helps,
Thank you for formally accepting my answer
Greg

More Answers (2)

Walter Roberson
Walter Roberson on 13 Nov 2013
Most of the NN routines initialize weights randomly. It is also possible to set particular weights, but I do not recall how to do that (Greg Heath has shown how to do it more than once.)

3 Comments

Thank you Walter! I sent him a link of my problem. Hopefully, he will answer, thank you again!
You could search... I would guess using the terms
NN random weight contributor:"greg heath"
might find relevant posts.
Hi Walther,
thank you for your nice suggestion!
All the best!

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Parwaz Ali
Parwaz Ali on 14 Nov 2013
Dear Mr. Gerg,
Thank you very much for your wonderful contribution. I could obtain the same results as those of nntool, using another code and your hint of [-1 1] data normalization.
I had no idea that MATLAB uses default scaling of [-1 1]. Thank you again!
All the best!

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on 13 Nov 2013

Commented:

on 14 Nov 2013

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