Asked by Parwaz Ali
on 13 Nov 2013

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?

*No products are associated with this question.*

Answer by Greg Heath
on 14 Nov 2013

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

Answer by 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.)

Parwaz Ali
on 13 Nov 2013

Thank you Walter! I sent him a link of my problem. Hopefully, he will answer, thank you again!

Walter Roberson
on 13 Nov 2013

You could search... I would guess using the terms

NN random weight contributor:"greg heath"

might find relevant posts.

Answer by 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!

Opportunities for recent engineering grads.

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