why I obtained different result from NN sim and manual calculation using the weights

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I use newff generate a very simple neural network with one input layer, one hidden layer (just one neuron for easy manual calculation), and one output layer like this:
>>x1=[-15:1:15];
>>y1=0.05*x1.^3-0.2*x1.^2-3*x1+20;
% the hidden layer and output layer both use "purelin" as transfer function.
>>net=newff(x1,y1, 1, {'purelin'});
%training the network
>>net=train(net,x1,y1);
% I use 15 as test, the output from sim is 49.8170
>> output=sim(net, 15)
output = 49.8170
%%%%%%and I also manually calculate the output using 15 as input,
%my calculation is like this , first, I obtained the input and layer weights, and the bias are as following:
>>IW=net.IW
IW =
[0.6972]
[]
>>LW=net.LW
LW =
[] []
[0.5009] []
>> b=net.b
b =
[-0.5228]
[ 0.5173]
% the my calculation like this
>> (15*0.6972-0.5228)*0.5009+0.5173
ans = 5.4938
Why the "sim" and my manual calculation is different? Where is wrong? ??

Respuestas (4)

Bing
Bing el 15 de Mayo de 2012
Thanks for the answer.
But the "a" of purelin function in matlab is 1. This is what I got from matlab
>> purelin((purelin(15*0.6972-0.5228))*0.5009+0.5173)
ans =
5.4938

Thijs
Thijs el 15 de Mayo de 2012
doing some matlab hackery i think i found the answer. both the inputs and the outputs of the neural network are mapped onto a minmax range, the inputs are mapped from -15 to 15 onto 0 to 1 (so 15 becomes 1). these mapped inputs are then passed through the neural network. purelin((purelin(1*0.6972-0.5228))*0.5009+0.5173)=0.8196. This output is then mapped from -1 to 1 onto ymin to ymax.
You can find ymin and ymax by entering: >>net.outputs{2}.range
conclusion: to manually duplicate the network performance:
input=15;
%find the input and output mapping
input_range=net.inputs{1}.range;
output_range=net.outputs{2}.range;
%perform the mapping
in=(input-input_range(1))/(input_range(2)-input_range(1));
%pass the input through the net
temp=(IW{1}*1+b{1})*LW{2,1}+b{2}
%map the output
y1=(temp--1)/2*(output_range(2)-output_range(1))+output_range(2)

Bing
Bing el 16 de Mayo de 2012
your anweer to input=15 is right. I also test input=-15. maunally, I got 6.604 but sim generate -36.6091
>> temp=(IW{1}*0+b{1})*LW{2,1}+b{2}
temp =
0.2554
y1=(temp+1)/2*(output_range(2)-output_range(1))-148.75
y1 =
6.6040
output=sim(net, -15)
output =
-36.6091

hassan khatir
hassan khatir el 19 de Jul. de 2023
Editada: hassan khatir el 19 de Jul. de 2023
use this function:
function y2=sim2(net,x)
xoffset=net.inputs{1}.processSettings{1}.xoffset;
gain=net.inputs{1}.processSettings{1}.gain;
ymin=net.inputs{1}.processSettings{1}.ymin;
w1 = net.IW{1}; % (10x6)
w2 = net.LW{2}; % (2x10)
b1 = net.b{1}; % (10x1)
b2 = net.b{2};
% Input 1
y1 = (x-xoffset).*gain+ymin;
% Layer 1
a1 = 2 ./ (1 + exp(-2*(repmat(b1,1,size(x,2)) + w1*y1))) - 1;
% output
outputs=repmat(b2,1,size(x,2)) + w2*a1;
gain = net.outputs{2}.processSettings{:}.gain;
ymin = net.outputs{2}.processSettings{:}.ymin;
xoffset = net.outputs{2}.processSettings{:}.xoffset;
y2 = (outputs-ymin)./gain + xoffset;
end

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