My neural network is getting trained but not yielding correct answers?plz help

 This code is written to form a relation between

• two actuator's movement z1 and z2 with
• laser points x,y coordinate.

basically aiming for two input,two output network. laser point is input and actuator position is output. i want to provide target to network that my laser needs to be directed to some particular location.The network must provide me actuator coordinates as output.

clear;

clc;

hold off;

format long;

filename = 'neural_network_model.xlsx';
%%input file contains laser coordinates and corresponding z1,z2 value.

point_Datar= xlsread(filename,'A1:Z300');

input =point_Datar(1:300,17:18);
%inputs in column 17,18,these are laser pointers x,y coordinates

ninput=(input(:,1) - min(input(:,1)))./(max(input(:,1)) - min(input(:,1)));
%%normalizing input 1

ninput(:,2)=(input(2) - min(input(2)))./(max(input(2)) - min(input(2)));
%normalizing input 2

input(:,1)-(ninput(:,1)*(max(input(:,1)) - min(input(:,1)))+min(input(:,1)))
%check

z1=point_Datar(1:300,7);
%read output 1,this is actuator 1s z coordinate

nz1=(z1 - min(z1))/(max(z1) - min(z1));
%normalize output 1

z1-(nz1*(max(z1) - min(z1))+min(z1))
%check output1 norm must result zero

z2=point_Datar(1:300,11);
%read output 2,this is actuator 2s z coordinate

nz2=(z2 - min(z2))/(max(z2) - min(z2));

z2-(nz2*(max(z2) - min(z2))+min(z2))
%check

output =[nz1 nz2];
%above line makes the output matrix

net = newff(ninput,output,[15 15],{'tansig','tansig'});

view(net);

net = train(net,ninput,output);
% Iterates gradient type of loop

trained_output= sim(net,ninput);

figure(1);plot(trained_output(:,1),trained_output(:,2),'*g');hold on;

plot(output(:,1),output(:,2),'sr');

input_zero=ninput;

input_zero(300,1)=0.0;
%making input 1 to be zero

input_zero(300,2)=0.0;
%making input 2 to be zero

trained_output1= sim(net,input_zero);

plot(trained_output(:,1),trained_output(:,2),'.r');

%%%check output..my laser lies on a triangular stand,two vertex with actuator and one vertex with ball-socket joint

param1=0.249977909490350;
%empirical value to measure a point inside a triangle

param2=0.500013614834919;
%empirical value to measure a point inside a triangle

a=point_Datar(1,1:3);   
%first socket joint

b=[point_Datar(1,5:6) trained_output1(1,1)*(max(z1) - min(z1))+min(z1)]
%for actuator b -x,y are constant values and z is network prediction for actuator 1s position

d=[point_Datar(1,9:10) trained_output1(1,2)*(max(z2) - min(z2))+min(z2)] 
%for actuator d-x,y are constant values and z is prediction for actuator 2s position 

l=b-a;
%finding pos vector for line ab

m=d-a;
%finding pos vector for line ad

c=a+param1*l+param2*m;
%this gives me laser position on  triangle abd

cros=cross(l,m);
%direction normal to ab and ad

syms p;
%line parameter

line=c+p*cros;
%line parameter p,laser is normal to two lines l,m at c

z=vpa(solve(line(3),p));

laserx=vpa(subs(line(1),p,z))
%Theo_X_Laser_point

lasery=vpa(subs(line(2),p,z))
%Theo_Y_Laser_point
%if neural network output is correct  laserx and lasery must be zero.

As i make bothinput columns 0,0 for row 1. i get all output columns equal. column 1 value becomes equal to column 2 for all rows.

I dont understant where am i wrong.

Plz spot my error in program or else give me 2 input 2 output network code. Should i process my data in any other way?

[Greg Heath]

Please format your code

[sameer d]

Please any one help me out??

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup

[sameer d]

Fitnet is not working .I am not able to predict output correctly.

[Greg Heath]

Apparently, you are not concerned about formatting your code so that someone will feel respected enough to help.

STRANGE!!!

[sameer d]

Sorry for delay.jus now viewed the tutorial.please any neural network expert show some way out of this problem.

Answer 1

[Greg Heath]

Your input and target matrices are transposed

Why Hmin=1 instead of 0?

Where did you get Hmax = 50?? Search greg fitnet Hub

I = O = 2

Unecessary weight initialization. That is what configure does

Only use rng once: before outer loop

Have 100 nets. Choose one with smallest h with Rsq >= Rsqgoal

After you succeed

DESIGN THE INVERSE NET:

INPUT = A

TARGET = L

HOPE THIS HELPS

GREG

[sameer d]

I HAVE CHANGED HMAX TO 39 BUT I HAVE GOT THE NEAREST ANSWER AT H=46.ANY OTHER SUGGESTIONS APART FROM FOLLOWING MODIFICATIONS.i GOT THESE FROM YOUR POST Asked by DemoiselX on 15 Feb 2013. IS MY CHOICE FOR H=1 CORRECT.I DONT UNDERSTAND LOGIC FOR THAT what is the difference between hmax and hub.plz state some reference code. It seems i am very near to results plz..little more help

input =point_Datar(1:100,17:18);
%target
z1=point_Datar(1:100,7);
z2=point_Datar(1:100,11);
ntarget=[z1 z2];
L = input';%Lx,Ly laser coordinates input
At= ntarget'; %Ax,Ay actuator coordinates target
%Ay= []%Ax,Ay actuator coordinates output
%%%FINDING HMAX
I=2;
O=2;
H=1;
Neq=200 %N*OUTPUT
Nw = (I+1)*H+(H+1)*O
% H =< Hub ==> Neq >= Nw
Hub = -1 + ceil( (Neq-O)/(I+O+1))  % 39

Hmin=0;
dH = 1;
Hmax =39;

j = 0;
I=2;
O=2;
 rng(0)
 k=0;
 for h = Hmin:dH:Hmax 
     j = j+1;
     if h == 0
        net = fitnet([]);
     else 
        net = fitnet(h);
     end
     for i = 1:10
         net = configure(net,L,At);

   % --------SNIP
   %view(net)
    net = train(net,L,At);
    Ay = net(L);
    NMSE = mse(Ay-At)/mean(var(At',1));
   Rsq = 1- NMSE

    L(1,59)=0.1;%i want to know actuator position when laser is near zero
       L(2,59)=0.1;%i want to know actuator position when laser is near zero
       Ay = net(L);%simulating with trained network
      az=Ay(1,59)%denormalizing does not yield correct result
      xlswrite(filename,az,'sheet1','A26');
      bz=Ay(2,59)%denormalizing does not correct result
      xlswrite(filename,az,'sheet1','A27');

[Greg Heath]

Sorry, I wrote my last reply a few minutes ago before seeing this.

Bottom line:

 You need to design another net to perform the inverse.

Hope this helps.

Greg

[sameer d]

Thanks a lot.let me give a last try

Answer 2

[Greg Heath]

Since normalizations and data-divisions are defaults, I really don't understand what you are doing. If you want a 2-input/2-output regression net you should use the CURRENT regression/curve-fitting net FITNET which calls the generic net FEEDFORWARDNET.

What version of MATLAB and NNToolbox do you have?

>> ver

If your versions of MATLAB and the NNToolbox are 2007 or older, you have to use the OBSOLETE regression/curve-fitting net NEWFIT which calls the generic net NEWFF.

Regardless of whether you use NEWFIT or FITNET, Read the relevant documentation and examples obtained from the help and doc commands. Foe example,

 help fitnet
 doc fitnet

Also, you can reference previous posts in BOTH NEWSGROUP and ANSWERS with search words like

 fitnet
 greg fitnet

or

 newfit
 greg newfit

Reply if you need more help.

Greg

[sameer d]

thanks,i would like to elaborate a little more.i train the network with laser lights movement on screen i.e x,y coordinare as input and actuator movement as output z1,z2 for two actuators.this data is obtained by experiment.basically i want actuator movement required to make the laser to reach any desired x,y coordinate on laser screen. The neural network i have used above is getting trainec but not yielding correct result.i have tried to predict actuator movement to reach 0,0 on laser screen.which is not working.ok i wd try with fitnet and come back. Thanks for reply. Based on above explanation if u could suggest some thing.it would be helpful.

Answer 3

[Greg Heath]

 L = Lx,Ly laser coordinates input
 At= Ax,Ay actuator coordinates target
 Ay= Ax,Ay actuator coordinates output

 DOCUMENTATION CODE(from help&doc fitnet)   
     [L,At] = laseractuator_dataset;
     net = fitnet(10);
     net = train(net,L,At);
     view(net)
     Ay = net(L);
     perf = perform(net,At,Ay)

SIMPLE REVISION FOR "INITIAL" INVESTIGATION

     [L,At] = laseractuator_dataset;
     plot(L(1),L(2),'k'), hold on 
     plot(At(1),At(2),'b')
     MSE00 = mean(var(At',1)) % REFERENCE MSE
     net = fitnet;    % H=10 DEFAULT
     rng('default')   % FOR REPRODUCIBILITY
     [ net tr Ay E] = train(net,L,At);
     % Ay = net(L); E = At-Ay
     plot(Ay(1),Ay(2),'ro-')
     Rsq = 1-mse(E)/MSE00
     %See training record tr for trn/val/tst details

 COMPREHENSIVE FOLLOWUP 

      Minimize H subject to performance constraint

           Rsqtrn <= 0.995

      using 

      net.trainParam.goal = 0.005*MSE00

For zillions of examples search both NEWSGROUP 
and ANSWERS using

      greg fitnet Hmin Hmax Ntrials

Hope this helps

Thank you for formally accepting my answer

Greg

[sameer d]

SIR,I WOULD TRY AGAIN BASED ON YOUR INSTRUCTIONS.THANKS

[sameer d]

NO CORRECT ANSWERS achieved ....i am using 100 data sets

%%

L = ninput;%Lx,Ly laser coordinates input
At= ntarget %Ax,Ay actuator coordinates target
Ay= []%Ax,Ay actuator coordinates output
%DOCUMENTATION CODE(from help&doc fitnet)   
   %[L,At] = laseractuator_dataset;
   net = fitnet(10);
   net = train(net,L,At);
   view(net)
   Ay = net(L);
  perf = perform(net,At,Ay)
    Ay
   L(100,1)=(0.1 - min(z1))./(max(z1) - min(z1))%i want to know actuator position when laser is near zero
   L(100,2)=(0.2 - min(z1))./(max(z1) - min(z1))%i want to know actuator position when laser is near zero
   Ay = net(L)%simulating with trained network
(Ay(100,1)*(max(z1) - min(z1))+min(z1))%denormalizing does not yield correct result
(Ay(100,2)*(max(z1) - min(z1))+min(z1))%denormalizing does not correct result

whats wrong i dont know please help???

[Greg Heath]

What are you doing??

Ay is your answer. How close is it to At?

 NMSE = mse(Ay-At)/mean(var(At',1))
 Rsq = 1- NMSE

Why are you fooling around with normalizations? MATLAB takes care of that automatically.

Plot At 'b.' and Ay 'ro' on the same 2-D plot

[sameer d]

i removed normalization as per your advise The values that you have asked for are as follows :

NMSE = 0.303196341847070 Rsq = 0.696803658152930

The network always provides correct result for the training data properly but it is not yielding vague result when input is (0,0). What else should i do...?

[Greg Heath]

sameer d about 24 hours ago

>i removed normalization as per your advise The >values that you have asked for are as follows :

>NMSE = 0.303196341847070 Rsq = 0.696803658152930

>The network always provides correct result for >the training data

What?! Those numbers indicate otherwise. Rsq ~ 0.7 means that you are only successfully modelling ~ 70% of the total target variance.

The training goal that I specified is 99.5% for the training data variance

What are your individual percentages for the training, validation and test subsets?

Why don't you try to obtain two good nets

net1: net that yields the direct transformation L ==> At

net2: net that yields the inverse trainsformation At ==> L

My method for optimizing net performance is to design ~100 nets in a double loop:

OUTER LOOP: ~ 10 different values for the number of inner nodes: h = Hmin:dH:Hmax

INNER LOOP: ~ 10 different trials with different random initial weights:

              ntrials = 1:Ntrials

I have posted zillions of examples in BOTH NEWSGROUP and ANSWERS.

For examples search with

            fitnet Hmin Hmax Ntrials

get 8 NEWSGROUP HITS and 35 ANSWERS HITS

 Hope this helps.

 Greg

[sameer d]

• Thanks for your guidance,plz go through once whether i applied everything you said!!But very perfect results are not achieved yet!!

   clear;
    clc;
    hold off;
    format long;
    filename = 'neural_network_model_bothvary.xlsx';
    point_Datar= xlsread(filename,'A1:Z400');
    %input
    input =point_Datar(1:100,17:18);
    %target
    z1=point_Datar(1:100,7);
    z2=point_Datar(1:100,11);
    ntarget=[z1 z2];
    L = input;%Lx,Ly laser coordinates input
    At= ntarget; %Ax,Ay actuator coordinates target
    %Ay= []%Ax,Ay actuator coordinates output
    Hmin=1;
    dH = 5;
    Hmax =50;
    j = 0;
    I=100;
    O=100;
     for h = Hmin:dH:Hmax 
         j = j+1;
         if h == 0
            net = fitnet([]);
         else 
            net = fitnet(h);
         end
         for i = 1:10
             net = configure(net,L,At);
             rng(0)
    %random weights
    H=h
    IW = 0.01*randn(H,I);
    b1 = 0.01*randn(H,1);
    LW = 0.01*randn(O,H);
    b2 = 0.01*randn(O,1);

     % --------SNIP
   %view(net)
    net = train(net,L,At);
    Ay = net(L);
    NMSE = mse(Ay-At)/mean(var(At',1));
   Rsq = 1- NMSE;
   L(59,1)=0.1;%i want to know actuator position when laser is near zero
       L(59,2)=0.2;%i want to know actuator position when laser is near zero
       Ay = net(L);%simulating with trained network
      az=Ay(59,1)
      xlswrite(filename,az,'sheet1','A26');
      bz=Ay(59,2)
      xlswrite(filename,az,'sheet1','A27');
      %%%%%%%%%%%%%%%%%%%%%check

       end
   end  
i wd be obilized for your help

[Greg Heath]

% x = input; % L1,L2 laser coordinates input

% t = target; % At1,At2 actuator coordinates target

% y = net(x) % Ay1,Ay2 actuator coordinates output

 % %     Hmin=1;
 % %     dH = 5;
 % %     Hmax =50;
 % %     j = 0;
 % %     I=100;
 % %     O=100;

     No, No, No.

 [ I N ] = size(x) % [ 2 100 ]
 [ O N ] = size(t) % [ 2 100 ]  
 vart1   = mean(var(t',1))

 % Using MATLAB defaults

 Ntrn   = N -2*round(0.15*N) % 70
 Ntrneq = Ntrn*O  % 140 No.training equations

 % With an I-H-O fitnet
 % Nw = (I+1)*H+(H+1)*O  No. of unknown weights
 % H <= Hub  <==> Ntrneq >= Nw

 Hub = (Ntrneq-O)/(I+O+1) % 27.6

 Hmin = 0       % H = 0 is Linear Model
 dH   = 3
 Hmax = 27
 Ntrials = 10

 j = 0;
 rng(0)
 for h = Hmin:dH:Hmax
    j = j+1;
    if h == 0
        net = fitnet([]);
    else
        net = fitnet(h);
    end
    for i = 1:Ntrials
        net = configure(net,x,y);
        [ net tr y e ] =  train(net,x,y);
        % y = net(x); e = t-y;
        NMSE(i,j) = mse(e)/vart1;
    end
 end
 Rsq = 1- NMSE

 % This gives you L -> Ay
 % To get Ay --> L design another net with x = Ay,y=L

Hope this helps.

Greg

[sameer d]

1. 1.as far as i can understand from matlab site,for configure->network inputs and targets must be passed as parameters.

net = configure(net,x,y); is a typo mistake

Correct code net = configure(net,x,t);

 i got all rsq as posted below.nmse which represents best fit are near to 1.0.so all these values are valid except those which are negetive are acceptable?
rsq values obtained are..
* 0.000000528492003   0.000000661112007   0.000000631230909   0.000000720541757   0.000000707908832   0.000000691816482   0.000000642306807   0.000000433481236
*    0.000000536647301   0.000000701196188  -0.000000159524352   0.000000713369036   0.000000706147932   0.000000600795553   0.000000651187772   0.000000497793156
*    0.000000538103236   0.000000659537954   0.000000709852542   0.000000700149325   0.000000665710296   0.000000722310671   0.000000671238742   0.000000724710070
*    0.000000538770321   0.000000685247350   0.000000605220294   0.000000398002043   0.000000723346418   0.000000353181603   0.000000700993397   0.000000728061392
*    0.000000536182654   0.000000723381590   0.000000707352394   0.000000383880954   0.000000713585918   0.000000672634837   0.000000714694274   0.000000047815638
*    0.000000525321485   0.000000640293008   0.000000378668754   0.000000750771920   0.000000720715242   0.000000722818263   0.000000721924453  -0.000004218926507
*    0.000000523946690   0.000000711628720   0.000000706103165   0.000000681407714   0.000000751282177   0.000000731264375   0.000000715997665   0.000000708396695
*    0.000000538561761   0.000000687434956   0.000000722182058   0.000000707577796   0.000000709714718   0.000000730765507  -0.000000643526857   0.000000067430577
*   -3.255484912792066   0.000000574725431   0.000000714843268   0.000000685423532   0.000000672429065   0.000000712010031   0.000000419654604   0.000000648057580
*   -0.000002281539920   0.000000662460079   0.000000491755093   0.000000738760154   0.000000718663787   0.000000362611300   0.000000652087198   0.000000746630981
* 
*   Columns 9 through 10
* 
*    0.000000500506832   0.000000734894341
*    0.000000651063187   0.000000350088217
*   -0.000002681740241   0.000000729831014
*    0.000000682435283  -0.000001167821722
*    0.000000720648986   0.000000715926733
*    0.000000082349783   0.000000705775590
*    0.000000733060004   0.000000702600612
*   -0.000001255602564   0.000000727928701
*    0.000000728074399  -0.000000722278677
*    0.000000102201376   0.000000723451092
but none of these is giving correct result.why so?

[Greg Heath]

I don't understand your problem..

I don't understand your posted output. It should be a 10x10 matrix.

All nets with NMSE < 0.005 are valid solutions.

Typically you choose one with the smallest number of hidden nodes.

Hope this helps.

Greg

[sameer d]

For configure->network inputs and targets must be passed as parameters.in the code you have given,u have passesed x,y as input for configure.

net = configure(net,x,y);

is this a typo mistake? Correct code

   net = configure(net,x,t);??

[sameer d]

I have posted rsq.why am i getting such low value for rsq??

[Greg Heath]

Your correction of configure is correct.

Your tabulation looks like NMSE, NOT Rsq.

Double check your code.

Greg

[sameer d]

okay i missed 10e6 coming above.BUT NMSE values are also not so impressive.should i try something more

>> min(min((NMSE)))

ans =

     0.248717823086038

and the best i am getting for dH=8

min(min((NMSE)))
ans =
     0.24707

ADDITIONALLY

mean(var(t',1))*0.005

ans =

     0.943076999999999

I would like to acknowledge you on my masters thesis,please permit me for same.

[sameer d]

also i am unable to repeat the results.how can i save the values initialized by configure? would saving the following be sufficient??

    bias(i+j,:)=net.b{1};
    weight(i+j:i+j+1,:)=net.iw{1,1}';

so that i could back calculate i+j value and trace back the weights and biases??

[Greg Heath]

In order to rerun any particular design, you have to know the initial RNG setting. Therefore save it like

for i = 1:Ntrials

    s(i,j) = rng;

    configure ...

Hope this helps.

Greg

[sameer d]

Thanks BUT NMSE values are also not so impressive.should i try something more

>> min(min((NMSE))) ans = 0.248717823086038 and the best i am getting for dH=8

min(min((NMSE))) ans = 0.24707 ADDITIONALLY

mean(var(t',1))*0.005 ans = 0.943076999999999

I would like to acknowledge you on my masters thesis,please permit me for same.

[sameer d]

THIS IS THE BEST RESULT I HAVE OBTAINED.THANKS row =

     7

col =

     4

>> min(min((NMSE)))

ans =

     4.775377230289548e-15

>>