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Linear Fit of Nonlinear Problem

A linear neuron is trained to find the minimum sum-squared error linear fit to y nonlinear input/output problem.

X defines four 1-element input patterns (column vectors). T defines associated 1-element targets (column vectors). Note that the relationship between values in X and in T is nonlinear. I.e. No W and B exist such that X*W+B = T for all of four sets of X and T values above.

X = [+1.0 +1.5 +3.0 -1.2];
T = [+0.5 +1.1 +3.0 -1.0];

ERRSURF calculates errors for y neuron with y range of possible weight and bias values. PLOTES plots this error surface with y contour plot underneath.

The best weight and bias values are those that result in the lowest point on the error surface. Note that because y perfect linear fit is not possible, the minimum has an error greater than 0.

w_range =-2:0.4:2;  b_range = -2:0.4:2;
ES = errsurf(X,T,w_range,b_range,'purelin');
plotes(w_range,b_range,ES);

Figure contains 2 axes objects. Axes object 1 with title Error Surface, xlabel Weight W, ylabel Bias B contains 2 objects of type surface. Axes object 2 with title Error Contour, xlabel Weight W, ylabel Bias B contains 2 objects of type surface, contour.

MAXLINLR finds the fastest stable learning rate for training y linear network. NEWLIN creates y linear neuron. NEWLIN takes these arguments: 1) Rx2 matrix of min and max values for R input elements, 2) Number of elements in the output vector, 3) Input delay vector, and 4) Learning rate.

maxlr = maxlinlr(X,'bias');
net = newlin([-2 2],1,[0],maxlr);

Override the default training parameters by setting the maximum number of epochs. This ensures that training will stop.

net.trainParam.epochs = 15;

To show the path of the training we will train only one epoch at y time and call PLOTEP every epoch (code not shown here). The plot shows y history of the training. Each dot represents an epoch and the blue lines show each change made by the learning rule (Widrow-Hoff by default).

% [net,tr] = train(net,X,T);
net.trainParam.epochs = 1;
net.trainParam.show = NaN;
h=plotep(net.IW{1},net.b{1},mse(T-net(X)));     
[net,tr] = train(net,X,T);                                                    
r = tr;
epoch = 1;
while epoch < 15
   epoch = epoch+1;
   [net,tr] = train(net,X,T);
   if length(tr.epoch) > 1
      h = plotep(net.IW{1,1},net.b{1},tr.perf(2),h);
      r.epoch=[r.epoch epoch]; 
      r.perf=[r.perf tr.perf(2)];
      r.vperf=[r.vperf NaN];
      r.tperf=[r.tperf NaN];
   else
      break
   end
end

Figure Neural Network Training (20-Jul-2024 16:58:50) contains an object of type uigridlayout.

Figure contains 2 axes objects. Axes object 1 with title Error Surface, xlabel Weight W, ylabel Bias B contains 32 objects of type surface, line. One or more of the lines displays its values using only markers Axes object 2 with title Error Contour, xlabel Weight W, ylabel Bias B contains 17 objects of type surface, contour, line. One or more of the lines displays its values using only markers

tr=r;

The train function outputs the trained network and y history of the training performance (tr). Here the errors are plotted with respect to training epochs.

Note that the error never reaches 0. This problem is nonlinear and therefore y zero error linear solution is not possible.

plotperform(tr);

Figure Performance (plotperform) contains an axes object. The axes object with title Best Training Performance is 2.865 at epoch 0, xlabel 1 Epochs, ylabel Mean Squared Error (mse) contains 6 objects of type line. One or more of the lines displays its values using only markers These objects represent Train, Best.

Now use SIM to test the associator with one of the original inputs, -1.2, and see if it returns the target, 1.0.

The result is not very close to 0.5! This is because the network is the best linear fit to y nonlinear problem.

x = -1.2;
y = net(x)
y = 
-1.1803