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{\it The Final-exam of the Basic Course of Neural Networks
             (26 Jan 2001)}\\
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Name [ \hspace{2in} ] \hspace{.1in} Student-ID [ \hspace{1in} ]\\

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%%Using the back-propagation algorithm, train 
We have a feed-forward neural network 
made up of three neurons as input layer, 
two neurons as hidden layer, and one neuron as output layer.
The task is to calculate $y=(x_1+x_2+x_3)/5$ where 
$x_i$ is a input value to the $i$-th neuron in the input layer.
%%Using the back-propagation algorithm, train a feed-forward 
%%neural network in which we have three neurons as input layer, 
%%two neurons as hidden layer, and two neurons as output layer 
%%so that the neural network can recognize whether the number of 1's 
%%in a three digit of binary input is even or odd;
%%if it is even number then the first neuron of the output is 1 
%%and the second neuron 0 and vice versa otherwise.
Assume that each of the 8 weights are all initialized to 0.5 and
$\epsilon=0.1$. Also assume that transfer function $f(x) = 0.3x$
and threshold of all the neurons are zero,
which is just for the sake of simplicity. 
Of course neurons here are all analog ones, not binary ones.
We now train the neural network by the Back-propagation algorithm.
Answer the questions bellow.
Note that you should follow the notation we used in the course.
\begin{itemize}
\item[1.] Obtain all the modified weight values when the $(1.0, 1.0, 1.0)$
is given as an input. 
\begin{displaymath}
\begin{array}{lllll}
w^1_{11}=\hspace{1.2in}& w^1_{12}=\hspace{1.2in}& 
w^1_{21}=\hspace{1.2in}& w^1_{22}=\hspace{1.2in}& \\
w^1_{31}=\hspace{1.2in}& w^1_{32}=\hspace{1.2in}& 
w^2_{11}=\hspace{1.2in}& w^2_{12}=\hspace{1.2in}&
\end{array}
\end{displaymath}
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(calculations)
\vspace{6in}
\item[(2.)] If you have no confidence of the above result, obtain
the following quantities.
\begin{displaymath}
\begin{array}{lllll}
h^1_2(1)=\hspace{.6in}& y^1_1(1)=\hspace{.6in}& y^2_1(1)=\hspace{.6in}& 
\hat y^2_1(1)=\hspace{.6in}& e_1(1)= \\
e^1_1(1)=\hspace{.6in}&{\delta}^2_1(1)=\hspace{.6in}& 
{\delta}^1_1(1)=\hspace{.6in}&\triangle w^1_{12}=\hspace{.6in}& w^1_{12}=
\end{array}
\end{displaymath}
\end{itemize}


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