Laboratory Task 2

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Instruction: Perform a single forward pass and compute for the error.

\( x = \begin{bmatrix} 1 \\ 0 \\ 1 \\ \end{bmatrix} \)

\( y = \begin{bmatrix} 1 \\ \end{bmatrix} \)

\( f(z) = max(0, Z_n) \)

\( \text{hidden unit weights} = \begin{bmatrix} w_{11} & = & 0.2 & w_{12} & = & -0.3 \\ w_{13} & = & 0.4 & w_{14} & = & 0.1 \\ w_{15} & = & -0.5 & w_{16} & = & 0.2 \end{bmatrix} \)

\( \text{output unit weights} = \begin{bmatrix} w_{21} & = & -0.3 \\ w_{22} & = & -0.2 \end{bmatrix} \)

\( \theta = \begin{bmatrix} \theta_1 & = & -0.4 \\ \theta_2 & = & 0.2 \\ \theta_3 & = & 0.1 \end{bmatrix} \)

Solution

Hidden Layer

Compute each hidden unit pre-activation:

\( z_1 = (1)(0.2) + (0)(0.4) + (1)(-0.5) + (-0.4) = -0.7 \)

\( z_2 = (1)(-0.3) + (0)(0.1) + (1)(0.2) + (0.2) = 0.1 \)

Apply activation \((a_i = f(z_i))\):

\( a_1 = f(z_1) = f(-0.7) = 0, \quad a_2 = f(z_2) = f(0.1) = 0.1 \)

So the hidden activations are:

\( \mathbf{a} = \begin{bmatrix} 0 \\ 0.1 \end{bmatrix} \)

Output Layer

Weights and bias:

\( \mathbf{w}^{(o)}= \begin{bmatrix} -0.3 \\ -0.2 \end{bmatrix}, \quad \theta^{(o)} = 0.1 \)

Pre-activation:

\( z_o = (a_1)(-0.3) + (a_2)(-0.2) + \theta^{(o)} \)

\( z_o = (0)(-0.3) + (0.1)(-0.2) + 0.1 = 0.08 \)

Activation:

\( \hat{y} = f(z_o) = f(0.08) = 0.08 \)

Error Calculation

Target:

\( y = \begin{bmatrix} 1 \end{bmatrix} \)

Error:

\( E = \tfrac{1}{2}(y - \hat{y})^2 \)

\( E = \tfrac{1}{2}(1 - 0.08)^2 = \tfrac{1}{2}(0.92^2) = 0.4232 \)

Final Result

\( \hat{y} = 0.08, \quad E = 0.4232 \)