Deep Learning · TEE · section

R25 Q6a · 3–2–1 forward pass and update w7,w8

February 2025 re-exam · complete solved paper

Question

Question. For inputs[1,0,1], weights w1…w8=[0.2,−0.3,0.4,0.1,−0.5,0.2,−0.3,−0.2], biases[−0.4,0.2,0.1], target1 and sigmoid at all neurons: find prediction,error,and update w7,w8 with η=0.9. (10 marks)

Answer — step by step

Given

Diagram edge mapping: x1→h1=w1,x1→h2=w2;x2→h1=w3,x2→h2=w4;x3→h1=w5,x3→h2=w6;h1→out=w7,h2→out=w8.

$\begin{bmatrix}0.2&0.4&-0.5\\-0.3&0.1&0.2\end{bmatrix}$ LaTeX: \begin{bmatrix}0.2&0.4&-0.5\\-0.3&0.1&0.2\end{bmatrix}

[1,0,1] → Two sigmoid hidden units → One sigmoid output → Target1

To find

Prediction, explicit error convention, two simultaneous output-weight updates.

Formula: which one and why

$E=\frac12(o-t)^2,\quad\delta_o=(o-t)o(1-o),\quad\frac{\partial E}{\partial w_k}=\delta_o h_j,\quad w_k^{new}=w_k-\eta\frac{\partial E}{\partial w_k}$ LaTeX:

E=\frac12(o-t)^2,\quad\delta_o=(o-t)o(1-o),\quad\frac{\partial E}{\partial w_k}=\delta_o h_j,\quad w_k^{new}=w_k-\eta\frac{\partial E}{\partial w_k}

The paper says error without naming its scaling. Use half squared error and also report residual/SSE, so the gradient convention is clear.

Step 1. Calculate each hidden weighted sum then sigmoid

Unit	Products +bias	Net	Activation
1	1×(0.2) + 0×(0.4) + 1×(-0.5) +(-0.4)	-0.7	0.331812227832
2	1×(-0.3) + 0×(0.1) + 1×(0.2) +(0.2)	0.10000000000000003	0.524979187479

Step 2. Output weighted sum and sigmoid

z=0.331812227832×(−0.3)+0.524979187479×(−0.2)+0.1=-0.104539505845; o=1/(1+e^(−z))=0.473888898824.

Step 3. State error and loss

Residual o−t=-0.526111101176; SSE=0.276792890781; halfSSE=0.138396445390.

Step 4. Multiply the output chain-rule factors

δ=(0.473888898824−1)×0.473888898824×(1−0.473888898824)=-0.131169078214.

Step 5. Gradient for each edge

Weight	δ × its source activation	Gradient
w7	-0.131169078214×0.331812227832	-0.043523504065
w8	-0.131169078214×0.524979187479	-0.068861036103

Step 6. Update from the old weights

Weight	Old − learning rate × gradient	New
w7	-0.3−0.9×(-0.043523504065)	-0.260828846342
w8	-0.2−0.9×(-0.068861036103)	-0.138025067507

Final answer

w7′=-0.260828846342; w8′=-0.138025067507. Only the requested two weights are updated; use the same old forward pass for both gradients. With unhalved SSE the gradients double, so numerical updates would differ.

Source / convention note. Original does not specify loss scaling; half-SSE is declared and alternatives distinguished.