Level 4 — ApplicationMachine Learning (Aerospace Applications)

Machine Learning (Aerospace Applications)

⏱ 60 minutes60 marksprintable — key stays hidden on paper

Level 4 — Application (novel problems, no hints) Time limit: 60 minutes Total marks: 60

Answer all questions. Show all working. Calculators permitted. Where numeric answers are requested, keep at least 4 significant figures unless told otherwise.

Question 1 — System Identification via Ridge Regression (12 marks)

A quadrotor's pitch-rate dynamics are approximated by the linear model $y = w_0 + w_1 x$ , where $x$ is elevator deflection (deg) and $y$ is measured pitch acceleration (deg/s²). You collect three flight-test samples:

$(x,y) = (1,\,2),\ (2,\,2),\ (3,\,4)$

(a) Write the ridge-regularized cost $J(\mathbf{w})$ that penalizes only $w_1$ (not the bias $w_0$ ) with strength $\lambda$ . (2 marks)

(b) Derive the closed-form normal equation for this ridge problem and state the modified matrix that must be inverted. (3 marks)

(d) Explain physically why regularizing $w_1$ but not $w_0$ is the standard convention. (2 marks)

Question 2 — Backpropagation Through a Tiny Network (14 marks)

Consider a single-input network for a fault-detection classifier:

$z_1 = w_1 x + b_1,\quad a_1 = \tanh(z_1),\quad z_2 = w_2 a_1 + b_2,\quad \hat{y} = \sigma(z_2)$

Loss is binary cross-entropy $L = -[y\ln\hat{y} + (1-y)\ln(1-\hat{y})]$ .

Given: $x = 1,\ w_1 = 0.5,\ b_1 = 0,\ w_2 = 1,\ b_2 = 0,\ y = 1$ .

(a) Perform the forward pass: compute $z_1, a_1, z_2, \hat{y}$ and the loss $L$ . (4 marks)

(b) Show that $\dfrac{\partial L}{\partial z_2} = \hat{y} - y$ and evaluate it. (3 marks)

(c) Using the chain rule, derive and evaluate $\dfrac{\partial L}{\partial w_1}$ . (Recall $\tanh'(z) = 1 - \tanh^2 z$ .) (5 marks)

(d) One gradient-descent step with learning rate $\eta = 0.1$ updates $w_1$ . Give the new $w_1$ . (2 marks)

Question 3 — Scaled Dot-Product Self-Attention (12 marks)

A transformer processes two token embeddings. After projection you obtain:

$Q = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\quad K = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\quad V = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}$

with key dimension $d_k = 2$ .

(a) Write the self-attention formula and compute the scaled score matrix $\dfrac{QK^\top}{\sqrt{d_k}}$ . (3 marks)

(b) Apply row-wise softmax to obtain the attention weight matrix. Give entries to 4 decimals. (5 marks)

(d) State one reason the $\sqrt{d_k}$ scaling is used. (1 mark)

Question 4 — Q-Learning for a Guidance MDP (12 marks)

An interceptor operates in a 2-state MDP (states $S_0, S_1$ ; $S_1$ is terminal). From $S_0$ , action fire gives reward $r=+10$ and transitions to $S_1$ ; action wait gives reward $r=-1$ and stays in $S_0$ . Discount $\gamma = 0.9$ . Learning rate $\alpha = 0.5$ . All Q-values initialised to 0.

(a) Write the Q-learning update rule. (2 marks)

(b) Starting from $Q=0$ , apply the update for the transition $(S_0,\text{wait},\,r=-1,\,S_0)$ , then for $(S_0,\text{fire},\,r=+10,\,S_1)$ where $S_1$ is terminal ( $\max_a Q(S_1,\cdot)=0$ ). Give the updated Q-values. (5 marks)

(c) Determine the exact converged optimal values $Q^*(S_0,\text{fire})$ and $Q^*(S_0,\text{wait})$ using the Bellman optimality equation, and state the optimal policy. (5 marks)

Question 5 — Bias–Variance and Cross-Validation (10 marks)

You fit polynomial models of degree $d = 1, 3, 9$ to noisy aerodynamic drag data using 5-fold cross-validation.

(a) Sketch (describe) the expected trend of training error and validation error as $d$ increases, and identify which regime each degree likely occupies (underfit / good / overfit). (4 marks)

(b) Define bias and variance in one sentence each, and state which dominates the error for $d=1$ and for $d=9$ . (3 marks)

(c) In 5-fold CV, a dataset of 200 samples is split. State how many samples are used for training and validation in each fold, and give one reason to prefer k-fold CV over a single train/test split. (3 marks)

Answer keyMark scheme & solutions

Question 1 (12 marks)

(a) Cost with bias unpenalized: (2) $J(\mathbf{w}) = \sum_{i=1}^{3}(w_0 + w_1 x_i - y_i)^2 + \lambda w_1^2$ (1 mark squared-error term, 1 mark for $\lambda w_1^2$ only on $w_1$ .)

(b) With design matrix $X = \begin{bmatrix}1&1\\1&2\\1&3\end{bmatrix}$ , the normal equation is $(X^\top X + \lambda R)\mathbf{w} = X^\top y$ , where the penalty matrix $R = \begin{bmatrix}0&0\\0&1\end{bmatrix}$ (zero in bias slot). (3) Solution: $\mathbf{w} = (X^\top X + \lambda R)^{-1}X^\top y$ .

(c) Compute: (5) $X^\top X = \begin{bmatrix}3 & 6\\ 6 & 14\end{bmatrix}$ , $X^\top y = \begin{bmatrix} 2+2+4\\ 2+4+12\end{bmatrix}=\begin{bmatrix}8\\18\end{bmatrix}$ .

With $\lambda=1$ : $A = X^\top X + R = \begin{bmatrix}3 & 6\\ 6 & 15\end{bmatrix}$ , $\det A = 45-36 = 9$ .

$A^{-1} = \frac{1}{9}\begin{bmatrix}15 & -6\\ -6 & 3\end{bmatrix}$ .

$\mathbf{w} = A^{-1}\begin{bmatrix}8\\18\end{bmatrix} = \frac{1}{9}\begin{bmatrix}15\cdot8 - 6\cdot18\\ -6\cdot8 + 3\cdot18\end{bmatrix} = \frac{1}{9}\begin{bmatrix}12\\ 6\end{bmatrix}$ .

$\boxed{w_0 = 12/9 = 1.3333,\quad w_1 = 6/9 = 0.6667}$

(2 for $X^\top X, X^\top y$ ; 1 for adding $R$ ; 2 for inversion & solve.)

(d) The bias $w_0$ merely sets the output offset (mean level); shrinking it toward zero would introduce systematic error and makes the fit depend on the arbitrary origin of $y$ . Penalizing only slopes keeps the model shift-invariant and controls model complexity without biasing the intercept. (2)

Question 2 (14 marks)

(a) Forward pass: (4) $z_1 = 0.5\cdot1 + 0 = 0.5$ $a_1 = \tanh(0.5) = 0.46212$ $z_2 = 1\cdot0.46212 + 0 = 0.46212$ $\hat{y} = \sigma(0.46212) = 1/(1+e^{-0.46212}) = 0.61350$ $L = -\ln(0.61350) = 0.48865$ (1 each for $a_1,\hat y, L$ + $z$ values.)

(b) $\dfrac{\partial L}{\partial \hat y} = \dfrac{-y}{\hat y}+\dfrac{1-y}{1-\hat y}=\dfrac{\hat y - y}{\hat y(1-\hat y)}$ ; and $\dfrac{\partial \hat y}{\partial z_2}=\sigma'(z_2)=\hat y(1-\hat y)$ . Product gives $\dfrac{\partial L}{\partial z_2}=\hat y - y$ . (3) Value: $0.61350 - 1 = -0.38650$ .

(c) Chain rule: (5) $\frac{\partial L}{\partial w_1} = \frac{\partial L}{\partial z_2}\cdot w_2 \cdot (1 - a_1^2)\cdot x$ $1 - a_1^2 = 1 - 0.46212^2 = 1 - 0.21356 = 0.78644$ . $\dfrac{\partial L}{\partial w_1} = (-0.38650)(1)(0.78644)(1) = -0.30396$ .

(d) Update: $w_1^{new} = 0.5 - 0.1(-0.30396) = 0.5 + 0.030396 = 0.53040$ . (2)

Question 3 (12 marks)

(a) $\text{Attn}(Q,K,V)=\text{softmax}\!\left(\dfrac{QK^\top}{\sqrt{d_k}}\right)V$ . (3) $QK^\top = I$ , so scaled scores $= \dfrac{1}{\sqrt2}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}0.7071&0\\0&0.7071\end{bmatrix}$ .

(b) Row softmax of $[0.7071, 0]$ : (5) $e^{0.7071}=2.0281,\ e^0=1$ , sum $=3.0281$ . Weight on diagonal $=2.0281/3.0281 = 0.6698$ ; off-diagonal $=1/3.0281 = 0.3302$ . $W = \begin{bmatrix}0.6698 & 0.3302\\ 0.3302 & 0.6698\end{bmatrix}$

(c) Output $= WV$ , $V=\begin{bmatrix}2&0\\0&3\end{bmatrix}$ : (3) Row 1: $[0.6698\cdot2,\ 0.3302\cdot3] = [1.3396,\ 0.9906]$ . Row 2: $[0.3302\cdot2,\ 0.6698\cdot3] = [0.6604,\ 2.0094]$ . $O = \begin{bmatrix}1.3396 & 0.9906\\ 0.6604 & 2.0094\end{bmatrix}$

(d) Scaling by $\sqrt{d_k}$ keeps the dot-product magnitudes from growing with dimension, preventing softmax saturation (tiny gradients). (1)

Question 4 (12 marks)

(a) (2) $Q(s,a)\leftarrow Q(s,a)+\alpha\Big[r+\gamma\max_{a'}Q(s',a')-Q(s,a)\Big]$

(b) (5) Update 1 — $(S_0,\text{wait},-1,S_0)$ , $\max_a Q(S_0,\cdot)=0$ : $Q(S_0,\text{wait}) = 0 + 0.5[-1 + 0.9\cdot0 - 0] = -0.5$ .

Update 2 — $(S_0,\text{fire},+10,S_1)$ , terminal $\Rightarrow \max=0$ : $Q(S_0,\text{fire}) = 0 + 0.5[10 + 0.9\cdot0 - 0] = 5.0$ .

Result: $Q(S_0,\text{wait})=-0.5$ , $Q(S_0,\text{fire})=5.0$ . (2.5 each)

(c) Bellman optimality. (5) Fire is terminal-transition: $Q^*(S_0,\text{fire}) = 10 + \gamma\cdot0 = 10$ . Wait: $Q^*(S_0,\text{wait}) = -1 + \gamma \max_a Q^*(S_0,a)$ . Since fire gives 10, $\max = 10$ : $Q^*(S_0,\text{wait}) = -1 + 0.9\cdot10 = 8$ . Optimal policy: choose fire in $S_0$ (10 > 8). (2 + 2 + 1)

Question 5 (10 marks)

(a) Training error decreases monotonically as $d$ increases (more flexibility fits data). Validation error is U-shaped: high at $d=1$ (underfit), minimum around $d=3$ (good fit), rising again at $d=9$ (overfit, high variance). (4)

$d=1$ : underfit; $d=3$ : good; $d=9$ : overfit.

(b) (3) Bias = error from wrong/too-simple model assumptions (systematic). Variance = sensitivity of the fit to particular training samples (fluctuation). $d=1$ : bias dominates; $d=9$ : variance dominates.

(c) (3) 5-fold on 200 samples: each fold uses $200/5 = 40$ for validation and $160$ for training. Prefer k-fold because every sample serves in validation exactly once, giving a lower-variance, more reliable performance estimate than one arbitrary split (and uses data efficiently).

[
  {"claim":"Ridge Q1 solution w0=4/3, w1=2/3","code":"X=Matrix([[1,1],[1,2],[1,3]]); y=Matrix([2,2,4]); R=Matrix([[0,0],[0,1]]); A=X.T*X+R; w=A.inv()*(X.T*y); result = (w[0]==Rational(4,3)) and (w[1]==Rational(2,3))"},
  {"claim":"Q2 dL/dw1 approx -0.30396","code":"import math; a1=math.tanh(0.5); z2=a1; yh=1/(1+math.exp(-z2)); g=(yh-1)*1*(1-a1**2)*1; result = abs(g-(-0.30396))<1e-3"},
  {"claim":"Q3 softmax diagonal weight approx 0.6698","code":"import math; e=math.exp(1/math.sqrt(2)); w=e/(e+1); result = abs(w-0.6698)<1e-3"},
  {"claim":"Q4 Bellman Q*(wait)=8, Q*(fire)=10","code":"g=Rational(9,10); qfire=10; qwait=-1+g*max(qfire,0); result = (qfire==10) and (qwait==8)"}
]