Mock midterm paper

(a) ID = log₂(800/50 + 1) = log₂(17) ≈ 4.09 bits.

• Device A: T = 0.20 + 0.10 × 4.09 ≈ 0.61 s; IP = 1/b = 10 bits/s.
• Device B: T = 0.10 + 0.15 × 4.09 ≈ 0.71 s; IP = 1/b ≈ 6.7 bits/s.

(b) The claim is wrong for this task. The intercept a is only the start-up cost; the slope b dominates as difficulty rises, and throughput (1/b) is the standard single-number comparison. A wins at ID ≈ 4.1 bits by about 0.10 s per movement. The crossover is where the two lines meet: 0.20 + 0.10·ID = 0.10 + 0.15·ID → ID = 2 bits. Below 2 bits B is faster (very close, very large targets); above 2 bits A is faster, and a control room's targets are far above 2 bits. Buy A — but note that the constants are population estimates, so test with the actual operators.

Marking note. ID correct = 1; two times = 2; IP = 1; crossover argument = 2. Saying “B is better because a is smaller” caps (b) at 1 — the whole point is that b scales with difficulty.

(a) Topology.

1. idle —(touch screen / idle timeout not reached)—▶ zone select
2. zone select —(choose zone 1|2|3)—▶ ticket type
3. ticket type —(single)—▶ payment; —(return)—▶ payment
4. payment —(card ok or octopus ok)—▶ printing
5. printing —(ticket ejected)—▶ idle
6. Failure arcs: card declined → payment (retry, max 2); no zone touch for 30 s → idle; printer jam → out of service.
7. Escape: one cancel state reachable from zone select and ticket type, returning to idle. Note the design decision: cancel during payment is only legal before the card is charged — after that, the flow must reach printing or a refund state.

(b) The escape is a single shared cancel state rather than a duplicated cancel arc in every state — that is what stops the diagram becoming spaghetti. The diagram guarantees that payment cannot be taken without an ensuing printing state, and that every non-terminal state is reachable and escapable. It does not guarantee that the user can read the screen, that the zones are correctly priced, or that the printer works.

Marking note. 7 for topology: happy path 3, failure arcs 2, labelled arcs 2 (unlabelled arcs lose a mark each, minimum 0). 3 for (b): escape stated, guarantee stated, limit stated. An escape arc out of printing is a semantic error — say so if you see yourself drawing it.

(a) Any three, applied to this sentence:

• Tokenisation — split into tokens; “five minutes” and “a timer” are multi-word units that must stay interpretable as one concept; the digit/word choice (“5” vs “five”) must map to the same duration.
• POS tagging — “timer” is a noun here, “stir” is a verb, “remind” takes a clausal complement; “set” is famously ambiguous (noun or verb).
• Parsing — attachment: does “in five minutes” modify stir (the reminder time) or set a timer (the timer duration)? Two trees, two different behaviours.
• Semantic analysis — “for the pasta” is a label, not a duration; the timer's duration is unstated and must be inferred or asked for.
• Coreference resolution — “it” in “stir it” refers to the pasta, not to the timer and not to the reminder.

(b) Coreference resolution failed: “it” was bound to the most recent salient noun — timer — rather than to pasta. The sentence provides no gender or number cue (both are singular, neuter), so a heuristic that prefers recency picks the wrong antecedent; the surface pattern “stir <noun>” then produces “stir the timer”. A semantic check would have caught it, because a timer is not the kind of thing one stirs.

(c) Because the failure is not a gap in the language model's vocabulary but a missing model of the situation: pasta is stirred, timers are not. Scale improves association, not reference — the same reason a bigger model still cannot tell you what “it” means without knowing what is in the kitchen.

Marking note. (a) 2 per stage: named + applied. A generic definition with no application scores 1. (b) failure + stage + mechanism. (c) must say “world knowledge / situation model”, not “more layers”.