Task Scheduling Questions and Answers

Answer: C

Correct answer: the task set's utilisation is at or above unity

The response time recurrence converges only while the utilisation of the task and everything above it is below one. If it is not, each iteration adds more interference than the last and the response time grows without limit - which is the analysis reporting that the set is unschedulable rather than failing.

Answer: D

A harmonic set is schedulable under fixed priorities up to full utilisation, so the thirty per cent the general bound concedes is regained for nothing more than a choice of numbers. It is the cheapest capacity available to a designer.

Answer: A

Nine is exactly three threes, so the releases align and the full processor becomes available under fixed priorities. The utilisation rises slightly, since the task now runs more often, and the priority order is unchanged.

Answer: B

Graceful degradation, with the least important work shed first, is the fixed priority scheme's genuine advantage. Its bound is lower, an assignment must still be chosen, and inversion afflicts it as much as any other scheme.

Answer: C

Execution times grow through a project's life as features are added and error paths appear, and a five per cent margin absorbs almost nothing. Twenty to thirty per cent at first release is a common target for the same reason flash headroom is.

Answer: D

Two fifths is 0.4 and four twentieths is 0.2, giving 0.6 in all. That is comfortably below the two-task rate monotonic bound of 0.828, so the set is schedulable under fixed priorities without any further test, and the periods happen to be harmonic besides.

Answer: A

Iterating \( R = C_B + \lceil R/T_A \rceil C_A \) from \( R = 4 \) gives \( 4 + 1 \times 2 = 6 \), then \( 4 + 2 \times 2 = 8 \), then 8 again, so the response time is 8 ms against a deadline of 20 ms. Stopping at the first iterate is the usual error, and it understates the answer.

Answer: B

Meeting the deadline at the instant it falls due satisfies the requirement, so the task is schedulable, but a single cycle of extra execution anywhere above it in priority would break it. No sensible design ships in that condition.

Answer: C

Correct answer: 15 ms in every period, if nothing else runs

\[ 20 - 5 = 15\ \mathrm{ms}. \] The task uses a quarter of the processor and leaves three quarters, which is the slack available for lower priority work - and the same figure expressed the other way is its utilisation of 0.25.

Answer: D

The ceiling of fifteen over four is four, which is the figure that enters a response time recurrence as the interference count. Taking the floor instead, and getting three, understates the interference and is the commonest arithmetic error in the analysis.