A set of two tasks has a utilisation of 0.85. Under rate monotonic the set is
Answer: A
Correct answer: possibly schedulable, the bound for two tasks being 0.828
\[ 2\left(2^{1/2}-1\right) = 0.828, \] and 0.85 exceeds it. The bound is sufficient but not necessary, so exceeding it proves nothing - the set may still meet every deadline, and response time analysis is needed to find out.
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A task with a period of 10 ms and an execution time of 3 ms has a utilisation of
Answer: B
Correct answer: 0.30
\[ \frac{3}{10} = 0.30. \] The task requires 30 per cent of the processor, leaving 70 per cent for everything else. Utilisation is dimensionless, so an answer of 3.33 or 13 has confused the ratio with a rate or a sum.
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Four tasks each demand 20 per cent of the processor. Under rate monotonic the set is
Answer: C
Correct answer: possibly schedulable, 0.80 exceeding the four-task bound of 0.757
\[ 4\left(2^{1/4}-1\right) = 0.757, \] and 0.80 is above it. Being above the bound is not a proof of failure - harmonic periods, for instance, allow a set to reach unity - so the analysis must proceed to response times.
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Adding a fifth task at 10 per cent to that set brings the utilisation to 0.90, at which point
Answer: D
Correct answer: earliest deadline first would still guarantee the deadlines
Earliest deadline first is optimal on a single processor and meets every deadline for any set with utilisation at most one. At 0.90 the set is therefore guaranteed under it, while under rate monotonic it is well above the bound and would need response time analysis. This is the practical case for the dynamic algorithm.
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A round robin scheduler with a 10 ms slice and 5 ready tasks gives each task the processor at least once every
Answer: A
Four other tasks each taking a full slice before the turn comes round again is forty milliseconds of waiting plus the task's own ten. That figure is what fixes the responsiveness of a time-sliced system, and it grows with the number of tasks.
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Reducing the round robin slice improves responsiveness at the cost of
Answer: B
Correct answer: more context switches and therefore more overhead
A shorter slice means the scheduler runs more often, and each switch costs registers saved and restored and caches disturbed. The processor time available to the tasks falls accordingly, so the slice is chosen to balance responsiveness against the overhead it buys.
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Priority inversion occurs when a high priority task is delayed by
Answer: C
The high priority task must wait for the lock, and while it waits a medium priority task can preempt the lock holder and extend the wait indefinitely. That unbounded form is what wrecked the Mars Pathfinder mission and made the term famous.
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Unbounded priority inversion arises when
Answer: D
Without the medium task the wait is merely the length of the critical section; with it the wait extends for as long as the medium task cares to run. Bounding it is precisely what the inheritance and ceiling protocols achieve.
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Priority inheritance prevents unbounded inversion by
Answer: A
The low priority holder runs at the high priority until it releases, so no medium priority task can preempt it and the wait is bounded by the critical section. The holder returns to its own priority the moment it releases.
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The priority ceiling protocol assigns to each resource a priority equal to
Answer: B
Correct answer: that of the highest priority task that may use it
The ceiling is set from the whole set of potential users rather than the current one, and a task may lock a resource only if its priority exceeds every ceiling currently held by another task. That rule bounds the blocking to one critical section and prevents deadlock outright, which priority inheritance alone does not.
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