Task Scheduling Questions and Answers

Answer: C

When each period divides the next exactly, the releases line up and no capacity is lost to awkward phasing, so the full processor is available. Choosing periods as 1, 2, 4, 8 and 16 milliseconds rather than 1, 3, 7 and 13 is therefore worth real capacity.

Answer: D

Correct answer: the nearest absolute deadline

Earliest deadline first is dynamic: the priority is recomputed at each release from the absolute deadline, so a task’s priority rises as its deadline approaches. That is what makes it optimal on one processor, meeting every deadline whenever the utilisation is at most unity.

Answer: A

Any task set with deadlines equal to periods and a utilisation at or below unity is schedulable, which is the whole attraction of the algorithm. The price is more run-time bookkeeping and much worse behaviour when the system is momentarily overloaded.

Answer: B

Under overload a fixed priority system sheds the lowest priority tasks predictably, whereas earliest deadline first can miss deadlines across the whole set as everything becomes urgent at once. Its utilisation bound is the higher of the two, at unity.

Answer: C

Correct answer: predictably, the lowest priority tasks missing first

A fixed priority scheme degrades in a known order, since the high priority tasks continue to preempt and the shortfall falls on those at the bottom. That predictability is a real advantage: the critical work can be placed high and will survive an overload that the rest does not.

Answer: D

Taking the remaining work into account makes it optimal in the same sense as earliest deadline first but causes frequent switching when two tasks have nearly equal laxity, since the ordering flips repeatedly. That thrashing is why it is little used in practice.

Answer: A

The worst interference a task can suffer arises when everything above it arrives at the same moment, so the response time computed for that phasing is the worst case for all phasings. This theorem is what makes the analysis tractable at all.

Answer: B

The recurrence adds the task's own execution to the interference from higher priority tasks, which itself depends on the response time being computed, so it is solved by iteration from an initial guess. It converges when the utilisation is below one.

Answer: C

Correct answer: 0.71

\[ \frac{1}{4} + \frac{2}{6} + \frac{1}{8} = 0.25 + 0.333 + 0.125 = 0.708. \] The set demands about 71 per cent of the processor, which must now be compared against the bound for three tasks rather than against unity.

Answer: D

Correct answer: 0.780

\[ n\left(2^{1/n}-1\right) = 3\left(2^{1/3}-1\right) = 3 \times 0.2599 = 0.780. \] The bound falls with the number of tasks, towards \( \ln 2 = 0.693 \) as n grows without limit. Since 0.708 is below 0.780, this set is guaranteed schedulable under rate monotonic.