Five periodic tasks together demand seventy per cent of the processor. Applying the fixed priority bound test to them shows the set to be
Answer: A
Substituting five into the formula gives 0.743, and being below it is a sufficient condition, so no further analysis is needed. The limiting figure of 0.693 applies only as the number of tasks grows without bound and is always pessimistic for a finite set.
Please log in or register to report this question.
The blocking a high priority task suffers under the priority ceiling protocol is bounded by
Answer: B
The protocol permits at most one such blocking per release, which is what makes the bound so attractive: the analysis adds a single term rather than a sum. Plain inheritance can in the worst case block a task once for each resource it needs.
Please log in or register to report this question.
Adding a blocking term \( B \) to a task's analysis is done by
Answer: C
Treating the blocking as though it were part of the task's own demand keeps the recurrence in its familiar form and is conservative. A task that becomes unschedulable only when the blocking is included points at the shared resource as the thing to redesign.
Please log in or register to report this question.
A task set is schedulable in analysis but misses deadlines on the target. The most likely omission is
Answer: D
Handlers run above every task and steal time that no task's execution figure includes, so a busy interrupt source can invalidate an otherwise sound analysis. Charging the handler load as a highest-priority pseudo-task is the standard correction.
Please log in or register to report this question.
Interrupt handlers are included in a fixed priority analysis by treating them as
Answer: A
Hardware priority puts every handler above every task whatever the kernel's own priorities say, so they enter the recurrence as the topmost interference. Their periods are the minimum interarrival times of the events that trigger them.
Please log in or register to report this question.
A designer moves work out of an interrupt handler into a task in order to
Answer: B
Inside a handler the work outranks every task regardless of its importance; as a task it takes its proper place in the priority order and can be preempted by more urgent work. The total time consumed is much the same, and it still has to be analysed.
Please log in or register to report this question.
Time slicing between tasks of equal priority is provided by a kernel to
Answer: C
Without it the first of several equal tasks to start would run until it blocked, which is sometimes wanted and often not. It guarantees nothing about deadlines: the wait grows with the number of equal tasks and the slice length.
Please log in or register to report this question.
Assigning every task the same priority in a preemptive kernel with time slicing gives behaviour like
Answer: D
Equal priorities and a slice is precisely round robin, and every task then waits for all the others before running again. It is easy to configure and throws away the whole benefit of the priority scheme, so it suits only sets with no timing requirements.
Please log in or register to report this question.
A task whose deadline equals its period and whose response time is 60 per cent of its period has a timing margin of
Answer: A
The response time is what matters and it leaves four tenths of the period unused, which is the cushion against growth in the execution times above it. Expressing margins as a fraction of the period rather than of the execution time keeps them comparable between tasks.
Please log in or register to report this question.
Comparing a 1 ms tick with a 10 ms tick for a set whose shortest period is 5 ms, the 1 ms tick
Answer: B
A five millisecond period cannot be expressed at all in ten millisecond ticks, and one millisecond ticks give five per period with the release jitter that implies. The overhead of ten times as many tick interrupts is what has to be weighed against it.
Please log in or register to report this question.
