Abstract Response-Time Analysis

In this module, we propose the general framework for response-time analysis (RTA) of uni-processor scheduling of real-time tasks with arbitrary arrival models. We prove that the maximum (with respect to the set of offsets) among the solutions of the response-time bound recurrence is a response time bound for tsk. Note that in this section we do not rely on any hypotheses about job sequentiality.

Consider any type of tasks ...

... and any type of jobs associated with these tasks.

Consider any kind of uni-service ideal processor state model.

Consider any valid arrival sequence with consistent, non-duplicate arrivals...

... and any schedule of this arrival sequence...

... where jobs do not execute before their arrival nor after completion.

Assume that the job costs are no larger than the task costs.

Consider a task set ts...

... and a task tsk of ts that is to be analyzed.

Consider a valid preemption model...

...and a valid task run-to-completion threshold function. That is, task_rtct tsk is (1) no bigger than tsk's cost, (2) for any job of task tsk job_rtct is bounded by task_rtct.

Let's define some local names for clarity.

Assume we are provided with abstract functions for interference and interfering workload.

We assume that the scheduler is work-conserving.

For simplicity, let's define some local names.

Let L be a constant which bounds any busy interval of task tsk.

Next, assume that interference_bound_function is a bound on the interference incurred by jobs of task tsk.

For simplicity, let's define a local name for the search space.

Consider any value R that upper-bounds the solution of each response-time recurrence, i.e., for any relative arrival time A in the search space, there exists a corresponding solution F such that R ≥ F + (task_cost tsk - task_rtct tsk).

In this section we show a detailed proof of the main theorem that establishes that R is a response-time bound of task tsk.

Consider any job j of tsk with positive cost.

Assume we have a busy interval [t1, t2) of job j that is bounded by L.

Let's define A as a relative arrival time of job j (with respect to time t1).

In order to prove that R is a response-time bound of job j, we use hypothesis H_R_is_maximum. Note that the relative arrival time (A) is not necessarily from the search space. However, earlier we have proven that for any A there exists another A_sp from the search space that shares the same IBF value. Moreover, we've also shown that there exists an F_sp such that F_sp is a solution of the response time recurrence for parameter A_sp. Thus, despite the fact that the relative arrival time may not lie in the search space, we can still use the assumption H_R_is_maximum. More formally, consider any A_sp and F_sp such that:..

(a) A_sp is less than or equal to A...

(b) interference_bound_function(A, x) is equal to interference_bound_function(A_sp, x) for all x less than L...

(c) A_sp is in the search space, ...

(d) A_sp + F_sp is a solution of the response time recurrence...

(e) and finally, F_sp + (task_last - ε) is no greater than R.

In this section, we consider the case where the solution is so large that the value of t1 + A_sp + F_sp goes beyond the busy interval. Although this case may be impossible in some scenarios, it can be easily proven, since any job that completes by the end of the busy interval remains completed.

By assumption, suppose that t2 is less than or equal to t1 + A_sp + F_sp.

Then we prove that job_arrival j + R is no less than t2.

But since we know that the job is completed by the end of its busy interval, we can show that job j is completed by job arrival j + R.

In this section, we consider the complementary case where t1 + A_sp + F_sp lies inside the busy interval.

So, assume that t1 + A_sp + F_sp is less than t2.

Next, let's consider two other cases: CASE 1: the value of the fix-point is no less than the relative arrival time of job j.

Suppose that A_sp + F_sp is no less than relative arrival of job j.

In this section, we prove that the fact that job j is not completed by time job_arrival j + R leads to a contradiction. Which in turn implies that the opposite is true -- job j completes by time job_arrival j + R.

Recall that by lemma "solution_for_A_exists" there is a solution F of the response-time recurrence for the given relative arrival time A (which is not necessarily from the search space). Thus, consider a constant F such that:..

(a) the sum of A_sp and F_sp is equal to the sum of A and F...

(b) F is at mo1st F_sp...

(c) and A + F is a solution for the response-time recurrence for A.

Next, we assume that job j is not completed by time job_arrival j + R.

Some additional reasoning is required since the term task_cost tsk - task_rtct tsk does not necessarily bound the term job_cost j - job_rtct j. That is, a job can have a small run-to-completion threshold, thereby becoming non-preemptive much earlier than guaranteed according to task run-to-completion threshold, while simultaneously executing the last non-preemptive segment that is longer than task_cost tsk - task_rtct tsk (e.g., this is possible in the case of floating non-preemptive sections). In this case we cannot directly apply lemma "j_receives_at_least_run_to_completion_threshold". Therefore we introduce two temporal notions of the last non-preemptive region of job j and an execution optimism. We use these notions inside this proof, so we define them only locally. Let the last non-preemptive region of job j (last) be the difference between the cost of the job and the j's run-to-completion threshold (i.e. job_cost j - job_rtct j). We know that after j has reached its run-to-completion threshold, it will additionally be executed job_last j units of time.

And let execution optimism (optimism) be the difference between the tsk's run-to-completion threshold and the j's run-to-completion threshold (i.e. task_rtct - job_rtct). Intuitively, optimism is how much earlier job j has received its run-to-completion threshold than it could at worst.

From lemma "j_receives_at_least_run_to_completion_threshold" with parameters progress_of_job := job_rtct j and delta := (A + F) - optimism) we know that service of j by time t1 + (A + F) - optimism is no less than job_rtct j. Hence, job j is completed by time t1 + (A + F) - optimism + last.

However, t1 + (A + F) - optimism + last ≤ job_arrival j + R! To prove this fact we need a few auxiliary inequalities that are needed because we use the truncated subtraction in our development. So, for example a + (b - c) = a + b - c only if b ≥ c.

Recall that we consider a busy interval of a job j, and j has arrived A time units after the beginning the busy interval. From basic properties of a busy interval it follows that job j incurs interference at any time instant t ∈ [t1, t1 + A). Therefore interference_bound_function(tsk, A, A + F) is at least A.

As two trivial corollaries, we show that tsk's run-to-completion threshold is at most F_sp...

... and optimism is at most F.

Next we show that t1 + (A + F) - optimism + last is at most job_arrival j + R, which is easy to see from the following sequence of inequalities: t1 + (A + F) - optimism + last ≤ job_arrival j + (F - optimism) + job_last ≤ job_arrival j + (F_sp - optimism) + job_last ≤ job_arrival j + F_sp + (job_last - optimism) ≤ job_arrival j + F_sp + job_cost j - task_rtct tsk ≤ job_arrival j + F_sp + task_cost tsk - task_rtct tsk ≤ job_arrival j + R.

... which contradicts the initial assumption about j is not completed by time job_arrival j + R.

Putting everything together, we conclude that j is completed by job_arrival j + R.

CASE 2: the value of the fix-point is less than the relative arrival time of job j (which turns out to be impossible, i.e. the solution of the response-time recurrence is always equal to or greater than the relative arrival time).

Assume that A_sp + F_sp is less than A.

Note that the relative arrival time of job j is less than L.

We can use j_receives_at_least_run_to_completion_threshold to prove that the service received by j by time t1 + (A_sp + F_sp) is no less than run-to-completion threshold.

However, this is a contradiction. Since job j has not yet arrived, its service is equal to 0. However, run-to-completion threshold is always positive.

Using the lemmas above, we prove that R is a response-time bound.