Library prosa.analysis.abstract.run_to_completion

Run-to-Completion Threshold of a job

In this module, we provide a sufficient condition under which a job receives enough service to become non-preemptive. Previously we defined the notion of run-to-completion threshold (see file abstract.run_to_completion_threshold.v). Run-to-completion threshold is the amount of service after which a job cannot be preempted until its completion. In this section we prove that if cumulative interference inside a busy interval is bounded by a certain constant then a job executes long enough to reach its run-to-completion threshold and become non-preemptive.
Consider any type of tasks ...
  Context {Task : TaskType}.
  Context `{TaskCost Task}.

... and any type of jobs associated with these tasks.
  Context {Job : JobType}.
  Context `{JobTask Job Task}.
  Context `{JobArrival Job}.
  Context `{JobCost Job}.

In addition, we assume existence of a function mapping jobs to their preemption points.
  Context `{JobPreemptable Job}.

Consider any arrival sequence with consistent arrivals...
Next, consider any ideal uni-processor schedule of this arrival sequence.
Assume that the job costs are no larger than the task costs.
Let tsk be any task that is to be analyzed.
  Variable tsk : Task.

Assume we are provided with abstract functions for interference and interfering workload.
For simplicity, let's define some local names.
We assume that the schedule is work-conserving.
Let j be any job of task tsk with positive cost.
  Variable j : Job.
  Hypothesis H_j_arrives : arrives_in arr_seq j.
  Hypothesis H_job_of_tsk : job_task j = tsk.
  Hypothesis H_job_cost_positive : job_cost_positive j.

Next, consider any busy interval t1, t2) of job [j].
  Variable t1 t2 : instant.
  Hypothesis H_busy_interval : busy_interval j t1 t2.

First, we prove that job j completes by the end of the busy interval. Note that the busy interval contains the execution of job j, in addition time instant t2 is a quiet time. Thus by the definition of a quiet time the job should be completed before time t2.
  Lemma job_completes_within_busy_interval:
    completed_by sched j t2.
    move: (H_busy_interval) ⇒ [[/andP [_ LT] [_ _]] [_ QT2]].
    unfold pending, has_arrived in QT2.
    move: QT2; rewrite /pending negb_and; move ⇒ /orP [QT2|QT2].
    { by move: QT2 ⇒ /negP QT2; exfalso; apply QT2, ltnW. }
      by rewrite Bool.negb_involutive in QT2.

In this section we show that the cumulative interference is a complement to the total time where job j is scheduled inside the busy interval.
Consider any sub-interval t, t + delta) inside the busy interval [t1, t2).
    Variables (t : instant) (delta : duration).
    Hypothesis H_greater_than_or_equal : t1 t.
    Hypothesis H_less_or_equal: t + delta t2.

We prove that sum of cumulative service and cumulative interference in the interval t, t + delta) is equal to delta.
    Lemma interference_is_complement_to_schedule:
      service_during sched j t (t + delta) + cumul_interference j t (t + delta) = delta.
      rewrite /service_during /cumul_interference/service_at.
      rewrite -big_split //=.
      rewrite -{2}(sum_of_ones t delta).
      rewrite big_nat [in RHS]big_nat.
      apply: eq_bigrx /andP[Lo Hi].
      move: (H_work_conserving j t1 t2 x) ⇒ Workj.
      feed_n 5 Workj; try done.
      { by apply/andP; split; eapply leq_trans; eauto 2. }
      rewrite -/(service_at sched j x) service_at_is_scheduled_at.
      have->: scheduled_at sched j x = ~~ interference j x.
      { by apply/idP/negP; intuition. }
      exact: addn_negb.

  End InterferenceIsComplement.

In this section, we prove a sufficient condition under which job j receives enough service.
Let progress_of_job be the desired service of job j.
    Variable progress_of_job : duration.
    Hypothesis H_progress_le_job_cost : progress_of_job job_cost j.

Assume that for some delta, the sum of desired progress and cumulative interference is bounded by delta (i.e., the supply).
    Variable delta : duration.
    Hypothesis H_total_workload_is_bounded:
      progress_of_job + cumul_interference j t1 (t1 + delta) delta.

Then, it must be the case that the job has received no less service than progress_of_job.
    Theorem j_receives_at_least_run_to_completion_threshold:
      service sched j (t1 + delta) progress_of_job.
      case NEQ: (t1 + delta t2); last first.
      { intros.
        have L8 := job_completes_within_busy_interval.
        apply leq_trans with (job_cost j); first by done.
        rewrite /service.
        rewrite -(service_during_cat _ _ _ t2).
        apply leq_trans with (service_during sched j 0 t2); [by done | by rewrite leq_addr].
          by apply/andP; split; last (apply negbT in NEQ; apply ltnW; rewrite ltnNge).
      { move: H_total_workload_is_boundedBOUND.
         apply subh3 in BOUND.
         apply leq_trans with (delta - cumul_interference j t1 (t1 + delta)); first by done.
         apply leq_trans with (service_during sched j t1 (t1 + delta)).
         { rewrite -{1}[delta](interference_is_complement_to_schedule t1) //.
           rewrite -addnBA // subnn addn0 //.
         { rewrite /service -[X in _ X](service_during_cat _ _ _ t1).
           rewrite leq_addl //.
             by apply/andP; split; last rewrite leq_addr.

  End InterferenceBoundedImpliesEnoughService.

In this section we prove a simple lemma about completion of a job after is reaches run-to-completion threshold.
Assume that completed jobs do not execute ...
.. and the preemption model is valid.
Then, job j must complete in job_cost j - job_run_to_completion_threshold j time units after it reaches run-to-completion threshold.
    Lemma job_completes_after_reaching_run_to_completion_threshold:
        job_run_to_completion_threshold j service sched j t
        completed_by sched j (t + (job_cost j - job_run_to_completion_threshold j)).
      movet ES.
      set (job_cost j - job_run_to_completion_threshold j) as job_last.
      have LSNP := @job_nonpreemptive_after_run_to_completion_threshold
                     Job H2 H3 _ _ arr_seq sched _ j _ t.
      apply negbNE; apply/negP; intros CONTR.
      have SCHED: t', t t' t + job_last scheduled_at sched j t'.
      { movet' /andP [GE LT].
        rewrite -[t'](@subnKC t) //.
        eapply LSNP; eauto 2; first by rewrite leq_addr.
        rewrite subnKC //.
        apply/negP; intros COMPL.
        move: CONTR ⇒ /negP Temp; apply: Temp.
        apply completion_monotonic with (t0 := t'); try done.
      have SERV: job_last + 1 \sum_(t t' < t + (job_last + 1)) service_at sched j t'.
      { rewrite -{1}[job_last + 1]addn0 -{2}(subnn t) addnBA // addnC.
        rewrite -{1}[_+_-_]addn0 -[_+_-_]mul1n -iter_addn -big_const_nat.
        rewrite big_nat_cond [in X in _ X]big_nat_cond.
        rewrite leq_sum //.
        movet' /andP [NEQ _].
        rewrite service_at_is_scheduled_at.
          by rewrite lt0b; apply SCHED; rewrite addn1 addnS ltnS in NEQ.
      eapply service_at_most_cost with (j0 := j) (t0 := t + job_last.+1) in H_completed_jobs_dont_execute; last first.
      { by apply ideal_proc_model_provides_unit_service. }
      move: H_completed_jobs_dont_execute; rewrite leqNgt; move ⇒ /negP T; apply: T.
      rewrite /service -(service_during_cat _ _ _ t); last by (apply/andP; split; last rewrite leq_addr).
      apply leq_trans with (job_run_to_completion_threshold j + service_during sched j t (t + job_last.+1));
        last by rewrite leq_add2r.
      apply leq_trans with (job_run_to_completion_threshold j + job_last.+1); last by rewrite leq_add2l /service_during -addn1.
        by rewrite addnS ltnS subnKC //; eapply job_run_to_completion_threshold_le_job_cost; eauto.

  End CompletionOfJobAfterRunToCompletionThreshold.

End AbstractRTARunToCompletionThreshold.