# Library prosa.analysis.facts.periodic.arrival_separation

Require Export prosa.model.task.arrival.periodic_as_sporadic.
Require Export prosa.analysis.facts.sporadic.arrival_times.
In this section we show that the separation between job arrivals of a periodic task is some multiple of their corresponding task's period.
Section JobArrivalSeparation.
Consider periodic tasks with an offset ...
Context {Task : TaskType}.
Context `{PeriodicModel Task}.
... and any type of jobs associated with these tasks.
Context {Job : JobType}.
Context `{JobTask Job Task}.
Context `{JobArrival Job}.
Consider any unique arrival sequence with consistent arrivals, ...
Variable arr_seq : arrival_sequence Job.
Hypothesis H_valid_arrival_sequence : valid_arrival_sequence arr_seq.
... and any task tsk that is to be analyzed.
Variable tsk : Task.
Assume all tasks have a valid period and respect the periodic task model.
Hypothesis H_periodic_model : respects_periodic_task_model arr_seq tsk.
Hypothesis H_valid_period : valid_period tsk.
In this section we show that two consecutive jobs of a periodic task have their arrival times separated by their task's period.
Section ConsecutiveJobSeparation.
Consider any two consecutive jobs j1 and j2 of task tsk.
Variable j1 : Job.
Variable j2 : Job.
Hypothesis H_j1_from_arr_seq : arrives_in arr_seq j1.
Hypothesis H_j2_from_arr_seq : arrives_in arr_seq j2.
Hypothesis H_j1_of_task : job_task j1 = tsk.
Hypothesis H_j2_of_task : job_task j2 = tsk.
Hypothesis H_consecutive_jobs : job_index arr_seq j2 = job_index arr_seq j1 + 1.
We show that if job j1 and j2 are consecutive jobs with j2 arriving after j1, then their arrival times are separated by their task's period.
Lemma consecutive_job_separation :
job_arrival j2 = job_arrival j1 + task_period tsk.
Proof.
move : (H_periodic_model j2) ⇒ PERIODIC.
feed_n 3 PERIODIC ⇒ //; first by rewrite H_consecutive_jobs; lia.
move : PERIODIC ⇒ [pj' [ARR_IN_PJ' [INDPJ'J' [TSKPJ' ARRPJ']]]].
rewrite H_consecutive_jobs addnK in INDPJ'J'.
apply equal_index_implies_equal_jobs in INDPJ'J' ⇒ //; last by rewrite TSKPJ'.
by rewrite INDPJ'J' in ARRPJ'; lia.
Qed.
End ConsecutiveJobSeparation.
In this section we show that for two unequal jobs of a task, there exists a non-zero multiple of their task's period which separates their arrival times.
Section ArrivalSeparationWithGivenIndexDifference.
Consider any two consecutive jobs j1 and j2 of task tsk that stem from the arrival sequence.
Variable j1 j2 : Job.
Hypothesis H_j1_from_arr_seq : arrives_in arr_seq j1.
Hypothesis H_j2_from_arr_seq : arrives_in arr_seq j2.
Hypothesis H_j1_of_task : job_task j1 = tsk.
Hypothesis H_j2_of_task : job_task j2 = tsk.
We'll assume that job j1 arrives before j2 and that their indices differ by an integer k.
Variable k : nat.
Hypothesis H_index_difference_k : job_index arr_seq j1 + k = job_index arr_seq j2 .
Hypothesis H_job_arrival_lt : job_arrival j1 < job_arrival j2.
We prove that arrival of unequal jobs of a task tsk are separated by a non-zero multiple of task_period tsk provided their index differs by a number k.
Lemma job_arrival_separation_when_index_diff_is_k :
∃ n,
n > 0
∧ job_arrival j2 = job_arrival j1 + n × task_period tsk.
Proof.
move: j1 j2 H_j1_of_task H_j2_of_task H_index_difference_k H_job_arrival_lt H_j2_from_arr_seq H_j1_from_arr_seq;
clear H_index_difference_k H_job_arrival_lt H_j2_from_arr_seq H_j1_from_arr_seq H_j1_of_task H_j2_of_task j1 j2.
move: k ⇒ s.
elim: s ⇒ [|s IHs].
{ intros j1 j2 TSKj1 TSKj2 STEP LT ARRj1 ARRj2; exfalso.
specialize (earlier_arrival_implies_lower_index arr_seq H_valid_arrival_sequence j1 j2) ⇒ LT_IND.
feed_n 4 LT_IND ⇒ //; first by rewrite TSKj2.
by lia.
}
{ intros j1 j2 TSKj1 TSKj2 STEP LT ARRj2 ARRj1.
specialize (exists_jobs_before_j
arr_seq H_valid_arrival_sequence j2 ARRj2 (job_index arr_seq j2 - s)) ⇒ BEFORE.
destruct s as [|s].
- ∃ 1; repeat split.
by rewrite (consecutive_job_separation j1) //; lia.
- feed BEFORE; first by lia.
move : BEFORE ⇒ [nj [NEQNJ [TSKNJ [ARRNJ INDNJ]]]]; rewrite TSKj2 in TSKNJ.
specialize (IHs nj j2); feed_n 6 IHs ⇒ //; first by lia.
{ by apply (lower_index_implies_earlier_arrival _ H_valid_arrival_sequence tsk) ⇒ //; lia.
}
move : IHs ⇒ [n [NZN ARRJ'NJ]].
move: (H_periodic_model nj) ⇒ PERIODIC; feed_n 3 PERIODIC ⇒ //; first by lia.
move : PERIODIC ⇒ [sj [ARR_IN_SJ [INDSJ [TSKSJ ARRSJ]]]]; rewrite ARRSJ in ARRJ'NJ.
have INDJ : (job_index arr_seq j1 = job_index arr_seq j2 - s.+2) by lia.
rewrite INDNJ -subnDA addn1 -INDJ in INDSJ.
apply equal_index_implies_equal_jobs in INDSJ ⇒ //; last by rewrite TSKj1 ⇒ //.
∃ (n.+1); split; try by lia.
rewrite INDSJ in ARRJ'NJ; rewrite mulSnr.
by lia.
}
Qed.
End ArrivalSeparationWithGivenIndexDifference.
Consider any two distinct jobs j1 and j2 of task tsk that stem from the arrival sequence.
Variable j1 j2 : Job.
Hypothesis H_j1_neq_j2 : j1 ≠ j2.
Hypothesis H_j1_from_arr_seq : arrives_in arr_seq j1.
Hypothesis H_j2_from_arr_seq : arrives_in arr_seq j2.
Hypothesis H_j1_of_task : job_task j1 = tsk.
Hypothesis H_j2_of_task : job_task j2 = tsk.
Without loss of generality, we assume that job j1 arrives before job j2.
Hypothesis H_j1_before_j2 : job_arrival j1 ≤ job_arrival j2.
We generalize the above lemma to show that any two unequal jobs of task tsk are separated by a non-zero multiple of task_period tsk.
Lemma job_sep_periodic :
∃ n,
n > 0
∧ job_arrival j2 = job_arrival j1 + n × task_period tsk.
Proof.
apply job_arrival_separation_when_index_diff_is_k with (k := (job_index arr_seq j2 - job_index arr_seq j1)) ⇒ //.
- apply subnKC.
move_neq_up IND.
eapply lower_index_implies_earlier_arrival in IND ⇒ //.
by move_neq_down IND.
- case: (boolP (job_index arr_seq j1 == job_index arr_seq j2)) ⇒ [/eqP EQ_IND|NEQ_IND].
+ by apply equal_index_implies_equal_jobs in EQ_IND ⇒ //; rewrite H_j1_of_task.
+ move: NEQ_IND; rewrite neq_ltn ⇒ /orP [LT|LT].
× by eapply (lower_index_implies_earlier_arrival) in LT.
× eapply (lower_index_implies_earlier_arrival) in LT ⇒ //.
by move_neq_down LT.
Qed.
End JobArrivalSeparation.