Library rt.analysis.basic.workload_bound
Require Import
rt.util.all.
Require Import rt.model.basic.task rt.model.basic.job rt.model.basic.schedule
rt.model.basic.task_arrival rt.model.basic.response_time
rt.model.basic.workload rt.model.basic.schedulability.
Module WorkloadBound.
Import Job SporadicTaskset Schedule ScheduleOfSporadicTask SporadicTaskArrival ResponseTime Schedulability Workload.
Section WorkloadBoundDef.
Context {sporadic_task: eqType}.
Variable task_cost: sporadic_task → time.
Variable task_period: sporadic_task → time.
(* Consider any task tsk with response-time bound R_tsk,
that is scheduled in an interval of length delta. *)
Variable tsk: sporadic_task.
Variable R_tsk: time.
Variable delta: time.
(* Based on the number of jobs that execute completely in the interval, ... *)
Definition max_jobs :=
div_floor (delta + R_tsk - task_cost tsk) (task_period tsk).
(* ... Bertogna and Cirinei's workload bound is defined as follows. *)
Definition W :=
let e_k := (task_cost tsk) in
let p_k := (task_period tsk) in
minn e_k (delta + R_tsk - e_k - max_jobs × p_k) + max_jobs × e_k.
End WorkloadBoundDef.
Section BasicLemmas.
Context {sporadic_task: eqType}.
Variable task_cost: sporadic_task → time.
Variable task_period: sporadic_task → time.
(* Let tsk be any task...*)
Variable tsk: sporadic_task.
(* ... with period > 0. *)
Hypothesis H_period_positive: task_period tsk > 0.
(* Let R1 <= R2 be two response-time bounds that
are larger than the cost of the tsk. *)
Variable R1 R2: time.
Hypothesis H_R_lower_bound: R1 ≥ task_cost tsk.
Hypothesis H_R1_le_R2: R1 ≤ R2.
Let workload_bound := W task_cost task_period tsk.
(* Then, Bertogna and Cirinei's workload bound is monotonically increasing. *)
Lemma W_monotonic :
∀ t1 t2,
t1 ≤ t2 →
workload_bound R1 t1 ≤ workload_bound R2 t2.
End BasicLemmas.
Section ProofWorkloadBound.
Context {sporadic_task: eqType}.
Variable task_cost: sporadic_task → time.
Variable task_period: sporadic_task → time.
Variable task_deadline: sporadic_task → time.
Context {Job: eqType}.
Variable job_cost: Job → time.
Variable job_task: Job → sporadic_task.
Variable job_deadline: Job → time.
Variable arr_seq: arrival_sequence Job.
(* Assume that all jobs have valid parameters *)
Hypothesis H_jobs_have_valid_parameters :
∀ (j: JobIn arr_seq),
valid_sporadic_job task_cost task_deadline job_cost job_deadline job_task j.
(* Consider any schedule. *)
Context {num_cpus: nat}.
Variable sched: schedule num_cpus arr_seq.
(* Assumption: jobs only execute if they arrived.
This is used to eliminate jobs that arrive after end of the interval t1 + delta. *)
Hypothesis H_jobs_must_arrive_to_execute:
jobs_must_arrive_to_execute sched.
(* Assumption: jobs do not execute after they completed.
This is used to eliminate jobs that complete before the start of the interval t1. *)
Hypothesis H_completed_jobs_dont_execute:
completed_jobs_dont_execute job_cost sched.
(* Assumption: Jobs are sequential.
This is required to use interval lengths as a measure of service. *)
Hypothesis H_sequential_jobs: sequential_jobs sched.
(* Assumption: sporadic task model.
This is necessary to conclude that consecutive jobs ordered by arrival times
are separated by at least 'period' times units. *)
Hypothesis H_sporadic_tasks: sporadic_task_model task_period arr_seq job_task.
(* Before starting the proof, let's give simpler names to the definitions. *)
Let job_has_completed_by := completed job_cost sched.
Let workload_of (tsk: sporadic_task) (t1 t2: time) :=
workload job_task sched tsk t1 t2.
(* Now we define the theorem. Let tsk be any task in the taskset. *)
Variable tsk: sporadic_task.
(* Assumption: the task must have valid parameters:
a) period > 0 (used in divisions)
b) deadline of the jobs = deadline of the task
c) cost <= period
(used to prove that the distance between the first and last
jobs is at least (cost + n*period), where n is the number
of middle jobs. If cost >> period, the claim does not hold
for every task set. *)
Hypothesis H_valid_task_parameters:
is_valid_sporadic_task task_cost task_period task_deadline tsk.
(* Assumption: the task must have a constrained deadline.
This is required to prove that n_k (max_jobs) from Bertogna
and Cirinei's formula accounts for at least the number of
middle jobs (i.e., number of jobs - 2 in the worst case). *)
Hypothesis H_constrained_deadline: task_deadline tsk ≤ task_period tsk.
(* Consider an interval [t1, t1 + delta). *)
Variable t1 delta: time.
(* Assume that a response-time bound R_tsk for that task in any
schedule of this processor platform is also given, ... *)
Variable R_tsk: time.
Hypothesis H_response_time_bound :
∀ (j: JobIn arr_seq),
job_task j = tsk →
job_arrival j + R_tsk < t1 + delta →
job_has_completed_by j (job_arrival j + R_tsk).
(* ... such that R_tsk >= task_cost tsk and R_tsk <= task_deadline tsk. *)
Hypothesis H_response_time_ge_cost: R_tsk ≥ task_cost tsk.
Hypothesis H_no_deadline_miss: R_tsk ≤ task_deadline tsk.
Section MainProof.
(* In this section, we prove that the workload of a task in the
interval t1, t1 + delta) is bounded by W. *)
(* Let's simplify the names a bit. *)
Let t2 := t1 + delta.
Let n_k := max_jobs task_cost task_period tsk R_tsk delta.
Let workload_bound := W task_cost task_period tsk R_tsk delta.
(* Since we only care about the workload of tsk, we restrict
our view to the set of jobs of tsk scheduled in [t1, t2). *)
Let scheduled_jobs :=
jobs_of_task_scheduled_between job_task sched tsk t1 t2.
(* Now, let's consider the list of interfering jobs sorted by arrival time. *)
Let earlier_arrival := fun (x y: JobIn arr_seq) ⇒ job_arrival x ≤ job_arrival y.
Let sorted_jobs := (sort earlier_arrival scheduled_jobs).
(* The first step consists in simplifying the sum corresponding
to the workload. *)
Section SimplifyJobSequence.
(* After switching to the definition of workload based on a list
of jobs, we show that sorting the list preserves the sum. *)
Lemma workload_bound_simpl_by_sorting_scheduled_jobs :
workload_joblist job_task sched tsk t1 t2 =
\sum_(i <- sorted_jobs) service_during sched i t1 t2.
(* Remember that both sequences have the same set of elements *)
Lemma workload_bound_job_in_same_sequence :
∀ j, (j \in scheduled_jobs) = (j \in sorted_jobs).
(* Remember that all jobs in the sorted sequence is an
interfering job of task tsk. *)
Lemma workload_bound_all_jobs_from_tsk :
∀ j_i,
j_i \in sorted_jobs →
job_task j_i = tsk ∧
service_during sched j_i t1 t2 != 0 ∧
j_i \in jobs_scheduled_between sched t1 t2.
(* Remember that consecutive jobs are ordered by arrival. *)
Lemma workload_bound_jobs_ordered_by_arrival :
∀ i elem,
i < (size sorted_jobs).-1 →
earlier_arrival (nth elem sorted_jobs i) (nth elem sorted_jobs i.+1).
End SimplifyJobSequence.
(* Next, we show that if the number of jobs is no larger than n_k,
the workload bound trivially holds. *)
Section WorkloadNotManyJobs.
Lemma workload_bound_holds_for_at_most_n_k_jobs :
size sorted_jobs ≤ n_k →
\sum_(i <- sorted_jobs) service_during sched i t1 t2 ≤
workload_bound.
End WorkloadNotManyJobs.
(* Otherwise, assume that the number of jobs is larger than n_k >= 0.
First, consider the simple case with only one job. *)
Section WorkloadSingleJob.
(* Assume that there's at least one job in the sorted list. *)
Hypothesis H_at_least_one_job: size sorted_jobs > 0.
Variable elem: JobIn arr_seq.
Let j_fst := nth elem sorted_jobs 0.
(* The first job is an interfering job of task tsk. *)
Lemma workload_bound_j_fst_is_job_of_tsk :
job_task j_fst = tsk ∧
service_during sched j_fst t1 t2 != 0 ∧
j_fst \in jobs_scheduled_between sched t1 t2.
(* The workload bound holds for the single job. *)
Lemma workload_bound_holds_for_a_single_job :
\sum_(0 ≤ i < 1) service_during sched (nth elem sorted_jobs i) t1 t2 ≤
workload_bound.
End WorkloadSingleJob.
(* Next, consider the last case where there are at least two jobs:
the first job j_fst, and the last job j_lst. *)
Section WorkloadTwoOrMoreJobs.
(* There are at least two jobs. *)
Variable num_mid_jobs: nat.
Hypothesis H_at_least_two_jobs : size sorted_jobs = num_mid_jobs.+2.
Variable elem: JobIn arr_seq.
Let j_fst := nth elem sorted_jobs 0.
Let j_lst := nth elem sorted_jobs num_mid_jobs.+1.
(* The last job is an interfering job of task tsk. *)
Lemma workload_bound_j_lst_is_job_of_tsk :
job_task j_lst = tsk ∧
service_during sched j_lst t1 t2 != 0 ∧
j_lst \in jobs_scheduled_between sched t1 t2.
(* The response time of the first job must fall inside the interval. *)
Lemma workload_bound_response_time_of_first_job_inside_interval :
t1 ≤ job_arrival j_fst + R_tsk.
(* The arrival of the last job must also fall inside the interval. *)
Lemma workload_bound_last_job_arrives_before_end_of_interval :
job_arrival j_lst < t2.
(* Next, we upper-bound the service of the first and last jobs using their arrival times. *)
Lemma workload_bound_service_of_first_and_last_jobs :
service_during sched j_fst t1 t2 +
service_during sched j_lst t1 t2 ≤
(job_arrival j_fst + R_tsk - t1) + (t2 - job_arrival j_lst).
(* Simplify the expression from the previous lemma. *)
Lemma workload_bound_simpl_expression_with_first_and_last :
job_arrival j_fst + R_tsk - t1 + (t2 - job_arrival j_lst) =
delta + R_tsk - (job_arrival j_lst - job_arrival j_fst).
(* Bound the service of the middle jobs. *)
Lemma workload_bound_service_of_middle_jobs :
\sum_(0 ≤ i < num_mid_jobs)
service_during sched (nth elem sorted_jobs i.+1) t1 t2 ≤
num_mid_jobs × task_cost tsk.
(* Conclude that the distance between first and last is at least num_mid_jobs + 1 periods. *)
Lemma workload_bound_many_periods_in_between :
job_arrival j_lst - job_arrival j_fst ≥ num_mid_jobs.+1 × (task_period tsk).
(* Prove that n_k is at least the number of the middle jobs *)
Lemma workload_bound_n_k_covers_middle_jobs :
n_k ≥ num_mid_jobs.
(* If n_k = num_mid_jobs, then the workload bound holds. *)
Lemma workload_bound_n_k_equals_num_mid_jobs :
num_mid_jobs = n_k →
service_during sched j_lst t1 t2 +
service_during sched j_fst t1 t2 +
\sum_(0 ≤ i < num_mid_jobs)
service_during sched (nth elem sorted_jobs i.+1) t1 t2
≤ workload_bound.
(* If n_k = num_mid_jobs + 1, then the workload bound holds. *)
Lemma workload_bound_n_k_equals_num_mid_jobs_plus_1 :
num_mid_jobs.+1 = n_k →
service_during sched j_lst t1 t2 +
service_during sched j_fst t1 t2 +
\sum_(0 ≤ i < num_mid_jobs)
service_during sched (nth elem sorted_jobs i.+1) t1 t2
≤ workload_bound.
End WorkloadTwoOrMoreJobs.
(* Using the lemmas above, we prove the main theorem about the workload bound. *)
Theorem workload_bounded_by_W :
workload_of tsk t1 (t1 + delta) ≤ workload_bound.
End MainProof.
End ProofWorkloadBound.
End WorkloadBound.
Require Import rt.model.basic.task rt.model.basic.job rt.model.basic.schedule
rt.model.basic.task_arrival rt.model.basic.response_time
rt.model.basic.workload rt.model.basic.schedulability.
Module WorkloadBound.
Import Job SporadicTaskset Schedule ScheduleOfSporadicTask SporadicTaskArrival ResponseTime Schedulability Workload.
Section WorkloadBoundDef.
Context {sporadic_task: eqType}.
Variable task_cost: sporadic_task → time.
Variable task_period: sporadic_task → time.
(* Consider any task tsk with response-time bound R_tsk,
that is scheduled in an interval of length delta. *)
Variable tsk: sporadic_task.
Variable R_tsk: time.
Variable delta: time.
(* Based on the number of jobs that execute completely in the interval, ... *)
Definition max_jobs :=
div_floor (delta + R_tsk - task_cost tsk) (task_period tsk).
(* ... Bertogna and Cirinei's workload bound is defined as follows. *)
Definition W :=
let e_k := (task_cost tsk) in
let p_k := (task_period tsk) in
minn e_k (delta + R_tsk - e_k - max_jobs × p_k) + max_jobs × e_k.
End WorkloadBoundDef.
Section BasicLemmas.
Context {sporadic_task: eqType}.
Variable task_cost: sporadic_task → time.
Variable task_period: sporadic_task → time.
(* Let tsk be any task...*)
Variable tsk: sporadic_task.
(* ... with period > 0. *)
Hypothesis H_period_positive: task_period tsk > 0.
(* Let R1 <= R2 be two response-time bounds that
are larger than the cost of the tsk. *)
Variable R1 R2: time.
Hypothesis H_R_lower_bound: R1 ≥ task_cost tsk.
Hypothesis H_R1_le_R2: R1 ≤ R2.
Let workload_bound := W task_cost task_period tsk.
(* Then, Bertogna and Cirinei's workload bound is monotonically increasing. *)
Lemma W_monotonic :
∀ t1 t2,
t1 ≤ t2 →
workload_bound R1 t1 ≤ workload_bound R2 t2.
End BasicLemmas.
Section ProofWorkloadBound.
Context {sporadic_task: eqType}.
Variable task_cost: sporadic_task → time.
Variable task_period: sporadic_task → time.
Variable task_deadline: sporadic_task → time.
Context {Job: eqType}.
Variable job_cost: Job → time.
Variable job_task: Job → sporadic_task.
Variable job_deadline: Job → time.
Variable arr_seq: arrival_sequence Job.
(* Assume that all jobs have valid parameters *)
Hypothesis H_jobs_have_valid_parameters :
∀ (j: JobIn arr_seq),
valid_sporadic_job task_cost task_deadline job_cost job_deadline job_task j.
(* Consider any schedule. *)
Context {num_cpus: nat}.
Variable sched: schedule num_cpus arr_seq.
(* Assumption: jobs only execute if they arrived.
This is used to eliminate jobs that arrive after end of the interval t1 + delta. *)
Hypothesis H_jobs_must_arrive_to_execute:
jobs_must_arrive_to_execute sched.
(* Assumption: jobs do not execute after they completed.
This is used to eliminate jobs that complete before the start of the interval t1. *)
Hypothesis H_completed_jobs_dont_execute:
completed_jobs_dont_execute job_cost sched.
(* Assumption: Jobs are sequential.
This is required to use interval lengths as a measure of service. *)
Hypothesis H_sequential_jobs: sequential_jobs sched.
(* Assumption: sporadic task model.
This is necessary to conclude that consecutive jobs ordered by arrival times
are separated by at least 'period' times units. *)
Hypothesis H_sporadic_tasks: sporadic_task_model task_period arr_seq job_task.
(* Before starting the proof, let's give simpler names to the definitions. *)
Let job_has_completed_by := completed job_cost sched.
Let workload_of (tsk: sporadic_task) (t1 t2: time) :=
workload job_task sched tsk t1 t2.
(* Now we define the theorem. Let tsk be any task in the taskset. *)
Variable tsk: sporadic_task.
(* Assumption: the task must have valid parameters:
a) period > 0 (used in divisions)
b) deadline of the jobs = deadline of the task
c) cost <= period
(used to prove that the distance between the first and last
jobs is at least (cost + n*period), where n is the number
of middle jobs. If cost >> period, the claim does not hold
for every task set. *)
Hypothesis H_valid_task_parameters:
is_valid_sporadic_task task_cost task_period task_deadline tsk.
(* Assumption: the task must have a constrained deadline.
This is required to prove that n_k (max_jobs) from Bertogna
and Cirinei's formula accounts for at least the number of
middle jobs (i.e., number of jobs - 2 in the worst case). *)
Hypothesis H_constrained_deadline: task_deadline tsk ≤ task_period tsk.
(* Consider an interval [t1, t1 + delta). *)
Variable t1 delta: time.
(* Assume that a response-time bound R_tsk for that task in any
schedule of this processor platform is also given, ... *)
Variable R_tsk: time.
Hypothesis H_response_time_bound :
∀ (j: JobIn arr_seq),
job_task j = tsk →
job_arrival j + R_tsk < t1 + delta →
job_has_completed_by j (job_arrival j + R_tsk).
(* ... such that R_tsk >= task_cost tsk and R_tsk <= task_deadline tsk. *)
Hypothesis H_response_time_ge_cost: R_tsk ≥ task_cost tsk.
Hypothesis H_no_deadline_miss: R_tsk ≤ task_deadline tsk.
Section MainProof.
(* In this section, we prove that the workload of a task in the
interval t1, t1 + delta) is bounded by W. *)
(* Let's simplify the names a bit. *)
Let t2 := t1 + delta.
Let n_k := max_jobs task_cost task_period tsk R_tsk delta.
Let workload_bound := W task_cost task_period tsk R_tsk delta.
(* Since we only care about the workload of tsk, we restrict
our view to the set of jobs of tsk scheduled in [t1, t2). *)
Let scheduled_jobs :=
jobs_of_task_scheduled_between job_task sched tsk t1 t2.
(* Now, let's consider the list of interfering jobs sorted by arrival time. *)
Let earlier_arrival := fun (x y: JobIn arr_seq) ⇒ job_arrival x ≤ job_arrival y.
Let sorted_jobs := (sort earlier_arrival scheduled_jobs).
(* The first step consists in simplifying the sum corresponding
to the workload. *)
Section SimplifyJobSequence.
(* After switching to the definition of workload based on a list
of jobs, we show that sorting the list preserves the sum. *)
Lemma workload_bound_simpl_by_sorting_scheduled_jobs :
workload_joblist job_task sched tsk t1 t2 =
\sum_(i <- sorted_jobs) service_during sched i t1 t2.
(* Remember that both sequences have the same set of elements *)
Lemma workload_bound_job_in_same_sequence :
∀ j, (j \in scheduled_jobs) = (j \in sorted_jobs).
(* Remember that all jobs in the sorted sequence is an
interfering job of task tsk. *)
Lemma workload_bound_all_jobs_from_tsk :
∀ j_i,
j_i \in sorted_jobs →
job_task j_i = tsk ∧
service_during sched j_i t1 t2 != 0 ∧
j_i \in jobs_scheduled_between sched t1 t2.
(* Remember that consecutive jobs are ordered by arrival. *)
Lemma workload_bound_jobs_ordered_by_arrival :
∀ i elem,
i < (size sorted_jobs).-1 →
earlier_arrival (nth elem sorted_jobs i) (nth elem sorted_jobs i.+1).
End SimplifyJobSequence.
(* Next, we show that if the number of jobs is no larger than n_k,
the workload bound trivially holds. *)
Section WorkloadNotManyJobs.
Lemma workload_bound_holds_for_at_most_n_k_jobs :
size sorted_jobs ≤ n_k →
\sum_(i <- sorted_jobs) service_during sched i t1 t2 ≤
workload_bound.
End WorkloadNotManyJobs.
(* Otherwise, assume that the number of jobs is larger than n_k >= 0.
First, consider the simple case with only one job. *)
Section WorkloadSingleJob.
(* Assume that there's at least one job in the sorted list. *)
Hypothesis H_at_least_one_job: size sorted_jobs > 0.
Variable elem: JobIn arr_seq.
Let j_fst := nth elem sorted_jobs 0.
(* The first job is an interfering job of task tsk. *)
Lemma workload_bound_j_fst_is_job_of_tsk :
job_task j_fst = tsk ∧
service_during sched j_fst t1 t2 != 0 ∧
j_fst \in jobs_scheduled_between sched t1 t2.
(* The workload bound holds for the single job. *)
Lemma workload_bound_holds_for_a_single_job :
\sum_(0 ≤ i < 1) service_during sched (nth elem sorted_jobs i) t1 t2 ≤
workload_bound.
End WorkloadSingleJob.
(* Next, consider the last case where there are at least two jobs:
the first job j_fst, and the last job j_lst. *)
Section WorkloadTwoOrMoreJobs.
(* There are at least two jobs. *)
Variable num_mid_jobs: nat.
Hypothesis H_at_least_two_jobs : size sorted_jobs = num_mid_jobs.+2.
Variable elem: JobIn arr_seq.
Let j_fst := nth elem sorted_jobs 0.
Let j_lst := nth elem sorted_jobs num_mid_jobs.+1.
(* The last job is an interfering job of task tsk. *)
Lemma workload_bound_j_lst_is_job_of_tsk :
job_task j_lst = tsk ∧
service_during sched j_lst t1 t2 != 0 ∧
j_lst \in jobs_scheduled_between sched t1 t2.
(* The response time of the first job must fall inside the interval. *)
Lemma workload_bound_response_time_of_first_job_inside_interval :
t1 ≤ job_arrival j_fst + R_tsk.
(* The arrival of the last job must also fall inside the interval. *)
Lemma workload_bound_last_job_arrives_before_end_of_interval :
job_arrival j_lst < t2.
(* Next, we upper-bound the service of the first and last jobs using their arrival times. *)
Lemma workload_bound_service_of_first_and_last_jobs :
service_during sched j_fst t1 t2 +
service_during sched j_lst t1 t2 ≤
(job_arrival j_fst + R_tsk - t1) + (t2 - job_arrival j_lst).
(* Simplify the expression from the previous lemma. *)
Lemma workload_bound_simpl_expression_with_first_and_last :
job_arrival j_fst + R_tsk - t1 + (t2 - job_arrival j_lst) =
delta + R_tsk - (job_arrival j_lst - job_arrival j_fst).
(* Bound the service of the middle jobs. *)
Lemma workload_bound_service_of_middle_jobs :
\sum_(0 ≤ i < num_mid_jobs)
service_during sched (nth elem sorted_jobs i.+1) t1 t2 ≤
num_mid_jobs × task_cost tsk.
(* Conclude that the distance between first and last is at least num_mid_jobs + 1 periods. *)
Lemma workload_bound_many_periods_in_between :
job_arrival j_lst - job_arrival j_fst ≥ num_mid_jobs.+1 × (task_period tsk).
(* Prove that n_k is at least the number of the middle jobs *)
Lemma workload_bound_n_k_covers_middle_jobs :
n_k ≥ num_mid_jobs.
(* If n_k = num_mid_jobs, then the workload bound holds. *)
Lemma workload_bound_n_k_equals_num_mid_jobs :
num_mid_jobs = n_k →
service_during sched j_lst t1 t2 +
service_during sched j_fst t1 t2 +
\sum_(0 ≤ i < num_mid_jobs)
service_during sched (nth elem sorted_jobs i.+1) t1 t2
≤ workload_bound.
(* If n_k = num_mid_jobs + 1, then the workload bound holds. *)
Lemma workload_bound_n_k_equals_num_mid_jobs_plus_1 :
num_mid_jobs.+1 = n_k →
service_during sched j_lst t1 t2 +
service_during sched j_fst t1 t2 +
\sum_(0 ≤ i < num_mid_jobs)
service_during sched (nth elem sorted_jobs i.+1) t1 t2
≤ workload_bound.
End WorkloadTwoOrMoreJobs.
(* Using the lemmas above, we prove the main theorem about the workload bound. *)
Theorem workload_bounded_by_W :
workload_of tsk t1 (t1 + delta) ≤ workload_bound.
End MainProof.
End ProofWorkloadBound.
End WorkloadBound.