linux-kernel - Re: [RFC 2/3] sched/fair: use util

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Date:   Mon, 4 Sep 2017 15:18:17 +0100
From:   Patrick Bellasi <patrick.bellasi@....com>
To:     Pavan Kondeti <pkondeti@...eaurora.org>
Cc:     LKML <linux-kernel@...r.kernel.org>, linux-pm@...r.kernel.org,
        Ingo Molnar <mingo@...hat.com>,
        Peter Zijlstra <peterz@...radead.org>,
        "Rafael J . Wysocki" <rafael.j.wysocki@...el.com>,
        Paul Turner <pjt@...gle.com>,
        Vincent Guittot <vincent.guittot@...aro.org>,
        John Stultz <john.stultz@...aro.org>,
        Morten Rasmussen <morten.rasmussen@....com>,
        Dietmar Eggemann <dietmar.eggemann@....com>,
        Juri Lelli <juri.lelli@....com>,
        Tim Murray <timmurray@...gle.com>,
        Todd Kjos <tkjos@...roid.com>,
        Andres Oportus <andresoportus@...gle.com>,
        Joel Fernandes <joelaf@...gle.com>,
        Viresh Kumar <viresh.kumar@...aro.org>
Subject: Re: [RFC 2/3] sched/fair: use util_est in LB

On 29-Aug 10:15, Pavan Kondeti wrote:
> On Fri, Aug 25, 2017 at 3:50 PM, Patrick Bellasi
> <patrick.bellasi@....com> wrote:
> > When the scheduler looks at the CPU utlization, the current PELT value
> > for a CPU is returned straight away. In certain scenarios this can have
> > undesired side effects on task placement.
> >
> 
> <snip>
> 
> > +/**
> > + * cpu_util_est: estimated utilization for the specified CPU
> > + * @cpu: the CPU to get the estimated utilization for
> > + *
> > + * The estimated utilization of a CPU is defined to be the maximum between its
> > + * PELT's utilization and the sum of the estimated utilization of the tasks
> > + * currently RUNNABLE on that CPU.
> > + *
> > + * This allows to properly represent the expected utilization of a CPU which
> > + * has just got a big task running since a long sleep period. At the same time
> > + * however it preserves the benefits of the "blocked load" in describing the
> > + * potential for other tasks waking up on the same CPU.
> > + *
> > + * Return: the estimated utlization for the specified CPU
> > + */
> > +static inline unsigned long cpu_util_est(int cpu)
> > +{
> > +       struct sched_avg *sa = &cpu_rq(cpu)->cfs.avg;
> > +       unsigned long util = cpu_util(cpu);
> > +
> > +       if (!sched_feat(UTIL_EST))
> > +               return util;
> > +
> > +       return max(util, util_est(sa, UTIL_EST_LAST));
> > +}
> > +
> >  static inline int task_util(struct task_struct *p)
> >  {
> >         return p->se.avg.util_avg;
> > @@ -6007,11 +6033,19 @@ static int cpu_util_wake(int cpu, struct task_struct *p)
> >
> >         /* Task has no contribution or is new */
> >         if (cpu != task_cpu(p) || !p->se.avg.last_update_time)
> > -               return cpu_util(cpu);
> > +               return cpu_util_est(cpu);
> >
> >         capacity = capacity_orig_of(cpu);
> >         util = max_t(long, cpu_rq(cpu)->cfs.avg.util_avg - task_util(p), 0);
> >
> > +       /*
> > +        * Estimated utilization tracks only tasks already enqueued, but still
> > +        * sometimes can return a bigger value than PELT, for example when the
> > +        * blocked load is negligible wrt the estimated utilization of the
> > +        * already enqueued tasks.
> > +        */
> > +       util = max_t(long, util, cpu_util_est(cpu));
> > +
> 
> We are supposed to discount the task's util from its CPU. But the
> cpu_util_est() can potentially return cpu_util() which includes the
> task's utilization.

You right, this instead should cover all the cases:

---8<---
 static int cpu_util_wake(int cpu, struct task_struct *p)
 {
-       unsigned long util, capacity;
+       unsigned long util_est = cpu_util_est(cpu);
+       unsigned long capacity;
 
        /* Task has no contribution or is new */
        if (cpu != task_cpu(p) || !p->se.avg.last_update_time)
-               return cpu_util(cpu);
+               return util_est;
 
        capacity = capacity_orig_of(cpu);
-       util = max_t(long, cpu_rq(cpu)->cfs.avg.util_avg - task_util(p), 0);
+       if (cpu_util(cpu) > util_est)
+               util = max_t(long, cpu_util(cpu) - task_util(p), 0);
+       else
+               util = util_est;
 
        return (util >= capacity) ? capacity : util;
 }
---8<---

Indeed:

- if *p is the only task sleeping on that CPU, then:
      (cpu_util == task_util) > (cpu_util_est == 0)
  and thus we return:
      (cpu_util - task_util) == 0

- if other tasks are SLEEPING on the same CPU, which however is IDLE, then:
      cpu_util > (cpu_util_est == 0)
  and thus we discount *p's blocked load by returning:
      (cpu_util - task_util) >= 0

- if other tasks are RUNNABLE on that CPU and
      (cpu_util_est > cpu_util)
  then we wanna use cpu_util_est since it returns a more restrictive
  estimation of the spare capacity on that CPU, by just considering
  the expected utilization of tasks already runnable on that CPU.

What do you think?

> Thanks,
> Pavan

Cheers Patrick

-- 
#include <best/regards.h>

Patrick Bellasi