Re: [RFC 2/3] sched/fair: use util_est in LB

From: Patrick Bellasi
Date: Mon Sep 04 2017 - 10:19:03 EST


On 29-Aug 10:15, Pavan Kondeti wrote:
> On Fri, Aug 25, 2017 at 3:50 PM, Patrick Bellasi
> <patrick.bellasi@xxxxxxx> 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