linux-kernel - Re: [PATCH 2/4] sched/fair: Drop out incomplete current period when sched averages accrue

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Message-ID: <570E61FE.4060000@arm.com>
Date:	Wed, 13 Apr 2016 16:13:02 +0100
From:	Dietmar Eggemann <dietmar.eggemann@....com>
To:	Yuyang Du <yuyang.du@...el.com>, peterz@...radead.org,
	mingo@...nel.org, linux-kernel@...r.kernel.org
Cc:	bsegall@...gle.com, pjt@...gle.com, morten.rasmussen@....com,
	vincent.guittot@...aro.org, juri.lelli@....com
Subject: Re: [PATCH 2/4] sched/fair: Drop out incomplete current period when
 sched averages accrue

On 10/04/16 23:36, Yuyang Du wrote:
> In __update_load_avg(), the current period is never complete. This
> basically leads to a slightly over-decayed average, say on average we
> have 50% current period, then we will lose 1.08%(=(1-0.5^(1/64)) of
> past avg. More importantly, the incomplete current period significantly
> complicates the avg computation, even a full period is only about 1ms.
> 
> So we attempt to drop it. The outcome is that for any x.y periods to
> update, we will either lose the .y period or unduely gain (1-.y) period.
> How big is the impact? For a large x (say 20ms), you barely notice the
> difference, which is plus/minus 1% (=(before-after)/before). Moreover,
> the aggregated losses and gains in the long run should statistically
> even out.
> 

For a periodic task, the signals really get much more unstable. Even for
a steady state (load/util related) periodic task there is a meander
pattern which depends on if we for instance hit a dequeue (decay +
accrue) or an enqueue (decay only) after the 1ms has elapsed.

IMHO, 1ms is too big to create signals describing task and cpu load/util
signals given the current scheduler dynamics. We simply see too many
signal driving points (e.g. enqueue/dequeue) bailing out of
__update_load_avg().

Examples of 1 periodic task pinned to a cpu on an ARM64 system, HZ=250
in steady state:

(1) task runtime = 100us period = 200us

  pelt		load/util signal

  1us: 		488-491

  1ms: 		483-534

We get ~2 dequeues (load/util example: 493->504) and ~2 enqueues
(load/util example: 496->483) in the meander pattern in the 1ms case.

(2) task runtime = 100us period = 1000us

  pelt	 	load/util signal

  1us: 		103-105

  1ms: 		 84-145

We get ~3-4 dequeues (load/util example: 104->124->134->140) and ~16-20
enqueues (load/util example: 137->134->...->99->97) in the meander
pattern in the 1ms case.

[...]