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Date: Thu, 14 Apr 2016 02:44:12 +0800 From: Yuyang Du <yuyang.du@...el.com> To: Vincent Guittot <vincent.guittot@...aro.org> Cc: Dietmar Eggemann <dietmar.eggemann@....com>, Peter Zijlstra <peterz@...radead.org>, Ingo Molnar <mingo@...nel.org>, linux-kernel <linux-kernel@...r.kernel.org>, Benjamin Segall <bsegall@...gle.com>, Paul Turner <pjt@...gle.com>, Morten Rasmussen <morten.rasmussen@....com>, Juri Lelli <juri.lelli@....com> Subject: Re: [PATCH 2/4] sched/fair: Drop out incomplete current period when sched averages accrue On Wed, Apr 13, 2016 at 05:28:18PM +0200, Vincent Guittot wrote: > > 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(). By "bailing out", you mean return without update because the delta is less than 1ms? > > 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 100us/200us = 50%, so the util should center around 512, it seems in this regard, it is better, but the variance is undesirable. > > 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. The same as above. > > yes, similarly i have some use cases with 2ms running task in a period > of 5.12ms. it will be seen either as a 1ms running task or a 2ms > running tasks depending on how the running is synced with the 1ms > boundary > > so the load will vary between 197-215 up to 396-423 depending of when > the 1ms boundary occurs in the 2ms running > Same as above, and this time, the util is "expected" to be 2/5.242*1024=391 of all the samples. We solve the problem of overly-decay, but the precision loss is a new problem. Let me see if we can get to a 2-level period scheme, :)
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