linux-kernel - Re: I.6 - Group scheduling

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Date:	Mon, 22 Jun 2009 13:53:10 +0200
From:	Ingo Molnar <mingo@...e.hu>
To:	eranian@...il.com
Cc:	LKML <linux-kernel@...r.kernel.org>,
	Andrew Morton <akpm@...ux-foundation.org>,
	Thomas Gleixner <tglx@...utronix.de>,
	Robert Richter <robert.richter@....com>,
	Peter Zijlstra <a.p.zijlstra@...llo.nl>,
	Paul Mackerras <paulus@...ba.org>,
	Andi Kleen <andi@...stfloor.org>,
	Maynard Johnson <mpjohn@...ibm.com>,
	Carl Love <cel@...ibm.com>,
	Corey J Ashford <cjashfor@...ibm.com>,
	Philip Mucci <mucci@...s.utk.edu>,
	Dan Terpstra <terpstra@...s.utk.edu>,
	perfmon2-devel <perfmon2-devel@...ts.sourceforge.net>
Subject: Re: I.6 - Group scheduling

> 6/ Group scheduling
>
> Looking at the existing code, it seems to me there is a risk of
> starvation for groups, i.e., groups never scheduled on the PMU.
>
> My understanding of the scheduling algorithm is:
>
> - first try to �schedule pinned groups. If a pinned group
>   fails, put it in error mode. read() will fail until the
>   group gets another chance at being scheduled.
>
> - then try to schedule the remaining groups. If a group fails
>   just skip it.
>
> If the group list does not change, then certain groups may always
> fail. However, the ordering of the list changes because at every
> tick, it is rotated. The head becomes the tail. Therefore, each
> group eventually gets the first position and therefore gets the
> full PMU to assign its events.
>
> This works as long as there is a guarantee the list will ALWAYS
> rotate. If a thread does not run long enough for a tick, it may
> never rotate.

You need to ensure the task never runs during the tick, this is a
statistical propery that is bound to untrue for any 'normal'
application.

This is similar to the old cputime hiding tricks. We don't think
this will be a problem, you really need to actively avoid the tick
to aggregate a significantly lower tick rate.

In any case this can also be excluded via a natural extension
mentioned above: scheduling based not on the scheduler tick but
based on one of the counters.
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