linux-kernel - Re: [RFC PATCH v3 00/16] Core scheduling v3

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Date:   Fri, 19 Jul 2019 13:52:38 +0800
From:   Aaron Lu <aaron.lu@...ux.alibaba.com>
To:     Tim Chen <tim.c.chen@...ux.intel.com>
Cc:     Julien Desfossez <jdesfossez@...italocean.com>,
        Aubrey Li <aubrey.intel@...il.com>,
        Subhra Mazumdar <subhra.mazumdar@...cle.com>,
        Vineeth Remanan Pillai <vpillai@...italocean.com>,
        Nishanth Aravamudan <naravamudan@...italocean.com>,
        Peter Zijlstra <peterz@...radead.org>,
        Ingo Molnar <mingo@...nel.org>,
        Thomas Gleixner <tglx@...utronix.de>,
        Paul Turner <pjt@...gle.com>,
        Linus Torvalds <torvalds@...ux-foundation.org>,
        Linux List Kernel Mailing <linux-kernel@...r.kernel.org>,
        Frédéric Weisbecker <fweisbec@...il.com>,
        Kees Cook <keescook@...omium.org>,
        Greg Kerr <kerrnel@...gle.com>, Phil Auld <pauld@...hat.com>,
        Valentin Schneider <valentin.schneider@....com>,
        Mel Gorman <mgorman@...hsingularity.net>,
        Pawan Gupta <pawan.kumar.gupta@...ux.intel.com>,
        Paolo Bonzini <pbonzini@...hat.com>
Subject: Re: [RFC PATCH v3 00/16] Core scheduling v3

On Thu, Jul 18, 2019 at 04:27:19PM -0700, Tim Chen wrote:
> 
> 
> On 7/18/19 3:07 AM, Aaron Lu wrote:
> > On Wed, Jun 19, 2019 at 02:33:02PM -0400, Julien Desfossez wrote:
> 
> > 
> > With the below patch on top of v3 that makes use of util_avg to decide
> > which task win, I can do all 8 steps and the final scores of the 2
> > workloads are: 1796191 and 2199586. The score number are not close,
> > suggesting some unfairness, but I can finish the test now...
> 
> Aaron,
> 
> Do you still see high variance in terms of workload throughput that
> was a problem with the previous version?

Any suggestion how to measure this?
It's not clear how Aubrey did his test, will need to take a look at
sysbench.

> >
> >  
> >  }
> > +
> > +bool cfs_prio_less(struct task_struct *a, struct task_struct *b)
> > +{
> > +	struct sched_entity *sea = &a->se;
> > +	struct sched_entity *seb = &b->se;
> > +	bool samecore = task_cpu(a) == task_cpu(b);
> 
> 
> Probably "samecpu" instead of "samecore" will be more accurate.
> I think task_cpu(a) and task_cpu(b)
> can be different, but still belong to the same cpu core.

Right, definitely, guess I'm brain damaged.

> 
> > +	struct task_struct *p;
> > +	s64 delta;
> > +
> > +	if (samecore) {
> > +		/* vruntime is per cfs_rq */
> > +		while (!is_same_group(sea, seb)) {
> > +			int sea_depth = sea->depth;
> > +			int seb_depth = seb->depth;
> > +
> > +			if (sea_depth >= seb_depth)
> 
> Should this be strictly ">" instead of ">=" ?

Same depth doesn't necessarily mean same group while the purpose here is
to make sure they are in the same cfs_rq. When they are of the same
depth but in different cfs_rqs, we will continue to go up till we reach
rq->cfs.

> 
> > +				sea = parent_entity(sea);
> > +			if (sea_depth <= seb_depth)
> 
> Should use "<" ?

Ditto here.
When they are of the same depth but no in the same cfs_rq, both se will
move up.

> > +				seb = parent_entity(seb);
> > +		}
> > +
> > +		delta = (s64)(sea->vruntime - seb->vruntime);
> > +	}
> > +

Thanks.