linux-kernel - Re: [RFC PATCH v2 11/17] sched: Basic tracking of matching tasks

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Message-ID: <20190510130624.GW2589@hirez.programming.kicks-ass.net>
Date:   Fri, 10 May 2019 15:06:24 +0200
From:   Peter Zijlstra <peterz@...radead.org>
To:     Aaron Lu <aaron.lu@...ux.alibaba.com>
Cc:     Vineeth Remanan Pillai <vpillai@...italocean.com>,
        Nishanth Aravamudan <naravamudan@...italocean.com>,
        Julien Desfossez <jdesfossez@...italocean.com>,
        Tim Chen <tim.c.chen@...ux.intel.com>, mingo@...nel.org,
        tglx@...utronix.de, pjt@...gle.com, torvalds@...ux-foundation.org,
        linux-kernel@...r.kernel.org, subhra.mazumdar@...cle.com,
        fweisbec@...il.com, keescook@...omium.org, kerrnel@...gle.com,
        Phil Auld <pauld@...hat.com>, Aaron Lu <aaron.lwe@...il.com>,
        Aubrey Li <aubrey.intel@...il.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 v2 11/17] sched: Basic tracking of matching tasks

On Mon, Apr 29, 2019 at 11:36:22AM +0800, Aaron Lu wrote:
> On Tue, Apr 23, 2019 at 04:18:16PM +0000, Vineeth Remanan Pillai wrote:
> > +/*
> > + * l(a,b)
> > + * le(a,b) := !l(b,a)
> > + * g(a,b)  := l(b,a)
> > + * ge(a,b) := !l(a,b)
> > + */
> > +
> > +/* real prio, less is less */
> > +static inline bool __prio_less(struct task_struct *a, struct task_struct *b, bool core_cmp)
> > +{
> > +	u64 vruntime;
> > +
> > +	int pa = __task_prio(a), pb = __task_prio(b);
> > +
> > +	if (-pa < -pb)
> > +		return true;
> > +
> > +	if (-pb < -pa)
> > +		return false;
> > +
> > +	if (pa == -1) /* dl_prio() doesn't work because of stop_class above */
> > +		return !dl_time_before(a->dl.deadline, b->dl.deadline);
> > +
> > +	vruntime = b->se.vruntime;
> > +	if (core_cmp) {
> > +		vruntime -= task_cfs_rq(b)->min_vruntime;
> > +		vruntime += task_cfs_rq(a)->min_vruntime;
> > +	}
> > +	if (pa == MAX_RT_PRIO + MAX_NICE) /* fair */
> > +		return !((s64)(a->se.vruntime - vruntime) <= 0);
> > +
> > +	return false;
> > +}
> 
> This unfortunately still doesn't work.
> 
> Consider the following task layout on two sibling CPUs(cpu0 and cpu1):
> 
>     rq0.cfs_rq    rq1.cfs_rq
>         |             |
>      se_bash        se_hog
> 
> se_hog is the sched_entity for a cpu intensive task and se_bash is the
> sched_entity for bash.
> 
> There are two problems:
> 1 SCHED_DEBIT
> when user execute some commands through bash, say ls, bash will fork.
> The newly forked ls' vruntime is set in the future due to SCHED_DEBIT.
> This made 'ls' lose in __prio_less() when compared with hog, whose
> vruntime may very likely be the same as its cfs_rq's min_vruntime.
> 
> This is OK since we do not want forked process to starve already running
> ones. The problem is, since hog keeps running, its vruntime will always
> sync with its cfs_rq's min_vruntime. OTOH, 'ls' can not run, its
> cfs_rq's min_vruntime doesn't proceed, making 'ls' always lose to hog.
> 
> 2 who schedules, who wins
> so I disabled SCHED_DEBIT, for testing's purpose. When cpu0 schedules,
> ls could win where both sched_entity's vruntime is the same as their
> cfs_rqs' min_vruntime. So does hog: when cpu1 schedules, hog can preempt
> ls in the same way. The end result is, interactive task can lose to cpu
> intensive task and ls can feel "dead".
> 
> I haven't figured out a way to solve this yet. A core wide cfs_rq's
> min_vruntime can probably solve this. Your suggestions are appreciated.

multi-queue virtual time is 'interesting'. I worked it out once and then
my head hurt, I've forgotten the details again. Esp. when combined with
affinity masks the simple things don't work right. For every
non-feasible weight scenario it comes apart.

I know pjt has an approximation somewhere that might work for us; but I
forgot those details again too.

On possible hack would be to allow min_vruntime to go backwards when
there is only a single task present; basically have min_vruntime =
p->vruntime when you enqueue the first task.