RE: [RFC PATCH v6 3/4] scheduler: scan idle cpu in cluster for tasks within one LLC

[Song Bao Hua (Barry Song)]

> -----Original Message-----
> From: Dietmar Eggemann [mailto:dietmar.eggemann@xxxxxxx]
> Sent: Friday, April 30, 2021 10:43 PM
> To: Song Bao Hua (Barry Song) <song.bao.hua@xxxxxxxxxxxxx>; Vincent Guittot
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> Subject: Re: [RFC PATCH v6 3/4] scheduler: scan idle cpu in cluster for tasks
> within one LLC
> 
> On 29/04/2021 00:41, Song Bao Hua (Barry Song) wrote:
> >
> >
> >> -----Original Message-----
> >> From: Dietmar Eggemann [mailto:dietmar.eggemann@xxxxxxx]
> 
> [...]
> 
> >>>>> From: Dietmar Eggemann [mailto:dietmar.eggemann@xxxxxxx]
> >>
> >> [...]
> >>
> >>>>> On 20/04/2021 02:18, Barry Song wrote:
> 
> [...]
> 
> > Though we will never go to slow path, wake_wide() will affect want_affine,
> > so eventually affect the "new_cpu"?
> 
> yes.
> 
> >
> > 	for_each_domain(cpu, tmp) {
> > 		/*
> > 		 * If both 'cpu' and 'prev_cpu' are part of this domain,
> > 		 * cpu is a valid SD_WAKE_AFFINE target.
> > 		 */
> > 		if (want_affine && (tmp->flags & SD_WAKE_AFFINE) &&
> > 		    cpumask_test_cpu(prev_cpu, sched_domain_span(tmp))) {
> > 			if (cpu != prev_cpu)
> > 				new_cpu = wake_affine(tmp, p, cpu, prev_cpu, sync);
> >
> > 			sd = NULL; /* Prefer wake_affine over balance flags */
> > 			break;
> > 		}
> >
> > 		if (tmp->flags & sd_flag)
> > 			sd = tmp;
> > 		else if (!want_affine)
> > 			break;
> > 	}
> >
> > If wake_affine is false, the above won't execute, new_cpu(target) will
> > always be "prev_cpu"? so when task size > cluster size in wake_wide(),
> > this means we won't pull the wakee to the cluster of waker? It seems
> > sensible.
> 
> What is `task size` here?
> 
> The criterion is `!(slave < factor || master < slave * factor)` or
> `slave >= factor && master >= slave * factor` to wake wide.
> 

Yes. For "task size", I actually mean a bundle of waker-wakee tasks
which can make "slave >= factor && master >= slave * factor" either
true or false, then change the target cpu where we are going to scan
from.
Now since I have moved to cluster level when tasks have been in same
LLC level, it seems it would be more sensible to use "cluster_size" as
factor?

> I see that since you effectively change the sched domain size from LLC
> to CLUSTER (e.g. 24->6) for wakeups with cpu and prev_cpu sharing LLC
> (hence the `numactl -N 0` in your workload), wake_wide() has to take
> CLUSTER size into consideration.
> 
> I was wondering if you saw wake_wide() returning 1 with your use cases:
> 
> numactl -N 0 /usr/lib/lmbench/bin/stream -P [6,12] -M 1024M -N 5

I couldn't make wake_wide return 1 by the above stream command.
And I can't reproduce it by a 1:1(monogamous) hackbench "-f 1".

But I am able to reproduce this issue by a M:N hackbench, for example:

numactl -N 0 hackbench -p -T -f 10 -l 20000 -g 1

hackbench will create 10 senders which will send messages to 10
receivers. (Each sender can send messages to all 10 receivers.)

I've often seen flips like:
waker wakee
1501  39
1509  17
11   1320
13   2016

11, 13, 17 is smaller than LLC but larger than cluster. So the wake_wide()
using cluster factor will return 1, on the other hand, if we always use
llc_size as factor, it will return 0.

However, it seems the change in wake_wide() could bring some negative
influence to M:N relationship(-f 10) according to tests made today by:

numactl -N 0 hackbench -p -T -f 10 -l 20000 -g $1

g             =      1     2       3       4
cluster_size     0.5768 0.6578  0.8117 1.0119
LLC_size         0.5479 0.6162  0.6922 0.7754

Always using llc_size as factor in wake_wide still shows better result
in the 10:10 polygamous hackbench.

So it seems the `slave >= factor && master >= slave * factor` isn't
a suitable criterion for cluster size?

Thanks
Barry