Re: [RFC PATCH v6 3/4] scheduler: scan idle cpu in cluster for tasks within one LLC
From: Dietmar Eggemann
Date: Fri Apr 30 2021 - 06:43:12 EST
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.
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