Re: [RFC PATCH] sched/fair: Bias runqueue selection towards almost idle prev CPU

[Vincent Guittot]

On Thu, 12 Oct 2023 at 17:56, Mathieu Desnoyers
<mathieu.desnoyers@xxxxxxxxxxxx> wrote:
>
> On 2023-10-12 11:01, Vincent Guittot wrote:
> > On Thu, 12 Oct 2023 at 16:33, Mathieu Desnoyers
> > <mathieu.desnoyers@xxxxxxxxxxxx> wrote:
> >>
> >> On 2023-10-11 06:16, Chen Yu wrote:
> >>> On 2023-10-10 at 09:49:54 -0400, Mathieu Desnoyers wrote:
> >>>> On 2023-10-09 01:14, Chen Yu wrote:
> >>>>> On 2023-09-30 at 07:45:38 -0400, Mathieu Desnoyers wrote:
> >>>>>> On 9/30/23 03:11, Chen Yu wrote:
> >>>>>>> Hi Mathieu,
> >>>>>>>
> >>>>>>> On 2023-09-29 at 14:33:50 -0400, Mathieu Desnoyers wrote:
> >>>>>>>> Introduce the WAKEUP_BIAS_PREV_IDLE scheduler feature. It biases
> >>>>>>>> select_task_rq towards the previous CPU if it was almost idle
> >>>>>>>> (avg_load <= 0.1%).
> >>>>>>>
> >>>>>>> Yes, this is a promising direction IMO. One question is that,
> >>>>>>> can cfs_rq->avg.load_avg be used for percentage comparison?
> >>>>>>> If I understand correctly, load_avg reflects that more than
> >>>>>>> 1 tasks could have been running this runqueue, and the
> >>>>>>> load_avg is the direct proportion to the load_weight of that
> >>>>>>> cfs_rq. Besides, LOAD_AVG_MAX seems to not be the max value
> >>>>>>> that load_avg can reach, it is the sum of
> >>>>>>> 1024 * (y + y^1 + y^2 ... )
> >>>>>>>
> >>>>>>> For example,
> >>>>>>> taskset -c 1 nice -n -20 stress -c 1
> >>>>>>> cat /sys/kernel/debug/sched/debug | grep 'cfs_rq\[1\]' -A 12 | grep "\.load_avg"
> >>>>>>>       .load_avg                      : 88763
> >>>>>>>       .load_avg                      : 1024
> >>>>>>>
> >>>>>>> 88763 is higher than LOAD_AVG_MAX=47742
> >>>>>>
> >>>>>> I would have expected the load_avg to be limited to LOAD_AVG_MAX somehow,
> >>>>>> but it appears that it does not happen in practice.
> >>>>>>
> >>>>>> That being said, if the cutoff is really at 0.1% or 0.2% of the real max,
> >>>>>> does it really matter ?
> >>>>>>
> >>>>>>> Maybe the util_avg can be used for precentage comparison I suppose?
> >>>>>> [...]
> >>>>>>> Or
> >>>>>>> return cpu_util_without(cpu_rq(cpu), p) * 1000 <= capacity_orig_of(cpu) ?
> >>>>>>
> >>>>>> Unfortunately using util_avg does not seem to work based on my testing.
> >>>>>> Even at utilization thresholds at 0.1%, 1% and 10%.
> >>>>>>
> >>>>>> Based on comments in fair.c:
> >>>>>>
> >>>>>>     * CPU utilization is the sum of running time of runnable tasks plus the
> >>>>>>     * recent utilization of currently non-runnable tasks on that CPU.
> >>>>>>
> >>>>>> I think we don't want to include currently non-runnable tasks in the
> >>>>>> statistics we use, because we are trying to figure out if the cpu is a
> >>>>>> idle-enough target based on the tasks which are currently running, for the
> >>>>>> purpose of runqueue selection when waking up a task which is considered at
> >>>>>> that point in time a non-runnable task on that cpu, and which is about to
> >>>>>> become runnable again.
> >>>>>>
> >>>>>
> >>>>> Although LOAD_AVG_MAX is not the max possible load_avg, we still want to find
> >>>>> a proper threshold to decide if the CPU is almost idle. The LOAD_AVG_MAX
> >>>>> based threshold is modified a little bit:
> >>>>>
> >>>>> The theory is, if there is only 1 task on the CPU, and that task has a nice
> >>>>> of 0, the task runs 50 us every 1000 us, then this CPU is regarded as almost
> >>>>> idle.
> >>>>>
> >>>>> The load_sum of the task is:
> >>>>> 50 * (1 + y + y^2 + ... + y^n)
> >>>>> The corresponding avg_load of the task is approximately
> >>>>> NICE_0_WEIGHT * load_sum / LOAD_AVG_MAX = 50.
> >>>>> So:
> >>>>>
> >>>>> /* which is close to LOAD_AVG_MAX/1000 = 47 */
> >>>>> #define ALMOST_IDLE_CPU_LOAD   50
> >>>>
> >>>> Sorry to be slow at understanding this concept, but this whole "load" value
> >>>> is still somewhat magic to me.
> >>>>
> >>>> Should it vary based on CONFIG_HZ_{100,250,300,1000}, or is it independent ?
> >>>> Where is it documented that the load is a value in "us" out of a window of
> >>>> 1000 us ?
> >>>>
> >>>
> >>> My understanding is that, the load_sum of a single task is a value in "us" out
> >>> of a window of 1000 us, while the load_avg of the task will multiply the weight
> >>> of the task. In this case a task with nice 0 is NICE_0_WEIGHT = 1024.
> >>>
> >>> __update_load_avg_se -> ___update_load_sum calculate the load_sum of a task(there
> >>> is comments around ___update_load_sum to describe the pelt calculation),
> >>> and ___update_load_avg() calculate the load_avg based on the task's weight.
> >>
> >> Thanks for your thorough explanation, now it makes sense.
> >>
> >> I understand as well that the cfs_rq->avg.load_sum is the result of summing
> >> each task load_sum multiplied by their weight:
> >
> > Please don't use load_sum but only *_avg.
> > As already said, util_avg or runnable_avg are better metrics for you
>
> I think I found out why using util_avg was not working for me.
>
> Considering this comment from cpu_util():
>
>   * CPU utilization is the sum of running time of runnable tasks plus the
>   * recent utilization of currently non-runnable tasks on that CPU.
>
> I don't want to include the recent utilization of currently non-runnable
> tasks on that CPU in order to choose that CPU to do task placement in a
> context where many tasks were recently running on that cpu (but are
> currently blocked). I do not want those blocked tasks to be part of the
> avg.

But you have the exact same behavior with load_sum/avg.

>
> So I think the issue here is that I was using the cpu_util() (and
> cpu_util_without()) helpers which are considering max(util, runnable),
> rather than just "util".

cpu_util_without() only use util_avg but not runnable_avg.
Nevertheless, cpu_util_without ans cpu_util uses util_est which is
used to predict the final utilization.

Let's take the example of task A running 20ms every 200ms on CPU0.
The util_avg of the cpu will vary in the range [7:365]. When task A
wakes up on CPU0, CPU0 util_avg = 7 (below 1%) but taskA will run for
20ms which is not really almost idle. On the other side, CPU0 util_est
will be 365 as soon as task A is enqueued (which will be the value of
CPU0 util_avg just before going idle)

Let's now take a task B running 100us every 1024us
The util_avg of the cpu should vary in the range [101:103] and once
task B is enqueued, CPU0 util_est will be 103

>
> Based on your comments, just doing this to match a rq util_avg <= 1% (10us of 1024us)

it's not 10us of 1024us

> seems to work fine:
>
>    return cpu_rq(cpu)->cfs.avg.util_avg <= 10 * capacity_of(cpu);
>
> Is this approach acceptable ?
>
> Thanks!
>
> Mathieu
>
> >
> >>
> >> static inline void
> >> enqueue_load_avg(struct cfs_rq *cfs_rq, struct sched_entity *se)
> >> {
> >>           cfs_rq->avg.load_avg += se->avg.load_avg;
> >>           cfs_rq->avg.load_sum += se_weight(se) * se->avg.load_sum;
> >> }
> >>
> >> Therefore I think we need to multiply the load_sum value we aim for by
> >> get_pelt_divider(&cpu_rq(cpu)->cfs.avg) to compare it to a rq load_sum.
> >>
> >> I plan to compare the rq load sum to "10 * get_pelt_divider(&cpu_rq(cpu)->cfs.avg)"
> >> to match runqueues which were previously idle (therefore with prior periods contribution
> >> to the rq->load_sum being pretty much zero), and which have a current period rq load_sum
> >> below or equal 10us per 1024us (<= 1%):
> >>
> >> static inline unsigned long cfs_rq_weighted_load_sum(struct cfs_rq *cfs_rq)
> >> {
> >>           return cfs_rq->avg.load_sum;
> >> }
> >>
> >> static unsigned long cpu_weighted_load_sum(struct rq *rq)
> >> {
> >>           return cfs_rq_weighted_load_sum(&rq->cfs);
> >> }
> >>
> >> /*
> >>    * A runqueue is considered almost idle if:
> >>    *
> >>    *   cfs_rq->avg.load_sum / get_pelt_divider(&cfs_rq->avg) / 1024 <= 1%
> >>    *
> >>    * This inequality is transformed as follows to minimize arithmetic:
> >>    *
> >>    *   cfs_rq->avg.load_sum <= get_pelt_divider(&cfs_rq->avg) * 10
> >>    */
> >> static bool
> >> almost_idle_cpu(int cpu, struct task_struct *p)
> >> {
> >>           if (!sched_feat(WAKEUP_BIAS_PREV_IDLE))
> >>                   return false;
> >>           return cpu_weighted_load_sum(cpu_rq(cpu)) <= 10 * get_pelt_divider(&cpu_rq(cpu)->cfs.avg);
> >> }
> >>
> >> Does it make sense ?
> >>
> >> Thanks,
> >>
> >> Mathieu
> >>
> >>
> >>>
> >>>> And with this value "50", it would cover the case where there is only a
> >>>> single task taking less than 50us per 1000us, and cases where the sum for
> >>>> the set of tasks on the runqueue is taking less than 50us per 1000us
> >>>> overall.
> >>>>
> >>>>>
> >>>>> static bool
> >>>>> almost_idle_cpu(int cpu, struct task_struct *p)
> >>>>> {
> >>>>>           if (!sched_feat(WAKEUP_BIAS_PREV_IDLE))
> >>>>>                   return false;
> >>>>>           return cpu_load_without(cpu_rq(cpu), p) <= ALMOST_IDLE_CPU_LOAD;
> >>>>> }
> >>>>>
> >>>>> Tested this on Intel Xeon Platinum 8360Y, Ice Lake server, 36 core/package,
> >>>>> total 72 core/144 CPUs. Slight improvement is observed in hackbench socket mode:
> >>>>>
> >>>>> socket mode:
> >>>>> hackbench -g 16 -f 20 -l 480000 -s 100
> >>>>>
> >>>>> Before patch:
> >>>>> Running in process mode with 16 groups using 40 file descriptors each (== 640 tasks)
> >>>>> Each sender will pass 480000 messages of 100 bytes
> >>>>> Time: 81.084
> >>>>>
> >>>>> After patch:
> >>>>> Running in process mode with 16 groups using 40 file descriptors each (== 640 tasks)
> >>>>> Each sender will pass 480000 messages of 100 bytes
> >>>>> Time: 78.083
> >>>>>
> >>>>>
> >>>>> pipe mode:
> >>>>> hackbench -g 16 -f 20 --pipe  -l 480000 -s 100
> >>>>>
> >>>>> Before patch:
> >>>>> Running in process mode with 16 groups using 40 file descriptors each (== 640 tasks)
> >>>>> Each sender will pass 480000 messages of 100 bytes
> >>>>> Time: 38.219
> >>>>>
> >>>>> After patch:
> >>>>> Running in process mode with 16 groups using 40 file descriptors each (== 640 tasks)
> >>>>> Each sender will pass 480000 messages of 100 bytes
> >>>>> Time: 38.348
> >>>>>
> >>>>> It suggests that, if the workload has larger working-set/cache footprint, waking up
> >>>>> the task on its previous CPU could get more benefit.
> >>>>
> >>>> In those tests, what is the average % of idleness of your cpus ?
> >>>>
> >>>
> >>> For hackbench -g 16 -f 20 --pipe  -l 480000 -s 100, it is around 8~10% idle
> >>> For hackbench -g 16 -f 20   -l 480000 -s 100, it is around 2~3% idle
> >>>
> >>> Then the CPUs in packge 1 are offlined to get stable result when the group number is low.
> >>> hackbench -g 1 -f 20 --pipe  -l 480000 -s 100
> >>> Some CPUs are busy, others are idle, and some are half-busy.
> >>> Core  CPU     Busy%
> >>> -     -       49.57
> >>> 0     0       1.89
> >>> 0     72      75.55
> >>> 1     1       100.00
> >>> 1     73      0.00
> >>> 2     2       100.00
> >>> 2     74      0.00
> >>> 3     3       100.00
> >>> 3     75      0.01
> >>> 4     4       78.29
> >>> 4     76      17.72
> >>> 5     5       100.00
> >>> 5     77      0.00
> >>>
> >>>
> >>> hackbench -g 1 -f 20  -l 480000 -s 100
> >>> Core  CPU     Busy%
> >>> -     -       48.29
> >>> 0     0       57.94
> >>> 0     72      21.41
> >>> 1     1       83.28
> >>> 1     73      0.00
> >>> 2     2       11.44
> >>> 2     74      83.38
> >>> 3     3       21.45
> >>> 3     75      77.27
> >>> 4     4       26.89
> >>> 4     76      80.95
> >>> 5     5       5.01
> >>> 5     77      83.09
> >>>
> >>>
> >>> echo NO_WAKEUP_BIAS_PREV_IDLE > /sys/kernel/debug/sched/features
> >>> hackbench -g 1 -f 20 --pipe  -l 480000 -s 100
> >>> Running in process mode with 1 groups using 40 file descriptors each (== 40 tasks)
> >>> Each sender will pass 480000 messages of 100 bytes
> >>> Time: 9.434
> >>>
> >>> echo WAKEUP_BIAS_PREV_IDLE > /sys/kernel/debug/sched/features
> >>> hackbench -g 1 -f 20 --pipe  -l 480000 -s 100
> >>> Running in process mode with 1 groups using 40 file descriptors each (== 40 tasks)
> >>> Each sender will pass 480000 messages of 100 bytes
> >>> Time: 9.373
> >>>
> >>> thanks,
> >>> Chenyu
> >>
> >> --
> >> Mathieu Desnoyers
> >> EfficiOS Inc.
> >> https://www.efficios.com
> >>
>
> --
> Mathieu Desnoyers
> EfficiOS Inc.
> https://www.efficios.com
>