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SubjectRe: [RFC PATCH] sched/fair: Bias runqueue selection towards almost idle prev CPU
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On 2023-10-12 12:24, Vincent Guittot wrote:
> On Thu, 12 Oct 2023 at 17:56, Mathieu Desnoyers
> <mathieu.desnoyers@efficios.com> wrote:
>>
>> On 2023-10-12 11:01, Vincent Guittot wrote:
>>> On Thu, 12 Oct 2023 at 16:33, Mathieu Desnoyers
>>> <mathieu.desnoyers@efficios.com> 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.

Ah, yes, @boost=0, which prevents it from using the runnable_avg.

> Nevertheless, cpu_util_without ans cpu_util uses util_est which is
> used to predict the final utilization.

Yes, I suspect it's the util_est which prevents me from getting
performance improvements when I use cpu_util_without to implement
almost-idle.

>
> 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)

If task A sleeps (becomes non-runnable) without being migrated, and therefore
still have CPU0 as its cpu, is it still considered as part of the util_est of
CPU0 while it is blocked ? If it is the case, then the util_est is preventing
rq selection from considering a rq almost idle when waking up sleeping tasks
due to taking into account the set of sleeping tasks in its utilization estimate.

>
> 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

Is the range of util_avg within [0..1024] * capacity_of(cpu), or am I missing something ?

Thanks,

Mathieu


>
>> 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
>>

--
Mathieu Desnoyers
EfficiOS Inc.
https://www.efficios.com

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