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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 13:00, Vincent Guittot wrote:
> On Thu, 12 Oct 2023 at 18:48, Mathieu Desnoyers
> <mathieu.desnoyers@efficios.com> wrote:
>>
>> 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
>
> No, the util_est of a cpu only accounts the runnable tasks not the sleeping one
>
> CPU's util_est = /Sum of the util_est of runnable tasks

OK, after further testing, it turns out that cpu_util_without() works
for my case now. I'm not sure what I got wrong in my past attempts.

>
>> 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 ?
>
> util_avg is in the range [0:1024] and CPU's capacity is in the range
> [0:1024] too. 1024 is the compute capacity of the most powerful CPU of
> the system. On SMP system, all CPUs have a capacity of 1024. On
> heterogeneous system, big core have a capacity of 1024 and others will
> have a lower capacity

Sounds good, I will prepare an updated patch.

Thanks,

Mathieu

>
>
>>
>> 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
>>

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

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