Messages in this thread | | | Date | Thu, 12 Oct 2023 10:33:56 -0400 | Subject | Re: [RFC PATCH] sched/fair: Bias runqueue selection towards almost idle prev CPU | From | Mathieu Desnoyers <> |
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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:
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
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