Messages in this thread Patch in this message | | | Subject | [PATCH] sched: fix sched-domain avg_load calculation. | Date | Thu, 7 Apr 2011 17:23:22 -0700 (PDT) | From | (Ken Chen) |
| |
In function find_busiest_group(), the sched-domain avg_load isn't calculated at all if there is a group imbalance within the domain. This will cause erroneous imbalance calculation. The reason is that calculate_imbalance() sees sds->avg_load = 0 and it will dump entire sds->max_load into imbalance variable, which is used later on to migrate entire load from busiest CPU to the puller CPU. It has two really bad effect:
1. stampede of task migration, and they won't be able to break out of the bad state because of positive feedback loop: large load delta -> heavier load migration -> larger imbalance and the cycle goes on.
2. severe imbalance in CPU queue depth. This causes really long scheduling latency blip which affects badly on application that has tight latency requirement.
The fix is to have kernel calculate domain avg_load in both cases. This will ensure that imbalance calculation is always sensible and the target is usually half way between busiest and puller CPU.
Signed-off-by: Ken Chen <kenchen@google.com> --- kernel/sched_fair.c | 3 ++- 1 files changed, 2 insertions(+), 1 deletions(-)
diff --git a/kernel/sched_fair.c b/kernel/sched_fair.c index c7ec5c8..c46568a 100644 --- a/kernel/sched_fair.c +++ b/kernel/sched_fair.c @@ -3127,6 +3127,8 @@ find_busiest_group( if (!sds.busiest || sds.busiest_nr_running == 0) goto out_balanced; + sds.avg_load = (SCHED_LOAD_SCALE * sds.total_load) / sds.total_pwr; + /* * If the busiest group is imbalanced the below checks don't * work because they assumes all things are equal, which typically @@ -3151,7 +3153,6 @@ find_busiest_group( * Don't pull any tasks if this group is already above the domain * average load. */ - sds.avg_load = (SCHED_LOAD_SCALE * sds.total_load) / sds.total_pwr; if (sds.this_load >= sds.avg_load) goto out_balanced; -- 1.7.3.1
| |