Thread (108 messages) 108 messages, 8 authors, 2014-11-24

[PATCH v2 08/11] sched: get CPU's activity statistic

From: Morten Rasmussen <hidden>
Date: 2014-06-04 10:26:07
Also in: lkml

On Wed, Jun 04, 2014 at 10:35:15AM +0100, Vincent Guittot wrote:
On 4 June 2014 11:23, Peter Zijlstra [off-list ref] wrote:
quoted
On Wed, Jun 04, 2014 at 09:55:42AM +0100, Morten Rasmussen wrote:
quoted
Both running_avg and runnable_avg are affected by other tasks on the
same cpus, but in different ways. They are equal if you only have one
task on a cpu. If you have more, running_avg will give you the true
requirement of the tasks until the cpu is fully utilized. At which point
the task running_avg will drop if you add more tasks (the unweighted sum
of task running_avgs remains constant).

runnable_avg on the other hand, might be affected as soon as you have
two task running on the same cpu if they are runnable at the same time.
That isn't necessarily a bad thing for load-balancing purposes, because
tasks that are runnable at the same time are likely to be run more
efficiently by placing them on different cpus. You might view as at sort
of built in concurrency factor, somewhat similar to what Yuyang is
proposing. runnable_avg increases rapidly when the cpu is over-utilized.
Agreed.
quoted
quoted
I'm not sure I see how 100% is possible, but yes I agree that runnable
can indeed be inflated due to this queueing effect.
You should only be able to get to 75% worst case for runnable_avg for
that example. The total running_avg is 50% no matter if the tasks
overlaps or not.
Yes, 75% is what I ended up with.
Can you explain how you reach 75% as it depends on the runtime and a
runtime longer than 345ms will end to a 100% load whatever the
idletime was previously ?
If the busy period of each task is long enough that the first one to run
runs to completetion before the other task is scheduled you get 25% and
50%. But as I said in my other reply, you can get higher task runnable
if the tasks busy period is long enough that you switch between them
before the first one completes.
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