Re: [patch v5 14/15] sched: power aware load balance

[Alex Shi]

On 03/22/2013 01:14 PM, Preeti U Murthy wrote:
>> > 
>> > the value get from decay_load():
>> >  sa->runnable_avg_sum = decay_load(sa->runnable_avg_sum,
>> > in decay_load it is possible to be set zero.
> Yes you are right, it is possible to be set to 0, but after a very long
> time, to be more precise, nearly 2 seconds. If you look at decay_load(),
> if the period between last update and now has crossed (32*63),only then
> will the runnable_avg_sum become 0, else it will simply decay.
> 
> This means that for nearly 2seconds,consolidation of loads may not be
> possible even after the runqueues have finished executing tasks running
> on them.

Look into the decay_load(), since the LOAD_AVG_MAX is about 47742, so
after 16 * 32ms, the maximum avg sum will be decay to zero. 2^16 = 65536

Yes, compare to accumulate time 345ms, the decay is not symmetry, and
not precise, seems it has space to tune well. But it is acceptable now.
> 
> The exact experiment that I performed was running ebizzy, with just two
> threads. My setup was 2 socket,2 cores each,4 threads each core. So a 16
> logical cpu machine.When I begin running ebizzy with balance policy, the
> 2 threads of ebizzy are found one on each socket, while I would expect
> them to be on the same socket. All other cpus, except the ones running
> ebizzy threads are idle and not running anything on either socket.
> I am not running any other processes.

did you try the simplest benchmark: while true; do :; done
I am writing the v6 version which include rt_util etc. you may test on
it after I send out. :)
> 
> You could run a similar experiment and let me know if you see otherwise.
> I am at a loss to understand why else would such a spreading of load
> occur, if not for the rq->util not becoming 0 quickly,when it is not
> running anything. I have used trace_printks to keep track of runqueue
> util of those runqueues not running anything after maybe some initial
> load and it does not become 0 till the end of the run.


-- 
Thanks Alex
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