Re: [PATCH 4/5] sched: don't consider upper se in sched_slice()

[Preeti U Murthy <preeti@linux.vnet.ibm.com>]

Hi Joonsoo,

On 03/28/2013 01:28 PM, Joonsoo Kim wrote:
> Following-up upper se in sched_slice() should not be done,
> because sched_slice() is used for checking that resched is needed
> whithin *this* cfs_rq and there is one problem related to this
> in current implementation.
> 
> The problem is that if we follow-up upper se in sched_slice(), it is
> possible that we get a ideal slice which is lower than
> sysctl_sched_min_granularity.
> 
> For example, we assume that we have 4 tg which is attached to root tg
> with same share and each one have 20 runnable tasks on cpu0, respectivly.
> In this case, __sched_period() return sysctl_sched_min_granularity * 20
> and then go into loop. At first iteration, we compute a portion of slice
> for this task on this cfs_rq, so get a slice, sysctl_sched_min_granularity.
> Afterward, we enter second iteration and get a slice which is a quarter of
> sysctl_sched_min_granularity, because there is 4 tgs with same share
> in that cfs_rq.
> 
> Ensuring slice larger than min_granularity is important for performance
> and there is no lower bound about this, except timer tick, we should
> fix it not to consider upper se when calculating sched_slice.
> 

I am assuming the sysctl_sched_latency = 20ms and
sysctl_sched_min_granularity = 4ms.
In that case:

With your patch, the sum of the sched_slice(runnable_task) on each_tg is
 40ms = sched_min_granularity * 10, while the parent tg has a
sched_slice of sysctl_sched_latency / 4 = (20 / 4) = 5ms.

Ideally the children's cpu share must add upto the parent's share.


Thank you

Regards
Preeti U Murthy