On Wed, Oct 31, 2007 at 10:10:32PM +0100, Peter Zijlstra wrote:
> Currently the ideal slice length does not take group scheduling into account.
> Change it so that it properly takes all the runnable tasks on this cpu into
> account and caluclate the weight according to the grouping hierarchy.
>
> Also fixes a bug in vslice which missed a factor NICE_0_LOAD.
>
> Signed-off-by: Peter Zijlstra <[EMAIL PROTECTED]>
> CC: Srivatsa Vaddagiri <[EMAIL PROTECTED]>
> ---
> kernel/sched_fair.c | 42 +++++++++++++++++++++++++++++++-----------
> 1 file changed, 31 insertions(+), 11 deletions(-)
>
> Index: linux-2.6/kernel/sched_fair.c
> ===================================================================
> --- linux-2.6.orig/kernel/sched_fair.c
> +++ linux-2.6/kernel/sched_fair.c
> @@ -331,10 +331,15 @@ static u64 __sched_period(unsigned long
> */
> static u64 sched_slice(struct cfs_rq *cfs_rq, struct sched_entity *se)
> {
> - u64 slice = __sched_period(cfs_rq->nr_running);
> + unsigned long nr_running = rq_of(cfs_rq)->nr_running;
> + u64 slice = __sched_period(nr_running);
>
> - slice *= se->load.weight;
> - do_div(slice, cfs_rq->load.weight);
> + for_each_sched_entity(se) {
> + cfs_rq = cfs_rq_of(se);
> +
> + slice *= se->load.weight;
> + do_div(slice, cfs_rq->load.weight);
> + }
>
> return slice;
Lets say we have two groups A and B on CPU0, of equal weight (1024).
Further,
A has 1 task (A0)
B has 1000 tasks (B0 .. B999)
Agreed its a extreme case, but illustrates the problem I have in mind
for this patch.
All tasks of same weight=1024.
Before this patch
=================
sched_slice(grp A) = 20ms * 1/2 = 10ms
sched_slice(A0) = 20ms
sched_slice(grp B) = 20ms * 1/2 = 10ms
sched_slice(B0) = (20ms * 1000/20) * 1 / 1000 = 1ms
sched_slice(B1) = ... = sched_slice(B99) = 1 ms
Fairness between groups and tasks would be obtained as below:
A0 B0-B9 A0 B10-B19 A0 B20-B29
|--------|--------|--------|--------|--------|--------|-----//--|
0 10ms 20ms 30ms 40ms 50ms 60ms
After this patch
================
sched_slice(grp A) = (20ms * 1001/20) * 1/2 ~= 500ms
sched_slice(A0) = 500ms
sched_slice(grp B) = 500ms
sched_slice(B0) = 0.5ms
Fairness between groups and tasks would be obtained as below:
A0 B0 - B99 A0
|-----------------------|-----------------------|-----------------------|
0 500ms 1000ms 1500ms
Did I get it right? If so, I don't like the fact that group A is allowed to run
for a long time (500ms) before giving chance to group B.
Can I know what real problem is being addressed by this change to
sched_slice()?
--
Regards,
vatsa
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