Messages in this thread | | | Date | Thu, 24 Apr 2014 19:14:53 +0200 | From | Peter Zijlstra <> | Subject | Re: [PATCH 1/3] sched, balancing: Update rq->max_idle_balance_cost whenever newidle balance is attempted |
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On Thu, Apr 24, 2014 at 09:53:37AM -0700, Jason Low wrote: > > So I thought that the original rationale (commit 1bd77f2d) behind > updating rq->next_balance in idle_balance() is that, if we are going > idle (!pulled_task), we want to ensure that the next_balance gets > calculated without the busy_factor. > > If the rq is busy, then rq->next_balance gets updated based on > sd->interval * busy_factor. However, when the rq goes from "busy" > to idle, rq->next_balance might still have been calculated under > the assumption that the rq is busy. Thus, if we are going idle, we > would then properly update next_balance without the busy factor > if we update when !pulled_task. >
Its late here and I'm confused!
So the for_each_domain() loop calculates a new next_balance based on ->balance_interval (which has that busy_factor on, right).
But if it fails to pull anything, we'll (potentially) iterate the entire tree up to the largest domain; and supposedly set next_balanced to the largest possible interval.
So when we go from busy to idle (!pulled_task), we actually set ->next_balance to the longest interval. Whereas the commit you referenced says it sets it to a shorter while.
Not seeing it.
So the code as modified by Ingo in one of the initial CFS commits, will move the ->next_balance time ahead if the balance succeeded (pulled_task), thereby reflecting that we are busy and we just did a balance so we need not do one again soon. (we might want to re-think this if we really make the idle balance only pull 1 task max).
Of course, I've now gone over this code 3 times today, so I'm terminally confused.
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