Re: [PATCH -next v2 2/5] block, bfq: add fake weight_counter for weight-raised queue

[Jan Kara]

On Tue 26-04-22 16:27:46, yukuai (C) wrote:
> 在 2022/04/26 15:40, Jan Kara 写道:
> > On Tue 26-04-22 09:49:04, yukuai (C) wrote:
> > > 在 2022/04/26 0:16, Jan Kara 写道:
> > > > Hello!
> > > > 
> > > > On Mon 25-04-22 21:34:16, yukuai (C) wrote:
> > > > > 在 2022/04/25 17:48, Jan Kara 写道:
> > > > > > On Sat 16-04-22 17:37:50, Yu Kuai wrote:
> > > > > > > Weight-raised queue is not inserted to weights_tree, which makes it
> > > > > > > impossible to track how many queues have pending requests through
> > > > > > > weights_tree insertion and removel. This patch add fake weight_counter
> > > > > > > for weight-raised queue to do that.
> > > > > > > 
> > > > > > > Signed-off-by: Yu Kuai <yukuai3@xxxxxxxxxx>
> > > > > > 
> > > > > > This is a bit hacky. I was looking into a better place where to hook to
> > > > > > count entities in a bfq_group with requests and I think bfq_add_bfqq_busy()
> > > > > > and bfq_del_bfqq_busy() are ideal for this. It also makes better sense
> > > > > > conceptually than hooking into weights tree handling.
> > > > > 
> > > > > bfq_del_bfqq_busy() will be called when all the reqs in the bfqq are
> > > > > dispatched, however there might still some reqs are't completed yet.
> > > > > 
> > > > > Here what we want to track is how many bfqqs have pending reqs,
> > > > > specifically if the bfqq have reqs are't complted.
> > > > > 
> > > > > Thus I think bfq_del_bfqq_busy() is not the right place to do that.
> > > > 
> > > > Yes, I'm aware there will be a difference. But note that bfqq can stay busy
> > > > with only dispatched requests because the logic in __bfq_bfqq_expire() will
> > > > not call bfq_del_bfqq_busy() if idling is needed for service guarantees. So
> > > > I think using bfq_add/del_bfqq_busy() would work OK.
> > > Hi,
> > > 
> > > I didn't think of that before. If bfqq stay busy after dispathing all
> > > the requests, there are two other places that bfqq can clear busy:
> > > 
> > > 1) bfq_remove_request(), bfqq has to insert a new req while it's not in
> > > service.
> > 
> > Yes and the request then would have to be dispatched or merged. Which
> > generally means another bfqq from the same bfqg is currently active and
> > thus this should have no impact on service guarantees we are interested in.
> > 
> > > 2) bfq_release_process_ref(), user thread is gone / moved, or old bfqq
> > > is gone due to merge / ioprio change.
> > 
> > Yes, here there's no new IO for the bfqq so no point in maintaining any
> > service guarantees to it.
> > 
> > > I wonder, will bfq_del_bfqq_busy() be called immediately when requests
> > > are completed? (It seems not to me...). For example, a user thread
> > > issue a sync io just once, and it keep running without issuing new io,
> > > then when does the bfqq clears the busy state?
> > 
> > No, when bfqq is kept busy, it will get scheduled as in-service queue in
> > the future. Then what happens depends on whether it will get more requests
> > or not. But generally its busy state will get cleared once it is expired
> > for other reason than preemption.
> 
> Thanks for your explanation.
> 
> I think in normal case using bfq_add/del_bfqq_busy() if fine.
> 
> There is one last situation that I'm worried: If some disk are very
> slow that the dispatched reqs are not completed when the bfqq is
> rescheduled as in-service queue, and thus busy state can be cleared
> while reqs are not completed.
> 
> Using bfq_del_bfqq_busy() will change behaviour in this specail case,
> do you think service guarantees will be broken?

Well, I don't think so. Because slow disks don't tend to do a lot of
internal scheduling (or have deep IO queues for that matter). Also note
that generally bfq_select_queue() will not even expire a queue (despite it
not having any requests to dispatch) when we should not dispatch other
requests to maintain service guarantees. So I think service guarantees will
be generally preserved. Obviously I could be wrong, we we will not know
until we try it :).

								Honza

-- 
Jan Kara <jack@xxxxxxxx>
SUSE Labs, CR