Re: [PATCH BUGFIX/IMPROVEMENT 2/6] block, bfq: put reqs of waker and woken in dispatch list

[Paolo Valente <paolo.valente@linaro.org>]

> Il giorno 3 feb 2021, alle ore 12:43, Jan Kara <jack@suse.cz> ha scritto:
> 
> On Thu 28-01-21 18:54:05, Paolo Valente wrote:
>> 
>> 
>>> Il giorno 26 gen 2021, alle ore 17:18, Jens Axboe <axboe@kernel.dk> ha scritto:
>>> 
>>> On 1/26/21 3:50 AM, Paolo Valente wrote:
>>>> Consider a new I/O request that arrives for a bfq_queue bfqq. If, when
>>>> this happens, the only active bfq_queues are bfqq and either its waker
>>>> bfq_queue or one of its woken bfq_queues, then there is no point in
>>>> queueing this new I/O request in bfqq for service. In fact, the
>>>> in-service queue and bfqq agree on serving this new I/O request as
>>>> soon as possible. So this commit puts this new I/O request directly
>>>> into the dispatch list.
>>>> 
>>>> Tested-by: Jan Kara <jack@suse.cz>
>>>> Signed-off-by: Paolo Valente <paolo.valente@linaro.org>
>>>> ---
>>>> block/bfq-iosched.c | 17 ++++++++++++++++-
>>>> 1 file changed, 16 insertions(+), 1 deletion(-)
>>>> 
>>>> diff --git a/block/bfq-iosched.c b/block/bfq-iosched.c
>>>> index a83149407336..e5b83910fbe0 100644
>>>> --- a/block/bfq-iosched.c
>>>> +++ b/block/bfq-iosched.c
>>>> @@ -5640,7 +5640,22 @@ static void bfq_insert_request(struct blk_mq_hw_ctx *hctx, struct request *rq,
>>>> 
>>>> 	spin_lock_irq(&bfqd->lock);
>>>> 	bfqq = bfq_init_rq(rq);
>>>> -	if (!bfqq || at_head || blk_rq_is_passthrough(rq)) {
>>>> +
>>>> +	/*
>>>> +	 * Additional case for putting rq directly into the dispatch
>>>> +	 * queue: the only active bfq_queues are bfqq and either its
>>>> +	 * waker bfq_queue or one of its woken bfq_queues. In this
>>>> +	 * case, there is no point in queueing rq in bfqq for
>>>> +	 * service. In fact, the in-service queue and bfqq agree on
>>>> +	 * serving this new I/O request as soon as possible.
>>>> +	 */
>>>> +	if (!bfqq ||
>>>> +	    (bfqq != bfqd->in_service_queue &&
>>>> +	     bfqd->in_service_queue != NULL &&
>>>> +	     bfq_tot_busy_queues(bfqd) == 1 + bfq_bfqq_busy(bfqq) &&
>>>> +	     (bfqq->waker_bfqq == bfqd->in_service_queue ||
>>>> +	      bfqd->in_service_queue->waker_bfqq == bfqq)) ||
>>>> +	    at_head || blk_rq_is_passthrough(rq)) {
>>>> 		if (at_head)
>>>> 			list_add(&rq->queuelist, &bfqd->dispatch);
>>>> 		else
>>>> 
>>> 
>>> This is unreadable... Just seems like you are piling heuristics in to
>>> catch some case, and it's neither readable nor clean.
>>> 
>> 
>> Yeah, these comments inappropriately assume that the reader knows the
>> waker mechanism in depth.  And they do not stress at all how important
>> this improvement is.
>> 
>> I'll do my best to improve these comments.
>> 
>> To try to do a better job, let me also explain the matter early here.
>> Maybe you or others can give me some early feedback (or just tell me
>> to proceed).
>> 
>> This change is one of the main improvements that boosted
>> throughput in Jan's tests.  Here is the rationale:
>> - consider a bfq_queue, say Q1, detected as a waker of another
>>  bfq_queue, say Q2
>> - by definition of a waker, Q1 blocks the I/O of Q2, i.e., some I/O of
>>  of Q1 needs to be completed for new I/O of Q1 to arrive.  A notable
> 					       ^^ Q2?
> 

Yes, thank you!

(after this interaction, I'll fix and improve all this description,
according to your comments)

>>  example is journald
>> - so, Q1 and Q2 are in any respect two cooperating processes: if the
>>  service of Q1's I/O is delayed, Q2 can only suffer from it.
>>  Conversely, if Q2's I/O is delayed, the purpose of Q1 is just defeated.
> 
> What do you exactly mean by this last sentence?

By definition of waker, the purpose of Q1's I/O is doing what needs to
be done, so that new Q2's I/O can finally be issued.  Delaying Q2's I/O
is the opposite of this goal.

> 
>> - as a consequence if some I/O of Q1/Q2 arrives while Q2/Q1 is the
>>  only queue in service, there is absolutely no point in delaying the
>>  service of such an I/O.  The only possible result is a throughput
>>  loss, detected by Jan's test
> 
> If we are idling at that moment waiting for more IO from in service queue,
> I agree.

And I agree too, if the drive has no internal queueing, has no
parallelism or pipeline, or is at least one order of magnitude slower
than the CPU is processing I/O.  In all other cases, serving the I/O
of only one queue at a time means throwing away throughput.  For
example, on a consumer SSD, moving from one to two I/O threads served
in parallel usually means doubling the throughput.

So, the best thing to do, if all the above conditions are met, is to
have this new I/O dispatched as soon as possible.

The most efficient way to attain this goal is to just put the new I/O
directly into the dispatch list.

> But that doesn't seem to be part of your condition above?
> 
>> - so, when the above condition holds, the most effective and efficient
>>  action is to put the new I/O directly in the dispatch list
>> - as an additional restriction, Q1 and Q2 must be the only busy queues
>>  for this commit to put the I/O of Q2/Q1 in the dispatch list.  This is
>>  necessary, because, if also other queues are waiting for service, then
>>  putting new I/O directly in the dispatch list may evidently cause a
>>  violation of service guarantees for the other queues
> 
> This last restriction is not ideal for cases like jbd2 thread since it may
> still lead to pointless idling but I understand that without some
> restriction like this several waking threads could just starve other ones.

Yeah, the goal here is to reduce a little bit false positives.

> So I guess it's fine for now.
> 

Yes, hopefully experience will lead us to even improvements or even
better solutions.

Thanks,
Paolo

> 								Honza
> -- 
> Jan Kara <jack@suse.com>
> SUSE Labs, CR