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

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