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

Paolo Valente
2021-01-28T18:01:35Z

> 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
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.
- 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
- 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

If these comments make things clearer, then I'll put them in the
commit message and the code, and I'll proceed with a V2.

Thanks,
Paolo

> --
> Jens Axboe
>