Re: [PATCH 0/4] locks: avoid thundering-herd wake-ups

[J. Bruce Fields]

On Wed, Aug 08, 2018 at 04:09:12PM -0400, J. Bruce Fields wrote:
> On Wed, Aug 08, 2018 at 03:54:45PM -0400, J. Bruce Fields wrote:
> > On Wed, Aug 08, 2018 at 11:51:07AM +1000, NeilBrown wrote:
> > > If you have a many-core machine, and have many threads all wanting to
> > > briefly lock a give file (udev is known to do this), you can get quite
> > > poor performance.
> > > 
> > > When one thread releases a lock, it wakes up all other threads that
> > > are waiting (classic thundering-herd) - one will get the lock and the
> > > others go to sleep.
> > > When you have few cores, this is not very noticeable: by the time the
> > > 4th or 5th thread gets enough CPU time to try to claim the lock, the
> > > earlier threads have claimed it, done what was needed, and released.
> > > With 50+ cores, the contention can easily be measured.
> > > 
> > > This patchset creates a tree of pending lock request in which siblings
> > > don't conflict and each lock request does conflict with its parent.
> > > When a lock is released, only requests which don't conflict with each
> > > other a woken.
> > 
> > Are you sure you aren't depending on the (incorrect) assumption that "X
> > blocks Y" is a transitive relation?
> > 
> > OK I should be able to answer that question myself, my patience for
> > code-reading is at a real low this afternoon....
> 
> In other words, is there the possibility of a tree of, say, exclusive
> locks with (offset, length) like:
> 
> 	(0, 2) waiting on (1, 2) waiting on (2, 2) waiting on (0, 4)
> 
> and when waking (0, 4) you could wake up (2, 2) but not (0, 2), leaving
> a process waiting without there being an actual conflict.

After batting it back and forth with Jeff on IRC....  So do I understand
right that when we wake a waiter, we leave its own tree of waiters
intact, and when it wakes if it finds a conflict it just adds it lock
(with tree of waiters) in to the tree of the conflicting lock?

If so then yes I think that depends on the transitivity
assumption--you're assuming that finding a conflict between the root of
the tree and a lock proves that all the other members of the tree also
conflict.

So maybe this example works.  (All locks are exclusive and written
(offset, length), X->Y means X is waiting on Y.)

	process acquires (0,3)
	2nd process requests (1,2), is put to sleep.
	3rd process requests (0,2), is put to sleep.

	The tree of waiters now looks like (0,2)->(1,2)->(0,3)

	(0,3) is unlocked.
	A 4th process races in and locks (2,2).
	The 2nd process wakes up, sees this new conflict, and waits on
	(2,2).  Now the tree looks like (0,2)->(1,2)->(2,2), and (0,2)
	is waiting for no reason.

?

--b.