On Mon, 8 May 2017, Loic Dachary wrote:
> On 05/08/2017 05:09 AM, Sage Weil wrote:
> > On Sun, 7 May 2017, Loic Dachary wrote:
> >> Hi Sage,
> >>
> >> When choosing the second replica, crush_bucket_choose picks an item and
> >> crush_choose_{indep,firstn} checks if it has already been chosen for the
> >> first replica. If so, it is discarded as a collision[1], r is modified
> >> and another attempt is made to get a different item. If no value is
> >> found after choose_tries attempt, the mapping is incomplete.
> >>
> >> Another way to do the same would be that crush_bucket_choose /
> >> bucket_straw2_choose[2] ignores the items that have already been chosen.
> >> It could be done by looping over the list but that would mean N (number
> >> of replicas) lookups for each item.
> >>
> >> The current implementation is optimized for the cases where collisions
> >> are rare. However, when the weights of the items are two order of
> >> magnitude appart or more, choose_tries has to be set to a very large
> >> value (more than 1000) for the mapping to succeed. In practice that is
> >> not a problem as it is unlikely that a host is 100 times bigger than
> >> another host ;-)
> >>
> >> When fixing an uneven distribution[3], CRUSH weights are sometime set to
> >> values with that kind of difference (1 against 100) to compensate for
> >> the probability bias and/or a small number of samples. For instance when
> >> there are 5 hosts with weights 5 1 1 1 1, modifying the weight set
> >> fails. It goes as far as 8.9, 0.01, 0.01, 0.01, 0.01 with choose_tries
> >> 2000. The value of choose_tries has to be increased a lot for a gain
> >> that is smaller and smaller and the CPU usage goes up. In this
> >> situation, an alternative implementation of the CRUSH collision seems a
> >> good idea.
> >>
> >> Instead of failing after choose_tries attempts, crush_bucket_choose
> >> could be called with the list of already chosen items and loop over them
> >> to ensure none of them are candidate. The result will always be correct
> >> but the implementation more expensive. The default choose_tries could
> >> even be set to a lower value (19 instead of 50 ?) since it is likely
> >> more expensive to handle 30 more collisions rather than looping over
> >> each item already selected.
> >>
> >> What do you think ?
> >
> > I think this direction is promising! The problem is that I think it's not
> > quite as simple as you suggest, since you may be choosing over multiple
> > levels of a hierarchy. If the weight tree is something like
> >
> > 4
> > / \
> > 2 2
> > / \ / \
> > 1 1 1 1
> > a b c d
> >
> > and you chose a, then yes, if you get back into the left branch you can
> > filter it out of the straw2 selections. And num_rep is usually small so
> > that won't be expensive. But you also need the first choice at the top
> > level of the hierarchy to weight the left *tree* with 1 instead of 2.
>
> I don't understand why this is necessary ? Here are the scenarios I have in mind:
>
>
> / \
> / \
> r1 4 r2 4 rack (failure domain)
> / \ / \
> 2 2 2 2 host
> / \ / \ / \ / \
> 1 1 1 1 1 1 1 1 device
> a b c d e f g h
>
> Say value 10 ends up in a the first time, it first went through rack
> r1 which is the failure domain. If value 10 also ends up in r1 the
> second time, straw2 will skip/collide it at that level because r1 is
> stored in out while a is stored in out2.
>
> There only case I can think of that requires collision to be resolved
> in a higher hierarchical level is when there is no alternative.
>
>
>
> / \
> / \
> r1 4 r2 4 rack
> / / \
> h1 2 2 2 host (failure domain)
> / / \ / \
> 1 1 1 1 1 device
> a e f g h
>
> If 10 ends up in h1 the first time and the second time, it will
> collide because there is no alternative. It will then retry_descent,
> ftotal increases which goes into r and it gets another chance at landing on a host
> that's not h1.
The problem isn't when choose[leaf] is specifying the intermediate
level (rack or host in your examples); it's when there is an intervening
level that a single choose is crossing. In your first tree,
root 6
> / \
> / \
> r1 4 r2 4 rack
> / \ / \
> h1 2 h2 2 h3 2 h4 2 host (failure domain)
> / \ / \ / \ / \
> 1 1 1 1 1 1 1 1 device
> a b c d e f g h
let's say the failure domain is the host instead of the rack. If we
choose a (h1) the first time, for subsequent descents from the root we
still pick r1 and r2 equally (4 vs 4) even though r1 only has 1 usable
host (h2). This normally is fine because we reject the entire descent for
50% of the r1 choices, so *effectively* r1's weight is only 2 (it's as if
the other attempts never happened). But with the smarter straw2 choose h2
100% for r1 and you'll end up with 2x more tiems for that second position
on h2 than you want.
Does that make sense?
sage
>
> I must be missing a use case :-)
>
>
> >
> > I think this could be done by adjusting the hierarchical weights as you go
> > (and I think one of Adam's early passes at the multipick problem did
> > something similar), but it's a bit more complex.
> >
> > It seems worth pursuing, though!
> >
> > And dynamically doing this only after the first N 'normal' attempts fail
> > seems like a good way to avoid slowing down the common path (no
> > collisions). I suspect the optimal N is probably closer to 5 than 19,
> > though?
> >
> > sage
> >
> >
> >>
> >> Cheers
> >>
> >> P.S. Note that even in this border case modifying the weights to 7.1,
> >> 0.5, 0.5, 0.4, 0.4 significantly improves the distribution (twice better
> >> instead of ten times better). Only it cannot do better because it hits a
> >> limitation of the current CRUSH implementation. But it looks like it is
> >> not too difficult to fix.
> >>
> >>
> >> [1] https://github.com/ceph/ceph/blob/master/src/crush/mapper.c#L541
> >> [2] https://github.com/ceph/ceph/blob/master/src/crush/mapper.c#L332
> >> [3] http://marc.info/?l=ceph-devel&m=149407691823750&w=2
> >> --
> >> Loïc Dachary, Artisan Logiciel Libre
> >>
>
> --
> Loïc Dachary, Artisan Logiciel Libre
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