Re: [PATCH 1/3] mm: memcontrol: fix memory.low proportional distribution

Roman Gushchin
2019-12-16T19:12:05Z

On Mon, Dec 16, 2019 at 01:25:18PM -0500, Johannes Weiner wrote:
> On Fri, Dec 13, 2019 at 08:40:31PM +0000, Roman Gushchin wrote:
> > On Fri, Dec 13, 2019 at 02:21:56PM -0500, Johannes Weiner wrote:
> > > When memory.low is overcommitted - i.e. the children claim more
> > > protection than their shared ancestor grants them - the allowance is
> > > distributed in proportion to each siblings's utilized protection:
> > >
> > > low_usage = min(low, usage)
> > > elow = parent_elow * (low_usage / siblings_low_usage)
> > >
> > > However, siblings_low_usage is not the sum of all low_usages. It sums
> > > up the usages of *only those cgroups that are within their memory.low*
> > > That means that low_usage can be *bigger* than siblings_low_usage, and
> > > consequently the total protection afforded to the children can be
> > > bigger than what the ancestor grants the subtree.
> > >
> > > Consider three groups where two are in excess of their protection:
> > >
> > > A/memory.low = 10G
> > > A/A1/memory.low = 10G, A/memory.current = 20G
> > > A/A2/memory.low = 10G, B/memory.current = 20G
> > > A/A3/memory.low = 10G, C/memory.current = 8G
> > >
> > > siblings_low_usage = 8G (only A3 contributes)
> > > A1/elow = parent_elow(10G) * low_usage(20G) / siblings_low_usage(8G) = 25G
> > >
> > > The 25G are then capped to A1's own memory.low setting, i.e. 10G. The
> > > same is true for A2. And A3 would also receive 10G. The combined
> > > protection of A1, A2 and A3 is 30G, when A limits the tree to 10G.
> > >
> > > What does this mean in practice? A1 and A2 would still be in excess of
> > > their 10G allowance and would be reclaimed, whereas A3 would not. As
> > > they eventually drop below their protection setting, they would be
> > > counted in siblings_low_usage again and the error would right itself.
> > >
> > > When reclaim is applied in a binary fashion - cgroup is reclaimed when
> > > it's above its protection, otherwise it's skipped - this could work
> > > actually work out just fine - although it's not quite clear to me why
> > > we'd introduce this error in the first place.
> >
> > This complication is not simple an error, it protects cgroups under
> > their low limits if there is unprotected memory.
> >
> > So, here is an example:
> >
> > A A/memory.low = 2G, A/memory.current = 4G
> > / \
> > B C B/memory.low = 3G B/memory.current = 2G
> > C/memory.low = 1G C/memory.current = 2G
> >
> > as now:
> >
> > B/elow = 2G * 2G / 2G = 2G == B/memory.current
> > C/elow = 2G * 1G / 2G = 1G < C/memory.current
> >
> > with this fix:
> >
> > B/elow = 2G * 2G / 3G = 4/3 G < B/memory.current
> > C/elow = 2G * 1G / 3G = 2/3 G < C/memory.current
> >
> > So in other words, currently B won't be scanned at all, because
> > there is 1G of unprotected memory in C. With your patch both B and C
> > will be scanned.
>
> Looking at the B and C numbers alone: C is bigger than what it claims
> for protection and B is smaller than what it claims for protection.
>
> However, A doesn't provide 4G to its children. It provides 2G to be
> distributed between the two. So how can B claim 3G and be exempted
> from reclaim?

First, what if the memory pressure comes from memory.high/max set on A?

Second, it's up to semantics we define. Looking at it from the other side:
there is clearly 1G of memory in C which is not protected no matter what.
B wants it's memory to be fully protected, but it's limited by the competition
on the parent level. Now we try to satisfy B's requirements until we can.
Should we treat B and C equally from scratch?

I think both approaches is acceptable, but if we're switching from one option
to another, let's make it clear.

>
> But more importantly, it isn't in either case! The end result is the
> same in both implementations. Because as soon as C is reclaimed down
> to below 1G, A is still in excess of its memory.low (because it's
> overcommitted!), and they both will be reclaimed proportionally.

I do not disagree: the introduction of the proportional reclaim
made this complication (partially?) obsolete. But originally it was
required to make target distribution correct.

>
> From the example in the current code:
>
> * For example, if there are memcgs A, A/B, A/C, A/D and A/E:
> *
> * A A/memory.low = 2G, A/memory.current = 6G
> * //\\
> * BC DE B/memory.low = 3G B/memory.current = 2G
> * C/memory.low = 1G C/memory.current = 2G
> * D/memory.low = 0 D/memory.current = 2G
> * E/memory.low = 10G E/memory.current = 0
> *
> * and the memory pressure is applied, the following memory distribution
> * is expected (approximately):
> *
> * A/memory.current = 2G
> *
> * B/memory.current = 1.3G
> * C/memory.current = 0.6G
> * D/memory.current = 0
> * E/memory.current = 0
>
> Even though B starts out within whatever it claims to be its
> protection, A is overcommitted and so B and C converge on their
> proportional share of the parent's allowance.
>
> So to go back to the example chosen above:
>
> > A A/memory.low = 2G, A/memory.current = 4G
> > / \
> > B C B/memory.low = 3G B/memory.current = 2G
> > C/memory.low = 1G C/memory.current = 2G
>
> With either implementation we'd expect the distribution to be about
> 1.5G and 0.5G for B and C, respectively.
>
> And they'd have to be, too. Otherwise the semantics would be
> completely unpredictable to anyone trying to configure this.
>
> So I think mixing proportional distribution with absolute thresholds
> like this makes the implementation unnecessarily hard to reason
> about. It's also clearly buggy as pointed out in the changelog.
>
> > > However, since
> > > 1bc63fb1272b ("mm, memcg: make scan aggression always exclude
> > > protection"), reclaim pressure is scaled to how much a cgroup is above
> > > its protection. As a result this calculation error unduly skews
> > > pressure away from A1 and A2 toward the rest of the system.
> >
> > It could be that with 1bc63fb1272b the target memory distribution
> > will be fine. However the patch will change the memory pressure in B and C
> > (in the example above). Maybe it's ok, but at least it should be discussed
> > and documented.
>
> I'll try to improve the changelog based on this, thanks for filling in
> the original motivation. But I do think it's a change we want to make.

Absolutely, I'm not against the change. I just want to make sure we will
put all details into the changelog.

Thanks!