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

Johannes Weiner
2019-12-16T18:25:22Z

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?

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.

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.