Date:   Wed, 6 Sep 2017 10:31:58 +0200
From:   Michal Hocko <mhocko@...nel.org>
To:     Roman Gushchin <guro@...com>
Cc:     linux-mm@...ck.org, Vladimir Davydov <vdavydov.dev@...il.com>,
        Johannes Weiner <hannes@...xchg.org>,
        Tetsuo Handa <penguin-kernel@...ove.SAKURA.ne.jp>,
        David Rientjes <rientjes@...gle.com>,
        Andrew Morton <akpm@...ux-foundation.org>,
        Tejun Heo <tj@...nel.org>, kernel-team@...com,
        cgroups@...r.kernel.org, linux-doc@...r.kernel.org,
        linux-kernel@...r.kernel.org
Subject: Re: [v7 2/5] mm, oom: cgroup-aware OOM killer

On Tue 05-09-17 21:23:57, Roman Gushchin wrote:
> On Tue, Sep 05, 2017 at 04:57:00PM +0200, Michal Hocko wrote:
[...]
> > Hmm. The changelog says "By default, it will look for the biggest leaf
> > cgroup, and kill the largest task inside." But you are accumulating
> > oom_score up the hierarchy and so parents will have higher score than
> > the layer of their children and the larger the sub-hierarchy the more
> > biased it will become. Say you have
> > 	root
> >          /\
> >         /  \
> >        A    D
> >       / \
> >      B   C
> > 
> > B (5), C(15) thus A(20) and D(20). Unless I am missing something we are
> > going to go down A path and then chose C even though D is the largest
> > leaf group, right?
> 
> You're right, changelog is not accurate, I'll fix it.
> The behavior is correct, IMO.

Please explain why. This is really a non-intuitive semantic. Why should
larger hierarchies be punished more than shallow ones? I would
completely agree if the whole hierarchy would be a killable entity (aka
A would be kill-all).
 
[...]
> > I do not understand why do we have to handle root cgroup specially here.
> > select_victim_memcg already iterates all memcgs in the oom hierarchy
> > (including root) so if the root memcg is the largest one then we
> > should simply consider it no?
> 
> We don't have necessary stats for the root cgroup, so we can't calculate
> it's oom_score.

We used to charge pages to the root memcg as well so we might resurrect
that idea. In any case this is something that could be hidden in
memcg_oom_badness rather then special cased somewhere else.

> > You are skipping root there because of
> > memcg_has_children but I suspect this and the whole accumulate up the
> > hierarchy approach just makes the whole thing more complex than necessary. With
> > "tasks only in leafs" cgroup policy we should only see any pages on LRUs
> > on the global root memcg and leaf cgroups. The same applies to memcg
> > stats. So why cannot we simply do the tree walk, calculate
> > badness/check the priority and select the largest memcg in one go?
> 
> We have to traverse from top to bottom to make priority-based decision,
> but size-based oom_score is calculated as sum of descending leaf cgroup scores.
> 
> For example:
>  	root
>           /\
>          /  \
>         A    D
>        / \
>       B   C
> A and D have same priorities, B has larger priority than C.
> 
> In this case we need to calculate size-based score for A, which requires
> summing oom_score of the sub-tree (B an C), despite we don't need it
> for choosing between B and C.
> 
> Maybe I don't see it, but I don't know how to implement it more optimal.

I have to think about the priority based oom killing some more to be
honest. Do we really want to allow setting a priority to non-leaf
memcgs? How are you going to manage the whole tree consistency? Say your
above example have prio(A) < prio(D) && prio(C) > prio(D). Your current
implementation would kill D, right? Isn't that counter intuitive
behavior again. If anything we should prio(A) = max(tree_prio(A)). Again
I could understand comparing priorities only on killable entities.
-- 
Michal Hocko
SUSE Labs

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