Re: [PATCH v5 01/16] mm: list_lru: optimize memory consumption of arrays of per cgroup lists
From: Muchun Song
Date: Sat Jan 08 2022 - 23:50:49 EST
On Fri, Jan 7, 2022 at 8:05 AM Roman Gushchin <guro@xxxxxx> wrote:
>
> On Mon, Dec 20, 2021 at 04:56:34PM +0800, Muchun Song wrote:
> > The list_lru uses an array (list_lru_memcg->lru) to store pointers
> > which point to the list_lru_one. And the array is per memcg per node.
> > Therefore, the size of the arrays will be 10K * number_of_node * 8 (
> > a pointer size on 64 bits system) when we run 10k containers in the
> > system. The memory consumption of the arrays becomes significant. The
> > more numa node, the more memory it consumes.
> >
> > I have done a simple test, which creates 10K memcg and mount point
> > each in a two-node system. The memory consumption of the list_lru
> > will be 24464MB. After converting the array from per memcg per node
> > to per memcg, the memory consumption is going to be 21957MB. It is
> > reduces by 2.5GB. In our AMD servers with 8 numa nodes in those
> > sysuem, the memory consumption could be more significant. The savings
> > come from the list_lru_one heads, that it also simplifies the
> > alloc/dealloc path.
> >
> > The new scheme looks like the following.
> >
> > +----------+ mlrus +----------------+ mlru +----------------------+
> > | list_lru +---------->| list_lru_memcg +--------->| list_lru_per_memcg |
> > +----------+ +----------------+ +----------------------+
> > | list_lru_per_memcg |
> > +----------------------+
> > | ... |
> > +--------------+ node +----------------------+
> > | list_lru_one |<----------+ list_lru_per_memcg |
> > +--------------+ +----------------------+
> > | list_lru_one |
> > +--------------+
> > | ... |
> > +--------------+
> > | list_lru_one |
> > +--------------+
> >
> > Signed-off-by: Muchun Song <songmuchun@xxxxxxxxxxxxx>
> > Acked-by: Johannes Weiner <hannes@xxxxxxxxxxx>
>
> As much as I like the code changes (there is indeed a significant simplification!),
> I don't like the commit message and title, because I wasn't able to understand
> what the patch is doing and some parts look simply questionable. Overall it
> sounds like you reduce the number of list_lru_one structures, which is not true.
>
> How about something like this?
>
> --
> mm: list_lru: transpose the array of per-node per-memcg lru lists
>
> The current scheme of maintaining per-node per-memcg lru lists looks like:
> struct list_lru {
> struct list_lru_node *node; (for each node)
> struct list_lru_memcg *memcg_lrus;
> struct list_lru_one *lru[]; (for each memcg)
> }
>
> By effectively transposing the two-dimension array of list_lru_one's structures
> (per-node per-memcg => per-memcg per-node) it's possible to save some memory
> and simplify alloc/dealloc paths. The new scheme looks like:
> struct list_lru {
> struct list_lru_memcg *mlrus;
> struct list_lru_per_memcg *mlru[]; (for each memcg)
> struct list_lru_one node[0]; (for each node)
> }
>
> Memory savings are coming from having fewer list_lru_memcg structures, which
> contain an extra struct rcu_head to handle the destruction process.
My bad English. Actually, the saving is coming from not only 'struct rcu_head'
but also some pointer arrays used to store the pointer to 'struct list_lru_one'.
The array is per node and its size is 8 (a pointer) * num_memcgs. So the total
size of the arrays is 8 * num_nodes * memcg_nr_cache_ids. After this patch,
the size becomes 8 * memcg_nr_cache_ids. So the saving is
8 * (num_nodes - 1) * memcg_nr_cache_ids.
> --
>
> But what worries me is that memory savings numbers you posted don't do up.
> In theory we can save
> 16 (size of struct rcu_head) * 10000 (number of cgroups) * 2 (number of numa nodes) = 320k
> per slab cache. Did you have a ton of mount points? Otherwise I don't understand
> where these 2.5Gb are coming from.
memcg_nr_cache_ids is 12286 when creating 10k memcgs. So the saving
of arrays of one list_lru is 8 * 1 (number of numa nodes - 1) * 12286 = 96k.
There will be 2 * 10k list_lru when mounting 10k points. So the total
saving is 96k * 2 * 10k = 1920 M.
Thanks Roman.