On 06/27/2013 11:12 AM, Nicolas Barbier wrote:
> Imagine that there are a lot of indexes, e.g., 50. Although a lookup
> (walking one index) is equally fast, an insertion must update al 50
> indexes. When each index requires one extra I/O (because each index is
> one level taller), that is 50 extra I/Os. In the partitioned case,
> each index would require the normal smaller amount of I/Os. Choosing
> which partition to use must only be done once: The result “counts” for
> all indexes that are to be updated.
I think you're underestimating the cost of partitioning. After all, the
lookup of what index to update for a given partition is a a lookup in
pg_index via pg_index_indrelid_index - a btree index.
Additionally, the depth of an index doesn't directly translate to the
number of I/O writes per insert (or delete). I'd rather expect the avg.
number of I/O writes per insert into a b-tree to be reasonably close to
one - depending mostly on the number of keys per page, not depth.
> Additionally: Imagine that the data can be partitioned along some
> column that makes sense for performance reasons (e.g., some “date”
> where most accesses are concentrated on rows with more recent dates).
> The other indexes will probably not have such a performance
> distribution. Using those other indexes (both for look-ups and
> updates) in the non-partitioned case, will therefore pull a huge
> portion of each index into cache (because of the “random distribution”
> of the non-date data). In the partitioned case, more cache can be
> spent on the indexes that correspond to the “hot partitions.”
That's a valid point, yes. I'd call this index partitioning. And with
partial indices, Postgres already has something that gets pretty close,
I think. Though, I don't consider this to be related to how the tuples
of the relation are laid out on disk.
Regards
Markus Wanner