On Thu, Jun 27, 2013 at 9:35 AM, Nicolas Barbier <nicolas.barb...@gmail.com>wrote:
> > My reasoning was: To determine which index block to update (typically > one in both the partitioned and non-partitioned cases), one needs to > walk the index first, which supposedly causes one additional (read) > I/O in the non-partitioned case on average, because there is one extra > level and the lower part of the index is not cached (because of the > size of the index). But the "extra level" is up at the top where it is well cached, not at the bottom where it is not. Cheers, Jeff
