I'm finally at the point of adding an SSD to my system, so I can get
reasonable dedup performance.
The question here goes to sizing of the SSD for use as an L2ARC device.
Noodling around, I found Richard's old posing on ARC->L2ARC memory
requirements, which is mighty helpful in making sure I don't overdo the
L2ARC side.
(http://www.mail-archive.com/zfs-discuss@opensolaris.org/msg34677.html)
What I haven't found is a reasonable way to determine how big I'll need
an L2ARC to fit all the relevant data for dedup. I've seen several
postings back in Jan about this, and there wasn't much help, as was
acknowledged at the time.
What I'm after is exactly what needs to be stored extra for DDT? I'm
looking at the 200-byte header in ARC per L2ARC entry, and assuming that
is for all relevant info stored in the L2ARC, whether it's actual data
or metadata. My question is this: the metadata for a slab (record)
takes up how much space? With DDT turned on, I'm assuming that this
metadata is larger than with it off (or, is it the same now for both)?
There has to be some way to do a back-of-the-envelope calc that says
(X) pool size = (Y) min L2ARC size = (Z) min ARC size
--
Erik Trimble
Java System Support
Mailstop: usca22-123
Phone: x17195
Santa Clara, CA
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