On Jun 15, 2010, at 6:40 AM, Roy Sigurd Karlsbakk wrote: >> I'm going to say something sacrilegious here: 128GB of RAM may be >> overkill. You have the SSDs for L2ARC - much of which will be the DDT, >> but, if I'm reading this correctly, even if you switch to the 160GB >> Intel X25-M, that give you 8 x 160GB = 1280GB of L2ARC, of which only >> half is in-use by the DDT. The rest is file cache. You'll need lots of >> RAM if you plan on storing lots of small files in the L2ARC (that is, >> if your workload is lots of small files). 200bytes/record needed in >> RAM for an L2ARC entry. >> >> I.e. >> >> if you have 1k average record size, for 600GB of L2ARC, you'll need >> 600GB / 1kb * 200B = 120GB RAM. >> >> if you have a more manageable 8k record size, then, 600GB / 8kB * 200B >> = 15GB > > Now I'm confused. First thing I heard, was about 160 bytes was needed per DDT > entry. Later, someone else told med 270. Then you, at 200. Also, there should > be a good way to list out a total of blocks (zdb just crashed with a full > memory on my 10TB test box). I tried browsing the source to see the size of > the ddt struct, but I got lost. Can someone with an osol development > environment please just check sizeof that struct?
Why read source when you can read the output of "zdb -D"? :-) -- richard -- Richard Elling rich...@nexenta.com +1-760-896-4422 ZFS and NexentaStor training, Rotterdam, July 13-15, 2010 http://nexenta-rotterdam.eventbrite.com/ _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss
