> 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? Vennlige hilsener / Best regards roy -- Roy Sigurd Karlsbakk (+47) 97542685 r...@karlsbakk.net http://blogg.karlsbakk.net/ -- I all pedagogikk er det essensielt at pensum presenteres intelligibelt. Det er et elementært imperativ for alle pedagoger å unngå eksessiv anvendelse av idiomer med fremmed opprinnelse. I de fleste tilfeller eksisterer adekvate og relevante synonymer på norsk. _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss
