ZFS datablocks are also a power of two what means, that if you have 1,2,4,8,16,32,.. datadisks, every write is evenly spread over all disks.
If you add one disk ex from 8 to 9 datadisks, any one disk is not used on a read/write. Does that means, 9 datadisks are slower than 8 disks? No, 9 disks are faster, maybee not 1/9 faster but faster. So think about more like a myth Add Raid redundancy disks to that count, example with 8 datadisks, add one disk for Z1 (9), 2 disks for Z2 (10) and 3 disks forZ3(11) for disks per vdev. Am 16.04.2013 um 23:37 schrieb Timothy Coalson: > On Tue, Apr 16, 2013 at 4:29 PM, Sašo Kiselkov <skiselkov...@gmail.com>wrote: > >> If you are IOPS constrained, then yes, raid-zn will be slower, simply >> because any read needs to hit all data drives in the stripe. This is >> even worse on writes if the raidz has bad geometry (number of data >> drives isn't a power of 2). >> > > Off topic slightly, but I have always wondered at this - what exactly > causes non-power of 2 plus number of parities geometries to be slower, and > by how much? I tested for this effect with some consumer drives, comparing > 8+2 and 10+2, and didn't see much of a penalty (though the only random test > I did was read, our workload is highly sequential so it wasn't important). > > Tim > _______________________________________________ > OpenIndiana-discuss mailing list > OpenIndiana-discuss@openindiana.org > http://openindiana.org/mailman/listinfo/openindiana-discuss -- _______________________________________________ OpenIndiana-discuss mailing list OpenIndiana-discuss@openindiana.org http://openindiana.org/mailman/listinfo/openindiana-discuss
