Question on RAID 5 and striping

Question

Hello, I have been doing some reading about how the striping works on the AX4-5 and would like to know if my interpretation is correct. The array has a stripe element size of 64K (not able to be changed acording to the info I read) so on a 5 drive RAID 5 array, that would mean the stripe size would be 256K On ESX we have a block size of 1MB. If we're using a Windows OS with a cluster size of 4K then is it right to say that the smallest read from the SAN would be 1MB and that this would require 16 reads and that each drive would receive 3 read commands (and 1 drive would need 4 reads) ? Thanks, Andrew

RRR · Accepted Answer

If ESX requires to read the whole 1MB block to satisfy the 4kB block read from a VM, then 1MB should be read from the Clariion. You are correct, you'd need to read 16 blocks of 64kB.

RRR · Answer

Hello AndrewC, It would be very bad that you'd need that many reads just for 1 read I/O. No, to read 1 block, you only need 1 read I/O on the back end.

AndrewC2 · Answer

Hello RRR,

Thanks for the reply.

Are you saying 1 read for each disk, but all disks would still have to be read in order to read the 1MB block. I'm trying to understand what is happening on each disk for a 1 MB read.

Regards,

Andrew

dysolution · Answer

My understanding is probably incomplete, but as far as I know, a VMFS formatted with an 8 MB block size does not necessarily mean that a 4 KB read or write from the guest will result in a 8 MB read or write to a LUN. (the horror!) VMFS uses sub-block addressing for smaller IOs and deals with 64 KB blocks even if the VMFS has a 8 MB block size.

http://www.yellow-bricks.com/2009/05/14/block-sizes-and-growing-your-vmfs/

You can see what ESX is actually doing to the disks by looking at the read and write IO sizes in Analyzer for a LUN and comparing those to what the guest reports it's doing (as in Perfmon Bytes/Read & Bytes/Write in Windows).

CLARiiON

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