May 16, 2005 at 5:09 am
Hopefully a quick and easy question.
In the situation of having a RAID 10 array with a total of 6 disks (3 striped and then mirrored) should I divide the Avg Disk Queue Length by 3 or by 6. I assume the physical Queue to the Windows server will the three striped disks and then the SAN will perform the mirror operation onto the other 3 mirrored disks.
Thanks,
TD
May 16, 2005 at 5:24 am
The average disk queue length shoud not be more than 6 in your case.
May 16, 2005 at 6:29 am
If you have a RAID array controller with a RAID 10 volume, Windows doesn't know anything about the underlying physical disks. To Windows, it just looks like one physical disk. You shouldn't have to worry about how many physical disks are in the array, since the Average Disk Queue Length basically represents the queue of disk i/o sent to the RAID controller. The RAID controller actually handles disk i/o to the individual disks in the array. You should be shooting for an average disk queue length of about 5.
May 16, 2005 at 6:30 am
OK.. so basically because it is Raid 10 I divide by the number of total spindles in the array by 2 e.g. 6/2 = 3... thus allowing for a queue of 2 per disk I should see more than 6 for that array.
If it was raid 5 I would divide by 6 instead.
Thanks,
TD
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