Each head or controller owns a set of disk, including spare(s).
When you write
2 Parity disks
1 Hotspare
3 System disks for the Passive head
and 6 unused disks this does not make sense to me.
So the layout should be:
controller/head 1 (active)
2 Parity disks
5 Data Disks
building the root-aggregate, containing the root-volume vol0 and data
1 Spare
controller/head 2 (passive)
2 Parity disks
1 Data disk
building the root-aggregate, containing the root volume vol0
1 Spare
I have also configured passive system with RAID4 instead of DP, giving one more disk as data disk for the active controller.
In the Storage FAQ you can find:
In an HA pair configuration NetApp recommends the following to maximize storage utilization:
Controller 1 (RAID-DP):
o
Number of Aggregates = 1
o
RAID Group Size = 8
Number of Data Drives = 6
Number of Parity Drives = 2
Number of Hot Spares = 1
Controller 2 (RAID-DP):
o
Number of Aggregates = 1
RAID Group Size = 3
Number of Data Drives = 1
Number of Parity Drives = 2
Number of Hot Spares = 0*
*As a result of assigning no spare drives to controller 2, warning messages will be generated that can be safely ignored (
raid.min_spare_count can be set to 0 to avoid warning messages).
so now we got the disk-layout.
How much do you get out of it, is the question. And the answer is as always: it depends.
There is a rule of thumb, stating that you get roughly 900-1200 IOPS out of 6 SAS disks. But what is the workload? Do you read 4kB or 64kB per IO? Which protocol (eg. CIFS has a lot more overhead than NFS)?
So first of all one should know your workload (protocols, IOPS, bandwidth, #clients etc) - or simply: what are you going to do? And what do you expect?
BTW there are lots of sizing tools from NetApp your consultant surely knows.
Mark