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June 2nd, 2013 03:00
Using QOS and the Background Class
A query about IOPS and QOS:
This is the scenario. I’ve a pool of 10 disks on a RAID 5 pool, comprising of 15K SAS disks. If I’ve understood how the IOPS are calculated, this pool can offer me a total of 230*8 IOPS. Now I am supposed to provision a LUN with a minimum of 1600 IOPS for a certain application, and plan to use the balance capacity of no less than 2.5 TB for other applications.
I’ve cut this LUN, and the system is running. I created a QOS policy to give the application an IOPs of 1600, +/- 10%. Then I did a storage V-motion to transfer the VMs from a slower disk to this new LUN for which I’ve created this policy. It showed me the IOPS of 1600 being granted, but the background class shows a value of 4000.
At present this pool has only this LUN for which I have a performance goal – no other LUNs are created. However when I go to the QoS control panel, I see that the IOPS are being granted for my goal. Yet the background class shows a value of near 4000 IOPS. What is this background class ? Is it the IOPS for other LUNs in this pool, or all other LUNs in the storage ? As far as I can understand the pool in question cannot grant me 4000+ IOPS for background class.
Could one who has a better knowledge of this, than myself please rid my ignorance, and help me in obtaining a clearer knowledge of this ?



RRR
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June 3rd, 2013 03:00
Where does your 230*8 come from? A 15k disk can do 180 IOps (rule of thumb). And what is this "8"?
Furthermore: when looking at IOps actually generated, bare in mind that small I/Os can have a much higher IOps than somewhat larger I/Os, so the rule of thumb (180) doesn't apply here. If all I/Os are 4 kB the 180 per disk is a good value to do the math. And then there's this thing called cache. The LUN might handle much more than 180 IOps per disk if there's a good locality of the data, I mean that the data is cached and since RAM is much faster than disk, you can get much higher IOps.
nwee75
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June 3rd, 2013 14:00
Hi RRR,
Thanks for responding.
230 comes from a document I've referred to in VMware. If it is wrong I stand corrected.
http://kb.vmware.com/selfservice/microsites/search.do?language=en_US&cmd=displayKC&externalId=1031773
The 8 comes from the pool using, in essence 8 for disks for writing, (the other two being for parity )
Please do point out if the math I did is wrong, so the I can learn of the correct one.
This is an environment with FAST cache; how does one integrate that to the equation ?
BTW, can you elaborate on the background class as well ?