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## How many IOPS can i get out of RAID 5 metaLUN?

Hi.

I have a metalun configured on cx4-480 array. How many total throughput (IOPS) can I expect on the LUN?

There are 6 component LUNs on different RGs R5(8+1) combination. MetaLUN is striped NOT concatenated.

Please suggest. Also if you can show how did you calculate the total IOPS that will be just awesume.

Many thanks!!!

Regards,

Nishant Kohli

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• ### CX-Series

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## Re: How many IOPS can i get out of RAID 5 metaLUN?

If you are talking about IOPS, then I would suppose the i/o pattern is [random, small], maybe 4KB. But you didn't give out the disk type, I would use SAS 15krpm drives as an example.

let's do the calculation:

=================================

SAS 15krpm drive supports 180 IOPS

You have 6*8 = 48 data disks -> 48*180 = 8640 IOPS.

Another key point is that you need to consider RAID 5 write penalty (1 app write = 4 disk writes) when calculating the final disk level I/O to compare with the the value we got just now (8640).

Le't assume your application level IOPS is = X, then:

60%X + 40%X * 4 must <= 8640 -> x<= 3927, which means your application level I/O shouldn't be greater than 3927 with R/W cache disabled.

In summary, if we don't consider the cache i/o absorption, for i/o pattern = [random, small], 60%R | 40%W, with 15krpm SAS, you can sustain application i/o <=3927 IOPS.

Again, this value is just a result from a rule of thumb, EMC professional services can give you an more accurate value through some of the internal tools.

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## Re: How many IOPS can i get out of RAID 5 metaLUN?

If you are talking about IOPS, then I would suppose the i/o pattern is [random, small], maybe 4KB. But you didn't give out the disk type, I would use SAS 15krpm drives as an example.

let's do the calculation:

=================================

SAS 15krpm drive supports 180 IOPS

You have 6*8 = 48 data disks -> 48*180 = 8640 IOPS.

Another key point is that you need to consider RAID 5 write penalty (1 app write = 4 disk writes) when calculating the final disk level I/O to compare with the the value we got just now (8640).

Le't assume your application level IOPS is = X, then:

60%X + 40%X * 4 must <= 8640 -> x<= 3927, which means your application level I/O shouldn't be greater than 3927 with R/W cache disabled.

In summary, if we don't consider the cache i/o absorption, for i/o pattern = [random, small], 60%R | 40%W, with 15krpm SAS, you can sustain application i/o <=3927 IOPS.

Again, this value is just a result from a rule of thumb, EMC professional services can give you an more accurate value through some of the internal tools.

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3 Silver

## Re: How many IOPS can i get out of RAID 5 metaLUN?

Hey Steve, that really opened up some clogs in my mind. This is what I needed to know.

I forgot to mention the drive type, though you were right on 180 IOPS per drive. Mine is 15k FC.

Thank you so much for your help!

Regards

Nishant Kohli

Storage - Guitar Center, ITO | Desk: (818) 735-8800 ext 2835

Mobile â€“ 818 584 5956

E: Nishant.Kohli@Guitarcenter.com<mailto:Nishant.Kohli@Guitarcenter.com>

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glen

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## Re: How many IOPS can i get out of RAID 5 metaLUN?

 é˜¿è¶…_SteveZhou wrote: let's do the calculation: ================================= SAS 15krpm drive supports 180 IOPS You have 6*8 = 48 data disks -> 48*180 = 8640 IOPS.

Remember, when calculating the back-end IOPs, we should take in consideration *all* drives and should not be deducting the (equivalent of) one drive for the distributed parity of RAID 5.  From a back-end perspective when considering IOPs (and also Bandwidth) we are calculating raw numbers regardless of what is written be it data or (distributed) parity.  On the other hand, the write penalty of RAID 5 is factored in when calculating the front-end IOPs as you later noted.

Therefore, for a RAID 5 (8+1) RG, we should be calculating: 180 IOPs * 9 (not 8 just because those are the data disks).

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## Re: How many IOPS can i get out of RAID 5 metaLUN?

you are correct, thanks for the reminder.

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## Re: How many IOPS can i get out of RAID 5 metaLUN?

Also consider the fact that response times increase exponentially when drives are utilized over Â±66%. This is called Little's law. Below Â±66% response times are good, but almost every tool will show disk utilization over 66% in a red color. Yes you can push them to 100% utilization, but response times are severe. So if you need to keep some decent response times, stick with 66% of what you just calculated.

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