IOPs calculation

It's the last step  - you take the Total IOPS (4500) and divide that be what each NL-SAA disk can handle - about 90 IOPS (and this depends on the IO size and randomness of the IO).

So divide 4500 Total IOPS by 90 IOPS per NL-SAS  (4500/90) = 50. So you need a total of 50 NL-SAS drives.

Now take the 50 disks and divide that by 8 (6+2 R6) and you get 6 raid groups that are each 6+2 and you have two disks left over. So add 6 more disks and create a 7th R6 group.

You're total would then be 56 drives in 7 6+2 R6 groups.

Now the question comes - to handle this 4500 IOPS, will you be creating a Pool or using Raid Groups? If Raid Groups then you'll need to create metaLUN (and you'll need to get the metaLUN configuration exactly right). If pools then you can create a single Pool with 56 NL-SAS drives. The extra 6 drives gives you some extra headroom for bursts, but if the response times are required to be 2ms, the only way you'll get that is with cache - Fast Cache or an EFD tier in the Pool. Remember these are spinning hard disks, 2ms is pretty hard to get as the drives are not this fast and 2ms is about what you can get with Writes and 100% System Cache hit. You may not get this NL-SAS and the requirement of 2ms seems like a critical application. I'd re-think the use of NL-SAS for this. Just my two cents and not anything official from EMC.

If you need a more detailed review of this, then I'd recommend you contact you EMC account team and see if they can put you in touch with a performance expert.

glen