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.  hope this helpful!
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. Read:60% Write: 40%  Please suggest. Also if you can show how did you calculate the total IOPS that will be just awesume. Many thanks!!!  Regards, Nishant Kohli
How many IOPS can i get out of RAID 5 metaLUN?