I am trying to study for my VNX TA Design certification and have hit a dead end while attempting to size disks according to IOPS for VNX.
I know the formula to calculate the number of disks is RG IOPS/Disk IOPS + Large I/O overhead + HotSpares + Vault Drives
However, what I do not get is how do I figure out the Disk IOPS.
There is a question in the E20-324 practice test regarding this
A customer has an application requirement of 5000 IOPS with 70:30 read/write ratio and 9.85 TB storage capacity. The I/O is small-block random. The customer is concerned about cost. You propose a VNX with 300 GB 15K SAS drives.
If spare drives are not configured, which configuration minimizes the costs while adhering to EMC best practices?
Which of these would you say it is correct and why ?
In terms of determining the number of the disks to handle a particular host workload, it is still accurate (not accounting for FAST cache). For VNX2, the white papers are not as detailed as some of the older white papers, but the disk speeds and raid calculations would be the same.
You would still use the R10 formula on page 117 for the question asked.
5000 IOPS at 70:30.
if looking at R5, then backend IOPS required would be 3500 reads and 1500 writes, so 1500*4 to take into account R5 overhead, so total disk IOps required would be 6000+3500 = 9500
Rule of thumb for 15K SAS disk is 180
9500/180 = 53, so best fit in your selection would be 55 drives.
Why did I start with R5 -- as better cost option. A higher write ration I would start thinking about R10, but with teh criteria proposed, R5 is the solution.
...and Glen was spot on -- recent best practices guides removed a lot of fundmental stuff to make it easier to read the whole thing and adhere to best practices. RAID calculations previoulsy covered in detail aren't necessarily best practices.
Small block random I/O means best will be suited with RAID-5.
1 disk of 300 GB SAS 15k rpm = 180 IOPS.
300*40 (drives)= 1200 GB.
300*55 (drives)= 16500 GB (This will not save his cost only, but will also give better performance).
Total requirement = 9.85 TB.
180(number of IOPS a drive can handle)*40(number of drives) = 7200 IOPS.
180(number of IOPS a drive can handle)*55(number of drives) = 9900 IOPS.
Capacity can be met with 40 drives but that will not make the sense in handling IOPS
Now 30 % write 1500*4=6000.
6000+3500 = 9500 IOPS should be addressed (55 drives 300G SAS 15K will easily be able to handle 9900 IOPS).
Goal is to acheive max ROI.
Both of C & D option come under EMC's best practices.
The answer is D with RAID-5 & 55 Drives.