IOPS (input/output operations per second) are still – maybe even more so than ever – the most prominent and important metric to measure storage performance. With SSD technology finding its way into affordable, mainstream server solutions, providers are eager to outdo each other offering ever higher IOPS dedicated servers and virtual private servers.
While SSD based servers will perform vastly better than SATA or SAS based ones, especially for random I/O, the type of storage alone isn’t everything. Vendors will often quote performance figures using lab conditions only, i.e. the best possible environment for their own technology. In reality, however, we are facing different conditions – several clients competing for I/O, as well as a wide ranging mix of random reads and writes along with sequential I/O (imagine 20 VPS doing dd bs=1M count=128 if=/dev/zero of=test conv=fdatasync).
Since most providers won’t offer their servers without RAID storage, let’s have a look at how RAID setups impact IOPS then. Read operations will usually not incur any penalty since they can use any disk in the array (total theoretical read IOPS available therefore being the sum of the individual disks’ read IOPS), whereas the same is not true for write operations as we can see from the following table:
||Backend Write IOPS per incoming write request
We can see that RAID 0 offers the best write IOPS performance – a single incoming write request will equate to a single backend write request – but we also know that RAID 0 bears the risk of total array loss in case a single disk fails. RAID 1 and 10, the latter being providers’ typical or most advertised choice, offers a decent tradeoff – 2 backend writes per single incoming write. RAID 5 and RAID 6, with their additional, robust setup, bear the largest penalty.
When calculating the effective IOPS, thus, keep in mind the write penalty individual RAID setups come with.
The effective IOPS performance of your array can be estimated using the following formula:
IOPSeff = n * IOPSdisk / ( R% + W% * FRAID )
with n being the number of disks in the array, R and W being the read and write percentage, and F being the RAID write factor tabled above.
We can also calculate the total IOPS performance needed based on an effective IOPS workload and a given RAID setup:
IOPStotal = ( IOPSeff * R% ) + ( IOPSeff * W% )
So if we need 500 effective IOPS, and expect around 25% read, and 75% write operations in a RAID 10 setup, we’d need:
500 * 0.25 + 500 * 0.75 * 2 = 875 total IOPS
i.e. our array would have to support at least 875 total, theoretical IOPS. How many disks/drives does this equate to? Today’s solid state drives will easily be able to handle that, but what about SATA or SAS based RAID arrays? A typical SAS 10k hard disk drive will give you around 100-140 IOPS. That means we will need 8 SAS 10k drives to achieve our desired IOPS performance.
All RAID levels except RAID 0 have significant impact on your storage array’s IOPS performance. The decision about which RAID level to use is therefore not only a question about redundancy or data protection, but also about resulting performance for your application’s needs:
- Evaluate your application’s performance requirements;
- Evaluate your application’s redundacy needs;
- Decide which RAID setup to use;
- Calculate the resulting IOPS performance necessary;
Calculate IOPS in a storage array by Scott Lowe, TechRepublic, 2/2010
Getting the hang of IOPS by Symantec, 6/2012