In my previous blog entry, I mentioned the write penalty for RAID-5/6. This factor will figure heavily in the way we size the RAID-level for performance capacity planning.
It is difficult to ascertain what kind of IOPS and throughput that are required for an application, especially a database, to run well with additional room to grow. From a DBA or an application developer, I believe they would have adequate information to tell what is the numbers of users that the application can support, both average and peak, transactions per second (TPS), block size required for logs, database files and so on.
But as we are all aware, most of the time, these types of information are not readily available. So, coming from a storage angle, the storage administrator can advise the DBA or the application developer that the configured RAID group or volume or LUN is capable of delivering a certain number of IOPS and is able to achieve a certain throughput MB/sec. These numbers will be off the box itself immediately. Of course, other factors such as HBA speed, the FC/iSCSI configurations, the network traffic and so on will affect the overall performance delivery to the application. But we can safely inform the DBA and/or the application developer that this is what the storage is delivering out of the box.
The building blocks of all storage RAID groups/volumes/LUNs are pretty much your hard disk drives (HDDs) and/or Solid State Drives (SSDs). The manufacturer of these disks will usually publish the IOPS and throughput of individual drives but if these information is not available, we can construct IOPS of an individual HDD from its seek and latency times.
For example, if the HDD’s
average latency = 2.8 ms; average read seek = 4.2 ms; average write seek = 4.8 ms
then the IOPS can be calculated as
1
IOPS = ---------------------------------------
(average latency) + (average seek time)
Therefore from the details above,
1
IOPS = ------------------- = 136.986 IOPS
(0.0028) + (0.0045)
That’s pretty simple, right? But of course, it is easier to just accept that a certain type of disk will have a range of IOPS as shown in the table below:
| Disk Type | RPM | IOPS Range |
| SATA | 5,400 | 50-75 |
| SATA | 7,200 | 75-100 |
| SAS/FC | 10,000 | 100-125 |
| SAS/FC | 15,000 | 175-200 |
| SSD | N/A | 5,000-10,000 |
The information from the table above is just for reference only and by no means a very accurate one but it is good enough for us to determine the IOPS of a RAID group/volume/LUN. Let’s look at the RAID write penalty again in the table below:
| RAID-level | Number of I/O Reads |
Number of I/O for Writes |
RAID Write Penalty |
| 0 | 1 | 1 | 1 |
| 1 (1+0, 0+1) | 1 | 2 | 2 |
| 5 | 1 | 4 | 4 |
| 6 | 1 | 6 | 6 |
Next, we need to know what is the ratio of Reads vs Writes for that particular database or application. I mentioned earlier that in OLTP-type of applications, we usually take a 2:1 or 3:1 ratio in favour of Reads.
To make things simpler, let’s assume we create a RAID-6 volume of 6 data disks and 2 parity disks in a RAID-6 (6+2) configuration. The disks used are SATA disks of 7,200 RPM, with each individual disk of 100 IOPS. Assume we are using a ratio of 2:1 in favour of Reads, which gives us 66.666% and 33.333% respectively for Reads and Writes.
Therefore, the combined IOPS of the 8 disks in the RAID-6 configuration is probably about 800 IOPS. However, because of the write penalty of RAID-6, the effective IOPS for the RAID-6 volume will be lower than that. Let’s do some calculation to see what happens:
1) Read IOPS + Write IOPS = 800 IOPS
2) (0.66666 x 800) + (0.33333 x 800) = 800 IOPS
3) Read IOPS will be 0.66666 x 800 = 533.328 IOPS
4) Write IOPS will be 0.33333 x 800 = 266.664 IOPS. However, since RAID-6 has a write penalty of 6, this number has to be divided by 6. 266.664/6 will be 44.444 IOPS for Writes
Therefore, what the RAID-6 volume is capable of is approximately 533 IOPS for Reads and 44 IOPS for Writes.
We have determined IOPS for the RAID volume but what about throughput. Throughput is determined by the block size used. Assume that our RAID-6 volume uses a 4-K block size. With a combined effective IOPS of 577 (533+44), we multiply the IOPS with the block size
Throughput = 577 IOPS x 4-KB
= 2308KB/sec
Therefore when I/O is sustained in a sequential manner, the effective throughput is 2308KB/sec.
On the other hand, we often were told to add more spindles to the volume to increase the IOPS. This is true, to a point, where the maximum amount of IOPS that can be delivered will taper into a flatline, because the I/O channel to the RAID volume has been saturated. Therefore, it is best to know that adding more spindles does not always equate to a higher IOPS.
Performance sizing for a database or an application is both a science and an art. Mathematically, we can prove things to a a certain amount of accuracy and confidence but each storage platform is very different in the way they handle RAID. Newer storage platforms have proprietary RAID that nowadays, it does not matter much what kind of RAID is best for the application. Vendors such as IBM XIV has RAID-X which both radical in design and implementation. NetApp will almost always say RAID-DP is the best no matter what, because RAID-DP is all NetApp.
So there is no right or wrong to choose the RAID-level for the application. But it is VERY important to know what are the best practice are and my advice is everyone is to do Proof-of-Concepts, and TEST, TEST, TEST! And ASK QUESTIONS!
