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What kind of IOPS and throughput do you get from RAID-5/6? – Part 2

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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!

SSDs coming into mainstream … be Ready!

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There has been a slew of SSD news in the storage blogosphere with the big one from eBay.

eBay has just announced that it has 100TB of SSDs from Nimbus Data Systems. On top of that, OCZ, SanDisk and STEC, all major SSD manufacturers, have announced a whole lot of new products with the PCIe SSD cards leading the way. The most interesting thing was the factor of $/GB has gone down significantly, getting very close to the $/GB of spinning disks. This is indeed good news to the industry because SSDs delivers low latency, high IOPS, low power consumption and many other new benefits.

Side note: As I am beginning to understand more about SSDs, I found out that NAND flash SSD has a latency in the microseconds compared to spinning HDDs, which has milliseconds latency range. In addition to that DRAM SSDs have latency that is in the range of nano seconds, which is basically memory type of access. DRAM SSDs are of course, more expensive. 

The SSDs are coming very soon into the mainstream, and this will inadvertently, drive a new generation of applications and accelerate growth in knowledge acquisition. We are already seeing the decline of Fibre Channel disks and the rise of SAS and SATA disks but SSDs in the enterprise storage, as far as I am concerned, brings forth 2 new challenges which we, as professionals and users in the storage networking environment, must address.

These challenges can be simplified to

  1. Are we ready?
  2. Where is the new bottleneck?

To address the first challenge, we must understand the second challenge first.

In system architectures, we know of various of performance bottlenecks that exist either in CPU, memory, bus, bridge, buffer, I/O devices and so on. In order to deliver the data to be process, we have to view the data block/byte service request in its entirety.

When a user request for a file, this is a service request. The end objective is the user is able to read and write the file he/she requested. The time taken from the beginning of the request to the end of it, is known as service time, which latency plays a big part of it. We assume that the file resides in a NAS system in the network.

The request for the file begins by going through the file system layer of the host the user is accessing, then to the user and kernel space, moving on through the device driver of the NIC card, through the TCP/IP stack (which has its own set of buffer overheads and so on), passing the request through the physical wire. From there it moves on through the NAS system with the RAID system, file system and so on until it reaches the file request. Note that I have shortened the entire process for simple explanation but it shows that the service request passes through a whole lot of things in order to complete the request.

Bottlenecks exist everywhere within the service request path and is also subjected to external factors related to that service request. For a long, long time, I/O has been biggest bottleneck to the processing of the service request because it is usually and almost always the slowest component in the entire scheme of things.

The introduction of SSDs will improve the I/O performance tremendously, into the micro- or even nano-seconds range, putting it in almost equal performance terms with other components in the system architecture. The buses and the bridges in the computer systems could be the new locations where the bottleneck of a service request exist. Hence we have use this understanding to change the modus operandi of the existing types of applications such as databases, email servers and file servers.

The usual tried-and-tested best practices may have to be changed to adapt to the shift of the bottleneck.

So, we have to equip ourselves with what SSDs is doing and will do to the industry. We have to be ready and take advantage of this “quiet” period to learn and know more about SSD technology and what the experts are saying. I found a great website that introduces and speaks about SSD in depth. It is called StorageSearch and it is what I consider the best treasure trove on the web right now for SSD information. It is run by a gentleman named Zsolt Kerekes. Go check it out.

Yup, we must be get ready when SSDs hit the mainstream, and ride the wave.