It has been a tough week for me and that’s why I haven’t been writing much this week. So, right now, right after dinner, I am back on keyboard again, continuing where I have left off with NetApp’s usable capacity.
A blog and a half ago, I wrote about the journey of getting NetApp’s usable capacity and stopping up to the point of the disk capacity after rightsizing. We ended with the table below.
|Manufacturer Marketing Capacity||NetApp Rightsized Capacity|
* The size of 34.5GB was for the Fibre Channel Zone Checksum mechanism employed prior to ONTAP version 6.5 of 512 bytes per sector. After ONTAP 6.5, block checksum of 520 bytes per sector was employed for greater data integrity protection and resiliency.
At this stage, the next variable to consider is RAID group sizing. NetApp’s ONTAP employs 2 types of RAID level – RAID-4 and the default RAID-DP (a unique implementation of RAID-6, employing 2 dedicated disks as double parity).
Before all the physical hard disk drives (HDDs) are pooled into a logical construct called an aggregate (which is what ONTAP’s FlexVol is about), the HDDs are grouped into a RAID group. A RAID group is also a logical construct, in which it combines all HDDs into data or parity disks. The RAID group is the building block of the Aggregate.
So why a RAID group? Well, first of all, (although likely possible), it is not prudent to group a large number of HDDs into a single group with only 2 parity drives supporting the RAID. Even though one can maximize the allowable, aggregated capacity from the HDDs, the data reconstruction or data resilvering operation following a HDD failure (disks are supposed to fail once in a while, remember?) would very much slow the RAID operations to a trickle because of the large number of HDDs the operation has to address. Therefore, it is best to spread them out into multiple RAID groups with a recommended fixed number of HDDs per RAID group.
RAID group is important because it is used to balance a few considerations
- Performance in recovery if there is a disk reconstruction or resilvering
- Combined RAID performance and availability through a Mean Time Between Data Loss (MTBDL) formula
Different ONTAP versions (and also different disk types) have different number of HDDs to constitute a RAID group. For ONTAP 8.0.1, the table below are its recommendation.
So, given a large pool of HDDs, the NetApp storage administrator has to figure out the best layout and the optimal number of HDDs to get to the capacity he/she wants. And there is also a best practice to set aside 2 HDDs for a RAID-DP configuration with every 30 or so HDDs. Also, it is best practice to take the default recommended RAID group size most of the time.
I would presume that this is all getting very confusing, so let me show that with an example. Let’s use the common 2TB SATA HDD and let’s assume the customer has just bought a 100 HDDs FAS6000. From the table above, the default (and recommended) RAID group size is 14. The customer wants to have maximum usable capacity as well. In a step-by-step guide,
- Consider the hot sparing best practice. The customer wants to ensure that there will always be enough spares, so using the rule-of-thumb of 2 HDDs per 30 HDDs, 6 disks are set aside as hot spares. That leaves 94 HDDs from the initial 100 HDDs.
- There is a root volume, rootvol, and it is recommended to put this into an aggregate of its own so that it gets maximum performance and availability. To standardize, the storage administrator configures 3 HDDs as 1 RAID group to create the rootvol aggregate, aggr0. Even though the total capacity used by the rootvol is just a few hundred GBs, it is not recommended to place data into rootvol. Of course, this situation cannot be avoided in most of the FAS2000 series, where a smaller HDDs count are sold and implemented. With 3 HDDs used up as rootvol, the customer now has 91 HDDs.
- With 91 HDDs, and using the default RAID group size of 14, for the next aggregate of aggr1, the storage administrator can configure 6 x full RAID group of 14 HDDs (6 x 14 = 84) and 1 x partial RAID group of 7. (91/14 = 6 remainder 7). And 84 + 7 = 91 HDDs.
- RAID-DP requires 2 disks per RAID group to be used as parity disks. Since there are a total of 7 RAID groups from the 91 HDDs, 14 HDDs are parity disks, leaving 77 HDDs as data disks.
This is where the rightsized capacity comes back into play again. 77 x 2TB HDDs is really 77 x 1.69TB = 130.13TB from an initial of 100 x 2TB = 200TB.
If you intend to create more aggregates (in our example here, we have only 2 aggregates – aggr0 and aggr1), there will be more consideration for RAID group sizing and parity disks, further reducing the usable capacity.
This is just part 2 of our “Playing with NetApp Capacity” series. We have not arrived at the final usable capacity yet and I will further share that with you over the weekend.
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|
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|
|1 (1+0, 0+1)||1||2||2|
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!
It’s a beautiful Saturday morning … the sun is out, and the birds are chirping … and here I am, thinking about RAID-5/6. What’s wrong with me?
Anyway, have you ever wondered almost all your volumes are in a RAID-5/6 configuration? Like an obedient child, the answer would probably be “Oh, my vendor said it is good for me …”
In storage, the rule is applications-read, applications-write. And different applications have different behaviors but typically, they fall under 2 categories:
- Random access
- Sequential access
The next question to ask is how much Read/Writes ratio (or percentage) is in that Random Access behavior and how much of Read/Write ratio in Sequential Access behavior.
We usually pigeonhole transactional databases such as SQL Server, Oracle into OLTP-type characteristics with random access being the dominant access method. Similarly, email applications such as Exchange, Lotus and even SMTP into similar OLTP-type characteristics as well. We typically do a 2:1 or 3:1 ratio for OLTP-type applications with Read heavy and less of Writes. Data warehouse type of databases tend to be more sequential.
However, even within these OLTP applications, there are also sequential access behaviors as well, as the following table for a database shows:
|Operation||Random or Sequential||Read/Write Heavy||Block Size|
|DB-Log||Random (Sequential in log recovery)||Write Heavy unless you are doing log recovery||1KB – 64KB|
|DB-Data Files||Random||Read/Write mix dependent on load||4KB – 32KB|
|Batch insert||Sequential||Write Heavy||8KB – 128KB|
|Index scan||Sequential||Read Heavy||8KB – 128KB|
We will look into 4 RAID-levels in this scenario and see how each RAID-level applies to an OLTP-type of environment. These RAID levels are RAID-0, RAID-1 (1+0, 0+1 included), RAID-5 and RAID-6.
RAID-0 is the baseline, with 1 x Read and 1 x Write being processed as per normal.
In RAID-1, it would require 2 x Writes and 1 x Read, because the write operation is mirrored. The RAID penalty is 2.
To avoid the cost of RAID-1, RAID-5 is almost always the RAID level of choice (unless you speak to those NetApp fellas). RAID-5 is a parity-based RAID and require 2 x Read (1 to read the data block and 1 to read the parity block) AND 2 x Write (1 to write the modified block and 1 to write the modified parity). Hence it has a RAID penalty of 4.
RAID-6 was to address the risk of RAID-5 because disk capacity are so freaking large now (3TB just came out). To rebuild a large-TB drive would take longer time and the RAID-5 volume is at risk if a second disk failure occurs. Hence, double parity RAID in RAID-6. But unfortunately, the RAID penalty for RAID-6 is 6!
To summarize the RAID write penalty,
|RAID-level||Number of I/O Reads
||Number of I/O for Writes
||RAID Write Penalty|
|1 (1+0, 0+1)||1||2||2|
So, it is well known that RAID 0 has good performance for reads and writes but with absolutely no protection. RAID-1 would be good for random reads and writes but it is costly. RAID-5 is good for applications with a high ratio of sequential reads vs writes (2:1, 3:1 as mentioned), and RAID-6, errr … should be taken similarly as RAID-5 with some additional performance penalty.
With that in mind, a storage administrator must question why a particular RAID-level was proposed to the database or any like-applications.
I am going out to enjoy the Saturday now … and today, August 13th is the World’s Left-Handed Day. More about this RAID penalty and IOPS in my next entry.