I have received a few questions relating to my previous post about NetApp VMware bootstorm results and want to answer them here.  I have also had a chance to look through the performance data gathered during the tests and have a few interesting data points to share. I also wanted to mention that I now have a pair of second generation Performance Accelerator Modules (PAM 2) in hand and will be publishing updated VMware boot storm results with the larger capacity cards.

What type of disk were the virtual machines stored on?

• The virtual machines were stored on a SATA RAID-DP aggregate.

What was the rate of data reduction through deduplication?

• The VMDK files were all fully provisioned at the time of creation. Each operating system type was placed on a different NFS datastore. This resulted in 50 virtual machines on each of 4 shares. The deduplication reduced the physical footprint of the data by 97%

A few interesting stats gathered during the testing. These numbers are not exact and due to the somewhat imprecise nature of starting and stopping statit in synchronization with the start and end of each test.

• The CPU utilization moved inversely with the boot time. The shorter the boot time, the higher the CPU utilization. This is not surprising as during the faster boots, the CPUs were not waiting around for disk drives to respond. More data was served from cache the the CPU could stay more utilized.
• The total NFS operations required for each test was 2.8 million.
• The total GB read by the VMware physical servers from the NetApp was roughly 49GB.
• The total GB read from disk trended down between cold and warm cache boots. This is what I expected and would be somewhat concerned if it was not true.
• The total GB read from disk trended down with the addition of each PAM. Again, I would be somewhat concerned if this was not the case.
• The total GB read from disk took a significant drop when the data was deduplicated. This helps to prove out the theory that NetApp is no longer going to disk for every read of a different logical block that points to the same physical block.

How much disk load was eliminated by the combination of dedup and PAM?

• The cold boots with no dedup and no PAM read about 67GB of data from disk. The cold boot with dedup and no PAM dropped that down to around 16GB. Adding 2 PAM (or 32GB of extended dedup aware cache) dropped the amount of data read from disk to less that 4GB.

I have migrated some data to ZFS filesystems recently and the capacity consumed has surprised me a couple times. In general, it has appeared that the data uses more capacity when stored on the ZFS filesystem. This prompted me to do a little investigating. Is ZFS using more capacity? Is it simply a reporting anomaly? Where is that space going? Does ZFS record size have a major impact? Does enabling compression have a significant impact?

In part, the extra space use is a result of ZFS reporting space utilization differently than other filesystems. When a ZFS filesystem is formatted, almost no capacity is used. A df command will show nearly the entire raw capacity. Many other filesystems take a portion of the raw capacity off the top and reserve it for metadata. This reserve will not show up in df. As data is added to the ZFS filesystem, blocks are allocated for both data and metadata. Both the data and metadata blocks will show up as used capacity. In many other filesystems, at least some of the metadata blocks will be taken from the reserve and only the data blocks will show as consumed capacity. For example, in Solaris, the du command will return the capacity used by the data blocks in a file. In ZFS, that du command returns the total space consumed by the file including metadata and compression. So the question at hand is, when storing a given set of files, does ZFS use more total space than other file systems? That one is difficult to test, given all the variables. But we can test various ZFS configuration options to determine the best settings for minimizing block use.

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After writing a couple of articles (here and here) about deduplication and how I think it should be implemented, I figured I would try it on a NetApp system I have in the lab. The goal of the testing here is to compare storage performance of a data set before and after deduplication. Sometimes capacity is the only factor, but sometimes performance matters. The test is random 4KB reads against a 100GB file. The 100GB file represents significantly more data than the test system can fit into its’ 16GB read cache. I am using 4KB because that is the natural block size for NetApp.

To maximize the observability of the results in this deduplication test, the 100GB file is completely full of duplicate data. For those who are interested, the data was created by doing a dd from /dev/zero. It does not get any more redundant than that. I am not suggesting this is representative of a real world deduplication scenario. It is simply the easiest way to observe the effect deduplication has on other aspects of the system.

This is the output from sysstat -x during the first test. The data is being transferred over NFS and the client system has caching disabled, so all reads are going to the storage device. (The command output below is truncated to the right, but the important data is all there.)

Random 4KB reads from a 100GB file – pre-deduplication:

 CPU   NFS  CIFS  HTTP   Total    Net kB/s   Disk kB/s     Tape kB/s Cache Cache  CP   CP Disk    FCP iSCSI   FCP  kB/s iSCSI  kB/s
                                  in   out   read  write  read write   age   hit time  ty util                 in   out    in   out
 19%  6572     0     0    6579  1423 27901  23104     11     0     0     7   16%   0%  -  100%      0     7     0     0     0     0
 19%  6542     0     0    6549  1367 27812  23265    726     0     0     7   17%   5%  T  100%      0     7     0     0     0     0
 19%  6550     0     0    6559  1305 27839  23146     11     0     0     7   15%   0%  -  100%      0     9     0     0     0     0
 19%  6569     0     0    6576  1362 27856  23247    442     0     0     7   16%   4%  T  100%      0     7     0     0     0     0
 19%  6484     0     0    6491  1357 27527  22870      6     0     0     7   16%   0%  -  100%      0     7     0     0     0     0
 19%  6500     0     0    6509  1300 27635  23102    442     0     0     7   17%   9%  T  100%      0     9     0     0     0     0

The system is delivering an average of 6536 NFS operations per second. The cache hit rate hovers around 16-17%. As you can see, the working set does not fit in primary cache. This makes sense. The 3170 has 16GB of primary cache and we are randomly reading from a 100GB file. Ideally, we would like to get a 16% cache hit rate (16GB cache / 100GB working set) and we are very close. The disks are running at 100% utilization and are clearly the bottleneck in this scenario. The spindles are delivering as many operations as the are capable of. So what happens if we deduplication this data?

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In a previous post I discussed the topic of deduplication for capacity optimization. Removing redundant data blocks on disk is the first, and most obvious, phase of deduplication in the marketplace. It helps to drive down the most obvious cost – the cost per GB of disk capacity. This market has grown quickly over the last few years. Both startups and established storage vendors have products that compete in the space. They are most commonly marketed as virtual tape library (VTL) or disk-to-disk backup solutions.

Does that mean that deduplication is a point solution for highly sequential workloads? No. There is another somewhat less obvious benefit of deduplication.

What storage administrator does not ask for more cache in the storage array? If I can afford 8GB, I want 16GB. If the system supports 16GB, I want 32GB. Whether it is for financial or technical reasons, cache is always limited. What about deduplicating the data in cache? When the workload is streaming sequential backup data from disk, this may not be very helpful. However, in a primary storage system with a more varied workload, this becomes very interesting.

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There is no debating that duplication is one of the hottest topics in IT. The question is if the hype has started to become bigger than the technology. Today, there are two primary use cases driving deduplication in the marketplace. The first is backup to disk and the second is virtual guest operating systems (VMware, Hyper-V, and Xen guests). (I will talk a bit about the disk to disk scenario in this article and the virtual guest topic in the next one.) These are both logical markets to adopt deduplication because they suffer from a common challenge. They both create a tremendous amount of redundant data on the disk array. The goal in both cases is to pack more data onto a disk drive and reduce the cost per GB. This is the first and most obvious use case for deduplication.

Disk drive capacity is growing exponentially, but disk performance is increasing at a much slower rate. In many cases, when helping customers size for their workload, performance drives the spindle count and not capacity. It is easy to meet the capacity needs with large drives, but will they meet the performance requirement? That is the problem. Often performance is what dictates the spindle count. It is no longer sufficient to size a storage device based solely on capacity requirements. This is a general challenge that must be taken into account when sizing a storage array.

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Last week I gave a presentation at the TechForum Roundtable in New York. Thank you to Priscilla Tate for running a great event.

Backup to disk is the number one application for deduplication today. My presentation covered the most common approaches vendors use to leverage deduplication in a backup environment. This includes client side, backup server based, inline processing, and post-processing.

  Demystifying Deduplication (7.9 MiB)

This topic was originally developed as part of a CTI Strategy Services consulting engagement. The customer credited us with saving them 6 months of effort.