IOPS Estimation for IBM V7000 Gen2

Written on:November 8, 2014
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In this article, we estimate the performance of the storage system controllers of the new IBM Storwize V7000 Gen2 for different workload profiles, as well as other parameters of IOPS per GHz, IOPS per GB.

In the article «IBM Storwize V7000 Gen2» it was noted that it was not necessary to consider 4 controller pairs, which are interrelated with common management, as a single storage system. Therefore, all our calculations are only applicable to one controller pair or IO-group in the terminology of IBM.

All the performed calculations are based on the methodology described in the article «Methodology of IOPS Estimation».

Let’s gather all necessary information needed for a calculation.

  • The frequency of processor cores – 1,9 GHz.
  • The number of cores on one controller – 8.
  • The sum frequency calculation – 8×1,9 GHz = 15,2 GHz.
  • At this stage we can get a rough estimation of IOPS for one controller. To do this, it is necessary to use the relation between the number of IOPS and sum frequency of the controller. This value is specified in the methodology (10000 IOPS / GHz). Let’s multiply 15.2 GHz x 10,000 IOPS / GHz, we obtain a rough estimation 152000 IOPS per controller. Thus, the performance of a pair of controllers is equal to 304000 IOPS. We can use these rough estimates for self-examination.
  • We take information about recommended configurations of RAID groups in «V7000 Implementation Guide»:
    • RAID 5 SSD are used in the configuration 7+1.
    • RAID 5 SAS are used in the configuration 7+1.
  • Max disk drive quantity for one pair of controllers is 504.

There is a recommendation to place Flash drives in the controllers shelf, cos it is connected with 2 SAS ports to  PM8xxx controller, in the article “IBM Storwize V7000 Gen2». Flash drives have 20 times more performance than general SAS drives.

Thus, one shelf of 24 Flash drives exceeds the total number of IOPS of 430 SAS drives.

In our configuration we see that 2 shelves of Flash drives (48 disks) give 2 times more IOPS performance than general SAS drives that can be installed in other disk slots (504 – 48 + 2 = 454 disks).

In this case, we get low load imbalance on SAS ports. The adding of more Flash drives will lead to long delays for SAS drives. Therefore, we will use the configuration with 48 Flash disks.

The estimation for max IOPS (Cache Read Miss)

Let’s set the values for the estimation:

  • Read operations – 99%.
  • The configuration for RAID groups of Flash drives – RAID5 7+1.
  • The quantity of RAID groups for Flash drives – 6.
  • The configuration for RAID groups of SAS 15K drives – RAID5 7+1.
  • The quantity of RAID groups for SAS 15K drives – 54.
  • We use one RAID5 6+1 group of SAS 15K drives to obtain 504 disk drives with hot swap.


Max IOPS for one IO-group is 239825.
Max IOPS for one controller is 119913.
The ratio of IOPS/GHz is 7889.
The ratio of IOPS/GB is 1,017.

Real Flash storage capacity is 29 TB.
Real SAS 15K storage capacity is 201 TB.

In general, we obtained  not a bad configuration with good values of IOPS and IOPS per GB.

Let’s compare this result with a rough estimation, which we did in the beginning of this article. Then we got 152000 IOPS for one controller. The difference from the new result is 21%.

That was a good estimation, taking into account that it had been made only on the basis of knowledge about the frequency of processor cores and their quantities on one controller.

The IOPS estimation for DB like workload (Cache Read Miss)

The Initial parameters are the same, but the percentage of read operations is different. A typical ratio between read and write operations is equal to 70/30 for databases (OLTP workload).


Max IOPS for one IO-group (DB like) is 130011.
Max IOPS for one controller (DB like) is 65006.
The ratio of IOPS per GHz is 4278.
The ratio of IOPS per GB is 1,017.

The real capacity of disk space on Flash and SAS drives is the same.

Note that the ratio of IOPS per GHz for more real workload profile is 2 times less than the characteristic value, which is specified in the methodology.

This indicates that the a half of cores is quite enough for this configuration, and the other part of cores can be used for other tasks.

In the article «IBM Storwize V7000 Gen2» it is told that “Up to 4 CPU cores on the controller can be used when RtC is active“.

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