IOPS Estimation for IBM V5000

Written on:January 1, 2015
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In this article, we estimate controller’s performance of midrange storage system IBM Storwize V5000 for different workload profiles, as well as other parameters of IOPS per GHz, IOPS per GB.

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 – 2,5 GHz.
  • The number of cores on one controller – 4.
  • The sum frequency calculation – 2,5×4 GHz = 10 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). Multiply 10 GHz x 10,000 IOPS / GHz, we obtain a rough estimation 100000 IOPS per controller. Thus, the performance of a pair of controllers is equal to 200000 IOPS. We can use these rough estimates for self-examination.
  • We take information about recommended configurations of RAID groups in «V5000 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 480.

There is a recommendation to place Flash drives in the controllers shelf, cos it is connected with 2 SAS ports to PM8018 controllers. In the article “IBM Storwize V3700: a Brief Overview” we used 8 Flash drives. Since V5000 system is not much different from V3700 in part of SAS chips, but they have markedly different processors (the difference in the sum frequency is more than 2.5 times), and the IO core in the new processors uses PCIE V3 standard, so we will use 3 times more Flash drives (whole disk shelf) than in V3700. Thus, we will use 24 Flash disks in the configuration RAID5 7+1 in accordance with the recommendation of IBM.

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 – 3.
  • The configuration for RAID groups of SAS 15K drives – RAID5 7+1.
  • The quantity of RAID groups for SAS 15K drives – 55.


Max IOPS for one IO-group is 158447.
Max IOPS for one controller is about 79224.
The ratio of IOPS/GHz is 7922.
The ratio of IOPS/GB is 0,716.

Real Flash storage capacity is about 15 TB.
Real SAS 15K storage capacity is about 202 TB.

Let’s compare this result with a rough estimation, which we did in the beginning of this article. Then we got 100000 IOPS for one controller. The difference from the new result is about 21%.

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 85895.
Max IOPS for one controller (DB like) is about 42948.
The ratio of IOPS per GHz is about 4295.
The ratio of IOPS per GB is about 0,388.

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.

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