BVQ Use Case: cost effective storage planning

This whitepaper shows how you can use BVQ analysis to exactly determine which kind of storage you need for a specific environment

This whitepaper is also available as PDF:
BVQ Use Case cost effective storage planning.pdf

Use case target

A 40 TB mail system has to be moved from an oversized 10k, 300GB storage to a new storage environment with lower cost 7.2k, 1TB disk drives and Raid 6. The question is: how many disks are needed in the new 7.2k disk arrays to deliver the same performance as before?

BVQ is able to analyze the aggregated performance values of all volumes, which make up the mail system.


Fact based knowledge about workload picture 2 (known by measurement)

  • 40TB capacity is needed
  • Mail system consumes 2500 IOPS
  • R/W distribution is 55% read
  • Diagram shows a constant read cache hit rate of 90% and write cache hit rate (overwrite in cache) of 15%. This will reduce the IOPS on disk by far.


  • Safety margins:
    There are big transfer rates, which will increase write penalties for RAID 6. Additionally a conservative IO estimation for 7k drives is used. It is ignored that the storage system behind SVC may further improve IO behavior by own caching.


Picture 1: BVQ Treemap with cost centers and applications:
the BVQ accounting package allows the grouping of volumes and to add them to applications or even cost centers. This is a good starting point for the investigation because not only single volumes are interesting but all volumes belonging to the mail system. Now a performance analysis can be started where all IOPS, transfer sizes and cache values are aggregated into single curves.


Picture 2: BVQ performance view screen
: the picture shows the aggregated values of IOPS, transfer size, cache hit read and cache hit write of all volumes. It is obvious that the backup IO patterns at night are completely different to the ones at business hours. But in both cases we have very stable cache hit results with 90% cache hit read and 15% cache hit write (overwrite in cache).



Picture 3: IOPS calculations
: this is a spreadsheet which is used to calculate the numbers of disks needed for specific IOPS scenarios with cache and RW distribution. It starts with the 2500 IOPS needed and the cache hits from read and write are subtracted. So it ends with only 1094 effective IOPS which have to be covered by the disks. Then more or less conservative IOPS/spindle constants and a RAID penalty are added and all this together leads to 48 spindles to cover the 1094 IOPSeff.

What makes this result particularly valuable?

With this based-facts approach we are able to find the most economic solution to meet the performance target. The hardware purchasing costs of today and the follow-up costs of tomorrow are reduced because fewer spindles, enclosures, systems, floor space, power and cooling are needed.

Ideas to improve even more!

  • Challenge backup duration – if it is critical – if not we can reduce the needed IOPS down to 2000 IOPS or even less. With 2000 IOPS we would end with only 38 disks to achieve the desired performance – (please keep in mind that we need 40TB. So 2TB disk drives could be used).
    One guess for this special example would be – it is critical because of the transfer sizes - backup starts at 9pm and finishes approx. 6am.
  • Having a look on Easy Tier–analysis of single disks to check whether we find IOP patterns which are handled preferred by Easy Tier.


Other scenarios without fact based knowledge
Without fact based knowledge about RW distribution and caches you are forced to work with assumptions.

  • Let’s assume RAID 6 and 50% read and 70% cache hit for read
      -> IOPSeff = 1625 -> 72 Disks
  • Let’s assume RAID 6 and nothing more
      -> IOPSeff = 2500 -> 201 Disks


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