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Consider an application that requires 1 TB of storage capacity and performs 4900 IOPS.

Application I/O size is 4KB. As it is business critical application, response time must be within an acceptable range. Specification of available disk drive:

Drive capacity = 73 GB

For rotational latency RPMs = 15,000 rpm

Average seek time = 5ms

Transfer rate: 40 MB/s

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Given : Application requires 1TB of storage capacity and performs 4900 IOPS

  • Application I/O size is 4 KB

  • Drive capacity = 73 GB

  • For Rotational latency RPM = 15000 rpm

  • Average seek time = 5ms

  • Transfer rate : 40 MB/s

Solution :

To calculate the time required to perform one I/O = seek time + (Rotational latency) + Transfer time.

Therefore

$5ms + \frac{30000}{15000} + \frac{4KB}{40 MB/s} = 7.1 ms$

Calculate the maximum number of IOPS a disk can perform as $\frac{1}{7.1 ms} = 0.140 \times 10^3 = 140 IOPS$

For acceptable response time, disk controller utilization must be less than 70% Therefore,

we get 140 x 0.7 = 98 IOPS

To meet the application performance requirement we need, 4900/98 = 50 disk

For capacity requirement, we need 1TB/73 GB

i.e. 14 disks.

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