To get the total amount of required IOPS, multiply the number of workloads by the number of functional application IOPS:

Total IOPS = (100 current workloads + 25 future workloads) * 65 IOPS = 8125 IOPS

To calculate the functional IOPS required for a specific workload, use the following formula:

Functional workload IOPS = (workload IOPS * %reads) + ((workload IOPS * %writes) * write penalty)

The write penalty is based on the number of IO operations a specific RAID configuration requires for a single write request. Writing data to multiple disks in mirror or parity calculations in a RAID5 or RAID6 configuration adds IO operations to the write request.
The write request is not completed until the data and parity are written to the disks.

The following table illustrates the write penalty based on the RAID levels:

Based on the requirements of 65 IOPS per workload with 90 percent reads and 10 percent writes on the storage configured in RAID5, the actual workload IOPS would be 85 IOPS:

Functional application IOPS = (65 * .90) + ((65* .10) * 4) = ~85 IOPS

Each disk in a storage array is able to provide a number of IOPS. The number of IOPS a single disk can deliver is calculated from the average latency and the average seek time of the disk. The formula to calculate disk performance is as follows:

IOPS = 1 / (average latency in milliseconds + average seek time in milliseconds)

The following table lists some approximate IOPS provided, based on the spindle speed and the drive type:

Based on the number of IOPS required, there will be a need of 47 15k SAS drives to support the workload:

8125 IOPS / 175 IOPS per drive = 46.4 or 47 15k SAS drives

The same workload on drives configured in RAID10 sets would require 52 15k SAS drives to provide the required IOPS:

Functional workload IOPS = (65 * .90) + ((65* .10) * 2) = 72 IOPS

Total IOPS = (100 current workloads + 25 future workloads) * 72 IOPS = 9,000 IOPS

9,000 IOPS / 175 IOPS per Drive = 51.4 or 52 15k SAS drives