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Exercise 6 - Disk Operation

The average I/O size of an application is 64 KB. The following specifications are available from the disk manufacturer: average seek time = 5 ms, 7,200 rpm, transfer rate = 40 MB/s. Determine Rotational latency and Seek time and number of Read/Write operations (IO Operations).

Solution

Given specification

Application size 64 KB = 65,536 bytes

T avg seek = 5 ms

Max rotation = 7200 rpm

Transfer rate = 40 MB/sec = 41,943,040 bytes/sec

Rotational Latency = 1/2 x Max rotation = 1/2 x (60 sec /7200 rpm) = 0.00416 sec = 4.16 ms

Avg Transfer time = 65536/41943040 = 0.00156 sec = 1.56 ms

Avg seek time = 5 ms

T access time = T avg seek + T avg rotation + T avg transfer = 5 + 4.16 + 1.56 = 10.72 ms

The average IO operation (IOPS) = 1/(Disk Average Latency in seconds + Disk Average Seek Time in seconds)

Average IOPS = 1/(0.00416+0.005) = 109

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