Question

Q1/A 7200 RPM drive has a full-stroke seek time of 18 ms. It transfers data internally at a rate of 40 MB s−1 and has a block size of 32 kB. Determine the IOPS for that drive.

Q2/The disk service time is 18 ms when the controller utilisation is 50 % . Calculate the expected disk service time when the controller utilisation is 70 %

Answer 1

Q1 . Given:

Rotational speed = 7200 RPM

full stroke seek time= 18 ms

transfer rate = 40 MB

Block size = 32 kB

Average latency = (1/Rotational speed / 60 ) *0.5 *1000

= (60/7200 ) * 500

= 30000/7200

= 4.17 ms = 0.00417 sec

Average seek time = (full stroke seek time /3)

= 18/3 = 6 ms = 0.006 sec

IOPS = 1/ (avg latency + avg seek time)

= 1/ (0.00417+0.006)

= 1/0.01017

= 98

That drive has about 98 IOPS

Q2. Given :

Disk service time (T_s)= 18 ms for 50% controller utilization

Response time (T_r) = T_s / 1- utilization

= 18/(1-0.5)

= 18/0.5

= 36 ms

At 70 % utilization,

Response time (T_r) = T_s / 1- utilization

Thus,

Disk service time (T_s)=  Response time (T_r) * (1- utilization)

= 36 * (1-0.7)

= 36*0.3

= 10.8 ms