Gate Question

Consider a disk pack with a seek time of 4 milliseconds and rotational speed of 10000 rotations per minute (RPM). It has 600 sectors per track and each sector can store 512 bytes of data. Consider a file stored in the disk. The file contains 2000 sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is _________. (A) 14020 (B) 14000 (C) 25030 (D) 15000 [GATE CS 2015]

Answer: (A)

Explanation

Seek time (given) = 4ms

RPM = 10000 rotation in 1 min [60 sec]

So, 1 rotation will be =60/10000 =6ms [rotation speed]

Rotation latency= 1/2 * 6ms=3ms

# To access a file,

total time includes =seek time + rot. latency +transfer time

To calc. transfer time, find transfer rate

Transfer rate = bytes on track /rotation speed

so, transfer rate = 600*512/6ms =51200 B/ms

transfer time= total bytes to be transferred/ transfer rate

so, Transfer time =2000*512/51200 = 20ms

Given as each sector requires seek tim + rot. latency

= 4ms+3ms =7ms

Total 2000 sector takes = 2000*7 ms =14000 ms

To read entire file ,total time = 14000 + 20(transfer time)

= 14020 ms

Seek Time is one of the key components of Disk Scheduling, Before going to Seek Time, let’s first discuss Disk Scheduling. Disk Scheduling is done by operating systems to schedule I/O requests arriving for the disk. Disk scheduling is also known as I/O Scheduling. Multiple I/O requests may arrive by different processes and only one I/O request can be served at a time by the disk controller. Thus other I/O requests need to wait in the waiting queue and need to be scheduled.