GATE | Gate IT 2007 | Question 44

A hard disk system has the following parameters :

• Number of tracks = 500
• Number of sectors/track = 100
• Number of bytes /sector = 500
• Time taken by the head to move from one track to adjacent track = 1 ms
• Rotation speed = 600 rpm.

What is the average time taken for transferring 250 bytes from the disk ?
(A) 300.5 ms
(B) 255.5 ms
(C) 255.0 ms
(D) 300.0 ms


Answer: (D)

Explanation: Avg. time to transfer = Avg. seek time + Avg. rotational delay + Data transfer time

• Avg Seek Time –  time taken  to move from 1st track to 1sr track : 0ms, 1st to 2nd : 1ms, 2ms, 3ms,….499ms
  Avg Seek time =( ∑0+1+2+3+…+499)/500 = 249.5 ms
• Avg Rotational Delay – RMP : 600 , 600 rotations in 60 sec (one Rotation = 60/600 sec = 0.1 sec) So, Avg Rotational Delay = 0.1/2= 50 ms
• Data Transfer Time: In One 1 Rotation we can read data on one track = 100 * 500 = 50,000 B data is read in one rotation. 250 bytes -> 0.1 * 250 / 50,000 = 0.5 ms

Therefore ATT = 249.5+50+0.5 = 300 ms

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