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GATE | GATE-CS-2015 (Set 1) | Question 65

  • Difficulty Level : Expert
  • Last Updated : 28 Jun, 2021

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


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


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