Loading libraries on the disk - Example - Operating System

Q.  An application loads 100 libraries at startup and loading each library exactly one disk access is required. 10ms is the seek time of the disk to a random location and 6000rpm is the rotational speed of the disk. Neglect the time to transfer data from the disk block once the head has been positioned at the start of the block. What will be the time taken by the application to load all the libraries, if all 100 libraries are loaded from random location on the disk?
- Published on 27 Aug 15

a. 1.50s
b. 1.00s
c. 1.87s
d. 0.74s

ANSWER: 1.50s

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    Discussion

  • Nirja Shah   -Posted on 18 Nov 15
    - As the transfer time can be neglected, the average access time is sum of average seek time and average rotational latency.

    - 10 ms is the average seek time for random location time.

    - The average rotational latency is half of the time needed for complete rotation.

    - It takes 1 minute to complete 6000 rotations.

    - So one rotation will take 60/6000 seconds which is 10 ms.

    - Therefore average rotational latency is half of 10 ms, which is 5ms.

    - Average disk access time = seek time + rotational latency
    = 10 ms + 5 ms
    = 15 ms
    For 100 libraries, the average disk access time will be 15∗100 ms

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