GATE CS Mastery
Cheat sheets

GATE CS 2011 · Operating Systems

Question 44

MCQPrinted as CS Q.442 marksWrong answer: −2/3Easy (estimate)

Question

An application loads 100 libraries at startup. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as 10 ms. Rotational speed of disk is 6000 rpm. If all 100 libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been positioned at the start of the block may be neglected.)

Choose one option
Official answer and solution
Official answer
B
From
Official GATE 2011 answer key

Solution

Calculate the average disk access time per library and multiply by 100 libraries.

  1. Given seek time to a random cylinder = 10 ms10\text{ ms}.
  1. Rotational speed = 6000 rpm. The time for one full rotation is:
Trot=60 s6000=0.01 s=10 msT_{\text{rot}} = \frac{60\text{ s}}{6000} = 0.01\text{ s} = 10\text{ ms}
  1. The average rotational latency is half of one full rotation:
Rotational Latency=10 ms2=5 ms\text{Rotational Latency} = \frac{10\text{ ms}}{2} = 5\text{ ms}
  1. Neglecting transfer time, total time per disk access is:
Taccess=10 ms+5 ms=15 ms=0.015 sT_{\text{access}} = 10\text{ ms} + 5\text{ ms} = 15\text{ ms} = 0.015\text{ s}
  1. For 100 independent random library loads:
Total Time=100×15 ms=1500 ms=1.50 s\text{Total Time} = 100 \times 15\text{ ms} = 1500\text{ ms} = 1.50\text{ s}
Final answer

B

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2011 Computer Science and Information Technology (CS)
Question
44 of 65, printed as CS Q.44
Section
Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.