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GATE CS 2015 Set 2 · Operating Systems

Question 59

NATPrinted as CS Q.492 marksNo negative markingMedium (estimate)

Question

Consider a typical disk that rotates at 15000 rotations per minute (RPM) and has a transfer rate of 50×10650 \times 10^6 bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller's transfer time is 10 times the disk transfer time, the average time (in milliseconds) to read or write a 512-byte sector of the disk is _____.

Official answer and solution
Accepted answers
6.1 to 6.2
From
Official GATE 2015 answer key

Solution

Sum seek time, rotational latency, disk transfer time, and controller transfer time.

  1. Rotational Delay:
Speed=15000 RPM=1500060 rotations/sec=250 rotations/sec.\text{Speed} = 15000\text{ RPM} = \frac{15000}{60}\text{ rotations/sec} = 250\text{ rotations/sec}.

Time for one full revolution:

Trev=1250 sec=4 ms.T_{\text{rev}} = \frac{1}{250}\text{ sec} = 4\text{ ms}.

Average rotational delay = Trev2=2 ms\dfrac{T_{\text{rev}}}{2} = 2\text{ ms}.

  1. Seek Time:
Average seek time=2×(average rotational delay)=2×2 ms=4 ms.\text{Average seek time} = 2 \times (\text{average rotational delay}) = 2 \times 2\text{ ms} = 4\text{ ms}.
  1. Disk Transfer Time:
Disk transfer time=512 bytes50×106 bytes/sec=10.24×10−6 sec=0.01024 ms.\text{Disk transfer time} = \frac{512\text{ bytes}}{50 \times 10^6\text{ bytes/sec}} = 10.24 \times 10^{-6}\text{ sec} = 0.01024\text{ ms}.
  1. Controller Transfer Time:
Controller transfer time=10×(disk transfer time)=10×0.01024 ms=0.1024 ms.\text{Controller transfer time} = 10 \times (\text{disk transfer time}) = 10 \times 0.01024\text{ ms} = 0.1024\text{ ms}.
  1. Total Average Access Time:
Tavg=4 ms+2 ms+0.01024 ms+0.1024 ms=6.11264 ms.T_{\text{avg}} = 4\text{ ms} + 2\text{ ms} + 0.01024\text{ ms} + 0.1024\text{ ms} = 6.11264\text{ ms}.
Final answer

6.11

Common trap

Omitting the controller transfer time or forgetting to divide the full rotation time by 2 to get the average rotational delay.

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

Source

Paper
GATE 2015 Computer Science and Information Technology (CS) Session 2
Question
59 of 65, printed as CS Q.49
Section
Computer Science and Information Technology

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