GATE CS 2015 Set 2 — Question 42
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Computer Organization & Architecture → I/O Organization → Disk Structure & Access Time
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Question
Consider a typical disk that rotates at 15000 rotations per minute (RPM) and has a transfer rate of 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 ________.
Correct answer
6.1 to 6.2
Solution
1.Calculate Average Rotational Delay:
- Rotational speed = 15000 RPM = rotations/sec.
- Time for one rotation = sec = 4 ms.
- Average rotational delay = rotation time = ms = 2 ms.
- Given as twice the average rotational delay.
- Average seek time = ms = 4 ms.
- Transfer rate = bytes/sec.
- Sector size = 512 bytes.
- Disk Transfer Time = sec = s s = 0.01024 ms.
- Given as 10 times the disk transfer time.
- Controller Transfer Time = ms = 0.1024 ms.
- Total Time = Seek Time + Rotational Delay + Disk Transfer Time + Controller Transfer Time
- Total Time = ms
- Total Time = ms.
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