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Consider a Disk I/O transfer, in which 1500 bytes are to be transferred, but number of bytes on a track is 1000, and rotation speed of disk is 1500 rps but the average time required to move the disk arm to the required track is 15 ms, then what will be total access time?
  • a)
    16.33 ms
  • b)
    15.33 ms
  • c)
    16.33 μs
  • d)
    15.33 μs
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Consider a Disk I/O transfer, in which 1500 bytes are to be transferre...

Given:
Ta→ transfer time T
s → average seek time =15 ms
r → rotation speed in rpms = 1500 rps
b → number of bytes to be transferred = 1500 bytes
N → num ber of bytes on a track = 1000 bytes

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Most Upvoted Answer
Consider a Disk I/O transfer, in which 1500 bytes are to be transferre...
To calculate the total access time, we need to consider three components: seek time, rotational delay, and transfer time.

1. Seek Time: The average time required to move the disk arm to the required track is given as 15 ms.

2. Rotational Delay: The rotation speed of the disk is given as 1500 rps. Since the average rotational delay is half a rotation, it will take (1/2 * 1/1500) seconds to reach the beginning of the desired sector.

3. Transfer Time: The number of bytes on a track is given as 1000. We want to transfer 1500 bytes, so we will need to read two tracks. The time required to transfer one track is (1000/1500) seconds.

Now, let's calculate the total access time:

Total Access Time = Seek Time + Rotational Delay + Transfer Time

Seek Time = 15 ms = 0.015 seconds
Rotational Delay = (1/2 * 1/1500) seconds
Transfer Time = (1000/1500) seconds

Total Access Time = 0.015 + (1/2 * 1/1500) + (1000/1500) seconds

Total Access Time = 0.015 + 0.000333333 + 0.666666667 seconds

Total Access Time = 0.681 seconds

Converting to milliseconds:

Total Access Time = 0.681 * 1000 ms

Total Access Time ≈ 681.33 ms

Therefore, the total access time will be approximately 681.33 ms, which can be rounded off to 681 ms.

Answer: c) 16.33 ms
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