Consider a hard disk with a rotational speed of 15000 rpm. The time to move…
2026
Consider a hard disk with a rotational speed of 15000 rpm. The time to move the read/write head from a track to its adjacent track is 1 millisecond. Initially, the head is on track 0. The number of sectors per track is 400. The sector size is 1024 bytes. It is necessary to transfer data from 10 randomly located sectors in each of the following tracks in the order: 5, 12 and 7.
The total time for the data transfer (in milliseconds) from the hard disk is _________. (rounded off to one decimal place)
Attempted by 3 students.
Show answer & explanation
Correct answer: 77.3
Solution
Given:
Rotational speed = 15000 rpm
Adjacent track seek time = 1 ms
Tracks accessed in order = 5, 12, 7
Sectors per track = 400
Sector size = 1024 bytes
10 randomly located sectors are accessed per track
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Step 1: Rotation Time
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Time for one complete rotation:
= 60 / 15000 seconds
= 0.004 s
= 4 ms
Average rotational latency:
= 4 / 2
= 2 ms
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Step 2: Seek Time
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Track movement:
0 -> 5 = 5 tracks
5 -> 12 = 7 tracks
12 -> 7 = 5 tracks
Total tracks moved:
= 5 + 7 + 5
= 17 tracks
Since adjacent-track seek time = 1 ms:
Total seek time:
= 17 ms
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Step 3: Rotational Latency
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Total sectors accessed:
= 10 sectors/track × 3 tracks
= 30 sectors
Since sectors are randomly located,
each sector requires average rotational latency.
Average rotational latency per sector:
= 2 ms
Total rotational latency:
= 30 × 2
= 60 ms
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Step 4: Transfer Time
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400 sectors are transferred in one rotation.
One rotation time = 4 ms
Time per sector:
= 4 / 400
= 0.01 ms
For 30 sectors:
= 30 × 0.01
= 0.3 ms
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Final Calculation
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Total time:
= Seek Time + Rotational Latency + Transfer Time
= 17 + 60 + 0.3
= 77.3 ms
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Final Answer
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77.3 ms
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