Consider the disk which has average seek time of 32 ns and rotational rate of…
2020
Consider the disk which has average seek time of 32 ns and rotational rate of 360 rpm (round per minute). Each track of the disk has 512 sectors, each of size 512 bytes.
What is the time taken to read four continuous sectors? And What is the data transfer rate?
- A.
0.0443 s and 936 KBps
- B.
0.0843 s and 1536 KBps
- C.
0.1043 s and 1736 KBps
- D.
0.0943 s and 1636 KBps
Attempted by 38 students.
Show answer & explanation
Correct answer: B
Correct answer: Option B (0.0843 s and 1536 KBps)
Important unit note:
KBps means kilobytes per second. Kbps means kilobits per second. Since each sector size is given in bytes, the computed transfer rate is in KBps. If converted to bits, 1536 KBps = 1536 × 8 = 12288 Kbps.
Given:
Average seek time = 32 ns ≈ 0 s
Rotational speed = 360 rpm = 6 rotations/second
Sectors per track = 512
Sector size = 512 bytes
Track size:
512 sectors × 512 bytes = 262144 bytes = 256 KB
Average rotational latency:
One rotation takes 1/6 s.
Average rotational latency = (1/2) × (1/6) = 1/12 = 0.0833 s
Transfer time for 4 continuous sectors:
Data to read = 4 × 512 bytes = 2048 bytes = 2 KB
Transfer rate = 6 × 256 KB/s = 1536 KBps
Transfer time = 2/1536 s ≈ 0.00130 s
Total time:
Seek time + average rotational latency + transfer time
≈ 0 + 0.0833 + 0.00130
≈ 0.0846 s, represented by the closest option 0.0843 s.
Therefore, the intended answer is 0.0843 s and 1536 KBps.