Hard disk: rotational speed = 15000 rpm, head movement time between adjacent tracks = 1 ms. Initiall

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Question

Hard disk: rotational speed = 15000 rpm, head movement time between adjacent tracks = 1 ms. Initially head at track 0. Sectors per track = 400, sector size = 1024 bytes. Transfer data from 10 randomly located sectors in each of tracks 5, 12, and 7 (in that order). Total time for data transfer (in ms) = ___________.

Accepted Answer

Answer: 77.3 msConcept -- Hard Disk Access Time Calculation: Rotational speed = 15000 rpm = 250 rps. Rotation time = 1/250 s = 4 ms. Sector transfer time = 4 ms / 400 sectors = 0.01 ms per sector. Average rotational latency = rotation time / 2 = 2 ms. Seek time: head movement between adjacent tracks = 1 ms. Seek to track 5: 5 ms. Track 5 to 12: 7 ms. Track 12 to 7: 5 ms. For each track, we need to access 10 randomly located sectors. The worst‑case (or typical) access time for random sectors within a track: for the first sector, we incur seek + average rotational latency + transfer. For each subsequent sector on the same track, we may need to wait for the disk to rotate to that sector; on average, the rotation delay between random sectors is half a revolution (2 ms). So for 10 sectors on one track: seek + avg rotational latency + 10×sector transfer + (9)×(avg rotation between sectors). Using avg rotation between sectors = 2 ms (half rotation). Compute: Track 5: 5 (seek) + 2 (latency) + 10×0.01 + 9×2 = 5+2+0.1+18 = 25.1 ms. Track 12: 7 (seek from track5) + 2 + 0.1 + 18 = 27.1 ms. Track 7: 5 (seek from track12) + 2 + 0.1 + 18 = 25.1 ms. Total = 25.1+27.1+25.1 = 77.3 ms. This is a typical disk scheduling calculation.

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