8.5 Assume that a disk drive is configured as follows. The total storage is appr

Question

8.5 Assume that a disk drive is configured as follows. The total storage is approximately 1033MB divided among 15 surfaces. Each surface has 2100 tracks, there are 64 sectors/track, 512 bytes/sector, and 8 sectors/cluster. The disk turns at 7200 rpm. The track-to-track seek time is 3 ms, and the average seek time is 20 ms. Now assume that there is a 512KB file on the disk. On average, how long does it take to read all of the data on the file? Assume that the first track of the file is randomly placed on the disk, that the entire file lies on contiguous tracks, and that the file completely fills each track on which it is found. Show your

Explanation / Answer

Size of a surface = 1033/15 = 68.86666 MB

Size of a track = (Number of sectors per track)*(Size of a sector) = 512*64 Bytes = 32.768 KB

Number of tracks where 512KB file is located = 512/32.768 = 15.625 (Approx 16 tracks)

So Seek time to go the first track of the file = 20ms

And seek time for track to track movement = 15*3 = 45 ms

We need 15.625 rotations to read whole file.

Time for 15.625 ratations = (60.0/7200)*15.625 seconds = 0.130208 seconds = 130.208 milli seconds

So total time to read whole file = 20 ms + 45 ms + 130.208 milli seconds

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