Question

3. Assume a disk drive from the late 1990s is configured as follows. The total storage is approximately 675MB divided among 15 surfaces. Each surface has 612 tracks; there are 144 sectors/track, 512 bytes/sector, and 8 sectors/cluster. The disk turns at 3600 rpm. The tract-to-track seek time is 20ms, and the average seek time is 80ms. Now assume that there is a 360KB file on the disk. On average, how long does it take to read all of the data in the file? Assume that the first track of the file is randomly placed on the disk, that the entire file lies on adjacent tracks, and that the file completely fills each track on which it is found. A seek must be performed each time the I/O head moves to a new track. Show your calculations

Answer 1

Answer:

total time to read whole file = ( 81.380208 mili seconds ) + 80ms + 80ms

Explanation:

Size of a surface = 675/15 = 45 MB

Size of a track = (Number of sectors per track)*(Size of a sector)

= (144*512)Bytes

= 73.728 KB

Number of tracks where 360KB file is located = 360/73.728 = 4.8828125 (Approx 5 tracks)

So Seek time to go the first track of the file = 80 ms

And after seek time for track to track movement = 4*20 = 80 ms

We need 4.8828125 rotations to read whole file.

Time for 4.8828125 ratations = (60.0/3600.0)*4.8828125 seconds = 0.081380208 seconds = 81.380208 mili seconds

So total time to read whole file = ( 81.380208 mili seconds ) + 80ms + 80ms