Suppose a disk with the following speci cation: 8 platters (16 surfaces) 1 surfa

Question

Suppose a disk with the following speci cation:
8 platters (16 surfaces)
1 surface = 10000 tracks
1 track = 256 sectors
1 sector = 512 bytes
Block size = 4096 bytes
Disk rotate rate = 5400 RPM
The time for the disk head to move from one cylinder to an adjacent cylinder is 0.001 ms.
Gaps consumes 10% of track.
A track has 256 gaps and 256 sectors.
Notice that the data transfer rate is not provided but this ratio depends upon the rotational
speed of the disks and the data recording density. Here, to compute the data transfer rate you
ignore the data recording density factor but consider the disk rotate rate and sector/gap structure.
Average seek distance is 1/3 maximum seek distance.

(a) What is the total capacity of this disk?
(b) How many bytes does one track hold?
(c) How many blocks are on one track?
(d) What is the minimum time (the best case) to access one disk block?
(e) What is the maximum time (the worst case) to access one disk block?
(f) What is the average time to access one disk block?

Explanation / Answer

(a) 16*10000*256*512 bytes

(b) 256*512 bytes

(c) (256*512)/4096 = 32 blocks

(d)

(e)5400 rev/min = 90 rev/sec = 11.11 msec/rev. Hence worst case time is full rotation (rotational latency)+0.001ms

i.e. 11.11 msec + 0.001 msec

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