Find the transfer rate of and capacity of a hard disk in DBMS, How to find the data transfer rate and the capacity of hard disk for efficient storage access in dbms?
Question:
5) Consider a disk with the following specifications:
512 bytes sector size, 8 double sided platters (2 surfaces per platter), 5000
tracks per surface, 100 sectors per track, average seek time of 10 msec, and
the disk rotates at 5400 revolutions per minute (rpm). Which of the following
is the correct combination of number of cylinders, average rotational delay,
and transfer rate (if an entire track of data can be transferred per revolution
of disk)?
a) 5000 cylinders, 5.55 milliseconds,
4.5 KB
b) 5000 cylinders, 4.5 milliseconds, 12
KB
c) 4000 cylinders, 5.55 milliseconds, 10
KB
d) 4000 cylinders, 2
milliseconds, 2.5 KBAnswer:
a) 5000
cylinders, 5.55 milliseconds, 4.5 KB
Discussion:
Given,
Number of platters - 8
Tracks per disk surface - 5000
Sectors per track - 100
Sector size - 512
bytes
Average seek time - 10 msec
rpm - 5400.
Number of cylinders is same as the
number of tracks. Hence, the number of cylinders is 5000.
The average rotational delay is half
of the rotation time. One complete rotation takes 1/5400 in a minute = (1/5400)
* 60 seconds = 0.0111 seconds = 11.1 milliseconds. The average rotational delay = 11.1/2 = 5.55
msec (approx).
We need the capacity of the track to
find the transfer rate. The capacity of the track can be found as follows;
Number of bytes per track = bytes per
sector * sectors per track
=
512 * 100 = 51200 bytes = 50 KB.
We know that, from above, the time for
one complete rotation is 11.1 milliseconds. Hence, the transfer rate is 50 KB/11.1 milliseconds
= 4.5 KB/msec (approx.).
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