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Capacity, Data and Transfer Rates, Access Time of Disk Drive

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A hard disk drive has 10 disks and 18 surfaces available for recording. Each surface is composed of 200 concentric tracks and the disks rotate at 7200 r.p.m. Each track is divided into 8 blocks of 256 32-bit words. There is one read/write head per surface and it is possible to read the 18 tracks of a given cylinder simultaneously. The time to step from track to track is 1ms. Between data transfers the head is parked at the outermost track of the disk.

Calculate:
a. The total capacity of the disk drive.
b. The maximum data rate in bits/second.
c. The average access time in milliseconds. (Neglect the head load time and head settling time).
d. The average transfer rate when reading 256 word blocks located randomly on the disk.
e. The recording density (bits/inch) of the innermost and outermost tracks if the disk has a 6 inch diameter and the outermost track comes to 1 inch from the edge of the disk. The track density is 200 tracks/inch.

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a. Total capacity of the disk drive = (Number of heads or surfaces available for recording) * (Number of tracks per surface) * (Number of sectors or blocks per track) * (Number of bits per block)
= 18 * 200 * 8 * (256 * 32)
= 235929600 bits or 29491200 bytes or 28800 KB or 28.125 MB
where 1 KB = 1024 bytes, and 1 MB = 1024 KB

b. Since the disk rotates at 7200 r.p.m. or (7200/60=) 120 r.p.s. (rotations per second), it will completely traverse a track in (1/120) seconds.
Maximum data that can be read during one rotation of the disk = (Number of surfaces available for recording) * (Number of blocks per track) * (Number of bits per block)
= 18 * 8 * (256 * 32)
= 1179648 bits

Hence, the maximum data rate in bits/second = 1179648 bits / (1/120) seconds
= 141557760 bits/second or 138240 Kb/second or 135 Mb/seconds
where 1 Kb = 1024 bits, and 1 Mb = 1024 Kb

Above computation assumes that it is possible to read data from all the surfaces simultaneously. However if disk allows to read the data from only one surface at a time, we will remove "Number of surfaces available for recording" from above computation, as shown below.

Maximum data that can be read during one rotation of the disk = (Number of blocks per track) * (Number of bits per block)
= 8 * (256 * 32)
= 65536 bits

Hence, the maximum data rate in bits/second = 65536 bits / ...

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