Consider a disc with the following characteristics. Answer questions A and B. -a

Question

Consider a disc with the following characteristics. Answer questions A and B.

-a sector size of 512 bytes
-2000 tracks per surface
-40 sectors per track
-10 one-sided platters
-Average seek-time=10 msec

A) If the disk platters rotate at 6400 rpm (revolution per minute), what is the maximum rotational latency (i.e., the maximum time (in seconds) required for a full rotation of platter)?

B) If one track of data can be transferred per revolution (i.e., 1 rotation of platter), what is the transfer rate (how many bytes are transferred per unit of time)?

Explanation / Answer

Answer:

a)

If the disk platters rotate at 6400 rpm (revolution per minute) the time required for onecomplete rotation, which is the maximum rotational latency, is given by

(1/6400) X 60 = 0.0093 seconds

The average rotational delay is half of the rotation time i.e 0.0046 seconds.

b)

Capacity of a track is given by

bytes/track = bytes/sector × sectors/track

= 512 × 40 = 20.48K

The capacity of a track is 20.48K bytes. Only one track of data can be transferred per revolution, the data transfer rate is   

  capacity of a track / maximum rotational latency

( 20.48K /0.009 ) = 2275 Kbytes/second

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