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Consider a disk with a sector size of $512$ bytes, $50$ sectors per track, $2000$ tracks per surface and five double-sided platters (i.e. $10$ surfaces). The disk platters rotate at $5400$ rpm. The average seek time is $10$ msec. A block size is chosen as $1024$ bytes.

A file containing $100,000$ records of $100$ bytes each is to be stored on this disk, and no record is allowed to span two blocks. Then, the rotational latency of the disk is ________ seconds.
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Formula

maximum rotational delay = (1/p) min = (60/p) sec

rotational delay  = ½ * cost of 1 revolution = ½ * (60/p) sec

If the disk platters rotate at 5400rpm, the time required for one complete rotation, which is the maximum rotational delay, is 1/5400 * 60 = 0.011 seconds

The average rotational delay is half of the rotation time 0.011/2 =  0.0055

Answer:

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