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Sonja [21]
3 years ago
6

A disk rotates at a rate of 7200 revolutions per minute. Seek operations (i.e., moving the access head to a desired track) take

on average 20 milli-seconds. Once the access head is on the appropriate track, it takes an average of one half revolution of the disk to get to the beginning of a requested sector. Each track contains 128 sectors and each sector contains 512 bytes of data. How long on average does it take to get the access head to the beginning of a randomly selected sector on a randomly selected track? Express your answer in micro-seconds (not seconds or milli-seconds).
Computers and Technology
1 answer:
Leno4ka [110]3 years ago
8 0

Answer:

24.167 micro seconds.

Explanation:

The given rotation rate = 7200 rpm = 7200 rounds per minute

Rotational latency is the average time taken for the head to reach starting of sector .

Rotational latency (in micro seconds) = (1 / (RPM / 60)) * 0.5 * 1000

(1/(7200/60))* 0.5 * 1000 = 4.167 ms

Thus, rotational latency = 4.167 ms.

Seek time = 20 ms

The average time taken for the access head to get to the beginning of randomly selected sector will be equal to the average time to first reach the random track plus the average time taken to reach random sector .

= 20 ms + 4.167  ms  = 24.167 micro seconds.

Thus, it would take 24.167 micro seconds to get the access head to the beginning of a randomly selected sector on a randomly selected track.

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Answer:

It throws an error.

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Answer:

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Required

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Explanation:

The explanation is provided in the attached document due to some errors in uploading the answer.

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