Computer Science, asked by rajesh702be, 1 year ago

consider following information about a disk. rotational speed=10000 RPM.seek time=average seek time=4.9ms.one stripe unit is SKB.each sector is 512 bytes.disk has 200 sectors per track. what is disk access time for a single unit on a single disk?
a)8.38 ms
b)11.38 ms
c)8.14 ms
d)11.14 ms

Answers

Answered by vimalkumarmdb
2

There are three components to any disk access. That is { seek time, rotational

delay and finally transfer time.

Rotational speed = 10000 RPM. Thus, the time for one rotation is 60seconds=10000rotations =

6ms per rotation. Now, average Rotational delay = 12 * Time of Rotation in milliseconds. So,

average rotational delay is 1=2 āˆ— 6ms = 3ms.

Seek time = average seek time = 4:9ms.

One stripe unit is 8KB. Each sector is 512 bytes. Thus, one stripe unit is 8192 / 512 = 16

sectors. Disk has 200 sectors per track, and performs one rotation in 6ms. Thus, the transfer

time is 16=200 āˆ— 6ms = 0:48ms.

Thus, the total disk access time = 3ms + 4:9ms + 0:48ms = 8:38ms

optation A is correct.

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