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
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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 ā 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 ā 6ms = 0:48ms.
Thus, the total disk access time = 3ms + 4:9ms + 0:48ms = 8:38ms
optation A is correct.
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