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# What is the total capacity of the disk?

Consider a two-platter disk with the following parameters:
Number of heads = 4 (one head for each surface of the platters)
Rotation speed = 7200 rpm
Number of tracks on one side of the platter = 30 000
Number of sectors per track = 600
Number of bytes per sector = 512 bytes
Seek time = one ms for every hundred tracks traversed.

Let the disk receive a request to access a random sector on a random track and assume that the head starts at track 0 (the outermost track of the disk).

a. What is the average access time?
b. What is the transfer time for a sector?
c. What is the total average time to satisfy a request?
d. What is the total capacity of the disk?

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a. Average access time will be sum of average seek time (time required to position the head on correct track) and average latency (rotating the disk to bring the required sector under the head).

Given that the head starts at track 0 (the outermost track of the disk) when the disk receives a request, the extreme cases of head positioning could be that the next request is for a sector on the outermost track, or the next request is for a sector on the innermost track.

When it is a request for a sector on the outermost track, head will not have to traverse any track and thus seek time would be 0.
When it is a request for a sector on the innermost track, head will have to traverse 30000 tracks ...

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