Question

Consider a typical disk that rotates at 15000 rotations per minute (rpm) and has a transfer rate of 50 106 bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller's transfer time is 10 times the disk transfer time, the average time (in milliseconds) to read or write a 512 byte sector of the disk is _____________

Answer 1

Answer:

6.1 milliseconds

Explanation:

Disk latency = Seek Time + Rotation Time + Transfer Time + Controller Overhead

Seek Time? Depends no. tracks the arm moves and seek speed of disk

Rotation Time? depends on rotational speed and how far the sector is from the head

Transfer Time? depends on data rate (bandwidth) of disk (bit density) and the size of request

Disk latency = Seek Time + Rotation Time +

Transfer Time + Controller Overhead

Average Rotational Time = (0.5)/(15000 / 60) = 2 miliseconds

[On average half rotation is made]

It is given that the average seek time is twice the average rotational delay

So Avg. Seek Time = 2 * 2 = 4 miliseconds.

Transfer Time = 512 / (50 × 106 bytes/sec)

= 10.24 microseconds

Given that controller time is 10 times the average transfer time

Controller Overhead = 10 * 10.24 microseconds

= 0.1 miliseconds

Disk latency = Seek Time + Rotation Time +

Transfer Time + Controller Overhead

= 4 + 2 + 10.24 * 10-3 + 0.1 miliseconds

= 6.1 miliseconds

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