GATE | GATE-CS-2015 (Set 2) | Question 65

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 _____________
(A) 6.1


Answer: (A)

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

Refer http://cse.unl.edu/~jiang/cse430/Lecture%20Notes/reference-ppt-slides/Disk_Storage_Systems_2.ppt


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