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 _____________
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 milliseconds [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 milliseconds. 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 milliseconds Disk latency = Seek Time + Rotation Time + Transfer Time + Controller Overhead = 4 + 2 + 10.24 * 10-3 + 0.1 milliseconds = 6.1 milliseconds
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