UGC-NET | UGC NET CS 2017 Jan – III | Question 50

Consider a disk queue with I/O requests on the following cylinders in their arriving order:

6, 10, 12, 54, 97, 73, 128, 15, 44, 110, 34, 45 

The disk head is assumed to be at cylinder 23 and moving in the direction of decreasing number of cylinders. Total number of cylinders in the disk is 150. The disk head movement using SCAN-scheduling algorithm is:
(A) 172
(B) 173
(C) 227
(D) 228


Answer:

Explanation: According to SCAN disk scheduling algorithm:

When it completed its last given sequence from input, it stops.
So, total disk head movement = 23+128 = 151.

So, none option matches.


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