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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:

1. $172$
2. $173$
3. $227$
4. $228$

@vijaycs If nothing is given(and suppose we have to fill in numerical) by default we have to take from 0 to 149. Isn't it??
ans : 151 . not mentionedin any option .

The R/W head current position is at cylinder 23 and moving inwards (low cylinder).

Range of cylinders [0,150].

So it will move 23 to 0 and then changes direction from 0 to largest cylinder of I/O request which is 128.

So, total head movement is = (23-0) + (128-0) = 151.
by
SCAN scheduling algorithm used therfore scans all the cylinders of the disk back and forth.

In this ques,  Head moves from 23 to 0 then from 0 to 128 servicing all the requests in between.

Therefore,  disk movement = (23-0)+(128-0) =151.
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