ISRO2014-14

Question

There are 200 tracks on a disc platter and the pending requests have come in the order - 36, 69, 167, 76, 42, 51, 126, 12 and 199. Assume the arm is located at the 100^th track and moving towards track 200. If sequence of disc access is 126, 167, 199, 12, 36, 42, 51, 69 and 76 then which disc access scheduling policy is used?

• A. Elevator
• B. Shortest seek-time first
• C. C-SCAN
• D. First Come First Served

Answer 1

It should be C-SCAN il  as the disc access has continued towards the nearest end from initial from 100 and

serviced126,167,199  untill it has hit end the top max

and then jumped to the other end and serviced 12,36,42,51,69,76 in similar manner .

Comments

if in option would C-LOOK SCAN then it will be more right na?

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ya .. kind of the  huge jump is not considered as head movement  

just i feel it adds time overhead . but C-look would have been appropriate

Answer 2

answer is (c). C-SCAN because it has been started from 100th and serving all the right hand side request till the last point then reverse the direction and starts from track 0 to serve further request.

SCAN will just reverse direction and service the next closest disk request and C-SCAN will always go back to track 0 and start going to higher track requests on reversal.

for more info@

http://www.cs.iit.edu/~cs561/cs450/disksched/disksched.html

Comments

In C-SCAN we move to both the extreme ends i.e 0,200 but here in question we didn't move to either 0 or 200 na

Answer 3

After 199 it is moving to 12 . so it should be elevator algo.and in circular scan it will move upto 0

Comments

I think it should be C-LOOK

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Yes , u r rgt but no option here

So we have to go with elevator

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but in elevator after 199 it should serve 76 instead of 12...76 is nearest to 100 which was not serviced but i think it i going to track 0 then servicing tract 12,36...like that

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Right @cse23 , but still a bit confusing, why not elevator ?

Answer 4

Answer is C-SCAN.

SCAN is wrong because it works like elevator servicing requests while going up and coming down, whereas in this case if you observe series [126, 167, 199, 12, 36, 42, 51, 69 and 76 ], after 199 it suppose to scan 76 but not 12.

If SCAN then answer would be in the series-126,167,199,76,69,51,42,36,12]

So in C-SCAN starts from 100, 126,167, and after reaching 199, without servicing any request during return it will reach 0 initially, then it start servicing requests on the way from 0 [12,36,42,51,69 and 76].till it reaches end of the disk.