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Net 28

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By Team EasyExamNotes

CBSE NET January 2017 PAPER II

OPERATING SYSTEM

Q. There are three processes P1, P2 and P3 sharing a semaphore for synchronising a variable. Initial value of semaphore is one. Assume that negative value of semaphore tells us how many processes are waiting in queue. Processes access the semaphore in following order:
(a) P2 needs to access
(b) P1 needs to access
(c) P3 needs to access
(d) P2 exits critical section
(e) Pi exits critical section
The final value of semaphore will be:

(A) 0
(B) 1
(C) -1
(D) -2

Ans :- (A)

Explanation:-

Given,Initial value of semaphore S = 1.
(a) When P2 needs to  access, S=0
(b) Now P1 needs to access, S=-1, which says 1 process P1 on wait, because P2 in section.
(c)  Now P3 needs to access, S= -2, which says 2 processes P1 and P2 on wait, because in P2 in section.
(d) Now P2 exits critical section, so P1 enter,S= -1, because only P2 on wait.
(e) Now P1 exits critical section, so P2 enter,S= 0, because no process on wait.
So, correct answer is A.
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