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Operating Systems | Set 9

  • Difficulty Level : Easy
  • Last Updated : 27 Mar, 2017

Following questions have been asked in GATE 2009 CS exam.

1) In the following process state transition diagram for a uniprocessor system, assume that there are always some processes in the ready state: Now consider the following statements:

I. If a process makes a transition D, it would result in another process making transition A immediately.
II. A process P2 in blocked state can make transition E while another process P1 is in running state.
III. The OS uses preemptive scheduling.
IV. The OS uses non-preemptive scheduling.
Which of the above statements are TRUE?

(A) I and II
(B) I and III
(C) II and III
(D) II and IV

Answer (C)
I is false. If a process makes a transition D, it would result in another process making transition B, not A.
II is true. A process can move to ready state when I/O completes irrespective of other process being in running state or not.
III is true because there is a transition from running to ready state.
IV is false as the OS uses preemptive scheduling.



2) The enter_CS() and leave_CS() functions to implement critical section of a process are realized using test-and-set instruction as follows:

void enter_CS(X)
{
    while test-and-set(X) ;
}
void leave_CS(X)
{
   X = 0;
}

In the above solution, X is a memory location associated with the CS and is initialized to 0. Now consider the following statements:
I. The above solution to CS problem is deadlock-free
II. The solution is starvation free.
III. The processes enter CS in FIFO order.
IV More than one process can enter CS at the same time.

Which of the above statements is TRUE?
(A) I only
(B) I and II
(C) II and III
(D) IV only



Answer (A)
The above solution is a simple test-and-set solution that makes sure that deadlock doesn’t occur, but it doesn’t use any queue to avoid starvation or to have FIFO order.



3) A multilevel page table is preferred in comparison to a single level page table for translating virtual address to physical address because
(A) It reduces the memory access time to read or write a memory location.
(B) It helps to reduce the size of page table needed to implement the virtual address space of a process.
(C) It is required by the translation lookaside buffer.
(D) It helps to reduce the number of page faults in page replacement algorithms.

Answer (B)
The size of page table may become too big (See this) to fit in contiguous space. That is why page tables are typically divided in levels.

Please see GATE Corner for all previous year paper/solutions/explanations, syllabus, important dates, notes, etc.

Please write comments if you find any of the answers/explanations incorrect, or you want to share more information about the topics discussed above.

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

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