Operating Systems | Set 7

27 Mar, 2017
Following questions have been asked in GATE CS exam.1) Let the time taken to switch between user and kernel modes of execution be t1 while the time taken to switch between two processes be t2. Which of the following is TRUE? (GATE CS 2011)(A) t1 > t2(B) t1 = t2(C) t1 < t2
(D) Nothing can be said about the relation between t1 and t2
Answer: - (C)
Process switching involves mode switch. Context switching can occur only in kernel mode.2) A system uses FIFO policy for page replacement. It has 4 page frames with no pages loaded to begin with. The system first accesses 100 distinct pages in some order and then accesses the same 100 pages but now in the reverse order. How many page faults will occur? (GATE CS 2010)(A) 196(B) 192(C) 197(D) 195
Answer (A)Access to 100 pages will cause 100 page faults. When these pages are accessed in reverse order, the first four accesses will not cause page fault. All other access to pages will cause page faults. So total number of page faults will be 100 + 96.
3) Which of the following statements are true? (GATE CS 2010)I. Shortest remaining time first scheduling may cause starvationII. Preemptive scheduling may cause starvationIII. Round robin is better than FCFS in terms of response time(A) I only(B) I and III only(C) II and III only(D) I, II and III
Answer (D)
I) Shortest remaining time first scheduling is a preemptive version of shortest job scheduling. It may cause starvation as shorter processes may keep coming and a long CPU burst process never gets CPU.II) Preemption may cause starvation. If priority based scheduling with preemption is used, then a low priority process may never get CPU.III) Round Robin Scheduling improves response time as all processes get CPU after a specified time.
4) Consider the methods used by processes P1 and P2 for accessing their critical sections whenever needed, as given below. The initial values of shared boolean variables S1 and S2 are randomly assigned.
Method Used by P1
while (S1 == S2) ;
Critica1 Section
S1 = S2;

Method Used by P2
while (S1 != S2) ;
Critica1 Section
S2 = not (S1);

Which one of the following statements describes the properties achieved? (GATE CS 2010)(A) Mutual exclusion but not progress(B) Progress but not mutual exclusion(C) Neither mutual exclusion nor progress(D) Both mutual exclusion and progress
Answer (A)It can be easily observed that the Mutual Exclusion requirement is satisfied by the above solution, P1 can enter critical section only if S1 is not equal to S2, and P2 can enter critical section only if S1 is equal to S2.Progress Requirement is not satisfied. Let us first see definition of Progress Requirement.Progress Requirement: If no process is executing in its critical section and there exist some processes that wishes to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely.If P1 or P2 want to re-enter the critical section, then they cannot even if there is other process running in critical section.
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.
GATE-CS-2010
GATE-CS-2011
GATE CS
MCQ
Operating Systems
Operating Systems
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.