UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 60)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 60)

Q 60. In which of the following scheduling criteria, context switching will never take place ?

(A) ROUND ROBIN
(B) Preemptive SJF
(C) Non-preemptive SJF
(D) Preemptive priority


Answer : (C) Non-preemptive SJF


In Non – preemptive algorithms context switching will never take place because it doesn’t allow to switch the process until it is completed fully.

Except Non-Preemptive SJF , All are Preemptive.
CPU will leave the Shotest Process only after its Full execution,Once process is started to execute.

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 59)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 59)

Q 59. Consider the following three processes with the arrival time and CPU burst time given in milliseconds:

Process	Arrival Time	Burst Time
P1	0	7
P2	1	4
P3	2	8

The Gantt Chart for preemptive SJF scheduling algorithm is _________.

(A)

P1

P2

P3

0                                           7                      13                        21

(B)

P1

P2

P1

P

0                      1                        5                      11                    19    

(C)

P1

P2

P3

0                                           7                      13                        19

(D)

P2

P3

P1

0                      4                       12                                            19

Answer : (B)

Explanation : 
A) Not preemptive. P1 continues execution without being preempted.
B) Preemptive SJF . P1 executes for 1 second followed by execution of P2 for 4 secs. Then P1 finishes execution followed by P3.
C) Not preemptive as each of the processes carries on execution without being preempted.
D)Preemptive but P2 starts execution even before it has arrived which can't happen in reality.

P₁ arrive at 0 it will be served by CPU for 1 unit. After 1 unit P₂ arrived ans it is shortest So it will execute for 4 unit, after its execution There are 2 processes out of which P₁ is shortest so it will execute for 6 unit after its execution P₁ will execute for 8 unit.

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 56-58)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 56-58)

Q 56. Which of the following statements are true ?
(a) External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is contiguous.
(b) Memory Fragmentation can be internal as well as external.
(c) One solution to external Fragmentation is compaction.

Code:
(A) (a) and (b) only
(B) (a) and (c) only
(C) (b) and (c) only
(D) (a), (b) and (c)

Answer : (C) (b) and (c) only

Explanation:

External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is contiguous. This is a false statement because external fragmentation occurs due to non contiguous available space.

Memory Fragmentation can be internal as well as external.True

One solution to external Fragmentation is compaction or shuffle memory contents.True

Q 57. Page information in memory is also called as Page Table. The essential contents in each entry of a page table is/are _________.

(A) Page Access information
(B) Virtual Page number
(C) Page Frame number
(D) Both virtual page number and Page Frame Number

Answer : (C) Page Frame number

Explanation: Page information in memory is also called as Page Table. The essential contents in each entry of a page table is page frame number. (i.e. Frame Number is essential in page table.)


Q 58. Consider a virtual page reference string 1, 2, 3, 2, 4, 2, 5, 2, 3, 4. Suppose LRU page replacement algorithm is implemented with 3 page frames in main memory. Then the number of page faults are_________.

(A) 5
(B) 7
(C) 9
(D) 10


Answer : (B) 7

Explanation: Page replacement diagram for LRU:

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UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 54-55)

UGC NET JULY 2018 (Computer Science And Applications Paper-II) (Question 54-55)

Q 54. Normally user programs are prevented from handling I/O directly by I/O instructions in them. For CPU's having explicit I/O instructions, such I/O protection is ensured by having the I/O instructions privileged. In a CPU with memory mapped I/O, there is no explicit I/O instruction. Which one of the following is true for a CPU with memory mapped I/O ?

(A) I/O protection is ensured by operating system routines.
(B) I/O protection is ensured by a hardware trap.
(C) I/O protection is ensured during system configuration.
(D) I/O protection is not possible.

Answer : (A) I/O protection is ensured by operating system routines.

Explanation: Memory mapped I/O means, accessing I/O via general memory access as opposed to specialized IO instructions.
The programmer can directly access any memory location directly. To prevent such an access, the OS (kernel) will divide the address space into kernel space and user space. An user application can easily access user application. To access kernel space, we need system calls (traps).


Q 55. Which UNIX/Linux command is used to make all files and sub-directories in the directory “progs” executable by all users ?

(A) chmod− R a+x progs
(B) chmod −R 222 progs
(C) chmod−X a+x progs
(D) chmod −X 222 progs

Answer : (A) chmod− R a+x progs

Solution : Option A will make all files and sub-directories in the directory (progs) executable by all users.
-R is for recurssive calls, so that sub-directories within the directories can be accessed.
a is for all users
+x is for making things executable

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