Operating Systems | Set 5

Following questions have been asked in GATE 2012 exam.

1. A process executes the code

  fork ();
  fork ();
  fork ();

The total number of child processes created is
(A) 3
(B) 4
(C) 7
(D) 8

Answer (C)

Let us put some label names for the three lines

  fork ();    // Line 1
  fork ();   // Line 2
  fork ();   // Line 3

       L1       // There will be 1 child process created by line 1
    /     \
  L2      L2    // There will be 2 child processes created by line 2
 /  \    /  \
L3  L3  L3  L3  // There will be 4 child processes created by line 3

We can also use direct formula to get the number of child processes. With n fork statements, there are always 2^n – 1 child processes. Also see this post for more details.

2. consider the 3 processes, P1, P2 and P3 shown in the table

Process     Arrival time    Time unit required
  P1                0                    5
  P2                1                    7
  P3                3                    4

The completion order of the 3 processes under the policies FCFS and RRS (round robin scheduling with CPU quantum of 2 time units) are
(A) FCFS: P1, P2, P3 RR2: P1, P2, P3
(B) FCFS: P1, P3, P2 RR2: P1, P3, P2
(C) FCFS: P1, P2, P3 RR2: P1, P3, P2
(D) FCFS: P1, P3, P2 RR2: P1, P2, P3

Answer (C)



3. Consider the virtual page reference string
1, 2, 3, 2, 4, 1, 3, 2, 4, 1
On a demand paged virtual memory system running on a computer system that main memory size of 3 pages frames which are initially empty. Let LRU, FIFO and OPTIMAL denote the number of page faults under the corresponding page replacements policy. Then
(A) OPTIMAL < LRU < FIFO (B) OPTIMAL < FIFO < LRU (C) OPTIMAL = LRU (D) OPTIMAL = FIFO Answer (B) The OPTIMAL will be 5, FIFO 6 and LRU 9.


4. A file system with 300 GByte uses a file descriptor with 8 direct block address. 1 indirect block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the size of each disk block address is 8 Bytes. The maximum possible file size in this file system is
(A) 3 Kbytes
(B) 35 Kbytes
(C) 280 Bytes
(D) Dependent on the size of the disk

Answer (B)

Total number of possible addresses stored in a disk block = 128/8 = 16

Maximum number of addressable bytes due to direct address block = 8*128
Maximum number of addressable bytes due to 1 single indirect address block = 16*128
Maximum number of addressable bytes due to 1 double indirect address block = 16*16*128

The maximum possible file size = 8*128 + 16*128 + 16*16*128 = 35KB

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