Operating Systems | Set 15

Following questions have been asked in GATE CS 2006 exam.

1) Consider three processes (process id 0, 1, 2 respectively) with compute time bursts 2, 4 and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF ties are broken by giving priority to the process with the lowest process id. The average turn around time is:
(A) 13 units
(B) 14 units
(C) 15 units
(D) 16 units

Answer (A)
Let the processes be p0, p1 and p2. These processes will be executed in following order.

  p2  p1  p2  p1  p2  p0  p1   p2   p0   p1   p2
0   4   5   6   7   8   9   10    11   12   13   14 

Turn around time of a process is total time between submission of the process and its completion.
Turn around time of p0 = 12 (12-0)
Turn around time of p1 = 13 (13-0)
Turn around time of p2 = 14 (14-0)

Average turn around time is (12+13+14)/3 = 13.

2) Consider three processes, all arriving at time zero, with total execution time of 10, 20 and 30 units, respectively. Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again. The operating system uses a shortest remaining compute time first scheduling algorithm and schedules a new process either when the running process gets blocked on I/O or when the running process finishes its compute burst. Assume that all I/O operations can be overlapped as much as possible. For what percentage of time does the CPU remain idle?(A) 0%
(B) 10.6%
(C) 30.0%
(D) 89.4%

Answer (B)
Let three processes be p0, p1 and p2. Their execution time is 10, 20 and 30 respectively. p0 spends first 2 time units in I/O, 7 units of CPU time and finally 1 unit in I/O. p1 spends first 4 units in I/O, 14 units of CPU time and finally 2 units in I/O. p2 spends first 6 units in I/O, 21 units of CPU time and finally 3 units in I/O.

 idle   p0    p1     p2    idle
0    2     9     23     44     47

Total time spent = 47
Idle time = 2 + 3 = 5
Percentage of idle time = (5/47)*100 = 10.6 %

3) The atomic fetch-and-set x, y instruction unconditionally sets the memory location x to 1 and fetches the old value of x in y without allowing any intervening access to the memory location x. consider the following implementation of P and V functions on a binary semaphore .

void P (binary_semaphore *s) {
  unsigned y;
  unsigned *x = &(s->value);
  do {
     fetch-and-set x, y;
  } while (y);

void V (binary_semaphore *s) {
  S->value = 0;

Which one of the following is true?
(A) The implementation may not work if context switching is disabled in P.
(B) Instead of using fetch-and-set, a pair of normal load/store can be used
(C) The implementation of V is wrong
(D) The code does not implement a binary semaphore

Answer (A)
Let us talk about the operation P(). It stores the value of s in x, then it fetches the old value of x, stores it in y and sets x as 1. The while loop of a process will continue forever if some other process doesn’t execute V() and sets the value of s as 0. If context switching is disabled in P, the while loop will run forever as no other process will be able to execute V().

4) Consider the following snapshot of a system running n processes. Process i is holding Xi instances of a resource R, 1 <= i <= n. currently, all instances of R are occupied. Further, for all i, process i has placed a request for an additional Yi instances while holding the Xi instances it already has. There are exactly two processes p and q such that Yp = Yq = 0. Which one of the following can serve as a necessary condition to guarantee that the system is not approaching a deadlock?
(A) min (Xp, Xq) < max (Yk) where k != p and k != q (B) Xp + Xq >= min (Yk) where k != p and k != q
(C) max (Xp, Xq) > 1
(D) min (Xp, Xq) > 1

Answer (B)
Since both p and q don’t need additional resources, they both can finish and release Xp + Xq resources without asking for any additional resource. If the resources released by p and q are sufficient for another process waiting for Yk resources, then system is not approaching deadlock.

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