Process Management

Question 1
Consider the following code fragment:
  if (fork() == 0)
  { a = a + 5; printf("%d,%d\n", a, &a); }
  else { a = a –5; printf("%d, %d\n", a, &a); } 
Let u, v be the values printed by the parent process, and x, y be the values printed by the child process. Which one of the following is TRUE?
A
u = x + 10 and v = y
B
u = x + 10 and v != y
C
u + 10 = x and v = y
D
u + 10 = x and v != y
Process Management    
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Question 1 Explanation: 
fork() returns 0 in child process and process ID of child process in parent process. In Child (x), a = a + 5 In Parent (u), a = a – 5; Therefore x = u + 10. The physical addresses of ‘a’ in parent and child must be different. But our program accesses virtual addresses (assuming we are running on an OS that uses virtual memory). The child process gets an exact copy of parent process and virtual address of ‘a’ doesn’t change in child process. Therefore, we get same addresses in both parent and child. See this run for example.
Question 2
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
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Question 2 Explanation: 
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().
Question 3
Three concurrent processes X, Y, and Z execute three different code segments that access and update certain shared variables. Process X executes the P operation (i.e., wait) on semaphores a, b and c; process Y executes the P operation on semaphores b, c and d; process Z executes the P operation on semaphores c, d, and a before entering the respective code segments. After completing the execution of its code segment, each process invokes the V operation (i.e., signal) on its three semaphores. All semaphores are binary semaphores initialized to one. Which one of the following represents a deadlockfree order of invoking the P operations by the processes? (GATE CS 2013)
A
X: P(a)P(b)P(c) Y:P(b)P(c)P(d) Z:P(c)P(d)P(a)
B
X: P(b)P(a)P(c) Y:P(b)P(c)P(d) Z:P(a)P(c)P(d)
C
X: P(b)P(a)P(c) Y:P(c)P(b)P(d) Z:P(a)P(c)P(d)
D
X: P(a)P(b)P(c) Y:P(c)P(b)P(d) Z:P(c)P(d)P(a)
Process Management    Deadlock    
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Question 3 Explanation: 
Option A can cause deadlock. Imagine a situation process X has acquired a, process Y has acquired b and process Z has acquired c and d. There is circular wait now. Option C can also cause deadlock. Imagine a situation process X has acquired b, process Y has acquired c and process Z has acquired a. There is circular wait now. Option D can also cause deadlock. Imagine a situation process X has acquired a and b, process Y has acquired c. X and Y circularly waiting for each other. See http://www.eee.metu.edu.tr/~halici/courses/442/Ch5%20Deadlocks.pdf Consider option A) for example here all 3 processes are concurrent so X will get semaphore a, Y will get b and Z will get c, now X is blocked for b, Y is blocked for c, Z gets d and blocked for a. Thus it will lead to deadlock. Similarly one can figure out that for B) completion order is Z,X then Y. This question is duplicate of http://geeksquiz.com/gate-gate-cs-2013-question-16/
Question 4
A shared variable x, initialized to zero, is operated on by four concurrent processes W, X, Y, Z as follows. Each of the processes W and X reads x from memory, increments by one, stores it to memory, and then terminates. Each of the processes Y and Z reads x from memory, decrements by two, stores it to memory, and then terminates. Each process before reading x invokes the P operation (i.e., wait) on a counting semaphore S and invokes the V operation (i.e., signal) on the semaphore S after storing x to memory. Semaphore S is initialized to two. What is the maximum possible value of x after all processes complete execution? (GATE CS 2013)
A
-2
B
-1
C
1
D
2
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Question 4 Explanation: 
Processes can run in many ways, below is one of the cases in which x attains max value
Semaphore S is initialized to 2

Process W executes S=1, x=1 but it doesn't update the x variable.

Then process Y executes S=0, it decrements x, now x= -2 and 
signal semaphore S=1

Now process Z executes s=0, x=-4, signal semaphore S=1
Now process W updates x=1, S=2

Then process X executes X=2 
So correct option is D
Question 5
A shared variable x, initialized to zero, is operated on by four concurrent processes W, X, Y, Z as follows. Each of the processes W and X reads x from memory, increments by one, stores it to memory, and then terminates. Each of the processes Y and Z reads x from memory, decrements by two, stores it to memory, and then terminates. Each process before reading x invokes the P operation (i.e., wait) on a counting semaphore S and invokes the V operation (i.e., signal) on the semaphore S after storing x to memory. Semaphore S is initialized to two. What is the maximum possible value of x after all processes complete execution? (GATE CS 2013)
A
-2
B
-1
C
1
D
2
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Question 5 Explanation: 
Question 6
A certain computation generates two arrays a and b such that a[i]=f(i) for 0 ≤ i < n and b[i]=g(a[i]) for 0 ≤ i < n. Suppose this computation is decomposed into two concurrent processes X and Y such that X computes the array a and Y computes the array b. The processes employ two binary semaphores R and S, both initialized to zero. The array a is shared by the two processes. The structures of the processes are shown below.
Process X:                         Process Y:
private i;                         private i;
for (i=0; i < n; i++) {            for (i=0; i < n; i++) {
   a[i] = f(i);                       EntryY(R, S);
   ExitX(R, S);                       b[i]=g(a[i]);
}                                 }
Which one of the following represents the CORRECT implementations of ExitX and EntryY? (A)
ExitX(R, S) {
  P(R);
  V(S);
}

EntryY (R, S) {
  P(S);
  V(R);
}
(B)
ExitX(R, S) {
  V(R);
  V(S);
}

EntryY(R, S) {
  P(R);
  P(S);
}
(C)
ExitX(R, S) {
  P(S);
  V(R);
}
EntryY(R, S) {
  V(S);
  P(R);
}
(D)
ExitX(R, S) {
  V(R);
  P(S);
}
EntryY(R, S) {
  V(S);
  P(R);
}
A
A
B
B
C
C
D
D
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Question 6 Explanation: 
The purpose here is neither the deadlock should occur
nor the binary semaphores be assigned value greater 
than one.
A leads to deadlock
B can increase value of semaphores b/w 1 to n
D may increase the value of semaphore R and S to
 2 in some cases
Question 7
Three concurrent processes X, Y, and Z execute three different code segments that access and update certain shared variables. Process X executes the P operation (i.e., wait) on semaphores a, b and c; process Y executes the P operation on semaphores b, c and d; process Z executes the P operation on semaphores c, d, and a before entering the respective code segments. After completing the execution of its code segment, each process invokes the V operation (i.e., signal) on its three semaphores. All semaphores are binary semaphores initialized to one. Which one of the following represents a deadlock-free order of invoking the P operations by the processes?
A
X: P(a)P(b)P(c) Y: P(b)P(c)P(d) Z: P(c)P(d)P(a)
B
X: P(b)P(a)P(c) Y: P(b)P(c)P(d) Z: P(a)P(c)P(d)
C
X: P(b)P(a)P(c) Y: P(c)P(b)P(d) Z: P(a)P(c)P(d)
D
X: P(a)P(b)P(c) Y: P(c)P(b)P(d) Z: P(c)P(d)P(a)
Process Management    GATE CS 2013    
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Question 7 Explanation: 
Option A can cause deadlock. Imagine a situation process X has acquired a, process Y has acquired b and process Z has acquired c and d. There is circular wait now. Option C can also cause deadlock. Imagine a situation process X has acquired b, process Y has acquired c and process Z has acquired a. There is circular wait now. Option D can also cause deadlock. Imagine a situation process X has acquired a and b, process Y has acquired c. X and Y circularly waiting for each other. See http://www.eee.metu.edu.tr/~halici/courses/442/Ch5%20Deadlocks.pdf   Consider option A) for example here all 3 processes are concurrent so X will get semaphore a, Y will get b and Z will get c, now X is blocked for b, Y is blocked for c, Z gets d and blocked for a. Thus it will lead to deadlock. Similarly one can figure out that for B) completion order is Z,X then Y. This question is duplicate of http://geeksquiz.com/operating-systems-process-management-question-8/
Question 8
A shared variable x, initialized to zero, is operated on by four concurrent processes W, X, Y, Z as follows. Each of the processes W and X reads x from memory, increments by one, stores it to memory, and then terminates. Each of the processes Y and Z reads x from memory, decrements by two, stores it to memory, and then terminates. Each process before reading x invokes the P operation (i.e., wait) on a counting semaphore S and invokes the V operation (i.e., signal) on the semaphore S after storing x to memory. Semaphore S is initialized to two. What is the maximum possible value of x after all processes complete execution?
A
-2
B
-1
C
1
D
2
Process Management    GATE CS 2013    
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Question 8 Explanation: 
Background Explanation: A critical section in which the process may be changing common variables, updating table, writing a file and perform another function. The important problem is that if one process is executing in its critical section, no other process is to be allowed to execute in its critical section. Each process much request permission to enter its critical section. A semaphore is a tool for synchronization and it is used to remove the critical section problem which is that no two processes can run simultaneously together so to remove this two signal operations are used named as wait and signal which is used to remove the mutual exclusion of the critical section. as an unsigned one of the most important synchronization primitives, because you can build many other Decrementing the semaphore is called acquiring or locking it, incrementing is called releasing or unlocking. Solution : Since initial value of semaphore is 2, two processes can enter critical section at a time- this is bad and we can see why. Say, X and Y be the processes.X increments x by 1 and Z decrements x by 2. Now, Z stores back and after this X stores back. So, final value of x is 1 and not -1 and two Signal operations make the semaphore value 2 again. So, now W and Z can also execute like this and the value of x can be 2 which is the maximum possible in any order of execution of the processes. (If the semaphore is initialized to 1, processed would execute correctly and we get the final value of x as -2.) Option (D) is the correct answer. Another Solution: Processes can run in many ways, below is one of the cases in which x attains max value Semaphore S is initialized to 2 Process W executes S=1, x=1 but it doesn't update the x variable. Then process Y executes S=0, it decrements x, now x= -2 and signal semaphore S=1 Now process Z executes s=0, x=-4, signal semaphore S=1 Now process W updates x=1, S=2 Then process X executes X=2 So correct option is D Another Solution: S is a counting semaphore initialized to 2 i.e., Two process can go inside a critical section protected by S. W, X read the variable, increment by 1 and write it back. Y, Z can read the variable, decrement by 2 and write it back. Whenever Y or Z runs the count gets decreased by 2. So, to have the maximum sum, we should copy the variable into one of the processes which increases the count, and at the same time the decrementing processed should run parallel, so that whatever they write back into memory can be overridden by incrementing process. So, in effect decrement would never happen.

Related Links: http://quiz.geeksforgeeks.org/process-synchronization-set-1/ http://geeksquiz.com/operating-systems-process-management-question-11/ for explanation This solution is contributed by Nitika Bansal
Question 9
A certain computation generates two arrays a and b such that a[i]=f(i) for 0 ≤ i < n and b[i]=g(a[i]) for 0 ≤ i < n. Suppose this computation is decomposed into two concurrent processes X and Y such that X computes the array a and Y computes the array b. The processes employ two binary semaphores R and S, both initialized to zero. The array a is shared by the two processes. The structures of the processes are shown below.
Process X:                         Process Y:
private i;                         private i;
for (i=0; i < n; i++) {            for (i=0; i < n; i++) {
   a[i] = f(i);                       EntryY(R, S);
   ExitX(R, S);                       b[i]=g(a[i]);
}                                 }
Which one of the following represents the CORRECT implementations of ExitX and EntryY? (A)
ExitX(R, S) {
  P(R);
  V(S);
}

EntryY (R, S) {
  P(S);
  V(R);
}
(B)
ExitX(R, S) {
  V(R);
  V(S);
}

EntryY(R, S) {
  P(R);
  P(S);
}
(C)
ExitX(R, S) {
  P(S);
  V(R);
}
EntryY(R, S) {
  V(S);
  P(R);
}
(D)
ExitX(R, S) {
  V(R);
  P(S);
}
EntryY(R, S) {
  V(S);
  P(R);
}
A
A
B
B
C
C
D
D
Process Management    GATE CS 2013    
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Question 9 Explanation: 
The purpose here is neither the deadlock should occur
nor the binary semaphores be assigned value greater 
than one.
A leads to deadlock
B can increase value of semaphores b/w 1 to n
D may increase the value of semaphore R and S to
  2 in some cases
See http://geeksquiz.com/operating-systems-process-management-question-13/
Question 10
A process executes the code
fork();
fork();
fork(); 
The total number of child processes created is
A
3
B
4
C
7
D
8
GATE CS 2012    Process Management    
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Question 10 Explanation: 
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.
There are 45 questions to complete.

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