Process-Synchronization

Question 1

Each of a set of n processes executes the following code using two semaphores a and b initialized to 1 and 0, respectively. Assume that count is a shared variable initialized to 0 and not used in CODE SECTION P.

wait (a); count = count+1;
if (count==n) signal (b);
signal (a); wait (b); signal (b);

What does the code achieve?

A
It ensures that all processes execute CODE SECTION P mutually exclusively.
B
It ensures that at most two processes are in CODE SECTION Q at any time.
C
It ensures that no process executes CODE SECTION Q before every process has finished CODE SECTION P.
D
It ensures that at most n-1 processes are in CODE SECTION P at any time.
Question 1 Explanation: 
The wait(a) ensures that the count value is correctly incremented (no race condition) if(count==n) signal (b); // This signal(b) statement is executed by the last (nth) process only. Rest of the n-1 processes are blocked on wait(b). Once the nth process makes signal(b) then the rest of the processes can proceed and enter Code section Q.
Question 2

Fetch_And_Add(X,i) is an atomic Read-Modify-Write instruction that reads the value of memory location X, increments it by the value i, and returns the old value of X. It is used in the pseudocode shown below to implement a busy-wait lock. L is an unsigned integer shared variable initialized to 0. The value of 0 corresponds to lock being available, while any non-zero value corresponds to the lock being not available.

  AcquireLock(L){
         while (Fetch_And_Add(L,1))
               L = 1;
   }
  ReleaseLock(L){
         L = 0;
   }

This implementation

A
fails as L can overflow
B
fails as L can take on a non-zero value when the lock is actually available
C
works correctly but may starve some processes
D
works correctly without starvation
Question 2 Explanation: 
Check the loop first:
while (Fetch_And_Add (L,1))
L = 1; // A waiting process can be here just after
// the lock is released, and can make L = 1.
Assume P1 executes until while condition and preempts before executing L =1. Now P2 executes all statements, hence L = 0. Then P1 without checking L it makes L = 1 by executing the statement where it was preempted.
It takes a non-zero value (L=1) when the lock is actually available (L = 0). So option B.
Question 3

A critical section is a program segment

A
which should run in a certain specified amount of time
B
which avoids deadlocks
C
where shared resources are accessed
D
which must be enclosed by a pair of semaphore operations, P and V
Question 3 Explanation: 
In CS, share resources are accessed.
Question 4

The concurrent programming constructs fork and join are as below:
fork

   N = 2
   M = 2
   fork L3
   fork L4
   S1
L1:join N
   S3
L2:join M
   S5
L3:S2
   goto L1
L4:S4
   goto L2
next: 
A
Theory Explanation.
Question 5

Each Process Pi, i = 1.......9 is coded as follows

    repeat 
           P(mutex)
           {Critical section}
           V(mutex)
    forever   

The code for P10 is identical except it uses V(mutex) in place of P(mutex). What is the largest number of processes that can be inside the critical section at any moment?

A
1
B
2
C
3
D
None of the above
Question 5 Explanation: 
Since the both code (i.e., P1 to P9 and P10) can be executed any number of times and code for P10 is
repeat
{
V(mutex)
C.S.
V(mutex)
}
forever
Now, let me say P1 is in CS then P1 is in CS
then P10 comes and executes CS (upon mutex).
Now, P2 comes (down on mutex).
Now P10 moves out of CS (again binary semaphore will be 1).
Now P3 comes (down on mutex).
Now P10 comes (upon mutex).
Now P4 comes (down mutex).

So, if we take P10 'in' and 'out' of CS recursively, all 10 processes can be in CS at same time using binary semaphore only.
Question 6

When  the  result  of  a  computation  depends  on  the  speed  of  the  processes involved there is said to be

A
cycle stealing
B
rare condition
C
a time lock
D
a deadlock
Question 6 Explanation: 
When first result depends on ordering of processes it is called race condition.
Speed of processes corresponds to ordering of processes.
Question 7

A counting semaphore was initialized to 10. Then 6P (wait) operations and 4V (signal) operations were completed on this semaphore. The resulting value of the semaphore is

A
0
B
8
C
10
D
12
Question 7 Explanation: 
Let the semaphore be S which is initially 10.
S = 10
Now 6P operations and uv operations were completed on this semaphore. So final value of S will be
S = 10 - 6 + 4 = 8
Question 8

(a) A certain processor provides a 'test and set' instruction that is used as follows:

 TEST register, flag 

This instruction atomically copies flag to register and sets flag to 1. Give pseudo-code for implementing the entry and exit code to a critical region using this instruction.

(b) Consider the following solution to the producer-consumer problem using a buffer of size 1. Assume that the initial value of count is 0. Also assume that the testing of count and assignment to count are atomic operations.

            Producer:      
                Repeat 
                         Produce an item;
                         if count = 1 then sleep;
                         place item in buffer.
                         Count = 1;
                         Wakeup(Consumer);
               Forever 

            Consumer:
               Repeat
                         if count = 0 then sleep;
                         Remove item from buffer;
                         Count = 0;
                         Wakeup(Producer);
                         Consume item;
              Forever; 

Show that in this solution it is possible that both the processes are sleeping at the same time.

A
Theory Explanation.
Question 9

Let m[0] … m[4] be mutexes (binary semaphores) and P[0] … P[4] be processes. Suppose each process P[i] executes the following:

  wait (m[i]); wait(m[(i+1) mode 4]);
  ------
  release (m[i]); release (m[(i+1)mod 4]);

This could cause:

A
Thrashing
B
Deadlock
C
Starvation, but not deadlock
D
None of the above
Question 9 Explanation: 
Question 10

(a) Fill in the boxes below to get a solution for the readers-writers problem, using a single binary semaphore, mutex (initialized to 1) and busy waiting. Write the box numbers (1, 2 and 3), and their contents in your answer book.

     int R = 0, W = 0;
     Reader () {
 L1:        wait (mutex);
            If (W ==0) {
                R = R +1; 
                ▭______(1)
     }
     else {
                ▭______(2)
                goto L1;
                }
                …./* do the read */
                wait (mutex)
                R = R – 1;
                signal (mutex);
          }
          Writer () {
 L2:            wait(mutex);
                If (▭) {______(3)
                    signal (mutex);
                    goto L2;
                }
                W=1;
                signal (mutex);
                …./*do the write*/
                wait(mutex)
                W = 0;
                signal (mutex); 

(b) Can the above solution lead to starvation of writers?

A
Theory Explanation is given below.
Question 10 Explanation: 
(a) (i) Signal (mutex)
(ii) Signal (mutex)
(iii) R ⇒ 1 (or) W == 1
(b) In CS, atleast one of the reader is always present. That means writer can't be enter into the critical section.
This leads to readers-writers problem may occur in the queue.
Question 11

Consider Peterson’s algorithm for mutual exclusion between two concurrent processes i and j. The program executed by process is shown below.

   repeat   
      flag[i] = true; 
      turn = j; 
      while (P) do no-op; 
      Enter critical section, perform actions, then exit critical 
      section 
      flag[i] = false; 
      Perform other non-critical section actions. 
   until false; 

For the program to guarantee mutual exclusion, the predicate P in the while loop should be.

A
flag[j]=true and turn=i
B
flag[j]=true and turn=j
C
flag[i]=true and turn=j
D
flag[i]=true and turn=i
Question 11 Explanation: 
To ensure the mutual exclusion, we have to take care that ‘turn’ value which is ‘j’ so we should not allow that process in CS which is having flag[j] = true.
Question 12

Suppose we want to synchronize two concurrent processes P and Q using binary semaphores S and T. The code for the processes P and Q is shown below.

Process P:
while (1) {
W:
   print '0';
   print '0';
X:
}
	
Process Q:
while (1) {
Y:
   print '1';
   print '1';
Z:
}

Synchronization statements can be inserted only at points W, X, Y and Z.

Which of the following will always lead to an output staring with '001100110011'?

A
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1
B
P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S initially 1, and T initially 0
C
P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1
D
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S initially 1, and T initially 0
Question 12 Explanation: 
Option B is correct.
Process P1 will be executed first and then Process P2 can be executed next.
At the process P: W=P(S)
X=V(T)
At the process Q: Y=P(T)
Z=V(S)
Here, S=1, T=0 then the process P executes first and then Q, and both can run on process alternate way start with P.
Question 13

Suppose we want to synchronize two concurrent processes P and Q using binary semaphores S and T. The code for the processes P and Q is shown below.

Process P:
while (1) {
W:
   print '0';
   print '0';
X:
}
	
Process Q:
while (1) {
Y:
   print '1';
   print '1';
Z:
}

Synchronization statements can be inserted only at points W, X, Y and Z.

Which of the following will ensure that the output string never contains a substring of the form 01n0 or 10n1 where n is odd?

A
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1
B
P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1
C
P(S) at W, V(S) at X, P(S) at Y, V(S) at Z, S initially 1
D
V(S) at W, V(T) at X, P(S) at Y, P(T) at Z, S and T initially 1
Question 13 Explanation: 
Here to print the required output only one semaphore is enough, if we initialize two at a time then they will run concurrently and leads the processing error.
Question 14

Consider the solution to the bounded buffer producer/consumer problem by using general semaphores S, F, and E. The semaphore S is the mutual exclusion semaphore initialized to 1. The semaphore F corresponds to the number of free slots in the buffer and is initialized to N. The semaphore E corresponds to the number of elements in the buffer and is initialized to 0.

Producer Process                Consumer Process
Produce an item;                Wait(E);
Wait(F);                        Wait(S);
Wait(S);
Append the items to the buffer; Signal(S);
Signal(S);                      Signal(F);
Signal(E);                      Consume the item;

Which of the following interchange operations may result in a deadlock? 1. Interchanging Wait (F) and Wait (S) in the Producer process 2. Interchanging Signal (S) and Signal (F) in the Consumer process

A
1 only
B
2 only
C
Neither 1 nor 2
D
Both 1 and 2
Question 14 Explanation: 
Suppose Wait(F) and Wait(S) are interchanged. And let the slots are full → F=0.
Now if Wait(S) in producer succeeds, then producer will wait for Wait(F) which is never going to succeed as consumer would be waiting for Wait(S). So deadlock, can happen.
If Signal(S) and Signal(F) are interchanged in consumer, deadlock won't happen. It will just give priority to a producer compared to the next consumer waiting.
Question 15

The wait and signal operations of a monitor are implemented using semaphores as follows. In the following,
x is a condition variable,
mutex is a semaphore initialized to 1,
x_sem is a semaphore initialized to 0,
x_count is the number of processes waiting on semaphore x_sem, initially 0, next is a semaphore initialized to 0,
next_count is the number of processes waiting on semaphore next, initially 0.
The body of each procedure that is visible outside the monitor is replaced with the following:

P(mutex); 
      ⋮
body of procedure 
      ⋮
if (next_count > 0)
    V(next);
else
    V(mutex); 
Each occurrence of x.wait is replaced with the following:
x_count = x_count + 1;
if (next_count > 0)
    V(next)
else
    V(mutex);
------------------------------------------------------------ E1;
x_count = x_count - 1; 
Each occurrence of x.signal is replaced with the following:
if (x_count > 0)
{
    next_count = next_count + 1;
    ------------------- E2;
    P(next),
    next_count = next_count - 1;
} 
For correct implementation of the monitor, statements E1 and E2 are, respectively,

A
P(x_sem), V(next)
B
V(next), P(x_sem)
C
P(next), V(x_sem)
D
P(x_sem), V(x_sem)
Question 15 Explanation: 
x_count is the no. of processes waiting on semaphore x_sem, initially 0.
x_count is incremented and decremented in x_wait, which shows that in between them wait(x_sem) must happen which is P(x_sem). Correspondingly V(x_sem) must happen in x_signal. So, D choice.
Question 16

Processes P1 and P2 use critical_flag in the following routine to achieve mutual exclusion. Assume that critical_flag is initialized to FALSE in the main program.

        get_exclusive_access ( ) 
{ 
  if (critical _flag == FALSE) { 
      critical_flag = TRUE ; 
      critical_region () ; 
      critical_flag = FALSE; 
      } 
}

Consider the following statements. i. It is possible for both P1 and P2 to access critical_region concurrently. ii. This may lead to a deadlock.

Which of the following holds?

A
(i) is false and (ii) is true
B
Both (i) and (ii) are false
C
(i) is true and (ii) is false
D
Both (i) and (ii) are true
Question 16 Explanation: 
Both the processes run concurrently if they run concurrently till the execution then there is no deadlock.
Question 17

Synchronization in the classical readers and writers problem can be achieved through use of semaphores. In the following incomplete code for readers-writers problem, two binary semaphores mutex and wrt are used to obtain synchronization

wait (wrt)
writing is performed
signal (wrt)
wait (mutex)  
  readcount = readcount + 1
  if readcount = 1 then S1
S2
reading is performed
S3
  readcount = readcount - 1
  if readcount = 0 then S4 
signal (mutex) 

The values of S1, S2, S3, S4, (in that order) are

A
signal (mutex), wait (wrt), signal (wrt), wait (mutex)
B
signal (wrt), signal (mutex), wait (mutex), wait (wrt)
C
wait (wrt), signal (mutex), wait (mutex), signal (wrt)
D
signal (mutex), wait (mutex), signal (mutex), wait (mutex)
Question 17 Explanation: 
S1: If readcount is 1, i.e., some reader is reading, DOWN on wrt so that no writer can write.
S2: After readcount has been updated, UP on mutex.
S3: DOWN on mutex to update readcount.
S4: If readcount is zero, i.e., no reader is reading, UP on wrt to allow some writer to write.
Question 18

The following is a code with two threads, producer and consumer, that can run in parallel. Further, S and Q are binary semaphores equipped with the standard P and V operations.

semaphore S = 1, Q = 0; 
integer x; 
producer:                            consumer:
while(true) do                      while(true) do 
     P(S);                               P(Q); 
     x = produce ();                     consume (x); 
     V(Q);                               V(S); 
done                                done
Which of the following is TRUE about the program above?
A
The process can deadlock
B
One of the threads can starve
C
Some of the items produced by the producer may be lost
D
Values generated and stored in ‘x’ by the producer will always be consumed before the producer can generate a new value
Question 18 Explanation: 
(A) Deadlock cannot happen as both producer and consumer are operating on different semaphores (No hold and wait).
(B) No starvation happen because there is alteration between Producer and Consumer.
(C) No items is lost.
Question 19
Consider the following threads, T 1 , T 2 , and T 3 executing on a single processor, synchronized using three binary semaphore variables, S 1 , S 2 , and S 3 , operated upon using standard wait() and signal(). The threads can be context switched in any order and at any time. Which initialization of the semaphores would print the sequence BCABCABCA….?
A
S 1 = 1; S 2 = 1; S 3 = 1
B
S 1 = 1; S 2 = 1; S 3 = 0
C
S 1 = 1; S 2 = 0; S 3 = 0
D
S 1 = 0; S 2 = 1; S 3 = 1
Question 19 Explanation: 
Explanation: Since B is to be printed first semaphore S1 has to be set to 1, so that the thread enters its critical section and prints B. Then C needs to be printed, this happens when Thread 1 is blocked on semaphore s3 (blocked, s3=0). Then Thread1 invokes Thread 3 to print A.
There are 19 questions to complete.

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