Process-Synchronization
Question 1 |

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

What does the code achieve?
It ensures that all processes execute CODE SECTION P mutually exclusively.
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It ensures that at most two processes are in CODE SECTION Q at any time. | |
It ensures that no process executes CODE SECTION Q before every process has finished CODE SECTION P.
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It ensures that at most n-1 processes are in CODE SECTION P at any time.
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Question 2 |
Consider the following solution to the producer-consumer synchronization problem. The shared buffer size is N. Three semaphores empty, full and mutex are defined with respective initial values of 0, N and 1. Semaphore empty denotes the number of available slots in the buffer, for the consumer to read from. Semaphore full denotes the number of available slots in the buffer, for the producer to write to. The placeholder variables, denoted by P, Q, R and S, in the code below can be assigned either empty or full. The valid semaphore operations are: wait() and sigmal().

Which one of the following assignments to P, Q, R and S will yield the correct solution?
P: full, Q: full, R: empty, S: empty
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P: empty, Q: empty, R: full, S: full
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P: full, Q: empty, R: empty, S: full
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P: empty, Q: full, R: full, S: empty
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Initial: mutex = 1
empty = 0
full = N

Question 3 |
A multithreaded program P executes with x number of threads and used y number of locks for ensuring mutual exclusion while operating on shared memory locations. All locks in the program are non-reentrant, i.e., if a thread holds a lock l, then it cannot re-acquire lock l without releasing it. If a thread is unable to acquire a lock, it blocks until the lock becomes available. The minimum value of x and the minimum value of y together for which execution of P can result in a deadlock are:
x = 1, y = 2 | |
x = 2, y = 1 | |
x = 2, y = 2 | |
x = 1, y = 1 |
Here non re-entrant process can’t own same lock multiple times, so if process tries to acquire already owned lock, will get blocked, and deadlock will happen.
From the above options x=1 (a single thread) and y=1 (a single lock) deadlock is possible when we consider given situations in question.
Question 4 |
Consider the following proposed solution for the critical section problem. There are n processes: P0...P(n-1). In the code, function pmax returns an integer not smaller than any of its arguments. For all i, t[i] is initialized to zero.
Code for Pi: do { c[i]=1; t[i] = pmax(t[0],...,t[n-1])+1; c[i]=0; for every j ≠ i in {0,...,n-1} { while (c[j]); while (t[j] != 0 && t[j]<=t[i]); } Critical Section; t[i]=0; Remainder Section; } while (true);
Which one of the following is TRUE about the above solution?
At most one process can be in the critical section at any time | |
The bounded wait condition is satisfied | |
The progress condition is satisfied | |
It cannot cause a deadlock |
Based on the above code option B, C and D are not satisfied.
We can see that while (t[j] != 0 && t[j] <= t[i]);
because of this condition deadlock is possible when t[j] = = t[i].
Because Progress == no deadlock as no one process is able to make progress by stopping other process.
Bounded waiting is also not satisfied.
In this case both deadlock and bounded waiting to be arising from the same reason as if t[j] = = t[i] is possible then starvation is possible means infinite waiting.
Mutual exclusion is satisfied.
All other processes j started before i must have value of t[j]) < t[i] as function pMax() return a integer not smaller than any of its arguments.
So if anyone out of the processes j have positive value will be executing in its critical section as long as the condition t[j] > 0 && t[j] <= t[i] within while will persist.
And when this j process comes out of its critical section, it sets t[j] = 0; and next process will be selected in for loop.
So, when i process reaches to its critical section none of the processes j which started earlier before process i is in its critical section.
This ensure that only one process is executing its critical section at a time.
So, A is the answer.
Question 5 |
Consider the following two-process synchronization solution.
Process 0 Process 1 --------- --------- Entry: loop while (turn == 1); Entry: loop while (turn == 0); (critical section) (critical section) Exit: turn = 1; Exit: turn = 0;
The shared variable turn is initialized to zero. Which one of the following is TRUE?
This is a correct two-process synchronization solution. | |
This solution violates mutual exclusion requirement. | |
This solution violates progress requirement. | |
This solution violates bounded wait requirement. |
So False.
C) Progress means if one process does not want to enter the critical section then it should not stop other process to enter the critical section.
But we can see that if process 0 will not enter the critical section then value of turn will not become 1 and process 1 will not be able to enter critical section.
So progress not satisfied. True.
D) Bounded waiting solution as there is a strict alteration.
So, False.
Question 6 |
Consider a non-negative counting semaphore S. The operation P(S) decrements S, and V(S) increments S. During an execution, 20 P(S) operations and 12 V(S) operations are issued in some order. The largest initial value of S for which at least one P(S) operation will remain blocked is _________.
7 | |
8 | |
9 | |
10 |
P(S) operation remain in blocked state therefore it will -1.
The negative value of the counting semaphore indicates the number of processes in suspended list (or blocked).
Take any sequence of 20P and 12V operations, at least one process will always remain blocked.
So, X - 20 + 12 = -1
Here P(S) = 20 and V(S) = 12
X = 7
Question 7 |
P1() { C = B – 1; B = 2*C; } P2() { D = 2 * B; B = D - 1; }The number of distinct values that B can possibly take after the execution is
3 | |
4 | |
5 | |
6 |
If we execute P1 process after P2 process, then B = 4
If we did preemption between P1 & P2 processes, then B = 2 (Preemption have done from P1 to P2) or B = 3 (Preemption have done from P2 to P1). So, among 2 & 3 values, only one value will be saved in B. So, total no. of distinct values that B can possibly take after the execution is 3.
Question 8 |

The producer will be able to add an item to the buffer, but the consumer can never consume it. | |
The consumer will remove no more than one item from the buffer. | |
Deadlock occurs if the consumer succeeds in acquiring semaphore s when the buffer is empty. | |
The starting value for the semaphore n must be 1 and not 0 for deadlock-free operation. |
Question 9 |
AcquireLock(L){ while (Fetch_And_Add(L,1)) L = 1; } ReleaseLock(L){ L = 0; }This implementation
fails as L can overflow | |
fails as L can take on a non-zero value when the lock is actually available | |
works correctly but may starve some processes | |
works correctly without starvation |
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 10 |
Which one of the following statements describes the properties achieved?
Mutual exclusion but not progress
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Progress but not mutual exclusion | |
Neither mutual exclusion nor progress
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Both mutual exclusion and progress |
Question 11 |

At least twice | |
Exactly twice | |
Exactly thrice | |
Exactly once |
S1=0
S2=0
P0 enters the critical section first,
prints (‘0’)
releases S1,S2(i.e., S1=1 & S2=1)
Now P1 & P2 both can enter critical section releases S0 & prints (‘0’)
This process continues, hence the number of zero’s printed ≥2.
Question 12 |
- The enter_CS() and leave_CS() functions to implement critical section of a process are realized using test-and-set instruction as follows:
I only
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I and II | |
II and III
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IV only |
It is not using any queue to avoid starvation and there is no use of FIFO.
In CS, only one process can enter.
So, Answer is option A.
Question 13 |
0 and 0
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0 and 1 | |
1 and 0 | |
1 and 1 |
Yb must be '0' initially, because if Yb is '1' initially then two process can be in critical section at the same time.
So answer is Option (C).
Question 14 |
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?The process can deadlock | |
One of the threads can starve | |
Some of the items produced by the producer may be lost | |
Values generated and stored in ‘x’ by the producer will always be consumed before the producer can generate a new value |
(B) No starvation happen because there is alteration between Producer and Consumer.
(C) No items is lost.
Question 15 |
/* P1 */ while (true) { wants1 = true; while (wants2 == true); /* Critical Section */ wants1=false; } /* Remainder section */ /* P2 */ while (true) { wants2 = true; while (wants1==true); /* Critical Section */ wants2 = false; } /* Remainder section */
It does not ensure mutual exclusion. | |
It does not ensure bounded waiting.
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It requires that processes enter the critical section in strict alternation.
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It does not prevent deadlocks, but ensures mutual exclusion. |
Now, if in P1
wants1 = true ;
executed and preempted and now if in P2
wants2 = true ;
executed.
Then the deadlock situation will be created because both will fall into infinite loop.
Question 16 |
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?
(i) is false and (ii) is true | |
Both (i) and (ii) are false | |
(i) is true and (ii) is false | |
Both (i) and (ii) are true |
Question 17 |
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
signal (mutex), wait (wrt), signal (wrt), wait (mutex) | |
signal (wrt), signal (mutex), wait (mutex), wait (wrt) | |
wait (wrt), signal (mutex), wait (mutex), signal (wrt) | |
signal (mutex), wait (mutex), signal (mutex), wait (mutex)
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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 |
void barrier ( void ) { 1: P(S); 2: process_arrived++; 3. V(S); 4: while (process_arrived !=3); 5: P(S); 6: process_left++; 7: if (process_left==3) { 8: process_arrived = 0; 9: process_left = 0; 10: } 11: V(S); } |
The barrier implementation is wrong due to the use of binary semaphore S
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The barrier implementation may lead to a deadlock if two barriers in invocations are
used in immediate succession
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Lines 6 to 10 need not be inside a critical section
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The barrier implementation is correct if there are only two processes instead of three
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Hence, it is leads to deadlock.
Question 19 |
void barrier ( void ) { 1: P(S); 2: process_arrived++; 3. V(S); 4: while (process_arrived !=3); 5: P(S); 6: process_left++; 7: if (process_left==3) { 8: process_arrived = 0; 9: process_left = 0; 10: } 11: V(S); } |
Lines 6 to 10 are simply replaced by process_arrived--
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At the beginning of the barrier the first process to enter the barrier waits until process_arrived becomes zero before proceeding to execute P(S).
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Context switch is disabled at the beginning of the barrier and re-enabled at the end.
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The variable process_left is made private instead of shared.
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Question 20 |
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?
The implementation may not work if context switching is disabled in P | |
Instead of using fetch-and-set, a pair of normal load/store can be used | |
The implementation of V is wrong
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The code does not implement a binary semaphore |
B) If we use normal load and store instead of Fetch and Set, then there can be chance that more than one process sees S.value as 0 and then mutual exclusion will not be satisfied. So wrong.
C) Here we are setting S→value to 0, which is correct. This option thats why wrong.
D) Only one process can be in critical section at any time. So this option is wrong.
Question 21 |
- Interchanging Wait (F) and Wait (S) in the Producer process
- Interchanging Signal (S) and Signal (F) in the Consumer process
1 only | |
2 only | |
Neither 1 nor 2 | |
Both 1 and 2 |
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 22 |
-
- 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);
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- 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;
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- 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,
P(x_sem), V(next) | |
V(next), P(x_sem) | |
P(next), V(x_sem) | |
P(x_sem), V(x_sem) |
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 23 |
P(Sy), P(Sx); P(Sx), P(Sy)
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P(Sx), P(Sy); P(Sy), P(Sx)
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P(Sx), P(Sx); P(Sy), P(Sy)
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P(Sx), P(Sy); P(Sx), P(Sy)
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P1 : line 1
P2 : line 3 (block require Sx)
P1 : line 2
P2 : line 4 (still in block state)
P1 : execute CS, the push up the value of Sx.
P2 : line 3 line 4 (block require Sy)
P1 : push up Sy
P2 : line 4 get Sy and enter into CS
P2 : start execution.
So option D is Answer.
Question 24 |
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' ?
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1
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P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S initially 1, and T initially 0
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P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1 | |
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S initially 1, and T initially 0
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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 25 |
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 01^n0 or 10^n1 where n is odd?
P(S) at W, V(S) at X, P(T) at Y, V(T) at Z, S and T initially 1 | |
P(S) at W, V(T) at X, P(T) at Y, V(S) at Z, S and T initially 1 | |
P(S) at W, V(S) at X, P(S) at Y, V(S) at Z, S initially 1
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V(S) at W, V(T) at X, P(S) at Y, P(T) at Z, S and T initially 1 |
Question 26 |
Thrashing | |
Deadlock | |
Starvation, but not deadlock | |
None of the above |

Question 27 |
Theory Explanation is given below. |
(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 28 |
cycle steating | |
rare condition | |
a time lock | |
a deadlock |
Speed of processes corresponds to ordering of processes.
Question 29 |
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
0 | |
8 | |
10 | |
12 |
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 30 |
1 | |
2 | |
3 | |
None of the above |
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 31 |
which should run in a certain specified amount of time | |
which avoids deadlocks | |
where shared resources are accessed | |
which must be enclosed by a pair of semaphore operations, P and V
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Question 32 |
flag[j]=true and turn=i | |
flag[j]=true and turn=j | |
flag[i]=true and turn=j | |
flag[i]=true and turn=i |
Question 33 |
One which is enclosed by a pair of P and V operations on semaphores. | |
A program segment that has not been proved bug-free. | |
A program segment that often causes unexpected system crashes. | |
A program segment where shared resources are accessed. |