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All questions of Concurrency & Deadlock for Computer Science Engineering (CSE) Exam
Which of the following statements is not true?- a)Deadlock can never occur if resources can be shared by competing processes.
- b)Deadlock can never occur if resources must be requested in the same order by processes.
- c)The Banker’s algorithm for avoiding deadlock requires knowing resource requirements in advance
- d)If the resource allocation graph depicts a cycle then deadlock has certainly occured.
Correct answer is option 'B'. Can you explain this answer?
Which of the following statements is not true?
a)
Deadlock can never occur if resources can be shared by competing processes.
b)
Deadlock can never occur if resources must be requested in the same order by processes.
c)
The Banker’s algorithm for avoiding deadlock requires knowing resource requirements in advance
d)
If the resource allocation graph depicts a cycle then deadlock has certainly occured.
|
Madhurima Chakraborty answered |
Deadlock can’t take place if resources must be requested in same order by process. For deadlock, circular wait is a must condition.
Suppose we have a system in which processes is in hold and wait condition then which of the following approach prevent the deadlock.- a)Request all resources initially
- b)Spool everything
- c)Take resources away
- d)Order resources numerically
Correct answer is option 'A'. Can you explain this answer?
Suppose we have a system in which processes is in hold and wait condition then which of the following approach prevent the deadlock.
a)
Request all resources initially
b)
Spool everything
c)
Take resources away
d)
Order resources numerically
|
Ruchi Sengupta answered |
Preventing Deadlock by Requesting All Resources Initially
Deadlock is a situation in which two or more processes are unable to proceed because each is waiting for the other to release resources. It can occur in a system with limited resources and processes that are holding resources while waiting for other resources to be released.
One approach to prevent deadlock is to request all resources initially. This means that when a process needs to execute, it requests all the resources it will need for its entire execution before it starts. This approach can help prevent deadlock by ensuring that a process has all the necessary resources before it begins execution.
Advantages of Requesting All Resources Initially:
-Preventing Resource Deadlock: By requesting all resources initially, a process can ensure that it has all the resources it needs for its execution. This prevents the situation where a process holds some resources and waits for others, leading to deadlock.
-Efficient Resource Allocation: Requesting all resources initially allows the system to allocate resources more efficiently. Since a process requests all the resources it needs at once, the system can determine if there are enough resources available to satisfy the request. If there are not enough resources, the system can allocate them to other processes that can make progress, avoiding resource wastage.
Disadvantages of Requesting All Resources Initially:
-Resource Overallocation: Requesting all resources initially may lead to resource overallocation, where a process requests more resources than it actually needs. This can result in resource wastage and inefficient resource utilization.
-Low Concurrency: Requesting all resources initially can reduce the level of concurrency in the system. Since a process needs to wait until it acquires all the resources it needs, other processes may have to wait for a long time before they can execute, leading to decreased system performance.
Overall, requesting all resources initially can be an effective approach to prevent deadlock in a system. However, it is important to carefully consider the resource requirements of each process and balance the need for preventing deadlock with the need for efficient resource utilization and system performance.
Deadlock is a situation in which two or more processes are unable to proceed because each is waiting for the other to release resources. It can occur in a system with limited resources and processes that are holding resources while waiting for other resources to be released.
One approach to prevent deadlock is to request all resources initially. This means that when a process needs to execute, it requests all the resources it will need for its entire execution before it starts. This approach can help prevent deadlock by ensuring that a process has all the necessary resources before it begins execution.
Advantages of Requesting All Resources Initially:
-Preventing Resource Deadlock: By requesting all resources initially, a process can ensure that it has all the resources it needs for its execution. This prevents the situation where a process holds some resources and waits for others, leading to deadlock.
-Efficient Resource Allocation: Requesting all resources initially allows the system to allocate resources more efficiently. Since a process requests all the resources it needs at once, the system can determine if there are enough resources available to satisfy the request. If there are not enough resources, the system can allocate them to other processes that can make progress, avoiding resource wastage.
Disadvantages of Requesting All Resources Initially:
-Resource Overallocation: Requesting all resources initially may lead to resource overallocation, where a process requests more resources than it actually needs. This can result in resource wastage and inefficient resource utilization.
-Low Concurrency: Requesting all resources initially can reduce the level of concurrency in the system. Since a process needs to wait until it acquires all the resources it needs, other processes may have to wait for a long time before they can execute, leading to decreased system performance.
Overall, requesting all resources initially can be an effective approach to prevent deadlock in a system. However, it is important to carefully consider the resource requirements of each process and balance the need for preventing deadlock with the need for efficient resource utilization and system performance.
Busy Waiting in process synchronisation is a term used when a process is in critical section and other processes are waiting to enter critical section. Busy Waiting is also called ___________- a)Bounded Waiting
- b)Busy processing
- c)Spin Lock
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Busy Waiting in process synchronisation is a term used when a process is in critical section and other processes are waiting to enter critical section. Busy Waiting is also called ___________
a)
Bounded Waiting
b)
Busy processing
c)
Spin Lock
d)
None of these
|
Crack Gate answered |
Bounded Waiting: This is a condition that needs to be fulfilled to achieve synchronisation.
- It says that no process has to wait forever to enter into a critical section.
- There should be a specific time after which it should get a chance to enter into the critical section.
- If bounded waiting is not fulfilled, it can suffer from starvation.
Busy processing: There is no such thing as busy processing.
Spin Lock: Also called Busy Waiting.
Spin Lock: Also called Busy Waiting.
- When a process is in the critical section and other processes are waiting to enter the critical section then it is called spinlock.
Consider a system having 'm' resources of the same type. These resources are shared by 3 processes A, B, C, which have peak time demands of 3, 4, 6 respectively. The minimum value of ‘m’ that ensures that deadlock will never occur is
- a)13
- b)12
- c)11
- d)14
Correct answer is option 'A'. Can you explain this answer?
Consider a system having 'm' resources of the same type. These resources are shared by 3 processes A, B, C, which have peak time demands of 3, 4, 6 respectively. The minimum value of ‘m’ that ensures that deadlock will never occur is
a)
13
b)
12
c)
11
d)
14
|
Anisha Chavan answered |
M for which the system is deadlock-free is 9.
To determine the minimum value of m, we need to consider the worst-case scenario where all processes are requesting their peak time demands simultaneously.
In this case, process A requires 3 resources, process B requires 4 resources, and process C requires 6 resources. Therefore, the total number of resources required in the system is 3 + 4 + 6 = 13.
To avoid deadlock, the system should have enough resources to satisfy the maximum resource demand of any process. Therefore, the minimum value of m should be equal to or greater than 13.
However, since all processes are requesting their peak time demands simultaneously, it is possible for the system to enter a deadlock state if the number of available resources is exactly equal to the total resource demand.
Therefore, the minimum value of m for which the system is deadlock-free is m = 13 + 1 = 14.
However, since all resources in the system are of the same type, it is not possible to allocate fractional resources. Therefore, the minimum value of m should be an integer.
Therefore, the minimum value of m for which the system is deadlock-free is m = 14.
To determine the minimum value of m, we need to consider the worst-case scenario where all processes are requesting their peak time demands simultaneously.
In this case, process A requires 3 resources, process B requires 4 resources, and process C requires 6 resources. Therefore, the total number of resources required in the system is 3 + 4 + 6 = 13.
To avoid deadlock, the system should have enough resources to satisfy the maximum resource demand of any process. Therefore, the minimum value of m should be equal to or greater than 13.
However, since all processes are requesting their peak time demands simultaneously, it is possible for the system to enter a deadlock state if the number of available resources is exactly equal to the total resource demand.
Therefore, the minimum value of m for which the system is deadlock-free is m = 13 + 1 = 14.
However, since all resources in the system are of the same type, it is not possible to allocate fractional resources. Therefore, the minimum value of m should be an integer.
Therefore, the minimum value of m for which the system is deadlock-free is m = 14.
The following two functions P1 and P2 that share a variable B with an initial value of 2 execute concurrently.
The number of distinct values that B can possibly take after the execution is __________.
Correct answer is '3'. Can you explain this answer?
The following two functions P1 and P2 that share a variable B with an initial value of 2 execute concurrently.
The number of distinct values that B can possibly take after the execution is __________.
|
|
Sudhir Patel answered |
Initial value of B is 2 //value stored in main memory
Value of B = 4
First execute P2()
P2 ( ) {
D = 2 * B; // 2 × 2 = 4
B = D – 1; // B = 4 – 1 = 3
}
write the value of B = 3 to memory
execute P1()
P1 ( ) {
C = B – 1; // C = 3 – 1 = 2
B = 2 * C; // B = 2 × 2 = 4
}
write the value of B = 4 is memory.
Distinct value of B is 4.
Value of B =3
First Execute P1()
P1 ( ) {
C = B – 1; // C = 2 – 1 = 1
B = 2 * C; // B = 2 × 1 =2
}
write the value of B = 2 to memory
then execute P2()
P2 ( ) {
D = 2 * B; // 2 × 2 = 4
B = D – 1; // B = 4 – 1 = 3
}
write the value of B = 3 to memory
One of the distinct values of B is 3.
Value of B = 2
First execute P2()
P2 ( ) {
D = 2 * B; // 2 × 2 = 4
B = D – 1; // B = 4 – 1 = 3
}
//don’t write to memory
execute P1() // with initial value of B = 2
P1 ( ) {
C = B – 1; // C = 2 – 1 = 1
B = 2 * C; // B = 2 × 1 = 2
}
write the value B = 3 by P2() in memory.
overwrite the value written by P2() as B = 2 by P1() in memory.
One of the distinct values of B is 2.
Therefore, the distinct values of B are 2, 3 and 4.
The number of distinct values that B can possibly take after the execution is 3.
Value of B = 4
First execute P2()
P2 ( ) {
D = 2 * B; // 2 × 2 = 4
B = D – 1; // B = 4 – 1 = 3
}
write the value of B = 3 to memory
execute P1()
P1 ( ) {
C = B – 1; // C = 3 – 1 = 2
B = 2 * C; // B = 2 × 2 = 4
}
write the value of B = 4 is memory.
Distinct value of B is 4.
Value of B =3
First Execute P1()
P1 ( ) {
C = B – 1; // C = 2 – 1 = 1
B = 2 * C; // B = 2 × 1 =2
}
write the value of B = 2 to memory
then execute P2()
P2 ( ) {
D = 2 * B; // 2 × 2 = 4
B = D – 1; // B = 4 – 1 = 3
}
write the value of B = 3 to memory
One of the distinct values of B is 3.
Value of B = 2
First execute P2()
P2 ( ) {
D = 2 * B; // 2 × 2 = 4
B = D – 1; // B = 4 – 1 = 3
}
//don’t write to memory
execute P1() // with initial value of B = 2
P1 ( ) {
C = B – 1; // C = 2 – 1 = 1
B = 2 * C; // B = 2 × 1 = 2
}
write the value B = 3 by P2() in memory.
overwrite the value written by P2() as B = 2 by P1() in memory.
One of the distinct values of B is 2.
Therefore, the distinct values of B are 2, 3 and 4.
The number of distinct values that B can possibly take after the execution is 3.
The following program consists of 3 concurrent processes and 3 binary semaphores. The semaphores are initialized as S0 = 1, S1 = 0, S2 = 0.
How many times will process P0 print ‘0’? - a)At least twice
- b)Exactly twice
- c)Exactly thrice
- d)Exactly once
Correct answer is option 'A'. Can you explain this answer?
The following program consists of 3 concurrent processes and 3 binary semaphores. The semaphores are initialized as S0 = 1, S1 = 0, S2 = 0.
How many times will process P0 print ‘0’?
a)
At least twice
b)
Exactly twice
c)
Exactly thrice
d)
Exactly once
|
|
Crack Gate answered |
Concept:
Binary semaphore which can take only two values 0 and 1 and ensure mutual exclusion.
A process is entered into the critical section only when the value of semaphore(s) = 1
Otherwise, it will wait until semaphore(s) >0
Binary semaphore which can take only two values 0 and 1 and ensure mutual exclusion.
A process is entered into the critical section only when the value of semaphore(s) = 1
Otherwise, it will wait until semaphore(s) >0
The semaphores are initialized as S0=1, S1=0, S2=0.
Because S0 =1 then P0 enter into the critical section and other processes will wait until either S1=1 or S2 =1
The minimum number of times 0 printed:
Because S0 =1 then P0 enter into the critical section and other processes will wait until either S1=1 or S2 =1
The minimum number of times 0 printed:
- S0 =1 then P0 enter into the critical section
- print '0'
- then release S1 and S2 means S1 =1 and s2 =1
- now either P1 or P2 can enter into the critical section
- if P1 enter into the critical section
- release S0
- then P2 enter into the critical section
- release S0
- P1 enter into the critical section
- print '0'
The minimum number of time 0 printed is twice when executing in this order (p0 -> p1 -> p2 -> p0)
The Maximum number of times 0 printed:
The Maximum number of times 0 printed:
- S0 =1 then P0 enter into the critical section
- print '0'
- Then release S1 and S2 means S1 =1 and s2 =1
- Now either P1 or P2 can enter into the critical section
- If P1 enter into the critical section
- Release S0 means S0 =1
- S0 =1 then P0 enter into the critical section
- print '0'
- Then P2 enter into the critical section
- Release S0 means S0 =1
- S0 =1 then P0 enter into the critical section
- print '0'
Maximum no. of time 0 printed is thrice when execute in this order (p0 -> p1 -> p0 -> p2 -> p0)
So, At least twice will process P0 print ‘0’
So, At least twice will process P0 print ‘0’
At a particular time of computation the value of a counting semaphore is 7. Then 20 P operations and 15 V operation were completed on this semaphore. The resulting value of the semaphore is- a)42
- b)2
- c)7
- d)12
Correct answer is option 'B'. Can you explain this answer?
At a particular time of computation the value of a counting semaphore is 7. Then 20 P operations and 15 V operation were completed on this semaphore. The resulting value of the semaphore is
a)
42
b)
2
c)
7
d)
12
|
|
Sudhir Patel answered |
Concepts:
V(S): Signal operation will increment the semaphore variable, that is, S++.
P(S): Wait operation will decrement the semaphore variable., that is, S--.
V(S): Signal operation will increment the semaphore variable, that is, S++.
P(S): Wait operation will decrement the semaphore variable., that is, S--.
Data:
Initial counting semaphore = I = 7
Wait operation = 20 P
Signal operation = 15 V
Final counting semaphore = F
Initial counting semaphore = I = 7
Wait operation = 20 P
Signal operation = 15 V
Final counting semaphore = F
Formula:
F = I + 15V + 20P
F = I + 15V + 20P
Calculation:
F = 7 + 15(+1) + 20(-1)
∴ F = 2
the resulting value of the semaphore is 2
F = 7 + 15(+1) + 20(-1)
∴ F = 2
the resulting value of the semaphore is 2
An operating system makes use of Banker’s algorithm to allocate 12 printers. Printers will be allocated'to a process requesting them only if - there are enough available printers to allow it to run to completion.
User 1 using 7 printers and will need at most a total of 10 printers.
User 2 is using 1 printer and will need at most 4 printers.
User 3 is using 2 printers and will need at most 4 printers.
Each user is currently requesting 1 more printer, Which of the following is true?- a)The operating system will grant a printer to User 1 only.
- b)The operating system will grant a printer to User 2 only.
- c)The operating system will grant a printer to User 3 only.
- d)The operating system will not grant any more printers until some are relinquished.
Correct answer is option 'C'. Can you explain this answer?
An operating system makes use of Banker’s algorithm to allocate 12 printers. Printers will be allocated'to a process requesting them only if - there are enough available printers to allow it to run to completion.
User 1 using 7 printers and will need at most a total of 10 printers.
User 2 is using 1 printer and will need at most 4 printers.
User 3 is using 2 printers and will need at most 4 printers.
Each user is currently requesting 1 more printer, Which of the following is true?
User 1 using 7 printers and will need at most a total of 10 printers.
User 2 is using 1 printer and will need at most 4 printers.
User 3 is using 2 printers and will need at most 4 printers.
Each user is currently requesting 1 more printer, Which of the following is true?
a)
The operating system will grant a printer to User 1 only.
b)
The operating system will grant a printer to User 2 only.
c)
The operating system will grant a printer to User 3 only.
d)
The operating system will not grant any more printers until some are relinquished.
|
Athira Choudhary answered |
Total printers = 12
Printers allocated to user 1, user 2 and user 3 are 7, 1 and 2 respectively
So, only safest sequence is V3 first then either or V3. So, operating system will grant printer to user 3 only.
Printers allocated to user 1, user 2 and user 3 are 7, 1 and 2 respectively
So, only safest sequence is V3 first then either or V3. So, operating system will grant printer to user 3 only.
Two atomic operations permissible on Semaphores are __________ and __________.- a)wait, stop
- b)wait, hold
- c)hold, signal
- d)wait, signal
Correct answer is option 'D'. Can you explain this answer?
Two atomic operations permissible on Semaphores are __________ and __________.
a)
wait, stop
b)
wait, hold
c)
hold, signal
d)
wait, signal
|
|
Sudhir Patel answered |
Semaphores are integer variables that are used to solve the critical section problem by using two atomic operations, wait and signal that are used for process synchronization.
Wait:
The wait operation decrements the value of its argument S if it is positive. If S is negative or zero, then no operation is performed.
The wait operation decrements the value of its argument S if it is positive. If S is negative or zero, then no operation is performed.
wait(S)
{
while (S<=0)
S--;
}
Signal:
The signal operation increments the value of its argument S.
signal(S)
{
S++;
}
What is a reusable resource? - a)that can be used by one process at a time and is not depleted by that use
- b)none of the mentioned
- c)that can be used by more than one process at a time
- d)that can be shared between various threads
Correct answer is option 'B'. Can you explain this answer?
What is a reusable resource?
a)
that can be used by one process at a time and is not depleted by that use
b)
none of the mentioned
c)
that can be used by more than one process at a time
d)
that can be shared between various threads
|
|
Sanya Agarwal answered |
Reusable resource: A resource, such as a CPU or tape transport, that is not rendered useless by being used. A magnetic disk or tape can be used often indefinitely and are to be regarded as reusable resources. Compare consumable resources.
Consider:
P = { P1, P2,P3}
R = {R1, R2, R3, R4}
One instance of resource type R1 two instances of resource type R2, one instances of resource type R3 and three instances of resource type R4. The situation is P1 is holding one instance of R2 and waiting for an instance R1, P2 is holding an instance of R1 one instance of R2 and waiting for an instance of R3 and P3 is holding an instance of R3.- a)Deadlock will not occur
- b)Deadlock will occur
- c)All the instances of R4 are held by the processes
- d)All the instances of R2 are not allotted
Correct answer is option 'A'. Can you explain this answer?
Consider:
P = { P1, P2,P3}
R = {R1, R2, R3, R4}
One instance of resource type R1 two instances of resource type R2, one instances of resource type R3 and three instances of resource type R4. The situation is P1 is holding one instance of R2 and waiting for an instance R1, P2 is holding an instance of R1 one instance of R2 and waiting for an instance of R3 and P3 is holding an instance of R3.
P = { P1, P2,P3}
R = {R1, R2, R3, R4}
One instance of resource type R1 two instances of resource type R2, one instances of resource type R3 and three instances of resource type R4. The situation is P1 is holding one instance of R2 and waiting for an instance R1, P2 is holding an instance of R1 one instance of R2 and waiting for an instance of R3 and P3 is holding an instance of R3.
a)
Deadlock will not occur
b)
Deadlock will occur
c)
All the instances of R4 are held by the processes
d)
All the instances of R2 are not allotted
|
|
Hiral Nair answered |
Dead lock can not occure since first P3 will complete then P2 then P1 will complete.
An operating system contains 3 user processes each requiring 2 units of resource R. The minimum number of units of R such that no deadlock will ever occur is- a)3
- b)4
- c)5
- d)6
Correct answer is option 'B'. Can you explain this answer?
An operating system contains 3 user processes each requiring 2 units of resource R. The minimum number of units of R such that no deadlock will ever occur is
a)
3
b)
4
c)
5
d)
6
|
Dishani Basu answered |
Dead lock occurs when each of the 3 user processes hold one resource and make simultaneous demand for another. If there are 4 resources one of the 3 user processes will get the fourth instance of the resource and release one or both of the resource it is currently holding after using.
Necessary conditions for deadlock are- a)Non-preemption and circular wait
- b)Mutual exclusion and partial allocation
- c)Both (a) and (b)
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
Necessary conditions for deadlock are
a)
Non-preemption and circular wait
b)
Mutual exclusion and partial allocation
c)
Both (a) and (b)
d)
None of the above
|
Vaishnavi Kaur answered |
For deadlock four necessary condition are to be met
(i) Mutual exclusion
(ii) Non-preemption
(iii) Circular wait
(iv) Partial allocation [Bounded waiting]
(i) Mutual exclusion
(ii) Non-preemption
(iii) Circular wait
(iv) Partial allocation [Bounded waiting]
A state is safe if the system can allocate resources to each process (up to its maximum) in some order and still avoid deadlock, which of the following is/are true
1. Deadlocked state is unsafe,
2. Unsafe state may lead to a deadlock situation.
3. Unsafe state must lead to a deadlock situation.
4. Deadlock state is a subset of unsafe state,- a)1, 2 and 3
- b)1 and 2 only
- c)1, 3 and 4
- d)1, 2 and 4
Correct answer is option 'D'. Can you explain this answer?
A state is safe if the system can allocate resources to each process (up to its maximum) in some order and still avoid deadlock, which of the following is/are true
1. Deadlocked state is unsafe,
2. Unsafe state may lead to a deadlock situation.
3. Unsafe state must lead to a deadlock situation.
4. Deadlock state is a subset of unsafe state,
1. Deadlocked state is unsafe,
2. Unsafe state may lead to a deadlock situation.
3. Unsafe state must lead to a deadlock situation.
4. Deadlock state is a subset of unsafe state,
a)
1, 2 and 3
b)
1 and 2 only
c)
1, 3 and 4
d)
1, 2 and 4
|
Rahul Chavan answered |
(i) Deadlock is unsafe state
(ii) The unsafe state may lead to deadlock.
(iii) Clearly deadlock state is subset of unsafe state.
Its not compulsory that unsafe state always result in deadlock
An operating system implements a policy that requires a process to release all resources before making a request for another resource.- a)Both starvation and deadlock can occur
- b)Starvation can occur but deadlock cannot occur
- c)Starvation cannot occur but deadlock can occur
- d)Neither starvation nor deadlock can occur
Correct answer is option 'D'. Can you explain this answer?
An operating system implements a policy that requires a process to release all resources before making a request for another resource.
a)
Both starvation and deadlock can occur
b)
Starvation can occur but deadlock cannot occur
c)
Starvation cannot occur but deadlock can occur
d)
Neither starvation nor deadlock can occur
|
|
Pritam Goyal answered |
The given operating system follows DEAD LOCK prevention policy which also ensures neither starvation nor deadlock can occur.
Which of the following is NOT true of deadlock prevention and deadlock avoidance schemes?- a)In deadlock prevention, the request for resources is always granted if the resulting state is. safe.
- b)In deadlock avoidance, the request for resources is always granted if the resulting state is safe.
- c)Deadlock avoidance is less restrictive than deadlock prevention.
- d)Deadlock avoidance requires knowledge of resource requirements a priori.
Correct answer is option 'A'. Can you explain this answer?
Which of the following is NOT true of deadlock prevention and deadlock avoidance schemes?
a)
In deadlock prevention, the request for resources is always granted if the resulting state is. safe.
b)
In deadlock avoidance, the request for resources is always granted if the resulting state is safe.
c)
Deadlock avoidance is less restrictive than deadlock prevention.
d)
Deadlock avoidance requires knowledge of resource requirements a priori.
|
|
Madhurima Bajaj answered |
Deadlock Prevention and Deadlock Avoidance
Deadlock prevention and deadlock avoidance are two different strategies used to handle the problem of deadlocks in a computer system.
Deadlock Prevention
In deadlock prevention, the system tries to prevent deadlocks from occurring by ensuring that the four necessary conditions for deadlock do not occur. These conditions are mutual exclusion, hold and wait, no preemption, and circular wait. If any of these conditions are violated, the request for resources is always granted if the resulting state is safe. Deadlock prevention is considered to be a more restrictive approach than deadlock avoidance.
Deadlock Avoidance
In deadlock avoidance, the system tries to avoid deadlocks by predicting and preventing the occurrence of a deadlock before it actually happens. This is done by using techniques such as resource allocation graphs or banker's algorithm. In this approach, the request for resources is always granted if the resulting state is safe. Deadlock avoidance is considered to be less restrictive than deadlock prevention.
The Answer
The statement that is not true of deadlock prevention and deadlock avoidance schemes is option A, which states that in deadlock prevention, the request for resources is always granted if the resulting state is safe. This statement is incorrect because in deadlock prevention, the request for resources is not always granted even if the resulting state is safe. Deadlock prevention may require the system to deny the request for resources in order to prevent the occurrence of a deadlock.
Deadlock prevention and deadlock avoidance are two different strategies used to handle the problem of deadlocks in a computer system.
Deadlock Prevention
In deadlock prevention, the system tries to prevent deadlocks from occurring by ensuring that the four necessary conditions for deadlock do not occur. These conditions are mutual exclusion, hold and wait, no preemption, and circular wait. If any of these conditions are violated, the request for resources is always granted if the resulting state is safe. Deadlock prevention is considered to be a more restrictive approach than deadlock avoidance.
Deadlock Avoidance
In deadlock avoidance, the system tries to avoid deadlocks by predicting and preventing the occurrence of a deadlock before it actually happens. This is done by using techniques such as resource allocation graphs or banker's algorithm. In this approach, the request for resources is always granted if the resulting state is safe. Deadlock avoidance is considered to be less restrictive than deadlock prevention.
The Answer
The statement that is not true of deadlock prevention and deadlock avoidance schemes is option A, which states that in deadlock prevention, the request for resources is always granted if the resulting state is safe. This statement is incorrect because in deadlock prevention, the request for resources is not always granted even if the resulting state is safe. Deadlock prevention may require the system to deny the request for resources in order to prevent the occurrence of a deadlock.
Implementation of ________ may waste CPU cycles.- a)Spinlock
- b)Sleep and Wake
- c)Both of the above
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Implementation of ________ may waste CPU cycles.
a)
Spinlock
b)
Sleep and Wake
c)
Both of the above
d)
None of the above
|
Aditi Pillai answered |
Introduction:
In computer systems, various mechanisms are used to synchronize processes and ensure proper execution. Two common mechanisms used for this purpose are spinlocks and sleep and wake operations. However, the implementation of spinlocks can lead to wastage of CPU cycles.
Spinlocks:
Spinlocks are synchronization mechanisms used to protect critical sections of code from being executed simultaneously by multiple processes or threads. When a process wants to enter a critical section, it first checks if the lock associated with that section is available. If the lock is not available, the process enters a tight loop known as "spinning," repeatedly checking the lock's status until it becomes available. Once the lock is acquired, the process can safely execute the critical section.
Waste of CPU cycles:
Implementing spinlocks can lead to wastage of CPU cycles under certain conditions. This occurs when a process continuously spins in a loop, repeatedly checking the lock's status even when it is not available. The wastage of CPU cycles can be problematic in scenarios where the lock is held for a prolonged period by another process.
Reasons for wastage:
1. Busy waiting: Spinlocks involve busy waiting, where a process continuously checks the lock's status in a loop. This constant checking consumes CPU cycles, even if the process is unable to acquire the lock. Thus, it results in wastage of CPU resources.
2. Delayed execution: If a process is spinning in a loop waiting for a lock, it prevents other processes from executing on the CPU during that time. This delay in execution affects the overall system performance and resource utilization.
Comparison with Sleep and Wake operations:
Sleep and wake operations provide an alternative mechanism for synchronization. When a process cannot acquire a lock, it can voluntarily yield the CPU and enter a sleep state, allowing other processes to execute. The process is then woken up when the lock becomes available, avoiding the wastage of CPU cycles.
Conclusion:
In conclusion, spinlocks can waste CPU cycles due to busy waiting and delayed execution. While they are useful in certain scenarios where lock acquisition times are expected to be short, in cases where locks are held for longer durations, sleep and wake operations or other synchronization mechanisms may be more efficient in terms of CPU utilization and overall system performance.
In computer systems, various mechanisms are used to synchronize processes and ensure proper execution. Two common mechanisms used for this purpose are spinlocks and sleep and wake operations. However, the implementation of spinlocks can lead to wastage of CPU cycles.
Spinlocks:
Spinlocks are synchronization mechanisms used to protect critical sections of code from being executed simultaneously by multiple processes or threads. When a process wants to enter a critical section, it first checks if the lock associated with that section is available. If the lock is not available, the process enters a tight loop known as "spinning," repeatedly checking the lock's status until it becomes available. Once the lock is acquired, the process can safely execute the critical section.
Waste of CPU cycles:
Implementing spinlocks can lead to wastage of CPU cycles under certain conditions. This occurs when a process continuously spins in a loop, repeatedly checking the lock's status even when it is not available. The wastage of CPU cycles can be problematic in scenarios where the lock is held for a prolonged period by another process.
Reasons for wastage:
1. Busy waiting: Spinlocks involve busy waiting, where a process continuously checks the lock's status in a loop. This constant checking consumes CPU cycles, even if the process is unable to acquire the lock. Thus, it results in wastage of CPU resources.
2. Delayed execution: If a process is spinning in a loop waiting for a lock, it prevents other processes from executing on the CPU during that time. This delay in execution affects the overall system performance and resource utilization.
Comparison with Sleep and Wake operations:
Sleep and wake operations provide an alternative mechanism for synchronization. When a process cannot acquire a lock, it can voluntarily yield the CPU and enter a sleep state, allowing other processes to execute. The process is then woken up when the lock becomes available, avoiding the wastage of CPU cycles.
Conclusion:
In conclusion, spinlocks can waste CPU cycles due to busy waiting and delayed execution. While they are useful in certain scenarios where lock acquisition times are expected to be short, in cases where locks are held for longer durations, sleep and wake operations or other synchronization mechanisms may be more efficient in terms of CPU utilization and overall system performance.
With a single resource, deadlock occurs- a)If there are more than two processes competing for that resource.
- b)If there are only two processes competing for that resource.
- c)If there is a single process competing for that resource.
- d)None of these.
Correct answer is option 'D'. Can you explain this answer?
With a single resource, deadlock occurs
a)
If there are more than two processes competing for that resource.
b)
If there are only two processes competing for that resource.
c)
If there is a single process competing for that resource.
d)
None of these.
|
Gargi Menon answered |
Deadlock will never occur for a single resource.
A system has 3 processes sharing 4 resources. If each process needs a maximum of 2 units then, deadlock- a)Can never occur
- b)May occur
- c)Has to occur
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
A system has 3 processes sharing 4 resources. If each process needs a maximum of 2 units then, deadlock
a)
Can never occur
b)
May occur
c)
Has to occur
d)
None of the above
|
Abhijeet Unni answered |
At least one process will be holding 2 resources in case of a simultaneous demand from, all the processes. That process will release the 2 resources, thereby avoiding any possible deadlock.
'm' processes share 'n' resources of the same type. The maximum need of each process does not exceed 'n' and the sum all their maximum needs is always less than m + n. In this set up deadlock- a)Has to occur
- b)May occur
- c)Can never occur
- d)None of the these
Correct answer is option 'C'. Can you explain this answer?
'm' processes share 'n' resources of the same type. The maximum need of each process does not exceed 'n' and the sum all their maximum needs is always less than m + n. In this set up deadlock
a)
Has to occur
b)
May occur
c)
Can never occur
d)
None of the these
|
|
Rajesh Malik answered |
Assume m = 5 and n = 10
That means, 5 processes are sharing 10 resources and worst case will be if everyone is demanding equal number of cases.
So, for deadlock to be there every process must be holding 2 resources and seeking 1 more resource. This will make total demand to 15, which is nothing but 10 + 5 (m + n).
However, as given maximum demand is always less than (m + n), so we can say there will never be a deadlock.
That means, 5 processes are sharing 10 resources and worst case will be if everyone is demanding equal number of cases.
So, for deadlock to be there every process must be holding 2 resources and seeking 1 more resource. This will make total demand to 15, which is nothing but 10 + 5 (m + n).
However, as given maximum demand is always less than (m + n), so we can say there will never be a deadlock.
A computer system has 6 tape drives, with 'n' processes competing for them. Each process may need 3 tape drives. The maximum value of ‘n’ for which the system is guaranteed to be deadlock free is- a)2
- b)3
- c)4
- d)1
Correct answer is option 'A'. Can you explain this answer?
A computer system has 6 tape drives, with 'n' processes competing for them. Each process may need 3 tape drives. The maximum value of ‘n’ for which the system is guaranteed to be deadlock free is
a)
2
b)
3
c)
4
d)
1
|
Harshad Mukherjee answered |
2 process can never lead to deadlock as the peak time demand of 6 (3 + 3) tape drives can be satisfied. But 3 processes can lead to a deadlock if each hold 2 drives and then demand one more.
The time complexity of Banker’s algorithm to avoid deadlock having n processes and m resources is- a)O( m x n )
- b)O(m + n)
- c)O( m2P x n )
- d)O( n2 x m )
Correct answer is option 'D'. Can you explain this answer?
The time complexity of Banker’s algorithm to avoid deadlock having n processes and m resources is
a)
O( m x n )
b)
O(m + n)
c)
O( m2P x n )
d)
O( n2 x m )
|
Krithika Kaur answered |
The time complexity of the banker’s algorithm as a function of number n of processes and m of resources is O(n2xm).
Consider a system with three resources, tape drive, printer and disk drive. Each one of them are assigned a unique number:
R1 = F(Printer) = 12
R2 = F (Tape Drive) = 1
R3 = F(Disk Drive) = 4
Which of the following access patterns necessarily avoids deadlock?- a)R1, R3, R2
- b)R2, R3, R1
- c)R1, R2, R3
- d)R3, R2, R1
Correct answer is option 'B'. Can you explain this answer?
Consider a system with three resources, tape drive, printer and disk drive. Each one of them are assigned a unique number:
R1 = F(Printer) = 12
R2 = F (Tape Drive) = 1
R3 = F(Disk Drive) = 4
Which of the following access patterns necessarily avoids deadlock?
R1 = F(Printer) = 12
R2 = F (Tape Drive) = 1
R3 = F(Disk Drive) = 4
Which of the following access patterns necessarily avoids deadlock?
a)
R1, R3, R2
b)
R2, R3, R1
c)
R1, R2, R3
d)
R3, R2, R1
|
Ayush Mehta answered |
To avoid deadlock, we must request the devices in ascending order of enumeration. This ensures that circular wait does not hold. The correct sequence is 1, 4, 12, i.e. R2, R3, R1.
Dijkstra’s banking algorithm in an operating system performs- a)Deadlock avoidance
- b)Deadlock recovery
- c)Mutual exclusion
- d)Context switching
Correct answer is option 'A'. Can you explain this answer?
Dijkstra’s banking algorithm in an operating system performs
a)
Deadlock avoidance
b)
Deadlock recovery
c)
Mutual exclusion
d)
Context switching
|
|
Sreemoyee Singh answered |
Banker’s algorithm resources will only be allocated if system will remain in SAFE state (DEADLOCK can’t occur if system is in SAFE state), so performing DEADLOCK AVOIDANCE.
Consider a counting Semaphore variable 'S'. Following are the semaphore operations performed: 18P, 3V, 7V, 2P, 6V. What can be the largest initial value of S to keep one process in suspended list?- a)3
- b)4
- c)5
- d)None of the these
Correct answer is option 'A'. Can you explain this answer?
Consider a counting Semaphore variable 'S'. Following are the semaphore operations performed: 18P, 3V, 7V, 2P, 6V. What can be the largest initial value of S to keep one process in suspended list?
a)
3
b)
4
c)
5
d)
None of the these
|
|
Mahesh Pillai answered |
To determine the largest initial value of S that keeps one process in the suspended list, let's analyze the given semaphore operations step by step.
Given semaphore operations:
- 18P: This operation tries to decrement the semaphore value by 18. If the semaphore value is already less than 18, the process will be suspended. Otherwise, it will continue. Let's assume the initial value of S is x, then after this operation, the new value of S will be (x - 18).
- 3V: This operation increments the semaphore value by 3. After this operation, the new value of S will be (x - 18 + 3) = (x - 15).
- 7V: This operation increments the semaphore value by 7. After this operation, the new value of S will be (x - 15 + 7) = (x - 8).
- 2P: This operation tries to decrement the semaphore value by 2. If the semaphore value is already less than 2, the process will be suspended. Otherwise, it will continue. After this operation, the new value of S will be (x - 8 - 2) = (x - 10).
- 6V: This operation increments the semaphore value by 6. After this operation, the new value of S will be (x - 10 + 6) = (x - 4).
Now, we need to find the largest initial value of S such that it keeps one process in the suspended list. This means that after all the operations, the semaphore value should be less than 0 to suspend a process.
So, we need to find the value of x that satisfies the following inequality: (x - 4) <>
Solving this inequality, we get x <>
Therefore, the largest initial value of S that keeps one process in the suspended list is 3.
Hence, the correct answer is option A.
Given semaphore operations:
- 18P: This operation tries to decrement the semaphore value by 18. If the semaphore value is already less than 18, the process will be suspended. Otherwise, it will continue. Let's assume the initial value of S is x, then after this operation, the new value of S will be (x - 18).
- 3V: This operation increments the semaphore value by 3. After this operation, the new value of S will be (x - 18 + 3) = (x - 15).
- 7V: This operation increments the semaphore value by 7. After this operation, the new value of S will be (x - 15 + 7) = (x - 8).
- 2P: This operation tries to decrement the semaphore value by 2. If the semaphore value is already less than 2, the process will be suspended. Otherwise, it will continue. After this operation, the new value of S will be (x - 8 - 2) = (x - 10).
- 6V: This operation increments the semaphore value by 6. After this operation, the new value of S will be (x - 10 + 6) = (x - 4).
Now, we need to find the largest initial value of S such that it keeps one process in the suspended list. This means that after all the operations, the semaphore value should be less than 0 to suspend a process.
So, we need to find the value of x that satisfies the following inequality: (x - 4) <>
Solving this inequality, we get x <>
Therefore, the largest initial value of S that keeps one process in the suspended list is 3.
Hence, the correct answer is option A.
Consider three concurrent processes P1, P2 and P3 as shown below, which access a shared variable D that has been initialized to 100.
The processes are executed on a uniprocessor system running a time-shared operating system. If the minimum and maximum possible values of D after the three processes have completed execution are X and Y respectively, then the value of Y - X is __________.
Correct answer is '80'. Can you explain this answer?
Consider three concurrent processes P1, P2 and P3 as shown below, which access a shared variable D that has been initialized to 100.
The processes are executed on a uniprocessor system running a time-shared operating system. If the minimum and maximum possible values of D after the three processes have completed execution are X and Y respectively, then the value of Y - X is __________.
|
|
Sudhir Patel answered |
Maximum value (Y):
- Process P1 reads the initial value 100 during execution P1 will update the value of D to (100 + 20 = 120) and writes it to memory.
- P2 and P3 read the value 120 from the memory.
- P2 will get executed before P1 and write the value (120 - 50 = 70) to memory. Since P3 is holding D = 120 after its execution D will be equal to (120 + 10 = 130) ∴ Y = 130
Minimum value (X):
- P1, P2 and P3 read the value 100 from the memory
- P1 will execute first, after its execution it will update (D = 100 + 20 = 120)
- After P1, P3 will execute and it will update the value to (D = 100 + 10 = 110)
- Finally, P2 will be execute and D will be equal to 100 - 50 = 50 ∴ X = 50
- Y - X = 130 - 50 = 80
Mutex can also be referred to as ___________- a)Counting Semaphore
- b)Binary Semaphore
- c)Monitor
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
Mutex can also be referred to as ___________
a)
Counting Semaphore
b)
Binary Semaphore
c)
Monitor
d)
None of the above
|
|
Sudhir Patel answered |
Semaphores: It is a non-negative integer value used by various processes mutually exclusive to achieve synchronization. A semaphore S can be accessed by only two operations which are atomic and mutually exclusive in nature - Wait (P) and Signal (V). There are two types of semaphores:
- Counting Semaphore: In this, the value of the semaphore is from an unrestricted integer domain, that is, it can have an integer value as high as we want.
- Binary Semaphore: In this, the value can range only between 0 and 1.
Monitor: It is a high-level construct implemented using a programming language that packages the data and procedures to modify the data. Only one processor is allowed in the monitor at a time.
Mutex: It is a locking mechanism used to synchronize access to a resource. Only one task can acquire the mutex, hence, ownership is associated with mutex and only the owner can release the lock. This whole system can be implemented using a binary semaphore hence, it is referred to as a binary semaphore.
Mutex: It is a locking mechanism used to synchronize access to a resource. Only one task can acquire the mutex, hence, ownership is associated with mutex and only the owner can release the lock. This whole system can be implemented using a binary semaphore hence, it is referred to as a binary semaphore.
Match List-I with List-II select the correct answer using the codes given below the lists:
List-I
A. Inconsistency
B. Deadlock resolution
C. Deadlock avoidance
D. Transit
List-II
1. Requires the recognition of unsafe states.
2. A message received by Pj from Pi that is not shown as sent by Pi
3. A message sent by Pi to Pj (mij) that does not belong to the set of messages received by Pj
4. Always requires the abortion of one of more executing processes.
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'B'. Can you explain this answer?
Match List-I with List-II select the correct answer using the codes given below the lists:
List-I
A. Inconsistency
B. Deadlock resolution
C. Deadlock avoidance
D. Transit
List-II
1. Requires the recognition of unsafe states.
2. A message received by Pj from Pi that is not shown as sent by Pi
3. A message sent by Pi to Pj (mij) that does not belong to the set of messages received by Pj
4. Always requires the abortion of one of more executing processes.
List-I
A. Inconsistency
B. Deadlock resolution
C. Deadlock avoidance
D. Transit
List-II
1. Requires the recognition of unsafe states.
2. A message received by Pj from Pi that is not shown as sent by Pi
3. A message sent by Pi to Pj (mij) that does not belong to the set of messages received by Pj
4. Always requires the abortion of one of more executing processes.
a)
a
b)
b
c)
c
d)
d
|
|
Nishanth Mehta answered |
Inconsistency: message received by Pj from Pi that is not shown as sent by Pi. Deadlock resolution: always require the abortion of one or more processes. Deadlock avoidance: requires the recognition of unsafe state.
Transit: message sent by Pi to Pj that does not belong to the set of messages received by Pj.
Transit: message sent by Pi to Pj that does not belong to the set of messages received by Pj.
Consider a sytem with five processes P0 through P4 and three resource types A, B, C. Resource type A has 10 instances, resource type B has 5 instances, and resource type C has 7 instances. Suppose that, at time to, the following snapshot of the system has been taken:
Which of the following is the safe sequence for the above system?- a)<P0, P2, P3, P1, P4>
- b)<P1, P2, P4, P0, P3>
- c)<P1, P3, P4, P2, P0>
- d)We can not find a safe sequence
Correct answer is option 'C'. Can you explain this answer?
Consider a sytem with five processes P0 through P4 and three resource types A, B, C. Resource type A has 10 instances, resource type B has 5 instances, and resource type C has 7 instances. Suppose that, at time to, the following snapshot of the system has been taken:
Which of the following is the safe sequence for the above system?
Which of the following is the safe sequence for the above system?
a)
<P0, P2, P3, P1, P4>
b)
<P1, P2, P4, P0, P3>
c)
<P1, P3, P4, P2, P0>
d)
We can not find a safe sequence
|
Gargi Sarkar answered |
<P1, P3, P4, P2, P0> is the safe sequence.
Consider the resource allocation graph in figure
This system is in a deadlock state. This remark is- a)True
- b)False
- c)Impossible to determine
- d)Unpredictable
Correct answer is option 'B'. Can you explain this answer?
Consider the resource allocation graph in figure
This system is in a deadlock state. This remark is
This system is in a deadlock state. This remark is
a)
True
b)
False
c)
Impossible to determine
d)
Unpredictable
|
|
Sagar Saha answered |
The system is not in deadlock state, if the following sequence of fulfillment of request of the processes is made.
P2→ Po→ P1→ P3
P2→ Po→ P1→ P3
Referring to below system state diagram how many resources are available?
1. One instance of resource R1.
2. Two instance of resource R2.- a)Only 1
- b)Only 2
- c)Both 1 and 2
- d)Neither 1 nor 2
Correct answer is option 'B'. Can you explain this answer?
Referring to below system state diagram how many resources are available?
1. One instance of resource R1.
2. Two instance of resource R2.
1. One instance of resource R1.
2. Two instance of resource R2.
a)
Only 1
b)
Only 2
c)
Both 1 and 2
d)
Neither 1 nor 2
|
|
Nilesh Mukherjee answered |
From the given RAG we can observe that there are 2 instance of R2 and a single instance process P1 holding resource R1, and requesting for resource R2. Process P2 is requesting for resource R1 and Resource R2.
Hence at the given time both the instances of resource R2 are available.
Hence at the given time both the instances of resource R2 are available.
Consider the following system state:
Total resources are 11. The system will be in a safe state if- a)Process P1 is allocated one additional resource.
- b)Process P2 is allocated two additional resources.
- c)Process P3 is allocated three additional resources.
- d)Process P2 is allocated one additional resource.
Correct answer is option 'C'. Can you explain this answer?
Consider the following system state:
Total resources are 11. The system will be in a safe state if
Total resources are 11. The system will be in a safe state if
a)
Process P1 is allocated one additional resource.
b)
Process P2 is allocated two additional resources.
c)
Process P3 is allocated three additional resources.
d)
Process P2 is allocated one additional resource.
|
|
Crack Gate answered |
If three additional resources are allocated to P3, its need becomes zero. This action allows the system to:
- Receive seven resources back from P3.
- Use these resources to allocate to either P1 or P2.
This ensures the system remains in a safe state.
Which of the following is not an optimization criterion in the design of a CPU scheduling algorithm?- a)Minimum CPU utilization
- b)Maximum throughput
- c)Minimum turnaround time
- d)Minimum waiting time
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not an optimization criterion in the design of a CPU scheduling algorithm?
a)
Minimum CPU utilization
b)
Maximum throughput
c)
Minimum turnaround time
d)
Minimum waiting time
|
|
Sudhir Patel answered |
Maximum CPU utilization, maximum throughput, minimum turnaround time, and Minimum waiting time are the optimization criterion in the design of a CPU scheduling algorithm
Which of the following is true?
- a)Deadlock occurs in both the cases
- b)Deadlock occurs in (i) but not in (ii)
- c)Deadlock occurs in (ii) but not in (i)
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
Which of the following is true?
a)
Deadlock occurs in both the cases
b)
Deadlock occurs in (i) but not in (ii)
c)
Deadlock occurs in (ii) but not in (i)
d)
None of the above
|
|
Anand Chatterjee answered |
In (a) the processes P1, P2 and P3 are in deadlock state. P2 waiting on P3. Which is held by P3. P1 is waiting on ft, which is held by P2. While P3 in waiting on R2 which is held by P1 and P2.
In (b) we can observe that P4 can release instance o f R2, which can be allocated to either P1 or P3, breaking the cycle. Deadlock does not occur in (b).
In (b) we can observe that P4 can release instance o f R2, which can be allocated to either P1 or P3, breaking the cycle. Deadlock does not occur in (b).
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