Operating System MCQ

Ans. (c)

Explanation – The average response time is the average of the difference between the arrival time and the time when the job gets the CPU for the first time.

Average response time = ∑ (Arrival time of each job – The time when they get the CPU for the first time) / Number of jobs

The shortest job first algorithm is the way to minimize the average response time:

Suppose all the possible values of x

If 0 < x <= 10, then the order of the jobs will be D, C, E, B, A

If 10 < x <= 40, then the order of the jobs will be C, D, E, B, A

If 40 < x <= 60, then the order will be C, E, D, B, A

If 60 < x <= 80, then the order will be C, E, B, D, A

If x > 80, then the order will be C, E, B, A, D

Hence, the answer is C, D, E, B, A

Ans. (d)

Explanation: Given, Time quantum (TQ) = 2 units,

Time in context switching = 1 unit

The Gant chart will be like this:

Here, ࢚ represents the context switch.

Completion Time of P5 = 23 units

Turnaround time of P5 = CT – AT = 23 – 5 = 18 units

Hence, the correct answer is 23 units and 18 units.

Ans. (a)

Types of CPU Scheduling algorithms

SRTF (Shortest remaining time first) – It minimizes the average waiting time and average turnaround time.

Round Robin – Each process is executed after another for a fixed interval of time called time quantum.

Priority Scheduler – The process of higher priority gets executed first.

First come first serve – As the name suggests, the process that arrives first gets executed first.

Hence, the correct answer is 1-C, 2-B, 3-D, 4-A.

Ans. (c)

Explanation: The shortest remaining time first algorithm is used, the Gant chart will look like this

From the above Gant chart, we can find the completion time (CT), turnaround time (TAT), and waiting time (WT).

Average waiting time = (9+0+14+2)/4 = 6.25 units

Ans. (a)

Explanation: The CPU utilization is equal to 1-p^N, where N is the number of processes and p is the waiting fraction.

Here, N = 10 and p = 0.80

CPU utilization = 1 – (0.80)^ 20 = 1 – 0.01 = 0.99

Hence, the correct answer is 99 percent.

And. (c)

Explanation: Shortest job first has the least average waiting time as it is the optimal scheduling algorithm.

Ans. (d)

Explanation:a) SRTF may cause starvation for the processes with high burst time

b) Preemptive may also cause starvation for the processes with low priority

c) Longest remaining time first does not cause starvation

d) Round-robin is better than the FCFS in terms of response time.

Ans. (b)

Explanation: Consider the three processes, P1, P2, and P3, and their BT is 32, 48, and 64 units, respectively.

Time spent by P1, P2, and P3 on input-output is 8, 12, and 16 units, respectively.

Hence, the correct answer is 8 units.

Ans. (d)

Explanation:Starvation is caused by SJF, SRTF, LJF, Priority, and Multilevel feedback queue.

Hence, the correct answer is Priority and Shortest Job First (SJF).

Ans. (d)

Explanation: Shortest remaining time first is used, the Gant chart will look like this:

Form the above Gant chart, we can find the CT, TAT, and WT.

Average Turnaround time = (19 + 12 + 4 + 1 + 5 + 2)/6 = 7.1

Average Waiting Time = (12 + 7 + 1 + 0 + 3 + 1)/6 = 4

Hence, the correct answer is 4 units and 7.1 units.

Ans. (d)

Explanation: Round Robin with a time quantum of 50 units is used, The Gant chart will look like this:

We can find the completion time and waiting time from the above Gant chart.

CT of P1 = 510,        Waiting time of P1 = CT – BT = 510 – 250 = 260

CT of P2 = 360,        Waiting time of P2 = CT – BT = 360–110 = 250

CT of P3 = 410,        Waiting time of P3 = CT – BT = 410 – 150 = 260

Average Waiting Time = (260+250+260)/3 = 256.66667 = 257

Ans. (b)

Explanation: a) In Round-Robin, Preemption occurs when the time quantum expires.

b) In the First come first server, preemption does not occur; context switching happens only when a process completes its execution.

c) In the Multilevel Queue Scheduling, preemption occurs when a process with higher priority arrives.

d) In the Multilevel Queue Scheduling with feedback, preemption occurs when a process with higher priority arrives or when the quantum of the high priority queue expires

Ans. (b)

Explanation: Response ratio = Burst time / Turnaround time

The Gant chart for the following will look like this:

We can find the completion and turnaround time for process C from the above Gant Chart.

CT = 24

TAT = CT – AT = 24 – 0 = 24

Response ratio = 8/24 = 0.25

Hence, the correct answer is 0.25.

Ans. (b)

Explanation: Given, priority scheduling is used. The Gant Chart for the same will look like this.

We can find the completion and turnaround time from the above Gant chart:

Average TAT = (10+16+3+14)/4 = 10.75

Average WT = (3+14+0+10)/4 = 6.75

Hence, the correct answer is 6.75 and 10.75.

Ans. (b)

Explanation: Given that, Round robin is used with a time quantum of 2 units. The Gant chart for the following will look like this.

The completion time of process C is 18.

Ans. (b)

Explanation: 1. Starvation could not occur on an OS with no-preemptive scheduling. (False)

2. Starvation could occur on an OS with preemptive scheduling (True)

3. The deadlock could not occur on an OS with non-preemptive scheduling (False)

Ans. (a)

Explanation: Given that, the system has four processes and three resources. Each process requires a maximum of 2 resources. Assume that each process holds one resource and request for the other one to start processing. Hence, the circular wait condition is satisfied, and the system turns to deadlock. Let P1, P2, P3, and P4 be the four processes.

Ans. (c)

Explanation: By dissatisfying hold and wait, the process must get all the resources before execution and must release all the resources after completion. This causes starvation for other processes.

Ans. (a)

Explanation: Given that initially R1 and R2 are free. Now, consider process A run the statement 1 and context switch occur process B run the statement S1 and S2 and set the R1 = Busy and again context switch occurs. Now, Process A run the statement S2 and set the R1 = Busy which was already set by the process B. Following the same steps for the statements S3 and S4 for both the processes. After this, both the process tries to run S5 i.e., use R1 and R2 at the same time. Hence, Mutual Exclusion is not guaranteed.

Hence, the correct answer is option a.

Ans. (d)

Explanation: Maximum requirements of each process are 4, 5, 3, and 7. Allocate the resources one less than the maximum requirement for each process.

Resources allocated to process

P1 = 3

P2 = 4

P3 = 2

P4 = 6 Allocated resource = 15, this may lead to deadlock. To prevent deadlock, we need to full fill the requirement of at least one process. For that, we need one more resource. Hence the value of n will be 16.

Ans. (b)

Explanation: Given that there are 15 processes and each requires a maximum of 4 resources.

For the system to be in deadlock, we need to allocate resources one less than the maximum requirements.

P = no. of processes * (maximum requirement – 1)

P = 15 * 3 = 45

For the system to be deadlock-free, we need to full fill the requirement of at least one process.

Q = P + 1 = 46

Hence, b is the correct answer.

Ans. (b, c, and d)

Explanation: From the given data, we can find the current need and instances available

Total instances = 16

Allocated = 12

Available = 16 – 12 = 4

At this stage, we can only full fill the requirement of process D. Allocating the available resources to process D after completing the execution it will release all the instances and we will have 6 instances. Here, we allocate it to process E and it will release all the instances after execution and we will have 8 instances. Here, we can allocate it to processes C and A both. On allocating the resources to process A, there will one safe sequence which is D, E, A, C, and B. On the other hand, when we allocate the resource to the process C, we will have two safe sequences which are D, E, C, A, and B and D, E, C, B, and A.

Hence, option b, c and d are correct.

Ans. (c)

Explanation: We can clearly see this in the resource allocation graph given below. Processes P2, P2, and P6 form a circular graph. Here, the condition of circular wait is satisfied. Hence, c is the correct answer.

Ans. (d)

Explanation: Taking the value of x = 50 and y = 30. Now, calculating the current need of each process and instances available.

Instances allocated = 60+40+50+30 = 180

Instances available = 200-180 = 20

Here, We can full fill the demand of any process for x = 50 and y = 30. Hence, d is the correct answer.

Ans. (b)

Explanation: The Banker’s algorithm is used for deadlock avoidance

Ans. (d)

Explanation: The minimum value of N = (7-1) + (5-1) + (9-1) + (12-1) + 1 = 30. Hence, the correct answer is option d.

Ans. (a) and (c)

Explanation: We can find the current need of each process from the above data by subtracting the allocated instances from maximum need.

Here, We can full fill the demand of both P2 and P4 processes. So, there will be two orders P2, P4 and P4, P2. We can allocate either to P2 first or P4 first.

On allocating the resource to P2:

After complete execution of process P2 it will release all the instances.

Available Resources would be:

Here we can only full fill the demand of process P4

After complete execution of P4. It will also release all the instances. And

Available resources would be

Here, We can full fill the demand of process p1 and P3 both. So, it will depend on us to which process we are allocating the resources. And we will have two orders here to P1, P3 and P3, P1.

The possible safe sequences are (P2, P4, P1, P3), (P2, P4, P3, P1), (P4, P2, P1, P3), and (P4, P2, P3, P1). Hence, options a and c both are correct.

Ans. (a)

Explanation: Calculating the current need of each process:

Possible values of a, b, and c are (3, 0, 1), (2, 0, 2), (2, 0, 3) and (3, 2, 4)

Taking a = 3, b = 0, c = 1.

We can full fill the need of process P4.

After execution it will release all the resources.

Current Available: a = 3+1 = 4, b = 0+1 = 1, c = 1+1 = 2.

Now, we can full fill the demand of process P2.

After execution P2 will release all the resources.

Current available: a = 4+0 = 4, b = 1+1 = 2, c = 2+1 = 3

Here, we can full fill the demand of process P3.

After execution P3 will release all the resources.

Current available: a = 4+1 = 5, b = 2+0 = 2, c = 3+1 = 4

Now, we can full fill the demand of process P4.

Hence the minimum value of X = 3+0+1 = 4.

Ans. (a) and (b)

Explanation:Calculating the current need of each process

Option (a) will give only one possible safe sequence which is P2, P3, P4, P1.

Option (b) will give only two possible safe sequences which are P2, P3, P1, P4 and P2, P3, P4, P1.

Option (c) will give only one possible safe sequence which is P4, P2, P3, P1.

Hence the correct answer is (a) and (b).

Ans. (d)

Explanation: To prevent deadlock allocate resources to each process one less than their maximum demand. After this full fill the demand of at least one process.

Minimum no. of resources required = (k-1) + (k-1) + (k-1) + (k-1) + 1 = 4k – 3. Hence, d is the correct answer.

Ans.  (b)

Explanation: The wait for graph algorithm is only applicable when al the resources have only single instances and it can not work when the resources have multiple instances.

Ans. (d)

Explanation:To prevent deadlock, following are the necessary condition for deadlock:

• Mutual Exclusion
• Hold and Wait
• Circular wait
• No preemption

“No Preemption” not occurs to prevent deadlock.

Hence, d is the correct answer.

Ans. (a)

Explanation:

Both the process will start running concurrently

On line number 3, process p1 will wait on x1, and will change the value of x1 = 0. Also process p2 will wait on x2, and will change the value of x2 = 0.

Both the process will enter in their critical section

On line number 4, both the processes p1 and p2 will infinitely wait on x2 and x1 respectively. Hence, the deadlock will occur.

Ans. (b)

Explanation: As the a and b both are initialized to zero, both the processes can enter in the critical section at the same time hence, there is no mutual exclusion but it satisfies the progress.

Ans. (a), (b), and (c)

Explanation: The above question is based on the Peterson’s solution, where Mutual exclusion, bounded wait and progress is achieved but it involved busy waiting. Hence, option a, b and c are correct.

Ans. (d)

Explanation: The Peterson’s Solution of process synchronization satisfies mutual exclusion, progress and bounded wait but also involves busy wait.

Ans. (d)

Explanation:Decker’s Algorithm satisfies mutual exclusion and bounded wait but not progress.

Ans. (d)

Explanation: The Lock Variable method only satisfies the condition of progress and do not satisfy mutual exclusion and bounded wait.

Ans. (c)

Explanation: The test and set lock method satisfies mutual exclusion and progress and do not satisfy bounded wait.

Ans. (b)

Explanation: a) Peterson’s solution satisfies all mutual exclusion, progress and bounded wait.

b) Decker’s Algorithm satisfies mutual exclusion and bounded wait but not progress.

c) Lock variable satisfies only progress not mutual exclusion and bounded wait.

d) Test and Set Lock variable satisfies mutual exclusion and progress but not bounded wait. Hence, the correct option is (b).

Ans. (a)

Explanation: In binary semaphore, only one process can enter in the critical section and there is not time limit on how long a process can be run. This makes other processes starve.

Ans. (a)

Explanation:

P() refers to wait() which reduces the value of semaphore by 1 and v() refers to signal() which increase the value of semaphore by 1.

S = 50 – 23 + 11 = 38

Hence, (d) is the correct answer.

Ans. (d)

Explanation: All the above methods are implemented in user mode as they do not require hardware support.

Ans. (a)

Explanation: Critical section is the program segment where the shared resources are accessed.

Ans. (a)

Explanation: It satisfies the mutual exclusion as only one process can enter in the critical section at a time. But due to strict alternation, it does not satisfy the progress.

Ans. (a)

Explanation: As both the processes can enter in their critical section at the same time. Hence, it does not satisfy mutual exclusion.

Ans. (b)

Explanation: The given statement clearly denotes that the number of blocked processes is equal to n. Hence, (b) is the correct answer.

Ans. (b)

Explanation:On line one, process A waits on s1 and make s1 = 0 and preempted. Now, the process B waits on s2 and make s2 = 0. Here, both the semaphores are equal to zero. Hence, no process can enter in their critical section and deadlock occur.

Ans. (c)

Explanation:Here, both the processes can run infinite number of times and make other program starve.

Ans. (b)

Explanation: Mutual exclusion – No two processes should be present in the critical section at the same time, i.e., only one process is allowed to enter in the critical section at a time. Hence, (b) is the correct answer.