My Report

Operating System Practice Test 3


Correct Answer: 2 points | Wrong: -1 point
Grades: A* (100% score) | A (80%-99%) | B (60%-80%) | C (40%-60%) | D (0%-40%)
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 5%

Question 1 of 20

1. Physical memory is broken into fixed-sized blocks called ________

Question 1 of 20

Question 2 of 20

2. The process to be aborted is chosen on the basis of the following factors?

Question 2 of 20

Question 3 of 20

3. The disadvantage of moving all process to one end of memory and all holes to the other direction, producing one large hole of available memory is ____________

Question 3 of 20

Question 4 of 20

4. If execution time binding is being used, then a process ______ be swapped to a different memory space.

Question 4 of 20

Question 5 of 20

5. The base register is also known as the ____________

Question 5 of 20

Question 6 of 20

6. Swapping requires a _________

Question 6 of 20

Question 7 of 20

7. Each entry in a translation lookaside buffer (TLB) consists of ____________

Question 7 of 20

Question 8 of 20

8. In contiguous memory allocation ____________

Question 8 of 20

Question 9 of 20

9. Smaller page tables are implemented as a set of _______

Question 9 of 20

Question 10 of 20

10. The content of the matrix Need is ____________

Question 10 of 20

Question 11 of 20

11. Sometimes the overhead of keeping track of a hole might be ____________

Question 11 of 20

Question 12 of 20

12. Memory management technique in which system stores and retrieves data from secondary storage for use in main memory is called?

Question 12 of 20

Question 13 of 20

13. Operating System maintains the page table for ____________

Question 13 of 20

Question 14 of 20

14. A system with 5 processes P0 through P4 and three resource types A, B, C have A with 10 instances, B with 5 instances, and C with 7 instances. At time t0, the following snapshot has been taken:

Process
P0               
P1               
P2           
P3           
P4       
    
Allocation (process-wise : P0 through P4 top to bottom) 
A   B   C
0   1   0
2   0   0
3   0   2
2   1   1
0   0   2

Max (process-wise: P0 through P4 top to bottom)
A   B   C
7   5   3
3   2   2
9   0   2
2   2   2
4   3   3

Available
A   B   C
3   3   2

The sequence <P1, P3, P4, P2, P0> leads the system to ____________

Question 14 of 20

Question 15 of 20

15. The main memory accommodates ____________

Question 15 of 20

Question 16 of 20

16. In a system that does not support swapping ____________

Question 16 of 20

Question 17 of 20

17. 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?

Question 17 of 20

Question 18 of 20

18. The wait-for graph is a deadlock detection algorithm that is applicable when ____________

Question 18 of 20

Question 19 of 20

19. A deadlock eventually cripples system throughput and will cause the CPU utilization to ______

Question 19 of 20

Question 20 of 20

20. 'm' processes share 'n' resources of the same type. The maximum need of each process doesn't exceed 'n' and the sum of all their maximum needs is always less than m+n. In this setup, deadlock ____________

Question 20 of 20


 

Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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