BN104 Operating Systems

Published : 15-Sep,2021 | Views : 10

Question:

Explain their functions. Students must answer Multiple Answer questions about OS operation, and report on selected OS aspects and algorithms such as virtual memory, process scheduling and mobile operating systems. After successful completion of this assignment, students should be able to:

Report on the basics, and provide examples, of operating systems structure and functionality, including memory allocation virtual memory, demand paging and process and device management.
Describe the integration of hardware, operating systems and application software.

Answer:

The comparison between the internal and external fragmentation:

EXTERNAL FRAGMENTATION	INTERNAL FRAGMENTATION
The occurrence of the external fragmentation takes place due to the allocation of the processes in the variable sized memory.	The internal fragmentations occurs as a result of the allocation of the processes in the fixed sized memory.
The external fragmentations occur due to the allocation of the process in the variable memory and processes are removed from the memory.	The internal fragmentation occur due to the allocation of the process in the fixed memory and some spaces are left in the memory.
For resolving the issue of external fragmentation compaction, paging and segmentation techniques are used.	For the internal fragmentations the best fit techniques are used.

The relocation techniques are started before the start of the running phase of the process. The lower addresses starts from zero. In addition to this, the files that are executable are generated by the assemblers and the compilers. The address are adjusted according to the runtime of the processes.

TIME

100

105

120

The order of the pages that are to be replaced are: 3,16,8,1,18,4,15,12.

The Least recently Used or the LRU techniques makes use of the status bits. The Status bit provides information about the entries that are new and also about the validation of the entries. The valid entries are also denoted by the status bit. The least recently used entries are removed from the system.

The order in which the pages would be replaced are: 18,12,8,3,0,16,14,6.

FCFS
SJN
SRT
Round Robin

Total time taken by a job is known as the turnaround time. The total time that the job has to wait before being executed is known as the total waiting time.

i) FCFS

A= 11-0=11

B=24-2=22

C=29-5=24

D=30-7=23

E=33-13=20

TOTAL TURNAROUND TIME= (11+22+24+23+20) = 100

AVERAGE TURNAROUND TIME=(100/4)=25

ii) SJN

A= 20-0=20

B=33-2=31

C=9-5=4

D=1=1

E=4=4

TOTAL TURNAROUND TIME= (20+31+4+1+4)= 60

AVERAGE TURNAROUND TIME=(60/4)=15

iii) SRT

A= 25-15+2-0=12

B=5-2=3

C=11-6-5=0

D=8-7=1

E=33-13=20

TOTAL TURNAROUND TIME= (12+3+0+20)= 35

AVERAGE TURNAROUND TIME=35/4)=8.75

iv) Round Robin

A= 29-0=29

B=33-2=31

C=21-5=24

D=10-7=3

E=13-13=0

TOTAL TURNAROUND TIME= (29+31+24+3+0)= 87

AVERAGE TURNAROUND TIME=(87/4)=21.75

Bibliography

[1] Moraru, I., Andersen, D.G., Kaminsky, M., Tolia, N., Ranganathan, P. and Binkert, N, November. Consistent, durable, and safe memory management for byte-addressable non volatile main memory. In Proceedings of the First ACM SIGOPS Conference on Timely Results in Operating Systems (p. 1). ACM., 2013.

[2] Sirohi, A., Pratap, A. and Aggarwal, M. Improvised Round Robin (CPU) Scheduling Algorithm. International Journal of Computer Applications, 99(18), pp.40-43., 2014.