SOLVED CS604 Assignment 2 Solution and Discussion

[zareen]

Operating Systems (CS604)
Assignment # 02
FALL 2019
Total marks = 20

Deadline Date
02.12.2019

Please carefully read the following instructions before attempting assignment.

RULES FOR MARKING
It should be clear that your assignment would not get any credit if:
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 The submitted assignment file does not open or it is corrupt.
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Objective:
• The objective of this assignment is to provide hands on experience of:
Different Process Scheduling Algorithms in Operating System

NOTE

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If you find any mistake or confusion in assignment (Question statement), please consult with your instructor before the deadline. After the deadline no queries will be entertained in this regard.

For any query, feel free to email at:
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Question No 01 4+3+3=10 marks
Assume you have to apply Shortest Job First (SJF) scheduling algorithms on the set of different processes given in the table below. The CPU burst time is also given for each process. Consider that all the processes arrive in the ready queue within time 0 seconds except P2 that arrive in ready queue within time 8 seconds. You are required to show the Gantt Chart to illustrate the execution sequence of these processes and calculate the Total Waiting Time and Average Waiting Time for the given processes by using SJF algorithm.

Process	CPU Burst Time (seconds)
P0	2
P1	6
P2	1
P3	4
P4	3
P5	8

Question No 02 4+3+3=10 marks
Consider a scenario where you have to apply Round Robin scheduling algorithm on the below given set of processes with each having a quantum size=8 milliseconds. The CPU burst time and arrival time for each process is also provided in the given table. You are required to show the Gantt Chart to illustrate the execution sequence of these processes. Moreover, calculate the Average Turnaround Time and Average Waiting Time for given processes by using round robin algorithm.

Process	CPU Burst Time (Milliseconds)	Arrival Time(Milliseconds)
P0	15	0
P1	8	4
P2	25	18
P3	18	5

Wish you very Best of Luck!

[zareen]

Solution:
Q. 1 Answer

P0	P4	P3	P2	P1	P5
0	2	5	9	10	16	24

Total Waiting Time = 0+2+5+1+10+16 = 34 Seconds
Average Waiting Time = 34/6 = 5.66 Seconds

Q.2 Answer:
Solution 1:

P0	P1	P3	P0	P2	P3	P2	P3	P2	P2
0	8	16	24	31	39	47	55	57	65	66

Turnaround time= Exit Time- Arrival Time

Turnaround time for P0 = 31-0 = 31
P1 = 16-4 = 12
P2 = 66-18 = 48
P3 = 57- 5= 52

Total Turnaround Time = 39+12+48+52 = 143
Average Turnaround Time = 143 / 4 =35.75 millisecond

Waiting Time= Turnaround Time- Burst Time

Waiting time for P0 = 31-15 = 16
P1 = 12-8 = 4
P2 = 48-25 = 23
P3 = 52-18 = 34

Total Waiting Time = 16+4+23+34 =77
Average Waiting Time = 77 / 4 =19.25 millisecond

Solution 2:

P0	P1	P3	P2	P0	P3	P2	P3	P2	P2
0	8	16	24	31	39	47	55	57	65	66

Turnaround time= Exit Time- Arrival Time

Turnaround time for P0 = 39-0 = 39
P1 = 16-4 = 12
P2 = 66-18 = 48
P3 = 57- 5= 52

Total Turnaround Time = 39+12+48+52 = 151
Average Turnaround Time = 151 / 4 =37.75 millisecond

Waiting Time= Turnaround Time- Burst Time

Waiting time for P0 = 39-15 = 24
P1 = 12-8= 4
P2 = 48-25= 23
P3 = 52-18 = 34

Total Waiting Time = 24+0+5+29 = 85
Average Waiting Time = 85 / 4 =21.25 millisecond

[zareen]

Solution:
Q. 1 Answer

P0	P4	P3	P2	P1	P5
0	2	5	9	10	16	24

Total Waiting Time = 0+2+5+1+10+16 = 34 Seconds
Average Waiting Time = 34/6 = 5.66 Seconds

Q.2 Answer:
Solution 1:

P0	P1	P3	P0	P2	P3	P2	P3	P2	P2
0	8	16	24	31	39	47	55	57	65	66

Turnaround time= Exit Time- Arrival Time

Turnaround time for P0 = 31-0 = 31
P1 = 16-4 = 12
P2 = 66-18 = 48
P3 = 57- 5= 52

Total Turnaround Time = 39+12+48+52 = 143
Average Turnaround Time = 143 / 4 =35.75 millisecond

Waiting Time= Turnaround Time- Burst Time

Waiting time for P0 = 31-15 = 16
P1 = 12-8 = 4
P2 = 48-25 = 23
P3 = 52-18 = 34

Total Waiting Time = 16+4+23+34 =77
Average Waiting Time = 77 / 4 =19.25 millisecond

Solution 2:

P0	P1	P3	P2	P0	P3	P2	P3	P2	P2
0	8	16	24	31	39	47	55	57	65	66

Turnaround time= Exit Time- Arrival Time

Turnaround time for P0 = 39-0 = 39
P1 = 16-4 = 12
P2 = 66-18 = 48
P3 = 57- 5= 52

Total Turnaround Time = 39+12+48+52 = 151
Average Turnaround Time = 151 / 4 =37.75 millisecond

Waiting Time= Turnaround Time- Burst Time

Waiting time for P0 = 39-15 = 24
P1 = 12-8= 4
P2 = 48-25= 23
P3 = 52-18 = 34

Total Waiting Time = 24+0+5+29 = 85
Average Waiting Time = 85 / 4 =21.25 millisecond

[zareen]

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[zareen]

@zareen said in CS604 Assignment 2 Solution and Discussion:

Assume you have to apply Shortest Job First (SJF) scheduling algorithms on the set of different processes given in the table below. The CPU burst time is also given for each process. Consider that all the processes arrive in the ready queue within time 0 seconds except P2 that arrive in ready queue within time 8 seconds. You are required to show the Gantt Chart to illustrate the execution sequence of these processes and calculate the Total Waiting Time and Average Waiting Time for the given processes by using SJF algorithm.

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