# SOLVED CS604 Assignment 2 Solution and Discussion

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

02.12.2019

RULES FOR MARKING
It should be clear that your assignment would not get any credit if:
 The assignment is submitted after the due date.
 The submitted assignment file does not open or it is corrupt.
 The submitted assignment solution is copied from any other student or from the internet.

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Any other formats like scan images, PDF, zip, rar, ppt and bmp etc will not be accepted.

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!

• Solution:

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

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

• Solution:

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

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

• 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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