Kingdom of Saudi Arabia
Ministry of Higher Education
Prince Sattam Bin Abdul-Aziz University
College of Computer
Engineering and sciences
‫المملكة العربية السعودية‬
‫وزارة التعليم العالي‬
‫جامـعـة األمير سطام بن عبدالعزيز‬
‫كلــيـــة هـنــدســـة وعلــــوم الحـاســب‬
2440 & 2544 :‫الشعبة‬
Course Title: CS3701-Operating System
CSC3701
Instructor Name: Mr. Abdul Haseeb Khan
Tutorial Lecture # 06
CH: CPU Scheduling part-II
Week# 06
Date: - - -2015
CPU Scheduling-Problem Solving Part: II
Q
Consider the following set of processes, with the length of CPU burst time
given in milliseconds:
Process
P1
P2
P3
P4
i.
Arrival Time
0
2
4
6
Burst Time
9
5
2
5
Draw the Gant charts illustrating the execution of these processes using non- preemptive
SJF and preemptive SJF?
Non- preemptive SJF:
P1
P3
0
9
P2
11
P4
16
21
Preemptive SJF:
P1
0
ii.
iii.
P2
2
P3
4
P2
6
P4
9
P1
14
21
What is the turnaround time of each process of the scheduling algorithms in part a?
Non- preemptive SJF: P1=
9
, P2= 14
, P3= 7
, P4= 15
Preemptive SJF:
21
, P2= 7
, P3= 2
, P4= 8
P1=
What is the waiting time of each process for each of the scheduling algorithms in part
a?
Non- preemptive SJF: P1=
0
, P2= 9
Preemptive SJF: P1= 12
, P2= 2
, P3= 5
, P3= 0
, P4= 10
, P4= 3
Round Robin
Q: Given in the following table are four processes with order P1, P2, P3, and P4
with their arrival time and CPU burst times shown. Draw and Find the following
CPU scheduling strategy is Round Robin with Quantum time: 2
i.
ii.
iii.
Process
Arrival Time
Burst Time
P1
1
9
P2
2
2
P3
3
1
P4
4
5
Draw the Gantt chart Round Robin
Waiting time for each process and Average waiting time
Turnaround time for each process and average turnaround time
Round Robin Quantum time: 2
i.
Gantt Chart:
ii.
Turn Around Time P1: 17
Average T.A.T: 8.5
iii.
Waiting time
P1:
Average W.T: 4.25
P2:
8
P2: 1
3
P3:
3
P4:
11
P3:
2
P4:
6
Q:
Consider the following set of processes, with the length of CPU burst time
given in milliseconds: for round robin algorithm (quantum = 5)
Process
Burst Time
P1
3
P2
7
P3
5
P4
2
P5
6
The processes are assumed to have arrived in the order P1, P2, P3, P4, P5,
all at time 0.
i.
Draw the Gantt charts illustrating the execution of these processes using, RR
scheduling.
P1
P2
P3
P4
P5
P2
P5
0
3
8
13
15
20
22 23
ii.
What is the turnaround time of each process?
P1=3,
P2= 22,
P3=13,
P4=15,
P5=23
iii.
What is the waiting time of each process?
P1=3-3=0,
P2=22-7= 15, P3=13-5=8, P4=15-2=13, P5=23-6=17
iv.
What is the average waiting time (over all processes)?
(0+15+8+13+17)/5=10.6
Q: Consider the following set of processes, with the length of the CPU-burst time
given in milliseconds:
Process
Burst Time
P1
3
P2
7
P3
5
P4
2
P5
6
The processes are assumed to have arrived in the order P1, P2, P3, P4, P5, all at time 0.
a.
Draw the Gantt charts using, RR (quantum = 5) scheduling.
P1
b.
c.
d.
P2
P3
P4
P5
P2
P5
3
8
13
15
20
22 23
What is the turnaround time of each process?
P1=3, P2= 22,
P3=13,
P4=15,
P5=23
What is the waiting time of each process?
P1=3-3=0,
P2=22-7= 15, P3=13-5=8, P4=15-2=13, P5=23-6=17
What is the average waiting time (over all processes)?
(0+15+8+13+17)/5=10.6
iv.
Q: Consider the following set of processes, with the length of the CPU-burst time
given in milliseconds:
Process
Arrival Time
Burst Time
P1
0
4
P2
1
4
P3
4
3
P4
8
3
Draw the Gant charts illustrating the execution of these processes using FCFS and
preemptive SJF?
FCFS
P1
P2
0
P3
4
P4
8
11
14
SJF:
P1
P3
0
4
P2
7
P4
11
14
RR:
P1
v.
P2
P1
P2
P3
P1
P2
P3
P4
P4
P4
0
1
2
3
4
5
6
7
8
9
10 11 12
14
What is the turnaround time of each process of the scheduling algorithms in part a?
13
FCFS: P1=
4
, P2= 7
, P3= 7
, P4= 6
SJF: P1=
4
, P2= 10
, P3= 3
, P4= 6
RR: P1=
10
, P2= 10
, P3= 8
, P4= 6
P1
P2 P3
vi.
What is the waiting time of each process for each of the scheduling algorithms in part a?
FCFS: P1= 0
, P2= 3
, P3= 4
, P4= 3
SJF: P1=
0
, P2= 0
, P3= 6
, P4= 3
RR: P1=
6
, P2= 6
, P3= 5
, P4= 3
vii.
Which of the schedules in part a results in the minimal average waiting time (over all
processes)?
FCFS Average waiting time= (0+3+4+3)/4 = 2.5
SJF Average waiting time= (0+0+6+3)/4 = 2.25
RR Average waiting time= (6+6+5+3)/4 = 5.0
SJF have minimal average waiting time.