Uniprocessor Scheduling II
Chapter 9
Shortest Process Next
0
5
10
15
20
1
2
3
4
5
P3: 4 time units
P4: 5 time units
P5 2 time units
z Nonpreemptive policy
z Process with shortest expected
processing time is selected next
z Short process jumps ahead of longer
processes
Shortest Process Next
z Predictability of longer processes is
reduced
z If estimated time for process not correct,
the operating system may abort it
z Possibility of starvation for longer
processes
Shortest Remaining Time
0
5
10
15
20
1
2
3
4
5
P2 Remaining 5 time units
P3 Remaining 4 time units
P2 Remaining 5 time units
P4 Remaining 5 time units
P5 Remaining 2 time units
z SRT is a preemptive version of shortest
process next (SPN) policy
z Must estimate processing time
Highest Response Ratio
Next (HRRN)
0
5
5
20
P4 RR = 2.4
2
4
15
P5 RR =3.5
1
3
10
P3 RR=2.25
P4 RR = 1.6
P5 RR = 1.5
z Choose next process with the greatest value of RR
z No Preemptive
RR =(w+s)/s =time spent waiting+expected service time
expected service time
P5 RR=( (13-8) + 2)/2 = 3.5 // P5 has been chosen at 13
P4 RR=( (13-6) + 5)/5 = 2.4
Feedback
z Scheduling is done on a preemptive basis,
and a dynamic priority mechanisms is
used.
z When s process first enters the system, it
is placed in RQ0. After its first execution,
when it returns to the Ready state, it is
placed in RQ1.
Feedback (q = 1)
5
0
1
2
0
1
0
RQ0
1
0
20
3
4
5
RQ5
3
2
1
0
4
15
2
1
0
3
5
10
2
3
RQ1
RQ2
1
4
RQ4
RQ3
z Penalize jobs that have been running longer.
Focus on the time spent in execution so far.
z Don’t know remaining time process needs to
execute
Feedback q=2i
5
0
1
2
3
4
5
0
1
0
10
15
20
RQ1,2 time units RQ2, 4 time units
1(2)
2(4)
0
1(2)
2
0
1(2)
2
0
1
z When a process first enters the system, it is placed
in RQ0. After its first execution, when it returns to
the Ready state, it is placed in RQ1 ……
z In general, a process scheduled from RQi is allowed
to execute 2 i time units (1, 2, 4, 8, …...) before
preemption.
Question - Case Study (1)
z Consider the following set of processes:
Process Name Arrival Time Processing Time
1
0
3
2
1
5
3
3
2
4
9
5
5
12
5
Perform the analysis of a comparison of scheduling
policies(FCFS, RR(q=1), RR(q=4), SPN, SRT,
HRRN, Feedback(q=1), and Feedback(q=2 i) )
Solution of Question
(2)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
FCFS 1 1 1 2 2 2 2 2 3 3 4 4 4 4 4 5 5 5 5 5
RR
1 2 1 2 3 1 2 3 2 4 2 4 5 4 5 4 5 4 5 5
RR
1 1 1 2 2 2 2 3 3 2 4 4 4 4 5 5 5 5 4 5
q=1
q=4
SPN 1 1 1 3 3 2 2 2 2 2 4 4 4 4 4 5 5 5 5 5
SRT 1 1 1 3 3 2 2 2 2 2 4 4 4 4 4 5 5 5 5 5
HRRN 1 1 1 2 2 2 2 2 3 3 4 4 4 4 4 5 5 5 5 5
FB
1 2 1 3 2 3 1 2 2 4 2 4 5 4 5 4 5 4 5 5
FB
1 2 1 1 3 2 2 3 2 2 4 4 5 4 4 5 5 4 5 5
q=1
q=2i
* Each square represents one time unit. The number refers to the running process
Solution of Question - FCFS
(3)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
FCFS 1 1 1 2 2 2 2 2 3 3 4 4 4 4 4 5 5 5 5 5
* Each square represents one time unit. The number refers to the running process
First Come First Served (FCFS)
Process
Pi
1
2 3 4 5
Arrival Time
Ta
0
1
3
9 12
Service Time
Ts
3
5
2
5
5
Finish Time
Tf
3
8 10 15 20
Turnaround Time Tq
3.0 7.0 7.0 6.0 8.0
Tq = Tf - Ta
Normalized
Tq/Ts 1.0 1.4 3.5 1.2 1.6
Turnaround Time
Average
6.20
1.74
Solution of Question Round Robin q=1
000000000011111111112
012345678901234567890
P1
P2
P3
P4
P5
1
0
1
Ready Time
3
2
4
3
7
9
5
9
10
12
12
14
16
13 15 17
12123123242454545455
Solution of Question - RR (q=1)
RR
q=1
(4)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 2 1 2 3 1 2 3 2 4 2 4 5 4 5 4 5 4 5 5
* Each square represents one time unit. The number refers to the running process
Round Robin (RR q=1)
Process
Pi
1
2 3 4
Arrival Time
Ta
0
1
3
9
Service Time
Ts
3
5
2
5
Finish Time
Tf
6 11 8 18
Turnaround Time Tq
6 10 5
9
Tq = Tf - Ta
Normalized
Tq/Ts 2.0 2.0 2.5 1.8
Turnaround Time
5
12
5
20
8
Average
1.6
1.98
7.6
Solution of Question Round Robin q=4
000000000011111111112
012345678901234567890
P1
P2
P3
P4
P5
Ready Time
0
1
7
3
9
14
12
18
11122223324444555545
Solution of Question - RR (q=4)
RR
q=4
(5)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 1 1 2 2 2 2 3 3 2 4 4 4 4 5 5 5 5 4 5
* Each square represents one time unit. The number refers to the running process
Round Robin (RR q=4)
Process
Pi
1
2 3 4 5 Average
Arrival Time
Ta
0
1
3
9 12
Service Time
Ts
3
5
2
5
5
Finish Time
Tf
3 10 9 19 20
Turnaround Time Tq
3 9 6 10 8
7.20
Tq = Tf - Ta
Normalized
Tq/Ts 1.0 1.8 3.0 2.0 1.6
1.88
Turnaround Time
Solution of Question -
SPN
(6)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
SPN 1 1 1 3 3 2 2 2 2 2 4 4 4 4 4 5 5 5 5 5
* Each square represents one time unit. The number refers to the running process
Shortest Process Next (SPN)
Process
Pi
1
2 3 4
Arrival Time
Ta
0
1
3
9
Service Time
Ts
3
5
2
5
Finish Time
Tf
3 10 5 15
Turnaround Time Tq
3 9 2
6
Tq = Tf - Ta
Normalized
Tq/Ts 1.0 1.8 1.0 1.2
Turnaround Time
5
12
5
20
8
Average
1.6
1.32
5.60
Solution of Question -
SRT
(7)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
SRT 1 1 1 3 3 2 2 2 2 2 4 4 4 4 4 5 5 5 5 5
* Each square represents one time unit. The number refers to the running process
Shortest Remaining Time (SRT)
Process
Pi
1
2 3 4 5 Average
Arrival Time
Ta
0
1
3
9 12
Service Time
Ts
3
5
2
5
5
Finish Time
Tf
3 10 5 15 20
Turnaround Time Tq
3 9 2
6 8
5.60
Tq = Tf - Ta
Normalized
Tq/Ts 1.0 1.8 1.0 1.2 1.6
1.32
Turnaround Time
Solution of Question -
HRRN
(8)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
HRRN 1 1 1 2 2 2 2 2 3 3 4 4 4 4 4 5 5 5 5 5
* Each square represents one time unit. The number refers to the running process
* At time 3: P2 RR (2+5)/5 = 1.4; P3 RR= (0+2)/2 = 1. So Choose P2
Highest Response Ratio Next (HRRN)
Process
Pi
1
2 3 4 5 Average
Arrival Time
Ta
0
1
3
9 12
Service Time
Ts
3
5
2
5
5
Finish Time
Tf
3
8 10 15 20
Turnaround Time Tq
3 7 7
6 8
6.20
Tq = Tf - Ta
Normalized
Tq/Ts 1.0 1.4 3.5 1.2 1.6 1.74
Turnaround Time
Solution of Question Feedback q=1
000000000011111111112
012345678901234567890
RQ1
P1
P2
P3
P4
P5
1
0
2
1
0
0
RQ0
RQ2
2
3
RQ3
1
3
0 1 2
0 1 2
4
3
RQ4
4
12132312244455524545
Solution of Question FB
q=1
FB q=1
(9)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 2 1 3 2 3 1 2 2 4 4 4 5 5 5 2 4 5 4 5
* Each square represents one time unit. The number refers to the running process
Feedback (FB q=1)
Process
Pi
1
2
Arrival Time
Ta
0
1
Service Time
Ts
3
5
Finish Time
Tf
7 16
Turnaround Time Tq
7 15
Tq = Tf - Ta
Normalized
Tq/Ts 2.3 3.0
Turnaround Time
3
3
2
6
3
4 5
9 12
5
5
19 20
10 8
Average
1.5
2.0 1.6
2.08
8.6
Solution of Question Feedback q= 2i
000000000011111111112
012345678901234567890
P1
P2
P3
P4
P5
RQ1
1
0
1
0
0
RQ0
RQ2
2
1
0
1
0
2
1
2
12113223224454455455
Solution of Question FB
q=2i
FB q= 2i (10)
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 2 1 1 3 2 2 3 2 2 4 4 5 4 4 5 5 4 5 5
* Each square represents one time unit. The number refers to the running process
Feedback (FB q = 2i)
Process
Pi
1
2 3
Arrival Time
Ta
0
1
3
Service Time
Ts
3
5
2
Finish Time
Tf
4 10 8
Turnaround Time Tq
4
9 5
Tq = Tf - Ta
Normalized
Tq/Ts 1.3 1.8 2.5
Turnaround Time
4 5
9 12
5
5
18 20
9 8
Average
1.8 1.6
1.81
7.00
Question 1 - Exercise/Home Work (1)
z Consider the following set of processes:
Process Name Arrival Time Processing Time
1
0
1
2
1
9
3
2
1
4
3
9
Perform the analysis of a comparison of scheduling
policies(FCFS, RR(q=1), RR(q=4), SPN, SRT,
HRRN, Feedback(q=1), and Feedback(q=2 i) )