Uniprocessor Scheduling
1
0
Priorities
• Scheduler will always choose a process of
higher priority over one of lower priority
• Have multiple ready queues to represent each
level of priority
• Lower-priority may suffer starvation
– To overcome that allow a process to change its
priority based on its age or execution history
2
3
Process Scheduling Example
4
1-
First-Come-First-Served
(FCFS)
• Each process joins the Ready queue
• When the current process finishes to execute, the
oldest process in the Ready queue is selected
• A short process may have to wait a very long
time before it can execute
• Favors CPU-bound processes
– I/O processes have to wait until CPU-bound
process completes
5
Simple Gantt Chart
6
Example (FIFO)
Job #
Jobtime
#
Arrival
1
2
3
4
5
85
Arrival
time
Finish
time
01
30
22
92
43
134
64
186
208
Servicetime
time
Response
33
76
94
125
122
Service time
3
6
4
5
2
Wait time
0
1
5
7
10
Average response time = (3+7+9+12+12)/5 = 8.6
Throughput = 5/20 = 1/4
Ave (RespTime/ServTime) = (1+7/6+9/4+12/5+12/2)/5 = 2.564
2-
Round-Robin
• Uses preemption based on a clock
• An amount of time is determined that allows each
process to use the processor for that length of time
• Clock interrupt is generated at periodic intervals
• When an interrupt occurs, the currently running
process is placed in the ready queue
– Next ready job is selected
• Known as time slicing
8
Example (Round Robin)
Job #
Arrival
time
Arrival time
1
00
2
22
3
44
4
66
55
88
Finish time
Service
time
Response time
43
4
18
6
16
17
4
13
20
5
14
15
2
7
Service time
3
6
4
5
2
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Round
Robin
T=1
Example (Round Robin)
Job #
Arrival time
1
0
2
2
3
4
44
66
55
88
Finish time
Service
time
Response time
53
5
17
6
15
13
4
9
20
5
14
15
2
7
Service time
3
6
4
5
2
5
Wait time
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timeQueueJ2,QueueJ:1QueueJ1,:
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Round
Robin
T=2
• Round-Robin
– Process take turns
– Each executes till its time-quantum expires
– Gives relatively reasonable turnaround time
• Better in standard deviation of turnaround time
– Useful when it is hard to estimate the execution time
• No need to know execution time in advance
– Overhead on process switching
• More process switch  more context switch
– I/O bound jobs may not use up their time quantum
• Not fair to I/O bound jobs
12
3- Shortest Process Next
• Nonpreemptive policy
• Process with shortest expected processing time is
selected next
• Short process jumps ahead of longer processes
• Predictability of longer processes is reduced
• If estimated time for process not correct, the operating
system may abort it
• Possibility of starvation for longer processes
13
Example (Shortest Job Next)
Job #
Arrival
Arrival time
time
11
00
22
22
3 3
4 4
4 4
6 6
5 5
8 8
Finish time
Service
time
Response time
33
3
69
7
4 15
11
5 20
14
2 11
3
Service time
3
6
4
5
2
511+14+
2Job
finishes
Average
respons e0time Job
= 1(3+7+
Wait time
7finishes 3)/5
9 = 7.16
Job 2Job
5 enters
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3 Job
enters
3,
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=
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isqueued
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Ave (RespTime/ServTime) = (1+7/6+11/4+14/5+3/2)/5 = 1.844
SJF
4- Shortest Remaining Time
• Preemptive version of shortest process
next policy
• Must estimate processing time
15
Example (Shortest Remaining
Time)
Job #
Arrival time
1
0
Finish time
Service
time
Response time
33
3
2
2
3
4
4
6
55
88
15
6
48
Job13
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20
5
14
10
2
2
Job
enters
3 4 Job
63 finishes4
5
2
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Job
5 finishes
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time
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time
Jobof2 Job
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enters
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0
7
0
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Service time
SRTF
5- Highest Response Ratio Next
(HRRN)
• Choose next process with the greatest ratio
(time spent waiting + expected service time)
expected service time
•The process with the greatest value will be scheduled to run.
17
Example (Highest Response Ratio)
RR = (waiting-time + service-time) / service-time
Job #
Arrival
Arrival time
time
11
00
22
22
33
44
4 4
6 6
5 5
8 8
Finish time
Service
time
Response time
33
3
69
7
4 13
9
5 20
14
2 15
7
Job62 finishes4
5
2
RR of JobJob
finishes = 2.25
3 =3((9–4)+4)/4
Average
respons
eJob
time
=50(3+7+
9+14+7
8= 0–6)+5)/5
Wait time
0 4Jenters
94 ==((13
Job 1 finishes
of JobRR
Job)/5
= 2.4
47=of
((9–6)+5)/5
1.6
obRR
enters
Throughput
=Job
5/20
= 1/4
JobJ2ob
Jobs
Job 5 = ((13–8)+2)/2
= 3.5
2 is
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enters
the
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5queue
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3,
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in
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enters
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= 2.144
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Service time
HRRF
3
• 6-
Multilevel Feedback Queue
– The system maintains multiple ready queues
• For example, N priority levels
– CPU always serves the jobs in the highest priority
queue
– A new process will be assigned a high priority
– If a process used up its time quantum
• priority decremented by one (placed in a lower level queue)
– Favor I/O bound jobs
• Not use up time quantum  stay at the same priority level
– Good for time sharing systems
• There are many variations of the algorithm
20
21
22
Example (Multi-level Queue)
Job #
Arrival time
11
00
22
22
33
44
44
66
55
88
Finish time
Service
time
311
620
4 16
5 19
210
Response
11
12
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