CPU SCHEDULING --> OPERSYS (MIDTERM) 3rd year 1st sem

1. Slide 1
CPU Scheduling
 Classification of CPU Scheduling
 Performance Criteria of Good Scheduler
 CPU Scheduling Algorithms
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Slide 2
*Property of STI J0024
CPU scheduling may be classified as:
 Non‐preemptive scheduling
CPU scheduling is non‐preemptive if once the CPU has been assigned to a
process and the process starts executing, the CPU cannot be taken away from
that process.
 Preemptive scheduling
In preemptive CPU scheduling, even though the CPU has been assigned to a
process and the process is already executing, the CPU scheduler may decide to
assign the CPU to another process in the ready queue.
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Slide 3
*Property of STI J0024
A good scheduler should optimize the following performance criteria:
 CPU utilization
 Throughput
 Turnaround time
 Response time
 Waiting time
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2. Slide 4
*Property of STI J0024
The different CPU scheduling algorithms are:
 First‐Come, First‐Served Algorithm
 Shortest Process First Algorithm
 Shortest Remaining Time First Algorithm
 Round Robin Algorithm
 Priority Scheduling
 Multilevel Feedback Queues
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Slide 5
*Property of STI J0024
First‐Come, First‐Served Algorithm (FCFS)
‐ the one that enters the Ready queue first gets to execute at the CPU first
a.
b.
c.
A
0 8
A
0
B
8 12
A
0
B C
8 12 17
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Slide 6
*Property of STI J0024
First‐Come, First‐Served Algorithm (FCFS)
d.
e.
WTA = 0 – 0 = 0 ms
WTB = 8 – 3 = 5 ms
WTC = 12 – 4 = 8 ms
WTD = 17 – 6 = 11 ms
WTE = 20 – 10 = 10 ms
The average waiting time is
(0 + 5 + 8 + 11 + 10)/5 = 34/5 = 6.8 ms
A
0
B C D
8 12 17 20
A
0
B C D
8 12 17 20
E
22
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3. Slide 7
*Property of STI J0024
First‐Come, First‐Served Algorithm (FCFS)
TAA = 8 ‐ 0= 8 ms
TAB = 12 ‐ 3 = 9 ms
TAC = 17 ‐ 4 = 13 ms
TAD = 20 ‐ 6 = 14 ms
TAE = 22 ‐ 10 = 12 ms
The average turnaround time is
(8 + 9 + 13 + 14 + 12)/5 = 56/5 = 11.2 ms
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Slide 8
*Property of STI J0024
Shortest Process First Algorithm (SPF)
‐ the process with the shortest CPU burst
a.
b.
c.
A
0 8
A
0
D
8 11
A
0
D
8 11 13
E
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Slide 9
*Property of STI J0024
Shortest Process First Algorithm (SPF)
d.
e.
WTA = 0 – 0 = 0 ms
WTB = 13 – 3 = 10 ms
WTC = 17 – 4 = 13 ms
WTD = 8 – 6 = 2 ms
WTE = 11 – 10 = 1 ms
The average waiting time is
(0 + 10 + 13 + 2 + 1)/5 = 26/5 = 5.2 ms
A
0
BD
8 11 13 17
E
A
0
B CD
8 11 13 17
E
22
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4. Slide 10
*Property of STI J0024
Shortest Process First Algorithm (SPF)
TAA = 8‐ 0 = 8 ms
TAB = 17 ‐ 3 = 14 ms
TAC = 22 ‐ 4 = 18 ms
TAD = 11 ‐ 6 = 5 ms
TAE = 13 ‐ 10 = 3 ms
The average turnaround time is
(8 + 14 + 18 + 5 + 3)/5 = 48/5 = 9.6 ms
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Slide 11
*Property of STI J0024
Shortest Remaining Time First Algorithm (SRTF)
‐ the preemptive version of SPF
a.
b.
c.
d.
Process ID Arrival Time CPU Burst
A 0 8
B 3 4
C 4 5
D 6 3
E 10 2
0
A
0
A
3
B
0
B
7
A
3
0
B
7
DA
3
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Slide 12
*Property of STI J0024
Shortest Remaining Time First Algorithm (SRTF)
e.
f.
g.
h.
WTA = (0 – 0) + (12 – 3) = 9 ms
WTB = 3 – 3 = 0 ms
WTC = 17 – 4 = 13 ms
WTD = 7 – 6 = 1 ms
WTE = 10 – 10 = 0 ms
The average waiting time is
(9 + 0 + 13 + 1 + 0)/5 = 23/5 = 4.6 ms
0
B D
7 10
A
3
0
B D
7 10 12
EA
3
A
0
B D
7 10 12 17
EA
3
A
0
B CD
7 10 12 17
E
22
A
3
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5. Slide 13
*Property of STI J0024
Shortest Remaining Time First Algorithm (SRTF)
TAA = 17 ‐ 0= 17 ms
TAB = 7 ‐ 3 = 4 ms
TAC = 22 ‐ 4 = 18 ms
TAD = 10 ‐ 6 = 4 ms
TAE = 12 ‐ 10 = 2 ms
The average turnaround time is
(17 + 4 + 18 + 4 + 2)/5 = 45/5 = 9.0 ms
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Slide 14
*Property of STI J0024
Round Robin Algorithm (RR)
‐ the process are selected on a first‐come, first‐served basis. However, each
process is given a time limit to execute at the CPU.
Assume a time quantum of 3 ms.
a.
b.
Process ID Arrival Time CPU Burst
A 0 8
B 3 4
C 4 5
D 6 3
E 10 2
0
A
3
0
B
6
A
3
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Slide 15
*Property of STI J0024
Round Robin Algorithm (RR)
c.
d.
e
f.
g.
h
i.
0
B A
6 9
A
3
0
B A
6 9 12
CA
3
D
0
B A
6 9 12 15
CA
3
D
0
B A
6 9 12 15
CA
3
B
16
D
0
B A
6 9 12 15
CA
3
B
16
A
18
D
0
B A
6 9 12 15
CA
3
B
16
A
18
E
20
D
0
B CA
6 9 12 15
C
22
A
3
B
16
A
18
E
20
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6. Slide 16
*Property of STI J0024
Round Robin Algorithm (RR)
WTA = (0 – 0) + (6 – 3) + (16 – 9) = 10 ms
WTB = (3 – 3) + (15 – 6) = 9 ms
WTC = (9 – 4) + (20 – 12) = 13 ms
WTD = 12 – 6 = 6 ms
WTE = 18 – 10 = 8 ms
The average waiting time is
(10 + 9 + 13 + 6 + 8)/5 = 46/5 = 9.2 ms
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Slide 17
*Property of STI J0024
Round Robin Algorithm (RR)
TAA = 18 ‐ 6= 18 ms
TAB = 16 ‐ 3 = 13 ms
TAC = 22 ‐ 4 = 18 ms
TAD = 15 ‐ 6 = 9 ms
TAE = 20 ‐ 10 = 10 ms
The average turnaround time is
(18 + 13 + 18 + 9 + 10)/5 = 68/5 = 13.6 ms
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Slide 18
*Property of STI J0024
Priority Scheduling
‐ Each process is assigned a priority and the CPU scheduler selects the process in
the ready queue with the highest priority to execute next.
Using the non‐preemptive priority algorithm:
a.
b.
Process ID Arrival
Time
CPU Burst Priority
A 0 8 4
B 3 4 3
C 4 5 1
D 6 3 2
E 10 2 2
A
0 8
A
0
C
8 13
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7. Slide 19
*Property of STI J0024
Priority Scheduling
c.
d.
e.
WTA = 0 – 0 = 0 ms
WTB = 18 – 3 = 15 ms
WTC = 8 – 4 = 4 ms
WTD = 13 – 6 = 7 ms
WTE = 16 – 10 = 6 ms
The average waiting time is
(0 + 15 + 4 + 7 + 6)/5 = 32/5 = 6.4 ms
A
0
C D
8 13 16
A
0
C D
8 13 16 18
E
A
0
BC D
8 13 16 18
E
22
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Slide 20
*Property of STI J0024
Priority Scheduling
TAA = 8 ‐ 0= 8 ms
TAB = 22 – 3 = 19 ms
TAC = 13 ‐ 4 = 9 ms
TAD = 16 ‐ 6 = 10 ms
TAE = 18 ‐ 10 = 8 ms
The average turnaround time is
(8 + 19 + 9 + 10 + 8)/5 = 54/5 = 10.8 ms
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Slide 21
*Property of STI J0024
Priority Scheduling
Using the preemptive priority algorithm:
a.
b.
c.
d.
Process ID Arrival
Time
CPU Burst Priority
A 0 8 4
B 3 4 3
C 4 5 1
D 6 3 2
E 10 2 2
0
A
0
BA
3
0
B C
4 9
A
3
0
B C
4 9
A
3
D
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8. Slide 22
*Property of STI J0024
Priority Scheduling
e.
f.
WTA = (0 – 0) + (17 – 3) = 14 ms
WTB = (3 – 3) + (14 ‐ 4) = 10 ms
WTC = (4 – 4) = 0 ms
WTD = (9 – 6) = 3 ms
WTE = (12 – 10) = 2 ms
The average waiting time is
(14 + 10 + 0 + 3 + 2)/5 = 29/5 = 5.8 ms
0
B C D
4 9 12
A
3
A
0
B C D
4 9 12 14
E
22
A
3
B
17
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Slide 23
*Property of STI J0024
Priority Scheduling
TAA = 22 – 0 = 22 ms
TAB = 17 ‐ 3 = 14 ms
TAC = 9 ‐ 4 = 5 ms
TAD = 12 ‐ 6 = 6 ms
TAE = 14 ‐ 10 = 4 ms
The average turnaround time is
(22 + 14 + 5 + 6 + 4)/5 = 51/5 = 10.2 ms
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Slide 24
*Property of STI J0024
Multilevel Feedback Queues
Q0All processes enter
the queuing network
at the rear of the
highest priority
queue
Processes that
finish executing or
request for an I/O
operation within the
given time slice
leave the queuing
network
FCFS (time slice = 10 ms)
Q1
FCFS (time slice = 20 ms)
Processes that do not finish executing
within the given time slice move to the next
lower-priority queue
Processes that
finish executing or
request for an I/O
operation within the
given time slice
leave the queuing
network
.
.
.
Qn
RR
Processes that do not finish executing
within the given time slice move to the next
lower-priority queue
Q2
FCFS (time slice = 30 ms)
Processes that
finish executing or
request for an I/O
operation within the
given time slice
leave the queuing
network
Processes that
finish executing or
request for an I/O
operation within the
given time slice
leave the queuing
network
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