Scheduling algorithms for embedded system

1. Nov 2nd 2012
Julian Ilham
Dept. of Electronic Engineering
Pukyong National University
Email: ilham.julian@gmail.com
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2.  Introduction
 Preemptive vs Non-Preemptive
 First Come First Served (FCFS)
 Shortest Job First (SJF)
 Priority Based Scheduling
 Round Robin (RR)
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3. What is scheduling?
Choosing which task to run and for how long
Why do we need it?
- Improve the system performance
- Support multitasking
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4. Running task can be forced to release even
Preemptive
though it is neither completed
Each running task keeps processing until it
Non - Preemptive
completed
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5.  Only available in Non-Preemptive mode
 Example:
Tasks Arrival Times (ms) Burst Time (ms)
P2 0.0 12
P3 3.0 8
P4 5.0 4
P1 10.0 10
P5 12.0 6
 Gantt chart
P2 P3 P4 P1 P5
0 20 24 34 40 (ms)
3 5 10 12
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6.  Time Calculation
P1 : 24 – 10 = 12 ms
P2 : 0 ms
P3 : 12 – 3 = 9 ms
P4 : 20 – 5 = 15 ms
P5 : 34 – 12 = 22 ms
Total waiting time : P1 + P2 + P3 + P4 + P5 = 58 ms
Average waiting time : 58 ms / 5 = 11.6 ms
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7.  Advantage: Very simple code and structure
 Disadvantage: Large average waiting time
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8.  Available in Preemptive and Non-Preemptive mode
 Example (Preemptive)
Tasks Arrival Times (ms) Burst Time (ms)
P2 0.0 12
P3 3.0 8
P4 5.0 4
P1 10.0 10
P5 12.0 6
 Gantt chart
P2 P3 P4 P3 P5 P2 P1
0 9 15 21 30 40 (ms)
3 5 10 12
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9.  Time Calculation (Preemptive)
P1 : 30 – 10 = 20 ms
P2 : 0 + (21 – 3) = 18 ms
P3 : (3 – 3) + (9 – 5) = 4 ms
P4 : 5 – 5 = 0 ms
P5 : 15 – 12 = 3 ms
Total waiting time : P1 + P2 + P3 + P4 + P5 = 45 ms
Average waiting time : 45 ms / 5 = 9 ms
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10.  Example (Non-Preemptive)
Tasks Arrival Times (ms) Burst Time (ms)
P2 0.0 12
P3 3.0 8
P4 5.0 4
P1 10.0 10
P5 12.0 6
 Gantt chart
P2 P4 P5 P3 P1
0 16 22 30 40 (ms)
3 5 10 12
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11.  Time Calculation (Non-Preemptive)
P1 : 30 – 10 = 20 ms
P2 : 0 ms
P3 : 22 – 3 = 19 ms
P4 : 12 – 5 = 7 ms
P5 : 16 – 12 = 4 ms
Total waiting time : P1 + P2 + P3 + P4 + P5 = 50 ms
Average waiting time : 50 ms / 5 = 10 ms
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12.  Advantage: Minimum average waiting time for a given set of
process
 Disadvantage: Difficult to estimate burst time, starvation
(indefinite blocking)
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13.  Available in Preemptive and Non-Preemptive mode
 Example (Preemptive)
Tasks Arrival Times (ms) Burst Time (ms) Priority
P2 0.0 12 2
P3 3.0 8 5
P4 5.0 4 1
P1 10.0 10 4
P5 12.0 6 3
 Gantt chart
P2 P4 P2 P5 P1 P3
0 9 16 22 32 40 (ms)
3 5 10 12
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14.  Time Calculation (Preemptive)
P1 : 22 – 10 = 12 ms
P2 : 9 – 5 = 4 ms
P3 : 32 – 3 = 29 ms
P4 : 5 – 5 = 0 ms
P5 : 16 – 12 = 4 ms
Total waiting time : P1 + P2 + P3 + P4 + P5 = 49 ms
Average waiting time : 49 ms / 5 = 9.8 ms
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15.  Example (Non-Preemptive)
Tasks Arrival Times (ms) Burst Time (ms) Priority
P2 0.0 12 2
P3 3.0 8 5
P4 5.0 4 1
P1 10.0 10 4
P5 12.0 6 3
 Gantt chart
P2 P4 P5 P1 P3
16 22 32 40 (ms)
3 5 10 12
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16.  Time Calculation (Non-Preemptive)
P1 : 22 – 10 = 12 ms
P2 : 0 ms
P3 : 32 – 3 = 29 ms
P4 : 12 – 5 = 7 ms
P5 : 16 – 12 = 4 ms
Total waiting time : P1 + P2 + P3 + P4 + P5 = 52 ms
Average waiting time : 52 ms / 5 = 10.4 ms
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17.  Advantage: Easier than SJF
 Disadvantage: Can cause indefinite blocking
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18.  Only available in Non-Preemptive mode
 FCFS is used in sequencing policy
 Need to determine time slice/ quantum as a unit of time for
each process
 Then continue to the next process after reaching maximum of
quantum
- Too small time quantum -> too many tasks switched
- Too large time quantum -> inherits the problem of FCFS
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19.  Example (assuming quantum is 5 ms):
Tasks Arrival Times (ms) Burst Time (ms)
P2 0.0 12
P3 3.0 8
P4 5.0 4
P1 10.0 10
P5 12.0 6
 Gantt chart
0 5 10 14 19 24 29 32 37 38 40
P2 P3 P4 P1 P5 P2 P3 P1 P5 P2
3 5 10 12
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20.  Time Calculation
P1 : (14 – 10) + (32 – 19) = 17 ms
P2 : 0 + (24 – 5) + (38 – 29) = 28 ms
P3 : (5 – 3) + (29 – 10) = 21 ms
P4 : 10 – 5 = 5 ms
P5 : (19 – 12) + (37 – 24) = 20 ms
Total waiting time : P1 + P2 + P3 + P4 + P5 = 91 ms
Average waiting time : 91 ms / 5 = 18.2 ms
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21.  Advantage: Guarantee fairness (bounded waiting time and no
starvation)
 Disadvantage: Average waiting time is long
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22. Q n A
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