algoritma fcfs
DESCRIPTION
ppt algoritma fcfsTRANSCRIPT
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-7
8
7
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-9
0:
97
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The final schedule (Gantt chart):
0
P1 (24)24 27
P2 (3) P3 (4)
P1 waiting time: 0P2 waiting time: 24P3 waiting time: 27
The average waiting time: (0+24+27)/3 = 17
What if P1 arrives at time 2
31
9
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6 445446
46$*416
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7@7
9
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0:
79
A-
0 3
P4 (3) P1 (6)9
P3 (7)16
P4 waiting time: 0P1 waiting time: 3P3 waiting time: 9P2 waiting time: 16
The total time is: 24The average waiting time (AWT):
(0+3+9+16)/4 = 7
P2 (8)24
Do it yourself
A
7@7
9
-
0:
79
A-
0 6
P4 (3)P1 (6)14
P3 (7)21
P1 waiting time: 0P2 waiting time: 6P3 waiting time: 14P4 waiting time: 21
The total time is the same.The average waiting time (AWT):
(0+6+14+21)/4 = 10.25(comparing to 7)
P2 (8)24
Do it yourself
-
>B
#25
#4
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$3 -
0
:
---
0 10
P1 (10)
P1 waiting time: 0P2 waiting time: 8
The average waiting time (AWT): (0+8)/2 = 4
P2 (2)2 (p2 arrives) 12
Do it yourself
#
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5
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#8
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0:
---
P1 waiting time: 4-2 =2P2 waiting time: 0
The average waiting time (AWT): (0+2)/2 = 1
0 122
P1 (8)P2 (2)4
P1 (2)
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7
7
0:
77
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Suppose time quantum is: 1 unit, P1, P2 & P3 never block
P1 P2 P3
10
P1 P2 P3 P1 P2 P3 P2
P1 waiting time: 4P2 waiting time: 6P3 waiting time: 6
The average waiting time (AWT): (4+6+6)/3 = 5.33
Do it yourself