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For task τ 2 we have
s (0)
2 , 1
=
B 2 + C 1 =5
B 2 + 5
T 1
+1 C 1 =5
s (1)
2 , 1
=
f 2 , 1
=
s 2 , 1 + C 2 =5+2=7
R 2
f 2 , 1 =7
D 2 .
=
For task τ 3 , the response time must be checked in the first three jobs.
For k =1:
s (0)
3 , 1
=
B 3 + C 2 + C 1 =7
B 3 + 7
T 1
+1 C 1 + 7
T 2
+1 C 2 =7
s (1)
3 , 1
=
f 3 , 1
=
s 3 , 1 + C 3 =7+3=10
R 3 , 1
=
f 3 , 1 =10 .
For k =2:
s (0)
3 , 2
=
B 3 + C 3 + C 1 + C 2 =10
B 3 + C 3 + 10
T 1
+1 C 1 + 10
T 2
+1 C 2 =16
s (1)
3 , 2
=
B 3 + C 3 + 16
T 1
+1 C 1 + 16
T 2
+1 C 2 =19
s (2)
3 , 2
=
B 3 + C 3 + 19
T 1
+1 C 1 + 19
T 2
+1 C 2 =22
s (3)
3 , 2
=
B 3 + C 3 + 22
T 1
+1 C 1 + 22
T 2
+1 C 2 =22
s (4)
3 , 2
=
f 3 , 2
=
s 3 , 2 + C 3 =22+3=25
R 3 , 2
=
f 3 , 2
T 3 =25
14 = 11 .
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