Hardware Reference
In-Depth Information
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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