Hardware Reference
In-Depth Information
For task τ 2 we have
S (0)
2
q last
2
=
B 2 + C 1 + C 2
=7+6+10
4=19
+ 19
T 1
+1 C 1 =19
S (1)
2
q last
2
=
B 2 + C 2
S 2 + q last
2
R 2
=
=19+4=23 .
For task τ 3 we have:
S (0)
3
B 3 + C 1 + C 2 + C 3 q last
=
3 =5+6+10+18 5=34
3 T 1
C 1 +
3 T 2
C 2 =40
S (1)
3
B 3 + C 3 q last
=
3 +
+1
+1
4 T 1
C 1 +
4 T 2
C 2 =50
S (2)
3
B 3 + C 3 q last
=
3 +
+1
+1
5 T 1
C 1 +
5 T 2
C 2 =56
S (3)
3
B 3 + C 3 q last
=
3 +
+1
+1
5 T 1
C 1 +
5 T 2
C 2 =56
S (4)
3
B 3 + C 3 q last
=
3 +
+1
+1
S 3 + q last
R 3 =
3 =56+5=61
For task τ 4 we have
S (0)
4
B 4 + C 1 + C 2 + C 3 + C 4 q last
=
4 =6+10+18+15 6=43
4 T 1
C 1 +
4 T 2
C 2 +
4 T 3
C 3 =59
S (1)
4
B 4 + C 4 q last
=
4 +
+1
+1
+1
5 T 1
C 1 +
5 T 2
C 2 +
5 T 3
C 3 =65
S (2)
4
B 4 + C 4 q last
=
4 +
+1
+1
+1
6 T 1
C 1 +
6 T 2
C 2 +
6 T 3
C 3 =65
S (3)
4
B 4 + C 4 q last
=
4 +
+1
+1
+1
S 4 + q last
R 4 =
4 =65+6=71 .
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