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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