Hardware Reference
In-Depth Information
Minimizing
U
over
G
we have
(1 +
G
2
)
(1 +
G
)
2
dU
dG
2
G
(1 +
G
)
−
=
=
G
2
+2
G
−
1
=
,
(1 +
G
)
2
and
dU/dG
=0for
G
2
+2
G
−
1=0, which has two solutions:
G
1
−
√
2
=
−
1
1+
√
2
.
G
2
=
−
Since 0
G<
1, the negative solution
G
=
G
1
is discarded. Thus, from equation
(4.7), the least upper bound of
U
is given for
G
=
G
2
:
≤
1+(
√
2
2
√
2
√
2
=2(
√
2
1)
2
−
4
−
1+(
√
2
U
lub
=
=
−
1)
.
−
1)
That is,
=2(2
1
/
2
U
lub
−
1)
0
.
83
.
(4.8)
Notice that, if
T
2
is a multiple of
T
1
,
G
=0and the processor utilization factor be-
comes 1. In general, the utilization factor for two tasks can be computed as a function
of the ratio
k
=
T
2
/T
1
. For a given
F
, from equation (4.5) we can write
F
)
2
F
+(
k
−
2
F
+
F
(
F
+1)
k
U
=
=
k
−
.
k
Minimizing
U
over
k
we have
dU
dk
F
(
F
+1)
k
2
=1
−
,
and
dU/dk
=0for
k
∗
=
F
(
F
+1). Hence, for a given
F
, the minimum value of
U
is
U
∗
=2(
F
(
F
+1)
F
)
.
Table 4.1 shows some values of
k
∗
and
U
∗
as a function of
F
, whereas Figure 4.10
shows the upper bound of
U
as a function of
k
.
−
Search WWH ::
Custom Search