Information Technology Reference
In-Depth Information
Proof
.
log
h
1
p
1
n
1
+
g
21
p
2
u
1
(
p
1
,
p
2
)
=
−
c
1
p
1
+
s
12
log
h
2
p
2
n
2
+
−
s
12
c
2
p
2
,
g
12
p
1
we have
∂u
1
(
p
1
,
p
2
)
∂p
1
1
p
1
−
s
12
g
12
n
2
+
=
g
12
p
1
−
c
1
.
Since
1
p
1
−
1
p
1
−
s
12
g
12
n
2
+
s
12
p
1
lim
p
1
ₒ
≥
lim
p
1
ₒ
=∞
g
12
p
1
0
0
and
1
p
1
−
s
12
g
12
n
2
+
lim
p
1
ₒ∞
=
0
g
12
p
1
and
∂
p
1
−
n
2
+
g
12
p
1
∂p
1
s
12
g
12
s
12
g
12
(
n
2
+
1
p
1
+
=−
g
12
p
1
)
2
1)
g
12
p
1
−
n
2
(
s
12
−
2
n
2
g
12
p
1
−
=
p
1
(
n
2
+
g
12
p
1
)
2
<
0,
there exists a unique value of
p
1
such that
1
p
1
−
s
12
g
12
n
2
+
g
12
p
1
−
c
1
=
0,
(3.11)
which is also the value of
p
SNE
1
. Solving (
3.11
), we obtain the desired result. Similarly,
we can obtain
p
SNE
2
.
Next we have the following result.
Corollary 3.2.
For the two-user SGUM-based power control game, each user's
transmit power at the SNE is decreasing as its social tie level with the other increases.
Proof
.
ʱ
1
+
p
SNE
1
=
ʲ
1
−
ʱ
1
and
s
12
g
12
+
c
1
n
2
−
g
12
2
c
1
g
12
n
2
c
1
g
12
ʱ
1
≡
,
ʲ
1
≡
>
0,
