Information Technology Reference
In-Depth Information
4.1 Channel Usage Analysis
Before discuss the channel usage, we make the following assumptions and def-
initions. There are
C
available channels in the network. Each node equips
K
network interfaces that work on
K
different channels.
K
satisfies the connectiv-
ity invariant condition, which is
K
2
+ 1. The transmission probability of
node
n
on channel
c
is
P
(
n,c
). The neighbors of node
n
is defined as
B
(
n
).
The node
n
detects no communication conflicts on channel
c
only if there is
no nodes transmitting or one node transmitting. The probability that only one
node is transmitting is
≥
⊛
⊞
⊝
P
(
i,c
)
×
⊠
(1
− P
(
j, c
))
i∈B
(
n
)
{n}
j∈B
(
n
)
{n}−{i}
=
j∈B
(
n
)
{n}
P
(
i,c
)
(1
−
P
(
j, c
))
×
P
(
i,c
)
.
1
−
i∈B
(
n
)
{n}
And the probability that no node is transmitting is
(1
−
P
(
i,c
))
.
i∈B
(
n
)
{n}
Then, the conflicting probability is 1 minus the above two parts. Thus:
P
conflict
(
n,c
)=1
−
⊛
⊞
P
(
i,c
)
⊝
⊠
×
P
(
i,c
)
+1
(1
−
P
(
j, c
))
.
(1)
−
1
i∈B
(
n
)
{n}
j∈B
(
c
)
{n}
Assume that
B
(
n
)
P
(
i,c
)=
q
and
|
{
n
}|
=
m.
i∈B
(
n
)
{n}
q
can be considered as the total channel usage. Usually, if all the nodes access the
channel based on CSMA/CA and there is no hidden terminal,
q
will be
1. If
there are hidden terminals,
q
can be
>
1. Because the transmission probability
of each node is
≤
1,
P
conflict
(
n,c
) reaches the maximum when
P
(
i,c
)=
m
for all
i
, using Lagrange
multiplier method. The maximum of
P
conflict
(
n,c
)is
≤
1,
q
is always
≤
m
, . We can prove that with 0
≤
q
≤
1
m−
1
.
m
+
q
1
q
m
q
m
1
−
−
−
For a fixed
m
, we define function
f
m
(
q
)=
1
m
+
q
1
m−
1
q
m
q
m
−
−
.