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首页DTMAC:基于收集者问题的水下传感器网络延迟容忍MAC协议
DTMAC:基于收集者问题的水下传感器网络延迟容忍MAC协议
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更新于2024-08-28
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本文档主要探讨了一种适用于水下无线传感器网络(Underwater Wireless Sensor Networks, UWSN)的延迟容忍性介质访问控制协议(Delay Tolerant MAC Protocol, DTMAC)。在水下通信环境中,与陆地无线传感器网络不同,其面临的挑战主要包括长传播延迟和群体移动性。这些特性导致网络吞吐量下降、公平性受损,以及不稳定的数据传输和频道预留问题。 DTMAC的设计灵感来源于“收集者彩票”问题,提出了一种分布式数据包重复传输策略。具体来说,当一个节点需要发送数据时,该数据包会被连续发送mtimes,每个传输尝试的概率设为p。为了优化协议性能,论文首先构建了一个基于干扰模型的理论框架,通过分析成功传输概率作为调整参数,确定了数据包重复次数m和传输概率p的最佳组合,从而最大化吞吐量。 由于DTMAC不依赖于确认机制或频道预订,它能够有效应对长传播延迟的影响,避免了由于延迟造成的性能损失。此外,由于协议设计的公平性,空间公平性问题也得到了解决,这意味着所有节点都能在一定程度上均匀地获取网络资源,即使在移动性的动态环境中也能维持稳定的数据交换。 然而,尽管DTMAC具有这些优点,它可能需要更复杂的路由算法来支持大规模的水下网络,并且对于网络规模、节点能量消耗和通信效率之间的平衡也需要深入研究。未来的研究方向可能包括改进的错误检测和纠正机制,以及针对特定应用场景的优化策略,以进一步提升水下无线传感器网络的性能和可靠性。
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Sensors-11696-2015
3
neighbors is given by neighbor discovery protocol which is an
efficient algorithm in a dynamic environment [24]. In addition,
the size of slot is equal to the maximum time required for
transmitting one data packet, and only one data packet can be
transmitted in one slot. Thirdly, there is collision between two
links, because their transmissions arrive at the receiver at the
same time result in incorrectly received packets. Lastly,
DTMAC requires time synchronization among the neighboring
nodes. DTMAC only requires information exchange between
neighboring nodes, and therefore it does not need time
synchronization of the whole network. The accuracy of time
synchronization is less than a slot, which is easy to achieve
using the time synchronization technology [25].
B. Algorithm description
In DTMAC, for all the nodes there are only two kinds of
states: sending state and receiving state and a transmission
process is divided into
m
slots. When a node needs to send a
data packet, it decides to send with a fixed probability
p
or to
receive with a fixed probability
1 p
in any slot of
m
slots.
After
m
slots, if the data packet is received at least one time by
the destination node, the transmission process is successful. In
the following, we prove the above conclusion and give the
value of
p
by theorem 1 and the value of
m
by theorem 2.
Theorem 1
In any single hop network, let
p
be the probability of a
node whether to send in a slot, and
1Nn
be the number of
nodes in the network. Furthermore, let
S
be a successful
transmission of a data packet after
m
slots, and
()PS
be the
probability of
S
. If
p
is set as
1/ 1n
, then
()PS
achieve
the maximum.
Proof.
In the network, one node successfully sends a data packet
to the destination node, if only this node is in sending state and
the other
n
nodes are in receiving state at the same slot. Hence,
we have
(1 )
n
P S p p
(1)
Let we denote
()PS
as the function of variable
p
, to obtain
the maximum of
()PS
, we calculate the first order derivation
of
()PS
as
1
'
(1 ) (1 ) ( 1)
nn
P S p p p
(2)
Let
'
0PS
, then we have
1/ 1pn
when
()PS
achieve
the maximum.
Due to possible collisions in shared channels, it is difficult to
confirm whether the data packet is successfully received by
other nodes in the network. Thus sending the same data packet
repeatedly can increase the probability of successful
transmissions. We then discuss how to find the minimum of
m
.
Lemma1
In any single hop network, let
( 1)nn
be the number of node
A’s neighbors, and
S
i
be the event that node A transmits
successfully at i-th slot. Furthermore, let
()PS
i
be the
probability of
S
i
, then
1
()
4
PS
i
n
.
Proof.
If
1n
, then we have
1
11
( ) (1 )
1
11
( 1)
1 1 1
1
(1 )
14
n
n
n
PS
i
n
nn
nn
n
n n n
(3)
Therefore, we have
1
()
4
PS
i
n
(4)
Theorem 2.
In any single hop network, let
( 1)nn
be the number of node
A’s neighbors, and
m
be the number of slots in a transmission
process. Furthermore, let
S
be a successful transmission of a
data packet after
m
slots, and
()PS
be the probability of
S
.
We set the successful transmission rate as
. If a node sends a
data packet and wants to make sure
()PS
, then
ln(1 )
1
ln(1 )
4
m
n
.
Proof.
Let
S
i
be the event that node A transmits successfully at i-th
attempt, and
()PS
i
be the probability of
S
i
. We know that,
1, ,S i m
i
is independent and identically distributed. For the
simplicity,
()PS
i
is abbreviated as
P
and
()PS
is abbreviated as
Q
.
Therefore, by Inclusion-Exclusion Principle [26], we have
( ) ( ) ( )
1 , ;
1
1
( ) ( 1) ( )
1
,,;
2 3 4
2 3 4
1
m
m
Q P S P S P S S
ii
ij
i i j i j
i
m
m
P S S S P S
i j k j
j
i j k i j k
m
m m m m
m
P P P P P
m
i
(5)
According to binomial theorem, and we have
(1 ) 1 ( )
0
23
1 ( ) ( ) ( ) ( )
23
23
1 ( ) ( ) ( )
23
m
m
m m i i
PP
i
i
m m m
m
m P P P P
m
m m m
m
mP P P P
m
(6)
Since
2 3 4
2 3 4
1
m
m m m m
m
Q P P P P P
m
i
(7)
,we have:
3
2
( ) ( )
23
m m m
m
P P P mP Q
m
(8)
Substitute (8) into (6), we have:
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