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1
Maximizing Lifetime of Sensor Surveillance Systems
IEEE/ACM TRANSACTIONS ON NETWORKING
Authors: Hai Liu, Xiaohua Jia, Peng-Jun Wan, Chih-Wei Yi, S. Kami Makki, and Niki Pissinou
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Outline
• Introductions
• System Model and Problem Statement
• Our Solutions– Find Maximal Lifetime– Decompose Workload Matrix– Determine Surveillance Tree
• Experiments and Simulations
• Conclusion
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Introductions
• Given a set of targets, a set of sensors (at most watch one target at a time) are used to watch the targets and collect sensed data to the BS.
• Lifetime– Duration until one target can no longer be watched by
any sensor or data can’t be forward to the BS.
• Problems– Schedule a subnet of sensors– Find the routes for the active sensor to send data
back to BS
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Introductions
BS
sensor
target
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System Model and Problem Statement
• B base station• S set of sensors, n = • T set of targets, m = • S(j) set of sensors that can watch target j• T(i) set of targets that are within the surveillance range
of sensor i• N(i) set of neighbors of sensor i• initial energy of sensor i• distance between sensor i and j• energy for transmitting and receiving one unit data• energy for watching a target per unit time• R data rate generated from sensors while watching t
argets
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System Model and Problem Statement
• S(1) = {S1, S2, S3}
• T(1) = {T1, T2, T3}
s1 s2
s3T1
s1T1 T2
T3
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System Model and Problem Statement
• Two requirements for sensors watching targets– Each sensor can watch at most one target at a
time– Each target should be watched by one sensor
at any time
• The problem is to find a schedule that meets the above two requirements for sensors watching targets, such that the lifetime of network is maximized.
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Our SolutionsFind Maximal Lifetime
• Linear Programming (LP)– total time sensor i watching target j– amount of data transmitted from sensor i to sensor j
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Our SolutionsFind Maximal Lifetime
• We call matrix the workload matrix– The sum of column is equal to L (each column)– The sum of row is less than or equal to L (each
row)
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Our SolutionsDecompose Workload Matrix
• are schedule matrices – Elements are either “0” or – Each column has exactly one non-zero element– Each row has at most one non-zero element
• The number of sensors is grater than or equal the number of targets ( n >= m)– n = m– n > m
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Our SolutionsSpecial Case n = m
• and denotes the sum of row I and the sum of column j in workload matrix
∵ and ∴
=>
=>
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Our SolutionsSpecial Case n = m
• Bipartite graph– Left hand side : sensors– Right hand side : targets– Edges :
• Since n = m, every sensor has a target to watch in each session
• Find perfect matchings…
S1
S2
Sn
T1
T2
Tm
……
….
……
….
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Our SolutionsSpecial Case n = m
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Our SolutionsSpecial Case n = m
2
1
0
1
1
1
0
1
2
0
1
0
1
0
0
0
0
1
2
0
0
0
1
1
0
1
1
s1
s1
s1
s2
s2
s2
s3
s3
s3
t1
t1
t1
t2
t2
t2
t3
t3
t3
1
2
21
111
12
11
1
111
G
P1
G
……
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Our SolutionsSpecial Case n = m
• Lema1: For of nonnegative real numbers, if = for 1 i,j n, =>exist perfect matching
• Proof: – – There doesn’t exist n positive entries in that no two
entries in the same column or row.– By Konig theorem, we can cover all the positive entries in
the matrix with e rows and f columns, such that e+f<n– The sum of all lines of is equal to 1, n e+f<n, =><
=
(L is sum of all elements in a row)
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Our SolutionsSpecial Case n = m
• Theorem 1: The DecomposeMatrix-nn algorithm can always find a perfect matching– From lemma 1
• Theorem 2: The time complexity of DecomposeMatrix-nn algorithm is O( ), where is the number of non-zero elements in– At most number of rounds to remove all edge
s in G– Find a perfect matching is O( )
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Our SolutionsSpecial Case n > m
• Let be the dummy matrix
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Our SolutionsSpecial Case n > m
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Our SolutionsSpecial Case n > m
=3, =5
=3, =2
=5, =3
=3, =2
= =3
=3
Let and record the sum of remaining undetermined elements of row i and column j
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Our SolutionsSpecial Case n > m
• Theorem 3: the FillMatrix Algorithm can compute the filled matrix– Given row sums and column sums of a matrix– Induction method– 1. n=1, m=1, since , we have
– 2. when n p-1, m q-1, can compute – 3. when n=p, m=q, we first compare with
• A. = , => , according 2.• B. > , => , monotonously decreases after ea
ch round and , there must exist in round l , => , according 2.
• C similar to B.
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Our SolutionsSpecial Case n > m
• Let denotes the matrix contains the first m columns in
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Our SolutionsSpecial Case n > m
• Theorem 4: The time complexity of FillMatrix Algorithm is O( )
• Theorem 5: The time complexity of DecomposeMatrix algorithm is O( )
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Our SolutionsDetermine Surveillance Tree
• Root : BS
• Leaf nodes : active sensors
• Suppose sensor i has l downstream nodes (i.e. are non-zero), let sensor i forward its outgoing data first to until is saturated, then switch to until the value of is met, and finally forward the last flow to .
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Experiments and SimulationsNumeric Example
• 10x10 region• Surveillance range: 0.4 *
10• Maximal transmission
range: 0.8*10• Initial energy is randomly
generated form [0, 100] with mean at 50
• =0.12, =0.1• =0.1 R=1
• α=2
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Experiments and SimulationsNumeric Example
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Experiments and SimulationsNumeric Example
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Experiments and SimulationsSimulations
• 1) Linear Growth of Decomposition Steps–
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Experiments and SimulationsSimulations
• 2) Comparison with Greedy Method:– Use the maximum matching algorithm in the
sensor-target bipartite graph to find the pairs of sensor and target
– For each active sensor we find the minimal energy cost path from it to the BS.
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Experiments and SimulationsSimulations
N=100, M=10
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Experiments and SimulationsSimulations
• The number of steps for decomposing the workload matrix is linear to the size of the system
• Our algorithm has better performance when– Large surveillance range– Large transmission range– Sensors are density deployed
• The increase of surveillance range is more effective than the increase of the maximal transmission range
M=50
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Conclusion
• We have presented the maximal lifetime scheduling problem in sensor surveillance systems.
• This is the first time in the literature that the problem of maximizing lifetime of sensor surveillance systems was formulated and the optimal solution was obtained.