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Data Transmission and Base Station Placement for Optimizing Network Lifetime.
E. Arkin, V. Polishchuk, A. Efrat, S. Ramasubramanian,J. Taheri, J. Mitchell, S. Sankararaman
1
• Motivation.• H & K algorithm for finding a Maximum
Matching.• Basic principles.• Computing an optimal forwarding protocol.• Optimizing the location of the base station.
2
Motivation:
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• Motivation.• Hopcroft & Karp algorithm for finding a
Maximum Matching.• Basic principles.• Computing an optimal forwarding protocol.• Optimizing the location of the base station.
4
Preliminaries
• A matching in a bipartite graph is a subset of edges such that every vertex is incident to at most one edge in the subset.
),( EUVG EM
5
Maximum matching
• A maximum cardinality matching in G can be found in time.)( nEO
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An augmenting path
Let M be a Matching, P is an augmenting path of M if:• P is an odd length path.• P starts and ends with an unmatched vertex.• Every other edge is not in M.
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Hopcroft & Karp algorithm U
V
8back
H & K First BFS stageU
V
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- A subset of unmatched vertices U
VVv '
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H & K First DFS stageU
V
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Second iteration stageU
V
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Second iteration stageU
V
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H & K Runtime
• Lemma 1: – The length of the shortest augmenting path
increases in each phase.
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H & K Runtime
• Lemma 2: – Let M* be a maximum matching.– Let M be a matching.– If M* have K edges more than M, there are K
vertex disjoint augmenting paths for M.
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Lemma 2 – cont.
– In the union of M* and M we can have:1. even length circles.2. even length paths.3. odd length paths.
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Even length circle and path.U
V
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An odd length pathU
V
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An odd length pathU
V
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Lemma 2- cont.
• We have to consider only the third group which are augmenting paths. – Each of the other groups doesn’t change the balance
between M and M* edges.• All these paths are vertex disjoint and contains
one more M*’s edge than M’s. – The degree of each vertex in the union of M and M* is
at most 2.– There is no augmenting path for M*.
• The number of such paths is k.20
The Runtime of H & K
• After phases:– By lemma 1- all the paths of length less than are already discovered.– By lemma 2- we have more vertex
disjoint augmenting paths.– Therefore, we get – In each step the remaining number of augmenting
paths is decreased by 1.– After at most phases we will get M*.
n
12 n
2/n
MMk *
)12/( nnk 2/n
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The Runtime of H & K
• Hence, there are at most phases.• Each phase takes . • In total we get .
)( nO
)( EO (BFS+DFS)
)( EnO
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-Matching
• Let be a vector of integers with a component, , associated with each vertex.
• A - matching is a subset of edge such that each vertex is incident to at most edges of .
b
b VU
b EM
Mb(v)
b(v)
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Extension of the H & K algorithm
• We extend the H & K algorithm for finding a maximum cardinality - matching.
• A vertex is “exposed” if it’s degree in M is less than .
• An “exposed” vertex is equivalent to unmatched vertex in the original H & K.
• The extended H & K takes also .
b
b(v)
)( EnO
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Maintaining a maximum - matching
• Let M be a maximum matching.• is increased by 1.• We can update M efficiently to get the
maximum - matching for the modified . • If an augmenting path exists, it must start
from v.• Therefore one BFS+DFS phase is enough.
b
bb
bb(v)
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Maintaining a maximum - matching
• Let M be a maximum matching.• is decreased by 1.• If M is still Maximum matching, we done.• Otherwise remove any edge (v,u) from M.• If an augmenting path exists, it must start
from u.• Therefore one BFS+DFS phase is enough.
b
bb(v)
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• Motivation.• H & K algorithm for finding a Maximum
Matching.• Basic principles.• Computing an optimal forwarding protocol.• Optimizing the location of the base station.
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•
Multi hop network topology
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Price-part 1
•
• Which is more expensive?
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•
• Which is more expensive?
Price-part 2
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More realistic model
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Simultaneously
• If we want that the cameras will be able to communicate they have to be awake in same time.
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Lifetime
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Definitions of lifetime
• The lifetime:– Starts from:
• The system starts operating.
– Until:• One sensor is dead.• Some percentage of the sensors is dead.• All sensors in some specific subset are dead.
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Definitions
• : a set of n points in the plane, each point represents a sensor.
• : a point in the plane,represents the base station.
2RS
2Rg
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The sensors
• The sensors are responsible for monitoring and gathering information.
• Each sensor can pass it’s information directly to the base station or vs. another sensor.
• The sensors can’t split their information.
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The base station
• The base station is responsible for processing the data gathered from the sensors.
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Energy
• Each sensor has a battery with limited capacity.
• For simplicity we assume that the capacity is 1.
• No cost for receiving.• For sending:
– Transmitting data a distance D requires- energy when . 42
D
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The followers
• Followers are sensors that passing their information vs. other sensors.
• The energy that required for such a sensor to pass it’s information defined by when is the Euclidian distance between the follower to the receiving sensor.
SF
fl fl
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The leaders
• Passes information directly to the base station.
• Can pass also information from other sensors.• Required energy:
– is the distance between to the base station . – is the degree.
SL
lg)(ld
lg l
)(ld
g
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The 2-tree
• g is the root.• Leaders are on the first level.• Followers are on the second.• The links from followers to leaders and from
leaders to the base station are the edges.
g
Level 1:
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• Motivation.• H & K algorithm for finding a Maximum
Matching.• Basic principles.• Computing an optimal forwarding protocol.• Optimizing the location of the base station.
42
Optimal forwarding protocol
• The problem: Find such that and
)max( FL 1lg)(, ldLl
1)(, ffLFf
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Splitting to followers and leaders
• Case 1: – S can’t transmit to g.– S have to be a follower.
• Case 2: – S can transmit directly to g.– S have to be a leader.
1sg
1sg
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Capacity
• Capacity of a leader- l:
• Capacity of a follower- f:
1.- }|lg|1/min{n, (l) b
1 (f) b
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The graph
• We define:
• We have an edge between and iff .
),( ELFG LlFf
1fl
46
Crucial observation
• Finding the optimal protocol is equivalent to computing maximum cardinality matching in G.
• A reminder.
b
47
Running time
• According to Hopcroft-Karp – • By exploiting the geometry of the problem -
)()( 5.2nOnEO
))log(( 5.1 nnO
48
An improved analysis
• We can reduce the time of the BFS to .• As before BFS+DFS phase invoked times.
– DFS takes time.– Improved BFS takes time.
• In total we have time.
))log(( nnO
)( nO)(nO
))log(( nnO
))log(( 5.1 nnO
49
The improved BFS
• From followers to leaders:– Start from all unmatched followers.– Find all leaders that these followers can transmit
to.– The method of finding the leaders contains:
• Defining a unit disk for each leader.• For each unmatched follower searching for the legal
leaders.
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The improved BFS
• For each unmatched follower, f:– Check whether the disk centered at one of the
leaders contains f.– If there is such disk:
• Add the corresponding edge to the matching and remove it’s disk.
• Go back to the first line.
– Otherwise, continue to the next unmatched follower.
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The improved BFS
• From leaders to followers :– Follow the matched edges from the reached
vertices of L.
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The improved BFS- Complexity
• Discovering and deleting each disk takes time (Efrat et al).• We can discover up to n disks, and each
discovered disk can be deleted only once.• Therefore, the total complexity of “BFS+DFS”
level is .
))(log(nO
))log(( nnO
53
The continues model
• Each sensor continuously transmitting data. • We will describe a subroutine for the decision
problem.• We use that subroutine for solving the
optimization problem.
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The decision problem-definitions
• We have:– t, the checked lifetime.– Location of g.
• If a sensor sends the data of k other sensors, it is spending per unit time.
sgk )1(
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The decision problem-constraints
• The problem: for the given check whether and .• If yes, there exists a protocol in which each
sensor continuously transmits data to g at a fixed rate.
• Since t is a constant the problem is analogues to the previous one and therefore can be solved in time.
1lg)(, ldtLl 1)(,
ffLtFf
t
))log(( 5.1 nnO
56
• Motivation.• H & K algorithm for finding a Maximum
Matching.• Basic principles.• Computing an optimal forwarding protocol.• Optimizing the location of the base station.
58
Optimal location of the base station
• The problem:– Finding a location g for the base station and the
tree T, maximizing the number of sensors that can transmit their data to g.
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Optimal location of the base station
• Another version: Finding a location g for the base station and the tree T, maximizing the network lifetime.
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The first version solution
• Observation:It is enough to consider only a polynomial number
of candidate locations.
We consider circles of Radii
centered at each sensor of S.The circle is the boundary of the region to which the sensor can transmit its data and the data from i-1 additional sensors.
/1/1/1 )1(,...,2,1 n
61
The first version solution – cont.
• g can be a vertex of the arrangement of the circles.
• If not, one can move g while not intersecting any circle without changing the transmission ability.
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The first version solution – cont.
• There are vertices in the arrangement.• For each vertex, we invoke the algorithm for
the Optimal forwarding protocol problem.• We find the vertex ,v*, maximizing the
number of sensors that can transmit their data to g.
• v* can be found in .))log(( 5.5 nnO
)( 4nO
63
An improved analysis
• One solution relays on the Previous one.• Start for an arbitrary vertex ,p.• Compute the optimal protocol Tp.(g=p)• Move g to an adjacent vertex ,q.
S1S2
S3S4
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An improved analysis – cont.
• As g is moved from p to q, the only changes in the graph concern to at most 2 sensors.
• They change their capacity by 1.• Thus, the optimal protocol can be obtained
from by one BFS+DFS phase, which takes .
),( EFLG
))log(( nnO
S1S2
S3S4
qT
pT
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The improved running time
• We can go over all the vertices of the arrangement by moving between adjacent vertices.
• Therefore, v* can be found in .))log(( 5 nnO
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The second version solution
• Given a lifetime t, we can decide in if there exists a g satisfying the problem conditions.• We assume that t* is known and refer to this
problem as a decision problem.– we can determine if a given t is lager, smaller or equal to t*.
))log(( 5 nnO
67
A solution using Parametric Search
• We increase t from 0 until it exceeds t*, while keeping track of the vertices of the arrangement using a parallel sorting networks.
• Using this technic, v* can be found in . ))(log( 25 nnO
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Approximation algorithm
• For maximization problems p < 1
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-approximation
• Let s’ be a sensor farthest from g.• We can set t to be so all the sensors
transmit their data directly to g.• Either s’ or its leader has to transmit to
distance at least . • Hence, .• Therefore, we get a approximation.
)2/1(
gs'/1
2/'gs
)2/1(
)2/'(* gst
70
-approximation)1(
a b
[a,b]- an interval for t.
a/b approximation
ab
approximationba /
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-approximation)1(
a b
[a,b]- an interval for t.ab
• We use a binary search.• We start from an - approximation• If we take enough steps of the binary
search we can get the needed approximation.
)2/1(
72
-approximation
• In we can get an - approximation for the maximum lifetime.
)1(
))/1log()log(( 5 nnO
)1(
73
Thank you
74
