Bridge Consensus: Ignoring Initial Inessentials

0.22 MB·English

by  David W. Casbeer

Bridge Consensus: Ignoring Initial Inessentials
David W. Casbeer1, Yongcan Cao1, Eloy Garcia3 and Dejan Milutinovic4
Abstract—In this paper, the problem of bridge consensus is A. Example
presented and solved. Bridge consensus consists of a network
Suppose four agents in a undirected graph, as pictured in
of nodes, some of whom are participating and others are
Fig.1.Agents1,3,and4haveinitialvaluesasdepictedinthe
non-participating. The objective is for all the agents to reach
average consensus of the participating nodes initial values in a figure.However,agent2doesnothaveaninitialopinionand,
distributedandscalablemanner.Todothis,thenodesmustuse therefore, does not have an initial value to contribute. The
the network connections of the non-participating nodes, which objective of the four agents is to come to average consensus
5
act as bridges for information and ignore the initial values
1 on the initial values of the participating agents (i.e., 1, 3,
of the non-participating nodes. The solution to this problem
0 and 4), while ignoring the non-value held initially by agent
is made by merging the ideas from estimation theory and
2 consensus theory. By considering the participating nodes has 2. Furthermore, we want the method to be scalable with

n having equal information and the non-participating nodes as the number of agents and degree of the graph. It is clear,
a havingnoinformation,thenodesinitialvaluesaretransformed that without the network connection that agent 2 provides,
J intoinformationspace.Twoconsensusfiltersareruninparallel
it would be impossible to reach a consensus. The next two
on the information state and information matrix. Conditions
2
paragraphs discuss two options that seem to be reasonable
ensuringthattheproductoftheinverseinformationmatrixand
1
the information state of each agent reaches average consensus approaches, but will not work. They are included to help

] of the participating agents’ initial values is given. better understand the difficulty of this problem.
Y
S
1 2 3 4
. I. INTRODUCTION
s
c x [0]=2 x [0]=∅ x [0]=4 x [0]=6
1 2 3 4
[ Suppose a scenario, where each node in a wireless sensor

network needs to estimate the state of the object/process
1 Fig.1. Anexamplegraphwherenodes1,3,and4areparticipatingand
v being observed. Furthermore, suppose that a number of node2isnotparticipating
3 sensors lose their sensor capability or due to limited sensor
1 capabilities they are unable to observe the object/process. For discussion’s sake, let us consider a discrete-time
7 In order for each sensor to maintain a local estimate, a implementation where each agent takes the average of its
2
distributed and scalable data fusion mechanism becomes neighbors.Supposethatagent2seeksto“initialize”itsvalue
0
. necessary. The sensors with no observation are still able withsomecombinationofitsneighborsvalues.Afterthefirst
1
0 to communicate with their neighbors and are most likely time step, the states of each agent would become: x1 = 2,
5 necessary to keep the network connected. x2 = 3, x3 = 5, x4 = 5; the average of which is 3.75.
1 This paper presents a distributed and scalable solution From this example, we see that a simple initialization of the
v: to this problem, which is aptly called bridge consensus, non-participatingnodesusingtheirneighborsvalueswillnot
work, because it biases the average towards the values held
i since the sensors without observations must act as “bridges”
X bythenon-participatingnodes’neighbors.Perhapsthereisa
to relay information, without themselves having an opin-
r way to perform a sophisticated initialization step where the
ion/informationtocontributetothedatafusionprocess.First
a
neighborsandthenon-participatingnodesadjusttheirvalues,
a simple example is shown to illustrate the difficulty of this
but at first glance a solution is not apparent.
problem. Then in Sections II-A and II-B we will briefly
Another possible avenue to attack this problem is to think
present the relevant ideas from estimation and consensus
ofthenon-participatingnodeasarelayofinformation.Here
theory that will be combined together in Section III to solve
agent 2 would store it’s neighbors values, and pass those
the bridge consensus problem. The paper is concluded with
alongtoitsneighbors.Figure2showsonepossibilescenario
a simulation to verify the results in this paper and a few
of how this could happen. At time 1, agent 2 stores the
closing statements.
informationitreceivesfromitsneighbors.Attime2,agent2
haspassedthisinformationtoitsneighbors,whilesimultane-
1D.Casbeer&Y.CaoarewithAFRL’sControlScienceCenterofExcel- ously storing the information it receives from it’s neighbors.
lence,Wright-PattersonAFB,OH,[email protected]&
[email protected] Withsuchascheme,theparticipatingagentswouldconverge
2E. Garcia is a contractor (Infoscitex Corp.) with AFRL’s tosomevaluethatisacombinationoftheirinitialvalues,and
Control Science Center of Excellence, Wright-Patterson AFB, OH, agent 2 would also know this value. At time 1, the average
[email protected]
of the participating agents is 4, which is the average of the
3D. Milutinovic is with the Department of Electrical Engineering, UC
SantaCruz,[email protected] initial conditions. However, at time 2, the average of the
Step 1 A. Maximum-likelihood Mean Estimate
1 2 3 4 Suppose that we have n independent normally distributed
random samples whose distributions are given by
x (1)=2 2→ x (1)=5 x (1)=5
1 3 4
←4 x ∼N(µ,R ), fori=1,...,n, (1)
i i
Step 2 thatis,eachrandomsamplehasthesamemean,butthevari-
ances differ. It is well known that the maximum likelihood
1 2 3 4
estimate of the mean, given these samples is given by
x (2)=3 3→ x (2)=3.5 x (2)=5
1 3 4 (cid:32) n (cid:33)−1 n
←3.5 (cid:88) (cid:88)
µˆ = R−1 R−1x . (2)
i i i
Fig.2. Examplewherethenon-participatingnode,agent2,actsasarelay. i=1 i=1
Rewriting Equation (2) in the information form yields
(cid:32) n (cid:33)−1 n
(cid:88) (cid:88)
participating agents is 3.8. The delay of information caused µˆ = Yi yi, (3)
the system to lose the average. Of course, for this simple i=1 i=1
example, if agent 2 were able to instantaneously relay its wheretheinformationmatrixandinformationstatearegiven,
neighbors’valuesthentherewouldbenoproblem.However, respectively, by Yi (cid:44) Ri−1 and yi (cid:44) Yixi. Equation (2) can
as the degree of the non-participating nodes increases or as be equivalently written as
the number of connected non-participating nodes increases. (cid:32) n (cid:33)−1 n
Theproblemofexactlyhowtorelayorroutetheinformation 1 (cid:88) 1 (cid:88)
µˆ = Y y , (4)
in a decentralized manner comes into question. n i n i
i=1 i=1
In summary, it is not clear how distributed averaging which shows that if one were able to compute the average
algorithms [5] or gossip algorithms [2], [1] can be adapted of the information state and information matrix, then the
to solve the average consensus problem with the specific maximum likelihood estimate of the mean would be easily
constraint that a subset of the nodes do not have an initial computed.
value to contribute, yet, they are necessary to maintain
B. Consensus Filter
connectivity of the communication graph. Obviously, the
applicationofgossipalgorithmsneedstorespectthephysical Consensus algorithms are decentralized methods for a
limitations of the graph. For instance, in the example shown team of agents to agree on specific consensus states. In
above,node1isneverabletoestablishdirectcommunication a consensus filter, each agent exchanges information with
with either node 3 or node 4 and the link that node 2 offers neighboring agents and not the entire team. Over time the
is essential to converge to a global common value. agents reach an agreement (or consensus) concerning the
consensus state [9], [7]. Furthermore, average consensus
occurs when the final consensus state is the average of the
initial values [8], [6].
II. BRIDGECONSENSUS
Before presenting the consensus algorithm used in this
paper, some graph theory terminology is needed. At any
Given and set of nodes, denoted by N, the bridge con- discrete-timeinstantτ,thecommunicationtopologybetween
sensus problem is for all of the nodes to reach average n agents can be described by the graph G[τ] = (N,E[τ])
consensus on the initial values of a subset of these nodes, where N ={1,...,n} is the vertex set and E[τ]⊆N ×N
in a distributed and scalable fashion. The subset of nodes is the edge set. The ijth element of the adjacency matrix
whose initial values are to contribute to the final average are A[τ] of graph G[τ] is Aij[τ] = 1 if i (cid:54)= j and the edge
denotedbytheparticipatingnodesetP ⊂N.Therestofthe (j,i) ∈ E[τ], otherwise Aij[τ] = 0. From the adjacency
nodesarecallednon-participating,andthissubsetisdenoted matrix one can construct the graph’s Laplacian matrix L[τ]
by P¯ = N \P. It is important that the non-participating as
(cid:40)
nodes share their communication links so that the entire set −A [τ] if i(cid:54)=j,
of nodes can reach a consensus on the average value of the Lij[τ]= (cid:80)nij A [τ] if i=j . (5)
participating nodes. j=1,j(cid:54)=i ij
In the consensus algorithm employed in this paper, each
To solve the bridge consensus problem, we combine the
agentinthenetworkmaintainsalocalcopyoftheconsensus
ideas from estimation theory and consensus literature. First,
state ξ ∈Rm. Each agent i updates ξ using its neighbors’
Section II-A briefly presents the maximum-likelihood mean i i
consensus states according to the rule
estimate of the of independent normally distributed random
variables.SectionII-Bfollowswiththenecessaryformalisms 1 (cid:88)n
inconsensustheory.Then,inSectionIII,thesetwoideasare ξi[τ +1]=ξi[τ]− d Aij[τ](ξi[τ]−ξj[τ]) (6)
τ
combined to solve the bridge consensus problem. j=1
where d ∈ [d ,∞) and d denotes the maximal tion (4)) yields
τ max,τ max,τ
degreeofG[τ].TheconsensusprotocolgiveninEq.(6)was
µ[∞] (cid:44) (Y [∞])−1y [∞] (13)
chosen by the need for discrete-time average consensus [6]. i i
Anicedistributedschemetochoosetheweightsdτ istouse (cid:32) 1 (cid:88)N (cid:33)−1 1 (cid:88)N
Metropolis weights [3]. = N Yi[0] N yi[0] (14)
Afterarrangingthelocalinformationstatesintothevector i=1 i=1
(cid:32) (cid:33)−1
ξ[τ]=[ξ1T[τ],...,ξnT[τ]]T, the update can be written as = (cid:88)C (cid:88)Cx [0] (15)
i
ξ[τ +1]=(Ψ[τ]⊗I)ξ[τ] (7) i∈P i∈P
1 (cid:88)
= C−1C x [0] (16)
where, |P| i
i∈P
1 1 (cid:88)
Ψ[τ]=I− L[τ], (8) = x [0] (17)
dτ |P| i
i∈P
I is the appropriate size identity matrix, and ⊗ denotes the
whichistheaverageoftheparticipatingnodes’initialvalues.
matrix Kronecker product. Note that Ψ[τ] defined in this
manner is a stochastic matrix. IV. SIMULATION
Lemma 1: [6] For a team of n agents whose states To verify the results of the paper, a simple example is
satisfy (6), ξi[τ]→ n1 (cid:80)ni=1ξi[0] as τ →∞ if presented.Inthisexample,6nodesareconnectedinagraph
1. G[τ] is balanced for every τ; as depicted in Figure 3. The white nodes, 2 and 4, are not-
2. For any τ , there exists T such that the union of the participating, while the gray nodes are participating. The
0
graphs over the time interval [τ ,τ + T] is strongly initial values at each node are given by:
0 0
connected.
x [0]=1 x [0]=9
1 4
x [0]=10 x [0]=4
2 5
III. BRIDGECONSENSUSSOLUTION
x [0]=0 x [0]=5
3 6
Consider the participating nodes in the bridge consensus
It can be seen in Figure 4 that the nodes converge to the
problem. These nodes can be thought of as being equally average of the participating nodes, namely 1(cid:80) x [0] =
important or containing the same information content, while 4 i∈P i
1(1+0+9+10)=2.5.Thenodesignoredtheinitialvalues
the non-participating nodes carry zero importance or infor- 4
of the participating nodes 2 and 4; the average if all the
mation.UsingtheinsightsfromSectionsII-AandII-B,sup-
nodes were included is 4.5.
pose that we initialize an information state and information
matrix for each agent as follows,
1
(cid:40)
0 if i∈P¯,
Y [0]= (9)
i
C if i∈P, and
3 2 6
(cid:40)
0 if i∈P¯,
y [0]= (10)
i
Y [0]x [0] if i∈P,
i i
4 5
where C is any positive-definite matrix.
Letthenetworkofagentsimplementtwoconsensusfilters Fig. 3. In this example, the nodes desire to reach average consensus
from the participating (gray) nodes (1, 3, 6,and 5) and where the white
according to Eq. (6), one for the information matrix (9)
nodes2and4arenot-participating.Withoutnode2and4’scommunication
and one for the information state (10). Assuming that the channels, average consensus is impossible, since node 3 would have no
conditions for average consensus are met (i.e., Lemma 1), incomingedges.
then the agent’s local values will converge asymptotically
towards
V. CONCLUSION
N
1 (cid:88) In this paper, the problem of bridge consensus was pre-
Y [∞] = Y [0], ∀i∈N, (11)
i N i sented and solved. Bridge consensus consists of a network
i=1 of nodes, some of whom are participating and others are
N
1 (cid:88) not-participating. The objective is for all the agents to reach
y [∞] = y [0], ∀i∈N, (12)
i N i averageconsensusoftheparticipatingnodesinitialvaluesin
i=1
a distributed and scalable manner. To solve this problem,
Dropping the node subscript, and using this converged value ideas from estimation theory and consensus theory were
to compute the maximum-likelihood mean estimate (Equa- merged. By considering the participating nodes has having
10
8
e
u
al 6 Average of all nodes
V
nt
e 4
r
r
u
C
2
Participating nodes’ average
0
0 2 4 6 8 10
Iteration
Fig. 4. Six agents running bridge consensus. Participating agents’ initial
values are 0, 1, 4, and 5. Non-participating agents’ initial values are 9
and10.Theagentconvergetoavalueof2.5,whichistheaverageofthe
participatingagents’initialvalues.
equal information and the non-participating nodes as having
no information, the nodes initial values are transformed into
information space. Two consensus filters are run in parallel
on the information state and information matrix. Conditions
ensuring that the product of the inverse information matrix
and the information state of each agent reaches average
consensus of the participating agents’ initial values was
given.
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