Sensor Location Problem - UNECE · 2013. 4. 30. · 1.1. Sensor Location Problem for graphs 5 If the set M includes the nodes from the set I∗, then we also know the values xi =

Sensor Location Problem

Minsk2013

3

CONTENTS

1 SENSOR LOCATION PROBLEM . . . . . . . . . . . 4

1.1 Sensor Location Problem for graphs . . . . . . . . . 41.2 Examples of Sensor Location Problem for graphs . . 6

1.2.1 Example 1 (Underdetermined system) . . . . . . . 61.2.2 Example 2 (Unique solution) . . . . . . . . . . . . 11

1. SENSOR LOCATION PROBLEM

One of applications of the given approach for solution of systems oflinear algebraic equations in the form (??) – (??) is the Sensor LocationProblem (SLP). That is the problem of location of the minimum numberof sensors in the nodes of the graph, in order to determine the arc flowsand node intensities. Let’s apply the decomposition algorithms for systemsof linear algebraic equations of type (??) – (??) to the Sensor LocationProblem for the graph

1.1. Sensor Location Problem for graphs

Let’s introduce the finite connected directed graph G = (I, U). The setU is defined on I × I (|I| < ∞,|U | < ∞). We assume, that the graph G

is symmetric: that is: if (i,j) ∈ U , then (j,i) ∈ U. We note that the graphG is not undirected: the flow on arc (i,j), in general, will not be the sameas the flow on arc (j,i). To designate this distinction, we refer to the graphG = (I, U) as a two way directed graph.

We represent the traffic flow by a network flow function x : U → R thatsatisfies the following system:X

j∈I+

i(U)

xi,j −X

j∈I−i

(U)

xj,i =

8<:xi, i ∈ I∗,

0, i ∈ I \ I∗(1.1.1)

where I∗ is the set of nodes with variable intensities, xi is the variableintensity of node i ∈ I∗, I+

i (U) = {j ∈ I : (i,j) ∈ U}, I−i (U) = {j ∈I : (j,i) ∈ U}. If the variable intensity xi of node i is positive, the node i isa source; if it is negative, this node i is a sink. For system (1.1.1) is true thefollowing condition:

Xi∈I

xi = 0. According to Theorem ?? if I∗ 6= ∅, then

the rank of the matrix of system (1.1.1) for a connected graph G = (I, U)is equal to |I|.

In order to obtain information about the network flow function x andvariables xi of nodes i ∈ I∗, sensors are placed at the nodes of the graphG = (I, U). The nodes in the graph G = (I, U) with sensors we callmonitored ones and denote the set of monitored nodes by M, M ⊆ I . Weassume that if a node i is monitored, we know the values of flows on alloutgoing and all incoming arcs for the node i ∈ M :

xij = fij, j ∈ I+i (U), xji = fji, j ∈ I−i (U), i ∈ M.

1.1. Sensor Location Problem for graphs 5

If the set M includes the nodes from the set I∗, then we also know thevalues xi = fi, i ∈ M

TI∗. So, we have

xij = fij, j ∈ I+i (U), xji = fji, j ∈ I−i (U), i ∈ M ;

xi = fi, i ∈ MT

I∗.(1.1.2)

Consider any node i of the network. For every outgoing arc (i,j) ∈ U

for this node i determine a real number pij ∈ (0,1] which denotes the partof the total outgoing flow

Xj∈I+

i(U)

xij from node i corresponding to the arc

(i,j). That is,xij = pij

Xj∈I+

i(U)

xij.

If |I+i (U)| > 2 for the node i ∈ I then we can write the arc flow along all

outgoing arcs from node i (except any selected arc) in terms of arc flow asingle outgoing arc, for example, (i,vi), vi ∈ I+

i (U):

xi,j =pi,j

pi,vi

xi,vi, j ∈ I+

i (U) \ vi. (1.1.3)

We continue this process for each node i ∈ I, if |I+i (U)| > 2.

Let |I+i (U)| > 2 for any node i ∈ I and xi,vi

is known for the arc (i,vi) andequal to fi,vi

. Then we can write the unknown arc flow along all outgoingarcs from node i (except any selected arc (i,vi)) in terms of arc flow for asingle outgoing arc (i,vi), where xi,vi

is known and equal to fi,vi.

So we shall formulate the Sensor Location Problem: what is the min�

imum number of monitored nodes |M | such that system (1.1.1) hasan unique solution?

Let’s substitute the calculated arc flows according to (1.1.2) and (1.1.3)in the equations of system (1.1.1). Let’s delete from graph G = (I, U) theset of the arcs on which the arc flow are known. Let’s delete from graphG the set of the nodes i ∈ M . Then we have a new graph G = (I, U).A new set of nodes with variable intensity for a new graph G is I

∗, where

I∗

= I∗ \ (MT

I∗). The new graph G can be non-connected. The graphG consists of connected components. Some connected components maycontain no nodes of the set I

∗. The system (1.1.1) for graph G = (I, U)

will be the following one:Xj∈I+

i(U)

xi,j −X

j∈I−i

(U)

xj,i =

8<:xi + bi, i ∈ I∗,

ai, i ∈ I \ I∗ (1.1.4)

6 1. SENSOR LOCATION PROBLEMX(i,j)∈U

lpijxij = 0, p = 1, q, (1.1.5)

where ai, bi, lpij – are constants.

Let’s state the steps of the algorithm for modeling the new graph G forthe given set M and the formation of set I \ M∗.

Step 1. Construct a cut CC(M) [2] with respect to the set of monitorednodes M .

Step 2. Find the nodes of the set I(CC(M)).Step 3. Construct the set M+ = I(CC(M)) \ M.

Step 4. Form sets M∗ = MS

M+ and I \ M∗.

The unknowns of the system (1.1.4) – (1.1.5) are the flows for outgoingarcs from the nodes of the set I \ M∗. Also the unknowns in the system(1.1.4) – (1.1.5) are the variable intensities xi, where i ∈ I

∗for the new

graph G = (I, U). The system (1.1.4) – (1.1.5) is a particular case of thesystem (??) – (??) if a connected component includes at least one nodefrom the set I

∗. The system (1.1.4) – (1.1.5) has an unique solution for

the given set M if and only if the rank of the matrix of system (1.1.4) –

(1.1.5) is equal to the number of unknowns of the system (1.1.4) – (1.1.5).According to Theorem ??, Theorem ?? computes the rank of the matrix ofsystem (1.1.4) for the given connected component for graph G = (I, U) ifthe connected component contains nodes of the set I

∗. We use the theory

of decomposition [!!!!!] if the connected components of the graph G don’tinclude the nodes of the set I

∗. For computing the rank of the matrix of

system (1.1.4) – (1.1.5) we use graph theoretical properties of the supportaccording to Theorem ??.

1.2. Examples of Sensor Location Problem for graphs

Consider the examples to the Sensor Location Problem in the graph G

for the case when the set M of monitored nodes contains a single node:|M | = 1.

1.2.1. Example 1 (Underdetermined system)

In Figure 1.1 we represent a finite connected directed symmetric graphG = (I, U), where

I = {1, 2, 3, 4, 5, 6},

U = {(1,2),(1,3),(2,1),(2,4),(2,6),(3,1),(3,5),(4,2),(4,6),(4,5),

1.2. Examples of Sensor Location Problem for graphs 7

(5,3),(5,4),(5,6),(6,2),(6,4),(6,5)},

I∗ = {2, 4, 5, 6}.

For the graph G = (I, U) (see Figure 1.1) we write the system of linearalgebraic equations of type (1.1.1).

x1,2 + x1,3 − x2,1 − x3,1 = 0

x2,4 + x2,6 + x2,1 − x4,2 − x1,2 − x6,2 = x2

x3,1 + x3,5 − x1,3 − x5,3 = 0

x4,2 + x4,5 + x4,6 − x2,4 − x5,4 − x6,4 = x4

x5,4 + x5,3 + x5,6 − x3,5 − x4,5 − x6,5 = x5

x6,2 + x6,4 + x6,5 − x2,6 − x4,6 − x5,6 = x6

(1.2.1)

12

3

4 5

6

Figure 1.1. Finite connected directed symmetric graph G

Suppose that the set M of monitored nodes is M = {1} for the graphshown in Figure 1.1.

We form the sets:

M+ = I(CC(M)) \ M = {2,3};

M∗ = M[

M+ = {1}[{2,3} = {1,2,3};

8 1. SENSOR LOCATION PROBLEM

M∗ = {1,2,3}, I \ M∗ = {4,5,6}.

2

3

4 5

6

Figure 1.2. Graph G′

In the Sensor Location Problem the values of flows on all incoming andoutgoing arcs for the each node i of the set M (monitored nodes) are knownand also we known the values xi = fi, i ∈ M

TI∗ :

x1,2 = f1,2, x2,1 = f2,1,

x1,3 = f1,3, x3,1 = f3,1.

(1.2.2)

We substitute the known values of the variables (1.2.2) to the system ofequations (1.2.1) and delete the corresponding arcs from the graph G. Wealso delete the nodes i ∈ M from the graph G. The graph G′ obtained aftersubstitution (1.2.2) is shown in Figure 1.2.

The rest of the flows for the outgoing arcs from the nodes of the setM+ = I(CC(M)) \ M = {2,3}, can be expressed from the flows of theoutgoing arcs for M+ by the following equations:

x2,4 =p2,4

p21f2,1; x2,6 =

p2,6

p21f2,1; x3,5 =

p3,5

p3,1f3,1. (1.2.3)

We substitute (1.2.3) into the system (1.2.1). We obtain the graph G =(I, U), by deleting from the graph G′ the arcs corresponding to variables(1.2.3). The graph G is shown in Figure 1.3.


2

3

4 5

6

Figure 1.3. Graph G

The system (1.2.1) for the graph G (see Fig.1.3) after substitution ofknown arc flows (1.2.2) and (1.2.3) transforms to the form (1.2.4):

f1,2 + f1,3 − f2,1 − f3,1 = 0

p2,4

p2,1f21 +

p2,6

p2,1f2,1 + f2,1 − x4,2 − f1,2 − x6,2 = x2

f3,1 +p3,5

p3,1f3,1 − f1,3 − x5,3 = 0

x4,2 + x4,5 + x4,6 −p2,4

p2,1f2,1 − x5,4 − x6,4 = x4

x5,4 + x5,3 + x5,6 −p3,5

p3,1f3,1 − x4,5 − x6,5 = x5

x6,2 + x6,4 + x6,5 −p2,6

p2,1f2,1 − x4,6 − x5,6 = x6

(1.2.4)

We form additional equations of type (??) for the system (1.2.4) for thegraph G (see Figure 1.3), where lp

i = 0, i ∈ I∗, p = 1,q. Part of the

unknowns of the system (1.2.4) constitute the arc flows for arcs, outgoingfrom nodes of the set I \ M∗ = {4,5,6}. For these arc flows we perform thefollowing steps.

• Choose arbitrary outgoing arc that starts from a node set I \M∗ ={4,5,6}, for example, the arc (4,5) is outgoing arc from node 4. Let us


express the arc flows to all other arcs outgoing from the node i = 4through the arc flow of x4,5 of selected arc (4,5).

x4,2 =p4,2

p4,5x4,5, x4,6 =

p4,6

p4,5x4,5.

• Choose any outgoing arc from the node i = 5, for example, (5,6).Let us express the arc flows to all other outgoing arcs from node i = 5through the arc flow of x5,6 outgoing arc (5,6).

x5,3 =p5,3

p5,6x5,6, x5,4 =

p5,4

p56x5,6.

• Choose any outgoing arc from the node i = 6, for example, (6,2).Let us express the arc flows to all other outgoing arcs from node i = 6through the arc flow of x6,2.

x64 =p64

p62x62,

x65 =p65

p62x62.

So, we have:

x4,2 =p4,2

p4,5x4,5, x4,6 =

p4,6

p4,5x4,5,

x5,3 =p5,3

p5,6x5,6, x5,4 =

p5,4

p5,6x5,6,

x6,4 =p6,4

p6,2x6,2, x6,5 =

p6,5

p6,2x6,2.

(1.2.5)

In this example, the additional equations of type (??) for the graph G areas follows:

x4,2 −p4,2

p4,5x4,5 = 0, x4,6 −

p4,6

p4,5x4,5 = 0,

x5,3 −p5,3

p5,6x5,6 = 0, x5,4 −

p5,4

p5,6x5,6 = 0,

x6,4 −p6,4

p6,2x6,2 = 0, x6,5 −

p6,5

p6,2x6,2 = 0.

(1.2.6)


We define the number of unknowns of the system (1.2.4), (1.2.6). Partof the unknowns of the system (1.2.4), (1.2.6) makes up outgoing arc flowsfor arcs outgoing from the set of nodes I \ M∗ = {4,5,6}:

x4,2, x4,5, x4,6, x5,3, x5,4, x5,6, x6,2, x6,4, x6,5.

The remaining part of the unknowns of the system (1.2.4), (1.2.6) de�

fines the values of the variables xi, i ∈ I∗, where I

∗= {2,4,5,6}:

x2, x4, x5, x6.

Thus, the number of unknowns of the system (1.2.4), (1.2.6) is equal to13. Number of equations in the system (1.2.4), (1.2.6) is equal to 11. Hencethe rank of the matrix of system (1.2.4), (1.2.6) is not greater than thenumber 11. System (1.2.4), (1.2.6) is a underdetermined system of linearalgebraic equations. System (1.2.4), (1.2.6) doesn’t have unique solutionfor given set of monitored nodes M = {1}. Consequently the system (1.2.1)for the set of monitored nodes M = {1} also has no unique solution. So,when we locate single sensor into the node 1, we cannot get unique solutionfor system (1.2.1).

1.2.2. Example 2 (Unique solution)

In Figure 1.4 represented by a finite connected directed symmetric graphG = (I, U), where

I∗ = {2,3,6,7},

I = {1,2,3,4,5,6,7,8},

U = {(1,2),(1,5),(2,1),(2,6),(3,4),(3,6),(4,3),(4,7),(4,8),(5,1),

(6,2),(6,3),(6,7),(7,4),(7,6),(8,4)}.


12

3

4

5 6

7

8

Figure 1.4. Finite connected directed symmetric graph G

For the graph G (see Figure 1.4) we consider the system of linear alge�

braic equations of the type (??):

x1,2 + x1,5 − x2,1 − x5,1 = 0x2,1 + x2,6 − x1,2 − x6,2 = x2

x3,4 + x3,6 − x4,3 − x6,3 = x3

x4,3 + x4,7 + x4,8 − x3,4 − x7,4 − x8,4 = 0x5,1 − x1,5 = 0

x6,2 + x6,3 + x6,7 − x2,6 − x3,6 − x7,6 = x6

x7,4 + x7,6 − x4,7 − x6,7 = x7

x8,4 − x4,8 = 0

(1.2.7)

Suppose that the set of monitored nodes is M = {6} for the graph G

shown in Figure 1.4. We denote a cut of the graph G with respect to agiven set M of nodes with CC(M). We construct the cut CC(M) for thegraph G respect to the set M = {6}.

We form the sets:

M+ = I(CC(M)) \ M = {2,3,7};

M∗ = M ∪ M+ = {6} ∪ {2,3,7} = {2,3,6,7};

I \ M∗ = {1,4,5,8}.


The values of arc flows on all incoming and outgoing arcs for the eachnode i of the set M = {6} (monitored node) are known:

x6,2 = f6,2, x6,3 = f6,3,

x6,7 = f6,7, x2,6 = f2,6,

x3,6 = f3,6, x7,6 = f7,6,

and also we know the values xi = fi, i ∈ MT

I∗ :

x6 = f6.

We have:x6,2 = f6,2, x6,3 = f6,3, x6,7 = f6,7,

x2,6 = f2,6, x3,6 = f3,6, x7,6 = f7,6,

x6 = f6.

(1.2.8)

Let us substitute (1.2.8) to the system of linear equations (1.2.7). Let’sdelete from graph G = (I, U) the set of the arcs on which the arc flow areknown according to (1.2.8). Also, we delete from graph G the set of thenodes i ∈ M , M = {6}. We obtain a graph G′ which is shown in Figure1.5.

The rest of the arcs flow for the outgoing arcs from the nodes of the setM+ = I(CC(M)) \ M = {2,3,7}, can be expressed from the arcs flow of


12

3

4

5

7

8

Figure 1.5. Graph G′

the outgoing arcs for M+ by the following equalities:

x2,1 =p2,1

p2,6x2,6,

x2,6 = f2,6;

x3,4 =p3,4

p3,6x3,6,

x3,6 = f3,6;

x7,4 =p7,4

p7,6x7,6,

x7,6 = f7,6.

(1.2.9)

In Figure 1.6 represented graph G = (I, U) obtained by deleting the setof the arcs on which the arc flow are known according to (1.2.9) from graphG′ (see Figure 1.5). The system (1.2.7) for the graph G (see. Figure 1.6)transforms to the form (1.2.10).


12

3

4

5

7

8

Figure 1.6. Graph G.

x1,2 + x1,5 −p2,1

p2,6f2,6 − x5,1 = 0

p2,1

p2,6f2,6 + f2,6 − x1,2 − f6,2 = x2

p3,4

p3,6f3,6 + f3,6 − x4,3 − f6,3 = x3

x4,3 + x4,7 + x4,8 −p3,4

p3,6f3,6 −

p7,4

p7,6f7,6 − x8,4 = 0

x5,1 − x1,5 = 0

f6,2 + f6,3 + f6,7 − f2,6 − f3,6 − f7,6 = f6

p7,4

p7,6f7,6 + f7,6 − x4,7 − f6,7 = x7

x8,4 − x4,8 = 0

(1.2.10)


Represent the system (1.2.10) in the form (1.2.11).

x1,2 + x1,5 − x5,1 = b1

−x1,2 = x2 + b2

−x4,3 = x3 + b3

x4,3 + x4,7 + x4,8 − x8,4 = b4

x5,1 − x1,5 = 0−x4,7 = x7 + b7

x8,4 − x4,8 = 0

(1.2.11)

where

b1 =p2,1

p2,6f2,6, b2 = −

p2,1

p2,6f2,6 − f2,6 + f6,2,

b3 = −p3,4

p3,6f3,6 − f3,6 + f6,3, b4 =

p3,4

p3,6f36 +

p7,4

p7,6f7,6

b5 = 0, b7 = −p7,4

p7,6f7,6 − f7,6 + f6,7, b8 = 0.

Arc flows xi,j,(i,j) ∈ U corresponding to the arcs outgoing from nodeset I \M∗ = {1,4,5,8} are unknown. For these unknown flows xi,j, (i,j) ∈U we form the additional equations of the type (??).

• Choose arbitrary outgoing arc that starts from a node i of set I \M∗ = {1,4,5,8}, if it is possible, for example, for the node i = 1 wechoose the arc (1,2). Then, for all outgoing arcs from a node i = 1except (1,2) the equality is true:

x1,5 =p1,5

p1,2x1,2 (1.2.12)

• Choose arbitrary outgoing arc that starts from a node i = 4 ofset I \ M∗ = {1,4,5,8}, for example, (4,8). Then, for the arc (4,3)outgoing from the node i = 4 and the arc (4,7) the equalities is true:

x4,3 =p4,3

p4,8x4,8, x4,7 =

p4,7

p4,8x4,8 (1.2.13)


As a result, we obtained the expressions (1.2.14) for unknown arc flowsfor the graph G:

x1,5 −p1,5

p1,2x1,2 = 0,

x4,3 −p4,3

p4,8x4,8 = 0,

x4,7 −p4,7

p4,8x4,8 = 0.

(1.2.14)

Note that for the nodes {5,8} ⊂ I \M∗ impossible to generate additionalequations of of type (??), because for each of these nodes has only oneoutgoing arc.

Join (1.2.12) and (1.2.13). Thus, the additional equations of type (??)for the graph G (see Figure 1.6) are (1.2.14).

Part of the unknowns of the system (1.2.11), (1.2.14) makes up outgoingarc flows for arcs from nodes of the set I \ I∗ = {1,4,5,8}:

x1,2, x1,5, x4,3, x4,7, x4,8, x5,1, x8,4.

The remaining part of the unknowns of the system (1.2.11), (1.2.14)defines the variables intensities xi, i ∈ I

∗, where I

∗= {2,3,7} :

x2, x3, x7.

New graph G = (I, U) (see Figure 1.6) I = {1,2,3,4,5,7,8}, U ={(1,2),(1,5),(4,3),(4,7),(4,8),(5,1),(8,4)}, I

∗= {2,3,7} consists of two con�

nected components: Gm = (Im, Um), m = 1,2.The connected component G1 = (I1, U 1) consists of nodes I1 = {1,2,5},

of arcs U 1 = {(1,2),(1,5),(5,1)}and nodes with variable intensities I∗1 = {2}.

The connected component G2 = (I2, U 2) consists of nodes I2 = {3,4,7,8},of arcs U 2 = {(4,3),(4,7),(4,8),(8,4)} and nodes with variable intensities I

∗2

= {3,7}.Choose the support R = {UR,I∗R} of the graph G for the system (1.2.11)

(see Figure 1.7): UR = {(1,2),(1,5),(4,3),(4,8)}, I∗R = {2,3,7}. [!!!!! 1, 3].After the support R = {UR,I∗R} of the graph G for the system (1.2.11) is

chosen (see Figure (1.7)), we determine what structures can be obtainedafter adding one no supporting element from U\UR or I

∗\I∗R to the support

R. We construct a system of characteristic vectors (basis of the solution


12

3

4

5

7

8

Figure 1.7. Support of the graph G for the system (1.2.11)

space) of the homogeneous system generated by the system (1.2.11). Thevectors d(4,7), d(5,1), d(8,4) entailed by arcs (4,7),(5,1),(8,4) respectivelyconstitute the system of characteristic vectors. We compute the charac�

teristic vectors d(4,7), d(5,1), d(8,4). We compute the characteristic vectord(4,7) = (d4,7i,j , (i,j) ∈ U ; d4,7

i , i ∈ I∗) entailed by arc (4,7) :d(4,7) = (d4,7

1,2 → 0, d4,71,5 → 0, d4,7

4,3 → −1, d4,74,7 → 1,d4,7

4,8 → 0, d4,78,4 → 0, d4,7

5,1 → 0, d4,72 → 0, d4,7

3 → 1, d4,77 → −1).

We compute the characteristic vector d(5,1) entailed by arc (5,1):d(5,1) = (d5,11,2 → 0, d5,1

1,5 → 1, d5,14,3 → 0, d5,1

4,7 → 0,d5,14,8 → 0, d5,1

8,4 → 0, d5,15,1 → 1, d5,1

2 → 0, d4,73 → 0, d4,7

7 → 0).

We compute the characteristic vector d(8,4) entailed by arc (8,4):d(8,4) = (d8,41,2 → 0, d8,4

1,5 → 0, d8,44,3 → 0, d8,4

4,7 → 0,d8,44,8 → 1, d8,4

8,4 → 1, d8,45,1 → 0, d8,4

2 → 0, d8,43 → 0, d8,4

7 → 0).

We enumerate the additional equations (1.2.14). For each addi�tional equation with number p we compute the determinants Lptr,


(t,r) ∈ U\UR = {(4,7),(5,1),(8,4)}, p = 1, 2, 3:Lptr =X

(i,j)∈UR

lpijdtrij + lptr, p = 1,q.

We form the matrix D. Matrix D consists of determinants of the struc�

tures, entailed by no supporting elements:L14,7 = 0,L1

5,1 = 1,L18,4 = 0,L2

4,7 = −1,L25,1 = 0,L2

8,4 = −p4,3

p4,8,L3

4,7 = 1,L35,1 = 0,L3

8,4 = −p4,7

p4,8.

We form the set W = {UW , I∗W}, |W | = q, UW ⊆ U \UR, I∗W ⊆ I∗\ I∗R.

As I∗W = ∅, then W = UW , UW = {(4,7),(5,1),(8,4)}}. We enumeratethe arcs of set UW , where t = t(t,r) – the number of arc (t,r) ∈ UW in theentered numbering:

t(4,7) = 1, t(5,1) = 2, t(8,4) = 3.

We form the matrix D:

D =

0BBBBBB� L14,7 L1

5,1 L18,4L2

4,7 L25,1 L2

8,4L34,7 L3

5,1 L38,4

1CCCCCCA .

So, the matrix D is:

D =

0BBBBBBBBBBBB�0 1 0

−1 0 −p4,3

p4,8

1 0 −p4,7

p4,8

1CCCCCCCCCCCCA .

Since the following relations are true: 0 < pi,j 6 1, (i,j) ∈ U , then thedeterminant of the matrix D is not equal to zero:


det D = −p4,3

p4,8−

p4,7

p4,86= 0.

Therefore, the system (1.2.11), (1.2.14) is system of full rank. The num�

ber of the unknown of system (1.2.11), (1.2.14) is equal to the rank of thematrix of system. Also, the number of the equations in the system (1.2.11),(1.2.14) is equal to the rank of the matrix. The system (1.2.11), (1.2.14)has the unique solution for given set M = {6}.

We use the theory of decomposition to build the unique solution of thesystem (1.2.11), (1.2.14) [!!!!!]. Apply the decomposition algorithms forsolution of the system (1.2.11), (1.2.14). To construct a general solutionof system (1.2.11) need build any particular solution x̃ = (x̃i,j,(i,j) ∈U, x̃i, i ∈ I

∗) of system (1.2.11). A particular solution x̃ of the system

(1.2.11) construct in according with the rules of remark ??. Non-support�ing components of a particular solution x̃ are equal to zeroes x̃t,r = 0,

(t,r) ∈ U\UR, x̃g = 0, g ∈ I∗\I∗R : x̃4,7 = 0, x̃5,1 = 0, x̃8,4 = 0. Using

the graph-theoretic properties of the support of the graph G for the system(1.2.11) support components of a particular solution of system (1.2.11) wefind in the time O(n), n = |I|. Support components of particular solutionx̃ are:

x̃1,2 =p2,1

p2,6f2,6, x̃2 = f2,6 − f6,2,

x̃4,3 =p3,4

p3,6f3,6 +

p7,4

p7,6f7,6,

x̃3 = f3,6 − f6,3 −p7,4

p7,6f76, x̃1,5 = 0,

x̃7 =p7,4

p7,6f7,6 + f7,6 − f6,7, x̃4,8 = 0.

Using (??), we compute the numbers A1, A2, A3:

A1 =f2,6 p1,5p2,1

p1,2 p2,6,

A2 = −p3,4

p3,6f3,6 −

p7,4

p7,6f7,6,

A3 = 0.


Since the matrix D is nonsingular, then we apply the formula (??) to findthe components xW = (xt,r,(t,r) ∈ UW ) = (x4,7, x5,1, x8,4) of the solutionof system in the form:

x5,1 = A1,

−x4,7 −p4,3

p4,8x8,4 = A2,

x4,7 −p4,7

p4,8x8,4 = A3,

(1.2.15)

where the vector b is the right-hand side of the system of type (??):

b =

0BBBBBB� b1b2b3

1CCCCCCA =

�A1

A2

A3

�.

We calculate the components xW = (x4,7, x5,1, x8,4) of vector xW fromsystem (1.2.15):

x5,1 =f2,6 p1,5 p2,1

p1,2 p2,6,

x4,7 =p4,7 (f7,6 p3,6 p7,4 + f3,6 p3,4 p7,6)

p3,6 (p4,3 + p4,7) p7,6,

x8,4 =p4,8 (f7,6 p3,6 p7,4 + f3,6 p3,4 p7,6)

p3,6 (p4,3 + p4,7) p7,6.

(1.2.16)

The remaining components of the desired unique solution of system(1.2.11), (1.2.14) we found from formulas (??) – (??), using network prop�

erties of support R = {UR,I∗R} of the graph G for system (1.2.11). So, wehave:


x1,2 =f2,6 p2,1

p2,6,

x2 = f2,6 − f6,2,

x3 = f3,6 − f6,3 −f7,6 p7,4

p7,6+

p4,7 (f7,6 p3,6 p7,4 + f3,6 p3,4 p7,6)

p3,6 (p4,3 + p4,7) p7,6,

x4,3 =p4,3 (f7,6 p3,6 p7,4 + f3,6 p3,4 p7,6)

p3,6 (p4,3 + p4,7) p7,6,

x1,5 =f2,6 p1,5 p2,1

p1,2 p2,6,

x7 = −f6,7 +−f3,6 p3,4 p4,7 p7,6 + f7,6 p3,6( p4,7 p7,6 + p4,3(p7,4 + p7,6))

p3,6 (p4,3 + p4,7) p7,6,

x4,8 =p4,8 (f7,6 p3,6 p7,4 + f3,6 p3,4 p7,6)

p3,6 (p4,3 + p4,7) p7,6.

We locate a single sensor to the node 6. The values of arc flows onall incoming and outgoing arcs for the each node i of the set M = {6}(monitored node) are known and also we know the values xi = fi, i ∈MT

I∗. We substitute the known values of the variables

f2,6 = 8, f3,6 = 3, f6,3 = 3, f6,2 = 6, f6,7 = 5, f7,6 = 2, f6 = 1

to the system of equations (1.2.7). Let’s delete from graph G = (I, U) theset of the arcs on which the arc flows are known:

(2,6), (3,6), (6,3), (6,2), (6,7), (7,6).

Also, we delete from graph G the set of the nodes i ∈ M , M = {6}. Weobtain a graph G′ which is shown in Figure 1.8.

The rest of the arc flows for the outgoing arcs from the nodes of the setM+ = I(CC(M)) \ M = {2,3,7}, can be expressed from the arc flows ofthe outgoing arcs for M+ by the following way:

x2,1 =p2,1

p2,6f2,6 = 8, x3,4 =

p3,4

p3,6f3,6 = 9,

x7,4 =p7,4

p7,6f7,6 = 1.

(1.2.17)


12

3

4

5 6

7

8

12

3

4

5

7

8

Figure 1.8. Graph G (on the left) and disconnected graph G′ (on the right)

Let’s delete from graph G′ (see Figure 1.8) the set of the arcs on whichthe arc flows are known according to (1.2.17).

In Figure 1.9 we represent graph G = (I, U) obtained by deleting theset of the arcs on which the arc flows are known according to (1.2.17) fromgraph G′ (see Figure 1.9).

The numerical values pi,j, (i,j) ∈ U are presented in Table 1.2.1.

Table 1.2.1Values of components of the vector p = (pi,j,(i,j) ∈ U)

(i,j) (1,2) (1,5) (2,1) (2,6) (3,4) (3,6) (4,3) (4,7)

pi,j

1

3

2

3

1

2

1

2

3

4

1

4

5

13

4

13

(i,j) (4,8) (5,1) (6,2) (6,3) (6,7) (7,4) (7,6) (8,4)

pi,j

4

131

5

10

2

10

3

10

1

3

2

31

Values of variable intensities of nodes from set I∗ = {2,3,6,7} are:

x2 = 2, x3 = 1, x6 = 1, x7 = −4.


12

3

4

5

7

8

12

3

4

5

7

8

Figure 1.9. Disconnected graph G (on the left) and support R of the graph G for thesystem (1.2.11) (on the right)

Table 1.2.2Values of known arc flows x = (xi,j ,(i,j) ∈ U)

(i,j) (1,2) (1,5) (2,1) (2,6) (3,4) (3,6) (4,3) (4,7)

xi,j 8 16 8 8 9 350

9

40

9

(i,j) (4,8) (5,1) (6,2) (6,3) (6,7) (7,4) (7,6) (8,4)

xi,j

40

916 6 3 5 1 2

40

9

As a result of the localization of a single sensor to the node 6 in the graphG which is represented in Figure 1.8 we have a unique solution of system(1.2.7). The arc flows of the unique solution of system (1.2.7) is presentedin Table 1.2.2. The numerical values of variable intensities of the uniquesolution of system (1.2.7) for the nodes of set I∗ are: x2 = 2, x3 = 1,x6 = 1, x7 = −4.

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Sensor Location Problem - UNECE · 2013. 4. 30. · 1.1. Sensor Location Problem for graphs 5 If the set M includes the nodes from the set I∗, then we also know the values xi =