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Distance Distributions and Boundary Effects in Finite Uniformly - - PowerPoint PPT Presentation

Background Distance Distributions Problem Formulation Results Conclusions Distance Distributions and Boundary Effects in Finite Uniformly Random Networks Salman Durrani Senior Lecturer Applied Signal Processing (ASP) Research Group


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Background Distance Distributions Problem Formulation Results Conclusions

Distance Distributions and Boundary Effects in Finite Uniformly Random Networks

Salman Durrani

Senior Lecturer Applied Signal Processing (ASP) Research Group Research School of Engineering, College of Engineering & Computer Science The Australian National University, Canberra, Australia http://users.cecs.anu.edu.au/~Salman.Durrani/

  • Jan. 2013

AusCTW 2013 1/35

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Background Distance Distributions Problem Formulation Results Conclusions

Outline

⋄ Motivation and Background

  • Spatial Point Processes
  • Distance Distributions

⋄ Prior Work

  • Poisson Point Process (PPP)
  • Binomial Point Process (BPP)

⋄ Problem Formulation

  • Modelling of Boundary Effects
  • Proposed Algorithm

⋄ Results ⋄ Conclusions

AusCTW 2013 2/35

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Background Distance Distributions Problem Formulation Results Conclusions

Background

⋄ Spatial point processes are used to model the locations of

  • bjects or events in a wide variety of scientific disciplines∗.
  • Forestry/Seismology/Geography/Astronomy
  • Locations of trees/earthquake epicenters/cities/galaxies
  • Medicine and Biology
  • Home locations of infected patients.
  • Spikes of neurons.
  • Microcalcifications in mammogram images.
  • Material Science
  • Positions of defects in industrial materials.

∗A. Baddeley, “Analysing spatial point patterns in R", CSIRO Workshop Notes, Feb. 2008. [Cited by 71]

AusCTW 2013 3/35

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Background Distance Distributions Problem Formulation Results Conclusions

Background

⋄ Spatial point processes are used to model the locations of

  • bjects or events in a wide variety of scientific disciplines∗.
  • Forestry/Seismology/Geography/Astronomy
  • Locations of trees/earthquake epicenters/cities/galaxies
  • Medicine and Biology
  • Home locations of infected patients.
  • Spikes of neurons.
  • Microcalcifications in mammogram images.
  • Material Science
  • Positions of defects in industrial materials.
  • Wireless Communications
  • A wireless network can be viewed as a collection of nodes,

where the location of nodes are seen as realizations of some spatial point process.

∗A. Baddeley, “Analysing spatial point patterns in R", CSIRO Workshop Notes, Feb. 2008. [Cited by 71]

AusCTW 2013 3/35

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Background Distance Distributions Problem Formulation Results Conclusions

Background - PPP

⋄ Popular model: infinite homogenous Poisson point process (PPP).

  • Rationale: Homogeneous PPP can be regarded as the limiting

case of a uniform distribution of N nodes on an area of size A, as N and A tend to infinity but their ratio ρ = N/A remains constant.

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Background Distance Distributions Problem Formulation Results Conclusions

Background - PPP

⋄ Popular model: infinite homogenous Poisson point process (PPP).

  • Rationale: Homogeneous PPP can be regarded as the limiting

case of a uniform distribution of N nodes on an area of size A, as N and A tend to infinity but their ratio ρ = N/A remains constant.

  • Advantage: Mathematical tractability - provides a model for

‘completely random’ distribution of points.

  • Main shortcoming: The number of nodes in disjoint areas is

independent.

AusCTW 2013 4/35

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Background Distance Distributions Problem Formulation Results Conclusions

Background - BPP

⋄ More realistic model: Finite number of nodes independently and uniformly distributed over a finite area (Binomial point process (BPP)).

  • Cellular networks: cells are hexagons.
  • Ad hoc and sensor networks: finite square region.

AusCTW 2013 5/35

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Background Distance Distributions Problem Formulation Results Conclusions

Background - BPP

⋄ More realistic model: Finite number of nodes independently and uniformly distributed over a finite area (Binomial point process (BPP)).

  • Cellular networks: cells are hexagons.
  • Ad hoc and sensor networks: finite square region.

⋄ Advantage: The number of nodes in disjoint areas is no longer independent: the more nodes in one sub-area, the fewer can fall in another.

AusCTW 2013 5/35

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Background Distance Distributions Problem Formulation Results Conclusions

Background - PPP & BPP

⋄ Illustration: Nodes distributed in 1×1 m2 area according to (a) PPP, 100 nodes/m2† and (b) BPP, N = 100.

PPP in 2D can be realized as a 1D PPP enriched by attaching to each

  • ne-dimensional point an

independent Uniform random variable to provide the second coordinate. BPP in 2D: »x=rand(1,100); »y=rand(1,100); »plot(x,y,‘r+’); 1 1 PPP,N=95 1 1 PPP,N=118 1 1 PPP,N=102 1 1 BPP,N=100 1 1 BPP,N=100 1 1 BPP,N=100

†Sheldon M. Ross, Simulation, 4th ed., Elsevier Inc., 2006.

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Background Distance Distributions Problem Formulation Results Conclusions

Background - Spatial Point Processes

⋄ Useful point processes for wireless network modeling‡:

‡J. G. Andrews et. al., “A primer on spatial modeling and analysis in wireless networks", IEEE

Communications Magazine, vol. 48, no. 9, pp. 156−163, Nov. 2010. [Cited by 42] AusCTW 2013 7/35

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Background Distance Distributions Problem Formulation Results Conclusions

Distance Distributions

⋄ The performance of wireless networks depends critically on the distances between the transmitters and receivers.

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Background Distance Distributions Problem Formulation Results Conclusions

Distance Distributions

⋄ The performance of wireless networks depends critically on the distances between the transmitters and receivers. ⋄ Euclidean distance to n-th neighbor from an arbitrarily chosen reference point.

  • n = 1 corresponds to nearest neighbour.
  • n = 2 corresponds to second nearest neighbour.
  • n = N corresponds to farthest neighbour.

R1 R2

AusCTW 2013 8/35

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Background Distance Distributions Problem Formulation Results Conclusions

n-th Neighbour PDF − PPP

⋄ PDF of Euclidean distance to n-th nearest neighbor in a homogeneous m-dimensional PPP: generalized Gamma distribution§

✎ ✍ ☞ ✌

fRn(r) = m(ρcmr m)n rΓ(n) e−ρcmrm where coefficients cm are given by cm =

    

π

m 2

( m

2 )!

for even m

π

m−1 2 ( m−1 2 )!

m!

for odd m (e.g., c1 = 2,c2 = π, c3 = 4π

3 )

§M. Haenggi, “On Distances in Uniformly Random Networks", IEEE Trans. Inf. Theory, vol. 51, no. 10, pp.

3584−3586, 2005. [Cited by 166] AusCTW 2013 9/35

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Background Distance Distributions Problem Formulation Results Conclusions

n-th Neighbour PDF − PPP

⋄ PDF of Euclidean distance to n-th nearest neighbor in a homogeneous m-dimensional PPP: generalized Gamma distribution§

✎ ✍ ☞ ✌

fRn(r) = m(ρcmr m)n rΓ(n) e−ρcmrm where coefficients cm are given by cm =

    

π

m 2

( m

2 )!

for even m

π

m−1 2 ( m−1 2 )!

m!

for odd m (e.g., c1 = 2,c2 = π, c3 = 4π

3 )

⋄ Special case (m = 2, n = 1):

✞ ✝ ☎ ✆

fR1(r) = 2πρre−ρπr2

§M. Haenggi, “On Distances in Uniformly Random Networks", IEEE Trans. Inf. Theory, vol. 51, no. 10, pp.

3584−3586, 2005. [Cited by 166] AusCTW 2013 9/35

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Background Distance Distributions Problem Formulation Results Conclusions

Distance Distributions − BPP

⋄ Distance distribution in a BPP with N nodes distributed inside a L-sided regular polygon (L-gon) with area A.

Triangle Square Hexagon Disk

Source: [Haenggi Paper] AusCTW 2013 10/35

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Background Distance Distributions Problem Formulation Results Conclusions

Distance Distributions − BPP

⋄ Distance distribution for BPP in a polygon (assuming center of polygon as reference point)¶

✛ ✚ ✘ ✙

fRn(r) =

      

2rπ A (1−p)N−npn−1 B(N−n+1,n)

0 < r ≤ Ri

2r(π−Lθ) A (1−q)N−nqN−n B(N−n+1,n)

Ri < r ≤ Rc Rc < r where

Ri =

  • A

L cot π L

  • ,

Rc =

  • 2A

L csc 2π L

  • ,

p = πr 2

A , q = πr 2−

  • Lr 2θ−LRi
  • r 2−R2

i

  • A

, θ = arccos(Ri/r), beta function B(a,b) = Γ(a)Γ(b)

Γ(a+b) . ¶S. Srinivasa and M. Haenggi, “Distance Distributions in Finite Uniformly Random Networks: Theory and

Applications", IEEE Trans. Veh. Tech., vol. 59, no. 2, pp. 940−949, Feb. 2010. [Cited by 34] AusCTW 2013 11/35

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Background Distance Distributions Problem Formulation Results Conclusions

Distance Distributions - Illustration

⋄ BPP with N = 5 nodes distributed inside a unit square (L = 4) (solid lines = BPP, dotted lines = PPP).

0.2 0.4 0.6 0.8 1 1 2 3 4 5 6 7 distance, r fk(r) k=1 k=2 k=3 k=4 k=5

AusCTW 2013 12/35

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Background Distance Distributions Problem Formulation Results Conclusions

Contribution of this Work

⋄ We derive the closed-form PDF of the distance between any arbitrary reference point and its n-th neighbour node, when N nodes are uniformly distributed inside a regular L-sided polygon.

AusCTW 2013 13/35

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Background Distance Distributions Problem Formulation Results Conclusions

Polygon Geometry

⋄ N nodes are independently and uniformly distributed inside a regular L-sided polygon A, inscribed in a circle of radius R centered at the origin. Let u = [x, y]T denote an arbitrary reference point. Circumradius: R Inradius: Ri = R cos(π/L) Area: A = |A| = 1

2LR2 sin

L

  • Side length: t = 2R sin

π

L

  • Interior angle: θ = π(L−2)

L

Central angle: ϑ = 2π

L

AusCTW 2013 14/35

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Background Distance Distributions Problem Formulation Results Conclusions

Problem Formulation

⋄ Define the cumulative density function (CDF) F(u;r), which is the probability that a random node falls inside a disk D(u;r) centered at the arbitrary reference point u, as

☛ ✡ ✟ ✠

F(u;r) = |D(u;r)∩A|

|A|

= O(u;r)

A

where O(u;r) = |D(u;r)∩A| is the overlap area. x S1 S2 S3 S4 V1 V2 V3 V4 r R y Ri

AusCTW 2013 15/35

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Background Distance Distributions Problem Formulation Results Conclusions

Problem Formulation

⋄ The CCDF expressing the probability that there are less than n nodes in the disk D is given by

✓ ✒ ✏ ✑

1−Fn(u;r) =

n−1

  • j=0
  • N

j

  • (F(u;r))j(1−F(u;r))N−j
  • S. Srinivasa and M. Haenggi, “Distance Distributions in Finite Uniformly Random Networks: Theory and

Applications", IEEE Trans. Veh. Tech., vol. 59, no. 2, pp. 940−949, Feb. 2010. [Cited by 34] AusCTW 2013 16/35

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Background Distance Distributions Problem Formulation Results Conclusions

Problem Formulation

⋄ The CCDF expressing the probability that there are less than n nodes in the disk D is given by

✓ ✒ ✏ ✑

1−Fn(u;r) =

n−1

  • j=0
  • N

j

  • (F(u;r))j(1−F(u;r))N−j

⋄ The corresponding PDF is

✎ ✍ ☞ ✌

fn(r) = (1−F(u;r))N−n(F(u;r))n−1 B(N −n +1,n) d dr F(u;r) where beta function B(a,b) = Γ(a)Γ(b)/Γ(a +b).

  • S. Srinivasa and M. Haenggi, “Distance Distributions in Finite Uniformly Random Networks: Theory and

Applications", IEEE Trans. Veh. Tech., vol. 59, no. 2, pp. 940−949, Feb. 2010. [Cited by 34] AusCTW 2013 16/35

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Background Distance Distributions Problem Formulation Results Conclusions

Why is the CDF F(u;r) so hard to find ?

⋄ Boundary effects: nodes located near the physical boundaries of the region have their coverage area reduced.∗∗

A B r r

∗∗C. Bettstetter, “On the minimum node degree and connectivity of a wireless multihop network", in Proc. 3rd

ACM international symposium on Mobile ad hoc networking & computing, 2002. [Cited by 727] AusCTW 2013 17/35

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Background Distance Distributions Problem Formulation Results Conclusions

Special case: reference point at the center

⋄ Boundary effects are easy to characterise: circular segment areas are symmetric, with no overlap. x S1 S2 S3 S4 V1 V2 V3 V4 r R x S1 S2 S3 S4 V1 V2 V3 V4 r R Ri Ri y y

AusCTW 2013 18/35

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Background Distance Distributions Problem Formulation Results Conclusions

General case: reference point at arbitrary location

⋄ Boundary effects are complicated to characterise:

  • Problem 1: circular segment areas are no longer symmetric

and they may have overlap. x S1 S2 S3 S4 V1 V2 V3 V4 r R Ri y

AusCTW 2013 19/35

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Background Distance Distributions Problem Formulation Results Conclusions

General case: reference point at arbitrary location

⋄ Boundary effects are complicated to characterise:

  • Problem 1: circular segment areas are no longer symmetric

and they may have overlap. x S1 S2 S3 S4 V1 V2 V3 V4 r R Ri y

  • Problem 2: since a L-gon has L sides and L vertices, there

can be 2∗L+1 different ranges for the distance r.

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Background Distance Distributions Problem Formulation Results Conclusions

Proposed Approach

⋄ We decompose the boundary effects into border and corner effects.

  • Let Bℓ = the area of the circular segment formed outside the

side Sℓ (ℓ = 1,2,...,L).

  • Let Cℓ = the corner overlap area formed at vertex Vℓ.

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Background Distance Distributions Problem Formulation Results Conclusions

Distances to Sides and Vertices

⋄ Distance d(u; V1) between the point u and the vertex V1 is

☛ ✡ ✟ ✠

d(u; V1) =

  • (x −R)2 +y2

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Background Distance Distributions Problem Formulation Results Conclusions

Distances to Sides and Vertices

⋄ Shortest distance d(u; S1) to the side S1 is

✎ ✍ ☞ ✌ d(u; S1) =

  • min(d(u; V1), d(u; V2)),

max(d(w;V1), d(w;V2)) > t; p(u; S1),

  • therwise;

where t is the side length and w =

  • R − 1

2(x −R)(cosϑ−1)+y sinϑ, sinϑ

  • (x −R)(cosϑ−1)+y sinϑ
  • 2(1−cosϑ)

T ✓ ✒ ✏ ✑ p(u; S1) = abs

  • y +tan

θ

2

  • x −R tan

θ

2

  • 1+tan2 θ

2

  • AusCTW 2013

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Background Distance Distributions Problem Formulation Results Conclusions

Corner Effects

⋄ Circular segment area formed outside side S1 of ℓ-gon is††

✗ ✖ ✔ ✕

B1(u;r) =

  

r2 arccos p(u;S1)

r

  • − (d(u;S1))2 arccos p(u;S1)

d(u;S1)

p(u;S1)

  • r2 −

p(u;S1)2 −

  • (d(u;S1))2 −

p(u;S1)2 , r ≥ d(u;S1); 0,

  • therwise;

⋄ Corner overlap area formed at vertex V1 of ℓ-gon is

✬ ✫ ✩ ✪

C1(u;r) =

                      

r2 2

  • arccos p(u;S1)

r

  • + arccos p(u;SL)

r

(d(u;V1))2 2

  • arccos p(u;S1)

d(u;V1)

  • + arccos p(u;SL)

d(u;V1)

  • +

p(u;S1) 2

  • d(u;V1)2

− p(u;S1)2 −

  • r2 −

p(u;S1)2

  • +

p(u;SL) 2

  • d(u;V1)2

− p(u;SL)2 −

  • r2 −

p(u;SL)2

π L

  • r2 −

d(u;V1)2 , r ≥ d(u;V1); 0,

  • therwise;

††Z. Khalid and S. Durrani, “Distance Distributions in Regular Polygons", IEEE Trans. Veh. Tech., 2013 (in

press: http://arxiv.org/abs/1207.5857). AusCTW 2013 23/35

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Background Distance Distributions Problem Formulation Results Conclusions

Rotation Operator

⋄ We define the rotation operator Rℓ which rotates an arbitrary point u = [x, y]T anti-clockwise around the origin by an angle ℓϑ. ⋄ The rotated point Rℓu can be expressed as

✞ ✝ ☎ ✆

(Rℓu) = Tu ⋄ The rotation matrix is given by

✎ ✍ ☞ ✌

T =

  • cos(ℓϑ)

−sin(ℓϑ) sin(ℓϑ) cos(ℓϑ)

  • AusCTW 2013

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Background Distance Distributions Problem Formulation Results Conclusions

Exploiting Rotational Symmetry

⋄ Solution to Problem 1: circular segment areas are no longer symmetric and they may have overlap: d(u; Vℓ) = d(R−(ℓ−1)u; V1) p(u; Sℓ) = p(R−(ℓ−1)u; S1) d(u; Sℓ) = d(R−(ℓ−1)u; S1) Bℓ(u;r) =

  • B1(R−(ℓ−1)u;r),

r ≥ d(u;Sℓ); 0,

  • therwise.

Cℓ(u;r) =

  • C1(R−(ℓ−1)u;r),

r ≥ d(u;Vℓ); 0,

  • therwise.

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Background Distance Distributions Problem Formulation Results Conclusions

Exploiting Rotational Symmetry

⋄ Define the distance vector d as

✞ ✝ ☎ ✆

d = [d(u;S1), ..., d(u;SL), d(u;V1), ..., d(u;VL)] and ´ d is the sorted distance vector in ascending order. k is the index vector that transforms d into ´ d. ⋄ We use the sorted distance vector ´ d and the index vector k to identify each unique range and to find the boundary effects for that range.

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Background Distance Distributions Problem Formulation Results Conclusions

Proposed Algorithm

⋄ Solution to Problem 2: there can be 2∗L+1 different ranges for the distance r:

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Background Distance Distributions Problem Formulation Results Conclusions

Results: Example 1

⋄ Arbitrary reference point: middle of side S4 for a square with R = 1.

d =

  • R

√ 2 , √ 2R, R √ 2 , 0, R √ 2 , √ 10R 2 , √ 10R 2 , R √ 2

  • ´

d =

  • 0,

R √ 2 , R √ 2 , R √ 2 , R √ 2 , √ 2R, √ 10R 2 , √ 10R 2

  • k = [4, 1, 3, 5, 8, 2, 6, 7]

x S1 S2 S3 S4 V1 V2 V3 V4 y

R

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Background Distance Distributions Problem Formulation Results Conclusions

Results: Example 1

⋄ Arbitrary reference point: middle of side S4 for a square with R = 1.

d =

  • R

√ 2 , √ 2R, R √ 2 , 0, R √ 2 , √ 10R 2 , √ 10R 2 , R √ 2

  • ´

d =

  • 0,

R √ 2 , R √ 2 , R √ 2 , R √ 2 , √ 2R, √ 10R 2 , √ 10R 2

  • k = [4, 1, 3, 5, 8, 2, 6, 7]

x S1 S2 S3 S4 V1 V2 V3 V4 y

R ⋄ The CDF is

Range F(u;r) 0 ≤ r ≤ R/ √ 2

πr2−(B4) A

R/ √ 2 ≤ r √ 2R

πr2−(B1+B3+B4−C1−C4) A

√ 2R ≤ r ≤ √ 10R/2

πr2−(B1+B2+B3+B4−C1−C2) A

r ≥ √ 10R/2 1

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Background Distance Distributions Problem Formulation Results Conclusions

Results: Example 1

⋄ Arbitrary reference point: middle of side S4 for a square with R = 1 and N = 5.

0.5 1 1.5 2 0.5 1 1.5 2 2.5 3 3.5 distance, r fn(r) Theory Simulation n=5 n=2 n=1 n=3 n=4

AusCTW 2013 29/35

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Background Distance Distributions Problem Formulation Results Conclusions

Results: Example 2

⋄ Arbitrary reference point located at vertex of a polygon with Area A = 100 and N = 10 nodes: PDF of nearest neighbour.

5 10 15 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 distance, r f1(r) L=3 L=4 L=6 L=100 AusCTW 2013 30/35

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Background Distance Distributions Problem Formulation Results Conclusions

Results: Example 3

⋄ Arbitrary reference point located at vertex of a polygon with Area A = 100 and N = 10 nodes: PDF of farthest neighbour.

5 10 15 20 0.1 0.2 0.3 0.4 0.5 0.6 distance, r f10(r) L=3 L=4 L=6 L=100 AusCTW 2013 31/35

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Background Distance Distributions Problem Formulation Results Conclusions

Conclusion and Future Work

⋄ In this work, we have derived the n-th neighbour distance distribution results in regular polygons. ⋄ The knowledge of these general distance distributions can be used to analyse the wireless network characteristics from the perspective of an arbitrary node located anywhere (i.e. not just the center) in the finite coverage area. ⋄ Applications:

  • Connectivity: S. Durrani, Z. Khalid and J. Guo, “A Tractable Framework for Exact

Probability of Node Isolation in Finite Wireless Sensor Networks", submitted to IEEE Trans. Veh. Tech., 2013 (http://arxiv.org/abs/1212.1283)

  • Interference and outage: work under progress.

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Background Distance Distributions Problem Formulation Results Conclusions

Thank you for your attention

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