Do Directional Antennas Facilitate In Reducing Interferences?

Transcription

1 Do Directional Antennas Facilitate In Reducing Interferences? Presented by Ury Matarazzo Ben-Gurion University, Beer-Sheva, Israel November 25, 2012

2 1 Introduction The question 2 The converge cast 3 Asymmetric model 4 Symmetric model

3 Introduction The converge cast Asymmetric model Symmetric The model question Definition of the problem The coverage area of a point P with angle α, radius r and direction.

4 Introduction The converge cast Asymmetric model Symmetric The model question Definition of the problem The interference of a point P. The interference of a graph G.

5 Introduction The converge cast Asymmetric model Symmetric The model question Definition of the problem P is a set of points in the plane and each point is a transceiver with circular antenna. UDG(P) is the unit disk graph of P.

6 Introduction The converge cast Asymmetric model Symmetric The model question Definition of the problem P is a set of points in the plane and each point is a transceiver with directional antenna. The induced symmetric communication graph (scg) of P.

7 Introduction The converge cast Asymmetric model Symmetric The model question Definition of the problem P is a set of points in the plane and each point is a transceiver with directional antenna. The induced directed communication graph (dcg) of P = The asymmetric communication graph of P.

8 Introduction The converge cast Asymmetric model Symmetric The model question Motivation of the problem The motivation to reduce interference is that high level of interferences affects the efficiency of a wireless network. This happens when two or more transmitters transmit simultaneously to a receiver placed in their coverage area.

9 How to construct a network with O(1) interference for α < 2π/3 Coverage-cast This is an operation in which data from many sources is routed towards a single base-station. For a fixed point s P, dcg(p) contains a path from each p P to s.

10 How to construct a network with O(1) interference for α < 2π/3 We use Euclidean MST, denoted MST (P). First, root MST (P) at s and orient every point p P to π(p) such that at the end, there will be a path from p to s.

11 How to construct a network with O(1) interference for α < 2π/3 We use Euclidean MST, denoted MST (P). Assign range (p, π(p)) to each p P {s}. The result is that there is a path from each point to s in dcg.

12 How to construct a network with O(1) interference for α < 2π/3 lune(p1, p2) lune(p1, p2) is the intersection of two discs centered at p1 and p2 of radius (p1, p2). lune(p1, p2) is the union of two wedges (Coverage area) of angle 2π/3 and range (p1, p2).

13 How to construct a network with O(1) interference for α < 2π/3 lune(p1, p2) L(MST (P)) is the set of all lunes corresponding to the edges of MST (P). It is enough to show that the depth of the arrangement of L(MST (P)) is at most 5.

14 Proof if x is a point in the plane and x / P, e is an edge such that e MST (P {x}) MST (P), then x is an endpoint of e.

15 Proof if x is a point in the plane then x belongs to at most 5 lunes in L(MST (P)).

16 Proof (p1, p2) / L(MST (P {x}))

17 Proof For each lune(p, q) L(MST (P)) such that x lune(p, Q), MST (P {x}) contains a unique edge that is connected to x and replaces the edge (p, q) of MST (P).

18 Proof x is a vertex of an Euclidean minimum spanning tree, and it s degree is at most 6, so x is covered by at most 6 lunes of L(MST (P)).

19 Theorem 1 Let P be a set of points representing directional antennas of angle α < 2π/3 and s P One can asign each p P an orientation θ and a range r such that from every p P there is a path from P to S in dcg(p), and I (dcg(p)) = O(1).

20 The asymmetric model The problem Assign to each antenna p P an orientation θ and a range r, such that dcg(p) is strongly connected and I (dcg(p)) = O(1).

21 Method overview Two steps Assign without bounding the maximum range of the antennas. Bound the maximum range by constant, assuming UDG(P) is connected.

22 Method details First step if P is sufficiently large, we can select a subset Q P of size l = 3.5k 6, where k = 2π/α, and assign to each antenna in Q an orientation θ and range r =, such that dcg(q) contains a tree T rooted at some point s Q whose leaves collectively cover the entire plane, and each point in Q T is oriented towards s. In particular, there exists a path from s to each point of P.

23 Method details second step if P is sufficiently large, we can select a subset Q P of size l = 3.5k 6, where k = 2π/α, and assign to each antenna in Q an orientation θ and range r =, such that dcg(q) contains a tree T rooted at some point s Q whose leaves collectively cover the entire plane, and each point in Q T is oriented towards s. In particular, there exists a path from s to each point of P.

24 Ensuring strong connectivity First step For each leave in T that do not cover any point in P T, we orient it towards s.

25 Ensuring strong connectivity Second step We build a converge-cast network for the set P Q {s}, with s as the base-station of the network. The orientation of s is not changed.

26 Ensuring strong connectivity Result The interference of the resulting dcg is bounded by the sum of Q and the interference of the converge-cast, which is 5, and Q + 5 = O(1).

27 The range of the antenna Without assumption The required range of the antennas can reach the diameter of P, which is not constant. Reducing range with the assumption Recall that we assumed that UDG(P) is connected, now we can bound the ranges of the antennas by O(1).

28 Partition with grid Partition We lay a grid G over P, such that the length of a cell side is 2l.

29 Partition with grid For each block For each block we construct converge-cast network and we connect the blocks.

30 Result The induced dcg is strongly connected The construction is similar to the previous construction and therefore it still is strongly connected. The interference at any point is O(1) The assigned ranges of the antennas is < 4 2l, so each point p P can be covered only by antennas from the 7 7 block around p s cell and this amount is constant, so I (dcg(p)) = O(1).

31 For 1D For α < 2π interference 3 Each odd point is oriented right and covers 3 points, except the last one. Each even point is oriented left and covers 3 points, except the first one. Result The graph is connected and I (scg(p)) = 3.

32 For 2D with α < π/2 For α < π/2 interference can be n 1 Consider the given example. n 1 points are located at vertical line and the n-th point p is far away so that any other point cannot cover other point and p. It can happen because α < π/2.

33 For 2D with α < π/2 For α < π/2 interference can be n 1 This way, all the points other than p must cover p and therefore the interference at p will be n 1.

34 How to construct a network with O( ) interference for π/3 α 3π/2 We sort the points by the x order First case - the order is montone.

35 First case - the order is montone First case - the order is montone We use the same orientations of the 1D case but we orient angles of π/4 and 5π/4 instead of 0 and π, respectively.

36 First case - the order is montone First case - the order is montone The interference of any point in the plane is at most 4.

37 How to construct a network with O( ) interference for π/3 α 3π/2 We sort the points by the x order General case - the order is not montone. We sort the points by the x order We partition the points to m = O( n) xy-monotone subsequences We connect each of the sequences separately We have O( n) sequences connected and the intererence is at most 4m = O( n) but the sequences are not connected.

38 How to connect the sequences Long sequences, length 8 We make the described construction and we connect each pair of sequences.

39 How to connect the points Long sequences, length > 8 We can separate the points by two lines such that for each side, we have 4 points of each sequence. Connecting the two sequences We assign orientations and radiuses such that the four points cover all the space and in particular the four points of the second sequence.

40 How to connect the sequences Short sequences, length < 8 For each point p in the sequence, we pick a point q that covers p and orient p to q. This way the sequence is connected to the other sequences.

41 The interference of the model Long sequences, length 8 The interference is bounded by 12m, where m = O( n) is the number of the sequences.

42 How to connect the short sequences Short sequences, length < 8 - connecting the points We pick an arbitrary point q from long sequence that covers p and orient p to q. Such q exists because a long sequence has 4 points that cover all plane, so one of them covers p. This way, scg(p) is connected.

43 How to connect the short sequences Short sequences, length < 8 - the interference The interference is increased only by a constant because it is easy to ensure that the amount of long sequences is constant.