Spatial Multiplexing Gains for Realistic Sized Ad Hoc Networks with Directional Antenna Arrays

We concentrate on an ad hoc network model with nodes on integer lattice points over a 2D plane. We examine the limits of ad hoc network performance for systems with antenna arrays capable of allowing both spatial multiplexing and directional processing. Two cases are considered. In the ﬁrst case, we consider “perfect” directional antenna arrays, in other words, each node can form beams of inﬁnitesimally narrow beamwidth. In this case, the throughput capacity of an ad hoc network is independent of the network size. In the second case, we consider a more practical system where each node can form a ﬁxed number of beams of ﬁnite beamwidth. Our results show that the spatial multiplexing gains depend on the system size, antenna beamwidth, and number of antenna beams. Furthermore, we show that spatial multiplexing gains o ﬀ setting the interference-related performance degradation can be achieved in ad hoc networks with thousands of nodes.


INTRODUCTION
The application of multiple antennas at both the transmitter and receiver sides of a wireless system for the purpose of spatial multiplexing (simply put, spatial multiplexing in this context means making use of multiple paths distinct in physical space to deliver information from a source to the corresponding destination) [1, 2] has been shown to have the potential of achieving extraordinary bit rates.As a result, this topic has received significant study recently [3-10].The issues of MAC/routing protocol design for ad hoc networks utilizing multiple antennas were also studied in [11-17].It should be noted that antenna arrays can implement directional processing and beamforming in addition to spatial multiplexing.When these approaches are suitably combined, good network performance is achieved.However, the majority of research has focused on point-to-point communications.Here we study spatial multiplexing at the network level.Further, we assume the antenna arrays used for the spatial multiplexing will also be used for beamforming.
We will study the uniform throughput capacity, or simply uniform throughput, which we define as the minimum long-term average rate at which every node in the network can transmit to its corresponding destination.The throughput in wireless ad hoc networks is inherently limited by in-terference since the nodes have to use the common wireless channel in order to transmit different information.The use of multiple directed beams and spatial multiplexing cannot completely eliminate the interference but, as we will see, can greatly alleviate it, even if a small number of beams at every node is used.
Previous results [18] have shown poor performance for large ad hoc networks without spatial multiplexing.In this paper, we will show that spatial multiplexing provides large gains in throughput for small networks, and that while these gains shrink for larger networks, there are still spatial multiplexing gains in networks with thousands of nodes.
In this paper, we consider a network consisting of n nodes located on a square grid with periodic boundary conditions.We begin by examining a simpler case of infinitely narrow beamwidth where every beam is just a zero-width ray with the origin at the transmitting node.We find that in this case, the uniform throughput is upper bounded by Wg/2, where W is the rate of point-to-point transmission along a single beam, and g is the number of beams each transmitter can form.Furthermore, we show that, under reasonable assumptions, the uniform throughput of Wg/2 can be achieved regardless of the network size (and the distance between sources and destinations).Next, we consider the case of a finite angular beamwidth D where the beams are infinite two-dimensional cones with vertices at the transmitter.We show that in this case the uniform throughput is bounded from above by a quantity, that is proportional to Wg, and for larger network sizes, proportional to l / √ n, where l is the average of the longest g hops possible from a given node without interference.Moreover, a fixed fraction of this upper bound can be shown to be achievable.The result is that, although the degradation of performance due to interference is still present for the finite beamwidth case, the spatial multiplexing allows one to "postpone" the throughput from falling below W (which is what the throughput would be for just a single source-destination pair) until fairly large network sizes (thousands of nodes for beamwidth of about 10 degrees and no more than 10 beams) which makes practically large network sizes entirely feasible.
The directional antenna assumptions used in this paper are consistent with accepted results [19] that imply that antenna arrays (smart antennas) can be used to form beams in n different directions if at least n antennas are available in an array.Further, by proper spacing of the antennas and by employing more antennas, these beams can be made more narrow.Therefore, the number of antennas limits the number of directional beams that each node may employ.
The rest of the paper is organized as follows.In Section 2, we formulate the model used in the paper.Section 3 is devoted to evaluating the uniform throughput for both cases of infinitely narrow beamwidth and finite beamwidth.Section 4 contains conclusions.

SYSTEM DESCRIPTION
In this paper, we evaluate the uniform throughput among n = m2 nodes with each node located at a unique integer point of the lattice Ω(m) = {(a, b) |a, b = 1, 2, . . ., m} covering an m-by-m square region with periodic boundary conditions (a torus). 1 We measure all the distances below in L 1 1 That is, as the coordinates are to be understood "module m."More precisely, the nodes with coordinates (x + m, y) are identified with nodes with coordinates (x, y) for y = 1, 2, . . ., m + 1 and nodes with coordinates (x, m + y) are identified with nodes with coordinates (x, y) for x = 1, 2, . . ., m + 1. metric ("Manhattan distance") in units of lattice space, unless noted otherwise.We assume that each transceiver node is equipped with an antenna array that can produce g antenna beams, each with angular width D (see Figure 1) such that gD ≤ 2π.Node-to-node transmissions on the torus are allowed only in the "shorter" direction, that is, the largest horizontal and vertical transmitting distance allowed by the model is m/2 .The latter requirement is used to imitate a real system with boundaries while disregarding boundary effects where they can lead to unwanted complications.
A transmission from node i to node j along a beam b i l , l = 1, 2, . . ., g, is assumed to be successful if 2 (1) node j lies inside the beam b i l , (2) node j does not lie inside any other beam b k l closer to the node k than the intended receiver.
If a transmission along a given beam is successful, the corresponding transmission rate is assumed to be equal to W. We assume that if a node-to-node transmission is successful, exactly one packet is transmitted. 3We use the full-duplex assumption: a node can both transmit and receive up to g packets simultaneously from different directions (along different beams).Thus the maximum number of packets that a node can simultaneously handle 4 (either transmit or receive) is equal to 2g.
Finally, we make the following assumption about the relative position of sources and destinations.Assumption 1.We assume each source and destination are separated by a distance m/2 lattice spaces in the horizontal and vertical directions (and thus are separated by a distance of m in L 1 metric).
This assumption is used to simplify calculations, and by removing the "randomness" from them, make it possible to obtain quantitative results for networks of finite size as opposed to asymptotic results valid only in the limit n→∞.
In the following, we measure the throughput in units of W. Thus in order to obtain the throughput in conventional units of bits/s, all the following results should be multiplied by W.

THROUGHPUT WITH SPATIAL MULTIPLEXING
In this section, we explore the uniform throughput for the model described above.

General bounds on throughput of ad hoc networks with spatial multiplexing
The uniform throughput of any network can be upper bounded on the basis of just the number of successful transmissions per time slot and the average number of nodeto-node hops necessary to complete a source-to-destination transmission.Assume that any successful node-to-node transmission happens at a rate of W. Let s be the expected number of distinct successful transmissions per node in a time slot for the given transmission scheme.Also, let h ia be the number of hops it takes to completely reach the destination from the source node i when using the path5 a. Define h i = min a h ia to be the length of the shortest path between node i and its destination.Finally, let h = (1/n) n i=1 h i be the mean hop count of the shortest source-to-destination path taken over all nodes in the network.Then a simple upper bound on the uniform throughput can be obtained.Theorem 1.For any transmission scheme, the uniform throughput per node, T , satisfies the inequality Proof.Let us consider a large number T of time slots.Then the total number of successful node-to-node transmissions over these T time slots is On the other hand, in order to obtain a throughput of at least T for the source node i, one would need at least T h i T successful node-to-node transmissions.Therefore, in order to obtain a throughput of at least T for all n sources, the corresponding count of successful node-to-node transmissions has to be at least It is clear then that we must have N T T ≤ N T , which implies, using ( 2) and ( 3), that In the following, we will be interested in transmission schemes which allow for all nodes to successfully transmit using all available beams in every time slot, 6 that is, schemes for which s = g.Consider a given node i.Let V i be the set of nodes such that if j ∈ V i , then g nonoverlapping beams b i l , l = 1, 2, . . ., g, originating at node i can be arranged in such a way that (i) j ∈ b i l for some value of l; (ii) if k ∈ b i l for k = j, then r i j < r ik (where r i j is the Euclidean distance between nodes i and j).
We will call nodes in V i visible without interference, or vwi, to node i.
Let L i be the distance between the source i and its destination.Also let l i1 , l i2 , . . ., l i|Vi| be the distances from node i to nodes in the set V i ordered so that l i1 ≥ l i2 ≥ •••l i|Vi| .Let l i = (1/g) g j=1 l i j be the mean of the g largest distances l i j , and let l = (1/n) n i=1 l i be the mean of the quantities l i .Then one can show that the uniform throughput T can be upper bounded as follows.(Here Theorem 2. The uniform throughput T satisfies the inequality Proof.Consider a long time period (measured in time slots) T.During this time, in order to have a throughput of T for all sources, the total information transport (i.e., information transmitted over distance, measured in bit • m) of would be needed.
On the other hand, let us compute the largest information transport that can be achieved during the same time T. If in every time slot every node uses all its beams for successful transmission (thus transmitting to vwi nodes only), the largest information transport would be Since C T ≤ C T , we see from ( 6) and ( 7) that Let us now explore the achievability of these upper bounds.Let Assumption 1 hold.In addition, let us assume that there exists a transmission scheme A such that (i) every node in the network successfully transmits r packets every time slot; (ii) each path from a source to the corresponding destination is at most h max hops long; (iii) the paths from each source to destination are the same (relative to the source and the corresponding destination) for every source node; (iv) every node uses the same directions (hops) for its transmissions in every time slot.
Then we can make the following claim.Proof.Due to symmetry between nodes (i.e., ensured by Assumption 1), the total number of source-destination paths passing through every node is the same.The total number of such paths is equal to rn, and the total number of links in them is at most rnh max .So, the number of paths passing through every node is at most rh max .Since every node can send and receive r packets in every time slot, there exists a schedule in which a node serves every path passing through it at least once in h max time slots.This means that every source node can send its own packets at least once in h max time slots using all r beams.By the definition of throughput, this implies that every source node can have a throughput of at least r/h max , which proves the theorem.

Infinitely narrow beamwidth
Let us first consider the case of infinitely narrow beamwidth, that is, D = 0.In this case, the beams are just straight lines.Given m and an arbitrary point ω ∈ Ω(m), let Ω(ω, m) be the set of lattice points in Ω(m) that are vwi to ω.If the beamwidth is infinitely small, a node is vwi to another node as long as no other nodes lie on the line segment between them.It is shown in [20] that, for a square (with boundaries) lattice, regardless of ω, This value can be thought of as the asymptotic fraction of nodes that are vwi to an arbitrary node on the grid.In the case of a torus, the number of nodes (lattice sites) that are vwi to a given node is obviously the same regardless of the lattice site ω.In fact, it is easy to see that for the network on an m × m torus, Figure 2 shows the minimum and maximum values for the quantity Ω(ω, m)/Ω(m) for the square grid (with boundaries).We see that for a torus the ratio |V i |/m 2 always stays above the limiting ratio 6/π 2 .
The following proposition shows that any node on the torus can communicate with any other node 7 in at most two hops.Proposition 1.Any node i can communicate with any other node j in two hops so that node-to-node transmissions are between vwi nodes.
Proof.In order to prove the proposition, we only need to show that the intersection of sets V i and V j is nonempty for any pair of nodes i and j.
Using the set-theoretic equality we can write which proves the lemma.
We can now state the upper bound on the throughput of a lattice ad hoc network with infinitely narrow beamwidth.
Theorem 4. The uniform throughput T for an ad hoc network on a square lattice with g beams of zero width for each node satisfies the inequality Proof.We use Theorem 1.In that theorem, s ≤ g, and, in order to take advantage of the spatial multiplexing, one needs at least 2 hops.So h ≥ 2, which proves the theorem.
As to the achievability of the upper bound, it could in principle depend on the location of sources and destinations.If, as we assumed, they all are separated by a distance of m/2 in both vertical and horizontal directions, then a transmission strategy employing 2 hops for all g paths from every source to destination can be used.Then, as is easy to see, a throughput of exactly g/2 for all source-destination pairs can be achieved whenever |V i ∩ V d(i) | ≥ g (where d(i) stands for the destination of the source i).In other words, the necessary and sufficient condition for the achievability of throughput g/2 is the existence of at least g nodes that are vwi to both the source and the destination.Any g of these nodes can be used as relays.Thus we have the following theorem.Theorem 5.For the lattice ad hoc network with the sourcedestination locations described in Assumption 1, the uniform throughput of g/2 is achievable provided On the other hand, from Lemma 1, we know that |V i ∩ V j | ≥ 2(α − 1)n for any pair of nodes i and j.It follows that the conditions of Theorem 5 are satisfied as long as (2α − 1)n ≥ g.Noting also that, since the source-todestination relative locations are the same for all sourcedestination pairs, all relay nodes will get the same number of packets to forward, we have the following corollary.

Finite beamwidth
Now, let the beamwidth be D > 0. As we will see, in this case the number of nodes that are vwi to a given node will not grow with the network size.Instead, it will be dependent on the beamwidth, resulting in the need for multiple hops in order to reach the destination.
The following theorem establishes a connection between the beamwidth D and the maximum distance H to a vwi node.

Theorem 6. The maximum beamwidth D that can be used to transmit without interference to a node a distance H away is
and the maximum distance for a transmission without interference given a beamwidth D is for any H ≥ 3 and D ≤ 90 • .
Proof.Equation ( 16) can easily be seen to be correct for H ≤ 4 by a straightforward enumeration of possibilities.In the following, we consider the case H ≥ 5. First, we show that D ≤ arctan (1/H − 2) for any vwi node at a distance H from the source.Let the source i be at the origin.Due to lattice symmetry, it is sufficient to only consider vwi nodes with coordinates (H − v, v) where v ≤ (H − 1/2) (the node with coordinates (H/2, H/2) cannot be vwi unles H = 2).We consider cases v ≤ H/3 and v > H/3 separately.

Case 1 ≤ v ≤ H/3
For the node (H−v, v) to be vwi, the corresponding beam has to "clear" the nodes with coordinates (H − v − 1, v) and (H − v − 2, v − 1).Let → u a and → u b be vectors with these coordinates, respectively (see Figure 3).Let us denote the angle between these vectors by θ 1 .Also let θ 0 = arctan (1/H − 2).Thus we need to show that θ 1 ≤ θ 0 .Using the standard vector algebra, we see that On the other hand, for the angle θ 0 we have Subtracting, we obtain where We see that d 1 (v) ≤ 0 for v ≤ H − 2 and v ≥ 1.This implies that cos 2 (θ 0 ) − cos 2 (θ 1 ) ≤ 0 in this range of v.We conclude that θ 1 ≤ θ 0 for all values of v not exceeding H/3.
In this case for the node (H − v, v) to be vwi to the source at the origin, the corresponding beam has to "clear" the nodes → u c be vectors with these respective coordinates (see Figure 4).Let θ 2 be the angle between these vectors.The use of standard vector algebra yields EURASIP Journal on Wireless Communications and Networking (0, 0) Subtracting cos 2 (θ 2 ) from cos 2 (θ 0 ), we obtain where The second derivative of d 2 (v) is It is easy to see that d 2 (v) < 0 everywhere as long as H ≥ 5.This implies that the first derivative d 2 (v) is monotonously decreasing everywhere and has one real root v 0 .Setting v = H − 1/2 and evaluating the first derivative, we obtain Finally, it follows directly from ( 16) that if the beamwidth D is given, then the maximum hop length H to a vwi node can be found as Theorem 6 described the largest L 1 distance for which a node is vwi to a source for a given beamwidth.It was also found that this maximum distance is achieved by a node whose position is one lattice point above the horizontal (v = 1).Numerical evaluation shows that for larger beamwidths, almost all the nodes within a certain distance can be vwi.As the beamwidth is decreased, the nodes along the horizontal and vertical directions can be vwi disproportionately more.Therefore, for larger beamwidths, the envelope of vwi nodes looks like a diamond, and as the beamwidth decreases, the envelope becomes more and more cross-like in appearance (see Figure 5).Some nodes within this envelope cannot be vwi since some nodes may directly block other nodes.For example, a node at lattice point (1, 1) cannot "see" a node at lattice point (3, 3) since a node at (2, 2) is blocking it.We found the total number of nodes vwi to a source node for various distances and for various beamwidths (Figure 6), using numerical evaluation.

Upper bound
Before stating the upper bound on the uniform throughput, let us define S max as the number of nodes that are vwi to a given node.This number depends on the beamwidth only.
Let α be an upper bound 8 on the fraction of nodes that are vwi for a given node in the zero beamwidth case.

Lemma 1.
In terms of the maximum hop size, Proof.For a system with nodes on a grid, it is obvious that the maximum number of nodes within a distance H can be found by counting the nodes within two squares of sides H and H + 1 surrounding the source.This is shown in Figure 7 for an H = 5.The total number of nodes within these squares is ( Removing the source node itself from the count results in a total of 2 H 2 + 2 H nodes within a distance H. Multiplying by the maximum fraction of nodes which are vwi α yields the statement of the lemma.
We can now obtain an upper bound on the throughput.

Theorem 7. The uniform throughput T satisfies the inequality
Proof.We know from Theorem 1 that T ≤ s/h, where h is the average length of the shortest source-to-destination path (measured in hops).It is clear that s ≤ min {g, S max , α n}.Also, for any transmission scheme, h ≥ max {2, √ n/ H} (the latter is because the distance between sources and destinations is equal to m, and the longest possible hop is equal to H).Noting that √ n/ H > 2 implies α n > S max , we obtain the statement of the theorem.

In case
√ n/ H > 2, that is, when it takes more than two hops to reach the destination from the corresponding source, the upper bound of Theorem 7 can be further tightened.

Theorem 8. If
√ n/ H > 2, the uniform throughput can be upper bounded as Proof.It follows from Theorem 2 by noting that all sourcedestination distances are equal to m = √ n, and therefore L = √ n.

Achievability
Let us assume, without loss of generality,9 that g < S max and g < α n.Under these assumptions, the upper bounds of Theorems 7 and 8 take the form Let us consider the two cases separately.
Consider the following transmission scheme.

Transmission Scheme 1
In this transmission scheme, the node-to-node transmissions are always to vwi nodes.The g successful transmissions from each node are possible in every time slot.Let us denote the possible hops to vwi nodes by the corresponding length in horizontal and vertical directions.Thus, if a transmission to a vwi node can be made in which a packet moves by k lattice space in horizontal direction and by l lattice space in the vertical direction, we denote such hop by (k, l).Because of system symmetry (no boundaries), all nodes have the same vwi hops available to them.Due to lattice symmetry, for every (k, l) vwi hop there is sgn(lk)(l, k) vwi hop. 10 Let us assume, for additional simplicity, that the number of beam g is divisible 11 by 4. Let (k 1 , l 1 ), (l 1 , k 1 ), . . ., (k g/2 , l g/2 ) be g hops listed from the largest value of the hop length |k| + |l| in a nonincreasing order.Our goal is to construct g paths from a source to the corresponding destination in such a way that (1) each node in the network is able to transmit to g vwi nodes in every time slot; (2) each node uses the same g hops in a given time slotthis ensures that every node receives exactly g transmissions.(5) Otherwise, choose sets of nodes 13 S (i) 1 ⊆ V ik 1 d(i) and Continue in the same way until either the required path continuations are found or declared to be impossible to find.In the former case the output of the algorithm will include sets The cardinalities of sets V id(i) for different values of D are shown in Table 1.
Thus we arrive at the second phase of source-todestination transmission.

Phase II
Forward the packets waiting at node i kj , j = 1, 2, . . ., r, to the destination via two hops: from i kj to one of the nodes in the set S (i) j and from that node to the destination.
We can now state the achievability result.Let l min be the smallest L 1 hop length used in Phase I of Transmission Scheme 1.
Theorem 9.The uniform throughput of at least g/ √ n/l min + 2 is achievable provided g ≤ |V i | and g path continuations can be found using Algorithm 1.
Proof.The proof follows directly from Theorem 3 where we set r = g and h max = √ n/l min + 2. It only remains to be noted that in order to make the overall transmission strategy (Transmission Scheme 1) satisfy the conditions of Theorem 3, we need to synchronize Phase I and Phase II.Namely, out of every √ n/l min + 2 time slots, √ n/l min are 13 Note that the set S (i) 1 chosen at this step may be different from S (i) 1 chosen at the previous step.dedicated to all nodes performing Phase I and 2 time slots to all nodes performing Phase II.
For some specific values of g and D we can actually establish the feasibility of Algorithm 1 and make more specific claims.For example, if g = 8, it is easy to see from Table 1 that finding the required path continuations are possible for any D ≤ 10 • .A brief consideration of the worst case scenario also shows that this is true for g = 16 as well.For larger values of g closer inspection would be needed.We can formulate these observations as a corollary.Let H = 2 + 1/ tan D as shown in Theorem 6.

Corollary 2. The uniform throughput of at least g/(
In this case, the inequality g H/ √ n > 1/2 holds as well, and, therefore, the source for every destination is located within the corresponding square S i .As we have already seen, twohop transmission is possible under these conditions.We have the following theorem. Theorem 10.The uniform throughput of at least g/2 is achievable provided V id(i) ≥ g.

Numerical results
The upper and lower bounds on uniform throughput per node for a system with beamwidth 25, 15, 10, and 5 degrees were numerically computed.The number of beams is set to g = 8.The throughput is measured in terms of W. Each beam is capable of sending W bits of data per second.The results are shown in Figures 10, 11, 12, and 13.These figures also show the size of the network required to bring the throughput per node down to one W, where there is no longer any spatial multiplexing gain.The results show that the network can be very large (thousands of nodes even for 10 degree beamwidth) before this occurs.Therefore, using spatial multiplexing with directional antennas, ad hoc networking can be implemented in practical sized systems without experiencing performance degradation (compared with the individual link rate) even for fairly wide beamwidths.

CONCLUSION AND DISCUSSION
We analyzed the throughput of ad hoc networks with nodes located on a square lattice with periodic boundary conditions.For the case of infinitely narrow beamwidth, we found that a uniform throughput proportional to the maximum number of beams a node can form and independent of the system size is achievable.
We also showed large gains (compared to systems without spatial multiplexing) for a practical system with a small number of antennas and a finite practically achievable beamwidth.These gains have been shown to offset the interference effect on throughput up to network sizes in the  thousands thus making network of such sizes effectively unaffected by interference-related throughput degradation.
Our results demonstrate that there is a strong incentive to design and deploy ad hoc networks with good directional antenna or beamforming capability in order to improve capacity or simplify the communication protocol design.
Obviously, one of the limitations of the proposed analysis approach is the assumption that the nodes are located on a regular square grid.While this assumption makes the analysis tractable, it does not fully reflect the topology of the ma-  jority of real networks.While the full consideration of more realistic models goes well beyond the scope of this paper, we will attempt to sketch an argument showing that the main results would likely not change much under a more realistic model.
The main result of the paper depends on the ability to simultaneously transmit along g beams over long distances (longer than the typical internode distance).To consider a different, perhaps more realistic model, let us assume that the nodes are placed randomly with a unit density inside a circle (or square) of area n 2 .The probability that a node at a distance r from a source is vwi to it can be approximately calculated (using the two-dimensional Poisson distribution) as e −r 2 D/2 (see Figure 14).The total number of nodes at a distance r from the source can be approximately calculated as 2πr (since the node density is unity).Therefore, the number of nodes at a distance r from the source that are vwi to it can be estimated as 2πre −r 2 D/2 .So for g = 8, one should be able to find 8 relays at a distance r ≈ 6 from the source for D = 5 • .For D = 10 • , one should be able to find 8 relays at a distance of r ≈ 3, and for D = 15 • , 8 relay nodes can be found at a distance of r ≈ 2. We see that these distances are roughly 2 times smaller than those found in this paper for the square grid model.This would result in the "efficient" network size (the size for which the throughput is no less than W) roughly 4 times smaller.Or, equivalently, the throughput for the same network size would be roughly 2 times smaller than the one found in this paper.It is also fairly easy to see that the total number of vwi nodes for both models is about the same for the same value of D, but the regular grid model has an advantage in the largest hop distance because of the presence of "preferred" directions along the grid's main axes.Also note that the real-life networks may well have such directions (i.e., along major streets in urban networks) and this would most likely lead to values of the achievable throughput between those found in this paper and those that can be obtained via a careful analysis of a random node placement model.

Figure 1 :
Figure 1: A node with 4 transmitting beams of angular width D.

Figure 3 :
Figure 3: In order for the node (H − v, v) to be vwi to the source at the origin, the corresponding beam has to "clear" nodes (H − v − 1, v) and (H − v − 2, v − 1).In this figure, H = 10 and v = 3 so that v < H/3.

Figure 4 :
Figure 4: In order for the node (H − v, v) to be vwi to the source at the origin, the corresponding beam has to "clear" nodes (H − v − 2, v − 1) and (H − v − 1, v − 1).In this figure, H = 11 and v = 4 so that v > H/3.

Figure 5 :Figure 6 :
Figure 5: Envelope of nodes vwi to a source for finite beamwidth.Only the nodes inside the envelope can be vwi to the node in the origin.

Figure 10 :
Figure 10: Throughput T versus n (upper and lower bounds) for beamwidth of 25 degrees.

Figure 11 :
Figure 11: Throughput T versus n (upper and lower bounds) for beamwidth of 15 degrees.

Figure 12 :
Figure 12: Throughput T versus n (upper and lower bounds) for beamwidth of 10 degrees.

Figure 13 :
Figure 13: Throughput T versus n (upper and lower bounds) for beamwidth of 5 degrees.

Figure 14 :
Figure14: The sector of area r 2 D/2 has to contain no nodes for the node j at a distance r from the source to be vwi to the latter.