A Novel Geometric Diagram and Its Applications in Wireless Networks

Transcription

1 A Novel Geometric Diagram and Its Applications in Wireless Networks Guangbin Fan * and Jingyuan Zhang * Department of Computer and Information Science, University of Mississippi University, MS 38677, gfan@cs.olemiss.edu Department of Computer Science, University of Alabama Tuscaloosa, AL 35487, zhang@cs.ua.edu Abstract Wireless networks have a lot of geometric properties since the signal strength corresponds directly to the distance. Earlier research findings in geometry, such as Voronoi diagram, have found a lot of applications in wireless networks. In this paper, a new geometric diagram, referred to as the umbrella diagram, is proposed. The umbrella diagram is comparable to the Voronoi diagram. The Voronoi diagram deals with a set of points, whereas the umbrella diagram deals with a set of different networks. The umbrella diagram of a set of hexagonal networks is to divide the plane into regions such that the points in a region have a larger distance to one network than to the other networks. Unlike the Voronoi diagram, the regions in the umbrella diagram are not necessarily convex. The paper gives an efficient solution to compute the umbrella diagram of n networks with the same cell size and orientation. Like the Voronoi diagram, the umbrella diagram has potential to be used in many areas. To illustrate its usefulness, the paper further gives two applications in wireless networks: the optimal network deployment and the maximum base station reuse. The optimal network deployment intends to find the best position to fix the network such that the network is closest to all existing base stations. The maximum base station reuse problem is to maximize the number of base stations that can be reused under a given bound. Keywords-Voronoi Diagram; Umbrella Diagram; Wireless Networks; Computational Geometry I. INTRODUCTION It is well known that there are two popular tessellations of a plane with regular polygons of the same kind: square and hexagonal [11]. In the square tessellation, each square has four (or eight) neighbors, and in the hexagonal tessellation, each hexagon has six neighbors. Fig. 1.1 illustrates both square and hexagonal tessellations. A tessellation can also be represented by the centers of its corresponding polygons, shown as the dots in Fig A lot of designs are based on those two popular tessellations. For example, mesh-connected computers are based on the square tessellation. The design of cellular networks is based on the hexagonal tessellation. In this paper we will consider the hexagonal tessellation, which closely approximates the circular radiation patterns and achieves the maximum coverage with a given number of nodes. Specifically, we will propose and compute the umbrella diagram of a set of different hexagonal tessellations, and show how to use the umbrella diagram to solve problems in the area of wireless networks. (The reason that the diagram is named as the umbrella diagram will be explained after its formal definition.) Figure 1.1 Square and hexagonal tessellations. To understand the umbrella diagram, we first introduce the well known closest-point and farthest-point Voronoi diagrams [9]. We assume a set S of n points in the plane, S = {P 0, P 1,,P n-1 }. The closest-point Voronoi diagram (or simply Voronoi diagram) of S is to find, for each point P i in S, the locus of points that are closer to P i than to any other points of S. The farthest-point Voronoi diagram of S is to find, for each point P i in S, the locus of points that are farther to P i than to any other points of S. The closest-point diagram of S divides the plane into n convex regions, each of which is associated with a point in S. Although the farthest-point diagram of S also divides the plane into convex regions (unbounded), each region is associated with a point in S that is located on the convex hull of S. A point in S that is within the convex hull of S does not affect the diagram [9]. Fig. 1.2 illustrates the closest-point and farthest-point Voronoi diagrams of 8 planar points. The left figure corresponds to the closest-point diagram that consists of 8 regions. The right figure, corresponding to the farthest-point diagram, consists of only 5 regions. The points located within the convex hull, P 5, P 6 and P 7, do not have an associated region. The Voronoi diagrams have found a lot of applications in areas ranging from archaeology to zoology [1, 4-8]. It can be used in graphics, image processing, wireless networks, robotics, motion planning, pattern recognition, etc. In [2], the author has presented a detailed survey of Voronoi diagrams and their applications. Voronoi diagrams are essential for

2 wireless networks. In a cellular network, the area covered by a base station can be viewed as a Voronoi region of the base station. Voronoi diagram has also been used in mobile ad hoc networks and sensor networks. In [8], a localized algorithm is developed based on the Voronoi diagram to solve the exposure problems in sensor networks. Figure 1.2 The closest-point and farthest-point Voronoi diagrams of a set of planar points. The proposed umbrella diagram is comparable to the Voronoi diagram. The Voronoi diagram deals with a set S of n planar points, whereas the umbrella diagram deals with n networks (patterns, or tessellations), each of which is determined by a point in S. In this paper, we assume honeycomb networks. It is easy to see that a honeycomb network is determined by three factors: cell size, network orientation, and the position of one cell center. Therefore, if the cell size and orientation is fixed, one single point will determine a network. A set of n points will determine n networks (or n tessellations). The umbrella diagram of a set S of n planar points divides the plane into regions. Each region is associated with a point in S. The points in the region associated with the point p in S have a larger distance to the network determined by p than to the networks determined by the other points in S. Here the distance from a point to a network is the distance between the point and its closest cell center in the network. Please refer to Fig. 2.2 for an illustration of the umbrella diagram of 3 planar points. We will give an efficient solution to compute the umbrella diagram of n planar points with a time complexity of O(n 3 ). The umbrella diagram can be used to solve problems in wireless networks. In this paper, we apply the diagram to solve two problems in wireless networks. One problem, referred to as the optimal network deployment, can be described as: given a set of existing base stations that are non-uniformly distributed, and a planned cellular network, find a deployment of the planned network such that the maximal distance between the existing base stations and their corresponding cell centers is minimized. The second application considers more constraints in network planning. In practice, most system designs permit a base station to be positioned up to one-fourth the cell radius away from the cell center to ensure an acceptable coverage [10]. Therefore the optimal network computed in the first application might not meet this constraint for all base stations. We assume the maximum allowed distance to be a constant d. So the problem, referred to as the maximum base station reuse, is to deploy the network such that number of nodes (old base stations) that can be located within distance d to their corresponding closest cell center is maximized. To the best of our knowledge, there is no solution to those problems reported in the literature. These problems can get solved utilizing the proposed umbrella diagram. One uses one application of the umbrella diagram of multiple nodes, and the other uses multiple applications of the umbrella diagram of two nodes. Although the solutions are given in cellular networks, the idea in general describes the important relationship between the maximum coverage (honeycomb) tessellation and the arbitrarily distributed nodes, which can be further used in the wireless ad hoc networks, image processing, etc. The rest of this paper is organized as follows. In Section 2, we define the umbrella diagram of a set of planar points. In Section 3, we show how to find the umbrella diagram. We further apply this diagram to solve two problems in wireless networks in Section 4. Finally we will summarize the results in Section 5. II. DEFINITION OF THE UMBRELLA DIAGRAM In this section, we will define the umbrella diagram of a set of planar points. The hexagonal plane tessellation divides the plane into hexagons as shown on the right side of Fig By using the hexagon geometry, the fewest number of cells can cover a geographic area, and the hexagon closely approximates a circular radiation pattern which would occur with an omni-directional antenna and free space propagation [10]. The hexagonal tessellation is also known as the honeycomb network [11]. In the following discussion, we will omit the word honeycomb when no confusion will arise. A network can be represented by a set of hexagons or a set of hexagon centers. A network is completely determined by three factors: the cell size, network orientation and the position of one cell center. If the cell size and orientation is fixed, the hexagon is solely determined by the position of one cell center. We define a network with a fixed cell size and orientation as movable. Definition 2.1 A network is movable if the cell size and orientation of the network is fixed but the position of the network is not fixed. In the following discussion, we will also refer to a movable network as a network pattern. Given a network pattern, if we fix the center of one cell at the position C, the network is completely fixed. We call the fixed network as the spreading network of C. Definition 2.2 Given a position C and a network pattern T, the spreading network of C with the pattern T, denoted as T(C), is the network with C being one of its cell centers. C is referred to as the starting point of T(C). Fig. 2.1 illustrates two spreading networks, T(P 0 ) and T(P 1 ), with a network pattern T. Next we will define the distance from a point to a network. It can be used to describe

3 the error between a base station and its ideal position in a newly planned network. Figure 2.2 The umbrella diagram of P 0, P 1, and P 2. Figure 2.1 Two spreading networks, T(P 0) and T(P 1), with a network pattern T. Definition 2.3 Given a point p and a network N, the distance between the point p and the network N, denoted by d(p,n), is defined as the distance between the point p and its closest cell center of N. d(p,n) is also called the distance from p to N. It is easy to verify that if the point p is located within a cell with its center being c, the network distance between p and N, d(p,n), is the distance between p and c since c is the closest cell center to p. Finally, we will define the umbrella diagram of a set of planar points. We will define the diagram of two points, followed by the diagram of multiple points. Definition 2.4 Given two planar points P 0 and P 1, and a network pattern T, the umbrella diagram of P 0 and P 1 consists of two regions, denoted as A T (P 0, P 1 ) and A T (P 1, P 0 ), where A T (P 0, P 1 ) is the locus of points x such that d(x, T(P 0 )) d(x, T(P 1 )), and A T (P 1, P 0 ) is the locus of points x such that d(x, T(P 1 )) d(x, T(P 0 )). The diagram is named as the umbrella diagram because of the umbrella shape of the 3d surface with equation z = x 2 + y 2, cut by a hexagon moving along the z axis. Definition 2.5 Given a set S of n planar points, S = {P 0, P 1,,P n-1 }, and a network pattern T, the umbrella diagram of S is a set of n regions, {A T (P i, S) 0 i n-1 }, where A T (P i, S) is the locus of points x such that d(x, T(P i )) d(x, T(P j )) for all j i. A T (P i, S) is called the umbrella region of P i. Fig. 2.2 illustrates the umbrella diagrams of 3 points. For clarity, we use U(P i ) instead of A T (P i, S) in the figure. The polygons of the same shape belong to the umbrella region of a point. We mark only a few polygons of each umbrella region in the figure. The umbrella diagram can be also defined from another point of view as follows: Definition 2.6 Given a set S of n planar points, S = {P 0, P 1,,P n-1 }, and a network pattern T, the umbrella diagram of S divides the plane into regions, each of which is associated with a point P i in S, such that the spreading network starting from a point in the umbrella region of P i is farther from P i than any other points in S. III. COMPUTING THE UMBRELLA DIAGRAM In this section, we will show how to compute the umbrella diagram of a set of planar points with a network pattern T. We first show how to find the umbrella diagram of two points. Then we extend our idea to find the umbrella diagram of a set of points. Finally we will present the algorithm and analyze its time complexity. A. Umbrella Diagram of Two Points Given two points, P 0 and P 1, and a network pattern T, we will show how to compute A T (P 0, P 1 ) and A T (P 1, P 0 ). A T (P 0, P 1 ) consists of all points that have a larger distance to T(P 0 ) than to T(P 1 ). Since the network T(P i ), i = 0 or 1, is the repeating of the hexagon centered at P i, the regions A T (P 0, P 1 ) and A T (P 1, P 0 ) are also repeatable. We will show how to compute A T (P 0, P 1 ). Without loss of generality, we will consider the points within the hexagon centered at P 0, referred to as H(P 0 ). Assume P is a point located within H(P 0 ). It is easy to verify that d(p, T(P 0 )) = PP 0. Next we will compute d(p, T(P 1 )). We assume H(P 0 ) is not identical to any hexagon of the network T(P 1 ). If they are identical, it is obvious that d(p, T(P 1 )) is always the same as d(p, T(P 0 )). Under that assumption, H(P 0 ) will intersect with either 3 or 4 hexagons of T(P 1 ) as shown in Fig. 3.1.

4 most five edges. Therefore, A T (P 0, P 1 ) within H(P 0 ) consists of 4 convex polygons, each of which has at most five edges. Figure 3.1 Two kinds of intersections between H(P 0) and T(P 1). We will consider the 4-intersection case with the 3- intersection case being similar. Assume the 4 hexagons of T(P 1 ) that are intersected with H(P 0 ) are centered at Q 0, Q 1, Q 2, and Q 3 as shown in Fig Since H(Q 0 ), H(Q 1 ), H(Q 2 ), and H(Q 3 ) completely cover H(P 0 ), we only need to consider H(Q 0 ), H(Q 1 ), H(Q 2 ), and H(Q 3 ) when computing d(p, T(P 1 )). Specifically we have the following. Lemma 3.1 Given two points, P 0 and P 1, and a network pattern T, assume Q 0, Q 1, Q 2, and Q 3 are the centers of the hexagons in T(P 1 ) that intersect with H(P 0 ). For any position P H(P 0 ), d(p, T(P 0 )) = PP 0 and d(p, T(P 1 )) = PQ i if P I i = H(P 0 ) I H(Q i ). Proof. If P I i = H(P 0 ) I H(Q i ), then P H(P 0 ) and P H(Q i ). Therefore d(p, T(P 0 )) = PP 0, and d(p, T(P 1 )) = PQ i. However, in order to compute the diagram of multiple points based on the diagram of two points, we will compute A T (P 0, P 1 ) and A T (P 1, P 0 ) a little bit differently. We will find the intersections of the bisector line segments (or their extensions). The bisector line segments l i and l i+1 (or their extensions) intersects on the boundary of cells H(Q i ) and H(Q i+1 ). This is because the intersection is the circumcenter of the triangle P 0 Q i Q i+1. If H(P 0 ) intersects with 4 hexagons in T(P 1 ), the polygon formed by the perpendicular bisectors (or their extensions) will be a quadrilateral; if H(P 0 ) intersects with 3 hexagons in T(P 1 ), the polygon formed will be a triangle as shown in Fig Denote the formed polygon by L(P 0,P 1 ) and the area outside L(P 0,P 1 ) by ~L(P 0,P 1 ). In addition, the contour of L(P 0,P 1 ) is visible from P 0. We summarize our finding in the following theorem. Theorem 3.2 Given two points, P 0 and P 1, and a network pattern T, A T (P 0,P 1 ) = ~L(P 0,P 1 ) I H(P 0 ) and A T (P 1,P 0 ) = L(P 0,P 1 ) I H(P 0 ) within H(P 0 ). Equivalently, if we compute A T (P 0, P 1 ) and A T (P 1, P 0 ) within H(P 1 ), then A T (P 0, P 1 ) = L(P 1,P 0 ) I H(P 1 ) and A T (P 1,P 0 ) = ~L(P 1,P 0 ) I H(P 1 ). If we put together the polygons of A T (P 0,P 1 ) I H(P 0 ), the resulting shape will be exactly the same as A T (P 0, P 1 ) I H(P 1 ). It is easy to see that A T (P 0, P 1 ) and A T (P 1, P 0 ) can be computed in constant time. Fig. 3.3 illustrates the umbrella diagrams of P 0 and P 1. For clarity, we use U(P 0 ) as A T (P 0, P 1 ) and U(P 1 ) as A T (P 1, P 0 ) in the figure. The polygons of the same shape belong to the umbrella region of the same point. We mark only a few polygons of each umbrella region in the figure. Figure 3.2 The construction of L(P 0, P 1). I i = H(P 0 ) I H(Q i ) is a convex polygon because both H(P 0 ) and H(Q i ) are convex polygons. It is easy to verify that I i consists of at most six edges. To compute the umbrella diagram of P 0 and P 1, we construct the perpendicular bisector l i of P 0 Q i in I i = H(P 0 ) I H(Q i ) as shown in Fig By Lemma 3.1, for any point P I i = H(P 0 ) I H(Q i ), if P is on the line segment l i, d(p, T(P 0 )) = PP 0 = PQ i = d(p, T(P 1 )); if P is on the side that is closer to P 0, d(p, T(P 0 )) = PP 0 < PQ i = d(p, T(P 1 )); if P is on the side that is farther to P 0, d(p, T(P 0 )) = PP 0 > PQ i = d(p, T(P 1 )). It is also easy to see that I i is divided by l i into two convex polygons, each of which has at Figure 3.3 The umbrella diagram of P 0 and P 1. B. Umbrella Diagram of a Set of Points In this subsection, we assume a set S of n planar points, S = {P 0, P 1,,P n-1 }. For each point P i, we will find A T (P i, S), the locus of points that have a larger distance to the spreading network of P i than to the spreading network of any other

5 points in S. In the previous subsection, we have shown how to compute the umbrella diagram of two points. Specifically for any two points P i and P j with j i, we know how to compute A T (P i ). By definition, A T (P i, S) = I AT ( Pi, Pj ). We will show how to compute A T (P i, S) within H(P i ). By Theorem 3.2, for j i, A T (P i,p j ) = ~L(P i,p j ) I H(P i ) within H(P i ), Therefore, within H(P i ), A T (P i, S) = I A ( P, P ) = (~ L( P, P ) I H ( P )) T = P ) I H ( i ( I = H ( P i ) I (~ U j i i j I Let L(P i, S) = U H P ) I ( i ~ L( P i )) L P i ) ( ). L P i ) i j (, then A T (P i, S) = ~L(P i, S). We summarize this in the following theorem. Theorem 3.3 Given a set S of n planar points, S = {P 0, P 1,,P n-1 }, and a network pattern T, A T (P i, S) within H(P i ) is equal to H ( P i ) I ~ L(P i, S) with L(P i, S) = U L( P i ). j i i The contour of L(P 0, S) is shown in a darker color in the figure. The umbrella region of P 0 within H(P 0 ), which is represented by the hatched part in the figure, consists of all points that are outside L(P 0, S) yet inside H(P 0 ). Once the umbrella regions of all P i S are computed, the umbrella diagram can be formed. As illustrated in Fig. 3.5, the umbrella regions are likely to be non-convex. Furthermore, the region of a point, e.g. U(P 0 ) in the figure, can consists of multiple polygons of different shape. C. The Algorithm and Its Time Complexity The algorithm to compute the umbrella diagram of n planar points can be described as follows. Algorithm: Umbrella Diagram Input: a set S of n planar points, S={P 0, P 1,,P n-1 }, and a network pattern T; Output: the umbrella diagram of S, that is, A T (P i, S) within H(P i ) for 0 i n-1. BEGIN FOR every point P i S FOR every point P j S ( j i ) Find the cell centers Q 0,,Q m-1 (m = 3 or 4) of T(P j ), the spreading network of P j, such that H(Q k ) intersects with H(P i ); Compute L(P i,p j ) formed by the perpendicular bisectors of P i Q k; END FOR; Compute L(P i, S) = U L( P i ) ; Compute A T (P i, S) = ~ L(P i, S) I H ( P i ) ; END FOR; END. Figure 3.4 Computing A T(P 0, S) within H(P 0) with S = {P 0, P 1, P 2, P 3}. Theorem 3.3 tells us how to compute the umbrella region of P i, A T (P i, S), within H(P i ). To compute A T (P i, S) within H(P i ), we first find L(P i,p j ) for every j i. Recall that L(P i,p j ) is a triangle or a quadrilateral whose contour is visible from P i. Then we compute L(P i, S) = U L( P i ). Finally find the part of H(P i ) that is located outside of L(P i, S). Fig. 3.4 illustrates how to compute A T (P 0, S) within H(P 0 ) with S = {P 0, P 1, P 2, P 3 }. L(P 0, S) is the union of L(P 0,P 1 ), L(P 0,P 2 ), and L(P 0,P 3 ). Figure 3.5 Umbrella diagram of S = {P 0, P 1, P 2, P 3}. Now we analyze the time complexity of the algorithm. Given two points P i and P j (j i), polygon L(P i,p j ) is either a triangle or a quadrilateral which can be computed in constant

6 time. Therefore, for a point P i, it takes O(n) time to compute all the polygons L(P i,p j ) for 0 j n-1 and j i. U Next we will show how long it takes to compute L(P i, S) = L(, ). Lemma 3.4 shows that L(P i, S) is a star-shaped P i P j polygon with O(n 2 ) edges. Lemma 3.4 The outer contour of L(P i, S) = U L( P i ) is a star-shaped polygon visible from P i with at most 4(n-1) 2 vertices. Proof. For simplicity, we call L(P i,p j ) as L j. We will prove L(P i, S) is a star-shaped polygon visible from P i by induction. Suppose C(k) is the union of k Ls. We will compute C(k+1) that is the union of C(k) and L k+1. We assume that p is an intersection of C(k) with L k+1. Because p is visible from P i in both C(k) and L k+1, p is visible from the P i in C(k+1). Therefore, all end-points in the union C(k+1) are visible from P i, that is, C(k+1) is a star-shaped polygon. That concludes L(P i, S) is a star-shaped polygon visible from P i. Now we compute the number of vertices on L(P i, S) in the worst case. An end-point on the outer contour of L(P i, S) is either the intersection of any two Ls, or an original vertex of L. There are three cases concerning the intersections for a pair of Ls, L(P i,p a ) and L(P i,p b ), as shown below. get the outer contour of the original n Ls. We first merge the angle list θ 1, θ 2,, θ u and the angle list θ 1, θ 2,, θ v. Note that C(A) and C(B) can have at most one intersection between any two consecutive angles in the resulting list. Without lost of generality, we only need to consider 6 cases for two consecutive angles α and β as shown in Figure 3.6. (Some symmetric cases have not been listed.) Case a: Keep the end points atα and β. Case b: Discard the end points atα and β. Case c: Find the intersection, and let γ be the angle for the intersection. Discard the end point at α, keep the end point at β, and add a new point at γ. Case d: Discard the end point at α, and keep the end point at β. Case e: Find the intersection, and let γ be the angle for the intersection. Keep the end points at α and β, add a new point at γ. Case f: Find the intersection, and let γ be the angle for the intersection. Discard the end points at α and β, and add a new point at γ. Number of L(P i,p a ) L(P i,p b ) intersections triangle quadrilateral 6 triangle triangle 6 quadrilateral quadrilateral 8 For each case, the number of intersections is no more than 8. There are (n-1)(n-2)/2 pairs of Ls. Therefore, the total number of end points on the outer contour of L(P i, S) is at most 8(n- 1)(n-2)/2 + 4(n-1) = 4(n-1) 2. Next we show that, given a point P i, it takes O(n 2 ) time to compute the outer contour of L(P i, S) = U L( P i ). We apply the divide and conquer approach. Again we denote L(P i,p j ) as L j, and assume P i is at the origin o. To find the outer contour of n Ls, we divide the n Ls into two groups, A and B, each of n/2 Ls. Assume the outer contours of A and B are C(A) and C(B) respectively. Realizing each end point p can be represented by the distance ρ between p and the origin o and the angle θ made by op with the x axis, the outer contour of k vertices can be represented by a list < ρ 1, θ 1 >, < ρ 2, θ 2 >,, < ρ k, θ k > with θ 1 < θ 2 < < θ k. Let C(A) = < ρ 1, θ 1 >, < ρ 2, θ 2 >,, < ρ u, θ u > and C(B) = < ρ 1, θ 1 >, < ρ 2, θ 2 >,, < ρ v, θ 2 v >. By Lemma 3.4, u ( n 2) 2 and v ( n 2). Now we will combine C(A) and C(B) to Figure 3.6 Six possible cases when merging two outer contours. It is easy to see that it takes O(n 2 ) time to find the union of C(A) and C(B) by Lemma 3.4. Let T(n) be the time complexity for computing the outer contour of n Ls, we have the following recursion. T(n) = 2T(n/2)+O(n 2 ) T(1) = O(1) Therefore T(n) = O(n 2 ). Once L(P i, S) = U L( P i ) is found, it takes O(n 2 ) time to compute the part of the L(P i, S) that is inside H(P i ). The part computed is, A T (P i, S) within H(P i ). Finally the body of the

7 outer loop will be executed for n times. Therefore, the total time complexity of the algorithm is O(n 3 ). We summarize our findings in the following theorem. Theorem 3.5 Given a set S of n points, S={P 0, P 1,,P n-1 }, and a network pattern T, the umbrella diagram of S with the pattern T can be solved in O(n 3 ) time. IV. APPLICATIONS OF THE UMBRELLA DIAGRAM IN WIRELESS NETWORKS The umbrella diagram in general characterizes the relationship between uniform patterns and arbitrarily distributed nodes, which exists in many areas such as wireless networks, graphics, image processing, pattern recognition, etc. In this paper, we give two applications of the umbrella diagram in wireless networks. To demonstrate the applications of the diagram, we only focus on the geometric constraints. The first problem is to find an optimal deployment of a planned network that reuses all the existing base stations. The second problem solves a constrained network deployment problem which maximizes the number of old base stations under a given bound. These design issues often exist in the network expansion, and the overlay design of integrated systems, where a new 3G system is superimposed over the existing 2G system. A. Optimal Cellular Network Deployment with the Reuse of All Existing Base Stations In a cellular network, a service area is divided into smaller areas called cells. Each cell is served by a base station. Cellular designs often use the honeycomb configuration because it is able to cover the maximum area with the minimum number of nodes [3, 10]. A uniform configuration is always used in the planning although the final base station placement may not be uniformly distributed because in reality an obstacle such as a river prevents a base station from being placed at a cell center. When the current cellular network can not meet the needs of rapidly growing subscribers, it is common to plan a new cellular network with a higher capacity, and it is wise to use existing base stations in the new network. While new base stations can be placed according to the cell centers based on the planned cellular network, an existing base station should be put as close to a cell center as possible in the planned cellular network. We assume the planned network is a uniform one, while the existing base stations are arbitrarily distributed. Our goal is to place the planed network such that the distance between existing base stations and their corresponding cell centers in the network is minimized. Some definitions are in order. First we will define the network difference between a base station and a cellular network. The network difference between a base station P and a cellular network N, denoted by nd(p,n), is defined as the distance between the base station P and the closest cell center of T. If a base station P is located within a cell with its center being C, the network difference between P and the cellular network N, nd(p,n), is the distance between P and C, i.e. PC. Second we will define the network difference between a group of base stations and a cellular network. The network difference between a set S of base stations and a cellular network N, denoted by nd(s,n), is the largest network difference between all the base stations and the network, that is, nd(s,n) = max{ nd(p i,n) P i S }. Figure 4.1 The network difference between a set of points and a network. For example, given the cellular network N as shown in Fig. 4.1, the base station P 0 is located within the cell centered at A. Therefore, the network difference between P 0 and the cellular network, nd(p 0,N), is the distance between P 0 and A, i.e. P 0 A. Similarly the network difference between P 1 (P 2, P 3 ) and network N is P 1 B ( P 2 C, P 3 D ). Suppose P 2 C is the largest among P 0 A, P 1 B, P 2 C and P 3 D, the network difference between the set S ={P 0, P 1, P 2, P 3 } and the network N is P 2 C. Finally we will define the problem of the optimal cellular network deployment. Given a set S of non-uniformly distributed base stations and a planned cellular network T (In fact, T is a network pattern because it is movable.), the optimal cellular network deployment is to find the position of one cell center for the planned network T such that the network difference between those base stations and the planned network T is minimized. Next we will show how to solve the problem of the optimal cellular network deployment using the umbrella diagram. Given a set S of n base stations S={P 0,P 1,,P n-1 }, and a planned network T, we are not sure that which base station has the largest network difference to the optimal deployment of T. Let us assume that P i has the largest network difference to the optimal deployment of T. We compute the umbrella region of P i under T, A T (P i, S), within H(P i ), which consists of all points that have a larger distance to T(P i ) than to T(P j ) with all j i. To solve the problem of the optimal cellular network deployment, we need to find the point, C i, on A T (P i, S) within H(P i ) such that C i has the smallest distance to

8 P i. By Lemma 3.4, A T (P i, S) within H(P i ) consists of polygons with a total of O(n 2 ) edges. Therefore it takes O(n 2 ) time to find C i. We find C i for 0 i n-1. Let d i be the distance between P i and C i. Therefore d i is the minimal distance between S and any network deployment of T under the condition that P i has a larger network difference to the deployment than any other base stations. Assume d k = min{d 0, d 1,,d n-1 }. Then T(C k ) is the optimal deployment of the planned network T that reuses P 0, P 1,,P n-1, and it can be found in O(n 3 ) time. We summarize our findings in the following. Theorem 4.1 Given a set of n planar points and a network pattern T, the optimal deployment of T that reuses all the n base stations can be found in O(n 3 ) time. Figure 4.2 shows the optimal deployment of the planned network T which reuses P 0, P 1, P 2, and P 3. In our example, d 1 = P 1 C 1 is the smallest one. Therefore C 1 will be selected as the center of one cell in the optimal deployment of the planned network T. T(C 1 ) is shown in the figure. maximum number of old base stations that can be reused for the given bound. The problem of the maximum base station reuse can be defined as follows: Given a set of n arbitrarily distributed old base stations, a network pattern and a maximum reuse distance d, deploy the network such that the maximum number of old base stations can be reused. Figure 4.3 Base Station Reuse with a Given Bound d. Figure 4.2 The optimal network deployment reusing P 0, P 1, P 2, P 3. B. Maximum Base Station Reuse with a Bound The algorithm given in the previous section can compute the optimal network deployment that minimizes the network difference between all the old base stations and the planned network. In practice, most system designs permit a base station to be positioned up to one-fourth the cell radius away from the cell center to ensure an acceptable coverage [10]. Therefore the optimal network deployment might not meet this physical constraint for all base stations. For example, as shown in Fig. 4.3, given a set of old base stations and a tolerance bound d, only P 0, P 1, P 5 and P 6 can be reused in this network deployment. For a given bound d, it is possible that no matter how we design the network, not all the old base stations can be located within the bound. Then the problem is how to find the To solve the problem, we begin with considering the umbrella diagram of two base stations. Given a new network pattern T, for a pair of old base stations, P 0 and P 1, we first construct the umbrella diagram of P 0 and P 1. As shown in Figure 3.2, L(P 0,P 1 ) is a polygon formed by the perpendicular bisectors (or their extensions) between P 0 and the centers of the cells in T(P 1 ) that intersect with H(P 0 ). The area that is outside L(P 0,P 1 ) but within H(P 0 ) represents the locus of all the starting points x such that the network difference d(p 0, T(x)) > d(p 1, T(x)). So for any network with a starting point located outside L(P 0,P 1 ) but within H(P 0 ), if P 0 can be reused, P 1 can be reused too because the network distance of P 0 is greater than that of P 1. Now we show how to incorporate the bound d into the contour. An old base station can be reused if and only if it is within the bound of a nearby cell center. If we draw a circle around the old base station P 0 with the radius being the bound d as shown in Fig. 4.4, the circled area stands for the reuse region of P 0. For any network with a starting point within the circle, it is obvious that P 0 can be reused. Therefore P 1 can be reused too because the network distance of P 0 is greater than that of P 1. Lemma 4.2 summarizes the above discussion. Lemma 4.2 Given two base stations P i and P j, the area that is outside L(P i,p j ) but inside the circle that centered at P i is the locus to deploy the cellular network such that both P i and P j can be reused subject to the condition that P i has an equal or larger network difference than P j.

9 difference. The following theorem summarizes our findings concerning our sub-optimal solutions. Figure 4.4 Bounded L(P 0, P 1) for P 0 and P 1. There could be three cases of intersection between L(P i,p j ) and the circle centered at P i : (a). L(P i,p j ) is totally outside the circle; (b). L(P i,p j ) is totally inside the circle; (c). L(P i,p j ) is intersecting with the circle. If L(P i,p j ) is totally outside the circle, it means P i cannot be reused in any network that has a larger distance to P i than to P j. If the L(P i,p j ) is totally inside the circle, it means that both P i and P j can reused in a network that has a larger distance to P i than to P j. In this case, the area that is outside of L(P i,p j ) but inside the circle is the locus to deploy a cellular network in which both old base stations can be reused subject to the condition that P i have a equal or larger network difference than P j. Finally, if L(P i,p j ) is intersecting with the circle as shown in Fig. 4.4, it means that the area that is outside L(P i,p j ) but inside the circle is the locus to deploy a cellular network in which both old base stations can be reused subject to the condition that P i has a equal or larger network difference than P j. Next, we will present our solution to the problem of maximum base station reuse. For each old base station P i, we will show how to get the circular arcs that will be used to solve the sub-maximum base station reuse problem that is subject to the condition that P i has the largest network difference. The maximum number of base stations that can be reused can be found by combining sub-optimal solutions. Suppose we have computed all the L(P 0,P j ) for all j 0, and draw them around P 0 as shown in Fig Next we add a circle around P 0 with the radius being the bound. Now if we walk from a point x on the circle straight towards the center P 0, the number of lines we encounter is one less than the number of base stations that can be reused if the new network is deployed at x. If the encountered line is a part of L(P 0,P j ) can be reused. Note P 0 can always be reused. By walking from all points on the circle, we can find the maximum number of base stations that can be reused. Of course all statements here are subject to the condition that P 0 has the largest network Figure 4.5 Inspecting maximum reused base stations. Theorem 4.3 Given a set S of n base stations, S={P 0, P 1,,P n- 1}, if the maximum number of contours L(P 0,P j ) with j 0 that intersect a radius starting from P 0 is k, the maximum number of base stations that can be reused is k+1 under the condition that P 0 has the largest network difference. Proof. Subject to the condition that P 0 has the largest network difference, we know that the network starting point must be located outside the contour L(P 0,P j ). In terms of P 0 if the starting point is inside the bounding circle centered at P 0, P 0 can be reused. Otherwise, P 0 cannot be reused. For a base station P j other than P 0, if the starting point is outside the contour L(P 0,P j ) can be reused as long as P 0 can be reused because, in this case, nd(p 0,T) is greater than nd(p j,t). Therefore, if we connect a point on the circle to the center P 0, the number of intersecting contours corresponds to the number of base stations that can be reused subject to the condition that P 0 is farther from the network than other base stations. It is easy to see that if the maximum number of contours that can be intersected with a radius is k, the maximum number of base stations that can be reused for the given bound and subject to the condition that P 0 has the largest network difference is k+1. For example, in Fig. 4.5, if we walk from point A towards the center P 0, only L(P 0, P 1 ) is encountered. Therefore, if the network is deployed at A, only old base stations P 0 and P 1 can be reused. However, if we walk from point B towards the center, L(P 0, P 1 ), L(P 0, P 2 ) and L(P 0, P 3 ) will be encountered. Therefore, in this case, the base stations, P 0, P 1, P 2 and P 3, can be reused when the network is deployed at B. However, it is impossible for us to inspect every radius to find the maximum number of base stations that can be reused. Next, we will use circular arcs to solve the maximum base station reuse problem. As shown in Fig. 4.5, we consider the intersections of the circle and L(P 0,P j ). Each intersection chord corresponds to a circular arc [12]. L(P 0,P j ) introduces at most 4 circular arcs. L(P 0,P 1 ), L(P 0,P 2 ),, L(P 0,P n-1 ) will introduce

10 a total of 4(n-1) circular arcs. To find the maximum number of circular arcs that intersect with a radius, we can first sort all the arc end points. Then we use a sweep line to get the maximum number of the arcs that intersect with the sweep line. The maximum number of overlapping arcs for n circular arcs can be found in O(nlogn) time. For each old base station P i, we can compute the maximum number of old base stations that can be reused under the condition that P i has the largest network difference. Next we will try to solve the maximum base station reuse problem without any condition. Assume T max is the deployed network in which the maximum number of base stations can be reused, and P k is the old base station that has the largest network difference to T max. Then the maximum number of old base stations can be reused is the same as the maximum number of old base stations can be reused under the condition that P k is farthest from the network. The following theorem summarizes how to find the solution to the maximum reuse problem. Theorem 4.4 Given a set S of n base stations, S={P 0, P 1,,P n- 1}, if k i (i=0..n-1) is the maximum number of old base stations that can be reused under the condition that P i has a larger network difference than all other old base stations, and k m is the maximum among all k i (i=0..n-1), k m is the maximum number of base stations that can be reused within the given bound. Next we consider the time complexity of the algorithm. As we know, given two points P i and P j (j i), polygon L(Pi,P j ) is either a triangle or quadrilateral which can be computed in constant time. For each P i, we have a circular arc graph of a maximum of 4(n-1) arcs. The maximum number of arcs that can intersect with a radius can be found in O(nlogn). There are n circular arc graphs, each of which corresponds to an old base station. Therefore it takes O(n 2 logn) time to compute the maximum number of old base stations that can be reused. V. SUMMARY Computational geometry has found a lot of applications in the area of wireless networks. In a cellular network, the area served by a base station can be viewed as a Voronoi region of that base station in the Voronoi diagram. In this paper, we have proposed a new geometric diagram, referred to as the umbrella diagram. The umbrella diagram is comparable to the famous Voronoi diagram. Whereas the Voronoi diagram deals with a set of points, the umbrella diagram deals with a set of different networks, each of which is derived from a point. The umbrella diagram of a set of hexagonal networks is to divide the plane into regions such that the points in a region have a larger distance to one network than to the other networks. The distance from a point to a network is the distance between the point and its closest cell center of the network. Unlike the Voronoi diagram, the regions in the umbrella diagram are not necessarily convex. The paper shows that the umbrella diagram of n networks with the same cell size and orientation (or the umbrella diagram of n points under a network pattern) can be computed in O(n 3 ) time. The umbrella diagram characterizes the relationship between uniform patterns and arbitrarily distributed nodes and it can be used to solve many problems in wireless networks. In this paper we show how to use the diagram to solve two problems: the problem of the optimal deployment of a planned network, and the problem of maximum base station reuse with a given bound. The optimal deployment of a planed network is to minimize the distance between the existing base stations and the deployed network if all existing base stations are to be reused. The maximum base station reuse problem is to find the maximum number of old base stations that can be reused under a given bound. 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