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1 Minimum-Energy Broadcast Routing in Static Ad Hoc Wireless Networks P.-J. Wan G. Calinescu.-Y. Li O. Frieder Abstract Energy conservation is a critical issue in ad hoc wireless networks for node and network life, as the nodes are owered by batteries only. One major aroach for energy conservation is to route a communication session along the routes which requires the lowest total energy consumtion. This otimization roblem is referred to as minimum-energy routing. While minimum-energy unicast routing can be solved in olynomial time by shortest-ath algorithms, it remains oen whether minimum-energy broadcast routing can be solved in olynomial time, desite of the NP-hardness of its general grah version. Recently three greedy heuristics were roosed in [8]: MST (minimum sanning tree), SPT (shortest-ath tree), and BIP (broadcasting incremental ower). They have been evaluated through simulations in [8], but little is known about their analytical erformance. The main contribution of this aer is the quantitative characterization of their erformances in terms of aroximation ratios. By exloring geometric structures of Euclidean MSTs, we have been able to rove that the aroximation ratio of MST is between 6 and, and the aroximation ratio of BIP is between and. On the other hand, the aroximation ratio of SPT is shown to be at least n,wherenis the number of receiving nodes. To our best knowledge, these are the rst analytical results for minimum-energy broadcasting. I. Introduction Ad hoc wireless networks have received signicant attention in recent years due to their otential alications in battleeld, emergency disaster relief and etc [7] [8]. Unlike wired networks or cellular networks, no wired backbone infrastructure is installed in ad hoc wireless networks. A communication session is achieved either through a gle-ho transmission if the communication arties are close enough, or through relaying by intermediate nodes otherwise. Omnidirectional antennas are used by all nodes to transmit and receive signals. They are attractive in their broadcast nature. A gle transmission by a node can be received by many nodes within its vicinity. This feature is extremely useful for multicasting/broadcasting communications. For the urose of energy conservation, each node can dynamically adjust its transmitting ower based on the distance of the receiving nodes and background noise. In the most common ower-attenuation model [6], the signal ower falls as r where r is the distance from the transmitter antenna and is the a constant between and 4 deendent on the wireless environment. All receivers have the same ower threshold for signal detection, which aretyically normalized to one. With these assumtions, the ower Deartment of Comuter Science, Illinois Institute of Technology, Chicago, IL s: fwan, calinescu, xli, ohirg@cs.iit.edu. required to suort a link between two nodes searated by ranger is r. Akey observation is that relaying signal between nodes may results in lower transmission than communicating over large distances due to the nonlinear ower attenuation. As a simle illustration, consider three nodes and with as k k > k k in Figure and assume =.Node wants to send a message to node. It has two otions. It can transmit the signal directly to node, with a energy consumtion of k k. Alternatively, it can relay the message through node and have it retransmit to node, with a total energy consumtion of k k + k k. Therefore if the angle is obtuse, the second otion consumes less total energy. A crucial issue is then how to nd a routing with minimum total energy consumtion for a given communication session. This roblem is referred to as the Minimum-Energy Routing [7] [8]. Fig.. Reduce energy consumtion through relaying. The minimum-energy broadcast/multicast routing in a simle ad hoc networking environment has been addressed by the ioneering work in [8]. To assess the comlex trade-os one at a time, the nodes in the network are assumed to be a oint set randomly distributed in a twodimensional lane and there is no mobility. Nevertheless, as argued in [8], the imact of mobility can be incororated into this static model because transmitter ower can be adjusted to accommodate the new locations of the nodes, as necessary. In other words, the caability to adjust transmission ower rovides considerable \elasticity" to the toological connectivity, and hence may reduce the The terms node, oint and vertex are interchangeable in this aer: node is a networking term, oint is a geometric term, and vertex is a grah-theoretic term.

2 need for hand-os and tracking. In addition, as assumed in [8], there are sucient bandwidth and transceiver resources. Under these assumtions, centralized (as oosed to distributed) algorithms were resented by [8] for minimum-energy broadcast/multicast routing. These centralized algorithms in this simle networking environment are exected to serve as the basis for further studies on the distributed algorithms in a more ractical network environment with limited bandwidth and transceiver resources, as well as node mobility. Three greedy heuristics were roosed in [8] for minimum-energy broadcast routing: MST (minimum sanning tree), SPT (shortest-ath tree), and BIP (broadcasting incremental ower). They have been evaluated through simulations in [8], but little is known about their analytical erformances in terms of the aroximation ratios. We believe that this analytical erformance is very essential and more convincing in evaluating these heuristics. For the minimum-energy broadcast routing, one may come u with many seemingly reasonable greedy criteria. But it's hard to tell from simulation oututs which one is better or oorer. Indeed, all the three heuristics roosed in [8] only have subtle dierences. For ure illustrative urose, another a slight variation of BIP, which is referred to Broadcast Average Incremental Power (BAIP), will be introduced in Section III. These subtle dierences, however, can have great imact on their erformance ratios. In fact, we will show thatthe aroximation ratios of MST and BIP are between 6 and and between and resectively on the other hand, the aroximation ratios of SPT and BAIP are at least n 4n and ln n ; o () resectively, where n is the number of nodes. To our best knowledge, these are the rst quantiative characterizations of heuristics for minimum-energy broadcast routing. The remaining of this aer is organized as follows. In Section II, we analyze the challenges for minimum-energy broadcast routing and briey overview the three greedy heuristics develoed in [8]. In Section III, we construct some bad instances to illustrate the oor erformance of both SPT and BAIP. These instances lead to the lower boundsonthearoximation ratios of SPT and BAIP. In Section IV, we construct instances to obtain lowerbounds on the aroximation ratios of MST and BIP. InSection V, we derive uer bounds on the aroximation ratios of MST and BIP. A cornerstone to the analysis of the uer bounds is an elegant structure roerty of Euclidean MST, which is exlored in Section VI. Finally in Section VIIwe summarize our results and oint outseveral future reseach roblems. II. Preliminaries In this aer, we assume the network nodes are given as a nite oint set P in the two-dimensional lane. For any,weuseg () to denote the weighted comlete grah over P in which the weight ofanedgee is equal to kek. The minimum-energy unicast routing is essentially a shortest-ath roblem in G (). Consider any for a unicast from a node P to another node q P can be easily obtained by alying a shortest-ath algorithm to G (). Consider any unicast ath from a node P to another node q P: = 0 m; m = q: In this ath, the transmission ower of node i for 0 i m; isk i i+ k and the transmission ower of m is zero. Thus the total transmission energy required by this ath is m; i= k i i+ k which isthetotalweight of this ath. So by alying any shortest-ath algorithm such as Dijkstra's algorithm [], one can get a minimum-energy unicast routing. However, for broadcast alications, and in general multicast alications, the minimum-energy routing is far more challenging. Any broadcast routing is an arborescence (a directed tree) T rooted at the source node of the broadcasting that sans all nodes. We usef T () to denote the transmission ower of the node required by T. Then for any leaf node of T, f T () = 0 and for any internal node of T, f T () = max qt kqk in other words, the -th ower of the longest distance between and its children in T P. The total energy required by T is then given by P f T (). Thus the minimum-energy broadcast routing is dierent from the conventional link-based minimum sanning tree (MST) roblem. Indeed, while the MST can be solved in olynomial time by algorithms such as Prim's algorithm [] and Kruskal's algorithm, it is still unknown whether minimum-energy broadcast routing can be solved in olynomial time. In its general grah version, the minimumenergy broadcast routing can be shown to be NP-hard [], and even worse, inaroxiable within a factor of ( ; ) log, where is the maximal degree and is any arbitrary small ositive constant unless NP DTIME n O(log log n) by an aroximation-reserving reduction from the Connected Dominating Set roblem [4]. However, this intractness of its general grah version does not necessarily imly the same hardness of its geometric version. In fact, as will be shown later in the aer, its geometric version can be aroximated within a constant factor. Nevertheless, this suggests that minimum-energy broadcast routing is considerably harder than the MST roblem. Three greedy heuristics have been roosed for the minimum-energy broadcast routing by [8]. The MST heuristic rst alies the Prim's algorithm to obtain a MST, and then orient it as an arborescence rooted at the source node. The SPT heuristic rst alies the Dijkstra's algorithm [] to obtain a SPT, and then orient it

3 as an arborescence rooted at the source node. The BIP heuristic is the node version of Dijkstra's algorithm for SPT. It maintains throughout its execution a gle arborescence rooted at the source node. The arborescence starts from the source node, and new nodes are added to the arborescence one at a time on a minimum incremental cost basis until all nodes are included in the arborescence. The incremental cost of adding a new node is the minimum additional ower increased by some node in the current arborescence so as to reach thisnew node. The imlementation of BIP is based on the standard Dijkstra's algorithm, with one fundamental dierence on the oeration whenever a new node is added. Whereas Dijkstra's algorithm udates the node weights (reresenting the distances), BIP udates the link costs (reresenting the incremental ower to reach the head node of the link). This udate is erformed by subtracting the cost of the added link from the cost of every link from the source node of the added link to a node not in new arborescence. The erformance of these three greedy heuristics have been evaluated in [8] by simulations studies. However, their analytic erformances in terms of the aroximation ratios remain oen. The subsequent sections of this aer will derive the bounds on their aroximation ratios. III. Greedy Is Not Always Good Greedy aroaches are the most natural and widely used techniques in design ractical heuristics for otimization roblems. For minimum-energy broadcast routing, one may think of many greedy heuristics, in addition to the three greedy heuristics roosed in [8]. The real challenge, however, is how to come u with a rovably good one. Two greedy criteria may only have slightdif- ference, but these small variation can have a great imact on their erformance. In addition, some criteria may erform quite well or even otimally in some situations, but may erform very oorly in some other situations. In this section, we describe two examles, one is SPT, another is a new one. The \bad" instance constructed in this section can not only lead to lower bounds on their aroximation ration, but also hel to design an overall good greedy criteria. For the simlicity, we only consider = in this section. We begin with the SPT algorithm. Let be a suciently small ositive number. Consider m nodes m evenly distributed on a cycle of radius centered at node o (see Figure ). For i m, let q i be the oint in the line segment o i with koq i k =. We consider a broadcasting from node o to these n =m nodes m q q q m : The SPT is the suerosition of the aths oq i i. Its total energy is + m ( ; ) : On the other hand, if the transmission ower of node o is, then the signal can reach all other oints. Thus the minimum energy is at most. So the aroximation ratio of SPT is + m ( ; ). As ;! 0, this ratio converges to n = m. ε ε o Fig.. q q q q m m A bad instance for SPT. The second greedy heuristic is very similar to Chvatal's algorithm [] for Set Cover Problem and is a variation of BIP. Like BIP, an arborescence, which starts with the source node, is maintained throughout the execution of algorithm. However, unlike BIP, many new nodes can be added one time. Similar to Chvatal's algorithm [], the new nodes added are chosen to have the minimal average incremental cost, which is dened as the ratio of the minimum addition ower increased by some node in the current arborescence so as to reach these new nodes to the number of these new nodes. We refer to this heuristic as the Broadcast Average Incremental Power, abbreviated by BAIP. In contrast to the + log m aroximation ratioofthechvatal's algorithm [] where m is the largest set size, we show that the aroximation ratio of BAIP is at least 4n ln n ; o () where n is the number of receiving nodes. Consider the following instance of minimum-energy broadcasting. All nodes are colinear with the source being the origin, the i-th receiving node at the osition i for i n ;, and the n-th receiving node at the osition n ; for some suciently small number >0. For any k n ;, the minimal transmission ower of the source to reach k receiving nodes is k = k, and thus the average ower eciency is k k =. On the other hand, the minimal transmission ower of the source to reach alln receiving nodes is ; n ; = n ;, and the thus the average ower eciency is n; n =; n. So BAIP will let the source to transmit at ower n ; to reach all nodes in a gle ste. However, the otimal routing is a directed ath consisting of all nodes from left

4 4 to right. So the minimum ower is n; i ; ; i ; + n ; ; n ; : i= < n i= i n; ; i ; =+ i= n; (n ; ) + + +ln 4i 4 = i= ln (n ; ) + 5 : 4 Thus the aroximation ratio is at least P n; i= n ; ; i ++ i ; i ; i ; + ; n ; ; n ; : As ;! 0, this ratio converges to P n i= = n ; = i ; i ; n + P n; i= 4i + P n; i= n + ln(n;)+ 4 4n ln (n ; ) + 5 = 4n ln n ; o () : n ( i++ i) Interestingly, SPT generates the otimal solution in the second bad instance, while BAIP can rovide nearotimal or otimal solution for the rst bad instance. On the other hand, MST and BIP have a lot similarities to SPT and BAIP, but have constant aroximation ratios as will be roved later. Thus one should carefully design and select greedy heuristics. IV. Lower Bounds on Aroximation Ratios In this section, we will derive lower bounds on aroximation ratios of MST and BIP. We begin with MST. Theorem : The aroximation ratio of MST is at least 6forany. Proof: Let be a suciently small ositive number. Consider seven nodes o 6 (see Figure ), which satises that ko k = Then for any i 5, and ko i k =+ i 6 k i i+ k = i 5: \ i o i+ < \ 6 o > : 4 5 Fig.. +ε +ε +ε +ε o 6 +ε A bad instance for MST. Consider the two triangles o and o 6. Since and by Law of Coe, we have ko k = ko 6 k \ 6 o > \ o k 6 k > k k =: We consider the broadcasting from the node o to nodes 6. Then the ath o 5 6 is the unique MST. It's total energy is 6. On the other hand, the otimal routing is the star centered at node o, whose total energy is ( + ). Thus the aroximation ratio is at 6 least (+), which converges to 6 as ;! 0. Now wedevelo the lower bound on the aroximation ratio of BIP. Theorem : The aroximation ratio of BIP is at least for any =. Proof: Let be a suciently small ositive number. Consider six oints 6 on a cycle of radius centered at node o (see Figure 4), with Then \ o = \ 5 o 6 = ; \ o = \ 4 o 5 = ; \ o 4 = ; \ 6 o = +: k k = k 5 6 k < k k = k 4 5 k < k 4 k < < k 6 k :

5 5 π/ θ Thus in the subsequent two stes, the oints and 5 are added, and the transmission ower of oints and 6 is k k = k 5 6 k. Similarly, in the last two stes, the oints and 4 are added, and the transmission ower of oints and 5 is k k = k 4 5 k. The total ower is 4 Fig. 4. π/ θ π/ θ 5 π/ θ q o q q q q m m+ π/ θ A bad instance for BIP. Let q be the oint in the erendicular bisector of 6 such that q is erendicular to : Choose a suciently large integer m such that koqk ; > k 4 k : m Let q q m+ be the m+ oints on the ray oq with koq i k = i m koqk for i m +. Then q m = q. We consider a broadcasting from oint o to oints q q m+ 6. The otimal solution is that the node o transmits at ower to reach all nodes. Now let's examine the outut of the BIP algorithm. As m is suciently large, in the rst m + stes, the oints q q m+ are sequentially added, and the transmission ower of the nodes o q q m all has the transmission ower Since the. angles koqk m \ q m+ q m = \ 6 q m+ q m > in the next two stes, the oints and 6 are added, and the transmission ower of oint q m+ is k q m+ k. At this moment, the incremental ower of all oints o q q m to reach any node i for i 5 is at least koqk ; > k 4 k > k k = k 5 6 k m and the incremental ower of oint q m+ to reach any node i for i 5 is also greater than k k = k 5 6 k as \ q m+ = \ 5 6 q m+ > \ q m = : 6 koqk (m +) + kqm+k +kk +kk m = m + m koqk + kqm+k +kk +kk : As ;! 0 and m ;!, the olygon converges to a regular hexagon, and the nodes q and q m+ converges to the center of the triangle o 6. Thus the total ower converges to +4=. Thus the aroximation ratio of BIP is at least 4:. V. Uer Bounds on Aroximation Ratios Our deriving of the uer bounds relies on extensively on the geometric structures of Euclidean MSTs. We rst observe that as long as the cost of a link is an increag function of the Euclidean length of the link, the set of MSTs of anyoint set coincides with the set of Euclidean MSTs of the same oint set. In fact, this can be followed from Prim's algorithm. In articular, for any sanning tree T of a (nite) oint setp, the arameter P et kek achieves its minimum if and only if T is an Euclidean MST of P. For any (nite) oint setp,weusemst (P ) to denote an arbitrary Euclidean MST of P. The radius of a oint set P is dened as inf su kqk : P qp Thus a oint set of radius one can be covered by a disk of the same radius. A key result P in this section is an uer bound on the arameter emst(p ) kek for any nite oint set P of radius P one. Note that the sureme of the length of mst (P ), emst(p ) kek, over all oint sets P of radius P one is innity. Amazingly however, the arameter emst(p ) kek is bounded by a constant for any oint set P of radius one, as shown P later. We use c to denote the sureme of the length of emst(p ) kek over all oint sets P of radius one. The next key theorem states that c is at most. Theorem : 6 c. The roof of this theorem involves comlicated geometric arguments, and therefore we ostone it in Section VI. Note that for any ointsetp of radius one, the length of each edge in mst (P ) isatmostone. Therefore, Theorem imlies that for any oint set P of radius one and any, emst(p ) kek emst(p ) kek c :

6 6 In the next, we exlore a relation between the minimum energy required by a broadcasting and the Euclidean MST of the corresonding oint set. Lemma 4: For any ointsetp in the lane, the total anergy required by any broadcasting among P is at least c P emst(p ) kek. Proof: Let T be an otimal arborescence for a broadcasting among P. For any none-leaf node in T, let T be an Euclidean MST of the oint set consisting and all children of in T. Suose that the longest Euclidean distance between and its children is r. Then the transmission ower of node is r, and all children of lie in the disk centered at with radius r. From the denition of c, we have which imlies that et r c kek c r et kek Let T denote the sanning tree obtained by suerog of all T 's for non-leaf nodes of PT. Then the total energy required by T is at least c et kek, which is P further no less than c emst(p ) kek. This comletes the roof. Consider any oint set P in the lane. Let T be an arborescence oriented from some mst P (P ). Then the total energy required by T is at most et kek. From Lemma 4, this total energy is at most c times the otimum cost. Thus the aroximation ratio of MST is at most c. Together with Theorem, this observation leads to the following theorem. Theorem 5: The aroximation ratio of MST is at most c, and therefore is at most. Finally, we derive the uer bound on the aroximation ratio of the BIP heuristic. Once again, the Euclidean MST will lay an imortant role. Lemma 6: For any broadcasting among a oint setp in the lane, the total energy required by P the arborescence generated by BIP algorithm is at most emst(p ) kek. Proof: Let G () be the comlete grah over the oint setp, in which theweight of an edge e is kek. Let T be the arborescence outut by the algorithm BIP. We construct another weighted grah H over the same oint set P according to the execution of BIP for generating T. Suose that during the execution of BIP the nodes are added in the order n where is the source node. Let T i be the arborescence just after node i is added. In H, the weight of the edge i i+ is equal to the incremental energy of the link from T i to i+ chosen during the execution of SPF and the weight ofany other edge is the same as that in G (). Note that for each edge i i+, its weight in H is not more than its weight in G (). Therefore, for any sanning tree, its weight inh is no more than its weight ing (). On the other hand, the execution of Prim's algorithm on H will emulate the algorithm BIP on G () in the sense that it will add the required nodes in the same order, and will outut the ath n. The weight of this ath in H is exactly the total energy required by T, but is at most the weight of any MSTin G (). This imlies that the total energy required by T is at most P emst(p ) kek. This comletes the roof. From the above lemma and Lemma 4, we can get the result for the BIP algorithm similar to Theorem 5. Theorem 7: The aroximation ratio of the BIP heuristic is at most c, and therefore is at most. VI. Proof of Theorem This section is devoted to the roof of Theorem. The lower bound is very trivial as it can follow from the instance consisting of seven oints: the center of a regular hexagon and its six vertices. However, the deriving of the uer bound is very challenging. We rst introduce some geometric structures and notations to be used in this section. All angles are measured in radians and take values in the range [0 ]. For any three oints and, the angle between the tworays and is denoted by \ or \. The closed innite area inside the angle \, also referred to as a sector, is denoted by ]. The triangle determined and is denoted by 4. The oen disk centered at with radius r, denoted by B ( r), is the set of oints whose distance from is less than r. The lune through oints and, denoted by L ( ), is the intersection of the two oen disks of radius k kcentered at and resectively (see Figure 5(a)). Thus it consists of oints whose distances from and are both less than k k. The oen diamond substended by a line segment, denoted by D ( ), is the rhombus with sides of length k k (see Figure 5(b)). Note that the interior angles at and within D ( )are equal to. Fig. 5. (a) (b) Illustration of (a) lune and (b) diamond. The Euclidean MSTs have many nice structure roerties [5]. Some basic roerties are listed blow. Any air of edges do not cross each other. The angles between any two edges incident to a common vertex is at least.

7 7 The length of each edge is at most the radius of the vertex set. The luna determined by each edge does not contain any other vertices. Let be any edge. Then the two endoints of any other edge are either both outside B ( k k), or both outside B ( k k). In this section, we willrstrove yet another structure roerty of the Euclidean MSTs, which isvery essential to bound the constant c: The diamonds of any two edges are disjoint. The roof of this roerty willmake use of the following lemma. Lemma 8: Let and be any three oints in the lane with \ = and k k = k k(see Figure 6). Let 4 be any oint in ] but outside 4 with \ 4 =. Then D ( 4 ) ] if and only if either 0 and k 4 k ( ;) k k or 5 6. On the other hand, as 4 and 4 4 x are similar, k 4 k k k = k xk k k = ; ; : Therefore, D ( 4 ) ] as long as k 4 k ( ;) k k. In the next we aly the above lemma to show that the diamond determined by any edge in an Euclidean MST is contained in some sector dened in the next lemma. Lemma 9: Let and be any three oints in the lane with being outside L ( ). Let 0 ( 0 resectively) be the vertex of D ( )(D ( ) resectively) which is lies on the oosite side of the line ( resectively) from ( resectively) (see Figure 7). Then D ( ) ] α α α π/ α x (a) 4 x y (a) x α α π/ α α x (b) y Fig. 6. (b) Two extreme cases for D (4) ]. Proof: Note that D ( 4 ) ] if > 5 6 and D ( 4 ) ] if 5 6. So we now assume 0. We x and calculate the maximum length of 4 such that D ( 4 ) ]. This haens when D ( 4 ) touches the ray,sayatx. We consider this extreme scenario. In this case, \ x = \ x = ; : Alying the Laws of Sine in 4 x,wehave k xk k k = ; ; : (c) Fig. 7. The three cases for Lemma 9 Proof: We assume by symmetry that is to the left of the line and to the right of the erendicular bisector of. Then k kk k. Since is outside L ( ), k kk k and \ <. x y

8 8 Therefore, q q \ 0 0 < = 5 6 \ q q q q y x \ < : Let x and y be the other twovertices of D ( )which lie between the left side and right side resectively of the line. It's sucient toshow that both x and y are within ] 0 0. This is true when \ 6 (see Figure 7(a)). So we assume that \ < 6. In this case x is within 4 0, and thus within ] 0 0, from Lemma 8 and k kk k. If \ 5 6, then y is within ] 0 ] 0 0 (see Figure 7(b)). If \ > 5 6,then \ y = ;\ ; \ y ; ; 6 = 5 6 which imlies that the ray y does not intersect with the ray 0 (see Figure 7(c)). So y is within ] 0 0. Therefore, in either case both x and y are within ] 0 0. This comletes the roof. Now we are ready to rove the \disjoint diamond" roerty of Euclidean MSTs. Lemma 0: In any Euclidean MST, the two diamonds determined by anytwo edges are disjoint. Proof: The lemma is true when two edges are incident to a common vertex as the angle between them is at least. So we consider two edges and q q with distinct endoints. We consider two cases. Case : At least one of and q q does not cross the erendicular bisector of the other. Without loss of generality, assume that q and q lie in the same side of the erendicular bisector of as (see Figure 8(a)). Let q 0 (q 0 resectively) be the vertex of D ( q ) (D ( q ) resectively) which is lies on the oosite side of the line q ( q resectively) from q (q resectively). Then from Lemma 9, D (q q ) ]q 0 q 0. On the other hand, ce both q and q are outside L ( ), D ( ) is outside ]q 0 q 0. Thus D ( ) and D (q q ) are disjoint. Case : Both and q q cross the erendicular bisector of the other. Without loss of generality, assume that q lies in the same side of the erendicular bisector of as (see Figure 8(b)). Then must lie in the same side of the erendicular bisector of q q as q, for otherwise k q k > k q k > k q k > k q k i.e., both and lie in the same side of the erendicular bisector of q q as q, which contradicts to the assumtion. Since q is outside L ( ) and (a) Fig. 8. Two cases for Lemma 0. k q k > k q k, we have k q k > k k. As k q k > k q k, q is outside 4q. Similarly, any of these four oints q and q is outside the triangle determined by theother three oints. This imlies that the convex hull determined by these four oints is a quadrilateral. Note that and q q can't be the two diagonals of the quadrilateral as they do not cross each other. Neither can be q and q as they are searated by the erendicular bisector of. Thus the two diagonals must be q and q, and consequently the boundary of the quadrilateral is q q. From the revious argument its four sides are all less than its two diagonals, and hence its four inner angles are all more than. Without loss of generality, we assume that k q kk q k. Then k q kk k, for otherwise q would be inside B ( k k) and q would be inside B ( k k), which is imossible. Similarly, k q kkq q k. Therefore, both \q q and \ q are less than. Since both \q and \ q q are less than,wehave \q \ q q 5 6 (b) : Let x be the oint inside ]q such that 4 q x is equilateral. Then both and q q are outside 4 q x. In addition, and \x \xq q 0 k xkk k kq xkkq q k : Let y be the center of 4 q x. Then from Lemma 8, D ( ) ] yx D (q q ) ]q yx: This imlies that D ( ) and D (q q ) are disjoint. Let P be any oint set of radius one. According to Lemma 0, the total area of the diamonds through the

9 edges in mst (P ) equals to 6 emst(p ) kek : Let be the oint in P such that all oints in P have distance of at most one from. Since all edges have lengths of at most one, all diamonds are contained in B. This imlies that 6 emst(p ) kek = 4 : q 9 Therefore emst(p ) kek 8 4:5: This estimation is quite loose and fails in getting the desired uer bound. In the next, we will rovide a tighter estimation which can lead to the uer bound. We observe that the total area of the diamonds is no more than the area of the disk B ( ) lus the sticking-out areas of these diamonds beyond B ( ). Let D ( )beany diamond which sticks out B ( ), and let q be its vertex which is outside B ( ) (see Figure 9). Let 0 ( 0 resectively) be the intersection between q ( q resectively) and the boundary of B ( ). Then the sticking-out area of D ( ) can be calculated by subtracting the area of the sector substended by 0 and 0 from the area of the quadrilateral 0 q0. The area of the quadrilateral 0 q0 can be further calculated by summing u the areas of and 4q 0 0. As \ 0 q0 is a constant, the area of 4q0 0 is maximized when kq 0 k kq 0 k. Let \ 0 0 =, then ; 0 and the sticking-out area of D ( )isaat most S () = + 6 ( ; cos ) ; : The area function S () has the following nice roerty. Lemma : For any ; 0,. if +, S ()+S() S ( + ). if +, S ()+S () S ; + ; + S ;. Proof: The lemma follows from the following two equality: for any and, S ( + ) ; S () ; S () = 4 6 ; + + ; + S S = 4 + ; 6 ; S () ; S () 6 ; 6 ; : Fig. 9. The calculation of the sticking-out area. We rst rove the rst equality. S ( + ) ; S () ; S () = ( ( + ) ; ; )+ ((cos +cos) ; (cos ( + )+)) 6 = + cos + ; + cos + = + cos + cos ; cos + = cos ; cos ; ; cos + + ; cos ; ; cos + cos + ; + = 4 6 ; + Now werove the second equality. S + ; + S ; S () ; S () =(S ( + ) ; S () ; S ()) ; S ( + ) ; S + ; ; S = 4 6 ; + ; 4 + ; 6 = 4 + ; 6 = + ; 6 = 4 + ; ; ; + cos 6 ; + ; ; 6 ; cos ; 6 ; :

10 0 Suose that there are k diamonds which stick out B ( ). For any i k, let i be the inner angle of the arc between the two intersection oints of the boundary B ( ) and the boundary of the i-th stickingout diamond. Then i ; 0 and k i= i < : By reeatedly alying the two inequalities in Lemma, the total sticking-out area of the diamonds is k i= &P k i= S ( i ) i 6S ' S = ; : Thus the total area of diamonds is at most Therefore, + ; = : emst(p ) kek 6 This comletes the roof Theorem. =: VII. Summary and Future Works In this aer, we have rovided theoretical erformance analysis the heuristics resented in [8]. The aroximation ratio of SPT is at least n, and thus less favorable from the theoretical ersective. The other two heuristics, MST and BIP, have constant aroximation ratios. Secically, the aroximation ratio of MST is between 6 and c, which is at most the aroximation ratio of the BIP heuristic is between and c. However, there are still several challenging issues for future research. First of all, the comutational comlexity of the Minimum-Energy Broadcasting remains unknown. As mentioned in Section I, the grah-version of this roblem is at least as hard as the Connected Dominate Set roblem. However, due to its geometric nature, this intractness of the grah version does not imly the same intractness of geometric version which isstudied in this aer. Indeed, while Connected Dominating Set roblem does not allow a constant-aroximation ratio, the geometric version does on the contrary, for examle, bymst or BIP. Secondly, the exact value of the constant c remains unsolved. A tighter uer bound on c can lead to tighter uer bounds on the aroximation ratios of both MST and BIP. From the deriving of the uer bound, we observe there are still rooms to imrove the uer bound. For examle, it's very unlikely for the diamonds to ll the unit disk fully. At least this is true for small number of nodes. However, the treatment of large number of nodes is quite challenging, and more geometric roerties of the Euclidean MSTs have to be exlored. The third interesting roblem is how to construct \worse" instances that can lead to better lower bounds on the aroximation ratios of both the MST and BIP. A major challenge, and a toic of continued research, is the develoment of distributed algorithms of MST and BIP. These algorithms should take advantage of the geometric roerties for fast imlementation. Furthermore, it is imortant to study the imact of limited bandwidth and transceiver resources, as well as to develo mechanisms to coe with node mobility [8]. References [] V. Chvatal, \A Greedy Heuristic for the Set-Covering Problem", Mathematics of Oerations Research, Vol. 4, No.,. {5, 979. [] T. J. Cormen, C. E. Leiserson, and R. L. Rivest, Introduction to Algorithms, MIT Press and MxGraw-Hill, 990. [] M. R. Garey, and D. S. Johnson, Comuters and Intractability: a Guide to the Theory of NP-Comleteness, W. H. Freeman and Comany, 979. [4] S. Guha, and S. Khuller, \Aroximation Algorithms for Connected Dominating Sets". Algorithmica 998, 0: [5] F. P. Prearata and M. I. Shamos, Comuatational Geometry: an Introduction, Sringer-Verlag, 985. [6] T. S. Raaort, Wireless Communications: Princiles and Practices, Prentice Hall, 996. [7] S. Singh, C. S. Raghavendra, and J. Steanek, \Power-Aware Broadcasting in Mobile Ad Hoc Networks", Proceedings of IEEE PIMRC'99, Osaka, Jaan, Se [8] J.E. Wieselthier, G.D. Nguyen, and A. Ehremides, \On the Construction of energy-ecient Broadcast and Multicast Trees in Wireless Networks". IEEE Infocom'000.