Flow Demands Oriented Node Placement in Multi-Hop Wireless Networks

Transcription

1 Flow Demands Oriented Node Plaement in Multi-Hop Wireless Networks Zimu Yuan Institute of Computing Tehnology, CAS, China arxiv: v1 [s.ni] 29 Mar 215 Abstrat In multi-hop wireless networks, flow demands mean that some nodes have routing demands of transmitting their data to other nodes with a ertain level of transmission rate. When a set of nodes have been deployed with flow demands, it is worth to know how to onstrut paths to satisfy these flow demands with nodes plaed as few as possible. In this paper, we study this flow demands oriented node plaement problem that has not been addressed before. In partiular, we divide and onquer the problem by three steps: alulating the maximal flow for single routing demand, alulating the maximal flow for multiple routing demands, and finding the minimal number of nodes for multiple routing demands with flow requirement. During the above solving proedure, we prove that the seond and third step are NP-hard and propose two algorithms that have polynomial-time omplexity. The proposed algorithms are evaluated under pratial senarios. The experiments show that the proposed algorithms an ahieve satisfatory results on both flow demands and total number of wireless nodes. I. INTRODUCTION Multi-hop wireless networks have gained a lot of attentions in the past few years. One of key design issues in multi-hop wireless networks is node plaement. By areful node plaement, we an make multi-hop wireless networks ahieve speial design goals. For example, studies on node plaement are related to wide topis suh as network traffi [5][18][21], network overage [15][16][22], network survivability [13][17][19], fault-tolerant [7][8][23], energy saving [2][9][1], and et. In this paper, we study the problem of flow demands oriented node plaement, i.e., how to use less wireless nodes to satisfy flow requirements for multi-hop wireless networks. Here, flow requirements indiate that the data should be transmitted at a ertain level of transmission rate. Existing studies on node plaement related to network traffi [5][18][21] mainly address the problem of optimizing the network throughput. However, the flow demands has signifiant differene from the throughput demands and new methods are needed to satisfy flow demands for node plaement. Atually, it is not straightforward to solve the problem of flow demands oriented node plaement. Therefore, we divide and onquer the problem by three steps: at first, we alulate the maximal flow for single routing demand, whih is the basis of analyzing multiple routing demand then, we alulate the maximal flow for multiple routing demands aording to the result on single routing demand finally, based on the maximal flow alulated for multiple routing demand, we try to merge routing paths to ahieve minimal number of wireless nodes. The above proedure involves several proofs and related algorithm design, i.e., two problems are proved as NP-hard and two heuristi algorithms are proposed. In evaluation, we verify the effiieny of the proposed algorithms by examining average satisfied rate (defined in Setion IV) of flow demands and the number of nodes used for plaement under the pratial senarios of data aggregation, demands with definite flow requirement and nodes with unknown flow requirement. To the best of our knowledge, the problem of satisfying the flow demands by node plaement in multi-hop wireless networks has not been studied yet. The ontributions are summarized as follows: 1) The omplexity of flow demands oriented node plaement is theoretially analyzed. For single routing demand, the theoretial maximal flow an be ahieved under the interferene model is onduted. For multiple routing demands, the proof of NP-hard to obtain the maximal flow under the interferene between routing paths, and the proof of NP-hard to minimize the number of nodes plaed by merging routing paths are given. 2) A novel approah is proposed to use less nodes to satisfy the flow requirements for node plaement. For multiple routing demands, a polynomial-time omplexity algorithm is given to ahieve larger flow as possible by finding the heaviest interferene node and assigning ativated time slots on eah routing path, and a polynomialtime omplexity algorithm is given to plae relay nodes as fewer as possible by prior to merge the onstruted routing paths with more exessive flow apaity. Both proposed algorithms are disussed with their worst ase. 3) The effiieny of the proposed algorithms are verified through several pratial senarios. Our experiments show that Average Satisfied Rate of flow demands redues slowly when inreasing the level of flow requirements, and the number of nodes for plaement is redued in average of 25.4%, 26.6% and 24.1%, both of whih prove the effiieny of proposed algorithms. The rest of the paper is organized as follows. Setion II introdues the model assumption. Setion III studies the flow demands oriented node plaement problem and proposes the algorithm. Setion IV evaluates the proposed algorithm. Setion V onludes the work. II. THE PROBLEM AND RELATED WORK Suppose that a set of nodes have been deployed in a plane. Some of these nodes may have routing demands of transmitting their data to other nodes with flow requirements that should be ahieved. Then, relay nodes are needed to be plaed to route and satisfy these flow demands. We assume that eah node is equipped with a radio, and the radio has the its maximal transmission range r. If two nodes, n 1 and

2 n 2, that are loated with their distane Dist(n 1, n 2 ) R an interfere the transmission between eah other. With respet to a suessful transmission of two nodes, we onsider the Protool Interferene Model. A transmission from node n 1 to n 2 is suessful if and only if 1) Dist(n 1, n 2 ) r 2) there does not exist a transmission node n 3 suh that Dist(n 1, n 3 ) R. This interferene model is widely used in referenes like [6][11]. Besides, we use a time slotted system. In the time slotted system, the time is divided into equal length slots, and the transmission between nodes are synhronized. We define f as the maximal flow (or maximal transmission rate) that an be transmitted in a single time slot. For the set of nodes with flow requirements, we model that there are total m pair of routing demands (sr q, dest q ) with flow requirement F A (sr q dest q ) should be satisfied, q = 1, 2,..., m. In this paper, we aim to plae relay nodes to onstrut paths to satisfy the flow requirement F A (sr q dest q ) between (sr q, dest q ), q = 1, 2,..., m, while use as fewer plaed nodes as possible. We named it as the flow demands oriented node plaement problem for short. As a matter of fat, the above problem has not been studied in previous researh. Some of existing studies [5][18][21] on node plaement address the throughput issue. These studies mainly fous on plaing nodes to optimize the network throughput. The senario of olleting all node transmissions towards sink nodes is onsidered in [5]. A grid-based relay nodes plaement method is used to optimize the network throughput in [18]. Two objetives are studied in [21]. The one is to maximize the minimum throughput for any relay node, and the other is to maximize the total throughput of the network. However, the throughput optimization of the whole network is not equal to the issue of satisfying the flow requirements of some nodes, and also, the flow requirements of different nodes may be different, whih should be with differential treatment. The flow requirements annot be satisfied within throughput optimization framework. Besides, the wireless interferene, an intrinsi harateristi of wireless hannel, among nodes is also seldom onsidered in these studies. We study the flow demands oriented node plaement problem to fill this blank. III. FLOW DEMANDS ORIENTED NODE PLACEMENT A. Methodology In this setion, we solve the problem of flow demands oriented node plaement in following three steps: 1) Calulating the maximal flow for single routing demand. At first, we find the maximal flow an be ahieved by onstruting paths for a single routing demand (sr 1, dest 1 ). 2) Calulating the maximal flow for multiple routing demands. Then, we prove that onstruting paths to maximize the flow for multiple routing demands (sr q, dest q ), q = 1, 2,..., m is NP-hard, and try to design an algorithm to ahieve greater flow as possible for multiple routing demands under the interferene between paths. 3) Finding the minimal number of nodes for multiple routing demands with the flow requirement. Finally, we also prove the problem of reduing the number of nodes plaed to the minimal with flow requirement F A (sr q dest q ) for (sr q, dest q ), q = 1, 2,..., m is NPhard, and try to propose our algorithm to redue the number of nodes plaed as possible while satisfying the flow requirement. B. Calulating The Maximal Flow for Single Routing Demand Suppose that there is a node pair (sr 1, dest 1 ) with routing demand from soure node sr 1 to destination node dest 1. Let p sr1dest 1,1 be the path onstruted between sr 1 and dest 1, and f sr1dest 1 be the flow between sr 1 and dest 1. Assume that the interferene range R is in [jr, (j + 1)r), j N +, and the distane Dist(sr 1, dest 1 ) between sr 1 and dest 1 has Dist(sr 1, dest 1 ) 2R 1. We give a theorem as follows: Theorem 1. With a single path onstruted, the flow f sr1dest 1 between sr 1 and dest 1 an reah F 1 = at most. f j+1 Proof: Assume that the distane between sr 1 and dest 1 has Dist(sr 1, dest 1 ) (ir, (i + 1)r]. At least i nodes should be plaed to onstrut a path p sr1dest 1,1 between sr 1 and dest 1 as shown in figure 1. With the ondition Fig. 1. A single path of (sr 1, dest 1 ) R [jr, (j + 1)r), j N +, we know that node n 11, n 12,..., n 1j are within interferene range of sr 1. The total j + 1 links of sr 1 n 11, n 11 n 12,..., n 1j 1 n 1j should be assigned with different time slots to avoid mutual interferene in transmission. The other links with transmission nodes out of the interferene range of sr 1 an be assigned with the time slots that have been used, i.e. Slot(n 1j+1 n 1j+2 ) = Slot(sr 1 n 11 ) = 1, Slot(n 1j+2 n 1j+3 ) = Slot(n 11 n 12 ) = 2,..., Slot(n 1i 1 n 1i ) = (i 1)%(j+1)+1 and Slot(n 1i dest 1 ) = i%(j + 1) + 1. There are total j + 1 time slots assigned to the path p sr1dest 1,1. As proved, the flow f sr1dest 1 between sr 1 and dest 1 an reah F 1 at most with one path p sr1dest 1,1 onstruted. With multiple paths onstruted, the flow f sr1dest 1 an reah a larger value. We have the following theorem: Theorem 2. With multiple paths onstruted, the flow f sr1dest 1 between sr 1 and dest 1 an reah F C at most. f F C = max (1) =1,2,...,C s Where s = max { q+1+ q=1,2,...,j m=2 { max{ floor( xm r ), }+ 1 } } and x m = q 2 r 2 os 2 2π(m 1) q 2 r 2 + R 2 + qr os 2π(m 1). 1 Node sr 1 and dest 1 an diretly ommuniate or ommuniate with only a few of relay nodes if we have Dist(sr 1, dest 1 ) < 2R, so it is meaningless to study the node plaement problem under the ondition of Dist(sr 1, dest 1 ) 2R. Also, it will be involved in omplex disussion of ases under this ondition

3 Proof: Suppose that there are paths onstruted between sr 1 and dest 1. To find the maximal flow that an be ahieved by the onstruted paths, the key is to seek out the heaviest interferene area that involves the maximum number of nodes in transmission. Obviously, the area surrounding sr 1 or dest 1 are inevitable to be the heaviest interferene area with the highest density of nodes plaed in the onstrution of paths, i.e. some nodes have to be plaed around sr 1 to reeive the transmission from sr 1 in figure 2, and thus, these nodes ould not be plaed far enough to avoid interferene. To find the maximal flow an be ahieved in the heaviest interferene area surrounding sr 1 or dest 1, it is neessary to find the heaviest interferene node on a onstruted path, whih is the node that interferes with the maximum number of nodes on other paths. Let S int (n) denote the interferene node set for a node n. Take the heaviest interferene area surrounding sr 1 for example. The heaviest interferene node n h has n h = arg max S int(n) (2) Dist(sr 1,n)<R Fig onstruted paths of (sr 1, dest 1 ) Next, we firstly try to minimize the interferes when onstruting paths so that the heaviest interferene node will interfere with the minimum number of nodes. With the path onstrution proess, the expression (2) an be rewritten as n h = arg min{ max S int(n) path onstrution} Dist(sr 1,n)<R (3) It an be proved that if these paths are onstruted with equal angle-interval of 2π from sr 1 and to dest 1 ( n 11 sr 1 n 21 = n 21 sr 1 n 31 = n 31 sr 1 n 11 = 2π 3 in figure 2), the interferene an be redued to the minimal between paths. It is not hard to finish this proof with the basi knowledge of analyti geometry. As spae is limited, we ignore the subordinate details here. After paths with equal angle-interval of 2π have been onstruted, the heaviest interferene node n h ould be found by expression 2. More speifi, we randomly selet a onstruted path and ompare all the nodes n having Dist(sr 1, n) < R to find the heaviest interferene node that determines the maximal flow an be ahieved. Without loss of generality, we set the seleted path as p sr1dest 1,1 and the seleted omparison nodes with the oordinates of (r, ), (2r, ),..., (jr, ). For a node (qr, ), q = 1, 2,..., j, the interferene length x m with another path p sr1dest 1,m an be alulated with the osine formula R 2 = q 2 r 2 + x 2 2π(m 1) m 2qrx m os (4) Solve this equation, we an get 2π(m 1) x m = qr os + q 2 r 2 2π(m 1) os2 q 2 r 2 + R 2 (5) Sine x m ould be negative value, we an alulate the number of interferene nodes on path p sr1dest 1,m as max{floor(x m /r), }, in whih floor() rounds down the frations. Thus, s m new time slots should be assigned to p sr1dest 1,m s m = max{ floor( x m ), } + 1 r (6) Add up all the new assigned time slots s(qr, ) = q s m (7) m=2 Find the maximal ounts of time slots assigned s = max s(qr, ) (8) q=1,2,...,j The node that introdues the maximal ounts of time slots s is the heaviest interferene node. We get the maximal ounts of time slots s for onstruted paths, and the flow ahieved an be denoted as f = f s (9) Let = 1, 2,..., C. We an obtain the maximal flow F C an be ahieved between sr 1 and dest 1 by 2 F C = max f (1) =1,2,...,C The proof is ompleted by ombining the expression (5)-(1). We have proved that the flow f sr1dest 1 between sr 1 and dest 1 an reah F C at most with multiple paths used. The One Soure Destination Pair with Multiple Paths (MP1) algorithm summarizes the proess to find F C (Algorithm 1). Paths are onstruted with equal angle-interval at their beginning and end part, and the onnetion paths between the beginning and end part are onstruted to avoid interferene mutually. Then, the maximal flow F C an be alulated by expression (1). MP1 algorithm exeutes with time omplexity of O(C 2 ) to onstrut paths. C. Calulating The Maximal Flow for Multiple Routing Demands Suppose that there are m routing demands (sr 1, dest 1 ), (sr 2, dest 2 ),...,(sr m, dest m ). As proved, the flow of a single demand (sr 1, dest 1 ) an reah the flow of F C by onstruting multiple paths. However, in m routing demands ase, not all these demands an reah F C. We have the following theorem: 2 The value of C is determined by R. When the number of paths inrease to a ertain value, the flow an be ahieved will derease then with too many nodes interfering in transmission.

4 Algorithm 1: Construt Multiple Paths for One Soure Destination Pair (MP1) Input: Soure destination pair (sr 1, dest 1 ), Flow f, Interferene range R, Maximal path ount C Output: The maximal flow F C and its onstruted paths p sr1dest 1 1 F C = 2 for = 1, 2,..., C do 3 for m = 1, 2,..., do 4 Plae nodes as the beginning of path p sr1dest 1,m with 2π(m 1) degrees to the diretion of sr 1 dest 1 from sr 1 until a plaed node on p sr1dest 1,m does not interfere other nodes in other already onstruted paths 5 Plae nodes as the end of path p sr1dest 1,m with π 2π(m 1) degrees to the diretion of sr 1 dest 1 to dest 1 until a plaed node on p sr1dest 1,m does not interfere other nodes in other already onstruted paths 6 Connet the nodes at the beginning and the end of p sr1dest 1,m and let the nodes at the onnetion path do not interfere the transmission of other already onstruted paths 7 end 8 s = max q=1,2,...,j { q m=2 { max{ floor( xm r 9 x m = q 2 r 2 os 2 2π(m 1) f 1 if max =1,2,...,C s f 11 F C = max =1,2,...,C ), } + 1 } } q 2 r 2 + R 2 + qr os 2π(m 1) > F C then s 12 Reord the onstruted paths p sr1dest 1,m, m = 1, 2,..., 13 end 14 end 15 return F C and p sr1dest 1 Theorem 3. Not all m routing demands of node pair (sr 1, dest 1 ), (sr 2, dest 2 ),...,(sr m, dest m ) an definitely reah the flow of F C by onstruting paths. For example, there are two routing demands (sr 1, dest 1 ) and (sr 2, dest 2 ) in figure 3. Both sr 1 and dest 1 interfere with sr 2 in transmission. When sr 2 does not transmit, the flow of (sr 1, dest 1 ) an reah F. When sr 2 transmits with interferene, the link sr 1 dest 1 annot be ativated all the time, and thus, annot reah the flow of F C. However, even when the transmission of a routing demand is under the interferene of onstruted paths of other demands, the flow of this routing demand an still reah F C by assigning time slots in some ases. In figure 4, there are two routing demands (sr 3, dest 3 ) and (sr 4, dest 4 ). The path p sr3dest 3,1 is assigned with time slots set of {1, 2, 3, 4, 5} to its links, and thus, an reah flow of f 5. The path p sr 4dest 4,2 is assigned with time slots {1, 2, 3, 4} and an reah flow of 4. Both node n 12 and n 13 on path p sr3dest 3,1 interfere with n 22 and n 23 Fig. 3. Two routing demands (sr 1, dest 1 ) and (sr 2, dest 2 ) Fig. 4. Two routing demands (sr 3, dest 3 ) and (sr 4, dest 4 ) in transmission. By assigning time slots of 4, 5, 2 and 3 to links n 12 n 13, n 13 n 14, n 22 n 23 and n 23 n 24 respetively, both path p sr3dest 3,1 and p sr4dest 4,2 an still reah the flow of f 5 and f 4 without adding new slot to existing time slots set to avoid interferene. As a summary, we give a theorem: Theorem 4. The flow of a routing demand an reah F C if time slots an be assigned to avoid interferene of other demands without adding new time slots. Sine not every routing demand an reah its maximal flow of F C, we onsider m routing demands as a whole, and try to answer the problem that how to maximize the sum of the flow of all these m routing demands. The problem is formalized as following objetive funtion: max m q=1 f sr qdest q s.t. f srqdest q > However, it is NP-hard to solve this objetive funtion. (11)

5 Algorithm 2: Construt Multiple Paths for Multiple Soure Destination Pairs (MPM) Input: Soure destination pairs (sr q, dest q ), q = 1, 2,..., m, Flow f, Interferene range R, Maximal path ount C Output: The flow F C (sr qdest q ) and path p srqdest q, q = 1, 2,..., m 1 for q = 1, 2,..., m do 2 Get the pair (sr q, dest q ) 3 [F C (sr q dest q ), p srqdest q ] = MP 1((sr q, dest q ), f, R, C) 4 end 5 for q = 1, 2,..., m do 6 Get the pair (sr q, dest q ) 7 Set F C (sr qdest q ) = 8 for Eah onstruted path p in p srqdest q do 9 Set n(p) as a randomly seleted node on path p 1 Set S int (p) = 11 for Eah plaed node n in p do 12 Find the nodes that interfere the transmission of node n, and reord these nodes into S int (n) (Count the soure and destination nodes with the times of the number of onstruted paths that onnet them in S int (n) 3 ) 13 if S int (n) > S int (p) then 14 Set n(p) = n 15 Set S int (p) = S int (n) 16 end 17 end 18 Assign time slots 1, 2,..., S int (p) to the links transmitted by the elements in S int (p) if the link has not been assigned with time slot, and other links in path p ould reuse these time slots without interferene in transmission 19 Set F C (sr qdest q )+ = 1 S int(p) 2 end 21 end 22 return The flow F C (sr qdest q ) and path p srqdest q, q = 1, 2,..., m Theorem 5. Construt paths to maximize the sum of the flow of m routing demands is NP-hard. Proof: This problem an be redued to the NP-hard problem of Theorem 1 in [11]. The NP-hard problem in [11] is to find the maximal sum of the flow for a group of soure and destination nodes on a given network with deployed nodes. Our problem is differene in that nodes are plaed to onstrut paths to maximize the sum of the flow for a group of soure and destination nodes. The node plaement an be seen as seleting a set of nodes from infinite andidate nodes from the plane. By onsidering these andidate nodes as deployed nodes in the network, our problem an be redued to the NPhard problem of Theorem 1 in [11]. It is NP-hard to maximize the sum of the flow of m routing demands by onstruting paths. Also, it has been proved in [11] that the maximum sum of the flow is NP-hard to be approximated. In other words, with the redution of our problem to this problem, we know that it is NP-hard to find a solution to onstrut paths that guarantee to approximate the maximal flow. Although it is NP-hard to approximate the maximal flow, the onstruted paths should try to avoid interferene between eah other. The interferene between paths depends on the position of the soure and destination nodes. Reonstrut the paths if interferene exists between them an make little effet beause reonstruted paths usually introdue new interferene, espeially when the reonstruted path goes through the soure or destination nodes, or intersets with other paths. So we design the MPM algorithm that does not try to reonstrut paths to avoid interferene, but try to find the heaviest interferene node and assign time slots based on its interferene set. MPM algorithm (Algorithm 2) find the heaviest interferene node n(p) for path p, and reord its interferene nodes to set S int (p). The flow F C (p) ahieved by 1 path p is S. The flow F int(p) C (sr qdest q ) between sr q and dest q an be alulated by summing up the flow of onstruted paths between them. MPM algorithm proesses eah path one, and returns the flow F C (sr qdest q ) and onstruted path p srqdest q, q = 1, 2,..., m. The ahieved sum of flow F C for all soure and destination pairs is m q=1 F C (sr qdest q ). D. Finding The Minimal Number of Nodes for Routing Demands with Flow Requirement m routing demands with flow requirement of F C has been disussed in last setion. Using MPM algorithm, m routing demands an ahieve flow of F C. Atually, the flow requirement of routing demand an be arbitrary in pratial. In this setion, we fous on the ase of m routing demands (sr 1, dest 1 ), (sr 2, dest 2 ),...,(sr m, dest m ) with flow requirement of F A (sr q dest q ), in whih F A (sr q dest q ) F C (sr qdest q ) for q = 1, 2,..., m. In this ase, the apaity of some paths may exeed the flow requirement of some routing demands. This will lower link utilization rate of these paths. It is a waste of resoure. For this reason, some paths an be merged to inrease the link utilization rate and derease the total number of nodes needed to be plaed. We have the following objetive funtion: min m q q=1 a=1 Length(p sr qdest q,a) (12) q s.t. a=1 f sr qdest q,a F A (sr q dest q ) As implied in the objetive funtion, two path, one for routing demand (sr q, dest q ) and the other one for (sr qk, dest qk ), an be merged if the length of the merged path is redued than the total length of these two paths, and the flow requirement, F A (sr q dest q ) and F A (sr qk dest qk ), an be met after these two path have been merged. I.e., in figure 5, let Length(d k1 d k2 ) denote the length of path p srqk dest qk,a k between two points, d k1 and d k2. We have Dist(d k1, p srqdest q,a) + Dist(d k1, p srqdest q,a) < Length(d k1 d k2 ), whih makes the merged path shorter than the total length of the original two paths and the flow requirement of F A (sr q dest q ) and F A (sr qk dest qk ) an still be met.

6 Algorithm 3: Merge Input: Paths p srqdest q with flow requirement F A (sr q dest q ), q = 1, 2,..., m, Flow F C Output: Merged paths p sr qdest q, q = 1, 2,..., m 1 for q = 1, 2,..., m do 2 Let q denote the number of paths for (sr q, dest q ) 3 while Delete the longest path from p srqdest q, still having the flow ahieved F q (sr q dest q ) > F A (sr q dest q ) do 4 Delete the longest path from p srqdest q 5 Set q = q 1 6 end 7 Set df (sr q dest q ) = F q (sr q dest q ) F A (sr q dest q ) 8 end 9 Sort the (sr q, dest q ) by the value of df (sr q dest q ) in desending order, q = 1, 2,..., m, and reord the result in array D 1 for q = 1, 2,..., m do 11 for u = q + 1, q + 2,..., m do 12 Get the pair (sr q, dest q ) and (sr u, dest u ) from array D 13 Equal-lengthly sample a set of points in all paths of p srqdest q and p srudest u 14 Calulate the distane from all paths of p srqdest q to all paths of p srudest u by averaging the distane between sample points, and reord the results into the q u matrix M 15 while 1 do 16 Selet the minimal value of distane from matrix M, denoted by M(a q, a u ) (M(a q, a u ) ) 17 Set the a q row M(a q, :) =, and the a u olumn M(:, a u ) = 18 if p srqdest q,a q and p srqdest q,a q is mergable then 19 Merge the path p srqdest q,a q and p srqdest q,a q 2 else 21 break 22 end 23 end 24 Reord the newly onstruted path p sr qdest q and p sr udest u 25 end 26 end 27 return paths p sr qdest q, q = 1, 2,..., m (a) Before merging (b) After merging Fig. 5. Merge path The path p srqdest q,a and p srqk dest qk,a k are mergable. We use the mergable element P to denote mergable paths, suh as P = (p srqdest q,a, p srqk dest qk,a k ). A mergable set S mer omposes of mergable elements, having p 1 p 2 for p 1 P 1, p 2 P 2, and P 1, P 2 S mer. Then, a maximum mergable set an be desribed as there are no other mergable paths for (sr q, dest q ), q = 1, 2,..., m after applying all the merging for mergable paths in the set. To solve the objetive funtion that minimize the total path length while still meet the flow demands, all the maximum mergable set should be found and ompared so that we an find the set minimize the total path length. However, it is NPhard to solve this objetive funtion. Theorem 6. Minimize the total length of all paths for m routing demands with flow requirement restrition by merging paths is NP-hard. Proof: Let eah mergable element represent a node in the network. Establish a link between nodes if two mergable elements ontain a same path. Then, our problem an be redued to the NP-hard problem of finding all maximal independent set in the network [12]. It is NP-hard to minimize the total length of all paths for m routing demands by merging paths. The Merge algorithm (Algorithm 3) is proposed to try to minimize the total length of onstruted paths while still meet the flow requirement F A of these m routing demands. In the Merge algorithm, extra paths with longer length of routing demands are firstly deleted, and the extra flows df (sr q dest q ), q = 1, 2,..., m that represents the exessive flow ompared to flow demand F A (sr q dest q ) are reorded. The paths of a routing pair with more exess flow are more likely to be merged with other paths, so we sort the routing pairs by df (sr q dest q ), q = 1, 2,..., m in desending order, and try to merge the paths with more exess flow (detailed in Algorithm 3). Finally, the Merge algorithm returns the newly merge path p sr qdest q, q = 1, 2,..., m with O(m 2 ) merge attempts between routing demands. The merging between paths depends on the positions

7 of (sr q, dest, q), q = 1, 2,..., m, and their flow requirement F A (sr q dest q ). In worst ase, no path an be merged to redue the total path length. E. Summary In this setion, we first analyze a single routing demand of (sr 1, dest 1 ) and prove that the flow an reah F 1, F C with one path and multiple paths onstruted respetively. Then, for multiple routing demands of (sr q, dest q ), q = 1, 2,..., m, we prove that it is NP-hard to maximize the sum of the flow by onstruting paths and show that the flow an ahieve F C by MPM algorithm. For multiple routing demands with flow requirement, we prove that it is NP-hard to minimize the total length of all paths by merging path and propose a greedy algorithm of merging paths. Algorithm 4: Flow Demands Oriented Node Plaement Input: Soure destination pairs (sr q, dest q ) with flow requirement F A (sr q dest q ), q = 1, 2,..., m, Flow f, Interferene range R, Maximal path ount C Output: The plaement position P of nodes 1 [F C, p sr qdest q ] = MP M((sr q, dest q ), f, R, C), q = 1, 2,..., m 2 p sr qdest q = Merge(F C, F A, p srqdest q ), q = 1, 2,..., m 3 Plae nodes along the onstruted paths p sr qdest q 4 Reord the position of nodes to P 5 return P The Flow Demands Oriented Node Plaement Algorithm is summarized in Algorithm 4. In line 1, the MPM algorithm (Algorithm 2) onstruts paths to find the maximal flow of multiple routing demands that an be supported by the network. The MPM algorithm try to satisfy the flow F C obtained by the MP1 algorithm (Algorithm 1) and an ahieve the flow of F C. In line 2, given flow requirement F A of routing demands, The Merge algorithm (Algorithm 3) try to minimize the total length of paths. Then, the algorithm plaes nodes along the merge paths and returns the position of plaed nodes. Following the analysis above, proposed Algorithm 4 proesses routing demands in O(m 2 ) times. The performane of the MPM algorithm depends on the positions of (sr q, dest, q), q = 1, 2,..., m. In worst ase, only one link an be ativated for transmission in eah time slot. The performane of the Merge algorithm depends on the positions of (sr q, dest, q), q = 1, 2,..., m, and their flow requirement F A (sr q dest q ). In worst ase, no path an be merged at all. We will examine the performane of proposed algorithm later in evaluation. IV. EVALUATION A. The Metri to Measure the Effiieny of Algorithms We evaluate algorithms proposed in Setion III through simulations. Three senarios, Data Aggregation, Demands with Definite Flow Requirement and Nodes with Unknown Flow Requirement, are used for evaluation. By default setting, nodes are plaed in a 2 2 square region. There are 1 routing demands. The transmission range r is set to 1 and the interferene range R is set to 2r. The flow f is set to 1. For eah senarios, the graph is randomly generated for 1 times. In evaluation of these senarios, the area size is hanged in 15 15, 2 2, and 3 3. The Interferene range is hanged R = r, R = 1.4r, R = 1.8r and R = 2.2r. The number of routing demands is hanged in 5, 1, 15 and 2. We define the Satisfied Rate as 1% f sr qdest q f srqdest q SR q = f sr qdest q (13) 1% f sr f qdest q < f srqdest q srqdest q where f srqdest q is the flow requirement of (sr q, dest q ) and f sr qdest q is the flow atually an be ahieved. Based on equation 13, we define the Average Satisfied Rate for m routing demands as m q=1 ASR = SR q (14) m The higher the average satisfied rate an be ahieved, the better the flow demands an be met. B. Experiments 1) Senario 1: Data Aggregation: In senario 1, data are gathered from a set of soure nodes and sent to a destination node. For this senario, m soures nodes and 1 destination node are randomly generated in graph. The flow requirement is randomly generated for eah soure. With the setting of flow f = 1, the total flow of all soure-destination pairs reahing the destination has no more than flow of 1 for eah time slot. Thus, the average flow requirement of eah pair is generated from.1 to.1 in evaluation. The impats of hanging area size, interferene range and routing demands on average satisfied rate are shown in figure 6, 7 and 8 respetively. The impats of hanging area size, interferene range and routing demands on the total number of nodes used for plaement are shown in figure 9, 1 and 11 respetively. The results are summarized later in setion IV-C. 2) Senario 2: Demands with Definite Flow Requirement: In senario 2, there is a set of m routing demands (sr q, dest q ) with definite flow requirement f srqdest q, q = 1, 2,..., m. For this senario, m routing demands are randomly generated in graph. The flow requirement is randomly generated for eah demand. The average flow requirement of eah pair is generated from.4 to.4 in evaluation. The impats of hanging area size, interferene range and routing demands on average satisfied rate are shown in figure 12, 13 and 14 respetively. The impats of hanging area size, interferene range and routing demands on the total number of nodes used for plaement are shown in figure 15, 16 and 17 respetively. The results are summarized later in setion IV-C. 3) Senario 3: Nodes with Unknown Flow Requirement: In senario 3, there are a set of nodes that have been deployed, but we do not know the definite routing demands and flow requirement. For this senario, a set of 2m nodes are randomly generated. m nodes out of these nodes are randomly seleted

8 *15 2*2 25*25 3* Demands 1 2 Demands Fig. 6. Senario 1: Area Size Fig. 7. Senario 1: Interferene Range Fig. 8. Senario 1: Routing Demands *15 2*2 25*25 3* Demands 1 2 Demands Fig. 9. Senario 1: Area Size Fig. 1. Senario 1: Interferene Range Fig. 11. Senario 1: Routing Demands *15 2*2 25*25 3* Demands 1 2 Demands Fig. 12. Senario 2/3: Area Size Fig. 13. Senario 2/3: Interferene Range Fig. 14. Senario 2/3: Routing Demands *15 2*2 25*25 3* Demands 1 2 Demands Fig. 15. Senario 2: Area Size Fig. 16. Senario 2: Interferene Range Fig. 17. Senario 2: Routing Demands *15 2*2 25*25 3* Demands 1 2 Demands Fig. 18. Senario 3: Area Size Fig. 19. Senario 3: Interferene Range Fig. 2. Senario 3: Routing Demands

9 as soure nodes and the other m nodes are seleted as destination nodes. As the set of nodes are with unknown flow requirement, the paths onstruted between soure-destination pairs try to fulfill the flow of F C (Theorem 2) using MPM algorithm (Algorithm 2). In evaluation, the average flow requirement of eah pair is generated from.4 to.4. Atually, the results on average satisfied rate is the same as the results in senario 2 (figure 12, 13 and 14). The differene between senario 3 and senario 2 is that the flow requirement is unknown in senario 3. Thus, the Merge funtion in algorithm 4 does not merge paths in senario 3 and more nodes need to be plaed than senario 2. The impats of hanging area size, interferene range and routing demands on the total number of nodes used for plaement are shown in figure 18, 19 and 2 respetively. C. Analysis The evaluation results are summarized as follows: 1) The result of average satisfied rate. The average satisfied rate is slightly higher with a larger area size, sine a larger area has lower probability of interferene between nodes. The average satisfied rate is lower with a larger interferene range, sine a larger interferene range has higher probability of interferene between nodes. The average satisfied rate is lower with more routing demands, sine more routing demands with more nodes ause higher probability of interferene between nodes. As an be seen from figure 6, 7, 8, 12, 13 and 14, the average satisfied rate of flow demands redues slowly when inreasing the level of flow requirements. 2) The result of total number of nodes used. Slightly less nodes needed to be plaed in senario 1 than senario 2, sine more paths an be merged in data aggregation senario. Less nodes needed to be plaed in senario 2 than senario 3, sine more nodes are needed to satisfy possible larger flow in senario 3. (Average 25.4%, 26.6% and 24.1% nodes haven been merged in figure 15, 16 and 17 than figure 18, 19 and 2 respetively.) The result verify the effiieny of merging routing paths. V. CONCLUSION AND DISCUSSION The flow demands oriented node plaement problem has been addressed in this paper. The problem is solved in three steps of alulating the maximal flow for single routing demand, alulating the maximal flow for multiple routing demands and finding the minimal number of nodes for routing demands with flow requirement. For single routing demand, we ondut its theoretial maximal flow an be ahieved. For multiple routing demands, we prove both the problem of alulating the maximal flow and finding the minimal number of nodes for plaement are NP-hard and propose polynomialtime omplexity algorithms. The proposed algorithms are extensively evaluated in the senarios of data aggregation, demands with definite flow requirement and nodes with unknown flow requirement, whih verify its effiieny. 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