CHAPTER 5 TASK ALLOCATION AND ROUTING IN WSN

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1 CHAPTER 5 TASK ALLOCATION AND ROUTING IN WSN 5.1 Introduction The lifetime of the sensor network can be improved by efficient assignment of task and routing. The task allocation inside the network plays a vital role in enhancement of lifetime because if the task distributed to different sensor nodes in WSN is unbalanced, it results in unbalanced use of energy. Due to the unbalanced utilization of energy inside the network, some sensor nodes come to end early as compared to others; it consequently breaks the communication links. As per the literature the task should be assigned to sensor nodes in some balanced manner which can increase the lifetime. One way to balance the task among the sensor nodes is schedule the sensor nodes according to some parameters and then assign the task. Due to the conditions that require the response from these sensor nodes in few seconds, some routing methods must be developed that respond in few seconds in distribution or collection of task. The key challenge to fulfill this goal is the battery power of sensor nodes and the presence of hurdles inside the region. The hurdles inside the WSN generate a problem of achieving a short length path for result collection. To overcome these problems, three methods are developed in this thesis. First method maintains clusters inside the WSN and assumes that the sensor nodes inside the clusters form a tree like structure. Second method, deploy and schedule the sensor nodes inside the region and on the basis of achieved scheduling distribute the task among the sensor nodes. Third method aims to provide the efficient route for collecting the task s results. 5.2 Material and Methodology In this section different methodologies are discussed that are developed in the present work Tree Based Methodology Before providing the steps that are used to elaborate the methodology this section gives some models that are applied to achieve the tasks allocation. This section will discuss the network model, remaining battery model, task distribution model etc. 60

2 i) Network System Model Network system consists of a set of clusters as N= {c 1, c 2, c 3, c n }and each cluster consists of many sensor nodes as C={n 1, n 2, n 3, n t }. The size of the cluster is variant and the battery power of these sensor nodes can be different but the computing power of the sensor nodes is same. Suppose the battery power of these sensor nodes are given by the set B= {b 1, b 2, b 3, b t } then the energy of the cluster can be given as. The network can be represented as a graph of many nodes, the upper layer of graph consists of clusters and lower layer consists of sensor nodes. The clusters are connected with each other by radio links and the sensor nodes inside the cluster forms a tree like structure. The link X uv means the cluster u can communicate with the cluster v. The graph for upper layer(l 1 ) can be modeled as G l1 ={N, X}, where X={X 12,X 34,X 35, X pn }. The edge(link) X bc represents that the cluster b can directly communicate with cluster c. The graph for lower layer(l 2 ) is modeled as G l2 = {C,Y}, where Y is a set {Y 12,Y 34,Y 35 Y qt }. The link Y mn represents that the sensor node m can communicate with sensor node n. An allocation matrix(a) mxn shows the amount of task assigned to a particular cluster. If an entry (A ij ) =0 is present in A, it shows that part of task i is not allocated to cluster j and if A ij = x, shows that x percent task of task i is assigned to the cluster j. ii) Task Model Suppose an application consists of many tasks such as {t 1, t 2, t 3, t n }. Each cluster inside the network is monitored by one head node (H). The structure of the network inside a cluster is like directed acyclic graph (DAG) and the root node of this DAG is the head node of that cluster. Each cluster is assigned to complete some application and the tasks that come under the application are managed by the head node of this cluster. iii) Cluster Formation Phase For constructing the cluster unsupervised learning is applied on a data set gathered from [92]. The original data set contains many fields such as date, time, epoch, moteid, temperature, humidity, light, voltage. The data set is preprocessed to obtain the fields of temperature (Celsius) and voltage (ma) and on this preprocessed data set k- means clustering is applied to form the cluster. Initially, 2 clusters are taken but it does not provide balanced density of sensor nodes. The clusters are increased from 2-3, 3-4 and 5. When the clusters are 5, it provides nearly equal density at each cluster as shown in figure

3 Fig. 5.1 Distribution of Sensor nodes iv) Task Distribution Model The application comprises of many tasks which are assigned to the sensor nodes inside the cluster according to their remaining battery power (R b ). Suppose the task t i is assigned to a sensor node (i) contains remaining battery power (R bi ) then the percentage of task assigned to i is given by. The assignment of the task to the sensor nodes at different level is decided by the upper level. E.g. the nodes at level i are n 1,n 2 ; n 1 contains x% of task and n 2 contains y% of task assigned by the sensor node present in level i-1, suppose that the nodes on level i+1 are n 3,n 4 and E is given as {(n 1,n 3 ),(n 1,n 4 )}. Now the division of x% of task is decided according to the remaining battery of the nodes n3(r bn3 ) and n4 (R bn4 ) as. and respectively. Figure 4.2 shows the task distribution process. 62

4 H x y n 1 n 2 (x*r bn3 ) /100 (x*rbn4)/ 100 n 3 n 4 Fig. 5.2 Task Distribution v) Remaining battery Model The remaining battery of the network can be represented as a set. So the remaining battery of the network becomes R BN =. The Cluster s battery power can be represented by a set. The nodes at level, i informs about their remaining battery to their parent level i-1. The remaining battery of the level i is calculated as, where k= {1,2,3,4, l}.where, total number of nodes at level i are l. The remaining battery power of each node at level i is calculated as, for x = i-1, i-2, i-3,, z and z is the level of the leaf node. At last, node H will calculate the total battery remaining inside the cluster after getting the remaining battery power of its sensor nodes. vi) Result Collection Model The results of the tasks distributed inside the cluster are collected by cluster head and cluster head send this result to base station (BS). The sensor nodes at lower level send its result to its parent at upper level. Suppose the nodes at level i are n i1, n i2, n i3, n i4, these sensor nodes send their results to sensor node at level i-1 without waiting for each other send process. The sensor node at level i-1 collects the result (Result i-1 ) and sends this result to sensor node at level i-2. (5.1) The whole process of result collection is shown in figure

5 n 5 n 4 n 2 n 6 n 3 n 1 n 3 n 6 n 1 n 5 BS n 2 n 4 n 1 n 5 n 1 n 2 n 3 n 4 n 2 n 3 n 4 n 6 n 7 n 5 n 6 Fig. 5.3 Result Collection Process vii) Task Allocation and Collection Algorithm Algorithm 5.1: Step 1: Step 2: Step 3: The Base Station, BS Allocates the task to each clusters on the basis of their remaining Battery Power(R BC ).The BS assign the task to cluster through their Cluster Head (CH). The task Assigned to the cluster T C is given as (5.2) Here α is a constant, it depend upon the battery utilization in communication (Cluster head communicates with all the nodes present in the cluster and α is the cost of communication in the form of battery utilization that incur in, communication of cluster head with other nodes present in the cluster ). H collect the information for the remaining battery of their child nodes; assign the task according to their remaining battery power (R NC ), the task assigned to a child node is calculated as (5.3) The task assigned to the CH is given as (5.4) R BH is the remaining battery power of H, Now the child node of CH assigns the task to their child node and itself according to their remaining battery power by using equation (5.3) and (5.4). 64

6 Step 4: Step 5: Step 6: Step 7: Repeat Step-3 until the leaf node of the DAG does not come. After the distribution of T total in the cluster the parent nodes collect the result from their child nodes. Each parent node waits for a threshold time for getting the result from the child node. Suppose the throughput of the processor of the nodes is η and the allocated task to the node is T NC then the threshold time (τ)., λ is the communication cost and ψ is the task partition cost. The cluster CH collects the task s results from their child nodes and replies the result to BS. viii) Relation between Battery and Temperature Due to the application of WSN in varying temperature regions, it needs to find the relation between battery and temperature and by getting this information it becomes able to decide the depletion rate of the battery. According to depletion rate of the battery the task assigned to cluster head can be calculated by the formula if the depletion rate of the battery is more than the standard depletion rate, depends upon the deletion rate. The relation between temperature (Celsius) and battery/ voltage (ma) is shown in figure 5.4. The distribution of voltage/battery and temperature are given in figure 5.5 and figure 5.6 respectively. Fig. 5.4 Relation between Battery/voltage and temperature 65

7 Fig. 5.5 Distribution of Battery/Voltage (ma) Fig. 5.6 Distribution of Temperature (Celsius) Task Distribution by Scheduling In this methodology the sensor nodes are deployed, scheduled and then tasks are distributed among the sensor nodes according to the schedule. The methodology applies the Teaching Learning Based optimization to find the deployment location and schedule and then provide a 66

8 task distribution algorithm to assign the task among the sensor nodes on the basis of some parameters. a) Objective Function Let the position of each Point of Interest (PoI i ) of Group G k is (x ki,y ki ) and Position of sensor (S j ) is. The batteries of sensor nodes are B j, efficiency of sensor nodes are E j then (5.5) ( ) (5.6) Lower value of objective function provides optimal location with good efficiency. b) Scheduling Algorithm Algorithm 5.2: Step 1: Apply the Grouping algorithm (Algorithm 5.4) to form the groups of POIs as (G 1, G 2, G 3, G 4, G 5, G n ).Suppose the G 1 = {PoI 1, PoI 5, PoI 8 }, G2= {PoI 2, PoI 9, PoI 20 } and so on. Step 2: Apply algorithm 5.3 in the result of step 1. C. Algorithm 5.3 Step 1: Step 2: Initialize the Optimization parameters such as Population size (P n ), Number of Generations (G n ), Number of Design variables (D n ) and limits of design variables (L, U). Initialize the Population. (Generate the random population according to the Population size (number of sensors) and number of design variables (Groups)). Sensors/Groups G 1 G 2 G G n S 1 V 11 V 12 V V 1N S 2 V 21 V 22 V V 2N S n V N1 V N2 V N3.... V NN 67

9 Step 3: Teacher Phase (a) Calculate the mean population column-wise, Which will give the mean for particular group as. (b) The best solution will act as the teacher for that iteration. (c) Teacher will try to shift the mean from, which will act as the new mean as (d) Find the difference between two means as Where is the teaching factor and selected as 1 or 2. r is the random number between 0 and 1. (e) Now add the Diff to current solution to get the updated solution as Step 4: Learner Phase for each i in Randomly select two groups as if ( ), where r i is in range [0,1]. else Accept if it gives better c) Algorithm 5.4 It explores the steps that are applied to form the group of sensor nodes. The Flow Chart for these steps is given in figure

10 Start Assign number of Groups (C n ) Calculate Centroid YES Distance from Sensor nodes to Centroid Change in any Group NO Exit Group the Sensor nodes based on minimum distance Fig. 5.7 Flow Chart for Grouping Algorithm Further text discusses each step that is applied to form the group of sensor nodes. i) Assignment of Number of Groups Many sensor nodes are deployed to cover the region of interest. In this step, the numbers of groups are provided that covers the whole area of interest. The pattern of group is assumed hexagon. Let the area of interest is A then the number of groups (C n ) is given as (5.7) ( ) ii) Centroid Calculation Centroids are the points that resemble as the center point of any group. Each group inside the region of interest consists of one centroid. Step 1: Assign the number of centroids needed 69

11 Step 2: Randomly select the sensor nodes that are deployed in the region as centroids Step 3: In this step the efficiency of different centroids in checked by the equation (5.5) and (5.6) iii) Calculation of Distance from Centroid to Sensor node Let the position of sensor node is (x i,y i ) and the location of centroid is (cx j, cy j ) then the distance(d i ) between sensor node and centroid is given as: (5.8) iv) Minimum Distance Calculation In this step, put the sensor nodes in different groups on the basis of distance given by equation (4.9). (5.9) Where is the mean of points in c) Task Distribution and Result Collection An application must be divided into tasks to perform the collaborative processing inside the WSN. There are many applications which require imprecise data and some applications of WSN which require precise data. Such as fire tracking inside forest requires imprecise data and traffic monitoring inside a city require precise data. Some applications also require priority so the task distribution algorithm must work on the basis of the parameters as priority and preciseness. As the scheduling algorithm provides a centroid node for each group which cover a region of interest. As these centroid nodes cover whole region of interest, these active centroid nodes only can provides the information about the event occur inside the region but if the application requires precise information about event it can utilize the other sensor nodes present inside the group after awakening them. This section provides an algorithm 5.5 which distribute the task on the basis of precise or non-precise and priority or non-priority. Suppose, the task (T) is to collect the information about the traffic of a city. T must be divided into zones as {T 1, T 2, T 3, T cn }. These tasks must be assigned among the zones. 70

12 Algorithm 5.5: Step 1: Step 2: Step 3: Divide the task into different zones according to number of clusters Decide the priority and accuracy of data for each zones if the priority of any zone is high then (a) Assign the sleep- sensor nodes in different clusters (Those comes in the path from zone to base station) for forwarding the data (i) If the accuracy required is low then No assignment for other sensor nodes present in the zone (ii) If accuracy required high then Assign more sensor nodes in the zone to collect the data from the environment. Step 4: If the priority of any zone is low then (a) Don t assign any extra sensor node for forwarding the data (i) If the accuracy required is low then No assignment for other sensor nodes present in the zone (ii) If accuracy required high then Assign more sensor nodes in the zone to collect the data from the environment Routing of Task/ Query In wireless sensor network many queries flow towards the region of interest for collecting some sensed data. Some assumptions are taken in this methodology as: The base station (BS) node is enabled with GPS system and by applying some localization strategy the position of other sensor nodes can be accomplished. The links between the sensor nodes are bidirectional. a) Network System The network consists of many sensor nodes which are able to communicate with others. The sensor nodes sense the data from environment and send this data to other sensor nodes by 71

13 messaging. The sensor nodes contain a small memory for storing some value. The network contains some regions which do not come under the coverage of any sensor nodes these regions are called as hole. The network can be represented by Euclidean space and the sensor nodes can be known as Euclidean points. b) Developed Methodology This section gives the different steps that are applied to fulfill the requirement of forwarding the task/ query. Step 1: Step 2: Step 3: Step 4: Step 5: Step 6: Step 7: Step 8: Select any exterior node p in the network. Perform the flooding from an exterior node p in the network. Each node records the minimum hop count from p. This can provide the shortest path tree from the node p. Determine the nodes that form the JOIN. The JOIN is a point where different homotopy type shortest path tree meets. If there are multiple JOINs inside a network then there is more than one hole inside the network. It can be found by the method given by Alstrup et al[93] for calculating the LCA. If there are two points(s,t) on the JOIN with same LCA(s,t). If we move from s or t (consider the edge between s and t) and go through LCA(s,t) with following the spanning tree, we will get a cycle. Determine the two nodes in the JOIN with minimum Euclidean distance from the node p. Determine the third node present in the inner boundary of the hole. It can be found by using the geometric rules and Euclidean distance. Determine the nodes present in the inner boundary of the hole. Determine the nodes present in the outer boundary of the region. Now it provides the set of cycles that give the inner and outer boundary. Now the medial axis between inner and outer boundary can be calculated. In the further text these steps are discussed in more detail. i) Shortest Path Tree Construction The hop count field for each node is initialized excluding node p. Now the node p flood the message which contains it ID and hop count (p s hop count is 0). The node (z) that receives the message increment its hop count by 1 and sets its ancestor node as p. Now z floods the 72

14 message to other nodes and other nodes increment their hop count by 1 and set the ancestor node as z. Any node (predecessor) gets the message from its ancestor, first it calculates the hop count value and compares new hop count value with its old hop count value if old hop count value is less than the new hop count value then it maintains its hop count value without any update. The process of construction of shortest path tree is given in figure 5.8 and message format is given in figure 5.9. Fig. 5.8 Construction of shortest path tree Sender Hop Count Node_ID Fig. 5.9 Message Format For example, in the figure 5.8 node p is flooding the message (1 0) to its neighbors (r,s,t) which are in its sensing range. After receiving the message from p, the neighbor nodes r, s and t calculate the hop count value, which becomes 0+1=1, since 1 is less than, the updated value for hop count in nodes r, s and t is stored as 1. Now the node r floods the message (2 1) to its neighbor s x, y, p and s the neighbor nodes calculates the hop count values as follows p s hop count value= r s hop count value(from received message) + 1= 1+1=2 s s hop count value= r s hop count value(from received message)+ 1= 1+1=2 x s hop count value= r s hop count value(from received message)+ 1=1+1=2 y s hop count value= r s hop count value(from received message)+ 1=1+1=2 Since the calculated hop count value for p is greater than the stored old value, so the hop count value for p will not be change, the calculated hop count value for s is also greater than 73

15 the stored old value so it will also be not change. The calculated value of hop count for nodes x and y will change because it is less than the stored old value. ii) Determine the Nodes that Form the JOIN The nodes where two branches of shortest path tree meets are known as JOIN nodes and the space that comes under these nodes is known as JOIN. Let the depth of the left shortest path tree branch (T L ) is d 1 and right shortest path tree branch (T R ) is d 2. Suppose there are n 1, n 2, n 3 nodes are present at depth d 1, nodes are present at depth d 2. The nodes that comes under each other sensing range are comes in the JOIN nodes. If the edges among these nodes are added, it provides a graph, because these nodes are present at the fork of either tree T L or T R. The Euclidean distance between p and n 1, n 2, n 3 or is β, then β will always be greater than zero. Generally the value of β is taken in some constant fraction of the diameter of the sensing range. Alstrup et al [93] have given the method for getting the Least Common Ancestor (LCA). The LCA of the two node in the (will be far from the nodes ) will be the path from to n 1 and path from to are well separated. This process can be applied to find more than one JOINs if present in the Euclidean space. The process of JOIN formation is given in figure Fig JOIN Formation iii) Nodes Present in the Inner Boundary of the Hole 74

16 Let the nodes that are present in the JOIN are n 1, n 2, n 3, n 4 Find the Euclidean distance between p and n 1, p and n 2, p and n 3, p and other nodes present in JOIN. Now find the two lowest Euclidean distance nodes in JOIN. These nodes become the nodes in the circumference of the hole. Now determine the node with highest depth that present in the line between p and a with lowest Euclidean distant node in first JOIN in the shortest path tree which becomes the third node on the circumference of the hole. If there are more than one hole in the Euclidean space then there will be more than one JOIN, if the Euclidean space contains two holes then there will be two JOINs. The two nodes on the circumference of the second hole can be determined by getting the two lowest Euclidean distance node from p in second JOIN and the third node in the circumference of the second hole can be determined by finding the node with highest Euclidean distance in first JOIN. The process is shown in figure Fig Euclidean Space with three JOINS Procedure for Determining the Third Node in JOIN (first) Step 1: Start from the root node p, take the depth value d=0 Step 2: Find the equation of line (L) between p and a Step 3: Find the next node node(l n ) in the shortest path tree that belongs in line L as follows (a) Put the coordinates value of L n in L and get the value(v) 75

17 Step 4: (b) if the value V becomes zero after +-C where C is a threshold (c) then increment d and mark L n as the next root (d) repeat step 3 until we get the child node Mark the node L n as the third node on the circumference of the first hole. Now two cases are considered about the shape of hole to find the nodes in inner boundary of hole. Case 1: Shape of hole is a circle Case 2: Shape of hole is any other polygon CASE 1 Let the equation of the circle with the center at (h, k) is (5.10) Calculating the center of the circle and radius given the three point on the circumference as Put x=x 1 and y=y 1 in equation (5.10) provides Put x=x 2 and y=y 2 in equation (5.10) Put x=x 3 and y=y 3 in equation (5.10) The equation (5.11), (5.12) and (5.13) gives Solve the equation (5.14) (5.11) (5.12) (5.13) (5.14) (5.15) Solve the equation (5.15) (5.16) Solve the equation (5.16) and (5.17) for the values of h and k (5.17) 76

18 (5.18) (5.19) Put the value of h and k from equation (5.18), (5.19) and x=x 1, y=y 1 in the circle equation (5.10). It provides the radius of the circle. Suppose the coverage radius of the sensor node is given as Fig Calculation of Rotational Angle In figure 5.12, suppose the distance, Then, (5.20) Calculation of the point from the point s(x 1, y 1 ) [ ] [ ] [ ] [ ] [ ] Same way, the point Fig Determine the point in the circumference of hole can be calculated as 77

19 [ ] [ ] [ ] [ ] [ ] CASE 2 Consider the figure 5.8 with one hole. // Input: Take a buffer as A Step 1.Consider the Left shortest path tree branch a. Find the value t 1 =d 1 -d -1. b. Put the coordinates of node n 1 in A. c. Find the parent (P) of parent node of n 1. d. While t 1 0 e. Find the right most child of P put the coordinate value of right most child in A f. t 1 =t 1-1 g. Assign node P as the parent of P Step 2.consider the right shortest path tree branch a. Find the value t 2 =d 2 -d -1. b. Put the coordinates of node n 1 in A. c. Find the parent (P) of parent node of n 1. d. While t 2 0 e. Find the left most child of P put the coordinate value of left most child in A f. t 2 =t 2-1 g. Assign node P as the parent of P A contains the coordinates of the nodes present in circumference of the hole. If more than one hole is present in the Euclidean space then apply the above algorithm for each pair of the lowest Euclidean distance at each hole. iv) Determine the Node Present in Exterior Boundary of Hole To find the exterior boundary it needs to find the outer node, outer node is a node which is far from the interior node in the Euclidean space. Following steps are applied to find the exterior boundary. // Input: take two buffer B and C Start from the root node p in the tree Find the val=max(d 1,d 2 ) for i=1 to val do 78

20 Find the left most child(l) and right most child(r) Put the Euclidean point of l in B and Euclidean point of r in C The buffer B and C provides the nodes present in the exterior boundary of the Euclidean space. v) Determine the Medial Axis The distance between nodes at interior boundary and exterior boundary is used to find the medial axes. On the basis of these distances different regions are maintained by using voronoi diagram. Let D i is the distance of a node from the inner boundary and D o is the distance of this node from the outer boundary. If D i is greater than D o then this node will belong to the region near to inner boundary. If D o is greater than the D i then this node will belong to the region near the outer boundary. The medial axis contains the nodes which have equal Euclidean distance from both inner and outer boundary. Only the mantissa part of a fractional number are considered for D i and D o. If the difference between D i and D o is C and C<1 then consider the D i and D o equal. This process provides the medial axis and the voronoi diagram for the considered Euclidean space. This voronoi diagram and medial axis can be used to perform the routing/ distribution inside the Euclidean space. c) Query/ Task Dissemination and Collection Each region inside the Euclidean space contains a master node (M); the master node receives the data from the nodes inside the region. BS generates the query like, Query: send the wind speed where the temperature is below 35. The node (K) that receives the query from BS inside a region R, transfer the query towards the master node in that particular region. The master node checks the condition in the query message and transfers the query towards the neighboring regions. Steps of Query/task dissemination and collection are given as follows: for each region R if condition==true (a) Check the timestamp(t) of the data (i) If t<threshold then Send the data towards the node K. Node K transfers the data to BS. (ii) Otherwise 79

21 Master node broadcast the message for gathering the required data from the region, master node collect the data and forwards the data towards the node K. Node K transfers the data to BS. The process of Query/ task dissemination and collection is shown in figure Fig Query/ Task Dissemination and Collection 5.3 Performance Results This section discusses the performance of different methodologies that are applied to allocate / route the task among the sensor nodes present in the WSN Results of Tree based Methodology To evaluate the performance of tree based methodology, throughput (number of task completed per seconds) is taken as the parameter. A scenario is created in OPNET which contains 64 sensor nodes and 5 clusters. The initial battery backups of clusters are 0.51, 0.21, 0.36, 0.14 and 0.04 joules respectively. To evaluate the throughput, the scenario is run for 600 seconds and provides the comparison between the throughput when developed method is applied and when the developed method is not applied. When developed method is applied it provides throughput as bps and when developed method is not applied it provides throughput as bps. As the results show developed method provides 32.51% increment in the throughput. 80

22 5.3.2 Results of Task Distribution by Scheduling To evaluate the performance of this methodology, four scenarios are created in MATLAB 2013b as scenario 1 with 5000 sensor nodes, scenario 2 with sensor nodes, scenario 3 with sensor nodes and scenario 4 with sensor nodes in 10 3 x10 3 m 2 area. Each scenario generates 100 tasks for each group of sensor nodes present in the region. Figure 5.15, 5.16, 5.17 and 5.18 show the results. Each figure contains four graphs which show Task Completed (%), Task Completion Time by Group, Battery used group-wise and Total Task allocated to each group for scenario. The graph that shows the Task Completed(%)(suppose in first iteration 20% task completed then in second iteration 80% task is given as input, suppose in second iteration 60% task completed then in third iteration 25% task will be given as input). Fig Scenario 1 81

23 Fig Scenario 2 Fig Scenario 3 82

24 Fig Scenario 4 These four scenarios are also created in an area of 10 2 x10 2 m 2 and compared the results (lifetime of network) when developed method is applied and when the deployment is random. The comparative results are given in table 5.1. Table 5.1 Lifetime of WSN (Seconds) Random deployment After Application of developed method Area of the WSN Improvement (%) Scenario 10 2 X10 2 m X10 3 m X10 2 m X10 3 m X10 2 m X10 3 m Results of Routing of Task /Query First, two analyses are presented to evaluate the performance of developed methodology as Traffic and Energy Analysis. 83

25 a) Traffic Analysis Let the scenario contains the regions {R 1, R 2, R 3, R 4 R n }. For the purpose of simplicity it is assumed that the each region contains maximum L levels of nodes and the total number of nodes in a region are Ln. Suppose each node generates p messages and each message has the same length as B bits. A region R have R neighboring regions. So the total number of messages generated within a region is given as follows: Case 1: if the current data is present at M (consider figure 5.14) Total Traffic of Region (TOTAL_Traff_RGN)= 2L*p*B+2LR *p*b Case 2: if the instant data is not present at M (Consider figure 5.14) Suppose, R k = kr 1 is the radius of k-hops coverage area of one sensor node. The area of 1-hop coverage (A 1 ) by a sensor node is πr 2 1. Suppose the total area of the region is S, since the total sensor nodes present in the region are Ln, so the number of sensor nodes per unit area is given as L n/s. So the number of sensor nodes present in k-hop distance (N k ) of one sensor node are Ln* πr 2 k /S = Ln* π k 2 R 2 1 /S. The area of k-hop ring (A k ) is given as A k = πr 2 k πr 2 k-1 = π(r 2 k R 2 k-1 ) = π(k 2 R (k-1) 2 R 2 1 ) = πr 2 1 (k 2 - k 2-1+2k) = πr 2 1 (2k-1) =(2k-1)A 1. Suppose the maximum hop count in a region is L then the area from the k-hop ring to the L- hop ring (A k-h ) is ( ) ( ) (( ) ( )) The traffic (TRAF_GEN k ) generated by each senor node present in the k-hop ring can be calculated as (The nodes in the k-hop ring forward the message of outer rings and generate the message itself). TRAF_GEN k = Traffic generated by each 1-hop distance node is given as TRAF_GEN 1 = 84

26 Traffic generated by each 2-hop distance node is given as So the total number of messages generated by 1-hop ring is given as (5.21) So the total number of messages generated by a region of maximum L leveled node is given as: Total Traffic of the Region (Total_Traffic_RGN)= (5.22) b) Energy Analysis Consider the following figure 5.19 for transmission and receiving the data. Suppose the energy requirement for transmission is E trans, energy for receiving is E receive and energy for amplification is E amp. So the energy requirement for transmitting the B bits of data for a distance of r is [ ] (5.23) The energy requirement for receiving B bits of data for distance r is [ ] (5.24) So the total energy requirement is given as [ ] [ [ ] [ ]] (5.25) 85

27 Fig Transmission and Receiving Model To compare the results of developed methodology, a scenario is created with 20 sensor nodes and 1 base station in OPNET and run for 3200 seconds and 7200 seconds. The developed method is compared with AODV routing protocol for the following parameters: Retransmission Attempts (Packets) Network Load(bits/sec) Data Dropped (bits/sec) Throughput(bits/sec) The comparative results are given in table 5.2. Table 5.2 Comparative Results S.No Parameters Retransmission Attempts Network Load Data Dropped Throughput 3600 seconds 7200 seconds AODV EECR AODV EECR Conclusion Efficient task allocation inside the WSN can improve the lifetime of network. As WSNs are established in disastrous area so it is impossible to recharge the battery of sensor nodes. This type of area provides a variation in temperature. The task inside the network must be assigned to sensor nodes in such a manner so that the lifetime of network enhanced. For the purpose of solving these problems three methods are given in this section. The tree based methodology 86

28 provides enhancement in the connectivity and removes the unbalanced assignment of the tasks/ query. It improves the throughput 32.51% as compared to the situation when this methodology is not applied. Second methodology enhances the lifetime of the network 4.62x for small scale problems and 16.19x for large scale problem when compared with random deployment of the sensor nodes inside the WSN. Third methodology provides the results as the density of the network increased by 33.33%, it provides 22.59% decrease in the throughput and 25.55% decrease in the end to end delay. The developed method is compared with AODV protocol as the results show it provides good results over AODV. The traffic increases in the network approximately linear with respect to increase in the density of the network. The growth of the power requirement increase in polynomial with increase in the density of the network. 87

Fig. 2: Architecture of sensor node

Fig. 2: Architecture of sensor node Volume 4, Issue 11, November 2014 ISSN: 2277 128X International Journal of Advanced Research in Computer Science and Software Engineering Research Paper Available online at: www.ijarcsse.com To Reduce

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