is always multiple by 2 in loop and in odd node is represented as n. We get the coordinates of the vertical and horizontal lines separately for each case. At the end, we put or apply the fitting function to get the minimum coordinates in the deployment area as shown in Figure 6.16.
6.4.6 Energy Calculation
Energy calculation is the amount of energy used from source to destination node. Read the energy from the excel file and it measures the energy from source node to zone scatter node (num1) then from zone scatter node to destination scatter node (num2) and at the end from destination scatter to destination node (num3). To find the final count of energy, we sum up all the forms of energy num = num1 + num2 + num3. To find the percentage of energy lost in the whole procedure is x = (num/N)*100. We also find out the energy used at each node for delivery packet from source to destination node.
Figure 6.15 Mesh formation.
Figure 6.16 Minimum spanning tree.
Figure 6.17 Throughput vs. delivery ratio.
6.4.7 Average Delay
Average delay is the delay in time (ms) from source to destination node by every node in the deployment area. We Enter the delay in time (ms) = 100. The average delay is 320.000 ms.
6.4.8 Throughput
It is defined as the total number of packets delivered over the total simulation time. We put the interval time 100 ms. Result of throughput is 420.000 kbps. The results are presented in Figure 6.17.
6.5 Simulation of VANET Network
For simulation of VANETs, we will first put the numbers of nodes in the area. First of all, we will set the coordinates of x and y. Coordinates of x and y is low = 0 (%lower bound to both the axis).
6.5.1 Placement of Nodes
Nodes value should be more than 80. Now, we will enter the sender node and the receiver node. Now, put the value of sensor nodes. In this case, we will put the value of number of sensors = 100 as shown in Figure 6.18.
6.5.2 Sender Node and Receiver Node
Next, we will put the value of sender node = 21 and receiver node = 57. The master nodes and receiver nodes in bold letters are presented in Figure 6.19
6.5.3 Euclidean Distance Between Two Coordinates
Now, we calculate the average distance between two nodes by the mean square root formula of the nodes with the help of coordinates of x and y. We calculate distance in upward, downward, leftward, and rightward direction by the average distance formula. To calculate the nodes in downward direction, select all the nodes down the current node and find out coordinates of node(x) and node(y) and find the down node index. Similarly, we will find out the upward by considering all the nodes up node from the current node, left node and right node index.
Figure 6.18 Ideal placement of nodes.
Figure 6.19 Sensor nodes with faulty points.
6.5.4 Separation of Faulty Nodes
Now, we will separate out the faulty nodes from the whole nodes. To separate out faulty nodes formula is p = 40 where p is percentage of faulty nodes out of the total number of nodes.
Types of Faulty Nodes
We have four types of faulty nodes
1 1. Permanent
2 2. Transient
3 3. Intermittent
4 4. Dynamic
Now, assign the x and y coordinates of faulty nodes and initiate the master node and calculate the distance of each faulty nodes from master nodes by mean square root formula, also the time from the master node to the faulty nodes. Plot the all four types (permanent, transient, intermittent, and dynamic) faulty nodes in the figure.
6.5.5 Best Match of the Node
By subtracting the faulty nodes from the equation, we will get the best match from up nodes, down nodes, left nodes and right nodes indexes. To get best match, we calculate the transfer information from sender to receiver node. All nodes toward the receiver, calculate the horizontal distance between “coor1” and “r_coor” and vertical distance between “coor1” and “r_coor”.
6.5.6 Cases of Simulation
Now, there are three conditions or cases. First two cases in which sender and receiver are outbound and third case is wen sender and receiver is inbound. But for third case, there are four scenarios.
1 1. Scenario with no error.
2 2. Scenario with error.
3 3. Scenario with path hoping.
4 4. Scenario with errors and correction.
But, here we will discuss three cases of scenario without and with errors and with errors and correction.
a. Case without errorsIn scenario without errors, we will first transfer the information of sender and reliever nodes. Now, check the range and distance between the nodes. Next step is to generate the blinking line and check the set of nodes (faulty nodes) and then send the information to the destination with the help of coordinates of x and y. In this figure, sender 21 is sending packet to the receiver through the minimum path and time delay. It passes through 3 nodes (72, 54, 65).
b. Case with errorIn scenario with error, loop will run until
