Featured Examples

Throughput and delay variation with distance

Open NetSim and Select Examples-> Underwater Acoustic Network ->Throughput and delay variation with distance and then click on the tile in the middle panel to load the example as Figure-1

Figure-1: List of scenarios for the example of Throughput and delay variation with distance

In this example, we understand how UWAN throughput and delay varies as the distance between 1 transmitter and 1 receiver is varied. Even with No pathloss the throughput in UWAN varies with Tx-Rx distance which is not the case in terrestrial RF based transmissions. The two parameters that affect throughput and delay are the speed of sound and the slot length of s-Aloha. The speed of sound in water is given by the formula

\[c_{sound} = 1449.05 + 45.7t - 5.21 \times t^{2} + \ 0.23 \times t^{3} + \left( \ 1.333 - 0.126t + 0.009 \times t^{2} \right)(S - 35) + 16.3 \times z + 0.18 \times z^{2}\]

where \(t\ \)is one-tenth of the temperature of the water in degrees Celsius, \(z\) is the depth in km and \(S\) is the salinity of the water in ppt. Then using \(t = \frac{25}{10} = 2.5,\) \(z = 50\), and \(S = 35\) - where t is one-tenth of the temperature of the water in degrees Celsius, z is the depth in meters and S is the salinity of the water - we get \(c_{sound} = 2799.33\ m/s\). When the transmitter receiver distance is \(d = 2km\), the propagation delay, \(\Delta = \frac{{2 \times 10}^{3}}{2799.33} = \ 714,456.4\ \mu s\)

Next, as explained in section 3.2.2, we consider ideal slot lengths for different transmitter receiver distances. In the case when \(d_{Rx}^{Tx} = 2\ km\) the slot length turns out as

\[L_{Slot} = \ T_{tx} + \Delta = 16,800 + 714,456.4 = 731,256.4\ \mu s = 0.73\]

Table-4 shows the ideal slot length settings for \(d_{Rx}^{Tx} = 4\ km\) and \(d_{Rx}^{Tx} = 6\ km.\)

Network setup:

The following network diagram illustrates what the NetSim UI displays when you open the example configuration file.

Figure-2: Network Scenario. Two underwater devices connected via an acoustic ad hoc link

In case #1, distance between the underwater devices is set to be 2km. In case #2 the distance is 4km, while in case #3 it is set to 6 km
Click on link, expand the property panel on the right and set the Channel characteristics as No pathloss.

Device Configuration:

Device > Interface (ACOUSTIC) > Datalink Layer
Slot Length(µs)	731257 for 2 km
Device> Interface (ACOUSTIC) > Physical Layer
Source Level (\(\mathbf{dB//1\mu Pa}\))	200
Modulation	QPSK
Data Rate (kbps)	20

Table-1: Device properties set for this example

Application Configuration:

We run simulations for different traffic generation rates. The generation rate depends on the inter arrival time – a GUI input in NetSim – in the following way

\[Generation\ Rate\ (Mbps) = \frac{Packet\ Size\ (Bytes)\ \times \ 8}{Interarrival\ Time\ (µs)}\]

Application Properties
Application Method	App1 CBR
Source ID	1
Destination ID	2
Packet Size (Bytes)	14
	Inter arrival Time (µs)	Generation rate (bps)
Case-1	4480000	25
	2240000	50
	1120000	100
	896000	125
	746666.6666	150
	640000	175
	560000	200
Case-2	5600000	20
	2800000	40
	1866666.6666	60
	1400000	80
	1120000	100
Case-3	5600000	20
	3733333.333	30
	2800000	40
	2240000	50
	1866666.6666	60
	1600000	70

Table-2: Application properties for the different samples in each case studied in this example

Enable packet trace and Acoustic Measurements Log under Configure report tab as shown below

Figure-3: Enabling the Acoustic Measurements Log

Run the Simulation for 100 sec.

Theoretical Predictions

The predicted propagation delay when the speed of sound \(c_{sound} = 2799.33\ m/s\) is

Distance between devices	Propagation delay (\(\mathbf{\Delta}\) in \(\mathbf{\mu s}\))
2km	714456.4
4km	1428912.7
6km	2143369.1

Table-3: Theoretically predicted propagation delay for different Tx-Rx distances

Transmission delay and Saturation Throughput

Considering a slot length of \(731,257\ \mu s\), we see that one packet exactly fits one slot and hence the predicted saturation throughput would be

\[\theta_{sat}^{2km} = \frac{{(L}_{pkt} \times 8)}{L_{slot}} = \frac{(14 \times 8)}{731256.4 \times 10^{- 6}} = 153\ bps\]

Proceeding similarly for 4 km and 6 km, the predictions for saturation throughput are

Distance between devices	Slot Length (\(\mathbf{L}_{\mathbf{slot}}\))	Saturation Throughput \(\mathbf{(}\mathbf{\theta}_{\mathbf{sat\ }}\)in bps)
2 km	731257	153
4 km	1445713	77
6 km	2160170	52

Table-4: Ideal slot lengths and theoretically predicted saturation throughputs (θ_sat) for different Tx-Rx distances.

Simulation results

We calculate of queuing delay, transmission delay, propagation delay from the packet trace. The steps are:

Open Packet Trace file using the Packet Trace option available in the Simulation Results window under traces.
The difference between the PHY LAYER ARRIVAL TIME(US) and the MAC LAYER ARRIVAL TIME(US) will give us the delay of a packet. (Refer )

\[Queuing\ Delay\ (\mu s) = PHYSICAL\ LAYER\ ARRIVAL\ TIME(\mu s) - MAC\ LAYER\ ARRIVAL\ TIME\ (\mu s)\]

Figure-4 : Screen shot of NetSim trace showing the Queuing Delay column

Now, calculate the mean queuing delay by taking the average of the queueing delays of all the packets. This is nothing but the column average. (Refer )
Similarly, users can get the Mean Transmission Delay and Mean Propagation Delay from the packet trace using the formulas

\[Transmission\ Delay\ (\mu s) = PHY\ LAYER\ START\ TIME(\mu s) - PHY\ LAYER\ ARRIVAL\ TIME(\mu s)\]

\[Propagation\ Delay\ (\mu s) = PHY\ LAYER\ END\ TIME(\mu s) - PHY\ LAYER\ START\ TIME(\mu s)\]

	Generation Rate (bps)	Throughput (bps)	Delay (µs)	Mean Propagation Delay(µs)	Mean Transmission Delay(µs)	Mean Queuing Delay(µs)
Case #1: Distance between underwater devices is 2km	25	26	1113144.789	714456.35	16800	381888.43
	50	50	1111731.732	714456.35	16800	380475.37
	100	100	1094568.647	714456.35	16800	363312.29
	125	124	1091089.643	714456.35	16800	359833.28
	150	149	1116252.457	714456.35	16800	384996.10
	175	152	6891103.855	714456.35	16800	6159847.5
	200	152	12291103.85	714456.35	16800	11559847.5
Case #2: Distance between underwater devices is 4km	20	20	2036146.042	1428912.7	16800	590433.33
	40	40	2081859.042	1428912.7	16800	636146.33
	60	59	2148454.189	1428912.7	16800	702741.47
	80	77	2999954.709	1428912.7	16800	1554242
	100	77	12519954.71	1428912.7	16800	11074242
Case #3: Distance between underwater devices is 6km	20	20	3163994.064	2143369.06	16800	1003825
	30	30	3230739.438	2143369.06	16800	1070570.37
	40	39	3194853.635	2143369.06	16800	1034684.57
	50	49	3340415.655	2143369.06	16800	1180246.59
	60	52	8763994.065	2143369.06	16800	6603825
	70	52	14763994.06	2143369.06	16800	12603825

Table-5: Tabulated results (throughput and delays) for 3 different Tx-Rx distances.

Figure-5: Throughput vs. Generation rate plotted for Tx-Rx distances of 2km, 4km and 6km based on earlier tables.

From Table-5, we see that the propagation delays from simulation match predictions in Table-3. Then from we observe that saturation throughput (the Y axis value once the curve flattens) matches prediction.

Distance between devices	Saturation Throughput. Predicted \(\mathbf{(}\mathbf{\theta}_{\mathbf{sat\ }}\)in bps)	Saturation Throughput. Simulation \(\mathbf{(}\mathbf{\theta}_{\mathbf{sat\ }}\)in bps)
2 km	153	152
4 km	77	77
6 km	52	52

Table-6: NetSim UWAN Simulation results vs. theoretical prediction of saturation throughput, for different Tx-Rx distances.

Underwater propagation losses and device range

Open NetSim and Select Examples-> Underwater Acoustic Network ->Underwater propagation losses and device range and then click on the tile in the middle panel to load the example as Figure-5.

Figure-6: List of scenarios for the example of Underwater propagation losses and device range.

In this example, we understand the Thorp propagation model, the sources of underwater noise, the passive sonar equation and how device range can be estimated based on received SNR. Refer to section 3.1 for the underlying theory on signal level, transmission losses, and the passive sonar equation.

In the NetSim GUI, we provide 3 samples per modulation scheme, totaling 18 samples for the 6 modulation techniques. The complete set of configuration files (scenario, settings and other related files) comprising of 621 samples, is available at https://github.com/NetSim-TETCOS/UWAN_Examples_v14.4/archive/refs/heads/main.zip.

Click on the link given and download UWAN Experiments

Extract the zip folder. The extracted project folder consists of Underwater propagation losses and device range example files.
How to import the workspace is explained in section 4.9.2 in NetSim User Manual.

Network setup

The following network diagram illustrates what the NetSim UI displays when you open the example configuration file.

Figure-7: Network Scenario

Click on link , expand the right-side property panel and set the Channel characteristics as Pathloss Only.
Click on Underwater Device 1, expand the right-side property panel change the following parameters,

Device Properties > Physical Layer
Source Level (\(\mathbf{dB//1\mu Pa}\))	190.8, 187.78,183.81
Data Rate (kbps)	20
Modulation Technique	QPSK, BPSK, FSK, 16QAM, 64QAM, 256QAM

Table-7: Device Properties

Configure a CBR Application from Source ID as 1 and Destination ID as 2 from set traffic window present in the ribbon at the top with Default Properties.
Enable packet trace and Acoustic Measurements Log under Configure report tab as shown in Figure-3.
Run the Simulation for 1000 sec.

Analytical computations

In the Thorp model, the \(db/km\) attenuation is given by.

\[\begin{split}10\log_{10}{\alpha(f)} = \left\{ \begin{array}{r} 0.11 \times \left( \frac{f^{2}}{1 + f^{2}} \right) + \ 44 \times \left( \frac{f^{2}}{4100 + f^{2}} \right) + 2.75 \times 10^{- 4} \times f^{2} + 0.003\ \ \ \ f \geq 0.4\ kHz \\ 0.002 + 0.11 \times \left( \frac{f^{2}}{1 + f^{2}} \right) + 0.011 \times f^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f < 0.4\ kHz \end{array}\ \ \right.\end{split}\]

For this example, substituting \(f = 20,\) we get \(10\log_{10}{\alpha(f)} = 4.133\ db/km\), we see that the total pathloss is

\[10\ log\ A(d,\ f) = \ k \times \ 10\ log\ (d_{m})\ + d_{km} \times 10\ log\ \alpha(f)\]

Using input parameters \(K\ (spread\ coefficient) = 2,\ f = 20\ kHz\ \)and distance between the source and destination, \(d = 18\ km\), and we get the total transmission loss, \(TL,\) as

\[TL = \ 10\ log\ A(d,\ f) = 159.51\ dB\]

Next, we turn to noise level \(NL.\ \) The turbulence, shipping, wind, and thermal, noise level in dB is given by

\[10\ logN_{t}(f) = \ 17\ - \ 30log(\ f)\]

\[10\ logN_{s}(f) = \ 40\ + \ 20 \times (s\ - \ 0.5) + \ 26\ log\ f\ - 60 \times log(f\ + \ 0.03)\]

\[10\ logN_{w}(f) = \ 50\ + \ 7.5 \times \sqrt{w}\ + \ 20\ log\ f - 40\ log(f\ + \ 0.4)\]

\[10\ logN_{th}(f)\ = \ - 15\ + \ 20\ log\ f\]

Substituting \(f = 20\ kHz,\) shipping factor \(s = 0.5,\ \)surface windspeed \(w = 0\ m/s,\ \) we get \(N_{t} = - 22.03\ dB\), \(N_{s} = - 4.27\ dB,\) \(N_{w} = 23.63\ dB\), and \(N_{th} = 11.02\ \ dB\). As explained in section 3.1.4 we see that wind noise has the most impact. After adding these noises in the linear scale and then converting back to \(dB\), Total noise, \(N_{Total}^{dB} = 23.87\). From the passive sonar equation

\[SNR = SL - TL - (NL - DI)\]

Substituting we get

\[SNR = 190.80 - 159.51 - (23.87 - 0) = 1.41\ dB\]

Results: Packet Error Rate vs Distance

For the above SNR, we plot PER vs. distance for different modulation schemes given default packet size of 14B.

\[PER = \frac{No.\ of\ errored\ packets}{(No.\ of\ errored\ packets + No.\ of\ received\ packets)}\]

No. of errored packets can be obtained from link metrics and No. of received packets can be obtained from application metrics of the results dashboard as shown in the image below.

Figure-8: Result Dashboard window.

Distance (m)	Source Level (dB/µPa)	Modulation Technique	Packets Received	Packets Errored	PER
18000	183.81	FSK	0	19	1
18000	183.81	BPSK	0	19	1
18000	183.81	QPSK	0	19	1
18000	187.78	FSK	0	19	1
18000	187.78	BPSK	0	19	1
18000	187.78	QPSK	0	19	1
18000	190.8	FSK	0	19	1
18000	190.8	BPSK	131	24	0.154839
18000	190.8	QPSK	0	19	1
23000	183.81	16QAM	119	2	0.016529
23000	183.81	64QAM	32	41	0.561644
23000	183.81	256QAM	0	13	1
23000	187.78	16QAM	121	0	0
23000	187.78	64QAM	119	2	0.016529
23000	187.78	256QAM	7	22	0.758621
23000	190.8	16QAM	121	0	0
23000	190.8	64QAM	121	0	0
23000	190.8	256QAM	113	8	0.066116

Table-8: PER values for different modulations and distances.

Generally, range is defined as the Tx-Rx distance at which the PER is 10 %. From these plots we can determine a device’s range. In summary, we see how the device range is dependent on Source Level, Noise, MCS and packet size.

s-Aloha performance with multiple transmit nodes

Open NetSim and Select Examples-> Underwater Acoustic Network -> s-Aloha performance with multiple transmit nodes and then click on the tile in the middle panel to load the example as Figure-9

Figure-9: List of scenarios for the example of s-Aloha performance with multiple transmit nodes.

Network setup

We consider three scenarios as shown in the figure below, with 2, 3 and 4 transmitting nodes.

Figure-10: Simulation scenarios with 2 transmitting nodes in (A), 3 transmitting nodes in (B) and 4 transmitting nodes in (C). In all cases there is a single receiver.

Properties

Then we set the UWAN device properties as shown below

Device Properties
Device > Interface (ACOUSTIC) > Datalink Layer
Retry Limit	0,4,6
Slot Length(µs)	741257
Device> Interface (ACOUSTIC) > Physical Layer
Source Level (\(\mathbf{dB//1\mu Pa}\))	190.8
Modulation	QPSK
Data Rate (kbps)	20

Table-9: UWAN Device Properties

Here, we set the Slot Time as 741257 μs, which is the ideal value of 731257 μs plus a guard interval of 10,000 μs
Configure a CBR Application from the source nodes (2, 3, 4 and 5 per the cases) to the destination (Node 1) with a packet size of 14 bytes and Inter arrival time as 560000μs from set traffic window present in the ribbon at the top.
Enable packet trace and Acoustic Measurements Log under Configure report tab as shown in Figure-3.
Run the Simulation for 10000sec.

Results

We observe throughputs from network metrics and packets transmitted and packets collided from the Link Metrics. We collision probability as \(P_{c} = \frac{Collision\ Count}{Packet\ Transmitted}\) and tabulate the results in the different cases.

Case #1: Two transmitting nodes
Retry Limit	Throughput N1 (bps)	Throughput N2 (bps)	Aggregate Throughput(bps)	Collision Count	Packet Transmitted	\[\mathbf{P}_{\mathbf{c}}\]
0	0	0	0	26980	26980	1
4	51	55	106	7104	16624	0.427
6	65	71	136	2548	14647	0.173

Table-10: Simulation Results with 2 transmitting nodes

Case #2: Three transmitting nodes
Retry Limit	Throughput N1 (bps)	Throughput N2 (bps)	Throughput N3 (bps)	Aggregate Throughput (bps)	Collision Count	Packet Transmitted	\[\mathbf{P}_{\mathbf{c}}\]
0	0	0	0	0	40470	40470	1
4	26	26	27	79	12234	19348	0.632
6	43	41	37	121	4969	15726	0.316

Table-11: Simulation Results with 3 transmitting nodes

Case #3: Four transmitting nodes
Retry Limit	Throughput N1 (bps)	Throughput N2 (bps)	Throughput N3 (bps)	Throughput N4 (bps)	Aggregate Throughput (bps)	Collision Count	Packet Transmitted	\[\mathbf{P}_{\mathbf{c}}\]
0	0	0	0	0	0	53960	53960	1
4	15	15	14	16	60	16942	22293	0.75
6	28	27	26	26	107	7202	16756	0.42

Table-12: Simulation Results with 4 transmitting nodes

We carry out simulations with different settings of Retry Count. The final results are plotted below. When Retry count is set to zero, all packets collide even when just two nodes are

Figure-11: Collision probability vs Number of Transmitting Nodes

transmitting. With retry count set to 0, the node attempts a packet transmission. If it fails, there is no retry and the packet is dropped. Recall, that in s-Aloha the transmitter does not back off before the first transmission attempt for a packet. With backlogged queues, the two transmitting nodes keep attempting at each slot. This leads to continuous collisions.

When the retry count is set to 4 (or 6) a transmitting node back off per the exponential backoff algorithm, before every retransmission. The back off algorithm is explained in section MAC Layer. Hence there is an element of randomness in packet transmissions at each slot. Nodes may or may not transmit. The probability of transmission at a particular slot reduces as the Retry Count is increased. Hence, we see lower collision probabilities for Retry count of 6.

Energy consumption analysis in underwater acoustic networks under varying traffic loads

Introduction

Efficient energy usage is important for underwater devices because the underwater environment imposes constraints on recharging options. Optimizing energy consumption is essential to ensure the longevity of the network

Consider a practical underwater acoustic network comprising sensor nodes deployed in the ocean, which collect data and relay it to a master node. This master node aggregates the data, transfers it to a shore-based control center, and controls the sensor nodes. The network traffic consists of packetized data delivery from the sensor nodes to the master node.

In our project, we model such a network in NetSim. The setup includes master and sensor nodes distributed across 16 underwater devices, organized into three clusters (A, B, and C). Each cluster contains sensor nodes responsible for data collection and transmission, with the master node managing data aggregation. The scenario is based on [9].

We analyse the energy consumption patterns of the sensors and the master node under different traffic loads.

Open NetSim and Select Examples-> Underwater Acoustic Network -> Energy consumption analysis in underwater acoustic networks under varying traffic loads and then click on the tile in the middle panel to load the example as shown in figure-12

Figure-12: List of scenarios for the example of Energy consumption analysis in underwater acoustic networks

Network setup

The scenario comprises of 16 underwater devices, organized into three clusters (A, B, and C) and a master node.

Figure-13: Scenario representing 3 different clusters and master node and data transmission from each cluster to the master node.

Cluster A: Underwater nodes 12, 13, 14, 15, 16.
Cluster B: Underwater nodes 11, 10, 9, 8, 7.
Cluster C: Underwater nodes 6, 5, 4, 3, 2.
Master Node: Node 1, responsible for collecting data from all the clusters.

The network is configured with static routing to ensure data transfer from sensor nodes to the master node. We assume an ideal channel with no pathloss.

Device Configuration

Device Properties
Mac Layer
Protocol	Slotted Aloha
Slot Length(\(\mu s\))	525420
Phy Layer
Source Level (\(dB//1\mu Pa\))	190
Modulation	QPSK
Data Rate (kbps)	20
Power
Power source	Battery
Initial energy (mAH)	10416
Transmitting current (mA)	6250
Idle mode current (mA)	1.6
Voltage (v)	48
Receiving current (mA)	37.5

Table-13: Device properties set for this example

In our project, we use a data rate of 20 Kbps, whereas at the time of publication of reference [9],the modems supported data rates of 10s to 100s of bits per second. Consequently, the network in the current NetSim simulation can support a much higher traffic load. Therefore, while we expect different numerical results when comparing the outcomes, we anticipate observing similar trends to those reported in [9].

Slot Length Calculation

This is a global parameter applicable to all UWAN devices. As a starting step, estimate the transmission time \(T_{tx}\), which would be

\[T_{tx}(\mu s) = \frac{\left( L_{pkt} + OH \right) \times 8}{PHYRate}\]

where \(L_{pkt}\ \)is the application layer packet size, \(OH\) is the overheads of all layers which is equal to 28B, and \(PHYRate\) is the data rate set in the PHY layer. Next, the propagation delay, \(\Delta\) is computed as \(\Delta = \frac{d}{c_{sound}},\) where \(d\) is the distance between the transmitter and the receiver. Thus, the ideal slot length should be

\[L_{slot} = \ \frac{\left( L_{pkt} + OH \right) \times 8}{PHYRate} + \frac{d}{C_{sound}}\]

In our scenario \(L_{pkt} = 14B\) and \(PHYRate = 20\ Kbps\) which leads to \(T_{tx} = 16,800\ \mu s\). Then using \(t = \frac{25}{10} = 2.5,\) \(z = 50\), and \(S = 35\) - where t is one-tenth of the temperature of the water in degrees Celsius, z is the depth in meters and S is the salinity of the water - we get \(c_{sound} = 2799.33\ m/s\).

When the transmitter receiver distance is \(d = 1423.79\ m\), the propagation delay,

\[\Delta = \frac{1423.79}{2799.33} = \ 508618.13\ \mu s.\]

Substituting all these, we see that the ideal slot length (when \(d = 1423.79\)) would be

\[L_{slot} = T_{tx} + \Delta = 16,800 + 508618.13 = 525420\mu s\]

Note: The slot length is set based on the largest Tx-Rx distance i.e., from node2 to Master node 1

Application Configuration

Create a three CBR Application from the source nodes (12, 11, 6) to the Destination (Node 1) with a packet size of 14 bytes each and we will vary the inter arrival according to the load

Load (Packet/sec)	Inter Arrival time \[(\mu s)\]	Load (Packet/sec)	Inter Arrival time \[(\mu s)\]
0.001	1000000000	0.0095	105263157
0.0015	666666666	0.01	100000000
0.002	500000000	0.02	50000000
0.0025	400000000	0.03	33333333
0.003	333333333	0.04	25000000
0.0035	285714285	0.05	20000000
0.004	250000000	0.06	16666666
0.0045	222222222	0.07	14285714
0.005	200000000	0.08	12500000
0.0055	181818181	0.09	11111111
0.006	166666666	0.1	10000000
0.0065	153846153	0.2	5000000
0.007	142857142	0.3	3333333
0.0075	133333333	0.4	2500000
0.008	125000000	0.5	2000000
0.0085	117647058	0.6	1666666
0.009	111111111

Table-14: Application properties for different loads

Figure-14: The network consists of 16 underwater devices connected via an acoustic link. Three applications are configured to send data from underwater sensors to the master node using static routes :App1 from Node 12 to Master Node 1 and App3 from Node 6 to Master Node 1.

Simulation results

Post simulation, click on the additional metrics in the simulation results window and scroll down for battery model metrics as shown below.

Figure-15: Battery model metrics

The transmitting energy, receiving energy, idle energy, and total consumed energy for the Master node,Layer1 Node 7, and Layer2 Node 15 are tabulated in individual tables for different loads

Load (Packet/sec)	Master Node 1
Load (Packet/sec)	Transmit Energy (mJ)	Receive Energy (mJ)	Idle Energy (mJ)	Total Consumed Energy (mJ)
0.001	0	26307	691807	718113
0.0015	0	40796	716045	756841
0.002	0	55483	728281	783764
0.0025	0	68962	735374	804336
0.003	0	85278	739642	824919
0.0035	0	98271	742889	841159
0.004	0	111749	744856	856605
0.0045	0	125491	746883	872373
0.005	0	140663	748698	889361
0.0055	0	153426	748517	901943
0.006	0	169059	749059	918118
0.0065	0	182570	749340	931910
0.007	0	198170	750409	948580
0.0075	0	211879	749936	961815
0.008	0	223927	751157	975083
0.0085	0	235974	750280	986253
0.009	0	249913	750492	1000405
0.0095	0	266096	750124	1016220
0.01	0	281333	749353	1030686
0.02	0	575793	740784	1316577
0.03	0	854663	731155	1585818
0.04	0	1202139	715724	1917863
0.05	0	1619240	698524	2317764
0.06	0	2258239	670898	2929137
0.07	0	3088371	636044	3724415
0.08	0	3558266	616165	4174430
0.09	0	3645710	612425	4258136
0.1	0	3724541	608901	4333442
0.2	0	3488490	619132	4107622
0.3	0	3471500	619817	4091317
0.4	0	3377058	622000	3999057
0.5	0	3479548	619362	4098910
0.6	0	3542675	616830	4159505

Table-15: Tabulated results for Master node

Load (Packet/sec)	Layer1 Node 7
Load (Packet/sec)	Transmit Energy (mJ)	Receive Energy (mJ)	Idle Energy (mJ)	Total Consumed Energy (mJ)
0.001	1312019	10995	691065	2014079
0.0015	1908392	15224	716275	2639890
0.002	2504764	20298	729141	3254203
0.0025	3339685	24527	735728	4099940
0.003	4174607	30447	740628	4945682
0.0035	4770979	35522	743810	5550311
0.004	5605901	42288	745971	6394159
0.0045	6202273	46516	748664	6997454
0.005	7037194	50745	749118	7837057
0.0055	7752841	54974	750409	8558223
0.006	8349214	60048	751573	9160835
0.0065	9064860	65969	751702	9882531
0.007	9780507	74426	752167	10607100
0.0075	10496154	78655	752933	11327742
0.008	11092527	89650	754046	11936222
0.0085	11688899	93878	753350	12536127
0.009	12285271	98953	753788	13138012
0.0095	12881644	103182	753777	13738603
0.01	13478016	107410	753323	14338750
0.02	27790954	219049	748891	28758894
0.03	42700264	317612	741797	43759672
0.04	59517965	434000	732687	60684652
0.05	76335667	584479	723131	77643277
0.06	108062678	712383	708987	109484048
0.07	147542531	1025897	686074	149254502
0.08	170443231	1274352	669363	172386946
0.09	201454596	1458335	653378	203566309
0.1	204078634	1622671	645366	206346672
0.2	236521293	2101952	617747	239240991
0.3	243319938	1984392	620982	245925312
0.4	229841922	1961101	625426	232428449
0.5	211473652	1905802	631998	214011452
0.6	255128111	2118476	612071	257858659

Table-16: Tabulated results for Layer 1, Node 7

Load (Packet/sec)	Layer 2 Node 15
Load (Packet/sec)	Transmit Energy (mJ)	Receive Energy (mJ)	Idle Energy (mJ)	Total Consumed Energy (mJ)
0.001	1360311	8456	690873	2059639
0.0015	1943301	14093	716389	2673783
0.002	2526291	17475	728847	3272613
0.0025	3303611	20858	736534	4061003
0.003	3789436	24240	741228	4554905
0.0035	4469592	27059	744525	5241176
0.004	4955417	29877	746823	5732117
0.0045	5441242	34387	748685	6224314
0.005	6121397	37206	751498	6910101
0.0055	6898718	40024	751542	7690284
0.006	7481708	43970	751749	8277427
0.0065	8161863	48480	752431	8962775
0.007	8647688	51862	753131	9452682
0.0075	9133513	55245	753387	9942145
0.008	9910834	58063	754561	10723458
0.0085	10493824	60882	754372	11309078
0.009	10979649	64828	754200	11798678
0.0095	11562639	68210	754552	12385402
0.01	12048465	71593	754728	12874785
0.02	23513939	145440	752394	24411774
0.03	35076578	212523	747783	36036884
0.04	48874014	303846	741121	49918981
0.05	58292600	395733	734991	59423323
0.06	77508797	546810	724029	78779637
0.07	106500128	751442	707997	107959567
0.08	123932855	968475	688867	125590197
0.09	144988766	1032739	686546	146708050
0.1	158314046	1169160	676869	160160075
0.2	129284179	1241880	681804	131207862
0.3	136798000	1231169	679611	138708780
0.4	140626797	1258228	678243	142563269
0.5	167477286	1303889	669477	169450652
0.6	141795989	1167468	681776	143645234

Table-17: Tabulated results for Layer 2, node15

Throughput and Packet collision count

The values for throughput of three applications and packets collided are listed below for different loads:

Load (Packet/sec)	Throughput-1 (Mbps)	Throughput-2 (Mbps)	Throughput-3 (Mbps)	Packets Collided
0.001	0	0	0	154
0.0015	0	0	0	233
0.002	0	0	0	306
0.0025	0	0	0	374
0.003	0	0	0	439
0.0035	0	0	0	511
0.004	0	0	0	578
0.0045	0.000001	0.000001	0.000001	642
0.005	0.000001	0.000001	0.000001	719
0.0055	0.000001	0.000001	0.000001	782
0.006	0.000001	0.000001	0.000001	843
0.0065	0.000001	0.000001	0.000001	913
0.007	0.000001	0.000001	0.000001	980
0.0075	0.000001	0.000001	0.000001	1052
0.008	0.000001	0.000001	0.000001	1125
0.0085	0.000001	0.000001	0.000001	1194
0.009	0.000001	0.000001	0.000001	1261
0.0095	0.000001	0.000001	0.000001	1331
0.01	0.000001	0.000001	0.000001	1396
0.02	0.000002	0.000002	0.000002	2805
0.03	0.000003	0.000003	0.000003	4167
0.04	0.000004	0.000004	0.000004	5747
0.05	0.000005	0.000005	0.000005	7542
0.06	0.000006	0.000006	0.000006	10088
0.07	0.000006	0.000006	0.000006	14228
0.08	0.000005	0.000006	0.000006	18428
0.09	0.000005	0.000006	0.000005	21346
0.1	0.000004	0.000005	0.000004	24555
0.2	0.000002	0.000004	0.000003	27718
0.3	0.000002	0.000004	0.000002	27771
0.4	0.000002	0.000003	0.000002	28000
0.5	0.000002	0.000003	0.000002	28074
0.6	0.000002	0.000004	0.000002	28143

Table-18: Tabulated results for throughput and packets collided.

Packets Collision Count

Figure-16: We observe the variation in collision count vs load, for this 16-node network running slotted aloha in the MAC layer.

As the load increases, the number of collisions rises sharply until it reaches a plateau. At low loads, the probability of collision is relatively low, and most packets are successfully transmitted. However, as the load increases, the probability of two or more packets being transmitted in the same time slot rises exponentially, leading to a rapid increase in collisions.

NetSim slotted Aloha implementation uses the exponential backoff algorithm when collisions occur. As collisions become frequent at high loads, nodes spend more time in backoff, effectively reducing their transmission attempts and stabilizing the collision rate.

Throughput

Figure-17: Throughput plots for all three applications.

The throughput behaviour can be explained by considering both the collision plot and the throughput graphs:

Initial increase: At low loads, throughput increases as more packets are successfully transmitted with relatively few collisions.
Peak throughput: The throughput reaches a maximum at an optimal load point (around 0.05-0.07 packets/sec). This represents the best balance between channel utilization and collision avoidance.
Sharp decline: As load increases beyond the optimal point, we see a sharp rise in collisions (from the collision plot). This leads to a rapid drop in throughput because:
- More transmission attempts result in collisions rather than successful transmissions.
- Colliding packets waste channel capacity without contributing to throughput.
- The exponential backoff algorithm causes nodes to wait longer before retransmitting, reducing overall transmission attempts.
Gradual stabilization: The collision plot shows a plateau at higher loads, but throughput continues to decrease slightly or stabilize at a lower level. This occurs because:
- The network is saturated with collisions.
- Most transmission attempts fail due to collisions.
- The backoff algorithm limits new transmission attempts.
- The actual number of successful transmissions becomes a small fraction of the total load.
Differences between applications:
- Applications 1 and 3 use routes with 4 hops and show similar throughput patterns. They experience more throughput degradation at high loads as compared to application 2 due to longer paths.
- Application 2 uses a route with only 3 hops and sees better throughput, at higher loads. This is because of the shorter path length, which reduces the overall collision probability because each additional hop increases the likelihood of collisions and packet loss
- The throughput differences among applications stem from cross layer interactions of routing path lengths and the slotted Aloha MAC layer's behaviour under varying loads.

Master Node 1 Energy Consumption

See Figure-18

The transmission energy from the master node will be zero because no transmission is occurring from the master node.
As the load increases, the number of collisions initially rises rapidly but flattens out. Although the number of successfully received packets decreases, the node continues to receive collided packets. In the NetSim energy model, note that energy is expended in receiving these collided packets. However, once received, the node cannot decode the packets that have undergone collisions.
The idle energy remains significant throughout all loads, though it slightly decreases at higher loads. This is because more energy is being consumed in receiving packets, leaving less time for the node to be idle.

Figure-18: Energy Consumption Plots for the Master Node.

The total consumed energy initially increases with network load, primarily due to the rise in receiving energy, and then flattens out.

Layer-1 Device - Node 7, Energy Consumption

See Figure-19

The transmitting energy for Layer 1 Node 7 increases significantly with the network load as it relays packets of Application 2 to the master node.
The receiving energy also increases with load. This reflects its role in receiving packets that it must then forward to the master node.
The idle energy remains relatively stable but shows a slight decrease at higher loads. This is due to the node spending more energy on transmission and reception rather than staying idle, at higher loads.

Figure-19: Energy Consumption Plots for the Layer1 Node 7

The total consumed energy increases with the network load, driven by the substantial rise in both transmitting and receiving energies.
These plots reflect Node 7's active role in relaying traffic from outer layers to the master node.

Layer-2 Device - Node 15 Energy Consumption

See Figure-20

The curves depicted in the four panels in Figure-20 closely resemble those in Figure-19 with the only distinction being slightly lower values. This difference arises because: Node 15 serves as a relay for Application 1, while Node 7 relays for Application 2. Application 1 sees lower throughput compared to Application 2 due to its longer path (4 hops versus 3 hops). Consequently, Node 15 relays fewer packets than Node 7, resulting in reduced transmit and receive energy consumption. This, in turn, leads to lower total energy consumption for Node 15.

Figure-20: Energy consumption plot for a Layer-2 device, Node 15