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, 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 4‑4 shows the ideal slot length settings for $d_{Rx}^{Tx} = 4\ km$ and $d_{Rx}^{Tx} = 6\ km.$
Network setup:#
- Create a scenario with 2 Underwater Devices
Figure 4‑1: 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 6km
- Channel characteristics as NO_PATHLOSS
Device Configuration:#
| Device>Interface | (ACOUSTIC)>Datalink Layer |
|---|---|
| Slot Length(µs) | 16800 |
| Device>Interface(ACOUSTIC)> | Physical Layer |
| Source Level($\mathbf{dB//1}\mathbf{\mu Pa}$) | 200 |
| Modulation | QPSK |
| Data Rate (kbps) | 20 |
Table 4‑1: Device properties set for this example
Application Configuration:#
We run simulations for different traffic generation rates. The generation rate depends on the inter-packet 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 | |
| Case-1 | Inter arrival Time (µs) | Generation rate (bps) |
| 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 4-2: Application properties for the different samples in each case studied in this example
- Click on Packet Trace option and select the Enable Packet Trace check box.
- 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 4‑3: Theoretically predicted propagation delay for different Tx-Rx distances
Transmission delay and Saturation Throughput#
Considering a slot length of $731,256.4\ \mu s$, we see that one packet exactly fits one slot and hence the predicted saturation throughput would be
$$\theta_{sat}^{2km} = \frac{{(L}{pkt} \times 8)}{L} = \frac{(14 \times 8)}{731256.4 \times 10^{- 6}} = 153\ bps$$
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 | 731256.4 | 153 |
| 4 km | 1445712.7 | 77 |
| 6 km | 2160169.1 | 52 |
Table 4‑4: Ideal slot lengths and theoretically predicted saturation throughputs ($\theta_{sat}$) 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 Open Packet Trace option available in the Simulation Results window.
-
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 Figure 4‑2)
$$Queuing\ Delay\ (\mu s) = PHYSICAL\ LAYER\ ARRIVAL\ TIME(\mu s) - MAC\ LAYER\ ARRIVAL\ TIME\ (\mu s)$$
Figure 4‑2 : 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 Figure 4‑2)
-
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 | 736612.9 | 714456.4 | 16800 | 5356.52 |
| 50 | 50 | 736856.4 | 714456.34 | 16800 | 5600 | |
| 100 | 100 | 736793.4 | 714456.4 | 16800 | 5537.08 | |
| 125 | 124 | 736856.4 | 714456.4 | 16800 | 5600 | |
| 150 | 149 | 738666.9 | 714456.4 | 16800 | 7410.53 | |
| 175 | 151 | 7377656.4 | 714456.4 | 16800 | 6646400 | |
| 200 | 151 | 12737656.4 | 714456.4 | 16800 | 12006400 | |
| Case #2: Distance between underwater devices is 4km |
20 | 20 | 1451313 | 1428912.7 | 16800 | 5600 |
| 40 | 40 | 1451313 | 1428912.7 | 16800 | 5600 | |
| 60 | 59 | 1453285 | 1428912.7 | 16800 | 7572.33 | |
| 80 | 76 | 3509313 | 1428912.7 | 16800 | 2063600 | |
| 100 | 76 | 12889313 | 1428912.7 | 16800 | 11443600 | |
| Case #3: Distance between underwater devices is 6km |
20 | 20 | 2165769 | 2143369.1 | 16800 | 5600 |
| 30 | 30 | 2167636 | 2143369.1 | 16800 | 7466.67 | |
| 40 | 39 | 2165609 | 2143369.1 | 16800 | 5440 | |
| 50 | 49 | 2165769 | 2143369.1 | 16800 | 5600 | |
| 60 | 52 | 8922169 | 2143369.1 | 16800 | 6762000 | |
| 70 | 52 | 14922169 | 2143369.1 | 16800 | 12762000 |
Table 4‑5: Tabulated results (throughput and delays) for 3 different Tx-Rx distances
Figure 4‑3: Throughput vs. Generation rate plotted for Tx-Rx distances of 2km, 4km and 6km based on earlier tables
From Table 4‑5, we see that the propagation delays from simulation match predictions in Table 4‑3. Then from Figure 4‑4 we observe that saturation throughput (the Y axis value once the curve flattens) matches prediction.
| Distance betweendevices | Saturation Throughput. Predicted $\mathbf{(}\mathbf{\theta}_{\mathbf{sat\ }}$in bps) | Saturation Throughput. Simulation $\mathbf{(}\mathbf{\theta}_{\mathbf{sat\ }}$in bps) |
|---|---|---|
| 2 km | 153 | 151 |
| 4 km | 77 | 76 |
| 6 km | 52 | 52 |
Table 4‑6: NetSim UWAN Simulation results vs. theoretical prediction of saturation throughput, for different Tx-Rx distances