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, 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:#

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