THE UNIVERSITY OF SYDNEY
⃝c MAHYAR SHIRVANIMOGHADDAM 1
Tutorial 2 – Solutions
ELEC5518: IoT for Critical Infrastructure Energy Efficiency and Multi-hop Networks
OBJECTIVES
In these examples, we practise how to measure the energy efficiency and life time of wireless networks.
EXAMPLES 1
A linear personal area network (PAN) is composed of 3 close-by motes and a distant Gateway (S). The communication is performed by electing a cluster head among the three motes which collects the traffic from the other motes and sends it remotely to the gateway. The energy required to operate the TX/RX circuitry, Ec = 50[nJ/bit]. The energy required to support sufficient transmission output power Etx(d) = kd2[nJ/bit], being k = 1[nJ/bit/m2]. The length of each packet to be exchanged is b = 2000 [bits]. The distance d is 5 [m] and all the three nodes A, B and C are in transmission range. Assume that the sensing energy is negligible and the exclude the energy consumed at the sink node from your calculation.
1. Write the expression of the energy consumed for sending one packet per node to the gateway when using A, B and C as cluster heads respectively.
2. Find the values of D for which the best solution is to elect B as cluster-head.
3. Write the energy consumption under direct transmission (no cluster-head).
AdBdC D
Fig. 1. Example Network 1
S
Solution:
Case 1: A is cluster-head
In this case node A needs to receive one packet from B, one packet from C and send the three packets
(B’s, C’s and its own one) to the gateway. Thus, the total energy consumed by node A in this case can be written as:
EA1 =2Ecb+3(Ecb+Etx(D+2d)b) The energy consumed by B and C in this very same case is:
EB1 = Ecb + Etx(d)b EC1 =Ecb+Etx(2d)b
THE UNIVERSITY OF SYDNEY ⃝c MAHYAR SHIRVANIMOGHADDAM 2
The total energy required to deliver the three packets to the central controller is:
E1 =E1+E1+E1 tot A B C
Case 2: B is cluster-head
In this case node B needs to receive one packet from A, one packet from C and send the three packets (A’s, C’s and its own one) to the gateway. Thus, the total energy consumed by node B in this case can be written as:
EB2 = 2Ecb + 3(Ecb + Etx(D + d)b) The energy consumed by A and C in this very same case is:
EA2 =EC2 =Ecb+Etx(d)b
The total energy required to deliver the three packets to the central controller is:
E2 =E2+E2+E2 tot A B C
Case 3: C is cluster-head
In this case node C needs to receive one packet from A, one packet from B and send the three packets (A’s, B’s and its own one) to the gateway. Thus, the total energy consumed by node C in this case can be written as:
EC3 =2Ecb+3(Ecb+Etx(D)b) The energy consumed by A and B in this very same case is:
EB3 = Ecb + Etx(d)b EA3 =Ecb+Etx(2d)b
The total energy required to deliver the three packets to the central controller is:
E3 =E3+E3+E3 tot A B C
We can safely say that E1 ≥ E3 (think about it). Thus, we have to find the conditions in which E2 ≤ E3. The energy consumption is case the three nodes perform direct transmission to the gateway the total energy consumption would be:
Edirect = 3Ecb + Etx(D + 2d)b + Etx(d + D)b + Etx(D)b
Note that we are not considering the energy for receiving the three packets by the gateway node.
THE UNIVERSITY OF SYDNEY
⃝c MAHYAR SHIRVANIMOGHADDAM 3
EXAMPLES 2
A sensor node generates a stream of 3 packets at a fixed rate of one packet every x [s]. The packets must be delivered to a sink node for further processing. The nominal data rate is R = 250[kb/s] and the length of each packet is L = 1000[bit]. The operating power level for TX circuitry is Ptx = 100[mW ]; the power emitted to the antenna is Po = 100[mW ]; the power consumed while in idle and sleep states are Pidle = 60[mW ] and Psleep = 10[mW ], respectively; In case the sensor goes to sleep, it needs a wake-up time of Tw = 500[μs]. Assume that the sensing energy is negligible.
• Write the energy consumption for transmitting the 3 packets in the 2 cases where
1) the sensor node goes to sleep after each transmission and wakes up when the following packet
is ready.
2) the sensor node is always active (assume that in both cases the sensor node is asleep at the very
beginning of the operations).
• Is there any value of x for which case (2) is more energy-efficient than case (1)? Solution:
Fig. 2. Reference case where mote switches on and off.
Fig. 3. Reference case where mote activates at the beginning and then stays active.
Figures 2 and 3 report the temporal evolution of the systems in the two cases where the mote switches on and off, and the mote stays active after first transmission, respectively.
Calling T the transmission time of one packet, with T = L = 4 ms. The energy consumed in the two R
cases can be written as follows:
Case 1 – mote switches on and off:
E1 = 3[TwPtx + (Ptx + Po)T + Psleep(x − T − Tw)] Case 2 – mote stays active after first transmission
E2 = PtxTw + (Ptx + Po)T + x Pidle − PidleTw − PidleT + 2[(Ptx + Po)T + Pidle(x − T )]
We have to check for which values of x it is E2 ≤ E1. By solving the inequality in x, we get: x ≤ 4.76 ms.
THE UNIVERSITY OF SYDNEY
⃝c MAHYAR SHIRVANIMOGHADDAM 4
EXAMPLES 3
The personal area network (PAN) is used to gather periodical information on the average temperature of a given area. N motes are geared with temperature sensors. Consider the following two scenarios:
1) The sensors periodically report their measures to the PAN Coordinator which then relays each report to the information sink which performs the average operation.
2) The sensors periodically report their measures to the PAN Coordinator which performs the average operation and then sends one averaged sample to the information sink.
d
Fig. 4. Example 3
Assuming that:
• Sensors are d = 5m away from the PAN Coordinator, The PAN Coordinator is D = 10m away from
sink.
• The length of each packet containing the temperature samples and the averaged temperature is 127
[byte].
• The energy for operating TX/RX circuitry is Ec = 50[nJ/bit];
• the energy required to support sufficient transmission output power Etx(d) = kd2[nJ/bit], being
k = 1[nJ/bit/m2];
• the energy required to perform the average operation is Ep = 4[μJ] for each temperature measure;
• the energy consumption for the averaging operation performed at the information sink is negligible.
The sensing energy is negligible.
i) Write the energy consumption to get one averaged temperature value at the sink when N = 4 in the
two aforementioned cases.
ii) Under the given parameters, is there any value of N beyond which performing the averaging operation
at the PAN Coordinator is less energy-efficient than at the sink? If so, find out this value.
Solution:
In Case 1, the overall energy consumption can be written as:
E1 =Nb(2E +E (d))+Nb(E +E (D)); tot c tx c tx
where the first term accounts for the total energy consumed by all the sensor nodes to deliver 1 packet to the PAN coordinator and for the energy consumed by the PAN coordinator for receiving one packet from
D
THE UNIVERSITY OF SYDNEY ⃝c MAHYAR SHIRVANIMOGHADDAM 5
all the sensor nodes; the second term accounts for the energy consumed by the PAN coordinator to deliver all the received packets to the SINK. Note that in this case there’s no energy consumed for performing the averaging operation across the temperature samples as this is performed by the SINK which is commonly expected to be attached to the mains.
In Case 2, the overall energy consumption can be written as:
E2 =Nb(2E +E (d))+NE +b(E +E (D));
tot c tx p c tx
where the first term accounts for the total energy consumed by all the sensor nodes to deliver 1 packet to the PAN coordinator and for the energy consumed by the PAN coordinator for receiving one packet from all the sensor nodes; the second term accounts for the energy consumed by the PAN coordinator to perform the averaging operation across the N temperature samples received and the last term accounts for the energy consumed by the PAN coordinator for to deliver the single packet containing the averaged measure to the SINK.
If N = 4, it turns out that E1 = 1.117mJ and E2 = 676.4μJ. tot tot
In order to verify if there exists any value of N for which Case 2 is less energy-efficient than Case 1, we have to solve (in N) the inequality E1 ≤ E2 , that is,
tot tot
Nb(2Ec + Etx(d)) + Nb(Ec + Etx(D)) ≤ Nb(2Ec + Etx(d)) + NEp + b(Ec + Etx(D))
After some maths, it turns out that N ≤ 1.01, that is, the only case when Case 2 is less energy efficient than Case 1 corresponds to a topology with one single sensor node and the PAN coordinator. Note that this case is not much practical in the reference scenario since no averaging operation would be required with one single temperature sample.