MAXIMUM THROUGHPUT OF 10 NODES FOR IEEE 802.15.4 WPAN

MAXIMUM THROUGHPUT OF 10 NODES FOR IEEE 802.15.4 WPAN

(WIRELESS PERSONAL AREA NETWORKS)

Dr. P. C. Jain1, Reena 2, Divya Gautam3 1.2.3School Of Electronics, Centre for Development of Advanced Computing, Noida, India

[email protected], [email protected], [email protected]

Abstract-ZigBee is the result of ZigBee Alliance and is also an open global standard providing wireless networking based on the IEEE 802.15.4 standard. IEEE 802.15.4 consists mainly of Physical and MAC layers. ZigBee and the underlying protocol IEEE 802.15.4 were developed with low data rate. Data rate requirements for audio will range between tens of Kbps and hundreds of Kbps. Compression methods could be applied to reduce the data rate. We can use IEEE 802.15.4 for this reduced data rate. This paper describes a five-phased approach for measuring and attaining the maximal throughput for 10 nodes in IEEE 802.15.4. The phases consist of: 1) theoretical calculations for single-hop 2) theoretical calculations for multi-hop 3) NS-2 simulations for single-hop 4) NS-2 simulations for multi-hop 5) NS-2 simulations for 10 nodes.

Keywords: Overhead, Payload, Single-hop, Multi-hop, Throughput

1. INTRODUCTION

In March 1999, the IEEE established the 802.15 working group as part of the IEEE Computer Society’s 802 Local and Metropolitan Area Standards Committee. The 802.15 working group was established with the specific purpose of developing standards for short distance wireless networks, otherwise known as wireless personal area networks (WPANs).The IEEE 802.15.4 standard defines the characteristics of the physical and MAC layers for Low-Rate Wireless Personal Area Networks (LR-WPAN).ZigBee and the underlying layers have been designed for low data rate applications, where low power consumption is of great importance. ZigBee/IEEE 802.15.4 can be used in applications like light switch panel, remote controlled devices, remote metering, medical healthcare, home automation, etc. The complex application becoming very popular is wireless sensor network (WSN).

Some of the important features of IEEE 802.15.4 are

– supports Guaranteed Time Slot (GTS) mechanism

– CSMA-CA access

– supports star-mesh network

– uses AES-128 security encryption

– uses DSSS

– works at 2.4GHz

– has 16 channels with 2MHz BW in 5MHz spacing

– data rate 250kb/s at 2.4GHz ( for all countries), channels used are 11-26

To measure the throughput for 10 nodes, the paper is described in following manner. Section 2 describes the ZigBee/IEEE 802.15.4 overhead and payload. Section 3 illuminates the theoretical calculations for single-hop. Section 4 highlights the upper-bound on multi-hop. Section 5 describes NS-2 simulations for single-hop. Section 6 describes NS-2 simulations for multi-hop and finally section 7 describes NS-2 simulations for 10 nodes.

2. ZigBee/IEEE 802.15.4 Overhead and Payload

The PHY/MAC layers of IEEE 802.15.4 facilitate transmission information, node addressing, network formation and scheduling. This can be done by concatenating headers with outgoing data frames. The contents of a data message that is being transmitted is called payload. The extra bytes of frame format information that is stored in the header at different layers, which combine with assembly and disassembly of packets, that reduces the overall transmission of payload is called overhead.

In IEEE 802.15.4 the maximum packet size is 133 bytes but the maximum physical service data unit length is 127 bytes where the remaining 6 bytes correspond to overhead of preamble, start-of-frame delimiter (SFD) and frame length field. The overhead added by MAC layer varies due to the variable length addressing fields (when a device is a member of multiple PANs or there are multiple PANs operating in a same region, the MAC layer defines two fields viz source PAN ID and destination PAN ID. If the source and destination PAN ID are equal

Dr. P. C. Jain et al. / International Journal of Engineering Science and Technology (IJEST)

ISSN : 0975-5462 Vol. 5 No.01 January 2013 54

and PAN ID compression sub-field of MAC header is set to 1 then the source PAN ID is dropped and the destination PAN ID is present.). The overheads can be shown in figure 1.

ZigBee AF – 1 byte

Transaction Count Frame Type

ZigBee APS – 6 bytes

Frame Control Destination Endpoint Cluster Identifier Profile Identifier Source Endpoint

Application Data – 101 bytes

Application data (payload)

ZigBee NWK – 8 bytes

Frame Control Destination Address Source Address Radius Sequence Number

IEEE 802.15.4 MAC – 9 bytes

Frame Control Sequence Number Destination PAN ID Destination Address Source Address

MAC FCS – 2 bytes

FCS

IEEE 802.15.4 PHY – 6 bytes

Preamble SFD Frame length (Reserved)

Fig. 1 - ZigBee/IEEE 802.15.4 frame format

ZigBee adds three additional headers to the outgoing data. These layers are : (i) Network (NWK) which takes care of routing of data. It can establish a new network and can achieve synchronization between two devices through beacon or poll. (ii) Application Services layer (APS) which provides data transmission service between 2 or more devices located on the same network, security service, binding of devices, establish and removal of group addresses (iii) Application Framework layer (AF) which provides a device’s potential services as dictated by a given AF profile. In total, the ZigBee headers are of 15 bytes for each data frame.

According to the 2003 standard, we define some constants used while calculating the maximum payload.

aMaxMACFrameOverhead represents the maximum MAC header length which is of 25 bytes. aMaxMACFrameSize represents the maximum payload supported by MAC layer which is to be determined.

aMaxMACFrameSize = aMaxPHYPacketSize – aMaxMACFrameOverhead = 127-25 = 102

This definition restricts the available payload in a packet to 102 bytes. In fig 1, 11 out of the maximum 25 (aMaxMACFrameOverhead) MAC header bytes are used and rest 14 are unused for application data. Actual packet transmitted = 102+6(PHY header) +9(MAC header) + 2 (MAC FCS)= 119 bytes instead of 133 bytes. If ZigBee headers are taken into account, the application payload = aMaxMACFrameSize – 15(8 bytes NWK+ 6 bytes APS + 1 byte AF). While the desired application payload is 101 bytes.

According to the revised version of IEEE 802.15.4 in 2006 published by IEEE, aMaxMACFrameSize variable is replaced by few new variables: aMaxMACPayloadSize represents minimum MAC header size, aMaxMACSafePayloadSize indicates that MAC payloads of length greater than this value may not be handled properly.

aMaxMACPayloadSize = aMaxPHYPacketSize – aMinMPDUOverhead = 127 -9 = 118

aMaxMACSafePayloadSize = aMaxPHYPacketSize – aMaxMPDUUnsecuredOverhead = 127 – 25=102

Where aMinMPDUOverhead represents the minimum number of octets added by the MAC sublayer to the PSDU having a value of 9 bytes and aMaxMPDUUnsecuredOverhead represents the maximum number of octets added by the MAC sublayer to the PSDU without security. Its length is 25 bytes.



Theoretical Calculations for Single-hop

For the calculations of throughput there are certain assumptions:

No loss or corruption of information due to transmission errors.

Acknowledgement is requested for reception of each data packet.

The application payload is 101 bytes.

There should be two nodes for transmission.

Time is considered in symbols, which can be further converted to seconds by dividing by symbol rate of 62,500 symbols per second.

Fig 2 shows two transmission phases consisting of various procedures:

1. Transmission Phase 1 {CSMA-CA (Carrier Sense Multiple Access with Collision Avoidance)}: shows how to gain access to wireless channel. The remaining steps are executed only if CSMA-CA procedure grants channel access and are error free.

2. Transmission Phase 2 : the procedures involved are:

i. TX Data: deals with transmission of data packet.

ii. ACK Turnaround: deals with turnaround time.

iii. TX ACK: deals with transmission of ACK (acknowledgement).

iv. IFS: deals with the interframe spacing.

2.1 Transmission Phase 1 (CSMA-CA )

In MAC layer, it dictates how the devices can gain transmit access to a wireless channel by first determining if another device is transmitting currently.

Fig 2 - IEEE 802.15.4 Packet Transmission Procedure

The unslotted CSMA-CA algorithm can be shown in fig 3. Steps (2) and (3) can be executed in variable amount of execution time depending on channel activity. The value of BE can be bounded by macMinBE (3) to macMaxBE (5) by IEEE 802.15.4 standard.

Fig 3- IEEE 802.15.4 CSMA-CA procedure for unslotted portion



macMaxCSMABackoffs (default 4) represents the maximum number of iterations to gain channel access. Channel access failure occurs if the algorithm does not gain access to the channel within macMaxCSMABackoffs. After channel access failure, transmission is cancelled. While traversing the Markov Chain from ‘TX Request’ to the ‘Access Failure’ probability of a channel access failure is obtained. This can be shown in fig 4.

Fig 4- FSM representing relation of CSMA-CA to packet transmit process

Here Pinactive is set to be 0.7 for throughput calculations.

Paccess_failure= (1-Pinactive) macMaxCSMABackoffs = (0.3)4 = 0.0081

The average symbol time for a given CCA stage, CCAi is defined as,

symbolsCCAi = backoff_durationi + cca_duration = backoff_durationi + 8 symbols.

backoff_durationi= E (Xi) * aUnitBackoffPeriod = E(Xi) * 20 symbols

Then for a given CCAi, according to the random function, xi is a random variable of standard uniform distribution with parameter b = 2BEi - 1. Therefore,

E (Xi) = (2BEi - 1) / 2.

The CCA durations are calculated by these means with the following results:

symbolsCCA1 = 78 symbols




By traversing the state machine transitions, the average time to complete one iteration, of unknown outcome, of the transmission process can be calculated as:

symbolsiteration = CCA1 + (1-Pinactive)*(CCA2 + (1-Pinactive)*(CCA3 + (1-Pinactive)*CCA4)) +

symbols TX-Phase2*Paccess_granted

The geometric distribution requires one parameter, the probability of a success for a given trial, which shall be Paccess_granted.

Therefore, based on the properties of the geometric distribution, the expected number of iterations required to gain channel access is,

Paccess_failure = (1-Pinactive) 4

Paccess_granted = 1 - Paccess_failure = 1 - (1-Pinactive) 4 = 1 – (0.3)4 = 0.9919

E (iteration_count) = 1/Paccess_granted. =1.0082



Finally, leaving the average symbol time required for a successful transmission of a single packet

symbolstransmit = symbolsiteration * E (iteration_count) = symbolsiteration/Paccess_granted

symbolstransmit = (CCA1 + (1-Pinactive)*(CCA2 + (1-Pinactive)*(CCA3 + (1-Pinactive)*CCA4 ))) / Paccess_granted

+ symbolsTX-Phase2.

This formula can be separated to represent the two distinct transmission phases with the introduction of a new variable, symbolsCSMA-CA.

symbolsCSMA-CA = (CCA1 + (1-Pinactive)*(CCA2 + (1-Pinactive)*(CCA3 + (1- Pinactive)*CCA4)))

/Paccess_granted

symbolstransmit = symbolsCSMA-CA +symbolsTX-Phase2

For Pinactive = 0.7, symbolsCSMA-CA evaluates to,

symbolsCSMA-CA = (78 + (0.3)*(158 + (0.3)*(318 +(0.3)*318)))/0.9919 = 163.934 symbols.

2.2 Transmission Phase 2

This phase is a conditional phase executed only if channel access is granted by the CSMA-CA procedure. To attain a maximum throughput the largest packet size is used. So each packet is assumed to contain 133 bytes (101 bytes application payload + 32 bytes overhead). The total execution time for phase 2 is calculated as

symbolsTX-Phase2 = symbolstx + symbolsturnaround + symbolstx-ACK + symbolsIFS.

3.2.1 Transmission Time (TX Data):

The packets are transmitted by the physical layer after assembling them into the PPDU structure. Thus, 6 additional bytes are concatenated by the physical layer for a total of 133 octets to be transmitted. The actual symbol time required for transmission is:

lengthPPDU = 133 bytes

lengthsymbol = 4 (bits/symbol)

symbolstx = lengthPPDU (bytes) * 8 (bits/byte) / lengthsymbol (bits/symbol)

symbolstx = 133*8/4 = 266 symbols.

3.2.2 Turnaround Time (ACKTurnaround):

After receiving the last octet of a packet at the destination, the IEEE 802.15.4 standard specifies that a node do not require more than aTurnaroundTime symbols to switch the RF transceiver from receive to transmit mode to transmit an acknowledgement. Thus for the practical case:

symbolsturnaround = aTurnaroundTime = 12 symbols.

3.2.3 Acknowledgment Transmission Time (TX ACK):

The transmission time required for an acknowledgement frame is calculated similarly to that of the data frame calculated in step (i) earlier. Only the parameter lengthPPDU is reduced to 11 bytes

symbolstx-ACK = 11*8/4 = 22 symbols.

3.2.4 Interframe Spacing Time (LIFS):

For the scenario under consideration,

symbolsIFS = aMinLIFSPeriod = 40 symbols.

3.3 Total Transmission Time:

After these calculations, the average time required for transmission of a single packet is computed in the given table:



Table 1 Packet Transmission Time for Pinactive = 0.7

Procedure Identifier Symbols Milliseconds

CSMA-CA symbols CSMA-CA 163.934 2.63

TX Packet symbolsTX 266 4.26

ACK Turnaround symbolsturnaround 12 0.19

TX ACK symbolsACK 22 0.35

LIFS symbolsIFS 40 0.64

Sum symbolssum 503.934 8.07

Finally, throughput is computed given the average number of packets transmitted per second.

packetssec = ratesymbol / symbolssum = 62,500/503.934 = 124.024 (packets/s)

throughputkbps = AppPayload (bytes/packet) *8 (bits/byte) * packetssec (packets/s)

throughputkbps = 101*8*124.024 = 100.21kbps

Thus maximum throughput for single hop transmission in a lightly loaded, non-beacon enabled PAN is approximately 100.21 kbps.

Table 2 Packet Transmission Time for Pinactive = 0.7 without ACK

Procedure Identifier Symbols Milliseconds

CSMA-CA symbols CSMA-CA 163.934 4.71

TX Packet symbolsTX 266 4.26

LIFS symbolsIFS 40 0.64

Sum symbolssum 469.934 9.61

packetssec = ratesymbol / symbolssum = 62,500/469.934 =132.997(packets/s)

throughputkbps = 101*8*132.997=107.46 kbps.

Table 3 Comparison of Throughput for Different Pinactive

Pinactive Throughputkbps (with ACK) Throughput kbps (without ACK)

0.25 40.76 41.9

0.5 79.57 84.07

0.7 100.21 107.4

0.9 115.5 125.2

1 120.8 131.5



Fig 5 - Graph Representing Throughput for Different Pinactive for ACK and without ACK

3. Upper Bound on Multi-hop

As compared to single-hop, additional limitations occur on throughput for multi-hop forwarding. Upper bound on multi-hop can be defined by the number of nodes within interference range that have to transmit the same radio packet. When all the nodes are active i.e. interfering with each other, resulting in n-hop transmission, the upper bound on throughput is

Tm (n) = Ts / n,

where Tm (n) is measured in Kbit/s.

4.1 Three -hop transmission:

Fig 6 - Three-hop transmission

When a sending node wants to transmit, it sends a request-to-send (RTS) packet to the receiver, who responses with a clear-to-send (CTS) packet if it receives the RTS packet correctly. Thereafter, the sending node can transmit data.

The figure above, illustrates how a source, node 0, can send data to the destination node 3, via node 1, node 2. In order to forward a data packet, node 1 needs to transmit control packet: a CTSl packet destined to node 0. Due to the broadcast characteristics of 802.15.4, CTSl packet is also sent to node 2 and it conveys the information of a data packet arrival at node 1 to node 2. After receiving CTS1, node 2 understands that node 1 is busy in communicating with node 1. So node 2 does not sent any data to node 1. Node 0 will transmit data to node 1. Now node 1 sends RTS2 packet to node 2. Node 2 invites node 1 to forward that data via CTS2 packet, and therefore RTS2 packet can be suppressed. The CTS2 sent to node 3 will also perform the same function as the CTS1 received at node 2. Node 1 transmits the data to node 2. Then the process continues in the same manner till the data reaches node 3 without affecting other nodes transmission.

In the scenario when only 4 nodes i. e. 3-hops are active at a time, the throughput is given as

Tm (3) = 130/3 = 40 kbps.

This concept can be used for any number of nodes. Thus the throughput for 10 nodes taken in the sets of 4 nodes, can be calculated as



Tm (3) = 130/3 = 40 kbps.

But if all the 10 nodes are active i.e. interfering with each other, the throughput is

Tm (9) = 130/9 = 14.33 kbps.

4. NS-2 Simulations for Single-hop

The position of the nodes is fixed. Both are separated by a distance of 25 meters. Node 0 is labeled as PAN Coordinator and node is labeled as Full-functional Device. The simulation time is taken as 90 seconds. Node 1 starts to constructs a constant bit rate (CBR) traffic flow to node 0 at a time of 30 seconds. CBR traffic in NS-2 requires two parameters, packet size and packet rate. The size of the packet is taken as 116bytes (101 bytes of payload and 15 bytes of ZigBee header). The packet rate set for simulation is a single packet every 3.0 milliseconds, less than half of the estimated 7.0 milliseconds required for complete transmission. This is shown in fig 7.

Fig 7 - Two nodes labeled as PAN Coor and FullFunctionDevice at simulation time=0 seconds

Fig 8 - Two nodes transmitting data at simulation time=30 seconds

The tracegraph snapshots have been taken with the simulation time of 90 seconds. In fig 9, the entire simulation scenario has been displayed



Fig 9 – Simulation Information

Fig 10 – throughput versus receiving bits for single-hop

The throughput of receiving bits is obtained as 130kbps.

Here the simulation time is taken as 90 seconds this is because when we take a small value (less than 10 seconds) of the simulation time proper tracegraph is not obtained i.e. there might only be some peaks not a proper wave this will lead to misinterpretation of result (sometimes there is a huge delay while transmitting data. If we have taken less simulation time there might be probability that the data transmission occurs after that simulation time and we obtain no result.).

The result can be better interpreted when simulation time is taken large (approximately greater than 1000 seconds). But because the increase in the simulation time leads to generation of a very large trace file and it will



take a large amount of time to load the trace file in the tool Tracegraph 2.02. Hence to avoid such conditions we take a reasonable simulation time.

The drop in the graph is because of the disassociation of node and/or packet drops. Sometimes we can see that there is certain fall in the number of packets received because of the disassociation of nodes, collision between data at a single node which results in packet drop.

5. NS-2 Simulations for Multi-hop

As discussed in section 4, the upper bound on throughput for 3-hop is given as

Tm (n) = Ts / n,

In section V, the throughput for single-hop is achieved as 130kbps. So the throughput for 3-hop is

Tm (4) = 1 * Tm (1) = 40 kbps

3

Fig 11 shows the simulation of 4 nodes i.e. 3-hop transmission. The simulation time is taken as 100 seconds. Node 0 is labeled as PAN Coordinator.

Fig 11- Data Transmission between 4 nodes

The throughput can be obtained by tracegraph. This can be shown as

Fig 12 – Throughput of Receiving Bits versus simulation time for 3-hop



Throughput for 4 nodes i.e. 3-hop transmission is 30 kbps. This value is approximately 1/3 of the throughput of single-hop transmission.

According to Section IV the throughput for single hop is 130 kbps and for n hop transmission throughput is 1/n of single hop. Throughput for 4 nodes i.e. 3-hop transmission is 30 kbps. This value is approximately 1/3 of the throughput of single-hop transmission and achieved value is 30 kbps approximately. The peaks show the data transmission when nodes are in association with PAN coordinator and no packets are dropped. The sudden fall in graph indicates packets dropping and/or node(s) getting dissociated from PAN coordinator.

The simulation has also been done when there are 8 nodes. The nodes are transmitting as 2 sets of 4 nodes each transmitting as 3-hop transmission. In this case, when only nodes 4, 5, 6, and 7 (where node 4 is labeled as PAN Coordinator) are in active state and nodes 0 to 3 are inactive then we obtained the throughput as 30 kbps. The simulation time in this case is also 100 seconds.

The simulation has also been done when both sets of nodes i.e. 0-3 and 4-7 are transmitting the result obtained can be shown below. Here both nodes 0 and 4 are PAN Coordinators. The simulation time in this case is also 100 seconds.

Fig 13 - Data Transmission between Nodes 0-3 and nodes 4-7



The tracegraph is shown in fig 14.

Fig 14 – throughput of receiving bits versus simulation time for 8 nodes

Throughput for 8 nodes when used as a set of two 3-hop transmission (node 0 to node 3, node 4 to node 7) if all the two sets are transmitting simultaneously is approximately 30kbps. This value is same as that for a single 3-hop transmission.

Fig 15 - Throughput of Receiving Bits versus Simulation Time for 8 nodes in sets of 4 nodes each set independent of transmission range of

other set



The throughput in this case is also 30kbps.

This graph is different than that of the previous graph in fig. 14. It can be see that in this graph there seems to be more amount of data flow after simulation time of 70 seconds as compared to previous graph. This indicates more data transmission and less packet loss.

This occurs because here the two sets of four nodes are out of each other’s transmission range. And because of this, the two PAN Coordinators work independently without interrupting each other’s data transmission. This leads to less collisions resulting in increased data flow.

The simulation is also done when there are 12 nodes. The nodes are transmitting as 3 sets of 4 nodes each transmitting as 3-hop transmission. In this case, when only nodes 8, 9, 10, and 11 (where node 8 is labeled as PAN Coordinator) are in active state and nodes 0 to 3 & nodes 4 to 7 are inactive then we obtained the throughput as 30 kbps. The simulation time in this case is also 100 seconds.

The simulation is also done when all 3 sets of nodes i.e. 0-3, 4-7 and 8-11 are transmitting the result obtained can be shown below. Here nodes 0, 4 and 8 are PAN Coordinators. The simulation time in this case is also 100 seconds.

Fig 16 - Data Transmission between Nodes 0-3, nodes 4-7 and nodes 8-11



The tracegraph is shown in figure 17 for 100 seconds of simulation time

Fig 17 - Throughput of Receiving Bits versus Simulation Time for 12 nodes

Throughput for 12 nodes when used as a set of three 3-hop transmission (node0 to node3, node4 to node7, node8 to node11) if all the three sets are transmitting simultaneously is approximately 30kbps.This value is same as that for a single 3-hop transmission. This indicates that whether it can be any number of nodes, the transmission occurs between 4 nodes i.e. 3-hop without interrupting the transmission of other nodes at a time.

6. NS-2 Simulation for 10 Nodes

Here all the 10 nodes are active. The simulation time is 100 seconds. Node 0 is the PAN Coordinator. The simulation is shown as

Fig 18 - Data Transmission between nodes 0 and 9



The tracegraph for 100 seconds is shown in fig 19.

Fig 19 – Throughput of receiving bits versus simulation time for 10 nodes in active state

Throughput for 10 nodes i.e. 9-hop transmission if all the 10 nodes are in working stage is 14 kbps. This value is approximately 1/9 of the throughput of single-hop transmission. Here the simulation time is taken as 100 seconds this is because when we a small value (less than 40 seconds) of the simulation time is taken, a proper tracegraph is not obtained i.e. there might only be some peaks not a proper wave this will lead to misinterpretation of result (sometimes there is a huge delay while transmitting data. If we have taken less simulation time there might be probability that the data transmission occurs after that simulation time and we obtain no result.).

The result can be better interpreted when simulation time is taken large (approximately greater than 1000 seconds). But because the increase in the simulation time leads to generation of a very large trace file and it will take a large amount of time to load the trace file in the tool Tracegraph 2.02. Hence to avoid such conditions a reasonable simulation time is taken.

The peaks in the graph are because of the association of node transmitting data. Sometimes we can see that there is certain fall in the number of packets received because of the disassociation of nodes, collision between data at a single node which results in packet drop.

The throughput in this case is very less. This is because of the increase in the collisions and delay. This also leads to loss of information. The throughput for the last case in section 6 is 30 kbps for 12 nodes and it is 14 kbps for 10 nodes. This is because while simulating the 10 nodes in series it takes a large amount of time to activate all the nodes. In this case all the 10 nodes are activated at simulation time of 40 seconds. This delay leads to decrease in throughput.

7. Summary and Conclusion

We limited our study only to calculate the maximum throughput. The throughput for 10 active nodes is achieved as 14 kbps. The comparison of throughput by varying the number of hops is shown in fig 20.

Fig 20 - Throughput of Receiving bits versus Number of Hops

0

20

40

60

80

100

120

140

1 2 3 4 5 6 7 8 9

throughput(kbps)

throughput(kbps)



By increasing the number of nodes the throughput decreases. This is because of the increase in the collisions and delay. This also leads to loss of information.

This throughput can further be increased by 1) using multi-channel approach for multi-hop, 2) using pipelining [1], 3) disabling CCA.

We can also measure certain other parameters despite of throughput. These can be 1) latency, 2) delivery order,

3) delay.

References [1] Fredrik Osterlind and Adam Dunkels “Approaching the Maximum 802.15.4 Multi-hop Throughput ” ,March 2008. [2] Institute of Electrical and Electronics Engineers, Inc., IEEE Std. 802.15.4- 2003, IEEE Standard for Information Technology —

Telecommunications and Information Exchange between Systems — Local and Metropolitan Area Networks — Specific Requirements — Part 15.4: Wireless Medium Access Control (MAC) and Physical Layer (PHY) Specifications for Low Rate Wireless Personal Area Networks (WPANs). New York: IEEE Press. 2003.

[3] Institute of Electrical and Electronics Engineers, Inc., IEEE Std. 802.15.4-2006, IEEE Standard for Information Technology — Telecommunications and Information Exchange between Systems — Local and Metropolitan Area Networks — Specific Requirements — Part 15.4: Wireless Medium Access Control (MAC) and Physical Layer (PHY) Specifications for Low Rate Wireless Personal Area Networks (WPANs). New York: IEEE Press. 2006.

[4] Jianliang Zheng and Myung J. Lee “A Comprehensive Performance Study of IEEE 802.15.4”. [5] Kevin Fall and Kannan Varadhan “The ns Manual”, Nov 2011. [6] Marc Greis.Ns Tutorial. http://www.isi.edu/nsnam /ns/tutorial/index.html [7] Ns2 tutorial-http://www.isi.edu/nsnam/ns [8] Sureshbabu Ramalingam, Subramanian Srinivasan, Lakshmikanth Yerradoddi “Performance analysis of Routing protocols in wireless

mesh and ad hoc networks”. [9] T. Ryan Burchfield and S. Venkatesan and Douglas Weiner “Maximizing Throughput in ZigBee Wireless Networks through Analysis,

Simulations and Implementations”. [10] Teerawat Issariyakul and Ekram Hossain “Introduction to Network Simulator NS2”, Springer. [11] Tracegraph http://www.tracegraph.com/download.html [12] ZigBee Alliance Document 053474r06: ZigBee Specification, December 2004



MAXIMUM THROUGHPUT OF 10 NODES FOR IEEE 802.15.4 WPAN

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