Muhammad Mahmudul Islam

Ronald Pose

Carlo Kopp

School of Computer Science & Software Engineering

Monash University, Australia

Problem Statement

Supporting real-time traffic in multi-hop ad-hoc network (e.g. a SAHN) with a contention based MAC protocol is a challenging task

In a previous paper in WOCN 2005

We have explained the challenges

&

Provided a solution with respect to a SAHN

using IEEE 802.11e operating in EDCA mode

In this paperExtend our previous work

SAHN: Suburban Ad-Hoc Network

EDCA: Enhanced Distributed Channel Access, an improved version of DCF (Distributed Coordination Function) of legacy 802.11

SAHN-MAC: EDCA of 802.11e + Proposed protocol

Topics Covered

SAHN

Challenges

Solution

Simulation results

SAHN (Suburban Ad-Hoc Network)

Multi-hop ad-hoc networkMulti-hop ad-hoc network Ideal for cooperative nodesIdeal for cooperative nodes,, ee.g. .g. connecting houses and businessconnecting houses and business Topology is quasiTopology is quasi--staticstatic Uses wireless technologyUses wireless technology Multi-hop QoS routing Multi-hop QoS routing DecentralizedDecentralized Multi MbpsMulti Mbps broadband broadband

serservicevice No charges for No charges for

SAHN trafficSAHN traffic Can run alonCan run alongside gside

TCP/IPTCP/IP Conceived byConceived by Ronald PoseRonald Pose && Carlo KoppCarlo Kopp inin 1997 1997 atat Monash University, A Monash University, Australiaustralia

Simulation Setup

Used GloMoSim (version 2.02)Nodes are separated by at most 240 meters, Nodes use same TX power with a TX range of 240 mUse EDCA of IEEE 802.11e in the link layerPhysical layer uses OFDM with a Physical layer operates at the TX rate of 54 MbpsSession consists of UDP type CBR trafficRouting is done with DSR

Default setup

Challenges (1/9)How to support QoS for real-time traffic?

Prevent network saturation

Why? (Explaining next)

Challenges (2/9) Effect of saturation in network performance (1/2)

B C D EA

For 512 bytes payload, max achievable throughput between AE 5.2 Mbps

Establish a 2.6 Mbps session between AE

Since below saturation

Achieved throughput = 2.6 Mbps

End-to-end delay = 0.9 ms

Challenges (3/9) Effect of saturation in network performance (2/2)

Throughput degraded by 35% (1.7 Mbps)

End-to-end delay increased by 550% (559 ms)At over saturation

1000

0.1

1

10

100

1

2

3

4

5

0

En

d-t

o-en

d d

elay

(m

s)

Th

rou

ghp

ut

(Mb

ps)

End-to-end delay

Throughput

Unsaturated[2.6 + 1.0]

NearSaturation[2.6 + 2.4]

Saturated[2.6 + 2.7]

OverSaturated[2.6 + 4.1]

Initial Load Added Load

Challenges (4/9)

Prevent network saturation

How?

Prevent adding new sessions if they saturate the network

How?

Reserve bandwidth

Bandwidth reservation for multi-hop ad-hoc network with contention based MAC protocol is not trivial

Why? (Explaining next)

Challenges (5/9) Throughput

Amount of data carried from one node to another in a given time period

Expressed in bps

Associated with the application layer

Bandwidth

Bandwidth and throughput are same at the application layer

For adding overheads of different layers

BW at the physical layer > Throughput

Bandwidth Utilization (U) = 100 %Bandwidth Consumed

Total Bandwidth

Challenges (6/9)

Adding

NW & MAC headers & RTS/CTS/ACK overheads

UA 18 % & UB 18 %

BA

B C D EA

Establish a 3.4 Mbps session between end nodes

Each packet = 512 bytes

Effect of multiple hops on U

Challenges (7/9)

Establish a 3.4 Mbps session between AE, Each packet = 512 bytes

B C D EA

G H I JF K Active Participant (α)

Passive Participant (ρ)

U of neighbors may be wasted

Challenges (8/9) Why we need to know U of passive participants?

Add another 3.4 Mbps

session GK

U of some of the nodes exceed their working limits

E.g. UC has to be

128.724%

(72.619 + 56.105)

B C D EA

G H I JF K

Challenges (9/9)

At each α measure U for itself (Usα) and for neighboring ρ (Us

ρ)

At each ρ measure U for itself (Usρ)

How?

How?

Support QoS for real-time traffic

Do not allow new session if it causes the network to get saturated

Allocate BW before establishing a session

Measure BW without choking ongoing sessions

How?

How?

Challenges (9/9)

How?

How?

Support QoS for real-time traffic

Do not allow new session if it causes the network to get saturated

Allocate BW before establishing a session

How?

How? Solution in previous work

At each α measure U for itself (Usα) and for neighboring ρ (Us

ρ)

At each ρ measure U for itself (Usρ)

Measure BW without choking ongoing sessions

Challenges (9/9)

How?

How?

Support QoS for real-time traffic

Do not allow new session if it causes the network to get saturated

Allocate BW before establishing a session

How?

How? In this work

At each α measure U for itself (Usα) and for neighboring ρ (Us

ρ)

At each ρ measure U for itself (Usρ)

Measure BW without choking ongoing sessions

Solution (1/14) Basics of the analytical model

RTS CTS

DATA ACK

A B C D

Interference Zone T1-T2

Step 1 starts at T1 Step 2

Step 3 Step 4 finishes at T2

A single transaction

A B C D A B C D

A B C D A B C D

Solution (2/14)

Base case (it consists of 2 nodes) UsA U of session s at node A

Us(b) base case U of session s

UsA= Us

B = Us(b)

ACKDATACTSRTS

ACK

CTSDATA

RTS

UsB = Us (b)Us

A = Us (b)

A B

Solution (3/14)

Other case (3 nodes)

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

UsB = 2 x Us (b)

UsC = 2 x Us (b)

UsA = 2 x Us (b)

A B C UsA= Us

B = UsC = 2 x Us(b)

Solution (4/14)

Other case (4 nodes)

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

UsB = 3 x Us (b)

UsC = 3 x Us (b)

UsA = 2 x Us (b)

UsD = 2 x Us (b)

A B C D

UsA= Us

D = 2 x Us(b)

UsB= Us

C = 3 x Us(b)

Solution (5/14)

Other case (5 nodes)

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

UsB = 3 x Us (b)

UsC = 4 x Us (b)

UsA = 2 x Us (b)

UsD = 3 x Us (b)

UsE = 2 x Us (b)

A B C D E

UsA= Us

E = 2 x Us(b)

UsB= Us

D = 3 x Us(b)

UsC = 4 x Us(b)

Solution (6/14)

Other case (6 nodes)

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

ACKDATACTSRTS

ACK

CTSDATA

RTS

UsB = 3 x Us (b)

UsC = 4 x Us (b)

UsA = 2 x Us (b)

UsD = 4 x Us (b)

UsE = 3 x Us (b)

A B C D E

ACKDATACTSRTS

ACK

CTSDATA

RTS

F

UsF = 2 x Us (b)

UsA= Us

F = 2 x Us(b)

UsB= Us

E = 3 x Us(b)

UsC= Us

D = 4 x Us(b)

Solution (7/14)

We can infer Us

α depends on the number of transactions α can hear transferring the same data packet for s

Each transaction involvesa specific link that joins the TX and the RX active participants

5 Nodes

UsA= Us

E = 2 x Us(b)

UsB= Us

D = 3 x Us(b)

UsC = 4 x Us(b)

6 Nodes

UsA= Us

F = 2 x Us(b)

UsB= Us

E = 3 x Us(b)

UsC= Us

D = 4 x Us(b)

2 Nodes

UsA= Us

B = Us(b)3 Nodes

UsA= Us

B = UsC = 2 x Us(b)

4 Nodes

UsA= Us

D = 2 x Us(b)

UsB= Us

C = 3 x Us(b)

Generalized formUs

α= n Us(b) n = number of links an α hears carrying the same data packet for s

Generalized form of the analytical model for Usα

Solution (9/14) Generalized form of the analytical model for Us

ρ (1/6)

G H I JF

A B C DE K

T1 T2 T3 T4

Consider the session s between AD

s consists of a single data packet

Total 3 transactions on links AB, BC & CD during T1-T2, T2-T3 & T3-T4 respectively

E-K are passive participants of s

Solution (9/14)

G H I JF

A B C DE K

T1 T2

T1-T2: UsE = Us

F = UsG = Us

H = Us(b)

Generalized form of the analytical model for Usρ (2/6)

Solution (9/14)

G H I JF

A B C DE K

T2 T3

T1-T2: UsE = Us

F = UsG = Us

H = Us(b)

T2-T3: UsG = Us

H = UsI = Us(b)

Generalized form of the analytical model for Usρ (3/6)

Solution (9/14)

G H I JF

A B C DE K

T3 T4

T1-T2: UsE = Us

F = UsG = Us

H = Us(b)

T2-T3: UsG = Us

H = UsI = Us(b)

T3-T4: UsH = Us

I = UsJ = Us

K = Us(b)

Generalized form of the analytical model for Usρ (4/6)

Solution (9/14)

G H I JF

A B C DE K

T1 T2 T3 T4

T1-T2: UsE = Us

F = UsG = Us

H = Us(b)

T2-T3: UsG = Us

H = UsI = Us(b)

T3-T4: UsH = Us

I = UsJ = Us

K = Us(b)

T1-T4: UsE = 1

Us(b)

UsF = 1

Us(b)

UsG = 2 Us(b) Us

H = 3 Us(b)

UsI = 2 Us(b) Us

J = 1 Us(b) UsK = 1 Us(b)

Generalized form of the analytical model for Usρ (5/6)

Solution (9/14)

G H I JF

A B C DE K

T1 T2 T3

The relationship between UsG & Us(b) depends on the number of links

carrying the same packet for s

Similar relation holds for other passive participants too

T1-T4: UsE = 1

Us(b)

UsF = 1

Us(b)

UsG = 2 Us(b) Us

H = 3 Us(b)

UsI = 2 Us(b) Us

J = 1 Us(b) UsK = 1 Us(b)

Generalized form of the analytical model for Usρ (7/8)

Solution (9/14)

Therefore we can write

Usα/ρ= n Us(b)

n = number of links an α/ρ hears carrying the same data packet for s

Generalized form of the analytical model for Usρ (8/8)

Solution (10/14)

Each active participant has to measure Us

α= n Us(b) Us

ρ= n Us(b)

Each passive participant has to measureUs

ρ= n Us(b)

A session initialization request packet (SIREQ) is sent before a session starts

SIREQ containsThroughput requirement

List of active participants in the route

EstimateUs(b) from Info_1

Estimate the value of n

What to measure?

……….Info_1……….Info_2

………Trivial ………Explaining next

Solution (11/14)

Let NH represent a neighbor of each node within 2–hop radius

Each node knows the following information about NH(1) NH’s geographical location(2) list of the NH’s neighbors and NH’s neighbors’ geographical locations(3) transmission ranges assigned to NH with its neighbors

Info_2 + Info_3

Estimate n

Estimate the value of n

Algorithm

……….Info_3

Solution (12/14)

Let α be the active participant estimating the value of n

Initialize n to 0Makes a list of all active participants (APs) within 2-hop radius. The list (α1, α 2….. α k) including αFor each α i (α 1, α 2….. α k), add 1 to n if

α i = α

α i is neighbor as in Fig (1)

α i is a 2-hop neighbor as in Fig (2) & the TX range of α i+1 α i reaches α

α i is a neighbor as in Fig (3) & the TX range of α i α i+1 reaches α

Estimate n of Usα by an α, i.e. an α is estimating n for itself

αiαi+1 α

RTS, DATA

CTS, ACK CTS, ACK

αi+1αiαRTS, DATA

CTS, ACKRTS, DATA

αi αRTS, DATA

CTS, ACK

(1)

(2)

(3)

Solution (13/14)

Let ρ be a passive participant of α

Initialize n to 0Makes a list of all APs within 2-hop radius of ρ The list (α1, α 2….. α k) including αFor each α i (α 1, α 2….. α k), add 1 to n if

α = α i & ρ is located as in Fig (1) & the TX range of α iα i+1 (α i+1α i) reaches ρ

α = α i & ρ is a 2-hop neighbor as in Fig (2) &the TX range of α i+1α i reaches ρ

α = α i & ρ is a 1-hop neighbor as in Fig (3) & the TX range of α i α i+1 reaches ρ

Estimate n of Usρ by an α, i.e. an α is estimating n for its passive neighbors

αi αi+1

ρ

RTS, DATA

CTS, ACK

CTS, ACK

αi+1αi

ρRTS, DATA

CTS, ACK

RTS, DATA

αi αi+1

RTS, DATA

CTS, ACK

(1)

(2)

(3)

ρCTS, ACKRTS, DATA

Solution (14/14)

Let ρ be a passive participant overhearing a SIREQ

Initialize n to 0Makes a list of all APs within 2-hop radius of ρ The list (α1, α 2….. α k)For each α i (α 1, α 2….. α k), add 1 to n if

ρ is located as in Fig (1) & the TX range of α iα i+1 (α i+1α i) reaches ρ

α i is a 2-hop neighbor of ρ as in Fig (2) &the TX range of α i+1α i reaches ρ

α i is a 1-hop neighbor as in Fig (3) & the TX range of α i α i+1 reaches ρ

Estimate n of Usρ by a ρ, i.e. a ρ is estimating n for itself

αi αi+1

ρ

RTS, DATA

CTS, ACK

CTS, ACK

αi+1αi

ρRTS, DATA

CTS, ACK

RTS, DATA

αi αi+1

RTS, DATA

CTS, ACK

(1)

(2)

(3)

ρCTS, ACKRTS, DATA

Simulation Setup

30 nodes on a 1500 × 1500 square meters flat terrainEach node had at most 6 neighborsEach simulation run consisted of at most 12 sessionsEach session was offered a load of 1 MbpsA new session was added every 2 secSimulation time for each test case was 50 secFor simplicity all sessions were of the same ACAll established sessions were executed till the end of the simulation runPath length of each route in each test case was fixedPath lengths among various test cases varied between 2-6The avg values of performance metrics were recorded at 1 sec interval

For performance evaluation

Simulation Result (1/5)

SAHN-MAC & 802.11e have been compared with respect to End-to-end delayThroughput Delivery ratio

Delivery ratio is the percentage of data received successfully at the final destination

With the given configurations both SAHN-MAC & 802.11e performed similarly up to path length 3

Simulation Result (2/5)

Path length = 5

Fairly stable

Not stable

Simulation Result (3/5)

Path length = 5

Not stable

Fairly stable

Simulation Result (4/5)

Path length = 5

Not stable

Fairly stable

Simulation Result (5/5)

Additions of new sessions increased network load802.11e cannot stop the network from overloading since it does not have any admission control mechanismSAHN-MAC did not allow any session to initiate if the new session could choke ongoing sessionsThus SAHN-MAC maintains fairly stable network performance compared to 802.11e

Summary of performance evaluation

Conclusion

Extended initial SAHN-MAC by considering neighboring nodes in bandwidth calculationSimulation results show SAHN-MAC can support fairly deterministic QoS which important for real-time trafficAt present we are extending SAHN-MAC for multiple frequency channels & directional antennasWe would also like to build a scheduling scheme at the MAC layer to handle different classes of traffic efficiently

Questions

Thank you