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 We explain the challenges & Provide a solution with respect to a SAHN using IEEE e operating in EDCA mode

SAHN: Suburban Ad-Hoc Network EDCA: Enhanced Distributed Channel Access, an improved version of DCF (Distributed Coordination Function) of legacy SAHN-MAC: EDCA of e + Proposed protocol

Topics Covered SAHN Challenges Solution Simulation results

SAHN (Suburban Ad-Hoc Network)  Multi-hop ad-hoc network  Ideal for cooperative nodes, e.g. connecting houses and business  Topology is quasi-static  Uses wireless technology  Multi-hop QoS routing  Decentralized  Multi Mbps broadband service  No charges for SAHN traffic  Can run alongside TCP/IP  Conceived by Ronald Pose & Carlo Kopp in 1997 at Monash University, Australia at Monash University, Australia

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 m  Use EDCA of IEEE e in the link layer  Physical layer uses OFDM with a  Physical layer operates at the TX rate of 54 Mbps  Session consists of UDP type CBR traffic  Routing is done with DSR Default setup

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

Challenges (2/9) Effect of saturation in network performance (1/2) For 512 bytes payload, max achievable throughput between A  E  5.2 Mbps Establish a 2.6 Mbps session between A  E 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

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

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 U A  18 % & U B  18 % 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 A  E, Each packet = 512 bytes 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 G  K U of some of the nodes exceed their working limits E.g. U C has to be  % ( )

Challenges (9/9) At each α measure U for itself and for neighboring ρ At each ρ measure U for itself 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? Measure BW without choking ongoing sessions

Challenges (9/9) At each α measure U for itself and for neighboring ρ At each ρ measure U for itself 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? Measure BW without choking ongoing sessions

Solution (1/10) Basics of the analytical model

Solution (2/10) U s A = U s B = U s(b) Base case (it consists of 2 nodes) U s A  U of session s at node A U s(b)  base case U of session s

Solution (3/10) Other case (3 nodes) U s A = U s B = U s C = 2 x U s(b)

Solution (4/10) Other case (4 nodes) U s A = U s D = 2 x U s(b) U s B = U s C = 3 x U s(b)

Solution (5/10) Other case (5 nodes) U s A = U s E = 2 x U s(b) U s B = U s D = 3 x U s(b) U s C = 4 x U s(b)

Solution (6/10) Other case (6 nodes) U s A = U s F = 2 x U s(b) U s B = U s E = 3 x U s(b) U s C = U s D = 4 x U s(b)

Solution (7/10) We can infer U s α depends on the number of transactions α can hear transferring the same data packet for s Each transaction involves a specific link that joins the TX and the RX active participants 5 Nodes U s A = U s E = 2 x U s(b) U s B = U s D = 3 x U s(b) U s C = 4 x U s(b) 6 Nodes U s A = U s F = 2 x U s(b) U s B = U s E = 3 x U s(b) U s C = U s D = 4 x U s(b) 2 Nodes U s A = U s B = U s(b) 3 Nodes U s A = U s B = U s C = 2 x U s(b) 4 Nodes U s A = U s D = 2 x U s(b) U s B = U s C = 3 x U s(b) Generalized form U s α = n  U s(b) n = number of links an α hears carrying the same data packet for s Generalized form of the analytical model

Solution (8/10) Each active participant has to measure U s α = n  U s(b) A session initialization request packet (SIREQ) is sent before a session starts SIREQ contains Throughput requirement List of active participants in the route Estimate U s(b)  from Info_1 Estimate the value of n What to measure? ……….Info_1 ……….Info_2 ………Trivial ………Explaining next

Solution (9/10) 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 (10/10) Let α be the active participant estimating the value of n  Initialize n to 0  Makes a list of all active participants (APs) within 2-hop radius. The list  (α 1, α 2 ….. α k ) including α  For 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 Algorithm to estimate n of U s α by an α

Simulation Result (1/3) We have validated the correctness of the proposed bandwidth estimation scheme using some simple test cases in 2 phases

Simulation Result (2/3) Phase1  Each test case dealt with a single session  All nodes were arranged in a straight line  Each session was established between end nodes  Each node could be within the TX range of at most 2 other nodes  For various test cases  the number of nodes varied between 2-11  traffic load of varied between 45 Kbps - 8 Mbps  We have logged U of each node every 1 min & averaged them at the end Result SAHN-MAC estimated U at each active participant with  93% accuracy

Simulation Result (3/3) Phase2  70 nodes placed in a 3000 m by 3000 m flat terrain  Each node had at most 6 neighbors  Each test case consisted of 20 sessions with same path length & AC but with different src & dst pairs  Path length varied various test cases between 3-5  We have logged U of each node every 1 min & averaged them at the end Result SAHN-MAC estimated U at each active participant with  85% accuracy Why SAHN-MAC is less accurate in Phase 2? At present SAHN-MAC does not consider passive participants for estimating utilization of each session

Conclusion  Discussed the challenges for supporting deterministic QoS for real-time traffic in multi-hop ad-hoc networks  Provided a solution  We are extending SAHN-MAC to estimate U s ρ and measure performance  We would like to  extend SAHN-MAC for multiple frequency channels & directional antennas  build a scheduling scheme at the MAC layer to handle different classes of traffic efficiently

Questions Thank you