1. Muhammad Mahmudul Islam Ronald Pose Carlo Kopp School of Computer Science & Software Engineering Monash University, Australia

2. 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 paper Extend our previous work

3. 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

4. Topics Covered SAHN Challenges Solution Simulation results

5. 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

6. 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 802.11e 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

7. Challenges (1/9) How to support QoS for real-time traffic? Prevent network saturation Why? (Explaining next)

8. 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

9. 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

10. 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)

11. 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

12. 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

13. 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

14. 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  128.724% (72.619 + 56.105)

15. Challenges (9/9) At each α measure U for itself (U s α ) and for neighboring ρ (U s ρ ) At each ρ measure U for itself (U s ρ ) 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?

16. Challenges (9/9) 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? Solution in previous work At each α measure U for itself (U s α ) and for neighboring ρ (U s ρ ) At each ρ measure U for itself (U s ρ ) Measure BW without choking ongoing sessions

17. Challenges (9/9) 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? In this work At each α measure U for itself (U s α ) and for neighboring ρ (U s ρ ) At each ρ measure U for itself (U s ρ ) Measure BW without choking ongoing sessions

18. Solution (1/14) Basics of the analytical model

19. Solution (2/14) 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 U s A = U s B = U s(b)

20. Solution (3/14) Other case (3 nodes) U s A = U s B = U s C = 2 x U s(b)

21. Solution (4/14) 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)

22. Solution (5/14) 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)

23. Solution (6/14) 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)

24. Solution (7/14) 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 for U s α

25. Solution (9/14) Generalized form of the analytical model for U s ρ (1/6) Consider the session s between A  D s consists of a single data packet Total 3 transactions on links AB, BC & CD during T 1 -T 2, T 2 -T 3 & T 3 -T 4 respectively E-K are passive participants of s

26. Solution (9/14) T 1 -T 2 : U s E = U s F = U s G = U s H = U s(b) Generalized form of the analytical model for U s ρ (2/6)

27. Solution (9/14) T 1 -T 2 : U s E = U s F = U s G = U s H = U s(b) T 2 -T 3 : U s G = U s H = U s I = U s(b) Generalized form of the analytical model for U s ρ (3/6)

28. Solution (9/14) T 1 -T 2 : U s E = U s F = U s G = U s H = U s(b) T 2 -T 3 : U s G = U s H = U s I = U s(b) T 3 -T 4 : U s H = U s I = U s J = U s K = U s(b) Generalized form of the analytical model for U s ρ (4/6)

29. Solution (9/14) T 1 -T 2 : U s E = U s F = U s G = U s H = U s(b) T 2 -T 3 : U s G = U s H = U s I = U s(b) T 3 -T 4 : U s H = U s I = U s J = U s K = U s(b) T 1 -T 4 : U s E = 1  U s(b) U s F = 1  U s(b) U s G = 2  U s(b) U s H = 3  U s(b) U s I = 2  U s(b) U s J = 1  U s(b) U s K = 1  U s(b) Generalized form of the analytical model for U s ρ (5/6)

30. Solution (9/14) The relationship between U s G & U s(b) depends on the number of links carrying the same packet for s Similar relation holds for other passive participants too T 1 -T 4 : U s E = 1  U s(b) U s F = 1  U s(b) U s G = 2  U s(b) U s H = 3  U s(b) U s I = 2  U s(b) U s J = 1  U s(b) U s K = 1  U s(b) Generalized form of the analytical model for U s ρ (7/8)

31. Solution (9/14) Therefore we can write U s α/ρ = n  U s(b) n = number of links an α/ρ hears carrying the same data packet for s Generalized form of the analytical model for U s ρ (8/8)

32. Solution (10/14) Each active participant has to measure U s α = n  U s(b) U s ρ = n  U s(b) Each passive participant has to measure U s ρ = 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

33. 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

34. Solution (12/14) 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 Estimate n of U s α by an α, i.e. an α is estimating n for itself

35. Solution (13/14) Let ρ be a passive participant of α  Initialize n to 0  Makes a list of all APs within 2-hop radius of ρ 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 = α i & ρ is located as in Fig (1) & the TX range of α i α i+1 (α i+1 α i ) reaches ρ  α = α i & ρ is a 2-hop neighbor as in Fig (2) & the TX range of α i+1 α 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 U s ρ by an α, i.e. an α is estimating n for its passive neighbors

36. Solution (14/14) Let ρ be a passive participant overhearing a SIREQ  Initialize n to 0  Makes a list of all APs within 2-hop radius of ρ The list  (α 1, 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 i α i+1 (α i+1 α i ) reaches ρ  α i is a 2-hop neighbor of ρ as in Fig (2) & the TX range of α i+1 α 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 U s ρ by a ρ, i.e. a ρ is estimating n for itself

37. Simulation Setup  30 nodes on a 1500 × 1500 square meters flat terrain  Each node had at most 6 neighbors  Each simulation run consisted of at most 12 sessions  Each session was offered a load of 1 Mbps  A new session was added every 2 sec  Simulation time for each test case was 50 sec  For simplicity all sessions were of the same AC  All established sessions were executed till the end of the simulation run  Path length of each route in each test case was fixed  Path lengths among various test cases varied between 2-6  The avg values of performance metrics were recorded at 1 sec interval For performance evaluation

38. Simulation Result (1/5) SAHN-MAC & 802.11e have been compared with respect to  End-to-end delay  Throughput  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

39. Simulation Result (2/5) Path length = 5 Fairly stable Not stable

40. Simulation Result (3/5) Path length = 5 Not stable Fairly stable

41. Simulation Result (4/5) Path length = 5 Not stable Fairly stable

42. Simulation Result (5/5)  Additions of new sessions increased network load  802.11e cannot stop the network from overloading since it does not have any admission control mechanism  SAHN-MAC did not allow any session to initiate if the new session could choke ongoing sessions  Thus SAHN-MAC maintains fairly stable network performance compared to 802.11e Summary of performance evaluation

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

44. Questions Thank you