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THESIS PRESENTATION

THESIS PRESENTATION. Radio Placement Algorithm and Dynamic Channel Allocation for Large Area Community Networks. By: Prashant Sharma Supervisor: Dr. Bhaskaran Raman. Community Network. What are community networks? What is Large Area Community Network?.

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THESIS PRESENTATION

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  1. THESIS PRESENTATION Radio Placement Algorithm and Dynamic Channel Allocation for Large Area Community Networks By: Prashant Sharma Supervisor: Dr. Bhaskaran Raman

  2. CommunityNetwork • What are community networks? • What is Large Area Community Network? Characteristics of Large Area Community Network Splitter Directional Antenna Single Radio Tree Topology (high quality links) (IITK Network) Cost of Radio Channel Allocation Algorithm NEXT

  3. Landline node is the node which provides services to other nodes. Node represented by IITK is the landline node in the network shown above. BACK

  4. Two–way Splitter BACK

  5. 1 3 2 Problems Addressed • Radio Placement Algorithm (RPA) • Dynamic Channel Allocation (DCA) Need of RPA ?? How to solve this problem ?? Interference 1 1 Adding New Radios

  6. Addition of new radio is not Trivial Local Optimal Solution: Node 4 Best Solution: Node 5 What needs to be done ?? See the Global Effect of connecting a new Radio

  7. Need of DCA Static Channel Allocation: Leads to Contention in some cases DCA: Effective and Efficient use of available channels

  8. Comparison with present Work Radio Placement Algorithm: Paper ‘Centralized channel assignment and routing algorithm for wireless mesh networks’ gives channel allocation algorithm, when all the nodes are equipped with equal k number of radios. We consider a general scenario in which network can have any number of radios. We not only finds the position of new radio but also determines channel allocation after addition of a new radio. Dynamic Channel Allocation All the existing community networks uses static channel allocation scheme. In comparison to present state of art static channel allocation scheme, we proposes a DCA scheme to use available channels in best possible way to minimize contention in the network.

  9. Design and Evaluation Methodology Design of RPA Revolves around a Contention Based Heuristic obtained by analysis of Simulations Target Metric Traffic Pattern Used Simulator used TeNs CBR UDP Throughput (The Enhanced Network Simulator) NEXT

  10. The Enhanced Network Simulator • Modification of ns-2 • Various Functionalities added in ns-2 are: • Support of Directional antenna (Antenna Radiation Pattern). • Support of 11 channels. • Support of multiple interfaces at a node. However, no support for splitter is present. … + Final Radiation Pattern Modification Done + Radiation pattern of antenna 1 Antenna 2 Antenna k Modification is done in following files: a) dir-antenna.c b) antenna.h c) wirelss-phy.c BACK

  11. Antenna Radiation Pattern BACK

  12. Why CBR UDP ?? 0 Rate Limiting Mechanism or Fair Queuing Mechanism 1 2 To Solve this problem 3 4 TCP : Not Possible UDP : Possible We uses, Equal Rate Traffic Pattern Each node sends ((T=6Mbps) * (Number of Channels)) / (number of nodes) Mbps of traffic. 6Mbps is the maximum rate attainable at application level, when 11 Mbps data rate is used, after discounting PHY/MAC/IP overheads. BACK

  13. Design and Evaluation Methodology (contd..) Design of DCA: Done using the Contention based heuristic. We uses this heuristic to determine which channel to assign to the link so that contention of the network is minimum. Evaluation: RPA: Evaluation is done by comparing RPA with a naïve but optimal exhaustive search method. We will show that throughput improvement achieved by RPA compares well with that of exhaustive search method. DCA: It is based on intuitive proof. We will show that no possible channel allocation can further increase the throughput.

  14. Preliminaries Basic assumptions and conditions behind RPA: • Radio addition leads to change in channel. • Usage of newly added radio. • Splitter Configuration. • Free Channel Available. NEXT

  15. Why channel should be changed ?? Consider the example: Simulations were carried out on both topologies. For a) UDP Traffic b) TCP Traffic TCP traffic is used as there are only two nodes 2 and 0, one hop distance from the landline node.

  16. Tables showing simulation results Reason: Even after adding a radio, there is no considerable decrease in contention of network. Results with UDP traffic Results with TCP traffic

  17. Why channel should be changed ?? (Contd..) Consider the previous example with channel changed after adding new radio Simulation Results Results with UDP traffic

  18. Simulation Results (Contd ..) Results with TCP traffic Reason: Contention at node 1 is eliminated. BACK

  19. Usage of newly added radio Various possibilities of adding a new radio: a) One link from 1-2, 1-3, 1-4, 1-5 can be used b) Two links from (1-2, 1-3), (1-2, 1-4) and so on However, splitter is also needed for case b and so on. BACK In RPA, one the case when one link is made to work through new radio is considered.

  20. Splitter Configuration Link 1-3 is made to work through new radio. Other possible changes: One of the links 1-2, 1-4, 1-5 can be shifted with 1-0. Suppose link 1-5 is shifted. BACK In RPA, we do not consider these changes.

  21. Free Channel Available 1 11 In this case, link can be made to work on any channel by combining it with the radio of link working on that channel through splitter. 6 New radio added to the link IDEA: Less costly splitter can be used. However, if cheap radios are being used then merit of this assumption weakens. BACK

  22. Design of Contention Based Heuristic (RPA) • RPA uses a heuristic to decide the position of new radio. • This heuristic determines Total Contention Value of Network (TCV). (TCV determines the amount of contention present in the network) We add new radios to decrease contention in the network. With this heuristic we tries to find out the position of new radio that leads to minimum contention for the network. This position is expected to give maximum throughput for network when provided with additional radio. IDEA ?? Determine TCV?? It is the sum of Contention Value (CV) of all the nodes in the network. CV of a node is the amount of contention faced by the traffic from given node to the landline node. CV of Node

  23. Significance of CV of a node By calculating CV of the node X for different positions of adding a new radio, we determine the position P which will lead to minimum contention for the node X. This position P is expected to give maximum throughput for node X when provided with a new radio. How to determine CV of a node ?? STEP 1: We will use some topologies and will determine position of a new radio in order to maximum throughput of some node X through simulations. STEP 2: We will show conclusion drawn from the simulation results regarding the determination of CV of node x. STEP 3: We then give our heuristic to determine CV of a node. STEP 4: This contention based heuristic will be substantiated using some examples.

  24. CV of a node Procedure given in the previous slide will be used to show how to determine CV of a node first for simple topologies and then we will move ahead with complex topologies. What are simple topologies?? Topology in which each node has at most 2 links. How is it helpful?? In simple topologies, CV calculation of a node is not complex. This is because contention faced by traffic from given node is only because of the nodes present in the path of given node to the landline node. No other node is present in the network which can cause contention to traffic from the given node.

  25. CV of a node (contd..) Consider the two topologies: Topology (i): Contention faced by node 3 is only because of nodes 2, 1 and 0 or nodes present in the path of node 3 to the landline node. Same is the case with other nodes. Topology (ii): Contention faced by node 4 is because of nodes 1, 2, 3 where 2 and 3 are not present in the path of node 4 to the landline node.

  26. CV of a node (contd..) We will now determine CV of a node in simple topology. Procedure described earlier: STEP1: Radio position to maximum throughput of node 5 is to be found. Simulation is done. Traffic source is connected at node 5. UDP traffic is used. Connecting a radio at a node leads to change in channel of more than one link.

  27. CV of a node (contd..) Simulation Results: Throughput maximum for node 2, minimum for node 1. STEP 2: Topology (i) has 4 links working on same channel which is causing more contention than the Topology (ii) having one set of 3 links working on same channel and one set of 2 links working on same channel. STEP 3: Heuristic we deduce is the different in the number of links causing contention to two links. A link x causing contention to two links y and z is represented as C[x->y,z]. CV of node 5 in terms of C[x->y,z] units in topology (i) and (ii) is: C[link(3-4)->link(4-5),link(2-3)] and C[link(2-3)->link(1-2),link(3-4)]) C[link(3-4)->link(4-5),link(2-3)]

  28. Another Example There can be some topologies in which there is one link causing contention to single link. In the previous example, new radio position to maximum throughput of node 4 is to be found. Traffic source is connected at node 4. UDP traffic is send (6 Mbps). Only node 4 is taken as active node. Link 4-5 is not taken as it is not causing any contention. Possible positions of adding a new radio. Simulation results

  29. Another Example (contd..) Throughput value is maximum at node 2 and minimum at node 1. Describing CV of node 4 in terms of C[x->y,z] units for both topologies: Topology (i) C[link(2-3)->link(1-2),link(3-4)], Topology (ii) No such cases. However, throughput of node 4 is almost same in both cases. This happens because even if topology (ii) has no C[x->y,z] units but it has two links with each link causing contention to single link. A link x causing contention to single link y is represented as C[x->y]. In Topology (ii), number of C[x->y] units are: C[link(0-1)->link(1-2)] and C[link(2-3)->link(3-4)]. Throughput of node 4 is almost same in both topologies. So we deduce that amount of contention caused by one C[x->y,z] is double that of C[x->y] or C[x->y,z] = 2* C[x->y]. We will use Cdouble and Csingle to represent C[x->y,z] and C[x->y]. CV of a node is calculated in terms of Cdouble and Csingle units.

  30. Complex Topologies We will see CV calculation of a node for complex topologies. The topologies having nodes with any number of links. Even in complex topologies, if new radio position to maximize throughput of node X is to be found, then only the path of node X to landline node is taken active. So any complex topology will translate to simple topology in which each node has at most two links active. However, in reality many nodes are active. In this topology all the nodes are taken active. New radio position to maximize throughput of node 6 is to be found. In this case, not only the nodes present in the path of node 6 are affecting traffic of node 6 but also the nodes 2 and 3 (because link 1-2 and 1-3 are working on same channel as link 0-1 and 1-4).

  31. Finding new radio position through CV heuristic

  32. Simulation Results Simulation is carried out using Equal Rate Traffic Pattern (1Mbps). UDP traffic source is used. Another Example Simulation results

  33. Radio Position through Contention Based Heuristic

  34. TCV Calculation (Sum of CV of all nodes) We will first find CV of node 4. Sub graph with respect to node 4. Topology CV of node 4 = C[link(7-8)->link(5-8),link(8-6)] + C[link(2-5)->link(0-5),link(5-8)] + C[link(5-8)->link(0-5),link(6-8)] + C[link(6-9)->link(9-4)] = 3Cdouble+Csingle = 7 units TCV = CV(node 4) + CV (node 9) + CV (node 6) and so on = 7 + 9 + 6 + 6 +6 + 2 + 6 + 6 +6 = 52

  35. Radio Placement Algorithm (RPA) Comparison with Naïve but Optimal Exhaustive Search Method • Connect given new radios in linear time. • Idea is to reduce the problem of connecting more than one new radios to a given network, to a problem of connecting a single new radio one by one. So RPA uses a sequential approach. • Suppose a network with n nodes and p possible positions of adding a new radio (p is O(n)). Let say k new radios are to be added to this network. Naïve exhaustive search method will try out all possibilities i.e c(p,k). Moreover, for each position of k new radios, link of each new k radios can work on all available channels. Let ch be the number of available channels. For each possible position of c(p,k), there are chk possibilities of allocating channels. Complexity of exhaustive search method is therefore chk*c(p,k) or O(chk * pk). However, with RPA position of k new radios is found sequentially. So RPA run k times and each time c(p,1) positions are checked. Complexity of RPA is therefore ch*k*c(p,1) or ch*k*p.

  36. Radio Placement Algorithm (Contd..) Procedure of RPA Contention based heuristic is used to find out position of single new radio in the network. Approach is to connect single new radio to all possible positions and to calculate TCV of the network for each possible position. TCV’s are compared to find out the position P for which TCV is minimum. Not only the position of single new radio is found but the channel of the corresponding link is also found. Once position of single new radio is found, it is added to the network and then same procedure is repeated to find out the position of next radio. Procedure finishes when the given number of radios are added sequentially to the network.

  37. Evaluation of RPA • Position of given k new radios are found using RPA. • Position of k new radios are then found using exhaustive method. • We will show that throughput improvement achieved by RPA compares well with the throughput improvement achieved by exhaustive search method. • In exhaustive search method, all possible position of k new radios are are considered and increase in throughput is found for each case. Increase in throughput is found using simulation. Equal rate traffic pattern is used. • Topologies used for RPA are generated as follows. • Given number of nodes are generated randomly in a area of size 8Km * 8Km. • Network is obtained by connecting them through minimum spanning tree. • Node 0 is taken as the landline node. Topologies has more than one radio at some nodes. Extra radios are connected to nodes in the chronological order.

  38. Addition of three radios In this topology, nodes 0 and 1 have two radios each. Using RPA, position of three new radios is found to be at node 6 (link 6-5, channel 11), node 8 (link 8-2, channel 1), node 2 (link 2-3, channel 1)

  39. Graph

  40. Addition of four radios Nodes 0, 1 and 3 have two radios each. Position of four new radios for this topology is found to be at node 11 (link 11-1, channel 6), node 6 (link 6-0, channel 11), node 8 (link 8-0, channel 1) and node 8 (link 8-7, channel 11)

  41. Graph

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