Peering in Infrastructure Ad hoc NetworksPresentation OutlineSlide 3Ad hoc Networks : Current Modes of OperationA Hybrid ApproachScenario : Reduced PowerSlide 7ObjectivesAssumptionsSlide 10Problem FormulationProblem Formulation (contd.)No Peering (Relaying)Node B Relays Node A’s trafficNode B relays Node A’s trafficProblem formulation (contd.)Slide 17Token Distribution StrategiesToken Distribution Strategies …Slide 20User Centric ApproachUser Centric Approach (contd.)Slide 23System Centric ApproachSystem Centric Approach (contd.)Slide 26Experimental ResultsSlide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36ConclusionsFuture WorkReferencesPowerPoint PresentationPeering in Infrastructure Ad hoc NetworksMentor : Linhai HeGroup : Matulya Bansal Sanjeev KohliEE 228a Course ProjectPresentation OutlineIntroduction to the problemObjectivesProblem FormulationAnalysis of approachesExperimental ResultsConclusionsPresentation OutlineIntroduction to the problemObjectivesProblem FormulationAnalysis of approachesExperimental ResultsConclusionsAd hoc Networks : Current Modes of OperationPeer-to-Peer ModeNodes relay each other’s trafficInfrastructure ModeNo relaying between nodesNodes directly communicate with Base StationA Hybrid Approach Does Peering in Infrastructure Mode make sense?A BBase stationScenario : Reduced PowerA and B both want to communicate with the base station.Using direct connections, A ends up using more power. With B agreeing to peer, A can reduce its power consumption while B can increase its throughput.ABBase StationPresentation OutlineIntroduction to the problemObjectivesProblem FormulationAnalysis of approachesExperimental ResultsConclusionsObjectivesAnalyze the advantages of peering in infrastructure mode based on two approaches:Individual User Centric: Each user tries to maximize its own performanceSystem Centric: Users collaborate to maximize overall system performanceShow improvement in network performance with experimental resultsAssumptionsBase Station distributes tokens to each user in every cycleThe number of tokens, T, distributed by BS in every cycle equals the no of transmission slots in each cycleA user can transmit in a slot only if it has a tokenUnderline MAC layer resolves contention for slotsPresentation OutlineIntroduction to the problemObjectivesProblem FormulationAnalysis of approachesExperimental ResultsConclusionsProblem FormulationTotal tokens in system - TNode A has TA tokensNode B has TB tokens TA + TB = TPower level of transmission is same for each user, PTA BBSProblem Formulation (contd.) Data rate/slot for ABS = rA 1/(dA) Data rate/slot for BBS = rB 1/(dB) Data rate/slot for AB = rAB 1/(dAB)A BBSNo Peering (Relaying)A BBS Throughput of node A = TA.rA Throughput of node B = TB.rB Throughput of the whole system = = TA.rA + TB.rB Power consumption for above throughput = (TA + TB)PTNode B Relays Node A’s traffic Node A sends a request to node B for relaying its data. Information of total data to be relayed is sent with this the request Node B analyzes the cost of relaying (in terms of power spent and throughput gained) and sends a response to node A asking for the no of tokens it wants in lieu of relaying Node A analyzes this response and decides to relay its traffic through node B if it can meet node B’s demand Assumption: Protocol setup time is negligibleNode B relays Node A’s trafficA BRequestResponseDataBSProblem formulation (contd.)Throughput of node A = TAB.rAB = TA.rANo of tokens available for distribution = TA – TABwhere TAB = TA(dAB/dA)Throughput of node B = (TB+TB’).rB where TB’ is the minimum no of tokens gained by node B to justify the relay i.e. to satisfy its utility functionProblem formulation (contd.)No of tokens saved in the system = TA – (TAB + TB’ + TB’’)where TB’’ is the no of tokens needed by node B to transmit node A’s data i.e. TB’’ = (TAB.rAB)/rB Power spent by the system for same throughput as in the case of no relay = (TAB + TB + TB’’)PTToken Distribution Strategies Equal tokensBoth nodes get half of the total tokens TA = TB = T/2 [ Not Fair ] Equal Bandwidth Both nodes get equal throughput TA.rA = TB.rB TA/TB = (dA/dB) [ Doesn’t optimize overall system throughput ]Token Distribution Strategies …• Equal normalized rate of change in throughput w.r.t. no of tokensThroughput of node A = TA.rA TA /(dA) d(TA.rA)/d(TA) 1/(dA)Similarly, d(TB.rB)/d(TB) 1/(dB) [d(TA.rA)/d(TA)]/TA = [d(TB.rB)/d(TB)]/TBTA/TB = (dB/dA)A BBS[Optimizes overall system throughput and seems fair]Presentation OutlineIntroduction to the problemObjectivesProblem FormulationAnalysis of approachesExperimental ResultsConclusionsUser Centric Approach In this system, each user tries to improve its own performance i.e. it doesn’t relay any data for social cause, it relays only to improve its own performanceWe define the utility function of relaying node as:U(T, P) = T[1 + C(log P)/P]where T and P are the no of tokens and battery power of relaying node (available for itself) respectivelyCaptures the token gain and energy payoff of relaying node very wellUser Centric Approach (contd.)Value of utility function of node B before relaying is:U(TB , PB) = TB[1 + C(log PB)/PB]If TB’ is the no of tokens gained by relaying the data and PB’ is its new residual power , then new value of node B’s utility function is:U((TB+TB’), PB’) = (TB + TB’)[1 + C(log PB’)/PB’]where PB’ = PB – (PT.TB’’) and TB’’ = TAB.rAB/rBUser Centric Approach (contd.)Since relaying node wants to maximize its performance, its utility value shouldn’t decrease after relaying the date i.e. :U((TB+TB’), PB’) - U(TB, PB) > 0 TB’ > TBC [(log PB/PB’)/ PB]TBC [(log PB/PB’)/ PB] is the minimum no of tokens needed by node B for its own usage in order to justify relaying node A’s data.System Centric ApproachThe users in this system try to enhance the overall system performance rather than their own.A user relays the traffic from another node if the ratio of residual battery power of relay node and source node is greater than the ratio of
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