Peering in Infrastructure Ad hoc NetworksMentor : Linhai HeGroup : Matulya BansalSanjeev KohliEE 228a Course ProjectPresentation Outline Introduction to the problem Objectives Problem Formulation Analysis of approaches Experimental Results ConclusionsPresentation OutlineIntroduction to the problem Objectives Problem Formulation Analysis of approaches Experimental Results ConclusionsAd hoc Networks : Current Modes of Operation Peer-to-Peer Mode Nodes relay each other’s traffic Infrastructure Mode No relaying between nodes Nodes directly communicate with Base StationA Hybrid Approach Does Peering in Infrastructure Mode make sense?ABBase stationScenario : Reduced Power A 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 Outline Introduction to the problemObjectives Problem Formulation Analysis of approaches Experimental Results ConclusionsObjectives Analyze the advantages of peering in infrastructure mode based on two approaches: Individual User Centric: Each user tries to maximize its own performance System Centric: Users collaborate to maximize overall system performance Show improvement in network performance with experimental resultsAssumptions Base Station distributes tokens to each user in every cycle The number of tokens, T, distributed by BS in every cycle equals the no of transmission slots in each cycle A user can transmit in a slot only if it has a token Underline MAC layer resolves contention for slotsPresentation Outline Introduction to the problem ObjectivesProblem Formulation Analysis of approaches Experimental Results ConclusionsProblem Formulation Total tokens in system - T Node A has TAtokens Node B has TBtokens TA+ TB= T Power level of transmission is same for each user, PTABBSProblem Formulation (contd.) Data rate/slot for ABS = rA∝ 1/(dA)α Data rate/slot for BBS = rB∝ 1/(dB)α Data rate/slot for AB = rAB∝ 1/(dAB)αABBSNo Peering (Relaying)ABBS Throughput of node A = TA.rA Throughput of node B = TB.rB Throughput of the whole system = τ = 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 trafficABRequestResponseDataBSProblem formulation (contd.) Throughput of node A = TAB.rAB =TA.rA No of tokens available for distribution = TA –TABwhere TAB = TA(dAB/dA)α 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 tokens Both 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)α [ 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)α d(TA.rA)/d(TA) ∝ 1/(dA)αSimilarly, d(TB.rB)/d(TB) ∝ 1/(dB)α[d(TA.rA)/d(TA)]/TA = [d(TB.rB)/d(TB)]/TB TA/TB =(dB/dA)αABBS[Optimizes overall system throughput and seems fair]Presentation Outline Introduction to the problem Objectives Problem FormulationAnalysis of approaches Experimental Results ConclusionsUser 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 performance We 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) respectively Captures 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 Approach The 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 energy spent per bit by the relay node and the source node for transmission i.e. :where EDirectand ERelay are the energies spent by source node and relay node to transmit one bit to the Base StationDirectRelayDirectRelayEEPP>System Centric Approach (contd.) In our 2-node case, Hence, node B will relay node A’s data if,α)(EEABABdd=ABBSα)(PPABABdd>Presentation Outline Introduction to the problem Objectives Problem Formulation Analysis of approachesExperimental Results ConclusionsExperimental ResultsApproach 1: Varying position of node B, power of B fixedExperimental ResultsApproach 1: Varying position of node B, power at B fixedExperimental ResultsApproach 1: Varying position of node A, power at B
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