June 2, Mobile Computing COE 446 Mobile Ad hoc Networks Tarek Sheltami KFUPM CCSE COE

June 2, Many Applications Personal area networking cell phone, laptop, ear phone, wrist watch Military environments soldiers, tanks, planes Civilian environments taxi cab network meeting rooms sports stadiums boats, small aircraft Emergency operations search-and-rescue policing and fire fighting

June 2, Why Ad hoc Networks? No infrastructure needed (no routing in fixed wireless) Can be deployed quickly, where there is no wireless communication infrastructure present Can act as an extension to existing networks enhances coverage Cost-effective – cellular spectrum costs $XX billion Adaptive computing and self-configuring Support for heterogeneous computational devices and OSs

June 2, Ad hoc Constraints Dynamic topologies Bandwidth-constrained Constraints on Tx power Infrastructure-less property, no central coordinators hidden terminal, exposed terminal No QoS preservation Load balancing Energy-constrained operation Limited physical security

June 2, Background Table-driven routing protocols Routing Disadvantages On-demand routing protocols Routing Disadvantages Cluster-based routing protocols (Laura Feeney, Infocom 2001) Routing Disadvantages

June 2, DSDV Routing Protocol

June 2, DSDV Routing Protocol (Cont d) Return

June 2, Disadvantages of Table Driven Protocols Routing is achieved by using routing tables maintained by each node The bulk of the complexity in generating and maintaining these routing tables If the topological changes are very frequent, incremental updates will grow in size Return

June 2, S D On-demand Routing Protocol (DSR) 1,8 1,3,8 1,2,8 1,3,8 1,3,4,7,8 1,2,5,8 1,3,4,8 1,3,6,8 8,5,2,1 5,2,1 2,1 X DATA

June 2, timeout AODV Routing Protocol D S X X DATA Return

June 2, Disadvantages of On-demand Protocols Not scalable to large networks, because of the source routing requirement. Furthermore, the need to place the entire route in both route replies and data packets causes a significant overhead. Some of them requires symmetric links between nodes, and hence cannot utilize routes with asymmetric links. Return

June 2, Cluster-based Routing Transmission Range of MT 1

June 2, Cluster-based Routing.. Range of MT 1

June 2, Cooperative Routing vs Direct Sending In simple radio model, a radio dissipates E ele = 50 nJ/bit at the sender and receiver sides. Let us assume the d is the distance between the source and destination, then, the energy loss is d 2. The transmit amplifier at the sender consumes E amp d 2, where E amp = 100 pJ/bit/m2. Therefore, from the sender side, to send one bit at distance d, the required power is E ele + E amp d 2, whereas at the receiver will need is E ele only. Normalizing both by dividing by E amp : P t = E + d 2 and P r = E, where P t and P r are the normalized transmission and reception power respectively, and E = E ele / E amp = 500m 2 At the HCB-model, the power needed for transmission and reception at distance d is: u(d) = P t + P r = 2E + d 2 u(d) = ad α + c Power-aware localized routing in wireless networks Stojmenovic, I.; Lin, X.; Parallel and Distributed Systems, IEEE Transactions on Volume 12, Issue 11, Nov Page(s):1122 – 1133

June 2, Where in HCB-model α = 2, a = 1, and c = 2E = 1,000 Let us assume that the source S can reach the destination D directly. Let us further assume that there is a middle node between the source and the destination. Let |SA| = x and |SD| = d as in the below Figure d > (c/(a(1-2 1-α ))) 1/α If d > (c/(a(1-2 1-α ))) 1/α, then there is an intermediate node A between the source and destination such that the retransmission of the packet through A will save the energy. Moreover, the greatest saving is achieved when A in the middle of SD. Cooperative Routing vs Direct Sending..

June 2,

June 2, d > (c/(a(1-2 1-α ))) 1/α d(a(α-1)/c) 1/α Also if d > (c/(a(1-2 1-α ))) 1/α, then the greatest power saving are obtained when the interval SD is divided into n > 1 equal subintervals, where n is the nearest integer to d(a(α-1)/c) 1/α. Cooperative Routing vs Direct Sending..

June 2,

June 2, u(d) = ad α + c d(c/(a(1-2 1-α ))) 1/α d>(c/(a(1-2 1-α ))) 1/α Now the power needed for direct transmission is u(d) = ad α + c, which is optimal when d(c/(a(1-2 1-α ))) 1/α, otherwise when d>(c/(a(1-2 1-α ))) 1/α, n-1 is equally spaced nodes can be selected for transmission, n = d(a(α-1)/c) 1/α Where, n = d(a(α-1)/c) 1/α The minimal power: v(d) = dc(a(α-1)/c) 1/α + da(a(α-1)/c) (1-α)/α Cooperative Routing vs Direct Sending..