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## Mobility Increase the Capacity of Ad-hoc Wireless Network

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**Mobility Increase the Capacity of Ad-hoc Wireless Network**Matthias Gossglauser / David Tse Infocom 2001**Outline**• Introduction • Model • Results • Numerical results • Conclusion**Introduction**• The time variation of the channel • Multipath fading • Path loss via distance attenuation • Interference from other users • Diversity • Used to cope with the time-variation of the channel**Introduction**• The main idea of [6] • On the uplink in a single cell, with multiple users communicating to the basestation via time-varying channels. • In order to maximize the total information theoretic capacity, the optimal strategy is to schedule at any one time only the user with the best channel to transmit to the basestation.**Introduction**• The main result in [4] • In the fixed ad-hoc networks • Even using optimal scheduling, routing and relaying of packet, the throughput per S-D pair decreases approximately like 1/√n (n: number of nodes)**Introduction**• Summary of this paper • Nodes are mobile • Focus on applications that can tolerate end-to-end delay • The result shows that the average long-term throughput per S-D pair can keep constant even as the number of node per unit area n increase**Model**• There are n nodes all lying in the open disk of unit area(π-1/2). • The location of the ith user at time t is given by Xi(t). The process {Xi(‧)} is independent, stationary and ergodic with stationary distribution uniform on the open disk. • This paper stipulate that the source node i has data intended for destination node d(i).**Model**• At time t, let Pi(t) be the transmit power of node i • γij(t) be the channel gain from node i to node j • α is a parameter greater than 2 • The received power at node j is Pi(t) γij(t).**Model**• At time t, node i transmits data at rate R packet/sec to node j if its SIR satisfied • β is the signal-to-interference ratio • N0 is the background noise power • L is the processing gain**Model**• Consider a scheduling and relay policy π • be the number of source node i packets that destination d(i) receives at time t under policy π • We shall say a long term throughput of λ(n) is feasible if there is a policy πsuch that for every S-D pair i,**Results**Fixed Nodes • Theorem 3.1 : There exists constants c and c’ such that and**Results**Mobile node without relaying • Lemma 3.2: Consider a scheduling policy that schedules direct transmissions only. Fix an arbitrary time t. Let S(t) be the set of source nodes that are scheduled successful transmission to their respective destinations. Then If β increases (SIR ↑) , then B will decrease. And the number of the long range communication will decrease.**Results**• Theorem 3.3: Assume that the policy is only allowed to schedule transmission between the source and destination nodes, and no relaying is permitted. If c is any constant satisfying. then for sufficiently large n. This result shows that without relaying, the achievable throughput per S-D pair goes to zero at least as fast as**Results**Mobile nodes with relaying • The idea now is to spread out packets to a large number of intermediate relay nodes that temporarily buffer packets until final delivery to the destination is possible. • How many times a packet has to be relayed? Answer :As the node location processes {Xi(t)} are independent, stationary, ergodic, it is actually sufficient to relay only once. As no packet is transmitted more than twice, the achievable total throughput is O(n).**Results**• Scheduling policy π • A sender density parameter θ distributed in (0,1) • nS=θn is the nodes as senders in each time slot, and the remaining nRnodes as potential receivers. • Each sender node transmits packets to its nearest neighbor among all nodes which are receivers, using unit transmission power (Pi=1). • Define Nt be the S-D pairs, whose transmission are possible.**Results**• Theorem 3.4: For the scheduling policy π, the expected number E[Nt] of feasible sender-receiver pairs is O(n), i.e. Furthermore, for two arbitrary nodes i and j, the probability that (i,j) is scheduled as a sender-receiver pair is O(1/n).**Results**• The over algorithm is divided into two phases: • Scheduling of packet transmissions form sources to relays (or the final destination),and • Scheduling of packet transmission from relays (or the source) to final destinations.**Results**Applying theorem 3.4, we see that both the arrival rate and the service rate of the queue is O(1/n). Summing over the throughputs of all the n-1 routes, it can be seen that the total average throughput per S-D pair is O(1). It has proved the theorem 3.5.**Results**• Theorem 3.5: The two-phased algorithm achieves a throughput per S-D pair of O(1), i.e. there exists a constant c > 0 such that Note that the largest possible throughput is c=ψ/2.**Conclusion**• This paper propose to spread traffic randomly to intermediate relay node and a single relay node is sufficient to use the entire throughput capacity of the network.