Communication Networks A Second Course Jean Walrand Department of EECS University of California at Berkeley

Ad-Hoc Networks: Capacity & Fairness Examples Random Network Mobility Effects Fair MAC?

Examples Links in Series Omni-directional antennas; r(interference) < 2 r(comm) link 4 interferes with 2, 3, 5 (4 sends  2 cannot be heard) => Independent sets: {1, 4, 7}, {2, 5, 8}, {3, 6} => R < 1/3 Omni-directional antennas; r(interference) = 2 r(comm) link 4 interferes with 1, 2, 3, 5 (4 sends  1 cannot be heard) => Independent sets: {1, 5}, {2, 6}, {3, 7}, {4, 8} => R < 1/4 Directional antennas; r(interference) = 2 r(comm) link 3 interferes with 2, 4 (3 sends  4 cannot be heard) => Independent sets: {1, 3, 5, 7}, {2, 4, 6, 8} => R < 1/2 Note: details of protocol matter …. 1 2 3 4 5 6 7 A B 8

Examples Links in Series 1 2 3 A B Links in Series Assume each node sends R/3 to every other node. What is the maximum possible value of R? Now we distinguish link 2 () and link 2’ () r(int) < 2r(com) {1’,3} do not conflict Same for {1, 3’} All other pairs conflict 1 2 3 1’ 2’ 3’ Independent sets: {1’, 3}, {1, 3’}, {2}, {2’} A, B, C, D = set rates 1, 3’: R; 2, 2’: 4R/3; 1’,3: R  B = A = f; C = D = 4f/3 2f + 8f/3 = 14f/3 = 1 => f = 3/14 R = 4f/3 = 2/17

Random Network Random network: Assume N nodes randomly placed in some unit square; they all transmit to a randomly picked destination at a total rate R. How large can R be? We show that R < O(1/N0.5). Each node is d = O(1/N0.5) away from neighbors Each transmission at rate R to a neighbor covers a disk of height R and area O(d2) = O(1/N) => volume R.O(1/N). To reach a random destination, we need to cover an average distance O(1) with hops of size O(1/N0.5), we need O(N0.5) hops. That transmission from source to destination covers a volume O(N0.5)R.O(1/N) = R.O(1/N0.5). Note: shortest hop minimizes volume: increasing hops sizes by k increases volume by k2/k = k There are N transmissions => They use volume R.O(N0.5) The total available volume is C.O(1) Hence, R.O(N0.5) < C.O(1)  R < O(1/N0.5) Note that this rate is achievable if the nodes are in a regular grid; they can use a simple horizontal, then vertical routing

Random Network S D About N0.5 hops of size 1/N0.5 They cover a volume equal to N0.5(R/N) => R/N0.5 per transmission There are N such transmissions Hence RN0.5 < C => R < C/N0.5

Random Network The capacity per node is O(1/N0.5) [Franceschetti et al, 2004] We have seen the upper bound The lower bound goes as follows. Consider square with sides n = N0.5. Divide into grid of squares with sides c. Consider nonempty squares to form bridges. If P(nonempty) > ½, then there are O(n) paths from left to right. Associate each path to a horizontal slab with O(1) height. Sources in that slab connect to path with O(log n) hop; same vertically.

Random Network There are A horizontal paths and A vertical paths. We associate a path to each horizontal slab of size O(1). O(n) paths through each cell. Hence, R.O(n) = C. However, we need to take Interference into account and the location of the paths b n = N0.5

Random Network Access to path: O(log n) Rate >> O(n) Paths: Divide square into square cells with side O(1). If square nonempty, it form a bridge. If P(bridge) > ½, one can go from left to right with a probability close to 1 as N increases. Moreover, there are many paths: O(n). Thus, one finds a path close to each slab. n = N0.5 Access to path: O(log n) Rate >> O(n) Bottleneck is path b O(log n)

Mobility Effects Basic idea: Nodes can wait to be closer to transmit If they move fast enough, the capacity increases However, this may increase delay Unfortunately, mobility assumptions are flaky.

Fair Ad Hoc MAC? Motivation Protocol Analysis Simulations Work with Rajarshi Gupta

Fair Ad Hoc MAC? Motivation  Protocol Analysis Simulations

Motivation: Exponential Backoff is Unfair Exponential backoff scheme (e.g. 802.11b) Nodes pick backoff uniformly in a backoff range If collision, double the backoff range Multiple interference domains Node in center sees more contention and collision It backs off more Gets lesser share of bandwidth Unfair towards middle nodes in network Active Link Rcvd on A Rcvd on B Rcvd on X A 6 A,B A,X 3 A,B,X 4 2 All rates in Mbps

Motivation: TCP cannot overcome x y 6 2 4 x - losses x, y - losses y - losses A1 A2 B1 B2 X1 X2 x y Simplified Model: WFQ 2 4 y x

Motivation: TCP cannot overcome In complex network, constraints are non-local: Independent Sets  Shadow prices cannot be computed locally Here, we have modest goals: Improve node-fairness (max-min)….

Fair Ad Hoc MAC? Motivation Protocol  Analysis Simulations

Protocol: Impatient Backoff Algorithm Approach: Nodes that face more contention should get higher priority Key Mechanism Upon collision, nodes decrease their backoff Need to worry about Stability Fairness Throughput

Protocol: Backoff Update If collision or quiet Decrease the mean backoff delay b := b/m, where m>1 If successful transmission Increase the mean backoff delay b := bm Note: Distributed reset mechanism When a node’s mean delay falls below threshold, node broadcasts “multiply by K” ….

Protocol: Simplified MAC Model All packet lengths are same Transmissions occur slot by slot Local synchronization is assumed Similar to any slotted protocol No RTS/CTS

Protocol: IBA Mechanism Backoff Contention Phase Each node has mean backoff b Picks backoff delay B using exponential variable with mean b Sends out Slot Capture Message after B backoff mini-slots If a node carrier senses another message sooner – it keeps quiet Packet Transmission Phase Starts after completion of Backoff Contention Phase Nodes with successful Slot Capture Messages transmit Constant packet length Transmission confirmed by ack Collision occurs if two neighbors pick same backoff Neither hears slot capture Both try to transmit Packet transmission wasted

Fair Ad Hoc MAC? Motivation Protocol Analysis  Simulations

Markov Chain Models Two extreme topologies interference Two extreme topologies Star Topology (unfair) Triangle Clique Topology (symmetric) Model ratio between mean backoffs Prove stability, fairness Throughput-fairness tradeoff (in Star) Max throughput = 0 + 41 = 4 But fair throughput = 0.5 + 40.5 = 2.5 interference

Star Topology: Birth-Death Chain Stable: Positive recurrent for m>1 Strong drifts towards stable state S0 Fair: Expected transmission rate for all nodes is 0.5 Note: Unstable for m < 1 (patient backoff) interference

Star Topology: Varying Neighbors Model sleeping nodes Every 100 slots, some nodes go to sleep Fairness = 1 Average success probability Middle Node(sX)= 0.473 Outer Nodes(sZ)= 0.470

Triangle Topology Markov Chain interference Prove positive recurrence using Lyapunov function Chain drifts towards bottom-left stable states Fairness is due to symmetry

Fair Ad Hoc MAC? Motivation Protocol Analysis Simulations 

Simulations on Random Topology Impatient Backoff Exponential Backoff Min Throughput = 9% of mean Jain’s Fairness Index = 0.58 Mean Throughput = 0.101 Min Throughput = 49% of mean Jain’s Fairness Index = 0.68 Mean Throughput = 0.102 Circle = Node : Center = Location, Area = Throughput

Variations in Simulation Nodes execute random walk Initial bias against selected nodes Nodes switch between active and sleep cycles Similar comparisons with exponential backoff Comparable throughput Significantly better fairness