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

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

3. 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

4. Examples Links in Series1 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

5. 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

6. 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

7. 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.

8. 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

9. 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)

10. 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.

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

12. Fair Ad Hoc MAC?
Motivation 
Protocol
Analysis
Simulations

13. Motivation: Exponential Backoff is Unfair
Exponential backoff scheme (e.g b)
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

14. Motivation: TCP cannot overcomex 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

15. 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)….

16. Fair Ad Hoc MAC?
Motivation
Protocol 
Analysis
Simulations

17. 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

18. 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” ….

19. 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

20. 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

21. Fair Ad Hoc MAC?
Motivation
Protocol
Analysis 
Simulations

22. 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 = 2.5
interference

23. 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 < (patient backoff)
interference

24. 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

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

26. Fair Ad Hoc MAC?
Motivation
Protocol
Analysis
Simulations 

27. 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

28. 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