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Page 1: Scheduling: Contention, Fairness and Throughput

Scheduling: Contention, Fairness and Throughput

University of California at Berkeley

Motivations for a Multi-Channel Wireless MAC

t=0 frequencySender 1

t=1frequency

Sender 2

t=2frequency

Sender 3

Today: Each wireless network uses 1 channel only

: :

PowerDensity

t=0frequency

PowerDensity

Sender 1 Sender 3

t=1frequency

Sender 2 Sender 1 Sender 4

Sender 4

t=2 frequencySender 3: :

Sender 2Sender 4

Why not: simultaneous sending on different channels?

Channel 1 Channel 2 Channel 3

Wasted spectru

m

Simple Rendezvous Scheme

1. Node i generates a pseudo random sequence X(Si,ti) i.i.d. ~uniform({1,2,... ,C}). Si is the seed. ti is a local version of the real time.

2. Each packet sent by i contains i, Si and ti.3. Idle nodes occasionally broadcast empty packets.4. Eventually each node hears every neighbor once.5. Node i listens on the default channel X(Si,ti) at time t

Ch2

Ch1

98765432t=1

node2node1 node3

Rendezvous: Sending Illustrated

Ch2

Ch1

98765432t=1

2. RTS/CTS/Data

1. waiting to send

4. Hopping returns to the original hopping schedule

3. Hopping stoppedduring data transfer

Some simple improvements:1. A sender checks if any receivers will be on its default channel

during the coming time slot.2. If not, the sender chooses a channel c’ with a receiver in it

uniformly at random.3. Sender transmits an RTS on channel c’ with probability 1/N(c’,t)

where N(c’,t) is the estimated number of nodes on channel c’ at time t.

Limitations of Local Scheduling

Optimal schedulers (Maximum Weight Matching) require global communication – not practical for MAC.

What is the throughput of local scheduling? No explicit exchange of scheduling information. No scheduling state other than the backlog.

Idealization: iterated Longest Queue First (iLQF): Nodes with longer backlog transmit first.

e.g., MAC with backlog dependent backoff time.

Results Partial Pooling: graph condition for optimality of iLQF. Instability in the case that Partial Pooling (P.P.) is not

satisfied.

Contention model

Incidence matrix A=(Ajk). Set of matches S[K]={m2{0,1}K s.t. Am· 1}. Maximal matches M[K] ½S[K]. Optimal throughput: < 2 Co(M[K]).

Classes: k2 K Resources: j2 J

Activities: k2 K

1

2

3

λ2

λ3

λ1

12 3

conflict graph

Captures dynamic contention for shared resources, e.g., wireless network, packet switch, distributed computation.

Optimality under Partial Pooling

Def: P.P. holds for A½K if 9 nonempty B ½A s.t.

8 m2 M[A], k2 Bmk=Const(A,B).

Def: P.P. holds if P.P. holds for all A ½K.

P.P. class strictly includes tree conflict graphs.

If system conflict graph satisfies P.P., then iLQF is throughput optimal. Approach: fluid limits, longest queue size is a Lyapunov

function.

Instability when P.P. fails

Assume load close to capacity (1/2). “Efficient” matches: {1,3,5,7},{2,4,6,8} “Inefficient”: {1,4,6},… all size 3 matches

8 3

4

6

7

1 2

5 “meta-stability” in 8-cycle:

Most states activate efficient matches.

But, inefficient states are attracting.

Mu

lti-

Ch

an

nel

Wir

ele

ss

MA

C

Imp

ati

en

t

Back

off

A

lgori

thm

Lim

itati

on

s of

Loca

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ched

ulin

g

Antonis Dimakis [email protected]

Rajarshi Gupta [email protected]

Wilson So [email protected]

2+: guaranteedsuccess

1:contentionsuccess

Idle: no suitablereceivers on samechannel

Quiet:everyone backs off

Receiver Absent:receiver stuck on another

channel

Time Slot Utilization Breakdown

2+: guaranteedsuccess

Idle: no suitablereceivers on samechannel

Quiet: everyone backs off

1:contentionsuccess

Avg

Per

Nod

e Q

ueu

e L e

ngth

[M

ini-Pa

c ket

s]

Time [slot]

(10,000 slots = 5 sec)

Collision: >1 sendsReceiver Absent:receiver on another

channel

Exp.1 : High (80% ) load, Long (5-slot) packets.

Per Node Queue Length vs. Time

x 104

Avg

Per

Nod

e Q

ueu

e L e

ngth

[M

ini-Pa

c ket

s]

(10,000 slots = 5 sec)

x 104Time [slot]Exp.2 : High (80% ) load, Short (1-slot) packets.

Collision:>1 sends

SmartNets Research Group

EECS, U C Berkeley Fall 2004

Key Idea In exponential backoff (e.g. 802.11)

Upon collision, nodes back off and become less aggressive Problem in networks spanning multiple interference domains

Nodes in the middle face more collisions They backoff more. Get lesser share of bandwidth This is unfair towards nodes in the middle

Idea: Give higher priority to nodes facing more contention Key: Upon collision, nodes decrease their backoff delay Characteristics

Achieves stable system Maintains throughput in random networks Significant improvement in fairness

EECS, U C Berkeley Fall 2004

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 mini-slots If a node carrier senses another

message sooner – it keeps quiet Packet Transmission Phase

Starts after completion ofBackoff Contention Phase

Nodes with successful Slot Capture Messages transmit

Constant packet length Confirmation using ack

If collision or quiet, b:= b/ m If successful transmission, b :=bm m > 1 Markov analysis indicates stability

and fairness

1 5432

interference

1

5

4

3

2

1's Packet Transmission

5's Packet Transmission

BackoffContentionPhase

PacketTransmissionPhase

backoff

slot capture

ack

ack

EECS, U C Berkeley Fall 2004

Resetting Rates Problem of MIMD scheme

When many congested neighbors all decrease backoffs

Resulting backoffs are small Many collisions

Solution: reset backoff delays when exceeds a limit

If any mean backoff goes below reset_limit

Multiply all mean backoffs by constant reset_factor

Simulation parameters m= 1.2 reset_limit = 16/5=3.2 reset_factor = 10

Reset propagation/loss Resets move hop-by-hop

across network May get lost Effect of lost reset

Keeps backoff low So node wins next few slots Increases mean backoff

Simulation Results Reset propagation Loss of up to 10% of resets Random walk movement

EECS, U C Berkeley Fall 2004

Simulations on Random TopologyExponential Backoff Impatient Backoff Algorithm

Jain’s Fairness Index = 0.58 Mean Throughput = 0.101

Jain’s Fairness Index = 0.68 Mean Throughput = 0.102

Circle = Node : Center = Location, Area = Throughput