stability and fairness of service networks jean walrand – u.c. berkeley joint work with a....
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Stability and Fairness of Service Networks
Jean Walrand – U.C. Berkeley
Joint work with A. Dimakis, R. Gupta, and J. Musacchio
Outline
Stability of Longest Queue First Fluctuations can stabilize
Fairness through flow control Control of long term rates
Fairness of multiple access Impatience may help in a crowd
Outline
Stability of Longest Queue First Fluctuations can stabilize
Fairness through flow control Control of long term rates
Fairness of multiple access Impatience may help in a crowd
LQF - Motivation Wireless:
Goals: Simple protocol, large throughputTransmission priority increases with backlog
2 3 4 5 1 6
LQF: Motivation
Iterated Longest Queue First (iLQF) [McKeown’95]:Queues are considered in decreasing queue size order.
Maximum throughput?
1112
22
21
1 1
2 2
input 1
input 2output 2
output 1
Q11(t)
Q12(t)
Q21(t)
Q22(t)
Switch:
Stability of LQF: Easy CaseExample:
LQF: 12 – 9 – 8
7 – 9 – 8
9 – 9 – 8
9 – 9 – 8
w.p. 1/2
1 2 3
1 2 3
service vectors
i.i.d. arrivals
w.p. 1/2
Stability of LQF: Easy Case Necessary:
1+2<1, 2+3<1.
Sufficient!
Under LQF, longest queues tend to decrease:Say, Q1¼ Q2>>Q3, for some time.
Then, Q1+Q2 decreases, and so do Q1,Q2.
Key: locally in time, service from common resource pool.
1 2 3
1 2 3
Stability of LQF: Easy Case
Local Pooling:
Assume {1, 2} are longest for some time
Note that {1, 2} are served at constant rate (1)
(We say that {1, 2} satisfies Local Pooling.)
{1, 2} must decrease (because 1+2<1)
longest queue must decrease
1 2 3
1 2 3
Stability of LQF: Easy Case
Local Pooling:
Assume {1, 2, 3} are longest for some time
Note that {1, 2} are served at constant rate (1)
(We say that {1, 2, 3} satisfies Local Pooling.)
{1, 2} must decrease (because 1+2<1)
longest queue must decrease
1 2 3
1 2 3
Stability of LQF: Easy Case
Local Pooling:
Assume {1, 3} are longest for some time
Note that {1, 3} are served at constant rate (2)
(We say that {1, 3} satisfies Local Pooling.)
{1, 3} must decrease (because 1+3<2)
longest queue must decrease
1 2 3
1 2 3
Stability of LQF: Easy Case Local Pooling: Set L satisfies LP if
it has a subset K that LQF serves at a constant rate
Theorem:If every set L satisfies LP and if the rates are feasible,then LQF makes system stable
Proof: Longest queue is a Lyapunov function(Consider fluid limit ….)
1 2 3
1 2 3
Stability of LQF: Subtle Effect Graph that does not satisfy Local Pooling:
3
6 5
1 4
2
{1, 2, 3, 4, 5, 6} has no subset served at constant rate {1, 2, 3, 4, 5, 6} does not satisfy LP
Every proper subset satisfies LP
E.g., {1, 2, 3, 5} longest serve {2, 3} at rate 1
Service Vectors: {1, 3, 5}, {2, 4, 6} {1, 4}, {2, 5}, {3, 6}
Stability of LQF: Subtle Effect Note: Deterministic inputs with
rate close to 0.5 unstable
(LQF serves 2/6 a positive fraction of time)
3
6 5
1 4
2
Theorem: LQF stable if i.i.d. arrivals with nonzero variance
Key Idea: {1, 2, 3, 4, 5, 6} cannot be set of longest queues for a positive fraction of time! LP holds most of the time Longest queue decreases
Stability of LQF: Subtle Effect
3
6 5
1 4
2Key Idea: {1, 2, 3, 4, 5, 6} cannot be set of longest queues for a positive fraction of time!
Assume all queues are longest for a while{2, 3} and {5, 6} served at same rate
Stability of LQF: Subtle Effect
(n)
(k + 1)(n)k(n)
L
Max – Min large at k(n) A subset L of queues dominates the others during intervalThis subset satisfies LP Longest queue decreases.
Stability of LQF: Subtle EffectTheorem:
Assume that whenever a set L does not satisfy LP, the corresponding service vectors have rank ≤ |L| - 2.
Assume also the arrivals are i.i.d. with positive variance(and satisfy a large deviation bound).
Then LQF is stable for any feasible arrival rates.
Outline
Stability of Longest Queue First Fluctuations can stabilize
Fairness through flow control Control of long term rates
Fairness of multiple access Impatience may help in a crowd
Motivation Example
11
12
21
22
11
21
12
22
a1
a2
b1
b2
h: discardthreshold
11
12
21
22
1
1
Intuitively: h large enough max – min fair
Long-term average rates max – min for h >> 1
h = discard threshold
Motivation
11
12
21
22
11
21
12
22
0.75
0.25
0.5
0.5
h: discardthreshold
1
1
0.75
0.25
0.2
0.7
h = discard threshold
Analysis
11
12
21
22
11
21
12
22
0.75
0.25
0.5
0.5
nh
n
n
n
n
n
n
Qn(nt): Scale thresholds and speed up
Analysis
11
12
21
22
11
21
12
22
0.75
0.25
0.5
0.5
h
1
1
Qn(nt)/n: Scale space
Qn(nt)/n fluid limit Q(t) with suitable rates….
AnalysisRoughly, x(n; t) := Qn(nt)/n uoc fluid limit
Q(t)For t ≥ t0, Q(t) = q* with suitable rates.This implies
Key argument: Most of the time t ≥ 0, x(n; t) ≈ q*
However, we want
AnalysisRoughly, x(n; t) := Qn(nt)/n uoc fluid limit
Q(t)For t ≥ t0, Q(t) = q* with suitable rates.This implies
However, we want
Key argument: Most of the time t ≥ 0, x(n; t) ≈ q*
To show this:1) Uniformly in |x(n; 0) – q*| ≤ , E[|x(n; t) – q*|] 0 for t ≤ t0
2) Uniformly in y = |x(n; 0) – q*| > , E[|x(n; yt0) – q*|] < y3) Expected time E() until |x(n; + t0) – q*| ≤ is small for n >> 1
Outline
Stability of Longest Queue First Fluctuations can stabilize
Fairness through flow control Control of long term rates
Fairness of multiple access Impatience may help in a crowd
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 6 6
A,X 3 3
A,B,X 4 4 2
A1
inte
rfer
ence
inte
rfer
ence
A2
B1
B2
X1
X2
Cory HallRoom 273
Cory HallRoom 264M
CoryHallway
All rates in Mbps
Protocol: Impatient Backoff Algorithm
Approach: Nodes that face more contention should get higher priority
Key MechanismUpon 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 mechanismWhen 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
1 5432
interference
1
5
4
3
2
1's Packet Transmission
5's Packet Transmission
BackoffContentionPhase
PacketTransmissionPhase
backoff
slot capture
ack
ack
Collision occurs if two neighbors pick same backoff
Neither hears slot capture
Both try to transmit Packet transmission
wasted
Markov Chain Models 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 + 41 = 4 But fair throughput = 0.5 + 40.5 = 2.5
interference
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
n n.m4
n.m2
n.m-2
n.m-4
S0
S1 S2S-1S-2
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
111
144
11616
114
1416
11664
1116
1464
116
256
14
256
1164
16464
1 1/9 1/33
8/9 32/33 128/1292/3
1/3 1/21
16/21 64/81
4/21 16/818/9
64/69 256/273
1/9 1/69
4/69 16/27332/33
1/33
256/261
interference
m=2
Prove positive recurrence using Lyapunov function Chain drifts towards bottom-left stable states Fairness is due to symmetry
Simulations on Random TopologyExponential 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
Impatient Backoff
Variations in Simulation Nodes execute random walk Initial bias against selected nodes Nodes switch between active and sleep cycles
Similar comparisons with exponential backoffComparable throughputSignificantly better fairness