# New Insights from a Fixed Point Analysis of Single

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```					New Insights from a Fixed
Point Analysis of Single Cell
IEEE 802.11 WLANs
A. Kumar, E. Altman, D. Miorandi and M. Goyal
INFOCOM 2005

Slides and Presented by Yong Yang
New Insights from a Fixed Point Analysis
of Single Cell IEEE 802.11 WLANs
   IEEE 802.11 – CSMA / Collision Avoidance
 Maintains a Contention Window (CW) and follows Binary Exponential Back-off

 Maintains a Back-off Counter (BC), uniformly sampled from [0, CW-1]
 BC is decremented by one for every idle slot with a pre-defined length

 Transmit only when BC reaches zero

   Single Cell
 Nodes are within a distance of each other such that concurrent transmissions
result in collision
t1              t2         t3       t4        t5   t6
1
BC1=0                    20                  15             12

2
BC2=8                    8                   3              24

3
BC3=5                    5                   25             22

4
BC4=8                    8                   3              18
New Insights from a Fixed Point Analysis
of Single Cell IEEE 802.11 WLANs
   1. Modeling IEEE 802.11
   Transmission attempt rate: related to back-off process
β = G (γ)
γ = Γ (G (γ))
   Collision probability γ = Γ (β)
New Insights from a Fixed Point Analysis
of Single Cell IEEE 802.11 WLANs
   1. Modeling IEEE 802.11
   Transmission attempt rate: related to back-off process
β = G (γ)
γ = Γ (G (γ))
   Collision probability γ = Γ (β)

   2. Insights for Network Throughput
   When traffic is open-loop (e.g. UDP)
   When traffic is closed-loop (e.g. TCP)
   When network size is large (closed form analysis)
New Insights from a Fixed Point Analysis
of Single Cell IEEE 802.11 WLANs
   1. Modeling IEEE 802.11
   Transmission attempt rate: related to back-off process
β = G (γ)
γ = Γ (G (γ))
   Collision probability γ = Γ (β)

   2. Insights for Network Throughput
   When traffic is open-loop (e.g. UDP)
   When traffic is closed-loop (e.g. TCP)
   When network size is large (closed form analysis)
Key Observation on IEEE 802.11
t1                t2           t3   t4        t5     t6
1
BC1=0                     20                15               12

2
BC2=8                     8                 3                24

3
BC3=5                     5                 25               22

4
BC4=8                     8                 3                18

Channel activities can be inferred from back-off processes
t1 t2     t3 t4        t5 t6
1                                                            Instants when a new
2                                                            back-off starts
3                                                            Back-off Rate = β
4

Back-off processes of nodes are correlated
Key Approximation of the Back-off Process

   Key Approximation: back-off processes are i.i.d.
   Identical: all nodes use the same back-off parameters, and
then achieve the same long-run average back-off rate β
   Independent: each transmission attempt experiences a
constant long-run collision probability γ

   Saturation assumption is needed in order to infer
channel activities from back-off processes
   Every node always has a packet to transmit
Focusing on the Back-offs of One Node
   Rj: # back-offs until success/discarded for packet j
   Xj: total back-off duration for packet j
R1=2       R2=1                  R3=4
Successes
Collisions
X1                                X3
X2
   Intuitively, the back-off rate: β = E[R] / E[X]
   Can be formally proved by renewal reward theorem

   K: maximum # retries for a packet
   bk: the mean back-off duration of the kth attempt for a packet, 0<=k<=K
E[ R]  k 0 P[# coll  k ] 1 1     2  ...  K
K

E[ X ]  k 0 P[# coll  k ]  bk  b0  b1   2b2  ...  k bk  ...  K bK
K
Focusing on the Back-offs of One Node
   Thus, the back-off rate (attempt rate):
1     2  ...  K
                                         : G( )
b0  b1   b2  ...  bk  ...  bK
2            k         K

   Note
   More general than Bianchi’s model (the standard Binary Exponential
Back-off (BEB))
 K = ∞ , and m = log2(CWmax / CWmin)

(2k CWmin + 1)/2            when 0<=k<m;
   bk =
(CWmax + 1)/2               when k>=m
   Substituting these into G(γ) yields Bianchi’s model:
2(1  2 )
G ( ) 
(1  2 )(CWm in  1)  CWm in (1  (2 ) m )
The Fixed Point Equation
   Under the i.i.d. assumption, the collision probability
γ = 1 – (1 – β)(n-1) := Γ (β)

   Finally, we obtain γ = Γ (G (γ))

Theorem 5.1: γ = Γ (G (γ)) has a unique fixed point if
0<= γ <=1 and bk, k>=0, is a non-decreasing sequence

    G (γ) is non-increasing in γ if bk, k>=0, is a non-decreasing sequence
(Lemma 5.1)
   Γ(β) is non-decreasing in β                   Γ(G(γ))
   Γ(G(γ)) is non-increasing in γ                        1

   Γ(G(γ)) must have a unique
intersecting point with “y=x” line

0
γ
1
Numerical and Simulation Results
   We know in practice, b0 = 16, bk = 2kb0

Γ(G(γ))                     K=7

   Verification by ns2
Part 2: Network Throughput
   Focusing on the aggregate attempt process

Renewal Time                 Renewal Time
Back-off           Success            Back-off   Collision      Back-off

Renewal Instants
   Under the i.i.d. assumption, the process is renewed at the end of every channel
activity (success or collision)

   Prob. that there is no channel activity: Pi = (1 – β)n
   Prob. that a channel activity is a success for node i: Ps|a = β(1 – β)n-1 / [1 – Pi ]

Ps|a  E[ Payload ]            Ps|a  E[ Payload ]
   Throughput of node i:          i                          
E[renewal _ time ]        E[back  off ]  E[ Activity]

   E[back-off] = 1 / (1-Pi )       (Expectation of a geometric distribution)
2.1 Open-loop Traffic
   Suppose each node i only has one flow
   Li : payload size
   Ci : transmission rate
   To : the fixed overhead with a packets transmission
     including RTS, CTS, DIFS, SIFS, Preamble and ACK
   Tc : the fixed overhead for a collision
     including RTS, DIFS, SIFS and CTS
   Thus, we have
   E[Payload] = Li
L
E[ Activity]  i 1 Ps|a ( i  To )  (1  nPs|a )  Tc
n

Ci
   Substituting all parameters in,
 (1   ) n1 Li
i 
L
1  i 1[  (1   ) n1 ( i  To )]  [1  (1   ) n  n(1   ) n1 ]Tc
n

Ci
   A simpler situation: all payloads have the same length L
n
  n i                min 1i  n Ci
1
i 1 C
n

i
   The total network throughput is bounded above by the minimum bit
rate of n flows
Conclusion & Comments
   Present a simpler and more general model for single cell IEEE
802.11 WLANs
   Assume collision always results in frame loss
   Capture Effect: signal with higher signal strength will capture the
transceiver when SINR is larger than a threshold

   Assume collision is the only cause of frame loss
   Channel condition is complicated (e.g. background noise)

   Provide a throughput formula for open-loop and closed-loop traffics
   Closed form analysis for K∞ and n∞

   Results can not be readily applied to multi-hop wireless networks
Conclusion & Comments (Contd.)
   In multi-hop WLANs
   Collision may happen in the middle of transmission
CS Range

A       B      C         D

   Overlapped transmissions may both succeed

A       B       C         D

Interference Range         Interference Range
Focusing on the Back-offs of One Node
   Rj: # back-offs (attempts) until success/discarded for packet j
   Bji: the sequence of back-offs (in slots) for packet j
X j  i 0 B j : total # back-offs (in slots) for packet j
R 1
j      i


R1=2         R2=1             R3=4

Successes        B1          B11   B2     B3   B3    B3    B3
0                 0      0    1     2     4
Collisions
X1          X2                X3

   Intuitively, the back-off rate: β = E[R] / E[X]
   This can be formally proved by renewal reward theorem
 Sequence Xj are renewal life times

 Rj is a reward associated with the renewal cycle of length Xj

 By renewal reward theorem: the back-off rate β = E[R] / E[X]
Detail about Back-off Process Evolution
Z(t): back-    0        1   0    0   1     2         3
off stage
Y(t): residual
back-off process

X1       X2             X3

   (Z(t), Y(t)) is a 2-dimensional Markov chain
   Bianchi’s method
   Another way
   Z(t) is a Markov chain embedded at every attempt instants
   Z(t) and the successive back-off intervals constitute a
Markov renewal process
   For a Markov renewal process, the event (back-off) rate is
insensitive to the distributions of life-times (Xj)
Proof of Lemma 5.1
1     2  ...  K
                                          : G( )
b0  b1   b2  ...  bk  ...  bK
2            k         K

is non-increasing in γ if bk, k>=0, is a non-decreasing sequence
   Derivative of G(γ) is non-positive                                                              1             k        2k

       b  k ( j 1 j             j 1
)  k 0  k ( j 1 jb j   j 1
K                      K                       K          K
k 0 k
)
j             n
                       jbk    k  j 1
 k 0  j 1 jb j   k  j 1
K          K                               K      K
k 0       j 1                                                                      1   n-k       k        2k

                 n 1

2K                     min{n , K }
n 1                   j  max{n  K ,1}       j (b j  bk )  0                                       n
k n j                                                           j

   Denote m(n) = |{(j,k) | j+k=n, 1<=j,k<K}|
max(n-K,1}                min{n, K}

m ax{n  K ,1} m / 2 1
j  m ax{n  K ,1}                   [(( n  j )  j )( bn  j  bk )]
k n j

 n(bn  b0 )1{1 n K }  0                                                               Floor{m/2}
Verification by Coupled Back-off DTMC

   Build a Discrete Time
Markov Chain
   State Space = {m0, m1, …mK
| ∑mk=n }
   The hold time (back-off) of a
node in stage k is
geometrically distributed with
parameter 1/bK
   bk = pk∙b0, where p>=1
   This DTMC is positive
recurrent
   Numerically compute the
equilibrium distribution
   Obtain γ
2.1 Open-loop Traffic (Multi Flows)

   Each node i may have mi connections with several
other nodes
   Flow (i, j) represents jth flow of node I
      Li,j: packet length of flow (i, j)
      Ci,j : transmission rate of flow (i, j)
      pi,j : probability the packet is from flow j on node I
   Throughput of each flow (i, j)
 (1   ) n1 pi , j Li , j
i , j 
Li ,k
1  i 1[  (1   )   n 1
((k 1 pi ,k       )  To )]  [1  (1   ) n  n(1   ) n 1 ]Tc
n                            mi

Ci ,k
1

2.2 Closed-loop Traffic
   TCP connections are closed-loop when
 Transferring long files                                 2
 No bit error in transmitting

 Buffers at wireless devices are large (no overflow)

 WLAN is the bottleneck

 TCP window sizes are large enough that buffers at wireless
devices never empty out
   Key observation of close loop: if a data packet (ACK) is
transmitted, then an ACK (data packet) is inserted into the
transmitter queue of the destination (source) node
   Queues are evolving only at successful transmitting instants, and
the process can be modeled by discrete-time Markov chain
2.2 Closed-loop Traffic (contd.)
      There are m connections
    s(j): source node of jth connection
    r(j): receiver node of jth connection
    hi,j : prob. that the HOL packet at node i is a data packet from
connection j
    hi,j(ack): prob. that the HOL packet at node i is an ACK for connection j
 (1   ) n 1 hs ( j ), j L j
j 
Lk                                   L(kack )
1  i 1[  (1   )   n 1
((k 1 hs ( k ),k                                                           )  To )]  [1  (1   ) n  n(1   ) n 1 ]Tc
n                             m                                      m    ( ack )
h
k 1 r ( k ),k
Cs ( k ),r ( k )                         Cr ( k ),s ( k )
      Let λj be the average number of packets of connect j that pass
through the node s(j) per packet served in the polling model
      Theorem 8.1: If at each success instant one of the nodes is polled
with equal probability, then hs(j),j =λjn
    Every HOL packet occupy the position for n time unites
    Same argument for ACK: hs(j),j(ack)=λj(ack)n

      Model analysis yields λj andλj(ack)
2.3 Asymptotic Analysis (K , n )
   Fixed points appear to be converging as K  and n 
   Suppose bk = pk∙b0, where p>=1
1 1  p
G ( )                        ( p  1)
b0   1 

1 1  p n 1           n  1 1  p
  1  (1            )  1  exp(              )
b0 1                   b0    1 

Theorem 7.2
1) γ < 1/p: justify the assumption pγ < 1
2) lim n   1 / p : collision probability only depends on p if K and n are large
p
3) lim n n  ln(        ) : the mean attempt rate of a node scales as O(1/n)
p 1
Proof of Theorem 7.1
    Theorem 7.1: the fixed point is of the form
LambertW ( ( p  1)ep )   ( p  1)                  n 1
 ( )                                                   (         )
LambertW ( ( p  1)ep )                            b0
    Proof
n  1 1  p                     p  p 1  p                          ( p  1)
  1  exp(                )  1  exp(   (             ))  1  exp( p) exp(            )
b0    1                        1    1                            1 

 ( p  1)
1    exp( p) exp(              )
1 
Multiply n(p-1)
 ( p  1)       ( p  1)
n( p  1) exp(p)               exp(            )
1             1 
Definition of LambertW(.)
 ( p  1)
 LambertW (n( p  1) exp(p))
1 
2.3 Asymptotic Analysis (contd.)
      Assuming all flows has the same payload L and transmission rate C
      The aggregate network throughput:               1
(1  ) L
ne  n L                                                  p
                                                                                                                  :  ( p)
L                                             1            1 L              1/ p          1
1  ne  n (      To )  (1  e  n    ne  n )Tc               (1  )(  To )  (             1  )Tc
C                                              p           p C                 p          p
ln(      )                      ln(      )
p 1                            p 1
Theorem 7.4
1) lim p   ( p )  0 : a collision will cause a drastic reduction in attempt rate
2) lim p 1  ( p )  0 : collision probability becomes large if no node increase back-off
intervals when collisions happen
3) τ(p) is maximized at                                     Tc /(Tc  1)
p
LambertW (Tc /(e(Tc  1))  Tc /(Tc  1)
where LambertW(x) is the inverse function of zez

In practice, throughput is
not very sensitive to p
unless p is very close to 1
θ

p
Relaxed Fixed Point Iteration

   Usual fixed point iteration γk+1=f(γk)
   Converge if f(γ) is a contraction: |f(a)-f(b)|<=c|a-b| where c<1
   Relaxed iteration: γk+1=(1-α)f(γk) + α γk
   When {γk} converge?
Γ(G(γ))
   Only if γk>=f(γk)
1
   Since                           n 1 p  2     ,
f ( )'|r 1/ p               : D
b0   p 1
f(γk+1)- f(γk) <=|D|(γk –γk+1 )
0
γ
   Only if γk+1- f(γk) >=|D|(γk –γk+1 )                            1/p   1

   Only if α(γk+1- f(γk)) >=|D|(1-α)(γk –γk+1 )
   Only if α>=|D|/(|D|+1)
Lemma 7.1
LambertW (axe x )
   For a > 0, lim x                     1
x
   Denote z(x) = LambertW(axex)
   Thus, z(x)ez(x) = axex                ln ax
1
z ( x)   x
   ln(z(x)) + z(x) = ln(ax) + x  x  ln z ( x)
1
x
   For a > 0, lim x  ( LambertW (axe x )  x)  ln a
   z(x) – x = ln(ax)-ln(z(x)) = ln(ax/z(x))

   For 0<a<=1, LambertW(axex) <=x
   LambertW(xex) is monotone increasing

   For 0<a<=1, the convergence in Eqn (9) is from below
Theorem 7.2
1) γ < 1/p: justify the assumption pγ < 1
2) lim n   1 / p : collision probability only depends on p if K and n are large
p
3) lim n n  ln(        ) : the mean attempt rate of a node scales as O(1/n)
p 1

LambertW ( ( p  1)ep )   ( p  1)
   1) and 2), we have                   ( ) 
LambertW ( ( p  1)ep )
and                                                     p 1
LambertW ( ( p  1)ep )  LambertW (                pep )
p

   3)                                (1   )  e  ( n 1) 

(n  1)   ln(1   )
n                             1             p
n         [( n  1)  ]  n ln(         )  ln(      )
n 1                         1 1/ p         p 1
Theorem 7.3

   Rate of convergence in Theorem 7.2
            LambertW ( ( p  1)ep )   ( p  1)
 ( ) 
LambertW ( ( p  1)ep )

   Denote x=ηp, a = (1-p)/p
1             x           1               x
 ( )       1              x
a   a(1               x
)
p      LambertW (axe )    p        LambertW (axe )
1          ax
p( ( )  )                       ( LambertW (axe x )  x)
p    LambertW (axe x )

1
lim n p( ( )  )  a ln a
p
Bianchi’s Model (1)
   Consider an ideal single-hop system with N users
   Let bi,k be the probability that a node is in state (i,k)
when time∞
(1-p)/W0                                    The backoff time is decremented
at the beginning of each slot time
1               1             1         1             1
0,0          0,1               0,2                    0,W0-2         0,W0-1
….
Successful
Transmission                     p/W1
….           ….                ….          ….                 ….             ….
i-1,0
p/Wi             Unsuccessful Transmission
p/Wi 1                1             1              1               1
i,0         i,1               i,2                          i,Wi-2         i,Wi-1
….

….           ….     p/Wm       ….          ….                 ….             ….
p/Wm 1                  1             1              1               1
m,0          m,1               m,2                         m,Wm-2         m,Wm-1
p/Wm
p/Wm               i,j       Stage i, Backoff Window j
Bianchi’s Model (2)
   Note
(1)

(2)

(3)

   By (1) and (2), we have                                         (4)
   So (3) can be written as                                        (5)

   Since               , by (1)(2)(5), we can get

   Thus, the probability that a station transmits in a randomly chosen
slot is

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