Transcript YNU-ANTL - Yeungnam Univ. Adavanced Networking Technology

2014 YU-ANTL Lab Seminar
Impact of Block ACK Window sliding on IEEE
802.11n throughput performance
June 7, 2014
Shinnazar Seytnazarov
Advanced Networking Technology Lab. (YU-ANTL)
Dept. of Information & Comm. Eng, Graduate School,
Yeungnam University, KOREA
(Tel : +82-53-810-3940; Fax : +82-53-810-4742
http://antl.yu.ac.kr/; E-mail : [email protected])
OUTLINE
 Introduction
 Frame aggregations
 BAW sliding
 The analytical model
 Expected A-MPDU length derivation
 Throughput derivation
 Analytical results
 Conclusion
 References
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Introduction (1)
 A-MPDU (Aggregation of MPDUs) - aggregation scheme [1]
 Sender can aggregate up to 64 MPDUs in A-MPDU frame
 If receiver receives at least one of the MPDUs successfully, it sends back
Block ACK (Block acknowledgement) frame informing about transmission
status MPDUs
MPDU
MAC
delimiter header
MAC payload
FCS
Pad
MPDU
PLCP header
MPDU1
MPDU2
...
MPDU64
Tail/Pad
A-MPDU
Fig. 1. Aggregation of MPDUs
Starting
Sequence
Number
1 2 3 ... 63 64
Bitmap
PLCP
header
Frame
Control
Receiver Transmitter
BA
BA
Address
Address
Control Info
FCS
Fig. 2. Block ACK frame format
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Introduction (2)
 Block ACK Window (BAW) sliding [1]
 BAW size is equal to 64 that is the maximum allowed A-MPDU length
 Sender can transmit the MPDUs that are within the BAW
 BAW continues sliding forward unless any of the MPDUs inside the BAW
fails
BAW sliding direction
Previous position of BAW
Current position of BAW
100 101 102 101 102 103
...
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Introduction (3)
 Simple example for BAW = 4
Sender’s window
Sender is sending SNs: 101~104
...
Transmitted and successfully received MPDU
...
Transmitted but failed MPDU
...
New MPDU in A-MPDU and BAW
...
MPDU outside of BAW
Receiver’s window
Receiver is anticipating SNs: 101~104
101
102
103
104
105
101
102
103
104
105
...
Erro
r
100
100
...
TX
104
103
102
101
A-MPDU1
Receiver is anticipating SNs: 103, 105, 106
TX
1
Sender is sending SNs: 103, 105, 106
103
104
105
106
107
0
1
102
103
104
105
106
107
...
Erro
r
102
1
BlockACK1
...
TX
106
105
103
A-MPDU2
Receiver is anticipating SN: 103
TX
0
Sender is sending SN: 103
102
103
104
105
106
107
1
1
0
102
BlockACK2
103
104
105
106
107
...
...
TX
103
Receiver is anticipating SNs: 107~110
TX
ACK
Sender is sending SNs: 107~110
106
107
108
109
110
111
106
107
108
109
110
111
...
...
TX
110
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109
108
A-MPDU3
5
107
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Expected A-MPDU length derivation (1)
 We introduce several random variables:





L – number of MPDUs in A-MPDU i.e. length of A-MPDU, L = 1, 2, . . , 64
N – number of new MPDUs in A-MPDU, N = 0, 1, 2, . . , L
S – number of successful MPDUs in A-MPDU, S = 0, 1, 2, . . , L
F – number of failed/erroneous MPDUs in A-MPDU, F = 0, 1, 2, . . , L
X – number of successful MPDUs until the first failure in A-MPDU, X = 1,
2, . . , L
 We need to find:
 Expected number of MPDUs in A-MPDU - E[L]
 Expected number of successful MPDUs in A-MPDU - E[S]
 Expected number of failed MPDUs in A-MPDU - E[F]
 Assumptions:
 Sender’s buffer always has enough number of MPDUs to fill the BAW
window
 MPDU errors occur independently and identically over MPDUs of A-MPDU
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Expected A-MPDU length derivation (2) [2]
 Considering assumption (2), the number of failed MPDUs has binomial distribution F ~ B(pe,
L), where pe is MPDU error probability and L is the number of MPDUs in A-MPDU:
P F=k =
L
k
pke (1 − pe )L−k
(1)
 So, the expected number of failed/erroneous MPDUs is:
L
k=0 k P
EF =
F=k =
L
L
k=0 k k
pke 1 − pe
L−k
(2)
= pe L
 Number of successfully transmitted MPDUs also has a binomial distribution S ~ B(1 - pe, L):
P S=n =
L
n
(1 − pe )n pL−n
e
(3)
 So, the expected number of successful MPDUs per A-MPDU is:
L
n=0 n P
ES =
S=n =
L
L
n=0 n n
(1 − pe )n pL−n
= 1 − pe L = L − E[F]
e
(4)
 PMF for the number of first successful MPDUs in A-MPDU can be written as:
P X=k =
(1 − pe )k pe , for 1 ≤ k ≤ L − 1
(1 − pe )k , for k = L
( 5)
 Using the above PMF we can calculate expected number of new MPDUs in A-MPDU; 𝑘
the expected window shift, where W depicts the window size which is 64:
E[N] =
L−1 W
k=1 k L (1 −
k
k+1 W
L−1
k=1 k(q
−q
)
L
pe )k pe + L
W
L
(1 − pe )L =
L−1 W k
k=1 k L q (1 −
W
L
+ WqL =
q−qL+1
1−q
𝐿
gives
q) + WqL =
+ WqL = q − q2 + 2q2 − 2q3 + ⋯ + L − 1 qL−1 − L − 1 qL
q + q2 + q3 + ⋯ + qL−1 + qL − LqL
𝑊
− LqL
W
L
+ WqL
W
L
+ WqL =
(6)
Here, q = 1 − pe
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Expected A-MPDU length derivation (3) [2]
 The length of A-MPDU – L is the composition of failed MPDUs of previous A-MPDU
and newly included MPDUs.
L(i) = F(i − 1) + N(i)
(7)
 It is obvious that under certain channel conditions, the expected length of AMPDU is the sum of the expectations of failed MPDUs and new MPDUs:
E L =E F +E N
(8)
 Thus, we will use the expected A-MPDU length instead of A-MPDU length for
Equations (1-6):
E L = E F + E N = pe E L +
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q−qE L +1
1−q
8
− E L qE L
W
EL
+ WqE L =
1−qE L
1−q
−
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Performance of BAW sliding under different
channel conditions (1)
 Expected length of A-MPDU for different window sizes
under different channel conditions
E[L](W=64)
E[L](W=128)
140
120
100
80
60
35.34
40
20.65
20
24.30
14.57
0
0
0.05
0.1
0.15
0.2
0.25
0.3
MPDU error probability
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Performance of BAW sliding under different
channel conditions (2)
 Expected length of A-MPDU, expected number of successful
and failed MPDUs under different channel conditions
E[L](W=64)
E[S](W=64)
E[F](W=64)
70
60
50
40
30
20
10
0
0
0.05
0.1
0.15
0.2
0.25
0.3
MPDU error probability
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Discrete time Markov chain [3]
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Transmission probability
 Transmission probability τ that a station transmits in a randomly chosen
slot time.
τ=
2(1−2p)
1−2p w+1 +pw 1− 2p m
(10)
 p is backoff stage increment probability due to either collision or A-MPDU
failure because of channel noise:
(11)
p = 1 − (1 − τ)𝑛−1 1 − pe E L
 Equations (10) and (11) can be solved using numerical method
and have unique solution for 𝛕.
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Slot durations
 Idle slot duration Ti: When all STAs are counting down, no station
transmits a frame and we have
(12)
Ti = σ
 Successful slot duration Ts: At least one MPDU in A-MPDU successfully
received by receiver, the slot duration is the sum of a A-MPDU, a SIFS
and an Block ACK duration
Ts = TPHY_hdr + TA−MPDU + TSIFS + TBlock_ACK
(13)
 Collision and ‘A-MPDU failure due to noise’ slot durations Tc and Tf:
Tc = Tf = TPHY_hdr + TA−MPDU + TEIFS
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(14)
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Probabilities of Time Slots
 Idle slot is observed if none of the stations transmits:
Pi = (1 − τ)n
(15)
 Successful slot is observed if only one station transmits and A-MPDU is not
fully failed
Ps =
n
1
τ 1−τ
n−1
1 − pe E L
= nτ(1 − τ)n−1 1 − pe E L
(16)
 Failure slot is observed if only one station transmits and A-MPDU is fully
failed
Pf =
n
1
τ 1−τ
n−1 p E L
e
= nτ(1 − τ)n−1 pe E L
(17)
 Collision slot is observed if none of other slots is observed:
(18)
Pc = 1 − Pi − Ps − Pf
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Network throughput
 Network throughput can be defined as:
S=
E[successful A−MPDU length]
E[slot duration]
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=
Ps E[S]
𝑘 Pk Tk
15
P 1−pe E[L]
s Ts +Pc Tc +Pf Tf
i i
= P T +Ps
(19)
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Parameters for numerical analysis
MAC header
MPDU payload
PHY header duration
Data transmission rate
Block ACK transmission rate
Maximum backoff stage (m)
CWmin
Slot duration (σ)
DIFS
SIFS
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34B
1000B
44us
300/600Mbps
24Mbps
5
15
9us
28us
10us
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Performance analysis of IEEE 802.11n
considering BAW sliding (1)
 Network throughput “with BAW” at R = 300Mbps
Pe = 0.0
Pe = 0.1
Pe = 0.3
275
225
175
125
75
1
10
20
30
Number of stations
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Performance analysis of IEEE 802.11n
considering BAW sliding (2)
 Network throughput “with BAW” at R = 600Mbps
Pe = 0.0
Pe = 0.1
Pe = 0.3
500
400
300
200
100
1
10
20
30
Number of stations
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Performance analysis of IEEE 802.11n
considering BAW sliding (3)
 Network throughput comparison “with and without BAW” at
R = 300Mbps
w/oBAW_Pe = 0.1
withBAW_Pe = 0.1
w/oBAW_Pe = 0.3
withBAW_Pe = 0.3
250
200
150
100
50
1
10
20
30
Number of stations
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Performance analysis of IEEE 802.11n
considering BAW sliding (4)
 Network throughput comparison “with and without BAW” at
R = 600Mbps
w/oBAW_Pe = 0.1
withBAW_Pe = 0.1
w/oBAW_Pe = 0.3
withBAW_Pe = 0.3
450
400
350
300
250
200
150
1
10
20
30
Number of stations
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Performance analysis of IEEE 802.11n
considering BAW sliding (5)
 Difference (%) between 'with BAW' and 'without BAW' at
different PHY rates
600M_Pe = 0.3
300M_Pe = 0.3
600M_Pe = 0.1
300M_Pe = 0.1
60
50
40
30
20
10
0
1
10
20
30
Number of stations
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Conclusion
 In this presentation
 We analyzed the BAW sliding effect on A-MPDU length under different
channel conditions
 When MPDU error probability increases from 0.0 to 0.3 BAW decreases the AMPDU length from
–
–
64 to 14.57 for window size of 64
128 to 20.65 for window size of 128
 BAW model was applied in DTMC model for IEEE 802.11n
 Network throughput was analyzed for different number of nodes and different
channel conditions
 Existing DTMC models for IEEE 802.11n performance have huge difference:
–
–
Over 20% when MPDU error probability 0.1 at 600Mbps PHY rate
Over 10% when MPDU error probability 0.1 at 300Mbps PHY rate
 Conclusion
 BAW sliding has significant impact on A-MPDU size and network performance
under erroneous channel conditions
 It is essential to consider BAW effect in order to have an accurate network
performance estimations
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References
[1] IEEE 802.11n, Part 11: Standard for Wireless LAN Medium Access Control (MAC) and Physical
Layer (PHY) Specifications Amendment 5: Enhancements for Higher Throughput, Sept. 2009.
[2] Ginzburg, Boris, and Alex Kesselman. "Performance analysis of A-MPDU and A-MSDU aggregation
in IEEE 802.11 n." In Sarnoff symposium, 2007 IEEE, pp. 1-5. IEEE, 2007.
[3] G. Bianchi, “Performance analysis of the IEEE 802.11 distributed coordination function,” IEEE
JSAC, vol. 18, no. 3, pp. 535–547, Mar. 2000.
[4] T. Li, Q. Ni, D. Malone, D. Leith, Y. Xiao, and R. Turletti, “Aggregation with fragment retransmission
for very high-speed WLANs,” IEEE/ACM Transactions on Networking, vol. 17, no. 2, pp. 591–604,
Apr. 2009.
[5] Chatzimisios, P., A. C. Boucouvalas, and V. Vitsas. "Influence of channel BER on IEEE 802.11 DCF."
Electronics letters 39.23 (2003): 1687-9.
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