Performance Analysis of IEEE 802.11 Random Access (Exponential Random Backoff) using Markov Chain

1. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
Probability and Random Process
Markov Chains (Advanced): IEEE 802.11
Exponential Random Backoff Modeling
Prof. Joongheon Kim
Korea University, School of Electrical Engineering
Artificial Intelligence and Mobility Laboratory
https://joongheon.github.io
joongheon@korea.ac.kr

2. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
Reference Paper
Giuseppe Bianchi, “Performance Analysis of the IEEE 802.11 Distributed
Coordination Function,” IEEE Journal on Selected Areas in Communications,
18(3):535-547, March 2000.
• This is the most famous paper which shows the analysis procedure of IEEE
802.11 distributed coordination function (DCF) random access.
• This paper presents the packet transmission probability and throughput of
IEEE 802.11 DCF random access, with backoff window modeling via a
discrete-time bidimensional Markov chain.
2

3. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
Discrete-Time Markov Chain for IEEE 802.11 DCF
• Basic Assumptions
• Finite and fixed number of contending stations, i.e., 𝑛.
• Saturation condition (i.e., each station is assumed to always have packets in its
own transmission buffer/queue).
• The packet collision probability, i.e., 𝑝, is constant.
• The IEEE 802.11 DCF modeling is defined as the bidimensional 𝑠(𝑡), 𝑏(𝑡) ,
• 𝑠(𝑡): stochastic process for the backoff stage 0, ⋯ , 𝑚 of the station at 𝑡.
• 𝑏(𝑡): stochastic process for the backoff time counter of the station at 𝑡.
3
𝑠(𝑡), 𝑏(𝑡)

4. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Analysis Procedure
• Modeling the Markov Chain (states and transition probabilities)
• Deriving the Stationary Distribution
• Computing the Probability that a Station Transmits in a Randomly Chosen Time (𝜏 )
• Computing the Throughput in Multiple Access (𝜏 vs 𝑝)
4

5. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
5
{0, 0} {0, 1} {0, W0-2} {0, W0-1}
{1, 0} {1, 1} {1, W1-2} {1, W1-1}
{m, 0} {m, 1} {m, Wm-2} {m, Wm-1}
……
……
……
…… …… …… ……
……

6. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 1: backoff counter is not expired
• Case 2: backoff counter is expired and the transmission is successful
• Case 3: backoff counter is expired and the transmission is collided
• Case 4: backoff stage is maximum and the transmission is collided
6
IEEE 802.11 DCF
Backoff Mechanism
Backoff Counter Expiration
Not Expired Expired
Backoff
Stage
Not Maximum Case 1 Case 2 if the transmission is successful
Case 3 if the transmission is collided
Maximum Case 1 Case 2 if the transmission is successful
Case 4 if the transmission is collided

7. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 1: backoff counter is not expired
• 𝑃({𝑖, 𝑘}|{𝑖, 𝑘 + 1}) = 1
7
Note that 𝑃 𝑖1, 𝑘1 | 𝑖0, 𝑘0 = 𝑃 𝑠(𝑡 + 1) = 𝑖1, 𝑏(𝑡 + 1) = 𝑘1|𝑠(𝑡) = 𝑖0, 𝑏(𝑡) = 𝑘0

8. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 2: backoff counter is expired and the transmission is successful
• 𝑃({0, 𝑘}|{𝑖, 0}) = (1 − 𝑝)/𝑊0
8

9. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 3: backoff counter is expired and the transmission is collided
• 𝑃({𝑖, 𝑘}|{𝑖 − 1,0}) = 𝑝/𝑊𝑖
9

10. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 4: backoff stage is maximum and the transmission is collided
• 𝑃({𝑚, 𝑘}|{𝑚, 0}) = 𝑝/𝑊
𝑚
10

11. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain (Final Form)
11
{0, 0} {0, 1} {0, W0-2} {0, W0-1}
{1, 0} {1, 1} {1, W1-2} {1, W1-1}
{m, 0} {m, 1} {m, Wm-2} {m, Wm-1}
……
……
……
1 1
1
1
1 1
1
1
1 1
1
1
…… …… …… ……
……
𝑃({𝑚, 𝑘}|{𝑚, 0}) = 𝑝/𝑊
𝑚
(1-p)/W0
p/W1
p/W2
p/Wm
p/Wm

12. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Analysis Procedure
• Modeling the Markov Chain (states and transition probabilities)
• Deriving the Stationary Distribution
• Computing the Probability that a Station Transmits in a Randomly Chosen Time (𝜏 )
• Computing the Throughput in Multiple Access (𝜏 vs 𝑝)
12

13. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• 𝑏𝑖,𝑘: stationary distribution for 𝑖 ∈ (0, 𝑚) and 𝑘 ∈ (0, 𝑊𝑖 − 1), i.e.,
• 𝑏𝑖−1,0 ∙ 𝑝 = 𝑏𝑖,0 → [A1]
Proof)
• For the state 𝑖 − 1,0 , 0 < 𝑖 < 𝑚, the next state should be selected from states 𝑖, 0 ,
𝑖, 1 , …, 𝑖, 𝑊𝑖 − 1 with probability 1/𝑊𝑖 for each state, if the transmission is collided
(with probability 𝑝)
• For the states 𝑖, 0 , 𝑖, 1 , …, 𝑖, 𝑊𝑖 − 1 , they will eventually converge to 𝑖, 0 with
probability 1 when 𝑡 → ∞
• Thus, it is obvious that the state 𝑖 − 1,0 goes to 𝑖, 0 with probability 𝑝, 0 < 𝑖 < 𝑚
• Then, the general form: 𝑏𝑖,0 = 𝑝 ∙ 𝑏𝑖−1,0 = 𝑝2
∙ 𝑏𝑖−2,0 = ⋯ = 𝑝𝑖
∙ 𝑏0,0
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𝑏𝑖,𝑘 = lim
𝑡→∞
𝑃 𝑠 𝑡 = 𝑖, 𝑏 𝑡 = 𝑗

14. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• 𝑏𝑚,0 =
𝑝𝑚
1−𝑝
𝑏0,0 → [A2] (when 𝑖 = 𝑚)
Proof) In state {𝑚, 0},
• Incoming Transition) This is the case where the transmission is collided with probability
𝑝 in the state {𝑚 − 1,0}, i.e., 𝑏𝑚−1,0 ∙ 𝑝
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Ignored because the transitions eventually
will be back to the origin {𝑚, 0}

15. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• 𝑏𝑚,0 =
𝑝𝑚
1−𝑝
𝑏0,0 → [A2] (when 𝑖 = 𝑚)
Proof, Continued) Note that the incoming probability is 𝑏𝑚−1,0 ∙ 𝑝. In state {𝑚, 0},
• Outgoing Transition) This is the case where the transmission is successful with
probability 1 − 𝑝 in the state {𝑚, 0}., i.e., 𝑏𝑚,0 ∙ (1 − 𝑝).
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𝑏𝑖−1,0 ∙ 𝑝 = 𝑏𝑖,0 → [A1]
Ignored because the transitions eventually
will be back to the origin {𝑚, 0}
Incoming = Outgoing, i.e., by [A1]
𝑏𝑚,0 =
𝑝
1 − 𝑝
𝑏𝑚−1,0 =
𝑝2
1 − 𝑝
𝑏𝑚−2,0 = ⋯ =
𝑝𝑚
1 − 𝑝
𝑏0,0 → [A2]

16. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• For each 𝑘 ∈ (0, 𝑊𝑖 − 1),
𝑏𝑖,𝑘 can be computed with three separated parts, i.e.,
• Case i) 𝑖 = 0 [initial backoff stage]
• Case ii) 0 < 𝑖 < 𝑚 [intermediate backoff stages]
• Case iii) 𝑖 = 𝑚 [maximum backoff stage]
16

17. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• Case i) For each 𝑘 ∈ (0, 𝑊𝑖 − 1), 𝑖 = 0 [initial backoff stage]
• The [Case 1) of the Modeling the Markov Chain] is not considered.
17

18. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• Case i) For each 𝑘 ∈ (0, 𝑊𝑖 − 1), 𝑖 = 0 [initial backoff stage]
• For the [Case 2) of the Modeling the Markov Chain], the 𝑘 number of transitions are not
considered among all 𝑊0 number of transitions. Thus, the 𝑊0 − 𝑘 number of transitions
are considered with probability (1 − 𝑝)/𝑊0
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𝑏0,𝑘 =
1 − 𝑝
𝑊0
𝑊0 − 𝑘 𝑏0,0 + 𝑏1,0 + ⋯ + 𝑏𝑚,0 =
𝑊0 − 𝑘
𝑊0
1 − 𝑝 ෍
𝑗=0
𝑚
𝑏𝑗,0
→ [B1]

19. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• Case ii) For each 𝑘 ∈ (0, 𝑊𝑖 − 1), 0 < 𝑖 < 𝑚 [intermediate backoff stages]
• The [Case 1) of the Modeling the Markov Chain] is not considered.
19

20. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• Case ii) For each 𝑘 ∈ (0, 𝑊𝑖 − 1), 0 < 𝑖 < 𝑚 [intermediate backoff stages]
• For the [Case 3) of the Modeling the Markov Chain], the 𝑘 number of transitions are not
considered among all 𝑊𝑖 number of transitions. Thus, the 𝑊𝑖 − 𝑘 number of transitions
are considered with probability 𝑝/𝑊𝑖
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𝑏𝑖,𝑘 =
𝑝
𝑊𝑖
𝑊𝑖 − 𝑘 𝑏𝑖−1,0 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝𝑏𝑖−1,0
→ [B2]

21. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• Case iii) For each 𝑘 ∈ (0, 𝑊𝑖 − 1), 𝑖 = 𝑚 [maximum backoff stage]
• The [Case 1) of the Modeling the Markov Chain] is not considered.
21

22. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• Case iii) For each 𝑘 ∈ (0, 𝑊𝑖 − 1), 𝑖 = 𝑚 [maximum backoff stage]
• For the [Case 3) of the Modeling the Markov Chain], two situations exist, i.e.,
• The state 𝑚 − 1,0 with probability 𝑝/𝑊
𝑚 for each transition
22

23. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• Case iii) For each 𝑘 ∈ (0, 𝑊𝑖 − 1), 𝑖 = 𝑚 [maximum backoff stage]
• For the [Case 4) of the Modeling the Markov Chain], two situations exist, i.e.,
• The state 𝑚, 0 with probability 𝑝/𝑊
𝑚 for each transition
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𝑏𝑚,𝑘 =
𝑝
𝑊
𝑚
𝑊
𝑚 − 𝑘 𝑏𝑚−1,0 + 𝑏𝑚,0 =
𝑊
𝑚 − 𝑘
𝑊
𝑚
𝑝 𝑏𝑚−1,0 + 𝑏𝑚,0
→ [B3]

24. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• Finally, for all Case i), Case ii) and Case iii),
• For 𝑖 = 0,
• For 0 < 𝑖 < 𝑚,
• For 𝑖 = 𝑚,
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𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝 𝑏𝑖−1,0 + 𝑏𝑖,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝𝑏𝑖−1,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
1 − 𝑝 ෍
𝑗=0
𝑚
𝑏𝑗,0
𝑏𝑖−1,0 ∙ 𝑝 = 𝑏𝑖,0 → [A1]
𝑏𝑚,0 =
𝑝𝑚
1−𝑝
𝑏0,0 → [A2]
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝 𝑏𝑖−1,0 + 𝑏𝑖,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝𝑏𝑖−1,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
1 − 𝑝 ෍
𝑗=0
𝑚
𝑏𝑗,0
For 𝑖 = 0,
For 0 < 𝑖 < 𝑚,
For 𝑚 < 𝑖,
→ [B1]
→ [B2]
→ [B3]

25. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• When the packet transmission is successful,
and suppose the backoff expiration states
(i.e., 0,0 , 1,0 , … , 𝑚, 0 ) go to 0, 𝑘
where 𝑘 ∈ 0, 𝑊0 − 1
with probability (1 − 𝑝)/𝑊0
for all states,
• And the 𝑊0 number of result states converge to 𝑏0,0
with probability 1, i.e.,
25
1 − 𝑝
𝑊0
෍
𝑗=0
𝑚
𝑏𝑗,0
𝑊0
1 − 𝑝
𝑊0
෍
𝑗=0
𝑚
𝑏𝑗,0 = 𝑏0,0 thus, ෍
𝑗=0
𝑚
𝑏𝑗,0 =
𝑏0,0
1 − 𝑝
→ [C]
𝑏𝑖−1,0 ∙ 𝑝 = 𝑏𝑖,0 → [A1]
𝑏𝑚,0 =
𝑝𝑚
1−𝑝
𝑏0,0 → [A2]
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝 𝑏𝑖−1,0 + 𝑏𝑖,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝𝑏𝑖−1,0
For 0 < 𝑖 < 𝑚,
For 𝑚 < 𝑖,
→ [B2]
→ [B3]
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
1 − 𝑝 ෍
𝑗=0
𝑚
𝑏𝑗,0
For 𝑖 = 0,
→ [B1]

26. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• For [B1] (for 𝑖 = 0),
• By applying [C] to [B1],
• In this case, 𝑏0,0 = 𝑏𝑖,0, i.e., 𝑏𝑖,𝑘 =
𝑊𝑖−𝑘
𝑊𝑖
𝑏𝑖,0
• For [B2] (for 0 < 𝑖 < 𝑚),
• By applying [A1] to [B2],
26
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
1 − 𝑝 ෍
𝑗=0
𝑚
𝑏𝑗,0 =
𝑊𝑖 − 𝑘
𝑊𝑖
1 − 𝑝
𝑏0,0
1 − 𝑝
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑏0,0
→[D]
𝑏𝑖−1,0 ∙ 𝑝 = 𝑏𝑖,0 → [A1]
𝑏𝑚,0 =
𝑝𝑚
1−𝑝
𝑏0,0 → [A2]
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝 𝑏𝑖−1,0 + 𝑏𝑖,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝𝑏𝑖−1,0
For 0 < 𝑖 < 𝑚,
For 𝑚 < 𝑖,
→ [B2]
→ [B3]
→ [C]
෍
𝑗=0
𝑚
𝑏𝑗,0 =
𝑏0,0
1 − 𝑝
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝𝑏𝑖−1,0 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑏𝑖,0
→[D]
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
1 − 𝑝 ෍
𝑗=0
𝑚
𝑏𝑗,0
For 𝑖 = 0,
→ [B1]

27. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• For [B1] and [B2], 𝑏𝑖,𝑘 =
𝑊𝑖−𝑘
𝑊𝑖
𝑏𝑖,0
• For [B3] (for 𝑖 = 𝑚),
• By [A2],
• Then, [B3] can be as follows:
27
→[D]
𝑏𝑖−1,0 ∙ 𝑝 = 𝑏𝑖,0 → [A1]
𝑏𝑚,0 =
𝑝𝑚
1−𝑝
𝑏0,0 → [A2]
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝 𝑏𝑖−1,0 + 𝑏𝑖,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝𝑏𝑖−1,0
For 0 < 𝑖 < 𝑚,
For 𝑚 < 𝑖,
→ [B2]
→ [B3]
→ [C]
෍
𝑗=0
𝑚
𝑏𝑗,0 =
𝑏0,0
1 − 𝑝
1 − 𝑝
𝑝
𝑏𝑚,0 = 𝑏𝑚−1,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝 𝑏𝑖−1,0 + 𝑏𝑖,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝
1 − 𝑝
𝑝
𝑏𝑖,0 + 𝑏𝑖,0 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑝
1
𝑝
𝑏𝑖,0
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑏𝑖,0
→[D] (General Form)
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
1 − 𝑝 ෍
𝑗=0
𝑚
𝑏𝑗,0
For 𝑖 = 0,
→ [B1]

28. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• The summation of all 𝑏𝑖,𝑘 should be 1, i.e.,
28
→ [D]
𝑏𝑖,𝑘 =
𝑊𝑖 − 𝑘
𝑊𝑖
𝑏𝑖,0
1 = ෍
𝑖=0
𝑚
෍
𝑘=0
𝑊𝑖−1
𝑏𝑖,𝑘 = ෍
𝑖=0
𝑚
෍
𝑘=0
𝑊𝑖−1
𝑊𝑖 − 𝑘
𝑊𝑖
𝑏𝑖,0 (by [D])
= ෍
𝑖=0
𝑚
𝑏𝑖,0 ෍
𝑘=0
𝑊𝑖−1
𝑊𝑖 − 𝑘
𝑊𝑖
= ෍
𝑖=0
𝑚
𝑏𝑖,0 ෍
𝑘=0
𝑊𝑖−1
1 −
𝑘
𝑊𝑖
= ෍
𝑖=0
𝑚
𝑏𝑖,0 𝑊𝑖 −
1
𝑊𝑖
෍
𝑘=0
𝑊𝑖−1
𝑘
= ෍
𝑖=0
𝑚
𝑏𝑖,0 𝑊𝑖 −
1
𝑊𝑖
𝑊𝑖 − 1 𝑊𝑖
2
= ෍
𝑖=0
𝑚
𝑏𝑖,0 𝑊𝑖 −
𝑊𝑖 − 1
2
= ෍
𝑖=0
𝑚
𝑏𝑖,0
𝑊𝑖 + 1
2
=
1
2
෍
𝑖=0
𝑚
𝑏𝑖,0 𝑊𝑖 + 1
[General Form of Backoff Mechanisms]
→ [E]

29. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• In IEEE 802.11 DCF
(Exponential Random Backoff), 𝑊𝑖 = 2𝑖
𝑊
where 𝑊 = 𝐶𝑊min and 𝐶𝑊
max = 𝑊
𝑚 = 2𝑚𝑊
• Therefore, [E] can be developed as follows:
29
→ [E]
1 =
1
2
෍
𝑖=0
𝑚
𝑏𝑖,0 𝑊𝑖 + 1
1 =
1
2
෍
𝑖=0
𝑚
𝑏𝑖,0 𝑊𝑖 + 1 =
1
2
𝑏0,0 𝑊0 + 1 + ⋯ + 𝑏𝑚−1,0 𝑊𝑚−1 + 1 + 𝑏𝑚,0 𝑊
𝑚 + 1
1 =
1
2
𝑏0,0 20
𝑊 + 1 + ⋯ + 𝑝𝑚−1
𝑏0,0 2𝑚−1
𝑊 + 1 +
𝑝𝑚
1 − 𝑝
𝑏0,0 2𝑚
𝑊 + 1
𝑏𝑖−1,0 ∙ 𝑝 = 𝑏𝑖,0 → [A1]
𝑏𝑚,0 =
𝑝𝑚
1−𝑝
𝑏0,0 → [A2]
(by [A1], [A2])
1 =
𝑏0,0
2
20
𝑊 + 1 + ⋯ + 𝑝𝑚−1
2𝑚−1
𝑊 + 1 +
𝑝𝑚
1 − 𝑝
2𝑚
𝑊 + 1
Continued…

30. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• In IEEE 802.11 DCF
(Exponential Random Backoff), Continued, …
30
1 =
𝑏0,0
2
20
𝑊 + 1 + ⋯ + 𝑝𝑚−1
2𝑚−1
𝑊 + 1 +
𝑝𝑚
1 − 𝑝
2𝑚
𝑊 + 1
1 =
𝑏0,0
2
𝑊 20
+ ⋯ + 𝑝𝑚−1
2𝑚−1
+
𝑝𝑚
1 − 𝑝
2𝑚
+ 1 + ⋯ + 𝑝𝑚−1
+
𝑝𝑚
1 − 𝑝
1 =
𝑏0,0
2
𝑊 ෍
𝑖=0
𝑚−1
(2𝑝)𝑖
+
(2𝑝)𝑚
1 − 𝑝
+
1(1 − 𝑝𝑚
)
1 − 𝑝
+
𝑝𝑚
1 − 𝑝
1 =
𝑏0,0
2
𝑊 ෍
𝑖=0
𝑚−1
(2𝑝)𝑖
+
(2𝑝)𝑚
1 − 𝑝
+
1
1 − 𝑝
→ [F]

31. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• In IEEE 802.11 DCF
(Exponential Random Backoff), Continued, …
• From [F],
31
𝑏0,0 =
2
𝑊 σ𝑖=0
𝑚−1
(2𝑝)𝑖 +
(2𝑝)𝑚
1 − 𝑝
+
1
1 − 𝑝
=
2
𝑊
2𝑝 0(1 − 2𝑝 𝑚)
1 − 2𝑝
+
(2𝑝)𝑚
1 − 𝑝
+
1
1 − 𝑝
→ [F]
1 =
𝑏0,0
2
𝑊 ෍
𝑖=0
𝑚−1
(2𝑝)𝑖
+
(2𝑝)𝑚
1 − 𝑝
+
1
1 − 𝑝
𝑏0,0 =
2
𝑊 1 − 2𝑝 𝑚 1 − 𝑝 + 𝑊 2𝑝 𝑚(1 − 2𝑝)
(1 − 2𝑝)(1 − 𝑝)
+
1
1 − 𝑝
𝑏0,0 =
2
𝑊 1 − 2𝑝 𝑚 1 − 𝑝 + 𝑊 2𝑝 𝑚(1 − 2𝑝)
(1 − 2𝑝)(1 − 𝑝)
+
(1 − 2𝑝)
(1 − 2𝑝)(1 − 𝑝)
𝑏0,0 =
2(1 − 2𝑝)(1 − 𝑝)
𝑊 1 − 2𝑝 𝑚 1 − 𝑝 + 𝑊 2𝑝 𝑚(1 − 2𝑝) + (1 − 2𝑝)

32. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Deriving the Stationary Distribution)
• Deriving the Stationary Distribution
• From
32
𝑏0,0 =
2(1 − 2𝑝)(1 − 𝑝)
𝑊 1 − 2𝑝 𝑚 1 − 𝑝 + 𝑊 2𝑝 𝑚(1 − 2𝑝) + (1 − 2𝑝)
𝑏0,0 =
2(1 − 2𝑝)(1 − 𝑝)
𝑊 − 𝑊 2𝑝 𝑚 − 𝑝𝑊 + 𝑝𝑊 2𝑝 𝑚 + 𝑊 2𝑝 𝑚 − 2𝑝𝑊 2𝑝 𝑚 + (1 − 2𝑝)
𝑏0,0 =
2(1 − 2𝑝)(1 − 𝑝)
𝑊 − 𝑝𝑊 − 𝑝𝑊 2𝑝 𝑚 + (1 − 2𝑝)
𝑏0,0 =
2(1 − 2𝑝)(1 − 𝑝)
(1 − 2𝑝)𝑊 + (1 − 2𝑝) + 𝑝𝑊(1 − 2𝑝 𝑚)
𝑏0,0 =
2(1 − 2𝑝)(1 − 𝑝)
(1 − 2𝑝)(𝑊 + 1) + 𝑝𝑊(1 − 2𝑝 𝑚)
𝑏0,0 =
2(1 − 2𝑝)(1 − 𝑝)
𝑊 − 2𝑝𝑊 + 𝑝𝑊 − 𝑝𝑊 2𝑝 𝑚 + (1 − 2𝑝)
→ [G]

33. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Probability to Transmit in a Randomly Chosen Time)
• Analysis Procedure
• Modeling the Markov Chain (states and transition probabilities)
• Deriving the Stationary Distribution
• Computing the Probability that a Station Transmits in a Randomly
Chosen Time (𝝉)
• Computing the Throughput in Multiple Access (𝜏 vs 𝑝)
33

34. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Probability to Transmit in a Randomly Chosen Time)
• Probability that a Station Transmits in a Randomly Chosen Time (𝜏)
• 𝜏: the probability that a node transmits in a randomly chosen slot time.
• As any transmission occurs when the backoff time counter is equal to zero, regardless of
the backoff stage, it is obvious that
• By applying [C] and [G],
34
𝜏 = ෍
𝑖=0
𝑚
𝑏𝑖,0
→ [C]
෍
𝑗=0
𝑚
𝑏𝑗,0 =
𝑏0,0
1 − 𝑝
𝑏0,0 =
2(1 − 2𝑝)(1 − 𝑝)
(1 − 2𝑝)(𝑊 + 1) + 𝑝𝑊(1 − 2𝑝 𝑚)
→ [G]
𝜏 = ෍
𝑖=0
𝑚
𝑏𝑖,0 =
𝑏0,0
1 − 𝑝
=
1
1 − 𝑝
2(1 − 2𝑝)(1 − 𝑝)
(1 − 2𝑝)(𝑊 + 1) + 𝑝𝑊(1 − 2𝑝 𝑚)
𝜏 =
2(1 − 2𝑝)
(1 − 2𝑝)(𝑊 + 1) + 𝑝𝑊(1 − 2𝑝 𝑚)
→ [H]

35. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Probability to Transmit in a Randomly Chosen Time)
• Probability that a Station Transmits in a Randomly Chosen Time (𝜏)
• From [H],
35
𝜏 =
2(1 − 2𝑝)
(1 − 2𝑝)(𝑊 + 1) + 𝑝𝑊(1 − 2𝑝 𝑚)
=
2(1 − 2𝑝)
(1 − 2𝑝)(𝑊 + 1) + 𝑝𝑊 1 − (2𝑝) σ𝑖=0
𝑚−1
2𝑝 𝑖
෍
𝑖=0
𝑚−1
(2𝑝)𝑖
=
(2𝑝)0
1 − (2𝑝)𝑚
1 − (2𝑝)
𝜏 =
2
(𝑊 + 1) + 𝑝𝑊 σ𝑖=0
𝑚−1
2𝑝 𝑖
(W: 2, m: 7)
Single User Access

36. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Probability to Transmit in a Randomly Chosen Time)
• Analysis Procedure
• Modeling the Markov Chain (states and transition probabilities)
• Deriving the Stationary Distribution
• Computing the Probability that a Station Transmits in a Randomly Chosen Time (𝜏 )
• Computing the Throughput in Multiple Access (𝝉 vs 𝒑)
36

37. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Computing the Throughput)
• Computing the Throughput in Multiple Access (𝜏 vs 𝑝)
• In general, 𝜏 (the probability that a node transmits in a randomly chosen slot time)
depends on the conditional collision probability 𝑝
• To find 𝑝, we have to note that 𝑝 is the probability that, in a time slot, at least
one of the (𝑛 − 1) remaining nodes transmit
• The fundamental independence assumption given above implies that each
transmission ”sees” the system in the same state, i.e., in steady state
• At steady state, each remaining station transmits a packet with probability 𝜏
• Thus, 𝑝 = 1 − (1 − 𝜏)𝑛−1
• By inverting this 𝑝 = 1 − (1 − 𝜏)𝑛−1, it is true that 1 − 𝜏 = (1 − 𝑝)
1
𝑛−1, finally,
where it is continuous monotonic increasing in 𝑝 ∈ (0,1),
that starts from 𝜏∗
0 = 0 to 𝜏∗
1 = 1
37
𝜏∗
𝑝 = 1 − (1 − 𝑝)
1
𝑛−1

38. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Computing the Throughput)
• Computing the Throughput in Multiple Access (𝜏 vs 𝑝)
38
𝜏∗
𝑝 = 1 − (1 − 𝑝)
1
𝑛−1
Relationship) If there is no collision
probability (𝑝 = 0), it is true that
there is no transmission
from the other stations in any
random slots

39. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
Q&A
Thank you for your attention!
• Questions?
• joongheon@korea.ac.kr
39

40. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
Appendix A.
Markov Chain Illustrations for IEEE
802.11 DCF
Prof. Joongheon Kim
Korea University, School of Electrical Engineering
Artificial Intelligence and Mobility Laboratory
https://joongheon.github.io
joongheon@korea.ac.kr

41. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
41
{0, 0} {0, 1} {0, W0-2} {0, W0-1}
{1, 0} {1, 1} {1, W1-2} {1, W1-1}
{m, 0} {m, 1} {m, Wm-2} {m, Wm-1}
……
……
……
…… …… …… ……
……

42. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 1: backoff counter is not expired
• Case 2: backoff counter is expired and the given packet is successfully
transmitted
• Case 3: backoff counter is expired and the given packet is collided
• Case 4: backoff stage is maximum and the given packet is collided
42

43. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 1: backoff counter is not expired
• For state {𝑖, 𝑗}, 𝑏(𝑡) = 𝑗 ≠ 0
• The corresponding states should eventually go to {𝑖, 0}
• Once the backoff counter is selected with the uniform distribution in the interval
0, 𝑊𝑖 − 1 , the counter definitely decreases toward 0 (i.e., expired) with probability 1
• 𝑃({𝑖, 𝑘}|{𝑖, 𝑘 + 1}) = 1
• Note that 𝑃 𝑖1, 𝑘1 | 𝑖0, 𝑘0 = 𝑃 𝑠(𝑡 + 1) = 𝑖1, 𝑏(𝑡 + 1) = 𝑘1|𝑠(𝑡) = 𝑖0, 𝑏(𝑡) = 𝑘0
43

44. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain (for Case 1)
44
{0, 0} {0, 1} {0, W0-2} {0, W0-1}
{1, 0} {1, 1} {1, W1-2} {1, W1-1}
{m, 0} {m, 1} {m, Wm-2} {m, Wm-1}
……
……
……
1 1
1
1
1 1
1
1
1 1
1
1
…… …… …… ……
……
𝑃({𝑖, 𝑘}|{𝑖, 𝑘 + 1}) = 1

45. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 2: backoff counter is expired and the given packet is successfully
transmitted
• For state {𝑖, 𝑗}, 𝑏(𝑡) = 𝑗 = 0
• When packet transmission is successful (probability: (1 − 𝑝)), the backoff stage goes to
initial setting (i.e., 𝑠(𝑡) = 𝑖 = 0)
• In the initial backoff stage, the backoff counter is selected in the interval 0, 𝑊0 − 1
• Within this interval, the backoff time is uniformly selected, i.e., the probability for
selecting one next stage is 1/𝑊0
• 𝑃({0, 𝑘}|{𝑖, 0}) = (1 − 𝑝)/𝑊0
45

46. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain (for Case 2)
46
{0, 0} {0, 1} {0, W0-2} {0, W0-1}
{1, 0} {1, 1} {1, W1-2} {1, W1-1}
{m, 0} {m, 1} {m, Wm-2} {m, Wm-1}
……
……
……
1 1
1
1
1 1
1
1
1 1
1
1
…… …… …… ……
……
𝑃({0, 𝑘}|{𝑖, 0} = (1 − 𝑝)/𝑊0
(1-p)/W0

47. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 3: backoff counter is expired and the given packet is collided
• For state {𝑖, 𝑗}, 𝑏(𝑡) = 𝑗 = 0
• When packet transmission is collided (probability: 𝑝), the backoff stage increases, one-
stage further (i.e., at 𝑠(𝑡) = 𝑖 − 1, the stage is updated to 𝑠(𝑡) = 𝑖)
• In the updated backoff stage, the backoff counter is selected in the interval 0, 𝑊𝑖 − 1
• Within this interval, the backoff time is uniformly selected, i.e., the probability for
selecting one next stage is 1/𝑊𝑖
• 𝑃({𝑖, 𝑘}|{𝑖 − 1,0}) = 𝑝/𝑊𝑖
47

48. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain (for Case 3)
48
{0, 0} {0, 1} {0, W0-2} {0, W0-1}
{1, 0} {1, 1} {1, W1-2} {1, W1-1}
{m, 0} {m, 1} {m, Wm-2} {m, Wm-1}
……
……
……
1 1
1
1
1 1
1
1
1 1
1
1
…… …… …… ……
……
𝑃({𝑖, 𝑘}|{𝑖 − 1,0}) = 𝑝/𝑊𝑖
(1-p)/W0
p/W1
p/W2
p/Wm

49. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain
• Case 4: backoff stage is maximum and the given packet is collided
• For state {𝑖, 𝑗}, 𝑠(𝑡) = 𝑚 and 𝑏(𝑡) = 𝑗 = 0
• When packet transmission is collided (probability: 𝑝), the backoff stage needs to
increase, one-stage further, however, it is impossible because the stage is already in
maximum. Thus, the stage cannot be updated, i.e., 𝑠(𝑡) = 𝑚
• In the backoff stage, the backoff counter is selected in the interval 0, 𝑊
𝑚 − 1
• Within this interval, the backoff time is uniformly selected, i.e., the probability for
selecting one next stage is 1/𝑊
𝑚
• 𝑃({𝑚, 𝑘}|{𝑚, 0}) = 𝑝/𝑊
𝑚
49

50. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain (for Case 4)
50
{0, 0} {0, 1} {0, W0-2} {0, W0-1}
{1, 0} {1, 1} {1, W1-2} {1, W1-1}
{m, 0} {m, 1} {m, Wm-2} {m, Wm-1}
……
……
……
1 1
1
1
1 1
1
1
1 1
1
1
…… …… …… ……
……
𝑃({𝑚, 𝑘}|{𝑚, 0}) = 𝑝/𝑊
𝑚
(1-p)/W0
p/W1
p/W2
p/Wm
p/Wm

51. Dr. Joongheon Kim
https://joongheon.github.io
Probability and Random Process
IEEE 802.11 DCF (Modeling the Markov Chain)
• Modeling the Markov Chain (Final Form)
51
{0, 0} {0, 1} {0, W0-2} {0, W0-1}
{1, 0} {1, 1} {1, W1-2} {1, W1-1}
{m, 0} {m, 1} {m, Wm-2} {m, Wm-1}
……
……
……
1 1
1
1
1 1
1
1
1 1
1
1
…… …… …… ……
……
𝑃({𝑚, 𝑘}|{𝑚, 0}) = 𝑝/𝑊
𝑚
(1-p)/W0
p/W1
p/W2
p/Wm
p/Wm