Presentation is loading. Please wait.

Presentation is loading. Please wait.

Presented by Hoang Nguyen

Similar presentations


Presentation on theme: "Presented by Hoang Nguyen"— Presentation transcript:

1 Presented by Hoang Nguyen
Performance Analysis of the IEEE Distributed Coordination Function Giuseppe Bianchi Presented by Hoang Nguyen CS598JH - Spring 06

2 Saturation Throughput
Problem Formulation Saturation Throughput Asymptotic throughput Assumptions Ideal Channel Condition (No Hidden Terminal, No Channel Capture) Finite Number of Stations Constant & Independent Collision Probability Overload Condition

3 Outline Analytical Model Saturation Throughput Analysis Maximum Saturation Throughput Performance Evaluation Conclusion

4 Analytical Model

5 Analytical Model b(t): stochastic process representing the backoff counter s(t): stochastic process representing the backoff stage (0..m) W = CWmin and CWmax = 2mW Wi = 2i W at stage i 2 (0,m) {s(t), b(t)}: bidimensional process with discrete-time Markov Chain Conditional Collision Probability p Transmission Probability 

6 Markov Model for Single Station
(1-p)/W0 Successful Transmission 1 1 …. 1 1 1 0,0 0,1 0,2 0,W0-2 0,W0-1 …. p/W_1 The backoff time is decremented at the beginning of each slot time i-1,0 p/Wi p/Wi 1 1 1 1 1 i,0 i,1 i,2 …. i,Wi-2 i,Wi-1 Unsuccessful Transmission …. 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

7 Stationary Distribution Derivation
Derivation for bi-1,0.p = bi,0 p/Wi p/Wi 1 1 …. 1 i,0 i,1 i,Wi-1 W_0. (1-p)/W_0 Flow-in Flow-out Wi+1. p/Wi+1 bi-1,0.p/Wi + bi+1,0.1 = bi,0 + bi-1,0.p/Wi + bi+2,0.1 = bi,1 …. bi-1,0.p/Wi = bi,Wi-1 bi-1 p = bi,0

8 Stationary Distribution Derivation
Use similar derivation, we get… Use the fact that We get Plus Therefore, Then,

9 Transmission Probability
Now, we have… And, p = 1 – (1-)n-1 Thus, p and  can be solve by numerical techniques

10 Saturation Throughput Analysis

11 Saturation Throughput
Consider a slot time… Probability at least one transmission Ptr = 1 – (1-)n Probability of successful transmission Ps = n(1-)n-1 / Ptr

12 Saturation Throughput (cont.)
S = E[payload information transmitted in a slot time] / E[length of a slot time] where Ts = average time channel sensed busy due to a successful transmission Tc = average time channel sensed busy due to a collision  = duration of empty slot time E[P] = average packet payload size Probability of a collision Probability of a successful transmission Probability of idle slot

13 Saturation Throughput (cont.)
Ts and Tc depend on channel access mechanism For basic access mechanism Similarly, for RTS/CTS mechanism

14 Model Validation Very accurate

15 Maximum Saturation Throughput

16 Maximum Saturation Throughput
Re-arrange S To maximize S, maximize this term… Take the derivation and impose to 0…

17 Approximate Solution More sensitive
n is unknown;  depends on m and W, which are fixed in hardware!

18 Performance Evaluation

19 Saturation Throughput vs. Initial Window Size
Almost independent when W · 64 Saturation throughput highly depends on W and n

20 Maximum Backoff Stage Greater than 4 is fine

21 Conclusions

22 Conclusions Pros Cons: Analytical Model
Accurate Simple Performance Evaluation on saturation Throughput Cons: Only Saturation Throughput Only Overload Condition Only Ideal Channel Condition

23 Thank You!

24 Backup Slides

25 Markov Model for Single Station
(1-p)/W0 1 1 1 …. 1 1 0,0 0,1 0,2 0,W0-2 0,W0-1 …. p/W_1 i-1,0 p/Wi 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


Download ppt "Presented by Hoang Nguyen"

Similar presentations


Ads by Google