Initial Markov under Bianchi s École Polytechnique, Paris, France June 20th, 2005
Outline Initial Markov under Bianchi s 1 2 Initial Markov under Bianchi s 3
Outline Initial Markov under Bianchi s 1 2 Initial Markov under Bianchi s 3
Single-cell WLAN Initial Markov under Bianchi s N stations share a wireless channel. Everybody hears everybody ( hypothesis). Ideal channel, no capture. Saturation hypothesis. Motivation : estimate the saturation throughput.
The MAC layer Initial Markov under Bianchi s 802.11 MAC (medium access control) layer is a collision avoidance scheme. Time is slotted. Every station s state is entirely determined two integers : its backoff level L i (t) (number of collisions for the current packet), and its backoff counter C i (t) (number of slots before the next attempt). At the beginning of a slot, station i : decrements its backoff counter C i (t) if C i (t) > 0. attempts to transmit if C i (t) = 0, and then chooses randomly its new backoff counter. The other stations freeze their counters during channel activity.
Station 1 transmits Initial Markov under Bianchi s When its backoff counter reaches 0, station 1 transmits and then chooses a new backoff counter. STA 1 c1=2 c1=1 c1=0 c1=9 CHANNEL ACTIVITY duration of a slot c1=8 Physical time
Station 2 freezes Station 2 listens to the channel. When channel activity is detected, station 2 freezes its backoff counter. Initial Markov under Bianchi s STA 1 c1=2 c1=0 c1=9 c1=1 CHANNEL ACTIVITY c1=8 Physical time duration of a slot c2=7 c2=5 c2=4 STA 2 c2=6 COUNTER IS FROZEN c2=3
Backoff time definition Initial Markov under Bianchi s So station 1 and station 2 decrement their counters at the same time : we can define a discrete so-called backoff time which corresponds to the instants when backoff counters are decremented. STA 1 STA 2 c1=2 c2=7 c1=1 c2=6 c1=0 c1=9 CHANNEL ACTIVITY duration of a slot c2=5 c2=4 COUNTER IS FROZEN c1=8 Physical time c2=3 BACKOFF TIME t=0 t=1 t=2 t=3 t=4
What happens during channel activity? Initial Markov under Bianchi s Two different schemes can be used, namely Basic and RTS/CTS. The Basic access mechanism is a two-way handshaking technique (DATA - ACK). The RTS/CTS access mechanism is a four-way handshaking technique (RTS - CTS - DATA - ACK). itself is independent of the access mechanism (the backoff time is frozen during channel activity).
Calculating the throughput Initial Markov under Bianchi s The throughput, which motivated our work, is defined as : E[payload information per slot] E[length of a slot] To calculate this, we need to know : the access mechanism used (Basic, RTS/CTS), to deal with channel activity periods, the attempt rate and collision rate, to deal with backoff periods. So the purpose of the following work is to calculate these two values : the attempt rate (mean number of attempts per slot for a given station) and the collision rate (mean number of collisions per attempt for a given station).
Remarks concerning the model... Initial Markov under Bianchi s Collisions are detected thanks to a ACK TIME OUT (or CTS TIME OUT), which are not taken into account. Backoff time is discrete and different from physical time. The physical duration of one slot is different depending on whether the slot is an idle one, a success or a collision. BACKOFF TIME t=0 t=1 t=2 t=3 t=4
Outline Initial Markov under Bianchi s 1 2 Initial Markov under Bianchi s 3
How backoff counters are chosen Initial Markov under Bianchi s Every station (i) has a backoff level L i (t) and a backoff counter C i (t). There are K + 1 backoff levels (L i (t) [0, K]). After K + 1 attempts, a packet either succeeds or is discarded. For each backoff level L [0, K], there is a given so-called contention window CW(L) N. Contention windows are non-decreasing (CW(L + 1) CW(L)). For instance, 802.11 contention windows are 16/ 32/ 64/ 128/ 256/ 1024/ 1024/ 1024.
How backoff counters are chosen Initial Markov under Bianchi s If C i (t) > 0 : C i (t + 1) = C i (t) 1 (the counter is decremented), L i (t + 1) = L i (t) (the level is unchanged). If C i (t) = 0 : station i attempts to transmit, a new level and a new counter are chosen depending on whether the attempt was successful or not.
How backoff counters are chosen Initial Markov under Bianchi s In case of success by station i at time t : L i (t + 1) = 0 (backoff level is brought back to 0), C i (t + 1) is chosen with uniform distribution in [0, CW(0) 1]. In case of collision by station i at time t : L i (t + 1) = L i (t) + 1 (unless L i (t) = K, in which case the packet is discarded and L i (t + 1) = 0), C i (t + 1) is chosen with uniform distribution in [0, CW(L i (t + 1)) 1].
The process is markovian Initial Markov under Bianchi s The joint process for the N stations is markovian : the state S := (L 1 (t), C 1 (t),..., L N (t), C N (t)) is an irreductible Markov. Let π be its stationary distribution. Calculating π would give us the attempt rate and the collision rate. Problem : the Markov is too complicated for π to be calculated explicitly.
Bianchi s Initial Markov under Bianchi s Let us consider the following (Bianchi, 2000) : 1. the collision probability for any station at any time is a constant p, 2. the collisions at different times are independent. This is similar to a very general method, called the mean field. It is very popular in Statistical Physics, Quantum Mechanics (adiabatic ), Mechanics, Fluid Mechanics... Under Bianchi s, each station s level and counter (L i (t), C i (t)) are a Markov independently of the other stations (hence the name, decoupling ).
Resulting simplified Markov (1-p)/(CW(0)) Initial Markov under Bianchi s... 0,0 0,1 0,2 0,CW(0)-1 p/(cw(1)) 1,0 1,1 1,2... p/(cw(m)) m,0 m,1 m,2 p/(cw(m))...... 1,CW(1)-1 m,cw(m)-1
Emission rate in stationary state Initial Markov It is possible to calculate the stationary distribution of the simplified Markov, ρ p (note that it depends on the value of p, which is unknown for the moment). The emission probability for a given station is then : under Bianchi s τ = K ρ p (l, 0) l=0 since ρ p is explicitly known, we find an explicit formula : τ = G(p)
Initial Markov under Bianchi s Assumed collision probability and fixed point equation Since all stations are supposed to be independent, at equilibrium, and they transmit on the channel with probability τ : The number of emissions at a given time t seen by a given station is binomial with parameters (N 1,τ). In particular, the collision probability seen by a given station is : p = 1 (1 τ) N 1 So we find a fixed point equation for p : p = 1 (1 G(p)) N 1
Outline Initial Markov under Bianchi s 1 2 Initial Markov under Bianchi s 3
Ex I : fixed point equation For 802.11 contention windows (16/ 32/ 64...) with N = 10 stations, the fixed point equation looks like : Initial Markov under Bianchi s 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 (There is always a unique balanced fixed point).
Ex. I : simulation results If we simulate the initial markov with the same contention windows and N = 10 stations, and if we plot the attempt and collision rates, we find : Initial Markov under Bianchi s 1 0.9 0.8 0.7 Markov simulator Attempt rate Collision rate fixed point analysis (gamma) fixed point analysis (tau) Rates 0.6 0.5 0.4 0.3 0.2 0.1 0 0 10000 20000 30000 40000 50000 60000 70000 80000 90000 100000 Time
Initial Markov under Bianchi s Ex. II : fixed point equation Let us now consider the following contention windows, with N = 10 stations : 1/ 3/ 9/ 27/ 81/ 243/ 729/ 2187. The fixed point equation looks like : 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Ex. II : simulation results If we simulate the initial markov with the same contention windows and N = 10 stations, and if we plot the collision rates, we find : Initial Markov 1 0.9 Example II Collision rate Fixed point analysis under Bianchi s 0.8 0.7 Rates 0.6 0.5 0.4 0.3 0.2 0.1 0 0 10000 20000 30000 40000 50000 60000 70000 80000 90000 100000 Time
Initial Markov under Bianchi s Unbalanced fixed points Bianchi s analysis assumes that all stations have the same emission probability τ, and see the same collision probability p = 1 (1 G(p)) N 1. Though, it is possible to consider unbalanced fixed points : p 1 = 1 (1 G(p 2 )) (1 G(p 3 )) p 2 = 1 (1 G(p 1 )) (1 G(p 3 ))... p N = 1 (1 G(p N 2 )) (1 G(p N 1 )) We already know that there is a balanced solution of the form p 1 = p 2 = = p N. An unbalanced solution for this system would correspond to a situation of unfairness between the stations (some stations experience a higher collision rate than others).
Resolution of the system Initial Markov under Bianchi s The fixed point system is non-linear and N-dimensional. Under specific assumptions on the contention windows (which are satisfied for 802.11 contention windows) it is possible to prove that there is no unbalanced fixed point. Otherwise, the system is difficult to solve.
Ex. II fixed points Initial Markov under Bianchi s The situation in example II was : N = 10, the contention windows are 1/ 3/ 9/ 27/ 81/ 243/ 729/ 2187. For this system it is very likely that the only fixed points are either of the form (p, p,..., p) or of the form (p 1, p 2, p 2,..., p 2 ). Under this assumption, and after some calculations, the N-dimensional system simplifies to a one-dimensional fixed point equation for p 2.
Graphical resolution for p 2 : Initial Markov under Bianchi s 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 We find three solutions for p 2, which corresponds to three values for p 1.
Simulation results Initial Markov under Bianchi s 1 0.9 0.8 0.7 0.6 Example II (unbalanced fixed points) Markov p1 (1) p1 (2) p1=p2 (balanced fixed point) p2 (2) p2 (1) Rates 0.5 0.4 0.3 0.2 0.1 0 0 10000 20000 30000 40000 50000 60000 70000 80000 90000 100000 Time
Cause and effects of the existence of multiple fixed points Initial Markov under Bianchi s Contention windows : 1/ 3/ 9/ 27/ 81/ 243/ 729/ 2187. Bianchi s model is not accurate at all in this case : the balanced fixed point is unstable, even the most unbalanced fixed point is a poor model since no equilibrium is reached. Short term unfairness between the stations occurs. The decoupling does not apply here.
Initial Markov under Bianchi s
Initial Markov under Bianchi s Mean-field (main idea) : replace a of interacting particles (here, stations) by a of independent particles in an external effective field. It is very accurate in most cases (N large enough, unique fixed point). Yet in specific cases (2 or 3 stations, or multistability), the can be very poor.
Open questions Initial Markov under Bianchi s Find a formal proof that the is accurate, solve the system of equations for unbalanced fixed points, or at least find an upper bound for the number of solutions, when several fixed points exist, study the stability of a fixed point.
The end... Initial Markov under Bianchi s Any questions?