Utilisation des m´ ethodes champ moyen pour l’´ evaluation de performance Nicolas Gast (Inria) Inria, Grenoble, France S´ eminaire de l’institut Fourier, Octobre 2016 Nicolas Gast (Inria) – 1 / 26
Models of interacting objects (in computer science) Wifi: object = device object = content Cluster: object = server Nicolas Gast (Inria) – 2 / 26
Models of interacting objects (in computer science) Wifi: object = device object = content Problem: state space explosion S states per object, N objects ⇒ S N states (and 4 20 = 10 12 ) Cluster: object = server Nicolas Gast (Inria) – 2 / 26
Mean-field model Population of N objects. X i ( t ) = fraction of objects in state i Nicolas Gast (Inria) – 3 / 26
Mean-field model Population of N objects. X i ( t ) = fraction of objects in state i Example: N servers 1 The state is ( X 0 , X 1 , X 2 . . . ). . . . . N ρ X i ( t ) = fraction of servers . . with i jobs 1 Randomly choose two, and select one Nicolas Gast (Inria) – 3 / 26
Some systems simplify as N grows 0.35 ODE ( N = ∞ ) 0.30 0.25 0.20 0.15 0.10 0.05 0.00 0 1 2 3 4 5 Time Example. Two-choice model Fraction of servers with 3 jobs At time 0: all servers have 1 jobs. Nicolas Gast (Inria) – 4 / 26
Some systems simplify as N grows 0.35 ODE ( N = ∞ ) 0.30 N=10 0.25 0.20 0.15 0.10 0.05 0.00 0 1 2 3 4 5 Time Example. Two-choice model Fraction of servers with 3 jobs At time 0: all servers have 1 jobs. Nicolas Gast (Inria) – 4 / 26
Some systems simplify as N grows 0.35 0.30 0.25 0.20 0.15 0.10 ODE ( N = ∞ ) N=10 0.05 N=100 0.00 0 1 2 3 4 5 Time Example. Two-choice model Fraction of servers with 3 jobs At time 0: all servers have 1 jobs. Nicolas Gast (Inria) – 4 / 26
Some systems simplify as N grows 0.35 0.30 0.25 0.20 0.15 ODE ( N = ∞ ) 0.10 N=10 N=100 0.05 N=1000 0.00 0 1 2 3 4 5 Time Example. Two-choice model Fraction of servers with 3 jobs At time 0: all servers have 1 jobs. Nicolas Gast (Inria) – 4 / 26
Some systems simplify as N grows 0.35 0.30 0.25 0.20 0.15 ODE ( N = ∞ ) N=10 0.10 N=100 N=1000 0.05 N=10000 0.00 0 1 2 3 4 5 Time Example. Two-choice model Fraction of servers with 3 jobs At time 0: all servers have 1 jobs. Nicolas Gast (Inria) – 4 / 26
Some systems simplify as N grows 0.35 0.30 0.25 0.20 Objective of this talk 0.15 ODE ( N = ∞ ) When is the ODE N=10 0.10 approximation N=100 N=1000 0.05 valid / not valid? N=10000 0.00 What is the 0 1 2 3 4 5 Time accuracy? Example. Two-choice model Fraction of servers with 3 jobs At time 0: all servers have 1 jobs. Nicolas Gast (Inria) – 4 / 26
Outline (Classical) Kurtz Population Model 1 Accuracy of the Approximation 2 Example: jobs allocation 3 Conclusion and recap 4 Nicolas Gast (Inria) – 5 / 26
Population CTMC A population process is a sequence of CTMC X N , indexed by the population size N , with state spaces E N ⊂ E such that the transitions are (for ℓ ∈ L ): X �→ X + ℓ at rate N β ℓ ( X ) . N � The drift is f ( x ) = ℓβ ℓ ( x ). ℓ We denote by x the solution of the associated ODE x = f ( x ) . ˙ Nicolas Gast (Inria) – 6 / 26
Transient regime Let Φ t denotes the (unique) solution of the ODE: � t Φ t x = x + Φ s xds . 0 Theorem (Kurtz 70s) If f is Lipschitz-continuous with constant L, then for any fixed T: � � � � X N ( t ) − x ( t ) � N →∞ sup lim � = 0 . t < T Proof. Martingale concentration + Gronwall. Nicolas Gast (Inria) – 7 / 26
The fixed point method Markov chain Transient regime p = pK ˙ t → ∞ Stationary π K = 0 Nicolas Gast (Inria) – 8 / 26
The fixed point method Markov chain Mean-field Transient regime p = pK ˙ x = f ( x ) ˙ N → ∞ t → ∞ f ( x ∗ ) = 0 Stationary π K = 0 ? fixed points Method was used in many papers: Bianchi 00, Performance analysis of the IEEE 802.11 distributed coordination function. Ramaiyan et al. 08, Fixed point analys is of single cell IEEE 802.11e WLANs: Uniqueness, multistability. Kwak et al. 05, Performance analysis of exponenetial backoff. Kumar et al 08, New insights from a fixed-point analysis of single cell IEEE 802.11 WLANs. Nicolas Gast (Inria) – 8 / 26
Does it always work? SIRS model: Markov chain is irreducible. Unique fixed point f ( x ∗ ) = 0. I Fixed point Stat. measure 1 + 10 x I N = 10 3 , 10 4 . . . f ( x ) = 0 5 x S + a π S π I x S x I a = . 3 0.209 0.234 0.209 0.234 10 x S + 10 − 3 S R Nicolas Gast (Inria) – 9 / 26
Does it always work? SIRS model: Markov chain is irreducible. Unique fixed point f ( x ∗ ) = 0. I Fixed point Stat. measure 1 + 10 x I N = 10 3 , 10 4 . . . f ( x ) = 0 5 x S + a π S π I x S x I a = . 3 0.209 0.234 0.209 0.234 10 x S + 10 − 3 S R a = . 1 0.078 0.126 0.11 0.13 Nicolas Gast (Inria) – 9 / 26
What happened? 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 limit cycle Fixed point x ∗ = π N true stationnary distribution Fixed point 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 0.0 0.0 0.0 0.0 1.0 1.0 1.0 1.0 a = . 1 a = . 3 Nicolas Gast (Inria) – 10 / 26
Fixed points? Markov chain Mean-field Transient regime p = pK ˙ x = f ( x ) ˙ N → ∞ t → ∞ f ( x ∗ ) = 0 Stationary π K = 0 ? fixed points Nicolas Gast (Inria) – 11 / 26
Fixed points? Markov chain Mean-field Transient regime p = pK ˙ x = f ( x ) ˙ N → ∞ t → ∞ t → ∞ f ( x ∗ ) = 0 f ( x ∗ ) = 0 Stationary π K = 0 N → ∞ fixed points Nicolas Gast (Inria) – 11 / 26
Fixed points? Markov chain Mean-field Transient regime p = pK ˙ x = f ( x ) ˙ N → ∞ t → ∞ if yes t → ∞ f ( x ∗ ) = 0 f ( x ∗ ) = 0 Stationary π K = 0 N → ∞ fixed points then yes Nicolas Gast (Inria) – 11 / 26
Fixed points? Markov chain Mean-field Transient regime p = pK ˙ x = f ( x ) ˙ N → ∞ t → ∞ if yes t → ∞ f ( x ∗ ) = 0 Stationary π K = 0 N → ∞ fixed points then yes Theorem (Benaim Le Boudec 08) If all trajectories of the ODE converges to the fixed points, the stationary distribution π N concentrates on the fixed points In that case, we also have: N →∞ P [ Z 1 = i 1 . . . Z k = i k ] = x ∗ 1 . . . x ∗ lim k . Nicolas Gast (Inria) – 11 / 26
Example of 802.11 1 1 Cho, Le Boudec, Jiang, On the Asymptotic Validity of the Decoupling Assumption for Analyzing 802.11 MAC Protoco. 2010 Nicolas Gast (Inria) – 12 / 26
Quiz Consider the 802.11 model: Under the stationary distribution π N : (A) P ( Z 1 = 0 , Z 2 = 0) ≈ P ( Z 1 = 0) P ( Z 2 = 0) (B) P ( Z 1 = 0 , Z 2 = 0) > P ( Z 1 = 0) P ( Z 2 = 0) (C) P ( Z 1 = 0 , Z 2 = 0) < P ( Z 1 = 0) P ( Z 2 = 0) (D) There is no stationary distribution (E) I do not know Nicolas Gast (Inria) – 13 / 26
Quiz Consider the 802.11 model: Under the stationary distribution π N : (A) P ( Z 1 = 0 , Z 2 = 0) ≈ P ( Z 1 = 0) P ( Z 2 = 0) (B) P ( Z 1 = 0 , Z 2 = 0) > P ( Z 1 = 0) P ( Z 2 = 0) (C) P ( Z 1 = 0 , Z 2 = 0) < P ( Z 1 = 0) P ( Z 2 = 0) (D) There is no stationary distribution (E) I do not know Answer: B P ( Z 1 ( t ) = 0 , Z 2 ( t ) = 0) = x 1 ( t ) 2 . Thus: positively correlated. Nicolas Gast (Inria) – 13 / 26
Outline (Classical) Kurtz Population Model 1 Accuracy of the Approximation 2 Example: jobs allocation 3 Conclusion and recap 4 Nicolas Gast (Inria) – 14 / 26
How accurate is mean-field approximation? X N i ( t ) = fraction of 0.35 objects in state i . 0.30 0.25 0.20 Theorem (Kurtz 70s’) 0.15 ODE ( N = ∞ ) When f is Lipschitz: 0.10 N=10 N=100 0.05 � 1 N=1000 X N ( t ) − x ( t ) = O √ ) 0.00 0 1 2 3 4 5 N Time Example. Two-choice model, Fraction of servers with 3 jobs In practice, we use mean-field for N ≥ 50. Are we wrong? Nicolas Gast (Inria) – 15 / 26
How accurate is mean-field approximation? X N i ( t ) = fraction of 0.35 objects in state i . 0.30 0.25 0.20 Theorem (Kurtz 70s’) 0.15 ODE ( N = ∞ ) When f is Lipschitz: 0.10 N=10 N=100 0.05 � 1 N=1000 X N ( t ) − x ( t ) = O √ ) 0.00 0 1 2 3 4 5 N Time Example. Two-choice model, Fraction of servers with 3 jobs In practice, we use mean-field for N ≥ 50. Are we wrong? 10 100 1000 + ∞ N Average queue length ( m N ) 3.81 3.39 3.36 3.35 Error ( m N − m ∞ ) 0.45 0.039 0.004 0 Nicolas Gast (Inria) – 15 / 26
0.20 0.15 Proba(3 jobs) 0.10 Limit ( N = ∞ ) 0.05 Steady-state probability for fixed N 0.00 0 50 100 150 200 N Where is the catch? √ O (1 / N ) X i = fraction of x i (mean-field servers with i jobs approx) Nicolas Gast (Inria) – 16 / 26
0.20 0.15 Proba(3 jobs) 0.10 Limit ( N = ∞ ) 0.05 Steady-state probability for fixed N 0.00 0 50 100 150 200 N Where is the catch? √ O (1 / N ) X i = fraction of Proba(one server x i (mean-field servers with i jobs has i jobs) = E [ X i ] approx) √ ? O (1 / N ) (CLT) Nicolas Gast (Inria) – 16 / 26
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