Diffusion approximations of queueing networks: a direct method for proving stationary distribution convergence

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1 Diffusion approximations of queueing networks: a direct method for proving stationary distribution convergence Anton Braverman 1, Jim Dai 2 and Masakiyo Miyazawa 3 1,2 School of Operations Research and Information Engineering Cornell University and 3 Department of Information Sciences Tokyo University of Science Stochastic Network Conference at UCSD June 24, / 29

2 Motivation: M/M/1 queue L In M/M/1, P{L( ) = k} = (1 ρ)ρ k for k Z +. As ρ = λ e /λ s 1, Set n = 1/(1 ρ). Then, (1 ρ)l ρ ( ) = Z( ) exp(1). (1) {L n (t)/ n, t 0} = Z = {Z(t), t 0}, (2) where Z is a one-dimensional reflecting Brownian motion, whose stationary distribution follows exp(1). (2) ought to imply (1). Can we prove (1) without referencing (2) in general? 2 / 29

3 Topics Single class queueing networks Braverman, Dai, and Miyazawa, Heavy traffic approximation for the stationary distribution of a generalized Jackson network: the BAR approach. August arxiv: Miyazawa (2015): Diffusion Approximation for Stationary Analysis of Queues and Their Networks: A Review, Journal of the Operations Research Society of Japan. Multiclass queueing networks? 3 / 29

4 An example of a generalized Jackson network (GJN) External Arrivals R e,1 L 1 R s, P = L 2 R s,2 L 3 R s, / 29

5 Model inputs External arrivals to station i follow a renewal process. Inter-arrival times are i.i.d T e,i, where E[T e,i ] = 1 λ e,i < and Var(T e,i ) = σ 2 e,i <. Service times at station i are i.i.d T s,i, where E[T s,i ] = 1 λ s,i < and Var(T s,i ) = σ 2 s,i <. After completing service at station i, a customer moves on to station j with probability p ij. Let P be the transition matrix defined by [P] ij = p ij. 5 / 29

6 System dynamics Markovian Representation: X = {X(t)} t 0 is Markov process, where state at time t is X(t) = (L(t), R e (t), R s (t)). L i (t) = number of customers at station i at time t R e,i (t) = residual time until next exogenous arrival to station i R s,i (t) = residual time until next service completion at station i The GJN in the figure has a 7-dimensional representation: ( L1 (t), R e,1 (t), R s,1 (t) ), (L 2 (t), R s,2 (t)), (L 3 (t), R s,3 (t)) 6 / 29

7 Stationary distribution Effective arrival rate to station i is λ a,i, which satisfies λ a,i = λ e,i + λ a,j p ji or λ a = λ e + P λ a. When λ a < λ s, Markov process X is positive Harris recurrent (Down-Meyn 1995): as t, L(t) = L( ) = (L 1 ( ), L 2 ( ), L 3 ( )). Understand the distribution of L( ). Computer simulations Heavy traffic asymptotics 7 / 29

8 Heavy traffic approximation Consider a sequence of systems with d stations, indexed by n 1. Heavy traffic condition: λ (n) s,i λ (n) a,i = b i n, b i > 0, i = 1,..., d. Theorem Assume {(T (n) s,i ) 2 } n 1 and {(T (n) e,i ) 2 } n 1 are uniformly integrable for each i. Then L (n) ( ) n Z( ) and ( 1 ) L n L (n) ( ) L (n) i ( ) = 0 L(Z( ) Z i ( ) = 0), where L(X) is the distribution of a random variable X and Z( )... 8 / 29

9 Semimartingale reflecting Brownian motion (SRBM) Z(t) = X(t) + RY (t) for all t 0, X is a (µ, Σ)-Brownian motion, Z(t) R d + for all t 0, Y ( ) is continuous and nondecreasing with Y (0) = 0, Y i ( ) only increases when Z i ( ) = 0, i = 1,..., d. Z 2 R (1) µ For d = 2, inputs are ( ) ( ) µ1 Σ11 Σ 12 µ =, Σ =, µ 2 Σ 21 Σ 22 ( ) R = (R (1), R (2) r11 r 12 ) =, r 21 r 22 R (2) Z 1 where R is assumed to be completely-s Z exists and is unique in distribution; (Harrison-Reiman 81, Taylor-Williams 92) 9 / 29

10 Characterization: Basic Adjoint Relationship (BAR) Harrison-Williams 1987 A (µ, Σ, R)-SRBM corresponding to a GJN has a stationary distribution iff R 1 µ < 0. Random variable Z( ) has the stationary distribution ν. Moment generating functions: for θ R d with θ 0 [ ] φ(θ) = Ee θ,z( ), φ i (θ) = E e θ,z( ) Z i ( ) = 0, i = 1,..., d. BAR (Dai-Kurtz 94, Kang-Ramanan 12, Dai-Miyazawa-Wu 14) A (µ, Σ, R)-SRBM has stationary distribution ν iff for θ 0 ( 1 2 θ, Σθ + µ, θ ) ϕ ν (θ) + d a i R (i), θ ϕ νi (θ) = 0 (3) i=1 for some constant a = (a i ) > / 29

11 Previous results: limit interchange L (n) (t) t L (n) ( ) Z(t) n t Z( ) n Convergence of steady-state queue lengths of a GJN already established: Gamarnik and Zeevi (2006): light-tailed inter-arrival and service times Budhiraja and Lee (2009): second moments and uniform integrability There is a growing literature: Tezcan (2008), Zhang-Zwart (2008), Katsuda (2010), Gamarnik-Stolyar (2012), D-Dieker-Gao (2014), Gurvich (2014), Ye-Yao (2015), and more... Both methods centered around justifying the limit interchange lim lim t n L(n) (t) = lim lim n t L(n) (t). 11 / 29

12 Proof outline Define the moment generating functions (mgf s) For i = 1,..., d, φ n (θ) := E [ e θ,l(n) ( )/ n ] for θ 0. φ n i (θ) := E [ e θ,l(n) ( )/ n L (n) i ( ) = 0 ] for θ 0. Intend to prove that (1) as n, the following limits exist φ n (θ) φ(θ) and φ n i (θ) φ i (θ) for θ 0, (2) the limits φ and φ i satisfy BAR (3), (3) the limits are mgf s of proper random variables. 12 / 29

13 Step 1: Preliminary 0 φ n (θ) 1 for each n 1 and each θ 0 Fixing an n, φ n is monotone in θ φ n (θ) φ n (θ ) if θ θ. For each subsequence {n }, there exists another subsequence {n } such that and φ i (θ) := are well defined. φ(θ) := lim E[ ) e θ,l(n ( )/ n ] for θ 0 n lim E[ ) e θ,l(n ( )/ n L (n ) n i ( ) = 0 ] for θ 0 13 / 29

14 Step 2: Limits satisfy BAR To prove (3), it suffices to prove that for each θ 0 ( 1 2 θ, Σθ + µ, θ )E [ e θ,l(n ) ( )/ n ] + d i=1 b i R (i), θ E [ e θ,l(n ) ( )/ n L (n ) i ( ) = 0 ] 0 as n. (4) Define L (n) (t) = L (n) (t)/ n. 14 / 29

15 Step 3: Continuity of φ s at θ = 0 Prove the continuity at θ = 0, namely, Namely, lim θ 0 φ(θ) = 1 and lim θ 0 φ i (θ) = 1 for all i = 1,..., d (5) φ(0 ) = 1 and φ i (0 ). (6) The continuity holds iff φ(θ) and φ i (θ) are mgf s of proper random variables. The continuity holds iff { L (n ) ( )} and { L (n ) ( ) L (n ) i ( ) = 0} are tight. The continuity (6) is shown to hold by algebraic manipulations of BAR (3) using the fact that R is an M-matrix. 15 / 29

16 Step 3: Tightness proof, tandem queue example L 1 L 2 BAR is ( 1 2 θ, Σθ + (b 1 b 2 )θ 2 b 1 θ 1 ) φ(θ) + b 1 (θ 1 θ 2 )φ 1 (θ 2 ) + b 2 θ 2 φ 2 (θ 1 ) = 0 Setting θ 2 = 0, one has ( 1 2 Σ 11θ 2 1 b 1 θ 1 )φ(θ 1, 0) + b 1 θ 1 φ 1 (0) = 0. Taking θ 1 0, we have φ(0, 0) = φ 1 (0) = 1, implying { L (n) 1 ( )} is tight. Setting θ = r(1, r ) and taking r 0, we have φ(0, 0 ) = φ 1 (0 ), Setting θ 1 = 0 and taking θ 2 0, we have which implies φ(0, 0 ) φ 1 (0 ). b 1 (φ(0, 0 ) φ 1 (0 )) + b 2 (1 φ(0, 0 )) = 0, which implies that φ(0, 0 ) = 1. Thus, { L (n) 2 ( )} is tight. Marginal tightness of { L (n) (n) 1 ( )} and { L 2 ( )} implies joint tightness. 16 / 29

17 Step 2: Piecewise deterministic Markov process (PDMP) L 1 L 2 The process X = (L 1, L 2, R e, R s,1, R s,2 ) is a piecewise deterministic Markov process (PDMP); Davis (1981) A sample path of a PDMP is composed of two parts, deterministic and continuous sections and (random) jumps due to expiration of remaining times. T s (1) T s (2) R s (t) R e (t) T s (3) T s (4) 0 L(t) T e (1) T e (2) T e (3) T e (4) T e (5) t Figure: A sample path of remaining times for GI/G/1 queue 17 / 29

18 Step 2: Change of variables for PDMP Define the jump size f (X(s)) = f (X(s)) f (X(s )), f (X(t)) f (X(0)) = t = 0 t 0 d dr e f (X(u)) d f (X(u)) du + f (X(s i )) ds s i (0,t] 2 i=1 [ t + f (X(u))dN A (u) + 0 1{L i (u) > 0} d dr s,i f (X(u))du 2 i=1 t 0 f (X(u))dN Di (u) ] ] (continuous) (jumps) N A arrival process. N Di station i departure process. 18 / 29

19 Step 2: Tandem queue BAR L 1 L 2 Basic Adjoint Relationship (BAR) 0 = E ν t 0 [ d dr e f (X(u)) + 2 i=1 [ t +E ν f (X(u))dN A (u) + 0 1{L i (u) > 0} d ] f (X(u)) du dr s,i 2 i=1 t 0 f (X(u))dN Di (u) jump terms intractable; getting rid of arrival-jump term requires ] (continuous) (jumps) E Te [ f (L1 + 1, L 2, T e, R s,1, R s,2 ) f (L 1, L 2, 0, R s,1, R s,2 ) ] = 0 (7) for any given L = (L 1, L 2 ) and R s = (R s,1, R s,2 ). 19 / 29

20 Step 2: Choosing test functions Choose f (X) = exp (θ 1 L 1 / n + θ 2 L 2 / ) n + ηr e + ζ 1 R s,1 + ζ 2 R s,2 To get rid of arrival jump, choose η such that [ E exp ( ) ] ηt e = e θ1/ n To get rid of station 2 service completion jump, choose ζ 2 such that [ E exp ( ) ] ζ 2 T s,2 = e θ2/ n To get rid of station 1 service completion jump, choose ζ 1 such that [ E exp ( ) ] ζ 1 T s,1 = e (θ1 θ2) n (8) (9) (10) 20 / 29

21 Step 2: Reduced BAR BAR becomes (η + ζ 1 + ζ 2 )E[f (X( ))] ζ 1 E[1 {L1( )=0}f (X( ))] ζ 2 E[1 {L2( )=0}f (X( ))] = 0. Or (η + ζ 1 + ζ 2 )E[f (X( ))] + ζ 1 (1 ρ 1 )E[f (X( )) L 1 ( ) = 0] + ζ 2 (1 ρ 2 )E[f (X( )) L 2 ( ) = 0] = 0. [ E[f (X( ))] = E exp(θ 1 L 1 / n + θ 2 L 2 / ] n) exp(ηr e ) exp(ζ 1 R s,1 ) exp(ζ 2 R s,2 ) Or φ n (θ 1, θ 2 ) E[f (X( )) L i ( ) = 0] φ n i (θ), i = 1, 2. (η + ζ 1 + ζ 2 )φ n (θ) + ζ 1 (1 ρ 1 )φ n 1(θ) + ζ 2 (1 ρ 2 )φ n 2(θ) / 29

22 Step 2: Taylor expansion Lemma (Taylor Expansion) Let H 0 be a bounded random variable with EH > 0. Set λ H = 1/EH and σh 2 = Var(H). Then the function f : R R satisfying E(e f (x)h ) = e x, x R is well defined and finite. Furthermore, f (x) satisfies the Taylor expansion f H (x) = λ H x 1 2 λ3 Hσ 2 Hx x 3 f (y), x R, where y is some number between x and zero. 22 / 29

23 Step 2: Expansions of η and ζ i η = f Te (θ 1 / θ 1 n) λ e 1 n 2 λ3 eσe 2 θ1 2 n ζ 1 = f Ts,1 ((θ 2 θ 1 )/ (θ 1 θ 2 ) n) λ s,1 1 n 2 λ3 s,1σs,1 2 (θ 1 θ 2 ) 2 n ζ 2 = f Ts,2 ( θ 2 / n) λ s,2 θ 2 n 1 2 λ3 s,2σ 2 s,2 θ 2 2 n. (η + ζ 1 + ζ 2 ) (λ s,1 λ e ) θ 1 (λ s,2 λ s,1 ) θ 2 n n + 1 (λ 3 2n eσe 2 θ1 2 + λ 3 s,2σs,2θ λ 3 s,1σs,1(θ 2 2 θ 1 ) 2) 23 / 29

24 Multiclass: a 2-station, 3-class reentrant-line L 1 L 2 FBFS policy L 3 Let f (x) = e θ,ln / n e ηte e ζ,rs Basic Adjoint Relationship (BAR) ) ] E ν [(η + ζ 1 1 {L1>0 + ζ 2 1 {L2>0} + ζ 3 1 {L1=0,L 3>0}} f (X) = 0 24 / 29

25 Reentrant line: BAR ρ 1,1 = λ e λ s,1, ρ 1 = λ e λ s,1 + λ e λ s,2, ρ 2 = λ e λ s,2. ( η + ζ1 + ζ 2 +(ζ 3 ζ 1 )(1 ρ 1,1 ) ) φ n (θ) ζ 2 (1 ρ 2 )φ n 2(θ) ζ 3 (1 ρ 1 )φ n 1(θ) [( ) ] + (ζ 3 ζ 1 )E 1 {L1=0} (1 ρ 1,1 ) f (X) 0 ( η + ζ1 + ζ 2 + (ζ 3 ζ 1 )(1 ρ 1,1 ) ) 1/n for all (θ 1, θ 2, θ 3 ) / 29

26 5-class reentrant line: state space collapse ζ 3 λ s,3 θ 3 n 1 2 λ3 s,3σ 2 s,3 θ 2 3 n (θ 1 θ 2 ) ζ 1 λ s,1 1 n 2 λ3 s,1σs,1 2 (θ 1 θ 2 ) 2 n For any (θ 2, θ 3 ) 0, by taking one has θ 1 = λ s,3 λ s,1 θ 3 + θ 2 0 (11) ζ 3 ζ 1 1/n. 26 / 29

27 5-class reentrant line: tightness The limit satisfies γ(θ 1, θ 2, θ 3 )φ(θ 1, θ 2, θ 3 ) + b 2 λ s,2 (θ 2 θ 3 )φ 2 (θ 1, θ 3 ) + b 1 λ s,3 θ 3 φ 1 (θ 2 ) = 0 for (θ 2, θ 3 ) 0 and θ 1 satisfying (11), where γ(θ 1, θ 2, θ 3 ) = λ e θ 1 + λ s,1 (θ 1 θ 2 ) + λ s,2 (θ 2 θ 3 ) + quadratic term Setting θ 3 = r 2 0, we have φ(0, 0, 0 ) = φ 2 (0, 0 ). Setting θ 2 = 0, φ(0, 0, 0 ) φ 2 (0, 0 ) + (1 φ(0, 0, 0 )) = 0, implying that φ(0, 0, 0 ) = / 29

28 Multiclass: 5-class reentrant-line (LBFS): challenge Station 1 Station 2 α 1 m 1 m 2 m 3 m 4 m 5 Does not work: choose θ 3 ( ) θ 4 θ1 = A, then θ θ 2 5 θ 3 θ 4 0 θ 5 for "too many" ( θ1 ) < 0. θ 2 One possible solution is to prove SSC: as n, but how? E [ Z n i ( )/ n ] 0 for i = 3, 4, 5, 28 / 29

29 Summary The direct ( BAR ) method works for GJNs; works for some multiclass priority networks including Rybko-Stolya network; does not work for the 2-station, 5-class reentrant line under LBFS policy. 29 / 29