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Perfect Sampling of Queuing Networks with Complex Routing - - PowerPoint PPT Presentation

System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis Perfect Sampling of Queuing Networks with Complex Routing complexity and computational aspects Jean-Marc Vincent


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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Perfect Sampling of Queuing Networks with Complex Routing

complexity and computational aspects Jean-Marc Vincent

MESCAL-INRIA Project Laboratoire d’Informatique de Grenoble Universities of Grenoble, France {Jean-Marc.Vincent}@imag.fr

This work was partially supported by ANR SETIN Checkbound

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Outline

1

System Simulation Application Problems Markovian Modelling Steady-state Sampling of Markov Models

2

Perfect Sampling General Description Coupling Inequalities Forward Coupling Backward Coupling

3

Discrete Time Markov Chain Transition Function Coupling Condition Doeblin Matrices Binary-Uniform Decomposition Examples

4

Event Driven Simulation Monotonicity of Events Acyclic Networks Coupling Time Ψ2 Software Modelling Queuing Systems

5

Case Studies Finite Queuing Network Communication Switch Grid Load Balancing

6

Advanced Topics Stochastic Automata Networks Non-Monotonic Events Variance Reduction

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Application Problems : Modeling and Analysis

  • f Complex Systems

Complex system

  • utput

Environment

Input of the system System

System

Basic model assumptions System :

  • automaton (discrete state space)
  • discrete or continuous time

Environment : non deterministic

  • time homogeneous
  • stochastically regular

Problem Generate “typical” states

  • steady-state sampling
  • ergodic simulation starting point
  • state space exploring techniques

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Queuing Networks with Finite Capacity

Network model Finite set of resources :

servers waiting rooms

Routing strategies :

state dependent

  • verflow strategy

blocking strategy ...

Average performance :

load of the system response time loss rate ...

Markov model

1−p λ µ ν C1 C2 Rejection Blocking p

Poisson arrival, exponential services distribution, probabilistic routing ⇒ continuous time Markov chain

C1 1 2 C2−1 C2 Queue 2 2 Queue 1 1 C1−1

Problem Computation of steady state distribution ⇒ state-space explosion

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Interconnexion Networks

Delta network

8 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 10 9

Input rates Service rates Homogeneous routing Overflow strategy Problem Loss probability at each level Analysis of hot spot ...

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Call centers

Multilevel Erlang model

Traffic 2 Overflow traffic 2 Overflow traffic 1 Type I servers Type II servers Type III servers Traffic 1

Types of requests Input rates Different service rates Overflow strategy Problem Optimization of resources Quality of service (waiting time, rejection probability,...) ...

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Resource Broker

Grid model

Q9 Resource Broker λ µ2 µ1 µ3 µ4 µ5 µ6 µ7 µ8 µ9 Overflow Q2 Q3 Q1 Q10 Q11 Q12 Q4 Q6 Q5 Q7 Q8

Input rates Allocation strategy State dependent allocation Index based routing : destination minimize a criteria Problem Optimization of throughput, response time,... Comparison of policies, analysis of heuristics ...

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Formalization : Markov Chain

Quantification

7 12 1 12 4 12 7 12 3 12 3 12 11 12 1 12 6 12 2 12 3 12

4 3 2 1

Stochastic matrix : transition probability

P = 1 12 2 6 6 4 2 3 7 1 11 3 6 3 4 7 1 3 7 7 5 Non-negative, if irreducible and aperiodic Unique probability vector π satisfying π = πP, π =

1 350 [46, 47, 142, 115]

1856-1922

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Solving methods

Solving π = πP

Analytical/approximatin methods Formal methods N 50 Maple, Sage,... Direct numerical methods N 1000 Mathematica, Scilab,... Iterative methods with preconditioning N 100, 000 Marca,... Adapted methods (structured Markov chains) N 1, 000, 000 PEPS,... Monte-Carlo simulation N 107

Postprocessing of the stationary distribution Computation of rewards (expected stationary functions) Utilization, response time,...

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Ergodic Sampling(1)

Ergodic sampling algorithm Representation : transition fonction Xn+1 = Φ(Xn, en+1). x ← x0 {choice of the initial state at time =0} n = 0; repeat n ← n + 1; e ← Random event(); x ← Φ(x, e); Store x {computation of the next state Xn+1} until some empirical criteria return the trajectory Problem : Stopping criteria

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Ergodic Sampling(2)

Start-up Convergence to stationary behavior lim

n→+∞ P(Xn = x) = πx.

Warm-up period : Avoid initial state dependence Estimation error : ||P(Xn = x) − πx|| Cλn

2.

λ2 second greatest eigenvalue of the transition matrix

  • bounds on C and λ2 (spectral gap)
  • cut-off phenomena

λ2 and C non reachable in practice (complexity equivalent to the computation of π) some known results (Birth and Death processes)

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Ergodic Sampling(3)

Estimation quality Ergodic theorem : lim

n→+∞

1 n

n

  • i=1

f (Xi) = Eπf . Length of the sampling : Error control (CLT theorem) CLT for additive functionals of Markov chains Complexity Complexity of the transition function evaluation (computation of Φ(x, .)) Related to the stabilization period + Estimation time

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Ergodic sampling(4)

Typical trajectory

States time Warm−up period Estimation period

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Replication Method

Typical trajectory

States time replication periods

Sample of independent states Drawback : length of the replication period (dependence from initial state)

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Regeneration Method

Typical trajectory

States time start−up period regeneration period R1 R2 R3 ....

Sample of independent trajectories Drawback : length of the regeneration period (choice of the regenerative state)

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Stochastic recursive sequences

Description [Borovkov et al]

Discrete state space X (usually lattice, product of intervals,...) Innovation state space, and an innovation process Dynamic of the system : transition function Φ : X × E − → X (x, ξ) − → y Trajectory given by x0 and {ξn} an innovation process X0 = x0; Xn+1 = Φ(Xn, ξn)

Discrete event systems

state space : usually lattice, product of intervals,... Innovations : usually a set of events E Independent innovation process : Poisson systems (uniformization)

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Markovian Modelling

Theorem (Markov process) If {ξn} is a sequence of iid random variables , the process {Xn} is a homogeneous discrete time Markov chain. Random Iterated system of functions The trajectory Xn is the successive application of random functions taken in the set {Φ(., ξ), ξ ∈ E} according a probability measure on E [Diaconis and Friedman 98]

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Coupling Inequality

Typical trajectory

Coupling time Stationary version τ time States

After τ the two processes are not distinguishable, then stationary Scheme used to prove Markov convergence (coupling inequality) |P(Xn ∈ A) − πA| P(τ n)

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Forward Sampling : avoid initial state dependence

Forward coupling

Steady-state ? f3 f4 f6 f7 f3 f1 Time State

Example

1 1 − p p

Always couple in the blue state Does not guarantee the steady state !

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Perfect Sampling : Backward Idea

Set dynamic In what state could I be at time n = 0 ? X0 ∈ X = Z0 ∈ Φ(X, e−1) = Z1 ∈ Φ(Φ(X, e−2), e−1) = Z2 . . . . . . ∈ Φ(Φ(· · · Φ(X, e−n), · · · ), e−2), e−1) = Zn

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Perfect sampling : Backward Idea

All the trajectories Time −i −j −τ ∗ S t a t i

  • n

a r y P r

  • c

e s s X X X X Zi Zj Z−τ ∗ = {X0} Z0 = X collapse 23 / 115 Perfect Sampling of Queuing Networks with Complex Routing

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Perfect Sampling : Convergence Theorem

Theorem Provided some condition on the events the sequence of sets {Zn}n∈N is decreasing to a single state, stationary distributed. τ ∗ = inf{n ∈ N; Card(Zn) = 1}. backward coupling time The set of possible states at time 0 is decreasing with regards to n

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Perfect Sampling : Coupling Condition

Theorem Suppose that the set of events is finite. Then the two conditions are equivalent:

τ ∗ < +∞ almost surely; There exist a finite sequence of events with positive probability S = {e1, · · · , eM} such that |Φ(X, e1→M)| = 1.

The sequence S is called a synchronizing pattern (synchronizing word, renovating event,...)

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Perfect sampling : Coupling Condition (proof)

Proof

⇒ If τ ∗ < +∞ almost surely there is a trajectory that couples in a finite

  • time. This finite trajectory is a synchronizing pattern.

⇐ Suppose there is a synchronizing pattern with length M. Because the sequence of events is iid, it occurs almost surely on every trajectory. Applying Borel-Cantelli lemma gives the result.

The forward and backward coupling time have the same distribution τ ∗ has an exponentially dominated distribution tail P(τ ∗ > M.n) (1 − P(e1→M))n. Practically efficient

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Perfect sampling : convergence theorem (proof 1)

Proof based on the shift property First, because τ < +∞ and the ergodicity of the chain there exists N0 s.t. |P(Φ(X, e1→n) = {x}) − πx| ǫ. But the sequence of events is iid (stationary) then P(Φ(X, e1→n) = {x}) = P(Φ(X, e−n+1→0) = {x}) τ ∗ < +∞ then there exists N1 such that P(τ ∗ N1) ǫ ; then P(Φ(X, e−n+1→0) = {x}) = P(Φ(X, e−n+1→0) = {x}, τ ∗ < N1) + P(Φ(X, e−n+1→0) = {x}, τ ∗ N1), = P(Φ(X, e−τ ∗→0) = {x}, τ ∗ < N1) + ǫ′, = P(X0 = x, τ ∗ < N1) + ǫ′ = P(X0 = x) + ǫ′′. By the dominated convergence theorem the result follows

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Perfect Sampling : Convergence Theorem (proof 2)

Proof based on the coupling property [Haggstrom] Consider N such that P(τ ∗ N) ǫ then consider a process {X n} with the same events e−N+1→0 but with X −N+1 generated according π. The process {X n} is stationary. On the event (τ ∗ < N) we have X0 = X 0 and P(X0 = X 0) P(τ ∗ N) ǫ (coupling inequality). Finally P(X0 = x) − πx = P(X0 = x) − P(X 0 = x) P(X0 = X 0) ǫ; πx − P(X0 = x) = P(X 0 = x) − P(X0 = x) P(X 0 = X0) ǫ; and the result follows.

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Perfect Sampling : Algorithm

Backward algorithm

Representation : transition fonction Xn+1 = Φ(Xn, en+1). for all x ∈ X do y(x) ← x end for repeat e ← Random event(); for all x ∈ X do z(x) ← y(Φ(x, e)); end for y ← z until All y(x) are equal return y(x) Convergence : If the algorithm stops, the returned value is steady state distributed Coupling time: τ < +∞, properties of Φ

Trajectories

Time States 0000 0001 0010 0011 0100 0101 0110 1000 1001 1010 1100 −4 −3 −2 −1 −5 −6 −7 −8 −9 −10 U1 U2 U3 U4 U5 U6 U7 U8 τ∗

Mean time complexity cΦ mean computation cost of Φ(x, e) C Card(X).Eτ.cΦ.

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Perfect Reward Sampling

Backward reward

Representation : transition fonction Xn+1 = Φ(Xn, en+1). Arbitrary reward function for all x ∈ X do y(x) ← x end for repeat e ← Random event(); for all x ∈ X do y(x) ← y(Φ(x, e)); end for until All Reward(y(x)) are equal return Reward(y(x)) Convergence : If the algorithm stops, the returned value is steady state reward distributed Coupling time: τr τ < +∞

Trajectories

Time States 0000 0001 0010 0011 0100 0101 0110 1000 1001 1010 1100 −4 −3 −2 −1 −5 −6 −7 −8 −9 −10 U1 U2 U3 U4 U5 U6 U7 U8 1 Cost

Mean time complexity CReward Card(X).Eτ.cΦDepends on the reward function.

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Inverse of PDF

P(X x) 1 1 2 3 K − 1 K Cumulative distribution function x p1 p2 p3 · · ·

Generation Divide [0, 1[ in intervals with length pk Find the interval in which Random falls Returns the index of the interval Computation cost : O(EX) steps Memory cost : O(1) Inverse function algorithm s=0; k=0; u=random() while u >s do k=k+1 s=s+pk end while return k

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Searching optimization

Optimization methods

pre-compute the pdf in a table rank objects by decreasing probability use a dichotomy algorithm use a tree searching algorithm (optimality = Huffmann coding tree)

Comments

  • Depends on the usage of the generator (repeated use or not)
  • pre-computation usually O(K) could be huge
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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Generation : Visual representation

[0, 1] partitionning

1

3 12 1 12 6 12 2 12 7 12 1 12 4 12 3 12 7 12 3 12 11 12

f1 f3 f4 f5 f6 f7 f8 f2 4 3 2 1

Random iterated system of functions

Function f1 f2 f3 f4 f5 f6 f7 f8 Probability

1 12 1 12 1 12 1 12 1 12 4 12 2 12 1 12

Stochastic matrix P = ⇒ simulation algorithm = RIFS

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

The coupling problem

τ estimation

Eτ = 2 1 Couples with probability 1

2

1 Never couples τ = ∞ 1 1 Couples with probability 1 τ = 1

1 2 1 2 1 2 1 2

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

General problem

Objective

Given a stochastic matrix P = ((pi,j)) build a system of function (fθ, θ ∈ Θ) and a probability distribution (pθ, θ ∈ Θ) such that :

1

the RIFS implements the transition matrix P,

2

ensures coupling in finite time

3

achieve the “best” mean coupling time : tradeoff between

  • choice of the transition function according to ((pθ)),
  • computation of the transition

Remarks

Usual method |Θ| = number of non-negative elements of P = O(n2)

choice in O(log n)

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Non sparse matrices

Rearranging the system

1 ”Synchronizing” transformation 1

3 12 1 12 4 12 7 12 3 12 3 12 11 12 1 12 6 12 2 12 7 12

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Non sparse matrices

Convergence rate When at least one column is non-negative ⇒ one step coupling. The RIFS ensures coupling and the coupling time τ is upper bounded by a geometric distribution with rate

  • j

min

i

pi,j number of transition functions : could be more than the number of non-negative elements at most n2

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Aliasing technique

Initialization K objects list L=∅,U=∅; for k=1; k K; k++ do P[k]=pk if P[k] 1

K then

U=U+{k}; else L=L+{k}; end if end for Alias and threshold tables while L = ∅ do Extract k ∈ L Extract i ∈ U S[k]=P[k] A[k]=i P[i] = P[i] - ( 1

K -P[k])

if P[i] 1

K then

U=U+{i}; else L=L+{i}; end if end while Combine uniform and alias value when rejection

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Aliasing technique : generation

1/8

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Aliasing technique : generation

Generation k=alea(K) if Random .

1 K S[k] then

return k else return A[k] end if Complexity Computation time :

  • O(K) for pre-computation
  • O(1) for generation

Memory :

  • threshold O(K) (real numbers as probability)
  • alias O(K) (integers indexes in a tables)

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Sparse matrices

Rearranging the system

1 Aliasing transformation 1

6 12 11 12 3 12 3 12 7 12 4 12 1 12 3 12 7 12 2 12 1 12

f1 f3 f4 f5 f6 f7 f8 f2 4 3 2 1

Complexity

M = maximum out degree of states pθ uniform on {1, · · · , M}, threshold comparison O(1) to compute one transition Combination with “Synchronizing” techniques 42 / 115 Perfect Sampling of Queuing Networks with Complex Routing

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Uniform-binary decomposition

Uniform superposition

1 1 1 1 1 1 3 4 2 A4 A4

1 3

1

2 3 1 3

1

2 2

1 2 3 4 A3 A3 1 1 1 1 4 A2 A2 2 1 3

1 3 2 3 1 3

1 1

2 3

4 3 2 1 A1 A1 1 Aliasing transformation

3 12 4 12 1 12 7 12 2 12 6 12 1 12 3 12 11 12 3 12 7 12

1 2 3 4

Decomposition

P = 1 M M X i=1 Pi , Pi : stochastic matrix with at most 2 non negative elements per row 43 / 115 Perfect Sampling of Queuing Networks with Complex Routing

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Coupling property

Exchange of columns or thresholds give an equivalent representative

representation

  • Equivalent

Spanning tree Irreducibility = ⇒ there is a spanning tree going to a single state where coupling occurs. P(τ ∗ < +∞) = 1. τ is geometrically bounded, so τ ∗ and τ ∗

C.

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Ψ software

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Example

Random transition coefficients: Number of states 10 100 500 1000 3000 Mean coupling time 3.1 4.5 5.3 5.7 6.1 Mean execution time µs 3 17 170 360 1100 Pentium III 700MHz and 256Mb memory. Sample size 10000. Remarks:

  • very small coupling time
  • Coefficients : same order of magnitude, aliasing enforces coupling

Comparison with birth and death process : Number of states 10 100 500 1000 3000 Mean coupling time 41 557 2850 5680 17000 Mean execution time µs 28 1800 88177 366000 3.5s Remarks:

  • large coupling time
  • sparse matrix, large graph diameter

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Overflow model

Model

K

Overflow rejection λ µ1 µ2 µ3 µ

Coupling time distribution

Coupling time (iterations) Density Sample size : 10000

0.05 0.1 0.15 0.2 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000

Parameters K servers, priority on overflows input rate λ, different service rate state (x1, · · · , xK), xi ∈ {0, 1}, size ∼ 130000 low diameter non product-form structure, Statistics Parameter Value minimum 113 maximum 1794 median 465 mean 498 Std 180 exponential tail, low mean value

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Overflow model (2)

Marginal distribution

Index of the server Utilization of server Sample size 10000

0.2 0.4 0.6 0.8 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 K

Marginal probability estimation P(Xi = 1) Occupied servers

Sample size = 10000 Number of occupied servers Probability

0.05 0.1 0.15 0.2 0.25 0.3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Marginal distribution of the occupied servers

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Overflow model (3)

Reward coupling

mean value Coupling time (iterations) Marginal law (server number) Sample size 10000 Maximum Quartile 3 Median Minimum Quartile 1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 500 1000 1500 2000 2500 load

  • ccupation

state

Reward coupling time

  • gain 20% for the first marginals
  • utilization : best reduction

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Monotonicity and perfect sampling : idea

(X, ≺) partially ordered set (lattice) Typically componentwise ordering on products of intervals min = (0, · · · , 0) and Max = (C1, · · · , Cn). An event e is monotone if Φ(., e) is monotone on X If all events are monotone then X0 ∈ Zn ⊂ [Φ(min, e−n→0), Φ(Max, e−n→0)] ⇒ 2 trajectories

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The Doubling Scheme

Complexity

Need to store the backward sequence of events Consider 2 trajectories issued from {min, Max} at time −n and test if coupling One step backward ⇒ 2.(1 + 2 + · · · + τ ∗) = τ ∗(τ ∗ + 1) = O(τ ∗2) calls to the transition function. Consider 2 trajectories issued from {min, Max} at time −2k and test if coupling Doubling step backward ⇒ 2.(1 + 2 + · · · + 2k) = 2k+2 − 2 calls to the transition function, with k such that 2k−1 < τ ∗ 2k, Number of calls : O(τ ∗)

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Monotonicity and Perfect Sampling

Monotone PS

Doubling scheme n=1;R[1]=Random event; repeat n=2.n; y(min) ← min y(Max) ← Max for i=n downto n/2+1 do R[i]=Random event; end for for i=n downto 1 do y(min) ← Φ(y(min), R[i]) y(Max) ← Φ(y(Max), R[i]) end for until y(min) = y(Max) return y(min)

Trajectories

State 2 1 M −1 −2 −4 −8 −16 −32 States : : Maximum : minimum Generated

Mean time complexity Cm 2.(2.Eτ).cΦ. Reduction factor :

4 Card(X).

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Event Modelling

Multidimensional state space : X = X1 × · · · × XK with Xi = {0, · · · , Ci}. Event e : ❀ transition function Φ(., e); (skip rule) ❀ Poisson process λe

Time States Events e1 e2 e3 e4

1781-1840

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Event modelling

Uniformization Λ =

  • e

λe and P(event e) = λe Λ ; Trajectory : {en}n∈Z i.i.d. sequence. ⇒ Homogeneous Discrete Time Markov Chain [Bremaud 99] Xn+1 = Φ(Xn, en+1). Generation among a small finite space E : O(1)

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Index Routing in Queuing Networks

Index functions for event e For queue i I e

i : {0, · · · , Ci} −

→ O (totally ordered set). Property : ∀xi, xj I e

i (xi) = I e j (xj).

ex: inverse of a priority,... Routing algorithm:

if xorigin >0 then { a client is available in the origin queue} xorigin = xorigin − 1; { the client is removed from the origin queue} j = argmini I e

i (xi); { computation of the destination}

if j = -1 then xj = xj+1; { arrival of the client in queue j } { in the other case, the client goes out of the network} end if end if

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Monotonicity of Index Routing Policies

Proposition If all index functions I e

i are monotone then event e is monotone.

Proof : Let x ≺ y two states and let be an index routing event. Let i be the

  • rigin queue for the event.

jx = argminjI e

j (xj) and jy = argminjI e j (yj)

Case 1 xi = yi = 0 nothing happens and Φ(x, e) = x ≺ y = Φ(y, e) Case 2 xi = 0, yi > 0 then Φ(x, e) = x ≺ y − ei + ejy = Φ(y, e) Case 3 xi > 0, yi > 0 then I e

jx(xjx) < I e jy (xjy ) I e jy (yjy ) < I e jx(yjx);

then xjx < yjx and Φ(x, e) = x − ei + ejx y − ei y − ei + ejy = Φ(y, e)

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Monotonicity of Routing

Examples [Glasserman and Yao] All of these events could be expressed as index based routing policies :

  • external arrival with overflow and rejection
  • routing with overflow and rejection or blocking
  • routing to the shortest available queue
  • routing to the shortest mean available response time
  • general index policies [Palmer-Mitrani]
  • rerouting inside queues

...

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Monotonicity of Routing : Examples

Stateless routing Overflow routing I e

j (xj) =

prio(j) if xj < Cj; +∞ elsewhere I e

−1 = max j

Cj. Routing with blocking I e

j (xj) =

  • prio(j)

if xj < Cj; +∞ elsewhere I e

i = max j

Cj. State dependent routing Join the shortest queue I e

j (xj) =

xj if xj < Cj; +∞ elsewhere; I e

−1 = max j

Cj. Join the shortest response time I e

j (xj) =

  • xj+1

µj

if xj < Cj; +∞ elsewhere; I e

−1 = max i

Ci.

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Coupling Experiment

Feed-forward queuing model

5

C C C C

1 2 3

λ λ λ λ λ λ

1 2 3 4

Estimation of Eτ :

50 100 150 200 250 300 350 400 τ 1 2 3 4 λ

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Main Result

Bound on coupling time Eτ

K

  • i=1

Λ Λi Ci + C 2

i

2 ,

  • Λ : global event rate in the network,
  • Λi the rate of events affecting Qi
  • Ci is the capacity of Queue i.

Sketch of the proof

  • Explicit computation for the M/M/1/C
  • Computable bounds for the M/M/1/C
  • Bound with isolated queues

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Explicit Computation for the M/M/1/C

Eτ b = E min(h0→C, hC→0) Absorbing time in a finite Markov chain; p =

λ λ+µ = 1 − q

1,C 1,C−1 0,C−2 1,C−2 2,C−1 2,C 3,C C−2,C−1 C−1,C C,C 0,0 0,1 1,2 0,C 0,C−1 0,C−3 p p p p p p p p p p p p p p p p q q q q q q q q q q q q q q q q Level 3 Level 4 Level 5 Level C+1 Level C+2 Level 2

Explicit recurrence equations Case λ = µ Eτ b = C+C 2

2

.

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Computable bounds for M/M/1/C

If the stationary distribution is concentrated on 0 (λ < µ), Eτ b Eh0→C is an accurate bound. Theorem The mean coupling time Eτ b of a M/M/1/C queue with arrival rate λ and service rate µ is bounded using p = λ/(λ + µ) = 1 − q. Critical bound: ∀p ∈ [0, 1], Eτ b C 2+C

2

. Heavy traffic Bound: if p > 1

2,

Eτ b

C p−q − q(1−( q

p) C )

(p−q)2

. Light traffic bound: if p < 1

2,

Eτ b

C q−p − p(1−( p

q) C )

(q−p)2

.

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Computable Bounds for M/M/1/C

Example with C = 10

20 40 60 80 100 120 0.2 0.4 0.6 0.8 1

Eτ b p heavy traffic Light traffic bound

C+C 2 2

C + C 2 bound

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Example for tandem queues

Coupling of Queue 0 Time X 0

0 = 5

5 4 2 1 3 = C1 6 = C0 −τ b Coupling of queue 1 conditionned by state of queue 0 4 3 = C1 1 −τ b

1 (s0 = 2)

Time X 1

1 = 2

X 1

0 = 3

5 = X 0 6 2 X 1

1 = 2

5 4 2 1 3 = C1 6 = C0 −τ b

0 − τ b 1 (s0 = 5)

X 0

0 = 5

τ b

1 (s0 = 5)

Time X 1

0 = 3

Then τ b st

∞τ b 1 + τ b 0 , normalized

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Bound with Isolated Queues

Theorem In an acyclic stable network of K M/M/1/Ci queues with Bernoulli routing and loss if overflow, the coupling time from the past satisfies in expectation, E[τ b]

  • K−1
  • i=0

Λ ℓi + µi    Ci qi − pi − pi(1 −

  • pi

qi

Ci ) (qi − pi)2   

  • K−1
  • i=0

Λ ℓi + µi (Ci + C 2

i ).

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Conjecture for General Networks

700 800 0.5 1 1.5 2 2.5 3 3.5 4 500 400 300 200 100 600

λ5

Eτ b B1 (proven) B1 ∧ B2 ∧ B3 B3 (conjecture) B2 (conjecture)

Extension to cyclic networks, Generalization to several types of events Application : Grid and call centers

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Software architecture

Aim of the software

finite capacity queuing network simulator rare events estimation (rejection, blocking,...) statistical guarantees (independence of samples)

⇒ Simulation kernel

  • pen source (C, GPL licence)

extensible library of events multiplatforms (Linux (Debian), mac OSX,...)

General architecture

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Queueing Network Description

Constrained communications

Capacity C λ ν µ2 µ1 Overflow

Events types

type action 1 Server departure 2 External arrival to the first empty room in the list DQ 3 Multi-server departure to DQ 4 Join the shortest queue in DQ 5 Index routing according an index table 6 Routing to the first empty room in the list DQ and overflow 7 Routing to the first empty room in the list DQ and blocking in the origin queue

Description file

# Number of queues 3 # Queues capacities 1 1 50 # queues minimal initial state 0 0 0 # queues maximal initial state 1 1 50 # Number of events 4 # Index file - N for No index file File: N # table of events # id type rate nbq origin d1 d2 d3 d4 0 2 0.8 5 -1 : 0 1 2 -1 1 1 0.6 2 0 :

  • 1

2 1 0.4 2 1 :

  • 1

3 7 2.0 5 2 : 0 1 2 -1

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Simulation control and output

Control parameters

# Sample number 10000 # Number of Antithetic variable 1 # Size of maximal trajectory 3000000 # Random generator seed 5 # Coupling file name File: No file

0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 10 20 30 40 50 0.01 0.02

Statistical analysis

Probability distribution in the buffer

Output

# P.S.I.2 version 4.4.4 # Data Network model # Number of queues ... # Parameters # Sample number # 10000 # Number of Antithetic variates ... # ============= 0 [ [ 0 1 10 ] ] 1 [ [ 1 1 13 ] ] 2 [ [ 1 1 2 ] ] 3 [ [ 1 1 33 ] ] ... 9999 [ [ 1 1 2 ] ] # Size 10000 Sampling time : 3809.202000 micro-seconds # Seed Value 5

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Example

Delta interconnection network, C = 10 ρ = 0.9

8 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 10 9

9999 [ [ 0 2 5 7 2 8 7 4 0 7 10 3 3 2 1 5 0 0 6 3 3 6 0 3 9 1 2 4 3 1 3 6 ] ] # Size 10000 Sampling time : 4302.413600 micro-seconds

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Monotonous reward sample

First server analysis 92 [ [ 0 ] ] 93 [ [ 1 ] ] 94 [ [ 1 ] ] 95 [ [ 1 ] ] 96 [ [ 1 ] ] 97 [ [ 1 ] ] 98 [ [ 1 ] ] 99 [ [ 1 ] ] # Size 100 Sampling time : 36.230000 micro-seconds Time reduction 99 [ [ 1 1 1 ] ] # Size 100 Sampling time : 308.100000 micro-seconds

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Coupling time study (doubling scheme)

Coupling time for each queue Perfect Simulation (with doubling period) started 0 6 2 1 6 [ 1 0 9 ] 1 7 5 1 7 [ 1 0 8 ] 2 8 2 6 8 [ 1 1 7 ] 3 8 1 2 8 [ 1 0 8 ] 4 8 1 1 8 [ 1 1 9 ] 5 7 7 4 7 [ 0 1 0 ] 6 6 1 2 6 [ 1 0 10 ] 7 8 1 1 8 [ 1 1 1 ] Carefull : number of steps to couple ( τ = 2nb steps) Last queue have the largest coupling time.

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Coupling Time Study (one step scheme)

Coupling time for each queue Perfect Simulation started 0 26 1 1 26 [ 1 0 10 ] 1 58 1 1 58 [ 0 0 10 ] 2 100 2 1 100 [ 1 0 8 ] 3 91 1 51 91 [ 1 1 2 ] 4 114 1 1 114 [ 1 1 6 ] 5 210 1 1 210 [ 1 1 9 ] Distribution of the rewards coupling time

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Download : http://gforge.inria.fr/projects/psi

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Synthesis

The psi2 unix command USAGE : psi2 unix [-ipo] argument [-hdtv]

  • i :

input file in ext directory

  • p :

parameter file in ext directory

  • o :
  • utput file in ext directory

By default, output file has outputtest.txt name in ext directory

  • h :

help.

  • d :

With details on output file

  • t :

Without doubling period, show coupling time of each queue

  • v :

version

Enjoy !

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Priority Servers

Erlang model

Output Arrivals Servers Overflow

  • n next free server

Rejection if all servers are buzy

X = {0, 1}3 E = {e0, e1, e2, e3} Card(X) = 2K Events Event type Rate Origin Destination list Arrival λ −1 Q1 ; Q2 ; Q3 ; −1 Departure µ1 Q1 −1 Departure µ2 Q2 −1 Departure µ3 Q3 −1 Results

Validation χ2 test K = 30 µi decreasing Saturation probability 0.0579 ± 4.710−4 Simulation time 0.4ms τ = 577

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Priority servers

Erlang model

Output Arrivals Servers Overflow

  • n next free server

Rejection if all servers are buzy

X = {0, 1}40 µ1 = 1, µ2 = 0.8, µ3 = 0.5 Sample size 5.106 Card(X) = 2K Saturation probability

Prob 0.1 0.2 0.3 0.4 0.5 0.6 10 20 30 40 50 60 70 λ

Proba

1e−06 1e−04 0.001 8 8.5 9 9.5 10 10.5 11 11.5

λ

1e−05

Coupling time

s Reward coupling time Coupling time

200 250 300 350 10 20 30 40 50 60 70 λ 100 50 µ 400 150

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Line of Servers

Tandem queues

reject µ2 C2 C3 µ3 C4 µ4 C5 µ5 C1 µ1 λ

X = {0, · · · , 100}5 E = {e0, · · · , e5} Card(X) = C K Events Event type Rate Origin

  • Dest. list

Arrival λ −1 Q1 ; −1 Routing/block µ1 Q1 Q2 ; Q1 Routing/block µ2 Q2 Q3 ; Q2 · · · · · · · · · · · · Departure µ5 Q5 −1 Results

C = 100 λ = 0.9; µ = 1 p = 1

2

Blocking probability b1 = 0.34, b2 = 0.02 b3 = 0.02, b4, = 0.02. Simulation time < 1ms

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Multistage network

Delta network

8 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 10 9

X = {0, · · · , 100}32 E = {e0, · · · , e64} Card(X) = C K Events Event type Rate Origin

  • Dest. list

Arrival λ −1 Qi ; −1 Routing/rejection

1 2µ

Qi Qj ; −1 · · · · · · · · · · · · Departure µ Qk −1 Results

C = 100 λ = 0.9; µ = 1 Loss rate Simulation time 135ms τ ≃ 400000

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Multistage network

Delta network

8 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 10 9

X = {0, · · · , 100}32 E = {e0, · · · , e64} Card(X) = C K Sample size 100000 Queue length and saturation proba at level 3

λ

1 2 3 4 5 6 7 E(N_34) 0.2 0.4 0.6 0.8

λ

0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0014 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 0.74 0.76 0.78

Coupling time

s

Global coupling Reward at least 1 queue saturated (3rd level) Reward queue 31 saturated

10000 12000 14000 16000 18000 0.2 0.4 0.6 0.8 6000 4000 2000

λ µ

8000

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Resource Broker

Grid model

Q9 Resource Broker λ µ2 µ1 µ3 µ4 µ5 µ6 µ7 µ8 µ9 Overflow Q2 Q3 Q1 Q10 Q11 Q12 Q4 Q6 Q5 Q7 Q8

Input rates Allocation strategy State dependent allocation Index based routing : destination minimize a criteria Problem Optimization of throughput, response time,... Comparison of policies, analysis of heuristics ...

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Routing Customers in Parallel Queues

The problem:

Find a routing policy maximizing the expected (discounted) throughput of the system. Several variations on this problem depend on the information available to the controller: current size of all queues (and size of the arriving batch).

The applications:

improve batch schedulers for cluster and grid infrastructures. Assert the value of information in such cases.

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Index policies for routing

Optimal routing policy problem is still open for n different M/M/1 Heuristic : index policy inspired from the Multi-Armed Bandit ⇒ free parameter and compute an equilibrium point. [Mitrani 2005] for routing and repair problems.

µ S servers Capacity C λ W b µ µ

W is the rejection cost (free parameter). Theorem There is an optimal policy of threshold type: there exists θ such that :Reject if x θ and accept otherwise.

  • θ does not depend on C as long as C > θ (including if C is infinite).
  • θ is a non-decreasing function of W .

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System Simulation Perfect Sampling Discrete Time Markov Chain Event Driven Simulation Case Studies Advanced Topics Synthesis

Index policies for Routing(II)

Computation of θ(W ) linear system of corresponding to Bellman’s equation, after uniformization.

5 10 15 20 25 30 2 4 6 8 10 12 14 16 18 20 BestThreshold(W) W ’TW.txt’

Index function I(x) = inf{W | θ(W ) = x}. Indifference case : when queue size is x, rejecting or accepting the next batch are both optimal choices if the rejection cost is I(x).

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Some numerical experiments(I)

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 1-9_1-9_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.5 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 1-9_9-9.95_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 9-1_1-9_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 9-1_9.0-9.95_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim*

Several cases with two queues with respective parameters (µ = 9, 1), C = 100

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Some numerical experiments(III)

0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 8-2_1-9_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 8-2_9.0-9.95_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1.18 1.2 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 8-2_1-9_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 8-2_9.0-9.95_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim*

Several cases with two queues with respective parameters (µ = 8, 2), C = 100.

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Some numerical experiments(IV)

0.95 1 1.05 1.1 1.15 1.2 1.25 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 8.1-1.9_1-9_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 8.1-1.9_9.0-9.95_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.99 1 1.01 1.02 1.03 1.04 1.05 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 7-3_1-9_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.995 1 1.005 1.01 1.015 1.02 1.025 1.03 1.035 1.04 1.045 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 7-3_9.0-9.95_100.txt JSQ JSQ-mu JSQ-mu2 Seq *optim*

Some other cases

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Some numerical experiments(V)

0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 8-1-1_1-9_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 8-1-1_9.0-9.95_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 7-2-1_1-9_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.95 1 1.05 1.1 1.15 1.2 1.25 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 7-2-1_9.0-9.95_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim*

Three queues. Respectively, µ = 8, 1, 1 and µ = 7, 2, 1.

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Numerical experiments(VI)

0.95 1 1.05 1.1 1.15 1.2 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 6-3-1_1-9_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 6-3-1_9.0-9.95_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.96 0.98 1 1.02 1.04 1.06 1.08 1.1 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 6-2-2_1-9_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.97 0.98 0.99 1 1.01 1.02 1.03 1.04 1.05 1.06 1.07 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 6-2-2_9.0-9.95_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim*

Now, µ = 6, 3, 1 and µ = 6, 2, 2.

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Numerical experiments(VII)

0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 1.14 1.16 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 5-4-1_1-9_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.98 1 1.02 1.04 1.06 1.08 1.1 1.12 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 5-4-1_9.0-9.95_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.97 0.975 0.98 0.985 0.99 0.995 1 1.005 1.01 1.015 1.02 1.025 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 5-3-2_1-9_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.97 0.975 0.98 0.985 0.99 0.995 1 1.005 1.01 1.015 1.02 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 5-3-2_9.0-9.95_50.txt JSQ JSQ-mu JSQ-mu2 Seq *optim*

Now, µ = 5, 4, 1 and µ = 5, 3, 2.

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Robustness of Index policies

The index policy was computed for λ = 5 or 9 and used over the whole range λ = 1 to 10.

0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 9-1_1-9_50.txt.1 mu : 9, 1 c : 1, 1 Nmax : 50 lambda index : 5 JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 9-1_9.0-9.95_50.txt.1 mu : 9, 1 c : 1, 1 Nmax : 50 lambda index : 5 JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1 2 3 4 5 6 7 8 9 Average waiting time (ratio with optim) Load 9-1_1-9_50.txt.2 mu : 9, 1 c : 1, 1 Nmax : 50 lambda index : 9 JSQ JSQ-mu JSQ-mu2 Seq *optim* 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 8.9 9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 10 Average waiting time (ratio with optim) Load 9-1_9.0-9.95_50.txt.2 mu : 9, 1 c : 1, 1 Nmax : 50 lambda index : 9 JSQ JSQ-mu JSQ-mu2 Seq *optim*

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Adaptation to Structured Models

Model Parameters

set of uniformized events E = {e1, .., ep} global states are tuples of local states ˜ s = (s1, . . . , sK) transition function: Φ(˜ s, ei) = ˜ r * each ˜ s ∈ X has a set of enabled events and its firing conditions and consequences

Constraints

Well-formed SAN models needed * exploring the subset X R (Reachable state space) State space explosion still a problem

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Analysis of Complex Discrete Systems

GRAPHICAL MODEL (STATES + TRANSITIONS) 0(1) 1(1) e1 e2

A(1)

e3

A(2)

0(2) 2(2) 1(2) e4 e5 e2 e5

˜ s ∈ X R ˜ r = Φ(˜ s, ep), ep ∈ ξ Φ(˜ s, e1) Φ(˜ s, e2) Φ(˜ s, e3) Φ(˜ s, e4) {0;0} {1;0} {0;0} {0;0} {0;0} {0;1} {1;1} {0;1} {0;2} {0;1} {0;2} {1;2} {0;2} {0;2} {0;0} {1;0} {1;0} {0,2} {1;0} {1;0} {1;1} {1;1} {1;1} {1;2} {1;1} {1;2} {1;2} {1;2} {1;2} {1;0}

ep ∈ ξ Rates Uniformized Rates e1 λ1 λ1/(λ1 + λ2 + λ3 + λ4 + λ5) e2 λ2 λ2/(λ1 + λ2 + λ3 + λ4 + λ5) e3 λ3 λ3/(λ1 + λ2 + λ3 + λ4 + λ5) e4 λ4 λ4/(λ1 + λ2 + λ3 + λ4 + λ5) e5 λ5 λ5/(λ1 + λ2 + λ3 + λ4 + λ5)

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New solutions for huge SAN models

Monotonicity and Perfect Simulation Idea

Monotonicity property for SAN related to the analysis of structural conditions * component-wise state space formation

Families of SAN models

SAN models with a natural partial order (canonical) * e.g. derived from Queueing systems models [Vincent 2005] SAN models with a given component-wise partial order (non-lattice)

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Partially Ordered State Spaces

Canonical component-wise ordering

Equivalent MC Queueing Network Model

λ µ K1

. . . Equivalent SAN Model

K1

A(1)

K2

. . . A(2)

K2

e2 e2 e12 e12 e1 e1 e12 e12 e12 e1 ν

Type Event Rate loc e1 λ syn e12 µ loc e2 ν

00 20 10 01 11 21 02 12 22 03 13 23 Event e12 Event e2 Event e1

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Partially Ordered State Spaces

Non-lattice component-wise ordering

Find a partial order of X demands a high c.c. Possible to find extremal global states in the underlying chain * |X M| states: more than two extremal states Complexity: related to τ, but also |X M|

Extremal states

Component-wise formation has ordered state indexes * consider an initial state composing X M * add to X M the states without transitions to states with greater indexes

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Non-lattice component-wise ordering

Resource sharing model with reservation (Dining Philosophers)

Type Event Rate loc lti µ syn tri λ syn rli λ loc rtK µ syn tlK λ syn lrK λ lrK T (0) R(0) L(0)

P(1)

rl2 tr1 lt1 rl1 lrK lrK rl0 tlK rtK T (K) trK−1 rl0

P(K) P(K−1)

T (K−1) L(K−1) R(K−1) ltK−1 rlK−1 trK−2 tlK trK−1 trK−2

P(i)

tri−1 tri rli rli+1 T (i) L(i) R(i)

i = 2 . . . (K − 2)

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Non-lattice component-wise ordering

e.g. three philosophers with resources reservation, graphical model of the underlying transition chain, extremal states identification

201 101 011 002 200 001 010 100 000 020 012 102 101 / 115 Perfect Sampling of Queuing Networks with Complex Routing

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SAN Perfect Simulation

Resource sharing model with reservation- K Philosophers

K X X R X M PEPS* (iteration) Perfect PEPS* (sample) 8 6,561 985 43 0.003185 sec. 0.032354 sec. 10 59,049 5,741 111 0.038100 sec. 0.111365 sec. 12 531,441 33,461 289 0.551290 sec. 0.689674 sec. 14 4,782,969 195,025 755 5.712210 sec. 2.686925 sec. 16 43,046,721 1,136,689 1,975 68.704325 sec. 15.793501 sec. 18 387,420,489 6,625,109 5,169 —- 83.287321 sec.

Numerical results

3.2 GHz Intel Xeon processor under Linux, 1 GByte RAM times: for one iteration on PEPS and for one sample generation on Perfect PEPS Remarks: X contraction in |X M| X limitation 6 × 107 onPEPS overcame

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Non Monotonic Systems

Classic non monotonous events

Batch arrivals Batch services Join procedures Negative customers · · ·

Almost monotonous events Monotonicity is not verified at the frontier

Batch arrivals : batch rejection (queue full) Batch services : system almost empty Join procedures : system almost empty Negative customers : system almost empty

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Envelopes

Aim : Build a monotonous upper and lower process Hypothesis : X is lattice, T = max X and B = min X U(M, m, e) = sup

msM

Φ(s, e); L(M, m, e) = inf

msM Φ(s, e).

Remark : if e is monotonous U(M, m, e) = Φ(M, e) and L(M, m, e) = Φ(m, e) Bounding process : Y0 = T and Z0 = B; Yn = U(Yn−1, Zn−1, e1→n) and Zn = L(Yn−1, Zn−1, e1→n). Remark : Yn and Zn are not Markov chains but the couple is Markov. Yn Xn Zn.

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Envelopes (2)

Theorem (Convergence) Either the process (Yn, Zn) hits the diagonal in finite time with probability 1 or nether hits the diagonal. When the process hits the diagonal the value is stationary distributed. Problems

(P1) The assumption that Sn hits the diagonal may not be verified. (P2) Even if convergence theorem, the coupling time could become prohibitively large. (P3) The time needed to compute U(M, m, e) and L(M, m, e) might depend

  • n the number of states between m and M

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Variance Reduction

Coupled trajectories

StatesReward RA/A R2/A StatesReward U4 U5 · · · Un StatesReward R1/A U0 U1 U2 U3

Coupled samples hope : negatively correlated

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Variance Reduction

3 coupled trajectories Perfect Simulation (with doubling period) -= Antithetic with 3 variables =- started 0 [ [ 1 0 10 ] [ 1 1 0 ] [ 1 1 7 ] ] 1 [ [ 1 1 10 ] [ 1 1 3 ] [ 1 0 0 ] ] 2 [ [ 1 0 9 ] [ 1 1 0 ] [ 1 1 10 ] ] 3 [ [ 1 1 0 ] [ 1 1 7 ] [ 1 1 0 ] ] 4 [ [ 1 1 9 ] [ 1 0 0 ] [ 0 1 3 ] ] 5 [ [ 1 0 6 ] [ 1 1 7 ] [ 0 1 2 ] ] 6 [ [ 1 1 4 ] [ 1 1 1 ] [ 0 0 10 ] ] 7 [ [ 1 1 4 ] [ 1 0 6 ] [ 1 1 6 ] ] 8 [ [ 1 0 8 ] [ 0 1 10 ] [ 1 1 9 ] ] 9 [ [ 1 1 0 ] [ 1 1 0 ] [ 1 1 8 ] ] 10 [ [ 1 1 6 ] [ 1 1 6 ] [ 1 1 8 ] ] 11 [ [ 0 1 3 ] [ 1 1 6 ] [ 1 1 5 ] ] Correlation analysis ⇒ variance reduction example VarX0 > Var(X0 + X ′

0 + X ′′ 0 )/3

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Synthesis

Exact sampling

Poisson systems (uniformization) independence of events (SRS) Reversed process ⇒ exact criteria (convergence proof) Polynomial coupling time in the size of the models Monotonicity and contraction on sets ⇒ RIFS and fractals

⇒ model structure Perfect samplers

General DTMC : O(Eτ ∗.|X|) Monotone DTMC : O(Eτ ∗.|Ext|) Monotone DTMC lattice : O(Eτ ∗)

⇒ numerically tractable

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Other approaches in perfect sampling

Algorithms

Forward/backward algorithm [Fill et al] Horizontal sampling [Foss et al] Read once samplers · · ·

Application contexts

Interacting particle systems (statistical physics) Stochastic geometry Networking [Le Boudec et al] Samplers of complex distributions · · ·

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Open problems

Method

Optimal coupling problem (general case : event decomposition) Infinite state space : coupling condition Non-monotone systems Transformation of generators

Models

Structured models : partial order construction Structured models : from description to efficient event decomposition (Max, +) dynamical systems (Petri nets) · · ·

Software Integration in general modeling framework

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Bibliography

Books

  • P. Br´

emaud Markov chains: Gibbs fields, Monte Carlo Simulation and

  • Queues. Springer-Verlag, 1999
  • O. Haggstrom. Finite Markov Chains and algorithmic applications. 2001.

Matematisk Statistik, Chalmers teknisk hogshola och Goteborgs universitet.

  • P. Glasserman and D. Yao. Monotone Structure in Discrete-Event
  • Systems. John Wiley and Sons, 1994.
  • T. Lindvall Lectures on the Coupling Method John Wiley and Sons 1992

Websites

David Wilson’s web site http://dimacs.rutgers.edu/∼dbwilson/exact

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Basic references and application papers

Fundamental papers among (in my opinion)

  • D. Propp, J.and Wilson. Exact sampling with coupled Markov chains and applications to statistical mechanics. Random

Structures and Algorithms, 9(1&2):223–252, 1996.

  • A. Borovkov and S. Foss. Two ergodicity criteria for stochastically recursive sequences. Acta Appl. Math., 34, 1994.
  • P. Diaconis and D. Freedman. Iterated random functions. SIAM Review, 41(1):45–76, 1999.
  • V. Craiu and X.-L. Meng. Multiprocess parallel antithetic coupling for backward and forward Markov chain Monte Carlo. Annals
  • f Statistics, 33(2):661–697, 2005.
  • J.A. Fill. An interruptible algorithm for perfect sampling via Markov chains, Annals of Applied Probability, 8:131–162,1998.
  • O. Stenflo. Ergodic theorems for markov chains represented by iterated function systems. Bull. Polish Acad. Sci. Math, 2001.
  • A. Walker. An efficient method for generating discrete random variables with general distributions. ACM Trans. Math. Software,

3:253–256, 1974.

Some application papers in computer science

  • B. Gaujal and F. Perronnin. Perfect simulation of stochastic hybrid systems with an application to peer to peer systems”. Journal
  • f Discrete Event Dynamic Systems, 2007. Special issue on hybrid systems.
  • F. Baccelli, B. Blaszczyszyn, and F. Tournois. Spatial averages of coverage characteristics in large cdma networks. Wirel. Netw.,

8(6):569–586, 2002.

  • V. Berten and B. Gaujal. Brokering strategies in computational grids using stochastic prediction models. Parallel Computing,
  • 2007. Special Issue on Large Scale Grids.
  • J.-Y. Le Boudec and M. Vojnovic. The Random Trip Model: Stability, Stationary Regime, and Perfect Simulation The Random

Trip Model: Stability, Stationary Regime, and Perfect Simulation. IEEE/ACM Transactions on Networking, 14(6):1153–1166, 2006.

  • A. Bouillard and B. Gaujal. Backward coupling for perfect simulation of free-choice nets. Journal of Discrete Event Dynamics

Systems, theory and applications, 2006. Special issue of selected papers from the Valuetools conference. 113 / 115 Perfect Sampling of Queuing Networks with Complex Routing

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Some references

Discrete time Markov chains

  • J.-M. Vincent and C. Marchand. On the exact simulation of functionals of stationary markov chains. Linear Algebra and its

Applications, 386:285–310, 2004.

  • J.-M. Vincent. Markov chains, iterated systems of functions and coupling time for perfect simulation. In Transgressive

Computing, pages 387–398, Grenada, Apr. 2006.

  • I. Y. Kadi, J.-M. Fourneau, N. Pekergin, J. Vienne, and J.-M. Vincent. Perfect simulation and monotone stochastic bounds. In

VALUETOOLS, Nantes, France, Oct. 2007

Finite queueing Networks

  • J.-M. Vincent. Perfect simulation of queueing networks with blocking and rejection. In Saint IEEE conference, pages 268–271,

Trento, 2005.

  • J.-M. Vincent and J. Vienne. Perfect simulation of index based routing queueing networks. Performance Evaluation Review,

34(2):24–25, 2006.

  • J. Dopper, B. Gaujal, and J.-M. Vincent. Bounds for the coupling time in queueing networks perfect simulation. In Numerical

Solutions for Markov Chain (NSMC06), pages 117–136, Charleston, June 2006.

Low probability events estimation

  • J.-M. Vincent. Perfect simulation of monotone systems for rare event probability estimation. In Winter Simulation Conference,

Orlando, Dec. 2005.

  • J.-M. Vincent and J. Vienne. Perfect simulation of monotone systems with variance reduction. In Proceedings of the 6th Int.

Workshop on Rare Event Simulation, pages 275–285, Bamberg, Oct. 2006.

Relaxing monotonicity properties

  • P. Fernandes, J.-M. Vincent, and T. Webber. Perfect simulation of stochastic automata networks. In ASMTA Conference,

Nicosie, jun 2008.

  • A. Busic, B. Gaujal and J.-M. Vincent Perfect simulation and non-monotone Markovian systems Valuetools 2008

Software tools

  • J.-M. Vincent and J. Vienne. Psi2 a software tool for the perfect simulation of finite queueing networks. In QEST, Edinburgh,
  • Sept. 2007.

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Thanks

Searchers and students Vandy Berten, Ana Busic, Jantien Dopper, Florentine Dubois, Paulo Fernandes, Bruno Gaujal, Ga¨ el Gorgo, Im` ene Kadi, Corine Marchand, Florent Morata, Nihal Pekerhin, No´ emie Sidaner, J´ erˆ

  • me Vienne, Tha¨

ıs Webber, Research groups ANR Checkbound, ACI SurePath, ANR SMS. Technical suggestions and software support Vincent Danjean, Guillaume Huard, Arnaud Legrand, Nicolas Maillard, gforge.inria.fr, Images http://www.gap-system.org/∼history/Biographies/

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