bounds for the coupling time in queueing networks perfect ... · outline 1 queueing networks with...

Bounds for the Coupling Time in QueueingNetworks Perfect Simulation

J.G. Dopper2, B. Gaujal and J.-M. Vincent1

1Laboratory ID-IMAGMESCAL Project

Universities of Grenoble, France{Bruno.Gaujal,Jean-Marc.Vincent}@imag.fr

2Mathematical Institute,Leiden University, [email protected]

J.G. Dopper, B. Gaujal and J.-M. Vincent (Universities of Grenoble)Bounds for the Coupling Time in Queueing Networks Perfect SimulationMAM 2006, june12 1 / 27

Outline

1 Queueing Networks with finite capacity

2 Event modelling and monotonicity

3 Perfect simulation and coupling time

4 Acyclic networks

5 Synthesis and future works

Outline

4 Acyclic networks

Queueing networks with finite capacity

Network modelFinite set of resources :

servers

waiting room

Routing strategies :

state dependent

overflow strategy

blocking strategy...

Average performance :

load of the system

response time

loss rate ...

Markov model

Assumptions :- Poisson arrival,- exponential distribution for service times,- probabilistic routing with overflow

⇒ continuous time Markov chain

ProblemComputation of the stationary distribution⇒ state space explosion

servers

waiting room

state dependent

overflow strategy

load of the system

response time

loss rate ...

Markov model

5

C

C0

1

2

3λ

λ

λλ

01

2

3

4

servers

waiting room

state dependent

overflow strategy

load of the system

response time

loss rate ...

Markov model

5

C

C0

1

2

3λ

λ

λλ

01

2

3

4

Related works

Non reversible systems (reverse event)Product form solution ??Widely studied domain

- Analytical solution [Perros 94]- specific cases- numerical computation of normalization constant

- Numerical computation [Stewart 94]

- Approximation techniques [Onvural 90, Perros 94,...]

- Simulation [Banks & al. 01,...]simulation of Markov modelssimulation of event graphsdiscrete event simulationperfect simulation [Mattson 04]

Outline

4 Acyclic networks

Event modelling

Queueing model :

5

C

C0

1

2

3λ

λ

λλ

01

2

3

4

Event description :rate origin destination enabling condition routing policy

e0 λ0 Q−1 Q0 none rejection if Q0 is fulle1 λ1 Q0 Q1 s0 > 0 rejection if Q1 is fulle2 λ2 Q0 Q2 s0 > 0 rejection if Q2 is fulle3 λ3 Q1 Q3 s1 > 0 rejection if Q3 is fulle4 λ4 Q2 Q3 s2 > 0 rejection if Q3 is fulle5 λ5 Q3 Q−1 s3 > 0 none

Event modelling

Multidimensional state spaceX = X0 × · · · × XK−1]

with Xi = {0, · · · , Ci}.Event e :; transition function Φ(., e);; Poisson process λe

Poisson driven system

Uniformization ⇒ GSMP representation

Λ =∑

e

λe and P(event e) =λeΛ

; Trajectory : {en}n∈Z i.i.d.

⇒ Homogeneous Discrete Time Markov Chain [Bremaud 99]

Xn+1 = Φ(Xn, en+1).J.G. Dopper, B. Gaujal and J.-M. Vincent (Universities of Grenoble)Bounds for the Coupling Time in Queueing Networks Perfect SimulationMAM 2006, june12 8 / 27

Event modelling

Time

States

Events

e1

e2

e3

e4

Λ =∑

e

Event modelling

Time

States

Events

e1

e2

e3

e4

Λ =∑

e

Monotonicity of routing strategy

(X ,≺) partially ordered set (componentwise)

x = [x0, x1, · · · , xK−1] ≺ y = [y0, y1, · · · , yK−1] iff ∀i , xi 6 yi .

An event e is said to be monotone if

x ≺ y ⇒ Φ(x , e) ≺ Φ(y , e).

Examples [Glasserman and Yao]All of these routing events are monotone:- 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...

Monotonicity of routing strategy

(X ,≺) partially ordered set (componentwise)

x = [x0, x1, · · · , xK−1] ≺ y = [y0, y1, · · · , yK−1] iff ∀i , xi 6 yi .

An event e is said to be monotone if

x ≺ y ⇒ Φ(x , e) ≺ Φ(y , e).

Examples [Glasserman and Yao]All of these routing events are monotone:- 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...

Outline

4 Acyclic networks

Classical forward simulation

ForwardRepresentation : 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);{computation of the next state Xn+1}

until some empirical criteriareturn x

Convergence : biased sampleSampling : Warm-up period

Trajectory

ComplexityRelated to the stabilization periodEstimation : replication or ergodic estimation

Xn+1 = Φ(Xn, en+1).

Trajectory

Xn+1 = Φ(Xn, en+1).

Trajectory

Initial state

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Time

1 2 3 4 5 5 6 7 8

States

0000

Xn+1 = Φ(Xn, en+1).

Trajectory

Initial state

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Time

1 2 3 4 5 5 6 7 8

States

0000

Xn+1 = Φ(Xn, en+1).

Trajectory

Initial state

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Time

1 2 3 4 5 5 6 7 8

States

0000

Xn+1 = Φ(Xn, en+1).

Trajectory

Initial state

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Time

1 2 3 4 5 5 6 7 8

States

0000

Xn+1 = Φ(Xn, en+1).

Trajectory

Initial state

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Time

1 2 3 4 5 5 6 7 8

States

0000

Xn+1 = Φ(Xn, en+1).

Trajectory

Initial state

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Time

1 2 3 4 5 5 6 7 8

States

0000

Xn+1 = Φ(Xn, en+1).

Trajectory

6

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Initial state

Time

1 2 3 4 5 7 8 9

States

0000

0001

Xn+1 = Φ(Xn, en+1).

Trajectory

6

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Initial state

Time

1 2 3 4 5 7 8 9

States

0000

0001

Xn+1 = Φ(Xn, en+1).

Trajectory

6

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Initial state

Time

1 2 3 4 5 7 8 9

States

0000

0001

Xn+1 = Φ(Xn, en+1).

Trajectory

6

0010

0011

0100

0101

0110

1000

1001

1010

1100

0

Initial state

Time

1 2 3 4 5 7 8 9

States

0000

0001

Xn+1 = Φ(Xn, en+1).

Trajectory

stabilization period

0101

0110

1000

1001

1010

1100

0

Initial state

6

Steady

state ?

Time

1 2 3 4 5 7 8 9

States

0000

0001

0010

0011

0100

Xn+1 = Φ(Xn, en+1).

Trajectory

stabilization period

0101

0110

1000

1001

1010

1100

0

Initial state

6

Steady

state ?

Time

1 2 3 4 5 7 8 9

States

0000

0001

0010

0011

0100

Perfect simulation : backward idea

Representation : transition fonction

Xn+1 = Φ(Xn, en+1), {en}n∈Z i.i.d. sequence.In what state could I be at time n = 0 ?

X0 ∈ X = Z0∈ Φ(X , e0) = Z1∈ Φ(Φ(X , e−1), e0) = Z2· · ·∈ Φ(Φ(· · ·Φ(X , e−n+1), · · · ), e0) = Zn

TheoremProvided some condition on the sequence of events, the sequence of sets

Z0 ⊇ Z1 ⊇ Z2 ⊇ · · · ⊇ Zn ⊇ · · · is decreasing to a single state.

The generated state is stationary distributed (steady state sample).

τb = inf{n ∈ N; Card(Zn) = 1}.

backward coupling time

Perfect simulation : backward idea

Representation : transition fonction

Xn+1 = Φ(Xn, en+1), {en}n∈Z i.i.d. sequence.In what state could I be at time n = 0 ?

X0 ∈ X = Z0∈ Φ(X , e0) = Z1∈ Φ(Φ(X , e−1), e0) = Z2· · ·∈ Φ(Φ(· · ·Φ(X , e−n+1), · · · ), e0) = Zn

TheoremProvided some condition on the sequence of events, the sequence of sets

Z0 ⊇ Z1 ⊇ Z2 ⊇ · · · ⊇ Zn ⊇ · · · is decreasing to a single state.

The generated state is stationary distributed (steady state sample).

τb = inf{n ∈ N; Card(Zn) = 1}.

backward coupling time

Perfect simulation

Backward algorithmRepresentation : transition fonction

Xn+1 = Φ(Xn, en+1).

for all x ∈ X doy(x) ← x

end forrepeat

u ← Random;for all x ∈ X do

e ← Random event();y(x) ← y(Φ(x, e));

end foruntil All y(x) are equalreturn y(x)

Convergence : If the algorithm stops, thereturned value is steady state distributedCoupling time: τ < +∞, properties of Φ

Trajectories

Mean time complexitycΦ mean computation cost of Φ(x , e)

C 6 Card(X ).Eτ.cΦ.J.G. Dopper, B. Gaujal and J.-M. Vincent (Universities of Grenoble)Bounds for the Coupling Time in Queueing Networks Perfect SimulationMAM 2006, june12 13 / 27

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6U7

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6U7

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6U7U8

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6U7U8

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6U7U8

τ∗

Perfect simulation

Xn+1 = Φ(Xn, en+1).

end forrepeat

Trajectories

Time

States

0000

0001

0010

0011

0100

0101

0110

1000

1001

1010

1100

−4 −3 −2 −1−5−6−7−8−9−10 0U1U2U3U4U5U6U7U8

τ∗

Monotonicity and perfect simulation : idea

min = (0, · · · , 0) and Max = (C1, · · · , Cn).

If all events are monotone then

X0 ∈ Zn ⊂ [Φ(min, e−n→0),Φ(Max , e−n→0)]

⇒ 2 trajectories

Monotonicity and perfect simulation

Monotone PSDoubling scheme

n=1;R[1]=Random event;repeat

n=2.n;y(min) ← miny(Max) ← Maxfor i=n downto n/2+1 do

R[i]=Random event;end forfor i=n downto 1 do

y(min) ← Φ(y(min), R[i])y(Max) ← Φ(y(Max), R[i])

end foruntil y(min) = y(Max)return y(min)

Trajectories

: minimum

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

0

Mean time complexity

Cm 6 2.(2.Eτ).cΦ. Reduction factor : 4Card(X ) .

Trajectories

: minimum

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

0

Trajectories

: minimum

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

0

Trajectories

: minimum

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

0

Trajectories

: minimum

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

0

Trajectories

: minimum

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

0

Trajectories

: minimum

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

0

Trajectories

: minimum

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

0

Trajectories

State

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

: minimum

Generated

0

Trajectories

State

2

1

M

−1−2−4−8−16−32 0

States

: : Maximum

: minimum

Generated

0

Coupling time

definition

τb = min{n ∈ N; Card(Zn) = 1};= min{n ∈ N; |Φ(X , e−n→0| = 1}.

Properties

- Backward τb and forward τ f coupling times have the sameprobability distribution;

- Marginal coupling : denote by τbi the backward coupling time forQi

τb = max τbi .

Problem : compute the mean coupling time

Coupling time

definition

Properties

τb = max τbi .

Coupling time

definition

Properties

τb = max τbi .

Outline

4 Acyclic networks

Coupling experiment

Queueing model :

5

C

C0

1

2

3λ

λ

λλ

01

2

3

4

Estimation of Eτ :

Coupling experiment

Queueing model :

5

C

C0

1

2

3λ

λ

λλ

01

2

3

4

Estimation of Eτ :

0

50

100

150

200

250

300

350

400

τ

0 1 2 3 4 λ

Main result

Theorem (Bound on coupling time)

Eτ 6K∑

i=1

Λ

Λi

Ci + C2i2

,

- 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

Main result

Theorem (Bound on coupling time)

Eτ 6K∑

i=1

Λ

Λi

Ci + C2i2

,

- 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

Explicit computation for the M/M/1/C

Eτb = E min(h0→C , hC→0)Absorbing time in a finite Markov chain; p = λλ+µ = 1− q

1,C

1,C−10,C−2

1,C−2 2,C−1

2,C

3,C

C−2,C−1 C−1,C

C,C0,0

0,1 1,2

0,C

0,C−1

0,C−3

p

pp p p p

ppp

p p

pq

q

q q q q q

qqq

q q

qLevel 3

Level 4

Level 5

Level C+1

Level C+2

Level 2

Explicit recurrence equationsCase λ = µ Eτb = C+C22 .

Computable bounds for M/M/1/C

If the stationary distribution is concentrated on 0 (λ < µ),

Eτb 6 Eh0→C is an accurate bound.

Theorem

The mean coupling time Eτb of a M/M/1/C queue with arrival rate λand service rate µ is bounded using p = λ/(λ + µ) = 1− q.

Critical bound: ∀p ∈ [0, 1], Eτb 6 C2+C2 .

Heavy traffic Bound: if p > 12 , Eτb 6 Cp−q −

q(1−“

qp

”C)

(p−q)2.

Light traffic bound: if p < 12 , Eτb 6 Cq−p −

p(1−“

pq

”C)

(q−p)2.

Computable bounds for M/M/1/C

Example with C = 10

0

20

40

60

80

100

120

0 0.2 0.4 0.6 0.8 1

Eτb

p

heavy trafficLight trafficbound

C+C2

2

C + C2

bound

Example for tandem queues

Coupling of Queue 0

Time

0

X 00 = 55

4

2

1

3 = C1

6 = C0

0

−τ b0Coupling of queue 1 conditionned by state of queue 0

4

3 = C1

1

0

−τ b1 (s0 = 2) 0Time

X 11 = 2

X 10 = 3

5 = X 00

6

2

X 11 = 2

5

4

2

1

3 = C1

6 = C0

0

−τ b0 − τ b1 (s0 = 5)

X 00 = 5

τ b1 (s0 = 5) 0Time

X 10 = 3

Then τb 6st ∞τb1 + τb0 , normalized

Bound with isolated queues

TheoremIn an acyclic stable network of K M/M/1/Ci queues with Bernoullirouting and losses in case of overflow, the coupling time from the pastsatisfies in expectation,

E[τb] 6K−1∑i=0

Λ

`i + µi

Ciqi − pi

−pi(1−

(piqi

)Ci)

(qi − pi)2

6

K−1∑i=0

Λ

`i + µi(Ci + C

2i ).

Outline

4 Acyclic networks

Synthesis

Computable bound for the mean coupling time :

- linear in the number of component of the model;

- at most quadratic in queues sizes;

- large capacity queues ( bound is accurate).

Practical impact

- Accurate bounds, dimensionning of trajectories length;

- Simulation useful even for low probability events;

- Coupling time is explained by the spread of the stationarydistribution.

Future works

Conjecture for general networks.

0

700

800

0 0.5 1 1.5 2 2.5 3 3.5 4

500

400

300

200

100

600

λ5

Eτb

B1 (proven)

B1 ∧ B2 ∧ B3

B3 (conjecture)

B2 (conjecture)

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

Queueing Networks with finite capacityEvent modelling and monotonicityPerfect simulation and coupling timeAcyclic networksSynthesis and future works

bounds for the coupling time in queueing networks perfect ... · outline 1 queueing networks with...

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