Analyzing the M/G/1 Queue with a Branching Process

Céline Comte Nokia Bell Labs France - Télécom ParisTech

Reading Group ”Network Theory” December 18, 2017 M/G/1 queue

Branching process

Stability Recurrence and transcience Positive recurrence

Related work and references

Branching process

Stability Recurrence and transcience Positive recurrence

Related work and references

λ µ

M The arrival process is Poisson → Arrival rate 0 < λ < +∞ G The service times are i.i.d. with a general distribution 1 → Mean service time 0 < µ < +∞ 1 A single server ? Infinite queue length

FCFS Customers leave in their arrival order

More on Poisson processes

tt1 Time

• Exponentially distributed inter-arrival times • The number of arrivals during a time interval of length t has a Poisson distribution with mean λt.

tt1 Time

• Exponentially distributed inter-arrival times • The number of arrivals during a time interval of length t has a Poisson distribution with mean λt. • The numbers of arrivals during two disjoint intervals are independent.

′ tt1 t Time

λ µ

FCFS Customers leave in their arrival order

• Xt = number of customers in the queue at time t

• (Xt)t≥0 is not a Markov process (in general)

Queue length 1

0 0 2 4 6 8 10 12 14 Time

• Dn = departure time of the n-th customer

• Dn is a regeneration point of (Xt)t≥0

Queue length 1

0 0 2 4 6 8 10 12 14 Time

• Dn = departure time of the n-th customer

• Dn is a regeneration point of (Xt)t≥0

Queue length 1

0 D0 D1 D2 D3 D4 D5 Time

• = Yn XDn number of customers left behind customer n • (Yn)n∈N is a Markov chain

Queue length 1

0 D0 D1 D2 D3 D4 D5 Time

• Because of the FCFS assumption, Yn = number of customers that arrived since customer n entered the queue

Queue length 1

0 D0 D1 D2 D3 D4 D5 Time

• Because of the FCFS assumption, Yn − Yn−1 + 1 = number of customers that arrived between the departures of customers n − 1 and n

Queue length 1

0 D0 D1 D2 D3 D4 D5 Time

• Because of the FCFS assumption, Yn − Yn−1 + 1 = number of customers that arrived during the service of customer n

Queue length 1

0 D0 D1 D2 D3 D4 D5 Time

• Because of the FCFS assumption, Yn − (Yn−1 − 1)+ = number of customers that arrived during the service of customer n

Queue length 1

0 D0 D1 D2 D3 D4 D5 Time

p0 p1 p0

• The number of arrivals during a service time of length t has a Poisson distribution with mean λt. • Letting F denote the CDF of the service time distribution, ∫ +∞ (λt)k = −λt pk e dF(t). 0 k! • Independence

Why is it a Markov chain ?

0 1 2 3 ...4

• = pk probabily that k customers arrive during the service = pk time of a customer

p0 p1 p0

• Letting F denote the CDF of the service time distribution, ∫ +∞ (λt)k = −λt pk e dF(t). 0 k! • Independence

Why is it a Markov chain ?

0 1 2 3 ...4

• = pk probabily that k customers arrive during the service = pk time of a customer • The number of arrivals during a service time of length t has a Poisson distribution with mean λt.

p0 p1 p0

• Independence

Why is it a Markov chain ?

0 1 2 3 ...4

• = pk probabily that k customers arrive during the service = pk time of a customer • The number of arrivals during a service time of length t has a Poisson distribution with mean λt. • Letting F denote the CDF of the service time distribution, ∫ +∞ (λt)k = −λt pk e dF(t). 0 k!

p0 p1 p0

Why is it a Markov chain ?

0 1 2 3 ...4

Why is it a Markov chain ?

p 0 1 2 2 3 ...4 p1 p0

p p2 p 0 1 1 2 2 3 ...4 p0 p1 p0

• = pk probabily that k customers arrive during the service = pk time of a customer • Transition matrix :   p p p p ... p ...  0 1 2 3 k  p p p p ... p ...   0 1 2 3 k     p0 p1 p2 ... pk ...    p0 p1 ... pk ... p0 ... pk ...

→ If (Yn)n∈N is positive recurrent, then it is also ergodic

Properties of (Yn)n∈N

p p2 p 0 1 1 2 2 3 ...4 p0 p1 p0

• Irreducible

→ All the states of (Yn)n∈N have the same nature (positive recurrent, null recurrent or transcient) → We just need to know the nature of state 0

p p2 p 0 1 1 2 2 3 ...4 p0 p1 p0

• Irreducible

→ All the states of (Yn)n∈N have the same nature (positive recurrent, null recurrent or transcient) → We just need to know the nature of state 0 • Aperiodic

→ If (Yn)n∈N is positive recurrent, then it is also ergodic

p p2 p 0 1 1 2 2 3 ...4 p0 p1 p0

• State 0 is recurrent ⇔ If the queue starts empty, then with probability 1 it empties out an infinite number of times ⇔ If the queue starts empty, then with probability 1 it eventually empties out

p p2 p 0 1 1 2 2 3 ...4 p0 p1 p0

Assume that state 0 is recurrent. • State 0 is positive recurrent ⇔ If the queue starts empty, then the mean time until it empties out again is finite • State 0 is null recurrent ⇔ If the queue starts empty, then the mean time until it empties out again is infinite

• We just need to look at one busy period of the queue.

Queue length 1

0 Start of a End of a busy period busy period

Branching process

Stability Recurrence and transcience Positive recurrence

Related work and references

• One node at generation 0 • The children of the nodes of generation n belong to generation n + 1 • The number of children of a node is

• Mean number of children per node ρ < +∞

Deﬁnition

• Random tree

A = 2

• The children of the nodes of generation n belong to generation n + 1 • The number of children of a node is

• Mean number of children per node ρ < +∞

Deﬁnition

• Random tree • One node at generation 0

A = 2

• The number of children of a node is

• Mean number of children per node ρ < +∞

Deﬁnition

• Random tree • One node at generation 0 • The children of the nodes of generation n belong to generation n + 1

A = 2

Deﬁnition

• Random tree • One node at generation 0 • The children of the nodes of generation n belong to generation n + 1 • The number of children of a node is

A = 2

Deﬁnition

• Random tree • One node at generation 0 • The children of the nodes of generation n belong to generation n + 1 • The number of children of a node is − random,

A = 2

Deﬁnition

• Random tree • One node at generation 0 • The children of the nodes of generation n belong to generation n + 1 • The number of children of a node is − random, − with a probability distribution that is the same for all nodes, − independent of the number of A = 2 children of other nodes,

B Given the number of children of the root, the subtrees rooted at these nodes are independent and have the same distribution as the initial tree

A = 2

B Given the number of children of the root, the subtrees rooted at these nodes are independent and have the same distribution as the initial tree

A = 2

2 Queue length

0 0 10 20 30 40 50 60 Time

Time

2 Queue length

0 0 10 20 30 40 50 60 Time

Time

27 © 2017 Nokia Public • = pk probability that k customers arrive = pk during a single service time = pk probability that a node has k children • The number of arrivals during a service time of length t has a Poisson distribution with mean λt. • Independence

t A = 2 Time

Why is it a branching process ?

t Time

Why is it a branching process ?

• = pk probability that k customers arrive = pk during a single service time = pk probability that a node has k children

A = 2

Why is it a branching process ?

• = pk probability that k customers arrive = pk during a single service time = pk probability that a node has k children • The number of arrivals during a service time of length t has a Poisson distribution with mean λt.

t A = 2 Time

• S = service time of a given customer. 1 S is a random variable with mean µ . • A = number of customers arrived during A = the service of this customer • Given S, A has a Poisson distribution with mean λS.

S A = 2 Time

• Mean number of children of a node E(A) = E(E(A|S)) = E(λS) = λE(S) λ = µ . λ ⇒ ρ = µ = load of the queue, λ ρ = µ = mean number of children of λ A = 2 ρ = µ = a node in the tree.

Branching process

Stability Recurrence and transcience Positive recurrence

Related work and references

Branching process

Stability Recurrence and transcience Positive recurrence

Related work and references

λ • ρ = µ = load of the queue, λ ρ = µ = mean number of children per node.

• Branching process result : − ρ ≤ 1 : the tree dies out with probability 1. ρ ≤ 1 : (assuming that p1 < 1 if ρ = 1) − ρ > 1 : the tree dies out with probability < 1. • Queueing reformulation :

− ρ ≤ 1 : (Yn)n∈N is recurrent. − ρ > 1 : (Yn)n∈N is transcient.

• P satisfies the fixed-point equation

+∞∑ = k P P pk. k=0

Sketch of proof

• = pk probability that k customers arrive = pk during a single service time = pk probability that a node has k children

N2 = 2

+∞∑ = k P P pk. k=0

Sketch of proof

• = pk probability that k customers arrive = pk during a single service time = pk probability that a node has k children

• P = probability that the queue empties P = probability that the tree is finite

N2 = 2

• = pk probability that k customers arrive = pk during a single service time = pk probability that a node has k children

• P = probability that the queue empties P = probability that the tree is finite

• P satisfies the fixed-point equation N2 = 2 +∞∑ = k P P pk. k=0

• = pk probability that k customers arrive = pk during a single service time = pk probability that a node has k children

• P = probability that the queue empties P = probability that the tree is finite

• P satisfies the fixed-point equation N2 = 2 +∞∑ = k P P pk. k=0

P = ϕ(P),

where +∞∑ ϕ = k (x) x pk k=0

is the generating function of (pk)k∈N

Sketch of proof

• P satisfies the fixed-point equation

+∞∑ = k P P pk k=0

N2 = 2

• P satisfies the fixed-point equation

+∞∑ = k P P pk k=0

• P satisfies the fixed-point equation

P = ϕ(P),

where +∞∑ N2 = 2 ϕ = k (x) x pk k=0

is the generating function of (pk)k∈N

0.6 y

0.4 +∞∑ ′ϕ(x) = xkp p0 k k=0 +∞∑ 0.2 ϕ′ = k−1 (x) kx pk k=1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x

• P is the smallest solution in [0,1] of the fixed-point equation

P = ϕ(P),

where +∞∑ ϕ = k (x) x pk k=0

is the generating function of (pk)k∈N N2 = 2

• (Pn)n∈N satisfies +∞∑ = k = ϕ Pn+1 Pn pk, that is, Pn+1 (Pn) k=0

• P0 = 0

Sketch of proof