Wi Wireless Access Graduate course in Communications Engineering - - PowerPoint PPT Presentation

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Wi Wireless Access Graduate course in Communications Engineering - - PowerPoint PPT Presentation

Lecture 7 November 11, 2020 Wi Wireless Access Graduate course in Communications Engineering University of Rome La Sapienza Rome, Italy 2020-2021 Poisson processes Poisson random process Reference tool in modelling the distribution


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Wi Wireless Access

Graduate course in Communications Engineering University of Rome La Sapienza Rome, Italy 2020-2021

Lecture 7 – November 11, 2020

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Poisson processes

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Poisson random process

  • Reference tool in modelling the distribution of

discrete events in time

  • Can be generalized to the birth and death process
  • Examples:

– Arrival of messages (packets) at a digital communication multiplexer – Arrival of photons in a light beam at an optical receiver (optical communications)

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Poisson random process

  • Definition of Random points in time: points in time

at which an event occurs

  • Let the time of the k-th arrival be denoted by tk

where tk≥ tj for k > j

  • Define a continuous-time random process N(t) equal

to the number of arrivals since starting time t0 up to current time t.

  • We call N(t) a counting process
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Poisson random process

  • N(t) has the following properties:

– it only assumes non-negative values; – it has initial condition N(t0) = 0; – it increases by 1 at each random point in time.

a) Only arrivals b)Arrivals and departures

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Poisson random process

  • Typical example: a queueing system
  • N(t) is the difference between number of arrivals and

number of departures accumulated over time

  • The model is adequate for example for a buffer in a packet-

switching network

b)Arrivals and departures

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Poisson random process

  • In most cases, the count N(t) is all we need to

predict the system evolution after t

  • When this is true, N(t) is the state of the system at

time t: we say that the system is in state j if N(t)=j

  • The system can change state at any time (differently

from what we saw for Markov chains)

  • Note that a sample N(ti) of the counting process at

time ti is a discrete-valued random variable

  • We define the probability of being in state j as:

qj t

( )=Pr N t=ti

( )= j

⎡ ⎣ ⎤ ⎦ ∀ti ∈R

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  • We need to model the evolution of the system from
  • ne state to the other
  • We cannot use the probability of transition from one

state to the other at a given time t, because it will be almost always equal to 0

  • We can define, however, the rate R of transitions

between two states

  • If R is a constant, in a time δt one has Rδt

transitions

  • For δt

0, we can neglect the possibility of two transitions in δt , i.e.:

– Rδt is the probability of one transition in δt, – (1 - Rδt) is the probability of no transition in δt.

Birth and death process

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Birth and death process

  • We can thus define the following state transition diagram for a birth

and death process: where:

  • are the transition rates from state i to state i+1
  • are the transition rates from state j to state j-1

λi t

( ),

i = 0,1, µ j t

( ),

j = 1, 2,

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Birth and death process

  • The evolution of the process is then governed by the following set
  • f equations:

with the conditions: dqj t

( )

dt = λ j−1 t

( )qj−1 t ( ) + µ j+1 t ( )qj+1 t ( ) − λ j t ( ) + µ j t ( )

( )qj t

( ),

j ≥ 0 q−1 t

( ) = 0

⎧ ⎨ ⎪ ⎩ ⎪ q0 t0

( ) = 1,

qj t0

( ) = 0 ∀j > 0

⎧ ⎨ ⎪ ⎩ ⎪

Birth from j-1 Death from j+1 Birth to j+1 Death to j-1

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Pure birth process

  • The special case of a birth and death process with:

is known as pure birth process or Poisson process with constant rate λ

  • For this process one has the following diagram:

λi t

( ) = λ,

i = 0,1, µ j t

( ) = 0,

j = 1, ⎧ ⎨ ⎪ ⎩ ⎪

  • and the following set of equations:

dqj t

( )

dt = λqj−1 t

( ) − λqj t ( ),

j ≥ 0 q0 0

( ) = 1

with

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Pure birth process

  • The first order differential equations can be solved easily
  • If we define the Laplace transform of the state probability:

and take the Laplace transform of both sides we get: Qj s

( ) =

qj t

( )

e−stdt

  • Taking into account the initial condition in t0=0 one has:

sQj s

( ) − qj 0 ( ) + λQj s ( ) = λQj−1 s ( )

Q0 s

( ) =

1 s + λ Qj s

( ) =

λ s + λ Qj−1 s

( ),

j > 0 and

  • Which can be solved by iteration leading to:

Qj s

( ) =

λ j s + λ

( )

j+1

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Pure birth process

  • And finally, by taking the inverse Laplace transform, one has,

for t ≥ 0 and j ≥ 0 :

qj t

( ) = Pr N t ( ) = j

⎡ ⎣ ⎤ ⎦ = λt

( )

j

j! e−λt

Poisson distribution

  • Thus a pure birth process is actually a Poisson process with

constant rate (but the result can be generalized for a variable rate as well)

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Poisson distribution properties

  • The Poisson distribution with parameter a

pN k

( ) = a ( )

k

k! e−a 1. A random variable N characterized by a p.d.f. equal to a Poisson distribution with parameter a has the following expectations: 2. The moment generating function for the random variable N takes the form: E N

[ ] = a

Mean loge ΦN s

( ) = a es −1

( )

E N 2 ⎡ ⎣ ⎤ ⎦ − E N

[ ]

2 = a

Variance has the following properties:

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Poisson process

  • Going back to the Poisson process, it can be proved that, for a

general initial condition N(t0)=k, one has: qj t

( ) = λ t − t0

( )

( )

j−k

j − k

( )!

e

−λ t−t0

( ), for j ≥ k, t ≥ t0

– which is a Poisson distribution with parameter λ(t-t0), that is the expected number of arrivals from instant t0.

  • (j-k) is the count since the starting time t0
  • IMPORTANT: the number of arrivals starting from t=t0 has a

distribution that does not depend in any way on what happened before t0

  • As a consequence, the number of arrivals in the interval (t0 ,t) is

statistically independent of the number of arrivals in any other non-overlapping interval of time

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  • The special case of a birth and death process with:

is known as a pure death process, and is characterized by the following transition diagram:

  • Under the assumption that the state of the system in t=0 is n, it can be

proved that for this process one has the following state probabilities:

Pure death process

λi t

( ) = 0

µ j t

( ) = jµ

⎧ ⎨ ⎪ ⎩ ⎪ qj t

( ) = n

j ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ p j t

( ) 1− p t ( )

( )

n− j

with

p t

( ) = e−µt

(Binomial distribution)

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Poisson process with variable rate

  • Let us consider a Poisson process with time-varying

rate, that is λ=λ(t)

  • The set of equations describing the system is:

dqj t

( )

dt + λ t

( )qj t ( ) = λ t ( )qj−1 t ( ),

j ≥ 0 q−1 t

( ) = 0

⎧ ⎨ ⎪ ⎩ ⎪ q0 t0

( ) = 1

qj t0

( ) = 0

j ≥ 1 ⎧ ⎨ ⎪ ⎩ ⎪

with: assuming thus that the system is in state j=0 at t=t0

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Poisson process with variable rate

  • Since λ (t) depends on time t the Laplace transform is of no help
  • One defines then:

Λ t

( )=

λ τ

( )dτ

t0 t

that is the total number of arrivals in the interval [t0, t]:

λ(t) λ λt

Λ(t)

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Poisson process with variable rate

  • The probability of n arrivals in the interval [t0, t] is

then determined by a Poisson distribution with parameter λ(t)

qn t

( ) = Λn t ( )

n! e

−Λ t

( )

in perfect analogy with the constant rate case

  • In this case one has for the number N(t) of arrivals during the

interval [t0, t] the following expectations:

E N t

( )

⎡ ⎣ ⎤ ⎦ = Λ t

( )

Mean σ N t

( )

2

= Λ t

( )

Variance

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Birth and death process: the queueing problem

  • Birth and death processes are particularly suitable for modelling

queueing systems

  • We will adopt the following convention/terminology:

– A queue is a buffer or memory which stores messages/packets – Messages in the queue are cleared by one or more servers (each server processes one packet at the time) – The queue can only keep a limited number of messages, referred to as the number of waiting positions – The state of the system (which tracks a counting process) is the sum of the packets waiting in the queue and of the packets currently being served – Messages arrive at the queue (births) at random times, and depart from the queue (deaths) due to the completion of service

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The queueing problem

Example: one server and no waiting positions

  • Constant arrival rate λ
  • Constant service (departure) rate μ
  • Since there is one server and no waiting positions, if a

message arrives while the server is busy the message is lost -> only two states, 0 and 1

  • It can be proved that the probability that the server is

not busy is:

q0 t

( ) =

µ λ + µ + q0 0

( )−

µ λ + µ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥e

− µ+λ

( )t

  • In such a system is usually more interesting to derive

the state probabilities at steady state, that is out of the initial transient