tcom 501: lecture 3

1

TCOM 501: Networking Theory & Fundamentals

Lecture 3 January 29, 2003

Prof. Yannis A. Korilis

3-2 Topics Markov Chains Discrete-Time Markov Chains Calculating Stationary Distribution Global Balance Equations Detailed Balance Equations Birth-Death Process Generalized Markov Chains Continuous-Time Markov Chains

3-3 Markov Chain Stochastic process that takes values in a

countable set Example: {0,1,2,…,m}, or {0,1,2,…} Elements represent possible “states” Chain “jumps” from state to state

Memoryless (Markov) Property: Given the present state, future jumps of the chain are independent of past history

Markov Chains: discrete- or continuous- time

3-4 Discrete-Time Markov Chain Discrete-time stochastic process {Xn: n = 0,1,2,

…} Takes values in {0,1,2,…} Memoryless property:

Transition probabilities Pij

Transition probability matrix P=[Pij]

1 1 1 0 0 1

1

{ | , ,..., } { | }{ | }

n n n n n n

ij n n

P X j X i X i X i P X j X iP P X j X i

0

0, 1ij ijj

P P

3-5

n step transition probabilities

Chapman-Kolmogorov equations

is element (i, j) in matrix Pn

Recursive computation of state probabilities

Chapman-Kolmogorov Equations

{ | }, , 0, , 0nij n m mP P X j X i n m i j

nijP

0

, , 0, , 0n m n mij ik kj

k

P P P n m i j

3-6

State Probabilities – Stationary Distribution State probabilities (time-dependent)

In matrix form:

If time-dependent distribution converges to a limit

is called the stationary distribution

Existence depends on the structure of Markov chain

11 1

0 0

{ } { } { | } π πn nn n n n j i ij

i i

P X j P X i P X j X i P

0 1π { }, π (π ,π ,...)n n n n

j nP X j

1 2 2 0π π π ... πn n n nP P P

π lim πn

n

π πP

3-7

Aperiodic: State i is periodic:

Aperiodic Markov chain: none of the states is periodic

Classification of Markov ChainsIrreducible: States i and j

communicate:

Irreducible Markov chain: all states communicate

, : 0, 0n mij jin m P P 1: 0n

iid P n d

0

3 4

21

0

3 4

21

3-8 Limit TheoremsTheorem 1: Irreducible aperiodic Markov chain For every state j, the following limit

exists and is independent of initial state i Nj(k): number of visits to state j up to time k

j: frequency the process visits state j

0π lim { | }, 0,1,2,...j nnP X j X i i

0

( )π lim 1j

j k

N kP X i

k

3-9

Existence of Stationary DistributionTheorem 2: Irreducible aperiodic Markov chain. Thereare two possibilities for scalars:

1. j = 0, for all states j No stationary distribution2. j > 0, for all states j is the unique stationary

distributionRemark: If the number of states is finite, case 2 is theonly possibility

0π lim { | } lim nj n ijn n

P X j X i P

3-10 Ergodic Markov Chains Markov chain with a stationary distribution

States are positive recurrent: The process returns to state j “infinitely often”

A positive recurrent and aperiodic Markov chain is called ergodic

Ergodic chains have a unique stationary distribution

Ergodicity ⇒ Time Averages = Stochastic Averages

π 0, 0,1,2,...j j

π lim nj ijn

P

3-11

Calculation of Stationary Distribution

A. Finite number of states Solve explicitly the

system of equations

Numerically from Pn which converges to a matrix with rows equal to Suitable for a small number of states

B. Infinite number of states Cannot apply previous

methods to problem of infinite dimension

Guess a solution to recurrence:

0

0

π π , 0,1,...,

π 1

m

j i iji

m

ii

P j m

0

0

π π , 0,1,...,

π 1

j i iji

ii

P j

3-12 Example: Finite Markov Chain

Markov chain formulation i is the number of umbrellas

available at her current location

Transition matrix

Absent-minded professor uses two umbrellas when commuting between home and office. If it rains and an umbrella is available at her location, she takes it. If it does not rain, she always forgets to take an umbrella. Let p be the probability of rain each time she commutes. What is the probability that she gets wet on any given day?

0 2 1 1 p

p1 p

1 p

0 0 10 1

1 0P p p

p p

3-13 Example: Finite Markov Chain

0 2 1 1 p

p1 p

1 p0 0 10 1

1 0P p p

p p

2 0

0 2

1 1 20 1 2

1 2

1

0

π π π

π (1 )ππ π π (1 )π π 1 1 1π , π , π

π 1 3 3 3π π π 1

i i

pP p p p

pp p p

01{gets wet} π3

pP p pp

3-14 Example: Finite Markov Chain Taking p = 0.1:

Numerically determine limit of Pn

Effectiveness depends on structure of P

0 0 10 0.9 0.1

0.9 0.1 0P

1 1 1π , , 0.310, 0.345, 0.3453 3 3

pp p p

0.310 0.345 0.345lim 0.310 0.345 0.345 ( 150)

0.310 0.345 0.345

n

nP n

3-15 Global Balance Equations Markov chain with infinite number of states Global Balance Equations (GBE)

is the frequency of transitions from j to i

Intuition: j visited infinitely often; for each transition out of j there must be a subsequent transition into j with probability 1

0 0

π π π π , 0j ji i ij j ji i iji i i j i j

P P P P j

π j jiP

Frequency of Frequency oftransitions out of transitions into j j

3-16 Global Balance Equations Alternative Form of GBE

If a probability distribution satisfies the GBE, then it is the unique stationary distribution of the Markov chain

Finding the stationary distribution: Guess distribution from properties of the system Verify that it satisfies the GBE Special structure of the Markov chain simplifies task

π π , 0,1,2,...j ji i ijj S i S i S j S

P P S

3-17 Global Balance Equations – Proof

0 0 0 0

π π π π

π π π

π π

j ji i ij j ji i iji i j S i j S i

j ji ji i ij i ijj S i S i S j S i S i S

j ji i ijj S i S i S j S

P P P P

P P P P

P P

0 0

0 0

π π and 1

π π π π

j i ij jii i

j ji i ij j ji i iji i i j i j

P P

P P P P

3-18 Birth-Death Process

One-dimensional Markov chain with transitions only between neighboring states: Pij=0, if |i-j|>1

Detailed Balance Equations (DBE)

Proof: GBE with S ={0,1,…,n} give:

0 1 n+1n2

, 1n nP

1,n nP ,n nP, 1n nP

1,n nP 01P

10P00P

S Sc

, 1 1 1,π π 0,1,...n n n n n nP P n

, 1 1 1,0 1 0 1

π π π πn n

j ji i ij n n n n n nj i n j i n

P P P P

3-19 Example: Discrete-Time Queue In a time-slot, one arrival with probability p or

zero arrivals with probability 1-p In a time-slot, the customer in service departs

with probability q or stays with probability 1-q Independent arrivals and service times State: number of customers in system

0 1 n+1n2

(1 )p

p (1 )p q

(1 )(1 )p q pq (1 )q p (1 )q p

p (1 )p q

(1 )(1 )p q pq (1 )q p

3-20 Example: Discrete-Time Queue

0 1 n+1n2

(1 )p

p (1 )p q

(1 )(1 )p q pq (1 )q p (1 )q p

(1 )p q

(1 )(1 )p q pq (1 )q p

(1 )p q

1 1(1 )

π (1 ) π (1 ) π π , 1(1 )n n n n

p qp q q p n

q p

0 1 1 0/

π π (1 ) π π1p q

p q pp

(1 )Define: / ,

(1 )p q

p qq p

1 0 10

1

π 1 π π , 1

1π π , 1

nn

n n

p np

n

3-21 Example: Discrete-Time Queue Have determined the distribution as a function of 0

How do we calculate the normalization constant 0? Probability conservation law:

Noting that

10π π , 1

1n

n np

1 11

00 1

1π 1 π 1 11 (1 ) (1 )

nnn np p

1

)1()1()1(111

qpq

pqqppqpp

01

π 1

π (1 ) , 1nn n

3-22 Detailed Balance Equations General case:

Imply the GBE Need not hold for a given Markov chain Greatly simplify the calculation of stationary distributionMethodology: Assume DBE hold – have to guess their form Solve the system defined by DBE and ii = 1

If system is inconsistent, then DBE do not hold If system has a solution {i: i=0,1,…}, then this is the unique

stationary distribution

π π , 0,1,...j ji i ijP P i j

3-23 Generalized Markov Chains Markov chain on a set of states {0,1,…}, that

whenever enters state i The next state that will be entered is j with probability Pij Given that the next state entered will be j, the time it spends

at state i until the transition occurs is a RV with distribution Fij

{Z(t): t ≥ 0} describing the state the chain is in at time t: Generalized Markov chain, or Semi-Markov processIt does not have the Markov property: future depends on

The present state, and The length of time the process has spent in this state

3-24 Generalized Markov Chains Ti: time process spends at state i, before making a

transition – holding time Probability distribution function of Ti

Tii: time between successive transitions to i Xn is the nth state visited. {Xn: n=0,1,…}

Is a Markov chain: embedded Markov chain Has transition probabilities Pij

Semi-Markov process irreducible: if its embedded Markov chain is irreducible

0 0

( ) { } { | next state } ( )i i i ij ij ijj j

H t P T t P T t j P F t P

0[ ] ( )i iE T t dH t

3-25 Limit TheoremsTheorem 3: Irreducible semi-Markov process, E[Tii] < ∞ For any state j, the following limit

exists and is independent of the initial state.

Tj(t): time spent at state j up to time t

pj is equal to the proportion of time spent at state j

lim { ( ) | (0) }, 0,1,2,...j tp P Z t j Z i i

[ ][ ]

jj

jj

E Tp

E T

( )lim (0) 1j

j t

T tP p Z i

t

3-26 Occupancy DistributionTheorem 4: Irreducible semi-Markov process; E[Tii] < ∞.Embedded Markov chain ergodic; stationary distribution π

Occupancy distribution of the semi-Markov process

πj proportion of transitions into state j E[Tj] mean time spent at j

Probability of being at j is proportional to πjE[Tj]

0 0

π π , 0; π 1j i ij ii i

P j

π [ ], 0,1,...

π [ ]j j

ji i

i

E Tp j

E T

3-27 Continuous-Time Markov ChainsContinuous-time process {X(t): t ≥ 0} taking

values in {0,1,2,…}. Whenever it enters state i Time it spends at state i is exponentially

distributed with parameter νi

When it leaves state i, it enters state j with probability Pij, where j≠i Pij = 1

Continuous-time Markov chain is a semi-Markov process with

Exponential holding times: a continuous-time Markov chain has the Markov property

( ) 1 , , 0,1,...itijF t e i j

3-28 Continuous-Time Markov Chains When at state i, the process makes

transitions to state j≠i with rate:

Total rate of transitions out of state i

Average time spent at state i before making a transition:

ij i ijq P

ij i ij ij i j i

q P

[ ] 1/i iE T

3-29 Occupancy Probability Irreducible and regular continuous-time Markov chain

Embedded Markov chain is irreducible Number of transitions in a finite time interval is finite with

probability 1 From Theorem 3: for any state j, the limit

exists and is independent of the initial state pj is the steady-state occupancy probability of state j pj is equal to the proportion of time spent at state j

[Why?]

lim { ( ) | (0) }, 0,1,2,...j tp P X t j X i i

3-30 Global Balance Equations Two possibilities for the occupancy probabilities:

pj= 0, for all j pj> 0, for all j, and j pj = 1

Global Balance Equations

Rate of transitions out of j = rate of transitions into j

If a distribution {pj: j = 0,1,…} satisfies GBE, then it is the unique occupancy distribution of the Markov chain

Alternative form of GBE:

, 0,1,...j ji i iji j i j

p q p q j

, {0,1,...}j ji i ijj S i S i S j S

p q p q S

3-31 Detailed Balance Equations Detailed Balance Equations

Simplify the calculation of the stationary distribution

Need not hold for any given Markov chain Examples: birth-death processes, and

reversible Markov chains

, , 0,1,...j ji i ijp q p q i j

3-32 Birth-Death Process

Transitions only between neighboring states

Detailed Balance Equations

Proof: GBE with S ={0,1,…,n} give:

0 1 n+1n2

0

1

n

1n

1

2

1n

n

, 1 , 1, , 0, | | 1i i i i i i ijq q q i j

1 1, 0,1,...n n n np p n

1 10 1 0 1

n n

j ji i ij n n n nj i n j i n

p q p q p p

S Sc

3-33 Birth-Death Process

Use DBE to determine state probabilities as a function of p0

Use the probability conservation law to find p0

Using DBE in problems: Prove that DBE hold, or Justify validity (e.g. reversible process), or Assume they hold – have to guess their form – and solve system

1 11

1 1 2 1 2 01 2 0 0

01 1 1 1

...

n n n nn

n n n n n in n n

in n n n n i

p p

p p p p p

11 1 1

0 00 1 1 10 0 01 1 1

1 1 1 1 , if n n n

i i in

n n n ni i ii i i

p p p

3-34 M/M/1 Queue Arrival process: Poisson with rate λ Service times: iid, exponential with parameter

µ Service times and interarrival times:

independent Single server Infinite waiting room N(t): Number of customers in system at time t

(state)0 1 n+1n2

3-35 M/M/1 Queue

Birth-death process → DBE

Normalization constant

Stationary distribution

0 1 n+1n2

1

1 1 0...n n

nn n n

p p

p p p p

0 00 1

1 1 1 1 , if 1nn

n n

p p p

(1 ), 0,1,...nnp n

3-36The M/M/1 Queue

Average number of customers

Applying Little’s Theorem, we have

Similarly, the average waiting time and number of customers in the queue is given by

1

0 0 0

2

(1 ) (1 )

1(1 )(1 ) 1

n nn

n n n

N np n n

N

11NT

1 and 1 2

WNTW Q