1 TCOM 501: Networking Theory & Fundamentals Lecture 3 January 29, 2003 Prof. Yannis A. Korilis.
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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 i
P 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 Chains
Irreducible: 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 Theorems
Theorem 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 distribution
2. j > 0, for all states j is the unique stationary distribution
Remark: If the number of states is finite, case 2 is theonly possibility
0π lim { | } lim nj n ij
n nP 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 1
0 1
1 0
P p p
p p
3-13 Example: Finite Markov Chain
0 2 1 1 p
p1 p
1 p0 0 1
0 1
1 0
P p p
p p
2 0
0 2
1 1 20 1 2
1 2
1
0
π π π
π (1 )π
π π π (1 )π π 1 1 1π , π , π
π 1 3 3 3
π π π 1i i
p
P p p p
pp p p
0
1{gets wet} π
3
pP p p
p
3-14 Example: Finite Markov Chain Taking p = 0.1:
Numerically determine limit of Pn
Effectiveness depends on structure of P
0 0 1
0 0.9 0.1
0.9 0.1 0
P
1 1 1π , , 0.310, 0.345, 0.345
3 3 3
p
p p p
0.310 0.345 0.345
lim 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 of
transitions 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 π
1
p qp q p
p
(1 )Define: / ,
(1 )
p qp q
q 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 1
100 1
1π 1 π 1 1
1 (1 ) (1 )n
nn np p
1
)1(
)1()1(111
q
pq
pq
qppqpp
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 distribution
Methodology: 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 Theorems
Theorem 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
[ ]
[ ]j
jjj
E Tp
E T
( )lim (0) 1j
j t
T tP p Z i
t
3-26 Occupancy Distribution
Theorem 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 Chains
Continuous-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
3-36
The 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
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