Discrete Time Markov Chains

EE384X Review 2

Winter 2006

Outline

• Some examples

• Definitions

• Stationary Distributions

References (on reserve in library):

1. Hoel, Port, and Stone: Introduction to Stochastic Processes

2. Wolff: Stochastic Modeling and the Theory of Queues

Simple DTMCs

• “States” can be labeled (0,)1,2,3,…

• At every time slot a “jump” decision is made randomly based on current state

•

10

p

q

1-q1-p

10

2

a

d fc b

e

(Sometimes the arrow pointing back to the same state is omitted)

1-D Random Walk

• Time is slotted

• The walker flips a coin every time slot to decide which way to go

•

X(t)

p1-p

Single Server Queue

• Consider a queue at a supermarket

• In every time slot:– A customer arrives with probability p– The HoL customer leaves with probability q

Bernoulli(p)Geom(q)

Birth-Death Chain

• Can be modeled by a Birth-Death Chain (aka. Geom/Geom/1 queue)

• Want to know: – Queue size distribution – Average waiting time, etc.

0 1 2 3

Markov Property

• “Future” is independent of “Past” given “Present”

• In other words: Memoryless

• We’ve seen memoryless distributions: Exponential and Geometric

• Useful for modeling and analyzing real systems

Discrete Time Markov Chains

• A sequence of random variables {Xn} is called a Markov chain if it has the Markov property:

• States are usually labeled {(0,)1,2,…}

• State space can be finite or infinite

Transition Probability

• Probability to jump from state i to state j

• Assume stationary: independent of time

• Transition probability matrix:

P = (pij)

• Two state MC:

Stationary Distribution

Define

Then k+1 = k P ( is a row vector)

Stationary Distribution:

if the limit exists.

If exists, we can solve it by

Balance Equations

• These are called balance equations– Transitions in and out of state i are balanced

In General

• If we partition all the states into two sets, then transitions between the two sets must be “balanced”.– Equivalent to a bi-section in the state transition

graph– This can be easily derived from the Balance

Equations

Conditions for to Exist (I)

• Definitions:– State j is reachable by state i if

– State i and j commute if they are reachable by each other

– The Markov chain is irreducible if all states commute

Conditions for to Exist (I) (cont’d)

• Condition: The Markov chain is irreducible

• Counter-examples:21 43

32p=1

1

Conditions for to Exist (II)

• The Markov chain is aperiodic:

• Counter-example:

10

2

1

0 01 1

0

Conditions for to Exist (III)

• The Markov chain is positive recurrent:– State i is recurrent if

– Otherwise transient– If recurrent

• State i is positive recurrent if E(Ti)<1, where Ti is time between visits to state i

• Otherwise null recurrent

Solving for

10

p

q

1-q1-p

Birth-Death Chain

• Arrival w.p. p ; departure w.p. q

• Let u = p(1-q), d = q(1-p), = u/d

• Balance equations:

u u u u

d d d d

0 1 2 31-u

1-u-d 1-u-d 1-u-d

Birth-Death Chain (cont’d)

• Continue like this, we can derive:

(i-1) u = (i) d

• Equivalently, we can draw a bi-section between state i and state i-1

• Therefore, we have

Birth-Death Chain (cont’d)

Any Problems?

• What if is greater than 1?– Then the stationary distribution does not exist

• Which condition does it violate?

2£2 Switch w/ HoL Blocking

• Packets arrive as Bernoulli iid uniform • Packets queued at inputs• Only one packet can leave an output every time slot

1

1

1

2

2

1

1

2

2£2 Switch (cont’d)

• If both HoL packets are destined to the same output– Only one of them is served (chosen randomly)– The other output is idle, as packets are blocked– This is called head of line blocking– HoL blocking reduces throughput

• Want to know: throughput of this switch

2£2 Switch - DTMC

• States are the number of HoL packets destined to output 1 and output 2

• But states (0,2) and (2,0) are the same– Can “collapse” them together

1,1 2,00,2 0.25 0.25

0.5 0.5

0.5

0.5 0.5

2£2 Switch – DTMC (cont’d)

• Now P{(0,2)} = P{(1,1)} = 0.5

• Switch throughput = 0.5£1+0.5£2 = 1.5

• Per output throughput = 1.5/2 = 0.75

1,10,2

0.5

0.5

0.5 0.5

Another Method to Find

• Sometimes the Markov chain is not easy to solve analytically

• Can run the Markov chain for a long time, then

{fraction of time spent in state i} ! (i)