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Continuous Time Markov Chains and Basic Queueing Theory

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These timers are independent of each other. Recall that Exponential distribution is memoryless. When the first timer expires, the MC makes the corresponding transition ... – PowerPoint PPT presentation

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Title: Continuous Time Markov Chains and Basic Queueing Theory


1
Continuous Time Markov Chainsand Basic Queueing
Theory
  • EE384X Review 4
  • Winter 2008

2
Review DTMC
  • pij is the transition probability from i to j
    over one time slot
  • The time spent in a state is geometrically
    distributed
  • Result of the Markov (memoryless) property
  • When there is a jump from state i, it goes to
    state j with probability

3
Continuous Time Version
  • qij is the transition rate from state i to state
    j

4
CTMC
  • Upon entering state i, a random timer
    TijExp(qij) is started for each potential
    transition i!j
  • These timers are independent of each other
  • Recall that Exponential distribution is
    memoryless
  • When the first timer expires, the MC makes the
    corresponding transition
  • Let Ti be the time spent in state i, and qiÃ¥j¹i
    qij, then Ti Exp(qi)
  • When there is a transition, the probability of
    jumping to state j is qij /qi

5
Definitions
  • X(t)t0 is a continuous time Markov chain
    ifPX(st)j X(u) us PX(st)j X(s)
  • Similar to Discrete Time MCs, Continuous Time MCs
    have stationary distribution p
  • Exists when Markov chain is positive recurrent
    and irreducible

6
Stationary Distribution
  • Balance equations
  • Transition rates in and out of state i are equal
  • Define matrix transition rate Q (qij) with qii
    -qi , then p Q 0, where p is a row vector
  • Together with Ã¥i p(i) 1, can solve for p

7
Queueing Theory Notation
  • A/S/s/k
  • A is the arrival process, e.g., Geometric,
    Poisson, Deterministic
  • S is the service distribution, e.g., Geometric,
    Exponential, Deterministic
  • s is the number of servers, e.g., 1, N, 1
  • k is the buffer size (if k is absent, then k 1)
  • E.g., Geom/M/1, M/M/1, M/D/1, M/M/1

8
M/M/1 Queue
  • Arrivals are Poisson with rate l
  • Inter-arrival times are exp(l)
  • Services are exponential with rate m
  • These are also transition rates for the Markov
    chain
  • This looks very similar to Geom/Geom/1 queue, but
    different

9
Solving M/M/1 Queue
  • We have pi l pi1 m
  • Let r l/m, then pi pi-1 r p0 ri
  • If r lt 1, the stationary distribution exists
    pi (1 - r) ri
  • Average Queue size

10
M/M/1 Queue
  • NQ is the queue size, excluding the one in
    service

11
M/M/1 Queue
  • Customer arrival process is Poisson(l)
  • All customers are served in parallel exp(m)
  • Departure rate proportional to of customers

12
Solving M/M/1 Queue
  • We have pi-1 l pi i m
  • Let r l/m, then
  • Thus

13
M/M/1 Queue
  • The queue size distribution of the M/M/1 queue is
    Poisson(r)
  • Therefore the average queue size is E(Q)r
  • Whats the condition for the queue to be
    recurrent?
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