QUEUING. CONTINUOUS TIME MARKOV CHAINS {X(t), t >= 0} is a continuous time process with > sojourn times S 0, S 1, S 2,... > embedded process X n = X(S.

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Presentation transcript:

QUEUING

CONTINUOUS TIME MARKOV CHAINS {X(t), t >= 0} is a continuous time process with > sojourn times S 0, S 1, S 2,... > embedded process X n = X(S n-1 +) X is a CTMC if S n ~ Exp(q i ) where i=X n

MATRICES Probability Transition Matrix

GENERATOR MATRIX GIVES RISE TO THE NUMERICAL METHODS INVOLVING RAISING A MATRIX TO A POWER -- ROW SUMS EQUAL ZERO -- DIAGONAL-DOMINATE

M/M/1 QUEUE  = rate of arrival (# per unit time)  = rate of service (1/  = avg serve time)

FIRST PASSAGE TIME Want to know how soon X(t) gets to a special state: m i = E[min t: X(t) is “special”|X(0) = i]

LIMITING DISTRIBUTION Corollary of the General Key Renewal Theorem m i,i = time to leave state i and return

TRANSITION DIAGRAM    .....  is the rate of transitioning from n to n+1  is the rate of transitioning from n+1 to n   is the rate of transitioning out

STEADY STATE TRANSITION BALANCE EQN let... then... the rate of transition into state i equals the rate of transition out long run probability we arrive to see j people waiting

law of total probability the ol’ summing trick  <1 required for queue stability

QUEUE LENGTH L L = long run avg number of people in the queue

QUEUE LENGTH L L = long run avg number of people in the queue

LITTLE’S LAW Imagine that a customer pays $1/min. to stand in line Let (0, T] be a long time interval Let N(t) be the number of customers arriving in (0, T] Let $ = proceeds in (0, T]

$ 1 = T * Avg system earnings per min system earning per min = length of waiting line (L) $ 2 = N(T) * Avg customer waiting cost customer waiting cost = his waiting time (W) $ 1 = $ 2

LITTLE’S LAW TRUE FOR ALL QUEUES!

SUMMARY Analytical approaches to M/M/1 queues Similar results for M/G/1, M/M/s Traffic intensity  < 1 for stability Little’s Law (L= W) holds in general

EXTENSION arrive ~ depart ~ Service~   We can prove the output from a M/M/1 queue is a Poisson Process!

EXTENSION arrive ~ depart ~ Service~    100 44 22  666  123 11 arrive ~ We can prove the output from a M/M/1 queue is a Poisson Process! the JACKSON NETWORK probabilistic routing is PP filtering