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2003 G.L. Li and V. O.K. Li, The University of Hong Kong 1 Networks of Queues: Myth and Reality Guang-Liang Li and Victor O.K. Li The University of Hong Kong { glli,vli}@eee.hku.hk

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  • 1

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 1

    Networks of Queues: Myth and Reality

    Guang-Liang Li and Victor O.K. Li

    The University of Hong Kong{glli,vli}@eee.hku.hk

  • 2

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 2

    Outline

    1. “How Networks of Queues Came About”2. Jackson Networks of Queues and Jackson’s

    Theorem3. Unsolved Mysteries4. Counterexample 1: M/M/1 Queue with

    Feedback5. Counterexample 2: Two M/M/1 queues in

    Tandem6. Possible Behavior of Networks of Queues7. Conclusion

  • 3

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 3

    1. “How Networks of Queues Came About”

    • 2002, J. Jackson, “How networks of queues came about,” Operations Research, vol. 50, no. 1, pp. 112-113.

    • 1957, J. Jackson, “Networks of waiting lines,”Operations Research, vol. 5, no. 4, pp. 518-521.

    • 1963, J. Jackson, “Jobshop-like queueing systems,”Management Science, vol. 10, no. 1, pp. 131-142.

    • After 1963, various generalizations and variations by others.

    In the 50th anniversary of Operations Research, Jackson reminisced on how he accidentally came up with the product form solution.

  • 4

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 4

    2. Jackson Networks of Queues and Jackson’s Theorem

    • Jackson Network of Queues• independent Poisson arrivals from outside• independent exponential service times, also

    independent of arrivals• first-come-first-served• once served at a queue, customer may either leave

    network, or go to the same or another queue in the network

    These are the basic assumptions of Jackson’s Network of Queues

  • 5

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 5

    m

    k

    M1

    mmθmλ*

    *1θ

    *kθ

    *Mθ

    11θ

    kkθ

    MMθkmθ

  • 6

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 6

    Jackson’s Theorem

    • Assumption: Network state (k1, k2, … , km) is a stationary Markov process

    • Theorem: In steady state, every queue in a Jackson network behaves as if it was an M/M/m queue in isolation, independent of all other queues in the network.

    Mkkkkkk MM

    PPPP ...21,...,, 2121 =

    We believe the assumption that (k1, k2, … km) is a stationary Markov Process is invalid.

  • 7

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 7

    3. Unsolved Mysteries• “product form solution”• tandem network

    – waiting times are dependent, cf., P.J. Burke, “The dependence of delays in tandem queues,” Ann. Math. Statist., vol. 35, no. 2, June, 1964, pp. 874-875.

    – but sojourn times are mutually independent, cf., E Reich, “Note on queues in tandem,” Ann. Math. Statist., 34 338-341, 1963.

    • M/M/1 with feedback behaves as if it was without feedback, but

    – with feedback: transition is impossible in small time interval if feedback occurs

    – without feedback: transition is always possible in any time interval

    These are some of the mysteries which make Jackson’s result doubtful.

    When a joint probability mass function (pmf) has a product form, i.e., it is equal to the product of marginal pmfs, it means that the underlying random variables, i.e., k1, k2, … km, are independent. But, based on the model described in slide 4, we know they are not.

    The tandem queue considered in Burke (1964) and Reich (1963) is the same. It is an M/M/1 queue feeding another queue with an exponentially distributed service time at the same rate, but independent of the service time, as the first queue.

    Sojourn time = service time + waiting time. How could the waiting times at the two queues be dependent, while the sojourn times are independent? This is a contradiction. We believe Reich is correct.

    Jackson claimed a feedback queue behaves as if it is a queue without feedback. We don’t believe it is possible.

  • 8

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 8

    4. Counterexample 1: M/M/1 Queue with Feedback

    Diagnosing Jackson’s Proof• m = 1, 2, … , M: labels of queues• nm: number of servers at queue m• µm: service rate at queue m• λm: arrival rate of customers at queue m from outside network• θkm: probability that customers go from queue m to queue k• = 1 - Σkθkm: probability that customers leave network from

    queue m• αi(k) = min{k, ni}, δi = min{k, 1}

    *mθ

    We are using the same notation as Jackson.

  • 9

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 9

    First Equation in Jackson’s Proof

    Pk1, … , kM(t+h) = {1-(Σλi)h – [Σαi(ki)µi]h}Pk1, … , kM(t)

    +Σαi(ki+1)µi hPk1, … , ki+1, … , kM(t)

    +ΣλiδihPk1, … , ki-1, … , kM(t)

    +ΣΣαj(kj+1)µjθijhPk1, … , kj+1, … , ki-1, … , kM(t)+o(h)

    *iθ

    This is the equation copied from Jackson’s paper.

  • 10

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 10

    • M=1, n1=1, k1=k, λ1=λ>0, µ1=µ>0, θ11=θ>0• For k>1

    Pk(t+h) = (1-λh-µh)Pk(t)+µ(1-θ)hPk+1(t)+λhPk-1(t) +o(h). (1)

    • X(t): number of customers waiting and being served in the single-server queue at time t.

    • Equation (1) is actuallyP{X(t+h) = k} = P{X(t+h) = k|X(t) = k}P{X(t) = k}+P{X(t+h) = k|X(t) = k+1}P{X(t) = k+1}+P{X(t+h) = k|X(t) = k-1}P{X(t) = k-1}+o(h) (2)

    We use Jackson’s equation to come up with the equation of a single server queue with feedback, i.e., Equ (1).

  • 11

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 11

    • Compare (1) with (2)P{X(t+h) = k|X(t) = k} = 1-λh-µh

    • Replace k by k-1 in (1) and (2), and compare the obtained equations.

    P{X(t+h) = k-1|X(t) = k} = µ(1-θ)h• Replace k by k+1 in (1) and (2), and compare the obtained

    equations.P{X(t+h) = k+1|X(t) = k} = λh

    • Contradiction: the sum of the above probabilities is not equal to one.

    • X(t) is not a Markov process. Jackson’s theorem does not hold.

  • 12

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 12

    Without Diagnosing Jackson’s Proof• τ(2): time spent by X(t) in state k=2 until a transition to k=1.• τ(2) is not exponential (to be shown) implies X(t) is not a

    Markov process.• actual service time of a customer: initial service time plus any

    extra service time due to feedback• τ(2) = τd+τp• τd>0: the (residual) actual service time of the departing

    customer• τp>0: part of the actual service time of the other customer• v(x): probability density function of τp• S: the (residual) exponential service time first expired in τ(2)

  • 13

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 13

    P{τ(2)

  • 14

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 14

    5. Counterexample 2: Two M/M/1 queues in Tandem

    • All customers arrive at the first queue, go to the second queue after service, and leave the network form there.

    • Jackson’s theorem in this case: corollary of Burke’s theorem:

    The output of the first queue is a Poisson Process at the same rate as that of the arrival process.

    The second queue is also an M/M/1 system.

  • 15

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 15

    Outline of Our Argument

    1. The output of the first queue has both a marginal version, and a non-marginal version (shall be demonstrated).

    2. The non-marginal version is neither a Poisson process nor a stationary process.

    3. If the two queues are considered jointly as a network, the arrival process at the second queue is the non-marginal, non-stationary version.

    4. The second queue is not an M/M/1 queue and is unstable.5. The state of this network is not stationary.6. So Jackson’s theorem does not hold.

    If the input of a queue is not stationary, the queue cannot be stable. We show that the output of an M/M/1 queue has a marginal, stationa ry version (Burke’s Thm), and a non-marginal, non-stationary version. Actually, the physical, observable output process is the non-marginal version, and the marginal version is obtained by averaging. (See next few slides.) If you look at each queue in isolation, we can use the marginal version. But if you look at the queues together, as in Jackson’s Thm, in which he looks at (k1, k2) together, you must deal with the non-marginal version. Since the non-marginal version is non-stationary, the second queue is not stable, and it does not make sense to talk about steady state any more.

  • 16

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 16

    Output of M/M/1 Queue

    Simulation (thought experiment) of stable M/M/1 queue in steady state

    • Inter-departure time (t-s) is sampled in either case below

    • Case (a): server is busy at time s- (t-s) is distributed as a service time- color a line segment of length (t-s) red- use “R” to represent the segment

    This thought experiment shows that the physical output of an M/M/1 queue is non-stationary.

  • 17

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 17

    • case (b): server is idle at time s

    (t-s) is distributed as the sum of an idle time of the server and a service time

    color a segment of length (t-s) blue

    use “B” to represent the segment

    • sample path of the inter-departure time sequence corresponds to a sequence of colored segments

  • 18

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 18

    Observation and Fact

    • sequence of colored segmentsRRRBRRRRBBBRRRBBBBBRR… .

    • segments of two colors: inter-departure times follow two different distributions

    • tendency for segments with the same color to aggregate: Markov dependence

    The output process is not independent. It also consists of two distinct distributions. It is NOT iid.

  • 19

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 19

    Non-Marginal Version of the Output

    • the inter-departure time sequence: not i.i.d., not stationary

    • the corresponding departure process: not Poisson process, not stationary

  • 20

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 20

    Marginal Version of the Output

    • obtained by averaging out the impact of the state of the queue

    • Experimental construction

    divide interval (0, H) into N (H) consecutive, disjoint subintervals of equal length

    for all segments with length less than H, calculate the frequencies that the lengths of the segments are in the small intervals, regardless of their colors.

    As H à ∞, an exponential pdf with parameter equal to the arrival rate is found

    The marginal version is obtained by averaging the state of the queue (or server). If you use the marginal version, you have to forget completely the state of the queue. You cannot say, “I know the server is busy, and the output of the queue is Poisson with rate equal to the arrival rate” because it is not correct. This is why Burke’s 1964 paper (referred to in slide 7) is incorrect. He considers the two queues in tandem together, and yet he uses the marginal version of the output of the first queue as input to the second queue.

  • 21

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 21

    Marginal Version of the output (cont’d.)

    • experimental construction (continued)

    sample random variables independently, regardless of the state of the queue, from the constructed pdf

    sampled random variables form an i.i.d. exponential sequence

    marginal version: the Poisson process corresponding to the exponential sequence

  • 22

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 22

    What Does Burke’s Theorem Really MeanSeparate queues in tandem based on the marginal version so

    as to treat them individually rather than jointly.“It is intuitively clear that, in tandem queuing processes of the

    type mentioned above, if the output distribution of each stage was of such character that the queuing system formed by the second stage was amenable to analysis, then the tandem queue could be analyzed stage-by-stage insofar as the separate delay and queue-length distributions are concerned. Such a stage-by-stage analysis can be expected to be considerably simpler than the simultaneous analysis heretofore necessary. Fortunately, under the conditions stated below, it is true that the output has the required simplicity for treating each stage individually.”

    P.J. Burke, “The output of a queueing system,” Operations Research, vol. 4, pp. 699-714, 1956.

    Actually Burke’s original paper advocates looking at each queue in tandem in isolation, rather than together. It seems he has forgotten about this in his 1964 paper.

  • 23

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 23

    What if two queues in tandem are considered jointly?

    • the output of the first queue is the non-marginal, non-stationary version

    • the second queue is not M/M/1, and is not stable• The state of the network is not a stationary process• Jackson’s theorem does not hold

    If you consider the two queues together, you must use the non-marginal version.

  • 24

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 24

    6. Possible Behavior of Networks of Queues

    • Jackson network without loops:

    not stationary if queues considered jointly

    after separation based on the marginal version, queues standing alone can be stable

    • Jackson network with loops: not stationary

    • network with renewal-type external arrivals and generally distributed service times: not stationary in general

    • tandem network with renewal-type external arrivals and generally distributed service times: can be isolated and isolated queues can be stable

  • 25

    2003 G.L. Li and V. O.K. Li, The University of Hong Kong 25

    7. Conclusion• Jackson’s theorem does not hold, as shown by the

    counterexamples.• The assumption (i.e., network state is a stationary

    Markov process) made by Jackson is invalid.• All known “proofs” are based on this invalid

    assumption.• Jackson network is not stationary, unless queues can

    be isolated.• Generalizations and variations of Jackson networks

    are questionable.• Re-investigation of related issues is necessary.