guang-liang li and victor o.k. li the university of hong kong … · 2006. 6. 1. · guang-liang li...

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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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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

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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.

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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

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m

k

M1

mmθmλ*

mθ

*1θ

*kθ

*Mθ

Mλ

kλ

1λ

11θ

kkθ

MMθkmθ

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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.

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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.

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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.

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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.

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• 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).

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• 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.

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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)

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P{τ(2)<t} = P{τ(2)<t|τp = x}v(x)dx (3)P{τp = 0} = 1 implies v(x) = δ(x) (Dirac delta function), τ(2) = S, (3) becomes

P{τ(2)<t} = P{S<t|τp = x}δ(x)dx= P{S<t|τp = 0} = P{S<t}

If 0<P{τp = 0}<1P{τ(2)<t} = P{τ(2)<t|τp = 0}P{τp = 0}+ P{τ(2)<t|τp = x}v(x)dx= P{S<t}P{τp = 0}+P{τ(2)<t|τp>0}P{τp>0}

For any 0<t<∞,P{S<t}>P{τ(2)<t|τp>0}

So τ(2) cannot be exponential.

∞∫0

∞∫0

0+∞∫

τ(2) is not exponentially distributed. Formal proof is in Appendix of paper, “Network of queues: myth and reality,”by GL Li and VOK Li.

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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.

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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.

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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.

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• 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

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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.

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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

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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.

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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

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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.

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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.

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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

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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.

guang-liang li and victor o.k. li the university of hong kong … · 2006. 6. 1. · guang-liang li...

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