2003 g.l. li and v. o.k. li, the university of hong kong 1 networks of queues: myth and reality...

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,[email protected]


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


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.


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


m

k

M1

mmm*

m

*1

*k

*M

M

k

1

11

kk

MMkm


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


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


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

m: service rate at queue m

m: arrival rate of customers at queue m from outside network

km: probability that customers go from queue m to queue k

• = 1 - kkm: probability that customers leave network from queue m

• i(k) = mink, ni, i = mink, 1

*m


First Equation in Jackson’s Proof

Pk1, …, kM(t+h) = 1-(i)h – [i(ki)i]hPk1, …, kM(t)

+i(ki+1)i hPk1, …, ki+1, …, kM(t)

+iihPk1, …, ki-1, …, kM(t)

+j(kj+1)jijhPk1, …, kj+1, …, ki-1, …, kM

(t)+o(h)

*i


• M=1, n1=1, k1=k, 1=>0, 1=>0, 11=>0

• For k>1

Pk(t+h) = (1-h-h)Pk(t)+(1-)hPk+1(t)+hPk-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 actually

P{X(t+h) = k} = P{X(t+h) = k|X(t) = k}P{X(t) = k}

+P{X(t+h) = kX(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)


• Compare (1) with (2)

PX(t+h) = kX(t) = k = 1-h-h

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

PX(t+h) = k-1X(t) = k = (1-)h

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

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

• 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.


Without Diagnosing Jackson’s Proof (2): time spent by X(t) in state k=2 until a transition to k=1. (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

(2) = d+p

d>0: the (residual) actual service time of the departing customer

p>0: part of the actual service time of the other customer

• v(x): probability density function of p

• S: the (residual) exponential service time first expired in (2)


P{(2)<t = P(2)<tp = x}v(x)dx (3)P{p = 0 = 1 implies v(x) = (x) (Dirac delta function), (2) = S, (3) becomes

P(2)<t} = P{S<t|p = x(x)dx= P{S<tp = 0} = PS<t}

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

For any 0<t<,P{S<t}>P{(2)<t|p>0}

So (2) cannot be exponential.

0

0

0+


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.


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.


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


• 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


Observation and Fact

• sequence of colored segments

RRRBRRRRBBBRRRBBBBBRR….• segments of two colors: inter-departure times

follow two different distributions• tendency for segments with the same color to

aggregate: Markov dependence


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


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 , an exponential pdf with parameter equal to the arrival rate is found


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


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.


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


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


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.

2003 g.l. li and v. o.k. li, the university of hong kong 1 networks of queues: myth and reality...

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