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10 The MlMl1||K q ueue or the terminal model

10 The MlMl1||K q ueue or the terminal model

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4.10 The



MIM(l(IK



queue



or the terminal



Figure 4.9: State transition



\



model



diagram for the MJM/lJJK



system



Figure 4.10: Simple terminal

Notice that this queueing station



diagram



=



po



ig



XW



k=O



-



w



is depicted.



large.



In



Again, we can use (4.8) to obtain the



i-l



= p0 n p(K - k) = popi n (K - k),



P



Using the normalisation



When all the jobs are in the



so the queue will not grow infinitely



i-l

pi



model



can never be overloaded.



no new jobs will arrive,



Figure 4.9 the state transition

following expression for pi:



model



K users



computer



queueing station,



87



k=O



i = 0, 1,. . . , K.



(4.51)



k=O



equation we find that

(4.52)



In Figure



4.10 we sketch this so-called



terminal



model.



It is an example of a closed



queueing network with a population of K customers circling between the terminals and the

processing system; note the special representation

we use to signify infinite server nodes.

One can think of closed queueing models as models in which a departing customer at one

station will result immediately

stations

through



in the arrival



of that customer



in the model. We will discuss queueing network

13.



at one of the other queueing



models at length in Chapter



10



88



4.11



4 MIMI 1 queueing



Mean



values



for the



terminal



models



model



Using the birth-death

model developed in Section 4.10 we can compute the steady-state

probabilities

p- and from those we can compute average performance measures such as

E[N] in a similar way to that done before. However, if we are only interested in average

performance values and not in the precise queue-length

method available. This method, known as mean-value

cases for more general queueing networks;



distribution

p,

- there is a simpler

analysis, can be applied in many



we will come back to it in Chapter



11 through



13. Here we develop the method for this special case only.

Let us first address the average response times at the terminals (E[&])

and at the

I n an infinite-server

system (E[R,(K)]).

queueing station, every job has its own server, so

no queueing or waiting occurs. Therefore: E[&] = E[Z], independently

of the number of

customers



K actually



present.



time depends on the number



For the processing

of customers



K.



system, however,

To compute



the average response



E[R,(K)],



we first have to



introduce the average cycle time as E[C(K)]

= E[&] + E[R,(K)]

= E[Z] + E[R,(K)].

E[C(K)]

expresses the mean time it takes for a customer to go once through the cycle

( a g ain note the dependence on K) can now be

“think-serve”

. The throughput

X(K)

expressed as K/E[C(K)],



th a t is, as the product of the frequency

is not equal to l/E[Z]).



and the number of jobs (notice that X(K)



with which jobs cycle

Combining the above



two results, we obtain

X(K)



K

= E[C(K)]



K

= E[Z] + E[R,(K)]’



(4.53)



from which we derive the response time law:



E[WWl = &j

Using Little’s



- E[Z].



law for the system, we have E[N,(K)]



for the terminals,



we have E[N,(K)]



two instances of Little’s

E[N,(K)]



= X(K)E[&].



law, we can eliminate

= f$z::#;K,



X(K),



(4.54)



= X(K)E[R,(K)].

Using X(K)



Using Little’s

= K/E[C(K)]



law



in the



as follows:



and E[N,(K)]



=



E’Z1



Jw(K)I



K.



(4.55)



As can be observed, the K customers spread themselves over the two nodes in the model

with ratios proportional

to the average time spend at the nodes divided by the time spend

on an average cycle.

Although the above equations give some insight,



they still do not yield us answers.



For



that purpose we need to know the throughput X(K) which equals the product of the serverbusy probability

and the service rate of the system: X(K)

= (1 -po)p = (1 -po)/E[S].



4.11 Mean



values



for the terminal



We can evaluate this throughput



model



89



by using the result of Section 4.10, thereby again noting



that if we change K, po will change as well (we therefore write po as a function

In summary, we have



of K below).



- E[Z].



(4.56)



Instead of computing PO(K) explicitly, let us first address two asymptotic results. For large

values of K, the idle fraction is very small so that the denominator 1 - po( K) will approach

1. For large K we therefore



have E[R,(K)]



E KE[S]



- E[Z].



For K = 1, the server-busy



probability

(p( 1)) th e utilisation

in case of 1 job) simply equals E[S]/(E[S]

+ E[Z]) (the

average time the job spends in the server divided by the time for an average cycle). This

is due to the fact that E[R,(l)]

= E[S] since queueing will not occur.

The above two limiting cases can be regarded as asymptotes for the actual curve of

E[&(K)1.

Th eir crossing point, that is, the value for K* such that E[S] = K*E[S]

is called the saturation point and computed as follows:

E[R,(l)]



= E[R,(oo)]



a



- E[Z],



K* = E’s~s;r,l.



(4.57)



Notice that K* = l/p(l).

Wh en the think and service times would have been constants

rather than random variables, K* would have been the maximum number of users that

could have been served before any queueing would



occur in the system.



Having



more



than K* customers present would for sure imply that at some place in the model queueing

would occur. Since the involved service times are not constants but random variables the

actual values for the response time at the system (E[R,(K)])



are of course larger than the



asymptotes and queueing will already occur for values of K smaller than K*. In Figure 4.11

E [R, (K)] is depicted as a function of K for the parameter values of the example to be

discussed below. The asymptotes are also indicated.

For the precise calculation

of E[R, (K)] we still



need to know



the value of po( K).



Although we can compute this value by using the summation as in Section 4.10 there

recursively from E [R, (K - l)] . To

is a smarter way to go. We can calculate E[R,(K)]

understand this, we need a result, which we will discuss in more detail in Chapters 11 and

12, known as the arrival



theorem which was proven in the late 1970s.



Theorem

4.2. Arrival

theorem.

A customer in a closed queueing network

queue in equilibrium

is one customer



(with average filling),



arriving



at a queue, will



see this



however, for the case in which there



less in the queueing network.



cl



90

According



4 MINI1 1 queueing

to this theorem,



models



the average response time at the system can be expressed as



follows:

E[R,(K)]

The first term represents

service)

arriving.



upon arrival,



= E[N,(K



the average waiting



- l)]E[S]



(4.58)



+ E[S].



time because of jobs already queued (or in



whereas the second term is the average service time of the job just



Now, using (4.55) we can write

(4.59)



so that



E[&(K - l)] (K - 1) E[S]+ E[S].

-q&(K)] = E[&(K

- l)] + q-1

)

(



To begin this recursion,

Example

Consider



we use E[R,(l)]



= E[S].



4.5. An MVA of the terminal

model.

a terminal model as described throughout



E[Z] = 10. The saturation



(4.60)



this section in which



point can easily be calculated



K*=EISl$-EIZlEM



as



6



2+10

2



E[S] = 2 and



(4.61)



*



are respectively

given as E[R, (Ii = l)] = E[S] = 2 and

The asymptotes of E[R,(K)]

E[R,(K

+ oo)] = KE[S] - E[Z] = 2K - 10. Using the MVA recursion we can now

compute E[R,(K)],

for K = 1, a.. ,12. Notice that we can also express p(K) and X(K)

directly



in terms of E[R,(K)]

dK)



as follows:



KWI



= E[R,(K)]



K

+ E[Z]



and X(K)



=



E[Rs(K)]



+ E[Z] ’



(4.62)



In Table 4.2 we present the values for E[R,(K)],

p(K) and X(K).

In Figure 4.11 we show

the average response time curve and its asymptotes (lower bounds).

In Figure 4.12 we

show the throughput



curve, again with its asymptotes



(upper bounds).



Note that in both



cases the asymptotes are very easy to compute and that their crossing points lie in both

cases at K*. It is clearly visible that adding more customers (adding more terminal users)

does increase the response times; however, it does not significantly

after a particular

point.



increase the throughput

0



4.11 Mean



values



for the terminal



16



model



91



I



I



I



I



I



2



4



6



8



10



14

12

10



WdK)18

6

4

2

0



Figure 4.11: E[R,(K)]



an d i t s t wo lower bounds as a function



12



of K



0.6

0.5

0.4

X(K)

0.3



Figure 4.12: X(K)



and its two upper bounds as a function



of K



92



4 MIMI 1 queueing



K E[WWl



P(K) X(K)



K wL(K)I



P(K) X(K)



1

2



2.00



0.167



0.084



7



5.92



0.880



0.440



2.33

2.76



0.324

0.470



0.162

0.235



8

9



7.21



3



0.930

0.963



0.465

0.482



4



3.30



0.602



0.301



10



0.982



0.492



5



3.98

4.85



0.715

0.808



0.358

0.404



11

12



0.992

0.997



0.496

0.497



6



Table 4.2: E[R,(K)],



4.12

Further



Further

information



p(K)



8.70

10.37

12.18

14.08



and X(E()



models



for K = 1,. .. ,12



reading

on birth-death



processes and their



models can be found in many performance



evaluation



application



to simple queueing



books. Early work related to birth-



death processes and blocking probabilities

is due to Erlang [84]. Background information

on the existence of local balance equations and their relation to the global balance equations

can be found in the books by Kelly [153] and Van Dijk [74]. The PASTA property has been

described by Wolff [291]. Th e arrival theorem and the associated mean-value analysis will

be discussed in more detail in Chapters 11 through 12; seminal papers in this area have

been published by Reiser and Lavenberg [245, 2431.



4.13



Exercises



4.1. Calculation

of B(k).

Compute the minimum number of jobs Ic such that for an M]M(l



queue with p = 0.8 the



value B(3c) < lo-“.

4.2. Waiting

time distribution.

The waiting time distribution

in the M]M]l

response time distribution.

and that with probability

Show that

Fw(t)



= Pr{W



queue can be derived in a similar



First notice that with



probability



pl, (k = 1,2, . * a) the waiting



5 t} = p. + 2 pl,&,(t)

k=l



p.



the



waiting



time



way as the

equals



time has an Erlang-Ic distribution.



= 1 - pe-(p-x)t



= 1 - pe-‘“(l-P)t.



0



4.13 Exercises

4.3. Variance



93

in the MIMI1



queue.



Show that the variance of the number of customers

to (4.18).

4.4. Multi-server



queueing



For the multiserver



M]M(m



1. the solutions



stations.

queue introduced



for pi do indeed fulfill



can indeed be computed



4.5. Comparing



average



Show that inequality

4.6. The MIMI1



in Section 4.5 show that:



for pi reduce to those for the MI M ] 1 queue when m = 1,



2. the expressions

3. E[N]



in an M ]M (1 queue indeed corresponds



the global balance equations,



using (4.27).



response



times



when



K = 2.



(4.36) holds for K = 2.

queue



with



server



breakdowns.



Consider an MIMI 1 queue in which the server has an exponentially

distributed

life-time

with mean l/Z. Once failed, the server is repaired; such a repair takes an exponentially

distributed amount of time, with mean l/r. Customer arrivals form a Poisson process with

rate X and services last a negative exponentially

distributed

time with mean l/p. For the

time being, assume that the arrival

1. Draw the state-transition



process is stopped as soon as the server breaks down.



diagram for this extended MIMI1 queue.



2. Derive the global balance equations for this CTMC.

3. Derive a formula



for E[N]



4. Derive a formula



for 0%.



(b e inspired



5. Now assume that arrivals



continue



by the normal M ]MI 1 queue).



to occur, even if the server has broken



down.



How does this change the state-transition

diagram and the global balance equations?

What is the stability condition in this situation?



4.7. Capacity



and arrival



rate



increase.



How does the average response time in an MIMI1 queue change, when both the arrival rate

X and the service rate p are increased by a multiplicative

factor Q, i.e., when p remains the

same?



4 M/MI



94



1 queueing



models



4.8. Erlang’s

B formula.

Calculate B(m, X/p) f or m = 5, 10, 15, 25 and 100 and X/,T.L= 0.1,. . . ,0.9.

4.9. Multi-server



queues.



Prove (4.26) and (4.27) using (4.24) and (4.25).

4.10.



MVA



Consider



and the terminal



a terminal



model with K customers,



1. Compute



the saturation



2. Compute

a graph.



the asymptotes



3. Compute the asymptotes

4. Compute



via (4.52)!



E[Z] = 15 and E[S] = 3.



point K*.

for the system response time E[R,(K)]



for the system throughput



the exact values of X(K)



5. Propose an MVA-based

pute pa(K)



model.



and E[R,(K)],



scheme to compute pa(K)



X(K)



and draw them in



and draw them in a graph.



for K = 1,. -. ,12, using MVA.

from p(K).



Do not directly



com-



Performance of Computer Communication Systems: A Model-Based Approach.

Boudewijn R. Haverkort

Copyright © 1998 John Wiley & Sons Ltd

ISBNs: 0-471-97228-2 (Hardback); 0-470-84192-3 (Electronic)



Chapter



5



MI G (LFCFS



I



queueing



models



N the previous chapter we have discussed a number of Markovian

shown various applications



of them.



In practice,



queueing models and



however, there are systems for which



the negative exponential service times that were assumed in these models are not realistic.

There exist, however, also single server models that require less strict assumptions regarding

model, the GIG11 model and

the used service time distributions.

Examples are the MjGll

the GiPHll

model. The analysis of these models is more complicated than that of the

simple birth-death

models encountered in Chapter 4.

In this chapter we focus on the MlGll

applicable



in environments



where multiple



are using a scarce resource,

distributed

periods of time.

This chapter is organised



q ueueing model. This model is rather generally

users (a large population



such as a transmission

as follows.



of potential



customers)



line or a central server, for generally



In Section 5.1 we present the well-known



for various mean performance measures of interest for the MlGll

attention to the impact of the general service time distribution.



results



queue. We pay special

The MIGIl result can



be proven in an intuitive fashion; we do so in Section 5.2. A rigorous proof based on a

embedded Markov chain is then presented in Section 5.3. In Section 5.4 we discuss an

extension



of the MlGll



model in which batches of jobs arrive simultaneously.



Finally,



in



Section 5.5, we discuss MI G 11 queueing models with server breakdowns.



5.1



The



MIGIl



result



Consider



a single server queueing station with unlimited



customer



population.



Jobs arriving



rate X. The service requirement



buffering



at the queueing station

of a job is a random



capacity



form.a



variable



and unlimited



Poisson process with



S, distributed



according



5 M/G1 l-FCFS



96

to the distribution



function



B(s),



i.e., B(s)



= Pr{S



5 s}.



queueing



S has expectation



models

E[S] (first



moment) and second moment E[S2].

Again we use the notation E[N] for the average number of jobs in the queueing system,

E[N,]



for the average number of customers in the queue, and E[N,] for the average number



of customers in the server. Applying Little’s

XE[S] and we assume p < 1 for stability.

complicated.



law for the server alone we have: p = E[N,] =

The derivation of E[N,] is somewhat more



At this stage we will only present and discuss the result.



Proofs will



be



postponed to later sections.

For the average number of jobs in the queue of an M ]G ] 1 queueing station the following

expression



has been derived:

E[N,]



= fl.



(5-l)



L\l - P)



Applying



Little’s



law (E[W]



= E[N,I/X),



we obtain



E[W]



= ?!!!!%



(5.2)



20 - P>



These two equations only address the queueing part of the overall queueing station.

including the service, we arrive at the following expressions:

E[N]



=



X2E[S2]

XE[S] + ~



=



XE[S2]

E[S] + ~



By



20 - P)’



E[R]

The M]G]l



result is presented mostly



w



in one of the four forms above. The form (5.3) is



often referred to as the Pollaczek-Khintchine

in more detail now.

Looking



- P>’



(or PK-) formula.



Let us discuss this equation



we observe that E[N] depends on the first and second

What does this imply?

From the first two

moment of the service time distribution.

moments of a distribution

its variance can be obtained as ai = E[(S - E[S])2] = E[S2] E[S12.



at the PK-formula



We thus see that a higher variance



implies



a higher average number



of jobs in



the system.



From a queueing point of view, exhibiting no variance in the service times

exists

(E[S2] = E[S12) 1s

* o pt imal. This is a very general observation: the more variability

With worse performance we of course mean

in the system, the worse the performance.

longer queues, longer waiting times etc.

Example

Consider



5.1. Influence

of variance.

two almost equivalent queueing stations.



In the first one the service requirement



5.1 The MIGII



result



97



is 1, deterministically,

the service

(with



i.e., E[Si]



requirement



probability



var[S2]



= 0.25.



higher



than



Example



0.5 each), i.e., E[Sz]

Consequently,



in the first



a queueing



The service

and fs(s)



station



requirement



= 0. In the second

are either



= 0.5(0.5)2+0.5(1.5)2

the average



the average



at which



jobs arrive



S has a probability



work



0.5 or 1.5



= 1.25, so that



waiting



times



requirements



,qs2]



will



be 25%



are the same!



process



time



service



•I



rate X = 0.4.



time,



s 2 1,



we obtain



= 2 (seconds).



station



is stable.



distribution,



= Srn 2s-‘ds



s2fs(s)ds



with



fs (s) = 2/s3, whenever



= Lrn 2~-~ds = (-2s-‘)lIT



of the service



= irn



function



the average



= 0.8 < 1 the queueing



moment



as a Poisson



density



s < 1. Calculating



since p = XE[S]

the second



= 1 and E[Sg]



although



E[S] = Am sfs(s)ds



lating



var[Si]

values



variance.



= 0, whenever



Clearly,



= 1 so that



but the actual



in the second system,



system,



5.2. Infinite



Consider



= 1 and E[SF]



is also 1 on average,



However,



when



calcu-



we obtain



= (2 Ins)::;”



= ~0.



1



Application



of the PK formula



thus reveals



that



E[N]



= 00, even though



the queue is not

q



overloaded!

It is important

is given

The



by the first two



effect



distribution



that



c;



= a;/E[X12,

Using



not the total



that



moments



The



becomes

squared



expresses



this notation



becomes



Before

bility



that



E[N]



distribution,



when



coefficient



of variation



the variance



variable



relative



JwS12(1+ G> = dwl(1+ c;>

2(1- P>

W-P) *

two



with



examples,



Cg. For E[W]



we make



X,



that



is,



to its (squared)



ww1)20 + G> = p + P2(1+ G>

w - P>

a(1-4 -



E[W]



with



variable



time



of variation



as:



linearly



=



of the service

coefficient



of a stochastic



can be rewritten



time behaviour



but only the averages.



we use the squared



of a random



the PK-formula



or waiting



increases



we end this section

of the PK-formula:



worse

when



the variance



q ueueing



time



clear nicely



E[N] = XE[S] +

We observe



MlGll



of the service



the performance



increases



in our formulae.

mean.



to note



we obtain



a few remarks



a similar



(5.5)

equation:



(5.6)

about



the applica-



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