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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

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

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

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

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

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)

the applica-

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