10 The MlMl1K 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
il
= 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
il
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 socalled
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 birthdeath
model developed in Section 4.10 we can compute the steadystate
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 queuelength
method available. This method, known as meanvalue
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 infiniteserver
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
“thinkserve”
. 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 serverbusy
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)] + q1
)
(
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 birthdeath
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 meanvalue 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 ErlangIc distribution.
= 1  pe(px)t
= 1  pe‘“(lP)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. Multiserver
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
lifetime
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 statetransition
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 statetransition
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. Multiserver
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 MVAbased
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 ModelBased Approach.
Boudewijn R. Haverkort
Copyright © 1998 John Wiley & Sons Ltd
ISBNs: 0471972282 (Hardback); 0470841923 (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 birthdeath
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 wellknown
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 lFCFS
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.
(5l)
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 PollaczekKhintchine
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 PKformula
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>
WP) *
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(14 
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 PKformula
or waiting
increases
we end this section
of the PKformula:
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