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10 The MlMl1||K q ueue or the terminal model
or the terminal
Figure 4.9: State transition
diagram for the MJM/lJJK
Figure 4.10: Simple terminal
Notice that this queueing station
Again, we can use (4.8) to obtain the
= p0 n p(K - k) = popi n (K - k),
Using the normalisation
When all the jobs are in the
so the queue will not grow infinitely
can never be overloaded.
no new jobs will arrive,
Figure 4.9 the state transition
following expression for pi:
i = 0, 1,. . . , K.
equation we find that
4.10 we sketch this so-called
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
in the arrival
of that customer
in the model. We will discuss queueing network
at one of the other queueing
models at length in Chapter
4 MIMI 1 queueing
Using the birth-death
model developed in Section 4.10 we can compute the steady-state
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;
- there is a simpler
analysis, can be applied in many
we will come back to it in Chapter
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
queueing station, every job has its own server, so
no queueing or waiting occurs. Therefore: E[&] = E[Z], independently
of the number of
time depends on the number
For the processing
the average response
we first have to
introduce the average cycle time as E[C(K)]
= E[&] + E[R,(K)]
= E[Z] + E[R,(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
. The throughput
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
= E[Z] + E[R,(K)]’
from which we derive the response time law:
E[WWl = &j
law for the system, we have E[N,(K)]
for the terminals,
we have E[N,(K)]
two instances of Little’s
law, we can eliminate
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.
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].
for the terminal
We can evaluate this throughput
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).
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
For K = 1, the server-busy
(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
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:
K* = E’s~s;r,l.
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.
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.
A customer in a closed queueing network
queue in equilibrium
is one customer
(with average filling),
at a queue, will
however, for the case in which there
less in the queueing network.
4 MINI1 1 queueing
to this theorem,
the average response time at the system can be expressed as
The first term represents
the average waiting
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
E[&(K - l)] (K - 1) E[S]+ E[S].
-q&(K)] = E[&(K
- l)] + q-1
To begin this recursion,
we use E[R,(l)]
4.5. An MVA of the terminal
a terminal model as described throughout
E[Z] = 10. The saturation
this section in which
point can easily be calculated
E[S] = 2 and
given as E[R, (Ii = l)] = E[S] = 2 and
The asymptotes of E[R,(K)]
+ oo)] = KE[S] - E[Z] = 2K - 10. Using the MVA recursion we can now
for K = 1, a.. ,12. Notice that we can also express p(K) and X(K)
in terms of E[R,(K)]
+ E[Z] ’
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
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
increase the throughput
for the terminal
Figure 4.11: E[R,(K)]
an d i t s t wo lower bounds as a function
Figure 4.12: X(K)
and its two upper bounds as a function
4 MIMI 1 queueing
Table 4.2: E[R,(K)],
for K = 1,. .. ,12
processes and their
models can be found in many performance
to simple queueing
books. Early work related to birth-
death processes and blocking probabilities
is due to Erlang . 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  and Van Dijk . The PASTA property has been
described by Wolff . 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.
Compute the minimum number of jobs Ic such that for an M]M(l
queue with p = 0.8 the
value B(3c) < lo-“.
The waiting time distribution
in the M]M]l
response time distribution.
and that with probability
queue can be derived in a similar
First notice that with
pl, (k = 1,2, . * a) the waiting
5 t} = p. + 2 pl,&,(t)
way as the
time has an Erlang-Ic distribution.
= 1 - pe-(p-x)t
= 1 - pe-‘“(l-P)t.
in the MIMI1
Show that the variance of the number of customers
For the multiserver
1. the solutions
for pi do indeed fulfill
can indeed be computed
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
in an M ]M (1 queue indeed corresponds
the global balance equations,
K = 2.
(4.36) holds for K = 2.
Consider an MIMI 1 queue in which the server has an exponentially
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
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
4. Derive a formula
(b e inspired
5. Now assume that arrivals
by the normal M ]MI 1 queue).
to occur, even if the server has broken
How does this change the state-transition
diagram and the global balance equations?
What is the stability condition in this situation?
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
Calculate B(m, X/p) f or m = 5, 10, 15, 25 and 100 and X/,T.L= 0.1,. . . ,0.9.
Prove (4.26) and (4.27) using (4.24) and (4.25).
and the terminal
model with K customers,
3. Compute the asymptotes
E[Z] = 15 and E[S] = 3.
for the system response time E[R,(K)]
for the system throughput
the exact values of X(K)
5. Propose an MVA-based
scheme to compute pa(K)
and draw them in
and draw them in a graph.
for K = 1,. -. ,12, using MVA.
Do not directly
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)
MI G (LFCFS
N the previous chapter we have discussed a number of Markovian
shown various applications
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
model. The analysis of these models is more complicated than that of the
models encountered in Chapter 4.
In this chapter we focus on the MlGll
are using a scarce resource,
periods of time.
This chapter is organised
q ueueing model. This model is rather generally
users (a large population
such as a transmission
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.
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
of the MlGll
model in which batches of jobs arrive simultaneously.
Section 5.5, we discuss MI G 11 queueing models with server breakdowns.
a single server queueing station with unlimited
rate X. The service requirement
at the queueing station
of a job is a random
Poisson process with
5 M/G1 l-FCFS
to the distribution
S has expectation
moment) and second moment E[S2].
Again we use the notation E[N] for the average number of jobs in the queueing system,
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.
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.
postponed to later sections.
For the average number of jobs in the queue of an M ]G ] 1 queueing station the following
has been derived:
L\l - P)
20 - P>
These two equations only address the queueing part of the overall queueing station.
including the service, we arrive at the following expressions:
XE[S] + ~
E[S] + ~
20 - P)’
result is presented mostly
in one of the four forms above. The form (5.3) is
often referred to as the Pollaczek-Khintchine
in more detail now.
(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
a higher average number
of jobs in
From a queueing point of view, exhibiting no variance in the service times
(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.
two almost equivalent queueing stations.
In the first one the service requirement
5.1 The MIGII
is 1, deterministically,
0.5 each), i.e., E[Sz]
in the first
= 0. In the second
S has a probability
0.5 or 1.5
= 1.25, so that
are the same!
rate X = 0.4.
s 2 1,
= 2 (seconds).
= Srn 2s-‘ds
fs (s) = 2/s3, whenever
= Lrn 2~-~ds = (-2s-‘)lIT
of the service
= 0.8 < 1 the queueing
as a Poisson
s < 1. Calculating
since p = XE[S]
= 1 and E[Sg]
E[S] = Am sfs(s)ds
= 0, whenever
= 1 so that
but the actual
in the second system,
= 1 and E[SF]
is also 1 on average,
= (2 Ins)::;”
of the PK formula
= 00, even though
the queue is not
It is important
by the first two
not the total
JwS12(1+ G> = dwl(1+ c;>
Cg. For E[W]
to its (squared)
ww1)20 + G> = p + P2(1+ G>
w - P>
of the service
of a stochastic
can be rewritten
but only the averages.
we use the squared
of a random
we end this section
of the PK-formula:
E[N] = XE[S] +
of the service
in our formulae.
a few remarks