INSTITUTE OF SfATl$TfCS
BOX 5457 .
STATE COU_EGF STATION
RALEIGH, NORTH CAROUNj~
UNIVERSITY OF NORTH CAROLINA
Department of
Statist~cs
Chapel Hil1 1 N. C.
ON :NNWORKS OF QUEUES
by
Frank Fish
August 1962
This research 1as supported by the Office of Naval
Research under contract No. Nonr-855(09) for research
in probability and statistics at the University of
North Carolina, Chapel Hill, N. C. Reproduction in
whole or in part is permitted for any purpose of the
United states Government •
.
~:
•.
Institute of Statistics
Mimeo Series No. 336
ii
...
ACKNOWLEDGEMENTS
•
I wish to e::pres.J rtlJ heartfelt thari!::; to Professor il. L.
Smith, "Tho suggested this topic, encouraged lile throughout,
meticulously read
t~e L~nuscript
and suggested improvements.
His aid was invalvuable.
I must also record :".1:y thanks to the Office of Naval Research
for financial assistance and to the "Fulbright" authorities who
made this year in the United states possible.
Finally, I am most grateful to Mrs. Doris Gardner for her care-
"
e
•
:':'111 typing of the manuscript and to Miss Martha JorCl.a.n for her aid.
in all things regula.tory •
/if
_All.
iii
T.l\BLE OF CONTENTS
Page
CHAPTER
----
~~o~amms
- - ii
- - ·v
INTRODUCTION
I
GENERAL QUb-,;EnIG THEORY
1.1 Introduction
1
.. - - -
- - - - -
1
1.2 HistOl7 - - - - -
1.3 Specification of a queue
1
- - - -
1.4 A single server queue - - - - - - -
2
3
1.5 Proof of the existence of a limiting
distribution of waiting-times - - - - -
4
1.6 Uniqueness of the limiting distribution
function of the waiting times - -
II
- - - - 9
1.7 The waiting time
- 11
GENERAL QUEtJEllTG SYSTEMS
- 16
2.1 Introduction - - - - - - - -
16
2.2 The output of a single multi-server queue - - 16
2.~
The independence of the interde]?arture
interval and the state of the system at
the depaTtuTe - - - - - - - -
- 20
2.4 The independence of inter-departure
intervals - - - - - - - - - - - - -
- 21
iv
Page
CHAPTER
2.5 A single servel' queue with l'andom arrivals a.nd
22
general service time - - - - - - - - - - -
2.6 The
III
TV
time at second multi-server queue
28
32
'.1
32
Introduction - - - -
3.2 A series of queues - - - - - -
32
3.3. The ergodicity of queues in series - - - - -
32
3.4 The steact" state solution
42
3.5 A network of queues
50
THE EFFEC'r ·OF LIHITING THE SIZE OF THE WAITING
54
- - - - - - - -
4.1 Introduction - - - - - - -
54
4.2 A waiting room of size N - - - -
54
4.3 Several special cases - - - -
61
FEEDBACK- - - - - -
67
5.1 Introduction-
67
5.2 Terminal feedback
VI
-
MlJILrI-STAGE (:),1JimEING SYSTEMS
ROcn1
v
v~iting
- - - - - - - - -
68
5.3 Single stage feedback
69
APPLICATIONS- - - - - - - - -
71
6.1 Introduction - -
71
6.2 The s:pare and standby machine problem
71
6.3 An inventoljT problem - - - - - - - - -
73
6.4 A generalisation of the
n~chine
interference
problem - - - - - - - - - - - - - - - - - -
6.5 A consideration of some practical problems BIBLIOGRAPHY - - - - - - - - - - - - - - - - -
v
INTRODUCTION
.
This M. S. thesis is devoted to an
e:~osition
of some of
the principle results in queueing theoI""J iThich serve to thrmv
light on the
queues.
under~rinc
theme of a discussion of a network of
Although a lot of 'tvork has been (tone on this subject,
it is felt that many aut:10rs seem'-una:vl8.::,e of other contributions
to this field, ovling, no doubt, to the some'tThat scattered nature
of the papers.
Thus, it appears convenient, at this time, to
collect some of this ,Tor:;: together.
The inJ;portant concept of a steady state distribution of cus··
tomers and waiting times is introduced, the conditions under which
it exists are found, for a variety of systems, and the solution of
the steady state equations is displayed in each case.
solutions are unique is shown.
That these
The account goes on to discuss
series of queues in the same way and leao.s naturally to our main
topic of a network of queues.
We discuss this, give an account of
some of the applications of the
theor~
and close the thesis with a
deduction of the solutions to tvTO connnon practical problems.
These
problems are
(i)
to find an optimum arrangement :?or a fixed amount of
service power, and
(ii)
to find, for c dixed imput rate, the best arrangement of
sel~ice
rate and size of waiting room in the case of two
IllD.chines in series, separated
~)y U
ilaiting room, vThere the
first machine is blocked if the waiting room overflows.
CHAPrER ONE
GENERAL QUEUEING TIillORy
1.1.
l
Introduction.
The theory of qv.eues has long interested rnathema.ticia.ns and
statisticians alike and, consequently, there exists a large literature on the subject.
Because of the astonishing range of applica-
tions this is dissipated through many journals on a wide variety of
topics.
In this chapter we will display some of the basic proper-
ties insofar as they affect the topic under consideration.
1.2.
History.
The earliest work of importance
viaS
toot of A. K. Erlang in
1908 under the auspices of the Copenhagen Telephone Company and his
results in Telephone Engineering include many of our basic formulae.
Great strides were then made by Pollaczec and IChintchine, working in
the early 1930' s.
Huch of this early worle 'Was the outcome of studies
of Telephone and Telegraphic problems, and an
exc~llent
survey of
some of the early congestion problems is to be found in many
..U1G. t e.\.(;i)O
,,.../-' ok ~.'"
include both
engip~e:i:-
Later workers, whose work we mainly reproduce here,
D. G. Kendall £1"2.7, and D. V. Lindley £l§}, both
writing in 1951.
~his
research 'Was supported by the Office of Naval Research under
contract No. Nonr-855(09) for research in probability and statistics
at the University of North Carolina, Chapel Hill, N. C. P.eproduction
in whole or in part is permitted for any purpose of the United states
Government.
2
1.3.
Specification o.f a queue.
FollOWing Kendall
[-rd.7,
we 'Will specify a Q.ueue by its three
basic properties.
(a)
The input process.
(b)
The queue-discipline.
(c)
The service mechanism.
The input process.
Throughout this account we will, in the main, restrict ourselves
to a consideration of a random (Poisson) input to the queues.
The
advantages of the Poisson input lie in the facts that it is the most
common, and theoretically the most simple type of input one can consider.
The queue-discipline.
With one or tvTo
e;~ceptions
we shall always assume the queue-dis-
cipline to be the usual first came-first served tY2e.
The service mechanism.
•
T
There are two special cases of importance.
(1) Negative exponential service tim.es.
(2)
Constant service times.
Most of our 1n.terest -will be
latter 'pill also receive some
cen~Gred
at~.:;eD:don.
on the former case, but the
Bot,h (1) and (2) were con-
sidered by Erlang) who also employed an intermadi~te hypoth~sis in
2
which the serv:!.ce time liaS of the X form: •
1
-f~
[\k,
•
e
.. uk/b
k-l
• u
d u )
(0 < u < 00)
3
(which has an interesting interpretation in terms of k
'stages' of service).
successive
It 'Will be noted that (l) and (2) are the
limiting forms of this distribution when k=l and k
=
00
respectively.
1.4. A single server queue.
Let Us consider a queue which has a single server.
We malte two
assumptions.
(l)
The time intervals between the arrivals of successive cus-
tomers are independent random variables with identical probability distributions, and finite means.
customers numbered
rand
He denote by
r+l.
t
r
the interval between
This specifies the imput.
If a customer arrives whilst the server is occupied, he takes
his place in the queue immediately behind anyone else who may be
waiting, 'Waits until they have been served, and then immediately begins his own service time.
Let s
r
be the service time of the
r-th
customer.
(2)
The
1 Sr 1 are
identically and independently distributed
with finite expectation, and the two sets of random variables
1tr~
(r
= 1,2, ••• )
are independent.
t sr J'
This completely specifies the sys-
tem.
We n$h to investigate the 'Waiting time of the customers, Le.
the time that the customer waits in line before he starts his service.
In particular, we are interested in whether this distribution tends
to a limiting form as the number of customers increases.
There are
several proofs in the literature that, under certain conditions this
is the case.
Kendall £~?7, utilizes the' 1'lork of Feller
£1/,
and
4
his general theory of recurrent events, to investigate the ergodic
properties of the queue after reducing the problem to one of a Markov
chain in 'discrete time'.
He notes that the epochs of departure are
points of regeneration, (i.e. what Feller
~-27
calls 'recurrent events.').
Yosida and Kakutani ~2~7 give a very elegant discussion of the limiting
behaviour in the case of a Markov Process with an enumerable, infinite
number of possible states (which we have here if we allow an infinite
queue) • However, they remark that their method is, in general, inade w
quate if the number of possible states is finite.
The proof we will follow is that of D. V. Lindley [~l§}.
1.5. Proof of the existence of a limiting distribution of waitingtimes.
Let
•
ur
= sr
(the difference between service and
- tr
and inter-arrival times.)
and let w
r
be the waiting time of the rwth customer, i.e. the length
of time he spends in the queue waiting to begin his service time.
vIe have, easily
wr+ l=wr +ur
IIIl
0
if wr + u r >
0
if w + u <
0
r
In virtue of our two assumptions the
r-
1 rJ
U
I _7 is finite. Also, ur
r
is necessarily positive and hence
independently distributed and Ef I u
independent of w • But,
r
F (x)
r
=0
for x <
0,
are identically and
w
r
where F (x)
r
= (=p(wr-< x»
is
is the distribution
5
function of the random variable w. But, F (0) is the probability
r
r
that the r-th customer 'Will not have to l're.it. Thus, in general,
F{X) has a discontinuity at the origin.
We assume that the first customer does not have to wait, that
is, the process starts 'With his entry into the service system.
Hence,
=1
and Fl(X)
N01'T, from (1.5.1) J we have, for any
Fr+lex)
x~
for
x?
0
•
0,
= pCwr+1 -< x)
= pewr+1 = 0)
< wr+1 -< x)
+ p(o
= pewr + ur -< 0) + p(o < wr + ur -< x)
= p(iV'r
+
=J
U
W +u <
r r-
< x)
r-
d F (w ) d G(u )
x
r
r
r
F (x - u ) d G(u ) •
r
where G(x)
r
r
is the distribution function of any u
r
•
Thus, knOi'Ting the distribution of G(x), I'Te can calculate F (]::)
r
successively for increasing
r.
We have then the result that the
i're.iting time distribution depends not on the individual distributions
of the service-times and inter-arrival-times, but only on the distribution of the difference between them.
6
sand t
NowlWre. have we used the fact that
r
r
are independent,
and hence we could use weaker assumptions, but this would not serve
any useful purpose.
tribution of the
Below, we utilise the factorisation of the dis-
u's in a special case.
Lindley shows that our queueing problem can be related to a random mlk "\>dth an absorbing barrier.
= p(us-<
F lex)
N
In fact, he shows that
s < r)
-
x for all
where, after replacine our original Us
by u + _ for 1 ~ s ~ r,
r l s
Le. inverting the series about its median, we define Us to equal
r
Z
s=l
u
s
The queueing problem thus corresponds to an infinity of random
mlks with different barriers, one for each value of x considered.
If E
r
is the event
[ Us
~
tor eo11
x
the s~Q.ue:oce of events
fEr)
the limit event E, as
r
f Us ~ x
s
(:
~
r
~
J
= 1,2, ••• )
is decreasing, tending to
increases where E is the event,
for all
s ~ 1
1
Hence, by a well-known property of a probability measure, there exists
lim
r -
> ()()
Fr+_lex)
= lim
r ->
If we denote this limit by
()()
P(E) = peE)
r
F(X), it clearly satisfiea, for x >
0 •
7
(1.5.4)
~
F(X) -=
F(x-u)d G(u)
u<x
or ,
(1.5.5)
F(X)
;:
~
F(Y) d G(x-y)
y~o
= p(Us
< x for all s > 1), F(x)
is a non-negative, non-decreasing function with F(x) -= 0 for x < o.
It follows that, since F(x)
Finally, we must discover when F(X) is a distribution function,
SO
that an equilibrium distribution of vlai ting times exists.
There
are three cases.
EL::]
(i)
>0 ,
By the strong law of large numbers
lim U /n
n
->00
n
= E(u),
with
probability one and thus, except for an event inth zero probability,
we can find, tor any sequence
U
n
>
1
-2 n
f u l ' u2 '
E~r u7 for all n > n.
0
••• l'!3-n
n
such that
TUus J for '.::Loy x~ qy chaos ing na
sufficiently large, ·Un > x for all n > no' with psotability one·
Thus, F(x) = 0 for all x. In other words, the distribution function of waiting times does not tend to a limiting distribution; the
waiting time increases beyond all bounds.
(ii)
E 1~7 <
0
•
As in the first case, the strong law of large numbers applies
and shows that for any positive
(1.,.6)
a
there exists an n
o
such that,
p(U < 0, for all n > 0) > 1 - 0/2
n-
By considering the distribution of U , ••• Un; a positive
l
x
8
can be found so that
p(u < x, for all n < n ) > 1 - 5/2
n -
-
0
(1.5.6) and (1.5.7) together show that
p(un..< x, for all n -> 1) > 1 - 0 ,
lim F(x) < live must have
and since
lim F(X)
x .......> Q(I
F(X) is a limiting distribution function.
x
->
00
(iii) E ~~7
= lj consequently,
-
=0
•
This case cannot be treated by the strong law which only shows
the.t the U In -> 0, and provides no infonnation about the probability
n
of the event E. More delicate methods are needed, and have been
supplied in a paper by Chung and FUchs ~~7, who show, as a special
case of some general results, that if EL~7
= 0,
and not all values
assumed by u are integral multiplies of a fixed number, then
(1.5.8)
p(
IUn
- x , < €, for an infinity of n)
=1
€
> o. If u only assumes integral multiplies
of a fixed number then
(1.5.8) only holds for x having these same
for any x and any
values, and the case vThere u
immediate consequence that
EO
F(x)
is excluded.
=0
for all
This result has the
x, since any value vdll
be exceeded by U, for some n, with probability one.
n
us with the case u EO, which is trivial.
We have, therefore, the following result:-
This leaves
9
A necessary and sufficient condition that the distribution fUnction of the 'Waiting time tends to a non-degenerate limit, as the number of customers increases, is that
If
EL'E:7 ~
0, and
El-":!:7 < 0
or u
= 0 certainly.
u :JO, the probability of 'Waiting a time not more
than x, tands to zero, for any x.
1.6. Uniqueness of the limiting distribution function of the 'Waiting
times.
Suppose the first customer 'Waits a time
y, and Je t
F
r
(x/y) de-
note the conditional distribution function of customer number
r.
Then a calculation along the lines of that used to establish
(1.5.3)
shows that,
Fr +l(x/y) = F(Us -< x , for 1 < s < rj Ur-< x-y)
> F(Us -< x, for 1 < s < r) - F(Ur > x-y) •
Hence,
(~.6.1)
inf F (x/y) > F(x),
lim
r
->
r
00
-
since P(U > x-y) t~ds to zero by the strong law of numbers whenever
r
E ~'E:7 < O.
But also,
and so,
(1.6.2)
lim
r
->
F(X) .
00
10
(1.6.1) and (1.6.2)
establish the independence of the waiting time,
y, of the first customer, and that of any subsequent customer.
y
has a distribution function
]'rex/HeY»~ =
If
H{y) ,
,f Fr(x/Y)d H(y) ,
and hence,
Fr( ~C/H( y »
lim
r->
co
=
J
lim
->
I'
= F(X)
Fr(x/y) d H(Y)
co
,
by Lebesgue's theorem on dominated convere;ence.
These results show
that there is one and only one distribution function satisfying (1.5.5)
whenever
shown.
J
y dG(Y) < 0'.
That there exists at least one has been
To see that it is unique, suppose there were another solution,
and that it was the waiting-time distribution of the first customer,
then it would be the waiting-time distribution of every customer, and
hence the limiting distribution, which is impossible.
Thus, we can damw the following conclusion.
The system has a unique stationary distribution if
1.1
== O.
ELE.? < 0,
or
Therefore, recalling the equivalence of our process and a ran-
dom walk, that the event A, that a customer does not have to wait, will
occur and that it will have finite mean recurrence time.
hand, if
EL"::.7
On the other
> 0, then A is a transient event, and with probability
one there is some point when every subsequent customer will have to
queue.
Finally, if
EL"::.7 = 0,
finite mean recurrenca time.
then A is a certain event with an inThis completes our study of the existence
of the distribution of 'tmiting times.
11
1.7 The waiting time
Let us now assume that,
ELi? < 0
,
i.e.
Thus, if we define the traffic intensity
then
p
p
E£s ~7/E'.LT t ~7,
'by
< 1. Suppose also that the input is random with mean rate
,,-, and that we have a general service time with mean rate
~.
Then
we are assl.mling that,
The epochs at which customers leave are points of regeneration,
as defined by Bartlett and Kendall
I"J:.I.
Consider one such and let
q be the size of the queue this customer leaves behind him.
Let the
service time of the next customer be
customers
arrive during this time.
tribution with mean
Denoting by
q'
v/"-,
v, and suppose that
Then, conditionally,
where
v
r
r
has a Poisson dis-
has the service time distribution.
the size of the queue left behind by the second cus-
tomer, we have in statistical equilibrium (if it can exist),
(1.7.1)
q' = max (q-l, 0) + r
,
or,
qf
where
8
= 8(q)
= q-l
+
8 + r
,
is zero for all non-zero
conseq'l.\e..'lCe of the definition of
q, and
8( q), we have
8(0)
= 1.
As a
12
2
6
=0
and
q(l-o)
=q
.
Assuming no'W that an equilibrium solution exists and that the
equilibrium values of
E.L9:7
and
EL q'2_7
are finitel then I on
forming the expectation of both sides of (1.7.2) 'We have l
orl
if.'2.7=
1 -
ELE7
= 1 - P
This is the probability that a departing customer leaves an empty
counter behind him.
Squaring both sides of (1.7.2) and using (1.7.3) 'We obtain
qf 2
=q _ 2
2q(1-r) +2
(r-l) + o(2r-l)
and on forming expectations I
= ~
>..
+
va-rev) + ~2
2A (>.. - ~)
Suppose that a departing customer has waiting-time and servicetime 'Wand v, reapectively I and leaves
Then
~
(1. 7.6)
' q customers behind him.
is the number of arrivals in a total time 'W + v ani so,
13
From this, and (1. 7•.5 ), it is easily deduced that
=
p
2(1-p)
f1
+ var (v/IJ.~
If the mean service time is kept constant and the frequency of
calls for service remains the same, then
A and IJ.
will be fixed
and in these circumstances it eollows f"rom (1.7.7) that maximum effid.~ncy
will be obtained if, and only if, there is no
the service time.
variation in
't'lith a negative-exponential distribution of service
times, however, the ratio (1. 7.7) is equal to t'Vr.Lce the minimum value.
For a fixed form of service time distribution the ratio is a
constant multiple of
reducing
p/(l-p)
p (i.e. by a reduction either in mean service time or in fre-
quency of calls for service).
of time
and consequently can only be reduced by
This implies an increase in the fraction
l-p (1.7.4) during "Which the counter is unused, and so an in-
crease of efficiency in one direction leads to a decrease in another.
Now let us consider the distribution of lvaiting times.
E.L!!7/Ef!:..7
If
is considerably less than unity an approximation to the
limiting distribution can be found by performing the necessary integrations of (1.5.2)
Consider no"W the most important situation, "When the arrival times
are random.
•
Let
t
G (y)
l
and
G (y)
2
be the distribution functions of
respectively, so that both vanish for negative arguments and
sand
14
If the arrivals are random, the distribution of t
is negative
exponential and in this case,
00
() = 1
Gy
e-7\.z G ( y+Z ) dz
l
7\.
o
ELt_7
where
Let
is
F*(x)
(1.7.8)
for all
,
l/7\..
be defined by the equation
=
F*{x)
J
F(X-y) d G(y)
y~x
x.
Then, in the special case of negative exponential arrivals, we
have, whenever x < 0, in virtue of the fact that
Gl(X)
= 0,
the
relations
=
/l.J
u>O
=
C e/l.x
,
J
e-/l.(u+v-x) F(U) d Gl(V) d U
v>O
where C is a constant.
Taking Fourier transforms of (1. 7•8) we have,
00
foo
•
e
iTx
dFl}(x)
00 00
= 100 100
e
iTX
d F(x-y) d.G(y)
hence, since
0
J-00
•
and
dF*(x)
=
F(O)
F*(x)
=
F(X)
for
x> 0
,
15
•
Letting ¢(T)
wand s
and
VeT)
denote the characteristic functions of
respectively, then
-c
= ¢( T)
+ ¢( T)
--.;.A~
A + i,.
and thus,
= c.
Now ¢(o)
= 1.
r l + A • 1 -. V( T)
L
J. T
_7
-1
Thus, as
-lim
e
,. ->
.
0
c
(1.1.11)
Lindley
flg,
gives tables of the values of the probability of
not having to 'Wait, mean 'Waiting time and variance of 'Waiting time in
the cases of random arrival and regular arrival when the service time
is of X2
distribution wi.th
2
or 3
degrees of freedom.
He also
showed that if service time were exponentially distributed and arrivals
were regular then the waiting time was aBain exponentially distributed,
and it has been knovm for a long time that the same is true if both
arrivals and service times were exponentially distributed.
£127,
W. L. Smith
however, has shown that if the service time is distributed eJ~
ponentially, then so is the waiting time, whatever the arrival-time distribution.
•
CHAPTER TWO
GENERAL QUEUEING SYSTEMS
2.1 Introduction.
In this chapter
1'1e
will consider only the simple arrangements
of one or two queues in series, either single, or multi-server in
either case.
We will discuss such topics of interest as, the out-
put; independence of interdeparture intervals and the state of the
system at those departures; mutual independence of interdeparture
intervals; and the distributions of waiting times at the various queues.
2.2 The output of a single multi-server queue.
We will consider, with P. J. Burke
arrivals and exponential service times.
["2},
a queue with random
Let there be
each 'With an exponential service time 'With mean
average interval bet1'1een arrivals be 1/11..
s
servers
l/IJ., and let the
Ttlen, it is well kno'WIl,
Kendall Lld..7, that if s IJ. > A there is an equilibrium distribution of states of the system, where by state of the system we mean
the number of customers queueing at any particular instant, under
general independence conditions between the service and inter-arrival times and the state of the system at any particular time.
vIe
shall assume that all customers remain in the system until they have
received service,
othel~lise
the queue discipline, or order of service,
is irrelevant, as, in this section, we are only interested in the
output.
17
The probability of the system being in some state k
when a
customer departs" is the same as that of being in the state k
a customer arrives.
"Then
For" the first case represents a transition from
(k+l) to k" whilst the latter represents a transition from k
to
(k+l).
Over a period of time" the number of transitions in each di-
rection
cannot differ by more than one" but the limit of the proba-
bi1ity in the latter case is known to be equal to the probab111ty of
the system bej.ng in the state k
at an arbitrary instant.
Hence"
so is the former limit.
Following Feller
state k
at any time
£27"
let the probability that the system is in
t" be
Pk(t)" and consider
Ik
(t + 5t).
In any small interval of time" tit" the system. is such that the
probability of more than one event happening is of the order O~ 5t2
and hence is negligible.
Thus" the probability of the system being
in the state k
t + 5t
at time
can be represented by"
(1 - A5t) + Pl(t) ~ 5t
pO(t+5t)
= poet)
~(t+5t)
= Pk(t)(l-k
~ 5t - A 5t)
~(t+5t)
= Pk(t)(l-S
~ 5t - A 6t)
"
18
From these equations, we can infer that
dt
d ~(t)
=
dt
(k+1)~ Pk+l(t) - (~+k~)pk(t)+ ~ pk_1(t) ,
(k+1 ~ s)
(k+1 > s) •
In the steady state our probabilities are independent of time, and
so we havel1'1
(2.2.1)
=
pk+1 =
Pk +1
= -
~
~
-
.
Po
.(~ + k~)
(k+1)~
(~
+
s~)
s I..l
,
~
+
1\
+
~ ~-l
(k+1) ~
~
s
These equations are obviously satisfied by,
k
,
(1;)
/ k~
I'k = Po ~
I'k
=
~ Ie
Po (-)
~
/ s~ sk-s
I'k_l
,
,
(k+1 ~ s)
,
(k
?
s) •
~
(0
~
k < s) •
(k~s) •
This solution is just a special case of the more general maltistage solution of R. R. P. Jackson ~1~7, which will be discussed
19
later and the uniqueness of which will be demonstrated.
the solution,
To complete
Po is determined by the nor:mali.ing condition
00
Let L denote the arbitrary inter-departure interval, and net)
the state of the system at a time
Let Fk(t)
t
after the previous departure.
be the joint probability that
net)
=k
and that
L > t.
The marginal distributions of L and the equilibrium probabilities are given by,
00
L:
Fk(t)
==
the distribution function of L,
F(t)
k=o
and
the equilibrium probability of being
in state k.
For an infinitesimal interval of length
Fo (t + dt)
= F0 (t)
2
neglecting terms of o(dt)
•
Similarly,
and
These equations lead to:-
(1 - ~ d t) ,
d t,
20
=
F'(t)
o
-~
F0 (t)
,
(k < s)
(k :: s)
subject to the initial conditions,
Fk(O)
= Pk •
The latter equations
yield,
as the unique solutions subject to the initial conditions.
Thus, we have that,
F(t)
=
and consequently, the marginal distribution of the interdeparture
interval is exponential with parameter A, i.e. the same as the
inter-arrival processes.
2.3 The independence of
~he
interdeparture interval and the atate
of the system at the departure.
We have,
p(t+dt > L > t, and n(L + 0)
=
i.e.
1
A
k~ (il)
k
= k)
[ Fk +l (t) (1\:+1)
s
Fk+1(t)
At A dt
Po e-
,
and,
1
A k
-At
,it-a . ("ii) Po e
s.s
A
dt ,
I..l.
I..l.
dt,
dt,
for k+l < a ,
for k+l > s ,
21
respectively, from (2.2.1) and (2.2.2).
These probabilities can be factored into the marginal distributions of net)
and
t, thus proving the independence of t
and
net) •
2.4.
The
inter-departure intervals.
independenc~of
tet A
represent the set of lengths of an arbitrary number of
inter-departure intervals subsequent to
t.
Because of the randomness
of the input, and because of the exponential service time distribution,
the output process is
tem.
~~rkovian
with respect to the state of the sys-
This implies that,
(2. .1)
p(.A /n(t»
1'( A /n(t), t)
=
can be taken to mean n(IrrO).
where to avoid ambiguity net)
independence of net)
and
t
=
p(n(t), t)
,
The
gives us,
p(n(t»
pet).
We can write,
) = p( A
peL, n(t), A
fr~n
and on substituting
peL, n(t), A
Whence,
)
/t, n(t»
p(t, n(t»
•
(2.4.1) amd (2.4.2), we have,
=
1'( A
=
p( .A , n(L»
/jn(t»
p(n(t»
pet)
pet) •
00
l' (t, ..:'\.
)
=
2:
n(t)=o
p( ..A, n(t»
pet)
= p(
j\.. )p(t) •
22
Thus, we have that all intervals are mutually independent.
2. 5 A single server queue with random
ar~i vale
and general service
time.
In this case, P. D. Finch~§7 considers a single server queue
with a Poisson input process and general service time.
Let the Poisson input process have parameter A and let the
service time of customer number r
be
s.
r
Suppose that the
'$ s r j)
are identically and independently distributed with density function
B(X),
such that
if!!.7 < 00, and suppose B(x) paeeesses a continuous
second derivative.
Define nr(t)
f~
as before,
departures of customers number
joint frequency function for
as the time interval between the
rand r+l,
nr+l and
fr
and H l(t,j)
r+
, that is
(j
we have for our marginal distributions of
Kr
as the
= 0,
1, ••• )
and the steady state
probabilities,
00
~
H let, j) 8t
r+
j=o
=
pet + 8t > (r > t) .
00
~
o
Hr+l(t, j) dt = p(nr +l = j),
(j = 0,1, ••• )
We consider only the case A E~!!.7 < 1, for, as we know, this is the
only case of practical importance which has a stationary distribution.
The number of custonlers left behind by the departure number (r+l)
'Will be ind.ependent of the time interval between departures numbers
rand
r -->
(r+l), in the limit as
=
R(t, j)
(2.5.1)
where,
=
R(t)
00
if, and only if,
R(t) p(j)
lim R (t)
_>
r
00
r
Now,
Hr+let) 5t
=
pet + 5t > Kr > tinr
= 0)
penr
= 0)
> 0) pet + 5t > Kr > tinr = 0)
+ penr
tr
but the conditional distribution of
given n > 0 is just that
r
of a service time, whilst the conditional distribution of
that
n
r
=0
Kr
given
is the sum of an arrival interval and a service time.
Thus,
1
t
Hr+ let)
Since,
=
p (0) A e-i\t
r
=
p(o)
lim
->
r
. eAX dB(X) +
o
Pr(o)
0
exists, it follows that
R(t)
=
Rr (t)
lim
r
->
00
exists and is given by
R{t)
If nr
t
= 0,
= p(o)
A e- At
~
ei\X
then there will be
after departure number r
dB(x) + (l-p(o»
B'(tz
o
wle re
j+l customers present at time
t + 5t >
f r > t,
if and only if
a customer arrives in some interval (T, T + 5T) after departure number
r(where
T
interval
< t), and if there are j sUbsequent arrivals in the time
( T, t) where the service time
r+
extends from T to
1
Similarly, if
n
are present at time
=m
r
t
it will be the case that (j+l) customers
after the departure number
> Kr > t, if and only if j + K -
the departure number
r, and that
m arrive in time
t
after
< s r < t + at.
rand t
Thus, as
= pen r+1 = j,
Hr+l(t,j)
we have,
t + at > ( r > t)
t
=p
j)
(0)
r
J,..
.
1
i\ e-
(:'.~')
i\(t--r)
f.!~_~"~_ e
B'(t.T)dT
i\t,
0
'J
J.
.
j+l
+
1::
p
m=l
r
e-i\tB'(t) •
(m)
Since, under our assumptions the following exist,
-lim
r->
t
t + at •
but not to
t + at
s
p (0)
lim
and
r
->
r
lX)
p
00
r
(m)
then the existence of the following limit is implied.
H(t,j)
=
r
H (t, j)
lim
_>
00
Moreover, we dan deduce that
r
25
j+l
(~t)j+l-m]
t
+ l3
p(m) ( (j+l-m)!
e-7\ B'(t)
m=l
I
In order to prove that the number of customers left by departure
number r
is not independent of the time interval between departures
number rand r+l,
in the limit as
r ->
co,
it is SUfficient to
prove that
#
H(t, j)
i'or some value of
Het) p(j)
j •
Let us consider the case
j
= O.
Using equations (2.5.2) and (2.5.3) we have that a necessary condition for
H(t, 0)
= H(t)
p(o)
is that,
P.( 1) B' ( t) e -At + p ()
0
B( t) A e -",t
Differentiating both sides of (2.5.5) with respect to t
we have,
(2.5.6) p(l) e-AtBIt(t) + ",(p(o) _ pel»~ B'{t) e-;...t _ ~2p(0)B(t) e-i-·t
:t
= p(o)(l-p(o»BfI(t)
+ A p(o)B'(t) -
A2p(0)e-~te/l.XdB(J:)
Eliminating the integral between equations (2.5.5) and (2.5.6)
26
gives, after simplification,
which has the solution
-At
l-p(O}
B'{t) = C e
£-
p(o)(l-p(o»)-p(l)e~
At
p(0)~I-p(0)1~p(1)
pel (I-p{o )
.]
p(o)(l-p(o»-r(i)
where C is a positive normalising constant.
B' (0+)
Hence,
=C > 0
but, by letting t--> 0+ in (2.5.5) we see that, since
B (0 +)
and B' (0+)
# 0,
= 0,
we must have,
pel)
= p(o)
(l-p(o»
Equation (2.5.6) then reduces to
(l-p(
0» B"(t) A B' (t) = 0
whence
B(t)
=1
_ e-At/l-p(o)
Thus the service time is exponential.
If we let P I (n/T) be the conditional probability that cusr+
tomer number r· leaves n customers behind him on departure, when
it is given that
Kr = T,
probability density for
and let Hr+ 2(t/T) be the conditional
f r +l , we have, by all argument similar to
that used in deriving (2.5.2)
27
t
H 2(t/T)
r+
= (l-p r+1(0/'1'»
r ->
Thus, as
B'(t) + p 1(0/'1') A a-At
r+
r eAxdB(x).
~
co ,
t
H(t/T)
= (l-p(o/T»B'(t)+p(o/,,)
A e-At! e
AX
dB(x)
However, if successive intervals were independent in the limit,
we would have,
H(t/T)
= H(t)
which would imply that,
t
(l-p(o/T»B'(t) + B(O/T)A e- At ]
e- Ax dB(x)
o
t
= (l-p(o» B'(t)
i.e.
Ax
+ p(o) A e-At] e
o
(p(O/T) - P(o»B'(t) - (p(O/T)-p(O»
/" e-
At
!
dB(x),
t
e
AX
dB(x)
=0
•
it follows that either (i) p(a/ '1')= 'p(O),
or
If (i)
(ii) B' ( t )
is the case it follows that
R(t, 0)
= H(t)
p(o)
If not (i) then (ii) must be true.
eliminating the integral, gives B"(t)
Differentiati.ng (ii) and
=0
for all t.
Hence,
B'(x) = C for all x, and on substituting this back in (ii) we find
that C
= O.
Hence B(::)
cannot be a distribution function.
Thus, (i) must be true) and we have, in conclusion, that the
interdeparture intervals are independent if and only if the service
time distribution is negative exponential.
28
2.6.
The 'Waiting time at a second multi-server queue.
E. Reich fl17~ uses a different approach to attack the problem
of 'Waiting times and independence of inter-departure times.
He
uses the reversibility of the Markov chain to study the distribution
of 'Waiting times wilen the customer proceeds to a second multi-server
queue after having been served at the first.
We restrict our attention to unsaturated queues in "eqUilibrium."
Definition:A stationary stochastic process
if N(t) and N(-t)
N(t)
is said to be
reversible~
have the same multivariate distributions.
This necessary and sufficient condition for reversibility becomes~
if N(t)
is a discrete or continuous parameter Markov chain
with a denumera.ble
state space,
Qi.(t) = Pi P .. (t) = P.Fj.(t) =
J
where
J
~J
~
Q ..
J~
(t) for all i,j=0,1,2 •••
Pi and pij(t) are, respectively, the stationary and transi-
tion probabilities of N(t).
Kolmogorovts criteriotl, £1'2.7, for the reversibility of Markov
chains with a finite state space, may, in a special case, which ineludes Reich IS, (14), "type A", be innnediately generalised to the
denumerable state space as follows.
Theorem.
Let N(k); k =
0,1,2~
••• ; be an irreducible stationary discrete
parameter Markov chain with the state space
stationary probabilities be
~,
0, 1, 2, ••• , let the
and let the single step transition
29
probabilities be
• A necessary and sufficient condition for the
ij
reversibility ot the Harkov chain is that,
1C
for every sequence of non-negative integers (ii J 1 , •••. ), be~
2
and end1Dg w:Lth the sane integer.
The proof of the theorem is to be found in Reichls paper, and
as it is of no interest as far as the general theme of this account
is concerned, it is
omit~ed
here.
By considering the epecial case he calls his "type A", Reich
•
e
shows easily that tbe
ble.
sta~ionary
-
Then, by noting its
birth and death process is revers i-
;~quivalence,
with constant birth probabili-
ties independent of the state of the system, and that of a
stationa~J
unsaturated queue with Poisson input, exponential service time, first
come-first served queue discipline and
things.
s
servers, he shows three
If customers arrivals correspond to births and departures to
deaths then,
a) the sequence of departure times form a Poisson process,
b) the value of n(t) is independent of all past departure times,
and,
c) if to
is a departure time, then n(to+o)iS independent of
all past departure times.
These results are equivalent to the theorem of Burke
1'27,
proved
30
earlier by a different method.
Suppose the customers, upon departure from the first queue,
immediately enter a second multi-server queue, where once again
service is first come-first served, with independent exponential service times.
It follows from result (b), that if nl(t) and n2 (t)
refer, respectively, to the first and second queues then nl(t) and
n2 (T) are independent,
special case s
= 1,
t
T
< t. This resUlt
=T
'ViaS
first proved in the
by R. R. P Jackson ~1~7, by a different
method which will be demonstrated later.
Here, and throughout the rest of this account, stage time will
be used to refer to the time spent by the customer between arrival ,
and departure, i.e. the sum of the Vlaiting time and service time.
Let
and Te represent a customer's stage times at the first and
1
denote the number
second queues respectively, and let n and n
l
2
of customers this customer leaves behind him at the first queue, and
T
finds at the second queue, respectively (customers being served ineluded) • As a corollary of (c ) above,
Let A(t, k)
n
l
= P(.rl < t/n 2 = k). If
the Poisson input process, then nl
is the
and n
are independent.
2
is the parameter of
A.
~umber
which occurred during the Vlaiting period fl.
that
p
(nl = j I Tl = t,
n2
= k) = e-~
of Poisson events
Therefore, we have
31
Hence,
= Et
r
_ e
AtZ
e
-At'
_7
co
=[
e
Atz
e-
At
dA(t, 1;:) •
But the left hand side of this last equation is independent of
k
(as
nl , n are independent),
2
on k, and, thus,
Therefore, A(t, k) does not depend
n
, and consequently T2 ,are independent of
2
In other words, the numbers of customers -waiting in the first
'
l
queue and the second queue at any time are independent, as are also
T
a customer's stage times in the first and second queues_
We are, of
course, assuming infinite queues as a possibility at either queue.
CHAPTER THREE
MUIJrI-STAGE QUEUEING SYSTEMS
3.1. Introduction.
In this chapter we define a series of queues, and then investigate the ergodicity of one such series.
After establishing the con-
ditions under which a steady state exists, we study the series in its
steady state, both as a whole and in its single queue elements.
Finally,
. we study the concept of a network of queu,ss.
3.2. A series of queues.
A series of queues is a system whereby the input to any queue is
.e
the output of the preceding one in the series, except for the initial
queue whose input is the input to the system as a whole.
3.3. The ergodicity of queues in series.
This topic has been treated thoroughly by J. Sachs
ing him, let us consider the case where we have
s
.L~ff/.
Follow'·
queues, each "tn th
one server.
(J'
For (J' = 1, 2, ••• s, let Rn be the service time of customer
o
number n in queue (J' . Denote by R + the time 1" 1 - 1" , where
n l
n+
n
T
is the time at which customer number n enters the system, i.e.
n
enters the first queue.
Let
wcrn denote the waiting time of the n-th
person in queue (J'.
It is immediately obvious, by induction, that
(3.3.1) wP
_ro, riPn + t Pn + ~E1)-1
(t(J' + If vf ) 7
(J'=l n
n - n+l-
n+l
= max
33
where
to"
n
= Rcrn
0"
Let Tk
where
ma.x*
- Rcr- l
n+ l
k
•
"£1
-T
... •• .. Tp"/
t~ and let D0" = ma",::*"
k
j
j J.
1
P
= r:
i=l
implies maximisation over all
0:5
jl
:5 ... :5
jp
:5 k
.:0
•
Also, let
cr
=
~
(t
n 0"=..1.~
n
HP
+ vf - vf 1 ) ,
n
n+
then, for all n, it is obvious that
Assume now that
.e
Then, from (3.3.1) and the induction hypothesis,
=
t P+l
max ( 0, up+l
I'Y
k + k
= max ( TkP+l
0:5j:5 R
Upon using this
t P+l + Wp+l _ vTP+l
n
n
n
_p+l + D.P - Dp) •
- ';,L....
1t
J
J
eA~ression
= t P+l
n
for k
+ max
=n
=n
- 1 and k
we obtain
('IJl+ l _ TJ?+l + DP - DP )
J
j
n-l
<. < n- 1 n-l
0_ J
(~+l .. TP+l +
n
j
nJ? _ nP)
J
n
=
DP+1 .. J!+1 + D'P .. J!
•
n-l
n
n
n-l
Thus,
HP
HP+1 = t'P+1 + W'P+1 _ W'P+1 +
n
n
n
n+l
n
=
D'P+1 ..
n-l
J!+1
n
'
and hance,
(
';l(
';l(
';l()
~.~.~
H'P _ D'P
n n-l -
0 ~
for all n and all
zn = (~,
_'0
~
,
'P
,
~
s •
W~) is not a Markov 'Process, but it is obvious
= (wl,
;wnS , Rnl , ••• RnS .. l ) is the. n-th random variable in
n
n
a stationary Markov process, for we have, (3.3.1),
tr~t
Y
'P..l
W'P 1 = max (0, H'P + t'P + !: (to" + vf ..
n+
n
n 0"=1 n
n
= max
if
n+
1»
(0 U'P + t'P + lI'P-1)
'n
n
n
but t'P and H'P- l are both functions of Y only. This leads us to:n
n
n
Lemma 1.
35
where
t.J .>
0 ,
-
i = 1, 2, ••• s
and
x.J --.
>0 ,
j = 1, 2, ••• s
•
Then,
for all n, x, t, r.
Proof of Lemma 1.
R2' ••• , R'
n'
Fix a point w in the sample space of
0
R ,
2
..., RnO,• ... ,. Rl ,
6
wi (lv, x) = ~
wi (w, x) x
=
2
S
••• , R
n
and let
,
,
Then,
for
It is clear that
2
~ j ~
lJ~(w, 0) ~ ~(w, x) for each
j.
n •
vTe then observe
that,
x) - max(o, wp-l(w, x) + tP-l(w)
n.
+ ~-2(w,
n
x»
36
Thus, we have
_ ( w, x ) + t P-l()
_ ( w, x »
+ min ( Wp--l
_ W , - Hp-2
j
l
j
l
j
l
and it followB easily that W~(w, 0) ~ 1'T~(w, x) for all 2 ~
j
<n ,
which completes the proof of Lemma 1.
Lemma. 2.
p(Zn ~ t/Z l = 0) --> F(t) as n --> ~, where F is an s dimensional distribution function whose variation over a-dimensional space
may be less than 1.
i.e. F may not be a probability distribution
function.
Proof of Lemma 2.
Then,
Since
l YnJ
by Lemma. 1.
is a stationary Markov process,
H* be the distribution function of Rl ,
Let
of
H •
2
Then, using
p(z
(3.3.5)
1 < t/zl=o)
n+ -
(3.3. 4,
in
we have,
<JP(Zn -< t/z 1 = 0,
-
=
J
p{Zn
and therefore
~ t/Z l = 0,
Rl
= r)
d H(x, r)
Rl
= r)
d H*(x, r)
= p(zn-< t/Xl = 0) •
Thus,
p(z
< t/z l = 0) is a monotone sequence and therefore con-
n-
verges to a limit which we call Fe t ) •
s
p(wl
n -< O:l' ••• 'Wn -< as /~l =
Repla cing
zi
~ = 1,2, ••• s) --->
0,
by O:i' we have,
F(al,···,o:s) •
Theorem.
Let
<
IJ.,
o
lJ.~
= E.rR~ _7,
for
~
= 0,
1, • u
Then, if
s.
max IJ.
1< ~ < ~
F is a probability distribution function.
Proof of Theorem 1.
Because of Lemma 2, we need only shaw that, under the conditions
stated here,
~znj
is
bounded in probability. i.e., for all
where
This is equivalent to proving that
probability.
Lindley
and Sachs proves that
rl6/
-
-
rr
n
'f;~,
showed that
l1(t) -> 0 as
••• ,
lrn
w:
t ->
n,
co •
are each bounded in
'tvas bounded in probability
is bounded in probability, but the proof is
long and of mainly intrinsic value only, and is omitted here.
This
38
~,
leaves us then,
W~
••• ,
and a trivial induction argument makes
it ob"'rious that we need only consider the
argument given is legitimate when
(3.3.6)
(3.3.2)
=
max
o ~ j 1:5
Let
So
= s,
Actually, the
is replaced by any
l
S l
s l
_D - < T + ••• + T - .
We observe that
Thus, fram
s
i Vl~ 1.
n
~th
k
.•..~
jS
and d.efine
-
n
= n,
l'
1
-< 'D- -<
s.
n
=s
- 1,
:5 n
si
to be the largest
er
< Si_l' (er
~ 0),
with the property that
Per - IJ. s . 1 > 0 •
~-
Notice that
Let
er
=0
satisfies
k be the first
i
(3.3.7),
such that
51 = o.
check that
IJ.
o
and that, for
= IJ. sk > ••• > IJ. So = IJ. s
si < er < Si_l '
(3.3.8)
Define, for
1
= 1, ••• ,
k ,
so that
'
8
1
is well defined.
Then it is easy to
39
(
<
max
o <-
j
1 < •••-< j si_l _.< n
si+ -
Because of (3.3.6), we have,
{H:}
and, therefore, in order to show that
is bounded in probability,
i
we have only to show that each U , (i = 1, ••• ), is bounded in proban
bility, which can be summarized in the following lemma.
Lemma 3
For m = 1, •.• , M, let
,Jll i
Ai'
= 1 , •••
be a sequence of identi-
cally and independently distributed random variables with
,
where Ao >
'A... _
>
OM -
max
,.g<a<M
~"i
assumed that
>
Aa
and
k
~
I
min
A > O.
(It is not
o<a<M a
are independent of one another).
M
~ X~ , and let W = max*' ( ~(~ - sU: ))
n
m=l n
Jm
i=l
Then, W is bounded in probability.
n
i
It is easily shovrn that the u can be taken as the wand
n
n
satisfy the conditions of the lemma, so that the proof of Lemma 3 gives
Also, let
(
I:
us our proof of the theorem.
Proof of Lemma 3
Let y
= min'("'i
- Aj ), where min'
implies minimisation over all
40
o < i, j ~ M with Ai - Aj > o.
strictly positive.
and let
€M+1
= 0,
For 2 ~ m ~ M,
and
=0
€1
= y;M. 0 is, of
define Em = ~ - Am-1
Let
0
course,
+(M-m+l)o,
•
Then,
€2' ••• ,
~1
are positive.
(3.3.13)
< (-l/M)(Ao - ~)
<
Letting
jM+l
= n,
0 •
~+1
and taking note of the fact that
= 0,
we
have
M
(3.3.1~)
E ~(n-jm + 1)
m=l
€
m+
where
1 - (n - j)€ 7
m m0
=
~ j 1 ~ ••• ~ jM+1
o ,
=n
•
Upon using (3.3.14), we find
'n
= maJ~~ ~
M
m
n
E (Sn - Sj ) + (n-j
m=l
m
m+
1)
€
m+
1 - (n-j )
ill
<
_
<
m
n
max ["Sn - Sj + (n-j m)(€m+1 - €)
m-7
m=l o<j < n
m
M
E
-
In-
€
7
ill·-
41
Each of the ter.ms in the summation of the right hand side of (3.3.15)
is bounded in probability, since
k
~km = E (j~ +
where,
and
i=l).
E;-xm.
-).
+
€
m+1 -
€
€
1 -
m+
7 < 0,
Dr-
€
m
),
owing to (3.3.12) and (3.3.13).
This concludes the proof of Lemma 3
and hence of Theorem 1.
Theorem 2
(J >
cr
<
s
-
If max
1< cr
-
then
(J.
0"
F sO.
Proof of Theorem 2
Let
that
p
cr > 1 with
be the first
Iln -->
00
~ !-L
!-L
cr
o
'
We need only show
in probability.
Using (3.3.2), we find
wP
n+1
=
;-TP - T~ + D~-l_
o.:s: j < n - n
J
J
if- 1 7
max
n-
=
..
(3.3.16)
=
max"
o~jl~ ...
.9;5 n
,
1
..
n
••• +T
... - T~J
1 -
+Tp-l - T1 - ••• -T.p-1
n
jl
7
J p _1 -
7
•
The last term on the right hand side of (::5.3.16) is bounded in probamax
I-tCT < 1J. ' by Theorem 10
p-l
Looking at the first term on the right halid side, we have
bility, because
l~ CT ~
...
n
=
max
0<
k
.L z
<n
j=k+l
.,. - T~J 1 - ]
.
(~- R~) +
J
The last term written is bounded in probability, whilst the preceding
term tends to infinity in probability as
It is then quite clear that
and hence
wPn+1
n
->
00
because
IJ.
> IJ.
•
P- 0
must tend to infinity in probability
F == 0 •
To sUllJIlJarize, vle have proved that if and only if
then the set of -we.1ting
times at the
S
max
~t
l<CT<sCT
<
IJ. ,
stages tends to a limit which
has a probability distribution function.
3.~.
The steady state solution.
Suppose that 't'le have a sequence of queues.
R. R. P. Jackson fll] does, that at queue
i
Suppose, further, as
there are
r
i
identi-
cal servers, each of which has associated 'tnth it an exponential service time distribution 'With mean
l/lJ.
i
0
units of time.
If a customer
arrives at the queue to find all
r.
servers busy, we assume that he
J.
takes his place in line behind any previous arrivals and waits until
a server becomes empty, service being first come-first served.
Suppose, further, that new customers arrive at the first queue
at random, 'With a mean rate of arrival
service at the k-th queue.
')-., and leave when they finish
We assume that
')-.
< r i Ili
for all i.
Hence from the previous section, the system has a stationary probabi1ity distribution function.
vie have sho'Wll in 2.2
to each queue will be random with a mean rate
that the input
')-..
P(n , n , .... , ~,t) be the probability that at time t
1
2
there are n
customers in the first stage, n
in the second, ••• ,
2
1
Let
and
!it
in the
If we consider the changes in the system over
k-th.
a small time interval 8t, we can obtain a set of differential difference equations in the usual manner.
Then, as before, we obtain a set
of steady state equations in which,
p(n1 , ••• ,~,t)
for all
t.
a(nj )
Define,
= nj ,
nj
< rJ
,
n.
> r.
1
,
nj
0
,
~ :J
=rj
8(n )
j
=
=
and
8(~ + 1)
=
1.
J -
n
j
j
j = 1, 2, ••• k •
J
j
= 1,
2, ••• k .
= l?{nl'''''~)
44
Then the steady state equations are,
k
=
j~l o(n j + 1) a(n j + 1) ~j p(nl,···,n j + 1, nj+l-l,nj+2""'~)
where all P's
containing a negative value of n are automatically
taken to be zero.
The last term in the summation on the right hand
side is
o(~ + 1) a(~ + 1) ~ p(nl , n2 , ••• , ~-l' ~ + 1) •
It can be shown that p(nl , n2 , ... ~)
k
11
= pCo)
j=l
bFn j _7
is a solution of the set of equations, where
r.
=
p(o, ••. ,o) = p(o)
equation,
A.
J
1·
- I (--)
r .• ~j
J
7
J-
-
series, equal to
n. > r j
J -
•
can be found by utilising the usual normalising
~ p(nl""'~{)
The b f-n.
,
= 1,
in the follolnng manner.
00
are all positive and
I:
njeo
b r nj-7 is a convergent
'L
rj
rj
r. !
lJ.
00
C--_"A_)
I-L
j rj
i:
nj=o
J
J
=
r.
r J
1
j
::-r
r j • 1 -C-_L)
I-L
j rj
(_1:-)
where
I-L
j
r
<
1 .
j
The positivity of the terms ensures that
00
00
,
i:
i:
n =0 n =0
1
2
and so, on writing
p(o)
k
= 1r
j=l
j
= 1, •••
k,
it follows that
-1
A.
J
The solution 'ive have given can be verified by sustitution.
It re-
mains to show that it is a unique solution.
Proof of Uniqueness of solution
Consider the associated Markov process generated by the states
(nl""'~)
of the real system, but with the transitions now supposed
to occur at discrete times separated by time intervals, all of length h.
If we choose the value of h
such that the quantities
"Ah, 1 - "Ah,
I-L h, 1 - k
i I-L i hj (k i ~ r i ) lie between 0 and 1, they can be
i i
regarded as transition probabilities. This process obviously differs
k
from the true process.
If we let A be the transition matrix of the associated process,
we can show that it is reducible, by the following arg'UD,le3it.
It is
clearly possible to reach the null state from any other state, merely
by successive departures without arrivals, for instance.
The proba-
bility of this occurring is clearly positive, as it is the product
(l-~h), k.~ih,
of such numbers as
J.
~.
(1 - k.J.
h).
J.
Similarly, it is
possible to reach any other state from the null state, again with positive probability.
Combining these two procedures, it is apparent
that we can reach any state from any other state, 'With positive probaThus, the matrix A is irreducible (Feller ~27, p.
bility.
and hence a unique solution exists to the e-quation
~
is the row vector of state probabilities.
~
exists a unique solution to the equation
~
A ==
~,
'18),
where
In other words, there
(A - I) == E .
The diagonal elements of A are of the form 1 - a h where a
n
is a function of
(A;
have the factor h.
n
~l""'~; kl""'~) whilst all other elements
Thus, we may write
(A - I)!h == B, and it follows
that there exists a unique solution to l:a == 2. •
This is the required result, for B is, in fact, the matrix of
the steady state equations of the true process.
p'istributions for any stage
Let the marginal probability that there are n customers at
stage
j
of service, i.e. in queue
j, be given by p.(n), where
J
00
•••
1:
I:
nj_l=o nj+l==o
•••
1:
~=o
p(n ,n , ••• ,n _ ,
j l
l 2
47
In the same way as we evaluated p(o) it can be shown that Pj(n) ~
brn.
-
J
~
n
-
7/A..
J
Thus, the probability distribution of customers
present in stage
is dependent only on
j
Il j
r j , ')..., and is the
'
same as that obtained from the system with one queue;
r j servers;
random input and a negative-exponential distribution of service times.
In the case where we have a single server system, we obtain
=
where
p(o)
=
We note that
xj
putting
k
(1 - ').../11.).
J
(3.4.1)
= 1,2, ••• ,k),
Pj(n), (j
.e
n
j";l
= ').../I-l j
p(o)
factorises, so that the distributions
are independent and we may write immediately,
•
Thus, the average nunfuer of customers in stage
co
L:
n xnj(l - x.) = x.(l -
n=o
J
J
X
j
co
L:
)
n=l
n
j
X
is
n-l
j
= Xj(l - Xj)(l -
=
.
xj(l - xj )
-1
.
The average number of customers being services is equal to the probability of one or more customers being in the stage
48
ro
= t
n=l
=
Thus, from
(3.4.2) - (3.4.3), the average number of customers
waiting in line is
To obtain the average values for the system as a whole, we must
sum the averages for the individual stages.
Let us now consider a customer arriving at stage
customers already there.
occurring.
to find n
j
n
This event has probability x (1 - x.)
.j
of
J
Now, knowing that the distribution of service times is
negative exponential with mean
pel departure in
l/~.,
J
g, g + ds) =
we have,
~j
e
-~. ~
J
ds
5, g + de) = de
1
Therefore,
P(second departure is in
0
~j e
-~.x
J'~<i(;!
-I-l.(g-:~)
J
At this point let us assume that the probability of departure number
(n-l) occurring in the interval
5' 5 + dg is,
n-l n-2 -~jS
g
e
ds
~j
(n - 2) ~
clx
Then the probability of departure number n occurring in this in-
•
terval is given by
n-l n-2
I-L j
x
e
-I-L x
j
I-L
(n-2) !
=
which is true for
Hence,
n
j
e
"I-L (~-x)
j
ax
n n-l
I-L j e
d~
(n-l) ~
= 1,2, ••• by induction.
the waiting time distribution is given by the probability
density function,
00
Z
(1 -
n=l
X )
j
n n n-l "l-LjS
.I-L j S
e
Xj
(n-l) :
and the probability of not writing at all is,
(:~.4.6)
The conditional probability density function of waiting time,
-(I-L.-7-..)g
given that there is some waiting, is (I-L j - 7-..) e J
•
In addition, it is easy to show that
a)
the mean waiting time is
b)
given that the mean service time is
efficiency defined to be
Xjl ll-Lj(l -
mean waiting time
mean service tiffie
X
j »)
,
l/I-L.' a measure of
J
is
xj
I( 1
- xj
),
and that
c)
Z
(7-..
j I-Lj I-L j
the average time spent waiting in the system as a whole is
7-..)' whilst the average service time is Z
-
j
1
I-L j
50
When the service times are all equal the system is somewhat sim..
plified, and it is easily seen that
P(n)
~ n
(-)
=
•••
=
p(o)
fJ.
n+k-l
C _
n l
;.5. A network of ~~~~
Let us consider now a more complicated arrangement, as discussed
by J. R. Jackson flQ/, in which we have, once again, a number of queues.
However, in this case, we assume that a customer finishing service at
queue k
.e
transfers to queue' j according to a fixed probability asso-
ciated with the queue it is leaving.
Call this probability Qjk'
In
addition, customers arrive randomly from outside the system and enter
queue
j
at a mean rate
A. ••
J
Specifically, our assumptions are these:Let there be stages
1,2, ••• ,M (each stage is a queue).
Then, for
m = 1, 2, •.• H.
1)
Stage m contains
r m servers.
2)
Customers from outside the system arrive in stage m in a
Poisson type series at a mean rate A.
m
;)
•
Customers arriving in stage m, from inside or outside the
system, are served on a first come-first served basis with an exponen..
tial distribution of service times
mean 1/11"
m
51
4)
Once served in stage m, a customer goes instantaneously
to stage k, (k
= 1,
2, ••• M), with probability Qkm'
the system with probability 1 -
He leaves
• It is on the basis of
km
assumption (4) that this system is termed a 'network' of 'Waiting
~
Q
k
lines.
For m = 1, 2, ••• , M, let r
be the mean arrival rate of
m
customers at stage m from any source. It is obvious that in a
steady state we must have
r
Let
m
=11.
m
+ ~
k
Qmkrk'
n denote the number of Gustomers waiting or in service in
m
stage m, and define the state of the system to be the vector (nl, ••• ,n~.
.e
Define P (n), (m
m
= 1,
2, ••• , M; n = 1, 2, ••• ) by the following
equations.
(3.5.2)
,
Pm(n)
= Pm(o)
(rm/~m)n/n!
Pm(n)
= Pm(0)
n-r
(rm/~m)n/.f m! r m m
n
= 0,
,
••• , r ,
m
n> r m
-
,
co
where the P (0) are determined by the equations
m
We then have:-
~
n=o
P (n)
m
= 1-
Theorem.
A steady state distribution of the state of the present system is
given by the products.
Provided r m < J..Lmr m
condition) •
for
m = 1, 2, ••• H, (the usual steady state
52
Proof.
The general approa.ch of Feller
£'2.7
gives us the steady state
equations,
The theorem can be proved by verifying that the given probabilities satisfy the steady-state equations.
Thus, we have shom'! that at
least insofar as the steady states are concerned, the system we are
discussing behaves as if each of 1ts stages
v~s
an independent system.
This conclusion is hardly surpising in view of the ea.rlier work of
Burke £~7, and Reich
1117.
Koenigsburg .Ll~7 independently discusses a cyclic queueing system, i.e. one in. Which operations are carried out in a fixed cycle, of
m stages.
This, of course, is just a special case of the above in
53
which
~i
= 0 for all i, and
0jk = 1
,
=0
,
Q
jk
j
=k+l
(mod m)
otherwise •
Koen1gsburg's steady state distributionsare the appropriate
special cases of Jackson's results, and his calculation of the amount
of time each server is busy (though in his case they are machines),
and the output of each machine (obviously equal)
work of
(:~,4),
ar~
covered by the
He gives tables of mean numbers of customers being
served and 'Waiting; the utilisation factor of the machines and the time
to complete a full circuit, as well as discussing the application of
.e
his results in the
coal~mining
industry•
CHAPTER FOUR
THE: EFFECT OF LIMITING THE SIZE OF THE
WAITING ROOl-1
4.1.
Introduction.
We have assumed, hitherto, that at any stage in a network, \Ve
can have an inf'inite queue length.
However, in most practical cases,
owing to lack of' storage space or invento!"'J costs, this assumption
is unrealistic.
In the case where we have a f'inite
queue, a new
customer who, on arrival, f'inds the queue is already full, may depart, never to return, or be may stop in the previous stage and thus
cause a blockage at that particular point, stopping production there.
The ef'f'ects of' this limitation of' the queue size have been discussed
in the literature, nota.bly by
4.2.
A -waiting room _~_size
Suppose that we have a
P. D. Finch
i§1
2/.
and G. C. Hunt.L
N.
~iting
room which will hold
N cus-
tomers, excluding the one being served, ana that if' a customer
arrives to f'ind the room full, he leaves, never to return.
pose that the arrivals are random, with mean rate
'A
We sup-
and we let the
distribution function of the service time of customer number
say, be
B(s) •
Then
k_~)
...
the probability of
r
00
(4.2.1)
k
r
= --rr
~
e-'As('As)r d B(s)
r
arrivals in a service
time, is given by
1-
r, s
.
55
Let p
r
n
denote the probability that customer number
customers behind him on completion of his service.
r
leaves
n
Then,
if n > N
and,
p r +l
(4.2.2)
0
pr+l
n
=
(pr + pr) k
100
k + Pnr k + ••• + P2r k _ + (p~+pr)k ,
= pr
n+l 0
o n
n l
l
(n < N)
pr+l
N
r
r
= PN Kl + PN-l K2+
+
P~ I~_l+(P~+P~) EW '
where Kj = k j + k + + •••
j l
The existence of a limiting distribution follows from the fact
that the process is a finite, irreducible; aperiodic Markov chain.
The equations which determine the lirdting distribution are
pn =Pn+lk 0 +Pk
(n=o,l, ••• N-l),
n l + ••• + P2kn- l+(Pl+P)k,
0
n
(k.2.3)
( P = PIfl + PN-l~+
N
llrite P ( x )
k(x)
Thus,
= Po +
PI + P~x + ••• + PN
)r-l
2,". .
= k o + k1x -:- ••• + k r xr + •••
'
56
P(x)k(x)
= (p o
7x
+ P1 )k 0 + -;-(p 0 + P1)k1 + P k 02
... + PN k 7xN- 1
N+j
. + ... + P It. 1 7
N J+ 0-
+ /-(p
'-
0
+ P )lL
1 -~+J
X
j
= 0,
,
1, 2, .•.
On substituting from (4.2.3), we have
rep
=--
where P~+j
0
+ P1 )k.-• + ••• + PN k.J+1:7
--N+J
-
.
Hence,
00
p(x) k(1:)
= P0(1
j
N
- x) + xP(x) + x ['Pir - PN + j:1 P + j x .
N
and so,
(4.2.4)
00
p(x)
P 0 (1 -
= -
k(x)
If
(1 - x)/(k(x) -
-, )
.'"
+
- ."
:~)
~ PN
+·
N- PN + j=l
J
xNfP
k(~~)
v
xj
..7
_7
- x
can be expanded us a power series in x for
some suitable region of convergence, only the first term in
contributes to the coefficient of xn for
n < N.
(4.2.4)
The P can
n
then be determined in tenns of P, which in turn can be found from
N
the normalizing equation
~
j=l
0
Pj
=1
•
57
For example, in the case when the service time has a negative
l/~l, k
exponential distribution v11 th mean value
a:
= t-./(A.+I-L), f3 = l.t/(I"
=
+ IJ.).
r
= r:l f3
where
We then have,
f3. •
l-a: x
Thus,
1 - x
k(x)-x
(1 - x)(l - a: x)
=
1
2
(3-J~+ax
and we find
p
n
1 - A./I-L
N 1
1 -(t-./J.l) +
=
1
= N+'i
A. n
• (-)
IJ.
. (n = 0,1, .•. ,N; A. ..l. II) ,
'
T ~
,
(n
= o,I, ••• ,N;
t-.
= J.l)
•
Therefore,
t-.<I-L
(4.2.6)
11m
N
->
P
n
=
00
These values agree with the 1Jell knovm solutions for the familiar
case
N
=
00
•
However,
,
if
A. > IJ. ,
,
if
A.:5 I-L ,
which elucidates the difference between the limiting solutions in the
58
two cases
A~
A>
~,
If we confine
•
~
oUl~selves
to the usual case where both the input
process and the service mechanism have a negative exponential distribution function, it is easy to find the distribution function for
the waiting time of a customer.
Denoting by
that a customer arrives to find n
custome~s
the one (if any) being served, and by
fr
(In
the probability
before him, including
the probability that
r
customers leave the system between consecutive arrivals, we have,
for
r
= 0,1, •••
H,
We can determine the
vTe find that
of
CL 1
l'if+ -r
~+l(I-x)! f fex.) -
in a similar manner to the Pn above and
. the coefficient of x r-l in the expansion
J.s
On
...
xl
the case of random arrivals vTe find,
1 - ~!A
Q
n
=
1_(~!A)H+2
1
N+2
,
,
The first eqlmtion of (4.2.8) can be rearranged to give,
~
=
=
=
In
59
Thus,
lim
N
->
~
00
=
,
l1.m
N
->
00
'a fact which we have assumed implicitly in 2.2.
m customers before him on arrival (m < N+l)
If a customer finds
then, he 'Waits a time
8
1
+ ••. + sm' where
sj
is a random variable
If S (x) is the distribution
n
service times the distribution function for
with negative exponential distribution.
function of the sum of
'Waiting times,
n
VlN(X), is given for
:~
>
0
by
We have, easily,
Therefore, for
~ ~ ~ ,
( ~)N
• ["1 + ~x + ••• + Nr-
11
~
(-)
N+l
~
( x)NJ
+ ~tX + ••• + ~!
_7
Let ,us denote the mean lJaiting time for a 'Waiting room of size
~(w).
~('W)
We have then that
for
~ ~ ~ •
N by
60
and, for A = ~
,
.
~(w)
1
=
In the case A <
• (N+l)
2~
~
and N = ~, When the
li~miting
distribution exists,
we have
E (w)
~
It
'WaS
=
1
as before.
proved earlier that if we have
( i ) an infinite vlaiting room,
(ii)
(iii)
a Poisson input process, parameter
a distribution function B(x)
possesses a continuous
(iv)
A
A,
for service time which
2nd order derivative,
EL!!.7 < 1
then
a)
The queue size left by a departing customer is independent
(in the limit) of the duration of the interval since the
previous departure.
b)
Two successive departure intervals are independent (in the
limit) •
It is of interest to see that these two properties do not hold when we
have a 'Waiting room of size N < ~ and A E£~7 <~.
For, we proved
in 2.5 that the queue size is independent of the interdeparture intervalif, and only if,
B(t)
=1
_ e-At!(l-P(O»
,
61
and we have just shown (4.2.5), that,
where
p
= ~/~ = ~
plies that
Thus, for
E~~7
N<
00,
1 - P
N+l
1 - p
=
p(o)
E~~7
= (1
1~1en
N<
00,
,
is the traffic intensity.
- p(o»/~, i.e. that p(o)
But (4.2.9) im-
= l-~
E~~7
=1
- p.
it cannot be true that the queue size is independent
of the interdeparture inte:rval.
We also showed that (b) is the case only when N
service time distribution 'Was negative e::ponential.
not hold when N <
00
=
00
and the
Thus, (b) does
•
To sum up, when we have a 'Waiting room of size N < 00, the queue
size is not independent of the duration of the interval since the
previous departure, nor are two successive departure intervals independent.
4.3. Several special cases.
Statement of the cases.
G. C. Hunt
£9_7
conl!iders a sequence of single-server queues 1nth
exponential service tiraes and four different
t~lpes
of 'Waiting room
where,
(i)
(ii)
infinite queues are allowed in front of each server,
no queues are allowed in front of any server, except the
first which may have an infinite queue,
(iii)
finite queues are allowed in front of each server, 'With the
exception of the first which may have an infinite queue, and
(iv)
no queues and no vacant facilities are allowed, 'With the exception of the queue before the first server which may be
62
infinite, the line moves all at once, as a unit.
Case (i) •
..
This we have already discussed
e~ct;el1sively
and so we will just
reiterate same of the results.
1)
The steady state exists if the mean service rate is greater
than the mean arrival rate.
2)
Assuming Poisson arrivals at the first stage, which with (1)
we recall, ensures Poisson arrivals at all subsequent stages,
ve have that the steady state solution for stage
j
is
given by,
=
P
n
where
p
= A/~j
(1 _ p) pn
is the utilisation factor or traffic intensity,
the mean arrival rate and.
3)
,
~j
the mean service rate of stage
A
j.
We can define the mean number of emits in the system as
00
L
Z
=
Z p.(l - p.r
Z
j n=o
j
Case 2.
n(l - p. )p~
=
JJ
l
J
J
No queues allowed.
For the two stage problem, in this case, we can define the
following time dependent state probabilities.
?l(n)
= pen, 0, 0):
(n-l) units in the initial queue and one
unit in service at the first stage (unless
n
= 0,
which means the system is empty),
P2(n) = pen,
0,
1):
(n-l) units in the initial queue, one unit
in service in the first stage, and one
unit in service in the second (unless n=o,
when we have just one unit, in the second
stage),
P3(n) = pen, 1, 1):
n units in the initial queue, 1 unit in
the first stage with service completed, and
one unit in service in the second stage,
'With the corresponding steady state probabilities,
pen,
1), p(n, 1, 1).
0,
We also define
,
_e
(i
thl~e
= 1,
pen, 0, 0),
new state probabilities,
2, 3) ,
These states are obviously unchanged by nell arrivals.
The only tran-
sitions between these states occur when some customer completes his
service.
The equations for these xi,folloidng Feller ~~7, are
given by
~Xl = -~ Xl + ~2x2 + ~l p(o,o,o) ,
x2
= -(1.!1+~l2)X2 + ~lxl + ~2~~ - ~l p(o,o,o)+~lP(o,o,l),
*3 =
where
-~2 ~ + ~l x2 - ~l p(o,o,l) ,
are the mean service rates of the first and second stages
~l' ~l2
respectively.
Let
~l
=
J.l.
2
for the moment.
Then, for the steady state, 1-1e have:-
- xl + x2 + p(o,o,o)
2x + J~
2
Xl -
-
=0
p(o,o,o) + p(o,o,l)
=0
x2 - J~
=0
As the utilisation increases,
-
p(o,o,l)
p(o,o,o) and P(o,o,l) became smaller
and smaller and eventually can be neglected.
In the limit,
Thus, the fraction of time that the first staGe is free to service
incoming units equals the fraction of time it is unblocked, which, in
the limit, is
2
= "3
_e
Thus, the maximum effective service rate of the first stage is
21J./3 and its maximum possible utilisation is
2/3.
It is obvious that, in the general N-stage system with no queues,
blocking monotonically increases as the ntunber of the stage decreases,
and the blocking is greater in the first stage than in any other.
Hence,
the maximum utilisation of the first staGe is the maximum utilisation
of the system as a whole.
The general steady state equations can be obtained, by the usual
method, from which we can dedtlce the mean number of units in the systern, which is used for comparison purposes, as
00
L
=
L:
n=o
l-n
PI(n) + (n+l) P2(n) + (n+2) P3(n)
_7
Case 3.
Finite queues allowed.
If we have a 'WaitinG room of size N in front of the second
stage, by defining N+3
state probabilities as above, we obtain
by a similar method the maximum possible utilisation for different
service rates
Ill'
P max
as
~~2
=
(N+2
N+2
11 2 ~~l
- 112 )
N+3
N+3
III
- 112
,
which reduces to
p
max =
N + 2
If + 3
,
for the same service rate in both stages.
Note tllat
P max ->
112
III
as
N -->
00,
for
III > 11
2
,
as is well known.
Case
4.
No queues, no vacancies.
This case can be treated exactly as the two previous ones.
How-
ever, there is an easier method.
For the steaa;y- state operation, the mean throughput rate must
be greater than the mean arrival rate.
Thus, if TN is the mean
throughput time for N stages, the maJ:im1.1!ll possible utilisation is
.
'gJ.ven
by
p max = ( Il TN )-1 •
Now,. let
~(t)
be the probability
that service in all N stages has been completed in a time
the line moves.
Then,
t
after
66
We must now find an ex;pression for
R2(t)
P'N(t).
For two stages,
is equal to the probability that service is completed in the
first stage in a time less than t
multiplied by the probability
that the service is completed in the second stage between t
t + dt, plus a similar quantity for the second stage.
and
Thus, the
mean throughput time for two stages 'With equal service rates is
00
21
IJ.
e-lJ.t(l - e-lJ.t) t dt =
3/21J. .
o
We can therefore deduce that the maximum possible utilisation is
p
max
= 2/3 •
HUnt then gives a table of L for cases
_e
1, 2 and
4
for
two stages 'With equal service rates in which Lease 1 < Lease 2
< Lease 3 for different
p'S, and also a table for the calculated
values of p max for different numbers of stages.
that
He also showed
CHAPTER FIVE
FEEDBACK
5.1. Introduction.
Two
particular cases of J. R. Jackson's
are worthy of special
P. D. Finch
117.
considerat~on and
11f.F network
model
have received such from
TheJr are both cases of 'feedback', namely termi-
nal and single-stage feedback.
Feedback occurs when a customer at same stage in a sequence of
queues completes his service at that stage and then rejoins the queue
at some previous stage or, possibly, the same one.
_e
Let there be m stages, each with a single server.
With ter-
minal feedback, there is feedback only from the m-th server to one of
the other servers with a finite probability p.(j
J
m
= 1, ••. ,
m), where
Z p. and q = 1 - P is the probability (supposed non-zero)
j=l J
of the customer not returning to the system. In the second case,
=
p
that of single stage feedback, the customer, upon completion of his
service at stage
rejoins the queue at that stage with probability
j
p. (not the same p
J
for
as in the terminal feedback) where
j
j
o < p. < 1
-
J
= 1, ... , m. The Pj are supposed to be independent of the
state of the system at the moment of retu::..: n and of the customer who
has just completed service.
In each case, customers are assumed to
arrive at the first stage at random, wtth a mean rate "A..
time of server
j
The service
is further assumed to be negative exponential with
68
mean value
~/~j.
No assumptions are rmade regarding the point oZ
the queue at which a
~eturning
customer joins it as we are interested
only in the number of customers in the different stages.
The constraint is imposed on the sustem that there must be no
110re than N customers present.
A new customer, arriving to finec
N already in the system, departs, never to return.
U
The existence 0_'
limiting distribution for the number of customers present follows
because the process is an irreducible, Ci.?eriodic Markov chain.
5.2.
Terminal feedbaclc.
Write p(nl, ... ,n)
111
n
_e
for the steady state probability of havinG
j, including the one (if any) then being
m
served, subj ect to t he condition En. < N. We then have, by the
j=l J usual method, the steady state equations.
j
customers in stage
Vlriting,
il,
-
= "
la,
(1,
k
these ,.re:-
=
to,
n >
0
=
0
j
n
j
E
j
n
I:
j
n.
j
J
,
,
<N,
=
N,
ll1
°=
- (/\ k + .2: IJ. j
J=1
ti )
j
p(n1 ,·· .,~)+ 'A. €1 p(nl -l, n2 ,·· .,nm)
TIl-l
+
L;
. 1
IJ.
J=
€. P . p(nl ,n2 , ••• ,n. "n.-l,n l,·.·,n 1,n +J.)
j+
J-- J
m- m
J
m J
m-l
IJ. P € p(n1 ,.··,n)+ L; €_IJ..p(n ,···,n. 1,n.+l,nj 1-1,
1
m m ill
m . 1 ...1 J
JJ
+
+
J=
••• ,n )
m
It can be verified oy
c.:L~t· subst:itution,th.3.t 'i;,~1ifi
n
( :~ ) ill
=
where
X =
j
and P
=
m
P ( 0,
set. of
°, ...,
eq\.,1.dt:i.OJlG
0)
/\(1 - P + Pl + P2 + ••• + Pj )/lJ. j (l - p), (j = l,2, ••• m),
m
~
j=l
pee, 0, ••• ,0)
p .•
J
can be determined by normalisation.
The uniqueness of the solution can be demonstrated in e. manner
similar to that of R. R. P. Jackson, explained in section
Indeed, when each P
j
vanishes, each /\/P' j < 1,
and
N
3.3 •
=
00.
Then
this is, exactly, Jackson's solution.
If each
_
as
N -->
,
00,
p(nl , ... ,nm)
x. < 1,
J
p(o, ••• ,o) tends to a finite non-zero limit
but when any x. > 1
J
for any
then p(o, ••• ,o), and hence
fnj~' tends to zero as
N _>
00.
70
5.;.
Single stage feedback.
In this case, the steady state equations are:-
+
f
•• ,
n
m
+ 1) ,
n
E n. < Nand Cl. = 1 - p .•
J J
J
j=l
n
n
m
l
It can be shown, as before, that p(nl,···,nm) =(xl )
••• (xm) X
p (0, ••• ,0), where xj = A/~lj Clj ' is the unique solution to these
where
equations.
p(o, ... ,0)
is again given by normalisation.
As in the previous section, it is necessary and sufficient, for
p(nl, ••• ,n ) to tend to a finite non-zero limit as
m
finity, that each x be less than unity.
j
N tends to in-
CHAPTER SIX
APPLICATIONS
6.1.
Introduction.
It is reasonable to ask,
'~fuat
practical situations can be met
by applications of the theory in this account?"
The answer is that there are many and varied industrial and
mechanical systems which satisfy the linrltations of this theory
or yield to slight variations in the application of the theory.
For instance, Koenigsburg
£1::'7
developed his theory whilst
working on the organisation of coal-mining equipment and personnel;
R. R. P. Jackson
i-Ii.!
was concerned with a stuo.y of the opeI"d.tions
of a machine shop repairing aircraft engines; likewise J. R. Jackson
£12,7 also received his
stimulus whilst studying machine shops, in
research for the British Navy and Burke ~;5.! uses as an eY~mple the
derivation of the optimun! number of sales-clerks and cashiers in a
shop.
Further applications by Koenigsburg
F¥!.7,
Karush
£J.3.7
and
Benson and Gregory £~7 are discussed, at length,and the solution
of two common problems is deduced.
6.2.
The spare and standby machine problem.
Koenigsburg
£1::'7
discusses ,.,hat may be termed "the spare and
standby machine problem" in which we have a closed system with two
stages, one a repair and the other a working stage.
There are
N
machines, of which a maximum of A can be working a.t any time (A
72
operators), and M re1?uirmen.
donly with a mean rate
;~2
BrealtdolmS are assumed to occur ran-
whereupon the machines enter the repair
stage, each repairman having an eJeponential distribution of service
times with mean rate
1J.1" The numbers
N, A, M and
1J. /1J.
2 1
com-
pletely determine the output of the system.
The problem, of course, is just the simplest example of a multiserver cyclic queue.
DenO~ing
by p(n , n ) the steady state probal
2
bility of the state (n , n ), where n
and n
are, respectively,
l
2
2
l
the number of machines in stages one and ti10 (Irepair l and Iservice
and witing l
),
we obtain, by the usual procedure, the steady state
or
•
(N-A
< M -< n)
•
1
'(:)
l{hen N = A we have a special case of N-A < M and the results
are identical to those for the Swedish Machine problem.
We then
have:n
lP(nl,N-nl ) =
(6.2.3)
(N~/(N-nl)~nl~)(f.!2/1J.1) ~(o,N),(nl <1.1) ,
.
" p(n1,N-nl )
=
n -Ii
n
(N~/(N-nl)~M!M 1 )(f.!2/f.!1) 1 p( o,N), (nl~ M).
From these results, it is easy to derive such quantities of interest as mean number of tillits; mean lenGth of the waiting line and
the mean number of units being served, at either stage, or the probabilities that the servers are fully occupied or that machines must
wait at either stage.
Koenigsburg
fl!:!:j
gives tables of these values and also the
utilisations of machines, operators and repair facilities for various
values of N, A, M and
~~2/f.!1
and shows how these can be used to
throw light onto an organisational problem in which five alternatives
for improving the system are considered.
(i)
Adding one repairman,
(ii)
adding one machine,
(iii)
adding one machine and one operator,
(iv)
adding one machine and one repairman,
(v)
6.3.
and
speeding up repair time.
An
inventory problem.
W. Karush applies the methods of this thesis to an inventory
problem in which custOl:ler demand for a given commodity is assumed to
•
74
be Poisson with a mean rate
The constant inventory is
'f...
nand
at any moment this may be split up into an in-stock amount nand
o
an in-replenishment amount n -n. If a customer arrives when
o
n o > 0 then the result is a unit sale and the initiation of replenishment of that unit, but if he arrives when n :::
o
0
then he leaves,
never to return, a lost sale is recorded and no replenishment is initiated.
Successive replenishment times are assumed to be independent and
random.
The replenishment cycle assumed is best explained by a simple
diagram.
J-O---
The
~A(.I(
~-
SOLD
UNf'l
o's
represent stages
Eij
which have mean service times
,0
S"fOCI'(,
l/Ilij'
in the replenishment path
The
r
i
paths are taken with
probabilities
a • Each path consists 01' a finite chain of expoi
nential stages. vlhen an exponential stage is completed, the next
is immediately begun.
The mean replenishment time for path
i
is
thus
1
:::
lJ.il
1
+ -- +
ll i2
and the mean replenishment time
...
l/Il
,
j.l
:::
+
+ ••• +
J
for the whole system is given
by
1
(r ::: 1, 2, .•• J r)
ar
fJ. r
75
Using this model and our past theory for finding the steady state
probabilities of the system, we obtain the following steady state
solution,
"There
,
I
(6.3.1)
C
g
= n - n,
o
P
= :-./fl and
C
is the normalis1ng lIractor given
by:C
+ P + P2/21• + ••• + Pn/n.1
=1
•
In our usual form:p(noIn)
=
,
• p{n/n)
and, in particular, we have
.e
(6.3.2)
p{o/n)
=
l+p+ •• • +pn In!
for the probability of being out of stock,
or the ratio of lost
sales to total demand.
We can also obtain the probability that there are exactly nIl
Ell; n
in stage E ; ••. ; nij
12
12
This is given by the product
replenishments in stage
E ;
ij
q
(6.3.3)
A
C-
r' I
i,j
n ·
a: i J
i
n
ij
1
nij • flij
,
"i'tlere
and summing these products over all sets
r
~
I
yields (6.3.1).
q
=
.
n .. (
~J j
n
Z
i,j
ij
in stage
,
with fixed
q
If, instead of ass'Luning that the mean arrival rate
constant, we assume that it is
A is a
A(n), a fl..u1ction of the stock on
.
0
hand, the formula (6.3.3) is modified by j:ep1acing the factor
by A.(n)A.(n-1) ••• A.(n +1)
for
n
= 1, ••• ,n-1
and
1
for
i\q
n = n.
0 0 0
Thus, for.mu1a
(6.'.1)
1
=C
becomes
...
•
,
q
and, of course, requires a modification of the normalising constant
c.
By showing that
L(N) = p(o,n) is a convex function of
u,
Karush goes on to determine the optimum allocation of inventory
dollars among several commodities, each lTi th
D.
replenishment cycle
like the above, so as to minimize the total lost sales for a constant
inventory N.
This, however, entails methods totally irrelevent to the
present discussion and will be omitted.
of the theory presentee. here, he
'WaS
The point is that by means
able to advance to a stage when
the solution of the inventory problem reduced to the standard problem
of
t
distribution of error'e J, which has been treated in the literature.
6.4. A
generalisa!ior:~f_. the machine. inter.:.ference model.
Benson and Gregory [~7 carry the analysis of the cyclic queue
a stage further When they consider a closed loop of queues in a
series, each vnth a single server, the customers of which travel betvreen consecutive service gates with a transit time which is a random variable.
Each server may also have customers from sources outside
77
the system which join the queue and are served with those from the
closed loop, on a first come-first served basis.
On completion of
their service, the customers from outside depart to a destination
outside the system.
Benson and Gregory quote as an example an air-
line, which operates a service betVl6en A and B.
are those for service facilities, Which
ma~T
The two queues
also have to cater to a
certain amount of casual service from other airports.
In the general case, suppose that the circuit has
r
service
gates in series, the service time of which and the transit times
between which are asstuned to have negative exponential distributions;
as have the inter-arrival ttmes of successive customers fram the
'open' or tfrom outside' system.
at gate
i
Suppose further that these times
have means
and l/v.
respectively, where
J.
is the mean transit time fram gate
i
refers to the transit time from gate
open queue has a traffic intensit Jr
Pi
to gate (i+l), and
1"
= vi
lJ.
-1
i
to gate 1.
'
Each
which is assumed
to be less than unity.
Let N be the total number of customers belonging solely to the
closed system (a constant) and denote
line at gate
i,
b~/
Yij
the customer
j-th in
including the one (if any) in service, where
Yi . = 0 if the custouer is from the open system and Yij = 1 if he
J
,.
is from the closed system. Then, if s. is the total number of cusJ.
tomers in transit frou Gate
•
r
L:
i=l
t.
J.
r
+
i
to
g<1te
s.
J.
Z
L:
i=l
j=l
=
!~.
i-:-l, we have
The state of the system may be described at any time by the values
of fl, ••• ,K
together vdth the r sequences Y·l'···'Yi
r
~
si •
The steady state eq,uations can be written dOi'm in the usual
nJalUler, and can be shown to have the solution:p
•••
r
=
C
'if
i=l
m
i
\)i
s.-7:
~
J.
Ki ! J.L i Ai
,
where Y(i, t)
- ni
is the number of customers from the open queue at gate i;
si
n = ~ Yi' is the m:unber of customers from the closed queue at
i
j=l J
gate i. It fol101'1S, because of the distinguishable arrangements
of customers, that
r
p(Kl ,·· •Krim. , ••• ,mri n1 ,·· .,nr )= C
J.
If
(m.+n.)~
:;
i=l mi·
n~i
i·
from which vTe can derive, by summing over al.+ arrangements of
n , .•• ,n ,; K1, ••• ,f , and all values of m , that the proba~ility
2
r
r
i
of exactly u customers from the closed queue at gate 1 is given
by:-
..
(N - u - n 2 - • •• - n r ) ~
79
'In their 1951 paper on machine interference, Benson and Cox
introduced the function
N
F(x, N)
= E
(N-n) ~
n=o
genera1ising this, they have
F(a1 , ••• ,a r ;n)
In fact,
=
n
E ••• E
n1+·· .+nr=o
F(al, ••• ,ar;n) is generated by the identity,
F(al, ••• ,a ;n)=F(a , ••• ,a l;n)+n a F(a , ••• ,a ;n-1),
r
1
rr
1
r
and is symmetrical in the variables
A -I
-1
al, ••• ,a • If we let
r
-r
= A. + ••• + A.
' the average total transit time per
1
r
circuit and write xi = ../1_ (I-li - \l i f 1, the servicing factor for
the closed queue at gate
i
then, we obtain
PI(u) = C j\.-N ;~~ F(X2 ,. •• ,Xr;N-U)
r (N-U)!~
N
Since
E P1(u) = 1, we have
u=o
However,
~ x?(x2 ,···,Xr jN-U)
u=o
=
(N-u) ~
N
""u
"'~1
(N-U)~ ~
v
"'.
N-u
E
u=o (N-u) ~
n
1
E .... E
n2 +·· .+nr=o
n
v
r
• •• -....:r
-1
r
---1
iI ( 1-\11..I-l.)
.1.=1
1.
-1
•
80
x ; N) •
r
Thus,
and, therefore,
N-u)
with corresponding eX'pressions for
The generating f'uncti. on of PI( u)
N
(Z) = E
1
u=o
,
Pi (u) •
is
NI (z xl) u F(xo<::. " " ,:~r
j
N-'ll.)
------- ---••• J
=
... ,
,
X . N)
r'
x ; N)
r
Therefore,
-1
El:3] = xl tF(Xl,···,xr ;
If
L
= f1
+
f2
+ ••• +
a;~l
N)}
f r'
F(X1, x2 ' ••• , xr ' N) •
the total number of customers in
transit, then
ErL7
=N - -"
=
r
-1
) F(X , ••• ,x ; N)
l
r
(
N'
F(:C , ... ,Xrj N)
l
. 1
})I
.~(.:~l' ."_',' ,Xrj lIT-I)
"
Z
~=
,
::~'''c
~
ux.~ F(xl" •• ,X;N),
r
after some reduction.
01
If a
the
r
is the average time required to make a full circuit of
gates and transits 1 then any customer spends a proportion
/'.... -1 a-I
of his til:le in transit.
Thus, as this equals the
fraction of customers in transit,
a
N
==
./\.
-1
E[i!7 ~-l
The rate of arrival of closed system customers at any gate is
!ITa
-1
•
Denote the proba.bility that there are no customers from either
system at gate
.e
1
by
c~
r.
1\
..•
I:
(1+·· .+(~+n2+·· .+nr=N
=
A
C·'
-N
(Nn
-1
i=2
-~.
(
-1 -1 l (
1-1
,i\l-v.Il.)
\F;;:2, ••• ,X;N)\
. 2
~=
~
~
r
a product of the folloving two factors:-
.
i)
The
probabilit~r
that there are no customers from an open
queue with the same input and service :mechanism as the open queue
here.
82
ii)
The probability Pl(o)
the closed system.
that there are no customers from
Alternatively,
Pl(o)
can be interpreted as the
probability that there are no customers froD a closed system with the
same conditions as the one here, neglecting the effect of the open
queue, after reducing the mean rate of
se~~ice
from
~"
~
to
~.-v"
~
~
•
Benson and Gregory point out that, because the distribution of
the sum of k variables, each with the smne negative exponential dis-
X2 , then, if we introduce service gates
2k
service times, these results hold for transit times
tribution, is knoiill to be
with zero
avel~ge
with X~I: distributions and, in particular, for consta.nt transit
times.
Limits of the system.
Benson and Gregory show that as
constant, the queue at stage
" tens~
" ty xi
J.n
which Xi
-1
.
,wnere
~
i
N
tends to infinity with ..../"l
acts like an open queue with traffic
\1 = max
( Xi'
) ( J."=1 , ••• 1)' .
Queues f or
= ~ expand without limit. Also, only at the gate(s) at
which the average service time is a maximum will reducing the service
time affect the number of customers in transit, where the effect 1'7111
be to increase the
nlli,ilie~.
In the limit, the rate of arrival at any
gate is equal to the minimum adjusted
When N tends to infinity and
.I
sel~ice
rate.
1_ tends to zero, i •e. the
average transit times between two consecutive gates tends to infinity,
we find that:i)
A.
-1
r ->
00
The arrival rate of closed queue customers is
A.
r
N, where
in such a way that 71.1' N = c, vThere c < min (~i - vi).
83
ii)
Gate
i
behaves) as far as the closed queue customers
are concerned, as an open queue with traffic intensity
-1
iii)
c ) ~Li
6.5. A consideration of
~ome
J.
for all
is the probability
' where
that there are no cust.omers in stage
-1
c S.
i.
practical problems.
Suppose we have a network subject to one restriction) that
M
L
~m =~.
Then, it is of practical interest to find the optimtM
m=l
distribution of our service facilities in terms of output or inven-
tory, i. e. how can we maximize output or minimize our total inventor:' •
Suppose that we have only one server at each stage; then we have,
from (3.5.2),
p
(n)
m
= l'm(0) (rm/~m)n
We know that, in the steady state, the output from any queue must
be random, the same as the input.
dom, parameter
Thus, for stage
m, output is ran-
r m. HO't'iever, only a proportion (1 - p)
of this
m
output actually leaves the system, the rest goes back into some other
stage of the system.
wi th parameter
Hence,
actua1~
output of the stage is random
M
r m(1-1'm).
which is independent of
Thus, total output = L r (1 - l' )
m=l m
m
!.lm' for all m. Thus, as long as ~ > r
m
m
for all m, ensuring a steady state distribution, the output is unchanged by any arrangement of the
's.
~
m
The mean nunlber of customers in staGe
m, n , is given by,
m
84
n P (n)
m
co
~ nCrm/~m)n ) from (6.5.1) .
n=1
Now .p (0)
m
is given by
~
n=o
p (n)
m
=1
Hence,
r m/IJ.m + (rmI~D1 )
2
p (0) (1 +
m
i.e.
P' (0)
=
ill
1 -
=
+ ...)
1
r D1I~m .
Thus, the mean n1.U1lber of customers :in stage
m
is given by,
00
= (1 - rr/IJ'm)(rm/~m)
nrn
~ nCr At )n-l
m m
n=l
rm
=
rm
Ilm -
Thus, the average number of customers in the system is,
M
~
m=l
rm
--_---rr
Ilm
-TIl
We now attempt to miniLtlze
(6.5.2)
subject to the restriction
M
~.
m=l
~
m
=~ •
Let
F
=
M
L
m=l
r I(l ~_ - r)
m
1.1
m
11
(L !l -~) •
1 Iil=l m
+ A
is a Lagrance multiplier.
Thus, we have, in the usual . . ray,
and
2 f
Thus,
oF
a }lm
=
rm
.e
i.e.
(Pm
-
(Ilm
-r
f )2
m
)2
m
m
for all m, when
0,
=
,
>--1
all m
,
= ,-,I_~l '-·41
1
or
}lm
= rm
/-
+
...,_.
rm
/1.
1
This root is chosen so that
But, we require
M
l::
ID=1
Hence,
1/2
>--1
IJ.
III
IJ.
>
f
m
m
= 1J.1 and so we must have
86
> r , for all
Recall that
~
ill
for all m.
Consequently our optimum
( It
r-
f""m
ill,
ill
= rm +
_
and we have that
>
0
ar~angeillent
~ r)
rl / 2
m m
m=l
M
/
E rl 2
m
m=l
This is the distribution of the
~
m
f
S
which minimizes the
average number of customers in the system, and, consequently, the
inventory.
Let us now consio.er a variation on this problem.
case where
Vie
Consider the
have R servers, each with :{;lean service time
Queue m is assumed to have
for stage m is given by
r
m
servers.
l/~.
The steady state solution
(3.5.2) as
Pm(n) = Pm(0) (rmAtm )n/n~
,
(n
= 0, ••• , r m) ,
n-:i."
P (n)
m
where P (0)
L:
n=o
are deterrlined by the usual normalising equations
m
00
= Pill,( 0) (1'mm
/~ )n/r l r i l l , (n > r ) ,
m
m
--m
P(n)=l,(m=l, ••• ,M).
m
Now,
Hence
P (0)
m
r
=-
r
00
n-r
m
n
L:
(r
/r
p.)n/r
~r
m 7- 1
L: (r /r ~) /n~ +
1,1 m
TIl
m
m m
n=o
n=rm+1
The average nt'lllber in stage m is given by
r
r
00
1n
E nP (n) =P (0)
E nCr /r Il)n/n~ +
n=l m
m
- n=l
m m
Hence, the average totc;l number is,
r
(6.5.3)
m
00
n-r
E nCr /r ~)n/n! + E nCr /r Il)n/r ~r m
M
n=o
m m
r +1 m m
m m _
Z
m
m=l - r m
00
n-r
E (r /r Il)n + E (r /r Il)n/r!r m
m m
r +1 m m
m m
n=o
m
r
Having solved the equations (3.5.1) for the
the arrangement of
state conditions
r
subject to
IS,
m
r m < rmll
r,
we have to find
m
= R and the steady
1-1
E r
m=l m
for m = 1, ••• ,M, (which, incidentally,
assure us once again that the output is
0.
constant, whatever the
arrangement), which miniLuze (6.5.3).
The easiest way to do this calculation is, probably, a graphical
method.
individual average
creasing
r mIl > r m, the
of each stage cun be calculated for in-
such that
starting wi tIl the szrallest r
m
r
m
n~tmbers
and plotted as a straight :I_ine graph.
calculations are wade on a modern
cOrrr.l?utel~,
directly as the output.
r
As each
Or, if the
lle can get the graphs
increases, the average number
m
of customers in its stage will decrease. This decrease will be a
convex function for increasing r • NOll, we consider our starting
m
values on the graphs, i.e. the smallest r 's such that r Il > r ,
m
m
m
and find the graph which has the steepest descent for an increase of
one server.
We then move along this graph to the point which corres-
88
ponds to adding one se:..--ver to the stage and. l'epeat the process, each
time picking the graph llith the steepest line of descent and moving
along it, until our resulting number of oervers is
R, Whereupon
we note our resulting number of servers for each stage.
our OptimtU11 distribution.
to a lot of '\vork.
This is then
Obviously, without a computer, this leads
IIovrever, the work can be kept to-a minimum by first
rand
m
servers only (m = 1, ••• ,M), and noting the diffe~ences. TIlen,
calculating the mean ntuuber of customers in each stage for
rnjl
picking the largest of these differences, in queue
late the average number for
servers is
j
say, we calcu-
r j+2, and so on, until our total number of ."
R.
The following problem frequently occurs both in industry and
.e
in businesses where a service is given, subject to a
~iting
roam
of limited size, e.g. a doctor's surgery or a hairdressers.
We have a service mechanism with a negative exponentially distributed service time
s, which has mean
l/~t.
There is a
~i ting
room of size N before it which is feo. by a random input from
another source with parameter 'f...
If we consio.er the industrial case,
the input and service mechanisms will be tl10 machines.
Now, the
second machine can have its mean service time reduced, but only at
the cost of producing nore defectives, or, the waiting room can be
enlarged, again at a certain expense.
Further, when the waiting
room is full, and another customer arrives, he blocks the first
machine, which again proves costly.
Let the cost of de?ectives per unit o? working time be distrit
buted as
f(x), where
..
is the traffic intensity> and let the
cost of storage sp~ce be distributed as
C(N).
Further, let the
cost per unit of time that the first machine is stopped be
C.
The
problem is to find the optimum working arl'angement in terms of cost.
We have from 4.2 tha.t, for a given
that there will be
N and ~, the probability
N custone rs in the Ylaiting room is
where
'A. n
,
(.- )
~
and, writing
K
r
ex
= 'A./'A.+I_t
n = o,l, ••• ,N-l ,
= ~/'A.+I..L ,
and (3
... )
r l
= (3(cl + ex + +
=
for
2
(3 Cr(l + a+ ex
+
... )
r
= 13
Ct
I-a:
=
a
r
Thus,
'A.
.'A.+fl
,
90
after some reduction.
Thus,
N
(2: )
l.l
Let A/l.l
(as in
= x,
and assume that we have
l.l
> A, so that x < I
4.2). Then,
p
N
=
1 -
2C
Now, if a customer arrives to find the waiting room full, he
blocks the previous machine.
room of size N+l.
This is the same as having a waiting
Thus, the probability that the first machine is
blocked, given N and x, PB(N,x)
.e
(6.5.4)
is given by
I-x
Hence,
= ( I-x )
x
N+lr
I
__ -N+2
= D(x, N)
I-x
say.
Thus, for a given x, as long as we have
c•
> g(N+I) - g(N) ,
our system is improving as we increase N.
x
- -l_xN+3 - 7
91
Consider the
s~)ecia1
= kN,
case g(N)
Then, we have an improvement as long as
where k
is a constant.
D(x,N) > k •
C
We can represent "Ghis on the following graph.
F - - - - - - - - - - - - - - - : ) )(
)(.;/
"Graph to show the
.e
J
iuprovem.ent region r
in the size of the waiting
inth respect to unit increases
room~"
Where the shaded recion is the region in vh1ich an improvement can be
Lade.
vIe also have,
1~J"(1_:~N+2)r.(N+1)_ (N+2)J~7 + xN(x_x2 )(N+2)xN+1
=
(1 _ xN+2 )2
}~!-(N+1) - (N+2)X + }J"+2
=
which is positive in
At
(1 _ xN+2 )
0
7
2
< x < 1 and singular at x
~c = i-../fl' 0X/O~l:':
-i-../fl2
and vTe have:-
=1
•
92
,
=
2)
which is negative in
Thus,
0
x
<x <1 •
as long as
c
>
o f(x)
our system inproves with increasing
!J..
Then we can
Let us consider the special case,
represent our arrangement on the following graph •
.e
,
•
_ _-i-.
0'
"Graph to display the 'i,-,1provement reG:i;.c:::~ .1'.ith respect to cha.nges
•
where the stippled t.'e{tion is the
~im.:Provement
region'.
To find the Optimum arrangement, we find an
are in the improvement reeions of both graphs.
:!:~. II
x
and an
N which
Then, on the first
graph, we find all gre<:;.ter N such that this
improvement region.
For each of these
to find the maximuu
~~
~T' s
'"
is still in the
we check on both graphs
lor which ,re are still in the improvement
region and take the md.ninUlll of these in each case.
particular case ve increase
It
C
until we find that
k
D(x, N) =
and stop there, notinG
Now, for each
k fl
l
our values of x and N.
e 1
c
We calculate our
profits for each of these points and find the maximum.
The corres-
ponding values of :c and N are our optmun arrangement.
S4
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fl7
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'>
')
122.7
,