TWO QUEUES IN PARALIEL
by
Jeffrey J. Hunter
Department of Statistics
University of North Carolina at Chapel Hill
Institute of Statistics Mimeo Series No. 601
November 1968
This research was supported by the National Science Foundation under Grant
No. GU-2059.
1
StJMMI\Ry
A queueing system consisting of two queues, each with single servers (independent negative exponential service time distributions) and correlated bivariate poisson input is discussed.
The joint stationary probabilities for
the respective numbers of customers in the two queues in equilibrium are found
in the case where we have finite waiting rooms of different sizes for each individual queue.
For the model where the waiting room capacities are unlimited,
we derive a functional relationship for the joint probability generating
function of the equilibrium probabilities.
queues in parallel is also presented.
A survey of pUblished research on
2
1.
Introduction
In the past, considerable attention has been given to the theory of simple
queues.
The easy problems have been solved and the methods that have been de-
veloped are sufficiently powerful to analyze simple mdels e.g. those which
consist of a group of servers with one waiting line and which have special
types of inputs and service mechanisms.
A study of tlle general problem of a
network of queues (an arrangement of queues in parallel and in series) is,
however, very much in its infancy.
To the present time, mst interest has
been centered around the study of queues in series.
In this paper we consider two queues in parallel, each waiting line having
a single server,both ass'llJred to be operating independently with negative exponential service time distributions.
The arrivals to the system are ass'llJred
to have a correlated bivariate Poisson input distribution.
We discuss the
problem of finding the joint equilibrium probabilities for the respective
number of customers in the two queues under the restriction of finite waiting
room capacities and also in the case of unlimited capacities.
Before describing the model in detail, we give a historical background to
the general problem of queues in parallel by presenting (Section 2) a survey
of the published research in this area.
Subsequent sections deal with the ob-
taining of relationships between the equilibrium probabilities, their solution
in the case of finite waiting rooms of different capacities, and a discussion
of the problem when we have unlimited capacities.
are found explicitly for certain special cases.
The equilibrium probabilities
3
2.
Historical Background
The first pUblished attempt at studying two queues in parallel was made
by Haight (1958).
He considered a model whereby arrivals to the system are
asswned to be in the form of a homogeneous Poisson process with paran:eter
A
and that the service time distributions for each queue are independent negative
exponential distributions, with paran:eters
2 respectively.
~
and
~2
for queue 1 and queue
Let Xl (t)
and X (t) denote the lengths at time t of
2
queue 1 and queue 2 respectively. An arrival will join queue 1, if and only
if, at his time of arrival Xl(t).:s X2 (t). If, on the other hand, an arrival
finds Xl(t) > X (t) , he will join queue 2. ThUS, queue 1 has a certain
2
advantage in absorbing arrivals in the equally advantageous case when Xl(t)
X (t)
2
and is called the "near" queue.
Let Pxy(t)
= Pr(Xl(t) = x,
X (t)
2
=
Queue 2 is called the "far" queue.
= y).
Haight is interested in deter-
mining the asymptotic state probabilities i.e. in finding p
xy
= lim p (t).
t ... oo xy
Using the method of differential-difference equations, he obtains relations
between the
p
but is unable to solve for them explicitly.
He also finds
xy
relationships between the marginal probabilities, the mean queue sizes, and
the Pxy.
In the event that we permit the queuers to change queues whenever it
seems advantageous to do so, the formulation is simplified, and Haight finds
explicit expressions.
Hilkins (1960) showed that Haight's results can be extended to the following
more general case.
X (t)
2
If, at the time of arrival of a customer,
Xl (t)
= y, the customer joins queue 1 with prob W(x, y) where
= x and
4
W(x, y)
=
1
x
<
w(x)
x
=y
Y ,
,
Ox> Y •
Wilkins proceeds to show that some of the relations between the
modified and some remain unchanged.
p
xy
are
Needless to say, this generalisation did
not lead to a solution for the equilibrium probabilities.
However, Kingman (1961)) upon making the simplification of symmetry
between the two queues (i.e.,
~. ~
and w(x) =~)
was able to determine
conditions underwhich a state of statistical equilibrium is reached and f'urthermore was able to express the equilibrium probabilities as an infinite sum
of geometric distributions.
Recently, Ghirtis (1966, 1968) considered the imposition of limited
waiting rooms for each queue and made certain modifications to the arrival
pattern.
M and
He assumed that, if queues 1 and 2 had limited waiting rooms of size
N respectively, an arrival to the system joined queue 1 if there was
Vlaiting room and otherwise was assigned to queue 2.
If there was no waiting
room in queue 2, the customer left the system without returning.
Ghirtis was
able to find the marginal distributions and indicated how to obtain the general
solution of the system of simultaneous equations for the equilibrium state
probabilities.
~ = ~.
In his earlier paper, (1966), he considered the case of
The more general case, when
latter paper (1968).
~
f
~,
is considered in his
5
3.
Description
£f the model
The system consists of two queues each having a single server.
The
service time distributions for each queue are independent negative exponential
distributions with
par~ters
'\
and
~
for queue 1 and queue 2 respectively.
Arrivals to each queue are assumed to be governed by a bivariate Poisson process, as follows.
Let A. (t)
~
Then
= Number
Pr{Al (t)
= m,
of arrivals in the i-th queue during (O, t].
= n)
~(t)
= e -('J..+~+A)t
~n{m,n)
k=O
{'J..t)m-k {~t)n-k {At)k
(m-k):
(n-k):
k:
This bivariate Poisson process can be characterised in the following
manner.
A.
We have three independent Poisson processes with rates
'J.., \ , and
The input to queue 1 can be regarded as being composed of the sum of Poisson
components, of rates
'J..
and
POisson component of rate
'2
A, while the input to queue 2 consists of the
plus the
SaJOO
Poisson component of rate
A that
lms present in the input to queue 1.
From equation (3 .1), or the above characterisation, it is easily seen that
the marginal inputs are each Poisson,
~
+ A for queue 2.
rate
'J.. + A for queue 1, and rate
Furthermore, it can be shown that the covariance between
the inputs to each queue is
A.
Haight (1967) presents a collection of useful results and applications
pertaining to this process.
6
One of the basic motivations behind this model, excluding any considerations
of its application, was to devise a queueing model where
service facilities
are correlated in some specific manner.
the inputs to the
It will be seen
in the course of the presentation of ruy results that there remain many unsolved
problems relating to this model.
I have not attempted to carry out any inves-
tigation to determine the transient solutions.
Concerning applications of this model, one could visualize the following
type of situation.
failure.
Consider two processes, each with components SUbject to
Items that fail in any specific process are serviced by an operator
assigned to that process.
First~,
secon~,
The failures of items can be of two possible types.
failures may occur at random in each process
independent~,
and
failures may occur according to some lbisson process in both pro-
cesses sinnlltaneous~ (e.g., a failure in both systems due to a power failure).
7
4.
~
steady state equations
We first consider our queueing model under the restriction of limited
waiting room capacities.
ASSUIIVa that queue 1 has a rna.ximum capacity of M
custorers and that queue 2 has a maximum capacity of N custorers.
In the
event that an arrival finds his assigned queue at its maxiIIllJn capacity, he
does not join any queue and is lost to the system.
Let Eij (t)
== Event that there are
i
custorers in queue 1 and
custorers in queue 2 at time
j
t.
In setting up the steady state equations, we shall make use of the method
of differential-difference equations.
This method is widely used when we have
a model consisting of poisson components.
of rate
Suppose we have a Poisson process
v, say, generating events; then the probability of one of these
events occurring in the time interval
(t, t+8t]
is
vBt + o(Bt).
IOOdel, we have 5 possible Poisson processes (parareters
In our
\ , \ , /\, ~ and
~2)
each acting independently of each other.
We shall, for the present, assUIIVa that
M 2: 2, N 2: 2 •
Consider the
possible state changes that may occur in our queueing system during the tire
interval
(t,
t+~t].
Suppose
I
:s i :s M-I,
I
:s j :s N-I
then Eij (t+6t)
could have arisen from one of the 6 possibilities (to o(Bt»
(i)
Eij (t)
,·
and in
no
(t, t + M]
arrival., no
and in
one \
(t, t + M]
and in
(t, t +
no
arrival., one
\
(ii)
(iii)
Ei_l'j(t)
Ei , j-l(t)
,·
,·
\
arrival., no
~t]
no
'2
no
'2
no
'2
•
service, no ~2 service,
~
arrival, no A arrival.
service, no 1-'2 service,
~
arrival, no A arrival..
service,
service, no
~
~
arrival, no A arrival.
8
(iv)
(v)
Ei _l , j-l(t)
Ei+l'j(t)
,.
(t, t +
no
arrival, no
Ei'j+l(t)
~t
pij(t)
,.
\
and in
no
(vi)
~t]
and in
\
(t, t +
~t],
arrival, no
ai:d in
(t, t+~t], no
no
arrival, no
\
1\ service, no ~ service,
no
arrival, one
\
A arrival.
I!:L
one
service, no ~ service,
arrival, no A arrival.
\
1\
service, one
arrival, no
\
~
service,
A arrival.
= Pr(Eij(t)} •
By the independence of the Poisson processes involved, we have that
Pij (t+8t)
=
Pij(t) [1 - ( \ + \
~t
+ Pi_l'j(t) [ \
+ A+
~ + ~}~t + o(~t) ]
+ o(Bt) ]
+ Pi'j_l(t) [ \ 8t + o(Bt) ]
+ Pi_l'j_l(t)
·
S ~nce
+ o(Bt) ]
+ Pi+l'j(t)
[~
Bt + o(Bt) ]
+ Pi'j+l(t)
[~
Bt + o(Bt)
J.
p .. (t+Bt) - P.j(t)
~J
~Bt
1·~m
~t...
[A ~t
O
we deduce that
-(\ + ~ + A +
1\ + ~)Pij(t)
+ \Pi_l'j(t) + ~Pi'j_l(t) + ~i_l'j_l(t)
+ ~Pi+l'j(t) + ~i'j+l(t) •
9
For statistical equilibrium,
as
t ...
01
,
pO j
~
=lim
t ....
Poj(t).
~
we obtain a second order difference equation for
dpo (t)
(4.5) below) by setting
~i
=0
for
i,j
P
ij
(equation
•
In a similar manner, considering possible state
(t, t + Bt]
Assuming equilibrium
cl~ges
in the interval
taking on the boundary values of 0, M or
0, N re-
spectively, we obtain tIle following system of difference equations for the
£:P ij) , (0 .:::: i .:::: M, 0 .:::: j .:::: N)
when M ~ 2 , N > 2 •
(4.1)
(4.3)
(":!.+~+A+I1l)Poj = Vo' j-l + ~Plj + '1lPo' j+l ' (1'::::
j .:::: N-l),
(4.4)
(":!.+~+A+~+I1l)Pij = ":!.Pi-l'j + ~i'j-l + ~i-l'j-l + ~Pi+l'j + ~i'j+l '
(1 .:::: i .:::: M-l, 1':::: j
(~+A+J.1:L+I1l)PMj
(":!.+A+I1l)P ON
S N-l)
,
(4.5)
= ":!.PM- 1 , j + (~+A)PM' j-l + ~M-l' j-1 + ~M' j+l '
(1 ~ j S N-l) ,
(4.6)
= ~O'N-1
+ ~P1N '
(4.7)
(":!.+A+~+I1l)PiN = (":!.+A)Pi-l'N + ~i'N-1 + ~i-l'N-1 + ~Pi+1'N '
(1 SiS M-l) ,
If
(4.8)
M = 1 and N > 2 , then equations (4.1), (4.3), (4.4), (4.6), (4.7),
and (4.9) hold.
10
If M >
and
2 and N = 1 , then equations (4.1), (4.2), (4.3), (4.7), (4.8),
(4.9) hold.
If M = 1 and N = 1
, then equations (4.1), (4.3), (4.7) and (4.9)
hold.
In the case of unlimited waiting rooms, the
isfY equation
(4.1), equations (4.2) for i
and equations
(4.5) for
i ~ 1 , j ~ 1 •
~
{Pij}
(i ~ 0, j ~ 0)
1 , equations (4.4) for
satj
>1 ,
II
5.
Joint probability generating fUnction equation
In an attempt to solve equations (4.1) to (4.9), we first obtain a func-
tional equation involving the joint probability generating function of the
equilibrium probabilities.
Preparatory to deriving an expression for
n(sl,s2)
we obtain a set of
difference equations for the following univariate generating functions:
M
1: POj s
1=0 ~
Assume
M2: 2 , N 2: 2.
~~l(S)
~Cflj+l(s)
-
-
i
•
Then
[~(l-S) + ~
+ A+
= ~(~l)poo
-
~(l-t)~o(s)
~sM(l-S)PMb
'
(5.1)
['1. (l-s) + '2 + A + ~ (l-t) + ~] Cflj(S) + [~+AsJt>j_l (s)
=
\.11 (t-l)poj
-
~ sM(l_S )PMj
M
- AS (l-s )PM' j -1 '
(j = 1, ••• , N-l) ,
-
[(~ +A){l-s) + ~ (l-t) + ~J CflN(S)
= ~(i-l)PON
+
(5.2)
[~+Asl4>N_l (s)
- (\+A)sM(l-S)P MN - AsM(l-S)PM'N_l •
(5.3)
Equation (5.1) is obtained from the set of equations (4.1), (4.2),
(l~.3) by llRlltiplying the i-th equation in the set
o~ i
~
M.
by
si
and sununing over
Similarly for equations (5.2) and (5.3) from the sets (4.4), (4.5),
(4.6) and (4.7), (4.8), (4.9) respectively.
12
=
j:CN CPj(Sl)S2
j
•
equation of the set (5.1), (5.2), (5.3) by s2
Thus by multiplying the j-th
j
and summing over 0 ~ j ~ N ,
we obtain upon simplification the following equation.
[~(l-~)+ ~(1-s2) + A(1-sls2) + ~(l~l) + ~(1~2) J~(sl'
=
1 1 M
s2)
j
N
~(1-- )~(0,s2) + ~(l:s )~(sl'O) + (~+~2)sl (l-S l )( E PMj s2 )
sl
2
j=O
In the case of eacll waiting line having unlinrl. ted waiting room equations
co
•
(5.1), (5.2) and (5.3) are modified to give for the cp .(s) = E P.jSl~
J
the
~
i=O
following equations.
~cpl(s)
2CPj+l (s)
J.L
-
[A.t (l-s) + '2
+ A+
~ (l-t) Jepo(s)
-
[A.t (l-s) + '2
+ A+
1\ (l-J.s>
=
With
~(sl'
1\( t-l)CP j (0),
s2)
= .~
.~
~=O J=O
(j =
=
~(·~l)CPo(O),
J
+ J.L2 CPj(s) +
j
=
.~
(5.6)
CPj(sl)s2
+
A(1-sls2)
= ~(1~1)~(0,
j
,
we obtain
J=O
from (5.5) and (5.6) the following functional equation for
[~(l-Sl) + ~(1-s2)
(~+~)CPj-l (s)
1, 2, ••• ) .
Pij sli s2
(5.5)
+
s2) +
~(sl' s2)'
~(1~1) + ~(1~2) ].(Sl'
~(1~2)~(sl'
0) •
s2)
(5.7)
13
In order to solve for tIle equilibrium probabilities
ficient to determine tIle appropriate j.p.g.f.
~(sl'
s2)
(Pij)
it is suf-
satisfying equation
(5.4) in the finite waiting room case; or satisfying equation (5.7) in the
infinite waiting room case.
Concerning the solution of the 'finite' case, we do not attempt to solve
for
~(sl'
Section
8).
s2)
directly but rather use a matrix theoretic approach (see
However, such an approach is not applicable for the 'infinite'
case and a detailed examination of the functional equation
desirable.
Any attempts to find a j.p.g.f.
(5.7) have failed (see Section 10).
~(sl'
s2)
(5.7) above seems
satisfying equation
14
6.
The marginal distributions - Finite waiting
~ ~
We wish to determine the marginal probabilities for the finite waiting
room case (maxinrom capacities of M and N for queue 1 and 2 respectively).
p.j =
Define
=
Pi·
Putting
C{Jj(l) = P.j
s =
M
I:
'=0
~N
I:
j=O
p. j
,
(j
= 0,
1,
••• J
p. j
,
(i
= 0,
1,
... ,
~
~
N)
M)
1 in equations (5.1), (5.2), (5.3) and noting that
we have
~.l - (~ + A)p. O
~.j+l - (~+A+~)P.j + ('~+A)P.j_l
(j
= 1,
- '¥.N + ( \ + A)P. N _I
By defining
P2
=
A+ ~
\l2
=
0,
=
0,
(6.2)
••• , N-l)
=
(6.1)
°.
(6.3)
, these equations can be rewritten as
=
0,
(6.4)
= 0,
P' j +l - (1 + P2 )P. j + P2P' j - l
(j = 1, ••• , N-l) ,
- P ·N +
Solving
P~'P
t:: ·N-l
=
°
(6.6)
(6.5) by standard difference equation techniques and determining
the constants by the boundary conditions we obtain the following expressions
for the P. j •
15
=
(1 - P )
2
P2 j , (j
(1 _ p N+l)
= 0,
1, ••• , N) •
(6.7)
= 0,
1, ••• , M) ;
(6.8)
2
Similarly, it can easily be shown that
=
(1-P1 )
i
---"":"M':":"":"l- P
l
(1 - P +)
'
(i
1
where
These results are to be expected.
Focus attention on queue 1, say.
Then
ire have a single server queue with negative service time distribution, parameter
~,Poisson
process input with parameter
imiting room of maximum capacity M.
\
+ A,
and a finite
The stationary distribution for such
a queueing situation is a truncated geometric distribution, parameter
as is well known.
P ,
l
16
7-
~
marginal distributions - Infinite waiting
~ ~_
To derive the marginal probabilities in this case, we shall use the
functional relationship given by equation
Putting s2 = 1
(5.7).
in this equation gives upon simplification
=
Using the fact that
1C
(1, 1)
= 1, since we have a probability generating
function, we deduce tl1&t
and hence
TllUS
the marginal distribution in geometric, parameter
PI _ Similarly
As for the finite waiting room case, these results for the marginal distributions are well known, Saaty, (1961).
17
8.
Solution for equilibrium probabilities - Finite waiting
~
In this section, we outline a procedure for solving equations
through (4.9).
(4.1)
Rather than attempt to solve for the joint probability gen-
erating fUnction we use a matrix theoretic approach.
M 2: 2 , N 2: 2 •
Asswne
Le t
n.
NJ
I
=
(j - ,
0 .,
[p oj' Pi j ' · • ., p
]j '
M
Define A(x, y, z) , an
(M+l)X (M+l)
••• , N) •
matrix, the values of the elements
depending on x, y, z, by
A(x, y, z)
=
x-z
-z
0
•
-y
X
-z
....
0
-y
X
•
..
•
..
...
•
...
...
...
...
...
...
...
...
...
...
...
....
....
...
...
....
•
...
..
...
•
....
•
0
•
•
0
•
•
•
•
...
•
x
-z
0
-y
X
-z
0
-y
x-y
Note that all the non-zero elements of A(x, y, z) appear on the diagonal
or the off-diagonals.
We define
Such a matrix is known as a "continuant".
18
A
1
(":L+\+"+~+~ \
~".
~
,~,~),
- A
~ -
A ( \ +,,+~ +1l2
~
2
B
1 == A
( ~'~,O
\ -r..
)
,
\
+
r..
~
,
~) ,
•
With these definitions, it is easy to show that equations (4.1), (4.2),
(4.3) can be expressed
by
the matrix equation (8.1) below.
(4.5), (4.6) can be expressed
by
equation (8.2) and
Similarly (4.4),
(4.7), (4.8), (4.9) by
(8.3).
Ao~
= £1 '
A1£j
=
(8.1)
£j+1 + B1£j_1 '
(j
= 1,
... , N-1) ,
(8.2)
(8.3)
~1d
Since A(~, Y1' zl) + A(x2 , Y2' z2) = A(~ + x2, Y1 + Y2, zl + z2) ,
noting that A(l, 0, 0) = I, we have the following relationships
(8.4)
for
~1e
problem now reduces to one of finding
~
is found, expressions for the £j (j
~,
= 1,
since once an expression
••• , N)
are readily deter-
mined by (8.2).
let us define
N
£=
I:12.j
j=O
=
(8.6)
19
are expressed by equation
where the
(6.8) •
nle technique we use to solve the second order difference equation
for the vector £j
(8.2)
is based upon a teChnique used for solving systems of
second order differential equations.
We reduce equation
(8.2) to a first
order difference equation in the following manner.
We define, for
j =
0, ••• , N-l , the
(2M + 2) X 1
vector
~
by
=
Then from equation
~j
wl1ere
D is a
=
(8.2) it is easy to see that
D ~j -1'
(j
= 1,
(2M + 2) X (2M + 2)
D
... , N-l) ;
matrix given by
=
(8.8)
(8.7) we obtain, by recursion,
From equation
j = 1, ••• ,
Let us consider solving for
z
""0
(8.7)
+ ••• + ~-l
=
z
""0
[£ -~ ]
}2, - ~
N-l •
• First note that
(8.10)
•
Also
[~ ~ ]
z
""0
=
[~ ]
(8.11)
20
(8.12)
Tlms adding equations
(8.10), (8.ll),
and
(8.12)
we lave
(8.13)
Using the results of equations
(8.9),
equation
(8.13)
can be rewritten as
(8.14)
where
(8.15)
Let us e:xamine
H in a little IOClre detail.
We shall see that
H is in fact
a singular matrix and hence we are unable to use matrix inversion to solve
(8.14).
However, in the course of our e:xamination
is possible to solve for
~
explicitly.
H
of
H,
we find that it
Let
=
k>l
Also, let
Thus, we first observe that
D =
(8.16)
21
k
Furthermore, since D +1
= Dk
D
= D Dk
, we obtain, for
k~l,
by equating
the respective matrices in the block multiplication,
k+1Dl
=
D
- k 2Bl
= kD3
(8.17)
k+1D2 = kDl + kDI'-l
= kD4
,
(8.18)
D
k+1 3
= - BlkDl + AlkD.3
,
(8.19)
= -kD4Bl
D
DA
D
k+1 4 = k 3 + k 4 l
=
- BlkD2 + AlkD4
(8.20)
By making use of equations (8.17) and (8.18), equation (8.15) can be
split into its component matrices to yield
N-l
IS. = H:3 = I + t kD3 =
k=l
N-l
~ = H4 = I + t
kD4 =
k=l
ThuS we observe that the matrix H is singular.
(8.21)
(8.22)
However, equation (8.14)
becores
(8.23)
Upon simplification, equation (8.23) gives two identical equations,
namely
IS.~ + ~l
=
£.
(8.24)
Using the boundary condition equation (8.1), equation (8.24) reduces to
(8.25)
where
Simplification of
C is possible as follows:
N
C
=
I +
=
I +
=
I +
l:
k=l
{kD1 + kD'Cfo}
using (8.21), (8.22),
N
l:
k=l
{kD1 + kD'Cf1 - kD2}
using (8.4),
N
l:
k=l
{k+1D2 - ~2}
using (8.18),
using (8.16).
Thus
We have not been able to show that
N+1D2
is non-singular for all
M, N.
However, for any simple case considered direct evaluation has shown the determinant of this matrix to be non-zero.
gular for all
x
>0
).
M; (2D2 =
At
A(x, y, z)
Note that
1D2
is non-singular When
Furthermore, computor studies have not shown
for moderate values of
1,1
and
N.
and 2D2
N+1D2
are non-sin-
x - y - z
> 0,
to be singular
Therefore, we may assu.me
(8.26)
Having determined ~, we can now find
From equation (8.9), (1
Sj
~,
(j = 1, ••• , N),
as follows.
~ N-1),
(8.27)
23
Thus, for
j =
1, ••• , N-l ,
l2.j
=
jDl ~ + jD2 1?1
=
(jD l + jD2 Ao ) ~
using (8.1)
=
CjD l + jD2 Al - jD2 ) ~
using
=
(j+lD2 - jD2 ) ~
(8.4)
using (8.18).
Also, from equation (8.27), (taking j = N-l) ,
l4J
Hence, for
j
= 1,
=
N_lD3 ~ + N_lD4 ~l
=
(N_lD3 + N_lD4 A o ) ~
=
C~l + ~2 Ao } ~
=
(N+lD2 - ~2) ~
using (8.1),
using (8.18),
•
2, ••• , N ,
(8.28)
In order that this procedure be useful in solving for the equilibrium
• Using equa2
tions (8.16) through (8.20), we have the following simple iterative procedure.
probabilities, we require an easy technique for obtaining
jD
Lemma. 8.1:
--
and for
j ~
2 ,
j+lD2
=
{jD2 ~ - j_ l D2 Bl '
Al jD2 - Bl j-ID2 •
(8.29)
24
Tile results of this section may be conveniently summarised by the following theorem.
For M ~ 2 , N ~ 2,
Theorem 8.2:
the matrix equations (8.1), (8.2) and
(8.3) have the following solution.
=
'~lere
the
jD2
(j
= 1,
••• , N) ,
are given by equation (8.29).
TL1is method of solving for the equilibrium probabilities in the finite
waiting room case has been programmed for an IBM System 360, model 75
Note that the recursive generation of the
jD
2
co~utor.
enables us to set up a very
efficient computor program.
In any numerical studies it is useful to have checks on computation
details.
Besides the checks provided by the marginal probabilities, we have
the following result that states, that for any given
j+1D2
the sum of the
elements of each column are constant.
Let ~'
Lemma 8.3:
= (1, 1, ••• ,
For
j
~
1), a
1 X (M+l)
0 ,
pJ'+1
2
- p
2
Proof:
First note that
~' A(x, y, z)
and thus
row vector.
=
(x - y - z),r ,
(8.30)
25
Furthermore, from equations (8.29),
Thus the lemma is true for
tion (8.20) is true for
j =
j =
0, 1.
0, 1,
...,
Using induction, we assume that equaThen by equations (8.29)
k •
Actually this is an indirect check on the marginals, since from equation
(8.28)
=
Also, since
.s,'
N+ID2 ~
D
N+I 2 1Zo
i'
~
2
(8.31)
·0
= £'
=
[ \1 -
Thus,
p j P
=
N+l
P2
- P2
P2
N+l
] l'
~
=
~r }2,
=
1 •
,
1 - P2
and from equation (8.31)
P.j
=
[
1 - P
2
N+l
1 - P
2
as already obtained (equation 6.7).
,.
j
= 0,
1, ••• , N •
26
9.
Special cases
In Section 8, we outlined a method for obtaining the equilibrium prob-
abilities in the finite 'faiting room case when M > 2 and N > 2.
In this
section, we look at some special cases when M and N take on specific
values (not necessarily
.w.
~
2) •
M=l,N=l.
For this case, we wish to solve equations (4.1), (4.3), (4.7) and (4.9)
for Pij
(i = 0, 1 ; j = 0, 1).
form as was done in Section 8.
A(x, y, z)
=
These equations can be reduced to a matrix
Let
[ x-z
-z
-y
x-y
].
Then
Ao ~
= ~l '
(9.1)
A2 ~l
= Bl
(9.2)
'Where A ' A , and B
l
o 2
y, z as in Section 8.
= Ao
Defining Al
are determined using the same substitutions for
x,
+ I , as in equation (8.4), we have from (9.1)
(~ - I)~
Thus
~,
= ~l
~~ = ~ + l2.1 = l2..
Hence
~
-1
= Al l2.,
where
po.
~
=
=
Pl·
e
e
(9.3)
1 - p~)
1 - P1
I-PI)
1 - P1
2
1
1 + PI
=
PI
PI
.
1 + PI
27
Let us define
T=A:t
TA
l
1"1.
Il:I.
.~
~
=
T . A:t
--
A:t
~
det A
l
= .!..
~
From (9.3), we can solve for ~
£i = (Pol'
Pll )
Note that when
Tl1en
+~+A+1l:I. +~.
1-1
2
(1 + P )
2
= (poo'
=
~,
plo ) .
From the relation ~ + ~
=~ ,
can then be determined to yield,
A= 0
~]
Poo
=
1
(l+P ) (1+P2 ) [1 +
l
Pol
=
1
(l+Pl ) (1+P2 )
Plo
= (1+P )(1+P ) [Pl·
l
2
Pu
=
[P2 -~] ,
1
(l+P
1
,
f] ,
lh+P:1l [":LP:1 + ~J
·
(two independent queues in parallel)
Pij
is given by
the product of the marginals, as one would expect.
M
M=1,N~2.
In this case, we vnsh to solve equations (4.1), (4.3), (4.4), (4.6), (4.7),
and (4.9).
These equations reduce to the same matrix equations (8.1), (8.2)
and (8.3) as in Section 8 but with the change that
28
A(x, y, z)
=
The ~ = (P Oj ' p{j) , j
z
X [
- z ]
x - y
•
- y
= 0, 1, ••• , N, are thence solved as in Section
8.
As a special case, we solve these equations explicitly for
w':1ere
-1
,
~
= ;D2
£1
=
~D2 - lD2 )
~
=
£-
lD2
=
I
£
~
=
120
- £1
,
[~ ~J
'
,. - '"1.
D
2 2
= A =
1
M = 1, N = 2.
~
~
.-~
,
--'"1.
~
or - ' \
,
~
Also
•
29
Since
£=
=
po·
1
1 + PI
,
PI
1 + PI
Pl.
we solve equation (9.5) to yield
1
FrJm equation (9.6) ,
£1
=
P
2
~
+~
- -\112
we solve for £1
Pol
=
Pll
=
- -11~2
P
2
~
,
~
+-
to yield,
1
2
(1+P l )(P2 + P2 + 1)
1
2
(1+P1)(P2 + P2 + 1)
~
t
- A«(l-P2) + A} ]
2
t
P1 P2 +
I- - 1l2\
A«(1-P2 ) + A} ]
2
or - 1l2
'2
Finally, from (9.7),
Po2
=
P12 =
[ 2
A( <P2 + \) ]
1
P2 - 2
'
2
(1+P 1 )(P2 + P + 1)
or - ~'2
2
1
2
(1+P1 )(P2 + P2 + 1)
[
+ \)
.(2.
P1 P22 + *12
- ~\
J•
•
30
(c)
M~ 2 , N = 1
The equations (4.1), (4.2), (4.3), (4.7), (4.8), and (4.9) that we vlis11
to solve can be expressed in matrix form by equations (9.1) and (9.2) vrlth
A(x, y, z)
the
(M+1) X (M+1)
matrix as defined in Section 8.
Consider determination of the
-1
~
=
Al
£1
=
£ -
£
=
Pij
for the case wllere
£,
~,
where
1
=
po.
Pl·
and
Al
=
T-
-~
~
~
~
-'1.
-T
~
~
-\
0
Thus
det Al
~
0
•
-~
-~
T- '1.
~
2
=
(P2
[T -2 ~\ ] =
+ 1)
~
~
,
M= 2 , N
=1
•
We now use equation (9.8) to find ~
to find
£i = (Pol'
Pi l, P21 )
and also equation (9.9)
P1o' P2o )
yielding:
P00
=
1
2
(PI + PI + 1)(P2 + 1)
P10
=
1
+ PI + 1)(P2 + 1)
(pi
= (poo'
P20
=
1
2
(PI + PI + 1)(P2 + 1)
Pol
=
(pi
1
+ PI + 1)(P + 1)
2
Pi l
=
1
2
(PI + PI + 1)(P2 + 1)
r
1 +
"J."J.)] '
2
or - ~\
h( T +
r1
h(T(l-"J.) - hI ]
~ _ ~\
r
,
•
1 + \)]
i - h(TP
; - '\\
Also
P21 =
1
2
(PI + PI + 1)(P2 + 1)
~P
2 -
h( T +
;
~P P
"J."J.)l
- '\\
,
h(T(1-P1) 1 2 +;
t
2
~~ +
-'\\
\)J
h( T"J. +
2
or - ' \ \
h~. ,
•
32
Note that we could have derived these same results from the earlier soM=1 , N
lution when
\
and
(d)
=2
by interchanging the roles of
1\
and
~;
and
~_
To conclude this section, we investigate the derivation of the equilibM= 2
rium probabilities in the case
To solve for
io' i.l '
Ali.l
=
~ + Blio '
A2~
=
Bli.l'
ie
and
and
where
where
x - z
- z
o
- y
x
- z
o
-y
x-y
=
A(x, y, z)
The solution is given by
rLA2 - B ] -1~ ,
-1
~
= 3D2
~l
= Ao~ ,
ie
=
£ =
l
l
i. - ~ - i.l '
where
1
Po-
£
=
Pl-
P2 -
2
~ + ~ + 1
I:
2
~
Pl+~+l
p2
1
~+Pl+l
-
N= 2 _
33
Now
(-r-I"J.)
2
1
= "2
A1 - Bl
~
2
+ I"J."\ - ~ ~
2
2
Ii.
- "\(2T-I":1.) - ~""
T + 2 Ii.\
ti
- \ (2-r-\) - ~""
4
-r ) that
It can be easily shown (by expanding
A - B
l
1
2
- Ii. (2-r-1i.)
t\
-
~\
(T_\)2 +
~\
-
~P2
is positive, and hence
is non-singular.
~'l?i.I.'
Because of the tedious algebra involved, we do not solve for
~.
,
- Ii. (2T-\)
However, theoretical expressions for the
can be derived :trom equations
(9.13), (9.14)
Pij (i
and
= 0,
1, 2; j
= 0,
or
1, 2)
(9.15).
Before concluding this section, we note that equation
(9.16)
can be
expressed as follows,
~
= -\~
(p~ + P2
+
l)[CT4 - 2~~T ""J
where
0"4
=
(T -
.JI"J.\ - .J~'2)(-r - .JI"J.\
It may be that this form of
t\
apPear in the determinant of N+1D2.
gated.
+
.J~~)(-r
+ .J1i.\ -
.J~\)(-r +.J~ \
+
.J~\).
is an exhibition of some trend that may
This possibility has not been investi-
34
10.
Solution for equilibrium probabilities - Infinite waiting room
As pointed out in Section 5, any attempts to find a joint probabUi ty
generating function
~(sl'
s2)
satisfying equation
(5.7) have failed.
Since the marginal distributions are both geometric (special case of a
negative binomial distribution), we have that
~(sl'
s2)
is the j.p.g.f. of
some bivariate negative binomial distribution.
We attempted to choose the constants
of the form
~(sl'
s2)'
a o' a l , a , a12
2
so that a j.p.g.f.
-1
[ao + als l + ~s2 + ~sls2 ]
would be a suitable candidate for
It was found that, in general, it is impossible to determine such
constants for equations
(5.7) to be satisfied.
Any approach at solving for the equilibrium probabilities using the matrix
t~leoretic
approach as in Section 8 does not appear to give any fruitfUl results.
One may be interested in holding M fixed, and letting N -.
apply the approach of Section 8.
difficulties.
011
,
and
However, in this case, we also run into
Firstly, sup-pose we consider equations. (8.14) and (8.15).
~
would suspect simplification using
k
-1
D = [I - D]
•
However, for this
k=O
to be true, we require
[I - D]
to be non-singular vnlich is not the case.
I
I - D
I
-. '"' . -
=
B
l
f
I
I
- I
•
I - Al
Note that
~' I
= !'
,
!' (-
I)
=
-~'
One
,
35
Thus the column sums of the first
... ,
(1,
and the column sums of the last
Hence
det(I - D) = 0
and
... ,
1, - 1,
(M+l)
... ,
(P2'
(M+l) rows are
- 1) ,
rows are
P2 , - P2'
I - D
... ,
- P2 ) •
is not of full rank.
Secondly, if one wishes to use Theorem 8.2 when
evalute
lim N+1D2
N-+oo
of
ooD
2
-1
•
= ooD2
N -+
00
one has to
,
(if it exists) and then investigate determination
However, we show that if
ooD
2
exists, then it is singular.
Under the assWlI.Ption of the existence of
ooD
2
,
we have from equation
From equation (8.5)
(10.1)
where
Now A = A(x, y, z)
2
Thus
x-l
I - ~
=
=
y+z
y=
•
-y
z
0
y
-y -z
z
0
y
-y -z
"
"
.......
0
"
z
"
"
"
"-
"
"
0
"
"
" "
"
"
y
-y -z
"
0
y
0
" z
-z
,
~
z =1-1
2
36
Let
for each
2 = [dijJ ; i, j = 1, ... , M+1.
ooD
From equation (10.1), we have
i = I, ••• , M+1,
- di1 Y + di2 Y = 0,
(j
The only solution for these equations is
thus
ooD
2
d
ij
= di
= 2,
(j
... , M) ,
= I,
... , M+1)
and
is singular (in fact rank 1 ).
To investigate this result, the computor program (mentioned in Section 8)
was constructed so tllat
N+1D2
was printed out for selected values of
No limiting tendencies apPeared evident.
(M, N).
37
REFERENCES
Ghirtis, G.e., (1966).
A system of two servers with limited waiting
rooms and certain order of visits.
Grece,
L
Bull.. Soci. Math.
Fasc. 2, 80-138.
Ghirtis, G.e., (1968).
A system of two servers with limited waiting
rooms and certain order of visits.
Haight, F .A., (1958).
Two queues in parallel.
Biometrika,
22J
223-228.
Biometrika, ~
401-410.
Haight, F.A., (1967).
Handbook of the Poisson Distribution.
New York:
Wiley.
Kingman, J.F.e., (1961).
Two similar queues in parallel.
Ann. Math.
Stat., ~ 1314-1323.
Saaty, T.L., (1961).
E1elOOnts of Queueing Theory.
New York: McGraw-
Hill.
Wilkins, e.A., (1960).
198-199.
On two queues in parallel.
BiolOOtrika, ~