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A single channel queue with bulk service subject to interruptions
M ~ t e c t r o ¢ Re/~ab~Vol. 29, No. 5, pp. 813-818, 1989. Printed in Great Britain.
0026-271418953.00+ .00 © 1989 Pergamon Press plc
A SINGLE CHANNEL QUEUEWITH BULK SERYICE SUBJECT TO INTERRUPTIONS K. C. HADAN* College of Arts and Science, Department of Mathematics, University of Bahrain, P.O. Box 32038, Manama, Bahrain (Received for publication
20 December 1988)
ABSTRACT This paper studies a single channel queueing system with Poisson arrivals and exponential service in batches of fixed size b ( > 1). However, the service channel is subject to occasional breakdowns occuring randomly in time. Both the operative times and the repair times of the service channel have been assumed to be exponential. The probability generating function of the queue length has been obtained for the steady state. For a particular case b = 1, the steady state solutions as well as the average queue length have been obtained explicitly. Also some known steady state results have been derived in another particular case. KEY WORDS
: Poisson arrivals, bulk service, exponential repair times, the
probability generating function, average queue length, steady state INTRODUCTION Interruptions in the customer service due to occasional breakdowns in the system are a common phenonmenon. This definitely affects not only the system efficiency but also the queue length and, for that matter, the customers' waiting time in the queue. Such systems with random breakdowns have been studied by many authors including Gaver [2], Avi-ltzhak and Naor [3] and Mitrany and Avi-ltzhak [4]. The present author, Madan [5,6] considered two such queueing systems with interrupted service. Both these models studied the time-dependent behaviour of the system. The present model, however, studies the steady state behaviour of a system with bulk service subject to interruptions and explicit steady state solutions and the average queue lengths have been obtained for the system. A brief description of the model is given below : 1. Arrivals occur one by one in Poisson stream with mean arrival rate ~.. (~. > 0) 2. The service of units is rendered in batches of fixed size b ( _> 1) and the service times of successive batches are distributed exponentially wilh mean service time llp. (I~ > O)
3. As soon as the service of a batch is completed, another batch of the size min(b,n), where n (> 0) is the queue length, is taken for service. However, if n = 0 at such an instant then the service channel remains idle till such time when a new unit arrives.
* Department of Mathematics, University of Bahrain (Arabian Gulf) 813
814
K.C. MADAN 4. Each time a breakdown is encountered, the service of the batch is suspended and the service channel is immediately sent for repairs. As soon as its repairs are complete, the service channel instantly resumes service of the same batch of units whose service was suspended. 5. The queue discipline is 'first come, first served'. 6. The operative times of the service channel i.e. the times for which it remains in operative state and its repair times are exponentially distributed with mean operative time 1/~ and the mean repair time 1/11. (~,, ~ > 0)
STEADY STATE EQUATIONS FOR THE SYSTEM Let W n denote the steady state probability thai the service channel is in working state and there are n (n > 0) units in the queue excluding a batch in service; Rn the steady state probability that the service channel is under repairs and there are n (n 0) units in the queue excluding the batch of suspended service units; Q the steady state probability that there is no unit either in the queue or in service i.e. the service channel, though operative, is idle. The system has the following set of steady state equations : (Z+p.+~) W n = ~,Wn. 1 +P. Wn+ b + q R n b (~,+p.+~) W0 = %Q + p. ,~_,Wk + "q R0 k=l (%+~) R n= %Rn. 1 +~,W n (n>0) (%+-q) R 0 = ~[O + W 0]
(n>0)
(1) (2) (3) (4)
(X+~) Q = I~ W 0
(5)
We define the probability generating functions : oo
¢~o
W(~.) = ' ~ , W n e.n ;
R(e.) = ~
n=0
Rno. n
(6)
n-0
Performing the operations octimes equation (2) + ~ Cn+b times equation (1), n=l using equation (6) and simplifying we have b-1 P.(o.b-o-k) Wk+ ocb{ n R (c¢)+%O-IJ,W O} k=O W(a) = {%(1-~)*P-+~} e¢b - I~
(7)
Similarly performing equation (4) + ~, c¢n times equation (3) and using n=0 equation (6) and simplifying we have
~{O.W(c,)}
(8)
R(o¢) = % (I-c¢) + "q Solving equations (7) and (8) simultaneously and simplifying, we have b-1 {~.(t-o.)+-q } { ~ p.(o.b-e.k)Wk+o.b(~.Q-P.W0)} +~,Tl(xbQ k=0
w(~) =
(9)
{ %(l-(z)+q } { [%(1-~z)+p.+~,].b-p. } _~.qo~b
R((z) =
b-1 ~,{ ~., p.(ecb-o~k)Wk*a.b{%Q-P.W0}+{[%(1-(z)*p.+~]ab-~}Q} k=0
{
}{
b
(10)
Single channel queue with bulk service
815
It can be shown by Rouche's Theorem that the denominator in the right hand side of equation (9) has b zeroes inside the unit circle I o¢1 = 1. Let these zeroes be denoted by o¢i, i = 1, 2, 3 .... b. Since W(a) is regular inside the contour, the numerator of
the right hand side of equation (9) must vanish for these zeroes giving b equations in b+l unknowns W k ( k - 0, 1, 2 . . . . b - l ) and Q. However, equation (5) enables us to determine Q in terms of W 0' thus reducing the total number of unknowns to b which can all be determined. Hence W(c¢) and, for that matter, R(c¢) can be completely determined. We would continue with the rest of the analysis for the particular case b = 1. In that case, equation (9) would become W(c~) -
{ ~.(l-c~)+-q } { p,(~.-1 )W0+ot(~.e-,w0) } +~-qo~Q
}{
Using the relation
(11 )
p,W 0 = (~+~) Q from equation (5) and simplifying , equation
(11) can be written as (¢¢-1) {-~.2o.+X(~.+~+~)+~'q } O W(c¢) =
(12)
{ ~.(1-a)+T I } { [Z(1 -a)+p.+F~] a--p, }--~TI O~
NOW the denominator of the right hand side of equation (12) can be factored as (c¢-1){~2c¢2-Z(~+P,+P,+TI)cc+P,(X+11)}
so that the factor (~-1) can be
cancelled from the numerator and the denominator of the right hand side of equation (12) which can then be written as
}e
W(c¢) =
(13)
{ X.2o¢2-Z(;~,+p,+P,+~)(z+p,(;~+'q)} Similarly equation (10) can be simplified to R(o¢) =
(14) { k2 o~2-X(k+p,+~+n) a.+p,(~.+n) }
We will now use the normalising condition W(1) + R(1) + Q = 1
(15)
On substituting the values of Q, W(1), and R(1) from equations (5), (13) and (14) respectively and simplifying equation (15) yields Q = 1- (;VP,)(I+(~/TI)) (1 +(P,/p.)) (1 +(~/'q))
(16)
This is the probability that the service channel, though operative, is idle. We can see from equation (16) that the necessary condition for the steady state to exist is that (X/p,)(l+(P.Jrl) < 1
(17)
In the particular case when the service channel is not subject to breakdowns, we see that on letting ~, --> 0 in equation (16), Q ~ 1 - (;Up.) which agrees with the probability of the idle state of the system M/M/I. Also letting ~ = 0 in equation (17) we see that the steady state condition reduces to ;up. < 1, the steady state condition for the system M/M/I. The probability that the service channel is under repairs irrespective of the number of units in the system is given by R(1) = p,.q_~,.(,~+.q) E,(ll+~,) Q = (~=+p,)(~+.q) ~(ll+~)
(18)
And the probability that the service channel is in working state is given by RR 2g,5-J
W(1) = ~.(~+-q)+c;qj. p,.q_~.(~,+q) Q = (~.+p,)(~,+.q) ~.(~+~)+~,n
(19)
816
K.C. M A D A N Clearly the expression in the right hand side of equation (19) is the utilisatlon factor of the system which can be denoted by p. Thus ~,(~.+-n)+~.q P = (~+p)(F.+~)
(20)
In the particular case when the service channel is not subject to breakdowns, we see that on letting ~, -) 0 in equation system M/M/1
(20),
p -)
which is the utilisation factor of
(;up)
THE GENERAL STEADY STATE SOLUTIOH
The denominator in the right hand side of equation (13) has 2 zeroes given by "1, "2
( ~'+P'+F'+TI)+'~(~'+P'+~+'q) 2 - 4 P'(~'+'rl)
=
2~.
°¢1+ "2 = ~.+p.+~+~q ~. and
such that
(21)
a1"2 = p.(X+11) Z2
(22 )
With these notations now the denominator of the right hand expression of equation (13) can be written as ~.2(._ ctI )(._ a2 )
or
~.2(al _.)( o{2 _.)
Also the numerator of the dght hand side of (13)can be written as ~2(.* -a) Q where
a* =
Z2
, so that equation (13) can be expressed as
W(a) - (a* - . ) ( a I _.)-1 ( "2 _.)-1 Q (7.~'
(1_~-;.),(1
(23)
- ~" - 1 ( , - ~¢¢ -I o
.*
.
"a--"~" (1-~'), 1 1 + (a--'~l÷ (~__)2+ ...} {1
(7.
+ (a--'~2+ ( 7" ) ÷2
, ( # c,.(#
Q
...]
[1 ÷(~
÷(~2
"-}
+( 7- ÷2 ...1+
...
} Q
Finally, equation (23) can be expressed as W(.)
=
ai.2
{ 1 + [ ~ 1 +~2"2- '"~ ; ' J 1 + [ - - ~1- - + - - - ~1-
_
+.2 2 .
+.lt~
1
1
a l 3 + e l 2.-~-'~ + a l "22 1 1
_ '_L(.. ~
!.
1
1
(~+71.2
+ I "23 1
+~-T~ + ~.22 )] .3+ ... } O
On the similar lines as above, we can express R(c¢) in equation (14) as
(2~)
817
Singlechannelqueue withbulk service R(a) = ~- (a** --a) ( a l _ . ) - 1 ( a2 _ a ) - I Q
#,o ~---~::.-{ 1+[~
f
'
. 1 +,.._~__ 1 _ _ ~
+[.1 2 "1a2 +.22 +[
1
+
"1 3
1
c¢12a2
+
1
"1 a22
+
1 1 (~+~1=2
1
a23
--'" (~-7 +~-T~ +.22 )] .3+... }o
(25)
where a** = The solutions W n and Rn for n= 0,1,2 .... can be obtained by picking up the coefficients of various powers of c¢ in the right hand expressions of equations (24) and (25) respectively. Thus the general solutions can be expressed as n n-1 Wn =
"1 °"" 2
{
ZLaln_i.2 i . Z aln_i_l.2i } Q 1
i = 0
X "~"** 1"2
1
(26)
i = 0
n Rn=
I
n-1
' { Z a 1n-i"21 - *a~t Z . 1 n - i -1l . 2 i = 0 i = 0
I. } Q
(27)
The solutions given in equations (16), (26) and (27) can be seen for at least n = O, 1, 2 etc to be satisfying the system's equations (1) through (5) for b = 1. It may be noted that the use of equations (22) would facilitate the verifications***. A PARTICULAR CASE If we suppose that the service channel is not subject to breakdowns, then letting ~---0 in equation (21) we have X+TI" =c¢* and "2 I1 at = X =X With these values of "1 and "2 and ~=0 equations (16) and (27) yield Q=I-~"X
and
Rn = 0
(28)
and equation (26) would give X n+l X Wn = ( ~ ' ) (1 - ~ ' ) n=0,1,2 .... (29) The result given in equation (29) agrees with thai of the corresponding M/M/1 system except for notations.
THE AVERAGEQUEUELENGTHS Let L1 and L2 respectively denote the average queue lengths in the steady state when the service channel is busy and when it is under repairs. Then LI=
Id-"~
W(-)l .=
1
(~.2/It2) {(1 +(~,))2 +(~/,)((it~)_~)} ÷~,it { (.~)+(1/It)(1 .(~,)) } _
(30)
(1 +(~p.)) (1 +(~./TI)){ 1- (Z/It)( 1+(~/q)) }
*** The author has verified the solutions for a few initial values of n
K.C. M,~AN
818
L2-
Id
(;~.2~,/p`2,1)(s-(p/,~))-(~.~2/,2~)+(~.~,/,~2)(1+(~,/p`))2 (31)
(1 +(rip`)) (1 +(~/'q)) { I -(ZAP`)(1+(~/'q)) }
•(n+(,•/'q))
< 1
Let L denote the average queue length under the steady state, irrespective of whether the service channel is busy or under repairs. Then L=
LI + L2
(Z 21p`2)( l+(~/.q))2+(;L~/p`)((ll.rl)+(llp`))+(~.~/,q2)(1
+(~/p`))2
=
(32)
(1 +(~/p.)) (1 +(,~/'q)) { 1-(2,./p`)(1 +(F..j-q)) } In the particular case when the service channel is not subject to failures, we have on letting ~ = 0 in equation (32), ~ 2/p.2 L = 1-(;vp`)
'
( ~'/ p`) < 1
(33)
The mean queue length in equation (33) differs from that in the classical M/M/1 system because the definition of queue length does not include the unit in service. To do so we define L" to be the mean queue length including a unit in service, then L" = L+ W(1) (34) where W(1) , the proportion of time the service channel remains busy can be obtained from equation (19) with ~,= 0. Thus W(1) = Z/p` and therefore, equation (34) becomes L" =
~2/p2 +Zip. 1-(Z/p.)
(35)
which simplifies to L" =
~./11 1-(Z/p`)
(36)
Equation (36) agrees with the corresponding result in the M/M/1 system. REFERENCES 1. Saaty, T. L. (1961). "Elements of Queueing Theory", Mc-Graw Hill Book Company, New York. 2. Gaver, D. P. (1962). 'A Waiting Line with Interrupted Service Including Priorities', J. Roy. Statist. B, 24, 73-90. 3. Avi-ltzhak, B and Naor, P. (1963). 'Some Queueing Problems with the Service Station Subject to Breakdowns', Opns. Res., 11,303-320. 4. Mitrany, I. L. and Avi-ltzhak, B. (1968). 'A many Server Queue with Service Interruptions', Opns. res., 16, 628-638. 5. Madan, K.C. (1973). 'A Priority Queueing System with Service Interruptions', Statistica Neerlandica,27, NR 3, 115-123 6. (1976). 'Interrupted Service Queueing with Arrivals and Departures in Batches of Variable Size', Math. Operationsforsch. U. Statist., Heft, S. 139149. 7. Chaudhry, M. L. and Templeton, J. G. C.(1983). ' A First Course in Bulk Queues', Wiley InterScience, U. KJU. S. A.
0026-271418953.00+ .00 © 1989 Pergamon Press plc
A SINGLE CHANNEL QUEUEWITH BULK SERYICE SUBJECT TO INTERRUPTIONS K. C. HADAN* College of Arts and Science, Department of Mathematics, University of Bahrain, P.O. Box 32038, Manama, Bahrain (Received for publication
20 December 1988)
ABSTRACT This paper studies a single channel queueing system with Poisson arrivals and exponential service in batches of fixed size b ( > 1). However, the service channel is subject to occasional breakdowns occuring randomly in time. Both the operative times and the repair times of the service channel have been assumed to be exponential. The probability generating function of the queue length has been obtained for the steady state. For a particular case b = 1, the steady state solutions as well as the average queue length have been obtained explicitly. Also some known steady state results have been derived in another particular case. KEY WORDS
: Poisson arrivals, bulk service, exponential repair times, the
probability generating function, average queue length, steady state INTRODUCTION Interruptions in the customer service due to occasional breakdowns in the system are a common phenonmenon. This definitely affects not only the system efficiency but also the queue length and, for that matter, the customers' waiting time in the queue. Such systems with random breakdowns have been studied by many authors including Gaver [2], Avi-ltzhak and Naor [3] and Mitrany and Avi-ltzhak [4]. The present author, Madan [5,6] considered two such queueing systems with interrupted service. Both these models studied the time-dependent behaviour of the system. The present model, however, studies the steady state behaviour of a system with bulk service subject to interruptions and explicit steady state solutions and the average queue lengths have been obtained for the system. A brief description of the model is given below : 1. Arrivals occur one by one in Poisson stream with mean arrival rate ~.. (~. > 0) 2. The service of units is rendered in batches of fixed size b ( _> 1) and the service times of successive batches are distributed exponentially wilh mean service time llp. (I~ > O)
3. As soon as the service of a batch is completed, another batch of the size min(b,n), where n (> 0) is the queue length, is taken for service. However, if n = 0 at such an instant then the service channel remains idle till such time when a new unit arrives.
* Department of Mathematics, University of Bahrain (Arabian Gulf) 813
814
K.C. MADAN 4. Each time a breakdown is encountered, the service of the batch is suspended and the service channel is immediately sent for repairs. As soon as its repairs are complete, the service channel instantly resumes service of the same batch of units whose service was suspended. 5. The queue discipline is 'first come, first served'. 6. The operative times of the service channel i.e. the times for which it remains in operative state and its repair times are exponentially distributed with mean operative time 1/~ and the mean repair time 1/11. (~,, ~ > 0)
STEADY STATE EQUATIONS FOR THE SYSTEM Let W n denote the steady state probability thai the service channel is in working state and there are n (n > 0) units in the queue excluding a batch in service; Rn the steady state probability that the service channel is under repairs and there are n (n 0) units in the queue excluding the batch of suspended service units; Q the steady state probability that there is no unit either in the queue or in service i.e. the service channel, though operative, is idle. The system has the following set of steady state equations : (Z+p.+~) W n = ~,Wn. 1 +P. Wn+ b + q R n b (~,+p.+~) W0 = %Q + p. ,~_,Wk + "q R0 k=l (%+~) R n= %Rn. 1 +~,W n (n>0) (%+-q) R 0 = ~[O + W 0]
(n>0)
(1) (2) (3) (4)
(X+~) Q = I~ W 0
(5)
We define the probability generating functions : oo
¢~o
W(~.) = ' ~ , W n e.n ;
R(e.) = ~
n=0
Rno. n
(6)
n-0
Performing the operations octimes equation (2) + ~ Cn+b times equation (1), n=l using equation (6) and simplifying we have b-1 P.(o.b-o-k) Wk+ ocb{ n R (c¢)+%O-IJ,W O} k=O W(a) = {%(1-~)*P-+~} e¢b - I~
(7)
Similarly performing equation (4) + ~, c¢n times equation (3) and using n=0 equation (6) and simplifying we have
~{O.W(c,)}
(8)
R(o¢) = % (I-c¢) + "q Solving equations (7) and (8) simultaneously and simplifying, we have b-1 {~.(t-o.)+-q } { ~ p.(o.b-e.k)Wk+o.b(~.Q-P.W0)} +~,Tl(xbQ k=0
w(~) =
(9)
{ %(l-(z)+q } { [%(1-~z)+p.+~,].b-p. } _~.qo~b
R((z) =
b-1 ~,{ ~., p.(ecb-o~k)Wk*a.b{%Q-P.W0}+{[%(1-(z)*p.+~]ab-~}Q} k=0
{
}{
b
(10)
Single channel queue with bulk service
815
It can be shown by Rouche's Theorem that the denominator in the right hand side of equation (9) has b zeroes inside the unit circle I o¢1 = 1. Let these zeroes be denoted by o¢i, i = 1, 2, 3 .... b. Since W(a) is regular inside the contour, the numerator of
the right hand side of equation (9) must vanish for these zeroes giving b equations in b+l unknowns W k ( k - 0, 1, 2 . . . . b - l ) and Q. However, equation (5) enables us to determine Q in terms of W 0' thus reducing the total number of unknowns to b which can all be determined. Hence W(c¢) and, for that matter, R(c¢) can be completely determined. We would continue with the rest of the analysis for the particular case b = 1. In that case, equation (9) would become W(c~) -
{ ~.(l-c~)+-q } { p,(~.-1 )W0+ot(~.e-,w0) } +~-qo~Q
}{
Using the relation
(11 )
p,W 0 = (~+~) Q from equation (5) and simplifying , equation
(11) can be written as (¢¢-1) {-~.2o.+X(~.+~+~)+~'q } O W(c¢) =
(12)
{ ~.(1-a)+T I } { [Z(1 -a)+p.+F~] a--p, }--~TI O~
NOW the denominator of the right hand side of equation (12) can be factored as (c¢-1){~2c¢2-Z(~+P,+P,+TI)cc+P,(X+11)}
so that the factor (~-1) can be
cancelled from the numerator and the denominator of the right hand side of equation (12) which can then be written as
}e
W(c¢) =
(13)
{ X.2o¢2-Z(;~,+p,+P,+~)(z+p,(;~+'q)} Similarly equation (10) can be simplified to R(o¢) =
(14) { k2 o~2-X(k+p,+~+n) a.+p,(~.+n) }
We will now use the normalising condition W(1) + R(1) + Q = 1
(15)
On substituting the values of Q, W(1), and R(1) from equations (5), (13) and (14) respectively and simplifying equation (15) yields Q = 1- (;VP,)(I+(~/TI)) (1 +(P,/p.)) (1 +(~/'q))
(16)
This is the probability that the service channel, though operative, is idle. We can see from equation (16) that the necessary condition for the steady state to exist is that (X/p,)(l+(P.Jrl) < 1
(17)
In the particular case when the service channel is not subject to breakdowns, we see that on letting ~, --> 0 in equation (16), Q ~ 1 - (;Up.) which agrees with the probability of the idle state of the system M/M/I. Also letting ~ = 0 in equation (17) we see that the steady state condition reduces to ;up. < 1, the steady state condition for the system M/M/I. The probability that the service channel is under repairs irrespective of the number of units in the system is given by R(1) = p,.q_~,.(,~+.q) E,(ll+~,) Q = (~=+p,)(~+.q) ~(ll+~)
(18)
And the probability that the service channel is in working state is given by RR 2g,5-J
W(1) = ~.(~+-q)+c;qj. p,.q_~.(~,+q) Q = (~.+p,)(~,+.q) ~.(~+~)+~,n
(19)
816
K.C. M A D A N Clearly the expression in the right hand side of equation (19) is the utilisatlon factor of the system which can be denoted by p. Thus ~,(~.+-n)+~.q P = (~+p)(F.+~)
(20)
In the particular case when the service channel is not subject to breakdowns, we see that on letting ~, -) 0 in equation system M/M/1
(20),
p -)
which is the utilisation factor of
(;up)
THE GENERAL STEADY STATE SOLUTIOH
The denominator in the right hand side of equation (13) has 2 zeroes given by "1, "2
( ~'+P'+F'+TI)+'~(~'+P'+~+'q) 2 - 4 P'(~'+'rl)
=
2~.
°¢1+ "2 = ~.+p.+~+~q ~. and
such that
(21)
a1"2 = p.(X+11) Z2
(22 )
With these notations now the denominator of the right hand expression of equation (13) can be written as ~.2(._ ctI )(._ a2 )
or
~.2(al _.)( o{2 _.)
Also the numerator of the dght hand side of (13)can be written as ~2(.* -a) Q where
a* =
Z2
, so that equation (13) can be expressed as
W(a) - (a* - . ) ( a I _.)-1 ( "2 _.)-1 Q (7.~'
(1_~-;.),(1
(23)
- ~" - 1 ( , - ~¢¢ -I o
.*
.
"a--"~" (1-~'), 1 1 + (a--'~l÷ (~__)2+ ...} {1
(7.
+ (a--'~2+ ( 7" ) ÷2
, ( # c,.(#
Q
...]
[1 ÷(~
÷(~2
"-}
+( 7- ÷2 ...1+
...
} Q
Finally, equation (23) can be expressed as W(.)
=
ai.2
{ 1 + [ ~ 1 +~2"2- '"~ ; ' J 1 + [ - - ~1- - + - - - ~1-
_
+.2 2 .
+.lt~
1
1
a l 3 + e l 2.-~-'~ + a l "22 1 1
_ '_L(.. ~
!.
1
1
(~+71.2
+ I "23 1
+~-T~ + ~.22 )] .3+ ... } O
On the similar lines as above, we can express R(c¢) in equation (14) as
(2~)
817
Singlechannelqueue withbulk service R(a) = ~- (a** --a) ( a l _ . ) - 1 ( a2 _ a ) - I Q
#,o ~---~::.-{ 1+[~
f
'
. 1 +,.._~__ 1 _ _ ~
+[.1 2 "1a2 +.22 +[
1
+
"1 3
1
c¢12a2
+
1
"1 a22
+
1 1 (~+~1=2
1
a23
--'" (~-7 +~-T~ +.22 )] .3+... }o
(25)
where a** = The solutions W n and Rn for n= 0,1,2 .... can be obtained by picking up the coefficients of various powers of c¢ in the right hand expressions of equations (24) and (25) respectively. Thus the general solutions can be expressed as n n-1 Wn =
"1 °"" 2
{
ZLaln_i.2 i . Z aln_i_l.2i } Q 1
i = 0
X "~"** 1"2
1
(26)
i = 0
n Rn=
I
n-1
' { Z a 1n-i"21 - *a~t Z . 1 n - i -1l . 2 i = 0 i = 0
I. } Q
(27)
The solutions given in equations (16), (26) and (27) can be seen for at least n = O, 1, 2 etc to be satisfying the system's equations (1) through (5) for b = 1. It may be noted that the use of equations (22) would facilitate the verifications***. A PARTICULAR CASE If we suppose that the service channel is not subject to breakdowns, then letting ~---0 in equation (21) we have X+TI" =c¢* and "2 I1 at = X =X With these values of "1 and "2 and ~=0 equations (16) and (27) yield Q=I-~"X
and
Rn = 0
(28)
and equation (26) would give X n+l X Wn = ( ~ ' ) (1 - ~ ' ) n=0,1,2 .... (29) The result given in equation (29) agrees with thai of the corresponding M/M/1 system except for notations.
THE AVERAGEQUEUELENGTHS Let L1 and L2 respectively denote the average queue lengths in the steady state when the service channel is busy and when it is under repairs. Then LI=
Id-"~
W(-)l .=
1
(~.2/It2) {(1 +(~,))2 +(~/,)((it~)_~)} ÷~,it { (.~)+(1/It)(1 .(~,)) } _
(30)
(1 +(~p.)) (1 +(~./TI)){ 1- (Z/It)( 1+(~/q)) }
*** The author has verified the solutions for a few initial values of n
K.C. M,~AN
818
L2-
Id
(;~.2~,/p`2,1)(s-(p/,~))-(~.~2/,2~)+(~.~,/,~2)(1+(~,/p`))2 (31)
(1 +(rip`)) (1 +(~/'q)) { I -(ZAP`)(1+(~/'q)) }
•(n+(,•/'q))
< 1
Let L denote the average queue length under the steady state, irrespective of whether the service channel is busy or under repairs. Then L=
LI + L2
(Z 21p`2)( l+(~/.q))2+(;L~/p`)((ll.rl)+(llp`))+(~.~/,q2)(1
+(~/p`))2
=
(32)
(1 +(~/p.)) (1 +(,~/'q)) { 1-(2,./p`)(1 +(F..j-q)) } In the particular case when the service channel is not subject to failures, we have on letting ~ = 0 in equation (32), ~ 2/p.2 L = 1-(;vp`)
'
( ~'/ p`) < 1
(33)
The mean queue length in equation (33) differs from that in the classical M/M/1 system because the definition of queue length does not include the unit in service. To do so we define L" to be the mean queue length including a unit in service, then L" = L+ W(1) (34) where W(1) , the proportion of time the service channel remains busy can be obtained from equation (19) with ~,= 0. Thus W(1) = Z/p` and therefore, equation (34) becomes L" =
~2/p2 +Zip. 1-(Z/p.)
(35)
which simplifies to L" =
~./11 1-(Z/p`)
(36)
Equation (36) agrees with the corresponding result in the M/M/1 system. REFERENCES 1. Saaty, T. L. (1961). "Elements of Queueing Theory", Mc-Graw Hill Book Company, New York. 2. Gaver, D. P. (1962). 'A Waiting Line with Interrupted Service Including Priorities', J. Roy. Statist. B, 24, 73-90. 3. Avi-ltzhak, B and Naor, P. (1963). 'Some Queueing Problems with the Service Station Subject to Breakdowns', Opns. Res., 11,303-320. 4. Mitrany, I. L. and Avi-ltzhak, B. (1968). 'A many Server Queue with Service Interruptions', Opns. res., 16, 628-638. 5. Madan, K.C. (1973). 'A Priority Queueing System with Service Interruptions', Statistica Neerlandica,27, NR 3, 115-123 6. (1976). 'Interrupted Service Queueing with Arrivals and Departures in Batches of Variable Size', Math. Operationsforsch. U. Statist., Heft, S. 139149. 7. Chaudhry, M. L. and Templeton, J. G. C.(1983). ' A First Course in Bulk Queues', Wiley InterScience, U. KJU. S. A.