Abstract— We study an M/G/1 retrial queue with negative arrivals and repeated attempts. This model is motivated by several practical applications. In multiprocessor computer systems, negative arrivals represent commands to delete some transactions. In Neural networks, primary and negative arrivals represent excitatory and inhibitory signals respectively. Such models can be used in relation with some problems of virus infection.
We obtain the generating function of the number of primary customers in the system in stationary regime.
Index Terms— Reliability, Security, Retrial Queues, Negative Arrivals.
I. INTRODUCTION
Consider an M/G/1 retrial queue with unreliable server and two types of arrivals, regular and negative.
Regular arrivals correspond to primary customers who join the system with the intention of getting served and then leaving the system. They are treated in the normal way: (i) If the server is free and available, an arriving primary customer (regular) begins to be served and leaves the system after service completion (if no breakdowns had occurred during his service time).
(ii) If the server is busy or out of order, the arriving customer joins the orbit and seeks service again at subsequent epochs until he finds the server free and available.
A negative arrival has the effect of remaining a customer from the orbit, if customers are present. This model is motivated by several applications. Negative arrivals can represent commands to delete some transactions as in distributed computer systems or databases, in which some operations become impossible because of locking of data or because of inconsistency. Negative and positive customers may also represent inhibitory and excitatory signals, respectively, in mathematical models of Neural Networks, while queue length represents the input potential to a neuron.
Manuscript received on March 24, 2010. This work was supported in part by the Ministery of Higher Education and Scientific Research under Grant B00220060089.
The author is with the University of Science & Technology Houari Boumediene (USTHB), Bab-Ez-Zouar, BP 32 Ela Alia, 16111, Algeria (e-mail: aaissani@ usthb.dz).
In this paper we consider an M/G1 retrial queueing model with negative arrivals when the server is subject to random breakdowns and repairs. In section 2 we give the mathematical formulation of the model. In section 3, we study the model with constant retrial policy without breakdowns. In section 4, we give an extension to the model with breakdowns and repairs.
II. MATHEMATICALMODEL
We assume that arrival times form two independent homogeneous Poisson streams with rate 0 and 0, corresponding to primary (regular) and negative arrivals respectively.
The constant retrial policy for access to the server from the orbit can be described as follows. If the orbit is not empty at time t and the next attempt finds the server free, then a random customer (or the customer at the head of the orbit) is chosen to occupy the server after an exponentially distributed amount of time with rate 0.
The source of customers sends negative signals which have the effect of deleting one customer of the orbit (if any), who is selected according to some specified killing strategy. The service time of primary or secondary orders is a nonnegative random variable with distribution functionH(x), H(0)0 and Laplace-Stieltjes transformh(s),Re(s)0. Denote by
k
h the kth order moment of the service time, k1. Let D0(x) be the probability of a server breaking down during the interval (t,tx) given that it is idle at time
t
and no arrivals (primary or secondary) during the period x. Similarly, let D1(x) be the probability of a server breaking down during (t,tx) given that it is rendering service at time t. In this note we assume that ( ) 1 0 ,0
x
e x
D
) (
1 x
D 1e1x,x0
After the occurrence of a breakdown, a random renewal period begins in which the service is interrupted. We denote by R0(x) and R1(x) the distribution functions of the
corresponding stationary renewal times, with Laplace-Stieltjes transformsr0(s) andr1(s),Re(s)0 and first order momentsri,i0,1. When a breakdown occurs during the service of a certain customer, then this customer enters orbit with probability 1pq or leaves the system without completion of the service with probabilityp.
An M/G/1 Retrial Queue with Negative Arrivals
and Unreliable Server
III. DISTRIBUTION OF THE ORBIT SIZE
In this section we assume that the server is absolutely reliable, so that 0 0,10.We first derive the joint distribution of the server state and the orbit size, from which we deduce the marginal distribution of the orbit size.
Define the indicator of activity of the server at time
t
: 0) (t
C if the server is free at timet;C(t)1, if it is busy at this time. At time t, let R(t) be the number of customers in orbit and N(t) , be the number of customers in the system. Next, define the continuous random variable (t) as the residual time of the current service at timet, ifC(t)0. Consider now the following random process:
(t){0,R(t)} If C(t)0, (t)
1,R(t);(t)
If C(t)0.Defined on the state-space
0,1 IN, where IN is the set of nonnegative integers and the set of nonnegative real numbers. If the stationary regime exists, we can introduce the steady-state probabilities of the process
t,t0
as follows:. 0 , } ) ( , 0 ) ( { )
(
0
m m t R t C P im l m P
t
C t Rt m t x
P m li x m P
t
() 1, ( ) ; ()
) , (
1
,
m0,x0.B
y considering all possible transitions from state to state over the interval
t,th
,h0, and lettingh0, we derive the system of Kolmogorov forward equations:[(10m)(10m)]P0(m)
( ,0) 0( 1)
1
m P dx
m
dP
, (1)
(1 )] ( , ) [ 0m P1 m x
( ,0) (1 ) ( 1, ) )
, (
1 0 1
1
x m P dx
m dP dx
x m dP
m
) ( ) 1 ( ) , 1 ( ) ( )
(x P0 m P1 m x P0 m H x
H
.
(2) Define the partial generating functions
, ) ( }
0 ) ( ; { )
(
0 0 )
(
0
m m t
R t
m P z t
C z E m li z
Q
) ) ( ; 1 ) ( ; ( )
,
( ()
1 z x limE z C t t x
Q Rt
t
0
1( , ),
m m
x m P
z
which converges at least in the disk z 1. From (1) and (2), we obtain the following system of equation
) 0 ( ) ( ) 0 , ( ) ( )
( 0
1
0 z P
x z Q z Q
z
, (3)
x z Q x
x z Q x z Q z
( , ) ( ,0) )
, ( )
( 1 1 1
) ( ) 0 ( ) , 0 ( ) ( ) ( ) ( )
( 0 1 P0 H x
z x P z x H z Q
z
, (4)
where(z)(z)
;
z
z
( ) ,
(z)z(z);
z
z
( ) ,
(z) (z).
If we introduce the Laplace transform
0 1
1(z,s) e Q (z,x)dx,
f sx z 1, Re(s)0, then
the equation (4) becomes
x z Q s z f z s s
( ,0) )
, ( )) (
( 1 1
). ( ) 0 ( )
( ) ( ) ( ) ( )
( 0 P0 h s
Z s a z s h z Q
z
( ) 0( ) ( ) ( ) ( ) P0(0)h(s).
Z s a z s h z Q
z
(5)
We denote
0 1(0, ).
)
(s e dP x
a sx
By assumptions, f1(z,s)is an analytic function for each 1
,z
z in the domainRe(s)0, so for s(z) the right hand side of equation (5) must be zero and then
( )
. ) 0 () ( ) ( ) ( ) ( ) ( ) 0 , (
0 0 1
z h P z
z a z z h z Q z x
z Q
(6)
Substituting (6) back into (5), we obtain:
) , (
1 z s
f
,) (
) ( ) ( ) ( ) ( ) ( ) (
z s s
s a z a z s h z h z
(7)
where ( ) ( ) 0( ) P0(0) z z Q z
z
.
In view of Tauberian theorems
( , ) 1( , ), 1,
0
1
m sf z s z
li z Q
s Then, we obtain from (7)
) , (
1 z
Q
. )
(
) ( ) 0 ( ) ( ) ( 1 ) 0 ( ) ( )
( 0 0
z
z a a z z h P
z z Q z
Substituting now (6) into (5), we get
) (
0 z
Q
. )( ) ( )
(
) ( )
( ) 0 ( ) ( )
( 0
z h z z
z h z z
P z a z
(9)
Now, denote the generating function of the number of customers in the system in stationary regime by
( )
() 0( ) 1( ,)
z zQ z Q z
E m li z
Q Rt
t
. Then after elementary computations, we obtain
) (
) ( 1 1
) (
z z h z
z
Q
) ( ) ( )
(
) ( )
( ) 0 ( ) ( )
( 0
z h z z
z h z z
P z a z
-
.) (
) ( ) 0 ( )
( 1 ) 0 (
0
z
z a a z z h P
(10) By using the normalization condition Q(1)1 and L’Hospital’s rule, whenever necessary we have
)
, 0( ) 0 ( ) 1 (
1 1 0
0
h h P
a Q
(11)
, 1) 0 ( 1
) 0 (
1 1
1
1 1
0
h h
h
a h h
P
(12)
where h1h'(0) is the mean service time of an arbitrary customer and
a(0)P1(0,)Q1(0,),
1( )h1
. (13)Recall that from (7)
0 1
1(0,s) s e Q (0,x)dx
sf sx
0 e P1(0,x)dx
s sx
0 e limP(C(t) 1,R(t) 0; (t) x)dx
s
t
sx
)
(
s
a
(14)Now, by arguments of Renewal Theory, the conditional limiting distribution of the residual service time provided that the server is busy and the orbit idle is equal to
mP (t) x/C(t) 1,R(t) 0
li t
.
0 0
) ( 1
) ( 1 0
) ( , 1 ) (
0 ) ( , 1 ) ( , ) ( lim
x H
x H
t R t C P
t R t C x t P
x
t
.
(15)
Taking the Laplace transform of (15), we have that
1 1
1
) ( 1 ) , 0 ( ) , 1 ( ) (
sh s h Q
s f s
a . (16) Now, the function Q(z) given by formula (10) is entirely determined.
IV. THE MODEL WITH BREAKDOWNS
We consider now the case when the server is subject to random breakdowns (active and/or passive) and repairs under the assumptions of section 2, (0 0,10).
Define the indicator of the availability of the server at time t : E(t)0, if the server is available at time t
andE(t)1, if it is out of order at this time. As before, let )
(t
R be the number of customers in orbit at timet, N(t) be the number of customers in the system at timet, and C(t)be the indicator of business of the server at this time.
In order to embed R(t) into a Markov process, let us introduce the continuous random variable (t) as follows. IfC(t)0andE(t)0, then (t) is the residual renewal time of the current breakdown at timet. IfC(t)0 and
0 ) (t
E , then (t)is the residual time of the current service at this time.
Consider now the following random Markov process: IfC(t)E(t)0, then (t){C(t),E(t),R(t)}. IfC(t)0 orE(t)0, then
(t)
C(t),E(t),R(t);(t)
. This process is defined on the state-space
0
,
1
0
,
1
IN
,where IN is the set of nonnegative integers and the set of nonnegative real numbers.
If the stationary regime exists, we can introduce the steady-state probabilities of the process
(t);t0
as follows:, } ) ( , 0 ) ( , 0 ) ( { )
(
00 m limP E t C t Rt m
P
t
i,j
0,1,m0,x0,
) , (m x Pij
limP{E(t) i,C(t) j,R(t) m; (t) x}. t
It is not difficult to show that the steady-state probabilitiesare solutions of the following system of ordinary differential equations
[(10m)(10m)0]P00(m)
dx x m dP10( , )
) 1 ( ) , (
00
01
m P dx
x m
dP
(18)
(1 )] ( )
[ 0m P10 m
dx x m dP10( , )
( ,0) (1 0 ) 10( 1, )
10 x m P dx m dP m
P10(m 1,x) 0P00(m)R0(x) 1
), ( ) , ( ) 1 )( ( ) , 1
( 1 1 01 1
01 m R x q qP m R x
P
(19)
(1 ) ] ( )
[ 0m 1 P01 m
dx x m dP01( , )
( ,0) (1 0 ) 01( 1, )
01 x m P dx m dP m
hH(x)P00(m)P01(m1,x)P00(m1)H(x). (20)
Define the partial generating functions:
Q00(z)
} 0 ) ( , 0 ) ( ;
{z () E t C t E
m
li Rt
t
000( )
m m
m P
z , (21)
Qij(z)E
zRt ;E(t)i,C(t) j, (t)x
)
(
( , ),(, ) (0,1),(1,0),
0
m ij m j i x m Pz (22)
From (18)-(20), we obtain the following system of equation
~( ) ( ) ( ,0) ( ,0) ( ) 00(0)
01 10
00 z P
x z Q x z Q z Q z
(23) x x z Q x x z Q x z Q z ( , ) ( , ) ) , ( )
( 10 10 10
0Q00(z)R0(x) 1[(1q)qz]
Q01(z, )R1(x) (z)P10(0,x), (24)
x z Q x x z Q x z Q z
( , ) ( ,0) )
, ( )
( 01 01 01
z
Q00(z)H(x) ( ) 01(0, ) P00(0)H(x),
z x P z (25)
where~(z)0(z), ~(z)1(z) Define now the Laplace transform
0 ( , ) , , 0,1,
) ,
(z s e F z xdx i j
f ij
sx
ij
z 1,Re(s)0 (26) Applying these operators to the system (24)-(25), we obtain
s(z)
f10(z,s)s ( ,0) 0 00( ) 0( )
10 s r z Q x z Q ) ( ) ( ) ( ) , ( )
( 01 1 10
1 p qz Q z r s z a s
(27)
( )) ( , ) (s 1 z f01 z s
s
01( ,0) Q00(z)h(s) z
x z
Q
), ( ) 0 ( ) ( )
( 01 P00 hs z s a z (28)
where
0 (0, ).
)
(s e dP x
a ij
sx ij
Since f10(z,s) is an analytic function in s in the domainRe(s)0, and for s(z) the right hand side of equation (27) vanishes, so
( )
. ) ( )) ( ( ) , ( ] [ ) ( ) ( ) 0 , ( 10 1 01 1 0 00 0 10 z a z z r z Q qz p z r z Q x z Q (29)Similarly, for s1(z), the right hand-side of (28) vanishes, and
~( ) ( )
~( ). ) ( ) 0 , ( 01 00 01 z a z z h z Q z x z Q
~( ), ) 0 (00 h z
P
z
(30) Substituting now (29), (30) in (27), (28) respectively, we obtain
( ( )) ( )
, ) ( ) ( ) ( ) , ( ) ( ) ( ) ( ) ( ) , ( ) ( 10 10 1 1 01 1 00 0 0 0 10 s a z a z s r z r z Q qz p z Q s r z r s z f z s s (31)
~( ) ( )
) ( ) ( ) ( ~ ) ( ) , ( ) ( ~ 01 01 00 01 s a z a z s h z h z Q z s z f z s s
h(s) h
~(z)
P00(0).z
(32) In view of Tauberian theorems, we obtain from (31) and (32):
) , ( ) ( ) ( 1 ) ( ) ( ) ( ) ( 1 ) , ( 01 1 1 00 0 0 10 z Q z z r qz p z Q z z r z Q
, ) ( ) ( 1 ) ( 10 z z a z (33)
1 1 01 00 1 1 01 ) ( ) ( ( 1 ) ( ) ( ) ( ) ( 1 ) , ( z z a z z Q z z h z z Q
). 0 ( ) ( 1 ) ( 00 1 1 P z z h z Substituting now (29), (30) in (23 and taking into account (34), we have:
~( )
( )
. ) ( ~ ) ( ) ( ~ ) ( ) ( ) 0 ( ) ( 0 0 0 01 10 00 00 z r z U z z U z z z U z a z P z Q (35) Note that the function
1 1 1 1 1 ) ( ) ( 1 ) ( ) ( ) ( ) ( z z h z r qz p z h z U (36)has a meaningful interpretation. It is the Laplace transform of the « blocking time » i.e. the elapsed time from the instant at which a customer begins his service until the time at which the server is available for beginning the service of each other customer.
A similar interpretation may be given to the function
. ) ( ) ( 1 ) ( ) ( ) ( ) ( 1 01 1 1 1 01 1 01 z z a z r qz p z a z U (37) Now denote by
) , ( ) , 1 ( ) ( ) ( 10 0100
m P m P m Q m t N P m li P t m
the stationary probability that there are m customers in the system. Passing to the generating function, we have
) , ( ) ( ) ( )
(z Q00 z Q10 z zQ01 z
Q .
Using the obtained relations we have
) ( ) ( ) ( ) ( 1 ) ( ) ( ( 1 ) ( ) ( ) ( 1 ) ( 00 1 1 1 1 0 0 0 z z Q z z h z z z r qz p z z z r z Q
( ) ) ( )) ( ( 1 1 10 qz p z z z a 1 1 01 1 ) ( ) ) ( ( 1 ) ( )) ( ( 1 z z a z z r ). 0 ( ) ( ) ( ( 1 ) ( ) ( ) ( ( 1 00 1 1 1 1 P z z r qz p z z z hz
(38)
By using normalization conditionQ(1)1, we obtain
) 1 ( 1 ) 1 ( ) )( 1 ( 1 ) 0 ( 1 1 1 00 1 1 1 0 00 r h Q h r P (39) From (35), we have
. ) ( ) ( ) )( ( ) ( ) 0 ( ) ( ) ( ) ( )) ( 1 ( ) 1 ( 0 0 1 , 1 00 1 , 1 1 01 00 r U U P U U U Q (40) From (39) and (40), we obtain finally
, ) ( ) ( ) ( ) 1 ( ) ( 1 ) )( 1 ( 1 ) 0 ( 1 1 1 1 1 1 01 1 1 1 0 00 U U A r h U h r P where ri ri'(0) and
0 0
1 ,
1) ( )( ) ( ) ( )
( U r
U
A .
In order to found the functions a10(s) anda01(s), we may
use an argumentation similar to that of section 3. More precisely, the conditional limiting distribution of the residual service time provided that the server is busy and available is equal to
0 0 ) ( 1 ) ( 1 1 ) ( , 0 ) ( / ) ( x H x H t C t E x t P m li x t Taking the Laplace transform, we have that ( ) (1, ) 1 ( ) 10(0, )
1 10 10 Q sh s h s sf s a ,
Similarly, the Laplace transform of the limiting distribution of the residual renewal time provided that the server is down
() / ()1, ()0
mP t x E t C t
li t
is equal to 1 2 1 1 0 2 0 00 0
0 1 ( )
) )( ( ) ( 1 ) 1 ( r s s r qz p h r s s r
Q
.
Then we have
. ) ( 1 ) , 1 ( ) )( ( ) ( 1 ) 1 ( ) , 0 ( ) ( 1 1 01 1 0 0 00 0 0 01 01 r s r Q qz p h r s r Q s Q s a
V. CONCLUSION
We have provided the study of the above defined retrial queue with negative arrivals and unreliable server. It will be interesting to provide a more detailed study by considering sample path properties as in [6,7].
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