ABSTRACT. In this paper, single server queue with poisson arrivals, triple stages of service with service interruption & compulsory server vacation is considered.
International Journal of Computer Applications (0975 – 8887) Volume 41– No.7, March 2012
Time dependent solution of a Non-Markovian Queue with Triple stages of service having Compulsory vacation and service interruptions S. Maragatha Sundari
S. Srinivasan
Asst.Prof. Dept.of Mathematics
Prof. & HOD, Dept.ofMathematics
DMI College of Engineering, Chennai.
BSA University, Chennai.
. ABSTRACT In this paper, single server queue with poisson arrivals, triple stages of service with service interruption & compulsory server vacation is considered. After the completion of first stage and second stage of service, the server must provide the third-stage of service. After the completion of each third stage of service, the server will take compulsory vacation. The vacation time is, exponentially distributed.The time dependent probability generating functions have been obtained in terms of Laplace transforms.
Key Words Poisson arrival, probability generating functions, time dependent solution, service interruption, server vacation.
2.MATHEMATICAL DESCRIPTION OF THE MODEL We have the following assumptions:
Customers arrive at a poison process with arrival rate λ. Service time of the three stage follows different general distributions with distribution function Mj (ν),and density function mj(v), j = 1,2,3… Let si (x) dx be the conditional probability of completion of the i th stage of service during (x, x+dx) given that elapsed time is x, So
1.INTRODUCTION In this paper we consider a qucueing system with compulsory server vacation and service interruptions. A qucueing system might be interrupted. Due to this, server may not be able to proceed with the process till the interruption is cleared. In this paper,each arriving customer had to undergo three stages of services provided by a single server. The service time follows general distribution. Once the system interrupts, then the customer whose service is interrupted goes back to the head of the queue where the arrivals are poisson. Many authors have discussed about Two stages of services. In this article, we have developed a three stages of services which will be more advantageous in large scale industries. Performance measures will be improved. Char et al. [3] chang and Takine [4] and Igaki [7] have studied queues with generalized vacations. Vacation queens with C servers have been studied by Tian et al [15] Madan & Marghi [14] have studied batch arrivals having general vacation time. Kulkarni and choi [9] have studied retrial queues, k.c.so [6] formed the optional maintenance policies for single server queue. Vacation queues have been studied extensively by numerous authors Levy and yechiali [10], Doshi [5] and Madan [11], [12], [13] due to their applications in computer network .V.Thangaraj formed a two stage service [2] Aissani and Artalejo [1] made a work on single server retrial qucueing system. Jayewardene and kella [8] have studied M/G/D queues with service interruptions.
Ki (x) =
mi ( x) 1 M i ( x)
, i = 1,2,3
v
mi (ν) =K i (v) exp (-
K
i
(x) dx ), i =1,2,3
0
Vacation time follows general (arbitrary) distribution with distribution function V(s) and the density function ν(s). The conditional probability distribution (x) dx of a completion of a vacation time during (x, x+dx) given that the elapsed vacation time is x, so
(x) =
ν(s) =
v( x) 1 V ( x) s
(s) exp [ - (
(x) dx )] 0
Interruption arrive with mean rate β>o.Let δ be the server rate of attending interruption.
The mean vacation time is
1
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International Journal of Computer Applications (0975 – 8887) Volume 41– No.7, March 2012
3. DEFINITIONS
4. EQUATIONS GOVERNING THE SYSTEM
We define
P n,1 (1) (x,t) = probability that at time ‘t’ there are ‘n’ customers in the queue excluding the one being served, the server is active providing first stage of service and the elapsed service time for this customer is x.
P n,1 (1) (x,t) =
P n,1 (1) (x,t)dx denotes the
0
probability that at time t there are n customers in the queue excluding the one customer in the first stage of service irrespective of the value of x. P n,1(2) (x,t) = probability that at time ‘t’ the server is active providing second stage of service and there are n ( ,0) customers in the queue excluding the one being served and the elapsed service time for this customer is x.
P n,1(2) (x,t) =
P n,1(2) (x,t)dx denotes the
Pn,1(1) (x,t) + P n,1 (1) (x,t)+ ( t x P n,1 (1) (x,t) =
+K1(x)+ γ)
P n-1,1(1) (x,t)___________ (1)
P0,1 (1) (x,t) + P0,1 (1)(x,t)+ ( t x
+K1(x)+ γ)
P0,1 (1) (x,t) = 0. ___________ (2)
P n,1(2) (x,t) + P n,1(2)(x,t)+ ( t x P n,1(2)(x,t) =
x
+K2(x)+ γ)
P n-1,1(2) (x,t). ___________ (3)
P 0,1 (2) (x,t) +
t
P 0,1 (2) (x,t)+(
+K2(x)+ γ)
P 0,1 (2) (x,t) =0 . ___________ (4)
0
probability that at time ‘t’ there are n customers in the queue excluding the one customer in the second stage of service irrespective of the value of x. P n,1(3) (x,t) = probability that at time t, the server is active providing third stage of service & there are n ( ,o) customers in the queue excluding the one customer being served and the elapsed service time foe this customer is x.
P n,1
(3)
(x,t) =
P n,1(3) (x,t) +
P n,1(3) (x,t) =
x
t
P n,1(3) (x,t)+(
+K3(x)+ γ)
P n-1,1(3) (x,t). ___________ (5)
P 0,1 (3) (x,t) +
t
P 0,1 (3) (x,t)+(
+K3(x)+ γ)
P 0,1 (3) (x,t) =0 ___________ (6) P n,1
(3)
(x,t)dx denotes the
0
x
probability that at time t there are n customer in the third stage of service customer in the third stage of service irrespective of the value of x. Vn (x,t) = probability that at time ‘t’ the server is under vacation with elapsed vacation time x and there are n ( ,o) customers waiting in the queue for service.
x = x
Vn(x,t) +
Vn(x,t)+( t
+ γ (x))Vn(x,t)
Vn-1(x,t)n=1,2…….___________ (7)
V0(x,t) +
V0(x,t)+ ( t
+ γ (x))V0(x,t)=0. ________
(8)
Vn (t) =
Vn (x,t)dx denotes the probability that
0
at time ‘t’ there are n customers in the queue and the server is under vacation irrespective of the value of x. Rn (t) = probability that at time ‘t’ the server is inactive due to arrival of service interruption. Q (t) = probability that at time ‘t’ there are no customers in the system and the server is idle but available in the system.
d dt
Rn(t)= - ( +β) Rn(t)+
Rn -1 (t)+
Pn-1(1) (x,t)dx
0
+
Pn-1(2) (x,t)dx +
0
Pn-1(3) (x,t)dx ,
0
n= 1,2…________ (9)
d dt
R0 (t) = - ( +β) Ro(t) ___________ (10)
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International Journal of Computer Applications (0975 – 8887) Volume 41– No.7, March 2012
d dt
1
Q(t) = -
Q (t) + R0(t)β+
V0 (x,t) γ (x)dx____ (11)
X Zn Pn,1(2)(t)
Pq,1 (2)(z,t) =
n0
0 1
X Zn Pn,1(3)(x,t)
Pq,1(3) (x,z,t) =
n0
By the usage of following boundary conditions, the above equations (1-11) to be solved. (1)
P0,1 (0,t) =Q (t)
1
X Zn Pn,1(3)(t)
Pq,1(3) (z,t) =
+ R1 (t) β
n0
+
1
V1(x,t)γ(x)dx___________(12)
X Zn Vn(x,t)
Vq (x,z,t) =
0
n0
Pn,1(1)(0,t) =Rn+1 (t)β +
1
Vn+1(x,t) γ (x) dx
Vq (z,t) =
n0
0
n=1,2…….__________ (13)
1
Rq (z,t) =
Pn,1(2)(0,t) =
X Zn Rn(t) n0
Pn,1(1) (x,t) K1(x) dx, n=0,1……
0
Define the Laplace transform of f(t) as
.________ (14) F(s) =
Pn,1(3)(0,t) =
X Zn Vn(t)
e-st f(t) dt, R (s) >o. __________(19)
0
Pn,1(2) (x,t) K2(x) dx, n=0,1……
0
Taking the Laplace transform of equations (1) to (16) & using (17), we obtain
.________ (15)
Vn (0,t) =
P n,1(3) (x,t) K3(x)dx, n=0,1……________(16)
=λ
0
Assume that initially there is no customer in the system and the server is idle. V0 (0) =0, Vn (0) =0, Q (0) =1 Pjn,1 (0) = 0 for n=0,1,2………….
___________ (17)
j=1,2,3…………. Now we shall find the transient solution for the above equations. First, let us define the probability generating functions.
n0
(z,t) =
X Z Pn,1 (t) n
p( 0,1)
1
X Zn Pn,1(2)(x,t) n0
(1)
(x,s), n = 1,2…….__________(20)
(x,s) + (s + λ + K1 (x) + β )
(x,s)=0_____________(21)
( 2) p x (n,1) =λ
p( n 1,1)
( 2)
(x,s)
( 2)
(x,s) + (s + λ + K2 (x) + β )
p(n,1)
( 2)
(x,s)
(x,s), n = 1,2………__________(22)
(x,s) + (s + λ + K2 (x) + β )
(x,s) = 0. __________(23)
(1)
n0
Pq,1(2) (x,z,t) =
(1)
(1) p( 0,1) x
p( 0,1)
1
Pq,1
p( n 1,1)
( 2) p x (n,1)
1
X Zn Pn,1(1)(x,t)
Pq,1(1) (x,z,t) =
(1)
(1) (1) (x,s) + (s + λ + K1 (x) + β ) p p (n ,1) x (n ,1)
__________ (18)
( 3) ( 3) p(n,1) (x,s) + (s + λ + K3 (x) + β ) p(n,1) (x,s) x ( 3) =λ p( n 1,1) (x,s) n=1,2…..__________(24) x
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International Journal of Computer Applications (0975 – 8887) Volume 41– No.7, March 2012
( 3) p( 0,1) x
p( 0,1)
( 3)
Vn x
(1) p(q ,1) x
(x,s) + (s + λ + K3 (x) + β )
p(q ,1)
(x,s) = 0. __________(25)
Vn
(x,s) + (s + λ + γ (x))
(x,s) = λ
Vn 1
(x,s),
V0
(x,s) + (s + λ + γ (x))
(s+ λ+δ) Rn (s) = λ Rn 1 (s) +
(x,s) = 0.__________(27)
p( n 1,1)
(1)
(x,s)dx+
0
p( n 1,1)
( 2)
(x,s)dx +
0
p( n 1,1)
( 3)
Similar operations are to be carried for the equations (22) to (29)
p(q ,1)
( 2)
( 3)
(x,z,s) + (s + λ - λz +K2 (x) + β )
(x,z,s)= 0. __________(37)
( 3) p(q ,1) x
p(q ,1)
(x,z,s) + (s + λ - λz +K3 (x) + β )
(x,z,s)= 0. __________(38)
(x,s)dx
0
n = 1,2…….__________(28) (s+ λ+δ)
(x,z,s)= 0. __________(36)
( 2) p x (q ,1)
n =1,2……__________(26)
V0 x
(1)
(x,z,s) + (s + λ - λz +K1 (x) + β )
Vq x
(x,z,s) + (s + λ - λz + γ (x))
Vq
(x,z,s)= 0.
__________(39)
R0
(s) = 0__________(29)
(s+ λ)
Q
(s) = 1+
R0
(s) +
V0
(x,s) γ (x)dx __(30)
(0,s)= λ Q (s) +
R1
(s) δ+
+
V1
(x,s) γ (x)dx
p(q ,1)
(1)
(x,z,s)dx
(1)
(z,s) = α z (,
0
0
p( 0,1)
Rq
(s+ λ- λz+ δ
p(q ,1)
( 2)
(x,z,s) dx+
0
p(q ,1)
( 3)
(x,z,s) dx )
0
__________ (40)
0
Multiply both sides of equation (31) by z, multiply both sides of equation (32) by zn+1, sum over n from 1 to ∞, add the two results, then using (18) we get.
__________(31)
p(n ,1)
(1)
(0,s) = Rn 1 (s) δ +
Vn 1
(x,s) γ (x) dx
0
n = 1,2……..__________(32)
p(n,1)
(o,s) =
p(n ,1)
(1)
(x,s) K1 (x) dx
n =0,1…..__________(33)
(o,s) =
p(n,1)
(o,z,s) = λz
( 2)
n =0,1…..__________(34)
Vq
(x,z,s) γ (x) dx –
0
p(n,1)
( 3)
Rq
(z,s) - δ R0 (s) +
V0
(x,s) γ (x) dx.
0
p(q ,1)
( 2)
(o,z,s)=
p(q ,1)
(1)
(x,z,s) K1 (x) dx. (42)
0
(o,s) =
(s)+ δ
(x,s) K2 (x) dx
0
Vn
Q
Performing similar operations on equations (33), (34) and (35), we get
( 3)
(1)
__________(41)
0
p(n,1)
p(q ,1)
( 2)
Z
(x,s) K3 (x) dx
0
n =0,1…..__________(35)
p(q ,1)
( 3)
(o,z,s)=
p(q ,1)
( 2)
(x,z,s) K2 (x) dx.
0
__________ (43)
Multiply equation (20) by zn & summing over n from 1 to ∞, adding to equation 21 & using the generating function, we have
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International Journal of Computer Applications (0975 – 8887) Volume 41– No.7, March 2012
Vq
(o,z,s) =
p(q ,1)
( 3)
(x,z,s) K3 (x) dx. _____ (44)
Vq
(x,z,s)=
0
-
Vq
(0,z,s) exp {- (s+λ-λz+β)x
(t) dt }__________(51)
0
Using equation (30), equation (41) becomes,
p(q ,1)
Z
(1)
(o,z,s) = (1-S Q (s) + λ (z-1)
Q
(s)+ δ Rq (z,s)
+
Vq
(x,z,s) γ (x) dx. __________(45)
Where
p(q ,1)
( 2)
(0,z,s),
p(q ,1)
( 3)
(0,z,s) and
Vq
(0,z,s) are
given by equations (42), (43) and (44). Again integrating (49),(50) and (51) by parts with respect to x, we get
0
p(q ,1)
Integrating equation (36) from o to x, we get
p(q ,1)
(1)
(x,z,s) =
p(q ,1)
(1)
(z,s) =
p(q ,1)
( 2)
(0,z,s) 1 M 2 ( s z ) s z
p(q ,1)
( 3)
(0,z,s) 1 M 3 ( s z ) s z
__________(52) (0,z,s)exp {- (s+λ-λz+β) x
p(q ,1)
-
( 2)
k1(t) dt }
__________(46)
( 3)
(z,s) =
_________(53)
0
Where
p(q ,1)
(1)
(0,z,s)is given by equation (45). Again
Vq
(z,s) =
Vq
(0,z,s) 1 V ( s z ) __________ (54) s z
integrating (46) by parts with respect to x, Where
p(q ,1)
(1)
(z,s) =
p(q ,1)
(1)
(0,z,s) 1 M 1 ( s z ) s z
M2
(s+λ-λz+β) =
M3
Where
(s+λ-λz+β) =
exp ( - (s+λ-λz+ )x) dM1(x)
is the Laplace stieltjes transform of the first stage service time M1 (x). Now multiply both sides of equation (46) by K1 (x) & integrating over x, we obtain
p(q ,1)
(1)
(x,z,s)K1 (x) dx =
p(q ,1)
(1)
(0,z,s)
M1
(s+λ-
0
exp ( - (s+λ-λz+β)x )dM3(x)
_____(56) is the Laplace - stieltjes transform of the second stage service time M2 (x) and third stage service time M3(x). Multiply both sides of equation (50) by K2 (x) and integrating over x & also multiply both sides of equation (51) by K3 (x) and integrating over x, we get
Similary integrating the equations (37) & (39) from 0 to x, we obtain.
p(q ,1)
( 2)
(x,z,s)=
p(q ,1)
( 2)
(x,z,s)K2 (x) dx =
p(q ,1)
( 2)
(0,z,s)
0
M 2 (s+λ-λz+β)__________(57)
λz+β)_______(48)
p(q ,1)
( 2)
p(q ,1)
( 3)
(x,z,s) K3 (x) dx =
p(q ,1)
( 3)
(0,z,s)
0
(0,z,s) exp { - (s+λ-λz+β)x
-
(s+λ-λz+β) =
0
0
exp ( - (s+λ-λz+β)x)dM2(x) ____(55)
0
__________ (47)
M1
M 3 (s+λ-λz+β) __________(58)
K2(t) dt }__________(49)
V
0
(s+ λ-λz) =
exp ( - (s+λ-λz)x)dV (x)
0
p(q ,1)
( 3)
(x,z,s)=
p(q ,1)
( 3)
(0,z,s) exp { - (s+λ-λz+β) x
-
__________(59) is the Laplace stieltjes transform of V (x).
K3(t) dt} __________(50)
0
41
International Journal of Computer Applications (0975 – 8887) Volume 41– No.7, March 2012 Multiply both sides of equation (51) by γ (x) & integrating over x, we get
(x,z,s) (x) dx =
Vq
Using equations (48), (49), (64) & (65) equations (40) becomes,
Vq
(0,z,s)
V
(s+λ- λz)
(s+λ-λz+δ) Rq (z,s) = βz [
0
p(q ,1)
(1)
(x,z,s)dx
0
+ Now using equations (42), (48), (57), (58) we can write equation (44) as,
(0,z,s) =
p(q ,1)
p(q ,1)
( 3)
( 2)
(x,z,s) dx+
0
p(q ,1)
( 3)
(x,z,s) dx ]
0
= βz [
Vq
__________(60)
p(q ,1)
(x,z,s) K3 (x) dx
(1)
(0,z,s) 1 M 1 ( s z ) + s z
0
=
p(q ,1)
( 3)
M3
(0,z,s)
p(q ,1) (s+λ-λz+β)
p(q ,1)
( 2)
(x,z,s) K2 (x) dx [ M 3 (s+λ-λz+β)]
0
=
p(q ,1)
( 2)
(0,z,s)
M2
(s+λ-λz+β)
M3
(0,z,s) 1 M 2 ( s z ) + s z
p(q ,1)
( 3)
(0,z,s)
1 M 3 ( s z ) ] s z
=
( 2)
(s+λ-λz+β)
(1) = βz p q ,1 (0, z, s) [1- M 1 (s+λ-λz+β)]+ s z
M1
λz+β) [1- M 2 (s+λ-λz+β)]+[1- M 3 (s+λ-λz+β)]
(s+λ-
M2
(s+λ-
=
p(q ,1)
(1)
λz+β) [ M 1 (s+λ-λz+β)]
(x,z,s) K1 (x) dx [ M 2 (s+λ-λz+β)]
0
(1) = βz p q ,1 (0, z, s) [1- M 1 (s+λ-λz+β ) s z
[ M 3 (s+λ-λz+β)]
Vq
p(q ,1)
(0,z,s) =
λz+β)
M3
(1)
(0,z,s)
M1
(s+λ-λz+β)
M2
(z,s) =
λz+β)][
z+β) M 2 (s+λ-λz+β) ]
λ
(s+λ-λz+β) __________(61)
p(q ,1)
M3
(1)
(0,z,s) [ M 1 (s+λ-λz+β) ][ M 2 (s+λ-
(s+λ-λz+β)] 1 V ( s z ) _______(62) s z
By using (61), equation (60) becomes,
Rq
(x,z,s) γ (x) dx =
p(q ,1)
(1)
(z,s) =
(1) βz pq ,1 (0, z, s) 1 M 1 (s z )M 2 (s z )M 3 (s z )
s z
s z
(66) Using equations (60) & (66) in eq (45), solving for
Vq
(s+λ-
(s+λ-
Using above equations, equation (54) becomes,
Vq
M3
(0,z,s)[ M 1 (s+λ-λz+β)]
p(q ,1)
(1)
(0,z,s), we get
p(q ,1)
(1)
(0,z,s)
0
[ M 2 (s+λ-λz+β)][
M3
(s+λ-λz+β)][
V
(s+λ-λz] ___(63) =
Now using (48) & (49) equation (42) & (43) reduces to
p(q ,1)
( 2)
(0,z,s) =
p(q ,1)
(1)
[V (s - z)] - z[1- M 1 f1 ( z ) M 2 f1 ( z ) M 3 f1 ( z )]
(0,z,s)
M1
(s+λ-λz+β) __________ (67)
____(64)
p(q ,1)
( 3)
f1 ( z ) f 2 ( z )[1 sQ( s ) ( z 1)Q( s )] f1 ( z ) f 2 ( z )[ z M 1 f1 ( z ) M 2 f1 ( z ) M 3 f1 ( z )]
(0,z,s) =
p(q ,1)
( 2)
(0,z,s)
M2
(s+λ-λz+β)
Where f1 (z) = s+λ-λz + β, f2 (z) = s + λ – λz + δ
___(65)
42
International Journal of Computer Applications (0975 – 8887) Volume 41– No.7, March 2012 Substituting the value of
p(q ,1)
(1)
(0,z,s) in equations (47),
(49), (50), (62) & (66) we get
p(q ,1)
(1)
f 2 ( z )[1 sQ(s) ( z 1)Q(s)][1 M 1 f1 ( z )] DR
(z,s) =
-__________ (68)
p(q ,1)
( 2)
(z,s) =
f 2 ( z )[1 sQ(s) ( z 1)Q(s)][1 M 2 f1 ( z )] DR
__________ (69)
p(q ,1)
( 3)
(z, s) =
f 2 ( z )[1 sQ(s) ( z 1)Q(s)][1 M 3 f1 ( z )] DR
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__________ (70)
[6] A. Federgruen and K.C. So, Optimal maintainence policies for single server queueing systems subject to breakdowns, Operation Research, 38 (1990), 330-343.
Vq (z,s) = f 1 ( z ) f 2 ( z )[1 sQ( s) ( z 1)Q( s)][M 1 f 1 ( z ) M 2 f 1 ( z ) M 3 f 1 ( z )] 1 V ( s z ) s z ) DR
__________ (71)
Rq (z,s)
7.REFERENCES
[7] N. Igaki, Exponential two server queue with N- policy and general vacation, Queueing Systems, 10 (1992), 279294. [8] A. K. Jayawardene and O.Kella, M/G/∞ with altering renewal break-downs, Queueing Systems, 22 (1996),7995.
=
z[1 sQ(s) ( z 1)Q(s)][1 M 1 f1 ( z ) M 2 f1 ( z )M 3 f1 ( z )] DR
[9] V.G. Kulkarni and B.D. Choi, Retrial queue with server subject to brerak-downs and repairs, Queueing Systems, 7 (1990), 191-209.
__________ (72)
[10] Y. Levi and U.Yechilai, An M/M/s queue with server vacations, Infor., 14 (1976), 153-163.
Where DR =
f1 ( z ) f 2 ( z ) z M 1 f1 ( z ) M 2 f1 ( z ) M 3 f1 ( z ) [V (s - z)] - z[1- M 1 f1 ( z ) M 2 f1 ( z ) M 3 f1 ( z )] Thus
p(q ,1)
Rq (z,s)
(1)
(z,s),
p(q ,1)
( 2)
(z,s),
p(q ,1)
( 3)
, Vq (z,s) and
are completely determined. Thus the transient
solution is obtained.
5.CONCLUSION This paper clearly analyses the transient solution of the queueing system with three stages of service, server provides essential service in all the stages to the arriving customers.This model can be utilized in Large scale manufacturing industries. As a future work, the steady state behaviour of the system and the other performance measures can be derived.
6.ACKNOWLEDGEMENT Our Sincere thanks to Dr.V.Thangaraj ,Director ,Ramanujam institute for advanced Study of Mathematics,University of Madras,Chennai for his support in development of this article.
[11] K.C. Madan and A.Z. Abu-Dayyeh, On a single server queue with optional phase type server vacations based on exhaustive deterministic service and a single vacation policy, Apllied Mathematics and Computation, 149 (2004), 723-734. [12] K.C. Madan, W.Sbu-Dayyeh and M.F. Saleh, An M/G/1 queue with second optional service and Bernoulli schedule server vacations,Systems Science,28(2002),5162. [13] K.C. Madan and R.F. Anabosi, A single server queue with two types of service, Bernoulli schedule server vacations and a single vacation policy, Pakistan Journal of Statistics, 19 (2003), 331-342. [14] F.A. Maraghi, K.C. Madan and K.D. Dowman, Batch arrival queueing system with random breakdowns and Bernoulli schedule server vacations having general time distribution, International Journal of Information and Management Sciences, 20 (2009), 55-70. [15] N.Tian, Q.Li and J. Cao, Conditional stochastic decomposition in M/M/c queue with server vacations, Stochastic Models, 15 (1999), 367-377.
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