 Open Access
 Total Downloads : 52
 Authors : M. Nila , Dr. D. Sumitha
 Paper ID : IJERTV8IS090233
 Volume & Issue : Volume 08, Issue 09 (September 2019)
 Published (First Online): 10102019
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Batch Arrival Retrial Queueing Model with Starting Failures, Customer Impatience, Multi Optional Second Phase and Orbital Search
M. Nila
Research Scholar, Department of Mathematics Avinashilingam Institute for Home science and Higher Education for Women, Coimbatore, India
Dr. D. Sumitha
Associate professor, Department of Science and Humanities,
Avinashilingam Institute for Home science and Higher Education for Women, Coimbatore, India
Abstract In this paper, we consider a single server batch arrival retrial queueing model with starting failures, customer impatience and multi optional second phase. Batch arrival of customers is in accordance with Poisson process. The server is subject to starting failure. The customers balk and renege at particular times. The server provides two phases of service, first essential and second multi optional. On each service completion, the server searches for customers in the orbit if any or remains idle. We derive the system performance measures for the prescribed model.
Keywords Retrial queue, Starting failure, Customer impatience, Multi optional second phase, Orbital search

INTRODUCTION
Retrial queueing system is characterized by the feature that the customers who find the server busy on arrival, join the retrial queue (orbit) and try again after some random time. It has been widely used in switching systems, telecommunication networks, local area networks and daily life situations. There are several contributions considering queueing system with two phases of service, customer impatience and starting failures. Ke and Chang (2009) studied the batch arrival retrial queue with starting failures, two phases of heterogeneous service and Bernoulli vacation. Sumitha and Udaya Chandrika (2012) discussed retrial queueing system with starting failure, single vacation and orbital search. Suganthi and Madheswari (2015) considered the retrial queueing systems with customer impatience. Madheswari et al. (2016) discussed the retrial queueing system with retention of reneging customers. Krishna kumar and Suganthi (2019) dealt with M/G/1 retrial queues with second optional service and customer balking under two types of Bernoulli vacation schedule. Varalakshmi and Chandrasekaran (2017) analyzed an unreliable two phase retrial queue with immediate Bernoulli feedbacks. Madhan (2018) considered optional deterministic server vacations in a single server queueing system providing two types of additional optional service. Vanitha (2018) studied the M/G/1 feedback queue with two stage heterogeneous service and deterministic server vacations. Rajadurai et al. (2015) have examined the M x/G/1 retrial queue with two phases of service with Bernoulli vacation schedule and random breakdown.
In this paper we discuss a single server retrial queue with starting failures, Customer impatience, Multi optional second phase and orbital search. In section II, we introduce the mathematical model of the system. In section III, we derive the main results of the paper.

DESCRIPTION OF THE MODEL
Consider a single server retrial queueing model with starting failures. Customers arrive in batches according to Poisson process with parameter . The batch size Y is
a random variable with P{Y = k} = Ck; k 1, ,
k 1
probability generating function C(z) and first two moments m1 and m2.
If the arriving batch finds the server free, one of the
customers start the server which takes negligible time. If the server is started successfully, the customer gets the service and all the other customers join the orbit. The probability of successful commencement of service is . Otherwise the repair of the server commences and the arriving batch joins the orbit and makes their retrials later. The retrial time follows general distribution with distribution function A(x). Let a(x), A*() denote the respective probability density function (pdf) and Laplace stieltjes transform (LST). Repair times follow general distribution with distribution function R(x). Let (x), R*() be the respective pdf and LST with first two .
If the server is busy or unavailable, the new arriving batch may join the orbit with probability p or leave the system with complementary probability. On account of arrival of primary customer, the retrial customer cancels his attempt for service and returns to the retrial queue with probability q or leaves the system with complementary probability. The server provides two phases of service .The first phase of service is essential to all customers whereas the second phase is multi optional. After completion of essential service, the customer may leave the system with probability r0 or choose any one of the multi optional service with probability ri where 1 i M.
The essential service time follows general distribution with distribution . Let be the respective pdf and LST with first two moments . The multi optional service time follows general distribution
with distribution function . Let be the
dx n
i n
K nk
respective pdf and LST with first two moments
where 1 i M. On each service completion, the server searches for the customers in the orbit with probability
d Qi x = – p + x Qi x + p
n
n
k 1
C Qi x ,
or remains idle with probability .
Define the Markov process
{N(t); }={X (t), C(t);
n 1, for 1 i M. (6)
d R (x) = – p + x R (x) (7)
X(t) – Orbit size at time t , C(t) – Server state. C(t)—0 if the server is idle at time t.
—1 if the server is busy in first essential service at time t.
dx 1 1
n
d
—2 if the server is busy in second multi optional service at time t.
—3 if the server is on repair at time t.
We define the respective hazard rate function for repeated
Rn (x) = – p + x Rn (x) + p
dx
dx
k 1
n 0 0 n 0
n 0 0 n 0
with boundary conditions
CK Rnk x , n 2 (8)
attempts, first essential service, second multi optional service, and repair as
I (0) = r 1 P x x dx + 1
M
=
; = 0 ;
Qin x i x dx + Rn x x dx; n 1 (9)
1 0
10
0 0
i1
=
, 1 ; =
1
1
Po (0) = c1I0 + 0
I1 x x dx + q 0
I1 x dx +

STEADY STATE DISTRIBUTION Define the following probability densities
M
r0 0 P1 x 0 x dx + 0
i1
Qi1 x i x dx (10)
0
0
n1
(x, t)dx = P{C(t)=0, X(t)=n, 0 < + };
Pn (0) = cn+1I0 + In+1 x x dx + q CK
t 0, 0, 1
(x, t)dx = P{C(t)=1, X(t)=n, 1 < + };
t 0, 0, 0
n
0 Ink+2 x dx + q
k 1
k 1
CK 0 Ink+1 x dx +
M
( x, t)dx = P{C(t)=2, X(t)=n, < + };
r0 Pn+1 x 0 x dx +
Qin+1 x i x dx
2 0
t 0, 0, 0, 1 i M
0
i1
n i 0 n 0
n i 0 n 0
n 1 (11)
(x, t)dx = P{C(t)=3, X(t)=n, 3 < + };
t 0, 0, 1
Qi 0 = r P x x dx , n 0 ; 1 i M (12)
R (0) = I
+ I x x dx
(13)
The system of steady state equations that govers the model is 1
given below
0 0 1
n
M Rn (0) =
CK Ink x dx + In x x dx,
I0 = r0 P0 x 0 x dx+
Qi0 x i x dx (1)
0 0
k 1
0 0
i1
n 2 (14)
d I (x) = – + x I (x), n 1 (2)
dx n n
The normalizing condition is
d
d
dx
dx
Po (x) = – p + 0 x Po (x) (3)
I0 +
n1
0 In x dx +
n0
0 Pn x dx +
n
n
d
d
dx Pn (x) = – p + 0 x Pn (x) + p
k 1
CK Pnk x , n 1 (4)
n0
M
M
0
0
i 1
Qi x dx +
n1
0 Rn
x dx =1 (15)
d Qi x = – p + x Qi x , for 1 i M. (5)
n
n
Define the probability generating functions
dx 0 i 0
I(x, z) =
n 1
In x zn , P x, z =
n 0
Pn x zn ,
I 0, z = I0 1 c z R p pc z z2 zT2 z T3 z z2g2 z R p pc(z)
zc(z)T2 z T3 z / D(z)T3 z (28)
n
n
Qi x, z =
n 0
Qi x zn , 1 i M
P(0,z) = I0
T1 z D(z)
(29)
R x, z =
n 1
Rn x zn
Qi (0, z) = ri I0 B(p pc z )T1 (z) D(z)
0
0
1 i M (30)
R(0,z) = I(0,z) [(c(z)+(1c(z)) ] + I0c z . (31)
Multiplying equations (1) (14) by and summing over n, n=0, 1, 2, 3,we obtain the following partial differential equations.
where
D(z)= T2 (z) zA() + c(z)(1 A() q + qz +
d + + x I x, z = 0 (16)
z2g z R p pc(z) + zT z T
z z2
dx 2 2 3
d + p(1 c z ) + x P x, z = 0 (17)
dx 0
i i
i i
d + p(1 c z ) + x Q x, z = 0, 1 i M (18)
dx
T1 (z) = zA + c z 1 A [1 R p pc z c(z)] q + qz +zc(z)2 R p pc(z) (1 A ) z c(z)
d + p(1 c z ) + x R x, z = 0 (19)
dx
M
M
T2 (z) = B p pc z [r0 +
T2 (z) = B p pc z [r0 +
i
i
0
0
ri B p pc z ]
I(0,z) = r0 P x, z 0 x dx+ Qi x, z i x
i1
0 0
i1
T (z) = [1 g z R p pc(z) ]
0
0
dx + R(x, z) x dx – I
3 2
(20)
0 g2 z = c(z)+(1c(z)) A
P(0,z) = c z I0 + I x, z x dx+ q
z z 0 z2
q
q
0 I x, z c z dx + z 0 I x, z c z dx
M
r
By substituting the expressions (28) – (31) in the equations
(24) – (27) we get the expressions for I (x, z), P(x, (x, z) and R(x, z). Integrating these with respect to x from 0 to we obtain I(z), P(z), (z), R(z), 1 i M.
I z = I0 1 A 1 c z R p pc z z2
+ o P x, z 0 (x)dx+ Qi x, z i x dx (21)
z 0 z 0
i1
zT2 z T3 z z2g2 z R p pc(z)
zc(z)T2 z T3 z / D(z)T3 z
(0, z) = ri 0 P x, z 0 x dx , for 1 i M (22)
P(z)= I0 T1(z) {1B p pc z } / D(z) p pc(z)
R (0,z) = I(x, z)c(z) dx + I(x, z)(x) dx + 0
0 0
I0c z . (23)
Solving the partial differential equations (16) – (19), we get
Qi(z)= I0T1 (z) ri B p pc z {1B p pc z }
0 i
0 i
/ D z p pc z , 1 i M
I (x, z) = I(0,z) (1A(x))
(24)
P (x, z) = P (0, z)ep[1c z ]x (1B0(x))
(25)
(x, z) = (0, z)ep[1c z ]x(1Bi(x)),
1 i M
(26)
R (x, z) = R (0, z)ep[1c z ]x(1R(x))
(27)
Solving equations (20) – (23) we get,
I (x, z) = I(0,z) (1A(x))
(24)
P (x, z) = P (0, z)ep[1c z ]x (1B0(x))
(25)
(x, z) = (0, z)ep[1c z ]x(1Bi(x)),
1 i M
(26)
R (x, z) = R (0, z)ep[1c z ]x(1R(x))
(27)
Solving equations (20) – (23) we get,
R(z) = I0(1 R p pc(z) ){ c(z)T1 z T2 z +
(1 c z )T2 (z) A c(z)[ q R p pc(z) 1 A (z c(z)c(z)) z] (1c(z)) A
{ z2 g z R p pc(z) + zT z T z z2}}
2 2 3
/D(z)T3 z (p pc(z))
Performance measures
The probability that the server is idle during the retrial time is
M
M
I = I0 (1A ) {[1 (pm1 01 +
i1
ripm1 i1 )]
Pq z = I0
+ I z + P z +
M
M
i1
Qi z + R(z)
pm1 1 m1 } / D1
The probability that the server is busy in essential service is given by
P = I0 01 [A + 1 A ) m1 + q +
= I0{g1 z + (1 A )(p pc z ){(1 c(z)
p pc z )[z2 zT2 z T3 z z2g2 ()
p pc(z) ] z c(z)T3 z T2 z } + T1 ()
(m1 + 1) D1
(1 T2
z )3 (z) + (1 R p pc(z) ){
The probability that the server is busy in multi optional service is given by
c(z)T1 (z)T2 z + c(z)(1 c z )A
T2 z qR p pc z 1 A zc z
Q = r I [A + 1 A ) m
+ q +
c z z (1 c z ) A { z2g2 z R p
i i 0 i1 1
(m1 + 1) D1, 1 i M
The probability that the server is under repair is given by
R = I01 [A + 1 A ) m1 + q + (m1 + 1) D1
pc(z) + zT2 (z)T3 z z2}}} g1 z
The probability generating function of the number of customers in the system is
M
where
Ps z = I0 + I z + zP z + z
i1
Qi z + R(z)
M
M
D1= [pm1 01 + ripm1 i1 ] +[A + 1
i1
A ) q + pm1 1 + m1 (1 A )
2 +
Substituting the expressions of I, P, , and R in the normalizing condition (15) and solving we get the analytical expression for as
= I0{(1 A )(p pc z ){(1 c(z)
p pc z )[z2 zT2 z T3 z z2g2 ()
p pc(z) ] z c(z)T3 z T2 z } +T1 ()
T3 z T1 z T2 z T3 (z) + (1 R p pc z )
{ c(z)T1 (z) T2 z + c(z)(1 c z )A T2 (z)
M
M
I0= D1 A {[pm1 01 +
i1
ripm1 i1 ] +
[ qR( p pc(z)) 1 A (z c(z)c(z)) z](1 c(z)) A { z2g2 z R p pc(z) +
M zT (z)T
z z2}}+ g z } g
(z)
pm1 1 } + { [01 +
i1
rii1 ] + 1 }
2 3 1 1
where
[A + 1 A ) m1 + q + (m1 + 1) +g z = D(z)( p pc z ) T (z)
1 3
[( q + )(1A ) 1]The mean number of customer in the orbit is
The probability generating function of the number of
L = lim
d P z = N3D2D3N2
customers in the orbit is
q z1 dz q
3(D2)2
N2 = I0 2pm1 (1 A ) T4 2A pm1 D1
22T4 T6 2 T4 pm1 1 + A pm1 1 D1
D2 = 2pm1 D1
D3 = 6pm1 D1 {m1 1 A ) + pm1 1 3[D1 pm2 + D2 pm1 ]
N3= I0 {(3p (1 A ) [2m1T4 T6 +m1 T5 +
m2 T4 ] +6A pm1 D1 pm1 1
3A p 2m1 D1 pm1 1 + m2 D1 +
L = lim d P z
s s
s s
z1 dz
M
m1 D2 +3 2T4 T6 T8 T7 T4 T5 T6 +
1
1
{6(1 A ) pm2 1 T4
= Lq + P (1) +
i1
Qi 1 .
1
1
3pm1 1(2T6 T4 + T5 ) 3 T4 [p2 2 m22 +
1
1
pm21 ]} + {4pm21 D1 +
1
1
pm1 1 D2 +2 D1(p22 m2 2 + pm2 1 )}}
where
T4 = [A + 1 A ) m1 + q + (m1 + 1)

CONCLUSION
In this paper, we analyzed a batch arrival retrial queueing system with starting failures, customer impatience, Multi optional second phase service and orbital search. Various performance measures like the probability that the server is idle, busy, repair in steady state and mean orbit size, mean system size are derived. The research on the present investigation can be further extended by including the concepts of working breakdown, Bernoulli feedback and vacation.
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T5 = (1 A ) 2qm1 1 2 p 1 + m2 +

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