ISSN 2319-8133 (Online)
(An International Research Journal), www.compmath-journal.org
A Bulk Arrival Retrial Queue with Feedback and Exponentially Distributed Multiple Working Vacation
S. Pazhani Bala Murugan1 and R. Vijaykrishnaraj2
1
Mathematics Section,
Faculty of Engineering and Technology,
Annamalai University, Annamalainagar, Tamilnadu, INDIA.
2
Department of Mathematics,
Annamalai University, Annamalainagar, Tamilnadu, INDIA.
email:[email protected]
1and [email protected]
2.
(Received on: January 2, 2019)
ABSTRACTWe consider a bulk arrival retrial queue with feedback and exponentially distributed multiple working vacation. The server provides service to customers, one by one, on a FCFS basis. Just after completion of his service a customer may leave the system or may opt to repeat his service, in which case this customer rejoins the head of the queue. After completion of customer's service there is no customer in the orbit the server may take a multiple working vacation. Using supplementary variable method we obtain the probability generating function for the number of customers in the orbit. Some particular cases are discussed.
MSC: 60K25, 60K30.
Keywords: Bulk arrival retrial queues, Feedback, Linear retrial policy, Working vacation and Supplementary variable method.
1. INTRODUCTION
Retrial queueing systems are described by the feature that the arriving customers who find the server busy join the retrial orbit to try their requests agian. Retrial queues are widely and successfully used as Mathematical models of several Computer systems and telecommunication networks.
Choi et al.
4analysed an π/π/1 retrial queue with general retrial times. Martin and
Gomez-Corral
12considered an π/πΊ/1 retrial queue with linear control policy, Lillo
11investigated a πΊ/π/1 retrial queue. Sherman and Kharoufeh
17studied an π/π/1 retrial queue with unreliable server.
One additional feature which has been widely discussed in retrial queueing system is the Bernoulli feedback of customers. The phenomena of feedback in retrial queueing systems occured in many practical situations. For example, the retrial queue with feedback can be used to model the Automative Repeat Request protocal in a high frequency communication network. Falin
16studied an π/π/1 retrial queue with feedback. Kumar et al.
8investigated an π/πΊ/1 retrial queue with feedback and starting failures. Ke and Chang
6considered a modified vacation policy for the π/πΊ/1 retrial queue with balking and feedback. Kumar et al.
9discussed an π/π/1 retrial queue with feedback and collisions. Kumar et al.
10analyzed an π/πΊ/1 retrial queue with feedback and negative customers.
In the queueing theory, vacation queues and retrial queues have been intensive research topics; we can find general models in Artalejo
2and Gomez-Corral
12. The study of queueing system with working vacations can also provide the theory and analysis method to design the optimal lower speed period.In 2002, Servi and Finn
16first introduced working vacation policy and studied an π/π/1/ππ queue.
Wu and Takagi
18extended the π/π/1/ππ queue to an π/πΊ/1/ππ queue using the matrix-analytic method, Baba
5considered a πΊπΌ/π/1 queue with working vacations.
Krishnamoorthy and Sreenivasan
7analysed an π/π/2 queue with working vacations.
Do
5studied an π/π/1 retrial queue with working vacation. Zhang and Xu
19considered an π/π/1/ππ queue with N-policy. Aftab Begum
1studied π
π/πΊ/1 queue with exponentially distributed Multiple working vacations and Santhi and Pazhani Bala Murugan
14analysed π
π/πΊ/1 queue with Multiple working vacation and with feedback. Pazhani Bala Murugan and Vijaykrishnaraj
15studied A bulk arrival retrial queue with exponentially distributed multiple working vacation.
In this paper, we study a bulk arrival retrial queueing system with feedback and exponentially distributed multiple working vacation. The organization of the paper is as follows. In section 2, we described the model. In section 3, we obtained the steady state probability generating function. In section 4 some particular cases have been discussed.
2. MODEL DESCRIPTION
We consider an π
π/πΊ/1 queueing system where the primary customers arrive according to a compound Poisson process with arrival rate π(β₯ 0). The batch size π is a random variable ππ(π = π) = π
π, π = 1,2,3, . .. with probability generating function π(π§) =
β
βπ=1π
ππ§
πand the first and second factorial moments of π are defined by π
(1)(1) = πΈ(π) and π
(2)(1) = πΈ(π(π β 1)).
We assume that there is no waiting space and therefore if an arriving customer
(external or repeated) finds the server occupied, he leaves the service area and joins a pool of
blocked customers called orbit. We will assume that only the customer at the head of the orbit
is allowed to reach the server at a service completion instant. The retrial time follows a general
distribution, with distibution functions π΅(π₯). Let π(π₯) and π΅
β(π) denote the probability
density function and Laplace Stieltjes Transform of π΅(π₯) respectively for regular service period and let π(π₯), π΄
β(π) denote the probability density function and Laplace Stieltjes Transform of π΄(π₯) respectively for working vacation period. Just after the completion of a service, if any customer is in orbit the next customer to gain service is determined by a competition between the primary customer and the orbit customer.
The service time is assumed to follow general distribution, with distribution function π
π(π₯) and density funciton π
π(π₯). Let π
πβ(π) be the Laplace Stieltjes Transform (LST) of the service time π
π.
Whenever the orbit becomes empty at a service completion instant the server starts a working vacation and the duration of the vacation time follows an exponential distribution with rate π. At a vacation completion instant if there are customers in the system the server will start a new busy period. Otherwise he takes another working vacation. This type of vacation policy is called multiple working vacation. During the working vacation period, the server provides service with service time π
π£which follows a general distribution with distribution function π
π£(π₯). Letπ
π£(π₯) be the probability density function and π
π£β(π) be the Laplace Stieltjes Transform of π
π£(π₯). Further, it is noted that the service interrupted at the end of a vacation is lost and it is restarted with different distribution at the beginning of the following service period.
After completion of each service a customer may like to repeat his service with probability π or may leave the system with probability π = 1 β π in both not working vacation period and working vacation period.
We assume that inter-arrival times, service times, working vacation times and a retrial times are mutually independent.
We define the following random variables N(t)-the orbit size at time t.
π΄
0(π‘)-the remaining retrial time in working vacation period.
π΅
0(π‘)-the remaining retrial time in regular service period.
π
π£0(π‘)-the remaining service time in working vacation period.
π
π0(π‘)-the remaining service time in regular service period.
0 if the server is on working vacation period at time t but not occupied 1 if the server is in regular service period at time t but not occupied
( ) 2 if the server is busy on working vacatio
Y t n period at time t
3 if the server is busy in regular service period at time t
so that the supplementary variables π΄
0(π‘), π΅
0(π‘), π
π£0(π‘) and π
π0(π‘) are introduced in order to obtain the bivaritate Markov Process {π(π‘), β(π‘); π‘ β₯ 0}, Where
0 0 0 0
( ) if ( ) 0 ( ) if ( ) 1 ( ) ( ) if ( ) 2
( ) if ( ) 3
v b
A t Y t B t Y t
t S t Y t
S t Y t
We define the following limiting probabilites:
π
0,0= lim
π‘ββ
ππ{π(π‘) = 0, π(π‘) = 0}
π
0,π= lim
π‘ββ
ππ{π(π‘) = π, π(π‘) = 0, π₯ < π΄
0(π‘) β€ π₯ + ππ₯}; π β₯ 1 π
0,π= lim
π‘ββ
ππ{π(π‘) = π, π(π‘) = 1, π₯ < π΅
0(π‘) β€ π₯ + ππ₯}; π β₯ 1 π
1,π= lim
π‘ββ
ππ{π(π‘) = π, π(π‘) = 2, π₯ < π
π£0(π‘) β€ π₯ + ππ₯}; π β₯ 0 π
1,π= lim
π‘ββ
ππ{π(π‘) = π, π(π‘) = 3, π₯ < π
π0(π‘) β€ π₯ + ππ₯}; π β₯ 0
We define the Laplace Stieltjes Transform and the probability generating functions as follows,
*
( )
0 x( ) ;
b b
S e s x dx
*( )
0 x( ) ;
v v
S e s x dx
*( )
0 x( )
A e a x dx
*
( )
0 x( ) ;
B e b x dx
0,* 0,( )
0 x( ) ;
n n
Q e Q x dx
0,* 0,(0)
0( ) ;
n n
Q Q x dx
*
1,n
( )
0 x 1,n( ) ;
Q e Q x dx
1,* 1,(0)
0( )
n n
Q Q x dx ,
0,* 0,( )
0 x( ) ;
n n
P e P x dx
*
0,n
(0)
0 0,n( ) ;
P P x dx
0* 0,*1
( , )
n( )
n;
n
Q z Q z
0* 0,*1
( , 0)
n(0)
n;
n
Q z Q z
0 0,
1
( , 0)
n(0)
n;
n
Q z Q z
1* 1,*0
( , )
n( )
n;
n
Q z Q z
1* 1,*0
( , 0)
n(0)
n;
n
Q z Q z
1 1,
0
( , 0)
n(0)
n;
n
Q z Q z
0* 0,*1
( , )
n( )
n;
n
P z P z
0* 0,*1
( , 0)
n(0)
n;
n
P z P z
0 0,
1
( , 0)
n(0)
n;
n
P z P z
1* 1,*0
( , )
n( )
n;
n
P z P z
1* 1,*0
( , 0)
n(0)
n;
n
P z P z
1 1,
0
( , 0)
n(0) .
nn
P z P z
3. THE ORBIT SIZE DISTRIBUTION
By assuming that the system is in steady state condition, the differential difference equations governing the systems are as follows:
ππ
0,0= ππ
1,0(0) + ππ
1,0(0) (1)
β
ππ₯ππ
0,π(π₯) = β(π + π)π
0,π(π₯) + ππ
1,π(0)π(π₯); π β₯ 1 (2)
β
πππ₯
π
1,0(π₯) = β(π + π)π
1,0(π₯) + π
0,1(0)π
π£(π₯) + ππ
0,0π
π£(π₯)π
1+ ππ
1,0(0)π
π£(π₯) (3)
β π
ππ₯ π
1,π(π₯) = β(π + π)π
1,π(π₯) + β
π
π=1
ππ
1,πβπ(π₯)π
π+ π
0,π+1(0)π
π£(π₯)
+ππ
π£(π₯) β«
0ββ
ππ=1π
0,πβπ+1(π₯)π
πππ₯ + ππ
π+1π
0,0π
π£(π₯) + ππ
1,π(0)π
π£(π₯) (4)
β
πππ₯
π
0,π(π₯) = βππ
0,π(π₯) + ππ
1,π(0)π(π₯) + ππ(π₯) β«
0βπ
0,π(π₯)ππ₯ (5)
β
ππ₯ππ
1,0(π₯) = βππ
1,0(π₯) + π
0,1(0)π
π(π₯) + ππ
1,0(0)π
π(π₯) + ππ
π(π₯) β«
0βπ
1,0(π₯)ππ₯ (6)
β π
ππ₯ π
1,π(π₯) = βππ
1,π(π₯) + β
π
π=1
ππ
1,πβπ(π₯)π
π+ π
0,π+1(0)π
π(π₯) + ππ
1,π(0)π
π(π₯)
+ππ
π(π₯) β«
0ββ
ππ=1π
0,πβπ+1(π₯)π
kππ₯ + ππ
π(π₯) β«
0βπ
1,π(π₯)ππ₯ (7) Taking LST on both sides of the equation from (2) to (7) we get,
ππ
0,πβ(π) β π
0,π(0) = (π + π)π
0,πβ(π) β ππ
1,π(0)π΄
β(π); π β₯ 1 (8)
ππ1,0β (π) β π1,0(0) = (π + π)π1,0β (π) β π0,1(0)ππ£β(π) β ππ1π0,0ππ£β(π) β ππ1,0(0)ππ£β(π)(9) ππ
1,πβ(π) β π
1,π(0) = (π + π)π
1,πβ(π) β β
π
π=1
ππ
ππ
1,πβπβ(π) β π
0,π+1(0)π
π£β(π)
βππ
π+1π
0,0π
π£β(π) β π β
ππ=1π
0,π+1β(0)π
ππ
π£β(π) β ππ
1,0(0)π
π£β(π) (10) ππ
0,πβ(π) β π
0,π(0) = ππ
0,πβ(π) β ππ
1,π(0)π΅
β(π) β ππ
0,πβ(0)π΅
β(π) (11) ππ
1,0β(π) β π
1,0(0) = ππ
1,0β(π) β π
0,1(0)π
πβ(π) β ππ
1,0(0)π
πβ(π) β ππ
1,0βπ
πβ(π) (12) ππ
1,πβ(π) β π
1,π(0) = ππ
1,πβ(π) β β
π
π=1
ππ
ππ
1,πβπβ(π) β π
0,π+1(0)π
πβ(π) β ππ
1,π(0)π
πβ(π) βππ
πβ(π) β
ππ=1π
0,πβπ+1β(0)π
πβ ππ
πβ(π)π
1,πβ(0) (13) Multiplying (8) with π§
πand summed over π from 1 to β, we get
π β
β
π=1
π
0,πβ(π)π§
πβ β
β
π=1
π
0,π(0)π§
π= (π + π) β
β
π=1
π
0,πβ(π)π§
πβ ππ΄
β(π) β
β
π=1
π
1,π(π)π§
π[π β (π + π)]π
0β(π§, π) = π
0(π§, 0) β ππ΄
β(π)π
1(π§, 0) + ππ΄
β(π)π
1,0(0) (14) π§
πtimes (10) summed over π from 1 to β and added up with (9) gives
π β
β
π=0
π1,πβ (π)π§πβ β
β
π=0
π1,π(π)π§π = (π + π) β
β
π=0
π1,πβ (π)π§πβ π β
β
π=1
β
π
π=1
πππ1,πβπβ (π)π§π βππ£β(π) β
β
π=0
π0,π+1(0)π§πβ ππ0,0ππ£β(π) β
β
π=0
ππ+1π§π βπππ£β(π) β
β
n=0
π1,π(0)π§πβ πππ£β(π) β
β
π=1
β
π
π=1
π0,π+1βπβ (0)πππ§π [π β (π β ππ(π§) + π)]π1β(π§, π) = [1 β πππ£β(π)]π1(π§, 0) βππ£β(π)
π§ π0(π§, 0) β ππ0,0π(π§)ππ£β(π) π§
β
πππ£βπ§(π)π(π§)π
0β(π§, 0) (15) Inserting π = π + π in (14), we get
π
0(π§, 0) = ππ΄
β(π + π)[π
1(π§, 0) β π
1,0(0)] (16)
Substituting π = 0 in (14) and using (16), we get
π
0β(π§, 0) =
π[1βπ΄β(π+π)][π1(π§,0)βπ1,0(0)]π+π
(17)
Inserting π = π β ππ(π§) + π in (15), we get
π
1(π§, 0) =
ππ£β(πβππ(π§)+π)[βπ[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]π1,0(0)+π(π+π)π(π§)π0,0](π+π)π§[1βπππ£β(πβππ(π§)+π)]βπππ£β(πβππ(π§)+π)[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]
(18) Substituting (18) in (16), we get
π0(π§, 0) = ππ΄β(π + π)[ ππ£β(πβππ(π§)+π)π(π+π)π(π§)π0,0β(π+π)π§[1βπππ£β(πβππ(π§)+π)]π1,0(0)
(π+π)π§[1βπππ£β(πβππ(π§)+π)]βπππ£β(πβππ(π§)+π)[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]
(19) Let as consider
π(π§) = (π + π)π§[1 β ππ
π£β(π β ππ(π§) + π)] β ππ
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) +π)π΄
β(π + π)],
we find π(0) < 0 and π(1) > 0.
This implies that there exist a real root π§
1β (0,1) for the equation π(π§) = 0.
Hence at π§ = π§
1the equation (19) becomes π
1,0(0) =
ππ(π§1)ππ£β(πβππ(π§1)+π)π0,0π§1[1βπππ£β(πβππ(π§1)+π)]
(20)
Substituting (20) in (18), we get
π
1(π§, 0) =
ππ£β(πβππ(π§)+π)[π(π+π)π(π§)βππ(π§1)π[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]]π0,0(π+π)π§[1βπππ£β(πβππ(π§)+π)]βπππ£β(πβππ(π§)+π)[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]
(21) Substituting (20) in (19), we get
π
0(π§, 0) =
ππ(π+π)π΄β(π+π)[π(π§)ππ£β(πβππ(π§)+π)βπ(π§1)[1βπππ£β(πβππ(π§)+π)]π§]π0,0(π+π)π§[1βπππ£β(πβππ(π§)+π)]βπππ£β(πβππ(π§)+π)[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]
(22) Where π(π§) =
π(π§)π§
π
π£β(π β ππ(π§) + π) Substituting (20) and (21) in (17), we get
π
0β(π§, 0) =
ππ[1βπ΄β(π+π)][π(π§)ππ£β(πβππ(π§)+π)βπ(π§1)[1βπππ£β(πβππ(π§)+π)]π§]π0,0(π+π)π§[1βπππ£β(πβππ(π§)+π)]βπππ£β(πβπX(π§)+π)[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]
(23) Inserting π = 0 and substituting (21), (22) and (23) in (15), we get
π1β(π§, 0) = [1βππ£β(πβππ(π§)+π)][(π+π)π(π§)βππ(π§1)[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]]ππ0,0
(πβππ(π§)+π)[(π+π)π§[1βπππ£β(πβππ(π§)+π)]βπππ£β(πβππ(π§)+π)[ππ(π§)+(πβππ(π§)+π)π΄β(π+π)]] (
24) Multiplying (11) with π§
πand summed over π from 1 to β, we get
π β
β
π=1
π
0,πβ(π)π§
πβ β
β
π=1
π
0,π(0)π§
π= π β
β
π=1
π
0,πβ(π)π§
πβ ππ΅
β(π) β
β
π=1
π
1,π(0)π§
πβππ΅
β(π) β
β
π=1
π
0,πβ(0)π§
π(π β π)π
0β(π§, π) = π
0(π§, 0) β ππ΅
β(π)π
1(π§, 0) + π΅
β(π)ππ
1,0(0) β ππ΅
β(π)π
0β(π§, 0) (25)
π§
πtimes (13) is summed over π from 1 to β and added up with (12) gives
π β
β
π=0
π1,πβ (π)π§πβ β
β
π=0
π1,π(0)π§π = π β
β
π=0
π1,πβ (π)π§πβ π β
β
π=1
β
π
π=1
πππ1,πβπβ (π)π§πβ ππβ(π) β
β
π=0
π0,π+1(0)π§π βπππβ(π) β
β
π=0
π1,π(0)π§πβ πππβ(π) β
β
π=1
β
π
π=1
π0,πβπ+1β (0)πππ§π βπππβ(π) β
β
π=0
π1,πβ (0)π§π
[π β (π β ππ(π§))]π
1β(π§, π)=[1 β ππ
πβ(π)]π
1(π§, 0) β
ππβ(π)π§π
0(π§, 0) β
ππ(π§)π§π
πβ(π)π
0β(π§, 0) βππ
πβ(π)π
1β(π§, 0) (26)
Inserting π = π and substituting ππ
1,0(0) = [1 β ππ(π§
1)]ππ
0,0in (25),we get
π
0(π§, 0) = ππ΅
β(π)π
1(π§, 0) β π΅
β(π)[1 β ππ(π§
1)]ππ
0,0+ ππ΅
β(π)π
0β(π§, 0) (27) Inserting π = π β ππ(π§) and substituting (27) in (26), we get
[1 β πππβ(π β ππ(π§))]π1(π§, 0) =ππβ(π β ππ(π§))
π§ [ππ΅β(π)π1(π§, 0) β π΅β(π)[1 β ππ(π§1)]ππ0,0
+ππ΅
β(π)π
0β(π§, 0)] + ππ(π§)π
πβ(π β ππ(π§))
π§ π
0β(π§, 0)
+ππ
πβ(π β ππ(π§))π
1β(π§, 0) (28) Inserting π = 0 and substituting (27) and ππ
1,0(0) = [1 β π(π§
1)]ππ
0,0in (25), we get π
0β(π§, 0) =
[1βπ΅β(π)]π
[ππ
1(π§, 0) β [1 β ππ(π§
1)]ππ
0,0+ ππ
0β(π§, 0)] (29) Substituting (29) and (27)in (28), we get
π1(π§, 0) = β[1βππ(π§1)][π΅β(π)+π(π§)(1βπ΅β(π))]ππβ(πβππ(π§))ππ0,0+ππ΅ππβ(πβππ(π§))π0β(π§,0)+ππ§ππβ(πβππ(π§))π1β(π§,0) [1βπππβ(πβππ(π§))]π§βπππβ(πβππ(π§))[π΅β(π)+π(π§)(1βπ΅β(π))]
(30) Substituting (30) in (27), we get
π
0(π§, 0) =
[1βπππβ(πβππ(π§))]π§π΅β(π)[ππ0β(π§,0)β[1βππ(π§1]ππ0,0]+π΅β(π)ππ§Q1β(π§,0)πππβ(πβππ(π§))[1βπππβ(πβππ(π§))]π§βπππβ(πβππ(π§))[π΅β(π)+(1βπ΅β(π))π(π§)]
(31) Substituting (30) in (29), we get
π0β(π§, 0) = [1βπ΅πβ(π)][ππ§πππβ(πβππ(π§))π1β(π§,0)+[1βπππβ(πβππ(π§))]π§ππ0β(π§,0)β[1βπππβ(πβππ(π§))]π§[1βππ(π§1)]ππ0,0 [1βπππβ(πβππ(π§))]π§βπππβ(πβππ(π§))[π΅β(π)+(1βπ΅β(π)π(π§))] ]
(32) Substituting (23) and (24)in (32), we get
π
0β(π§, 0) =
ππ·1(π§)1(π§)
. π
0,0(33)
Where,
π
1(π§) = [1 β π΅
β(π)]ππ§ππ
πβ(π β ππ(π§))[1 β π
π£β(π β ππ(π§) + π)][(π + π)π(π§)
β ππ(π§
1)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]] + [1 β π΅
β(π)][π β ππ(π§) + π][1 β ππ
πβ(π β ππ(π§))]ππ§π[1 β π΄
β(π + π)][π(π§)π
π£β(π β ππ(π§) + π)
β π(π§
1)[1 β ππ
π£β(π β ππ(π§) + π)]π§] β [1 β π΅
β(π)][π β ππ(π§) + π][1
β ππ
πβ(π β ππ(π§))]π§[1 β ππ(π§
1)][(π + π)[1 β ππ
π£β(π β ππ(π§) + π]π§
β ππ
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]]
π·
1(π§) = [[1 β ππ
πβ(π β ππ(π§))]π§ β ππ
πβ(π β ππ(π§))[π΅
β(π) + (1 β π΅
β(π))π(π§)]][π
β ππ(π§)
+ π][(π + π)[1 β ππ
π£β(π β ππ(π§) + π]π§ β ππ
π£β(π β ππ(π§) + π)[ππ(π§) +(π β ππ(π§) + π)π΄
β(π + π)]]
Substituting π = 0 and Substituting (24), (27), (30) and (33) in (26) we get π
1β(π§, 0) =
ππ·2(π§)2(π§)
. ππ
0,0(34)
Where,
π2(π§) = ππ§[1 β ππβ(π β ππ(π§)][1 β ππ£β(π β ππ(π§) + π)][(π + π)π(π§) β ππ(π§1)[ππ(π§) + (π
β ππ(π§) + π)π΄β(π + π)]] + ππ[π΅β(π) + (1 β π΅β(π)π(π§)][1 β ππβ(π
β ππ(π§)][π β ππ(π§) + π][1 β π΄β(π + π)][π(π§)ππ£β(π β ππ(π§) + π) β π(π§1)[1
β πππ£β(π β ππ(π§) + π]π§]
π·
2(π§) = [π β ππ(π§)][π β ππ(π§) + π][[1 β ππ
πβ(π β ππ(π§))]π§ β ππ
πβ(π β ππ(π§))[π΅
β(π) + (1 β π΅
β(π)π(π§)]][(π + π)[1 β ππ
π£β(π β ππ(π§) + π)]π§ β ππ
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]]
We define π
π(π§) = π
0β(π§, 0) + π
1β(π§, 0) + π
(0,0)π
π(π§) =
ππ(π§)π·π(π§)
. π
0,0(35)
as the probability generating function for the number of customers in the orbit when the server is on working vacation period.
Where,
ππ(π§) = ππ[1 β π΄β(π + π)][π β ππ(π§) + π)][π(π§)ππ£β(π β ππ(π§) + π) β π(π§1)[1 β πππ£β(π
β ππ(π§) + π)]π§] + π[1 β ππ£β(π β ππ(π§) + π)][(π + π)π(π§) β ππ(π§1)[ππ(π§) + (π β ππ(π§) + π)π΄β(π + π)]] + [π β ππ(π§) + π)][(π + π)[1 β πππ£β(π
β ππ(π§) + π)]π§ β πππ£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄β(π + π)]
π·
π(π§) = [π β ππ(π§) + π][(π + π)[1 β ππ
π£β(π β ππ(π§) + π)]π§ β ππ
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]]
and
π
π΅(π§) = π
0β(π§, 0) + π
1β(π§, 0) π
π΅(π§) =
ππ΅(π§)π·π΅(π§)
π
0,0(36)
as the probability generating function for the number of customers in the orbit when the server is on not working vacation (normal busy) period.
Where,
π
π΅(π§) = πππ§[1 β π
π£β(π β ππ(π§) + π)][(π + π)π(π§) β ππ(π§
1)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]][[1 β ππ
πβ(π β ππ(π§)] β ππ
πβ(π β ππ(π§))[π΅
β(π) + (1
β π΅
β(π))π(π§)]] + πππ[1 β π΄
β(π + π)][π β ππ(π§) + π][π(π§)π
π£β(π β ππ(π§) + π) β π(π§
1)[1 β ππ
π£β(π β ππ(π§) + π)]π§][π§[1 β ππ
πβ(π β ππ(π§))] + [1
β π§ β π
πβ(π β ππ(π§)) + π§ππ
πβ(π β ππ(π§))][π΅
β(π) + (1 β π΅
β(π)π(π§)]]
β π[1 β ππ(π§
1)][π β ππ(π§) + π][(π + π)[1 β ππ
π£β(π β ππ(π§) + π)]π§
β ππ
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)][π§[1 β ππ
πβ(π
β ππ(π§))] + [1 β π§ β π
πβ(π β ππ(π§)) + π§ππ
πβ(π β ππ(π§))][π΅
β(π) + (1
β π΅
β(π))π(π§)]]
π·
π΅(π§) = [π β ππ(π§)][π β ππ(π§) + π][[1 β ππ
πβ(π β ππ(π§)]π§ β ππ
πβ(π β ππ(π§))[π΅
β(π) + (1 β π΅
β(π)π(π§)]][(π + π)[1 β ππ
π£β(π β ππ(π§) + π)]π§ β ππ
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]]
We define
π(π§) = π
π΅(π§) + π
π(π§) (37)
as the probability generating function for the number of customers in the orbit irrespective of the server state.
Where π
π(π§) and π
π΅(π§) are given in equation (35) and (36). We shall now use the normalizing condition π(1) = 1 to determine the unknown π
0,0which appears in (37). Substituting π§ = 1 in (37) and using L'Hospitals rule, we obtain
π0,0 = 1β[π+ππΈ(π)πΈ[ππ]+π(1βπ΅β(π))πΈ(π)]
[1βπ]
[
πβπππ(π§1)+ππ β[π+πβππ(π§1)[π+ππ΄β(π+π)]][ππΈ[ππ]ππ£β(π)+(1βπ΅β(π))[1βπππ£β(π)]]
(π+π)[1βπππ£β(π)]βπππ£β(π)[π+ππ΄β(π+π)]
] (38) 4. PARTICULAR CASES
Case (i):
If no customer receives the feedback service then by setting π = 0 in (37),
we get π(π§) = π
π(π§) + π
π΅(π§) (39)
π
π(π§) =
ππ(π§)π·π(π§)
π
0,0and π
π΅(π§) =
ππ΅(π§)π·π΅(π§)
π
0,0π
π(π§) = π[1 β π΄
β(π + π)][π β ππ(π§) + π][π(π§)π
π£β(π β ππ(π§) + π) β π(π§
1)π§] + π[1
β π
π£β(π β ππ(π§) + π)][(π + π)π(π§) β π(π§
1)[ππ(π§) + (π β ππ(π§)
+ π)π΄
β(π + π)]] + [π β ππ(π§) + π][(π + π)π§ β π
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]
π·
π(π§) = [π β ππ(π§) + π][(π + π)π§ β S
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]
π
π΅(π§) = πππ§[1 β π
π£β(π β ππ(π§) + π)][(π + π)π(π§) β π(π§
1)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]][1 β π
πβ(π β ππ(π§))[π΅
β(π) + (1 β π΅
β(π)π(π§)]] + ππ[1
β π΄
β(π + π)][π β ππ(π§) + π][π(π§)π
π£β(π β ππ(π§) + π) β π(π§
1)π§][π§ + [1
β π§ β π
π£β(π β ππ(π§))][π΅
β(π) + (1 β π΅
β(π)π(π§)] β π[1 β π(π§
1)][π
β ππ(π§) + π][(π + π)π§ β π
π£β(π β ππ(π§))[ππ(π§) + (π β ππ(π§) + π)]][π§ + [1 β π§ β π
πβ(π β ππ(π§))][π΅
β(π) + (1 β π΅
β(π)π(π§)]]
π·
π΅(π§) = [π β ππ(π§)][π β ππ(π§) + π][π§ β π
πβ(π β ππ(π§))[π΅
β(π) + (1 β π΅
β(π)π(π§)]][(π + π)π§ β π
π£β(π β ππ(π§) + π)[ππ(π§) + (π β ππ(π§) + π)π΄
β(π + π)]]
and
π
0,0=
1βππΈ(π)πΈ[ππ]β(1βπ΅β(π))πΈ(π)πβππ(π§1)+π
π β
[
π+πβπ(π§1)[π+ππ΄β(π+π)]π+πβππ£β(π)[π+ππ΄β(π+π)ππΈ(ππ)ππ£β(π)+(1βπ΅β(π))
] (40)
Equation (39) is well known generating function of the orbit size distribution of A bulk arrival
retrial queue with exponentially distributed multiple working vacation studied by S.Pazhani
Bala Murugan and R.Vijaykrishnaraj
15 irrespective of the notation.
Case (ii):
If there is no retrial then on setting π΅
β(π) = 1, π΄
β(π + π) = 1 in (37), we get
π(π§) = π
π(π§) + π
π΅(π§) (41)
π
π(π§) =
ππ(π§)π·π(π§)
π
0,0and π
π΅(π§) =
ππ΅(π§)π·π΅(π§)
π
0,0ππ(π§) = π[1 β ππ£β(π β ππ(π) + π)][π(π§) β ππ(π§1)π§] + [π β ππ(π§) + π][π§ β (π + ππ§)ππ£β(π
β ππ(π§) + π)]
π·
π(π§) = [π β ππ(π§) + π][π§ β (π + ππ§)π
π£β(π β ππ(π§) + π)]
ππ΅(π§) = πππ§[1 β ππ£β(π β ππ(π§) + π)][π(π§) β ππ(π§1)π§][1 β ππβ(π β ππ(π§))] β π[1 β ππ(π§1)][π
β ππ(π§) + π][π§ β (π + ππ§)ππ£β(π β ππ(π§) + π)][1 β ππβ(π β ππ(π§))]
π·π΅(π§) = [π β ππ(π§)][π β ππ(π§) + π][π§ β (π + ππ§)ππβ(π β ππ(π§))][π§ β (π + ππ§)ππβ(π β ππ(π§) + π)]
π
0,0=
1β[π+ππΈ(π)πΈ(ππ)][1βπ]
[
πβπππ(π§1)+ππ β1βππ(π§1)
1βππ£β(π)[Ξ»πΈ(ππ)ππ£β(π)]
] (42)
Equation (41) is well known generating function of the queue size distribution of an π
π/πΊ/1 queue with feedback and multiple working vacation studied by K.Santhi and S.Pazhani Bala Murugan
14irrespective of the notation.
Case(iii):
If no customer receives the feedback service and no retrial then on setting π΄
β(π + π) = 1, π΅
β(π) and π = 0 in (37), we get
π(π§) = π
π(π§) + π
π΅(π§) (43)
π
π(π§) =
ππ(π§)π·π(π§)
π
0,0and π
π΅(π§) =
ππ΅(π§)π·π΅(π§)
π
0,0ππ(π§) = π[1 β ππ£β(π β ππ(π§) + π)][π(π§) β π(π§1)π§] + [π β ππ(π§) + π][π§ β ππ£β(π β ππ(π§) + π)]
π·
π(π§) = [π β ππ(π§) + π][π§ β π
π£β(π β ππ(π§) + π)]
π
π΅(π§) = πππ§[1 β π
π£β(π β ππ(π§) + π)][π(π§) β π(π§
1)π§][1 β π
πβ(π β ππ(π§))]
π·
π΅(π§) = [π β ππ(π§)][π§ β π
πβ(π β ππ(π§)][π β ππ(π§) + π][π§ β π
π£β(π β ππ(π§) + π)]
π
0,0=
1βππΈ(π)πΈ(ππ)[
πβππ(π§1)+ππ β1βπ(π§1)1βππ£β(π)[Ξ»πΈ(ππ)ππ£β(π)]] (44) Equation (43) is well known generating function of the queue size distribution of an ππ/πΊ/1 queue with multiple working vacation studied by M.I.Aftab Begum (2011) irrespective of the notation.
ACKNOWLEDGMENT
The authors would like to thank UGC-RGNF Award No: F1-17.1/2016-17/RGNF- 2015-17-SC-TAM-26968/(SA-III/Website).
REFERENCES