• No results found

A Bulk Arrival Retrial Queue with Feedback and Exponentially Distributed Multiple Working Vacation

N/A
N/A
Protected

Academic year: 2022

Share "A Bulk Arrival Retrial Queue with Feedback and Exponentially Distributed Multiple Working Vacation"

Copied!
11
0
0

Loading.... (view fulltext now)

Full text

(1)

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]

1

and [email protected]

2

.

(Received on: January 2, 2019)

ABSTRACT

We 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.

4

analysed an 𝑀/𝑀/1 retrial queue with general retrial times. Martin and

Gomez-Corral

12

considered an 𝑀/𝐺/1 retrial queue with linear control policy, Lillo

11

(2)

investigated a 𝐺/𝑀/1 retrial queue. Sherman and Kharoufeh

17

studied an 𝑀/𝑀/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

16

studied an 𝑀/𝑀/1 retrial queue with feedback. Kumar et al.

8

investigated an 𝑀/𝐺/1 retrial queue with feedback and starting failures. Ke and Chang

6

considered a modified vacation policy for the 𝑀/𝐺/1 retrial queue with balking and feedback. Kumar et al.

9

discussed an 𝑀/𝑀/1 retrial queue with feedback and collisions. Kumar et al.

10

analyzed an 𝑀/𝐺/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

2

and 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

16

first introduced working vacation policy and studied an 𝑀/𝑀/1/π‘Šπ‘‰ queue.

Wu and Takagi

18

extended the 𝑀/𝑀/1/π‘Šπ‘‰ queue to an 𝑀/𝐺/1/π‘Šπ‘‰ queue using the matrix-analytic method, Baba

5

considered a 𝐺𝐼/𝑀/1 queue with working vacations.

Krishnamoorthy and Sreenivasan

7

analysed an 𝑀/𝑀/2 queue with working vacations.

Do

5

studied an 𝑀/𝑀/1 retrial queue with working vacation. Zhang and Xu

19

considered an 𝑀/𝑀/1/π‘Šπ‘‰ queue with N-policy. Aftab Begum

1

studied 𝑀

𝑋

/𝐺/1 queue with exponentially distributed Multiple working vacations and Santhi and Pazhani Bala Murugan

14

analysed 𝑀

𝑋

/𝐺/1 queue with Multiple working vacation and with feedback. Pazhani Bala Murugan and Vijaykrishnaraj

15

studied 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 𝑀

𝑋

/𝐺/1 queueing system where the primary customers arrive according to a compound Poisson process with arrival rate πœ†(β‰₯ 0). The batch size 𝑋 is a random variable π‘ƒπ‘Ÿ(𝑋 = 𝑛) = 𝑔

𝑛

, 𝑛 = 1,2,3, . .. with probability generating function 𝑋(𝑧) =

βˆ‘

βˆžπ‘˜=1

𝑔

π‘˜

𝑧

π‘˜

and the first and second factorial moments of 𝑋 are defined by 𝑋

(1)

(1) = 𝐸(𝑋) and 𝑋

(2)

(1) = 𝐸(𝑋(𝑋 βˆ’ 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 𝐡(π‘₯). Let 𝑏(π‘₯) and 𝐡

βˆ—

(πœƒ) denote the probability

(3)

density function and Laplace Stieltjes Transform of 𝐡(π‘₯) respectively for regular service period and let π‘Ž(π‘₯), 𝐴

βˆ—

(πœƒ) denote the probability density function and Laplace Stieltjes Transform of 𝐴(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 𝑆

𝑏

(π‘₯) and density funciton 𝑠

𝑏

(π‘₯). Let 𝑆

π‘βˆ—

(πœƒ) be the Laplace Stieltjes Transform (LST) of the service time 𝑆

𝑏

.

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œ‚. 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 𝑆

𝑣

which follows a general distribution with distribution function 𝑆

𝑣

(π‘₯). Let𝑠

𝑣

(π‘₯) be the probability density function and 𝑆

π‘£βˆ—

(πœƒ) be the Laplace Stieltjes Transform of 𝑆

𝑣

(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 𝑝 or may leave the system with probability π‘ž = 1 βˆ’ 𝑝 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.

𝐴

0

(𝑑)-the remaining retrial time in working vacation period.

𝐡

0

(𝑑)-the remaining retrial time in regular service period.

𝑆

𝑣0

(𝑑)-the remaining service time in working vacation period.

𝑆

𝑏0

(𝑑)-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 𝐴

0

(𝑑), 𝐡

0

(𝑑), 𝑆

𝑣0

(𝑑) and 𝑆

𝑏0

(𝑑) are introduced in order to obtain the bivaritate Markov Process {𝑁(𝑑), βˆ‚(𝑑); 𝑑 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

(4)

We define the following limiting probabilites:

𝑄

0,0

= lim

π‘‘β†’βˆž

π‘ƒπ‘Ÿ{𝑁(𝑑) = 0, π‘Œ(𝑑) = 0}

𝑄

0,𝑛

= lim

π‘‘β†’βˆž

π‘ƒπ‘Ÿ{𝑁(𝑑) = 𝑛, π‘Œ(𝑑) = 0, π‘₯ < 𝐴

0

(𝑑) ≀ π‘₯ + 𝑑π‘₯}; 𝑛 β‰₯ 1 𝑃

0,𝑛

= lim

π‘‘β†’βˆž

π‘ƒπ‘Ÿ{𝑁(𝑑) = 𝑛, π‘Œ(𝑑) = 1, π‘₯ < 𝐡

0

(𝑑) ≀ π‘₯ + 𝑑π‘₯}; 𝑛 β‰₯ 1 𝑄

1,𝑛

= lim

π‘‘β†’βˆž

π‘ƒπ‘Ÿ{𝑁(𝑑) = 𝑛, π‘Œ(𝑑) = 2, π‘₯ < 𝑆

𝑣0

(𝑑) ≀ π‘₯ + 𝑑π‘₯}; 𝑛 β‰₯ 0 𝑃

1,𝑛

= lim

π‘‘β†’βˆž

π‘ƒπ‘Ÿ{𝑁(𝑑) = 𝑛, π‘Œ(𝑑) = 3, π‘₯ < 𝑆

𝑏0

(𝑑) ≀ π‘₯ + 𝑑π‘₯}; 𝑛 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) .

n

n

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:

πœ†π‘„

0,0

= π‘žπ‘ƒ

1,0

(0) + π‘žπ‘„

1,0

(0) (1)

βˆ’

𝑑π‘₯𝑑

𝑄

0,𝑛

(π‘₯) = βˆ’(πœ† + πœ‚)𝑄

0,𝑛

(π‘₯) + π‘žπ‘„

1,𝑛

(0)π‘Ž(π‘₯); 𝑛 β‰₯ 1 (2)

βˆ’

𝑑

𝑑π‘₯

𝑄

1,0

(π‘₯) = βˆ’(πœ† + πœ‚)𝑄

1,0

(π‘₯) + 𝑄

0,1

(0)𝑠

𝑣

(π‘₯) + πœ†π‘„

0,0

𝑠

𝑣

(π‘₯)𝑔

1

+ 𝑝𝑄

1,0

(0)𝑠

𝑣

(π‘₯) (3)

βˆ’ 𝑑

𝑑π‘₯ 𝑄

1,𝑛

(π‘₯) = βˆ’(πœ† + πœ‚)𝑄

1,𝑛

(π‘₯) + βˆ‘

𝑛

π‘˜=1

πœ†π‘„

1,π‘›βˆ’π‘˜

(π‘₯)𝑔

π‘˜

+ 𝑄

0,𝑛+1

(0)𝑠

𝑣

(π‘₯)

+πœ†π‘ 

𝑣

(π‘₯) ∫

0∞

βˆ‘

π‘›π‘˜=1

𝑄

0,π‘›βˆ’π‘˜+1

(π‘₯)𝑔

π‘˜

𝑑π‘₯ + πœ†π‘”

𝑛+1

𝑄

0,0

𝑠

𝑣

(π‘₯) + 𝑝𝑄

1,𝑛

(0)𝑠

𝑣

(π‘₯) (4)

(5)

βˆ’

𝑑

𝑑π‘₯

𝑃

0,𝑛

(π‘₯) = βˆ’πœ†π‘ƒ

0,𝑛

(π‘₯) + π‘žπ‘ƒ

1,𝑛

(0)𝑏(π‘₯) + πœ‚π‘(π‘₯) ∫

0∞

𝑄

0,𝑛

(π‘₯)𝑑π‘₯ (5)

βˆ’

𝑑π‘₯𝑑

𝑃

1,0

(π‘₯) = βˆ’πœ†π‘ƒ

1,0

(π‘₯) + 𝑃

0,1

(0)𝑠

𝑏

(π‘₯) + 𝑝𝑃

1,0

(0)𝑠

𝑏

(π‘₯) + πœ‚π‘ 

𝑏

(π‘₯) ∫

0∞

𝑄

1,0

(π‘₯)𝑑π‘₯ (6)

βˆ’ 𝑑

𝑑π‘₯ 𝑃

1,𝑛

(π‘₯) = βˆ’πœ†π‘ƒ

1,𝑛

(π‘₯) + βˆ‘

𝑛

π‘˜=1

πœ†π‘ƒ

1,π‘›βˆ’π‘˜

(π‘₯)𝑔

π‘˜

+ 𝑃

0,𝑛+1

(0)𝑠

𝑏

(π‘₯) + 𝑝𝑃

1,𝑛

(0)𝑠

𝑏

(π‘₯)

+πœ†π‘ 

𝑏

(π‘₯) ∫

0∞

βˆ‘

π‘›π‘˜=1

𝑃

0,π‘›βˆ’π‘˜+1

(π‘₯)𝑔

k

𝑑π‘₯ + πœ‚π‘ 

𝑏

(π‘₯) ∫

0∞

𝑄

1,𝑛

(π‘₯)𝑑π‘₯ (7) Taking LST on both sides of the equation from (2) to (7) we get,

πœƒπ‘„

0,π‘›βˆ—

(πœƒ) βˆ’ 𝑄

0,𝑛

(0) = (πœ† + πœ‚)𝑄

0,π‘›βˆ—

(πœƒ) βˆ’ π‘žπ‘„

1,𝑛

(0)𝐴

βˆ—

(πœƒ); 𝑛 β‰₯ 1 (8)

πœƒπ‘„1,0βˆ— (πœƒ) βˆ’ 𝑄1,0(0) = (πœ† + πœ‚)𝑄1,0βˆ— (πœƒ) βˆ’ 𝑄0,1(0)π‘†π‘£βˆ—(πœƒ) βˆ’ πœ†π‘”1𝑄0,0π‘†π‘£βˆ—(πœƒ) βˆ’ 𝑝𝑄1,0(0)π‘†π‘£βˆ—(πœƒ)

(9) πœƒπ‘„

1,π‘›βˆ—

(πœƒ) βˆ’ 𝑄

1,𝑛

(0) = (πœ† + πœ‚)𝑄

1,π‘›βˆ—

(πœƒ) βˆ’ βˆ‘

𝑛

π‘˜=1

πœ†π‘”

π‘˜

𝑄

1,π‘›βˆ’π‘˜βˆ—

(πœƒ) βˆ’ 𝑄

0,𝑛+1

(0)𝑆

π‘£βˆ—

(πœƒ)

βˆ’πœ†π‘”

𝑛+1

𝑄

0,0

𝑆

π‘£βˆ—

(πœƒ) βˆ’ πœ† βˆ‘

π‘›π‘˜=1

𝑄

0,𝑛+1βˆ—

(0)𝑔

π‘˜

𝑆

π‘£βˆ—

(πœƒ) βˆ’ 𝑝𝑄

1,0

(0)𝑆

π‘£βˆ—

(πœƒ) (10) πœƒπ‘ƒ

0,π‘›βˆ—

(πœƒ) βˆ’ 𝑃

0,𝑛

(0) = πœ†π‘ƒ

0,π‘›βˆ—

(πœƒ) βˆ’ π‘žπ‘ƒ

1,𝑛

(0)𝐡

βˆ—

(πœƒ) βˆ’ πœ‚π‘„

0,π‘›βˆ—

(0)𝐡

βˆ—

(πœƒ) (11) πœƒπ‘ƒ

1,0βˆ—

(πœƒ) βˆ’ 𝑃

1,0

(0) = πœ†π‘ƒ

1,0βˆ—

(πœƒ) βˆ’ 𝑃

0,1

(0)𝑆

π‘βˆ—

(πœƒ) βˆ’ 𝑝𝑃

1,0

(0)𝑆

π‘βˆ—

(πœƒ) βˆ’ πœ‚π‘„

1,0βˆ—

𝑆

π‘βˆ—

(πœƒ) (12) πœƒπ‘ƒ

1,π‘›βˆ—

(πœƒ) βˆ’ 𝑃

1,𝑛

(0) = πœ†π‘ƒ

1,π‘›βˆ—

(πœƒ) βˆ’ βˆ‘

𝑛

π‘˜=1

πœ†π‘”

π‘˜

𝑃

1,π‘›βˆ’π‘˜βˆ—

(πœƒ) βˆ’ 𝑃

0,𝑛+1

(0)𝑆

π‘βˆ—

(πœƒ) βˆ’ 𝑝𝑃

1,𝑛

(0)𝑆

π‘βˆ—

(πœƒ) βˆ’πœ†π‘†

π‘βˆ—

(πœƒ) βˆ‘

π‘›π‘˜=1

𝑃

0,π‘›βˆ’π‘˜+1βˆ—

(0)𝑔

π‘˜

βˆ’ πœ‚π‘†

π‘βˆ—

(πœƒ)𝑄

1,π‘›βˆ—

(0) (13) Multiplying (8) with 𝑧

𝑛

and summed over 𝑛 from 1 to ∞, we get

πœƒ βˆ‘

∞

𝑛=1

𝑄

0,π‘›βˆ—

(πœƒ)𝑧

𝑛

βˆ’ βˆ‘

∞

𝑛=1

𝑄

0,𝑛

(0)𝑧

𝑛

= (πœ† + πœ‚) βˆ‘

∞

𝑛=1

𝑄

0,π‘›βˆ—

(πœƒ)𝑧

𝑛

βˆ’ π‘žπ΄

βˆ—

(πœƒ) βˆ‘

∞

𝑛=1

𝑄

1,𝑛

(πœƒ)𝑧

𝑛

[πœƒ βˆ’ (πœ† + πœ‚)]𝑄

0βˆ—

(𝑧, πœƒ) = 𝑄

0

(𝑧, 0) βˆ’ π‘žπ΄

βˆ—

(πœƒ)𝑄

1

(𝑧, 0) + π‘žπ΄

βˆ—

(πœƒ)𝑄

1,0

(0) (14) 𝑧

𝑛

times (10) summed over 𝑛 from 1 to ∞ and added up with (9) gives

πœƒ βˆ‘

∞

𝑛=0

𝑄1,π‘›βˆ— (πœƒ)π‘§π‘›βˆ’ βˆ‘

∞

𝑛=0

𝑄1,𝑛(πœƒ)𝑧𝑛 = (πœ† + πœ‚) βˆ‘

∞

𝑛=0

𝑄1,π‘›βˆ— (πœƒ)π‘§π‘›βˆ’ πœ† βˆ‘

∞

𝑛=1

βˆ‘

𝑛

π‘˜=1

π‘”π‘˜π‘„1,π‘›βˆ’π‘˜βˆ— (πœƒ)𝑧𝑛 βˆ’π‘†π‘£βˆ—(πœƒ) βˆ‘

∞

𝑛=0

𝑄0,𝑛+1(0)π‘§π‘›βˆ’ πœ†π‘„0,0π‘†π‘£βˆ—(πœƒ) βˆ‘

∞

𝑛=0

𝑔𝑛+1𝑧𝑛 βˆ’π‘π‘†π‘£βˆ—(πœƒ) βˆ‘

∞

n=0

𝑄1,𝑛(0)π‘§π‘›βˆ’ πœ†π‘†π‘£βˆ—(πœƒ) βˆ‘

∞

𝑛=1

βˆ‘

𝑛

π‘˜=1

𝑄0,𝑛+1βˆ’π‘˜βˆ— (0)π‘”π‘˜π‘§π‘› [πœƒ βˆ’ (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑄1βˆ—(𝑧, πœƒ) = [1 βˆ’ π‘π‘†π‘£βˆ—(πœƒ)]𝑄1(𝑧, 0) βˆ’π‘†π‘£βˆ—(πœƒ)

𝑧 𝑄0(𝑧, 0) βˆ’ πœ†π‘„0,0𝑋(𝑧)π‘†π‘£βˆ—(πœƒ) 𝑧

βˆ’

πœ†π‘†π‘£βˆ—π‘§(πœƒ)

𝑋(𝑧)𝑄

0βˆ—

(𝑧, 0) (15) Inserting πœƒ = πœ† + πœ‚ in (14), we get

𝑄

0

(𝑧, 0) = π‘žπ΄

βˆ—

(πœ† + πœ‚)[𝑄

1

(𝑧, 0) βˆ’ 𝑄

1,0

(0)] (16)

Substituting πœƒ = 0 in (14) and using (16), we get

(6)

𝑄

0βˆ—

(𝑧, 0) =

π‘ž[1βˆ’π΄βˆ—(πœ†+πœ‚)][𝑄1(𝑧,0)βˆ’π‘„1,0(0)]

πœ†+πœ‚

(17)

Inserting πœƒ = πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚ in (15), we get

𝑄

1

(𝑧, 0) =

π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)[βˆ’π‘ž[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]𝑄1,0(0)+πœ†(πœ†+πœ‚)𝑋(𝑧)𝑄0,0]

(πœ†+πœ‚)𝑧[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]βˆ’π‘žπ‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]

(18) Substituting (18) in (16), we get

𝑄0(𝑧, 0) = π‘žπ΄βˆ—(πœ† + πœ‚)[ π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)πœ†(πœ†+πœ‚)𝑋(𝑧)𝑄0,0βˆ’(πœ†+πœ‚)𝑧[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]𝑄1,0(0)

(πœ†+πœ‚)𝑧[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]βˆ’π‘žπ‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]

(19) Let as consider

𝑓(𝑧) = (πœ† + πœ‚)𝑧[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)] βˆ’ π‘žπ‘†

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) +πœ‚)𝐴

βˆ—

(πœ† + πœ‚)],

we find 𝑓(0) < 0 and 𝑓(1) > 0.

This implies that there exist a real root 𝑧

1

∈ (0,1) for the equation 𝑓(𝑧) = 0.

Hence at 𝑧 = 𝑧

1

the equation (19) becomes 𝑄

1,0

(0) =

πœ†π‘‹(𝑧1)π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧1)+πœ‚)𝑄0,0

𝑧1[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧1)+πœ‚)]

(20)

Substituting (20) in (18), we get

𝑄

1

(𝑧, 0) =

π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)[πœ†(πœ†+πœ‚)𝑋(𝑧)βˆ’πœ†π‘ˆ(𝑧1)π‘ž[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]]𝑄0,0

(πœ†+πœ‚)𝑧[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]βˆ’π‘žπ‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]

(21) Substituting (20) in (19), we get

𝑄

0

(𝑧, 0) =

π‘žπœ†(πœ†+πœ‚)π΄βˆ—(πœ†+πœ‚)[𝑋(𝑧)π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)βˆ’π‘ˆ(𝑧1)[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]𝑧]𝑄0,0

(πœ†+πœ‚)𝑧[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]βˆ’π‘žπ‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]

(22) Where π‘ˆ(𝑧) =

𝑋(𝑧)

𝑧

𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚) Substituting (20) and (21) in (17), we get

𝑄

0βˆ—

(𝑧, 0) =

πœ†π‘ž[1βˆ’π΄βˆ—(πœ†+πœ‚)][𝑋(𝑧)π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)βˆ’π‘ˆ(𝑧1)[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]𝑧]𝑄0,0

(πœ†+πœ‚)𝑧[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]βˆ’π‘žπ‘†π‘£βˆ—(πœ†βˆ’πœ†X(𝑧)+πœ‚)[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]

(23) Inserting πœƒ = 0 and substituting (21), (22) and (23) in (15), we get

𝑄1βˆ—(𝑧, 0) = [1βˆ’π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)][(πœ†+πœ‚)𝑋(𝑧)βˆ’π‘žπ‘ˆ(𝑧1)[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]]πœ†π‘„0,0

(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)[(πœ†+πœ‚)𝑧[1βˆ’π‘π‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)]βˆ’π‘žπ‘†π‘£βˆ—(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)[πœ†π‘‹(𝑧)+(πœ†βˆ’πœ†π‘‹(𝑧)+πœ‚)π΄βˆ—(πœ†+πœ‚)]] (

24) Multiplying (11) with 𝑧

𝑛

and summed over 𝑛 from 1 to ∞, we get

πœƒ βˆ‘

∞

𝑛=1

𝑃

0,π‘›βˆ—

(πœƒ)𝑧

𝑛

βˆ’ βˆ‘

∞

𝑛=1

𝑃

0,𝑛

(0)𝑧

𝑛

= πœ† βˆ‘

∞

𝑛=1

𝑃

0,π‘›βˆ—

(πœƒ)𝑧

𝑛

βˆ’ π‘žπ΅

βˆ—

(πœƒ) βˆ‘

∞

𝑛=1

𝑃

1,𝑛

(0)𝑧

𝑛

βˆ’πœ‚π΅

βˆ—

(πœƒ) βˆ‘

∞

𝑛=1

𝑄

0,π‘›βˆ—

(0)𝑧

𝑛

(πœƒ βˆ’ πœ†)𝑃

0βˆ—

(𝑧, πœƒ) = 𝑃

0

(𝑧, 0) βˆ’ π‘žπ΅

βˆ—

(πœƒ)𝑃

1

(𝑧, 0) + 𝐡

βˆ—

(πœƒ)π‘žπ‘ƒ

1,0

(0) βˆ’ πœ‚π΅

βˆ—

(πœƒ)𝑄

0βˆ—

(𝑧, 0) (25)

𝑧

𝑛

times (13) is summed over 𝑛 from 1 to ∞ and added up with (12) gives

(7)

πœƒ βˆ‘

∞

𝑛=0

𝑃1,π‘›βˆ— (πœƒ)π‘§π‘›βˆ’ βˆ‘

∞

𝑛=0

𝑃1,𝑛(0)𝑧𝑛 = πœ† βˆ‘

∞

𝑛=0

𝑃1,π‘›βˆ— (πœƒ)π‘§π‘›βˆ’ πœ† βˆ‘

∞

𝑛=1

βˆ‘

𝑛

π‘˜=1

π‘”π‘˜π‘ƒ1,π‘›βˆ’π‘˜βˆ— (πœƒ)π‘§π‘›βˆ’ π‘†π‘βˆ—(πœƒ) βˆ‘

∞

𝑛=0

𝑃0,𝑛+1(0)𝑧𝑛 βˆ’π‘π‘†π‘βˆ—(πœƒ) βˆ‘

∞

𝑛=0

𝑃1,𝑛(0)π‘§π‘›βˆ’ πœ†π‘†π‘βˆ—(πœƒ) βˆ‘

∞

𝑛=1

βˆ‘

𝑛

π‘˜=1

𝑃0,π‘›βˆ’π‘˜+1βˆ— (0)π‘”π‘˜π‘§π‘› βˆ’πœ‚π‘†π‘βˆ—(πœƒ) βˆ‘

∞

𝑛=0

𝑄1,π‘›βˆ— (0)𝑧𝑛

[πœƒ βˆ’ (πœ† βˆ’ πœ†π‘‹(𝑧))]𝑃

1βˆ—

(𝑧, πœƒ)=[1 βˆ’ 𝑝𝑆

π‘βˆ—

(πœƒ)]𝑃

1

(𝑧, 0) βˆ’

π‘†π‘βˆ—(πœƒ)𝑧

𝑃

0

(𝑧, 0) βˆ’

πœ†π‘‹(𝑧)𝑧

𝑆

π‘βˆ—

(πœƒ)𝑃

0βˆ—

(𝑧, 0) βˆ’πœ‚π‘†

π‘βˆ—

(πœƒ)𝑄

1βˆ—

(𝑧, 0) (26)

Inserting πœƒ = πœ† and substituting π‘žπ‘ƒ

1,0

(0) = [1 βˆ’ π‘žπ‘ˆ(𝑧

1

)]πœ†π‘„

0,0

in (25),we get

𝑃

0

(𝑧, 0) = π‘žπ΅

βˆ—

(πœ†)𝑃

1

(𝑧, 0) βˆ’ 𝐡

βˆ—

(πœ†)[1 βˆ’ π‘žπ‘ˆ(𝑧

1

)]πœ†π‘„

0,0

+ πœ‚π΅

βˆ—

(πœ†)𝑄

0βˆ—

(𝑧, 0) (27) Inserting πœƒ = πœ† βˆ’ πœ†π‘‹(𝑧) and substituting (27) in (26), we get

[1 βˆ’ π‘π‘†π‘βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧))]𝑃1(𝑧, 0) =π‘†π‘βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧))

𝑧 [π‘žπ΅βˆ—(πœ†)𝑃1(𝑧, 0) βˆ’ π΅βˆ—(πœ†)[1 βˆ’ π‘žπ‘ˆ(𝑧1)]πœ†π‘„0,0

+πœ‚π΅

βˆ—

(πœ†)𝑄

0βˆ—

(𝑧, 0)] + πœ†π‘‹(𝑧)𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))

𝑧 𝑃

0βˆ—

(𝑧, 0)

+πœ‚π‘†

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))𝑄

1βˆ—

(𝑧, 0) (28) Inserting πœƒ = 0 and substituting (27) and π‘žπ‘ƒ

1,0

(0) = [1 βˆ’ π‘ˆ(𝑧

1

)]πœ†π‘„

0,0

in (25), we get 𝑃

0βˆ—

(𝑧, 0) =

[1βˆ’π΅βˆ—(πœ†)]

πœ†

[π‘žπ‘ƒ

1

(𝑧, 0) βˆ’ [1 βˆ’ π‘žπ‘ˆ(𝑧

1

)]πœ†π‘„

0,0

+ πœ‚π‘„

0βˆ—

(𝑧, 0)] (29) Substituting (29) and (27)in (28), we get

𝑃1(𝑧, 0) = βˆ’[1βˆ’π‘žπ‘ˆ(𝑧1)][π΅βˆ—(πœ†)+𝑋(𝑧)(1βˆ’π΅βˆ—(πœ†))]π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))πœ†π‘„0,0+πœ‚π΅π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))𝑄0βˆ—(𝑧,0)+πœ‚π‘§π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))𝑄1βˆ—(𝑧,0) [1βˆ’π‘π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))]π‘§βˆ’π‘žπ‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))[π΅βˆ—(πœ†)+𝑋(𝑧)(1βˆ’π΅βˆ—(πœ†))]

(30) Substituting (30) in (27), we get

𝑃

0

(𝑧, 0) =

[1βˆ’π‘π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))]π‘§π΅βˆ—(πœ†)[πœ‚π‘„0βˆ—(𝑧,0)βˆ’[1βˆ’π‘žπ‘ˆ(𝑧1]πœ†π‘„0,0]+π΅βˆ—(πœ†)πœ‚π‘§Q1βˆ—(𝑧,0)π‘žπ‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))

[1βˆ’π‘π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))]π‘§βˆ’π‘žπ‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))[π΅βˆ—(πœ†)+(1βˆ’π΅βˆ—(πœ†))𝑋(𝑧)]

(31) Substituting (30) in (29), we get

𝑃0βˆ—(𝑧, 0) = [1βˆ’π΅πœ†βˆ—(πœ†)][πœ‚π‘§π‘žπ‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))𝑄1βˆ—(𝑧,0)+[1βˆ’π‘π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))]π‘§πœ‚π‘„0βˆ—(𝑧,0)βˆ’[1βˆ’π‘π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))]𝑧[1βˆ’π‘žπ‘ˆ(𝑧1)]πœ†π‘„0,0 [1βˆ’π‘π‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))]π‘§βˆ’π‘žπ‘†π‘βˆ—(πœ†βˆ’πœ†π‘‹(𝑧))[π΅βˆ—(πœ†)+(1βˆ’π΅βˆ—(πœ†)𝑋(𝑧))] ]

(32) Substituting (23) and (24)in (32), we get

𝑃

0βˆ—

(𝑧, 0) =

𝑁𝐷1(𝑧)

1(𝑧)

. 𝑄

0,0

(33)

Where,

𝑁

1

(𝑧) = [1 βˆ’ 𝐡

βˆ—

(πœ†)]πœ‚π‘§π‘žπ‘†

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))[1 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][(πœ† + πœ‚)𝑋(𝑧)

βˆ’ π‘žπ‘ˆ(𝑧

1

)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]] + [1 βˆ’ 𝐡

βˆ—

(πœ†)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][1 βˆ’ 𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))]πœ‚π‘§π‘ž[1 βˆ’ 𝐴

βˆ—

(πœ† + πœ‚)][𝑋(𝑧)𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)

βˆ’ π‘ˆ(𝑧

1

)[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑧] βˆ’ [1 βˆ’ 𝐡

βˆ—

(πœ†)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][1

βˆ’ 𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))]𝑧[1 βˆ’ π‘žπ‘ˆ(𝑧

1

)][(πœ† + πœ‚)[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚]𝑧

βˆ’ π‘žπ‘†

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]]

𝐷

1

(𝑧) = [[1 βˆ’ 𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))]𝑧 βˆ’ π‘žπ‘†

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))[𝐡

βˆ—

(πœ†) + (1 βˆ’ 𝐡

βˆ—

(πœ†))𝑋(𝑧)]][πœ†

βˆ’ πœ†π‘‹(𝑧)

(8)

+ πœ‚][(πœ† + πœ‚)[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚]𝑧 βˆ’ π‘žπ‘†

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) +(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]]

Substituting πœƒ = 0 and Substituting (24), (27), (30) and (33) in (26) we get 𝑃

1βˆ—

(𝑧, 0) =

𝑁𝐷2(𝑧)

2(𝑧)

. πœ†π‘„

0,0

(34)

Where,

𝑁2(𝑧) = πœ‚π‘§[1 βˆ’ π‘†π‘βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧)][1 βˆ’ π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][(πœ† + πœ‚)𝑋(𝑧) βˆ’ π‘žπ‘ˆ(𝑧1)[πœ†π‘‹(𝑧) + (πœ†

βˆ’ πœ†π‘‹(𝑧) + πœ‚)π΄βˆ—(πœ† + πœ‚)]] + πœ‚π‘ž[π΅βˆ—(πœ†) + (1 βˆ’ π΅βˆ—(πœ†)𝑋(𝑧)][1 βˆ’ π‘†π‘βˆ—(πœ†

βˆ’ πœ†π‘‹(𝑧)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][1 βˆ’ π΄βˆ—(πœ† + πœ‚)][𝑋(𝑧)π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚) βˆ’ π‘ˆ(𝑧1)[1

βˆ’ π‘π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚]𝑧]

𝐷

2

(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][[1 βˆ’ 𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))]𝑧 βˆ’ π‘žπ‘†

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))[𝐡

βˆ—

(πœ†) + (1 βˆ’ 𝐡

βˆ—

(πœ†)𝑋(𝑧)]][(πœ† + πœ‚)[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑧 βˆ’ π‘žπ‘†

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]]

We define 𝑃

𝑉

(𝑧) = 𝑄

0βˆ—

(𝑧, 0) + 𝑄

1βˆ—

(𝑧, 0) + 𝑄

(0,0)

𝑃

𝑉

(𝑧) =

𝑁𝑉(𝑧)

𝐷𝑉(𝑧)

. 𝑄

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,

𝑁𝑉(𝑧) = πœ†π‘ž[1 βˆ’ π΄βˆ—(πœ† + πœ‚)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][𝑋(𝑧)π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚) βˆ’ π‘ˆ(𝑧1)[1 βˆ’ π‘π‘†π‘£βˆ—(πœ†

βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑧] + πœ†[1 βˆ’ π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][(πœ† + πœ‚)𝑋(𝑧) βˆ’ π‘žπ‘ˆ(𝑧1)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)π΄βˆ—(πœ† + πœ‚)]] + [πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][(πœ† + πœ‚)[1 βˆ’ π‘π‘†π‘£βˆ—(πœ†

βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑧 βˆ’ π‘žπ‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)π΄βˆ—(πœ† + πœ‚)]

𝐷

𝑉

(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][(πœ† + πœ‚)[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑧 βˆ’ π‘žπ‘†

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]]

and

𝑃

𝐡

(𝑧) = 𝑃

0βˆ—

(𝑧, 0) + 𝑃

1βˆ—

(𝑧, 0) 𝑃

𝐡

(𝑧) =

𝑁𝐡(𝑧)

𝐷𝐡(𝑧)

𝑄

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,

𝑁

𝐡

(𝑧) = πœ†πœ‚π‘§[1 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][(πœ† + πœ‚)𝑋(𝑧) βˆ’ π‘žπ‘ˆ(𝑧

1

)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]][[1 βˆ’ 𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧)] βˆ’ π‘žπ‘†

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))[𝐡

βˆ—

(πœ†) + (1

βˆ’ 𝐡

βˆ—

(πœ†))𝑋(𝑧)]] + πœ†πœ‚π‘ž[1 βˆ’ 𝐴

βˆ—

(πœ† + πœ‚)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑋(𝑧)𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚) βˆ’ π‘ˆ(𝑧

1

)[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑧][𝑧[1 βˆ’ 𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))] + [1

βˆ’ 𝑧 βˆ’ 𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧)) + 𝑧𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))][𝐡

βˆ—

(πœ†) + (1 βˆ’ 𝐡

βˆ—

(πœ†)𝑋(𝑧)]]

βˆ’ πœ†[1 βˆ’ π‘žπ‘ˆ(𝑧

1

)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][(πœ† + πœ‚)[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑧

βˆ’ π‘žπ‘†

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)][𝑧[1 βˆ’ 𝑝𝑆

π‘βˆ—

(πœ†

βˆ’ πœ†π‘‹(𝑧))] + [1 βˆ’ 𝑧 βˆ’ 𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧)) + 𝑧𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))][𝐡

βˆ—

(πœ†) + (1

βˆ’ 𝐡

βˆ—

(πœ†))𝑋(𝑧)]]

(9)

𝐷

𝐡

(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][[1 βˆ’ 𝑝𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧)]𝑧 βˆ’ π‘žπ‘†

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))[𝐡

βˆ—

(πœ†) + (1 βˆ’ 𝐡

βˆ—

(πœ†)𝑋(𝑧)]][(πœ† + πœ‚)[1 βˆ’ 𝑝𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]𝑧 βˆ’ π‘žπ‘†

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]]

We define

𝑃(𝑧) = 𝑃

𝐡

(𝑧) + 𝑃

𝑉

(𝑧) (37)

as the probability generating function for the number of customers in the orbit irrespective of the server state.

Where 𝑃

𝑉

(𝑧) and 𝑃

𝐡

(𝑧) are given in equation (35) and (36). We shall now use the normalizing condition 𝑃(1) = 1 to determine the unknown 𝑄

0,0

which appears in (37). Substituting 𝑧 = 1 in (37) and using L'Hospitals rule, we obtain

𝑄0,0 = 1βˆ’[𝑝+πœ†πΈ(𝑋)𝐸[𝑆𝑏]+π‘ž(1βˆ’π΅βˆ—(πœ†))𝐸(𝑋)]

[1βˆ’π‘]

[

πœ†βˆ’πœ†π‘žπ‘ˆ(𝑧1)+πœ‚

πœ‚ βˆ’[πœ†+πœ‚βˆ’π‘žπ‘ˆ(𝑧1)[πœ†+πœ‚π΄βˆ—(πœ†+πœ‚)]][πœ†πΈ[𝑆𝑏]π‘†π‘£βˆ—(πœ‚)+(1βˆ’π΅βˆ—(πœ†))[1βˆ’π‘π‘†π‘£βˆ—(πœ‚)]]

(πœ†+πœ‚)[1βˆ’π‘π‘†π‘£βˆ—(πœ‚)]βˆ’π‘žπ‘†π‘£βˆ—(πœ‚)[πœ†+πœ‚π΄βˆ—(πœ†+πœ‚)]

] (38)

4. PARTICULAR CASES

Case (i):

If no customer receives the feedback service then by setting 𝑝 = 0 in (37),

we get 𝑃(𝑧) = 𝑃

𝑉

(𝑧) + 𝑃

𝐡

(𝑧) (39)

𝑃

𝑉

(𝑧) =

𝑁𝑉(𝑧)

𝐷𝑉(𝑧)

𝑄

0,0

and 𝑃

𝐡

(𝑧) =

𝑁𝐡(𝑧)

𝐷𝐡(𝑧)

𝑄

0,0

𝑁

𝑉

(𝑧) = πœ†[1 βˆ’ 𝐴

βˆ—

(πœ† + πœ‚)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑋(𝑧)𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚) βˆ’ π‘ˆ(𝑧

1

)𝑧] + πœ†[1

βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][(πœ† + πœ‚)𝑋(𝑧) βˆ’ π‘ˆ(𝑧

1

)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧)

+ πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]] + [πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][(πœ† + πœ‚)𝑧 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]

𝐷

𝑉

(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][(πœ† + πœ‚)𝑧 βˆ’ S

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]

𝑁

𝐡

(𝑧) = πœ†πœ‚π‘§[1 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][(πœ† + πœ‚)𝑋(𝑧) βˆ’ π‘ˆ(𝑧

1

)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]][1 βˆ’ 𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))[𝐡

βˆ—

(πœ†) + (1 βˆ’ 𝐡

βˆ—

(πœ†)𝑋(𝑧)]] + πœ†πœ‚[1

βˆ’ 𝐴

βˆ—

(πœ† + πœ‚)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑋(𝑧)𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚) βˆ’ π‘ˆ(𝑧

1

)𝑧][𝑧 + [1

βˆ’ 𝑧 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))][𝐡

βˆ—

(πœ†) + (1 βˆ’ 𝐡

βˆ—

(πœ†)𝑋(𝑧)] βˆ’ πœ†[1 βˆ’ π‘ˆ(𝑧

1

)][πœ†

βˆ’ πœ†π‘‹(𝑧) + πœ‚][(πœ† + πœ‚)𝑧 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]][𝑧 + [1 βˆ’ 𝑧 βˆ’ 𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))][𝐡

βˆ—

(πœ†) + (1 βˆ’ 𝐡

βˆ—

(πœ†)𝑋(𝑧)]]

𝐷

𝐡

(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑧 βˆ’ 𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))[𝐡

βˆ—

(πœ†) + (1 βˆ’ 𝐡

βˆ—

(πœ†)𝑋(𝑧)]][(πœ† + πœ‚)𝑧 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)[πœ†π‘‹(𝑧) + (πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)𝐴

βˆ—

(πœ† + πœ‚)]]

and

𝑄

0,0

=

1βˆ’πœ†πΈ(𝑋)𝐸[𝑆𝑏]βˆ’(1βˆ’π΅βˆ—(πœ†))𝐸(𝑋)

πœ†βˆ’πœ†π‘ˆ(𝑧1)+πœ‚

πœ‚ βˆ’

[

πœ†+πœ‚βˆ’π‘ˆ(𝑧1)[πœ†+πœ‚π΄βˆ—(πœ†+πœ‚)]

πœ†+πœ‚βˆ’π‘†π‘£βˆ—(πœ‚)[πœ†+πœ‚π΄βˆ—(πœ†+πœ‚)πœ†πΈ(𝑆𝑏)π‘†π‘£βˆ—(πœ‚)+(1βˆ’π΅βˆ—(πœ†))

] (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.

(10)

Case (ii):

If there is no retrial then on setting 𝐡

βˆ—

(πœ†) = 1, 𝐴

βˆ—

(πœ† + πœ‚) = 1 in (37), we get

𝑃(𝑧) = 𝑃

𝑉

(𝑧) + 𝑃

𝐡

(𝑧) (41)

𝑃

𝑉

(𝑧) =

𝑁𝑉(𝑧)

𝐷𝑉(𝑧)

𝑄

0,0

and 𝑃

𝐡

(𝑧) =

𝑁𝐡(𝑧)

𝐷𝐡(𝑧)

𝑄

0,0

𝑁𝑉(𝑧) = πœ†[1 βˆ’ π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑍) + πœ‚)][𝑋(𝑧) βˆ’ π‘žπ‘ˆ(𝑧1)𝑧] + [πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑧 βˆ’ (π‘ž + 𝑝𝑧)π‘†π‘£βˆ—(πœ†

βˆ’ πœ†π‘‹(𝑧) + πœ‚)]

𝐷

𝑉

(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑧 βˆ’ (π‘ž + 𝑝𝑧)𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]

𝑁𝐡(𝑧) = πœ†πœ‚π‘§[1 βˆ’ π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][𝑋(𝑧) βˆ’ π‘žπ‘ˆ(𝑧1)𝑧][1 βˆ’ π‘†π‘βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧))] βˆ’ πœ†[1 βˆ’ π‘žπ‘ˆ(𝑧1)][πœ†

βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑧 βˆ’ (π‘ž + 𝑝𝑧)π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][1 βˆ’ π‘†π‘βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧))]

𝐷𝐡(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑧 βˆ’ (π‘ž + 𝑝𝑧)π‘†π‘βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧))][𝑧 βˆ’ (π‘ž + 𝑝𝑧)π‘†π‘βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]

𝑄

0,0

=

1βˆ’[𝑝+πœ†πΈ(𝑋)𝐸(𝑆𝑏)]

[1βˆ’π‘]

[

πœ†βˆ’πœ†π‘žπ‘ˆ(𝑧1)+πœ‚

πœ‚ βˆ’1βˆ’π‘žπ‘ˆ(𝑧1)

1βˆ’π‘†π‘£βˆ—(πœ‚)[λ𝐸(𝑆𝑏)π‘†π‘£βˆ—(πœ‚)]

] (42)

Equation (41) is well known generating function of the queue size distribution of an 𝑀

𝑋

/𝐺/1 queue with feedback and multiple working vacation studied by K.Santhi and S.Pazhani Bala Murugan

14

irrespective of the notation.

Case(iii):

If no customer receives the feedback service and no retrial then on setting 𝐴

βˆ—

(πœ† + πœ‚) = 1, 𝐡

βˆ—

(πœ†) and 𝑝 = 0 in (37), we get

𝑃(𝑧) = 𝑃

𝑉

(𝑧) + 𝑃

𝐡

(𝑧) (43)

𝑃

𝑉

(𝑧) =

𝑁𝑉(𝑧)

𝐷𝑉(𝑧)

𝑄

0,0

and 𝑃

𝐡

(𝑧) =

𝑁𝐡(𝑧)

𝐷𝐡(𝑧)

𝑄

0,0

𝑁𝑉(𝑧) = πœ†[1 βˆ’ π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][𝑋(𝑧) βˆ’ π‘ˆ(𝑧1)𝑧] + [πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑧 βˆ’ π‘†π‘£βˆ—(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]

𝐷

𝑉

(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑧 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]

𝑁

𝐡

(𝑧) = πœ†πœ‚π‘§[1 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)][𝑋(𝑧) βˆ’ π‘ˆ(𝑧

1

)𝑧][1 βˆ’ 𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧))]

𝐷

𝐡

(𝑧) = [πœ† βˆ’ πœ†π‘‹(𝑧)][𝑧 βˆ’ 𝑆

π‘βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧)][πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚][𝑧 βˆ’ 𝑆

π‘£βˆ—

(πœ† βˆ’ πœ†π‘‹(𝑧) + πœ‚)]

𝑄

0,0

=

1βˆ’πœ†πΈ(𝑋)𝐸(𝑆𝑏)

[

πœ†βˆ’πœ†π‘ˆ(𝑧1)+πœ‚πœ‚ βˆ’1βˆ’π‘ˆ(𝑧1)1βˆ’π‘†π‘£βˆ—(πœ‚)[λ𝐸(𝑆𝑏)π‘†π‘£βˆ—(πœ‚)]

] (44) Equation (43) is well known generating function of the queue size distribution of an 𝑀

𝑋

/𝐺/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

1. M.I.Aftab Begum, Analysis of the batch arrival 𝑀

𝑋

/𝐺/1 queue with exponentially

distributed multiple working vacations, International Journal of Mathematical Sciences

and Applications, 1(2), 865-880 (2011).

(11)

2. J.R.Artalejo and A.Gemoz-Corral, Retrial Queueing Systems, Springer, Berlin, Germany, (2008).

3. Y.Baba, Analysis of a 𝐺𝐼/𝑀/1 queue with multiple working vacations, Operations Research Letters, Vol. 33, No.7, pp-654 - 681.

4. B.D.Choi, K.K.Park and C.E.M.Pearce, An 𝑀/𝑀/1 retrial queue with control policy and general retrial times, Queueing systems: Theory and Applications, Vol.14, No.3-4, pp.275- 292, (1993).

5. T.V.Do, 𝑀/𝑀/1 retrial queue with working vacations, Acta Informatica, Vol.47, No.1, pp.67 - 75, (2010).

6. J.Ke and F.Chang, Modified vacation policy for 𝑀/𝐺/1 retrial queue with balking and feedback, Computers and Industrial Engineering, Vol.57, No.1, pp. 433 - 443, (2009).

7. A.Krishnamoorthy and C.Sreenivasan, An 𝑀/𝑀/2 queueing system with hetrogeneous servers including one with working vacation, International Journal of Stochastic Analysis, Vol. 2012, Article ID 145867, 16 pages, (2012).

8. B.K.Kumar, S.P.Madheswari and A.Vijaykumar, The 𝑀/𝐺/1 retrial queue with feedback and starting failures, Applied Mathematical Modelling, Vol.26,No.11, pp.1057-1075, (2002).

9. B.K.Kumar, G.Vijayalakshmi, A.Krishnamoorthy and S.S.Basha, A Single server feedback retrial queue with collisions, Computers Operations Research, Vol.37, No.7, pp.1247 - 1255, (2010).

10. B.K.Kumar, S.P.Madheswari and S.R.A.Lakshmi, An 𝑀/𝐺/1 Bernoulli feedback retrial queueing system with negative customers, Operational Research, Vol.13, No.2, pp.187 - 210, (2013).

11. R.E.Lillo, A 𝐺/𝑀/1 queue with exponential retrial, Top, Vol.4, No.1, pp.99-120, (1996).

12. M.Martin and A.Gemoz-Corral, On the 𝑀/𝐺/1 retrial queueing system with linear control policy, Top, Vol.3, No.2, pp.285 - 305, 1995.

13. J.Medhi, Stochastic Processes, Wiley Eastern, (1982).

14. K.Santhi and S.Pazhani Bala Murugan, A Bulk Input Queueing System with Feedback and Single Working Vacation, International Journal of Scientific Research and Management Studies, 1, 168-176 (2014).

15. S.Pazhani Bala Murugan and R.Vijaykrishnaraj, A bulk arrival retrial queue with exponentially distributed multiple working vacation, Journal of Emerging Technologies and Innovative Research, 5, 1-9 (2018).

16. L.D.Servi and S.G.Finn, 𝑀/𝑀/1 queue with working vacations (𝑀/𝑀/1/π‘Šπ‘‰), Performance Evaluation, vol. 50, No 1, pp. 42-52, (2002).

17. N.P.Sherman and J.P.Kharoufeh, An 𝑀/𝑀/1 retrial queue with unreliable server, Operations Research Letters, Vol.34, No.6, pp.697-705, (2006).

18. D.Wu and H.Takagi, The 𝑀/𝐺/1 queue with multiple working vacation, Performance Evaluations, 63, 654-681 (2006).

19. Z.Zhang and X.Xu, Analysis for the 𝑀/𝑀/1 queue with mulitple working vacations and

N-policy, International Journal of Information and Management Sciences, Vol.19, No.3,

pp.495-506, (2008).

References

Related documents