A Bulk Arrival Non-Markovian Queueing System with Two types of Second Optional Services and with Second Optional Vacation

(1)

Vol. 6, No. 1, 2015,91-98

ISSN: 2320 –3242 (P), 2320 –3250 (online) Published on 22 January 2015

www.researchmathsci.org

91

International Journal of

A Bulk Arrival Non-Markovian Queueing System with

Two types of Second Optional Services and with Second

Optional Vacation

P.Manoharan1 and K.Sankara Sasi2 1

Department of Mathematics, Annamalai University, Annamalainagar– 608 002, India e-mail: [email protected]

Corresponding Author 2

Department of Mathematics, Annamalai University, Annamalainagar– 608 002, India e-mail: [email protected]

Received 2 October 2014; accepted 29 November 2014

Abstract. We study anMX/G/1queueing system where the arrival follows a compound Poisson process. The server provides the first service to all customers which is essential. Only some of the customers demand second service which is optional. The second optional service is given in two types. Soon after the system becomes empty the server takes a vacation and after returning from vacation the server may take second vacation which is optional. Using supplementary variable technique we derive the probability generating function for the number of customers in the system. Some performance measures are calculated. Some particular cases are derived.

Keywords: First essential service, Second optional service, Regular vacation, Second optional vacation and Supplementary variable technique.

AMS Mathematics Subject Classification (2010): 60K25, 60K30 1. Introduction

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92 2. The model

The following assumptions briefly describe the mathematical model of our study.

• Customers arrive at the system in batches of variable size in a compound Poisson process.

• There is a single server who provides the first essential service (FES) to all customers. The service time for FES follows a general distribution. Let B₁(x)and

) ( 1 x

b respectively denote the distribution function and the density function of the

FES times.

• As soon as the first service is completed the customer can exercise any one of the following three options.

i. may leave the system with probability 1− p=q

ii. may opt for type 1 optional service with probability pp₁ iii. may opt for type 2 optional service with probability pp₂ wherep+q=1 and p1+p2=1

The type i (i=1, 2) optional service times are assumed to be negative exponential with mean service times 1 ₀ ₁₂

, i ;

µ

,

µi i

=

> . Let B_i(ν)and b_i(ν);i=1,2be the

distribution and density function of type 1 and type 2 services respectively. • Further it is assumed thatµi(x)dxis the conditional probability of the completion

of the ith service given that the elapsed service time is x so that

(x) B

(x) b (x)dx

µ

i i

i = ₋

1

and therefore b(x) µ(x)e ;

x i(t)dt µ

i i

∫

= −0 i∈{1,2,3}

• We assume that the FES and SOS are mutually independent of each other. Let

} 3 , 2 , 1 { ), 1 ( ) ( ), (

* _s _E_B _k≥ _i∈

B k

i

i denote the LST and finite moments of two service

times respectively. Thus the total service time required by the server to complete the service cycle which may be called as modified service period is given by

  

− +

=

p 1 y probabilit with

B

p y probabilit with

B B

1 2 1

B

• Whenever the system becomes empty, the server goes for a first regular vacation (FRV) of random lengthV₁. Let V1(x)and v1(x) respectively be the distribution function and density function of the first vacation times.

• At the end of FRV, the server may take the second optional vacation SOV with probability θ. Otherwise he remains in the system with probability (1−θ)until a

new customer arrives. Let V2(x)andv2(x) respectively be the distribution function and density function for the SOV times. Further it is assumed that

dx x

i( )

ν is the conditional probability of the completion of the ݅௧௛vacation given that the elapsed vacation time is x so that

) ( 1

) ( )

(

x V

x v dx x

i i i

− =

ν and therefore

; )

( )

( 0

) (

∫

= −

x

i tdt

i

i x x e

v

ν

(3)

93

• It is also assumed that the vacation times V and ₁ V are mutually independent of ₂ each other having general distribution function V_i(x), LST *

i

V and finite moments

} 2 , 1 { ), 1 ( )

(V k≥ i∈

E ik . Thus the total vacation time required to complete the

vacation cycle, which may be called as modified vacation period is given by

  

− +

=

θ θ 1 y probabilit with

V

y probabilit with

V V

1 2 1

V

3. Queue size distribution at a random epoch

Here we first set up the steady state equations for the stationary queue size distribution by treating elapsed service time, FES time, SOS time, FRV time and SOV time as supplementary variables. Then we solve these equations and derive the PGF’s, assuming that the system is in steady state condition. Let N(t) be the queue size (including one being served, if any),B1(0)(t) be the elapsed service time at tfor FES, ()

) 0 ( 2 t

B be the elapsed service time at tfor the SOS, V1(0)(t)be the elapsed vacation time at t for the FRV, V2(0)(t)be the elapsed vacation time at t for the SOV. For further development of this model let us introduce the random variable Y(t) as follows.

  

  

 

=

t time at SOS 2 type giving busy is server the if 4

t time at SOS 1 type giving busy is server the if 3

t time at FES giving busy is server the if 2

t time at SOV on is server the if 1

t time at FRV on is server the if 0

Y(t)

The supplementary variables (), (); 1(0)() )

0 ( 2 ) 0 (

1 t V t B t

V ,B₂(0)(t) andB₃(0)(t) are introduced in order to obtain a bivariate Markov process

{

N(t);∂(t);t≥ 0

}

where

  

   

 

= = = = =

= ∂

4 Y(t) if (t) B

3 Y(t) if (t) B

2 Y(t) if (t) B

1 Y(t) if (t) V

0 Y(t) if (t) V

(t)

(0) 3

(0) 2

(0) 1

(0) 2

(0) 1

We define the limiting probabilities as follows.

{

() ; ( ) ( ); ( )

}

; 0; 0

Pr lim )

( (0) ₁(0)

1 ,

1 x dx= _→_∞ N t =n ∂t =V t x<V t ≤x+dx n≥ x> Q

t n

{

( ) ; ( ) ( ); ( )

}

; 0; 0

Pr lim )

( (0) ₂(0)

2 ,

2 x dx= _→_∞ N t =n ∂ t =V t x<V t ≤x+dx n≥ x> Q

t n

{

() ; ( ) ( ); ()

}

; 0; 0

Pr lim )

( (0) ₁(0)

1 ,

1 x dx= _→_∞ N t =n ∂ t =B t x<B t ≤x+dx n≥ x> P

t n

{

( ) ; () (); ( )

}

; 0; 0

Pr lim )

( ₂(0) ₂(0)

,

t n

{

( ) ; ( ) ( ); ( )

}

; 0; 0

Pr lim )

( ₃(0) ₃(0)

,

(4)

94

Further it is assumed thatBi(0)(0)=0;Bi(0)(∞)=1 for i∈{1,2,3}and (0) 0; ( ) 1 (0)

(0) ₌ _∞ ₌

i

i V

V

for i∈{1,2}and are continuous at x=0.The differential difference equations governing the system are

) ( )

( )) ( ( )

( ₁_, ₁

1 ,

1 1 ,

1 x x P x c P x

P dx

d

n n

k k n

n −

=

∑

= +

+ λ µ λ x>0n≥1

( )

1

0 ) ( )) ( ( )

( 1 1,0

0 ,

1 x + + x P x =

P dx

d _λ _µ _x_>₀

( )

2

) ( )

( )) ( ( )

( ₂_, ₁

1 ,

2 2 ,

2 x x P x c P x

P dx

d

n n

k k n

n −

=

∑

= +

+ λ µ λ x>0n≥1

( )

3

0 ) ( )) ( ( )

( ₂ ₂_,₀

0 ,

2 x + + x P x =

P dx

d

µ

λ x>0

( )

4

) ( )

( )) ( ( )

( ₃_, ₁

1 ,

3 3 ,

3 x x P x c P x

P dx

d

n n

k k n

n −

=

∑

= +

+ λ µ λ x>0 n≥1

( )

5 0

) ( )) ( ( )

( 3 3,0

0 ,

3 x + + x P x =

P dx

d

µ

λ x>0

( )

6

) ( )

( )) ( ( )

( ₁_, ₁

1 ,

1 1 ,

1 x x Q x c Q x

Q dx

d

n n

k k n

n −

=

∑

= +

+ λ ν λ x>0 n ≥1

( )

7

0 ) ( )) ( ( )

( ₁ ₁_,₀

0 ,

1 x + + x Q x =

Q dx

d

ν

λ x>0

( )

8

) ( )

( )) ( ( )

( ₂_, ₁

1 ,

2 2 ,

2 x x Q x c Q x

Q dx

d

n n

k k n

n −

=

∑

= +

+ λ ν λ x> 0n≥1

( )

9

0 ) ( )) ( ( )

( ₂ ₂_,₀

0 ,

2 x + + x Q x =

Q dx

d

ν

λ x>0

( )

10

dx x x P dx x x P dx x x P p

Q (1 ) ( ) ( ) ( ) ( ) ( ) ₃( )

0 3,0 2

0 2,0 1

0 1,0 0

,

1 µ µ µ

λ = −

∫

∞ +

∫

∞ +

∫

∞

dx x x Q dx x x

Q ( ) ( ) ( ) ( ) )

1

( ₂

0 2,0 1

0 1,0 ν ν

θ

_∫

∞ +

_∫

∞

− +

(11) whereQ Q (x)dx

0 1,0 0

,

1

∫

∞

=

The boundary conditions are

0 , 1 0

,

1 (0) Q

Q =λ

(12) 0

) 0 ( , 1n =

Q n≥1(13)

dx x x Q

Q _n(0) _n( ) ₁( )

0 1, ,

2 θ

∫

ν

∞

= n≥0(14)

P (0) (1 ) ( ) ( ) ( ) ( ) ( ) ₃( )

0 3,1 2

0 2,1 1

0 1,1 0

,

1

∫

µ

∫

µ

∫

µ

∞ ∞

∞

+ +

− =

dx x x Q dx x x

Q ( ) ( ) ( ) ( )

) 1

( 2

0 2,1 1

0 1,1 ν ν

θ

∫

∞ +

∫

∞

− +

(15)

P_n(0) (1 ) _n ( ) ( ) _n ( ) ( ) _n ( ) ₃( )

0 3, 1 2

0 2, 1 1

0 1, 1 ,

1

∫

µ

∫

µ

∫

µ

∞ + ∞

+ ∞

+ + +

− =

dx x x Q dx x x

Q _n ( ) ( ) _n ( ) ( ) )

1

( ₂

0 2, 1 1

0 1, 1 ν ν

θ

_∫

∞ ₊ +

_∫

∞ ₊

− +

(16)

dx x x P pp

P _n(0) _n( ) ₁( )

0 1, 1 ,

2

∫

µ

∞

(5)

95 dx

x x P pp

P3,n(0) 2

∫

₀ 1,n( )µ1( )

∞

= n≥0(18) and the normalizing condition is

1 ) ( )

(

1 2

1 0 , 1

3

1 0

, +

∑ ∑ ∫

=

∑ ∑ ∫

∞

= = ∞ ∞

= = ∞

dx x Q dx

x P

n i n i n i

n i

) 19 (

Now let us define the following PGF’s

); ( )

,

( ,

0

x P z z x

P in

n n

i

∑

∞ =

= x≥0; z≤1, i∈

{

1,2,3

}

(20)

); 0 ( )

, 0

( _,

0

n i n

n

i z z P

P

∑

∞

=

= z ≤1, i∈

{

1,2,3

}

(21)

); ( )

,

( _,

0

x Q z z x

Q _i_n

n n

i

∑

∞

=

= x≥0; z ≤1, i∈

{

1,2

}

(22)

); 0 ( )

, 0

( ,

0

n i n

n

i z z Q

Q

∑

∞

=

= z≤1, i∈

{

1,2

}

(23)

, ) , ( )

(

0 P x z dx z

Pi

∫

i

∞

= i∈

{

1,2,3

}

(24)

, ) , ( )

(

0 Q x z dx z

Qi

∫

i

∞

= i∈

{

1,2

}

(25)

Multiplying (1) by

z

n and summing over n=0 to

∞

and adding with (2), we get

(

( ) ( )

)

( , ) 0

) ,

( 1 1

1 x z + − X z + x P x z =

P dx

d _λ _λ _µ

[

]

X z x

e x B z P z x

P1( , ) 1(0, )1 1( ) (1 ( ))

− − −

= λ

(26)

where k

k kz

c z

X

∑

∞

=

1

) (

Multiplying (3) by

z

∞

(

( ) ( )

)

( , ) 0

) ,

( 2 2

2 x z + − X z + x P x z = P

dx d

µ λ

λ

[

]

X z x

e x B z P z x

P₂( , )= ₂(0, )1− ₂( ) −λ(1− ( )) (27)

Multiplying (5) by

z

∞

(

( ) ( )

)

( , ) 0

) ,

( 3 3

3 x z + − X z + x P x z = P

dx

d _λ _λ _µ

[

_B _x

]

_e X z x

z P z x

P (1 ( ))

3 3

3( , ) (0, )1 ( )

− − −

= λ

(28) Multiplying (7) by

z

n and summing over n=0to

∞

(

( ) ( )

)

( , ) 0 )

,

( 1 1

1 x z + − X z + x Q x z = Q

dx

d _λ _λ _ν

[

]

Xz x

e x V z Q z x

Q1( , ) 1(0, )1 1( ) (1 ( ))

− − −

= λ

(29) Multiplying (9) by z୬ and summing over n = 0 to∞ and adding with (10), we get

(

( ) ( )

)

( , ) 0 )

,

( 2 2

2 x z + − X z + x Q x z =

Q dx

d _λ _λ _ν

[

]

Xz x

e x V z Q z x

Q₂( , )= ₂(0, )1− ₂( ) −λ(1− ()) (30)

Multiplying (16) by

z

n+1 and summing over n=0to

∞

and adding with z times (15), we get

dx x z x P dx x z x P dx x z x P p z

zP(0, ) (1 ) ( , ) ( ) ( , ) ( ) ( , ) 3( )

0 3 2

0 2 1

0 1

1

∫

µ

∫

µ

∫

µ

∞ ∞

∞

+ +

− =

0 , 1 2

0 2 1

0 1( , ) ( ) ( , ) ( ) )

1

( −θ Q x zν x dx+ Q x zν x dx−λQ

+

_∫

∞

_∫

∞

(6)

96

From equation (17), we get P₂(0,z)=pp₁P₁(0,z)B₁*(λ−λX(z))(32) From equation (18), we get

)) ( ( ) , 0 ( ) , 0

( ₂ ₁ ₁*

3 z pp P z B X z

P = λ−λ (33)

From equation (26), we get ( , ) 1( ) 1(0, ) 1*( ( )) 0 P1 x z µ x dx=P z B λ−λX z

∫

∞ (34) From equation (27), we get

)) ( ( )) ( ( ) , 0 ( ) ( ) ,

( 2*

* 1 1 1 2

0 P2 x z µ x dx= ppP z B λ−λX z B λ−λX z

∫

)) ( ( )) ( ( ) , 0 ( ) ( ) , ( * 3 * 1 1 2 3

0 P3 x z µ x dx=pp P z B λ−λX z B λ−λX z

∫

∞

) 36 (

From equation (29), we get ( , ) ( ) (0, ) *( ( )) 1

1 1

0 Q1 x zν x dx=Q zV λ−λX z

∫

)) ( ( )) ( ( ) , 0 ( ) ( ) ,

( 2*

* 1 1 2

0 Q2 x zν x dx=θQ z V λ−λX z V λ−λX z

∫

∞ (38) From equation (13) and (14), we get Q₁(0,z)=λQ₁_,₀

(39) )) ( ( ) , 0 ( ) , 0

( 1 1*

2 z Q zV X z

Q =θ λ−λ (40) Using (34) to (40) in (31), we get

(

)

{

}

0 , 1 1 * 1 * 2 1 ) ( )) ( ( ))] ( ( 1 [ ) , 0 ( Q z D z X V z X V z P       ₋ ₊ ₋ ₋ = λ θ θ λ λ λ λ ) 41 (

whereD₁(z)=

{

z−[(1−p)+pp₁B₂*(λ−λX(z))+pp₂B₃*(λ−λX(z))]B₁*(λ−λX(z))

}

Using (41) in (32), we get

(

)

[

]

{

}

0 , 1 1 * 1 * 1 * 2 1 2 ) ( )) ( ( )) ( ( )) ( ( 1 ) , 0 ( Q z D z X B z X V z X V pp z P       ₋ ₊ ₋ ₋ ₋ = λ θ θ λ λ λ λ λ λ ) 42 (

Using (41) in (33), we get

(

)

[

]

{

}

0 , 1 1 * 1 * 1 * 2 2 3 ) ( )) ( ( )) ( ( )) ( ( 1 ) , 0 ( Q z D z X B z X V z X V pp z P _      ₋ ₊ ₋ ₋ ₋ = λ θ θ λ λ λ λ λ λ ) 43 (

Integrating (26) to (30) between 0 and ∞, we get

) , 0 ( )) ( ( )) ( ( 1 ) , ( ) ( ₁ * 1 0 1

1 P z

z X z X B dx z x P z P       − − − = =

_∫

∞ λ λ λ λ ) 44 ( ) , 0 ( )) ( ( )) ( ( )) ( ( 1 ) , ( )

( * ₁

1 *

2 1 0 2

2 B X z P z

z X z X B pp dx z x P z

P λ λ

λ λ λ λ ₋       − − − = =

_∫

∞ ) 45 ( ) , 0 ( )) ( ( )) ( ( )) ( ( 1 ) , ( ) ( 1 * 1 * 3 2 0 3

3 B X z P z

z X z X B pp dx z x P z

P λ λ

λ λ λ λ ₋       − − − = =

_∫

∞ ) 46 ( ) , 0 ( )) ( ( )) ( ( 1 ) , ( ) ( 1 * 1 0 1

1 Q z

z X z X V dx z x Q z Q       − − − = =

_∫

∞ λ λ λ λ ) 47 ( ) , 0 ( )) ( ( ))] ( ( 1 ))[ ( ( ) , ( ) ( 1 * 2 * 1 0 2

2 Q z

z X z X V z X V dx z x Q z Q       − − − − = =

_∫

∞ λ λ λ λ λ λ θ ) 48 (

(7)

97

[

]

{

}(

)

0 . 1 1 * 1 * 1 * 2 1 )) ( 1 )( ( )) ( ( 1 )) ( ( )) ( ( ) 1 ( ) , 0 ( Q z X z D z X B z X V z X V z P       − − − − − + − = θ θ λ λ λ λ λ λ ) 49 (

[

]

{

}(

)

0 . 1 1 * 1 * 2 * 1 * 2 1 2 )) ( 1 )( ( )) ( ( )) ( ( 1 )) ( ( )) ( ( ) 1 ( ) , 0 ( Q z X z D z X B z X B z X V z X V pp z P       − − − − − − + − = θ θ λ λ λ λ λ λ λ λ ) 50 (

[

]

{

}(

)

0 . 1 1 * 1 * 3 * 1 * 2 2 3 )) ( 1 )( ( )) ( ( )) ( ( 1 )) ( ( )) ( ( ) 1 ( ) , 0 ( Q z X z D z X B z X B z X V z X V pp z P       − − − − − − + − = θ θ λ λ λ λ λ λ λ λ ) 51 (

Using (39) in (47) and (48), we get

(

)

1,0 * 1 1 ) ( 1 )) ( ( 1 ) ( Q z X z X V z Q       − − − = λ λ ) 52 (

[

]

(

)

1,0 * 2 * 1 2 ) ( 1 )) ( ( 1 )) ( ( ) ( Q z X z X V z X V z Q       − − − − = θ λ λ λ λ ) 53 (

From the fact that P₁(1)+P₂(1)+P₃(1)+Q₁(1)+Q₂(1)=1 , we arrived

[

( ) ( )

]

1 2 1 0 , 1 V E V E Q θ λ ρ + −

= , where ρ=λX′(z)

[

E(B₁)+pp₁E(B₂)+pp₂E(B₃)

]

( )

₅₄

and (0) ( ), (0) ( ), *(0) ( ₃) 3 2 * 2 1 *

1 E B B E B B E B

B =− =− =− are the mean of service times of FES, type1 SOS time and type2 SOS time respectively, (0) ( 1)

*

1 EV

V =− and (0) ( 2)

*

2 EV

V =− are

the mean of vacation times of FRV and SOV respectively. Therefore

) ( ) ( ) ( Z D Z N z P =

( )

55

where

{

(

)

*

}

( )

₁_,₀

1 *

2( ( ))] ( ( )) 1 1

1 [ )

(z V X z V X z zQ

N = −θ +θ λ−λ λ−λ − −

(

)

{

[1 ( ( )) ( ( ))] ( ( ))

}

(1 ( ))

) ( * 1 * 3 2 * 2

1B X z ppB X z B X z X z

pp p z

z

D = − − + λ−λ + λ−λ λ−λ −

3.1. Performance measures

Let L_qand Ldenote the steady state average queue size and system size respectively.

Then

1 1 ₍ ₎

) ( ) ( = = = = z z q z D z N dz d z P dz d

L Using the L’Hospital rule twice we obtain

2 ' '' ' '' ' )) 1 ( ( 2 ) 1 ( ) 1 ( ) 1 ( ) 1 ( D D N N D

L_q = −

) 56 (

where ' ₁ ₂ ₁_,₀

)) ( ) ( )( ( 2 ) 1

( X z EV EV Q

N =− λ ′ +θ

(

)

{

1 2

}

1,0

2 2 2 1 2 2 2 1

''₍₁₎ ₃ _X ₍_z₎ _E₍_V₎ _E₍_V ₎ ₍_X ₍_z₎₎_[_E₍_V ₎ _E₍_V ₎ ₂ _E₍_V₎_E₍_V_)]_Q

N =− λ ′′ +θ +λ ′ +θ + θ

(

)

(

( ) ( ) ( ) ( ) 1

)

) ( 2 ) 1

( 1 1 2 2 3

' = ′ ′ + + − B E pp B E pp B E z X z X D

λ

(

)

{

2 ( ) ( ) ( ) ( ) ( ) ( )

3 ) 1

( 1 1 2 2 3

'' z X B E pp B E pp B E z X z X

D = λ ′ ′′ + + − ′′

(

)

{

( ) ( ) ( ) 2 ( ) ( ) ( )

}}

)

( 1 1 2 2 3

2 3 2 2 2 1 2 1 3 2 B E p B E p B pE B E pp B E pp B E z

X′ + + + +

+λ

where ( ), ( ), ( ), ( ), ( 2) 2 2 1 2 3 2 2 2

1 E B E B EV E V

B

E second moment of FES, type 1 SOS, type 2 SOS, FRV and SOV time respectively. Now, we can obtainL=L_q +ρ, where L_q

and ρhave been found in (56) and (54) respectively. Then using Little’s formulae, we obtain Wq,the average waiting time in the queue and ܹ, the average waiting time in the

system, as

λ

q q

L W = and

λ L

(8)

98 3.2. Particular cases

Case 1: If there is no one opting for type 2 second optional service then by setting p2 =0, then (55) takes the form

( )

[ ]

{ }

( )

[ ]

{ * }( ) 1,0

1 *

2

* 1 *

2

) ( 1 )) ( ( )) ( ( 1

) 1 ( 1 )) ( ( )) ( ( 1 )

( Q

z X z X B z X pB p z

z z X V z X V z

P

− − −

+ − −

− − − −

+ −

= θ θ λ_λ λ_λ λ_λλ_λ (57)

which coincides the model studied by Manoharan [8].

Case 2: If there is no one opting for second optional service (both type 1 and type 2) and if there is a single arrival then by setting݌ଵ= 0, ݌ଶ= 0 and X(z)=z (55) takes the

form

(

)

[

]

{

}

[

*

]

1,0

1

* 1 *

2

) (

1 ) ( ) ( 1

)

( Q

z B

z

z V

z P

λ λ

λ λ λ λ θ θ

− −

− − −

+ − =

(58)

which coincides with the PGF of Choudhury [4] irrespective of the notations used.

6. Conclusion

The analysis carried out in“Abulk arrival non-markovian queueing system with two types of second optional services and with second optional vacation” is to obtain the probability generating function for the number of customers in the system and also to obtain waiting time of a customer in the system.

REFERENCES

1. Y. Baba, On theMX/G/1queue with vacation time, Oper. Res. Let., 5 (1986) 93-98. 2. B.D.Choi and K.K.Park, The M/G/1 retrial queue with Bernoulli schedule, Queueing

System, 7 (1990) 219-228,.

3. G.Choudhury, Analysis of the MX/G/1 queueing system with vacation times, Sankhya, 64-B Pt. 1 (2002) 37-49.

4. G.Choudhury, An M/G/1 queue with an optional second vacation, Information and Management Sciences, 17(3) (2006) 19-30.

5. J.W.Cohen, The Single Server Queue, North-Holland, Amsterdam, 1969.

6. R.Kalyanaraman and S.Pazhani Bala Murugan, An M/G/1queue with second optional service and with server vacation, Annamalai University Science Journal, 45 (2008) 129-134.

7. C.Madan Kailash, AnM/G/1queue with second optional service, Queueing Systems, 34 (2000) 37-46.

8. P.Manoharan and K.Sankara Sasi, An MX/G/1 queue with Second optional service and with optional second vacation. Proceedings of ICMMA2012, Department of Mathematics, Annnamalai University, Annamalainagar-608 002, Tamilnadu, India, December 22-24, 2012.

9. J.Medhi, Stochastic Processes, Wiley Eastern, 1982.

10. V.Thangaraj and S.Vanitha, A single server M/G/1feedback queue with two type of service having general distribution, International Mathematical Forum, 5(1) (2010) 15-33.