A study on the behaviour of the server breakdown without interruption
in a
M
[X]/G
(
a,
b
)
/1
queueing system with Multiple vacations, Accessible
batches and Closedown time
Karpagam. S
a,∗and G.Ayyappan
ba,bDepartment of Mathematics, Pondicherry Engineering College, Puducherry-605014, India.
In this paper we consider M[X]/G(a,b)/1 queueing system with accessible service, server breakdown without interruption, multiple vacation and closedown time. The service rule is as follows: the server starts service only when a minimum number of customers0a0is available in the queue and the maximum service capacity is0b0 units such that late entries are allowed to join a service batch, without affecting the service time, if the size of the batch being served is less than0d0 (maximum accessible limit). After completing a batch of service, if the server is breakdown with probability(π)then the renovation of service station will
be considered. After completing the renovation of service station or if there is no breakdown of the server with probability(1−π), if the queue lengthδ < a, then the server performs a closedown work. Following
closedown work, the server leaves for a vacation of random length irrespective of queue length. When the server returns from a vacation and if the queue length is still less than0a0he leaves for another vacation and so on until he finds minimum0a0customers waiting for service in the queue. The probability generating function of the queue size at an arbitrary time, some important characteristic of the queueing system and cost model are derived.
Keywords: Bulk service, Accessible batches, Server breakdown, Multiple vacation, Closedown time.
c
2015 MJM. All rights reserved.
1
Introduction
In this paper we considerM[X]/G(a,b)/1 queueing system with accessible service, server breakdown without interruption, multiple vacation and closedown time.The service starts only if minimum of0a0customers are available in the queue. At the service completion epoch, if the number of customers is0δ0, wherea≤δ≤d−1
(d ≤b)then the server takes the entire queue for batch service and admits arrivals for service till the service of the current batch is over or, the accessible limit d, is reached, whichever occurs first. At the service initiation epoch, if the number of customers waiting in the queueδ(δ ≥ d)then the server takes min(δ,b)customers
for service and does not allow further arrival into service batch. On completion of a service, if the server is breakdown with probability(π)then the renovation of service station will be considered.After completing the
renovation of service station or if there is no breakdown of the server with probability(1−π), if the queue
lengthδ < a, then the server performs a closedown work such as, shutting down the machine, removing
the tools etc. Following closedown work, the server leaves for a vacation of random length irrespective of queue length. When the server returns from a vacation and if the queue length is still less than0a0he leaves for another vacation and so on until he finds minimum0a0 customers waiting for service in the queue. The concept of accessibility in Markovian queues with bulk service has been discussed by Medhi [5] and Gross and Harris [3]. Sivasamy[6] has analysed a M/M(a,d,b)/1 with accessible and non-accessible service batches. Ganesan [2] has analysed Erlangian queueing system with accessible batch service.
∗Corresponding author.
E-mail address:[email protected](Karpagam),[email protected](Ayyappan).
2
Notations
Let X be the group size random variable of the arrival, gk be the probability that ’k’customers arrive in a batch and X(z) be its probability generating function. Let S(.), V(.),C(.),and R(.) be the cumulative distributions of the service time,vacation time, closedown time, renovation time respectively. s(x), v(x), c(x),andr(x)be their corresponding probability density functions. At an arbitrary time, S0(t) denotes the remaining service time at time’t’ andV0(t),C0(t),andR0(t)denotes the remaining vacation time, closedown
time and renovation time at time ’t’ respectively. Let us denote the Laplace-Stieltjes transform s(x), v(x), c(x),andr(x)asS(˜θ)V˜(θ)C(˜θ)andR(˜θ).
ε(t) = 0, 1, 2 and 3 if the server is on busy, closedown, vacation and renovation
Z(t) =j, if the server is onjthvacation
Ns(t)=Number of customers in service at time t Nq(t)=Number of customers in queue at time t
Define the probabilities as,
Pij(x,t)dt= Pr{Ns(t) =i,Nq(t) =j,x≤S0≤x+dx,ε(t) =0},a≤i≤b,j≥0,
Qji(x,t)dt= Pr{Nq(t) =i,x≤V0≤x+dx,ε(t) =2,Z(t) =j}, i≥0 ,j≥1.
Cn(x,t)dt= Pr{Nq(t) =n,x≤C0≤x+dx,ε(t) =1},n≥0 .
Rn(x,t)dt= Pr{Nq(t) =n,x≤R0≤x+dx,ε(t) =3},n≥0.
In steady state, let us define forx >0, Pij(x) = lim
t→∞Pij(x,t)fora≤i≤b,j≥0
Qij(x) = lim
t→∞Qij(x,t)forl=1to∞,j≥1,
Cn(x) = lim
t→∞Cn(x,t)forn≥0 and
Rn(x)lim
t→∞Rn(x,t)forn≥0
Following Cox [1], the equations govern the system under steady state conditions can be obtained as follows:
−Pa00(x) =−λPa0(x) + (1−π)
b
∑
m=d
Pma(0)s(x) +
∞
∑
l=1
Qla(0)s(x) +Ra(0)s(x) (2.1)
−Pi00(x) =−λPi0(x) + (1−π)
b
∑
m=d
Pmi(0)s(x) +
∞
∑
l=1
Qli(0)s(x)
+ i−a
∑
k=1
Pi−k,0(x)λgk+ Ri(0)s(x), a+1≤i≤d (2.2)
−Pi00(x) =−λPi0(x) + (1−π)
b
∑
m=d
Pmi(0)s(x) +
∞
∑
l=1
Qli(0)s(x)
+Ri(0)s(x), d+1≤i≤b (2.3)
−Pdn0 (x) =−λPdn(x) + n
∑
k=1
Pd,n−k(x)λgk + d−a
∑
k=1
Pd−k,0(x)λgk+n, n≥1 (2.4)
−Pin0 (x) =−λPin(x) + n
∑
k=1
Pi,n−k(x)λgk, d<i<b, n≥1 (2.5)
−Pbn0 (x) =−λPbn(x) + (1−π)
b
∑
m=d
Pm,b+n(0)s(x) +
∞
∑
l=1
Ql,b+n(0)s(x)
+ n
∑
k=1
−C00(x) =−λC0(x) + (1−π)
b
∑
m=a
Pm0(0)c(x) +R0(0)c(x), (2.7)
−Cn0(x) =−λCn(x) + (1−π) b
∑
m=d
Pmn(0)c(x) + n
∑
k=1
Cn−k(x)λgk
+ Rn(0)c(x), 1≤n≤a−1, (2.8)
−Cn0(x) =−λCn(x) +
n
∑
k=1
Cn−k(x)λgk, n≥a (2.9)
−Q010(x) =−λQ10(x) +C0(0)v(x), (2.10)
−Q01n(x) =−λQ1n(x) + n
∑
k=1
Q1,n−k(x)λgk+Cn(0)v(x), n≥1 (2.11)
−Q0j0(x) =−λQj0(x) +Qj−1,0(0)v(x),j≥2 (2.12)
−Q0jn(x) =−λQjn(x) +Qj−1,n(0)v(x) + n
∑
k=1
Qj,n−k(x)λgk,j≥2, 1≤n≤a−1 (2.13)
−Q0jn(x) =−λQjn(x) + n
∑
k=1
Qj,n−k(x)λgk, n≥a, j≥2, (2.14)
−R00(x) =−λR0(x) +π
b
∑
m=d
Pm0(0)r(x) (2.15)
−R0n(x) =−λRn(x) +π
b
∑
m=d
Pmn(0)r(x) + n
∑
k=1
Rn−k(x)λgk, n≥1 (2.16)
3
Queue size distribution
The Laplace-Stieltjes transform ofPij(x), Qjn(x), Rn(x), Cn(x)are defined as follows:
˜ Pij(θ) =
Z ∞
0 e
−θxP
ij(x)dx, Q˜jn(θ) =
Z ∞
0 e
−θxQ
jn(x)dx,
˜ Cn(θ) =
Z ∞
0 e
−θxCn(x)dx, Rn˜ ( θ) =
Z ∞
0 e
−θxRn(x)dx,
(3.17)
Taking Laplace-Stieltjes transform on both sides of equations(2.1)to(2.16), we get
˜
θPa0(θ)−Pa0(0) =λPa˜0(θ)−(1−π)
b
∑
m=d
Pma(0)S(˜ θ)−
∞
∑
l=1
Qla(0)S(˜ θ)−Ra(0)S(˜ θ) (3.18)
˜
θPi0(θ)−Pi0(0) =λP˜i0(θ)−(1−π)
b
∑
m=d
Pmi(0)S(˜ θ)−
∞
∑
l=1
Qli(0)S(˜ θ)
− i−a
∑
k=1
˜
Pi−k,0(θ)λgk−Ri(0)S(˜ θ), a+1≤i≤d (3.19)
˜
θPi0(θ)−Pi0(0) =λPi˜0(θ)−(1−π)
b
∑
m=d
Pmi(0)S(˜ θ)−
∞
∑
l=1
Qli(0)S(˜ θ)
−Ri(0)S(˜ θ), d+1≤i≤b (3.20)
˜
θPdn(θ)−Pdn(0) =λP˜dn(θ)−
n
∑
k=1
˜
Pd,n−k(θ)λgk− d−a
∑
k=1
˜
θPin˜ (θ)−Pin(0) =λPin˜ (θ)−
n
∑
k=1
˜
Pi,n−k(θ)λgk, d<i<b, n≥1, (3.22)
θPbn˜ (θ)−Pbn(0) =λPbn˜ (θ)−(1−π)
b
∑
m=d
Pm,b+n(0)S(˜ θ)−Rb+n(0)S(˜ θ)
−
∑
∞l=1
Ql,b+n(0)S(˜ θ)−
n
∑
k=1
˜
Pb,n−k(x)λgk, n≥1 (3.23)
θC˜0(θ)−C0(0) =λC˜0(θ) − (1−π)
b
∑
m=a
Pm0(0)C(˜ θ) − R0(0)C(˜ θ), (3.24)
θCn˜ (θ)−Cn(0) = λCn˜ (θ)−(1−π)
b
∑
m=d
Pmn(0)C(˜ θ)−Rn(0)C(˜ θ)
− n
∑
k=1
˜
Cn−k(θ)λgk, 1≤n≤a−1, (3.25)
θCn˜ (θ)−Cn(0) =λCn˜ (θ)−
n
∑
k=1
˜
Cn−k(θ)λgk, n≥a (3.26)
θQ˜10(θ)−Q10(0) =λQ˜10(θ)−C0(0)V(˜ θ), (3.27)
θQ˜1n(θ)−Q1n(0) =λQ˜1n(θ)−
n
∑
k=1
˜
Q1,n−k(θ)λgk − Cn(0)V(˜ θ), n≥1 (3.28)
θQj˜ 0(θ)−Qj0(0) =λQj˜ 0(θ)−Qj−1,0(0)V(˜ θ), j≥2, (3.29) θQjn˜ (θ)−Qjn(0) =λQjn˜ (θ)−Qj−1,n(0)V(˜ θ)−
n
∑
k=1
˜
Qj,n−k(θ)λgk, j≥2, 1≤n≤a−1, (3.30)
θQ˜jn(θ) − Qjn(0) =λQ˜jn(θ)−
n
∑
k=1
˜
Qj,n−k(θ)λgk, n≥a, j≥2, (3.31)
θR˜0(θ)−R0(0) =λR˜0(0)−π
b
∑
m=d
Pm0(0)R(˜ θ) (3.32)
θRn˜ (θ)−Rn(0) =λRn˜ (θ)−π
b
∑
m=d
Pmn(0)R(˜ θ)−
n
∑
k=1
˜
Rn−k(θ)λgk, n≥1 (3.33)
To obtain the probability generating function of the queue size at an arbitrary time epoch, the following probability generating functions are defined.
˜
Pi(z,θ) =
∞
∑
j=0
˜
Pij(θ)zj Pi(z, 0) =
∞
∑
j=0
Pij(0)zj,(d≤i≤b)
˜
Qj(z,θ) =
∞
∑
n=0
˜
Qjn(θ)zn Qj(z, 0) =
∞
∑
n=0
Qjn(0)zn, j≥1
˜
C(z,θ) =
∞
∑
n=0
˜
Cn(θ)zn C(z, 0) =
∞
∑
n=0
Pn(0)zn
˜
R(z,θ) =
∞
∑
n=0
˜
Rn(θ)zn R(z, 0) =
∞
∑
n=0
Rn(0)zn
(3.34)
By multiplying the equations(3.18)to(3.33)with suitable powers ofzn and summing over n, (n = 0to∞) and using equation(3.34)
(θ−λ+λX(z))Pd˜ (z,θ) =Pd(z, 0)−S(˜ θ){[(1−π)
b
∑
m=d
Pmd(0) +
∞
∑
l=1
Qld(0) +Rd(0)]}
− λ
zd d−1
∑
i=a ˜
whereGi(z) =∑dn−=0i−1gnzn
(θ−λ+λX(z))Pi˜(z,θ) =Pi(z, 0)−S(˜ θ)
(1−π)
b
∑
m=d Pmi(0)
+
∞
∑
l=1
Qli(0) +Ri(0),d+1≤i≤b−1
(3.36)
(θ−λ+λX(z))P˜b(z,θ) =Pb(z, 0)− ˜ S(θ)
zb
(1−π)
b
∑
m=d
[Pm(z, 0)− b−1
∑
n=0
Pmn(0)zn]
+
∞
∑
l=1
[Ql(z, 0)− b−1
∑
n=0
Qln(0)zn]
+ [R(z, 0)− b−1
∑
n=0
Rn(0)zn]
(3.37)
(θ−λ+λX(z))Q˜1(z,θ) =Q1(z, 0)−C(z, 0)V(˜ θ) (3.38)
(θ−λ+λX(z))Q˜j(z,θ) =Qj(z, 0)−V(˜ θ)
a−1
∑
n=0
Qj−1,n(0)zn, j≥2 (3.39)
(θ−λ+λX(z))C(z,˜ θ) =C(z, 0)−
a−1
∑
n=0
Rn(0)znC(˜ θ)
−(1−π)
a−1
∑
n=1
b
∑
m=d
Pmn(0)znC(˜ θ)
−(1−π)
b
∑
m=a
Pm0(0)C(˜ θ)
(3.40)
(θ−λ+λX(z))R(z,˜ θ) =R(z, 0)−π
b
∑
m=d
Pm(z, 0)R(˜ θ)zn (3.41)
substitutingθ=λ−λX(z)in equations(3.35)to(3.41), we get
Pd(z, 0) = S(˜ λ−λX(z))
[(1−π)
b
∑
m=d
Pmd(0) +
∞
∑
l=1
Qld(0) +Rd(0)]
+ λ
zd d−1
∑
i=a ˜
Pi0(λ−λX(z))zi×[X(z)−Gi(z)] (3.42)
Pi(z, 0) = S(˜ λ−λX(z))
(1−π)
b
∑
m=d
Pmi(0) +
∞
∑
l=1
Qli(0) +Ri(0)
, d<i<b (3.43)
Q1(z, 0) = V(˜ λ−λX(z))C(z, 0) (3.44)
Qj(z, 0) = V(˜ λ−λX(z))
a−1
∑
n=0
Qj−1,n(0)zn, j≥2 (3.45)
C(z, 0) = C(˜ λ−λX(z))
a−1
∑
n=0
Rn(0)zn+ (1−π)
a−1
∑
n=1
b
∑
m=d
Pmn(0)zn
+ (1−π)
b
∑
m=a Pm0(0)
(3.46)
R(z, 0) = πR(˜ λ−λX(z))
b
∑
m=d
Pm(z, 0)R(˜ θ) (3.47)
Pb(z, 0) =
˜
S(λ−λX(z))f(z)
wheref(z) = [(1−π) +πR(˜ λ−λX(z))]
b−1
∑
m=d
Pm(z, 0) +
∞
∑
l=1
Ql(z, 0)
− b−1
∑
i=0
(1−π)
b
∑
m=d
Pmi(0)zi+Ri(0)zi+
∞
∑
l=1
Qli(0)zi
using the equations (3.35) to (3.41) in equations (3.42) to (3.48) after simplification, we get
˜
Q1(z,θ) = [
˜
V(λ−λX(z))−V(˜ θ)]C(z, 0)
(θ−λ+λX(z)) (3.49)
˜
Qj(z,θ) =
[V(˜ λ−λX(z))−V(˜ θ)]∑na−=01Qj−1(0)zn
(θ−λ+λX(z)) ,j≥2 (3.50)
˜
C(z,θ) =
˜
C(λ−λX(z))−C(˜ θ)
a−1
∑
n=0
Rn(0)zn
+ (1−π)[
a−1
∑
n=1
b
∑
m=d
Pmn(0)zn+ b
∑
m=a
Pm0(0)]
(θ−λ+λX(z))
(3.51)
˜
Pd(z,θ) =
˜
S(λ−λX(z))−S(˜ θ)×
[(1−π)
b
∑
m=d
Pmd(0) +
∞
∑
l=1
Qld(0) +Rd(0)]
+ λ
zd d−1
∑
i=a
[P˜i0(λ−λX(z))−Pi˜0(θ)]zi
×[X(z)−Gi(z)]
(−λ+λX(z)) (3.52)
˜
Pi(z,θ) =
˜
S(λ−λX(z))−S(˜ θ)×
(1−π)
b
∑
m=d
Pmi(0) +
∞
∑
l=1
Qli(0) +Ri(0)
(θ−λ+λX(z)) , d+1≤i≤b−1
(3.53)
˜
Pb(z,θ) = [
˜
S(λ−λX(z))−S(˜ θ)]f(z)
[θ−λ+λX(z)]
zb−(1−π)S(˜ λ−λX(z))
−πS˜(λ−λX(z))R˜(λ−λX(z))
(3.54)
˜
R(z,θ) = π[
˜
R(λ−λX(z))−R(˜ θ)]∑bm=dPm(z, 0)
(θ−λ+λX(z)) (3.55)
4
Probability Generating Function of Queue Size
4.1
PGF of Queue Size at an arbitrary epoch is obtained as,
P(z) =P˜d(z, 0) + b−1
∑
i=d+1
˜
Pi(z, 0) +P˜b(z, 0) +C(z, 0) +˜
∞
∑
l=1
˜
Ql(z, 0) +R(z, 0)˜ (4.56)
By substitutingθ=0 on the equations(3.49)to(3.55), then the equation(4.56)becomes
Nr(z) =(1−π)S(˜ λ−λX(z)) +πR(˜ λ−λX(z))S(˜ λ−λX(z))−1
× b−1
∑
i=d
(zb−zi)ci
+(zb−1)[C(˜ λ−λX(z))V(˜ λ−λX(z))−1]
a−1
∑
i=0
dizi
+(zb−1)[V(˜ λ−λX(z))−1]
a−1
∑
i=0
qizi (4.57)
Dr(z) =h(−λ+λX(z)) i
[zb−(1−π)S(˜ λ−λX(z))−πR(˜ λ−λX(z))S(˜ λ−λX(z))]
(4.58)
4.2
Steady state condition
The probability generating function has to satisfy P(1)=1. In order to satisfy this condition, apply L’Hospital rules and equating this expression to 1. Consequently,
[E(S) +πE(R)]
b−1
∑
i=d
(b−i)ci+b[E(C) +E(V)] a−1
∑
i=0
dizi+bE(V) a−1
∑
i=0
qizi
=b−λ.E(X).E(S)−π.λ.E(X).E(R)
since pi,qi are probabilities of i customers being in the queue, it follows that left hand side of the above expression must be positive. ThusP(1) =1 is satisfied ifzb−(1−π)S(˜ λ−λX(z))−πS(˜ λ−λX(z))R(˜ λ−
λX(z)) >0, ifρ = λ.E(X).[E(bS)+πE(R)] thenρ<1 is the condition to be satisfied for the existence of the steady
state for the model under consideration. Equation(58)hasb+aunknownsd0,d1,d2, ...,da−1,q0,q1,q2, ...,qa−1,
and ca,ca+1,ca+2, ...,cd−1,cd, ..,cb−1. using the followin result, express qi interms of di in such a way that numerator have onlybconstants. Now equation(57)gives the probability generating function of the number of customers involving onlybunknowns. By Rouche’s theorem of complex variables, it can be proved that zb−(1−π)S(˜ λ−λX(z))−πS(˜ λ−λX(z))R(˜ λ−λX(z))hasb−1 zeros inside and one on the unit circle
|z| = 1. Since P(z) is analytic within and on the unit circle, the numerator must vanishes at these points, which givesbequations inbunknowns. These equations can be solved by any suitable numerical technique.
5
Some important performance measures
5.1
Expected queue length
The expected queue length E(Q) at an arbitrary time epoch is obtained by differentiating P(z) at z=1 and is given by
E(Q) =
f1(X,S,R)
b−1
∑
i=d
(b(b−1)−i(i−1))ci
+f2(X,S,R)
b−1
∑
i=d
(b−i)ci+ f3(X,S,R,V)
a−1
∑
i=0
ci
+f4(X,S,R,V,C)
a−1
∑
i=0
di+f5(X,S,R,V)
a−1
∑
i=0
iqi
+f6(X,S,R,V,C)
a−1
∑
i=0
idi
2.hλ.X1.(b−S1−πR1) i2
The functions f1to f6are given by
f1(X,S,R) =T3T1,
f2(X,S,R) =T4T1 − T3T2
f3(X,S,R,V) =bV2T1+b(b−1)V1T1−bV1T2
f4(X,S,R,V,C) =b(b−1)C1T1+bC2T1+2bT1V1C1−bC1T2
f5(X,S,R,V) =2bV1T1, f6(X,S,R,V,C) =2bV1T1+2bC1T1
where
T1=λX1(b−S1−πR1)
T2=λX2(b−S1−πR1) +λX1[b(b−1)−S2−πR2−2πR1S1]
T3=S1+πR1, T4=S2+πR2+2πR1S1
and
X1 = E(X),S1=λX1.E(S),R1=λX1.E(R),
C1=λX1E(C),V1=λX1E(V),
S2 = λX2E(S) +λ2(E(X))2E(S2),
R2 = λX2E(R) +λ2(E(X))2E(R2)
C2 = λX2E(C) +λ2(E(X))2E(C2),
V2 = λX2E(V) +λ2(E(X))2E(V2)
5.2
Expected waiting time
The expected waiting time is obtained by the Little’s formula as;
E(W) = E(Q)
λE(X) (5.60)
whereE(Q)is given in equation(5.59)
References
[1] Cox, D.R. (1965). ”The analyses of non-markovian stochastic processes by the inclusion of supplementary variables”.Mathematical Proceedings of the Cambridge Philosophical Society, 51, 433-441.
[2] Ganesan,V.(2002).”Erlangian queueing system with accessible and non-accessible batch service”. Presented conference-Cochin University,India
[3] Gross,D. and C.M. Harris (1985). ”Fundamentals of Queueing theory”.,John WileyNew York.
[4] Jeyakumar,S. and B., Senthilnathan (2012). ”A Study on the behavoiur of the server breakdown without interruption in aM[X]/G(a,b)/1 queueing system with multiple vacations and closedown time”.Applied Mathematics and Computations,Vol. 219, 2618-2633.
[5] Medhi,J.(1984).”Recent developments in bulk queueing models”.John Wiley Eastern Ltd., NewDelhi
[6] Sivasamy,R.(1990). ”A bulk service queue with accessible and non-accessible service batches”.Opsearch, Vol.27. N0.1, 46-54
Received: August 12, 2015;Accepted: September 128, 2015
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