Consider a GI M 1 queue with set up period and working vacations. During the working vacation period, customers can be served at a lower rate, if there are customers at a service completion instant, the vacation can be interrupted and the server will come back to a set up period with probability or continue the working vacation with probability , and when the set up period ends, the server will switch to the normal working level. Using the matrix analytic method, we obtain the steady state distributions for the queue length at arrival epochs. Li Tao "The Queue Length of a GI/M/1 Queue with Set-Up Period and Bernoulli Working Vacation Interruption" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-5 | Issue-5 , August 2021, URL: https://www.ijtsrd.com/papers/ijtsrd43743.pdf Paper URL: https://www.ijtsrd.com/mathemetics/other/43743/the-queue-length-of-a-gim1-queue-with-setup-period-and-bernoulli-working-vacation-interruption/li-tao
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p ( 1
p p
= − ), and when the set-up period ends, the
server will switch to the normal working level.
Otherwise, the server continues the vacation.
Meanwhile, if there is no customer when a vacation
ends, the server begins another vacation, otherwise,
he switches to the set-up period, and after the set-up
period, the server switches to the normal busy period.
Suppose n
τ be the arrival epoch of nth customers with
0 0
τ = . The inter-arrival times { , 1
n n
τ ≥ } are
independent and identically distributed with a general
distribution function, denoted by ( )
A t with a mean
1/ λ and a Laplace Stieltjes transform (LST), denoted
by *
( )
A s . The service times during a normal service
period, the service times during a working vacation
period, the set-up times and the working vacation
times are exponentially distributed with rate µ ,η ,
β andθ , respectively.
Let ( )
L t be the number of customers in the system at
time t and ( 0)
n n
L L τ
= − be the number of the customers
before the nth arrival. Define 0
n
J = , the nth arrival
occurs during a working vacation period; 1
n
J = , the
nth arrival occurs during a set-up period; 2
n
J = , the
nth arrival occurs during a normal service period.
Then, the process {( , ), 1}
n n
L J n ≥ is an embedded
Markov chain with state space
{0,0} {( , ), 1, 0,1,2}.
k j k j
Ω = ∪ ≥ =
In order to express the transition matrix of ( , )
n n
L J , let
( , ),( , ) 1 1
( , , ).
i j k l n n n n
p P L k J l L i J j
+ +
= = = = =
∣
Meanwhile, we introduce the expressions below
0
( )
( ), 0,
!
k
t
k
t
a e dA t k
k
µ
µ
∞
−
= ≥
∫
( )
0 0
( ( ))
( ), 0,
!
k
t
x t x
k
t x
b e e dxdA t k
k
β µ
µ
β
∞
− − −
−
= ≥
∫ ∫
0
( )
( ), 0,
!
k
k t t
k
t
c p e e dA t k
k
η θ
η
∞
− −
= ≥
∫
( )
0 0
0
( )
( )
!
( ( ))
( ), 0,
( )!
l
k t t
l x x y x
k x
l
k l
t y
x
d p e e e
l
t y
e dydxdA t k
k l
η θ β
µ
η
θ β
µ
∞
− − − −
=
−
− −
=
−
≥
×
−
∑
∫ ∫ ∫
1
1 ( )
0 0
1
( )
( )
( 1)!
( ( ))
( ), 1,
( )!
l
k t t
l x x y x
k x
l
k l
t y
x
e p p e e e
l
t y
e dydxdA t k
k l
η θ β
µ
η η
β
µ
−
∞
− − − − −
=
−
− −
=
−
−
−
× ≥
∑
∫ ∫ ∫
( )
0 0
( )
( ), 0,
!
k
t
k x x t x
k
x
f p e e e dxdA t k
k
η θ β
η
θ
∞
− − − −
= ≥
∫ ∫
1
1 ( )
0 0
( )
( ), 1.
( 1)!
k
t
k x x t x
k
x
g p p e e e dxdA t k
k
η θ β
η η −
∞
− − − − −
= ≥
−
∫ ∫
Using the lexicographic sequence for the states, the
transition probability matrix of ( , )
n n
L J can be written
as the Block-Jacobi matrix
00 01
1 1 0
2 2 1 0
3 3 2 1 0
,
B A
B A A
P B A A A
B A A A A
=
M M M M M O
where
00 0 0 0 01 0 0 0
1 ; ( , , );
B c d f A c f d
= − − − =
0 0 0
*
0 0
0
0 ( ) ;
0 0
c f d
A A b
a
β
=
0 0 ,
0 0
k k k k k
k k
k
c f g d e
A b
a
+ +
=
0 0 0
1
*
0
0
1 ( )
1 ( ) , 1.
1
k
i i i i i
i
k
k i
i
k
i
i
c d e f g c d f
B b A k
a
β
=
=
=
− + + + + − − −
= − − ≥
−
∑
∑
∑
3. Steady-state distribution at arrival epochs
We first define
0 0 0
0 1 0 1
( ) , ( ) , ( ) ,
( ) , ( ) , ( ) , ( .
)
k k k
k k k
k k k
k k k k
k k k k
k k k k
A z a z B z b z C z c z
D z d z E z e z F z f z G z g z
∞ ∞ ∞
= = =
∞ ∞ ∞ ∞
= = = =
= = =
= = = =
∑ ∑ ∑
∑ ∑ ∑ ∑
In this section, we derive the steady-state distribution
for ( , )
n n
L J at arrival epochs using matrix-geometric
approach. In order to derive the steady-state
distribution, we need the following three lemmas.
Lemma 3.1.
*
( ) ( ),
A z A z
µ µ
= −
* *
[ ( ) ( )]
( ) ,
(1 )
A z A
B z
z
β µ µ β
β µ
− −
=
− −
*
( ) ( ),
C z A p z
θ η η
= + −
* *
* *
[ ( ) ( )]
( )
(1 )
[ ( ) ( )]
,
(1 ) ( )(1 )
A p z A
D z
z p z
A p z A z
z p z z
θβ θ η η β
β µ θ η η β
θβ θ η η µ µ
β µ θ η µ η
+ − −
=
− − + − −
+ − − −
−
− − + − − −
* *
* *
[ ( ) ( )]
( )
(1 )
[ ( ) ( )]
,
(1 ) ( )(1 )
p z A p z A
E z
z p z
p z A p z A z
z p z z
η β θ η η β
β µ θ η η β
η β θ η η µ µ
β µ θ η µ η
+ − −
=
− − + − −
+ − − −
−
− − + − − −
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* *
[ ( ) ( )]
( ) ,
A A p z
F z
p z
θ β θ η η
θ η η β
− + −
=
+ − −
* *
[ ( ) ( )]
( ) .
p z A A p z
G z
p z
η β θ η η
θ η η β
− + −
=
+ − −
Lemma 3.2. If 0
θ > , the
equation *
( )
z A p z
θ η η
= + − has a unique root in the
range 0<z<1.
Lemma 3.3. If 0
θ > , 0
β > and / 1
ρ λ µ
= < , then the
matrix equation
0
k
k
k
R R A
∞
=
= ∑ has the minimal
nonnegative solution
1 2 1 3 2 3 1
2 3 2
3
( ) ( ) ( )
0 ( ) ,
0 0
r r r r r r r
R r r r
r
δ δ γδ
− ∆ − − −
= ∆ −
where *
2 ( )
r A β
= , 1
r and 3
r are the unique roots in the
range 0<z<1 of equations
*
( )
z A p z
θ η η
= + − and *
( )
z A z
µ µ
= − , respectively, and
1 1
( ) / ( )
p r p r
δ θ η θ η η β
= + + − − , 2
/[ (1 )]
r
β β µ
∆ = − − ,
1 1
/[ ( )(1 )]
p r r
γ β θ η µ η
= + − − − .
Moreover, we can easily verify that the Markov
chain P is positive recurrent if and only if
0
θ > , 0
β > and 1
ρ < . And the matrix
00 01
1 1
1 1
0 0 0 0 0 0
0 0 0 0
2 1 2 1
1 1 1 1 1
0 3 2 0 0 3 2 0
2 3 2 2 3 2
0 0
3 3
[ ]
1
( ) ( )
1
( ) ( )
1 0 0
0 0 1
k k
k k
k k
B A
B R
R B R A
c d f c f d
c f c f
r r r r
r r r r r
a r r b a r r b
r r r r r r
a a
r r
δ δ
ω ω
∞ ∞
− −
= =
=
− − −
+ − −
− − − −
= ∆ − ∆ −
− + −
−
∑ ∑
with 3 2 3 1 0
2 1
0 0
2 3 1 3 1 2 1
( ) ( ) ( )
[ ]
r r r r d
r r
a b
r r r r r r r
δ γδ δ
ω
∆ − − −
= − + − has a
positive left invariant vector
1 2 1 3 2 3 1
(1, , ( ), ( ) ( ))
K r r r r r r r
δ δ γδ
− ∆ − − − (1)
where K ia a random positive real number.
Let ( , )
L J be the stationary limit of the process ( , )
n n
L J ,
and denote
0 00 0 1 2
; ( , , ), 1,
k k k k k
π π π π π π
= = ≥
{ , } lim { , }, ( , ) .
kj n n
n
P L k J j P L k J j k j
π
→∞
= = = = = = ∈Ω
Theorem 3.4. If 0
θ > , 0
β > and 1
ρ < , the stationary
probability distribution of ( , )
L J is given by
0 3 1
1 3 2 1
2 3 3 2 3 1
(1 ) , 0,
(1 ) ( ), 0,
(1 ) [ ( ) ( )], 1,
k
k
k k
k
k k k k
k
r r k
r r r k
r r r r r k
π σ
π σδ
π σ δ γδ
= − ≥
= − − ≥
= − ∆ − − − ≥
where
1 2
3 2 2 1 1 3 2 2 3 1
(1 )(1 )
.
(1 )[(1 ) ( )] (1 )( ) (1 )( )
r r
r r r r r r r r r r
σ
δ δ γδ
− −
=
− − + − + ∆ − − − − −
Proof. With the Theorem 1.5.1 in [8],
00 10 11 12
( , , , )
π π π π is given by the positive left invariant
vector Eq. (1), and satisfies the normalizing condition
1
00 10 11 12
( , , )( ) 1,
I R e
π π π π −
+ − =
where e is a column vector with all elements equal to
one. Substituting R into the above relationship, we
can get
1 2 3
3
3 2 2 1 1 3 2 2 3 1
(1 )(1 )(1 )
(1 ) .
(1 )[(1 ) ( )] (1 )( ) (1 )( )
r r r
K r
r r r r r r r r r r
σ
δ δ γδ
− − −
= = −
− − + − + ∆ − − − − −
Therefore, we have
10 11 12 3 1 2 1 3 2 3 1
( , , ) (1 ) ( , ( ), ( ) ( ))
r r r r r r r r
π π π σ δ δ γδ
= − − ∆ − − − .
Using the Theorem 1.5.1 of Neuts [8], we can obtain
1
0 1 2 10 11 12
( , , ) ( , , ) , 1.
k
k k k k R k
π π π π π π π −
= = ≥ (2)
Taking 10 11 12
( , , )
π π π and 1
k
R −
into Eq. (2), the theorem
can be derived.
Then, we discuss the distribution of the queue length
L at the arrival epochs. From Theorem 3.4, we have
0 00 3
{ 0} (1 ) ,
P L r
π π σ
= = = = −
0 1 2
3 1 2 3 2 3 1
{ }
(1 ) [(1 ) ( ) ( )], 1.
k k k k
k k k k k k
P L k
r r r r r r r k
π π π π
σ δ δ δ γδ
= = = + +
= − − + + ∆ − − − ≥
The state probability of a server in the steady-state is
given by
3
0
0 1
(1 )
{ 0} ,
1
k
k
r
P J
r
σ
π
∞
=
−
= = =
−
∑
3 2 1
1
1 1 2
(1 )( )
{ 1} ,
(1 )(1 )
k
k
r r r
P J
r r
σδ
π
∞
=
− −
= = =
− −
∑
1 3 2 2 3 1
2
1 1 2
(1 )( ) (1 )( )
{ 2} .
(1 )(1 )
k
k
r r r r r r
P J
r r
σ δ σγδ
π
∞
=
∆ − − − − −
= = =
− −
∑
Theorem 3.5. If 0
θ > , 0
β > and 1
ρ < , the stationary
queue length L can be decomposed as 0 d
L L L
= + ,
where 0
L is the stationary queue length of a classical
GI/M/1 queue without vacation, and follows a
geometric distribution with parameter 3
r . Additional
queue length d
L has a distribution
{ 0} ,
d
P L σ
= =
1
3 1 1
1
3 2 2
{ } ( 1 )( )
( 1)( ) 1
, .
k
d
k
P L k r r r
r r r k
σ δ γδ
σδ
−
−
= = − −
+ ∆ − ≥
−
−
Proof. The probability generating function of L is as
follows:
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0 1 2
0 1 1
( ) k k k
k k k
k k k
L z z z z
π π π
∞ ∞ ∞
= = =
= + +
∑ ∑ ∑
3 2 3 1
2 1
3
1 1 2 2 3 1 3
( ) ( )
( )
1
(1 ) [ ]
1 (1 )(1 ) (1 )(1 ) (1 )(1 )
r r z r r z
r r z
r
rz rz r z r z r z rz r z
σ δ δ γδ
− −
−
= − + +∆ −
− − − − − − −
3 3 2 1 3 3 2 3 1
3 1 1 2 2 1
1 1 ( )(1 ) ( ) ( )
[ ]
1 1 (1 )(1 ) 1 1
r r z r r r z z r r z r r z
r z r z r z r z r z r z
σ δ δ γδ
− − − − − −
= + + ∆ −
− − − − − −
3 3 1 3 1 3 2 3 2 3 1
3 1 1 2 2 1
1 ( ) ( ) ( ) ( ) ( )
[1 ]
1 1 1 1 1 1
r r r z r r z r r z r r z r r z
r z rz rz r z r z rz
σ δ δ δ γδ
− − − − − −
= − + − +∆ −
− − − − − −
3 3 1 3 2
3 1 2
1 ( ) ( )
[ ( 1 ) ( 1) ]
1 1 1
r r r z r r z
r z r z r z
σ σ δ γδ σδ
− − −
= + − − + ∆ −
− − −
0 ( ) ( ).
d
L z L z
=
which completes the proof.
Thus, the mean queue length at the arrival epoch is
given by
3 3 1 3 2
2 2
3 1 2
( 1 )( ) ( 1)( )
[ ] .
1 (1 ) (1 )
r r r r r
E L
r r r
σ δ γδ σδ
− − − ∆ − −
= + +
− − −
4. Steady-state distribution at arbitrary epochs
Now we consider the steady-state distribution for the
queue length at arbitrary epochs. And, denote the
limiting distribution of ( )
L t : lim { ( ) }
k
t
p P L t k
→∞
= = .
Theorem 4.1. If 0
θ > , 0
β > and 1
ρ < , the limiting
distribution of ( )
L t exists. And, we obtain
3 3
0
1
1 1 1
2 1
3 3 2 1
1
(1 ) (1 ) (1 )(1 )
1 [ ],
(1 ) (1 ) (1 )(1 )
(1 ) [ ], 1.
k k k
k
r r
p
p r
r r
p r r r r k
p r
δ δ γδ
δ γδ
σλ
µ β θ η η
δ δ γδ
δ γδ
σλ
µ β θ η η
− − −
− ∆ − − + −
∆ −
= − + +
+ −
− ∆ − − + −
∆ −
= − + + ≥
+ −
Proof. From the theory of SMP, the limiting
distribution of ( )
L t has the following expression (see
[9]):
1
2 0
1
( )
(1 ( ))
( 1 )!
i k
t
k i
i k
t
p e A t dt
i k
µ
µ
π λ
+ −
∞ ∞
−
= −
= −
+ −
∑ ∫
1
( )
1 0 0
1
( ( ))
(1 ( ))
( 1 )!
i k
t
x t x
i
i k
t x
e e dx A t dt
i k
β µ
µ
π β λ
+ −
∞ ∞
− − −
= −
−
+ −
+ −
∑ ∫ ∫
1,1 0
(1 ( ))
t
k e A t dt
β
π λ
∞
−
−
+ −
∫
1
1
0 0
1
( )
(1 ( ))
( 1 )!
i k
i k t t
i
i k
t
p e e A t dt
i k
η θ
η
π λ
+ −
∞ ∞
+ − − −
= −
+ −
+ −
∑ ∫
1
( )
0 0 0
1 0
1
( )
( )
!
( ( ))
(1 ( ))
( 1 )!
l
i k t t
l x x y x
i x
i k l
i k l
t y
x
p e e e
l
t y
e dydx A t dt
i k l
η θ β
µ
η
π θ β
µ
λ
∞ + −
∞
− − − −
= − =
+ − −
− −
+
−
−
+
×
− −
∑ ∑
∫ ∫ ∫
1
1
1 ( )
0 0 0
1
1
( )
( )
( 1)!
( ( ))
(1 ( ))
( 1 )!
l
i k t t
l x x y x
i x
i k l
i k l
t y
x
p p e e e
l
t y
e dydx A t dt
i k l
η θ β
µ
η η
π β
µ
λ
−
∞ + −
∞
− − − − −
= =
+ − −
− −
+
−
×
−
−
+ − −
∑ ∑
∫ ∫ ∫
1
1 ( )
0 0 0
1
( )
(1 ( ))
( 1 )!
i k
t
i k x x t x
i
i k
x
p e e e dx A t dt
i k
η θ β
η
π θ λ
+ −
∞ ∞
+ − − − − −
= −
+ −
+ −
∑ ∫ ∫
( )
0 0 0
( )
(1 ( ))
( )!
i k
t
i k x x t x
i
i k
x
p p e e e dx A t dt
i k
η θ β
η η
π λ
−
∞ ∞
− − − − −
=
+ −
−
∑ ∫ ∫
1 2 3 4 5 6 7 8.
b b b b b b b b
= + + + + + + +
We compute each part of the equation and have
*
1 1
2
3 3 2
2
*
1
1
1
1
1 ( )
1 (1 ) [
(1 )
1 ( )
],
(1 )
k k
k
A r
b r r r
r
A r
r
r
µ µ
δ γδ
σλ δ
µ µ
µ µ
γδ
µ
− −
−
− −
∆ −
= − − ∆
−
− −
+
−
*
1 1
2 2
3 2 2
2
*
1 1
1 2
1 1
1 1 1
1 ( ) 1
2 (1 ) [
(1 )
(1 ( )) (1 )
],
(1 )
k k
k k
A r r
b r r r
r
A r r
r r
r r r
µ µ
σλ δ δ
µ β
µ µ δ
βδ
β µ µ µ β µ µ
− −
− −
− − −
= − ∆ − ∆
−
− − −
− +
− + − − +
1 1
2 2
3 2 1
(1 ) (1 )
3 (1 ) [ ],
k k
r r
b r r r
δ δ
σλ
β β
− −
− −
= − −
1
1
3 1
1
1
4 (1 ) ,
k
r
b r r
p r
σλ
θ η η
−
−
= −
+ −
1
3 1 1 2
3
1 1 1
1 *
3 1 1 1
1 1 1
(1 ) 1 1
5 (1 ) ( )
(1 ) 1 1 ( )
[ ],
(1 )
k
k
r r r r
b r
r p r p r
r r r A r
r p r r
σλ θβ
σ
β µ µ θ η η β θ η η β
σλθγ µ µ
β µ µ θ η η µ
−
−
− − −
= − −
− + + − − + −
− − − −
− −
− + + − −
3 1 1 2
1 1 1
*
3 1 1 1
1 1 1
(1 ) 1 1
6 ( )
(1 ) 1 1 ( )
[ ],
(1 )
k
k
r p r r r
b
r p r p r
r p r r A r
r p r r
σλ ηβ
β µ µ θ η η β θ η η β
σλ ηγ µ µ
β µ µ θ η η µ
− − −
= −
− + + − − + −
− − − −
− −
− + + − −
1
3 1 2 1
1 1
(1 ) 1 1
7 ( ),
k
r r r r
b
p r p r
σλθ
θ η η β β θ η η
−
− − −
= −
+ − − + −
Then, using these expressions, the theorem can be
obtained by some computation.
Let L denote the steady-state system size at an
arbitrary epoch, the mean of L can be given by
3 2
0 2 1 1
3
(1 ) 1
[ ] (1 ) [ ].
(1 ) (1 )( )
(1 )
k
k
E L kp r
r r p r
r
δ γδ δ δ γδ
σλ
β θ η η
µ
∞
=
∆ − − ∆ − +
= = − + +
− − + −
−
∑
Remark 4.2. If 1
p = , the system reduces to the model
described in [4], and if 0
p = , the system becomes a
GI/M/1 queue with set-up period and multiple
working vacations [6].
References
[1] L. Servi and S. Finn, “M/M/1 queue with
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