On an
M
/
G
/1 Queueing Model with
k
-Phase Optional
Services and Bernoulli Feedback
Sayeedeh Abdollahi1, Mohammad Reza Salehi Rad2
Department of Statistics, Allameh Tabataba’i University, Tehran, Iran. Email: [email protected], [email protected]
Received May 16th,2012; revised June 18th, 2012; accepted July 2nd, 2012
ABSTRACT
In this article an M/G/1 queueing model with single server, Poisson input, k-phases of heterogeneous services and Ber-noulli feedback design has been considered. For this model, we derive the steady-state probability generating function (PGF) of queue size at the random epoch and at the service completion epoch. Then, we derive the Laplace-Stieltjes Transform (LST) of the distribution of response time, the means of response time, number of customers in the system and busy period.
Keywords:M/G/1 Queue; Bernoulli Feedback; Queue Size; Waiting Time and Busy Period
1. Introduction
The M/G/1 queueing model is one of the famous and
applied models in which the distribution of service times is unknown. For this reason, many of real models could be considered by this model. Multi-phase service is im-portant, because some of systems have more than one phase service, for example manufacture production lines. Feedback is also important, because in some queueing models, some customers, after completion the service, may need to go back to the end of queue to take the ser-vice again. It means that the serser-vice customer is not ac-ceptable and must go to the end of queue. According to these conditions, we have considered an M/G/1 queueing model with k-Phase Optional Services and Bernoulli feed- back.
For this model, first we find the steady-state probabil-ity generating function (PGF) of queue size at the ran-dom epoch and at the service completion epoch. Then, we derive the Laplace-Stieltjes Transform (LST) of the distribution of response time. The means of response time, number of customers in the system and busy period will be derived by using the PGF and LST.
In relation of this model, [1-9,11] have derived some results. The model that they have considered is two phases. But, in this article, we will consider a k-phase queue with optional service and Bernoulli feedback in all
phases. Of course, [10] studied an M/G/1 queue with
k-phase services and vacation, but without feedback that is different from this paper.
Following, in Section 2 we describe the model and
give some definitions. In Section 3, the PGF of the sys-tem size will be derived. In Section 4, we will find some measures of effectiveness. At the final section, we pro-vide a conclusion.
2. The Mathematical Model and Definitions
In this model, the server provides first phase of regular service to all the customers. As soon as the i-th (for i = 1, ···, k − 1) phase of service of a customer is completed , it
may leave the system or immediately go for (i + 1)-th
phase of optional service. However, after receiving each phase of unsuccessful service by a unit, then it may im-mediately join to the end of tail of the original queue as feedback customer to take service again. Thus, the as-sumptions of the model are:
1) Customers arrive at the system to a Poisson process with rate .
2) The service discipline of the system is FCFS1.
3) The server provides k-phases of heterogeneous ser-vice for any customer. The serser-vice times for k-phases are
independent random variable that denoted by i with
distribution functions
B
, 1, ,i
B x i k
* , 1, ,
i
B x i
l iE B
and LST of
these distributions are . These
vari-ables have finite moments, that is for .
k
l1
4) As soon as the i-th phase of service of a customer is completed, the customer may go to the (i + 1)-th phase of service with probability i,i1, , k.
5) After completion of the i-th phase, if the customer is dissatisfied with its service for certain reason or it re-
ceived unsuccessful service, in this case the customer may immediately joins the end of the original queue as a feedback customer for receiving the service again with
probability i, for , otherwise the customer
may depart the system with probability .
p i1, , k
k
, 1
i i i
q q p
Definition 2.1. The modified service time or the time required by a customer to complete the service cycle is given by
1 1
1 2 1 2
1 2 3 1 2 3
1
1 2 2 1
1
1 2 1
1
with probabaility 1 with probabaility 1 with probabaility 1
with probabaility 1
with probabaility
k
i k
i k
i k
i
B
B
B B
B B B
B
B
(2.1) where 01 and k 0. Then the LST of B is given
by
* *
1 1 1
1
j k
i i j j i
B s B s
(2.2)and
1
1 1
j k
i j i
E B E B
j
(2.3)As we know, in the queueing systems the utilization factor is E B
and denoted by . This measure says if the system is in steady state or not. In this article, we study the model in equiblirium. It happens when 1.Definition 2.2. The elapsed of i-th phases service
i at time “t” is denoted by for .
We introduce the random variable
as followps Bi0
t
Y t
1, 2, , i k
k
],
0 when theservser isidle at time t.when theserver isin thei-th phase at time t
Y t i
Thus, we have a bivariate Markov process
, where if and
if , for . Now, we
define probabilities as
NQ t ,L t
0
iL t B t
0 L t i
0Y t
1, 2, ,
i
Y t
,
0 0
, [ ,
; d
0, 0
i n Q i i
P x t P N t n L t
B t x B t x x
x n
for i1, 2, , k and
0 Q 0, 0
R t P N t L t
We know that and , for
. Also
is continuous at0 0
i
B
i
B x
1i
B
1, 2, ,
i k x0. Then
we have the hazard rate functions of Bi as
d d
1i i
i
B x x x
B x
(2.4)
where i1, 2, , k,and i
x is the conditionalprob-ability of completion of i-th phase of service during the time interval
x x, dx
, given that the elapsed servicetime is x.
By assuming that the system is in the steady state, we let
0 0
, ,
lim
d lim , d , 0,
t
i n i n t
R R t
P x x P x t x x n
0
(2.5)
for i1, 2, , k.
In the next section, we find the PGF of these prob-abilities.
3. The PGF of the System Size
Now, for i1, 2, , k, the PGF of the probabilities that
explained by (2-5), are defined as
,
0
, n
i i n n
P x z z P x z x
1, 0
(3.1)
,0
0, n (0) 1
i i n n
P z z P z
(3.2)For finding the steady-state PGF from Kolmogorov forward equations, for i1, 2, , k, we can write the
steady-state equations as
, ,
d d
1, 0
i n i i n i n
P x x P x P x
x
n x
, 1 (3.3)
,0 ,0
d
0
d Pi x i x Pi x
x
0 x x
(3.4)
Also
0 ,
0 0
1 d
k
i i i i i
R q x P
(3.5)It is clear that Pi, 1 0, for i1, 2, , k. Now, at
0
x , the boundary conditions are
1,0 0
1
,0 ,1
0 0
0 1
d d
k i i
i i i i i i
P R
p x P x x q x P x
x
(3.6)
1, ,
1 0
, 1 0
0 1
d
k
n i i i i n i
i i i n
P p x P
q x P x x
d
x x
,
P x
(3.7)
, 0 1 0 1 1 d
i n i i i n
P
x (3.8)
0 0 ,
1 0
d 1
k
i n i n
R P x
x
which we use this condition to find the relation between
and .
0
R Pi
1In the next Lemma, we derive the relation between
and .
,i
P x z Pi
0,zLemma 3.1. From relation (3-3), we have
,
0, 1
(1 z x)i i i
P x z P z B x e x0
x
(3.9)
Proof. See [4]. □
Proposition 3.2. By the z-transform of Bi that is
* (1 )
0 d
z
i i
B z
e Bwe have
1 0
* 1
0, 1
k i 0, 1 i i i i i
zP z R z
P z zp q B z
k
(3.10)
Proof. By multiplying the relation (3.7) in and
summation from to , and using (3.6), the proof
is completed. □
n
z 1
n
Proposition 3.3. For i1, , , we have
*
1 1 1
0, 0,
i i i i
P z P z B z (3.11)
Proof. By multiplying (3.8) in and summation on
to , we can obtain (3.11). □
n
z 0
n
Corollary 3.4. By proposition 3.3, we have
*
1 1
0, 0,
i i
P z P z A z
(3.12) where
* *
1
j
j l
l
A z B z
(3.13)Now, by (3.10) and (3.13), we obtain
0
1 1
*
1 0
1 0,
1
i k
l i i i i i l
z R
P z
z A z zp q
k
(3.14)
Corollary 3.5. If , for
, we have
0
, d
i i
P z
P x z x 1, ,i
*
0 1
1 1
*
0
1
1
i k
l i i i i i l
R B z
P z
A z zp q z
1
(3.15) and for i2, , k
1
* *
0 1
0 1
*
0
1
1
i
i i l
l i
i k
l i i i i i l
R B z A z
P z
A z zp q z
1
(3.16)
Now, if *
0 1
A , 01 and
is the1
k i i
P z P z
PGF of queue size distribution at a random epoch, then from (3.15) and (3.16), we have
1 2
1
* *
0 1
1 0
1 *
0
1
1
k i i
i k
l i i
i l i k
l i i i i i l
P z P z P z
R B z A z
A z zp q z
1
(3.17) and the PGF of the queue size at the departure epoch is
0
1
* *
1
0 0
0 1
*
1 0
1 1
1
s
i k
l i i
i l i k
l i i i i i l
P z
R zP z
z B z A
R
z
A z zp q z
(3.18) In the next section, we find some measures of effec-tiveness as the means system size, response time and busy period.
4. The Measures of Effectiveness
This section includes three sub-sections. In the sub-Sec- tion 4.1, we find the mean system size. In the sub-Section 4.2, the mean response time is obtained and in the last sub-section we calculate the mean busy period.
4.1. The Mean System Size
If Q be the mean number of customers in the queue,
then we have
L
d1 d
s Q
P z
L z
z
For calculating Q we use the following lemma.
Lemma 4.1.1. By (3.13) and for , we
have
L
1, 2, , j k
1) *
1 1
lim
j
j i
z
i
d
A z E
dz
B (4.1.1)2)
2 2
* 2
2
1 1 1
d
lim var
d
j j
j i i
z
i i
A z E B
z
B
(4.1.2)
Now, we write P zS
in form of
0
S
f z
P z R
g z
where
1 *
*
1 *
1
1 0 1 0
1 1
i i
k k
l i i l i i i i
i l i l
f z z B z A z A z zp q z
and
1 *
1 0
1
i k
l i i i i i l
g z A z zp q
z
Since
, then by using the L’Hopital rule , we have1 1
lim lim 0
z f z z g z
0 2
1 1 1 1
2 1
Q
f g f g
L R
g
where
1 1
* * * *
1 1
1 0 1 0
1 1
* * *
1
1 0 1 0
1 * 1 0
d 1
d
d d
1 1
d d
1 1
i i
k k
l i i l i i
i l i l
i i
k k
l i i l i i
i l i l
i k
l i i i i l
h s B s A s s B s A s
s
s B s A s A s sp
s s
A s p
i
1 1
1
1
1
1 0 1 0 1 1 0
0 k i l i k i l i j 1 i k i l i i
i l i l j i l
h E B E B p
1 1
* * * *
1 1
1 0 1 0
2
1 1
* * * *
1 2 1
1 0 1 0
d d
1
d d
d d
d d
i i
k k
l i i l i i
i l i l
i i
k k
l i i l i i
i l i l
l l
f z B z A z B z A z
z z
B z A z z B z A z
z z
z
1 1
* * * *
1 1
1 0 1 0
2
1 1
* * * *
1 2 1
1 0 1 0
1
0
d d d
1
d d d
d d d
1
d d d
i i
k k
i i l i i
i i l
i i
k k
l i i l i i
i l i l
i l l
B z A z B z A z
z z z
z B z A z B z A
z z z
z
2 1
* *
2
1 1 0
1
*
1 0
d d
1 1
d d
d 1
d
i
k k
i i i i l i i i
i i l
i k
l i i i i l
A z zp q A z p
z z
A z p
z
1 1 1
2 2 2
1 0 1 0 1 0 1
2
1 1
2
1 0 1 1 1 0
1 2 2
var 1 2 1
i i
k k k i
l i l i l i j
i l i l i l j
i i
k i i k
l j j i l i i
i l j j i l
f E B E B E B E B
E B B E B p
i
1 1
* *
1 0 1 0
1 1
1 0 1 1 0
d 1 1
d
1 1 1 1
i i
k k
l i i i i l i i i
i l i l
i i
k i k
l j i l i i
i l j i l
g z A z zp q A z p
z
g E B p
1
2
1 1
* *
2
1 0 1 0
1
* 1 0
d d
1 1
d d
d
1 d
i i
k k
l i i i i l i i
i l i l
i k
l i i i i l
g z A z zp q A z p
z z
A z p
z
i
1 i
p
1 2
2
1
1 0 1 1 1 0
1 k i l i j i var j 1 i 2k i l i i
i l j j i l
g E B B E B
hence 1
0 2
Q
L
L R
L
, in which
1 1 1 1
2 2 2
1
1 0 1 0 1 0 0
2 1
2
1 0 0 0
2
var 1
i i i
k k k i
l i l i l i j
i l i l i l j
i
k i i
l j j i
i l j j
L E B E B E B E B
E B B
1
1 0 0
1 1
1 0 0 1 0
1 1
1 0 0
2 1
1 1 1
i
k i
l i i j
i l j
i i
k i k
l j i l i i
i l j i l
i i
k
l j l
i l i l
E B
E B
E B
p
p
1
1 0 1 0
2
1 1
2
1 0 0 0 1 0 0
var 2
1 1
1 1
i
k i k
j i l i i
j i l
i i
k i i k i
l j j i l i i j
i l j j i l j
E B
E B B E B
p
p
1
and
2
1 1
2
1 0 1 1 0
2 k i l i j 1 i k i l] i i
i l j i l
L E B p
1 14.2. The Mean Response Time
Here, we show the response time variable with WR. For
finding the mean of WR, first we need to obtain the LST
of the distribution of waiting time in the queue, then by using this, we find the LST of the response time distribu-tion. Then we can find the mean response time.
From classical formula in the queueing theory, we
know that *
*
Q S
W z B z P z
z, where
1
* *
1 1 0
1
i k
l i i i l
B z A
Now, if we put s
1z
and using the (3.18), we have
1
* *
1 0 0
* *
0
* 1
* 1 0
1
* *
1 0 0
0
0
1 1
1
1
( ) 1
1
i k
l i i S
i l
Q k i
i
l i i i i i l
i k
l i i i l
l l
s
B s A s
P
W s R B s
B s p s s
A s p q
s B s A s
R
s
* 1
* 1
1 ( )
i k
i i i i
B s
A s p s s
On the other hand, the response time is defined by
, where is queueing time and is the
service time. Then, by using the convolution property, the LST of
R Q
* * * 1
R Q S
W
s
Ws
Bs
Ps
in which by some simple computation we find that
* 0
R
h s
W s R
k s
where
1 *
*
1 *
1
1 0 1 0
1 1
i i
k k
l i i l i i i
i l i l
h s s B s A s A s sp s
and
1 *
1 0
1
i k
l i i i i l
k s A s sp s
The mean response time, for an arbitrary customer, is
*
0
d d
R R
W
E W
s
s
s
But h
0 k
0 0, then, by using the L’Hopital rule, we have
0 2
0 0 0 0
2 0
R
h k k h
E W R
k
(4.2.1)
On the other hand, d *
d Bi s E Bi
s and
2
* 2
2
d
d Bi s E Bi
s , then
1 1
* * * *
1 1
1 0 1 0
1 1
* * *
1
1 0 1 0
1 * 1 0
d 1
d
d d
1 1
d d
1 1 ,
i i
k k
l i i l i i
i l i l
i i
k k
l i i l i i
i l i l
i k
l i i i i l
h s B s A s s B s A s
s
s B s A s A s sp
s s
A s p
i
1 1 ,
1
1
1
1 0 1 0 1 1 0
0 k i l i k i l i j 1 i k i l i i
i l i l j i l
h E B E B p
1 1
* * * *
1 1
1 0 1 0
2
1 1
* * * *
1 2 1
1 0 1 0
1
1 0
d d
1
d d
d d
d d
i i
k k
l i i l i i
i l i l
i i
k k
l i i l i i
i l i l
i k
l i l
h s B s A s B s A s
s s
B s A s s B s A s
s
s
1
* * * *
1 1
1 0
2
1 1
* * * *
1 2 1
1 0 1 0
2 1
* 2 0
d d ( ) 1 d
d d d
d d d
1
d d d
d
d
i k
i i l i i
i l
i i
k k
l i i l i i
i l i l
i
l i l
B s A s B s A s
s s s
s B s A s s B s A s
s s s
A s s
1
*
1 1 0
1
*
1 0
d
1 .
d
d 1 ,
d
i
k k
i i l i i i
i i l
i k
l i i i i l
sp A s p
s
A s p
s
1 1 1 1
2
1 0 1 0 1 0 0
2 1
1 0 1 1
1
1 0
0 = 2 2 [
var 1
2 1 ,
i i i
k k k i
l i l i l i j
i l i l i l j
i
k i i
l i j j i
i l j j
i i
l i i j j l
h E B E B E B
E B E B B
p E B
E B
1
k i
1 1
* *
1 0 1 0
1 1
1 0 1 1 0
2 1
* 2 0
d 1 1 1
d
0 1 1 1 ,
d
1 d
i i
k k
l i i i l i i i
i l i l
i i
k i k
l j i l i i
i l j i l
i
l i i
l
k s A s sp A s p
s
k E B p
k s A s sp
s
1
*
1 1 0
1
*
1 0
2
1 1
1 0 1 1 0
d
. 1
d d
[ 1 ] ,
d
(0) var 1
i
k k
i l i i i
i i l
i k
l i i i i l
i i
k i i
l j j i l
i l j j l
A s p
s
A s p
s
k E B B E B
1 1
1
1 0 1
1
1 .
k i
j i i
i j
i
k i
l j i i i l j
p
E B p
with replacing in (4.2.6), it results that
10 2
R
W
E W R
W
where
1 1 1 1
2 1
1 0 1 0 1 0 0
2 1
1 0 1 1 0
2 2
var 1 2
i i i
k k k i
l i l i l i j
i l i l i l j
i i
k i i
l j j i l
i l j j l
W E B E B E B E B
E B B
1
1 1
1 1
1 0 1 1 0
2 1
1 0
1
1 1 1
var
k i
j i i
i j
i i
k i k
l j i i l i i
i l j i l
i i
l j j
j j
l
E B p
E B p p
E B B
1
1 1 1 0 1
1 1
1 0 1 0 1
1 2 1
* 1
i
k i k i
i l j i
i i l j
i i
k k i
l i l j i l
i l i l j
E B p
E B E B
i
1
1 0
1 1
i k
i i i l
p
and
2
1 1
2
1 0 1 1 0
2 k i l i j 1 i k i l i i
i l j i l
W E B p
