International Journal of Research in Engineering & Applied Sciences
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84The Study Of Serial Queuing Processes With Feedback
RajkalaAssist. Prof. Dr. BRA. Govt. College, Kaithal-136027(Haryana), India
Abstract
O’Brien (1954), Jackson (1954) and Hunt (1955) studied the problems of serial queues in the steady state with Poisson assumptions. In these studies, it is assumed that the unit must go through each service channel without leaving the system. Barrer (1955) obtained the steady – state solution of a single channel queuing model having Poisson input, exponential holding time, random selection where impatient customers leave the service facility after a wait of certain time. Finch (1959) studied simple queues with customers at random for service at a number of service stations in series where the arrival from outside was considered at the initial stage. Feedback is permitted either from the terminal server or from each server of the series to the queue waiting for service at that stage by imposing an upper limit on the number of customers in the system at any time. Singh(1984) studied the problem of serial queues introducing the concept of reneging. Singh and Umed (1994) worked on the network of queuing processes with impatient customers. Punam (2011) found the steady-state solution of serial queuing processes where feedback is not permitted.
In this paper, the steady-state solutions are obtained for general queuing process in which
(i) The number of serial service channel is M
(ii) A customer may join any channel from outside and leave the system at any stage after getting service.
(iii) Feedback is permitted from any channel to its previous channel. (iv) Poisson arrivals and exponential service times are followed. (v) The queue disciple is random selection for service
(vi) Waiting space is infinite
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851. Formulation of Model
The system consists of the queues
Q
j
j
1,2,3,....,
M
with respective servers
1,2,3,....,
j
S
j
M
. Customers demanding different types of service arrive from outside the system in Poisson stream with parameters
j
j
1,2,...,
M
atQ
j
j
1,2,3,....,
M
respectively. After the completion of service atS
j, the customer either leave the system with probability pj or join the next channel with probabilityq
j or join back the previous channel with probabilityr
j such that1
j j j
p
q
r
j
1,2,3,...
M
. It is being mentioned here thatr
j
0
whenj
1
as there is no previous channel of the first channel andq
j
0
whenj
M
since there is no next chance after Mth channel. The service time distribution for seversS
j are mutually independent-negative distribution with parameters
j
j
1,2,3,....,
M
The applications of such models are of common occurrence. For example, consider the administration of a particular district in a particular state at the level of district head quarter consisting of Block development officer, Tehsildar, Sub-divisional magistrate, District magistrate etc. Here, the officers of the district correspond to the servers of the model. The people meet the officers of the district in connection with their problems. It is also a common practice that officers call the customers (people) for hearing randomly. The senior officer may send any customer to his junior if some information regarding the customer’s problem is lacking.
2. Formulation of Equations:
Define:
P n n n
1,
2, ,...
3n
M1,
n
M;
t
= the probability that at time ‘t
’ there aren
j customers (which may leave the system after service or join the next phase or join back the previous channel)waiting before
S
j
ˆ 1,2,3,....,
j
M
1,
M
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86
1 2 3
.
,
, ,...,
1,....
i i M
T n
n n n
n
n
1 2 3
.
i,
,
,...,
i1,....,
MT
n
n n n
n
n
1 1 2 3 1
. , .
i i,
,
,...,
i1,
i1,....,
MT
n
n n n
n
n
n
1
.,.
1,
2, ,...,
3 11,
1....,
i i i i M
T
n
n n n
n
n
n
Following the procedure given by Kelly (1979), we write the difference – differential equations
1 1
;
;
M M
i i i i
i i
d
P n t
n
P n t
dt
1.
;
M i i iP T n t
1.
;
Mi i i i
p P T
n t
1 1 1. , .
;
Mi i i i i
q P T
n t
1
1
.,.
;
M
i i i i i
r P T
n t
. (2.1)Where
n
i
0
i
1,2,3,....,
M
and
1
0
0
0
if x
x
if x
and
P n t
;
0
if any of the arguments in negative.3. Steady – State
Equations:-We write the following Steady–state equations of the queuing model by equating the time derivates to zero in the equation (2.1)
1 1 1 1
.
.
M M M M
i i i i i i i i
i i i i
n
P n
P T n
p P T
n
1 1 1 1 1. , .
.,.
M Mi i i i i i i i
i i
q P T
n
r P T
n
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87For
n
i
0
i
1,2,3,....,
M
4. Steady-State
Solutions:-The solutions of the Steady-State equations (3.1) can be verified to be
1 2 2 2 1 2 1 1 1 3 3 3 2 3 2 2 2 4 4 4 31 2 3
0
...
n
n n
r
q
r
q
r
P n
P
1
1 2 2 2 1 1 1
1
...
M M
n n
M M M M M M M M M M M
M M
q
r
q
(4.1)Where
1 2 2 2 1
1
r
2 1 1 1 3 3 3 2
2
q
r
3 2 2 2 4 4 4 3
3
q
r
(4.2)……….
……….
……….
1 2 2 2
1
1
M M M M M M M
M
M
q
r
1 1 1
M M M M
M
M
q
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88Solving these (4.2) M-equations for
M with the help of determinants, we get
M
M1
q
M1
M1
M2
q
M1q
M2
M2
M3
....
q
M1q
M2....
q
3 3
2
1 2 3 2 2 1 1 2 3 2 2 1 1
1 1 2
....
....
M M M M
M
M M M M
q
q
q q
q
q
q q q q
q
r
(4.3)where
M M1
q r
n1 M
M21 2 2 1 3
M M
q
Mr
M M
(4.4)Continuing in this way
3 2
q r
2 3 1
where
2
1 3
2 4
2 1
1
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
1
0
0
0
0
0
1
0
0
0
0
M
M M
M
r
q
r
q
r
q
r
q
2
2 1 2
1
1
1
1
1
1
1
r
q r
q
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89Since
M is obtained, we can get
M1by putting the value of
M in the last equation of
4.2 ,
M2by putting the values of
M1and
M in the last but one equation of
4, 2
continuing inthis way, we shall obtain
M3,
M4,
,
3,
2,
and
1,.
Thus, we write
4,1
as under
1 2 3
1 2 3
0
n n n nM;
0,
1,2,
,
,
M i
P n
P
for n
i
M
(4.5)We obtain
P
0
from the normalizing conditions.
01
nP n
and with the restriction that traffic intensity of each service channel of the system isless than unity,
3
1 2 1
3
1 2
1 2 3
1 1
1 2 3 1 0
1 2 3
0 0 0
1
0 0
1
0
....
1
0
....
...
M M
M M
M M
n
n n n n
M M
n
n
n n
n n n
n n M M n n
P
P
1 2 3 1
1
1
1
1
1
1
1
1
0
...
1
1
1
1
1
1
0
(4.7)
1
M M M i iP
P
Thus
P n
is completely determined.5. Steady-State Marginal Probability
Let
P n
1 be the steady-state marginal probability that there aren
1 units in the queue before the first phase. This is determined as
2 3 0
1
, ,.... M
n n n
P n
P n
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90Using (4.5) and (4.7)
11
(1
1)
1n
P n
Similarly
22
(1
2)
2n
P n
---
---
1
nMM M M
P n
6. Mean Queue Length
Mean queue length before the server
S
1 is determined by
1
1 1
1 1 1
0
1 1 1
0
2 3 4
1 1 1 1 1
1
1
(1
)
(1
)
2
3
4
....
(1
)
n
n n
L
n P n
n
Similarly
;
2,3,....
(1
)
k k
k
L
k
M
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91 11
(1
)
M k k M
k
k k
L
L
6. Corollary If our queuing system is restricted to 5 servers, then the Steady-State solution
4.5
can be written as
1 2 3 4 51
,
2,
3,
4,
50,0,0,0,0
1 2 3 4 5n n
n n n
P n n n n n
P
p
p
Where 1 1 2 2 2
1
r
2 1 1 1 3 3 3 2
2
q
r
5.1
3 2 2 2 4 4 4 3
3
q
r
4 3 3 3 5 5 5 4
4
r q
r
4 4 4 4 5
5
r q
Using (4.3),
5
1
q r
1 2
q r
2 3
q r
3 4
q q r r
1 3 2 4
q
4
4
1
q r
1 2
q r
2 3
4 3 3 1 2 4 3 2
2 4 3 2 1 1
5
5 1 2 2 3 3 4 1 3 2 4 4 5 1 2 2 3
1
1
1
q q
q r
q q q
q q q q
q r
q r
q r
q q r r
q r
q r
q r
(5.2)International Journal of Research in Engineering & Applied Sciences
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92References:
Finch, P.D (1959) : Cyclic queues. Opns. Res. Quart, Vol. 9, No. 1.
Jackson, R.R. P(1954) : Queuing system with phase type service. operat. Res. Quart.
109-120.
Hunt, G.C. (1955) : Sequential arrays of waiting lines. opps. Res. Vol. 4, No. 6.
Kelly, F.P (1976) : Network of a queues. Adv. Prob. 8,416-432.
Kelly, F.P (1979) : Reversibility and stochastic networks. Wiley, New York.
O’Brien, G.G. (1954) : The solution of some queuing problems. J.Soc. Ind. Appl. Math.
Vol. 2, 132-142.
Singh.M. (1984) : Steady-state behaviour of serial queuing processes with impatient customers. Math. operations forsch, U.statist. Ser. Statist.
Vol. 15.No 2, 289-298
Singh, M and Singh, Umed (1994): Network of serial and non-serial queuing processes with impatient customers. JISSOR Vol. 15, 1-4.
Punam; Singh.M; and Ashok (2011): The Steady-State solution of serial queuing processes with reneging due to long queue and urgent message.
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