http://www.scirp.org/journal/ajor ISSN Online: 2160-8849 ISSN Print: 2160-8830
DOI: 10.4236/ajor.2018.81003 Jan. 17, 2018 33 American Journal of Operations Research
Minimizing the Loss Probability in M/M/2/1
Queueing System with Ordered Entry
Eman Ahmed Kamel
Department of Mathematic, Faculty of Science, EL-Minia University, EL-Minia City, Egypt
Abstract
This study analyzed the M/M/2/1 queueing model with queue of length one (waiting room of capacity just one), heterogeneous servers and ordered entry using the method of semi-Markov process. The customers who arrive in the system enter the free server; if the two servers are free, the customers enter the first server. If the two servers are busy, just one customer can wait at the waiting room. If the two servers are busy and the waiting room has a custom-er, the following customers will leave the system without receiving any service. Such a customer is called LOST COSTOMER. The probability of lost custom-ers in the queueing system under examination was computed. Furthermore, by using inequality f s
( )
≥e−as obtained from Jensen’s inequality, it was shown that the loss probability was minimum when inter-arrival times fit de-terministic distribution [1][2].Keywords
Loss Probability, Heterogeneous Servers, Semi-Markov Process, Laplace-Stieltjes Transform
1. Introduction
The fundamental M/M/2/1 queueing models with waiting room of capacity one and with identical two servers have been examined. The most effective mea-surement of the system is the loss probability, and the loss probability occurs when the two servers are busy and the waiting room has a customer. The steady state probabilities of this system are obtained by the formula: [3][4]
0
!
, 0 , !
k k n k
k
k
P k n
k
ρ ρ λ
µ ρ
=
= ≤ ≤ =
∑
(1)How to cite this paper: Kamel, E.A. (2018) Minimizing the Loss Probability in M/M/2/1 Queueing System with Ordered Entry. American Journal of Operations Research, 8, 33-41.
https://doi.org/10.4236/ajor.2018.81003
Received: October 25, 2017 Accepted: January 14, 2018 Published: January 17, 2018
Copyright © 2018 by author and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).
http://creativecommons.org/licenses/by/4.0/
DOI: 10.4236/ajor.2018.81003 34 American Journal of Operations Research
where k is number of the occupied servers in the system; n is the number of servers in the system, in this model n = 2; λ is the inter-arrival rate µ is the
service rate; 1 1
,
λ µ− − are the mean of the inter-arrival time and the mean of the service time respectively.The probability of lost customers in the system can be computed by the following formula: [5]
3 3 0 3
1
k k k C h
P =
∑
= (2)( )
( )
2 1
1
k k
f k h
f k
µ µ
=
−
=
∏
(3)( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
0 1
2 3
1 1 2
1; ;
2
1 2 1 2
;
2 2
f f
h h
f f
f f f
h
f h f f
µ µ
µ µ
µ µ λ
µ µ λ
− −
= = ⋅
− − +
= =
+
Methods: In this paper, we get the steady state probabilities of this system by formula (1), get the total probability by formula (7), get Laplace-Stieltjes trans-form by trans-formula (8), get steady state probabilities πj by formula j = 0, 1, 2, 3 by formulas at (IV).
2. M/M/2/1 Queueing Model with Ordered Entry [6]
M/M/2/1 Queueing system with finite capacity and heterogeneous servers were analyzed in this study. In this model, inter-arrival times with a finite expected value
( )
0 1 d
a=
∫
∞ −F t t (4)inter-arrival times are independent of each other and have distribution function
F(t).
There are two servers in the system. Their mean service times are assumed to be different from each other. The service time of each customer in server k is a random variable bk and has an exponential distribution with parameter µk; k = 1, 2
(
)
1 e kt, 0 kP b ≤ = −t −µ t> (5)
The service discipline takes place with the principle of “ordered entry”. That is, the customer, who arrives in the system, enters the server with the lowest index among the free server. If the two servers are busy, he wait in the waiting room, if the two servers are busy and the waiting room not empty the customer leaves the system without receiving any service. Such customers are termed “lost custom-ers”. Thus, the main problem herein is to compute the probability of lost cus-tomers in the system and minimize this probability.
3. Analyzing the Model with Semi-Markov Process [7] [8]
Let t t t0, , ,1 2,tn be the arrival time of the customers in the system and0 1 2
DOI: 10.4236/ajor.2018.81003 35 American Journal of Operations Research
Let X(t) be the number of customers in the system at time t and Xn =X t
( )
n ,1
n≥ ; where the number of customers staying in the system immediately before
the time of arrival of the nth customer in the system is denoted by X
n. Let the
semi-Markov process representing the system be defined as follows
( )
t Xn,tn t tn1,n 1η = ≤ < + ≥ (6)
The function defined as:
( )
{
(
1 , 1)
}
ij n n n n
Q x =P X + = j t + − <t x X =i (7)
For x > 0 and 0 ≤ i, j ≤ 3 is called the kernel of process. Based on assumption of Formula (6) and the total probability formula, functions of Formula (7) are computed as follows: [9]
( )
( )
( )
00 01
Q x =F x −Q x
( )
(
1 2 ( 1 2))
( )
01 0 e e e d
x t t t
Q x =
∫
−µ + −µ − −µ µ+ F t ,( )
0( )
e dstF s =
∫
∞ f t − t( )
( )
02 03 0
Q x =Q x =
( )
(
1 2 (1 2))
( )
10 0 1 e e e d
x t t t
Q x =
∫
− −µ + −µ − −µ µ+ F t( )
( )
( )
( )
11 10 12
Q x =F x −Q x −Q x
( )
(1 2)( )
12 0e d
x t
Q x =
∫
−µ µ+ F t( )
13 0
Q x =
( )
(1 2)( )
20 0 1 e d
x t
Q x =
∫
− −µ µ+ F t( )
(
1 2 ( 1 2))
( )
21 0 e e e d
x t t t
Q x =
∫
−µ + −µ − −µ µ+ F t( )
(1 2)( )
22 0e d
x t
Q x =
∫
−µ µ+ F t( )
( 1 2 )( )
23 0e d
x t
Q x =
∫
−µ µ λ+ + F t( )
30 0
Q x =
( )
(
(1 2))
( )
31 0 1 e d
x t
Q x =
∫
− −µ µ+ F t( )
(
1 2 ( 1 2))
( )
32 0 1 e e e d
x t t t
Q x =
∫
− −µ + −µ − −µ µ+ F tWaiting room λ
µ1
DOI: 10.4236/ajor.2018.81003 36 American Journal of Operations Research
( )
(1 2 )( )
33 0e d
x t
Q x =
∫
−µ µ λ+ + F t( )
ijq s
: is the LS transforms of functionsQ
ij( )
x
( )
0 e sxd( )
; 0 , 3ij ij
q s =
∫
∞ − Q x ≤i j≤ ,( )
( )
0 e d
sx
f s =
∫
∞ − F x (8)( )
( )
(
)
(
)
(
)
00 1 2 1 2
q s = f s − f s+µ − f s+µ + f s+µ µ+ (9)
( )
(
)
(
)
(
)
01 1 2 1 2
q s = f s+µ + f s+µ − f s+µ µ+ (10)
( )
( )
02 03 0
q s =q s = (11)
( )
( )
(
)
(
)
(
)
10 1 2 1 2
q s = f s − f s+µ − f s+µ + f s+µ µ+ (12)
( )
(
)
(
)
(
)
11 1 2 2 1 2
q s = f s+µ + f s+µ − f s+µ µ+ (13)
( )
(
)
12 1 2
q s = f s+µ µ+ (14)
( )
13 0
q s = (15)
( )
( )
(
)
20 1 2
q s = f s − f s+µ µ+ (16)
( )
(
)
(
)
(
)
21 1 2 1 2
q s = f s+µ + f s+µ − f s+µ µ+ (17)
( )
(
)
22 1 2
q s = f s+µ µ+ (18)
( )
(
)
23 1 2
q s = f s+µ µ+ +λ (19)
( )
30 0
q s = (20)
( )
( )
(
)
31 1 2
q s = f s −f s+µ µ+ (21)
( )
( )
(
)
(
)
(
)
32 1 2 1 2
q s = f s − f s+µ − f s+µ + f s+µ µ+ (22)
( )
(
)
33 1 2
q s = f s+µ µ+ +λ (23)
(
1)
;( )
0 ; , 1, 2, 3ij n n ij ij
p =P X + = j X =i p =q i j= (24)
The probabilities are obtained from (17) for the model M/M/2/1 with hetero-geneous as the follows: [10][11]
( )
( )
(
)
(
)
(
)
00 0 00 0 0 1 0 2 0 1 2
q =p = f − f +µ − f +µ + f +µ µ+
( )
( )
(
)
00 1 1 2 1 2 1 1 2 12
p = − f µ − f µ + f µ µ+ = − −f f + f (25)
01 1 2 12
p = +f f −f (26)
02 03 0
p =p = (27)
10 1 1 2 12
p = − −f f +f (28)
11 1 2 2 12
p = +f f − f (29)
12 12
p = f (30)
13 0
p = (31)
20 1 12
p = − f (32)
21 1 2 12
p = +f f −f (33)
22 12
DOI: 10.4236/ajor.2018.81003 37 American Journal of Operations Research
(
)
23 1 2 12
p
=
f
µ µ λ
+
+
=
f
λ (35)30 0
p = (36)
31 1 12
p = −f (37)
32 1 2 12
p = +f f − f (38)
33 12
p = f λ (39)
4. Steady State Probabilities
π
jSatisfy the Following
Equations [12] [13] [14]
3
0 ; 1
j i ipij j j
π =
∑
=π∑
π =3
0 i0 ipi0 0p00 1p10 2p20 3p30
π =
∑
= π =π +π +π +π (40)(
p
00−
1
)
π π
0+
1 10p
+
π
2p
20=
0;
p
30=
0
0 0 1 1 2 2 0
aπ +πa +π a = (40.1)
(
)
0 00
1
1
1 2 121
1 2 12a
=
p
− = − − +
f
f
f
− = − − +
f
f
f
1 10 1 1 2 12
a =p = − −f f + f
2 20 1 12
a = p = − f
3
1 i0 ipi1 0p01 1p11 2p21 3p31
π =
∑
= π =π +π +π +π (41)(
)
0p01 1 p11 1 2p21 3p31 0
π
+π
− +π
+π
=0 0b 1 1b 2 2b 3 3b 0 π +π +π +π =
0 01 1 2 12
b =p = +f f − f
(
)
1 11
1
1 22
121
b
=
p
− = + −
f
f
f
−
2 21 1 2 12
b = p = +f f − f
3 31 1 12
b =p = − f
3
2 i0 ipi2 0p02 1p12 2p22 3p32
π =
∑
= π =π +π +π +π (42)(
)
1 12p 2 p22 1 3p32 0; p02 0
π
+π
− +π
= =1 1c 2 2c 3 3c 0
π +π +π = (42.1)
1 12 12
c = p = f
(
)
2 22 1 12 1
c = p − = f −
3 32 1 1 2 12
c = p = − −f f + f
3
3 i0 ipi3 0p03 1p13 2p23 3p33
π =
∑
= π =π +π +π +π (43)(
)
2
p
23 3p
331
0;
p
03p
130
π
+
π
− =
=
=
2d2 3d3 0
π +π = (43.1)
2 23 12
d =p = f λ
(
)
3 33 1 12 1
DOI: 10.4236/ajor.2018.81003 38 American Journal of Operations Research
1
j j
π =
∑
0 1 2 3 1
π +π π+ +π = (44)
From Equation (43.1):
2d2 3d3 0
π +π =
12
2 3
12
1
f f
λ
λ
π =− − π (45)
From Equation (42.1):
1 1c 2 2c 3 3c 0 π +π +π =
3
1 1 3 2 3 3 2
0
d
c c c
d
π − π +π =
2 3 3 2
1 3
1 2
c d c d
c d
π = − π (46)
From Equation (40.1):
0 0 1 1 2 2 0
aπ +πa +π a =
2 3 3 2 3
0 0 3 1 3 2
1 2 2
0
c d c d d
a a a
c d d
π + − π + − π =
2 3 3 2 3 2 1 3 1 2 3 1 3 2
0 0 3 1 3 2 3
1 2 2 1 2
c d c d d a c d a c d a c d
a a a
c d d c d
π = − − π + π = − + π
2 1 3 1 2 3 1 3 2
0 3
0 1 2
a c d a c d a c d
a c d
π = − + π (47)
Use Equations (45:47) in Equation (44):
0 1 2 3
1 1 3 1 2 3 1 1 2 0 2 3 0 1 3
a c c
a c d a c d a c c a c d a c d
π = −
− − + −
(
) (
)
(
)(
)
0 1 2 1 2 12 12 12
1
121
12 1 2 12a c c
= − − +
f
f
f
f
f
− =
f
−
f
f
+ −
f
f
(
)(
)
1 1 2 12 1 12 1 2 12
a c c = f − f f + −f f
(
)(
) (
)
3 1 2 1 0 12 1
d c −c a −a = f λ−
0
π : the probability that the system’s probability of being free
1
π : the probability that only one server is busy in the system
2
π : the probability that the two servers are busy in the system
3
π : the probability that the two servers are busy in the system and the waiting
room has a customer.
(
)(
)
(
12)
12(
1)(
2 12)
3
12 12 12 1 2 12
1 1 1
f f f f f
f λ f f f f f
π = − + −
− − − + − (48)
Loss probability and its minimization [15][16][17][18]
DOI: 10.4236/ajor.2018.81003 39 American Journal of Operations Research
equivalent to the probability of loss of customers in the system. That is, Formula (48) is equal to the loss probability. If we symbolize the probability of loss of customers in the system with PLOSS and is written as
LOSS 3
P =π
Under µ1=µ2= ≥ >µ λ 0 the formula of loss probability presented by
Formula (48) satisfies Palm’s loss Formula (2) for n = 3. Optimization of the queueing system given above to create more efficient systems in real life appears an important problem. For the M/M/2/1 queueing model with a waiting line of just one place and with heterogeneous servers, and µ µ1+ 2 = ≥µ 0 under, Nath
and Enns (1981) assigned the arriving customer to the server with the lowest mean service time among the free servers, thereby minimizing the probability of lost customers in the system.
Let Am be a class of distribution functions F of the inter-arrival times, the mean of which is a constant m, 0<m<t. Let PLOSS(F)be the loss probability for
the M/M/2/1 queueing system with heterogeneous servers and ordered entry, and F∈Am. Assume that D(t) is the deterministic distribution, in which
( )
1D t = for t≤m and D t
( )
=0 for t>m.The Laplace transform of Dirace Delta :
(
)
{
δ t−m}
= 0∞ −e stδ(
t−m)
dt=e−mt; m>0∫
(49)
Theorem:
When the distribution of inter-arrival times fits the deterministic distribution D among all distribution functions included in class Am, loss probability
( )
LOSS
P F becomes minimum, that is, min LOSS
( )
LOSS( )
m
F A∈ P F =P D ]
19
[
Proof:
For minimizing the loss probability, let Formula (48) be :
( )
(
12)
(
(
12)(
1)(
2 12)
)
LOSS 3
12 12 12 1 2 12
1 1 1
f f f f f
P F
f λ f f f f f
π − − + −
= =
− − − + − (50)
By using Jensen’s inequality f s
( )
≥e−ms which mean that the probability of lost customers in the system is minimum when under-arrival times fit the de-terministic distribution.The numerator:
(
)(
)
( 1 2)
(
(1 2))
(
1 2 (1 2))
12 1 12 1 2 12
e m 1 e m e m e m e m
f f f f f
µ µ µ µ µ µ µ µ
− + − + − − − +
− + −
≥ − + −
(
)(
)
(1 2)
(
(1 2))
(
1 2 ( 1 2))
12 12 1 1 2 12
e m 1 e m e m e m e m
f f f f f
µ µ µ µ µ µ µ µ
− + − + − − − +
− + −
≤ − − + −
The denominator:
(
)
(
)(
)
( )
(
1 2)
(1 2)(
( 1 2))
(
1 2 (1 2))
12 1 12 1 12 1 2 12
e m 1 em 1 e m e m e m em
f λ f f f f f
µ µ λ µ µ µ µ µ µ µ µ
− + + − + − + − − − +
− − − + −
DOI: 10.4236/ajor.2018.81003 40 American Journal of Operations Research
From (48) and use (50):
( )
(
)
(
(
)(
)(
)
)
( )
(
( ))
( )( ) ( ) ( ) ( )
1 2 1 2 1 2 1 2
1 2 1 2 1 2 1 2 1 2
12 12 1 2 12
LOSS
12 12 12 1 2 12
1 1 1
e e 1 e e e
e 1 e 1 e e e e
m m m m m
m m m m m m
f f f f f
P F
f λ f f f f f
µ µ µ µ µ µ µ µ
µ µ λ µ µ µ µ µ µ µ µ
− + − + − − − +
− + + − + − + − − − +
− + −
≥
− − − + −
− + −
∗ ∗
≥
− − − + −
(51)
The right side of (51) has the value of PLOSS
( )
D thus( )
( )
LOSS LOSS
min m
F A∈ P F =P D (Q.E.D) Numerical example:
Use Equation (51)
Case 1: At m=1;µ1=1;µ2=2;λ=1
( )
LOSS
0.0505
P
D
=
(52) Case 2: At m=1;µ1=2;µ2=2;λ=2( )
LOSS 0.0159
P D = (53)
Case 3: At m=1;µ1=2;µ2=2;λ=1
( )
LOSS
0.0164
P
D
=
(54) It is cleared that case 2 is the smallest probability of losing customers.5. Discussion
In this study, the M/M/2/1 model with recurrent entries, finite capacity and or-dered entry was analyzed and the steady-state probabilities of the system and the probability of lost customers in the system were obtained. Optimization was performed according to the arrival processes and it was shown that the loss probability was minimum with probability 0.0159 in the M/M/2/1 queueing sys-tem with heterogeneous servers and ordered entry when the deterministic dis-tribution was selected among the disdis-tributions of inter-arrival times with iden-tical means.
6. Conclusion
The loss probability is minimum in M/M/2/1 model where µ1=µ2 =λ.
7. Recommendations
It can be studied that the model M/M/n/r, 1≤ ≤r n and the distribution of
stream of overflows can be obtained by analyzing the stream of overflows. With the distribution to be obtained, the loss probability can be generalized for n ser-vices. Similar analyses can be made for “Random” service discipline or for dif-ferent suggested disciplines instead of “Ordered Entry” discipline.
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DOI: 10.4236/ajor.2018.81003 41 American Journal of Operations Research
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