ISSN 2319-8133 (Online)
(An International Research Journal), www.compmath-journal.org
Impatient Customers in an M/M/1 Working Vacation Queue with a Waiting Server and Setup Time
P. Manoharan* and T. Jeeva Annamalai University, INDIA.
email:[email protected], [email protected]
(Received on: May 14, 2019) ABSTRACT
Using probability generating function method, Impatient customers in an M/M/1 working vacation queue with a waiting server and set-up time is discussed in this paper. Customers impatience is due to the servers vacation. We obtain the probability generating functions of the stationary state probabilities, performance measures, sojourn time of a customer and stochastic decomposition of the queue length and waiting time.
Keywords: M/M/1 queue; Impatient customers; Working vacation; Setup time;
Probability generating function.
1. INTRODUCTION
Server vacations were introduced first by Levy and Yechiali
8. Substantial work in this area has been conducted during the last few decades while Tian and Zhang
13, Doshi
6, Baba
2and Takagi
12surveyed the vacation models. Recently, Ke, Wu and Zhang
7did a survey about the vacation models. These studies help understanding about a variety of vacation models.
Queueing systems with impatient customers appear in many real life situations such as those involving impatient telephone switchboard customers, hospital emergency rooms handling critical patients, and inventory systems that store perishable goods
9,1. There is growing interest in the analysis of queueing systems with impatient customers. This is due to their potential application in communication systems, call centers, Production-inventory systems and many other related areas, see for
4,3.
Queueing systems with impatient customers have been studied by a number of
authors. We mention a few of the more significant works below. Palm's pioneering work
11seems to be the first to analyze queueing systems with impatient customers by considering the
M/M/C queue, where the customers have independent exponentially distributed waiting times.
The situation of impatient customers in a server vacations period was investigated by Altman and Yechialli, Yue et al.
14have analyzed an M/M/1 queueing system with working vacation and impatient customers and obtained the pgf of the number of units in the model when the server is in a working vacation period and in a service period. Padmavathy et al.
10have studied the steady state behaviour of vacation queues with impatient customers and a waiting server.
In this paper we analyze the impatient customers in an M/M/1 working vacation queue with waiting server and setup times. In this model, if the system becomes empty, then the server waits for a random period of time. This option of the server waits reflects many real life queueing systems. It was first introduced by Boxma et al.
5for the ordinary M/G/1 queue with server vacations.
The rest of the paper is organized as follows. Section 2, gives a description of the Impatient customers in an Markovian working vacation queue with waiting server and setup times. Section 3, gives the analysis of the M/M/1 working vacation model with the stochastic decomposition property. In section 4, gives the numerical results.
2. MODEL DESCRIPTION
We consider a Impatient customers in an M/M/1 working vacation queue with a waiting server and setup time. The assumption of the model are as follows: Customers arrive according to a poisson process with arrival rate 𝜆. The service rate is 𝜇
𝑏during a normal service period. When the busy period is ended the server waits a random duration of time before beginning a vacation. This waiting time duration follows the exponential distribution with rate 𝜂 is the waiting rate of a server. The server takes a working vacation as soon as the system becomes empty. During the vacation period, the server serves the customers at an exponential rate 𝜇
𝑣, where 𝜇
𝑣< 𝜇
𝑏. The duration of a vacation is also exponential with rate 𝛾.
The customers in this system are assumed to be impatient as described below.
Whenever arriving customer has to join a queue and finds the system in working vacation, the customer activates an impatient timer 'T', which is exponential with rate 𝛼 and is independent of the number of customers in the system at that moment. If the server ends the vacation before the time T expires, the customer stays in the system till his service is completed, otherwise the customer leaves the system and will never return. When a vacation ends, if there are customers in the queue, the server changes his service rate from 𝜇
𝑣to 𝜇
𝑏and a regular busy period starts, otherwise the server begins a closed-down period. However, a server needs some setup time to be active so as to serve waiting customers. We assume that the setup time follows the exponential distribution with mean
1𝛽.
We assume that inter arrival times, service times, working vacation times, impatient times and setup times are mutually independent. In addition the service order is First in First out(FIFO). At time t, let Q(t) denote the total number of customers in the system and J(t) denotes the state of the server with,
J(t)=
0, at time t the server is busy in working vacation period, 1, at time t the server is free during in setup period, 2, at time t the server is busy in normal service period.
Then the pair {(𝑄(𝑡), 𝐽(𝑡)); 𝑡 ≥ 0} defines a two dimensional continuous time Markov chain with state space 𝐸 = {{(0,0), (0,1), (0,2)} ∪ {𝑖, 𝑗}, 𝑖 = 1,2, . . . , 𝑗 = 0,1,2}. Let 𝑃
𝑖,𝑗= lim
𝑡→∞
𝑃{𝑄(𝑡) = 𝑖, 𝐽(𝑡) = 𝑗}, (𝑖, 𝑗) ∈ 𝐸. Let 𝑃
𝑖,0, 𝑖 ≥ 0, be the probability that there are i customers in the system when the server is in working vacation period, 𝑃
𝑖,1, 𝑖 ≥ 0, be the probability that there are i customers in the system when the server is in setup period and 𝑃
𝑖,2, 𝑖 ≥ 0, be the probability that there are i customers in the system when the server is in normal busy period. Therefore, using the theory of Markov process, we obtain the following set of balance equations.
(𝜆 + 𝛾 ) 𝑝
0,0= 𝜂𝑝
0,2+ 𝜇
𝑣𝑝
1,0 ,(1)
(𝜆 + (𝑛 − 1)𝛼 + 𝜇
𝑣+ 𝛾) 𝑝
𝑛,0= 𝜆𝑝
𝑛−1,0+ (𝜇
𝑣+ 𝑛𝛼)𝑝
𝑛+1,0, 𝑛 ≥ 1, (2)
𝜆𝑝
0,1= 𝛾𝑝
0,0 ,(3)
(𝜆 + 𝛽)𝑝
𝑛,1= 𝜆𝑝
𝑛−1,1, 𝑛 ≥ 1, (4)
(𝜆 + 𝜂)𝑝
0,2= 𝜇
𝑏𝑝
1,2+ 𝛽 𝑃
0,1, (5)
(𝜆 + 𝜇
𝑏)𝑝
𝑛,2= 𝜆𝑝
𝑛−1,2+ 𝛽𝑝
𝑛,1+ 𝛾𝑝
𝑛,0+ 𝜇
𝑏𝑝
𝑛+1,2, 𝑛 ≥ 1 . (6)
Define the Probability generating functions (PGFs) 𝑄
0(𝑧) =
0 0 , n n np z , 𝑄
1(𝑧) =
1 0 , n n np z , 𝑄
2(𝑧) =
2 0 , n n np z (7)
with 𝑄
0(1) + 𝑄
1(1) + 𝑄
2(1) = 1 and 𝑄
0′(𝑧) =
1 1 ,0 n nnp z
nMultiplying (2) by 𝑧
𝑛and summing over n and using (1) , we get
(𝜆 + 𝜇𝑣+ 𝛾) 0 1 , n n np z
+ 𝛼 0 1 , n n np z
(𝑛 − 1) + (𝜆 + 𝛾)𝑝00= 𝜆𝑧 ,0 1 1 n np
nz
+ 𝜇𝑣 ,0 1 1 n np
nz
+𝛼 1,0 1 n nnp
nz
+ 𝜂𝑝0,2+ 𝜇𝑣𝑝1,0 𝛼𝑧(1 − 𝑧)𝑄0′(𝑧) − 𝑄0(𝑧)[(𝜆𝑧 − 𝜇𝑣+ 𝛼)(1 − 𝑧) + 𝛾𝑧] + (𝛼 − 𝜇𝑣)(1 − 𝑧)𝑝0,0+ 𝜂𝑧𝑝0,2= 0(8) Putting 𝑧 = 1 in (8) we get
𝛾𝑄
0(1) = 𝜂𝑝
0,2−𝑄
0(𝑧)[(𝜆𝑧 − 𝜇
𝑣)(1 − 𝑧) + 𝛾𝑧] − 𝜇
𝑣(1 − 𝑧)𝑝
0,0+ 𝜂𝑧𝑝
0,2= 0 𝑄
0(𝑧) = 𝜇
𝑣(1 − 𝑧)𝑝
0,0+ 𝜂𝑧𝑝
0,2𝜆𝑧 − 𝜆
2− 𝜇
𝑣+ 𝜇
𝑣𝑧 + 𝛾𝑧 Equation (8) can be written as 𝑄
0′(𝑧) − (
𝜆𝑧−𝜇𝑣+𝛼𝛼𝑧
+
𝛾𝛼(1−𝑧)
)𝑄
0(𝑧) +
(𝛼−𝜇𝑣)𝛼𝑧
𝑝
0,0+
𝜂𝛼(1−𝑧)
𝑝
0,2= 0
Integrating Factor
𝑒
− ∫ (𝜆𝑧−𝜇𝑣+𝛼𝛼𝑧 +𝛼(1−𝑧)𝛾 )𝑑𝑧= 𝑒
−𝜆𝛼𝑧𝑧
𝜇𝑣𝛼−1(1 − 𝑧)
𝛾𝛼The general solution to the differential equation is given by
𝑄0(𝑧) = 𝑒𝛼𝜆𝑧𝑧−𝜇𝑣𝛼−1(1 − 𝑧)−𝛾𝛼((𝜇𝑣𝛼 − 1)𝑝00∫
𝑧 0
𝑒−𝜆𝛼𝑥𝑥𝜇𝑣𝛼−2(1 − 𝑥)𝛾𝛼𝑑𝑥 −𝜂
𝛼𝑝0,2∫𝑧
0
𝑒−𝜆𝛼𝑥𝑥𝜇𝑣𝛼−1(1 − 𝑥)𝛾𝛼−1𝑑𝑥)
𝑄
0(𝑧) = 𝑧
−𝜇𝑣𝛼−1(1 − 𝑧)
−𝛼𝛾((
𝜇𝛼𝑣− 1)𝑝
0,0𝑘
1(𝑧) −
𝜂𝛼𝑝
0,2𝑘
2(𝑧)) (9) 𝑘
1(𝑧) = ∫
0𝑧𝑒
𝜆𝛼(𝑧−𝑥)𝑥
−𝜇𝑣𝛼−2(1 − 𝑥)
𝛾𝛼𝑑𝑥
𝑘
2(𝑧) = ∫
0𝑧𝑒
𝛼𝜆(𝑧−𝑥)𝑥
𝜇𝑣𝛼−1(1 − 𝑥)
𝛾𝛼−1𝑑𝑥 Taking
1
lim
z
in equation (8) we get
1
lim
z
𝑄
0(𝑧) = 𝑄
0(1) = [( 𝜇
𝑣𝛼 − 1)𝑝
0,0𝑘
1(𝑧) − 𝜂
𝛼 𝑝
0,2𝑘
2(𝑧)]
1
lim
z
(1 − 𝑧)
−𝛾𝛼(
𝜇𝛼𝑣− 1) 𝑝
0,0𝑘
1(1) =
𝛼𝜂𝑝
0,2𝑘
2(1)
𝑝
02= (
𝜇𝑣𝛼−𝛼𝛼𝜂𝑝
0,0𝑘𝑘1(1)2(1)
) 𝑝
02= (
𝜇𝑣𝜂−𝛼𝑝
0,0𝑘𝑘1(1)2(1)
)
(10)
Multiplying (3)and (4) by 𝑧
𝑛and summing over n, we get
(𝜆 + 𝛽)
01 ,
n n n
p z = 𝜆
,11 1 n
n
p
nz (𝜆 + 𝛽)𝑄
1(𝑧) − (𝜆 + 𝛽)𝑝
0,1= 𝜆𝑧𝑄
1(𝑧)
(𝜆 + 𝛽 − 𝜆𝑧)𝑄
1(𝑧) = (𝜆 + 𝛽)𝑝
0,1(11)
Similar way (5) and (6) with power of 𝑧
𝑛and summing over all possible values of n, we get
(𝜆 + 𝜇𝑏)(𝑄2(𝑧) − 𝑝0,2) + (𝜆 + 𝜂)𝑝0,2 = 𝜆𝑧𝑄2(𝑧) + 𝛽(𝑄1(𝑧) − 𝑝0,1) +𝜇𝑏𝑧 (𝑄2(𝑧) − 𝑃1,2𝑧 −𝑃0,2) + 𝛾(𝑄0(𝑧) − 𝑃0,0) + 𝜇𝑏𝑃1,2+ 𝛽𝑃0,1
(1 − 𝑧)(𝜆𝑧 − 𝜇
𝑏)𝑄
2(𝑧) = 𝛾𝑧𝑄
0(𝑧) + 𝛽𝑧𝑄
1(𝑧) − 𝜇
𝑏(1 − 𝑧)𝑃
0,2− (𝜂𝑝
02+ 𝛾𝑝
0,0)𝑧 𝑄
2(𝑧) =
𝛽𝑧𝑄1(𝑧)+𝛾𝑧𝑄0(𝑧)−(𝜂𝑝(𝜆𝑧−𝜇02+𝛾𝑝0,0)𝑧−𝜇𝑏(1−𝑧)𝑝0,2𝑏)(1−𝑧)
(12) Adding (5) and (6) over all possible values of n, we get
𝛽𝑄
1(1) + 𝛾𝑄
0(1) = 𝜂𝑝
0,2+ 𝛾𝑝
0,0(13) Equation (12) can be written as,
𝑄
2(1) =
𝛽𝑄1′(1)+𝛾𝑄𝜇 0′(1)𝑏−𝜆
+
𝜇𝜇𝑏𝑝0 2𝑏−𝜆
(14)
where, 𝑄
0′(1) = 𝐸[𝐿
0],𝑄
2(1) = 1 − 𝑄
0(1) − 𝑄
1(1) substituting in(13), we gives
1 − 𝑄0(1) − 𝑄1(1) = 𝛽𝐸[𝐿𝜇1]+𝛾𝐸[𝐿0]
𝑏−𝜆 +𝜇𝜇𝑏𝑝0,2
𝑏−𝜆
𝐸[𝐿0] =(𝜇𝑏𝛾−𝜆)(1 − 𝑄0(1) − 𝑄1(1)) −𝛽𝛾𝐸[𝐿1] −𝜇𝛾𝑏 𝑝0,2 (15)
So that 𝐸[𝐿
0] can be obtained.Adding (2), (4), (6) and rearranging the terms, we get
𝜆𝑃𝑛,0+ 𝜆𝑃𝑛,1+ 𝜆𝑃𝑛,2− [(𝜇𝑣+ 𝑛𝛼)𝑃𝑛+1,0+ 𝜇𝑏𝑃𝑛+1,2] =−[(𝜇𝑣 + (𝑛 − 1)𝛼)𝑃𝑛,0+ 𝜇𝑏𝑃𝑛,2], 𝑛 ≥ 1
𝜆𝑃𝑛−1,0+ 𝜆𝑃𝑛−1,1+ 𝜆𝑃𝑛−1,2
(16) Using (16) in (2),(4),(6), we get
𝜆𝑝
𝑛,0+ 𝜆𝑝
𝑛,1+ 𝜆𝑝
𝑛,2= (𝜇
𝑣+ 𝑛𝛼)𝑝
𝑛+1,0+ 𝜇
𝑏𝑝
𝑛+1,2, 𝑛 ≥ 0 (17) Adding over all possible values n in (17), we get
𝜆𝑄0(1) + 𝜆𝑄1(1) + 𝜆𝑄2(1) = 𝜇𝑏(𝑄2(1) − 𝑝0,2) + 𝜇𝑣(𝑄0(1) − 𝑝0,0) + 𝛼
∑∞𝑛=0(𝑛 + 1)𝑃𝑛+1,0− 𝛼(𝑄0(1) − 𝑝0,0), 𝑛 ≥ 0
𝐸[𝐿0] = ,
0
( 1)
n 1 0 n nn p z and
𝑄2(1) = 1 − 𝑄0(1) − 𝑄1(1), 𝐸[𝐿
0] in (15) and we get
𝜆𝑄0(1) + 𝜆𝑄1(1) + 𝜆 − 𝜆𝑄0(1) − 𝜆𝑄1(1) = 𝜇𝑏− 𝜇𝑏𝑄0(1)−𝜇𝑏𝑄1(1) + 𝜇𝑣𝑄0(1) − 𝜇𝑣𝑃0 ,0− 𝜇𝑏𝑝0,2− 𝛼𝑄0(1) + 𝛼𝑃0,0+ 𝛼𝐸[𝐿0]
𝜆 = 𝜇𝑏− 𝜇𝑏𝑄0(1) − 𝜇𝑏𝑄1(1) − 𝜇𝑏𝑝0,2+ 𝜇𝑣𝑄0(1) − 𝜇𝑣𝑝0,0− 𝛼𝑄0(1) + 𝛼𝑝0,0
+𝛼(𝜇𝑏𝛾−𝜆)(1 − 𝑄0(1) − 𝑄1(1)) −𝛼𝛽𝛾 𝐸[𝐿1] −𝛼𝜇𝛾𝑏𝑝0,2
𝜆𝛾 − 𝛼(𝜇
𝑏− 𝜆) = ((𝜇
𝑣− 𝜇
𝑏)𝛾 − 𝛼(𝛾 + 𝜇
𝑏− 𝜆))𝑄
0(1) − (𝛼(𝜇
𝑏− 𝜆) + 𝜇
𝑏𝛾) 𝑄
1(1) − 𝛾(𝜇
𝑣− 𝛼)𝑝
0,0− 𝜇
𝑏(𝛾 + 𝛼)𝑝
0,2− 𝛼𝛽𝐸[𝐿
1] (18) For z=1, from equation (9), we get
𝑝
0,0=
(𝜇𝛾𝑣−𝛼) 𝑘2(1)
𝑘1(1)
𝑄
0(1) (19)
(𝜆𝛾 − 𝛼(𝜇𝑏− 𝜆)) = ((𝜇𝑣− 𝜇𝑏)𝛾 − 𝛼(𝛾 + 𝜇𝑏− 𝜆))𝑄0(1) − (𝛼(𝜇𝑏− 𝜆) +𝜇𝑏𝛾)𝛾
𝛽𝑝0,0−𝛼
𝛽𝑝0,0𝜆𝛾 − 𝛾(𝜇𝑣− 𝛼)𝑝0,0− 𝜇𝑏(𝛾 + 𝛼)𝛾 𝜂𝑄0(1) 𝑄0(1) = 𝜆𝛾 − 𝛼(𝜇𝑏− 𝜆)(((𝜇𝑣− 𝜇𝑏)𝛾 − 𝛼(𝛾 + 𝜇𝑏− 𝜆) − 𝜇𝑏(𝛾 + 𝛼)𝛾𝜂)
−(𝜇𝛾
𝑣−𝛼) 𝑘2(1)
𝑘1(1)(𝜆𝛾𝛼𝛽 + 𝛾(𝜇𝑣− 𝛼) + (𝛼(𝜇𝑏− 𝜆) + 𝜇𝑏𝛾)𝛾𝛽))−1)
(20) From equation (11), we get
𝐸[𝐿1] = 𝑄1′(1) =𝛾𝜆𝛽2𝑝0,0
𝑄1′′(1) =𝛾𝜆𝛽32𝑝0,0
(21) We find the values 𝑝
0,1, 𝑝
0,2and 𝑝
1,0in the equation (3), (5) and (1) we get
𝑝0,1=𝜆+𝛽𝛾 𝑝0,0
(𝜆 + 𝜂)𝑝0,2= 𝜇𝑏𝑝1,2+𝜆+𝛽𝛽𝛾 𝑝0,0((𝜆 + 𝜂)(𝜇𝑣𝜂−𝛼)𝑘𝑘1(1)
2(1)−𝜆+𝛽𝛽𝛾 𝑝0,0= 𝜇𝑏𝑝1,2) 𝑝1,2=𝜇1
𝑏((𝜆 + 𝜂)(𝜇𝑣𝜂−𝛼)𝑘𝑘1(1)
2(1)−𝜆+𝛽𝛽𝛾) 𝑝1,0=𝜇1
𝑣((𝜆 + 𝛾) − (𝜇𝑣− 𝛼)𝑘𝑘1(1)
2(1)𝑝0,0)
(22)
Differentiating (12) with respect to z and using L' Hospital's rule as follows 𝐸(𝐿
2) = 𝑄
2′(1) =
[(𝜇𝑏−𝜆)(2𝛾𝑄0′(1)+2𝛽𝑄1′(1)+𝛾𝑄0′′(1)+𝛽𝑄1′′(1)) +2𝜆(𝛾𝑄0′(1)+𝛽𝑄1′(1)+𝜇𝑏𝑃02) ]
2(𝜇𝑏−𝜆)2
(23) Differentiating equation (8) twice and putting z=1, we get
𝑄
0′′(1) =
𝜆−𝛾−2𝛼2𝛼
𝑄
0′(1) +
𝜆𝛾
𝑄
0(1) (24) Substituting (15), (21) and (24) in (23) , we get 𝐸[L(2)] The expected number of customers in the system can be computed as 𝐸[𝐿] = 𝐸[𝐿
0] + 𝐸[𝐿
1] + 𝐸[𝐿
2] Let W denote the total sojourn time of a customer in the system.
𝐸[𝑊] = 1
𝜆 (𝐸[𝐿
0] + 𝐸[𝐿
1] + 𝐸[𝐿
2])
3. STOCHASTIC DECOMPOSITION RESULTS
Theorem: If 𝜌 < 1, the stationary queue length Q can be decomposed into the sum of two independent random variables: 𝑄 = 𝑄
0+ 𝑄
𝑑, where 𝑄
0is the stationary queue length of a 𝑀/𝑀/1 queue without vacation and the additional stationary queue length 𝑄
𝑑.
𝑄
𝑑(𝑧) =
11−𝜌
[ 𝑄
0(𝑧) (1 − 𝜌𝑧 −
𝜇 𝛾𝑧𝑏(1−𝑧)
) + 𝑄
1(𝑧) (1 − 𝜌𝑧 −
𝜇 𝛽𝑧𝑏(1−𝑧)
) +
𝑧(𝛾𝑄0(1)+𝛽𝑄1(1))𝜇𝑏(1−𝑧)
+
𝜇𝑣−𝛼𝜂 𝑘1(1)
𝑘2(1)
𝑝
00] (25)
Proof: consider
=
1−𝜌1−𝜌𝑧
[
𝜇𝜇𝑏𝑏−𝜆
( 𝑄
0(𝑧) (1 − 𝜌𝑧 −
𝛾𝑧𝜇𝑏(1−𝑧)
) + 𝑄
1(𝑧) (1 − 𝜌𝑧 −
𝛽𝑧𝜇𝑏(1−𝑧)
) +
𝑧(𝛾𝑄0𝜇(1)+𝛽𝑄1(1))𝑏(1−𝑧)
+
𝜇𝑣𝜂−𝛼𝑘𝑘1(1)2(1)
𝑝
0,0)]
=
1−𝜌1−𝜌𝑧
𝑄
𝑑(𝑧)
(26)
E(Q)=
1−𝜌𝜌+𝐸(𝑄
𝑑)
Theorem: If 𝜌 < 1, the stationary waiting time can be decomposed into the sum pf two independent random variables: 𝑊 = 𝑊
0+ 𝑊
𝑑, where 𝑊
0is the stationary waiting time of a 𝑀/𝑀/1 queue without vacations which has an exponential distribution with the parameter 𝜇
𝑏(1 − 𝑃) and the additional stationary waiting time 𝑊
𝑑due to the effect of working vacation and set up period with its Laplace Stieltjes transform(LST):
𝑊𝑑∗ (𝑠) =(𝜇 1
𝑏−𝜆)𝑠[[(𝜇𝑏− 𝜆 + 𝑠)𝑠 − 𝛾(𝜆 − 𝑠)]𝑄0(1 −𝜆𝑠) + [(𝜇𝑏− 𝜆 + 𝑠)𝑠 − 𝛽(𝜆 − 𝑠)]
𝑄1(1 −𝑠𝜆) + (𝛾𝑄0(1) + 𝛽𝑄1(1))(𝜆 − 𝑠) +𝜇𝑣𝜂−𝛼𝑘𝑘1(1)
2(1)𝜇𝑏𝑠𝑝0,0]
(27) Proof: The relationship between the Probability generating function L and LST of waiting time is given by
𝑄(𝑧) = 𝑊 ∗ (𝜆(1 − 𝑧))
Assume that 𝑠 = 𝜆(1 − 𝑧); so 𝑧 = 1 −
𝑠𝜆
and 1 − 𝑧 =
𝑠𝜆
. Applying the relations in (25), we obtain the desired result.
𝑊𝑑∗ (𝑠) = 1
1 − 𝜌[𝑄0(1 −𝑠
𝜆) (1 − 𝜌(1 −𝑠
𝜆) −𝛾(1 −𝜆𝑠) 𝜇𝑏𝑠
𝜆
) + 𝑄1(1 −𝑠
𝜆) (1 − 𝜌(1 −𝑠
𝜆) −𝛽(1 −𝜆𝑠) 𝜇𝑏𝑠
𝜆
)
+(𝛾𝑄0(1) + 𝛽𝑄1(1)) 𝜇𝑏𝜇𝑏
𝜆
+𝜇𝑣− 𝛼 𝜂
𝑘1(1) 𝑘2(1)𝑝0,0]
Differentiating above equation w.r.to s, we get
𝑊𝑑= 1(𝜇𝑏− 𝜆)[𝑄0(1 −𝑠
𝜆)((𝜇𝑏− 𝜆 + 𝑠) + 𝑠 + 𝛾) −1
𝜆𝑄0′(1 −𝑠
𝜆)((𝜇𝑏− 𝜆 + 𝑠)𝑠 − 𝛾(𝜆 − 𝑠) + 𝑄1(1 −𝑠 𝜆) ((𝜇𝑏− 𝜆 + 𝑠) + 𝑠 + 𝛽) −1
𝜆𝑄1′(1 −𝑠
𝜆) ((𝜇𝑏− 𝜆 + 𝑠)𝑠 − 𝛽(𝜆 − 𝑠))– (𝛾𝑄0(1) + 𝛽𝑄1(1)) +𝜇𝑣− 𝛼
𝜂
𝑘1(1) 𝑘2(1)𝜇𝑏𝑝0,0]
Put 𝑠 = 0 in above equation, we get
E(W) =(𝜇1𝑏−𝜆)[𝑄0(1)(𝜇𝑏− 𝜆 + 𝛾) + 𝑄1(1)(𝜇𝑏− 𝜆 + 𝛾) − (𝛾𝑄0(1) + 𝛽𝑄1(1)) + 𝑄0′(1)𝛾 + 𝑄1′(1)𝛽 +𝜇𝑣𝜂−𝛼𝑘𝑘1(1)
2(1)𝜇𝑏𝑝0,0]
Substituting the value of 𝑄
0′(1), 𝑄
1′(1), we get
0
lim
s 𝑊𝑑∗ (𝑠) = 1
. 𝑊
𝑑is a random variable of the additional waiting time.
4. NUMERICAL RESULTS
In this section, we obtain the mean system length and the mean waiting time of an arbitrary customer. In this section, we illustrate the results obtained above numerically and discuss the effect of system parameters on system performance indices. We assume that the service rate 𝜇
𝑏in a regular busy period equals 4.0, arrival rate 𝜇
𝑣equals 2.0, at the same time, we assume that working vacation is an exponential distributed random variable with mean 𝛾
=0.8, 𝛽 =1.0, 𝜂 =0.4. We plot the values of mean queue length E(Q) and mean waiting time E(W) by fixed the arrival rate 𝜆 and impatient rate 𝛼 increases.
Figure.1 The changing curve of E(Q) Figure.2 .The changing curve of E(W)