DOI 10.1007/s10957-007-9253-6
Optimal Control of an M/G/1/K Queueing System
with Combined F Policy and Startup Time
K.-H. Wang· C.-C. Kuo · W.L. Pearn
Published online: 24 July 2007
© Springer Science+Business Media, LLC 2007
Abstract We investigate the optimal management problem of an M/G/1/K
queue-ing system with combined F policy and an exponential startup time. The F policy queueing problem investigates the most common issue of controlling the arrival to a queueing system. We present a recursive method, using the supplementary vari-able technique and treating the supplementary varivari-able as the remaining service time, to obtain the steady state probability distribution of the number of customers in the system. The method is illustrated analytically for exponential service time distribu-tion. A cost model is established to determine the optimal management F policy at minimum cost. We use an efficient Maple computer program to calculate the opti-mal value of F and some system performance measures. Sensitivity analysis is also investigated.
Keywords F policy, M/G/1/K queue· Optimization · Recursive methods ·
Sensitivity analyses· Startup times · Supplementary variables
1 Introduction
A supplementary variable technique is used to study the optimal management prob-lem of the F policy M/G/1/K queueing system where the server needs a startup time before allowing customers in the system and K <∞ denotes the maximum number
Communicated by Y.C. Ho. K.-H. Wang (
)Department of Applied Mathematics, National Chung-Hsing University, Taichung, Taiwan, ROC e-mail: khwang@amath.nchu.edu.tw
C.-C. Kuo· W.L. Pearn
Department of Industrial Engineering and Management, National Chiao Tung University, Hsin Chu, Taiwan, ROC
of customers in the system. The method of controlling arrivals focuses on reducing the number of customers in the system. The model presented in this paper is very useful in real-life situations since the control of arriving customers is considered.
Gupta [1] developed analytic closed-form solutions for the F policy M/M/1/K queueing system with an exponential startup time. However, steady-state analytic solutions of the F policy queueing systems with interarrival times or service times distribution of the general type have rarely been found. It is extremely difficult, if not possible, to develop an explicit expression for the probability distributions of the number of customers in the system. However, it will become particularly helpful to use the supplementary variable technique for the non-Markovian queueing system having general interarrival times or general service times. The supplementary variable technique was first introduced by Cox [2]. Based on this technique, Gupta and Rao [3,
4] provided a recursive method to develop the steady-state probability distribution of the number of failed machines in the M/G/1 machine repair problem with no spares and the cold-standby M/G/1 machine repair problem, respectively.
Past work regarding queues may be divided into two categories: (i) the case of controlling the service and (ii) the case of controlling the arrivals. In the case of con-trolling the service, the N policy M/M/1 queueing system without startup was first in-troduced by Yadin and Naor [5]. This model was extended by Bell [6,7], Heyman [8], Kimura [9], Teghem [10], Wang and Ke [11], and others. Wang and Ke [11] presented a recursive method and applied the supplementary variable technique to develop the steady-state probability distributions of the number of customers for the N policy M/G/1 queueing system. Recently, Ke and Wang [12] developed a recursive method and used the supplementary variable technique to compute the steady-state probabil-ity distributions of the number of customers for the N policy G/M/1 queueing system. The server startup corresponds to the preparatory work of the server before beginning the service. In some real-life situations, the server often needs a startup time before starting the service. The research of several authors on queueing systems with startup time focus mainly on the N policy M/G/1 queues. The N policy M/M/1 queueing system with exponential startup time was first proposed by Baker [13]. Borthakur et al. [14] extended the Baker model to the general startup time. The N policy M/G/1 queueing system with startup time was analyzed by Medhi and Templeton [15], Tak-agi [16], Lee and Park [17], Hur and Paik [18], and so on. Recently, Ke [19] presented a recursive method and used the supplementary variable technique to investigate the operating characteristics for the N policy G/M/1 queueing system with exponential startup time. In the case of controlling the arrivals, so far very few researchers have examined the F policy M/G/1/K queueing system with server startup or the F policy G/M/1/K queueing system with server startup. Steady-state analytic solutions for the
F policy M/M/1/K queueing system with an exponential startup time was first de-rived by Gupta [1]. Through a series of propositions, the relationship between the N policy and the F policy are established by Gupta [1].
The primary objective of this paper is threefold. First, we develop a recursive method, using the supplementary variable technique and treating the supplementary variable as the remaining service time, to develop the steady-state probability dis-tributions of the number of customers for the F policy M/G/1/K queueing system. Second, to illustrate a recursive method, we present one simple example for expo-nential service time distribution. Third, we use an efficient Maple computer program
to determine the optimal management F policy to minimize the total expected cost per customer per unit time.
2 Description of the System
We consider controlling the arrivals to a finite capacity M/G/1/K queueing system with combined F policy exponential startup time. It is assumed that customers ar-rive according to a Poisson process with parameter λ, and the service time of the successive customers are independent and identically distributed (i.i.d.) random vari-ables having a distribution S(u) (u≥ 0), a probability density function s(u) (u ≥ 0) and mean service time s1. The arrival process is independent of the service process.
We assume that arriving customers form a single waiting line based on the order of their arrivals; that is, the first-come, first-served discipline is followed. Suppose that the server can serve only one customer at a time. Customers entering into the ser-vice facility and finding that the server is busy have to wait in the queue until the server is available. Gupta [1] first introduced the concept of a F policy. The defin-ition of a F policy is described as follows: When the number of customers in the system reaches its capacity K (i.e. the system becomes full), no further arriving cus-tomers are allowed to enter the system until enough cuscus-tomers in the system have been served so that the number of customers in the system decreases to a threshold value F (0≤ F < K − 1). At that time, the server requires to take an exponential startup time with parameter β to start allowing customers in the system. Thus, the system operates normally until the number of customers in the system reaches its capacity at which time the above process is repeated all over again.
3 Steady State Results
We use the following supplementary variable: U≡ remaining service time for the customer in service. The state of the system at time t is given by
N (t )≡ number of customers in the system,
U (t )≡ remaining service time for the customer being served.
Let us define
P0,n(u, t )du= Pr{N(t) = n, u < U(t) ≤ u + du}, u ≥ 0, n = 0, 1, . . . , K,
P1,n(u, t )du= Pr{N(t) = n, u < U(t) ≤ u + du}, u ≥ 0, n = 0, 1, . . . , K − 1, P0,n(t )= ∞ 0 P0,n(u, t )du, n= 0, 1, . . . , K, P1,n(t )= ∞ 0 P1,n(u, t )du, n= 0, 1, . . . , K − 1.
Relating the state of the system at time t and t+ dt, we obtain
(d/dt)P0,0(t )= −βP0,0(t )+ P0,1(0, t), (1)
(∂/∂t− ∂/∂u)P0,n(u, t )= −βP0,n(u, t )+ P0,n+1(0, t)s(u), 1≤ n ≤ F, (2)
(∂/∂t− ∂/∂u)P0,n(u, t )= P0,n+1(0, t)s(u), F + 1 ≤ n ≤ K − 1, (3)
(∂/∂t− ∂/∂u)P0,K(u, t )= λP1,K−1(u, t ), (4)
(d/dt)P1,0(t )= −λP1,0(t )+ βP0,0(t )+ P1,1(0, t), (5)
(∂/∂t− ∂/∂u)P1,1(u, t )= −λP1,1(u, t )+ βP0,1(u, t )
+ λP1,0(t )s(u)+ P1,2(0, t)s(u), (6)
(∂/∂t− ∂/∂u)P1,n(u, t )= −λP1,n(u, t )+ βP0,n(u, t )+ λP1,n−1(u, t )s(u) + P1,n+1(0, t)s(u), 2≤ n ≤ F, (7)
(∂/∂t− ∂/∂u)P1,n(u, t )= −λP1,n(u, t )+ λP1,n−1(u, t )s(u)
+ P1,n+1(0, t)s(u), F+ 1 ≤ n ≤ K − 2, (8)
(∂/∂t− ∂/∂u)P1,K−1(u, t )= −λP1,K−1(u, t )+ λP1,K−2(u, t ). (9) In steady state, let us define
P0,n= lim t→∞P0,n(t ), n= 0, 1, . . . , K, P1,n= lim t→∞P0,n(t ), n= 0, 1, . . . , K − 1, P0,n(u)= lim t→∞P0,n(u, t ), n= 1, 2, . . . , F, P1,n(u)= lim t→∞P1,n(u, t ), n= 0, 1, . . . , K − 1 and further define
P0,n(u)= P0,ns(u), n= 1, 2, . . . , F. (10)
From (1–10), we can obtain easily the following steady state equations:
0= −βP0,0+ P0,1(0), (11)
−(d/du)P0,n(u)= −βP0,ns(u)+ P0,n+1(0)s(u), 1≤ n ≤ F, (12)
−(d/du)P0,n(u)= P0,n+1(0)s(u), F+ 1 ≤ n ≤ K − 1, (13)
−(d/du)P0,K(u)= λP1,K−1(u), (14)
0= −λP1,0+ βP0,0+ P1,1(0), (15)
−(d/du)P1,1(u)= −λP1,1(u)+ βP0,1s(u)+ λP1,0s(u)+ P1,2(0)s(u), (16)
−(d/du)P1,n(u)= −λP1,n(u)+ βP0,ns(u)+ λP1,n−1(u)+ P1,n+1(0)s(u),
−(d/du)P1,n(u)= −λP1,n(u)+ λP1,n−1(u)+ P1,n+1(0)s(u),
F+ 1 ≤ n ≤ K − 2, (18)
−(d/du)P1,K−1(u)= −λP1,K−1(u)+ λP1,K−2(u). (19) Further define S∗(θ )= ∞ 0 e−θudS(u)= ∞ 0
e−θus(u)du,
P0,n∗ (θ )= ∞
0
e−θuP0,n(u)du,
P1,n∗ (θ )= ∞
0
e−θuP1,n(u)du,
P0,n= P0,n∗ (0)= ∞ 0 P0,n(u)du, P1,n= P1,n∗ (0)= ∞ 0 P1,n(u)du, ∞ 0 e−θu ∂ ∂uP0,n(u)du= θP ∗ 0,n(θ )− P0,n(0), ∞ 0 e−θu ∂ ∂uP1,n(u)du= θP ∗ 1,n(θ )− P1,n(0).
Therefore, if the LST is taken of both sides of (12–14) and (16–19), it is found that
−θP0,n∗ (θ )= −βP0,nS∗(θ )+ P0,n+1(0)S∗(θ )− P0,n(0), 1≤ n ≤ F, (20) −θP∗ 0,n(θ )= P0,n+1(0)S∗(θ )− P0,n(0), F+ 1 ≤ n ≤ K − 1, (21) −θP0,K∗ (θ )= λP1,K∗ (θ )− P0,K(0), (22) (λ− θ)P1,1∗ (θ )= βP0,1S∗(θ )+ λP1,0S∗(θ )+ P1,2(0)S∗(θ )− P1,1(0), (23) (λ− θ)P1,n∗ (θ )= βP0,nS∗(θ )+ λP1,n∗ −1(θ )+ P1,n+1(0)S∗(θ )− P1,n(0), 2≤ n ≤ F, (24) (λ− θ)P1,n∗ (θ )= λP1,n∗ −1(θ )+ P1,n+1(0)S∗(θ )− P1,n(0), F+ 1 ≤ n ≤ K − 2, (25) (λ− θ)P1,K∗ −1(θ )= λP1,K∗ −2(θ )− P1,K−1(0). (26) 3.1 Recursive Method
A recursive method is developed to obtain P0,n∗ (0) and P1,n∗ (0). Our solution algo-rithm will first obtain P0,n(0) (1≤ n ≤ K) which are then used for finding P0,n∗ (0).
Using (11) and setting θ= 0 in (20) and (21), we get P0,n(0)= β ζn−1 i=0 P0,i, 1≤ n ≤ K, ζn= n, 0≤ n ≤ F − 1, F, F≤ n ≤ K, (27) and P0,n+1(0)= −βϕn,FP0,n+ P0,n(0), 1≤ n ≤ K − 1, ϕn,F = 1, 1≤ n ≤ F, 0, otherwise. (28)
Using (28) in (20) and (21), we get
P0,n∗ (θ )= {[1 − S∗(θ )]/θ}P0,n(0), 1≤ n ≤ K − 1. (29) Taking limθ→0in (29) and using L’Hôspital’s rule once gives
P0,n∗ (0)= s1P0,n(0), 1≤ n ≤ K − 1, (30)
where s1= −S∗(1)(0) is the mean service time.
Using (27) in (30), we have P0,n∗ (0)= φnP0,0, 1≤ n ≤ K − 1, (31) φn= 1, n= 0, s1β(1+ s1β)ζn−1, 1≤ n ≤ K. (32) Thus, P0,1∗ (0), P0,2∗ (0), . . . , P0,K∗ −1(0) can be obtained by using (31).
Next, we derive the expressions of P1,n(0) (1≤ n ≤ K) in terms of P1,0and P0,0.
Using (31) in (23–24) and then setting θ= λ in (23–26), we finally obtain
P1,2(0)= [P1,1(0)− βφ1P0,0S∗(λ)− λP1,0S∗(λ)]/S∗(λ), (33) P1,n+1(0)= [P1,n(0)− βϕn,FφnP0,0S∗(λ)− λP1,n∗ −1(λ)]/S∗(λ),
2≤ n ≤ K − 2, (34)
P1,K−1(0)= λP1,K∗ −2(λ). (35)
To obtain P1,n∗ −1(λ) (1≤ n ≤ K − 1) in (34–35), using (31) in (23–24) again, differentiating (23–26) (l− 1) times with respect to θ and setting θ = λ, we finally get
P1,1∗(l−1)(λ)= −(S∗(l)(λ)/ l)[λP1,0+ βφ1P0,0+ λP1,2(0)],
P1,n∗(l−1)(λ)= −(1/l)[P1,n+1(0)S∗(l)(λ)+ βϕn,FφnP0,0S∗(l)(λ)+ λP1,n∗(l)−1(λ)], 2≤ n ≤ K − 2, l = 1, . . . , K − n − 1, (37)
P1,K∗ −1(λ)= −λP1,K∗(1)−2(λ), (38)
where P1,n∗(0)(λ)= P1,n∗ (λ) and S∗(l)(θ )= (dl/dθl)S∗(θ ) denote the lth derivative of S∗(θ ).
Solving (36–38) recursively, we obtain
P1,n∗ (λ)= − nS∗(λ)P1,0− ζn i=1 [β n−i+1φiS∗(λ)/λ]P0,0 − n i=1 [ n−i+1S∗(λ)/λ]P1,i+1(0), 1≤ n ≤ K − 1, (39) where n= −[(−λ)nS∗(n)(λ)/n!S∗(λ)], 1 ≤ n ≤ K − 1, 0, otherwise. (40) Using (39) in (34), we obtain P1,n(0)= [1/S∗(λ)]P1,n−1(0)+ n−2 i=1 n−i−1P1,i+1(0) + β ζn−2 i=1 n−i−1φi− ϕn−1,Fφn−1 P0,0 + λ n−2P1,0, 3≤ n ≤ K − 1. (41) We further define n= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 1, n= 0, 1≤k≤n τ1+τ2+···+τk=n κτ1κτ2· · · κτk, n= 1, 2, . . . , K − 3, τ1, τ2, . . . , τk∈ {1, 2, · · · , n} 0, otherwise, (42) where κn= ⎧ ⎪ ⎨ ⎪ ⎩ [1/S∗(λ)] + 1, n= 1, n, n= 2, 3, . . . , K − 3, 0, otherwise. (43)
Remark 3.1 The representative meaning of the above formulation (42) is to sum up all possible products of k κs in which the total of subscript values of κ equals n. We give an easily understood example for n= 4:
4= κ4+ κ3κ1+ κ2κ2+ κ1κ3+ κ1κ1κ2+ κ1κ2κ1+ κ2κ1κ1+ κ1κ1κ1κ1 = κ4+ 2κ3κ1+ κ2
2+ 3κ12κ2+ κ14.
Using (42) and (43) to solve (41) recursively, and including (15) and (33), we finally get P1,1(0)= A(1)P1,0+ B(1)P0,0, (44) P1,n(0)= n i=2
n−i[A(i)P1,0+ B(i)P0,0], 2 ≤ n ≤ K − 1, (45) where A(n)= ⎧ ⎪ ⎨ ⎪ ⎩ λ, n= 1, λ[1 − S∗(λ)]/S∗(λ), n= 2, λ n−2, 3≤ n ≤ K − 1, (46) B(n)= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −β, n= 1, −β[1 + ϕ1,Fφ1S∗(λ)]/S∗(λ), n= 2, β ζn−2 i=1 n−i−1φi− βϕn−1,Fφn−1, 3≤ n ≤ K − 1. (47)
Substituting (45), (44), and (35) into (39) finally yields
P1,0= − K−2 i=1 K−i−1 i+1 j=2 (i− j + 1)B(j) + K−1 i=2 [(K − i − 1)B(i)/S∗(λ)] + ζK−2 i=1 β K−i−1φi K−2 i=1 K−i−1 i+1 j=2 (i− j + 1)A(j) + K−1 i=2 [(K − i − 1)A(i)/S∗(λ)] + λ K−2 P0,0. (48)
Finally, we develop the steady-state probabilities P1,n∗ (0) in terms of P0,0. Setting θ= 0 in (23–26) we have P1,n∗ (0)= (1/λ) β ζn i=0 φiP0,0+ P1,n+1(0) , 0≤ n ≤ K − 2, (49) P1,K∗ −1(0)= (β/λ) F i=0 φiP0,0. (50)
As P1,1(0), P1,2(0), . . . , P1,K−1(0) and P1,0 are known, P1,1∗ (0), P1,2∗ (0), . . . , P1,K∗ −1(0) can be determined recursively using (49) and (50) in terms of P0,0.
Now the only unknown quantity is P0,K∗ (0), which can be obtained from (22). To find it, differentiating (22) with respect to θ and setting θ= 0, we have
P0,K∗ (0)= −λP1,K∗(1)−1(0). (51) To find λP1,K∗(1)−1(0), differentiating (23–26) with respect to θ and setting θ= 0, we finally obtain
P1,1∗(1)(0)= [P1,1+ βφ1P0,0S∗(1)(0)+ λP1,0S∗(1)(0)+ P1,2(0)S∗(1)(0)]/λ, (52)
P1,n∗(1)(0)= [P1,n+ βϕn,FφnP0,0S∗(1)(0)+ λP1,n∗(1)−1(0)+ P1,n+1(0)S∗(1)(0)]/λ,
2≤ n ≤ K − 2, (53)
P1,K∗(1)−1(0)= [P1,K−1+ λP1,K∗(1)−2(0)]/λ. (54) As P1,1∗(1)(0) is known completely from (52), the values P1,n∗(1)(0) for n= 2, 3, . . . ,
K− 1 can be found recursively from (53) and (54). Therefore, we obtain
P1,K∗(1)−1(0)= (1/λ) K−1 i=1 P1,i+ βS∗(1)(0) F i=1 φnP0,0 + S∗(1)(0)K−1 i=2 P1,i(0)+ λP1,0S∗(1)(0) . (55)
Substituting (55) into (51), we have
P0,K∗ (0)= − K−1 i=1 P1,i+ βS∗(1)(0) F i=1 φnP0,0+ S∗(1)(0) × K−1 i=2 P1,i(0)+ λP1,0S∗(1)(0) . (56)
So P0,1∗ (0), P0,2∗ (0), . . . , P0,K∗ (0) is known in terms of P0,0, which can be determined
using the normalizing condition K i=0 P0,i+ K−1 i=0 P1,i= 1. (57)
To demonstrate the working of the recursive method, we first describe the solution algorithm for calculating the steady state probabilities P0,n∗ (0) (0≤ n ≤ K) and
P1,n∗ (0) (0≤ n ≤ K − 1). Next, to illustrate the solution algorithm, we provide one simple example for the exponential service time distribution.
3.2 Solution Algorithm
Let F be the threshold, let K be the maximum capacity of the system, and let S∗(l)(θ )
be the lth derivative of S∗(θ ), where l= 1, 2, . . . , K. We set the values of F , K, and the LST expression of the service time distribution, namely S∗(θ ). The steps of the solution algorithm are stated as follows:
Step 1. For each n= 0, 1, . . . , K, compute φnusing (32).
Step 2. For each n= 1, 2, . . . , K − 1, compute P0,n∗ (0) using (31) in terms of P0,0.
Step 3. Compute n(1≤ n ≤ K − 2) and κn(1≤ n ≤ K − 3) using (40) and (43), respectively.
Step 4. For each n= 0, 1, . . . , K − 3, compute nusing (42).
Step 5. For each n= 1, 2, . . . , K − 1, compute A(n) and B(n) using (46) and (47). Step 6. For each n= 1, 2, . . . , K − 1, compute P1,n(0) using (44) and (45) in terms
of P1,0and P0,0.
Step 7. Compute P1,0using (48) in terms of P0,0. Thus, P1,n(0)(1≤ n ≤ K − 1) are
determined from Step 6.
Step 8. For each n= 1, 2, . . . , K − 1, compute P1,n∗ (0) using (49) and (50) in terms of P0,0.
Step 9. For n= K, compute P0,n∗ (0) using (56) in terms of P0,0.
Step 10. Determine P0,0 using (57). Thus P0,n∗ (0)(n= 1, 2, . . . , K) are obtained
from Steps 2 and 9, and P1,n∗ (0)(n= 0, 1, . . . , K − 1) are obtained from Steps 7 to 8.
3.3 Simple Example
For the F policy M/M/1/K queueing system, we set the mean service time s1= 1/μ,
where μ is the service rate. Assume that F= 1 and K = 4. In this case, we have
S∗(θ )= μ/(μ + θ).
Step 1. For each n= 0, 1, . . . , 4, compute φn. Using (32), we obtain
φ0= 1, φ1= (1 − α)/α, φ2= φ3= φ4= (1 − α)/α2,
where α= μ/(μ + β).
Step 2. For each n= 1, 2, 3, compute P0,n∗ (0) using (31) in terms of P0,0.
From (31), we finally get
P0,1∗ (0)= φ1P0,0= [(1 − α)/α]P0,0,
P0,2∗ (0)= P0,3∗ (0)= φ2P0,0= [(1 − α)/α2]P0,0.
Step 3. For each n= 1, 2, compute nand κnusing (40) and (43), respectively. For each n= 1, 2, using (40) yields 1= −1/(1 + σ ) and 2= −1/(1 + σ )2,
where σ = μ/λ.
Step 4. For each n= 0, 1, compute n.
It implies from (42) that 0= 1 and 1= (1 + σ + σ2)/σ (1+ σ ).
Step 5. For each n= 1, 2, 3, compute A(n) and B(n). Using (46) and (47), it follows that
A(1)= μ/σ, A(2)= μ/σ2, A(3)= −μ/σ (1 + σ ).
B(1)= −(1 − α)μ/α, B(2)= −(α + σ )(1 − α)μ/σ α2, B(3)= −(1 − α)2μ/(1+ σ )α2.
Step 6. For n= 1, 2, 3, using (44–45), we compute P1,n(0) in terms of P1,0and P0,0.
It yields from (44) and (45) that
P1,1(0)= A(1)P1,0+ B(1)P0,0,
P1,2(0)= 0[A(2)P1,0+ B(2)P0,0],
P1,3(0)= 1[A(2)P1,0+ B(2)P0,0] + 0[A(3)P1,0+ B(3)P0,0].
Step 7. Compute P1,0 using (48) in terms of P0,0. Thus, P1,n(0)(1≤ n ≤ 3) are
ob-tained from Step 6.
From (48), we finally have
P1,0= [σ (1 − α)(α + σ + σ2+ σ3)/α2]P0,0 (P1,0∗ (0)= P1,0), P1,1(0)= [σ μ(1 − α)(1 + σ + σ2)/α2]P0,0,
P1,2(0)= [σ μ(1 − α)(1 + σ )/α2]P0,0, P1,3(0)= [σ μ(1 − α)/α2]P0,0.
Step 8. For each n= 1, 2, 3, compute P1,n∗ (0) using (49) and (50) in terms of P0,0.
Using (49) and (50) yields
P1,1∗ (0)= [σ (1 − α)(1 + σ + σ2)/α2]P0,0, P1,2∗ (0)= [σ (1 − α)(1 + σ )/α2]P0,0, P1,3∗ (0)= [σ (1 − α)/α2]P0,0.
Step 9. For n= 4, compute P0,n∗ (0) using (56) in terms of P0,0.
Using (56), it follows that
P0,4∗ (0)= [(1 − α)/α2]P0,0.
Step 10. Determine P0,0using (57). Thus P0,n∗ (0) (n= 0, 1, . . . , 4) are obtained from
Steps 2 and 9, and P1,n∗ (0) (n= 0, 1, 2, 3) are obtained from Steps 7 to 8.
P0,0= α2/[α2+ α(1 − α) + 3(1 − α) + σ (1 − α)(3 + α + 3σ + 2σ2+ σ3)].
4 System Performance Measures
Our analysis is based on the following system performance measures of the F policy M/G/1 queueing system with exponential startup time. Let
Ls≡ average number of customers in the system;
Pb≡ probability that the server is busy;
Ps≡ probability that the server requires a startup time before starting service;
Pbl≡ probability that the server is blocked.
The expressions for Ls, Pb, Ps, Pblare given by
Ls= K n=1 nP0,n+ K−1 n=1 nP1,n, Pb= K n=0 P0,n+ K−1 n=0 P1,n, Ps= F n=0 P0,n, Pbl= K n=0 P0,n. 5 Optimal F Policy
We develop the total expected cost function per unit time for the F policy M/G/1 queueing system with startup times, in which F is a management decision variable. The main purpose of this paper is to determine the optimum management F policy so as to minimize this total expected cost function. Let
Ch≡ holding cost per unit time for each customer present in the system;
Cb≡ busy cost per unit time for a busy server;
Cs≡ startup cost per unit time for the preparatory work of the server before starting the service;
Cbl≡ fixed cost for every lost customer when the system is blocked.
Utilizing the definitions of each cost element listed above, the total expected cost function per unit time is given by
T C(F )= ChLs+ CbPb+ CsPs+ CblλPbl. (58)
The optimal value of F , F∗is determined by satisfying the following inequality:
Table 1 The optimal value of F and its minimum expected cost (the F policy M/M/1 queueing system) λ(μ, β)= (1.0, 3.0) μ(λ, β)= (0.8, 3.0) β(λ, μ)= (0.8, 1.0) 0.5 0.6 0.7 1.0 1.1 1.2 2.0 4.0 5.0 Case1 F∗ 9 7 5 4 7 10 5 4 4 T C(F∗) 105.00 127.49 151.42 177.60 158.45 143.31 177.68 177.56 177.54 Case2 F∗ 12 11 9 6 10 12 6 6 6 T C(F∗) 105.00 127.50 151.55 178.29 158.66 143.37 178.36 178.25 178.23 Case3 F∗ 12 11 8 6 10 12 6 6 6 T C(F∗) 105.00 127.50 151.56 178.31 158.67 143.37 178.40 178.27 178.24 Case4 F∗ 11 9 7 4 8 11 5 5 5 T C(F∗) 117.50 142.50 169.00 197.99 176.77 160.02 198.07 197.94 197.92 Case5 F∗ 5 4 3 2 4 6 2 2 2 T C(F∗) 122.47 149.93 180.05 213.87 189.10 169.77 213.96 213.83 213.80 6 Numerical Results
We now perform a sensitivity analysis on the optimum value F∗ based on changes in the specific values of the system parameter and fix the system capacity at K= 15. We present two simple examples for two different service time distributions such as exponential and 3-stage Erlang. We employ the following cost elements:
Case 1: Ch= 5, Cb= 200, Cs= 250, Cbl= 300.
Case 2: Ch= 5, Cb= 200, Cs= 250, Cbl= 350.
Case 3: Ch= 5, Cb= 200, Cs= 300, Cbl= 350.
Case 4: Ch= 5, Cb= 225, Cs= 300, Cbl= 350.
Case 5: Ch= 10, Cb= 225, Cs= 300, Cbl= 350.
The optimal value of F , F∗, and its minimum expected cost T C(F∗) for the above five cases are shown in Tables 1, 2. We first fix (μ, β) and vary the val-ues of λ. Tables1,2reveal that: (i) T C(F∗)increases as λ increases for any cases; (ii) F∗decreases as λ increases for any cases. Next, we fix (λ, β) and vary the values of μ. We observe from Tables1,2that: (i) T C(F∗)decreases as μ increases for any cases; (ii) F∗increases as μ increases for any cases. Finally, we fix (λ, μ) and vary the values of β. It appears from Tables1,2that: (i) T C(F∗)slightly decreases as β increases for any cases; (ii) F∗does not change at all when β changes from 2.0 to 5.0 for any cases. Intuitively, F∗is insensitive to changes in β.
It can be easily seen from Tables1,2that (i) F∗increases as Chdecreases or Cbl
increases (see Case 4–5 and Case 1–2) and that (ii) Chand Cblhave a larger effect on
F∗than Cband Cs(see Case 3–4 and Case 2–3).
7 Conclusions
In this paper, we have provided a recursive method for computing the steady state probability distribution of the number of customers in a finite system. We also
illus-Table 2 The optimal value of F and its minimum expected cost (the F policy M/E3/1 queueing system) λ(μ, β)= (1.0, 3.0) μ(λ, β)= (0.8, 3.0) β(λ, μ)= (0.8, 1.0) 0.5 0.6 0.7 1.0 1.1 1.2 2.0 4.0 5.0 Case1 F∗ 9 7 6 4 7 10 4 4 4 T C(F∗) 104.17 126.00 148.91 174.00 155.50 141.11 174.02 173.99 173.98 Case2 F∗ 12 11 9 6 10 12 6 6 6 T C(F∗) 104.17 126.00 148.93 174.22 155.54 141.12 174.24 174.20 174.20 Case3 F∗ 12 11 9 6 10 12 6 6 6 T C(F∗) 104.17 126.00 148.93 174.23 155.54 141.12 174.26 174.21 174.20 Case4 F∗ 11 9 7 5 9 12 5 5 5 T C(F∗) 116.67 141.00 166.42 194.12 173.71 157.78 194.15 194.11 194.10 Case5 F∗ 6 4 3 2 4 6 2 2 2 T C(F∗) 120.83 147.00 175.28 207.57 183.63 165.53 207.60 207.56 207.55
trated a recursive method by a study of the exponential service time distribution. In addition, we derived the optimum value of the control parameter F so as to minimize an expected cost function. We performed a sensitivity analysis among the optimal value of F , specific values of system parameters, and the cost elements. Based on the numerical results, we could make an effective decision based on exact solutions for practical and general queueing system with quantitative measurement.
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