Consider a two-stage queueing model. This model is for a selective inspection system that often is in operation for an international border-crossing station or a security-check point at airport. In such a system, there are two stages for inspection: the waiting space at the first stage is fairly and is assumed to be infinite; but at the second stage is the waiting space is quite limited due to the service capacity B which is expandable with a significant cost per spot. In the first stage service, customers after initial screen are either transferred immediately to the second stage further inspection with probability q or continues to the routine check and leaves the system. The first year provides a system description and model formulation with analytic solution. The second year will focus on determining the optimal staffing and inspection strategy for a given security-check level.
We consider the queueing system for the security-check waiting lines, where customers arrive according to a Poisson process with mean arrival rate λ. There are c
1
servers in this system, and each one provides a preliminary first phase of the first-stage inspection (Phase 1) to all arriving customers. The average service time of first-phase inspection is exponentially distributed with rate µ1
of each server. As soon as the first phase is completed, the customer may be provided with a second-phase service at the first-stage inspection with probability 1−q (0 ≤ q ≤ 1), where the average service rate of each server is µ 2
, or may enter the second-stage inspection with probability q. Assuming that the average service rate of the second-second-stage inspection is ν of each server. Those service times are mutually independent and follow exponential distributions.For constructing a queueing model, consider a two-stage M /Cox(2)/c
1 → /M/c 2
/B system. First stage has c1
servers and a buffer of infinite capacity and second stage has c2
servers and a buffer of finite capacity of B− c 2
. First, we define the system state as (n, i, j, m), where n denotes the number of customers at the first stage, m denotes the number of customers at the second stage, i and j denote the total number of customers in phase 1 and in phase 2 at the first stage, respectively. Then we have the state spaceS = {(n, i, j, m)| i + j = n, if n < c 1
; i + j = c1 , if n ≥ c 1 , i, j, n ∈ {0} ∪ N, m ∈ {0, 1, 2, . . . , B}}.
Denote n as the system state vector for n = 1, 2, . . ., where state
1 =
{(1, 0, 1, 0), (1, 1, 0, 0), (1, 0, 1, 1), (1, 1, 0, 1), . . . , (1, 0, 1, B), (1, 1, 0, B)}
2 =
{(2, 0, 2, 0), (2, 1, 1, 0), (2, 2, 0, 0), (2, 0, 2, 1), (2, 1, 1, 1), (2, 2, 0, 1), . . . , (2, 0, 2, B), (2, 1, 1, B), (2, 2, 0, B)}
1 is of dimension 2(B + 1). Given n < c
1
, any combination of n servers gives the same pattern of service speed since the servers are homogeneous, which is independent of m. For example, the service rate at state (1, 1, 0, B) is the same as at (1, 0, 1, B), while that at (1, 1, 0, 2) is the same as that at (1, 0, 1, 3). thus, staten 1
is of dimension (n + 1)(B + 1), when 0 < n < c1
, and n is of (c1
+ 1)(B + 1) when n≥ c 1
. A picture of possible transitions is drawn in Figure 1.3
The infinitesimal generater Q is of the block-tridiagonal form and written as follows
Define
C n
=
¯
c
¯
c
. ..¯
c
where ¯
c = λ[I n , 0] is the size of (n + 1) × (n + 2). The solution of this model is not easy to obtain
because of its highly dependent state relation and complicated stationary probability structure resulting ill-conditioned matrices. Due to the nature of the Coxian service, any sub-outflows of c1
can be considered.With probability 1
− q, the customer leaves the system after two phases of the service time (stage 1). This
output flow becomes the input flow of the second stage inspection (service). With probability q, the customer continues to complete the second phase of the service time. However, the service at stage 2 is independently carried out for its security check by each continuing customer which makes the computational procedure intricate and intractable.To simplify the analysis, we first investigate a stylized one stage queueing model with c
1
servers but of which an additional phase of service rate ν is made similarly as a security check in phase 2. It is certainly an approximation of the proposed model (a multi-server system for both stages).Once we have our Q matrix, the next step is to obtain some steady state results. The steady-state probabilities for this queue satisfy πQ = 0 and π1 = 1, where π
≥ 0 is partitioned into blocks corresponding
to the states for 0 customer, 1 customer, 2 customers, …, etc. That is, π = (P0 , P 1 , P 2 , . . .). Using the block
probabilities and the elements of the Q matrix, we need to find the Pn
’s that satisfy:P 0 B 0
+ P1 A 1
= 0, (3.1)P n C n
+ Pn+1 B n+1
+ Pn+2 A n+2
= 0, for n = 0, 1, . . . , c− 1,
(3.2)P n C + P n+1 B + P n+2 A = 0, n = c, c + 1, . . . .
(3.3) From (3.3), the matrix geometric procedure gives the vector solutionP n
= Pc R n −c , n = c, c + 1, . . . ,
(3.4)where R is the matrix solution of the equation
C + RB + R 2 A = 0.
Nuets [9] showed that the iteration
R i
=−(C + R 2 i −1 A)B −1 , i = 1, 2, . . . ,
converges to the solution R starting with R
0
= 0. Using the recurrence relation (3.4) in equations (3.1), (3.2) and the normalization equation π1 = 1, we can determine the steady state probability vector π.For example, we solve a special case of c
1
= 2, i.e., M /Cox(2)/2 queueing model for security-check waiting lines. Suppose that customers arrive the system according to a Poisson process with mean arrival rate λ.There are two servers (named server I and server II) in this system, and each one provides a preliminary first phase of the first-stage inspection (Status 1) to all arriving customers. The average service time of first-phase inspection is exponentially distributed with rate µ
1
. As soon as the first phase is completed, the customer may be provided with a second-phase service of the first-stage inspection (Status 2) with probability 1− q
(0≤ q ≤ 1) or may enter the second stage (Status 3) with probability q. Assuming that the average service
rate of the second-phase service is µ2
, and the average service rate of the second-stage inspection is ν. Those service times are mutually independent and follow exponential distributions.First, we define the system state as (n, i, j), where n represents the number of customers in the system,
i represents the status of service in the server I, j represents the status of service in the server II. Then we
have the state spaceS = {(n, i, j)| n ∈ {0} ∪ N, i ∈ {0, 1, 2, 3}, j ∈ {0, 1, 2, 3}}.
5
Denote n as the system state vector, where
0 =
{(0, 0, 0)},
1 =
{(1, 1, 0), (1, 2, 0), (1, 3, 0), (1, 0, 1), (1, 0, 2), (1, 0, 3)},
n = {(n, 1, 1), (n, 2, 1), (n, 3, 1), (n, 1, 2), (n, 2, 2), (n, 3, 2), (n, 1, 3), (n, 2, 3), (n, 3, 3)},
for n = 2, 3, . . .. The multi-dimensional states n correspond to the number of customers in the system, the status of server I and the status of server II.
The infinitesimal generator matrix has the following structure:
Q =
where those sub-matrices are