# Analysis of a Multiserver Queueing-Inventory System

Authors: A. Krishnamoorthy; R. Manikandan; Dhanya Shajin
Journal: Advances in Operations Research
 (2015)

---

## Abstract

We attempt to derive the steady-state distribution of theM
/
M
/
c queueing-inventory system with positive service time. First we analyze the case of c
=
2 servers which are assumed to be homogeneous and that the service time follows exponential distribution. The inventory replenishment follows the (
s
,
Q
) policy. We obtain a product form solution of the steady-state distribution under the assumption that customers do not join the system when the inventory level is zero. An average system cost function is constructed and the optimal pair (
s
,
Q
) and the corresponding expected minimum cost are computed. As in the case of M
/
M
/
c retrial queue with c
≥
3, we conjecture that M
/
M
/
c for c
≥
3, queueing-inventory problems, do not have analytical solution. So we proceed to analyze such cases using algorithmic approach. We derive an explicit expression for the stability condition of the system. Conditional distribution of the inventory level, conditioned on the number of customers in the system, and conditional distribution of the number of customers, conditioned on the inventory level, are derived. The distribution of two consecutive s to s transitions of the inventory level (i.e., the first return time to s) is computed. We also obtain several system performance measures.


---

## Body

## 1. Introduction

The notion of inventory with positive service time was first introduced by Sigman and Simchi-Levi [1]. They assumed arbitrarily distributed service time, exponentially distributed replenishment lead time with customer arrival forming a Poisson process. Under the condition of stability of the system, they investigate several performance characteristics. In the context of arbitrarily distributed lead time the readers attention is invited to a very recent paper by Saffari et al. [2] where the authors provide a product form solution for system state probability distribution under the assumption thatno customer joins the system when inventory level is zero.Reference [1] by Sigman and Simchi-Levi was followed by [3] of Berman et al. with deterministic service time wherein they formulated the model as a dynamic programming problem. A review paper by Krishnamoorthy et al. [4] provides the details of the research developments on queueing theory with positive service time. Schwarz et al. [5] were the first to produce product form solutions for single server queueing-inventory problem with exponentially distributed service time as well as lead time and Poisson input of customers. They arrived at product form solution for the system state distribution. Nevertheless this is achieved under the assumption that customers do not join when the inventory level is zero (of course, [2] of Saffari et al. is the extension of this to arbitrary distributed lead time). This is despite the strong correlation between the lead time and the number of customers joining the system during that time. Subsequently several authors made the above assumption in their investigations to come up with product form solution, the details of which could be seen below. Krishnamoorthy and Viswanath [6] subsume Schwarz et al. [5] by extending the latter to production inventory with positive service time. References [7] of Sivakumar and Arivarignan, [8] of Krishnamoorthy and Narayanan, [9] of Deepak et al., [10] of Schwarz and Daduna, [11] of Schwarz et al., and [12] of Krenzler and Daduna are a few other significant contributions to inventory with positive service time. Protection of production and service stages in a queueing-inventory model, with Erlang distributed service and interproduction time, is analyzed by Krishnamoorthy et al. [13].Classical queue with inventoried items for service is also studied by Saffari et al. [14] where the control policy followed is (
s
,
Q
) and lead time is mixed exponential distribution. Customers arriving during zero inventory are lost forever. This leads to a product form solution for the system sate probability. Schwarz et al. [11] consider queueing networks with attached inventory. They consider rerouting of customers served out from a particular station when the immediately following station has zero inventory. Thus no customer is lost to the system. The authors derive joint stationary distribution of queue length and on-hand inventory at various stations in explicit product form. A recent contribution of interest to inventory with positive service time involving a random environment is by Krenzler and Daduna [15] wherein also a stochastic decomposition of the system is established. They prove a necessary and sufficient condition for a product form steady-state distribution of the joint queueing-environment process to exist. A still more recent paper by Krenzler and Daduna [12] investigates inventory with positive service time in a random environment embedded in a Markov chain. They provide a counter example to show that the steady-state distribution of an M
/
G
/
1
/
∞ system with (
s
,
S
) policy and lost sales need not have a product form. Nevertheless, in general, loss systems in a random environment have a product form steady-state distribution. They also introduce a blocking set where all activities other than replenishment stay suspended whenever the Markov chain is in that set. This resulted in arriving at a product form solution to the system state distribution.The work on multiserver queueing-inventory systems is scarce. Nair et al. [16] consider an inventory system with number of servers varying from s
+
1 to S, depending on the inventory position. Another contribution is by Yadavalli et al. [17] wherein the authors consider a finite customer source system (this paper contains a few additional references to multiserver inventory system).In all work quoted above, customers are provided an item from the inventory on completion of service. Nevertheless, there are several situations where a customer may not be served/may not purchase the item with probability one at the end of his service. For example, customers who may buy an item arrive at a retail shop where there are one or more (finite number) servers (sales executives). The servers explain to each customer the features of product. The time required for this may be regarded as the service time. After listening to the server each customer, independently of the others, decides whether to buy the item (probabilityγ) or leaves the system without purchasing the item. A less realistic example is as follows: a candidate appears for an interview against a position. At the end of the interview the candidate decides to accept the offer of job with probability γ and with complementary probability rejects it. In this case the job is taken as an inventory. In this connection one may refer to Krishnamoorthy et al. [18] for some recent developments.We arrange the presentation of this paper as indicated below: in Section2 the M
/
M
/
2 queueing-inventory problem is mathematically formulated. The product form solution of the steady-state probability distribution, including some important performance measures, is obtained in Section 3. Further we numerically investigate the optimal (
s
,
Q
) pair values and the minimal cost for different values of γ. Section 5 discusses the M
/
M
/
c with c (greater than or equal to 3 but less than s) queueing-inventory problems by using algorithmic approach. Section 6 gives some conditional probability distributions and a few performance measures for the c(≥3) server case. Section 7 analyzes the distribution of the inventory cycle time. In Section 8 the optimal c and the corresponding minimal cost for different values of γ are investigated. Further we look for the optimal (
s
,
Q
) pair values that would result in cost minimization for different pairs of values of γ and c.


## 2. Mathematical Modelling of theM
/
M
/
2 Queueing-Inventory Problem

First we consider anM
/
M
/
2 queueing-inventory system with positive service time. Customer arrival process is assumed to be Poisson with rate λ. Each customer requires a single item having random duration of service which follows exponential distribution with parameter μ. However, it is not essential that inventory is provided to the customer at the end of his service. More precisely, the item is served with probability γ at the end of a service or else it is not provided with probability 1
-
γ. A crucial assumption of this model is that customers do not join the system when the inventory level is zero. When the number of customers is at least two and not less than two items are in inventory, the service rate is 2
μ. When the inventory level reaches a prespecified value s
>
0, a replenishment order is placed for Q units with Q
>
s. We fix S
=
Q
+
s as the maximum number of items that could be held in the system at any given time. The lead time follows exponential distribution with parameter β. Then X
(
t
)
∣
t
≥
0
=
{
(
N
(
t
)
,
I
(
t
)
)
∣
t
≥
0
} is a CTMC with state space Ω
1
=
⋃
i
=
0
∞
L
(
i
), where L
(
i
) is called the ith level. In each of the levels the number of items in the inventory can be anything from 0 to S. Accordingly we write L
(
i
)
=
(
i
,
0
)
,
…
,
(
i
,
Q
+
s
). The infinitesimal generator W
1 of this CTMC {
X
(
t
)
∣
t
≥
0
} is
(1)
W
1
=
B
00
A
0
B
20
B
10
A
0
A
2
A
1
A
0
A
2
A
1
A
0
⋯
⋱
⋱
⋱
,
where B
00 contains transition rates within L
(
0
); B
20 represents the transitions from level 1 to level 0; B
10 contains the transitions within level 1; A
0 represents the transition from level i to level i
+
1,  i
≥
1; A
1 represents the transitions within L
(
i
) for i
≥
2; and A
2 represents transitions from L
(
i
) to L
(
i
-
1
),  i
≥
2. The transition rates are
(2)
B
00
k
l
=
-
β
,
for
l
=
k
=
0
,
-
(
λ
+
β
)
,
for
l
=
k
;
k
=
1,2
,
…
,
s
,
-
λ
,
for
l
=
k
;
k
=
s
+
1
,
s
+
2
,
…
,
S
,
β
,
for
l
=
k
+
Q
;
k
=
0,1
,
…
,
s
,
0
,
otherwise
,
B
20
k
l
=
γ
μ
,
for
l
=
k
-
1
;
k
=
1,2
,
…
,
S
,
(
1
-
γ
)
μ
,
for
l
=
k
;
k
=
1,2
,
…
,
S
,
0
,
otherwise
,
B
10
k
l
=
-
β
,
for
l
=
k
=
0
,
-
(
λ
+
β
+
μ
)
,
for
l
=
k
;
k
=
1,2
,
…
,
s
,
-
(
λ
+
μ
)
,
for
l
=
k
;
k
=
s
+
1
,
s
+
2
,
…
,
S
,
β
,
for
l
=
k
+
Q
;
k
=
0,1
,
…
,
s
,
0
,
otherwise
,
A
0
k
l
=
λ
,
for
l
=
k
;
k
=
1,2
,
…
,
S
,
0
,
otherwise
,
A
1
k
l
=
-
β
,
for
l
=
k
=
0
,
-
(
λ
+
β
+
μ
)
,
for
l
=
k
=
1
,
-
(
λ
+
β
+
2
μ
)
,
for
l
=
k
;
k
=
2,3
,
…
,
s
,
-
(
λ
+
2
μ
)
,
for
l
=
k
;
k
=
s
+
1
,
s
+
2
,
…
,
S
,
β
,
for
l
=
k
+
Q
;
k
=
0,1
,
…
,
s
,
0
,
otherwise
,
A
2
k
l
=
γ
μ
,
for
l
=
k
-
1
;
k
=
1
,
(
1
-
γ
)
μ
,
for
l
=
k
=
1
,
2
γ
μ
,
for
l
=
k
-
1
;
k
=
2,3
,
…
,
S
,
2
(
1
-
γ
)
μ
,
for
l
=
k
;
k
=
2,3
,
…
,
S
,
0
,
otherwise
.
Note that all entries (block matrices) in W
1 are of the same order, namely, S
+
1, and these matrices contain transition rates within level (in the case of diagonal entries) and between levels (in the case of off-diagonal entries).


### 2.1. Analysis of the System

In this section we perform the steady-state analysis of the queueing-inventory model under study by first establishing the stability condition of the queueing-inventory system. DefineA
=
A
0
+
A
1
+
A
2. This is the infinitesimal generator of the finite state CTMC corresponding to the inventory level 0,1
,
2
,
…
,
S for any level i (≥1). Let ζ denote the steady-state probability vector of A. That is,
(3)
ζ
A
=
0
,
ζ
e
=
1
.
Write
(4)
ζ
=
ζ
0
,
ζ
1
,
…
,
ζ
s
,
…
,
ζ
Q
,
…
,
ζ
S
.
We have
(5)
A
=
-
β
β
γ
μ
-
(
β
+
γ
μ
)
2
γ
μ
-
(
β
+
2
γ
μ
)
⋱
⋱
⋱
2
γ
μ
-
(
β
+
2
γ
μ
)
β
2
γ
μ
-
2
γ
μ
⋱
⋱
2
γ
μ
-
2
γ
μ
.
Then using (3) we get the components of the vector ζ explicitly as
(6)
ζ
0
=
1
+
β
γ
μ
1
+
β
+
γ
μ
γ
μ
∑
i
=
0
s
‍
β
+
2
γ
μ
2
γ
μ
i
-
2
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
+
(
Q
-
s
-
2
)
β
+
2
γ
μ
2
γ
μ
s
-
1
1
+
β
+
γ
μ
γ
μ
∑
i
=
0
s
‍
β
+
2
γ
μ
2
γ
μ
i
-
2
llllll
+
β
2
γ
μ
β
+
γ
μ
γ
μ
llllll
×
∑
i
=
0
s
‍
β
+
2
γ
μ
2
γ
μ
i
-
2
β
+
2
γ
μ
2
γ
μ
s
-
i
+
1
-
1
β
+
2
γ
μ
2
γ
μ
s
-
1
-
γ
μ
β
+
γ
μ
l
l
l
l
l
l
l
l
l
l
+
∑
i
=
0
s
‍
β
+
2
γ
μ
2
γ
μ
i
-
2
β
+
2
γ
μ
2
γ
μ
s
-
i
+
1
-
1
-
1
,
ζ
i
=
β
γ
μ
ζ
0
,
for
i
=
1
,
β
γ
μ
β
+
γ
μ
2
γ
μ
×
β
+
2
γ
μ
2
γ
μ
i
-
2
ζ
0
,
for
i
=
2,3
,
…
,
s
+
1
,
ζ
i
+
1
,
for
i
=
s
+
1
,
s
+
2
,
…
,
Q
-
1
,
β
2
γ
μ
β
+
2
γ
μ
2
γ
μ
s
-
1
-
1
β
+
γ
μ
γ
μ
l
l
l
l
l
l
l
×
β
+
2
γ
μ
2
γ
μ
s
-
1

-
1
ζ
0
,
for
i
=
Q
+
1
,
ζ
Q
+
i
=
β
2
γ
μ
β
+
γ
μ
γ
μ
β
+
2
γ
μ
2
γ
μ
i
-
2
×
β
+
2
γ
μ
2
γ
μ
s
-
i
-
1
-
1
ζ
0
,
i
=
2,3
,
…
,
s
.
Since the Markov chain X
(
t
)
∣
t
≥
0 is an LIQBD, it is stable if and only if the left drift rate exceeds the right drift rate. That is,
(7)
ζ
A
0
e
<
ζ
A
2
e
.
Thus, we have the following lemma for stability of the system under study.Lemma 1.
The stability condition of theM
/
M
/
2 queueing-inventory system under consideration is given by λ
<
μ
2
-
β
ζ
0
/
γ
μ
(
1
-
ζ
0
).Proof.
From the well-known result by Neuts [19] on the positive recurrence of the Markov chain associated with A, we have ζ
A
0
e
<
ζ
A
2
e for the Markov chain to be stable. With a bit of algebra, this simplifies to λ
<
μ
2
-
β
ζ
0
/
γ
μ
(
1
-
ζ
0
).For future reference we defineρ
1 as
(8)
ρ
1
=
λ
μ
2
-
β
ζ
0
/
γ
μ
(
1
-
ζ
0
)
.


## 2.1. Analysis of the System

In this section we perform the steady-state analysis of the queueing-inventory model under study by first establishing the stability condition of the queueing-inventory system. DefineA
=
A
0
+
A
1
+
A
2. This is the infinitesimal generator of the finite state CTMC corresponding to the inventory level 0,1
,
2
,
…
,
S for any level i (≥1). Let ζ denote the steady-state probability vector of A. That is,
(3)
ζ
A
=
0
,
ζ
e
=
1
.
Write
(4)
ζ
=
ζ
0
,
ζ
1
,
…
,
ζ
s
,
…
,
ζ
Q
,
…
,
ζ
S
.
We have
(5)
A
=
-
β
β
γ
μ
-
(
β
+
γ
μ
)
2
γ
μ
-
(
β
+
2
γ
μ
)
⋱
⋱
⋱
2
γ
μ
-
(
β
+
2
γ
μ
)
β
2
γ
μ
-
2
γ
μ
⋱
⋱
2
γ
μ
-
2
γ
μ
.
Then using (3) we get the components of the vector ζ explicitly as
(6)
ζ
0
=
1
+
β
γ
μ
1
+
β
+
γ
μ
γ
μ
∑
i
=
0
s
‍
β
+
2
γ
μ
2
γ
μ
i
-
2
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
+
(
Q
-
s
-
2
)
β
+
2
γ
μ
2
γ
μ
s
-
1
1
+
β
+
γ
μ
γ
μ
∑
i
=
0
s
‍
β
+
2
γ
μ
2
γ
μ
i
-
2
llllll
+
β
2
γ
μ
β
+
γ
μ
γ
μ
llllll
×
∑
i
=
0
s
‍
β
+
2
γ
μ
2
γ
μ
i
-
2
β
+
2
γ
μ
2
γ
μ
s
-
i
+
1
-
1
β
+
2
γ
μ
2
γ
μ
s
-
1
-
γ
μ
β
+
γ
μ
l
l
l
l
l
l
l
l
l
l
+
∑
i
=
0
s
‍
β
+
2
γ
μ
2
γ
μ
i
-
2
β
+
2
γ
μ
2
γ
μ
s
-
i
+
1
-
1
-
1
,
ζ
i
=
β
γ
μ
ζ
0
,
for
i
=
1
,
β
γ
μ
β
+
γ
μ
2
γ
μ
×
β
+
2
γ
μ
2
γ
μ
i
-
2
ζ
0
,
for
i
=
2,3
,
…
,
s
+
1
,
ζ
i
+
1
,
for
i
=
s
+
1
,
s
+
2
,
…
,
Q
-
1
,
β
2
γ
μ
β
+
2
γ
μ
2
γ
μ
s
-
1
-
1
β
+
γ
μ
γ
μ
l
l
l
l
l
l
l
×
β
+
2
γ
μ
2
γ
μ
s
-
1

-
1
ζ
0
,
for
i
=
Q
+
1
,
ζ
Q
+
i
=
β
2
γ
μ
β
+
γ
μ
γ
μ
β
+
2
γ
μ
2
γ
μ
i
-
2
×
β
+
2
γ
μ
2
γ
μ
s
-
i
-
1
-
1
ζ
0
,
i
=
2,3
,
…
,
s
.
Since the Markov chain X
(
t
)
∣
t
≥
0 is an LIQBD, it is stable if and only if the left drift rate exceeds the right drift rate. That is,
(7)
ζ
A
0
e
<
ζ
A
2
e
.
Thus, we have the following lemma for stability of the system under study.Lemma 1.
The stability condition of theM
/
M
/
2 queueing-inventory system under consideration is given by λ
<
μ
2
-
β
ζ
0
/
γ
μ
(
1
-
ζ
0
).Proof.
From the well-known result by Neuts [19] on the positive recurrence of the Markov chain associated with A, we have ζ
A
0
e
<
ζ
A
2
e for the Markov chain to be stable. With a bit of algebra, this simplifies to λ
<
μ
2
-
β
ζ
0
/
γ
μ
(
1
-
ζ
0
).For future reference we defineρ
1 as
(8)
ρ
1
=
λ
μ
2
-
β
ζ
0
/
γ
μ
(
1
-
ζ
0
)
.


## 3. Computation of the Steady-State Probability

For computing the steady-state probability vector of the process{
X
(
t
)
∣
t
≥
0
}, we first consider a queueing-inventory system with unlimited supply of inventory items (i.e., classical M
/
M
/
2 queueing system). The rest of the assumptions such as those on the arrival process and lead time are the same as given earlier. Designate the Markov chain so obtained as {
N
(
t
)
∣
t
≥
0
}, where N
(
t
) is the number of customers in the system at time t. Its infinitesimal generator G
1 is given by
(9)
G
1
=
-
λ
λ
μ
-
(
λ
+
μ
)
λ
2
μ
-
(
λ
+
2
μ
)
λ
2
μ
-
(
λ
+
2
μ
)
λ
…
⋱
⋱
⋱
.
Let π be the steady-state probability vector of G
1. Partitioning π by levels we write π as
(10)
π
=
(
π
0
,
π
1
,
π
2
,
…
)
.
Then the steady-state vector must satisfy
(11)
π
G
1
=
0
,
π
e
=
1
.
From the relation (11) we get the vector π explicitly as follows:
(12)
π
i
=
1
+
λ
μ
1
-
λ
2
μ
-
1
-
1
for
i
=
0
,
λ
μ
π
0
for
i
=
1
,
1
2
i
-
1
λ
μ
i
π
0
for
i
≥
2
.
Further we consider an inventory system with negligible service time and no backlog of demands. The assumptions such as those on the arrival process and lead time are the same as given in the description of the model. Denote this Markov chain by {
I
(
t
)
∣
t
≥
0
}. Here I
(
t
) is the inventory level at time t. Its infinitesimal generator G
2 is given by(13)
G
2
=
0
1
⋮
s
⋮
Q
⋮
S
0
a
aaaa
1
aaa
⋯
a
aa
s
⋯
a
aa
Q
a
⋯
a
aaaa
S
-
β
β
γ
λ
-
(
γ
λ
+
β
)
⋱
⋱
⋱
γ
λ
-
(
γ
λ
+
β
)
β
γ
λ
-
γ
λ
⋱
⋱
γ
λ
-
γ
λ
γ
λ
-
γ
λ
.Letψ
=
(
ψ
0
,
ψ
1
,
…
,
ψ
S
) be the steady-state probability vector of the process {
I
(
t
)
∣
t
≥
0
}. Then ψ satisfies the relations
(14)
ψ
G
2
=
0
,
ψ
e
=
1
.
That is, at arbitrary epochs the inventory level distribution ψ
j is given by
(15)
ψ
j
=
1
+
Q
β
γ
λ
β
+
γ
λ
γ
λ
s
-
1
,
j
=
0
,
β
γ
λ
β
+
γ
λ
γ
λ
j
-
1
ψ
0
,
j
=
1,2
,
…
,
s
,
β
γ
λ
β
+
γ
λ
γ
λ
s
ψ
0
,
j
=
s
+
1
,
s
+
2
,
…
,
Q
,
β
γ
λ
β
+
γ
λ
γ
λ
j
-
Q
-
1
×
β
+
γ
λ
γ
λ
s
-
(
j
-
Q
-
1
)
-
1
ψ
0
,
j
=
Q
+
1
,
Q
+
2
,
…
,
S
.
Using the components of the probability vector ψ, we will find the steady-state probability vector of the original system. Let x be the steady-state probability vector of the original system. Then the steady-state vector must satisfy the set of equations
(16)
x
W
1
=
0
,
x
e
=
1
.
Partition x by levels as
(17)
x
=
x
0
,
x
1
,
x
2
,
…
,
where the subvectors of x are further partitioned as
(18)
x
i
=
(
x
i
(
0
)
,
x
i
(
1
)
,
x
i
(
2
)
,
x
i
(
3
)
,
…
,
x
i
(
S
)
)
,
i
≥
0
.
Then by using the relation x
W
1
=
0, we get
(19)
-
β
x
i
0
+
γ
μ
x
i
+
1
1
=
0
,
i
≥
0
,
λ
x
i
j
-
λ
+
2
μ
+
β
x
i
+
1
j
+
2
1
-
γ
μ
x
i
+
2
j
+
2
γ
μ
x
i
+
2
j
+
1
=
0
,
i
≥
1
,
2
≤
j
≤
Q
-
1
,
λ
x
i
j
+
β
x
i
+
1
j
-
Q
-
λ
+
2
μ
x
i
+
1
j
+
2
1
-
γ
μ
x
i
+
2
j
+
2
γ
μ
x
i
+
2
j
+
1
=
0
,
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
i
≥
1
,
Q
≤
j
≤
S
-
1
,
λ
x
i
S
+
β
x
i
+
1
s
-
λ
+
2
μ
x
i
+
1
(
S
)
+
2
1
-
γ
μ
x
i
+
2
S
=
0
,
i
≥
1
,
-
λ
+
β
x
0
j
+
1
-
γ
μ
x
1
j
+
γ
μ
x
1
j
+
1
=
0
,
lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll
1
≤
j
≤
s
,
-
λ
x
0
(
j
)
+
1
-
γ
μ
x
1
j
+
γ
μ
x
1
j
+
1
=
0
,
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
s
+
1
≤
j
≤
Q
-
1
,
β
x
0
(
j
-
Q
)
-
λ
x
0
(
j
)
+
(
1
-
γ
)
μ
x
1
(
j
)
+
γ
μ
x
1
j
+
1
=
0
,
lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll
Q
≤
j
≤
S
-
1
,
β
x
0
(
s
)
-
λ
x
0
(
S
)
+
(
1
-
γ
)
μ
x
1
(
S
)
=
0
,
λ
x
0
(
j
)
-
(
λ
+
β
+
μ
)
x
1
(
j
)
+
2
(
1
-
γ
)
μ
x
2
(
j
)
+
2
γ
μ
x
2
(
j
+
1
)
=
0
,
2
≤
j
≤
s
,
λ
x
0
(
j
)
-
(
λ
+
μ
)
x
1
(
j
)
+
2
(
1
-
γ
)
μ
x
2
(
j
)
+
2
γ
μ
x
2
(
j
+
1
)
=
0
,
s
+
1
≤
j
≤
Q
-
1
,
λ
x
0
(
j
)
+
β
x
1
(
j
-
Q
)
-
(
λ
+
μ
)
x
1
(
j
)
+
2
(
1
-
γ
)
μ
x
2
(
j
)
+
2
γ
μ
x
2
(
j
+
1
)
=
0
,
Q
≤
j
≤
S
-
1
,
λ
x
0
(
S
)
+
β
x
1
(
s
)
-
λ
+
μ
x
1
(
S
)
+
2
(
1
-
γ
)
μ
x
2
(
S
)
=
0
.
We assume a solution of the form
(20)
x
i
j
=
α
-
1
Θ
j
i
π
i
ψ
j
,
i
≥
0
,
0
≤
j
≤
S
,
for constants Θ
j
i, and then verify that the system of equations given in (16) is satisfied.The constantsΘ
j
i’s are given by
(21)
Θ
0
i
=
1
,
i
≥
0
,
Θ
1
i
=
1
γ
,
i
=
1
,
2
γ
,
i
≥
2
,
Θ
j
0
=
1
γ
j
,
1
≤
j
≤
S
-
1
,
Θ
2
i
=
β
+
γ
λ
β
+
λ
1
γ
2
,
i
=
1,2
,
2
β
+
(
1
+
γ
)
λ
β
+
λ
1
γ
2
,
i
≥
3
,
Θ
j
i
=
1
γ
(
β
+
λ
)
δ
j
-
1
i
,
3
≤
i
≤
2
(
j
-
1
)
,
3
≤
j
≤
s
+
1
,
(
β
+
γ
λ
)
γ
(
β
+
λ
)
Θ
j
-
1
i
-
2
,
i
≥
2
j
-
1
,
3
≤
j
≤
s
+
1
,
1
γ
λ
δ
j
-
1
i
,
3
≤
i
≤
2
(
j
-
1
)
,
s
+
2
≤
j
≤
Q
,
β
+
γ
λ
γ
λ
Θ
j
-
1
i
-
2
,
i
≥
2
j
-
1
,
s
+
2
≤
j
≤
Q
,
where δ
j
-
1
i
=
(
λ
+
2
μ
+
β
)
Θ
j
-
1
i
-
1
-
2
μ
Θ
j
-
1
i
-
2
-
(
1
-
γ
)
λ
Θ
j
-
1
i.Consider(22)
Θ
Q
+
k
i
=
1
γ
λ
β
+
λ
λ
s
-
1
-
1
×
ξ
Q
+
k
-
1
i
β
+
λ
λ
s
-
λ
,
3
≤
i
≤
2
Q
,
k
=
1
,
1
γ
λ
β
+
λ
λ
s
-
1
-
1
×
β
+
λ
λ
s
-
λ
ξ
Q
+
k
-
1
i
-
2
γ
λ
llll
×
β
+
λ
λ
s
-
λ
,
i
≥
2
Q
+
1
,
k
=
1
,
1
γ
λ
(
β
+
λ
)
×
β
+
λ
λ
s
-
(
k
-
1
)
-
1
-
1
×
β
+
λ
λ
s
-
(
k
-
2
)
-
1
ξ
Q
+
k
-
1
i
llll
×
β
+
λ
λ
s
-
(
k
-
2
)
-
1
llll
-
β
Θ
k
-
1
i
-
1
β
+
λ
λ
s
-
k
-
2
-
1
,
3
≤
i
≤
2
Q
+
k
-
1
,
2
≤
k
≤
s
,
1
γ
λ
(
β
+
λ
)
×
β
+
λ
λ
s
-
(
k
-
1
)
-
1
-
1
×
γ
λ
β
+
λ
λ
s
-
(
k
-
2
)
-
1
l
l
l
l
×
Θ
Q
+
k
-
1
i
-
2
-
β
Θ
k
-
1
i
-
1
β
+
λ
λ
s
-
(
k
-
2
)
-
1
,
i
≥
2
Q
+
k
-
1
,
2
≤
k
≤
s
,
where ξ
Q
+
k
-
1
i
=
(
λ
+
2
μ
)
Θ
Q
+
k
-
1
i
-
1
-
2
μ
Θ
Q
+
k
-
1
i
-
2
-
(
1
-
γ
)
λ
Θ
Q
+
k
-
1
i.Consider(23)
Θ
j
1
=
1
γ
(
β
+
λ
)
×
(
λ
+
β
)
Θ
j
-
1
0
l
l
l
-
(
1
-
γ
)
λ
Θ
j
-
1
1
,
3
≤
j
≤
s
+
1
,
1
γ
Θ
j
-
1
0
-
(
1
-
γ
)
Θ
j
-
1
1
,
s
+
2
≤
j
≤
Q
,
Θ
S
0
=
Θ
s
0
-
1
-
γ
Θ
S
1
,
Θ
j
2
=
1
γ
(
β
+
λ
)
×
(
λ
+
β
+
μ
)
Θ
j
-
1
1
l
l
l
-
μ
Θ
j
-
1
0
-
(
1
-
γ
)
λ
Θ
j
-
1
2
,
3
≤
j
≤
s
+
1
,
1
γ
λ
ϑ
j
-
1
,
s
+
2
≤
j
≤
Q
,
Θ
Q
+
k
2
=
1
γ
λ
β
+
λ
λ
s
-
1
-
1
×
ϑ
Q
β
+
μ
μ
s
-
β
,
k
=
1
,
1
γ
(
β
+
λ
)
β
+
λ
λ
s
-
(
k
-
1
)
-
1
-
1
×
ϑ
j
-
1
β
+
λ
λ
s
-
(
k
-
2
)
-
1
l
l
l
l
-
β
Θ
k
-
1
1
β
+
λ
λ
s
-
(
k
-
2
)
-
1
,
2
≤
k
≤
s
,
where ϑ
j
=
λ
+
μ
Θ
j
1
-
μ
Θ
j
0
-
1
-
γ
λ
Θ
j
2,  s
-
1
≤
j
≤
S.Thus we have(24)
∑
i
=
0
∞
‍
∑
j
=
0
Q
+
s
‍
Θ
j
i
π
i
ψ
j
=
1
+
Q
β
γ
λ
β
+
γ
λ
γ
λ
s
.
If we note x
e
=
1 and (20) we have
(25)
α
-
1
∑
i
=
0
∞
‍
∑
j
=
0
Q
+
s
‍
Θ
j
i
π
i
ψ
j
=
1
.
Write α
=
1
+
Q
β
/
γ
λ
β
+
γ
λ
/
γ
λ
s. Then dividing each Θ
j
i
π
i
ψ
j by α we get the steady-state probability vector of the original system.Thus we arrive at our main theorem.Theorem 2.
Suppose that the conditionρ
1
<
1 holds. Then the components of the steady-state probability vector of the process X
(
t
)
∣
t
≥
0 with generator matrix W
1 are x
i
(
j
)
=
α
-
1
Θ
j
i
π
i
ψ
j,  i
≥
0,  0
≤
j
≤
S, the probabilities π
i correspond to the distribution of number of customers in the system as given in (12), and the probabilities ψ
j are obtained in (15).The consequence of Theorem2 is that the two-dimensional system can be decomposed into two distinct one-dimensional objects one of which corresponds to the number of customers in an M
/
M
/
2 queue and the other to the number of items in the inventory.


### 3.1. Performance Measures

(i)
Mean number of customers in the system is as follows:(26)
L
s
=
α
-
1
∑
i
=
1
∞
‍
∑
j
=
0
Q
+
s
‍
i
Θ
j
i
π
i
ψ
j
.
(ii)
Mean number of customers in the queue is as follows:(27)
L
q
=
α
-
1
∑
i
=
2
∞
‍
∑
j
=
2
Q
+
s
‍
(
i
-
2
)
Θ
j
i
π
i
ψ
j
.
(iii)
Mean inventory level in the system is as follows:(28)
I
m
=
α
-
1
∑
i
=
0
∞
∑
j
=
1
Q
+
s
j
Θ
j
i
π
i
ψ
j
.
(iv)
Mean number of busy servers is as follows:(29)
P
BS
=
α
-
1
∑
j
=
3
Q
+
s
‍
∑
0
∑
i
=
2
∞
‍
Θ
1
i
π
i
ψ
1
+
∑
j
=
2
Q
+
s
‍
Θ
j
1
π
1
ψ
j
+
Θ
1
1
π
1
ψ
1
l
l
l
l
l
l
l
l
l
l
l
l
+
2
∑
i
=
3
∞
‍
Θ
2
i
π
i
ψ
2
+
∑
j
=
3
Q
+
s
‍
Θ
j
2
π
2
ψ
j
+
Θ
2
2
π
2
ψ
2
∑
j
=
3
Q
+
s
‍
∑
0
.
(v)
Depletion rate of inventory is as follows:(30)
D
inv
=
γ
μ
α
-
1
∑
i
=
1
∞
∑
j
=
1
Q
+
s
Θ
j
i
π
i
ψ
j
.
(vi)
Mean number of replenishments per time unit is as follows:(31)
R
r
=
β
α
-
1
∑
i
=
0
∞
‍
∑
j
=
0
s
‍
Θ
j
i
π
i
ψ
j
.
(vii)
Mean number of departures per unit time is as follows:(32)
D
m
=
μ
α
-
1
∑
i
=
1
∞
‍
Θ
1
i
π
i
ψ
1
+
∑
j
=
1
Q
+
s
‍
Θ
j
1
π
1
ψ
j
+
2
μ
α
-
1
∑
i
=
2
∞
‍
∑
j
=
2
Q
+
s
‍
Θ
j
i
π
i
ψ
j
.
(viii)
Expected loss rate of customers is as follows:(33)
E
loss
=
λ
α
-
1
∑
i
=
0
∞
Θ
0
i
π
i
ψ
0
.
(ix)
Expected loss rate of customers when the inventory level is zero per cycle isE
loss
c
=
E
loss
/
R
r.
(x)
Effective arrival rate is as follows:(34)
λ
A
=
λ
α
-
1
∑
i
=
0
∞
‍
∑
j
=
1
Q
+
s
‍
Θ
j
i
π
i
ψ
j
.
(xi)
Mean sojourn time of the customers in the system isW
s
=
L
s
/
λ
A.
(xii)
Mean waiting time of a customer in the queue isW
q
=
L
q
/
λ
A.
(xiii)
Mean number of customers waiting in the system when inventory is available is as follows:(35)
W
~
=
α
-
1
∑
i
=
0
∞
∑
j
=
1
Q
+
s
i
Θ
j
i
π
i
ψ
j
.
(xiv)
Mean number of customers waiting in the system during the stock out period is as follows:(36)
W
~
~
=
α
-
1
∑
i
=
0
∞
‍
i
Θ
0
i
π
i
ψ
0
.


## 3.1. Performance Measures

(i)
Mean number of customers in the system is as follows:(26)
L
s
=
α
-
1
∑
i
=
1
∞
‍
∑
j
=
0
Q
+
s
‍
i
Θ
j
i
π
i
ψ
j
.
(ii)
Mean number of customers in the queue is as follows:(27)
L
q
=
α
-
1
∑
i
=
2
∞
‍
∑
j
=
2
Q
+
s
‍
(
i
-
2
)
Θ
j
i
π
i
ψ
j
.
(iii)
Mean inventory level in the system is as follows:(28)
I
m
=
α
-
1
∑
i
=
0
∞
∑
j
=
1
Q
+
s
j
Θ
j
i
π
i
ψ
j
.
(iv)
Mean number of busy servers is as follows:(29)
P
BS
=
α
-
1
∑
j
=
3
Q
+
s
‍
∑
0
∑
i
=
2
∞
‍
Θ
1
i
π
i
ψ
1
+
∑
j
=
2
Q
+
s
‍
Θ
j
1
π
1
ψ
j
+
Θ
1
1
π
1
ψ
1
l
l
l
l
l
l
l
l
l
l
l
l
+
2
∑
i
=
3
∞
‍
Θ
2
i
π
i
ψ
2
+
∑
j
=
3
Q
+
s
‍
Θ
j
2
π
2
ψ
j
+
Θ
2
2
π
2
ψ
2
∑
j
=
3
Q
+
s
‍
∑
0
.
(v)
Depletion rate of inventory is as follows:(30)
D
inv
=
γ
μ
α
-
1
∑
i
=
1
∞
∑
j
=
1
Q
+
s
Θ
j
i
π
i
ψ
j
.
(vi)
Mean number of replenishments per time unit is as follows:(31)
R
r
=
β
α
-
1
∑
i
=
0
∞
‍
∑
j
=
0
s
‍
Θ
j
i
π
i
ψ
j
.
(vii)
Mean number of departures per unit time is as follows:(32)
D
m
=
μ
α
-
1
∑
i
=
1
∞
‍
Θ
1
i
π
i
ψ
1
+
∑
j
=
1
Q
+
s
‍
Θ
j
1
π
1
ψ
j
+
2
μ
α
-
1
∑
i
=
2
∞
‍
∑
j
=
2
Q
+
s
‍
Θ
j
i
π
i
ψ
j
.
(viii)
Expected loss rate of customers is as follows:(33)
E
loss
=
λ
α
-
1
∑
i
=
0
∞
Θ
0
i
π
i
ψ
0
.
(ix)
Expected loss rate of customers when the inventory level is zero per cycle isE
loss
c
=
E
loss
/
R
r.
(x)
Effective arrival rate is as follows:(34)
λ
A
=
λ
α
-
1
∑
i
=
0
∞
‍
∑
j
=
1
Q
+
s
‍
Θ
j
i
π
i
ψ
j
.
(xi)
Mean sojourn time of the customers in the system isW
s
=
L
s
/
λ
A.
(xii)
Mean waiting time of a customer in the queue isW
q
=
L
q
/
λ
A.
(xiii)
Mean number of customers waiting in the system when inventory is available is as follows:(35)
W
~
=
α
-
1
∑
i
=
0
∞
∑
j
=
1
Q
+
s
i
Θ
j
i
π
i
ψ
j
.
(xiv)
Mean number of customers waiting in the system during the stock out period is as follows:(36)
W
~
~
=
α
-
1
∑
i
=
0
∞
‍
i
Θ
0
i
π
i
ψ
0
.


## 4. Optimization Problem I

In this section we provide the optimal values of the inventory levels and the fixed order quantity Q. Now for computing the minimal costs of M
/
M
/
2 queueing-inventory model we introduce the cost function F
(
2
,
s
,
Q
) defined by
(37)
F
2
,
s
,
Q
=
h
·
I
m
+
c
1
·
E
loss
+
c
2
·
W
~
~
+
K
+
Q
·
c
3
·
R
r
+
c
4
·
P
BS
+
c
5
·
c
-
P
BS
,
where K is fixed cost for placing an order, c
1 is the cost incurred due to loss per customer, c
2 is waiting cost per unit time per customer during the stock out period, c
3 is variable procurement cost per item, c
4 is the cost incurred per busy server, c
5 is the cost incurred per idle server, and h is unit holding cost of inventory per unit per unit time. We assign the following values to the parameters: λ
=
5,  μ
=
3,  β
=
1,  K
=
$
500,  c
1
=
$
100,  c
2
=
$
50,  c
3
=
$
25,  c
4
=
$
10,  c
5
=
$
20,  and h
=
$
2. Using MATLAB program we computed the optimal pairs (
s
,
Q
) and also the corresponding minimum cost (in Dollars). Here γ is varied from 0.1 to 1 each time increasing it by 0.1 unit. The optimal pair (
s
,
Q
) and the corresponding cost (minimum) are given in Table 1.Table 1
Optimal(
s
,
Q
) pair and minimum cost.
γ
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Optimal(
s
,
Q
) pair and  minimum cost
(3, 15)
(3, 21)
(3, 27)
(3, 33)
(3, 39)
(3, 43)
(5, 46)
(5, 53)
(6, 53)
(6, 58)
82.684
106.87
130.57
153.76
176.29
198.10
219.04
239.03
258.26
277.18


## 5.M
/
M
/
c
(
c
≥
3
) Queueing-Inventory System

Next we considerM
/
M
/
c queueing-inventory system with positive service time for 3
≤
c
≤
s. We keep the model assumptions the same as in Section 2. Hence the service rate is i
μ, for i varying from 0 to c, depending on the availability of the inventory and customers. When the number of customers is at least c and not less than c items are in the inventory, the service rate is c
μ. Write {
Y
(
t
)
∣
t
≥
0
}
=
{
(
N
(
t
)
,
I
(
t
)
)
∣
t
≥
0
}. Then {
Y
(
t
)
∣
t
≥
0
} is a CTMC with state space Ω
2
=
⋃
i
=
0
∞
L
(
i
), where L
(
i
) is the collection of states L
(
i
)
=
(
i
,
0
)
,
…
,
(
i
,
Q
+
s
) as defined in Section 2. The infinitesimal generator W
2 of the CTMC {
Y
(
t
)
∣
t
≥
0
} is
(38)
W
2
=
B
A
-
0
A
2
1
A
1
1
A
-
0
A
2
2
A
1
2
A
-
0
⋱
⋱
⋱
A
2
c
-
2
A
1
c
-
2
A
-
0
A
2
c
-
1
A
1
c
-
1
A
-
0
A
-
2
A
-
1
A
-
0
A
-
2
A
-
1
A
-
0
⋯
⋱
⋱
⋱
,
and the transition rates are
(39)
B
k
l
=
-
β
,
for
l
=
k
=
0
,
-
(
λ
+
β
)
,
for
l
=
k
;
k
=
1,2
,
…
,
s
,
-
λ
,
for
l
=
k
;
k
=
s
+
1
,
s
+
2
,
…
,
S
,
β
,
for
l
=
k
+
Q
;
k
=
0,1
,
…
,
s
,
0
,
otherwise
,
A
-
0
k
l
=
λ
,
for
l
=
k
;
k
=
1,2
,
…
,
S
,
0
,
otherwise
,
A
-
1
k
l
=
-
β
,
for
l
=
k
=
0
,
-
λ
+
β
+
i
μ
,
for
l
=
k
;
k
=
1,2
,
…
,
c
,
-
(
λ
+
β
+
c
μ
)
,
for
l
=
k
;
k
=
c
+
1
,
c
+
2
,
…
,
s
,
-
(
λ
+
c
μ
)
,
for
l
=
k
,
k
=
s
+
1
,
s
+
2
,
…
,
S
,
β
,
for
l
=
k
+
Q
;
k
=
0,1
,
…
,
s
,
0
,
otherwise
,
A
-
2
k
l
=
i
γ
μ
,
for
l
=
k
-
1
;
k
=
c
,
c
+
1
,
c
+
2
,
…
,
S
,
i
γ
μ
,
for
l
=
k
-
1
;
k
=
1,2
,
…
,
c
-
1
,
c
(
1
-
γ
)
μ
,
for
l
=
k
;
k
=
c
,
c
+
1
,
c
+
2
,
…
,
S
,
i
(
1
-
γ
)
μ
,
for
l
=
k
;
k
=
1,2
,
…
,
c
-
1
,
0
,
otherwise
.Form
=
1,2
,
…
,
c
-
1,
(40)
A
2
m
k
l
=
m
γ
μ
,
for
l
=
k
-
1
;
m
≤
k
;
k
=
1,2
,
…
,
S
,
k
γ
μ
,
for
l
=
k
-
1
;
m
>
k
;
k
=
1,2
,
…
,
S
,
m
(
1
-
γ
)
μ
,
for
l
=
k
;
m
≤
k
;
k
=
1,2
,
…
,
S
,
k
γ
μ
,
for
l
=
k
;
m
>
k
;
k
=
1,2
,
…
,
S
,
0
,
otherwise
,
A
1
m
k
l
=
-
β
,
for
l
=
k
=
0
,
-
(
λ
+
β
+
m
μ
)
,
for
l
=
k
;
m
≤
k
;
k
=
1,2
,
…
,
s
,
-
(
λ
+
β
+
k
μ
)
,
for
l
=
k
;
m
>
k
≥
1
,
-
(
λ
+
m
μ
)
,
for
l
=
k
;
k
=
s
+
1
,
s
+
2
,
…
,
S
,
β
,
for
l
=
k
+
Q
;
k
=
0,1
,
…
,
s
,
0
,
otherwise
.


### 5.1. System Stability and Computation of Steady-State Probability Vector

The Markov chain under consideration is a LIQBD process. For this chain to be stable it is necessary and sufficient that(41)
ξ
A
-
0
e
<
ξ
A
-
2
e
,
where ξ is the unique nonnegative vector satisfying
(42)
ξ
A
-
=
0
,
ξ
e
=
1
,
and A
-
=
A
-
0  +  A
-
1  +  A
-
2 is the infinitesimal generator of the finite state CTMC on the set 0,1
,
…
,
S. Write ξ as ξ
0
,
ξ
1
,
…
,
ξ
S. Then we get from (42) the components of the probability vector ξ explicitly as
(43)
ξ
0
=
1
+
∑
i
=
1
c
-
1
‍
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
∏
k
=
1
c
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
-
1
llllll
+
Q
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
-
s
β
c
γ
μ
1
+
∑
i
=
0
s
-
2
‍
∏
k
=
0
i
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
β
2
c
γ
μ
1
+
∑
i
=
1
s
-
2
‍
∏
k
=
1
i
‍
β
+
k
γ
μ
k
γ
μ
-
1
,
ξ
i
=
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
ξ
0
,
for
1
≤
i
≤
c
,
β
+
c
γ
μ
c
γ
μ
i
-
c
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
ξ
0
,
for
c
+
1
≤
i
≤
s
+
1
,
ξ
i
+
1
,
for
s
+
1
≤
i
≤
Q
-
1
,
ξ
Q
+
i
=

∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
ξ
0
,
for
i
=
1
,

1
+
∑
j
=
0
i
-
2
‍
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
l
×
1
+
∑
j
=
0
i
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
ξ
0
,
for
2
≤
i
≤
s
.
From the relation (41) we have the following.Lemma 3.
The stability condition of the queueing-inventory system under study is given byρ
2
<
1, where ρ
2
=
λ
1
-
ξ
0
/
μ
∑
j
=
1
c
-
1
j
ξ
j
+
c
∑
j
=
c
Q
+
s
ξ
j.Proof.
The proof is on the same lines as that of Lemma1.Next we compute the steady-state probability vector ofW
2 under the stability condition. Let y denote the steady-state probability vector of the generator W
2. So y must satisfy the relations
(44)
y
W
2
=
0
,
y
e
=
1
.
Let us partition y by levels as
(45)
y
=
y
0
,
y
1
,
y
2
,
…
,
where the subvectors of y are further partitioned as
(46)
y
i
=
y
i
0
,
y
i
1
,
y
i
2
,
…
,
y
i
S
,
i
≥
0
.
The steady-state probability vector y is obtained as
(47)
y
i
+
c
-
1
=
y
c
-
1
R
i
,
i
≥
1
,
where R is the minimal nonnegative solution to the matrix quadratic equation
(48)
R
2
A
-
2
+
R
A
-
1
+
A
-
0
=
0
,
and the vectors y
0
,
y
1
,
…
,
y
c
-
1 can be obtained by solving the following equations:
(49)
y
0
B
+
y
1
A
2
1
=
0
,
y
i
-
1
A
-
0
+
y
i
A
1
i
+
y
i
+
1
A
2
i
+
1
=
0
,
1
≤
i
≤
c
-
1
,
y
c
-
2
A
-
0
+
y
c
-
1
A
1
c
-
1
+
R
A
-
2
=
0
.
Now from (49), we get
(50)
y
0
=
y
1
A
2
1
-
B
-
1
=
y
1
A
2
1
-
A
-
0
′
-
1
,
y
1
=
-
y
2
A
2
2
A
1
1
+
A
2
1
-
A
-
0
′
-
1
A
-
0
-
1
=
y
2
A
2
2
-
A
-
0
′
-
1
,
y
i
=
y
i
+
1
A
2
i
+
1
-
A
-
0
′
-
1
,
0
≤
i
≤
c
-
1
,
where
(51)
A
-
i
′
=
B
,
i
=
0
,
A
1
i
+
A
2
i
-
A
-
i
-
1
′
-
1
A
-
0
,
1
≤
i
≤
c
,
subject to normalizing condition
(52)
∑
i
=
1
c
-
2
y
i
+
y
c
-
1
I
-
R
-
1
e
=
1
.
Since R cannot be computed explicitly we explore the possibility of algorithmic computation. Thus, one can use logarithmic reduction algorithm as given by Latouche and Ramaswami [20] for computing R. We list here only the main steps involved in logarithmic reduction algorithm for computation of R.Logarithmic Reduction Algorithm for
RStep 0.  H
←
(
-
A
-
1
)
-
1
A
-
0, L
←
(
-
A
-
1
)
-
1
A
-
2, G
=
L, and T
=
H.Step 1. Consider
(53)
U
=
H
L
+
L
H
,
M
=
H
2
,
H
⟵
I
-
U
-
1
M
,
M
⟵
L
2
,
L
⟵
I
-
U
-
1
M
,
G
⟵
G
+
T
L
,
T
⟵
T
H
.
Continue Step 1 until e
-
G
e
∞
<
ϵ.Step 2. R
=
-
A
-
0
A
-
1
+
A
-
0
G
-
1.


## 5.1. System Stability and Computation of Steady-State Probability Vector

The Markov chain under consideration is a LIQBD process. For this chain to be stable it is necessary and sufficient that(41)
ξ
A
-
0
e
<
ξ
A
-
2
e
,
where ξ is the unique nonnegative vector satisfying
(42)
ξ
A
-
=
0
,
ξ
e
=
1
,
and A
-
=
A
-
0  +  A
-
1  +  A
-
2 is the infinitesimal generator of the finite state CTMC on the set 0,1
,
…
,
S. Write ξ as ξ
0
,
ξ
1
,
…
,
ξ
S. Then we get from (42) the components of the probability vector ξ explicitly as
(43)
ξ
0
=
1
+
∑
i
=
1
c
-
1
‍
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
∏
k
=
1
c
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
-
1
llllll
+
Q
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
-
s
β
c
γ
μ
1
+
∑
i
=
0
s
-
2
‍
∏
k
=
0
i
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
β
2
c
γ
μ
1
+
∑
i
=
1
s
-
2
‍
∏
k
=
1
i
‍
β
+
k
γ
μ
k
γ
μ
-
1
,
ξ
i
=
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
ξ
0
,
for
1
≤
i
≤
c
,
β
+
c
γ
μ
c
γ
μ
i
-
c
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
ξ
0
,
for
c
+
1
≤
i
≤
s
+
1
,
ξ
i
+
1
,
for
s
+
1
≤
i
≤
Q
-
1
,
ξ
Q
+
i
=

∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
ξ
0
,
for
i
=
1
,

1
+
∑
j
=
0
i
-
2
‍
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
l
×
1
+
∑
j
=
0
i
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
ξ
0
,
for
2
≤
i
≤
s
.
From the relation (41) we have the following.Lemma 3.
The stability condition of the queueing-inventory system under study is given byρ
2
<
1, where ρ
2
=
λ
1
-
ξ
0
/
μ
∑
j
=
1
c
-
1
j
ξ
j
+
c
∑
j
=
c
Q
+
s
ξ
j.Proof.
The proof is on the same lines as that of Lemma1.Next we compute the steady-state probability vector ofW
2 under the stability condition. Let y denote the steady-state probability vector of the generator W
2. So y must satisfy the relations
(44)
y
W
2
=
0
,
y
e
=
1
.
Let us partition y by levels as
(45)
y
=
y
0
,
y
1
,
y
2
,
…
,
where the subvectors of y are further partitioned as
(46)
y
i
=
y
i
0
,
y
i
1
,
y
i
2
,
…
,
y
i
S
,
i
≥
0
.
The steady-state probability vector y is obtained as
(47)
y
i
+
c
-
1
=
y
c
-
1
R
i
,
i
≥
1
,
where R is the minimal nonnegative solution to the matrix quadratic equation
(48)
R
2
A
-
2
+
R
A
-
1
+
A
-
0
=
0
,
and the vectors y
0
,
y
1
,
…
,
y
c
-
1 can be obtained by solving the following equations:
(49)
y
0
B
+
y
1
A
2
1
=
0
,
y
i
-
1
A
-
0
+
y
i
A
1
i
+
y
i
+
1
A
2
i
+
1
=
0
,
1
≤
i
≤
c
-
1
,
y
c
-
2
A
-
0
+
y
c
-
1
A
1
c
-
1
+
R
A
-
2
=
0
.
Now from (49), we get
(50)
y
0
=
y
1
A
2
1
-
B
-
1
=
y
1
A
2
1
-
A
-
0
′
-
1
,
y
1
=
-
y
2
A
2
2
A
1
1
+
A
2
1
-
A
-
0
′
-
1
A
-
0
-
1
=
y
2
A
2
2
-
A
-
0
′
-
1
,
y
i
=
y
i
+
1
A
2
i
+
1
-
A
-
0
′
-
1
,
0
≤
i
≤
c
-
1
,
where
(51)
A
-
i
′
=
B
,
i
=
0
,
A
1
i
+
A
2
i
-
A
-
i
-
1
′
-
1
A
-
0
,
1
≤
i
≤
c
,
subject to normalizing condition
(52)
∑
i
=
1
c
-
2
y
i
+
y
c
-
1
I
-
R
-
1
e
=
1
.
Since R cannot be computed explicitly we explore the possibility of algorithmic computation. Thus, one can use logarithmic reduction algorithm as given by Latouche and Ramaswami [20] for computing R. We list here only the main steps involved in logarithmic reduction algorithm for computation of R.Logarithmic Reduction Algorithm for
RStep 0.  H
←
(
-
A
-
1
)
-
1
A
-
0, L
←
(
-
A
-
1
)
-
1
A
-
2, G
=
L, and T
=
H.Step 1. Consider
(53)
U
=
H
L
+
L
H
,
M
=
H
2
,
H
⟵
I
-
U
-
1
M
,
M
⟵
L
2
,
L
⟵
I
-
U
-
1
M
,
G
⟵
G
+
T
L
,
T
⟵
T
H
.
Continue Step 1 until e
-
G
e
∞
<
ϵ.Step 2. R
=
-
A
-
0
A
-
1
+
A
-
0
G
-
1.


## 6. Conditional Probability Distributions

We could arrive at an analytical expression for system state probabilities ofM
/
M
/
2 queueing-inventory system. However for the M
/
M
/
c queueing-inventory system with c
≥
3, the system state distribution does not seem to have closed form owing to the strong dependence between the inventory level, number of customers, and the number of servers in the system. In this section we provide conditional probabilities of the number of items in the inventory, given the number of customers in the system and also that of the number of customers in the system conditioned on the number of items in the inventory.


### 6.1. Conditional Probability Distribution of the Inventory Level Conditioned on the Number of Customers in the System

Letη
=
(
η
0
,
η
1
,
…
,
η
S
) be the probability distribution of the inventory level conditioned on the number of customers in the system. Then we get explicit form for the conditional probability distribution of the inventory level conditioned on the number of customers in the system. We formulate the result in the following lemma.Lemma 4.
Assume thati is the number of customers in the system at some point of time. Conditional on this we compute the inventory level distribution η
j where there are j items in the inventory. We consider two cases as follows. (i)
Wheni
<
c, the inventory level probability distribution is given by
(54)
η
0
=
1
+
∑
j
=
1
i
-
1
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
∏
k
=
1
i
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
-
1
llllll
+
Q
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
-
s
β
i
γ
μ
1
+
∑
j
=
0
s
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
lllllll
+
β
2
i
γ
μ
1
+
∑
j
=
1
s
-
2
∏
k
=
1
j
‍
β
+
k
γ
μ
k
γ
μ
-
1
,
η
j
=
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
f
o
r
1
≤
j
≤
i
,
β
+
i
γ
μ
i
γ
μ
j
-
i
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
f
o
r
i
+
1
≤
j
≤
s
+
1
,
η
j
+
1
,
f
o
r
s
+
1
≤
j
≤
Q
-
1
,
η
Q
+
j
=

∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
l
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
η
0
,
f
o
r
j
=
1
,
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
l
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
l
×
1
+
∑
i
=
0
j
-
2
∏
k
=
0
i
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
f
o
r
2
≤
j
≤
s
.
(ii)
Wheni
≥
c, the inventory level probability distribution is derived by
(55)
η
0
=
1
+
∑
j
=
1
c
-
1
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
k
+
1
γ
μ
llllll
+
∏
k
=
1
c
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
-
1
llllll
+
Q
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
k
+
1
γ
μ
llllll
-
s
β
c
γ
μ
1
+
∑
j
=
0
s
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
k
+
1
γ
μ
lllll
+
β
2
c
γ
μ
1
+
∑
j
=
1
s
-
2
∏
k
=
1
j
‍
β
+
k
γ
μ
k
γ
μ
-
1
,
η
j
=
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
k
+
1
γ
μ
η
0
,
f
o
r
1
≤
j
≤
c
,
β
+
c
γ
μ
c
γ
μ
j
-
c
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
k
+
1
γ
μ
η
0
,
f
o
r
c
+
1
≤
j
≤
s
+
1
,
η
j
+
1
,
f
o
r
s
+
1
≤
j
≤
Q
-
1
,
η
Q
+
j
=
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
η
0
,
f
o
r
j
=
1
,

1
+
∑
c
=
0
j
-
2
∏
k
=
0
c
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
l
×
1
+
∑
c
=
0
j
-
2
∏
k
=
0
c
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
f
o
r
2
≤
j
≤
s
.Proof.
LetΓ
1 be the infinitesimal generator of the corresponding Markov chain.(i)
Case ofi
<
c.The infinitesimal generator Γ
1 is given by(56)
Γ
1
=
0
1
2
⋮
i
⋮
c
⋮
s
⋮
Q
⋮
S
aaa
0
a
aaaa
1
aaaa
aaa
⋯
aaaaaa
aaaa
i
aaaa
⋯
aaaa
c
aaa
aaaa
⋯
aaa
aa
s
⋯
a
Q
aa
⋯
a
S
aaaa
-
β
β
γ
μ
-
(
γ
μ
+
β
)
2
γ
μ
-
(
2
γ
μ
+
β
)
⋱
⋱
⋱
i
γ
μ
-
(
i
γ
μ
+
β
)
⋱
⋱
i
γ
μ
-
(
i
γ
μ
+
β
)
⋱
⋱
i
γ
μ
-
(
i
γ
μ
+
β
)
β
i
γ
μ
-
i
γ
μ
⋱
⋱
i
γ
μ
-
i
γ
μ
i
γ
μ
-
i
γ
μ
,and the inventory level distribution η can be obtained from the equations η
Γ
1
=
0 and η
e
=
1, and we get(57)
η
j
=
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
1
≤
j
≤
i
,
β
+
i
γ
μ
i
γ
μ
j
-
i
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
i
+
1
≤
j
≤
s
+
1
,
η
j
+
1
,
for
s
+
1
≤
j
≤
Q
-
1
,
η
Q
+
j
=
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
l
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
η
0
,
for
j
=
1
,

1
+
∑
i
=
0
j
-
2
∏
k
=
0
i
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
l
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
l
×
1
+
∑
i
=
0
j
-
2
∏
k
=
0
i
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
2
≤
j
≤
s
,
where
(58)
η
0
=
1
+
∑
j
=
1
i
-
1
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
∏
k
=
1
i
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
-
1
llllll
+
Q
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
-
s
β
i
γ
μ
1
+
∑
j
=
0
s
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
lllllll
+
β
2
i
γ
μ
1
+
∑
j
=
1
s
-
2
∏
k
=
1
j
‍
β
+
k
γ
μ
k
γ
μ
-
1
.
(ii)
Case ofi
≥
c.The infinitesimal generator Γ
2 is given by(59)
Γ
2
=
0
1
2
⋮
c
⋮
i
⋮
s
⋮
Q
⋮
S
aaa
0
a
aaaa
1
aaaa
aaa
⋯
aaaaaa
aaaa
c
aaaa
⋯
aaaa
i
aaa
aaaa
⋯
aaa
aa
s
⋯
a
Q
aa
⋯
a
S
aaaa
-
β
β
γ
μ
-
(
γ
μ
+
β
)
2
γ
μ
-
(
2
γ
μ
+
β
)
⋱
⋱
⋱
c
γ
μ
-
(
c
γ
μ
+
β
)
⋱
⋱
c
γ
μ
-
(
c
γ
μ
+
β
)
⋱
⋱
c
γ
μ
-
(
c
γ
μ
+
β
)
β
c
γ
μ
-
c
γ
μ
⋱
⋱
c
γ
μ
-
c
γ
μ
c
γ
μ
-
c
γ
μ
.By solving the equations η
Γ
2
=
0 and η
e
=
1, we get
(60)
η
j
=
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
1
≤
j
≤
c
,
β
+
c
γ
μ
c
γ
μ
j
-
c
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
c
+
1
≤
j
≤
s
+
1
,
η
j
+
1
,
for
s
+
1
≤
j
≤
Q
-
1
,
η
Q
+
j
=

∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
η
0
,
for
j
=
1
,
1
+
∑
c
=
0
j
-
2
∏
k
=
0
c
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
l
×
1
+
∑
c
=
0
j
-
2
∏
k
=
0
c
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
2
≤
j
≤
s
,
where
(61)
η
0
=
1
+
∑
j
=
1
c
-
1
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
∏
k
=
1
c
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
-
1
llllll
+
Q
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
-
s
β
c
γ
μ
1
+
∑
j
=
0
s
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
β
2
c
γ
μ
1
+
∑
j
=
1
s
-
2
∏
k
=
1
j
‍
β
+
k
γ
μ
k
γ
μ
-
1
.


### 6.2. Conditional Probability Distribution of the Number of Customers Given the Number of Items in the Inventory

Letp
i,  i
≥
0, denote the probability that there are i customers in the system conditioned on the inventory level at j. We have three different cases.(i)
Whenj
=
0,
(62)
p
i
=
μ
γ
λ
+
μ
+
β
×
Prob
1
item
in
inventory
,
1
customer
×
Prob
and
the
inventory
is
served
+
Prob
No
item
in
inventory
,
×
Prob
i
customers
in
the
system
=
μ
γ
λ
+
μ
+
β
y
i
+
1
1
+
y
i
0
,
i
≥
0
.
(ii)
When0
<
j
<
c,
(63)
p
i
=
μ
γ
λ
+
μ
+
β
1
[
j
+
1
≤
s
]
y
1
(
j
+
1
)
+
μ
(
1
-
γ
)
λ
+
μ
+
β
1
[
j
≤
s
]
y
1
(
j
)
+
1
-
λ
λ
+
β
1
[
j
≤
s
]
y
0
(
j
)
,
i
=
0
,
λ
λ
+
(
i
-
1
)
μ
+
β
1
[
j
≤
s
]
y
i
-
1
(
j
)
+
(
i
+
1
)
μ
γ
λ
+
(
i
+
1
)
μ
+
β
1
[
j
+
1
≤
s
]
y
i
+
1
(
j
+
1
)
+
(
i
+
1
)
μ
(
1
-
γ
)
λ
+
(
i
+
1
)
μ
+
β
1
[
j
≤
s
]
y
i
+
1
(
j
)
+
1
-
λ
+
i
μ
λ
+
i
μ
+
β
1
[
j
≤
s
]
y
i
(
j
)
,
1
≤
i
<
j
,
λ
λ
+
j
μ
+
β
1
[
j
≤
s
]
y
i
-
1
(
j
)
+
(
j
+
1
)
μ
γ
λ
+
(
j
+
1
)
μ
+
β
1
[
j
+
1
≤
s
]
y
i
+
1
(
j
+
1
)
+
j
μ
(
1
-
γ
)
λ
+
j
μ
+
β
1
[
j
≤
s
]
y
i
+
1
(
j
)
+
1
-
λ
+
j
μ
λ
+
j
μ
+
β
1
[
j
≤
s
]
y
i
(
j
)
,
i
≥
j
.The first term on the right hand side of the case of1
≤
i
≤
j in (63) has two factors; the former represent probability of an arrival before service completion as well as replenishment when there were i
-
1 customers and j inventory in the system. Similar explanations stand for the remaining terms and also for other expressions for p
i.(iii)
Whenj
≥
c,
(64)
p
i
=
μ
γ
λ
+
μ
+
β
1
[
j
+
1
≤
s
]
y
1
(
j
+
1
)
+
μ
(
1
-
γ
)
λ
+
μ
+
β
1
[
j
≤
s
]
y
1
(
j
)
+
1
-
λ
λ
+
β
1
[
j
≤
s
]
y
0
(
j
)
+
β
1
[
j
>
Q
]
λ
+
β
1
[
j
>
Q
]
y
0
(
j
-
Q
)
,
i
=
0
,
λ
λ
+
(
i
-
1
)
μ
+
β
1
[
j
≤
s
]
y
i
-
1
(
j
)
+
(
i
+
1
)
μ
γ
λ
+
(
i
+
1
)
μ
+
β
1
[
j
+
1
≤
s
]
y
i
+
1
(
j
+
1
)
+
(
i
+
1
)
μ
(
1
-
γ
)
λ
+
(
i
+
1
)
μ
+
β
1
[
j
≤
s
]
y
i
+
1
(
j
)
+
1
-
λ
+
i
μ
λ
+
i
μ
+
β
1
[
j
≤
s
]
y
i
(
j
)
+
β
1
[
j
>
Q
]
λ
+
i
μ
+
β
1
[
j
>
Q
]
y
i
(
j
-
Q
)
,
1
≤
i
<
c
,
λ
λ
+
c
μ
+
β
1
[
j
≤
s
]
y
i
-
1
(
j
)
+
c
μ
γ
λ
+
c
μ
+
β
1
[
j
+
1
≤
s
]
y
i
+
1
(
j
+
1
)
+
c
μ
(
1
-
γ
)
λ
+
c
μ
+
β
1
[
j
≤
s
]
y
i
+
1
(
j
)
+
1
-
λ
+
c
μ
λ
+
c
μ
+
β
1
[
j
≤
s
]
y
i
(
j
)
+
β
1
[
j
>
Q
]
λ
+
c
μ
+
β
1
[
j
>
Q
]
y
i
(
j
-
Q
)
,
i
≥
c
,where 1
[
j
≤
s
] and 1
[
j
>
Q
] indicate whether the replenishment process is on.


### 6.3. Performance Measures

(i)
Mean number of customers in the system isL
s
=
∑
i
=
1
∞
∑
j
=
0
Q
+
s
i
y
i
(
j
).
(ii)
Mean number of customers in the queue isL
q
=
∑
i
=
c
+
1
∞
∑
j
=
0
Q
+
s
(
i
-
c
)
y
i
(
j
).
(iii)
Mean inventory level in the system isI
m
=
∑
i
=
0
∞
∑
j
=
1
Q
+
s
j
y
i
(
j
).
(iv)
Mean number of busy servers is as follows:(65)
P
BS
=
∑
k
=
1
c
‍
k
∑
i
=
k
+
1
∞
y
i
(
k
)
+
∑
j
=
k
+
1
Q
+
s
y
k
(
j
)
+
y
k
(
k
)
.
(v)
Mean number of idle servers isP
IS
=
c
-
∑
i
=
0
∞
y
i
(
0
).
(vi)
Depletion rate of inventory isD
inv
=
γ
μ
∑
i
=
1
∞
∑
j
=
1
Q
+
s
y
i
(
j
).
(vii)
Mean number of replenishments per time unit isR
r
=
β
∑
i
=
0
∞
∑
j
=
0
s
y
i
(
j
).
(viii)
Mean number of departures per unit time is as follows:(66)
D
m
=
∑
k
=
1
c
-
1
‍
k
μ
∑
i
=
k
∞
y
i
(
k
)
+
∑
j
=
k
Q
+
s
y
k
(
j
)
+
c
μ
∑
i
=
c
∞
‍
∑
j
=
c
Q
+
s
y
i
(
j
)
.
(ix)
Expected loss rate of customers isE
loss
=
λ
∑
i
=
0
∞
y
i
(
0
).
(x)
Expected loss rate of customers when the inventory level is zero per cycle isE
loss
c
=
E
loss
/
R
r.
(xi)
Mean number of customers arriving per unit time is as follows:(67)
λ
A
=
λ
∑
i
=
0
∞
∑
j
=
1
Q
+
s
y
i
(
j
)
.
(xii)
Mean sojourn time of the customers in the system isW
s
=
L
s
/
λ
A.
(xiii)
Mean waiting time of a customer in the queue isW
q
=
L
q
/
λ
A.
(xiv)
Mean number of customers waiting in the system when inventory is available isW
~
=
∑
i
=
1
∞
∑
j
=
1
Q
+
s
i
y
i
(
j
).
(xv)
Mean number of customers waiting in the system during the stock out period isW
~
~
=
∑
i
=
1
∞
i
y
i
(
0
).


## 6.1. Conditional Probability Distribution of the Inventory Level Conditioned on the Number of Customers in the System

Letη
=
(
η
0
,
η
1
,
…
,
η
S
) be the probability distribution of the inventory level conditioned on the number of customers in the system. Then we get explicit form for the conditional probability distribution of the inventory level conditioned on the number of customers in the system. We formulate the result in the following lemma.Lemma 4.
Assume thati is the number of customers in the system at some point of time. Conditional on this we compute the inventory level distribution η
j where there are j items in the inventory. We consider two cases as follows. (i)
Wheni
<
c, the inventory level probability distribution is given by
(54)
η
0
=
1
+
∑
j
=
1
i
-
1
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
∏
k
=
1
i
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
-
1
llllll
+
Q
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
-
s
β
i
γ
μ
1
+
∑
j
=
0
s
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
lllllll
+
β
2
i
γ
μ
1
+
∑
j
=
1
s
-
2
∏
k
=
1
j
‍
β
+
k
γ
μ
k
γ
μ
-
1
,
η
j
=
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
f
o
r
1
≤
j
≤
i
,
β
+
i
γ
μ
i
γ
μ
j
-
i
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
f
o
r
i
+
1
≤
j
≤
s
+
1
,
η
j
+
1
,
f
o
r
s
+
1
≤
j
≤
Q
-
1
,
η
Q
+
j
=

∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
l
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
η
0
,
f
o
r
j
=
1
,
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
l
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
l
×
1
+
∑
i
=
0
j
-
2
∏
k
=
0
i
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
f
o
r
2
≤
j
≤
s
.
(ii)
Wheni
≥
c, the inventory level probability distribution is derived by
(55)
η
0
=
1
+
∑
j
=
1
c
-
1
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
k
+
1
γ
μ
llllll
+
∏
k
=
1
c
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
-
1
llllll
+
Q
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
k
+
1
γ
μ
llllll
-
s
β
c
γ
μ
1
+
∑
j
=
0
s
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
k
+
1
γ
μ
lllll
+
β
2
c
γ
μ
1
+
∑
j
=
1
s
-
2
∏
k
=
1
j
‍
β
+
k
γ
μ
k
γ
μ
-
1
,
η
j
=
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
k
+
1
γ
μ
η
0
,
f
o
r
1
≤
j
≤
c
,
β
+
c
γ
μ
c
γ
μ
j
-
c
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
k
+
1
γ
μ
η
0
,
f
o
r
c
+
1
≤
j
≤
s
+
1
,
η
j
+
1
,
f
o
r
s
+
1
≤
j
≤
Q
-
1
,
η
Q
+
j
=
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
η
0
,
f
o
r
j
=
1
,

1
+
∑
c
=
0
j
-
2
∏
k
=
0
c
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
l
×
1
+
∑
c
=
0
j
-
2
∏
k
=
0
c
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
f
o
r
2
≤
j
≤
s
.Proof.
LetΓ
1 be the infinitesimal generator of the corresponding Markov chain.(i)
Case ofi
<
c.The infinitesimal generator Γ
1 is given by(56)
Γ
1
=
0
1
2
⋮
i
⋮
c
⋮
s
⋮
Q
⋮
S
aaa
0
a
aaaa
1
aaaa
aaa
⋯
aaaaaa
aaaa
i
aaaa
⋯
aaaa
c
aaa
aaaa
⋯
aaa
aa
s
⋯
a
Q
aa
⋯
a
S
aaaa
-
β
β
γ
μ
-
(
γ
μ
+
β
)
2
γ
μ
-
(
2
γ
μ
+
β
)
⋱
⋱
⋱
i
γ
μ
-
(
i
γ
μ
+
β
)
⋱
⋱
i
γ
μ
-
(
i
γ
μ
+
β
)
⋱
⋱
i
γ
μ
-
(
i
γ
μ
+
β
)
β
i
γ
μ
-
i
γ
μ
⋱
⋱
i
γ
μ
-
i
γ
μ
i
γ
μ
-
i
γ
μ
,and the inventory level distribution η can be obtained from the equations η
Γ
1
=
0 and η
e
=
1, and we get(57)
η
j
=
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
1
≤
j
≤
i
,
β
+
i
γ
μ
i
γ
μ
j
-
i
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
i
+
1
≤
j
≤
s
+
1
,
η
j
+
1
,
for
s
+
1
≤
j
≤
Q
-
1
,
η
Q
+
j
=
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
l
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
η
0
,
for
j
=
1
,

1
+
∑
i
=
0
j
-
2
∏
k
=
0
i
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
l
×
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
i
γ
μ
l
×
1
+
∑
i
=
0
j
-
2
∏
k
=
0
i
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
2
≤
j
≤
s
,
where
(58)
η
0
=
1
+
∑
j
=
1
i
-
1
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
∏
k
=
1
i
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
-
1
llllll
+
Q
β
+
i
γ
μ
i
γ
μ
s
+
1
-
i
∏
k
=
0
i
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
-
s
β
i
γ
μ
1
+
∑
j
=
0
s
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
lllllll
+
β
2
i
γ
μ
1
+
∑
j
=
1
s
-
2
∏
k
=
1
j
‍
β
+
k
γ
μ
k
γ
μ
-
1
.
(ii)
Case ofi
≥
c.The infinitesimal generator Γ
2 is given by(59)
Γ
2
=
0
1
2
⋮
c
⋮
i
⋮
s
⋮
Q
⋮
S
aaa
0
a
aaaa
1
aaaa
aaa
⋯
aaaaaa
aaaa
c
aaaa
⋯
aaaa
i
aaa
aaaa
⋯
aaa
aa
s
⋯
a
Q
aa
⋯
a
S
aaaa
-
β
β
γ
μ
-
(
γ
μ
+
β
)
2
γ
μ
-
(
2
γ
μ
+
β
)
⋱
⋱
⋱
c
γ
μ
-
(
c
γ
μ
+
β
)
⋱
⋱
c
γ
μ
-
(
c
γ
μ
+
β
)
⋱
⋱
c
γ
μ
-
(
c
γ
μ
+
β
)
β
c
γ
μ
-
c
γ
μ
⋱
⋱
c
γ
μ
-
c
γ
μ
c
γ
μ
-
c
γ
μ
.By solving the equations η
Γ
2
=
0 and η
e
=
1, we get
(60)
η
j
=
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
1
≤
j
≤
c
,
β
+
c
γ
μ
c
γ
μ
j
-
c
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
for
c
+
1
≤
j
≤
s
+
1
,
η
j
+
1
,
for
s
+
1
≤
j
≤
Q
-
1
,
η
Q
+
j
=

∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
η
0
,
for
j
=
1
,
1
+
∑
c
=
0
j
-
2
∏
k
=
0
c
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
l
×
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
-
β
c
γ
μ
l
×
1
+
∑
c
=
0
j
-
2
∏
k
=
0
c
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
η
0
,
2
≤
j
≤
s
,
where
(61)
η
0
=
1
+
∑
j
=
1
c
-
1
∏
k
=
0
j
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
∏
k
=
1
c
-
1
‍
β
+
k
γ
μ
k
γ
μ
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
-
1
llllll
+
Q
β
+
c
γ
μ
c
γ
μ
s
+
1
-
c
∏
k
=
0
c
-
1
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
-
s
β
c
γ
μ
1
+
∑
j
=
0
s
-
2
∏
k
=
0
j
‍
β
+
k
γ
μ
(
k
+
1
)
γ
μ
llllll
+
β
2
c
γ
μ
1
+
∑
j
=
1
s
-
2
∏
k
=
1
j
‍
β
+
k
γ
μ
k
γ
μ
-
1
.


## 6.2. Conditional Probability Distribution of the Number of Customers Given the Number of Items in the Inventory

Letp
i,  i
≥
0, denote the probability that there are i customers in the system conditioned on the inventory level at j. We have three different cases.(i)
Whenj
=
0,
(62)
p
i
=
μ
γ
λ
+
μ
+
β
×
Prob
1
item
in
inventory
,
1
customer
×
Prob
and
the
inventory
is
served
+
Prob
No
item
in
inventory
,
×
Prob
i
customers
in
the
system
=
μ
γ
λ
+
μ
+
β
y
i
+
1
1
+
y
i
0
,
i
≥
0
.
(ii)
When0
<
j
<
c,
(63)
p
i
=
μ
γ
λ
+
μ
+
β
1
[
j
+
1
≤
s
]
y
1
(
j
+
1
)
+
μ
(
1
-
γ
)
λ
+
μ
+
β
1
[
j
≤
s
]
y
1
(
j
)
+
1
-
λ
λ
+
β
1
[
j
≤
s
]
y
0
(
j
)
,
i
=
0
,
λ
λ
+
(
i
-
1
)
μ
+
β
1
[
j
≤
s
]
y
i
-
1
(
j
)
+
(
i
+
1
)
μ
γ
λ
+
(
i
+
1
)
μ
+
β
1
[
j
+
1
≤
s
]
y
i
+
1
(
j
+
1
)
+
(
i
+
1
)
μ
(
1
-
γ
)
λ
+
(
i
+
1
)
μ
+
β
1
[
j
≤
s
]
y
i
+
1
(
j
)
+
1
-
λ
+
i
μ
λ
+
i
μ
+
β
1
[
j
≤
s
]
y
i
(
j
)
,
1
≤
i
<
j
,
λ
λ
+
j
μ
+
β
1
[
j
≤
s
]
y
i
-
1
(
j
)
+
(
j
+
1
)
μ
γ
λ
+
(
j
+
1
)
μ
+
β
1
[
j
+
1
≤
s
]
y
i
+
1
(
j
+
1
)
+
j
μ
(
1
-
γ
)
λ
+
j
μ
+
β
1
[
j
≤
s
]
y
i
+
1
(
j
)
+
1
-
λ
+
j
μ
λ
+
j
μ
+
β
1
[
j
≤
s
]
y
i
(
j
)
,
i
≥
j
.The first term on the right hand side of the case of1
≤
i
≤
j in (63) has two factors; the former represent probability of an arrival before service completion as well as replenishment when there were i
-
1 customers and j inventory in the system. Similar explanations stand for the remaining terms and also for other expressions for p
i.(iii)
Whenj
≥
c,
(64)
p
i
=
μ
γ
λ
+
μ
+
β
1
[
j
+
1
≤
s
]
y
1
(
j
+
1
)
+
μ
(
1
-
γ
)
λ
+
μ
+
β
1
[
j
≤
s
]
y
1
(
j
)
+
1
-
λ
λ
+
β
1
[
j
≤
s
]
y
0
(
j
)
+
β
1
[
j
>
Q
]
λ
+
β
1
[
j
>
Q
]
y
0
(
j
-
Q
)
,
i
=
0
,
λ
λ
+
(
i
-
1
)
μ
+
β
1
[
j
≤
s
]
y
i
-
1
(
j
)
+
(
i
+
1
)
μ
γ
λ
+
(
i
+
1
)
μ
+
β
1
[
j
+
1
≤
s
]
y
i
+
1
(
j
+
1
)
+
(
i
+
1
)
μ
(
1
-
γ
)
λ
+
(
i
+
1
)
μ
+
β
1
[
j
≤
s
]
y
i
+
1
(
j
)
+
1
-
λ
+
i
μ
λ
+
i
μ
+
β
1
[
j
≤
s
]
y
i
(
j
)
+
β
1
[
j
>
Q
]
λ
+
i
μ
+
β
1
[
j
>
Q
]
y
i
(
j
-
Q
)
,
1
≤
i
<
c
,
λ
λ
+
c
μ
+
β
1
[
j
≤
s
]
y
i
-
1
(
j
)
+
c
μ
γ
λ
+
c
μ
+
β
1
[
j
+
1
≤
s
]
y
i
+
1
(
j
+
1
)
+
c
μ
(
1
-
γ
)
λ
+
c
μ
+
β
1
[
j
≤
s
]
y
i
+
1
(
j
)
+
1
-
λ
+
c
μ
λ
+
c
μ
+
β
1
[
j
≤
s
]
y
i
(
j
)
+
β
1
[
j
>
Q
]
λ
+
c
μ
+
β
1
[
j
>
Q
]
y
i
(
j
-
Q
)
,
i
≥
c
,where 1
[
j
≤
s
] and 1
[
j
>
Q
] indicate whether the replenishment process is on.


## 6.3. Performance Measures

(i)
Mean number of customers in the system isL
s
=
∑
i
=
1
∞
∑
j
=
0
Q
+
s
i
y
i
(
j
).
(ii)
Mean number of customers in the queue isL
q
=
∑
i
=
c
+
1
∞
∑
j
=
0
Q
+
s
(
i
-
c
)
y
i
(
j
).
(iii)
Mean inventory level in the system isI
m
=
∑
i
=
0
∞
∑
j
=
1
Q
+
s
j
y
i
(
j
).
(iv)
Mean number of busy servers is as follows:(65)
P
BS
=
∑
k
=
1
c
‍
k
∑
i
=
k
+
1
∞
y
i
(
k
)
+
∑
j
=
k
+
1
Q
+
s
y
k
(
j
)
+
y
k
(
k
)
.
(v)
Mean number of idle servers isP
IS
=
c
-
∑
i
=
0
∞
y
i
(
0
).
(vi)
Depletion rate of inventory isD
inv
=
γ
μ
∑
i
=
1
∞
∑
j
=
1
Q
+
s
y
i
(
j
).
(vii)
Mean number of replenishments per time unit isR
r
=
β
∑
i
=
0
∞
∑
j
=
0
s
y
i
(
j
).
(viii)
Mean number of departures per unit time is as follows:(66)
D
m
=
∑
k
=
1
c
-
1
‍
k
μ
∑
i
=
k
∞
y
i
(
k
)
+
∑
j
=
k
Q
+
s
y
k
(
j
)
+
c
μ
∑
i
=
c
∞
‍
∑
j
=
c
Q
+
s
y
i
(
j
)
.
(ix)
Expected loss rate of customers isE
loss
=
λ
∑
i
=
0
∞
y
i
(
0
).
(x)
Expected loss rate of customers when the inventory level is zero per cycle isE
loss
c
=
E
loss
/
R
r.
(xi)
Mean number of customers arriving per unit time is as follows:(67)
λ
A
=
λ
∑
i
=
0
∞
∑
j
=
1
Q
+
s
y
i
(
j
)
.
(xii)
Mean sojourn time of the customers in the system isW
s
=
L
s
/
λ
A.
(xiii)
Mean waiting time of a customer in the queue isW
q
=
L
q
/
λ
A.
(xiv)
Mean number of customers waiting in the system when inventory is available isW
~
=
∑
i
=
1
∞
∑
j
=
1
Q
+
s
i
y
i
(
j
).
(xv)
Mean number of customers waiting in the system during the stock out period isW
~
~
=
∑
i
=
1
∞
i
y
i
(
0
).


## 7. Analysis of Inventory Cycle Time

We define the inventory cycle time random variable,Γ
cycle
⁡, as the time interval between two consecutive instants at which the inventory level drops to s. Thus Γ
cycle
⁡ is a random variable whose distribution depends on the number of customers at the time when inventory level dropped to s at the beginning of the cycle and the inventory level process prior to replenishment.We proceed with the assumption that
γ
=
1. If the number of customers present in the system is at least Q
+
c when the order for replenishment is placed, then we need not have to look at future arrivals to get a nice form for the cycle time distribution. In fact it is sufficient that there are at least Q customers at that epoch. However in this case the service rate during lead time may drop below c
μ even when there are at least c items in the inventory. This is so since number of customers may go below c. Thus we look at various possibilities below.


### 7.1. When the Number of Customersl
≥
Q
+
c

When the number of customers is at leastQ
+
c, future arrivals need not be considered. The service rate of the M
/
M
/
c queueing-inventory system depends on the number of customers, number of servers and number of items in the inventory. Thus we consider the following cases.Case 1 (replenishment occurs before inventory level hitsc
-
1).
We consider the state(
l
,
s
) as the starting state; thus the inventory level decreases from s to a particular level s
-
k, for k varying from 0 to s
-
c due to service completion at rate c
μ, during the lead time. At level s
-
k, the replenishment occurs and it is absorbed to Δ
1, where the absorbing state is defined as Δ
1
=
(
l
-
k
,
Q
+
s
-
k
)
∣
0
≤
k
≤
s
-
c. Therefore the time until absorption to Δ
1 follows Erlang distribution of order k with parameter c
μ, and it is denoted by E
(
c
μ
;
k
). Now, the number of customers in the system is l
-
k or larger with the corresponding inventory level Q
+
s
-
k, for k varying from 0 to s
-
c. Similarly, the inventory level reaches s from Q
+
s
-
k with Q
-
k service completions all of which have rate c
μ. This time duration also follows Erlang distribution of order Q
-
k. Write this as E
(
c
μ
;
Q
-
k
). Thus under the condition that there are at least Q
+
c customers at the beginning of the cycle and that the inventory level does not fall below c, the inventory cycle time, Γ
cycle
⁡, has Erlang distribution of order Q with parameter c
μ. That is,
(68)
Γ
cycle
⁡
~
E
(
c
μ
;
k
)
*
E
(
c
μ
;
Q
-
k
)
~
E
c
μ
;
Q
,
where the symbol “~” stands for “having distribution.” The probability of replenishment taking place before inventory level that drops to c
-
1 is given by ∫
0
∞
∑
k
=
0
s
-
c
e
-
μ
v
μ
v
k
β
e
-
β
v
/
k
!
d
v.Case 2 (replenishment after hittingc
-
1 but not zero).
The inventory level decreases froms to k, when k varies from 1 to c
-
1. The first s
-
c
+
1 services are at the same rate c
μ. Thereafter it slows down to (
c
-
1
)
μ and finally to k
μ, when replenishment occurs. Consequently the inventory level rises to Q
+
k. Now here onwards the service rate stays at c
μ. Thus in the cycle, the distribution of the time until replenishment takes place is the convolution of generalized Erlang distribution and that of an Erlang distribution E
(
Q
+
k
-
s
;
c
μ
). The conditional distribution of replenishment realization, after s
-
k
-
1 service is completed, but before (
s
-
k
)th is completed, can be computed as in Case 1. At the same level s
-
k, the replenishment will occur and it is absorbed to Δ
2, where the absorbing state is defined as Δ
2
=
(
l
-
(
s
-
k
)
,
Q
+
k
)
∣
1
≤
k
≤
c
-
1. Thus, the time until absorption to Δ
2 follows generalized Erlang distribution with parameters c
μ
,
(
c
-
1
)
μ
,
…
,
(
k
+
1
)
μ of order s
-
k and k vary from 1 to c
-
1. It is denoted by G
E
c
μ
,
(
c
-
1
)
μ
,
…
,
(
k
+
1
)
μ
;
s
-
k. Then from Δ
2 the inventory level reaches s due to service completion with parameter c
μ. Thus the time duration follows Erlang distribution with parameter c
μ of order Q
+
k
-
s, with k varying from 1 to c
-
1. That is, E
c
μ
;
Q
+
k
-
s. Hence the inventory cycle time, Γ
cycle
⁡, follows generalized Erlang distribution of order Q. Therefore, Γ
cycle
⁡ is defined as
(69)
Γ
cycle
⁡
~
G
E
c
μ
,
…
,
c
μ
︸
s
-
c
+
1
times
,
c
-
1
μ
,
…
,
k
+
1
μ
;
s
-
k
*
E
c
μ
;
Q
+
k
-
s
,
where G
E
(
·
) stands for generalized Erlang distribution.Case 3 (replenishment after inventory level reaching zero).
Then the inventory level reaches0 from the level s due to service completion with parameters c
μ (repeated s
-
c
+
1 times). Thus the time until absorption to Δ
3
=
(
l
-
s
,
Q
) follows generalized Erlang distribution of order s and parameters c
μ
,
(
c
-
1
)
μ
,
…
,
μ. When the inventory level hits 0, the system becomes idle for a random duration of time which follows exponential distribution with parameter β. After replenishment, the system starts service and consequently the inventory level reaches s from Q due to service completion with parameter c
μ. This part has Erlang distribution with parameter c
μ and order Q
-
s. Thus, Γ
cycle
⁡ follows generalized Erlang distribution of order Q. That is,
(70)
Γ
cycle
⁡
~
G
E
c
μ
,
…
,
c
μ
︸
s
-
c
+
1
times
,
(
c
-
1
)
μ
,
…
,
μ
;
s
*
exp
⁡
β
*
E
c
μ
;
Q
-
s
.
The cases we are going to consider hereafter result in cycle time distribution that are phase type with not necessarily unique representation. However, one can sort out the problem of minimal representation. Obviously this is the one which considers that many arrivals are needed to have exactly
Q
services in this cycle.


### 7.2. When the Number of Customersl
<
Q
+
c

In this case we may have to consider future arrivals as well, since number of customers available at the start of the cycle may be such that the service rate falls belowc
μ. Thus the cycle time will have more general distribution, namely, the phase type. We go about doing this. Our procedure is such that the moment we have enough customers to serve during the remaining part of the cycle, we stop considering future arrivals. Thus consider a Markov chain on the state space
(71)
A
s
,
l
,
s
-
1
,
l
-
1
,
…
,
0
,
l
-
s
,
s
,
l
+
1
,
l
l
s
-
1
,
l
,
…
,
0
,
l
-
s
+
1
,
…
,
s
+
Q
,
l
,
…
,
l
l
s
+
Q
-
l
,
0
,
s
+
Q
,
l
+
1
,
…
,
s
,
l
,
…
,
l
l
s
+
Q
,
s
+
Q
-
l
-
1
,
…
,
s
+
Q
-
1
,
l
-
1
,
l
s
+
Q
,
s
+
Q
-
l
,
…
,
s
,
s
-
l
-
1
A
.
The initial state (
s
,
l
): thus the initial probability vector will have one at the position corresponding to (
s
,
l
) and the rest of the elements zero. The absorption state in this Markov chain is (
s
,
*
), where * belongs to 0,1
,
2
,
…
,
Q
+
l
-
s and is a departure epoch. Let T be the block with transitions among transient states and let T
* be the column vector with transition rates to the absorbing states as elements. Then the cycle time has distribution 1
-
α
e
T
t
e where α is the initial probability vector with 1 at the position indicating the inventory level s as first coordinate and the number of customers (=l) at the beginning of the cycle as second coordinate. Note that the phase type representation obtained is not unique since the service rate strongly depends on both inventory level and number of customers in the system. The case of l
<
s: here again the procedure is similar to that corresponding to l
≥
s, but less than Q
+
c. The initial state is (
s
,
l
). After exactly Q service completions with a replenishment within this cycle and with arrivals truncated at that epoch which ensure rate c
μ for as many services as possible, the absorption state of the Markov chain generated corresponds to a departure epoch with s items in the inventory. Here again the cycle time has a PH distribution with representation which is not unique because the service rates may change depending on the number of customers in the system and the number of items in the inventory.


## 7.1. When the Number of Customersl
≥
Q
+
c

When the number of customers is at leastQ
+
c, future arrivals need not be considered. The service rate of the M
/
M
/
c queueing-inventory system depends on the number of customers, number of servers and number of items in the inventory. Thus we consider the following cases.Case 1 (replenishment occurs before inventory level hitsc
-
1).
We consider the state(
l
,
s
) as the starting state; thus the inventory level decreases from s to a particular level s
-
k, for k varying from 0 to s
-
c due to service completion at rate c
μ, during the lead time. At level s
-
k, the replenishment occurs and it is absorbed to Δ
1, where the absorbing state is defined as Δ
1
=
(
l
-
k
,
Q
+
s
-
k
)
∣
0
≤
k
≤
s
-
c. Therefore the time until absorption to Δ
1 follows Erlang distribution of order k with parameter c
μ, and it is denoted by E
(
c
μ
;
k
). Now, the number of customers in the system is l
-
k or larger with the corresponding inventory level Q
+
s
-
k, for k varying from 0 to s
-
c. Similarly, the inventory level reaches s from Q
+
s
-
k with Q
-
k service completions all of which have rate c
μ. This time duration also follows Erlang distribution of order Q
-
k. Write this as E
(
c
μ
;
Q
-
k
). Thus under the condition that there are at least Q
+
c customers at the beginning of the cycle and that the inventory level does not fall below c, the inventory cycle time, Γ
cycle
⁡, has Erlang distribution of order Q with parameter c
μ. That is,
(68)
Γ
cycle
⁡
~
E
(
c
μ
;
k
)
*
E
(
c
μ
;
Q
-
k
)
~
E
c
μ
;
Q
,
where the symbol “~” stands for “having distribution.” The probability of replenishment taking place before inventory level that drops to c
-
1 is given by ∫
0
∞
∑
k
=
0
s
-
c
e
-
μ
v
μ
v
k
β
e
-
β
v
/
k
!
d
v.Case 2 (replenishment after hittingc
-
1 but not zero).
The inventory level decreases froms to k, when k varies from 1 to c
-
1. The first s
-
c
+
1 services are at the same rate c
μ. Thereafter it slows down to (
c
-
1
)
μ and finally to k
μ, when replenishment occurs. Consequently the inventory level rises to Q
+
k. Now here onwards the service rate stays at c
μ. Thus in the cycle, the distribution of the time until replenishment takes place is the convolution of generalized Erlang distribution and that of an Erlang distribution E
(
Q
+
k
-
s
;
c
μ
). The conditional distribution of replenishment realization, after s
-
k
-
1 service is completed, but before (
s
-
k
)th is completed, can be computed as in Case 1. At the same level s
-
k, the replenishment will occur and it is absorbed to Δ
2, where the absorbing state is defined as Δ
2
=
(
l
-
(
s
-
k
)
,
Q
+
k
)
∣
1
≤
k
≤
c
-
1. Thus, the time until absorption to Δ
2 follows generalized Erlang distribution with parameters c
μ
,
(
c
-
1
)
μ
,
…
,
(
k
+
1
)
μ of order s
-
k and k vary from 1 to c
-
1. It is denoted by G
E
c
μ
,
(
c
-
1
)
μ
,
…
,
(
k
+
1
)
μ
;
s
-
k. Then from Δ
2 the inventory level reaches s due to service completion with parameter c
μ. Thus the time duration follows Erlang distribution with parameter c
μ of order Q
+
k
-
s, with k varying from 1 to c
-
1. That is, E
c
μ
;
Q
+
k
-
s. Hence the inventory cycle time, Γ
cycle
⁡, follows generalized Erlang distribution of order Q. Therefore, Γ
cycle
⁡ is defined as
(69)
Γ
cycle
⁡
~
G
E
c
μ
,
…
,
c
μ
︸
s
-
c
+
1
times
,
c
-
1
μ
,
…
,
k
+
1
μ
;
s
-
k
*
E
c
μ
;
Q
+
k
-
s
,
where G
E
(
·
) stands for generalized Erlang distribution.Case 3 (replenishment after inventory level reaching zero).
Then the inventory level reaches0 from the level s due to service completion with parameters c
μ (repeated s
-
c
+
1 times). Thus the time until absorption to Δ
3
=
(
l
-
s
,
Q
) follows generalized Erlang distribution of order s and parameters c
μ
,
(
c
-
1
)
μ
,
…
,
μ. When the inventory level hits 0, the system becomes idle for a random duration of time which follows exponential distribution with parameter β. After replenishment, the system starts service and consequently the inventory level reaches s from Q due to service completion with parameter c
μ. This part has Erlang distribution with parameter c
μ and order Q
-
s. Thus, Γ
cycle
⁡ follows generalized Erlang distribution of order Q. That is,
(70)
Γ
cycle
⁡
~
G
E
c
μ
,
…
,
c
μ
︸
s
-
c
+
1
times
,
(
c
-
1
)
μ
,
…
,
μ
;
s
*
exp
⁡
β
*
E
c
μ
;
Q
-
s
.
The cases we are going to consider hereafter result in cycle time distribution that are phase type with not necessarily unique representation. However, one can sort out the problem of minimal representation. Obviously this is the one which considers that many arrivals are needed to have exactly
Q
services in this cycle.


## 7.2. When the Number of Customersl
<
Q
+
c

In this case we may have to consider future arrivals as well, since number of customers available at the start of the cycle may be such that the service rate falls belowc
μ. Thus the cycle time will have more general distribution, namely, the phase type. We go about doing this. Our procedure is such that the moment we have enough customers to serve during the remaining part of the cycle, we stop considering future arrivals. Thus consider a Markov chain on the state space
(71)
A
s
,
l
,
s
-
1
,
l
-
1
,
…
,
0
,
l
-
s
,
s
,
l
+
1
,
l
l
s
-
1
,
l
,
…
,
0
,
l
-
s
+
1
,
…
,
s
+
Q
,
l
,
…
,
l
l
s
+
Q
-
l
,
0
,
s
+
Q
,
l
+
1
,
…
,
s
,
l
,
…
,
l
l
s
+
Q
,
s
+
Q
-
l
-
1
,
…
,
s
+
Q
-
1
,
l
-
1
,
l
s
+
Q
,
s
+
Q
-
l
,
…
,
s
,
s
-
l
-
1
A
.
The initial state (
s
,
l
): thus the initial probability vector will have one at the position corresponding to (
s
,
l
) and the rest of the elements zero. The absorption state in this Markov chain is (
s
,
*
), where * belongs to 0,1
,
2
,
…
,
Q
+
l
-
s and is a departure epoch. Let T be the block with transitions among transient states and let T
* be the column vector with transition rates to the absorbing states as elements. Then the cycle time has distribution 1
-
α
e
T
t
e where α is the initial probability vector with 1 at the position indicating the inventory level s as first coordinate and the number of customers (=l) at the beginning of the cycle as second coordinate. Note that the phase type representation obtained is not unique since the service rate strongly depends on both inventory level and number of customers in the system. The case of l
<
s: here again the procedure is similar to that corresponding to l
≥
s, but less than Q
+
c. The initial state is (
s
,
l
). After exactly Q service completions with a replenishment within this cycle and with arrivals truncated at that epoch which ensure rate c
μ for as many services as possible, the absorption state of the Markov chain generated corresponds to a departure epoch with s items in the inventory. Here again the cycle time has a PH distribution with representation which is not unique because the service rates may change depending on the number of customers in the system and the number of items in the inventory.


## 8. Optimization Problem II

We look for the optimal pair of control variables in the model discussed above. Now for computing the minimal cost of(
s
,
Q
) model we introduce the cost function: F
(
c
,
s
,
Q
), which is defined by
(72)
F
c
,
s
,
Q
=
h
·
I
m
+
c
1
·
E
loss
+
c
2
·
W
~
~
+
K
+
Q
·
c
3
·
R
r
+
c
4
·
P
BS
+
c
5
·
c
-
P
BS
,
where s
=
40, S
=
81, and K, c
1, c
2, c
3, c
4, c
5, and h are the same input parameters as described in Section 4. We provide optimal c and corresponding minimum cost for various γ values. From Table 2 we notice that the optimal value of c is 6 for various γ values, presumably because of the high holding cost.Table 2
Optimal serverc and minimum cost.
γ
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Optimalc and minimum cost
6
6
6
6
6
6
6
6
6
6
148.78
166.36
183.93
201.51
219.09
236.66
254.24
271.82
289.40
306.98In Table3, we examine the optimal pair (
s
,
Q
) and the corresponding minimum cost for various γ and c, keeping other parameters fixed (as in Section 4).Table 3
Optimal(
s
,
Q
) values and minimum cost.
c
γ
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
3
(4, 12)
(4, 15)
(4, 19)
(4, 23)
(4, 27)
(4, 31)
(4, 37)
(4, 39)
(4, 42)
(4, 45)
95.857
113.57
131.86
151.49
171.55
191.59
211.50
230.82
249.87
268.55
4
(5, 15)
(5, 19)
(5, 23)
(5, 27)
(5, 31)
(5, 34)
(5, 38)
(5, 41)
(5, 45)
(5, 48)
126.43
147.25
166.70
186.06
205.42
224.69
243.74
262.55
281.05
299.25
5
(6, 16)
(6, 22)
(6, 26)
(6, 31)
(6, 34)
(6, 38)
(6, 42)
(6, 45)
(6, 48)
(6, 52)
154.11
177.53
198.43
218.40
237.91
257.07
275.93
294.49
312.75
330.72
6
(7, 16)
(7, 23)
(7, 28)
(7, 32)
(7, 36)
(7, 40)
(7, 43)
(7, 46)
(7, 49)
(7, 53)
177.90
202.28
223.79
244.06
263.65
282.79
301.56
320.00
338.13
355.97
7
(8, 16)
(8, 23)
(8, 28)
(8, 32)
(8, 36)
(8, 40)
(8, 43)
(8, 46)
(8, 49)
(8, 53)
200.30
224.90
246.58
266.91
286.47
305.52
324.18
342.49
360.51
378.24
8
(9, 16)
(9, 23)
(9, 28)
(9, 32)
(9, 36)
(9, 40)
(9, 43)
(9, 46)
(9, 49)
(9, 53)
222.30
246.98
268.68
288.97
308.45
329.01
345.92
364.12
382.02
399.65
9
(10, 16)
(10, 23)
(10, 28)
(10, 32)
(10, 36)
(10, 40)
(10, 43)
(10, 46)
(10, 49)
(10, 53)
244.29
268.97
290.65
310.87
330.25
349.08
367.49
385.57
403.36
420.89
10
(11, 16)
(11, 23)
(11, 28)
(11, 32)
(11, 36)
(11, 40)
(11, 43)
(11, 46)
(11, 49)
(11, 53)
266.27
290.94
312.60
332.76
352.06
370.78
389.08
407.05
424.73
442.17


## 9. Conclusions

In this paper we studied multiserver queueing-inventory system with positive service time. First we considered two server queueing-inventory systems, where the steady-state distributions are obtained in product form. Further, we analyse queueing-inventory system with more than two servers. As observed in [21] by Falin and Templeton, we conjecture that M
/
M
/
c for c
≥
3, queueing-inventory system, does not have analytical solution. So such cases are analyzed by algorithmic approach. Conditional distribution of the inventory level, conditioned on the number of customers in the system, and conditional distribution of the number of customers, conditioned on the inventory level, are derived. We also provided the cycle time distribution of two consecutive s to s transitions of the inventory level (i.e., the first return time to s). We have computed the optimal number of servers to be employed and also computed optimal (
s
,
Q
) pair values and the corresponding minimum cost.



---

*Source: 747328-2015-01-20.xml*