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Proceedings of the Eighteenth Ramanujan Symposium On
Recent Trends in Dynamical Systems and Mathematical Modelling,
University of Madras, Chennai,
25–27, September, 2013 Volume 18 (2014), 57–64
Production Inventory with Positive Service Time and Loss
A. Krishnamoorthy, R. Manikandan, and B. Lakshmi
Abstract. A production inventory system, with the item produced being admitted (added to
the inventory) with probability δas well as an item from the inventory supplied to the customer
with probability γat the end of a service, is considered in this paper. The control policy followed
is of the (s, S) type. We obtain joint distribution of the number of customers and the number of
items in the inventory as the product of their marginals under the assumption that customers do
not join when inventory level is zero. Performance measures that impact the system are obtained.
Keywords. Production inventory, Positive service time, Stochastic decomposition.
2010 MSC. 60J10, 60K20, 60K25, 90B22, 90B05.
1. Introduction
Sigman and Levi [8] were the first to introduce inventory with arbitrarily distributed service time.
The lead time for inventory replenishment is exponentially distributed. This was followed by Berman
et al. [1] with deterministic service time wherein they formulated the model as a dynamic programming
problem, and a few other papers by various researchers. The results in Schwarz et al. [7] requires
special mention since under exponentially distributed service time as well as lead time and Poisson
input of customers, the authors come up with product form solution for the system state distribution
where customers are not allowed to join when inventory level is zero (of course Saffari et al. [6] is the
extension of this to arbitrary distributed lead time). This is despite the strong correlation between the
number of customers joining the system and the lead time. Their work is subsumed in Krishnamoorthy
and Viswanath [5]. It may be noted that very few work has been so far reported in production inventory
with positive service time. The first work on production inventory with service time could be atributed
to Krishnamoorthy and Viswanath [2]. A brief survey of inventory with positive service time is given
in Krishnamoorthy et al. [3]. Protection of production and service stages in a queueing-inventory
model, with Erlang distributed service and inter-production time is analyzed in Krishnamoorthy et al.
[4].
In all work quoted above, customers are provided an item from the inventory on completion of
service. Nevertheless, in reality a customer need not be served the item with probability one at the
end of his service. For example a candidate appears for an interview against a position. At the end
of the interview he/she may not be offered the position. In other words he is offered the job with
probability γ. Also there are cases where the candidate declines the offer of the job. In this case the
job is taken as an inventory. In this paper we analyze such type of situations under Poisson demand,
ISBN 978-81-929362-0-8
58 A. Krishnamoorthy, R. Manikandan, and B. Lakshmi
exponentially distributed service time and the time for producing an item has exponential distribution.
We further impose the condition that no customer joins when the on-hand inventory is zero. Further
an item produced is accepted with probability δand with complementary probability it is rejected.
Thus this paper further generalizes the work reported in Krishnamoorthy and Vishwanath [5].
We arrange the presentation of this paper as indicated below: Section 2 provides the mathematical
formulation of the problem under study. The analysis of the system is carried out in Section 3. In
particular, we derive the long run stability of the system. Then, under this condition we show that the
system state can be decomposed. Next we compute important system performance measures. Further,
in order to construct an appropriate cost function, we provide the expected length of a production cycle
in Section 3.3. Finally we discuss the first emptiness time distribution for the M/M/1/1 queueing-
inventory system with production.
Notations:
The following notations and abbreviations are used in the sequel.
N(t): Number of customers in the system at time t.
I(t): Inventory level in the system at time t.
P(t): Status of the production process:
That is, P(t) = (0,if production is off at time t.
1,if production is on at time t.
Ik: Identity matrix of order k.
e: (1,1, ..., 1)′a column vector of 1’s of appropriate order.
CTMC: Continuous time Markov chain.
LIQBD: Level independent quasi birth and death process.
2. Description of the model
We consider an (s, S) production inventory system with a single server. Demands by customers for
the item occur according to a Poisson process of rate λ. Processing of the customer request requires a
random amount of time, which is exponentially distributed with parameter µ. However, as assumed in
the previous chapter it is not essential that the item from inventory is provided to the customer at the
end of a service. More precisely, an item from inventory is provided to a customer with probability γat
the end of his service and with probability 1 −γthe customer leaves the system empty handed. When
the inventory level depletes to s, the production process is immediately switched on. Each production
is of 1 unit and the production process is kept in the on mode until inventory level becomes S. To
produce an item it takes an amount of time which is exponentially distributed with parameter β. A
produced item is not necessarily added to the inventory due to manufacturing defect: with probability
δit is accepted and with probability 1 −δthe item is rejected. We assume that no customer is
allowed to join the queue when the inventory level is zero; such demands are considered as lost. It is
assumed that the amount of time for the item produced to reach the retail shop is negligible. Thus
the system is a CTMC {X (t); t≥0}={(N(t),I(t),P(t)) ; t≥0}.The production process is in on
mode if 0 ≤ I(t)≤sand it is in off mode if I(t) = S; but when the inventory level lies between
Production Inventory with Positive Service Time and Loss 59
s+ 1 and S−1, P(t) is either 0 or 1 according as the production is in off or in on mode, respectively.
Thus to describe the status of the process we need write P(t) = 0 or 1 only when I(t) takes values
s+ 1,...,S −1. Thus the state space of the CTMC is Ω=
∞
S
i=0
L(i),where L(i), called level iof the
CTMC, is given by, {(i, j); 0 ≤j≤s}S{(i, j, k); s+ 1 ≤j≤S−1, k = 0,1}S{(i, S )},∀i≥0.The
number of states within ith level is 2S−s. This determines the order of the matrix entries in the
infinitesimal generator given below:
W=
B1A0
A2A1A0
A2A1A0. . .
.........
.
The block matrices appearing on the right side above are explained below:
B1=
−δβ δβ
−(λ+δβ)δβ
...
...
−(λ+δβ)δβ
−(λ+δβ)V1
U V2
U V2
...
...
U V2
U V3
−λ
with U="−λ0
0−(λ+δβ)#, V1=h0δ β i, V2="0 0
0δβ #, V3="0
δβ #; entries of B1correspond-
ing to transition rates within level 0.
A0="00
0λI(2S−s)−1#, A1=B1−µ
λA0
and
A2=
0. . . . . . 0
γµ (1 −γ)µ
...
...
γµ (1 −γ)µ
F1F2
F3F2
F3F2
...
...
F3F20
F4(1 −γ)µ
with F1="γµ
γµ #, F2="(1 −γ)µ0
0 (1 −γ)µ#, F3="γµ 0
0γµ #
and F4=h0γµ i.
60 A. Krishnamoorthy, R. Manikandan, and B. Lakshmi
3. Analysis of the system
We first establish the stability condition of the system. Define A=A0+A1+A2. Let ϕdenote the
steady-state probability vector of A. That is ϕsatisfies
ϕA= 0,ϕe = 1.(3.1)
Using the above relations, we get the components of the probability vector ϕexplicitly as:
ϕ(s−i) = ϕ(S)γµ
δβ −γµ 1−γµ
δβ S−s!γµ
δβ i
,0≤i≤s,
ϕ(i, 0) = ϕ(S), s + 1 ≤i≤S−1,
ϕ(i, 1) = ϕ(S)γµ
δβ −γµ 1−γµ
δβ S−i!, s + 1 ≤i≤S−1.
and the unknown probability
ϕ(S) = γµ
δβ −12
γµ
δβ S+2 −γµ
δβ s+2 −(S−s)γµ
δβ −1.
Since the Markov chain under study is an LIQBD process, it is stable if and only if the left drift rate
exceeds the right drift rate. That is,
ϕA0e<ϕA2e.(3.2)
We have the following lemma:
Lemma 3.1. The stability condition of the (s, S)production inventory model is given by λ < µ.
Proof. From the well known result in Neuts [9] on the positive recurrence of A, we have ϕA0e<
ϕA2e.With a bit of computation, this simplifies to the result λ < µ. For future reference we define ρ
as
ρ=λ
µ.(3.3)
3.1. Steady-state analysis. For computing the steady-state probability vector of the process {X (t); t≥
0}, consider a production inventory system with negligible service time with no backlog of customers.
The rest of the assumptions such as those on the arrival process and lead time are the same as given
earlier. Designate this Markov chain so obtained as {e
X(t); t≥0}={(I(t),P(t)) ; t≥0}. Its infinitesi-
mal generator f
Wis given by,
g
W=
−δβ δβ
γλ −(γλ +δ β)δβ
...
...
...
γλ −(γλ +δ β)δβ
γλ −(γλ +δ β)V1
ˆ
F1U V2
ˆ
F3U V2
...
...
...
ˆ
F3U V2
ˆ
F3U V3
ˆ
F4−γλ
,
Production Inventory with Positive Service Time and Loss 61
where ˆ
F1=λ
µF1,ˆ
F3=λ
µF3,ˆ
F4=λ
µF4,and all other sub matrices are as defined previously for
matrix .
Let π=(π(0), π(1), . . . , π(s), π (s+1,1), . . . , π (S−1,1), π(s+ 1,0), . . . , π(S−1,0), π(S)) be the steady-
state probability vector of the process e
X(t)={I(t); t≥0}. Then πsatisfies the relations
πf
W= 0,πe= 1 (3.4)
That is, at arbitrary epochs the components of the inventory level probability distribution πis given by:
π(s−i) = π(S)γλ
δβ −γλ 1−γλ
δβ S−s!γλ
δβ i
,0≤i≤s,
π(i, 0) = π(S), s + 1 ≤i≤S−1,
π(i, 1) = π(S)γλ
δβ −γλ 1−γλ
δβ S−i!, s + 1 ≤i≤S−1.
and the unknown probability
π(S) = γλ
δβ −12
γλ
δβ S+2 −γλ
δβ s+2 −(S−s)γλ
δβ −1.
Using the components of the probability vector π, we shall find the steady-state probability vector of
the CTMC {X (t); t≥0}. For this, let xbe the steady-state probability vector of the original system.
Then the steady-state vector must satisfy the set of equations
xW= 0,xe= 1.(3.5)
partition xby levels as
x= (x0, x1, x2, . . . )(3.6)
where the sub-vectors of xare further partitioned as, xi= (xi(0), xi(1),...,xi(s),
xi(s+1,1), . . . xi(S−1,1), xi(s+1,0), . . . xi(S−1,0), xi(S)), i ≥0.Then the above system of equations
reduces to
x0B1+x1A2= 0 (3.7)
xiA0+xi+1A1+xi+2 A2= 0, i ≥0 (3.8)
Assume that
x0=ξπ(3.9)
xi=ξλ
µi
π, i ≥1 (3.10)
where ξis a constant to be determined. We verify that the equations (3.7) and (3.8) are satisfied by
(3.9) and (3.10). For (3.7), we have
x0B1+x1A2=ξπB1+λ
µA2(3.11)
62 A. Krishnamoorthy, R. Manikandan, and B. Lakshmi
and from relation (3.8), we have,
xiA0+xi+1A1+xi+2 A2=ξλ
µi+1
πB1+λ
µA2(3.12)
Now from the matrices B1, A2and f
W, it follows that
B1+λ
µA2=f
W(3.13)
Also from (3.4) we have πf
W= 0. Hence the right hand side of the equation (3.11) and (3.12) are
zero. Hence if we take the vector xas given by (3.6), it follows that (3.7) and (3.8) are satisfied. Now
applying the normalizing condition xe=1, we get
ξ"1 + λ
µ+λ
µ2
+λ
µ3
+· · ·#= 1
Hence under the condition that λ < µ, we have
ξ= 1 −λ
µ.(3.14)
Thus we arrive at
Theorem 3.2. Under the necessary and sufficient condition λ < µ for stability, the components of
the steady-state probability vector of the process {X (t); t≥0}with generator matrix W,is given by
(3.9),(3.10) and (3.14). That is, x0= (1 −ρ)π,xi= (1 −ρ)ρiπ, i ≥1where ρis as defined in (3.3)
and πis the inventory level probability vector.
3.2. Performance measures. We enumerate below the long run system performance characteristics
that are useful in formulating an optimization problem.
•Mean number of customers in the system, Ls=λ
µ−λ.
•Mean number of customers waiting in the system during the stock out period,
Ws=Lsπ(0) = λ
µ−λ γλ
δβ s+1 1−γλ
δβ S−s
1−γλ
δβ !π(S)!.
•Mean number of customers waiting in the system when inventory is available,
f
Ws=Ls(1 −π(0)) = λ
µ−λ 1−γλ
δβ s+1 1−γλ
δβ S−s
1−γλ
δβ !π(S)!.
•Mean number of items in the inventory,
Einv =
s
P
i=0
iπ(i) +
S−1
P
i=s+1
i(π(i, 0) + π(i, 1)) = 2−(S−s)(S+s+3)
2γλ
δβ −1π(S).
•Mean rate at which the production process is switched on,
Eon =γµ ∞
P
i=1
ξλ
µiπ(s+ 1,0)=γλπ(S).
•Expected rate at which items are added to the inventory,
Erp =δβ s
P
i=0
π(i) +
S−1
P
i=s+1
+π(i, 1)=δβ (1 −(S−s)π(S)) .
•Expected loss rate of the manufactured item due to rejection,
Mloss = (1 −δ)βs
P
i=0
π(i) +
S−1
P
i=s+1
π(i, 1)= (1 −δ)β(1 −(S−s)π(S)).
Production Inventory with Positive Service Time and Loss 63
•Expected loss rate of customers (customers not joining the system for want of inventory),
Closs =λπ(0) = λ γ λ
δβ s+1 1−γλ
δβ S−s
1−γλ
δβ !π(S)!.
3.3. Expected cycle time of the production process. The production process is switched on
at a service completion epoch t0, which started with s+ 1 items in the inventory with one item from
inventory supplied to the customer and the production process being in off mode. The production
process, once turned on, is turned off only at an epoch t1at which the inventory level in the system
reaches S. A production cycle starts with the switching on of the production process as inventory
level drops progressively to sfrom Sand terminates with the inventory level reaching S. Thus the
expected cycle time of the production process Ecycle is given by
Ecycle =1
δβ (S−s)
s
X
j=0 γλ
δβ j
+
S−1
X
j=s+1
(S−j)γλ
δβ j!.
4. Computing optimal (s, S)pair and the minimum cost
We look for the optimal values of s(the level, reaching at which the production process is switched
on) and the maximum inventory level Sof the production inventory model under discussion. Now for
checking the optimality of sand S, the following cost function is constructed. Define F(s, S) as the
expected cost per unit time in the long run. Then
F(s, S) = K.Eon +hinv .Einv +c1.Closs +c2.Mloss +c3.Erp +c4.Ws+c5.f
Ws
where Kis the fixed cost for starting a production run, hinv is the cost per unit time per inventory
towards holding, c1is the cost incurred due to loss per customer when the inventory is out of stock,
c2is the cost incurred due to rejection per unit manufactured item, c3is the cost of production per
unit time, c4is the waiting cost per unit time per customer during the stock out period and c5is
the waiting cost per unit time per customer when inventory is available. Though we are not able to
compute explicitly the optimal values of sand S, due to the highly complex form of the cost function,
we arrive at these using numerical techniques. For the input values λ= 2, µ = 3, β = 2.5, K =
$5000, hinv = $20, c1= $400, c2= $100, c3= $200, c4= $300, c5= $100 and varying δand γwe arrive
at the optimal pair (1,11) and the minimum cost $461.02 which correspond to δ= 1 and γ= 0.1.
5. Emptiness time distribution for M/M/1/1production inventory system
We now compute the distribution of the time till the items in the inventory becomes empty (zero)
starting from the epoch at which the production is switched on reaching level S. Let χrepresent this
random variable. Now we specialize to the case of M/M/1/1 production inventory with positive service
time. This will enable us to deal with a finite state space CTMC with 3S−selements in its state space:
ℑ={(0,0,1) ,(1,0,1) , . . . , (s, 0,1) ,(s, 1,1) ,(s+ 1,0,0) ,(s+ 1,0,1) , . . . ,
(S−1,0,0) ,(S−1,0,1),(S−1,1,0),(S−1,1,1),(S, 0,0),(S, 1,0)},
The state ((0,0,1)) is regarded as absorbing. Thus the infinitesimal generator Qof the Markov
64 A. Krishnamoorthy, R. Manikandan, and B. Lakshmi
chain {(I(t),C(t),P(t))|t≥0}is of the form Q="T T 0
00#with initial probability vector α=
(0,0, . . . , 1,0) where 1 is in the (3S−s−2)th position; Tis of order 3S−s−1; T0is a 3S−s
component column vector such that Te+T0=0. Let χrepresent the random variable “time till the
items in the inventory becomes zero”. This time duration follows PH distribution with representation
(α,T). Therefore the expected time for the inventory level to reach zero is,
E(χ) = −α$T−1e.
Acknowledgments. Manikandan’s research is supported by the University Grants Commission,
Govt.of India, under RGNF Scheme (No.F.16-1041 (SC)/2008(SA-III)).
References
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[6] M. Saffari, S. Asmussen and R. Haji (2013). The M /M/1 queue with inventory, lost sale, and general lead times,
Queueing Syst DOI 10. 1007/s11134-012-9337-3.
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Queueing Syst 54:55-78.
[8] K. Sigman and D. Simchi-Levi (1992). Light traffic heuristic for an M/G/1 queue with limited inventory. Ann
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[9] M. F. Neuts (1994). Matrix-Geometric Solutions in Stochastic Models - An Algorithmic Approach, 2nd ed., Dover
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A. Krishnamoorthy E-mail:achyuthacusat@gmail.com
Address:
Department of Mathematics, Cochin University of Science and Technology,
Cochin-682 022, India
R. Manikandan E-mail:rangaswamy.mani@gmail.com
Address:
Department of Mathematics, Cochin University of Science and Technology,
Cochin-682 022, India
B. Lakshmi E-mail:luck@cusat.ac.in
Address:
Department of Mathematics, Cochin University of Science and Technology,
Cochin-682 022, India