ON A QUEUEING INVENTORY WITH COMMON LIFE
TIME AND REDUCTION SALE CONSEQUENT TO IN-
CREASE IN AGE
Abdul Rof V1
Assistant Professor, KAHM Unity Women’s College, Manjeri (India). Part-time Research Scholar, Department of
Mathematics, Cochin University of Science and Technology. Cochin (India).
abdulrof@cusat.ac.in
https://orcid.org/0000-0002-7479-3651
Achyutha Krishnamoorthy2
Centre for Research in Mathematics, CMS College, Kottayam (India). Department of Mathematics, Central University
of Kerala, Kasaragod, Kerala, India.
achyuthacusat@gmail.com
https://orcid.org/0000-0002-5911-1325
Reception: 01/08/2022 Acceptance: 16/08/2022 Publication: 29/12/2022
Suggested citation:
Abdul R. V. and A. Krishnamoorthy (2022). On a Queueing Inventory with Common Life Time and Reduction Sale
Consequent to Increase in Age. 3C Empresa. Investigación y pensamiento crítico,11 (2), 15-31.
https://doi.org/10.17993/3cemp.2022.110250.15-31
https://doi.org/10.17993/3cemp.2022.110250.15-31
15
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
ABSTRACT
Consider an inventoried item for which reduction in sales price is declared as the age of the item increases.
Decision to maintain sales price at the same level/reduce, is taken at stages 2
,···,k−
1
,k
. On the
items attaining CLT, they are sold at scrap value, provided items are still left in stock. Customer arrival
forms a non-homogenous Poisson process, with rate increasing with each sales price reduction. Service
time follows exponential distribution.The items are replenished according to (
S, s
)policy with positive
lead time. Each stage of CLT is iid which follows a Phase type distribution with representation(
α, S
)of
order
m
. The k-fold convolution of this distribution is the CLT of the inventoried items. The problem
is modelled as a queueing-inventory problem which is a continuous time Markov chain (CTMC). The
stationary distribution of this CTMC is computed and various performance measures are discussed.
A cost function is constructed to compute the optimal order quantity and reorder level.The model is
compared with queueing inventory model in which the CLT follows Erlang Distribution of order k.
KEYWORDS
Inventory, Lead time, Common Life Time, Reduction Sale
https://doi.org/10.17993/3cemp.2022.110250.15-31
1 INTRODUCTION
The quality of inventory items, especially perishable decreases gradually as their age increases.Seasonal
items also have this property.However for that price can be very high during the time which is not the
season for that item. It is very common to declare reduction in sales price while the quality of the
item decreases. Textile items are real life examples for this phenomenon. The quality of textile items
decreases gradually and finally it becomes unusable. Price reduction is an effective strategy to increase
the sales volume and to settle the items before they perish.
In this paper we consider a queueing inventory system with
S
items which has a common life time of
k
stages. In each stage it is phase type distributed. When it is absorbed from one stage the items reach
the next stage and finally reach the
kth
stage. When it is absorbed from
kth
stage the items attain
CLT and become scrapped. The arriving customer gets the inventory in the present stage. The arrival
rate of customers is assumed to be increasing in each stage due price reduction. The customers are not
allowed in the absence of the inventory. Here we seek answers to two questions. Firstly, how much stock
we need to maximize profits, and secondly, when to place an order to deliver items on time to waiting
customers.
The common life time inventory models were first introduced by Lian, Z and Neuts, M.F (2005).
The common life time was assumed to be of discrete phase type distribution. A perishable queueing
inventory system with (
Q, r
)policy, Poisson demands, identical life times, constant lead times and lost
sales with fixed ordering costs was studied by Berke, E and Gurler, L (2008). A cost function is used
to find the optimal (
Q, r
)policy. Krishnamoorthy et. al (2006) describes a queueing inventory with
reservation, cancellation, common life-time and retrial. Broekmeulen, R and Donselaar, K (2009) discuss
a perishable queueing inventory with batch ordering, positive lead time and time varying demand. The
replenishment is done using EWA policy (Estimated Withdrawal and Aging) in which the replenishment
is done considering the items that will be out-dated within the lead time. Williams, C and Patuwo, B
(1999) deal with a perishable inventory queueing system with positive lead time. The optimal reorder
quantity is found using numerical computations. Schwarz, M. (2006) derive stationary distributions of
joint queue length and inventory processes in explicit product form for various M/M/1-systems with
inventory under continuous review and different inventory management policies, and with lost sales.
Krishnamoorthy et al (2021) provides an overview of the work done in mathematical inventory models
with positive service time. Kirubhashankar.et. al (2021) analyzes a production inventory model for
deteriorating products with constant demand and backlogged shortages which depends on the length
of the waiting time for the next order level. Optimum shortage time and total cycle length with the
objective of optimizing the total cost are found.
Mehrez, A and Ben-Arieh, D (1991) describes a model that combines several dichotomies, into a
single model. It has the objective of deciding the optimal order quantities for a multi-item inventory
system over a finite horizon. The demand is probabilistic with service level constraints, and there is an
all-unit price break, for orders that exceed a given size. Amirthakodi et. al (2015) consider a continuous
review perishable inventory system with service facility consisting of finite waiting hall and a single
server. The items are replenished based on variable ordering policy. The lead time is assumed to have
phase type distribution. develop a deterministic inventory model with quantity discount, pricing and
partial back ordering when the product in stock deteriorates with time. Wee, H. M (1999) investigate a
base-stock perishable inventory system with Markovian demand and general lead-time and lifetime
distributions. Janssen, L et. al (2016) presents a review of literature of deteriorating inventory models.
The rest of the paper is distributed as follows. In the second section model description is given. The
steady state analysis, stability condition and various performance measures are discussed in the third
section. The numerical illustration is given in the fourth section.
2 MODEL DESCRIPTION
Consider a queuing-inventory which has S items. Initially all items are in stage 1. As the items getting
older the stages are changed to 2
,···,k−
1
,k
and finally the CLT of the items is reached and the
https://doi.org/10.17993/3cemp.2022.110250.15-31
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
16
ABSTRACT
Consider an inventoried item for which reduction in sales price is declared as the age of the item increases.
Decision to maintain sales price at the same level/reduce, is taken at stages 2
,···,k−
1
,k
. On the
items attaining CLT, they are sold at scrap value, provided items are still left in stock. Customer arrival
forms a non-homogenous Poisson process, with rate increasing with each sales price reduction. Service
time follows exponential distribution.The items are replenished according to (
S, s
)policy with positive
lead time. Each stage of CLT is iid which follows a Phase type distribution with representation(
α, S
)of
order
m
. The k-fold convolution of this distribution is the CLT of the inventoried items. The problem
is modelled as a queueing-inventory problem which is a continuous time Markov chain (CTMC). The
stationary distribution of this CTMC is computed and various performance measures are discussed.
A cost function is constructed to compute the optimal order quantity and reorder level.The model is
compared with queueing inventory model in which the CLT follows Erlang Distribution of order k.
KEYWORDS
Inventory, Lead time, Common Life Time, Reduction Sale
https://doi.org/10.17993/3cemp.2022.110250.15-31
1 INTRODUCTION
The quality of inventory items, especially perishable decreases gradually as their age increases.Seasonal
items also have this property.However for that price can be very high during the time which is not the
season for that item. It is very common to declare reduction in sales price while the quality of the
item decreases. Textile items are real life examples for this phenomenon. The quality of textile items
decreases gradually and finally it becomes unusable. Price reduction is an effective strategy to increase
the sales volume and to settle the items before they perish.
In this paper we consider a queueing inventory system with
S
items which has a common life time of
k
stages. In each stage it is phase type distributed. When it is absorbed from one stage the items reach
the next stage and finally reach the
kth
stage. When it is absorbed from
kth
stage the items attain
CLT and become scrapped. The arriving customer gets the inventory in the present stage. The arrival
rate of customers is assumed to be increasing in each stage due price reduction. The customers are not
allowed in the absence of the inventory. Here we seek answers to two questions. Firstly, how much stock
we need to maximize profits, and secondly, when to place an order to deliver items on time to waiting
customers.
The common life time inventory models were first introduced by Lian, Z and Neuts, M.F (2005).
The common life time was assumed to be of discrete phase type distribution. A perishable queueing
inventory system with (
Q, r
)policy, Poisson demands, identical life times, constant lead times and lost
sales with fixed ordering costs was studied by Berke, E and Gurler, L (2008). A cost function is used
to find the optimal (
Q, r
)policy. Krishnamoorthy et. al (2006) describes a queueing inventory with
reservation, cancellation, common life-time and retrial. Broekmeulen, R and Donselaar, K (2009) discuss
a perishable queueing inventory with batch ordering, positive lead time and time varying demand. The
replenishment is done using EWA policy (Estimated Withdrawal and Aging) in which the replenishment
is done considering the items that will be out-dated within the lead time. Williams, C and Patuwo, B
(1999) deal with a perishable inventory queueing system with positive lead time. The optimal reorder
quantity is found using numerical computations. Schwarz, M. (2006) derive stationary distributions of
joint queue length and inventory processes in explicit product form for various M/M/1-systems with
inventory under continuous review and different inventory management policies, and with lost sales.
Krishnamoorthy et al (2021) provides an overview of the work done in mathematical inventory models
with positive service time. Kirubhashankar.et. al (2021) analyzes a production inventory model for
deteriorating products with constant demand and backlogged shortages which depends on the length
of the waiting time for the next order level. Optimum shortage time and total cycle length with the
objective of optimizing the total cost are found.
Mehrez, A and Ben-Arieh, D (1991) describes a model that combines several dichotomies, into a
single model. It has the objective of deciding the optimal order quantities for a multi-item inventory
system over a finite horizon. The demand is probabilistic with service level constraints, and there is an
all-unit price break, for orders that exceed a given size. Amirthakodi et. al (2015) consider a continuous
review perishable inventory system with service facility consisting of finite waiting hall and a single
server. The items are replenished based on variable ordering policy. The lead time is assumed to have
phase type distribution. develop a deterministic inventory model with quantity discount, pricing and
partial back ordering when the product in stock deteriorates with time. Wee, H. M (1999) investigate a
base-stock perishable inventory system with Markovian demand and general lead-time and lifetime
distributions. Janssen, L et. al (2016) presents a review of literature of deteriorating inventory models.
The rest of the paper is distributed as follows. In the second section model description is given. The
steady state analysis, stability condition and various performance measures are discussed in the third
section. The numerical illustration is given in the fourth section.
2 MODEL DESCRIPTION
Consider a queuing-inventory which has S items. Initially all items are in stage 1. As the items getting
older the stages are changed to 2
,···,k−
1
,k
and finally the CLT of the items is reached and the
https://doi.org/10.17993/3cemp.2022.110250.15-31
17
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
items remain if any, are scrapped. In each stage the CLT is iid that has a phase type distribution
with representation (
α, S
)of order
m
. In the ith stage the distribution is the i-fold convolution of
the PH distribution. The inventory is replenished with (S,s) policy with positive lead time which
follows an exponential distribution with parameter
β
.The replenishment order is placed when either the
inventory level is dropped to s or CLT of the stocked items attains the level
q
where 1
≤q≤k
. The
customers arrive to the system according to a Poisson process of rate
λi,i
=1
,
2
,···,k
. A reduction
in price is declared when the stage is changed from 1
→
2
,
2
→
3
,
3
→
4
,........k −
1
→k
. The
service of customers is exponentially distributed with rate
µi,i
=1
,
2
,···,k
. The service of customers
is exponentially distributed with rate
µi,i
=1
,
2
,···,k
. Customers are not allowed to enterwhen there
is no inventory in the system.
Let
N
(
t
)be the number of customers,
I
(
t
)be the inventory level,
C
(
t
), the stage of CLT and
J
(
t
)is
the phase of CLT at time t.
Then the process {(N(t),I(t),C(t),J(t)) : t≥0} is a continuous time Markov chain with state space
{(n, 0) : n≥0}U{(n,i,r,j):n≥0,1≤i≤S, 1≤r≤k, 1≤j≤m}U{ˆ
0}
Transition due to arrival:
(n,i,r,j)→(n+1,i,r,j)with rate λr,n≥0,1≤i≤S, 1≤r≤k, 1≤j≤m
Transition due to service:
(n, 1, r, j)→(n−1,0) with rate µr,n≥1,1≤r≤k, 1≤j≤m
(n,i,r,j)→(n−1,i−1,r,j)with rate µr,n≥1,2≤i≤S, 1≤r≤k, 1≤j≤m
Transition due to replenishment:
(n, 0) →(n, S, 1,j)with rate βαj,n≥0,1≤j≤m
(n,i,r,j)→(n, S, 1,j)with rate β, n ≥0,1≤i≤s, 1≤r≤k, 1≤j≤m
(n,i,r,j)→(n, S, 1,j)with rate β, n ≥0,s+1≤i≤S, c ≤r≤k, 1≤j≤m
Transition due to phase change of Common Life Time:
(n,i,k,j)→(n, 0) with rate tj,n≥0,1≤i≤S, 1≤j≤m
(n,i,r,j)→(n,i,r,p)with rate tjp,n≥0,1≤i≤S, 1≤r≤k−1,1≤j, p ≤m
Transition due to stage change of Common Life Time:
(n,i,r,j)→(n, i, r +1,p)with rate αptj,n≥0,1≤i≤S, 1≤r≤k−1,1≤j, p ≤m
(n,i,k,j)→(n, 0) with rate tj,n≥0,1≤i≤S, 1≤j≤m
The infinitesimal generator matrix of the process is given by,
Q=
B1A0
A2A1A0
A2A1A0
.........
3 STEADY STATE ANALYSIS
Let
A
=
A0
+
A1
+
A2
and let Π=(
π0,π
1, ..., πS
)be the steady state probability vector where
πi= ((πi)1,(πi)2,···,(πi)k). Here (πi)r=(πi(r, 1),π
i(r, 2), ....πi(r, m)),1≤i≤S, 1≤r≤k
The condtions
https://doi.org/10.17993/3cemp.2022.110250.15-31
πA =0and πe =1, can be re-written as
−π0β+
k
r=1
(π1)rµrem+
S
i=1
(πi)kT0=0
(πi)r−1(T0⊗α)(1 −δir)+(πi)r−1[(T−(µr+β)I]+(πi+1)rµr=0,1≤i≤s, 1≤r≤k
(πi)r−1(T0⊗α)(1 −δir)+(πi)r−1(T−µrI)+(πi+1)rµr=0,s+1≤i≤S−1,1≤r≤q−1
(πi)r−1(T0⊗α)(1 −δir)+(πi)r−1(T−(µr+β)I)+(πi+1)rµr=0,s+1≤i≤S−1,q ≤r≤S
π0α+(
S
i=1
k
r=1
(πi)r)T=0
S
i=1
k
r=1
m
j=1
πi(r, j)=1
Simplifying the last equation, π−1
0=1−αT −1e
Solving the equations we get, πi, for each i=0,1,2, ..., S.
3.1 STABILITY CONDITION
Theorem:The system is stable if and only if
ie,
S
i=1
k
r=1
m
j=1
πi(r, j)λr<
S
i=1
k
r=1
m
j=1
πi(r, j)µr
Proof The system is stable if πA0e<πA
2e
ie,
S
i=1
k
r=1
m
j=1
πi(r, j)λr<
S
i=1
k
r=1
m
j=1
πi(r, j)µr
3.2 STATIONARY PROBABILITIES
Let x=(x0,x
1, ....)be the stationary probability vector of the system where
xi={xi(0) : i≥0}U{xi(p, r, j):i≥0,1≤p≤S, 1≤r≤k, 1≤j≤m}
Then xQ =0
, xe =1which results in the following equations.
x0B1+x1A2=0
xi−1A0+xiA1+xi+1A2=0,i≥1
Let xi=x0Ri,i≥1. Substituting in xi−1A0+xiA1+xi+1A2=0, we get,
R2A2+RA1+A0=0
Let R(0)=0be the initial solution.
From the recurrence relation
R(n)=(R2(n−1)A2+A0)(−A−1
1),n≥1
we find R.
https://doi.org/10.17993/3cemp.2022.110250.15-31
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
18
items remain if any, are scrapped. In each stage the CLT is iid that has a phase type distribution
with representation (
α, S
)of order
m
. In the ith stage the distribution is the i-fold convolution of
the PH distribution. The inventory is replenished with (S,s) policy with positive lead time which
follows an exponential distribution with parameter
β
.The replenishment order is placed when either the
inventory level is dropped to s or CLT of the stocked items attains the level
q
where 1
≤q≤k
. The
customers arrive to the system according to a Poisson process of rate
λi,i
=1
,
2
,···,k
. A reduction
in price is declared when the stage is changed from 1
→
2
,
2
→
3
,
3
→
4
,........k −
1
→k
. The
service of customers is exponentially distributed with rate
µi,i
=1
,
2
,···,k
. The service of customers
is exponentially distributed with rate
µi,i
=1
,
2
,···,k
. Customers are not allowed to enterwhen there
is no inventory in the system.
Let
N
(
t
)be the number of customers,
I
(
t
)be the inventory level,
C
(
t
), the stage of CLT and
J
(
t
)is
the phase of CLT at time t.
Then the process {(N(t),I(t),C(t),J(t)) : t≥0} is a continuous time Markov chain with state space
{(n, 0) : n≥0}U{(n,i,r,j):n≥0,1≤i≤S, 1≤r≤k, 1≤j≤m}U{ˆ
0}
Transition due to arrival:
(n,i,r,j)→(n+1,i,r,j)with rate λr,n≥0,1≤i≤S, 1≤r≤k, 1≤j≤m
Transition due to service:
(n, 1, r, j)→(n−1,0) with rate µr,n≥1,1≤r≤k, 1≤j≤m
(n,i,r,j)→(n−1,i−1,r,j)with rate µr,n≥1,2≤i≤S, 1≤r≤k, 1≤j≤m
Transition due to replenishment:
(n, 0) →(n, S, 1,j)with rate βαj,n≥0,1≤j≤m
(n,i,r,j)→(n, S, 1,j)with rate β, n ≥0,1≤i≤s, 1≤r≤k, 1≤j≤m
(n,i,r,j)→(n, S, 1,j)with rate β, n ≥0,s+1≤i≤S, c ≤r≤k, 1≤j≤m
Transition due to phase change of Common Life Time:
(n,i,k,j)→(n, 0) with rate tj,n≥0,1≤i≤S, 1≤j≤m
(n,i,r,j)→(n,i,r,p)with rate tjp,n≥0,1≤i≤S, 1≤r≤k−1,1≤j, p ≤m
Transition due to stage change of Common Life Time:
(n,i,r,j)→(n, i, r +1,p)with rate αptj,n≥0,1≤i≤S, 1≤r≤k−1,1≤j, p ≤m
(n,i,k,j)→(n, 0) with rate tj,n≥0,1≤i≤S, 1≤j≤m
The infinitesimal generator matrix of the process is given by,
Q=
B1A0
A2A1A0
A2A1A0
.........
3 STEADY STATE ANALYSIS
Let
A
=
A0
+
A1
+
A2
and let Π=(
π0,π
1, ..., πS
)be the steady state probability vector where
πi= ((πi)1,(πi)2,···,(πi)k). Here (πi)r=(πi(r, 1),π
i(r, 2), ....πi(r, m)),1≤i≤S, 1≤r≤k
The condtions
https://doi.org/10.17993/3cemp.2022.110250.15-31
πA =0and πe =1, can be re-written as
−π0β+
k
r=1
(π1)rµrem+
S
i=1
(πi)kT0=0
(πi)r−1(T0⊗α)(1 −δir)+(πi)r−1[(T−(µr+β)I]+(πi+1)rµr=0,1≤i≤s, 1≤r≤k
(πi)r−1(T0⊗α)(1 −δir)+(πi)r−1(T−µrI)+(πi+1)rµr=0,s+1≤i≤S−1,1≤r≤q−1
(πi)r−1(T0⊗α)(1 −δir)+(πi)r−1(T−(µr+β)I)+(πi+1)rµr=0,s+1≤i≤S−1,q ≤r≤S
π0α+(
S
i=1
k
r=1
(πi)r)T=0
S
i=1
k
r=1
m
j=1
πi(r, j)=1
Simplifying the last equation, π−1
0=1−αT −1e
Solving the equations we get, πi, for each i=0,1,2, ..., S.
3.1 STABILITY CONDITION
Theorem:The system is stable if and only if
ie,
S
i=1
k
r=1
m
j=1
πi(r, j)λr<
S
i=1
k
r=1
m
j=1
πi(r, j)µr
Proof The system is stable if πA0e<πA
2e
ie,
S
i=1
k
r=1
m
j=1
πi(r, j)λr<
S
i=1
k
r=1
m
j=1
πi(r, j)µr
3.2 STATIONARY PROBABILITIES
Let x=(x0,x
1, ....)be the stationary probability vector of the system where
xi={xi(0) : i≥0}U{xi(p, r, j):i≥0,1≤p≤S, 1≤r≤k, 1≤j≤m}
Then xQ =0
, xe =1which results in the following equations.
x0B1+x1A2=0
xi−1A0+xiA1+xi+1A2=0,i≥1
Let xi=x0Ri,i≥1. Substituting in xi−1A0+xiA1+xi+1A2=0, we get,
R2A2+RA1+A0=0
Let R(0)=0be the initial solution.
From the recurrence relation
R(n)=(R2(n−1)A2+A0)(−A−1
1),n≥1
we find R.
https://doi.org/10.17993/3cemp.2022.110250.15-31
19
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
3.3 PERFORMANCE MEASURES
1. Expected number of customers in the system =∞
i=0 S
p=1 k
r=1 m
j=1 ixi(p, r, j)
2. Expected number of items in the stage r
Er=∞
i=0 S
p=1 m
j=1 pxi(p, r, j),r =1,2,···,k
3. Expected number of scrapped items Es=∞
i=0 S
p=1 m
j=1 pxi(p,k,j)tj.
4. Probability that a newly joining customer is served instantaneously
=S
p=1 k
r=1 m
j=1 x0(p, r, j).
5. Probability that the server is busy=∞
i=1 S
p=1 k
r=1 m
j=1 xi(p,r,j)
=1−x0e−∞
i=1 xi(0)
6. Probability that all items are sold before the realization of CLT =∞
i=0 xi(0)
7. Probability that all items in a cycle are scrapped =∞
i=0 m
j=1 xi(S,k,j)tj
8. Probability that there are no customers in the system =S
p=1 k
r=1 m
j=1 xi(p,r,j)
3.4 DISTRIDUTION OF WAITING TIME OF A TAGGED CUSTOMER
Consider a customer who joins the system as
nth
tagged customerand waiting for service where
n>
0.
The rank of the customer reduces to
n−
1
,n−
2
,···,
2
,
1as the customers in the queue join for service.
Let
M
(
t
)be the rank of the customer,
I
(
t
)be the inventory level, C(t) the stage of CLT and J(t) the
phase of CLT at time
t
. Then the process
{
(
M
(
t
)
,I
(
t
)
,C
(
t
)
,J
(
t
)) :
t≥
0
}
is a Markov chain with state
space
{n, n −1,···,2,1}×{0}∪{n, n −1,···,2,1}×{1,2,···,S}×{1,2,···,k}×{1,2,···,m}∪{
ˆ
0}
where
ˆ
0
is the absorbing state of the Markov Chain which denotes the service epoch of the
nth
tagged
customer.
Let qr=Pr{N(t)=r, I (t)=i, C(t)=p, J(t)=j}=xr(0) + S
i=1 k
p=1 m
j=1 xr(i, p, j)
Let ωr=qr
∞
r=1 qrThen ω=(ωr,ω
r−1,........ω
1)is the initial probability vector of the distribution.
The infinitesimal generator matrix of the process is,
ˆ
P=WW
0
00
where
W=
Wn,n Wn,n−1
Wn−1,n−1Wn−1,n−2
W2,1
W1,1
Wi,i =A0+A1,i=1,···rand Wi,i−1=A2,i=2,···,r
and
W0=
0
A2
The waiting time distribution is
w
(
t
)=
ωeWxW0
The average waiting time of a tagged customer=
−ω
(
W−
1)
e
https://doi.org/10.17993/3cemp.2022.110250.15-31
3.5 DISTRIBUTION OF THE DURATION OF A CYCLE
The expected cycle length is the duration of time elapsed between two consecutive replenishment
orders.Starting from any system state, with the number of items in inventory S (an order has been
placed and got materialised), we look at the evolution until the next replenishment takes place,
either through realization of CLT or by reaching the inventory level to s. Consider the Markov chain
{
(
I
(
t
)
,N
(
t
)
,S
(
t
)
,J
(
t
)) :
t≥
0
}
with the state space
{
(0
,r
):
r≥
0
}∪{
(
r,i,p,j
):1
≤r≤S,
0
≤i≤
H, 1≤p≤k, 1≤j≤m}∪{∆}, where{∆}is the absorbing state of the Markov chain which denotes
the realization of the next replenishment by completing one cycle.
Let pr=Pr{N(t)=i, I(t)=r, S(t)=1}
i e:pr=H
i=1 m
j=1 xi(r, 1,j)
Let
γr
=
pr/S
r=0 pr
Then
γ
=(
γS,γ
S−1,........γ
0
)is the initial probability vector of the distribution.
The infinitesimal generator matrix of the process is,
ˆ
Q=UU0
00
where
U=
US,S US,S−10··· 0US,0
US−1,S US−1,S−1US−1,S−20··· US−1,0
.
.
..............
.
.
Us,S 0··· Us,s ...Us,0
.
.
..
.
...........
.
.
U0,S 0··· ··· ··· −βIS+1
(S+1)×(S+1)
Here,
for s+1≤i≤S
Ui,i =
K0,0K0,10··· 0
0K1,1K0,1··· 0
.
.
..
.
........
.
.
0··· ··· K1,1K0,1
0··· ··· 0K1,1
(S+1)×(S+1)
and for 1≤i≤s
Ui,i =
L0,0K0,10··· 0
0L1,1K0,1··· 0
.
.
...........
.
.
0··· ··· L1,1K0,1
0··· ··· 0L1,1
(S+1)×(S+1)
K0,0=
L1M10··· 0
0L2M1··· 0
.
.
..
.
........
.
.
0··· ··· Lk−1M1
0··· ··· 0Lk
k×k
https://doi.org/10.17993/3cemp.2022.110250.15-31
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
20
3.3 PERFORMANCE MEASURES
1. Expected number of customers in the system =∞
i=0 S
p=1 k
r=1 m
j=1 ixi(p, r, j)
2. Expected number of items in the stage r
Er=∞
i=0 S
p=1 m
j=1 pxi(p, r, j),r =1,2,···,k
3. Expected number of scrapped items Es=∞
i=0 S
p=1 m
j=1 pxi(p,k,j)tj.
4. Probability that a newly joining customer is served instantaneously
=S
p=1 k
r=1 m
j=1 x0(p, r, j).
5. Probability that the server is busy=∞
i=1 S
p=1 k
r=1 m
j=1 xi(p,r,j)
=1−x0e−∞
i=1 xi(0)
6. Probability that all items are sold before the realization of CLT =∞
i=0 xi(0)
7. Probability that all items in a cycle are scrapped =∞
i=0 m
j=1 xi(S,k,j)tj
8. Probability that there are no customers in the system =S
p=1 k
r=1 m
j=1 xi(p,r,j)
3.4 DISTRIDUTION OF WAITING TIME OF A TAGGED CUSTOMER
Consider a customer who joins the system as
nth
tagged customerand waiting for service where
n>
0.
The rank of the customer reduces to
n−
1
,n−
2
,···,
2
,
1as the customers in the queue join for service.
Let
M
(
t
)be the rank of the customer,
I
(
t
)be the inventory level, C(t) the stage of CLT and J(t) the
phase of CLT at time
t
. Then the process
{
(
M
(
t
)
,I
(
t
)
,C
(
t
)
,J
(
t
)) :
t≥
0
}
is a Markov chain with state
space
{n, n −1,···,2,1}×{0}∪{n, n −1,···,2,1}×{1,2,···,S}×{1,2,···,k}×{1,2,···,m}∪{
ˆ
0}
where
ˆ
0
is the absorbing state of the Markov Chain which denotes the service epoch of the
nth
tagged
customer.
Let qr=Pr{N(t)=r, I (t)=i, C(t)=p, J(t)=j}=xr(0) + S
i=1 k
p=1 m
j=1 xr(i, p, j)
Let ωr=qr
∞
r=1 qrThen ω=(ωr,ω
r−1,........ω
1)is the initial probability vector of the distribution.
The infinitesimal generator matrix of the process is,
ˆ
P=WW
0
00
where
W=
Wn,n Wn,n−1
Wn−1,n−1Wn−1,n−2
W2,1
W1,1
Wi,i =A0+A1,i=1,···rand Wi,i−1=A2,i=2,···,r
and
W0=
0
A2
The waiting time distribution is
w
(
t
)=
ωeWxW0
The average waiting time of a tagged customer=
−ω
(
W−
1)
e
https://doi.org/10.17993/3cemp.2022.110250.15-31
3.5 DISTRIBUTION OF THE DURATION OF A CYCLE
The expected cycle length is the duration of time elapsed between two consecutive replenishment
orders.Starting from any system state, with the number of items in inventory S (an order has been
placed and got materialised), we look at the evolution until the next replenishment takes place,
either through realization of CLT or by reaching the inventory level to s. Consider the Markov chain
{
(
I
(
t
)
,N
(
t
)
,S
(
t
)
,J
(
t
)) :
t≥
0
}
with the state space
{
(0
,r
):
r≥
0
}∪{
(
r,i,p,j
):1
≤r≤S,
0
≤i≤
H, 1≤p≤k, 1≤j≤m}∪{∆}, where{∆}is the absorbing state of the Markov chain which denotes
the realization of the next replenishment by completing one cycle.
Let pr=Pr{N(t)=i, I(t)=r, S(t)=1}
i e:pr=H
i=1 m
j=1 xi(r, 1,j)
Let
γr
=
pr/S
r=0 pr
Then
γ
=(
γS,γ
S−1,........γ
0
)is the initial probability vector of the distribution.
The infinitesimal generator matrix of the process is,
ˆ
Q=UU0
00
where
U=
US,S US,S−10··· 0US,0
US−1,S US−1,S−1US−1,S−20··· US−1,0
.
.
..............
.
.
Us,S 0··· Us,s ...Us,0
.
.
..
.
...........
.
.
U0,S 0··· ··· ··· −βIS+1
(S+1)×(S+1)
Here,
for s+1≤i≤S
Ui,i =
K0,0K0,10··· 0
0K1,1K0,1··· 0
.
.
..
.
........
.
.
0··· ··· K1,1K0,1
0··· ··· 0K1,1
(S+1)×(S+1)
and for 1≤i≤s
Ui,i =
L0,0K0,10··· 0
0L1,1K0,1··· 0
.
.
...........
.
.
0··· ··· L1,1K0,1
0··· ··· 0L1,1
(S+1)×(S+1)
K0,0=
L1M10··· 0
0L2M1··· 0
.
.
..
.
........
.
.
0··· ··· Lk−1M1
0··· ··· 0Lk
k×k
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21
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
K1,1=
P1M10··· 0
0P2M1··· 0
.
.
..
.
........
.
.
0··· ··· Pk−1M1
0··· ··· 0Pk
k×k
K0,1=
N10··· ··· 0
0N20··· 0
.
.
..
.
........
.
.
0··· ··· Nk−10
0··· ··· 0Nk
k×k
L0,0=
U1M10··· 0
0U2M1··· 0
.
.
..
.
........
.
.
0··· ··· Uk−1M1
0··· ··· 0Uk
k×k
L1,1=
V1M10··· 0
0V2M1··· 0
.
.
..
.
........
.
.
0··· ··· Vk−1M1
0··· ··· 0Vk
k×k
For 1≤i≤q−1,L
i=T−λiIm,U
i=T−(λi+µi)Im
For q≤i≤k, Li=T−(λi+β)Im,U
i=T−(λi+β+µi)Im
For 1≤i≤k, Pi=T−(λi+β)Im,V
i=T−(λi+β+µi)Im,N
i=λiImand M1=T0⊗α.
For 1≤i≤S
Ui,i−1=
0··· ··· ··· 0
J1,00··· ··· 0
.
.
...........
.
.
0··· J1,000
0··· ··· J1,00
(S+1)×(S+1)
where
J1,0=
H10··· ··· 0
0H20··· 0
.
.
..
.
........
.
.
0··· ··· Hk−10
0··· ··· 0Hk
k×k
For 1≤i≤k, Hi=µiIm
For s+1≤i≤S, Wi,S =0
https://doi.org/10.17993/3cemp.2022.110250.15-31
U0,S =
M0,00··· ··· 0
0M0,00··· 0
.
.
..
.
........
.
.
0··· ··· M0,00
0··· ··· 0M0,0
(S+1)×(S+1)
M0,0=βα 0··· ··· 01×k
For 1≤i≤s
Ui,S =
M1,10··· ··· 0
0M1,10··· 0
.
.
..
.
........
.
.
0··· ··· M1,10
0··· ··· 0M1,1
(S+1)×(S+1)
M1,1=
βT 0··· ··· 0
0βT 0··· 0
.
.
..
.
........
.
.
0··· ··· βT 0
0··· ··· 0βT
k×k
For 1≤i≤S
Ui,0=
G1,10··· ··· 0
0G1,10··· 0
.
.
..
.
........
.
.
0··· ··· G1,10
0··· ··· 0G1,1
(S+1)×(S+1)
G1,1=
0
···
0
T0
k×1
The probability density function of the distribution is, f(x)=γeUxU0
So the mean of the distribution is the first raw moment, (µ1)′=−γU−1e
So expected cycle length=−γU−1e
4 SPECIAL CASE: ERLANG CLT DISTRIBUTION
Consider an inventory item that has a common life time(CLT) with
k
stages. At the beginning of stage
1,the age of the item is zero. As the item gets older the age of the item changes to 2
,
3
, ..., ···,k−
1
,k
.Each
stage duration is iid exponention random variable. After
kth
stage the items remaining, if any, are
scrapped.The CLT is modelled as an Erlang Distribution of order
k
. Customers arrive according to a
Poisson process with rates depending on the stage of the CLT,
λi,i
=1
,
2
,···, k.
The service rate in
each stage is exponential with rate
µi,i
=1
,
2
,···, k.
A reduction in price is declared when the stage is
changed from 1
→
2
,
2
→
3
,
3
→
4
,........k−
1
→k.
The stages are changed according to exponential
https://doi.org/10.17993/3cemp.2022.110250.15-31
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
22
K1,1=
P1M10··· 0
0P2M1··· 0
.
.
..
.
........
.
.
0··· ··· Pk−1M1
0··· ··· 0Pk
k×k
K0,1=
N10··· ··· 0
0N20··· 0
.
.
..
.
........
.
.
0··· ··· Nk−10
0··· ··· 0Nk
k×k
L0,0=
U1M10··· 0
0U2M1··· 0
.
.
..
.
........
.
.
0··· ··· Uk−1M1
0··· ··· 0Uk
k×k
L1,1=
V1M10··· 0
0V2M1··· 0
.
.
..
.
........
.
.
0··· ··· Vk−1M1
0··· ··· 0Vk
k×k
For 1≤i≤q−1,L
i=T−λiIm,U
i=T−(λi+µi)Im
For q≤i≤k, Li=T−(λi+β)Im,U
i=T−(λi+β+µi)Im
For 1≤i≤k, Pi=T−(λi+β)Im,V
i=T−(λi+β+µi)Im,N
i=λiImand M1=T0⊗α.
For 1≤i≤S
Ui,i−1=
0··· ··· ··· 0
J1,00··· ··· 0
.
.
...........
.
.
0··· J1,000
0··· ··· J1,00
(S+1)×(S+1)
where
J1,0=
H10··· ··· 0
0H20··· 0
.
.
..
.
........
.
.
0··· ··· Hk−10
0··· ··· 0Hk
k×k
For 1≤i≤k, Hi=µiIm
For s+1≤i≤S, Wi,S =0
https://doi.org/10.17993/3cemp.2022.110250.15-31
U0,S =
M0,00··· ··· 0
0M0,00··· 0
.
.
..
.
........
.
.
0··· ··· M0,00
0··· ··· 0M0,0
(S+1)×(S+1)
M0,0=βα 0··· ··· 01×k
For 1≤i≤s
Ui,S =
M1,10··· ··· 0
0M1,10··· 0
.
.
..
.
........
.
.
0··· ··· M1,10
0··· ··· 0M1,1
(S+1)×(S+1)
M1,1=
βT 0··· ··· 0
0βT 0··· 0
.
.
..
.
........
.
.
0··· ··· βT 0
0··· ··· 0βT
k×k
For 1≤i≤S
Ui,0=
G1,10··· ··· 0
0G1,10··· 0
.
.
..
.
........
.
.
0··· ··· G1,10
0··· ··· 0G1,1
(S+1)×(S+1)
G1,1=
0
···
0
T0
k×1
The probability density function of the distribution is, f(x)=γeUxU0
So the mean of the distribution is the first raw moment, (µ1)′=−γU−1e
So expected cycle length=−γU−1e
4 SPECIAL CASE: ERLANG CLT DISTRIBUTION
Consider an inventory item that has a common life time(CLT) with
k
stages. At the beginning of stage
1,the age of the item is zero. As the item gets older the age of the item changes to 2
,
3
, ..., ···,k−
1
,k
.Each
stage duration is iid exponention random variable. After
kth
stage the items remaining, if any, are
scrapped.The CLT is modelled as an Erlang Distribution of order
k
. Customers arrive according to a
Poisson process with rates depending on the stage of the CLT,
λi,i
=1
,
2
,···, k.
The service rate in
each stage is exponential with rate
µi,i
=1
,
2
,···, k.
A reduction in price is declared when the stage is
changed from 1
→
2
,
2
→
3
,
3
→
4
,........k−
1
→k.
The stages are changed according to exponential
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3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
distribution with rate
α
. When the inventory level drops to
s
or the item reaches the stage
q,
1
≤q≤k
an order is placed.he inventory is replenished with positive lead time which is exponentially distributed
with rate β.The customers are blocked in the absence of inventory to reduce the system cost.
Let
N
(
t
)=Number of customers in the system,
I
(
t
)=Inventory level,
C
(
t
)is the stage of CLT
at time
t
. Then {
X
(
t
):
t≥
0}={(
N
(
t
)
,I
(
t
)
,S
(
t
)) :
t≥
0} is a Markov process with state space
{(0,0)}U{0,1,2,3,...}×{1,2,...,S}×{1,2,···,k}.
4.1 INFINITESIMAL GENERATOR MATRIX
Q=
B1A0
A2A1A0
A2A1A0
.........
Where, each block has order (kS + 1) ×(kS + 1) and,
B1=
−β0B0,S
B1,0B1,1B1,S
Bs,0Bs,s Bs,S
BS,00BS,S
where, for 1≤i≤s
Bi,i =
−(λ1+β+α)α
−(λ2+β+α)α
......
−(λk+β+α)
, for s+1≤i≤S
Bi,i =
−(λ1+α)α
......
−(λq+β+α)α
−(λk+β+α)
, for 1≤i≤S,
Bi,0=
0
0
.
.
.
α
,
B0,S =β0··· 0
https://doi.org/10.17993/3cemp.2022.110250.15-31
for 1≤i≤s
Bi,S =
β0··· 0
β0··· 0
.
.
.··· ....
.
.
β0··· 0
,
and for s+1≤i≤S
Bi,S =
00··· 0
.
.
.··· ....
.
..
.
.
β0··· 0
.
.
.··· ....
.
.
β0··· 0
,
A0=
00··· ··· 0
0L...··· 0
.
.
....L··· 0
.
.
...........
.
.
0··· ··· 0L
L=
λ10··· 0
0λ2··· .
.
.
.
.
........
.
.
0··· ··· λk
A2=
00··· ··· 0
M10...··· .
.
.
0M.......
.
.
.
.
...........
.
.
0··· ··· M0
where,
M1=
µ1
µ2
.
.
.
µk
,
M=
µ10··· 0
0µ2....
.
.
.
.
........
.
.
0··· ··· µk
https://doi.org/10.17993/3cemp.2022.110250.15-31
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
24
distribution with rate
α
. When the inventory level drops to
s
or the item reaches the stage
q,
1
≤q≤k
an order is placed.he inventory is replenished with positive lead time which is exponentially distributed
with rate β.The customers are blocked in the absence of inventory to reduce the system cost.
Let
N
(
t
)=Number of customers in the system,
I
(
t
)=Inventory level,
C
(
t
)is the stage of CLT
at time
t
. Then {
X
(
t
):
t≥
0}={(
N
(
t
)
,I
(
t
)
,S
(
t
)) :
t≥
0} is a Markov process with state space
{(0,0)}U{0,1,2,3,...}×{1,2,...,S}×{1,2,···,k}.
4.1 INFINITESIMAL GENERATOR MATRIX
Q=
B1A0
A2A1A0
A2A1A0
.........
Where, each block has order (kS + 1) ×(kS + 1) and,
B1=
−β0B0,S
B1,0B1,1B1,S
Bs,0Bs,s Bs,S
BS,00BS,S
where, for 1≤i≤s
Bi,i =
−(λ1+β+α)α
−(λ2+β+α)α
......
−(λk+β+α)
, for s+1≤i≤S
Bi,i =
−(λ1+α)α
......
−(λq+β+α)α
−(λk+β+α)
, for 1≤i≤S,
Bi,0=
0
0
.
.
.
α
,
B0,S =β0··· 0
https://doi.org/10.17993/3cemp.2022.110250.15-31
for 1≤i≤s
Bi,S =
β0··· 0
β0··· 0
.
.
.··· ....
.
.
β0··· 0
,
and for s+1≤i≤S
Bi,S =
00··· 0
.
.
.··· ....
.
..
.
.
β0··· 0
.
.
.··· ....
.
.
β0··· 0
,
A0=
00··· ··· 0
0L...··· 0
.
.
....L··· 0
.
.
...........
.
.
0··· ··· 0L
L=
λ10··· 0
0λ2··· .
.
.
.
.
........
.
.
0··· ··· λk
A2=
00··· ··· 0
M10...··· .
.
.
0M.......
.
.
.
.
...........
.
.
0··· ··· M0
where,
M1=
µ1
µ2
.
.
.
µk
,
M=
µ10··· 0
0µ2....
.
.
.
.
........
.
.
0··· ··· µk
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3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
A1=
−β00··· ··· A0,S
A1,0A1,10...··· A1,S
.
.
..............
.
.
As,0··· 0As,s ...As,S
.
.
..............
.
.
AS,00··· ··· 0AS,S
where, for 1≤i≤s
Ai,i =
−(λ1+µ1+β+α)α··· 0
0−(λ2+µ2+β+α)....
.
.
.
.
........
.
.
.
.
.......α
0··· 0−(λk+µk+β+α)
, for s+1≤i≤S
Ai,i =
−(λ1+µ1+α)α··· 0
.
.
........
.
.
0−(λq+µq+β+α)....
.
.
.
.
........
.
.
0··· 0−(λk+µk+β+α)
, for 1≤i≤S,
Ai,0=
0
0
.
.
.
α
,
A0,S =β0··· 0
for 1≤i≤s
Ai,S =
β0··· 0
β0··· 0
.
.
.··· ....
.
.
β0··· 0
and
Ai,S =
00··· 0
.
.
.··· ....
.
..
.
.
β0··· 0
.
.
.··· ....
.
.
β0··· 0
https://doi.org/10.17993/3cemp.2022.110250.15-31
4.2 STEADY STATE ANALYSIS
Let A=A0+A1+A2and π=(π0,π
1,π
2, ..., πS)be the steady state probability distribution of
{(I(t),S(t)) : t≥0}, where πi=(πi1,π
i2, ..., πik),1≤i≤S.
Then we have,
−π0β+π1(M1+A1,0)+···+πSA1,0=0
πi(A1,1+L)+πi+1M=0,1≤i≤s
πi(AS,S +L)+πi+1M=0,s+1≤i≤S−1and
π0A0,S
+(
π1
+
π2
+
···
+
πS−1
)
A1,S
+
πS
(
AS,S
+
L
)=0. Simplifying,
πi
=
π1
[(
A1,1
+
L
)
M−1
]
i,
1
≤i≤s
πi=π1[(A1,1+L)M−1]s[(AS,S +L)M−1]i−s−1,s+1≤i≤S
Here π1and π0can be obtained from the normalizing condition and the first equation.
4.3 STABILITY CONDITION
Theorem: The system is stable if and only if,
S
i=1 k
j=1 πijλj<S
i=1 k
j=1 πijµj
Proof: The system is stable if and only if
πA0e<πA
2e
Substituting we get,
[0 π1Lπ
2L··· πSL]e<
[0 π1M1π2M··· πSM0]e
Simplifying, S
i=1 k
j=1 πijλj<S
i=1 k
j=1 πijµj
Stationary Distribution
Let
x
=(
x0,x
1,x
2,···
)be the steady state probability vector of the Makov process
X
(
t
)=(
N
(
t
)
,I
(
t
)
,S
(
t
))
,t≥
0.Then, xi=xi−1Ri,where Ris the minimal non negative solution of the matrix equation,
R2A2+RA1+A0=0
Here, xi=xi(j, m),1≤j≤S, 1≤m≤k.
4.4 PERFORMANCE MEASURES
1. Expected number of customers in the queue, Ec=∞
i=0 ixi(0) + ∞
i=0 S
j=1 k
m=0 ixi(j, m).
2. Expected number of items in the in the stage m, Em=∞
i=0 S
j=1 jxi(j, m).
3. Expected number of items scrapped, Es=∞
i=0 S
j=1 αjxi(j, k).
4. Probability that all items are sold in the first stage, Es=∞
i=1 µ1xi(j, 1).
5. Probability that all items are scrapped, x0(S, k).
4.5 DISTRIBUTION OF THE DURATION OF A CYCLE
The expected cycle length is the duration of time elapsed between two consecutive replenishment orders.
To compute this we proceed as follows.
Choose an
ϵ
and a
H
(
ϵ
)such that the probability of the number of customers in system exceeding
H
has probability
<ϵ
. Take
ϵ
to be quite small (of the order 10
−5
). Accordingly
H
will be large. Yet
we have a finite system. We have the corresponding system state probability vector; the sum of these
probabilities is very close to 1. Starting from any system state, with the number of items in inventory S
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3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
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26
A1=
−β00··· ··· A0,S
A1,0A1,10...··· A1,S
.
.
..............
.
.
As,0··· 0As,s ...As,S
.
.
..............
.
.
AS,00··· ··· 0AS,S
where, for 1≤i≤s
Ai,i =
−(λ1+µ1+β+α)α··· 0
0−(λ2+µ2+β+α)....
.
.
.
.
........
.
.
.
.
.......α
0··· 0−(λk+µk+β+α)
, for s+1≤i≤S
Ai,i =
−(λ1+µ1+α)α··· 0
.
.
........
.
.
0−(λq+µq+β+α)....
.
.
.
.
........
.
.
0··· 0−(λk+µk+β+α)
, for 1≤i≤S,
Ai,0=
0
0
.
.
.
α
,
A0,S =β0··· 0
for 1≤i≤s
Ai,S =
β0··· 0
β0··· 0
.
.
.··· ....
.
.
β0··· 0
and
Ai,S =
00··· 0
.
.
.··· ....
.
..
.
.
β0··· 0
.
.
.··· ....
.
.
β0··· 0
https://doi.org/10.17993/3cemp.2022.110250.15-31
4.2 STEADY STATE ANALYSIS
Let A=A0+A1+A2and π=(π0,π
1,π
2, ..., πS)be the steady state probability distribution of
{(I(t),S(t)) : t≥0}, where πi=(πi1,π
i2, ..., πik),1≤i≤S.
Then we have,
−π0β+π1(M1+A1,0)+···+πSA1,0=0
πi(A1,1+L)+πi+1M=0,1≤i≤s
πi(AS,S +L)+πi+1M=0,s+1≤i≤S−1and
π0A0,S
+(
π1
+
π2
+
···
+
πS−1
)
A1,S
+
πS
(
AS,S
+
L
)=0. Simplifying,
πi
=
π1
[(
A1,1
+
L
)
M−1
]
i,
1
≤i≤s
πi=π1[(A1,1+L)M−1]s[(AS,S +L)M−1]i−s−1,s+1≤i≤S
Here π1and π0can be obtained from the normalizing condition and the first equation.
4.3 STABILITY CONDITION
Theorem: The system is stable if and only if,
S
i=1 k
j=1 πijλj<S
i=1 k
j=1 πijµj
Proof: The system is stable if and only if
πA0e<πA
2e
Substituting we get,
[0 π1Lπ
2L··· πSL]e<
[0 π1M1π2M··· πSM0]e
Simplifying, S
i=1 k
j=1 πijλj<S
i=1 k
j=1 πijµj
Stationary Distribution
Let
x
=(
x0,x
1,x
2,···
)be the steady state probability vector of the Makov process
X
(
t
)=(
N
(
t
)
,I
(
t
)
,S
(
t
))
,t≥
0.Then, xi=xi−1Ri,where Ris the minimal non negative solution of the matrix equation,
R2A2+RA1+A0=0
Here, xi=xi(j, m),1≤j≤S, 1≤m≤k.
4.4 PERFORMANCE MEASURES
1. Expected number of customers in the queue, Ec=∞
i=0 ixi(0) + ∞
i=0 S
j=1 k
m=0 ixi(j, m).
2. Expected number of items in the in the stage m, Em=∞
i=0 S
j=1 jxi(j, m).
3. Expected number of items scrapped, Es=∞
i=0 S
j=1 αjxi(j, k).
4. Probability that all items are sold in the first stage, Es=∞
i=1 µ1xi(j, 1).
5. Probability that all items are scrapped, x0(S, k).
4.5 DISTRIBUTION OF THE DURATION OF A CYCLE
The expected cycle length is the duration of time elapsed between two consecutive replenishment orders.
To compute this we proceed as follows.
Choose an
ϵ
and a
H
(
ϵ
)such that the probability of the number of customers in system exceeding
H
has probability
<ϵ
. Take
ϵ
to be quite small (of the order 10
−5
). Accordingly
H
will be large. Yet
we have a finite system. We have the corresponding system state probability vector; the sum of these
probabilities is very close to 1. Starting from any system state, with the number of items in inventory S
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27
3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
(an order has been placed and got materialised), we look at the evolution until the next replenishment
takes place, either through realization of CLT or by the sales of the last item left in the inventory in
that cycle.
Consider the Markov chain X={X(t):t≥0}={(N(t),I(t),S(t)) : t≥0}
with the state space
{
(
i,j,l
):0
≤i≤H,
1
≤j≤S,
1
≤l≤k}U{
∆
}
, where
{
∆
}
is the absorbing
state of the Markov chain which denotes the realization of CLT.
Let qi=Pr{N(t)=i, I(t)=S, S(t)=1}
and
ηi
=
qi/H
i=0 qi
Then
η
=(
η0,η
1,........η
H
)is the initial probability vector of the distribution.
The infinitesimal generator of the process is
ˆ
U=VV0
00
where,
V=
VS,S VS,S−10··· 0VS,0
0VS−1,S−1VS−1,S−20··· VS−1,0
.
.
..............
.
.
0··· Vs,s Vs,s−1...Vs,0
.
.
..
.
...........
.
.
00 ··· ...V1,1V1,0
00 ··· ··· ··· V0,0
(S+1)×(S+1)
Here,
for s+1≤i≤S
Vi,i =
BS,S L0··· 0
0AS,S L··· 0
.
.
..
.
........
.
.
0··· ··· AS,S L
0··· ··· 0AS,S
(H+1)×(H+1)
and for 1≤i≤s
Vi,i =
B1,1L0··· 0
0A1,1L··· 0
.
.
..
.
........
.
.
0··· ··· A1,1L
0··· ··· 0A1,1
(H+1)×(H+1)
For 2≤i≤S
Vi,i−1=
0··· ··· 0
M...··· 0
.
.
........
.
.
0··· M0
(H+1)×(H+1)
https://doi.org/10.17993/3cemp.2022.110250.15-31
V1,0=
A1,00··· 0
M1A1,0...0
.
.
........
.
.
0··· M1A1,0
(H+1)×(H+1)
For 2≤i≤S
Vi,0=
A1,00··· 0
0A1,0...0
.
.
........
.
.
0··· ··· A1,0
(H+1)×(H+1)
V0,0=−βIH+1 and
V0=
0
.
.
.
0
B
.
.
.
B
(S+1)×1
B=
β
.
.
.
β
k×1
The probability density function of the distribution is, f(x)=ηeVx
V0
So the mean of the distribution is the first raw moment, µ′
1=−ηV −1e
Therefore expected cycle length=−ηV −1e
5 OPTIMIZATION PROBLEM
Based on the above performance measures we construct a cost function to find the profit of the retailer.
Let
Cr
be the price per unit of the item in the
rth
stage for
r
=1
,···,k
. Assume that the holding
cost increases with the age of the stocked item and let
Hr
be the holding cost per unit of the item
in the
rth
stage. Let
K
be the fixed ordering cost while placing a replenishment order and
c
be the
variable procurement cost (per unit) of the item to the store.
The total revenue =
k
r=1 (Cr−Hr)Er
+
CscrapEscrap −
(
K
+
cS
)
/cycle length
, where
Cscrap
is the
tagged price for scrapped item and Escrap is the expected number of scrapped items.
6 NUMERICAL ILLUSTRATION
In this section we provide numerical illustration of the profit function with variation in the values of
s
and Sin both Phase Type and Erlang cases.
Table 1: Effect of S and s on Profit in Phase-Type Case We fixthe parameters as
k
=4
,m
=
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3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
28
(an order has been placed and got materialised), we look at the evolution until the next replenishment
takes place, either through realization of CLT or by the sales of the last item left in the inventory in
that cycle.
Consider the Markov chain X={X(t):t≥0}={(N(t),I(t),S(t)) : t≥0}
with the state space
{
(
i,j,l
):0
≤i≤H,
1
≤j≤S,
1
≤l≤k}U{
∆
}
, where
{
∆
}
is the absorbing
state of the Markov chain which denotes the realization of CLT.
Let qi=Pr{N(t)=i, I(t)=S, S(t)=1}
and
ηi
=
qi/H
i=0 qi
Then
η
=(
η0,η
1,........η
H
)is the initial probability vector of the distribution.
The infinitesimal generator of the process is
ˆ
U=VV0
00
where,
V=
VS,S VS,S−10··· 0VS,0
0VS−1,S−1VS−1,S−20··· VS−1,0
.
.
..............
.
.
0··· Vs,s Vs,s−1...Vs,0
.
.
..
.
...........
.
.
00 ··· ...V1,1V1,0
00 ··· ··· ··· V0,0
(S+1)×(S+1)
Here,
for s+1≤i≤S
Vi,i =
BS,S L0··· 0
0AS,S L··· 0
.
.
..
.
........
.
.
0··· ··· AS,S L
0··· ··· 0AS,S
(H+1)×(H+1)
and for 1≤i≤s
Vi,i =
B1,1L0··· 0
0A1,1L··· 0
.
.
..
.
........
.
.
0··· ··· A1,1L
0··· ··· 0A1,1
(H+1)×(H+1)
For 2≤i≤S
Vi,i−1=
0··· ··· 0
M...··· 0
.
.
........
.
.
0··· M0
(H+1)×(H+1)
https://doi.org/10.17993/3cemp.2022.110250.15-31
V1,0=
A1,00··· 0
M1A1,0...0
.
.
........
.
.
0··· M1A1,0
(H+1)×(H+1)
For 2≤i≤S
Vi,0=
A1,00··· 0
0A1,0...0
.
.
........
.
.
0··· ··· A1,0
(H+1)×(H+1)
V0,0=−βIH+1 and
V0=
0
.
.
.
0
B
.
.
.
B
(S+1)×1
B=
β
.
.
.
β
k×1
The probability density function of the distribution is, f(x)=ηeVx
V0
So the mean of the distribution is the first raw moment, µ′
1=−ηV −1e
Therefore expected cycle length=−ηV −1e
5 OPTIMIZATION PROBLEM
Based on the above performance measures we construct a cost function to find the profit of the retailer.
Let
Cr
be the price per unit of the item in the
rth
stage for
r
=1
,···,k
. Assume that the holding
cost increases with the age of the stocked item and let
Hr
be the holding cost per unit of the item
in the
rth
stage. Let
K
be the fixed ordering cost while placing a replenishment order and
c
be the
variable procurement cost (per unit) of the item to the store.
The total revenue =
k
r=1 (Cr−Hr)Er
+
CscrapEscrap −
(
K
+
cS
)
/cycle length
, where
Cscrap
is the
tagged price for scrapped item and Escrap is the expected number of scrapped items.
6 NUMERICAL ILLUSTRATION
In this section we provide numerical illustration of the profit function with variation in the values of
s
and Sin both Phase Type and Erlang cases.
Table 1: Effect of S and s on Profit in Phase-Type Case We fixthe parameters as
k
=4
,m
=
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3C Empresa. Investigación y pensamiento crítico. ISSN: 2254-3376
Ed. 50 Vol. 11 N.º 2 August - December 2022
3,λ
1=4,λ
2=4.5,λ
3=5,λ
4=6,µ
1=7,µ
2=7.5,µ
3=8,µ
4=8.5,β =5,α= [0.4,0.4,0.2]
T=
−15 4 6
1−13 5
34−17
and costs as
C1
= 10
,C
2
=8
,C −
3=6
,C
4
=2
,C
scrap
=1
.
5
,H
1
=2
,H
2
=1
.
5
,H
3
=1
,H
4
=
1,c=5,K =2
S↓s→1234567
6 105.91 104.67 103.04 99.02 96.26
7 112.82 112.40 111.67 110.60 107.32 105.32
8 121.60 121.38 121.01 120.32 119.20 116.07 113.18
9 131.24 131.09 130.86 130.50 129.80 128.57 125.47
10 141.34 141.22 141.06 140.84 140.47 139.77 138.50
11 151.73 151.62 151.49 151.33 151.10 150.72 149.98
12 162.32 162.21 162.01 161.97 161.80 161.57 161.17
13 173.06 172.94 172.95 172.71 172.58 172.42 172.19
14 183.89 183.78 183.67 183.55 183.44 183.31 183.15
15 194.81 194.69 194.58 194.46 194.35 194.23 194.11
Table 2: Effect of S and s on Profit in Erlang Case We fix the parameters as
k
=4
,m
=
3,λ
1=4,λ
2=4.5,λ
3=5,λ
4=6,µ
1=7,µ
2=7.5,µ
3=8,µ
4=8.5,β =5,α=5
and the costs as
C1
= 10
,C
2
=8
,C −
3=6
,C
4
=2
,C
scrap
=1
.
5
,H
1
=2
,H
2
=1
.
5
,H
3
=1
,H
4
=
1,c=5,K =2
S↓s→1234567
6 75.09 64.82 53.67 41.60 29.13
7 87.14 77.71 67.49 56.25 43.83 30.73
8 98.51 89.98 80.77 70.57 59.13 46.30 32.52
9 109.19 101.58 93.35 84.23 73.92 62.19 48.88
10 119.28 112.55 105.31 97.24 88.08 77.58 65.51
11 128.74 122.86 116.52 109.42 101.34 92.04 81.27
12 137.66 132.57 127.05 120.86 113.78 105.60 96.10
13 146.07 141.69 136.94 131.58 125.42 118.27 109.93
14 154.08 150.34 146.27 141.66 146.35 130.14 122.88
15 161.66 158.49 155.03 151.10 146.53 141.17 134.89
From the table it is clear that the profit increses as
S
increases but it decreases as
s
increases. The
change of profit when sis changed is very sensitive in Erlang case compared to Phase Type case.
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Ed. 50 Vol. 11 N.º 2 August - December 2022
30
3,λ
1=4,λ
2=4.5,λ
3=5,λ
4=6,µ
1=7,µ
2=7.5,µ
3=8,µ
4=8.5,β =5,α= [0.4,0.4,0.2]
T=
−15 4 6
1−13 5
34−17
and costs as
C1
= 10
,C
2
=8
,C −
3=6
,C
4
=2
,C
scrap
=1
.
5
,H
1
=2
,H
2
=1
.
5
,H
3
=1
,H
4
=
1,c=5,K =2
S↓s→1234567
6 105.91 104.67 103.04 99.02 96.26
7 112.82 112.40 111.67 110.60 107.32 105.32
8 121.60 121.38 121.01 120.32 119.20 116.07 113.18
9 131.24 131.09 130.86 130.50 129.80 128.57 125.47
10 141.34 141.22 141.06 140.84 140.47 139.77 138.50
11 151.73 151.62 151.49 151.33 151.10 150.72 149.98
12 162.32 162.21 162.01 161.97 161.80 161.57 161.17
13 173.06 172.94 172.95 172.71 172.58 172.42 172.19
14 183.89 183.78 183.67 183.55 183.44 183.31 183.15
15 194.81 194.69 194.58 194.46 194.35 194.23 194.11
Table 2: Effect of S and s on Profit in Erlang Case We fix the parameters as
k
=4
,m
=
3,λ
1=4,λ
2=4.5,λ
3=5,λ
4=6,µ
1=7,µ
2=7.5,µ
3=8,µ
4=8.5,β =5,α=5
and the costs as
C1
= 10
,C
2
=8
,C −
3=6
,C
4
=2
,C
scrap
=1
.
5
,H
1
=2
,H
2
=1
.
5
,H
3
=1
,H
4
=
1,c=5,K =2
S↓s→1234567
6 75.09 64.82 53.67 41.60 29.13
7 87.14 77.71 67.49 56.25 43.83 30.73
8 98.51 89.98 80.77 70.57 59.13 46.30 32.52
9 109.19 101.58 93.35 84.23 73.92 62.19 48.88
10 119.28 112.55 105.31 97.24 88.08 77.58 65.51
11 128.74 122.86 116.52 109.42 101.34 92.04 81.27
12 137.66 132.57 127.05 120.86 113.78 105.60 96.10
13 146.07 141.69 136.94 131.58 125.42 118.27 109.93
14 154.08 150.34 146.27 141.66 146.35 130.14 122.88
15 161.66 158.49 155.03 151.10 146.53 141.17 134.89
From the table it is clear that the profit increses as
S
increases but it decreases as
s
increases. The
change of profit when sis changed is very sensitive in Erlang case compared to Phase Type case.
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Ed. 50 Vol. 11 N.º 2 August - December 2022