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One for one period policy for perishable inventory
Accepted Manuscript One for One Period Policy for Perishable Inventory Anwar Mahmoodi, Alireza Haji, Rasoul Haji PII: DOI: Reference:
S0360-8352(14)00343-X http://dx.doi.org/10.1016/j.cie.2014.10.012 CAIE 3835
To appear in:
Computers & Industrial Engineering
Received Date: Revised Date: Accepted Date:
17 August 2013 12 March 2014 20 October 2014
Please cite this article as: Mahmoodi, A., Haji, A., Haji, R., One for One Period Policy for Perishable Inventory, Computers & Industrial Engineering (2014), doi: http://dx.doi.org/10.1016/j.cie.2014.10.012
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Title One for One Period Policy for Perishable Inventory
Abstract Recently, for zero ordering cost a new ordering policy named (1, T), in which the time interval between two consecutive orders and the value of the order size are both constant, have been developed for nonperishable products. In this paper, the (1, T) policy is developed for perishable products. Using an analogy among this inventory model, a queueing model with impatient customers, and a finite dam model, the long-run average total cost function of the inventory system is derived. It is observed that the total cost rate is independent from the lead time as is for nonperishable products. Since analyzing the convexity of the model is extremely complicated, a proposition is proved to define a domain for the optimal solution, and then a search algorithm is presented to obtain the optimal solution. Furthermore, a numerical analysis is carried out to examine the sensitivity of optimal T with respect to system parameters and to compare the performance of (1, T) policy with the well known (S–1, S) policy. This analysis shows that for fixed values of system parameters, there is a fixed value of lead time for which the performance of (1, T) policy is better than (S–1, S) policy. Further as the lead time increases this superiority is more pronounced. Keywords: Inventory Control, Perishable Products, Queueing Theory, Impatient Customers, Finite Dam Model
1. Introduction Considering nonperishable products, Haji and Haji (2007) introduce a new inventory control policy called (1, T), which is different from the classical inventory policy used in the literature of inventory and production control systems. The (1, T) policy is to order one unit at each fixed time period T. For the case of stochastic demand, the (1, T) inventory policy is the one in which the time interval between two consecutive orders and the value of the order size are both constant. Therefore, when (1, T) is employed in the first level of a supply chain, it prevents
1
expanding the demand uncertainty for other levels and makes their demand deterministic, one unit every T units of time. Hence, the following advantages are obtained using the (1, T) policy in retailers’ level of a supply chain (Haji and Haji (2007)): 1) The safety stock in supplier is eliminated (cost reduction). 2) Shortage cost in supplier due to uncertainty in demand is eliminated. 3) Information exchange cost for supplier due to the elimination of uncertainty of its demand is eliminated. 4) Inventory control and production planning in supplier are simplified. And 5) This policy is very easy to apply. Following Haji and Haji (2007), Haji et al. (2009) apply the (1, T) policy to a two-echelon inventory system with nonperishable products. In this paper, we consider the (1, T) policy for perishable products in a single stage inventory system. The assumption of infinite lifetime is common in most of the inventory models. However, the perishability of products is a major problem of some industry sectors in which by disregarding the finite lifetime of their products the resulting model may give inaccurate results. Hence, significant studies have been carried out for inventory control of perishable products. Nahmias (1982) and Karaesman et al. (2011) present two comprehensive reviews focusing on inventory control of perishable products. Schmidt and Nahmias (1985) for perishable items consider a system operating under (S–1, S) policy with lost sales, Poisson demand, outdating costs, purchase costs, and per unit per period holding costs. They derive the cost function by solving partial differential equations for the Sdimensional stochastic process corresponding to the time elapsed since the last S orders were placed. Perry and Posner (1998) generalize Schmidt and Nahmias (1985) to allow for general types of customer impatience behavior. Olsson and Tydesjo (2010) extend Schmidt and Nahmias (1985) by allowing backorders. In this study, we consider an inventory system under Schmidt and Nahmias (1985) assumptions but employ the (1, T) policy instead of the (S–1, S) policy. Using some concepts from queueing theory, it is shown that the considered model is similar to a D/M/1 queue with impatient customer. The long-run average total cost function including the total outdating, holding, shortage and purchase costs is derived utilizing the analogy of the D/M/1 queue with impatient customer and the finite dam model of Moran (1954). Since analyzing the convexity of the model is extremely complicated, a proposition is presented to introduce a domain for optimal T. Therefore, the optimal solution can be obtained using a search algorithm. 2
Haji and Haji (2007) prove the independency of the total cost function of (1, T) policy from the lead time for nonperishable products. In this study, we show that this result is also valid for the case of perishable products. Further, for the case of perishable products we have compared (1, T) and (S–1, S) policies. The results show that for fixed values of system parameters, there is a fixed value of lead time for which the performance of (1, T) policy is better than (S–1, S) policy. Furthermore, thorough a numerical example we show as the lead time increases this superiority is more pronounced. The proceeding parts of this paper are organized as follow: In section 2, the model and its analogy to other models is described. In section 3, the cost function of the model is derived and an algorithm is presented to obtain the optimal solution. In section 4, a numerical analysis is carried out. Finally in section 5, the conclusion is presented.
2. Problem Description and mathematical formulation 2.1 Problem Description An inventory system with lost sales, Poisson demand, outdating costs, purchase costs, and per unit per period holding costs is considered. Using (1, T) policy in which an order constantly is placed for one unit of product in each constant time interval, the long-run average total cost function including the total outdating, holding, shortage and purchase costs is derived. The objective is to determine the optimal time interval between any two consecutive orders which minimizes the long-run average total cost. It is assumed that the product shelf life is finite and constant, the lead time for an order is constant, and the fixed ordering cost is zero or negligible. General Notations μ
The demand rate.
π
Cost of a lost sale.
h
Rate of holding cost.
p
Cost of a perished product.
c
Unit price.
τ
Lead time.
m
Product shelf life. 3
T
Time interval between any two consecutive orders in (1, T) policy.
L
The long-run average number of units in system.
HC
Average holding cost per time unit.
Π
Average shortage cost per time unit.
OC
Average perishing cost per time unit.
PC
Average purchase cost per time unit.
C(T) Average total cost rate, for the (1, T) policy. C(S) Average total cost rate, for the (S–1, S) policy. wqn
The queue waiting time of the nth customer.
sn
The required service time of the nth customer.
Wq
The random variable of the queue waiting time of an arriving customer (product) in steady state.
S
The random variable of the required service time of a customer (product) in steady state.
S
The random variable of the occurred service time of a customer (product) in steady state.
Ws
A customer (product) average waiting time in the queueing system in steady state.
2.2. Methodology To obtain the total cost function of the system, one can resort to some concepts of queueing theory. To do this, consider a D/M/1 queueing system with impatient customers in which: 1) The arrival process is the arrival units of product to the system and the inter-arrival times are constant, equal to T, 2) The customers’ (product units’) patience time is the product shelf life. That is, the customer gives up whenever his sojourn time is larger than m, and 3) The inter-demand times, exponentially distributed with mean 1/μ, are the required service times of these units. Hence, the inventory problem can be interpreted as a D/M/1 queue with impatient customers, a single channel queueing system in which the arrival process is deterministic with rate 1/T, the 4
service times have exponential distribution with mean 1/μ, and the customer leaves the system whenever his sojourn time is larger than m. It is clear that the number of units in this queueing system is equal to the inventory on hand in the above inventory system. In the considered queueing system, customers (products) arrive at the instances 0, T, 2T, …, nT, … and are served in the order of their arrival. These customers require service for times s0, s1, …, sn, … . A customer will wait in the system only for a time not exceeding a fixed time m. We number the customers (arriving units to the inventory system) according to their arrival times, and define wqn as the waiting time in queue of the nth customer (product). The following relation can be obtained between wq(n+1) and wqn using a manner similar to that of Lindley (1952).
w q ( n +1)
⎧0 ⎪ = ⎨w qn + s n −T ⎪ ⎩ m −T
w qn + s n ≤ T
if if
T < w qn + s n < m
if
; n = 0,1,...
m ≤ w qn + s n
Or equivalently w q ( n +1) = [w qn + min (s n , m − w qn ) −T ]+ = [min(w qn + s n , m ) −T ]+ ; n = 0,1,...
(1)
Where [ x ]+ = max (0, x ) . Thus, the sequence of { wqn } forms a Markov chain with the state space [0, m − T]. The probability of a product perishing is Pr( S + Wq > m ) . Hence, to estimate the perishing costs, we have to obtain the distribution function of Wq . Define Fn ( x ) = Pr(Wqn ≤ x ) as the distribution function of Wqn , and let F ( x ) = lim n→∞ Fn ( x ) when the limit exists. To obtain F (x ) , we use the analogy of the finite dam model and D/M/1 queues with impatient customers.
2.3 The finite dam model Daley (1964) clarifies the analogy of the queueing system with impatient customers and the finite dam model. Assume a dam with finite capacity m, and consider its storage Zt for discrete time t = 0,1, 2,... . Suppose in the interval (t , t + 1) , an amount of Xt flows into the dam, filling it 5
to the level of min ( Z t + X t , m ) and any excess water being lost over the spillway. A fixed demand for an amount T occurs just before the instant t + 1 , and is met as fully as possible by the release of the quantity min[T , min ( Z t + X t , m)] = min (T , Z t + X t , m ) . Thus an amount Z t +1 leaves in the dam, where Z t +1 = min ( Z t + X t , m ) − min (T , Z t + X t , m ) ⇒ Z t +1 = [min ( Z t + X t , m ) − T ]+
(2)
If { X t } is assumed to be a sequence of independent non-negative random variables from an exponential distribution with parameter μ, equation (2) is of the same form as (1). Therefore, the distribution functions of {Z t } and {Wqn } are the same. Moran (1955) shows that h( y ) , the density function of Y = Z t + X t , satisfies the following equation (Prabhu (1958)):
⎧αμe μ y ⎪ ⎪ ⎪ h (m − y ) = ⎨ μy ⎪αμe ⎪ ⎪⎩
; − ∞ < y ≤T ⎛ n (− μ )i −i μT ⎞ e ( y − iT )i ⎜ μ∑ ⎟ ⎜ i =0 i ! ⎟ n ⎜ ⎟ (− μ )i −i μT e ( y − iT )i −1 ⎟ ⎜ +∑ ⎝ i =1 (i − 1)! ⎠
; nT < y ≤ (n + 1)T ; n = 1,..., N
(3)
m ⎡m⎤ Where N = floor ( ) = ⎢ ⎥ , and α is the normalizing constant which is obtained from Prabhu T ⎣T ⎦
(1958) as below:
α=
e −μm N
1 + ∑ (− μ ) e i
i =1
− i μT
(4)
[m − iT ] i i!
Using (3), the distribution function of Y = Z t + X t , H ( y ) is
6
n ⎧ ⎞ m − ( n + 1)T ≤ y < m − nT , ( − μ)i − iµT µ( m− y ) ⎛ − 1 αe e ( m − y − iT )i ⎟ ; ⎪ ⎜∑ H ( y) = ⎨ ⎝ i =0 i ! ⎠ and n = 1,..., N . ⎪ µ( m− y ) ;m − T ≤ y < ∞ ⎩1 − αe
(5)
For the dam storage, Z t , we have the relations Pr( Z t +1 = 0) = Pr( Z t + X t ≤ T ) = H (T ) Pr( Z t +1 < z ) = Pr( Z t + X t ≤ T + z ) = H (T + z ) ; 0 < z < m − T Pr( Z t +1 = m − T ) = Pr( Z t + X t > m ) = 1 − H ( m) Using these relations, Prabhu (1958) obtain the distribution function of Z t in steady state, F (z ) , with the assumption that m = N ⋅ T . Let this assumption to be released for T ≤ m , then F (z ) can be obtained as follows: N −1 ⎧ ⎞ ( − μ ) i − i μT μ ( m −T ) ⎛ = − (0) 1 α F e e ( m − (i + 1)T )i ⎟ ⎪ ⎜∑ ⎝ i =0 i ! ⎠ ⎪ i ⎪ N −1 ⎛ ( − μ ) − i μT ⎞ ⎪ F ( z ) = 1 − α e μ ( m − z −T ) ⎜ ∑ e ( m − z − (i + 1)T )i ⎟ ; ⎪ ⎝ i =0 i ! ⎠ ⎨ i n −1 ⎪ F ( z ) = 1 − α e μ ( m − z −T ) ⎛ ( − μ ) e − iμT ( m − z − (i + 1)T )i ⎞ ; ∑ ⎜ ⎟ ⎪ i! i =0 ⎝ ⎠ ⎪ ; ⎪ F ( z) = 1 ⎪ ⎩
0 ≤ z < m − NT m − ( n + 1)T ≤ z < m − nT ,
(6)
and n = 1, 2,..., N − 1 m −T ≤ z
Due to the similarity of (1) and (2), (6) is also the distribution function of Wq . So this distribution function is used to estimate the perishing costs.
7
3. Cost evaluation 3.1 Perishing cost Pr(Wq + S > m ) is the probability that a product is outdated, which is obtained from the relation
between the queueing system and the finite dam model as follows. Pr(Wq + S > m ) = Pr( X t + Z t > m ) = 1 − H ( m ) = α
(7)
Therefore, the proportion of products that is perished is
OC =
1 ⋅ Pr(Wq + S > m) . Thus T
p pα ⋅ Pr(Wq + S > m) = T T
(8)
3.2 Holding cost Let L denote the long-run average number of units in the system, then by Little’s formula
L=
1 Ws , where Ws is a product average waiting time in the system. Ws can be obtained by T
conditioning on S + Wq. So Ws = mα + (1 − α ) E (Wq + S | S + Wq < m )
(9)
Where α is the proportion of customers that leave the queueing system without complete service according to (7) and (4). Also, m
E (Wq + S | S + Wq < m) = E (Y | Y < m ) =
∫ yh( y )dy 0
H (m)
According to (5), h( y ) can be written as
8
m
=
∫ yh( y )dy 0
1− α
(10)
⎧ ⎛ N ( − μ)i −iµT ⎞ e (m − y − iT )i + ⎪ ⎜ μ∑ ⎟ i ! ⎪αe µ ( m− y ) ⎜ i =0 ⎟ ;0 ≤ y < m − NT N ⎪ ⎜ ⎟ ( − μ)i − iµT i −1 e (m − y − iT ) ⎟ ⎪ ⎜ ∑ i =1 (i − 1)! ⎝ ⎠ ⎪ ⎪⎪ h( y ) = ⎨ ⎛ n ( − μ)i −iµT ⎞ ⎪ μ e (m − y − iT )i + ⎟ m − (n + 1) T ≤ y < m − nT , ⎪ µ ( m− y ) ⎜ ∑ ! i ⎜ i =0 ⎟ ; ⎪αe n ⎜ ⎟ and n = 1,..., N − 1 ( − μ)i − iµT ⎪ i −1 e (m − y − iT ) ⎟ ⎜ ∑ ⎪ i =1 (i − 1)! ⎝ ⎠ ⎪ µ ( m− y ) ;m − T ≤ y ⎩⎪αμe
(11)
m
Let Θ = ∫ yh( y )dy , thus from (9), (10) and Little’s formula, we have 0
L=
Ws 1 = (mα + Θ ) T T
(12)
Hence,
HC = hL = h
1 (mα + Θ ) T
(13)
3.3 Shortage cost In order to calculate the shortage cost, the proportion of time that the system is out of stock, P0 , is required. Stanford (1979) considers the customer impatience only on the queue time and provides a relation to obtain P0 . He also presents a conjecture for P0 , when the customer impatience is considered on the sojourn time, but our simulation shows that his conjecture is not valid for our case (see Appendix A for more details). Therefore, we provide a relation to obtain P0 , when the customer impatience is on the sojourn time. Theorem 1: For D/M/1 queues with customer impatience on the sojourn time, P0 is obtained from the following relation.
P0 = 1 −
1− α μT
(14)
9
Proof: See Appendix B. Since P0 is the proportion of time that the system is out of stock, the proportion of demand that is lost is µP0 . Therefore, the average shortage cost per unit time is
Π = πμP0 = πμ(1 −
1− α ) μT
(15)
3.4 Purchase cost The amount of products purchased per time unit is 1 T , Thus
PC =
c T
(16)
3.5 System total cost The system total cost is the summation of the perishing, holding, shortage, and purchase costs. Form (8), (13), (15) and (16), the total cost rate of the system can be written as:
C (T ) = OC + HC + Π + PC =
pα c 1 1− α + h (mα + Θ ) + πμ(1 − )+ T T μT T
(17)
m
Where α is given in (4) and Θ = ∫ yh( y )dy in which h( y ) is according to (11). 0
From (17), C (T ) is independent of lead time as it is the case for nonperishable products [see Haji and Haji (2007)]. Due to the existence of α and Θ in (17), analyzing the convexity of C (T ) is extremely complicated. Therefore, the following proposition is presented to introduce a domain for the optimal value of T denoted by T*. Proposition 1: If the establishment of the inventory system is affordable, say T * ≠ ∞ , then T* will be in the interval (0,m]. Proof: See Appendix C.
10
When the establishment of the inventory system is not affordable, say T * = ∞ , then total lost sales is the optimal solution and the average per time unit system total cost is πμ . Therefore, by using a search algorithm to find the best amount of T in (0, m] and compare its total cost rate with πμ , one can find the optimal solution of (1, T) inventory system. Let ε be a small value, the optimal solution of (1, T) can be found with the accuracy of ε using the following algorithm. Algorithm 1. Step 1: Set T = ε . Calculate C (T ) using (17). Set T * = T and C * = C (T ) . Step 2: If T + ε ≤ m , set T = T + ε , calculate C (T ) using (17), and go to step 3. Otherwise, go to step 4. Step 3: If C (T ) < C * , set T * = T and C * = C (T ) . Go to step 2. Step 4: If πμ < C * , set T * = ∞ and C * = πμ . Go to step 5. Step 5: The algorithm is finished and (T * , C * ) is the best solution of (1, T) inventory system.
4. Numerical Analysis This section is devoted to compare the performance of (1, T) and (S–1, S) policies in terms of the total cost rate, and also to examine the relation between T * and the system parameters. The Matlab software is used for coding both policies. For all problems in this section, we assume the unit of the product shelf life and lead time is year, and the unit of costs is a cost unit, say dollar. Schmidt and Nahmias (1985) obtain the total cost function for (S–1, S) policy, C (S ) , as follow: ⎛ e − μτ τ S e − μτ τ S e − μ ( τ + m ) ( τ + m ) S −1 ⎞ ) − Kτ + h ⎜ S − μτ (1 − K ⎟ S! S! ( S − 1)! ⎝ ⎠ − μ ( τ +m ) S −1 e ( τ + m) + (c + p ) K ( S − 1)!
C ( S ) = cμ + ( π − c ) μK
11
(18)
Where K =
1 e
− μτ
S −1 − μx τ +m x τ e dx −∫ τ S! ( S − 1)! S
, and τ stands for the lead time.
From (17) one can see that the total cost of (1, T) policy is independent from the lead time. On the contrary, the total cost of (S–1, S) policy depends on this parameter which could be seen from (18). Therefore, we are interested in comparison between (1, T) and (S–1, S) policies when the lead time is varied. To do this, let µ = 10 , c = 100 , h = 20 , m = 0.1 , p = 50 , π = 300 and C ( S * ) − C (T * ) ε = 1 400 . Also, let S represent the optimal value of S, and % ΔC = × 100 show C(S * ) *
how much in percent the (1, T) policy performs better than the (S–1, S) policy. For various amounts of lead time the optimal solution is obtained for both policies. The results are shown in Table 1 and Figure 1. While in (1, T) policy the total cost is not sensitive to lead time, the total cost for the case of (S–1, S) policy is. As τ increases, the total cost of (S–1, S) policy increases but the total cost of (1, T) policy remains constant. This means that for fixed values of other parameters, there is a fixed value of τ for which the performance of (1, T) policy is better than (S–1, S) policy. Moreover as τ increases this superiority is more pronounced. Table 1: Comparison between (S–1, S) policy and (1, T) policy when the lead time is varied τ 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15
T* 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085
C(T*) 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281
S* 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
[Figure 1is here] 12
C(S*) 2044 2159 2249 2322 2382 2432 2475 2511 2519 2523 2530 2538 2546 2556 2565
%ΔC -11.6% -5.7% -1.4% 1.8% 4.2% 6.2% 7.8% 9.2% 9.4% 9.6% 9.8% 10.1% 10.4% 10.8% 11.1%
We also are interested in the interaction effects of the lead time, the product shelf life and the cost parameters. So we consider all combinations of τ ∈ {0.1, 0.2, 0.3} , m ∈{0.1, 0.2, 0.3} ,
π ∈ {200,300,400} , p ∈ {25,50,75} and h ∈ {15,30} . The fixed parameters are µ = 10 and c = 100 . We also set ε = 1 400 in algorithm 1. The results for h = 15 and h = 30 are presented in Table 2 and Table 3, respectively. Since the result of (1, T) policy is not sensitive to the lead time, it has just one solution for any arbitrary lead time. Therefore, in these tables, for any combinations of other parameters we have presented one solution for (1, T) policy but three solutions for (S–1, S) policy related to three different lead times. Table 2: Numerical results for all combination of varied parameters, µ=10, c=100, and h=15 Problem No.
m
π
p
T*
C(T*)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3
200 200 200 300 300 300 400 400 400 200 200 200 300 300 300 400 400 400 200 200 200 300 300 300 400 400 400
25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75
0.1 0.1 ∞ 0.0775 0.085 0.0925 0.065 0.07 0.075 0.1025 0.11 0.115 0.085 0.09 0.0925 0.0775 0.08 0.0825 0.105 0.1075 0.1125 0.09 0.0925 0.095 0.0825 0.085 0.0875
1837 1929 2000 2154 2278 2386 2345 2509 2656 1498 1545 1587 1671 1745 1811 1780 1871 1956 1361 1392 1420 1487 1537 1583 1567 1630 1688
*
S 1 1 0 2 2 1 3 2 2 2 2 2 3 2 2 3 3 3 2 2 2 3 3 3 3 3 3
τ=0.1 %ΔC C(S*) 1900 3.3% 1957 1.4% 2000 0.0% 2387 9.8% 2519 9.6% 2626 9.1% 2736 14.3% 2877 12.8% 3009 11.7% 1530 2.1% 1582 2.3% 1634 2.9% 1748 4.4% 1837 5.0% 1890 4.2% 1857 4.1% 1958 4.4% 2058 5.0% 1350 -0.8% 1372 -1.5% 1394 -1.9% 1440 -3.3% 1489 -3.2% 1539 -2.9% 1522 -3.0% 1572 -3.7% 1621 -4.1%
13
*
S 1 1 0 3 2 2 4 3 3 3 2 2 4 4 3 5 4 4 3 3 3 4 4 4 5 5 4
τ=0.2 %ΔC C(S*) 1938 5.2% 1973 2.2% 2000 0.0% 2489 13.5% 2610 12.7% 2689 11.3% 2852 17.8% 3034 17.3% 3163 16.0% 1604 6.6% 1661 7.0% 1692 6.2% 1850 9.7% 1945 10.3% 2014 10.1% 2026 12.1% 2121 11.8% 2216 11.7% 1409 3.4% 1438 3.2% 1467 3.2% 1549 4.0% 1601 4.0% 1652 4.2% 1641 4.5% 1722 5.3% 1785 5.4%
*
S 1 1 0 4 3 3 5 5 4 4 3 3 5 5 4 6 6 5 4 4 4 5 5 5 6 6 6
τ=0.3 %ΔC C(S*) 1955 6.0% 1981 2.6% 2000 0.0% 2532 14.9% 2641 13.7% 2731 12.6% 2918 19.6% 3088 18.8% 3228 17.7% 1642 8.8% 1683 8.2% 1723 7.9% 1908 12.4% 2000 12.8% 2075 12.7% 2086 14.7% 2211 15.4% 2305 15.1% 1441 5.6% 1475 5.6% 1509 5.9% 1612 7.8% 1665 7.7% 1718 7.9% 1711 8.4% 1789 8.9% 1866 9.5%
Table 3: Numerical results for all combination of varied parameters, µ=10, c=100, and h=30 Problem No.
m
π
p
T*
C(T*)
28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3
200 200 200 300 300 300 400 400 400 200 200 200 300 300 300 400 400 400 200 200 200 300 300 300 400 400 400
25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75
0.1 0.1 ∞ 0.0775 0.085 0.095 0.065 0.07 0.075 0.105 0.11 0.1175 0.085 0.09 0.0925 0.0775 0.08 0.085 0.105 0.11 0.115 0.09 0.095 0.0975 0.085 0.0875 0.09
1847 1939 2000 2167 2289 2396 2362 2524 2670 1514 1560 1601 1694 1766 1831 1806 1897 1980 1383 1413 1440 1518 1566 1610 1602 1664 1720
*
S 1 1 0 2 2 1 3 2 2 2 2 2 3 2 2 3 3 3 2 2 2 3 3 3 3 3 3
τ=0.1 %ΔC C(S*) 1906 3.1% 1962 1.2% 2000 0.0% 2400 9.7% 2531 9.6% 2632 9.0% 2756 14.3% 2890 12.7% 3021 11.6% 1546 2.1% 1598 2.4% 1650 3.0% 1773 4.5% 1853 4.7% 1905 3.9% 1883 4.1% 1984 4.4% 2084 5.0% 1367 -1.2% 1389 -1.7% 1411 -2.1% 1468 -3.4% 1518 -3.2% 1567 -2.7% 1551 -3.3% 1600 -4.0% 1650 -4.2%
*
S 1 1 0 3 2 2 4 3 3 3 2 2 4 4 3 4 4 4 3 3 3 4 4 4 5 5 4
τ=0.2 %ΔC C(S*) 1942 4.9% 1977 1.9% 2000 0.0% 2501 13.4% 2618 12.6% 2696 11.1% 2869 17.7% 3046 17.1% 3175 15.9% 1621 6.6% 1671 6.6% 1702 5.9% 1874 9.6% 1970 10.4% 2031 9.8% 2050 11.9% 2145 11.6% 2240 11.6% 1428 3.2% 1457 3.0% 1486 3.1% 1577 3.7% 1629 3.9% 1680 4.2% 1678 4.5% 1759 5.4% 1812 5.1%
*
S 1 1 0 4 3 3 5 5 4 3 3 3 5 5 4 6 5 5 4 4 4 5 5 5 6 6 6
τ=0.3 C(S*) 1958 1983 2000 2543 2650 2739 2934 3104 3239 1654 1695 1735 1931 2023 2092 2116 2235 2328 1462 1495 1529 1640 1693 1746 1747 1824 1901
%ΔC 5.7% 2.2% 0.0% 14.8% 13.6% 12.5% 19.5% 18.7% 17.6% 8.5% 8.0% 7.7% 12.3% 12.7% 12.5% 14.7% 15.1% 14.9% 5.4% 5.5% 5.8% 7.4% 7.5% 7.8% 8.3% 8.8% 9.5%
Most of the observed results agree with what one would intuitively expect. For example, as the perishing cost, p, increases, T* increases in order to decrease the average number of perished items. In addition, when the shortage cost, π, increases, T* decreases in order to decrease the average number of lost sales. Furthermore, the total cost rate increases as p or π increases, but it does decrease as m increases. To explain the sensitivity of T* with respect to p, consider Problems 13, 14 and 15 of Table 2 in which all parameters except p are fixed. The obtained values for T* are 0.085, 0.09 and 0.0925 14
for p = 25,50 and 75, respectively. Therefore, by 50% decreasing in p = 50 , T* decreases about 6%, and by 50% increasing in p = 50 , it increases about 3%. So with respect to p, T * is more sensitive at left. Also, consider Problems 11, 14 and 17 to see the sensitivity of T* with respect to π. The presented values for T* are 0.11, 0.09 and 0.08 for π = 200,300 and 400, respectively. Thus, by 50% decreasing in π = 300 , T* increases about 22%, and by 50% increasing in π = 300, it decreases about 11%. In addition, see Problems 5, 14 and 23 to find the sensitivity of T* with respect to m. For m = 0.1,0.2 and 0.3, respectively, T* is 0.085, 0.09 and 0.0925. Hence, 50% decreasing in m = 0.2 leads T* to be decreased about 6%, and 50% increasing in m = 0.2 leads it to be increased about 3%. Finally, to show the sensitivity of T* with respect to h, we can compare the results of table 2 and 3. In a few problems, T* increases when h increases, but in most of problems it remains constant. This may be due to the fact that h has a less amount than the other cost parameters and so a less impact on total cost. The positive amount of %ΔC shows that the (1, T) policy outperforms the (S–1, S) policy. From the results of Table 1, we know %ΔC increases as the lead time increases. We use the results of Table 2 and 3 to depict some diagrams which could help us to analyze the sensitivity of %ΔC with respect to the cost parameters and the product shelf life. Figures 2-5 present such diagrams which illustrate %ΔC versus π, p, m, and h, respectively, for different values of τ. From Figure 2 one can see that %ΔC is not a monotonic function of π. For τ = 0.1 , %ΔC at first is increasing and then decreasing with respect to π which means that the total cost of (1, T) policy is increased less than the total cost of (S–1, S) policy when π increases from 200 to 300, but it increases more when π increases from 300 to 400. However, for τ = 0.2 , and τ = 0.3 the total cost of (S–1, S) policy is more sensitive than the case of (1, T) policy. Thus %ΔC is an increasing function of π in these cases. Also, the sensitivity of %ΔC decreases as π increases. For example, in the case of τ = 0.2 , when π increases from 200 to 300, %ΔC increases about 3.3%, but it increases about 1.5% for the increasing of π from 300 to 400. [Figure 2 is here]
15
Figure 3 illustrates the behavior of %ΔC versus the changes in p. It shows that %ΔC is not a monotonic function of p. For all three cases of τ, %ΔC at first is increasing and then decreasing with respect to p. Furthermore, while its sensitivity does increase for τ = 0.1 , it decreases for the case of τ = 0.2 and τ = 0.3 . Namely, as p increases from 25 to 50, %ΔC increases by 0.6%, 0.6% and 0.4% for three lead times, respectively, but it decreases about 0.8%, 0.2% and 0.1% for increasing of p from 50 to 75. [Figure 3 is here] Figure 4 depicts %ΔC with respect to m for different values of lead time. It seems that %ΔC is a decreasing function of shelf life. As m increases the total cost of both (1, T) and (S–1, S) policies does decrease. But the decreasing intensity of total cost for the (S–1, S) policy is more than that of the (1, T) policy. Furthermore, as it can be seen, %ΔC is highly sensitive to variations in m, and it is much more when the lead time is small. In fact, for the case of τ = 0.1 , %ΔC decreases about 4.6%, when m increases from 0.1 to 0.2., and it decreases about 8.2% when m increases from 0.2 to 0.3. [Figure 4 is here] In Figure 5, the behavior of %ΔC versus the changes in h is shown. As it can be seen, the sensitivity of %ΔC to h is lower than its sensitivity to other parameters. When h increases from 15 to 30, for the case of τ = 0.1 and τ = 0.3 , %ΔC decreases about 0.3% and 0.1%, respectively, but for the case of τ = 0.2 , it increases about 0.1%. Thus, %ΔC is not a monotonic function of h. [Figure 5 is here] We have also performed a separate set of calculations to examine the sensitivity of the total cost of (1, T) model to the value of T in a neighborhood of the optimum. Fixed parameters are
p = 50, h = 15 , µ = 10 and c = 100 . The results are shown in Table 4. Each entry in this table represents the percentage increase in the total cost when T is set to 50% of its optimal value followed by the percentage increase when T is set to 150% of its optimal value. As it can be seen, the total cost is much more sensitive at left. For example, in the case of m = 0.2 and 16
π = 300 , when T is increased from the optimal by 50%, the total cost increases about 12%, but when it is decreased from its optimal by 50%, the total cost increases about 66%. Also as m or π increases, the percentage increase of total cost is more pronounced. Table 4: Sensitivity analysis on T in a neighborhood of the optimum π=200 π=300 π=400 m 38 , 1.3 38 , 8 53 , 9 0.1 49 , 5 66 , 12 77, 17 0.2 70 , 8 84 , 16 92 , 22 0.3 Finally, the computational efficiency of Algorithm 1 might be of interest in the sense that one might be curious to know that how much time is needed for finding the optimal policy of (1,T) model. To achieve this purpose, all problems of Table 2 and 3 are considered and the average time needed for these problems are presented in Table 5. The Matlab codes are performed on an Intel 2.53GHz core i3 CPU with 4.00 GB of RAM. Since in Algorithm 1, we search the interval of (0, m] by the step of ε, the required time depends on the value of m and ε. For the smaller value of ε, the more time is needed, but the more accurate result would be obtained. We believe, the needed accuracy in practical application does not require a very small value of ε. For example, when the shelf life of an item is 0.2 year (73 days), the accuracy of 1 day (ε = 1/365 year) is satisfactory in determining the optimal value of T. Therefore, based on the presented results, Algorithm 1 works rather well. Table 5: Average needed time (in second) for considered problems m 0.1 0.2 0.3
ε = 1 100
ε = 1 200
ε = 1 300
ε = 1 400
ε = 1 500
0.1577 0.5087 1.0513
0.5226 1.8365 3.9789
1.0400 3.9668 8.8479
1.8876 7.0651 15.6625
2.8158 10.9599 24.4376
5. Conclusion For the case of stochastic demand, the (1, T) inventory policy is the one in which the time interval between two consecutive orders and the value of the order size are both constant. In this study, the (1, T) policy was developed for perishable products. This inventory problem was interpreted as a D/M/1 queueing system with impatient customers in which if a customer’s sojourn time exceeds a predetermined constant value then he leaves the system. 17
The analogy between the considered queueing model and the finite dam model was used to develop the (1, T) cost function. Since it is extremely difficult to prove the convexity of the cost function, a proposition was presented to introduce a domain for optimal T. Furthermore, a search algorithm was developed to obtain the optimal solution. In addition, a numerical analysis was carried out to compare (1, T) policy with the well known (S–1, S) policy. According to this comparison, for fixed values of system parameters, there is a fixed value of lead time for which the (1, T) policy is better than the (S–1, S) policy. Further as the lead time increases this superiority is more pronounced. Moreover, it was observed that the (1, T) is independent from the lead time. Further, as the lost sale and single stage system was considered in this paper, developing the (1, T) policy when the shortages are backordered is introduced for future research. Also, considering a multi echelon inventory system with (1, T) policy can be an attractive direction.
References Daley, D. J., 1964. Single server queueing systems with uniformly limited queueing time. Journal of the Australian Mathematical Society. 4, 489–505. Haji, R., Haji, A., 2007. One-for-one period policy and its optimal solution. Journal of Industrial and Systems Engineering. 1, 200–217. Haji, R., Pirayesh Neghab, M., Baboli, A., 2009. Introducing a new ordering policy in a twoechelon inventory system with Poisson demand. International Journal of Production Economics. 117, 212–218 Karaesman, I., Scheller-Wolf, A., Deniz, B., 2011. Managing Perishable and Aging Inventories, Review and Future Research Directions, in: Kempf, K., Keskinocak, P., Uzsoy, P. (Eds.), Handbook of Production Planning, Kluwer International Series in Operations Research and Management Science, Advancing the State-of-the-Art Subseries. Lindley, D. V., 1952. The theory of queues with a single server. Proceedings of the Cambridge Philosophical Society. 48, 277–289. Moran, P. A. P., 1955. A probability theory of dams and storage systems: modifications of the release rules. Ibid. 6, 117–130. 18
Nahmias, S., 1982. Perishable inventory theory: a review. Operations Research. 30, 680–708. Olsson, F., 2010. Inventory Problems with perishable items: Fixed lifetime and backlogging. European Journal of Operational Research. 202, 131–137 Perry, D., Posner, M., 1998. An (S–1, S) inventory system with fixed shelf life and constant lead times. Operations Research. 46, 65–71. Prabhu, N. U., 1958. On the integral equation for the finite dam. Quarterly Journal of Mathematics. 2, 183–188. Schmidt, C. P., Nahmias, S., 1985. (S–1, S) policies for perishable inventory. Management Science. 31, 719–728. Stanford, R. E., 1979. Reneging phenomena in single channel queues. Mathematics of Operations Research. 4, 162–178
Appendix A. The comparison of Stanford’s Conjecture with simulation For the case in which the customer impatience is on system waiting time, the conjecture of Stanford (1979) with our notations could be written as P0 = 1 −
(1 − α ) E ( S | S + Wq < m )
(A.1)
T
To show whether or not (A.1) is valid in our case, we compare its result with simulation for 4 problems with different parameter settings. Each simulation consists of 10 runs, each with a run length of 10000 time units. We present the mean and the standard deviation of 10 runs for each problem. The results are presented in Table A.1. As it can be seen the result of Stanford’s conjecture is not compatible with simulation results. Table A.1: Numerical results of Stanford’s conjecture and simulation
Problem 1 2 3 4
M
μ
0.2 0.5 1 2 5 2 0.2 2
5 5 1 5
T
P0 Stanford’s conjecture 0.4660 0.8001 0.8586 0.1470
19
Mean of 10 simulation runs (Standard dev.) 0.1767 (0.0022) 0.7996 (0.0011) 0.8276 (0.0011) 0.0497 (0.0025)
Appendix B. Proof of Theorem 1. Since S is exponential, it is clear that E ( S | S + Wq > m ) = m − E (Wq ) + E ( R ) , where R is the remaining time of an exponential random variable and E ( R ) = 1 µ . Thus
E ( S | S + Wq > m) = m +
1 − E (Wq ) µ
(B.1)
Also,
E ( S ) = E ( S | S + Wq < m) ⋅ Pr( S + Wq < m) + E ( S | S + Wq > m) ⋅ Pr( S + Wq > m) = (1 − α ) E ( S | S + Wq < m) + αE ( S | S + Wq > m) Therefore, we can write
E ( S | S + Wq < m) =
1 α (m − E (Wq )) − μ 1− α
(B.2)
Clearly, E ( S + Wq | S + Wq > m ) = m + E ( R ) . So
E ( S + Wq | S + Wq > m) = m +
1 µ
(B.3)
In addition, E ( S + Wq | S + Wq > m ) = E (Wq | S + Wq > m ) + E ( S | S + Wq > m )
(B.4)
Furthermore, From (B.1), (B.3), and (B.4) we have E (Wq | S + Wq > m ) = E (Wq )
(B.5)
Also, equation (B.5) implies that E (Wq | S + Wq ≤ m ) = E (Wq )
(B.6)
Let S be the occurred service time of a customer. The expected value of S can be obtained as follows. 20
E ( S ) = (1 − α ) E ( S | S + Wq ≤ m ) + αE ( S | S + Wq > m )
(B.7)
Clearly, for S + Wq ≤ m we can write E ( S | S + Wq ≤ m ) = E ( S | S + Wq ≤ m )
(B.8)
And also for S + Wq > m we have E ( S | S + Wq > m ) = m − E (Wq )
(B.9)
Moreover, from (B.8) and (B.9) we can write (B.7) as follows. E ( S ) = (1 − α ) E ( S | S + Wq ≤ m ) + α (m − E (Wq ) )
(B.10)
Now from (B.2) and (B.10) we have
⎛1 α (m − E (Wq ))⎟⎟⎞ + α(m − E (Wq )) E ( S ) = (1 − α )⎜⎜ − ⎝ μ 1− α ⎠ Or equivalently,
E(S ) =
1− α μ
(B.11)
Furthermore, it is known from Little’s formula that P0 = 1 − λE ( S )
(B.12)
Where λ = 1 T is the arrival rate. Therefore, from (B.11) and (B.12) P0 = 1 −
E(S ) 1− α =1− T μT
(B.13)
■ Appendix C. Proof of Proposition 1. We obtain the total cost rate for m ≤ T . Since for this case the value of Wq is zero, we have
⎧S if S ≤ m S =⎨ , ⎩m if S > m
(C.1)
From (C.1), it is clear that Ws = E (S )
(C.2) 21
Using Little’s formula ( L = λWs ) and (C.2), the long-run average number of units in the system is obtained as follows. L=
E(S ) T
Hence, HC = hL = h
E(S ) T
(C.3)
Since Wq = 0 , the probability that a product is outdated is obtained as follows.
α = Pr(S > m) = e − µm Therefore, the proportion of products that is perished is
OC =
α . Thus T
pα T
(C.4)
Furthermore, the proportion of time that the system is out of stock can be obtained from (B.12) as follows. P0 = 1 −
E(S ) T
(C.5)
P0 is the proportion of time that the system is out of stock, so the proportion of demand that is lost is µP0 . Therefore, Π = πμP0 = πμ (1 −
E(S ) ) T
(C.6)
In addition, the amount of products purchased per time unit is 1 T , Thus
PC =
c T
(C.7)
22
Form (C.3), (C.4), (C.6), and (C.7), the total cost rate of the system for m ≤ T can be written as:
C (T ) = HC + OC + Π + PC = hL +
pα c + πμP0 + ⇒ T T
E ( S ) pe − µm E(S ) c C (T ) = h + + πμ(1 − )+ ⇒ T T T T C (T ) =
E (S ) pe − μ m c + + πμ (h − πμ ) + T T T
From (C.8), if
(C.8)
E(S ) pe− µm c (h − πμ) + + ≥ 0 , It is clear that T * = ∞ . Moreover, if T T T
E(S ) pe − µm c (h − πμ) + + < 0 , clearly C(T) is increasing with respect to T, thus T * = m . T T T Therefore, if T * ≠ ∞ , T* will be in the interval of (0, m]■
Caption of figures Figure 1: Total cost versus the lead time for both policies. Figure 2: %ΔC versus π (p=50, m=0.2, h=15). Figure 3: %ΔC versus p (π =300, m=0.2, h=15). Figure 4: %ΔC versus m (π =300, p=50, h=15). Figure 5: %ΔC versus h (π =300, p=50, m=0.2).
23
3000 2500
Total Cost
2000 1500
(1,T) Policy (S-1,S) Policy
1000 500 0
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 Lead time
24
%ΔC
18.0% 16.0% 14.0% 12.0% 10.0% 8.0% 6.0% 4.0% 2.0% 0.0%
τ=0.1 τ=0.2 τ=0.3
200
300
25
400
π
14.0% 12.0%
%ΔC
10.0% τ=0.1
8.0%
τ=0.2
6.0%
τ=0.3
4.0% 2.0% 0.0%
p 25
50
26
75
16.0% 14.0% 12.0% 10.0%
%ΔC
8.0%
τ=0.1
6.0%
τ=0.2
4.0%
τ=0.3
2.0% 0.0% -2.0%
0.1
0.2
-4.0% -6.0%
27
0.3
m
14.0% 12.0%
%ΔC
10.0% 8.0%
τ=0.1
6.0%
τ=0.2 τ=0.3
4.0% 2.0% 0.0%
h 15
30
28
Title One for One Period Policy for Perishable Inventory
Anwar Mahmoodi1, Alireza Haji1,*, Rasoul Haji1 1
Department of Industrial Engineering, Sharif University of Technology, Tehran, Iran
*
Corresponding Author; Email: [email protected], Tel.: +982166165704, Fax: +982166022702
29
Highlights The (1,T) policy is developed for perishable inventories. The system cost rate is provided using an analogy from queueing and finite dam models The (1,T) policy outperforms the (S–1,S) policy when the lead time is long enough. The (1,T) policy is independent from the lead time.
30
S0360-8352(14)00343-X http://dx.doi.org/10.1016/j.cie.2014.10.012 CAIE 3835
To appear in:
Computers & Industrial Engineering
Received Date: Revised Date: Accepted Date:
17 August 2013 12 March 2014 20 October 2014
Please cite this article as: Mahmoodi, A., Haji, A., Haji, R., One for One Period Policy for Perishable Inventory, Computers & Industrial Engineering (2014), doi: http://dx.doi.org/10.1016/j.cie.2014.10.012
This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
Title One for One Period Policy for Perishable Inventory
Abstract Recently, for zero ordering cost a new ordering policy named (1, T), in which the time interval between two consecutive orders and the value of the order size are both constant, have been developed for nonperishable products. In this paper, the (1, T) policy is developed for perishable products. Using an analogy among this inventory model, a queueing model with impatient customers, and a finite dam model, the long-run average total cost function of the inventory system is derived. It is observed that the total cost rate is independent from the lead time as is for nonperishable products. Since analyzing the convexity of the model is extremely complicated, a proposition is proved to define a domain for the optimal solution, and then a search algorithm is presented to obtain the optimal solution. Furthermore, a numerical analysis is carried out to examine the sensitivity of optimal T with respect to system parameters and to compare the performance of (1, T) policy with the well known (S–1, S) policy. This analysis shows that for fixed values of system parameters, there is a fixed value of lead time for which the performance of (1, T) policy is better than (S–1, S) policy. Further as the lead time increases this superiority is more pronounced. Keywords: Inventory Control, Perishable Products, Queueing Theory, Impatient Customers, Finite Dam Model
1. Introduction Considering nonperishable products, Haji and Haji (2007) introduce a new inventory control policy called (1, T), which is different from the classical inventory policy used in the literature of inventory and production control systems. The (1, T) policy is to order one unit at each fixed time period T. For the case of stochastic demand, the (1, T) inventory policy is the one in which the time interval between two consecutive orders and the value of the order size are both constant. Therefore, when (1, T) is employed in the first level of a supply chain, it prevents
1
expanding the demand uncertainty for other levels and makes their demand deterministic, one unit every T units of time. Hence, the following advantages are obtained using the (1, T) policy in retailers’ level of a supply chain (Haji and Haji (2007)): 1) The safety stock in supplier is eliminated (cost reduction). 2) Shortage cost in supplier due to uncertainty in demand is eliminated. 3) Information exchange cost for supplier due to the elimination of uncertainty of its demand is eliminated. 4) Inventory control and production planning in supplier are simplified. And 5) This policy is very easy to apply. Following Haji and Haji (2007), Haji et al. (2009) apply the (1, T) policy to a two-echelon inventory system with nonperishable products. In this paper, we consider the (1, T) policy for perishable products in a single stage inventory system. The assumption of infinite lifetime is common in most of the inventory models. However, the perishability of products is a major problem of some industry sectors in which by disregarding the finite lifetime of their products the resulting model may give inaccurate results. Hence, significant studies have been carried out for inventory control of perishable products. Nahmias (1982) and Karaesman et al. (2011) present two comprehensive reviews focusing on inventory control of perishable products. Schmidt and Nahmias (1985) for perishable items consider a system operating under (S–1, S) policy with lost sales, Poisson demand, outdating costs, purchase costs, and per unit per period holding costs. They derive the cost function by solving partial differential equations for the Sdimensional stochastic process corresponding to the time elapsed since the last S orders were placed. Perry and Posner (1998) generalize Schmidt and Nahmias (1985) to allow for general types of customer impatience behavior. Olsson and Tydesjo (2010) extend Schmidt and Nahmias (1985) by allowing backorders. In this study, we consider an inventory system under Schmidt and Nahmias (1985) assumptions but employ the (1, T) policy instead of the (S–1, S) policy. Using some concepts from queueing theory, it is shown that the considered model is similar to a D/M/1 queue with impatient customer. The long-run average total cost function including the total outdating, holding, shortage and purchase costs is derived utilizing the analogy of the D/M/1 queue with impatient customer and the finite dam model of Moran (1954). Since analyzing the convexity of the model is extremely complicated, a proposition is presented to introduce a domain for optimal T. Therefore, the optimal solution can be obtained using a search algorithm. 2
Haji and Haji (2007) prove the independency of the total cost function of (1, T) policy from the lead time for nonperishable products. In this study, we show that this result is also valid for the case of perishable products. Further, for the case of perishable products we have compared (1, T) and (S–1, S) policies. The results show that for fixed values of system parameters, there is a fixed value of lead time for which the performance of (1, T) policy is better than (S–1, S) policy. Furthermore, thorough a numerical example we show as the lead time increases this superiority is more pronounced. The proceeding parts of this paper are organized as follow: In section 2, the model and its analogy to other models is described. In section 3, the cost function of the model is derived and an algorithm is presented to obtain the optimal solution. In section 4, a numerical analysis is carried out. Finally in section 5, the conclusion is presented.
2. Problem Description and mathematical formulation 2.1 Problem Description An inventory system with lost sales, Poisson demand, outdating costs, purchase costs, and per unit per period holding costs is considered. Using (1, T) policy in which an order constantly is placed for one unit of product in each constant time interval, the long-run average total cost function including the total outdating, holding, shortage and purchase costs is derived. The objective is to determine the optimal time interval between any two consecutive orders which minimizes the long-run average total cost. It is assumed that the product shelf life is finite and constant, the lead time for an order is constant, and the fixed ordering cost is zero or negligible. General Notations μ
The demand rate.
π
Cost of a lost sale.
h
Rate of holding cost.
p
Cost of a perished product.
c
Unit price.
τ
Lead time.
m
Product shelf life. 3
T
Time interval between any two consecutive orders in (1, T) policy.
L
The long-run average number of units in system.
HC
Average holding cost per time unit.
Π
Average shortage cost per time unit.
OC
Average perishing cost per time unit.
PC
Average purchase cost per time unit.
C(T) Average total cost rate, for the (1, T) policy. C(S) Average total cost rate, for the (S–1, S) policy. wqn
The queue waiting time of the nth customer.
sn
The required service time of the nth customer.
Wq
The random variable of the queue waiting time of an arriving customer (product) in steady state.
S
The random variable of the required service time of a customer (product) in steady state.
S
The random variable of the occurred service time of a customer (product) in steady state.
Ws
A customer (product) average waiting time in the queueing system in steady state.
2.2. Methodology To obtain the total cost function of the system, one can resort to some concepts of queueing theory. To do this, consider a D/M/1 queueing system with impatient customers in which: 1) The arrival process is the arrival units of product to the system and the inter-arrival times are constant, equal to T, 2) The customers’ (product units’) patience time is the product shelf life. That is, the customer gives up whenever his sojourn time is larger than m, and 3) The inter-demand times, exponentially distributed with mean 1/μ, are the required service times of these units. Hence, the inventory problem can be interpreted as a D/M/1 queue with impatient customers, a single channel queueing system in which the arrival process is deterministic with rate 1/T, the 4
service times have exponential distribution with mean 1/μ, and the customer leaves the system whenever his sojourn time is larger than m. It is clear that the number of units in this queueing system is equal to the inventory on hand in the above inventory system. In the considered queueing system, customers (products) arrive at the instances 0, T, 2T, …, nT, … and are served in the order of their arrival. These customers require service for times s0, s1, …, sn, … . A customer will wait in the system only for a time not exceeding a fixed time m. We number the customers (arriving units to the inventory system) according to their arrival times, and define wqn as the waiting time in queue of the nth customer (product). The following relation can be obtained between wq(n+1) and wqn using a manner similar to that of Lindley (1952).
w q ( n +1)
⎧0 ⎪ = ⎨w qn + s n −T ⎪ ⎩ m −T
w qn + s n ≤ T
if if
T < w qn + s n < m
if
; n = 0,1,...
m ≤ w qn + s n
Or equivalently w q ( n +1) = [w qn + min (s n , m − w qn ) −T ]+ = [min(w qn + s n , m ) −T ]+ ; n = 0,1,...
(1)
Where [ x ]+ = max (0, x ) . Thus, the sequence of { wqn } forms a Markov chain with the state space [0, m − T]. The probability of a product perishing is Pr( S + Wq > m ) . Hence, to estimate the perishing costs, we have to obtain the distribution function of Wq . Define Fn ( x ) = Pr(Wqn ≤ x ) as the distribution function of Wqn , and let F ( x ) = lim n→∞ Fn ( x ) when the limit exists. To obtain F (x ) , we use the analogy of the finite dam model and D/M/1 queues with impatient customers.
2.3 The finite dam model Daley (1964) clarifies the analogy of the queueing system with impatient customers and the finite dam model. Assume a dam with finite capacity m, and consider its storage Zt for discrete time t = 0,1, 2,... . Suppose in the interval (t , t + 1) , an amount of Xt flows into the dam, filling it 5
to the level of min ( Z t + X t , m ) and any excess water being lost over the spillway. A fixed demand for an amount T occurs just before the instant t + 1 , and is met as fully as possible by the release of the quantity min[T , min ( Z t + X t , m)] = min (T , Z t + X t , m ) . Thus an amount Z t +1 leaves in the dam, where Z t +1 = min ( Z t + X t , m ) − min (T , Z t + X t , m ) ⇒ Z t +1 = [min ( Z t + X t , m ) − T ]+
(2)
If { X t } is assumed to be a sequence of independent non-negative random variables from an exponential distribution with parameter μ, equation (2) is of the same form as (1). Therefore, the distribution functions of {Z t } and {Wqn } are the same. Moran (1955) shows that h( y ) , the density function of Y = Z t + X t , satisfies the following equation (Prabhu (1958)):
⎧αμe μ y ⎪ ⎪ ⎪ h (m − y ) = ⎨ μy ⎪αμe ⎪ ⎪⎩
; − ∞ < y ≤T ⎛ n (− μ )i −i μT ⎞ e ( y − iT )i ⎜ μ∑ ⎟ ⎜ i =0 i ! ⎟ n ⎜ ⎟ (− μ )i −i μT e ( y − iT )i −1 ⎟ ⎜ +∑ ⎝ i =1 (i − 1)! ⎠
; nT < y ≤ (n + 1)T ; n = 1,..., N
(3)
m ⎡m⎤ Where N = floor ( ) = ⎢ ⎥ , and α is the normalizing constant which is obtained from Prabhu T ⎣T ⎦
(1958) as below:
α=
e −μm N
1 + ∑ (− μ ) e i
i =1
− i μT
(4)
[m − iT ] i i!
Using (3), the distribution function of Y = Z t + X t , H ( y ) is
6
n ⎧ ⎞ m − ( n + 1)T ≤ y < m − nT , ( − μ)i − iµT µ( m− y ) ⎛ − 1 αe e ( m − y − iT )i ⎟ ; ⎪ ⎜∑ H ( y) = ⎨ ⎝ i =0 i ! ⎠ and n = 1,..., N . ⎪ µ( m− y ) ;m − T ≤ y < ∞ ⎩1 − αe
(5)
For the dam storage, Z t , we have the relations Pr( Z t +1 = 0) = Pr( Z t + X t ≤ T ) = H (T ) Pr( Z t +1 < z ) = Pr( Z t + X t ≤ T + z ) = H (T + z ) ; 0 < z < m − T Pr( Z t +1 = m − T ) = Pr( Z t + X t > m ) = 1 − H ( m) Using these relations, Prabhu (1958) obtain the distribution function of Z t in steady state, F (z ) , with the assumption that m = N ⋅ T . Let this assumption to be released for T ≤ m , then F (z ) can be obtained as follows: N −1 ⎧ ⎞ ( − μ ) i − i μT μ ( m −T ) ⎛ = − (0) 1 α F e e ( m − (i + 1)T )i ⎟ ⎪ ⎜∑ ⎝ i =0 i ! ⎠ ⎪ i ⎪ N −1 ⎛ ( − μ ) − i μT ⎞ ⎪ F ( z ) = 1 − α e μ ( m − z −T ) ⎜ ∑ e ( m − z − (i + 1)T )i ⎟ ; ⎪ ⎝ i =0 i ! ⎠ ⎨ i n −1 ⎪ F ( z ) = 1 − α e μ ( m − z −T ) ⎛ ( − μ ) e − iμT ( m − z − (i + 1)T )i ⎞ ; ∑ ⎜ ⎟ ⎪ i! i =0 ⎝ ⎠ ⎪ ; ⎪ F ( z) = 1 ⎪ ⎩
0 ≤ z < m − NT m − ( n + 1)T ≤ z < m − nT ,
(6)
and n = 1, 2,..., N − 1 m −T ≤ z
Due to the similarity of (1) and (2), (6) is also the distribution function of Wq . So this distribution function is used to estimate the perishing costs.
7
3. Cost evaluation 3.1 Perishing cost Pr(Wq + S > m ) is the probability that a product is outdated, which is obtained from the relation
between the queueing system and the finite dam model as follows. Pr(Wq + S > m ) = Pr( X t + Z t > m ) = 1 − H ( m ) = α
(7)
Therefore, the proportion of products that is perished is
OC =
1 ⋅ Pr(Wq + S > m) . Thus T
p pα ⋅ Pr(Wq + S > m) = T T
(8)
3.2 Holding cost Let L denote the long-run average number of units in the system, then by Little’s formula
L=
1 Ws , where Ws is a product average waiting time in the system. Ws can be obtained by T
conditioning on S + Wq. So Ws = mα + (1 − α ) E (Wq + S | S + Wq < m )
(9)
Where α is the proportion of customers that leave the queueing system without complete service according to (7) and (4). Also, m
E (Wq + S | S + Wq < m) = E (Y | Y < m ) =
∫ yh( y )dy 0
H (m)
According to (5), h( y ) can be written as
8
m
=
∫ yh( y )dy 0
1− α
(10)
⎧ ⎛ N ( − μ)i −iµT ⎞ e (m − y − iT )i + ⎪ ⎜ μ∑ ⎟ i ! ⎪αe µ ( m− y ) ⎜ i =0 ⎟ ;0 ≤ y < m − NT N ⎪ ⎜ ⎟ ( − μ)i − iµT i −1 e (m − y − iT ) ⎟ ⎪ ⎜ ∑ i =1 (i − 1)! ⎝ ⎠ ⎪ ⎪⎪ h( y ) = ⎨ ⎛ n ( − μ)i −iµT ⎞ ⎪ μ e (m − y − iT )i + ⎟ m − (n + 1) T ≤ y < m − nT , ⎪ µ ( m− y ) ⎜ ∑ ! i ⎜ i =0 ⎟ ; ⎪αe n ⎜ ⎟ and n = 1,..., N − 1 ( − μ)i − iµT ⎪ i −1 e (m − y − iT ) ⎟ ⎜ ∑ ⎪ i =1 (i − 1)! ⎝ ⎠ ⎪ µ ( m− y ) ;m − T ≤ y ⎩⎪αμe
(11)
m
Let Θ = ∫ yh( y )dy , thus from (9), (10) and Little’s formula, we have 0
L=
Ws 1 = (mα + Θ ) T T
(12)
Hence,
HC = hL = h
1 (mα + Θ ) T
(13)
3.3 Shortage cost In order to calculate the shortage cost, the proportion of time that the system is out of stock, P0 , is required. Stanford (1979) considers the customer impatience only on the queue time and provides a relation to obtain P0 . He also presents a conjecture for P0 , when the customer impatience is considered on the sojourn time, but our simulation shows that his conjecture is not valid for our case (see Appendix A for more details). Therefore, we provide a relation to obtain P0 , when the customer impatience is on the sojourn time. Theorem 1: For D/M/1 queues with customer impatience on the sojourn time, P0 is obtained from the following relation.
P0 = 1 −
1− α μT
(14)
9
Proof: See Appendix B. Since P0 is the proportion of time that the system is out of stock, the proportion of demand that is lost is µP0 . Therefore, the average shortage cost per unit time is
Π = πμP0 = πμ(1 −
1− α ) μT
(15)
3.4 Purchase cost The amount of products purchased per time unit is 1 T , Thus
PC =
c T
(16)
3.5 System total cost The system total cost is the summation of the perishing, holding, shortage, and purchase costs. Form (8), (13), (15) and (16), the total cost rate of the system can be written as:
C (T ) = OC + HC + Π + PC =
pα c 1 1− α + h (mα + Θ ) + πμ(1 − )+ T T μT T
(17)
m
Where α is given in (4) and Θ = ∫ yh( y )dy in which h( y ) is according to (11). 0
From (17), C (T ) is independent of lead time as it is the case for nonperishable products [see Haji and Haji (2007)]. Due to the existence of α and Θ in (17), analyzing the convexity of C (T ) is extremely complicated. Therefore, the following proposition is presented to introduce a domain for the optimal value of T denoted by T*. Proposition 1: If the establishment of the inventory system is affordable, say T * ≠ ∞ , then T* will be in the interval (0,m]. Proof: See Appendix C.
10
When the establishment of the inventory system is not affordable, say T * = ∞ , then total lost sales is the optimal solution and the average per time unit system total cost is πμ . Therefore, by using a search algorithm to find the best amount of T in (0, m] and compare its total cost rate with πμ , one can find the optimal solution of (1, T) inventory system. Let ε be a small value, the optimal solution of (1, T) can be found with the accuracy of ε using the following algorithm. Algorithm 1. Step 1: Set T = ε . Calculate C (T ) using (17). Set T * = T and C * = C (T ) . Step 2: If T + ε ≤ m , set T = T + ε , calculate C (T ) using (17), and go to step 3. Otherwise, go to step 4. Step 3: If C (T ) < C * , set T * = T and C * = C (T ) . Go to step 2. Step 4: If πμ < C * , set T * = ∞ and C * = πμ . Go to step 5. Step 5: The algorithm is finished and (T * , C * ) is the best solution of (1, T) inventory system.
4. Numerical Analysis This section is devoted to compare the performance of (1, T) and (S–1, S) policies in terms of the total cost rate, and also to examine the relation between T * and the system parameters. The Matlab software is used for coding both policies. For all problems in this section, we assume the unit of the product shelf life and lead time is year, and the unit of costs is a cost unit, say dollar. Schmidt and Nahmias (1985) obtain the total cost function for (S–1, S) policy, C (S ) , as follow: ⎛ e − μτ τ S e − μτ τ S e − μ ( τ + m ) ( τ + m ) S −1 ⎞ ) − Kτ + h ⎜ S − μτ (1 − K ⎟ S! S! ( S − 1)! ⎝ ⎠ − μ ( τ +m ) S −1 e ( τ + m) + (c + p ) K ( S − 1)!
C ( S ) = cμ + ( π − c ) μK
11
(18)
Where K =
1 e
− μτ
S −1 − μx τ +m x τ e dx −∫ τ S! ( S − 1)! S
, and τ stands for the lead time.
From (17) one can see that the total cost of (1, T) policy is independent from the lead time. On the contrary, the total cost of (S–1, S) policy depends on this parameter which could be seen from (18). Therefore, we are interested in comparison between (1, T) and (S–1, S) policies when the lead time is varied. To do this, let µ = 10 , c = 100 , h = 20 , m = 0.1 , p = 50 , π = 300 and C ( S * ) − C (T * ) ε = 1 400 . Also, let S represent the optimal value of S, and % ΔC = × 100 show C(S * ) *
how much in percent the (1, T) policy performs better than the (S–1, S) policy. For various amounts of lead time the optimal solution is obtained for both policies. The results are shown in Table 1 and Figure 1. While in (1, T) policy the total cost is not sensitive to lead time, the total cost for the case of (S–1, S) policy is. As τ increases, the total cost of (S–1, S) policy increases but the total cost of (1, T) policy remains constant. This means that for fixed values of other parameters, there is a fixed value of τ for which the performance of (1, T) policy is better than (S–1, S) policy. Moreover as τ increases this superiority is more pronounced. Table 1: Comparison between (S–1, S) policy and (1, T) policy when the lead time is varied τ 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15
T* 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085 0.085
C(T*) 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281 2281
S* 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
[Figure 1is here] 12
C(S*) 2044 2159 2249 2322 2382 2432 2475 2511 2519 2523 2530 2538 2546 2556 2565
%ΔC -11.6% -5.7% -1.4% 1.8% 4.2% 6.2% 7.8% 9.2% 9.4% 9.6% 9.8% 10.1% 10.4% 10.8% 11.1%
We also are interested in the interaction effects of the lead time, the product shelf life and the cost parameters. So we consider all combinations of τ ∈ {0.1, 0.2, 0.3} , m ∈{0.1, 0.2, 0.3} ,
π ∈ {200,300,400} , p ∈ {25,50,75} and h ∈ {15,30} . The fixed parameters are µ = 10 and c = 100 . We also set ε = 1 400 in algorithm 1. The results for h = 15 and h = 30 are presented in Table 2 and Table 3, respectively. Since the result of (1, T) policy is not sensitive to the lead time, it has just one solution for any arbitrary lead time. Therefore, in these tables, for any combinations of other parameters we have presented one solution for (1, T) policy but three solutions for (S–1, S) policy related to three different lead times. Table 2: Numerical results for all combination of varied parameters, µ=10, c=100, and h=15 Problem No.
m
π
p
T*
C(T*)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3
200 200 200 300 300 300 400 400 400 200 200 200 300 300 300 400 400 400 200 200 200 300 300 300 400 400 400
25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75
0.1 0.1 ∞ 0.0775 0.085 0.0925 0.065 0.07 0.075 0.1025 0.11 0.115 0.085 0.09 0.0925 0.0775 0.08 0.0825 0.105 0.1075 0.1125 0.09 0.0925 0.095 0.0825 0.085 0.0875
1837 1929 2000 2154 2278 2386 2345 2509 2656 1498 1545 1587 1671 1745 1811 1780 1871 1956 1361 1392 1420 1487 1537 1583 1567 1630 1688
*
S 1 1 0 2 2 1 3 2 2 2 2 2 3 2 2 3 3 3 2 2 2 3 3 3 3 3 3
τ=0.1 %ΔC C(S*) 1900 3.3% 1957 1.4% 2000 0.0% 2387 9.8% 2519 9.6% 2626 9.1% 2736 14.3% 2877 12.8% 3009 11.7% 1530 2.1% 1582 2.3% 1634 2.9% 1748 4.4% 1837 5.0% 1890 4.2% 1857 4.1% 1958 4.4% 2058 5.0% 1350 -0.8% 1372 -1.5% 1394 -1.9% 1440 -3.3% 1489 -3.2% 1539 -2.9% 1522 -3.0% 1572 -3.7% 1621 -4.1%
13
*
S 1 1 0 3 2 2 4 3 3 3 2 2 4 4 3 5 4 4 3 3 3 4 4 4 5 5 4
τ=0.2 %ΔC C(S*) 1938 5.2% 1973 2.2% 2000 0.0% 2489 13.5% 2610 12.7% 2689 11.3% 2852 17.8% 3034 17.3% 3163 16.0% 1604 6.6% 1661 7.0% 1692 6.2% 1850 9.7% 1945 10.3% 2014 10.1% 2026 12.1% 2121 11.8% 2216 11.7% 1409 3.4% 1438 3.2% 1467 3.2% 1549 4.0% 1601 4.0% 1652 4.2% 1641 4.5% 1722 5.3% 1785 5.4%
*
S 1 1 0 4 3 3 5 5 4 4 3 3 5 5 4 6 6 5 4 4 4 5 5 5 6 6 6
τ=0.3 %ΔC C(S*) 1955 6.0% 1981 2.6% 2000 0.0% 2532 14.9% 2641 13.7% 2731 12.6% 2918 19.6% 3088 18.8% 3228 17.7% 1642 8.8% 1683 8.2% 1723 7.9% 1908 12.4% 2000 12.8% 2075 12.7% 2086 14.7% 2211 15.4% 2305 15.1% 1441 5.6% 1475 5.6% 1509 5.9% 1612 7.8% 1665 7.7% 1718 7.9% 1711 8.4% 1789 8.9% 1866 9.5%
Table 3: Numerical results for all combination of varied parameters, µ=10, c=100, and h=30 Problem No.
m
π
p
T*
C(T*)
28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3 0.3
200 200 200 300 300 300 400 400 400 200 200 200 300 300 300 400 400 400 200 200 200 300 300 300 400 400 400
25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75 25 50 75
0.1 0.1 ∞ 0.0775 0.085 0.095 0.065 0.07 0.075 0.105 0.11 0.1175 0.085 0.09 0.0925 0.0775 0.08 0.085 0.105 0.11 0.115 0.09 0.095 0.0975 0.085 0.0875 0.09
1847 1939 2000 2167 2289 2396 2362 2524 2670 1514 1560 1601 1694 1766 1831 1806 1897 1980 1383 1413 1440 1518 1566 1610 1602 1664 1720
*
S 1 1 0 2 2 1 3 2 2 2 2 2 3 2 2 3 3 3 2 2 2 3 3 3 3 3 3
τ=0.1 %ΔC C(S*) 1906 3.1% 1962 1.2% 2000 0.0% 2400 9.7% 2531 9.6% 2632 9.0% 2756 14.3% 2890 12.7% 3021 11.6% 1546 2.1% 1598 2.4% 1650 3.0% 1773 4.5% 1853 4.7% 1905 3.9% 1883 4.1% 1984 4.4% 2084 5.0% 1367 -1.2% 1389 -1.7% 1411 -2.1% 1468 -3.4% 1518 -3.2% 1567 -2.7% 1551 -3.3% 1600 -4.0% 1650 -4.2%
*
S 1 1 0 3 2 2 4 3 3 3 2 2 4 4 3 4 4 4 3 3 3 4 4 4 5 5 4
τ=0.2 %ΔC C(S*) 1942 4.9% 1977 1.9% 2000 0.0% 2501 13.4% 2618 12.6% 2696 11.1% 2869 17.7% 3046 17.1% 3175 15.9% 1621 6.6% 1671 6.6% 1702 5.9% 1874 9.6% 1970 10.4% 2031 9.8% 2050 11.9% 2145 11.6% 2240 11.6% 1428 3.2% 1457 3.0% 1486 3.1% 1577 3.7% 1629 3.9% 1680 4.2% 1678 4.5% 1759 5.4% 1812 5.1%
*
S 1 1 0 4 3 3 5 5 4 3 3 3 5 5 4 6 5 5 4 4 4 5 5 5 6 6 6
τ=0.3 C(S*) 1958 1983 2000 2543 2650 2739 2934 3104 3239 1654 1695 1735 1931 2023 2092 2116 2235 2328 1462 1495 1529 1640 1693 1746 1747 1824 1901
%ΔC 5.7% 2.2% 0.0% 14.8% 13.6% 12.5% 19.5% 18.7% 17.6% 8.5% 8.0% 7.7% 12.3% 12.7% 12.5% 14.7% 15.1% 14.9% 5.4% 5.5% 5.8% 7.4% 7.5% 7.8% 8.3% 8.8% 9.5%
Most of the observed results agree with what one would intuitively expect. For example, as the perishing cost, p, increases, T* increases in order to decrease the average number of perished items. In addition, when the shortage cost, π, increases, T* decreases in order to decrease the average number of lost sales. Furthermore, the total cost rate increases as p or π increases, but it does decrease as m increases. To explain the sensitivity of T* with respect to p, consider Problems 13, 14 and 15 of Table 2 in which all parameters except p are fixed. The obtained values for T* are 0.085, 0.09 and 0.0925 14
for p = 25,50 and 75, respectively. Therefore, by 50% decreasing in p = 50 , T* decreases about 6%, and by 50% increasing in p = 50 , it increases about 3%. So with respect to p, T * is more sensitive at left. Also, consider Problems 11, 14 and 17 to see the sensitivity of T* with respect to π. The presented values for T* are 0.11, 0.09 and 0.08 for π = 200,300 and 400, respectively. Thus, by 50% decreasing in π = 300 , T* increases about 22%, and by 50% increasing in π = 300, it decreases about 11%. In addition, see Problems 5, 14 and 23 to find the sensitivity of T* with respect to m. For m = 0.1,0.2 and 0.3, respectively, T* is 0.085, 0.09 and 0.0925. Hence, 50% decreasing in m = 0.2 leads T* to be decreased about 6%, and 50% increasing in m = 0.2 leads it to be increased about 3%. Finally, to show the sensitivity of T* with respect to h, we can compare the results of table 2 and 3. In a few problems, T* increases when h increases, but in most of problems it remains constant. This may be due to the fact that h has a less amount than the other cost parameters and so a less impact on total cost. The positive amount of %ΔC shows that the (1, T) policy outperforms the (S–1, S) policy. From the results of Table 1, we know %ΔC increases as the lead time increases. We use the results of Table 2 and 3 to depict some diagrams which could help us to analyze the sensitivity of %ΔC with respect to the cost parameters and the product shelf life. Figures 2-5 present such diagrams which illustrate %ΔC versus π, p, m, and h, respectively, for different values of τ. From Figure 2 one can see that %ΔC is not a monotonic function of π. For τ = 0.1 , %ΔC at first is increasing and then decreasing with respect to π which means that the total cost of (1, T) policy is increased less than the total cost of (S–1, S) policy when π increases from 200 to 300, but it increases more when π increases from 300 to 400. However, for τ = 0.2 , and τ = 0.3 the total cost of (S–1, S) policy is more sensitive than the case of (1, T) policy. Thus %ΔC is an increasing function of π in these cases. Also, the sensitivity of %ΔC decreases as π increases. For example, in the case of τ = 0.2 , when π increases from 200 to 300, %ΔC increases about 3.3%, but it increases about 1.5% for the increasing of π from 300 to 400. [Figure 2 is here]
15
Figure 3 illustrates the behavior of %ΔC versus the changes in p. It shows that %ΔC is not a monotonic function of p. For all three cases of τ, %ΔC at first is increasing and then decreasing with respect to p. Furthermore, while its sensitivity does increase for τ = 0.1 , it decreases for the case of τ = 0.2 and τ = 0.3 . Namely, as p increases from 25 to 50, %ΔC increases by 0.6%, 0.6% and 0.4% for three lead times, respectively, but it decreases about 0.8%, 0.2% and 0.1% for increasing of p from 50 to 75. [Figure 3 is here] Figure 4 depicts %ΔC with respect to m for different values of lead time. It seems that %ΔC is a decreasing function of shelf life. As m increases the total cost of both (1, T) and (S–1, S) policies does decrease. But the decreasing intensity of total cost for the (S–1, S) policy is more than that of the (1, T) policy. Furthermore, as it can be seen, %ΔC is highly sensitive to variations in m, and it is much more when the lead time is small. In fact, for the case of τ = 0.1 , %ΔC decreases about 4.6%, when m increases from 0.1 to 0.2., and it decreases about 8.2% when m increases from 0.2 to 0.3. [Figure 4 is here] In Figure 5, the behavior of %ΔC versus the changes in h is shown. As it can be seen, the sensitivity of %ΔC to h is lower than its sensitivity to other parameters. When h increases from 15 to 30, for the case of τ = 0.1 and τ = 0.3 , %ΔC decreases about 0.3% and 0.1%, respectively, but for the case of τ = 0.2 , it increases about 0.1%. Thus, %ΔC is not a monotonic function of h. [Figure 5 is here] We have also performed a separate set of calculations to examine the sensitivity of the total cost of (1, T) model to the value of T in a neighborhood of the optimum. Fixed parameters are
p = 50, h = 15 , µ = 10 and c = 100 . The results are shown in Table 4. Each entry in this table represents the percentage increase in the total cost when T is set to 50% of its optimal value followed by the percentage increase when T is set to 150% of its optimal value. As it can be seen, the total cost is much more sensitive at left. For example, in the case of m = 0.2 and 16
π = 300 , when T is increased from the optimal by 50%, the total cost increases about 12%, but when it is decreased from its optimal by 50%, the total cost increases about 66%. Also as m or π increases, the percentage increase of total cost is more pronounced. Table 4: Sensitivity analysis on T in a neighborhood of the optimum π=200 π=300 π=400 m 38 , 1.3 38 , 8 53 , 9 0.1 49 , 5 66 , 12 77, 17 0.2 70 , 8 84 , 16 92 , 22 0.3 Finally, the computational efficiency of Algorithm 1 might be of interest in the sense that one might be curious to know that how much time is needed for finding the optimal policy of (1,T) model. To achieve this purpose, all problems of Table 2 and 3 are considered and the average time needed for these problems are presented in Table 5. The Matlab codes are performed on an Intel 2.53GHz core i3 CPU with 4.00 GB of RAM. Since in Algorithm 1, we search the interval of (0, m] by the step of ε, the required time depends on the value of m and ε. For the smaller value of ε, the more time is needed, but the more accurate result would be obtained. We believe, the needed accuracy in practical application does not require a very small value of ε. For example, when the shelf life of an item is 0.2 year (73 days), the accuracy of 1 day (ε = 1/365 year) is satisfactory in determining the optimal value of T. Therefore, based on the presented results, Algorithm 1 works rather well. Table 5: Average needed time (in second) for considered problems m 0.1 0.2 0.3
ε = 1 100
ε = 1 200
ε = 1 300
ε = 1 400
ε = 1 500
0.1577 0.5087 1.0513
0.5226 1.8365 3.9789
1.0400 3.9668 8.8479
1.8876 7.0651 15.6625
2.8158 10.9599 24.4376
5. Conclusion For the case of stochastic demand, the (1, T) inventory policy is the one in which the time interval between two consecutive orders and the value of the order size are both constant. In this study, the (1, T) policy was developed for perishable products. This inventory problem was interpreted as a D/M/1 queueing system with impatient customers in which if a customer’s sojourn time exceeds a predetermined constant value then he leaves the system. 17
The analogy between the considered queueing model and the finite dam model was used to develop the (1, T) cost function. Since it is extremely difficult to prove the convexity of the cost function, a proposition was presented to introduce a domain for optimal T. Furthermore, a search algorithm was developed to obtain the optimal solution. In addition, a numerical analysis was carried out to compare (1, T) policy with the well known (S–1, S) policy. According to this comparison, for fixed values of system parameters, there is a fixed value of lead time for which the (1, T) policy is better than the (S–1, S) policy. Further as the lead time increases this superiority is more pronounced. Moreover, it was observed that the (1, T) is independent from the lead time. Further, as the lost sale and single stage system was considered in this paper, developing the (1, T) policy when the shortages are backordered is introduced for future research. Also, considering a multi echelon inventory system with (1, T) policy can be an attractive direction.
References Daley, D. J., 1964. Single server queueing systems with uniformly limited queueing time. Journal of the Australian Mathematical Society. 4, 489–505. Haji, R., Haji, A., 2007. One-for-one period policy and its optimal solution. Journal of Industrial and Systems Engineering. 1, 200–217. Haji, R., Pirayesh Neghab, M., Baboli, A., 2009. Introducing a new ordering policy in a twoechelon inventory system with Poisson demand. International Journal of Production Economics. 117, 212–218 Karaesman, I., Scheller-Wolf, A., Deniz, B., 2011. Managing Perishable and Aging Inventories, Review and Future Research Directions, in: Kempf, K., Keskinocak, P., Uzsoy, P. (Eds.), Handbook of Production Planning, Kluwer International Series in Operations Research and Management Science, Advancing the State-of-the-Art Subseries. Lindley, D. V., 1952. The theory of queues with a single server. Proceedings of the Cambridge Philosophical Society. 48, 277–289. Moran, P. A. P., 1955. A probability theory of dams and storage systems: modifications of the release rules. Ibid. 6, 117–130. 18
Nahmias, S., 1982. Perishable inventory theory: a review. Operations Research. 30, 680–708. Olsson, F., 2010. Inventory Problems with perishable items: Fixed lifetime and backlogging. European Journal of Operational Research. 202, 131–137 Perry, D., Posner, M., 1998. An (S–1, S) inventory system with fixed shelf life and constant lead times. Operations Research. 46, 65–71. Prabhu, N. U., 1958. On the integral equation for the finite dam. Quarterly Journal of Mathematics. 2, 183–188. Schmidt, C. P., Nahmias, S., 1985. (S–1, S) policies for perishable inventory. Management Science. 31, 719–728. Stanford, R. E., 1979. Reneging phenomena in single channel queues. Mathematics of Operations Research. 4, 162–178
Appendix A. The comparison of Stanford’s Conjecture with simulation For the case in which the customer impatience is on system waiting time, the conjecture of Stanford (1979) with our notations could be written as P0 = 1 −
(1 − α ) E ( S | S + Wq < m )
(A.1)
T
To show whether or not (A.1) is valid in our case, we compare its result with simulation for 4 problems with different parameter settings. Each simulation consists of 10 runs, each with a run length of 10000 time units. We present the mean and the standard deviation of 10 runs for each problem. The results are presented in Table A.1. As it can be seen the result of Stanford’s conjecture is not compatible with simulation results. Table A.1: Numerical results of Stanford’s conjecture and simulation
Problem 1 2 3 4
M
μ
0.2 0.5 1 2 5 2 0.2 2
5 5 1 5
T
P0 Stanford’s conjecture 0.4660 0.8001 0.8586 0.1470
19
Mean of 10 simulation runs (Standard dev.) 0.1767 (0.0022) 0.7996 (0.0011) 0.8276 (0.0011) 0.0497 (0.0025)
Appendix B. Proof of Theorem 1. Since S is exponential, it is clear that E ( S | S + Wq > m ) = m − E (Wq ) + E ( R ) , where R is the remaining time of an exponential random variable and E ( R ) = 1 µ . Thus
E ( S | S + Wq > m) = m +
1 − E (Wq ) µ
(B.1)
Also,
E ( S ) = E ( S | S + Wq < m) ⋅ Pr( S + Wq < m) + E ( S | S + Wq > m) ⋅ Pr( S + Wq > m) = (1 − α ) E ( S | S + Wq < m) + αE ( S | S + Wq > m) Therefore, we can write
E ( S | S + Wq < m) =
1 α (m − E (Wq )) − μ 1− α
(B.2)
Clearly, E ( S + Wq | S + Wq > m ) = m + E ( R ) . So
E ( S + Wq | S + Wq > m) = m +
1 µ
(B.3)
In addition, E ( S + Wq | S + Wq > m ) = E (Wq | S + Wq > m ) + E ( S | S + Wq > m )
(B.4)
Furthermore, From (B.1), (B.3), and (B.4) we have E (Wq | S + Wq > m ) = E (Wq )
(B.5)
Also, equation (B.5) implies that E (Wq | S + Wq ≤ m ) = E (Wq )
(B.6)
Let S be the occurred service time of a customer. The expected value of S can be obtained as follows. 20
E ( S ) = (1 − α ) E ( S | S + Wq ≤ m ) + αE ( S | S + Wq > m )
(B.7)
Clearly, for S + Wq ≤ m we can write E ( S | S + Wq ≤ m ) = E ( S | S + Wq ≤ m )
(B.8)
And also for S + Wq > m we have E ( S | S + Wq > m ) = m − E (Wq )
(B.9)
Moreover, from (B.8) and (B.9) we can write (B.7) as follows. E ( S ) = (1 − α ) E ( S | S + Wq ≤ m ) + α (m − E (Wq ) )
(B.10)
Now from (B.2) and (B.10) we have
⎛1 α (m − E (Wq ))⎟⎟⎞ + α(m − E (Wq )) E ( S ) = (1 − α )⎜⎜ − ⎝ μ 1− α ⎠ Or equivalently,
E(S ) =
1− α μ
(B.11)
Furthermore, it is known from Little’s formula that P0 = 1 − λE ( S )
(B.12)
Where λ = 1 T is the arrival rate. Therefore, from (B.11) and (B.12) P0 = 1 −
E(S ) 1− α =1− T μT
(B.13)
■ Appendix C. Proof of Proposition 1. We obtain the total cost rate for m ≤ T . Since for this case the value of Wq is zero, we have
⎧S if S ≤ m S =⎨ , ⎩m if S > m
(C.1)
From (C.1), it is clear that Ws = E (S )
(C.2) 21
Using Little’s formula ( L = λWs ) and (C.2), the long-run average number of units in the system is obtained as follows. L=
E(S ) T
Hence, HC = hL = h
E(S ) T
(C.3)
Since Wq = 0 , the probability that a product is outdated is obtained as follows.
α = Pr(S > m) = e − µm Therefore, the proportion of products that is perished is
OC =
α . Thus T
pα T
(C.4)
Furthermore, the proportion of time that the system is out of stock can be obtained from (B.12) as follows. P0 = 1 −
E(S ) T
(C.5)
P0 is the proportion of time that the system is out of stock, so the proportion of demand that is lost is µP0 . Therefore, Π = πμP0 = πμ (1 −
E(S ) ) T
(C.6)
In addition, the amount of products purchased per time unit is 1 T , Thus
PC =
c T
(C.7)
22
Form (C.3), (C.4), (C.6), and (C.7), the total cost rate of the system for m ≤ T can be written as:
C (T ) = HC + OC + Π + PC = hL +
pα c + πμP0 + ⇒ T T
E ( S ) pe − µm E(S ) c C (T ) = h + + πμ(1 − )+ ⇒ T T T T C (T ) =
E (S ) pe − μ m c + + πμ (h − πμ ) + T T T
From (C.8), if
(C.8)
E(S ) pe− µm c (h − πμ) + + ≥ 0 , It is clear that T * = ∞ . Moreover, if T T T
E(S ) pe − µm c (h − πμ) + + < 0 , clearly C(T) is increasing with respect to T, thus T * = m . T T T Therefore, if T * ≠ ∞ , T* will be in the interval of (0, m]■
Caption of figures Figure 1: Total cost versus the lead time for both policies. Figure 2: %ΔC versus π (p=50, m=0.2, h=15). Figure 3: %ΔC versus p (π =300, m=0.2, h=15). Figure 4: %ΔC versus m (π =300, p=50, h=15). Figure 5: %ΔC versus h (π =300, p=50, m=0.2).
23
3000 2500
Total Cost
2000 1500
(1,T) Policy (S-1,S) Policy
1000 500 0
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 Lead time
24
%ΔC
18.0% 16.0% 14.0% 12.0% 10.0% 8.0% 6.0% 4.0% 2.0% 0.0%
τ=0.1 τ=0.2 τ=0.3
200
300
25
400
π
14.0% 12.0%
%ΔC
10.0% τ=0.1
8.0%
τ=0.2
6.0%
τ=0.3
4.0% 2.0% 0.0%
p 25
50
26
75
16.0% 14.0% 12.0% 10.0%
%ΔC
8.0%
τ=0.1
6.0%
τ=0.2
4.0%
τ=0.3
2.0% 0.0% -2.0%
0.1
0.2
-4.0% -6.0%
27
0.3
m
14.0% 12.0%
%ΔC
10.0% 8.0%
τ=0.1
6.0%
τ=0.2 τ=0.3
4.0% 2.0% 0.0%
h 15
30
28
Title One for One Period Policy for Perishable Inventory
Anwar Mahmoodi1, Alireza Haji1,*, Rasoul Haji1 1
Department of Industrial Engineering, Sharif University of Technology, Tehran, Iran
*
Corresponding Author; Email: [email protected], Tel.: +982166165704, Fax: +982166022702
29
Highlights The (1,T) policy is developed for perishable inventories. The system cost rate is provided using an analogy from queueing and finite dam models The (1,T) policy outperforms the (S–1,S) policy when the lead time is long enough. The (1,T) policy is independent from the lead time.
30