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A generalized ordering and recovery policy for reusable items
European Journal of Operational Research 182 (2007) 764–774 www.elsevier.com/locate/ejor
Production, Manufacturing and Logistics
A generalized ordering and recovery policy for reusable items Dae-Won Choi a, Hark Hwang
a,*
, Shie-Gheun Koh
b
a b
Department of Industrial Engineering, KAIST, 373-1 Guseong-dong, Yuseong-gu, Daejeon 305-701, South Korea Department of Systems Management and Engineering, Pukyong National University, San 100, Yongdang, Namgu, Busan 608-739, South Korea Received 29 August 2005; accepted 17 August 2006 Available online 22 November 2006
Abstract This paper deals with a joint EOQ and EPQ model in which a stationary demand is satisfied by recovered products as well as newly purchased products. It is assumed that a fixed proportion of the used products are collected from customers and later recovered for reuse. We generalize the (P, R) policy in the literature by treating the sequence of orders for newly purchasing products and setups for recovery process within a cycle as a decision variable. Through example problems we illustrate the validity of the model and solution procedure developed. 2006 Elsevier B.V. All rights reserved. Keywords: Inventory management; Recycling; EOQ; EPQ
1. Introduction The recovery of used products is known to be economically more attractive than disposal. Metal scrap brokers, waste paper recycling and deposit systems for soft drink bottles are typical examples of product recovery that have been around for a long time. Also, the growth of environmental concerns has given an increasing attention to reuse (Fleischmann et al., 1997). It is reported that the environmental costs during the whole lifecycle of industrial products play an important role in the * Corresponding author. Tel.: +82 42 869 3113; fax: +82 42 869 3110. E-mail addresses: [email protected] (D.-W. Choi), [email protected] (H. Hwang), [email protected] (S.-G. Koh).
calculation of total production costs (Spengler et al., 1997). As a result of economic and social concerns, the manufacturers tend to make effort to minimize environmental impacts and energy consumption by adopting one of the following recovery options: repair, refurbishing, remanufacturing, cannibalization and recycling (Thierry et al., 1995). These options can be explained in terms of the required degree of disassembly. First three options upgrade used products in quality and/or technology. In repair, used products are returned to ‘‘working order’’ with a minimum level of disassembly. Quality standards, which are less rigorous than those for new products, are achieved through refurbishing. With complete disassemblies and extensive inspections, the remanufactured products have the same quality as rigorous as those for new products. The purpose of cannibalization is to
0377-2217/$ - see front matter 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2006.08.048
D.-W. Choi et al. / European Journal of Operational Research 182 (2007) 764–774
recover a limited set of reusable parts from used products or components. In recycling, after removing the identity and functionality of used products and components, materials from those are reused. In this study we focus on items that can be brought up to an ‘‘as-new’’ quality by adopting one of the first three recovery options. Fig. 1 shows the general framework of the situation under study. The supplier satisfies a stationary demand for an item, and the used products are returned from the customers. To satisfy the demand, one of two alternatives is used; either new products are ordered externally or used products are recovered to ‘‘asnew’’ condition. The objective is to control the external orders as well as the internal product recovery process while meeting the stationary demand with the minimum fixed and variable costs. A number of authors proposed inventory models and policies for the recovery systems. Review papers were provided by Guide et al. (1997), Fleischmann et al. (1997), Van der Laan et al. (1996), Guide (2000). Recently, Dyckhoff et al. (2003) reported major trends of the recent decade in supply chain management (SCM) putting a special emphasis on reverse logistics. The first deterministic model was reported by Schrady (1967). Assuming constant demand and return rates, and fixed lead times for external orders and recovery, he proposed a control policy with fixed lot sizes serving demand as far as possible from recovered products. Expressions for the optimal lot sizes for order and recovery were derived similar to the classical EOQ formula. Nahmias and Rivera (1979) extended the Schrady’s model to the case that the recovery rate is finite. Later, Mabini et al. (1992) modified the model by considering service level constraints and a multiitem system where items share the same repair facil-
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ity. Richter (1996a,b) proposed a slightly different model. He assumed that used items are collected in a location and brought back to the recovery shop at the end of each collection interval which coincides with a recovery cycle. Teunter (2001) also generalized the Schrady’s model in two ways. First, he considered more general policies. These policies alternated a number of manufacturing batches with a number of recovery batches. Second, he distinguished between the holding cost rates for manufactured and recovered items, whereas Schrady used the same rate for both items. Koh et al. (2002) generalized the work of Nahmias and Rivera (1979). They allowed the recovery rate to be both smaller and larger than the demand rate, and then considered the system that has one setup for recovery (or one order for new products) and many orders for new products (or many setups for recovery) in a cycle. For all four combinations, they derived closed-form expressions for the average total cost from which the optimal lot sizes are determined numerically. Later, Dobos and Richter (2004) discussed a similar system in which the disposal option of the returned items is allowed and all the recycling batches follow the production batches. Teunter (2004) pointed out the sub-optimality of Koh et al. (2002) when the inventory holding cost for serviceable products is greater than that of recoverable products. The existing deterministic models can be classified in terms of the order and recovery policies. Let P denote the number of orders for newly purchasing (or producing) items and R the number of recovery setups in a cycle. Most of existing studies adopted (P = 1, R) and/or (P, R = 1) policies. Richter (1996a,b), Teunter (2001), Dobos and Richter (2004) adopted (P, R) policy in which P orders for
Ordering new ones [Our system]
Recoverable Inventory
Recovery process
Serviceable Inventory
Customer
Return Fig. 1. The recovery system.
Scrap
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new products are followed by R recovery setups. Richter (1996a,b) assumed that all the recoverable products collected in a cycle are brought back to the recovery shop at the end of the cycle and so (P, R) seems a reasonable policy. The other two studies did not investigate the effects of the sequencing in (P, R) policy. Fig. 2 shows inventory levels of recoverable and serviceable products with (2, 3) policy in a cycle. As shown in Fig. 2(b), changing the sequence of the orders for new products and recovery setups in a cycle can reduce the inventory holding cost for recoverable products. Therefore, (2, 3) policy is no longer optimal.
Motivated by the above observations, we treat in this study the sequence of P orders and R setups as a decision variable. For the recovery system in Fig. 1, we find simultaneously (i) the economic order quantity for newly procured products, (ii) the optimal recovery lot size for the recoverable products, and (iii) the optimal sequence of orders and setups. This paper is organized as follows. In the following section the cost model is developed. And then a solution procedure to find an optimal ordering and recovery policy is proposed in Section 3. Section 4 reports the results of computational experiments
Recoverable Inventory
Serviceable Inventory
Order
Order
Recovery Recovery Recovery
Recoverable Inventory
Serviceable Inventory
Order
Recovery
Order
Recovery Recovery
Fig. 2. Inventory level with 2 orders and 3 recovery setups in a cycle: (a) 2 orders followed by 3 recovery setups and (b) the optimal sequence of orders and recovery setups.
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to validate the models and the solution procedure. Finally, we conclude the paper with summary in Section 5.
n
767
the number of setups in the recovery shop during a cycle (n P 1)
2. Model development
Dependent variables: The following variables can be expressed by the decision variables
For the development of the model the following assumptions and notations are used.
T Qr
2.1. Assumptions and notations
Ro
• Production rate of recovery process is a known constant (p). • Demand rate is deterministic and known (d). • Products used by customers are collected at a fixed and known rate (r). • All the cost parameters are known constants. • Purchase and recovery lead times are known. • Shortages are not allowed. • It is more economical to recover than to purchase. Therefore, the inventory holding cost for serviceable products (hs) is greater than that for the recoverable ones (hr). • Demand rate is greater than collection rate of recoverable products and less than the production rate of recovery process (r < d < p). • There are m orders for new products and n setups for recovery in a cycle. • The ordering lots are of equal size (Qo) through time. • The recovery lots are of equal size (Qr) through time. Known parameters p production rate of recovery process, [units]/ [time] d demand rate of the serviceable product, [units]/[time] r collection rate of recoverable products from customers, [units]/[time] cs setup cost for recovery process, [$]/[setup] co ordering cost for new products, [$]/ [order] hs inventory holding cost for the serviceable products, [$]/[unit]/[time] hr inventory holding cost for the recoverable products, [$]/[unit]/[time] Decision variables Qo ordering lot size for new products m the number of orders for serviceable products during a cycle (m P 1)
Rr
Rs
o cycle time, T ¼ mQodþnQr ¼ mQ dr recovery lot size in the recovery shop, r o ðmQo : nQr ¼ d r : rÞ ! Qr ¼ dr mQ n the amount of recoverable products collected during the time period in which demand is satisfied by one lot of new products (Qo), Ro ¼ r Qdo the net decrease of recoverable products during the time period in which demand is satisfied by one lot of recovered products (Qr), Rr ¼ Qr r Qdr ¼ Qr 1 dr the inventory level of recoverable products at the time when the first purchase order is made in a cycle, Rs ¼ Qr d1 1p r
2.2. Model Let T denote the cycle time of the system under study. It is defined as the time interval between two adjacent points where the inventory level of recoverable products is zero. Suppose we have m orders for new products and n setups for recovery processes in a cycle. For a time being, we will assume that m and n are relative prime and positive integers. Fig. 3 shows that for a given set of values of m, n and Qo the total inventory quantities of serviceable products in a cycle do not depend on the sequence of orders and recovery setups. Thus the total inventory related cost for serviceable products in a cycle can be obtained as follows: m co þ n cs þ r T hs : d
Qo d r Q ðp dÞ T hs þ r d 2 2 p ð1Þ
Note that the first term is the ordering cost for newly purchased products, second term the setup cost for recovery processes, and the final two terms inventory holding costs for serviceable products. The remaining term needed to build the total cost model is the inventory holding cost for recoverable products in a cycle. As shown in Fig. 4, the recoverable inventory can be decomposed into four types of polygons, A1 to A4. And the total recoverable inven-
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r Rr
Ro
r-p
Recoverable Inventory
Rs
Rs
Serviceable Inventory
Qo -d p-d Order
Recovery
Order
Recovery Recovery
Fig. 3. Inventory level when m = 2, n = 3.
tory during a cycle equals Area(A1) + Area(A2) + m Æ Area(A3) n Æ Area(A4). With notations in Fig. 4(c) each area can be expressed as follows: AreaðA1 Þ ¼
Qr r ðp dÞ T ; d p
Qo Qo r ; d d 2 2 2 Qr ðp rÞ r Q2r 1 1 AreaðA4 Þ ¼ þ 2 d p p 2
AreaðA3 Þ ¼
Qr Qr ðp rÞ: d p
ð2Þ ð3Þ
ð4Þ
However, the area of A2 depends on the sequence of purchasing orders and recovery setups, which implies the sequence is needed in advance to calculate the total costs per cycle. Let epoch be a point in time where the serviceable inventory level reaches zero. Thus there are (m + n) number of epochs in a cycle, at each of which either purchase order or recovery setup has to be made. The inventory level of recoverable products is increasing (decreasing) when the demand is being satisfied by purchased products (recovered products). From the definition of cycle, it has to start with an order for new products, while it has to end up with a recovery setup to run out of the recoverable inventory. To minimize the inventory of recoverable products in a cycle, a recovery setup is preferable at an epoch if the inventory level of recoverable products is greater than or equal to Rr + Rs. In other words, if the recoverable
inventory is less than Rr + Rs at an epoch, an order for new products has to be released to avoid shortage. Let ILk be the inventory level of recoverable products at the kth epoch. Based on the above observations, we propose ‘Sequencing Algorithm with inventory level (SAIL)’ that generates an optimal sequence of orders and recovery setups for given values of m and n. Sequencing Algorithm with inventory level (SAIL) Set k = 1 and ILk = Rs. Do If ILk < Rs + Rr, kth decision is an ordering for new products and set ILk+1 = ILk + Ro, Otherwise, kth decision is a recovery setup and ILk+1 = ILk Rr. Set k++. While k 6 m + n. The system parameters in SAIL have the following characteristics: (1) The initial value of IL1 does not have any effects on the results of SAIL. Neither does linear transformation of m/Rr of the system parameters. (2) During a cycle, the net increase in recoverable product inventory by m orders for new products and the net decrease in the inventory by n recovery setups must be the same, i.e., m Æ Ro = n Æ Rr.
D.-W. Choi et al. / European Journal of Operational Research 182 (2007) 764–774
Order
Order
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Recovery Recovery Recovery A4 A4 A4
A3 A3
A2 A1
Order
Order
Recovery
Recovery Recovery A4
A4
A3
A4
A3
A2 A1 Qr /d
r A3
A4
Ro /d r- p
Qo /d
Rr
Rs
A1
r T
Qo /p
Fig. 4. Decomposition of recoverable inventory: (a) all orders are followed by recovery setups, (b) optimal sequence and (c) heights and lengths of A1, A3 and A4.
Based on (1), we introduce a new variable, epoch index (EI), EIk = m/Rr Æ (ILk Rs). From (1) and (2) Ro and Rr becomes n and m, respectively. With the newly introduced EI, SAIL can be rewritten as follows:
Sequencing (SAEI)
Algorithm
Set k = 1 and EIk = 0. Do
with
Epoch
Indices
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If EIk < m, kth decision is an ordering for new products and set EIk+1 = EIk + n, Otherwise, kth decision is a recovery setup and set EIk+1 = EIk m. Set k++. While k 6 m + n SAEI issues an order of new products at epoch k if EIk is smaller than m and to make a recovery setup if EIk is greater than or equal to m. Let SEI = {EIk, k = 1, 2, . . . , m + n} and SIL = {ILk = EIk Æ Rr/m + Rs, k = 1, 2, . . . , m + n} be a sequence of epoch indices obtained from SAEI and corresponding sequence of recoverable inventories, respectively. SAEI is applied to a case with m = 2 and n = 3 and the results are listed in Table 1 for SEI, SIL and order/recovery decision sequence. Now, we will find the area of A2 utilizing the following property. Property 1. Let m and n be relative prime. Then SEI is a permutation of {0, 1, 2, 3, . . . , m + n 1}. Proof. In SAEI, EIk starts with zero, increases by n when it is less then m, and decreases by m when it is greater than or equals to m, which results in 0 6 EIk < m + n. Thus each EIk can be expressed as (i Æ n j Æ m) where i is the number of purchasing orders released (i = 0, 1, . . . , m) and j is the number of recovery setups made (j = 0, 1, . . . , n 1). Suppose EIk1 = (i1 Æ n j1 Æ m) and EIk2 = (i2 Æ n j2 Æ m) have a same value where k1 < k2. From EIk1 = EIk2, we have (i2 i1)n = (j2 j1)m and from k1 < k2, we have i1 < i2 and j1 < j2. Thus the relation of m:n = (i2 i1):(j2 j1) is obtained where (i2 i1) < m and (j2 j1) < n. This is contradiction to the assumption that m and n are relative prime integers. Therefore, EIk and EIi, i 5 k, have to be two different integers and consequently, SEI is a permutation of {0, 1, 2, 3, . . . , m + n 1}. h
As shown in Fig. 4, A2 consists of m + n rectangles. The height of each rectangle is found to be ILk Rs, k = 1, 2, . . . , m + n with the first one being zero. Rearranging them in ascending order of height of each rectangle, the area of A2 is easily obtained. An example for the case of m = 2 and n = 3 is shown in Fig. 5. The first m rectangles including zero height have the length of Qo/d and the other n rectangles have the length of Qr/d (Fig. 5(b)). The area of A2 is easily obtained and Q AreaðA2 Þ ¼ o f0 þ 1 þ þ ðm 1Þg d Qr Rr þ fm þ ðm þ 1Þ þ þðm þ n 1Þg d m ðm 1Þðd rÞT n 1 rT rQo þ mþ : ¼ 2d 2 d nd ð5Þ Dividing the total cost by the cycle time and utilizing the relationships of mQo = (d r)T and nQr = rT, the total average cost is obtained as follows:
RIL
RIL
3
1
4
2
Qo /d
Qr /d
Qo /d
Qr /d
Qr /d
0
1
2
3
4
Qo /d
Qr /d
Qr /d
Qr /d
0
Qo /d
Fig. 5. Splitting and rearranging A2 when m = 2 and n = 3: (a) splitting and (b) rearranging.
Table 1 Order/recovery sequence when m = 2 and n = 3 Epoch no. (k)
1
2
3
4
5
EIk ILk Condition Decision Change
0 Rs EI1 < 2 Order EI2 = EI1 + 3
3
1
4
2
3Rr 2
Rr 2
4Rr 2
2Rr 2
þ Rs EI2 P 2 Recovery EI3 = EI2 2
þ Rs EI3 < 2 Order EI4 = EI3 + 3
þ Rs EI4 P 2 Recovery EI5 = EI4 2
þ Rs EI5 P 2 Recovery –
D.-W. Choi et al. / European Journal of Operational Research 182 (2007) 764–774
n no 1 TCðm; n; Qo Þ ¼ co ðd rÞ þ cs ðd rÞ m Qo
1 Q þ ðd rÞhs þ 1 rhr o n 2d
ðp dÞr hs þ ðp rÞhr þ d r m Q : ð6Þ 2pðd rÞ n o Note that Eq. (6) becomes identical with the objective function of Koh et al. (2002) when there is only one recovery setup during a cycle, i.e. n = 1. 3. Solution procedure Eq. (6) can be rewritten as
n no 1 1 TCðm; n; Qo Þ ¼ A þ B þ C D Qo m Qo n m þ E Qo ; n
ð7Þ
where A ¼ co ðd rÞ; B ¼ cs ðd rÞ; ðd rÞhs þ rhr ; C¼ 2d
rhr ðp dÞr r hs þ ðp rÞhr : D ¼ ; and E ¼ d 2pðd rÞ 2d Suppose m and n are given. Then Eq. (7) becomes a convex function of Qo and from the first order necessary condition, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ Bn=m : ð8Þ Qo ¼ C D=n þ Em=n Since A, B, C, D and E are positive and C is larger than D, Qo always exists. Substituting (8) into (7) we have rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m n BD AD þ AC þ BE: TCAðm; nÞ ¼ 2 AE þ BC n m m n ð9Þ With a given value of n, TCA(m, n) is convex on m, and the optimal value of m becomes rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi BðC D=nÞ 0 m ¼n ; ð10Þ AE m 0 may not be an integer and thus two costs resulted from bm 0 c and dm 0 e need to be compared where
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bm 0 c is the greatest integer less than or equal to m 0 and dm 0 e the smallest integer greater than or equal to m 0 . For example, if the value of m 0 is 3.4 when n is 2, TCA(3, 2) and TCA(4, 2) need to be examined. To complete our discussion the following property is introduced. Property 2. If m and n are relative prime integers and l is an integer larger than one, the total average cost incurred by l Æ m purchasing orders and l Æ n recovery setups in a cycle is the same as that with m orders and n setups in a cycle. Proof. Suppose l Æ m orders and l Æ n setups are required in a cycle. From SAEI, it can be observed that EIk becomes zero when km orders and kn recovery setups, k = 0, 1, 2, . . ., l 1, are made. In other words, EI1+k(m+n) = km Æ ln kn Æ lm = 0, k = 0, 1, 2, . . . , l 1. Thus the given cycle is equivalent to l number of small cycles, each of which consists of m purchasing orders and n recovery setups. h Property 2 ensures that it is sufficient to search over (m, n) policy in which m and n are relative prime integers. Thus the assumption we made about m and n at the beginning of 2.2 can be disregarded at this point. Suppose m 0 = 3.4 and n = 2. From Property 2, TCA(5, 2) rather than TCA(4, 2) needs to be compared with TCA(3, 2). Let LB(n0) be a lower bound of Eq. (9) for a given value of n0. With m 0 found from Eq. (10), LB(n0) = TCA(m 0 , n0) and LBðn0 Þ ¼ 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ABEðC D=n0 Þ AD=n0 þ AC þ BE: ð11Þ
Note that Eq. (11) is increasing ffi pffiffiffiffiffiffiin n and pffiffiffiffiffiffifunction has the maximum value of 2 AC þ BE . Now, we suggest a solution algorithm, a type of enumeration method, as follows: [Step 1] Initialization (1) Set nmax = K(integer). (2) Set n = 1. (3) Find m1 = max(1, bm 0 c) and m2 = dm 0 e from Eq. (10). (4) Find TCA1 = TCA(m1, n) and TCA2 = TCA(m2, n) from Eq. (9). (5) If TCA1 < TCA2, then n* = n, m* = m1 and TCA* = TCA 0 = TCA1. Otherwise, n* = n, m* = m2 and TCA* = TCA 0 = TCA2.
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(6) If Min{TCA1, TCA2} < TCA*, a. Set TCA* = Min{TCA1, TCA2} and n* = n. b. If TCA1 < TCA2, then m* = m1, otherwise m* = m2.
[Step 2] Iterations and stopping rule (1) Set n = n + 1. (2) Find m1 = max(1, bm 0 c) and m2 = dm 0 e from Eq. (10). (3) If m1 and n are not relative prime, do m1 = m1 1 until m1 and n are relative prime. (4) If m2 and n are not relative prime, do m2 = m2 + 1 until m2 and n are relative prime. (5) Find TCA1 = TCA(m1, n) and TCA2 = TCA(m2, n) from Eq. (9).
(7) If TCA* > LB(n) and n < nmax, then go to (1). Otherwise, find Qo from Eq. (8) and terminate the algorithm.
Table 2 Parameter values and solutions of problems with e = 105 Problems
Parameter values
Optimal solutions
p
d
r
co
cs
hs
hr
m
n
Qo
TC
P1
300
200
100
10
20
2
1
P2
300
150
105
40
50
2
0.2
1 2 2
1 1 3
54.7723 36.5148 72.2189
109.545 109.545 143.314
60
120 118
50
116
m
114 112
30
110
20 10 0
m
108
TCA
106
LB
1
5
9
13
17
21
25
Total average cost
40
104 29
102
n 25
146 145
20
144
m
142 141
10
140 m
5
139
TCA
138
LB
0
137 1
5
9
13
17
21
25
29
n Fig. 6. Trend of optimal solutions: (a) problem P1 and (b) problem P2.
Total average cost
143 15
D.-W. Choi et al. / European Journal of Operational Research 182 (2007) 764–774
To validate the solution procedure, two example problems are solved. The system parameters and results are listed in Table 2. Note that two different optimal solutions exist for the problem P1, one of which is the same as that of Koh et al. (2002). In the optimal solution of the problem P2, both m and n are greater than one. Fig. 6 shows the trend of optimal solutions. The total average cost, TC, is shown to be fluctuating for small values of n, especially for P2 but eventually converging to LB(n) as n becomes larger. The total average costs are minimized when n equals 1 and 3 for problem P1 and P2, respectively. 4. Computational results Through sensitivity analysis, we studied how the system characteristics, i.e., the total average cost and optimal numbers of recovery setups and purchasing orders in each cycle, respond to changes of the parameter values for the analysis, 9 levels for d and r, and 10 levels for the remaining system parameters were set to generate the test problems (Table 3), which result in 8,100,000 problems. From the test results the followings observation can be made which are consistent with our expectation:
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(1) The number of orders for new products decreases and the order size increases as the ordering cost increase. (2) The number of orders for new products and recovery lot size increases (decreases) as the recovery setup (holding) cost increases. (3) With n fixed, the number of orders for new products increases while the order size decreases as the holding cost for serviceable items increases. One interesting question is how many problems out of 8,100,000 have optimal solutions in which both m and n are greater than one. Let NP be the number of those problems. Surprisingly, the value of NP is very small with NP = 16,161 (about 0.2%). Table 4 shows the distribution of these Table 3 Parameter values tested Parameters
Upper bound Lower bound Amount of increment
p
d
r
co
cs
hs
hr
100 1000 100
0.1p 0.9p 0.1p
0.1d 0.9d 0.1d
100 1000 100
100 1000 100
10 100 10
0.1hs hs 0.1hs
Table 4 Distribution of the problems in NP p
Value % D
100 9.99 0.165
200 10 0.165
300 9.97 0.165
400 9.99 0.165
500 10.04 0.164
600 9.97 0.165
700 10.01 0.165
800 10 0.165
900 9.99 0.165
1000 10.04 0.164
d
Value % D
0.1p 35.27 0.192
0.2p 27.31 0.178
0.3p 16.71 0.164
0.4p 14.85 0.113
0.5p 5.86 0.071
0.6p 0.00 –
0.7p 0.00 –
0.8p 0.00 –
0.9p 0.00 –
r
Value % D
0.1d 0.00 –
0.2d 0.00 –
0.3d 9.28 0.055
0.4d 25.99 0.201
0.5d 22.03 0.185
0.6d 25.37 0.199
0.7d 16.09 0.1
0.8d 1.24 0.011
0.9d 0.00 –
co
Value % D
100 5.57 0.168
200 8.66 0.136
300 9.9 0.153
400 10.66 0.185
500 13.08 0.156
600 9.28 0.166
700 9.9 0.177
800 10.05 0.169
900 11.14 0.17
1000 11.76 0.165
cs
Value % D
100 4.39 0.153
200 9.35 0.153
300 11.14 0.147
400 11.76 0.174
500 9.98 0.159
600 11.14 0.18
700 11.14 0.169
800 11.14 0.169
900 10.61 0.179
1000 9.36 0.152
hs
Value % D
10 9.99 0.165
20 9.99 0.165
30 10.01 0.164
40 9.99 0.165
50 999 0.165
60 10.01 0.164
70 10.02 0.164
80 9.99 0.165
90 10.01 0.164
100 9.99 0.165
hr
Value % D
0.1hs 100 0.165
0.2hs 0.00 –
0.3hs 0.00 –
0.4hs 0.00 –
0.5hs 0.00 –
0.6hs 0.00 –
0.7hs 0.00 –
0.8hs 0.00 –
0.9hs 0.00 –
hs 0.00 –
% = (the number of problems in NP with the associated parameter value)/16161 · 100%. D (deviation) = jA Bj/A * 100, where A: average of the total average costs of the optimal solutions, B: average of the total average costs when either m or n is set to one.
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16,161 problems over the range of each parameter. It also shows the degree of deviation from the optimal solution if the same problem is solved with (m = 1, n) and (m, n = 1) policy. For example, from the third row of the table we observe that the amount of average deviation is approximately 0.165% regardless of p value. The maximum value of the average deviation is 0.201%. Also, the number of problems in NP and amount of average deviation are not sensitive to changing values of the four parameters, p, co, cs and hs, while sensitive to the value of d, r and hr. Moreover, as either d decrease or r comes close to a half of d, they become much sensitive. 5. Conclusions This paper studies an inventory system through the mathematical model where the stationary demand can be satisfied by recovered products and newly purchased products. We first develop the objective cost function and then present the solution algorithm. We generalizes (P, R) policy of Teunter (2001) by treating the sequence of orders for newly purchasing items and setups for recovery process within a cycle as a decision variable. Through the sensitivity analysis we find that only 0.2% out of the 8,100,000 tested problems has an optimal solution in which both m and n are greater than one. In other words, with the maximum deviation of 0.2% from the optimal solution one may as well opt for (1, n) and (m, 1) policy rather than (m, n) policy. The solution obtained from the algorithm is confirmed to be always dominant to those from the previous models. The results in this paper could be extended to the following cases: (1) the recoverable items are deteriorating as seen from certain raw materials in newspaper manufacturing company, (2) the collection rate and/or demand rate are random variables and (3) the collection rate of recoverable items and demand rate of serviceable items are known but not constant through time. Acknowledgements This work was supported by the Korea Research Foundation Grant funded by the Korean Government (MOEHRD) (KRF-2006-311-D00249).
References Dobos, I., Richter, K., 2004. An extended production/recycling model with stationary demand and return rates. International Journal of Production Economics 90, 311–323. Dyckhoff, H., Lackes, R., Reese, J., 2003. Supply Chain Management and Reverse Logistics. Springer-Verlag. Fleischmann, M., Bloemhof-Ruwaard, J.M., Dekker, R., Van der Laan, E., Van Nunen, J.A., Van Wassenhove, L.N., 1997. Quantitative models for reverse logistics: A review. European Journal of Operational Research 103, 1–17. Guide, V.D.R., 2000. Production planning and control for remanufacturing: Industry practice and research needs. Journal of Operations Management 18, 467–483. Guide, V.D.R., Jayaraman, V., Srivastava, R., 1997. Repairable inventory theory: Models and applications. European Journal of Operational Research 102, 1–20. Koh, S.G., Hwang, H., Sohn, K.I., Ko, C.S., 2002. An optimal ordering and recovery policy for reusable items. Computers & Industrial Engineering 43, 59–73. Mabini, M.C., Pintelon, L.M., Gelders, L.F., 1992. EOQ type formulation for controlling repairable inventories. International Journal of Production Economics 28, 21–33. Nahmias, N., Rivera, H., 1979. A deterministic model for a repairable item inventory system with finite repair rate. International Journal of Production Research 17, 215–221. Richter, K., 1996a. The EOQ repair and waste disposal model with variable setup numbers. European Journal of Operational Research 95, 313–324. Richter, K., 1996b. The extended EOQ repair and waste disposal model. International Journal of Production Economics 45, 443–447. Schrady, D.A., 1967. A deterministic inventory model for repairable items. Naval Research Logistics Quarterly 14, 391–398. Spengler, Th., Puchert, H., Penkuhm, T., Rentz, O., 1997. Environmental integrated production and recycling management. European Journal of Operational Research 97, 308–326. Teunter, R.H., 2001. Economic order quantities for recoverable item inventory systems. Naval Research Logistics 48, 484–495. Teunter, R.H., 2004. Lot-sizing for inventory systems with product recovery. Computers & Industrial Engineering 46, 431–441. Thierry, M.C., Salomon, M., Van Nunen, J., Van Wassenhove, L., 1995. Strategic issues in product recovery management. California Management Review 37, 114–135. Van der Laan, E., Dekker, R., Salomon, M., Ridder, A., 1996. An (s, Q) inventory model with remanufacturing and disposal. International Journal of Production Economics 46, 339–350.
Production, Manufacturing and Logistics
A generalized ordering and recovery policy for reusable items Dae-Won Choi a, Hark Hwang
a,*
, Shie-Gheun Koh
b
a b
Department of Industrial Engineering, KAIST, 373-1 Guseong-dong, Yuseong-gu, Daejeon 305-701, South Korea Department of Systems Management and Engineering, Pukyong National University, San 100, Yongdang, Namgu, Busan 608-739, South Korea Received 29 August 2005; accepted 17 August 2006 Available online 22 November 2006
Abstract This paper deals with a joint EOQ and EPQ model in which a stationary demand is satisfied by recovered products as well as newly purchased products. It is assumed that a fixed proportion of the used products are collected from customers and later recovered for reuse. We generalize the (P, R) policy in the literature by treating the sequence of orders for newly purchasing products and setups for recovery process within a cycle as a decision variable. Through example problems we illustrate the validity of the model and solution procedure developed. 2006 Elsevier B.V. All rights reserved. Keywords: Inventory management; Recycling; EOQ; EPQ
1. Introduction The recovery of used products is known to be economically more attractive than disposal. Metal scrap brokers, waste paper recycling and deposit systems for soft drink bottles are typical examples of product recovery that have been around for a long time. Also, the growth of environmental concerns has given an increasing attention to reuse (Fleischmann et al., 1997). It is reported that the environmental costs during the whole lifecycle of industrial products play an important role in the * Corresponding author. Tel.: +82 42 869 3113; fax: +82 42 869 3110. E-mail addresses: [email protected] (D.-W. Choi), [email protected] (H. Hwang), [email protected] (S.-G. Koh).
calculation of total production costs (Spengler et al., 1997). As a result of economic and social concerns, the manufacturers tend to make effort to minimize environmental impacts and energy consumption by adopting one of the following recovery options: repair, refurbishing, remanufacturing, cannibalization and recycling (Thierry et al., 1995). These options can be explained in terms of the required degree of disassembly. First three options upgrade used products in quality and/or technology. In repair, used products are returned to ‘‘working order’’ with a minimum level of disassembly. Quality standards, which are less rigorous than those for new products, are achieved through refurbishing. With complete disassemblies and extensive inspections, the remanufactured products have the same quality as rigorous as those for new products. The purpose of cannibalization is to
0377-2217/$ - see front matter 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2006.08.048
D.-W. Choi et al. / European Journal of Operational Research 182 (2007) 764–774
recover a limited set of reusable parts from used products or components. In recycling, after removing the identity and functionality of used products and components, materials from those are reused. In this study we focus on items that can be brought up to an ‘‘as-new’’ quality by adopting one of the first three recovery options. Fig. 1 shows the general framework of the situation under study. The supplier satisfies a stationary demand for an item, and the used products are returned from the customers. To satisfy the demand, one of two alternatives is used; either new products are ordered externally or used products are recovered to ‘‘asnew’’ condition. The objective is to control the external orders as well as the internal product recovery process while meeting the stationary demand with the minimum fixed and variable costs. A number of authors proposed inventory models and policies for the recovery systems. Review papers were provided by Guide et al. (1997), Fleischmann et al. (1997), Van der Laan et al. (1996), Guide (2000). Recently, Dyckhoff et al. (2003) reported major trends of the recent decade in supply chain management (SCM) putting a special emphasis on reverse logistics. The first deterministic model was reported by Schrady (1967). Assuming constant demand and return rates, and fixed lead times for external orders and recovery, he proposed a control policy with fixed lot sizes serving demand as far as possible from recovered products. Expressions for the optimal lot sizes for order and recovery were derived similar to the classical EOQ formula. Nahmias and Rivera (1979) extended the Schrady’s model to the case that the recovery rate is finite. Later, Mabini et al. (1992) modified the model by considering service level constraints and a multiitem system where items share the same repair facil-
765
ity. Richter (1996a,b) proposed a slightly different model. He assumed that used items are collected in a location and brought back to the recovery shop at the end of each collection interval which coincides with a recovery cycle. Teunter (2001) also generalized the Schrady’s model in two ways. First, he considered more general policies. These policies alternated a number of manufacturing batches with a number of recovery batches. Second, he distinguished between the holding cost rates for manufactured and recovered items, whereas Schrady used the same rate for both items. Koh et al. (2002) generalized the work of Nahmias and Rivera (1979). They allowed the recovery rate to be both smaller and larger than the demand rate, and then considered the system that has one setup for recovery (or one order for new products) and many orders for new products (or many setups for recovery) in a cycle. For all four combinations, they derived closed-form expressions for the average total cost from which the optimal lot sizes are determined numerically. Later, Dobos and Richter (2004) discussed a similar system in which the disposal option of the returned items is allowed and all the recycling batches follow the production batches. Teunter (2004) pointed out the sub-optimality of Koh et al. (2002) when the inventory holding cost for serviceable products is greater than that of recoverable products. The existing deterministic models can be classified in terms of the order and recovery policies. Let P denote the number of orders for newly purchasing (or producing) items and R the number of recovery setups in a cycle. Most of existing studies adopted (P = 1, R) and/or (P, R = 1) policies. Richter (1996a,b), Teunter (2001), Dobos and Richter (2004) adopted (P, R) policy in which P orders for
Ordering new ones [Our system]
Recoverable Inventory
Recovery process
Serviceable Inventory
Customer
Return Fig. 1. The recovery system.
Scrap
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new products are followed by R recovery setups. Richter (1996a,b) assumed that all the recoverable products collected in a cycle are brought back to the recovery shop at the end of the cycle and so (P, R) seems a reasonable policy. The other two studies did not investigate the effects of the sequencing in (P, R) policy. Fig. 2 shows inventory levels of recoverable and serviceable products with (2, 3) policy in a cycle. As shown in Fig. 2(b), changing the sequence of the orders for new products and recovery setups in a cycle can reduce the inventory holding cost for recoverable products. Therefore, (2, 3) policy is no longer optimal.
Motivated by the above observations, we treat in this study the sequence of P orders and R setups as a decision variable. For the recovery system in Fig. 1, we find simultaneously (i) the economic order quantity for newly procured products, (ii) the optimal recovery lot size for the recoverable products, and (iii) the optimal sequence of orders and setups. This paper is organized as follows. In the following section the cost model is developed. And then a solution procedure to find an optimal ordering and recovery policy is proposed in Section 3. Section 4 reports the results of computational experiments
Recoverable Inventory
Serviceable Inventory
Order
Order
Recovery Recovery Recovery
Recoverable Inventory
Serviceable Inventory
Order
Recovery
Order
Recovery Recovery
Fig. 2. Inventory level with 2 orders and 3 recovery setups in a cycle: (a) 2 orders followed by 3 recovery setups and (b) the optimal sequence of orders and recovery setups.
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to validate the models and the solution procedure. Finally, we conclude the paper with summary in Section 5.
n
767
the number of setups in the recovery shop during a cycle (n P 1)
2. Model development
Dependent variables: The following variables can be expressed by the decision variables
For the development of the model the following assumptions and notations are used.
T Qr
2.1. Assumptions and notations
Ro
• Production rate of recovery process is a known constant (p). • Demand rate is deterministic and known (d). • Products used by customers are collected at a fixed and known rate (r). • All the cost parameters are known constants. • Purchase and recovery lead times are known. • Shortages are not allowed. • It is more economical to recover than to purchase. Therefore, the inventory holding cost for serviceable products (hs) is greater than that for the recoverable ones (hr). • Demand rate is greater than collection rate of recoverable products and less than the production rate of recovery process (r < d < p). • There are m orders for new products and n setups for recovery in a cycle. • The ordering lots are of equal size (Qo) through time. • The recovery lots are of equal size (Qr) through time. Known parameters p production rate of recovery process, [units]/ [time] d demand rate of the serviceable product, [units]/[time] r collection rate of recoverable products from customers, [units]/[time] cs setup cost for recovery process, [$]/[setup] co ordering cost for new products, [$]/ [order] hs inventory holding cost for the serviceable products, [$]/[unit]/[time] hr inventory holding cost for the recoverable products, [$]/[unit]/[time] Decision variables Qo ordering lot size for new products m the number of orders for serviceable products during a cycle (m P 1)
Rr
Rs
o cycle time, T ¼ mQodþnQr ¼ mQ dr recovery lot size in the recovery shop, r o ðmQo : nQr ¼ d r : rÞ ! Qr ¼ dr mQ n the amount of recoverable products collected during the time period in which demand is satisfied by one lot of new products (Qo), Ro ¼ r Qdo the net decrease of recoverable products during the time period in which demand is satisfied by one lot of recovered products (Qr), Rr ¼ Qr r Qdr ¼ Qr 1 dr the inventory level of recoverable products at the time when the first purchase order is made in a cycle, Rs ¼ Qr d1 1p r
2.2. Model Let T denote the cycle time of the system under study. It is defined as the time interval between two adjacent points where the inventory level of recoverable products is zero. Suppose we have m orders for new products and n setups for recovery processes in a cycle. For a time being, we will assume that m and n are relative prime and positive integers. Fig. 3 shows that for a given set of values of m, n and Qo the total inventory quantities of serviceable products in a cycle do not depend on the sequence of orders and recovery setups. Thus the total inventory related cost for serviceable products in a cycle can be obtained as follows: m co þ n cs þ r T hs : d
Qo d r Q ðp dÞ T hs þ r d 2 2 p ð1Þ
Note that the first term is the ordering cost for newly purchased products, second term the setup cost for recovery processes, and the final two terms inventory holding costs for serviceable products. The remaining term needed to build the total cost model is the inventory holding cost for recoverable products in a cycle. As shown in Fig. 4, the recoverable inventory can be decomposed into four types of polygons, A1 to A4. And the total recoverable inven-
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r Rr
Ro
r-p
Recoverable Inventory
Rs
Rs
Serviceable Inventory
Qo -d p-d Order
Recovery
Order
Recovery Recovery
Fig. 3. Inventory level when m = 2, n = 3.
tory during a cycle equals Area(A1) + Area(A2) + m Æ Area(A3) n Æ Area(A4). With notations in Fig. 4(c) each area can be expressed as follows: AreaðA1 Þ ¼
Qr r ðp dÞ T ; d p
Qo Qo r ; d d 2 2 2 Qr ðp rÞ r Q2r 1 1 AreaðA4 Þ ¼ þ 2 d p p 2
AreaðA3 Þ ¼
Qr Qr ðp rÞ: d p
ð2Þ ð3Þ
ð4Þ
However, the area of A2 depends on the sequence of purchasing orders and recovery setups, which implies the sequence is needed in advance to calculate the total costs per cycle. Let epoch be a point in time where the serviceable inventory level reaches zero. Thus there are (m + n) number of epochs in a cycle, at each of which either purchase order or recovery setup has to be made. The inventory level of recoverable products is increasing (decreasing) when the demand is being satisfied by purchased products (recovered products). From the definition of cycle, it has to start with an order for new products, while it has to end up with a recovery setup to run out of the recoverable inventory. To minimize the inventory of recoverable products in a cycle, a recovery setup is preferable at an epoch if the inventory level of recoverable products is greater than or equal to Rr + Rs. In other words, if the recoverable
inventory is less than Rr + Rs at an epoch, an order for new products has to be released to avoid shortage. Let ILk be the inventory level of recoverable products at the kth epoch. Based on the above observations, we propose ‘Sequencing Algorithm with inventory level (SAIL)’ that generates an optimal sequence of orders and recovery setups for given values of m and n. Sequencing Algorithm with inventory level (SAIL) Set k = 1 and ILk = Rs. Do If ILk < Rs + Rr, kth decision is an ordering for new products and set ILk+1 = ILk + Ro, Otherwise, kth decision is a recovery setup and ILk+1 = ILk Rr. Set k++. While k 6 m + n. The system parameters in SAIL have the following characteristics: (1) The initial value of IL1 does not have any effects on the results of SAIL. Neither does linear transformation of m/Rr of the system parameters. (2) During a cycle, the net increase in recoverable product inventory by m orders for new products and the net decrease in the inventory by n recovery setups must be the same, i.e., m Æ Ro = n Æ Rr.
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Order
Order
769
Recovery Recovery Recovery A4 A4 A4
A3 A3
A2 A1
Order
Order
Recovery
Recovery Recovery A4
A4
A3
A4
A3
A2 A1 Qr /d
r A3
A4
Ro /d r- p
Qo /d
Rr
Rs
A1
r T
Qo /p
Fig. 4. Decomposition of recoverable inventory: (a) all orders are followed by recovery setups, (b) optimal sequence and (c) heights and lengths of A1, A3 and A4.
Based on (1), we introduce a new variable, epoch index (EI), EIk = m/Rr Æ (ILk Rs). From (1) and (2) Ro and Rr becomes n and m, respectively. With the newly introduced EI, SAIL can be rewritten as follows:
Sequencing (SAEI)
Algorithm
Set k = 1 and EIk = 0. Do
with
Epoch
Indices
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If EIk < m, kth decision is an ordering for new products and set EIk+1 = EIk + n, Otherwise, kth decision is a recovery setup and set EIk+1 = EIk m. Set k++. While k 6 m + n SAEI issues an order of new products at epoch k if EIk is smaller than m and to make a recovery setup if EIk is greater than or equal to m. Let SEI = {EIk, k = 1, 2, . . . , m + n} and SIL = {ILk = EIk Æ Rr/m + Rs, k = 1, 2, . . . , m + n} be a sequence of epoch indices obtained from SAEI and corresponding sequence of recoverable inventories, respectively. SAEI is applied to a case with m = 2 and n = 3 and the results are listed in Table 1 for SEI, SIL and order/recovery decision sequence. Now, we will find the area of A2 utilizing the following property. Property 1. Let m and n be relative prime. Then SEI is a permutation of {0, 1, 2, 3, . . . , m + n 1}. Proof. In SAEI, EIk starts with zero, increases by n when it is less then m, and decreases by m when it is greater than or equals to m, which results in 0 6 EIk < m + n. Thus each EIk can be expressed as (i Æ n j Æ m) where i is the number of purchasing orders released (i = 0, 1, . . . , m) and j is the number of recovery setups made (j = 0, 1, . . . , n 1). Suppose EIk1 = (i1 Æ n j1 Æ m) and EIk2 = (i2 Æ n j2 Æ m) have a same value where k1 < k2. From EIk1 = EIk2, we have (i2 i1)n = (j2 j1)m and from k1 < k2, we have i1 < i2 and j1 < j2. Thus the relation of m:n = (i2 i1):(j2 j1) is obtained where (i2 i1) < m and (j2 j1) < n. This is contradiction to the assumption that m and n are relative prime integers. Therefore, EIk and EIi, i 5 k, have to be two different integers and consequently, SEI is a permutation of {0, 1, 2, 3, . . . , m + n 1}. h
As shown in Fig. 4, A2 consists of m + n rectangles. The height of each rectangle is found to be ILk Rs, k = 1, 2, . . . , m + n with the first one being zero. Rearranging them in ascending order of height of each rectangle, the area of A2 is easily obtained. An example for the case of m = 2 and n = 3 is shown in Fig. 5. The first m rectangles including zero height have the length of Qo/d and the other n rectangles have the length of Qr/d (Fig. 5(b)). The area of A2 is easily obtained and Q AreaðA2 Þ ¼ o f0 þ 1 þ þ ðm 1Þg d Qr Rr þ fm þ ðm þ 1Þ þ þðm þ n 1Þg d m ðm 1Þðd rÞT n 1 rT rQo þ mþ : ¼ 2d 2 d nd ð5Þ Dividing the total cost by the cycle time and utilizing the relationships of mQo = (d r)T and nQr = rT, the total average cost is obtained as follows:
RIL
RIL
3
1
4
2
Qo /d
Qr /d
Qo /d
Qr /d
Qr /d
0
1
2
3
4
Qo /d
Qr /d
Qr /d
Qr /d
0
Qo /d
Fig. 5. Splitting and rearranging A2 when m = 2 and n = 3: (a) splitting and (b) rearranging.
Table 1 Order/recovery sequence when m = 2 and n = 3 Epoch no. (k)
1
2
3
4
5
EIk ILk Condition Decision Change
0 Rs EI1 < 2 Order EI2 = EI1 + 3
3
1
4
2
3Rr 2
Rr 2
4Rr 2
2Rr 2
þ Rs EI2 P 2 Recovery EI3 = EI2 2
þ Rs EI3 < 2 Order EI4 = EI3 + 3
þ Rs EI4 P 2 Recovery EI5 = EI4 2
þ Rs EI5 P 2 Recovery –
D.-W. Choi et al. / European Journal of Operational Research 182 (2007) 764–774
n no 1 TCðm; n; Qo Þ ¼ co ðd rÞ þ cs ðd rÞ m Qo
1 Q þ ðd rÞhs þ 1 rhr o n 2d
ðp dÞr hs þ ðp rÞhr þ d r m Q : ð6Þ 2pðd rÞ n o Note that Eq. (6) becomes identical with the objective function of Koh et al. (2002) when there is only one recovery setup during a cycle, i.e. n = 1. 3. Solution procedure Eq. (6) can be rewritten as
n no 1 1 TCðm; n; Qo Þ ¼ A þ B þ C D Qo m Qo n m þ E Qo ; n
ð7Þ
where A ¼ co ðd rÞ; B ¼ cs ðd rÞ; ðd rÞhs þ rhr ; C¼ 2d
rhr ðp dÞr r hs þ ðp rÞhr : D ¼ ; and E ¼ d 2pðd rÞ 2d Suppose m and n are given. Then Eq. (7) becomes a convex function of Qo and from the first order necessary condition, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A þ Bn=m : ð8Þ Qo ¼ C D=n þ Em=n Since A, B, C, D and E are positive and C is larger than D, Qo always exists. Substituting (8) into (7) we have rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m n BD AD þ AC þ BE: TCAðm; nÞ ¼ 2 AE þ BC n m m n ð9Þ With a given value of n, TCA(m, n) is convex on m, and the optimal value of m becomes rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi BðC D=nÞ 0 m ¼n ; ð10Þ AE m 0 may not be an integer and thus two costs resulted from bm 0 c and dm 0 e need to be compared where
771
bm 0 c is the greatest integer less than or equal to m 0 and dm 0 e the smallest integer greater than or equal to m 0 . For example, if the value of m 0 is 3.4 when n is 2, TCA(3, 2) and TCA(4, 2) need to be examined. To complete our discussion the following property is introduced. Property 2. If m and n are relative prime integers and l is an integer larger than one, the total average cost incurred by l Æ m purchasing orders and l Æ n recovery setups in a cycle is the same as that with m orders and n setups in a cycle. Proof. Suppose l Æ m orders and l Æ n setups are required in a cycle. From SAEI, it can be observed that EIk becomes zero when km orders and kn recovery setups, k = 0, 1, 2, . . ., l 1, are made. In other words, EI1+k(m+n) = km Æ ln kn Æ lm = 0, k = 0, 1, 2, . . . , l 1. Thus the given cycle is equivalent to l number of small cycles, each of which consists of m purchasing orders and n recovery setups. h Property 2 ensures that it is sufficient to search over (m, n) policy in which m and n are relative prime integers. Thus the assumption we made about m and n at the beginning of 2.2 can be disregarded at this point. Suppose m 0 = 3.4 and n = 2. From Property 2, TCA(5, 2) rather than TCA(4, 2) needs to be compared with TCA(3, 2). Let LB(n0) be a lower bound of Eq. (9) for a given value of n0. With m 0 found from Eq. (10), LB(n0) = TCA(m 0 , n0) and LBðn0 Þ ¼ 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ABEðC D=n0 Þ AD=n0 þ AC þ BE: ð11Þ
Note that Eq. (11) is increasing ffi pffiffiffiffiffiffiin n and pffiffiffiffiffiffifunction has the maximum value of 2 AC þ BE . Now, we suggest a solution algorithm, a type of enumeration method, as follows: [Step 1] Initialization (1) Set nmax = K(integer). (2) Set n = 1. (3) Find m1 = max(1, bm 0 c) and m2 = dm 0 e from Eq. (10). (4) Find TCA1 = TCA(m1, n) and TCA2 = TCA(m2, n) from Eq. (9). (5) If TCA1 < TCA2, then n* = n, m* = m1 and TCA* = TCA 0 = TCA1. Otherwise, n* = n, m* = m2 and TCA* = TCA 0 = TCA2.
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(6) If Min{TCA1, TCA2} < TCA*, a. Set TCA* = Min{TCA1, TCA2} and n* = n. b. If TCA1 < TCA2, then m* = m1, otherwise m* = m2.
[Step 2] Iterations and stopping rule (1) Set n = n + 1. (2) Find m1 = max(1, bm 0 c) and m2 = dm 0 e from Eq. (10). (3) If m1 and n are not relative prime, do m1 = m1 1 until m1 and n are relative prime. (4) If m2 and n are not relative prime, do m2 = m2 + 1 until m2 and n are relative prime. (5) Find TCA1 = TCA(m1, n) and TCA2 = TCA(m2, n) from Eq. (9).
(7) If TCA* > LB(n) and n < nmax, then go to (1). Otherwise, find Qo from Eq. (8) and terminate the algorithm.
Table 2 Parameter values and solutions of problems with e = 105 Problems
Parameter values
Optimal solutions
p
d
r
co
cs
hs
hr
m
n
Qo
TC
P1
300
200
100
10
20
2
1
P2
300
150
105
40
50
2
0.2
1 2 2
1 1 3
54.7723 36.5148 72.2189
109.545 109.545 143.314
60
120 118
50
116
m
114 112
30
110
20 10 0
m
108
TCA
106
LB
1
5
9
13
17
21
25
Total average cost
40
104 29
102
n 25
146 145
20
144
m
142 141
10
140 m
5
139
TCA
138
LB
0
137 1
5
9
13
17
21
25
29
n Fig. 6. Trend of optimal solutions: (a) problem P1 and (b) problem P2.
Total average cost
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To validate the solution procedure, two example problems are solved. The system parameters and results are listed in Table 2. Note that two different optimal solutions exist for the problem P1, one of which is the same as that of Koh et al. (2002). In the optimal solution of the problem P2, both m and n are greater than one. Fig. 6 shows the trend of optimal solutions. The total average cost, TC, is shown to be fluctuating for small values of n, especially for P2 but eventually converging to LB(n) as n becomes larger. The total average costs are minimized when n equals 1 and 3 for problem P1 and P2, respectively. 4. Computational results Through sensitivity analysis, we studied how the system characteristics, i.e., the total average cost and optimal numbers of recovery setups and purchasing orders in each cycle, respond to changes of the parameter values for the analysis, 9 levels for d and r, and 10 levels for the remaining system parameters were set to generate the test problems (Table 3), which result in 8,100,000 problems. From the test results the followings observation can be made which are consistent with our expectation:
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(1) The number of orders for new products decreases and the order size increases as the ordering cost increase. (2) The number of orders for new products and recovery lot size increases (decreases) as the recovery setup (holding) cost increases. (3) With n fixed, the number of orders for new products increases while the order size decreases as the holding cost for serviceable items increases. One interesting question is how many problems out of 8,100,000 have optimal solutions in which both m and n are greater than one. Let NP be the number of those problems. Surprisingly, the value of NP is very small with NP = 16,161 (about 0.2%). Table 4 shows the distribution of these Table 3 Parameter values tested Parameters
Upper bound Lower bound Amount of increment
p
d
r
co
cs
hs
hr
100 1000 100
0.1p 0.9p 0.1p
0.1d 0.9d 0.1d
100 1000 100
100 1000 100
10 100 10
0.1hs hs 0.1hs
Table 4 Distribution of the problems in NP p
Value % D
100 9.99 0.165
200 10 0.165
300 9.97 0.165
400 9.99 0.165
500 10.04 0.164
600 9.97 0.165
700 10.01 0.165
800 10 0.165
900 9.99 0.165
1000 10.04 0.164
d
Value % D
0.1p 35.27 0.192
0.2p 27.31 0.178
0.3p 16.71 0.164
0.4p 14.85 0.113
0.5p 5.86 0.071
0.6p 0.00 –
0.7p 0.00 –
0.8p 0.00 –
0.9p 0.00 –
r
Value % D
0.1d 0.00 –
0.2d 0.00 –
0.3d 9.28 0.055
0.4d 25.99 0.201
0.5d 22.03 0.185
0.6d 25.37 0.199
0.7d 16.09 0.1
0.8d 1.24 0.011
0.9d 0.00 –
co
Value % D
100 5.57 0.168
200 8.66 0.136
300 9.9 0.153
400 10.66 0.185
500 13.08 0.156
600 9.28 0.166
700 9.9 0.177
800 10.05 0.169
900 11.14 0.17
1000 11.76 0.165
cs
Value % D
100 4.39 0.153
200 9.35 0.153
300 11.14 0.147
400 11.76 0.174
500 9.98 0.159
600 11.14 0.18
700 11.14 0.169
800 11.14 0.169
900 10.61 0.179
1000 9.36 0.152
hs
Value % D
10 9.99 0.165
20 9.99 0.165
30 10.01 0.164
40 9.99 0.165
50 999 0.165
60 10.01 0.164
70 10.02 0.164
80 9.99 0.165
90 10.01 0.164
100 9.99 0.165
hr
Value % D
0.1hs 100 0.165
0.2hs 0.00 –
0.3hs 0.00 –
0.4hs 0.00 –
0.5hs 0.00 –
0.6hs 0.00 –
0.7hs 0.00 –
0.8hs 0.00 –
0.9hs 0.00 –
hs 0.00 –
% = (the number of problems in NP with the associated parameter value)/16161 · 100%. D (deviation) = jA Bj/A * 100, where A: average of the total average costs of the optimal solutions, B: average of the total average costs when either m or n is set to one.
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16,161 problems over the range of each parameter. It also shows the degree of deviation from the optimal solution if the same problem is solved with (m = 1, n) and (m, n = 1) policy. For example, from the third row of the table we observe that the amount of average deviation is approximately 0.165% regardless of p value. The maximum value of the average deviation is 0.201%. Also, the number of problems in NP and amount of average deviation are not sensitive to changing values of the four parameters, p, co, cs and hs, while sensitive to the value of d, r and hr. Moreover, as either d decrease or r comes close to a half of d, they become much sensitive. 5. Conclusions This paper studies an inventory system through the mathematical model where the stationary demand can be satisfied by recovered products and newly purchased products. We first develop the objective cost function and then present the solution algorithm. We generalizes (P, R) policy of Teunter (2001) by treating the sequence of orders for newly purchasing items and setups for recovery process within a cycle as a decision variable. Through the sensitivity analysis we find that only 0.2% out of the 8,100,000 tested problems has an optimal solution in which both m and n are greater than one. In other words, with the maximum deviation of 0.2% from the optimal solution one may as well opt for (1, n) and (m, 1) policy rather than (m, n) policy. The solution obtained from the algorithm is confirmed to be always dominant to those from the previous models. The results in this paper could be extended to the following cases: (1) the recoverable items are deteriorating as seen from certain raw materials in newspaper manufacturing company, (2) the collection rate and/or demand rate are random variables and (3) the collection rate of recoverable items and demand rate of serviceable items are known but not constant through time. Acknowledgements This work was supported by the Korea Research Foundation Grant funded by the Korean Government (MOEHRD) (KRF-2006-311-D00249).
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