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Sequential and global optimization for a closed-loop deteriorating inventory supply chain
Mathematical and Computer Modelling 52 (2010) 161–176
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Sequential and global optimization for a closed-loop deteriorating inventory supply chain P.C. Yang a , H.M. Wee b,∗ , S.L. Chung c , P.C. Ho a a
Industrial Engineering & Management Department, St. John’s University, Tamsui, Taipei 25135, Taiwan, ROC
b
Industrial Engineering Department, Chung Yuan Christian University, Chungli 32023, Taiwan, ROC
c
Information Management Department, St. John’s University, Tamsui, Taipei 25135, Taiwan, ROC
article
info
Article history: Received 19 September 2009 Received in revised form 1 February 2010 Accepted 9 February 2010 Keywords: Supply chain management Closed-loop supply chain Multi-echelon inventory Deteriorating items
abstract Products such as IC chips, computers and mobile phones can become out of date due to technological innovation. However, these outdated products can be remanufactured and resold to market. In this paper, a closed-loop supply chain inventory system with multimanufacturing cycles and multi-remanufacturing cycles is analyzed using sequential and global optimization. In the case of sequential optimization, the decision is made initially by the down-stream player, then by the up-stream player. In the case of global optimization, the decision is made jointly by all the players. The supply chain includes a manufacturer, a retailer, a collector and a material supplier. This paper considers out of date as a type of deterioration and generalizes the following three cases: (i) single manufacturing cycle and single remanufacturing cycle, (ii) single manufacturing cycle and multi-remanufacturing cycles, and (iii) multi-manufacturing cycles and single remanufacturing cycle. The analytical results of this study show that a significant increase in the joint profit will result when the integrated policy is adopted. © 2010 Elsevier Ltd. All rights reserved.
1. Introduction One of the keys to successful supply chain management is the vendor–buyer inventory system integration. Clark and Scarf [1] presented a concept of serial multi-echelon structures to determine the optimal policy. Banerjee [2] derived a joint economic lot size model for a single vendor, single buyer system when the vendor has a finite production rate. Goyal [3] extended Banerjee’s model by relaxing the lot-for-lot production assumption. Ha and Kim [4] analyzed the integrated mathematical model between the buyer and the producer. Wee and Jong [5] studied the integration between parts and finished product with multi-lot size and deterioration. Yang et al. [6] derived a collaborative vendor–buyer inventory system with declining market. Wee considers product outdate or deterioration as decay, damage, spoilage, evaporation, obsolescence, pilferage, loss of utility or loss of marginal value of a commodity that results in decreasing usefulness from the original one [7] Ghare and Schrader [8] were the first authors to consider the on-going deterioration of inventory. Other authors such as Kang and Kim [9] and Raafat et al. [10] assumed either instantaneous or finite production with different assumptions on the patterns of deterioration. Yang and Wee [11] used a heuristic approach to develop an integrated vendor–buyer inventory model for deteriorating items. Recently many enterprises have focused their attention on reverse supply chain to meet environmental concerns/ regulations and social liability. Product remanufacturing such as transforming used items into marketable products through
∗
Corresponding author. Tel.: +886 3 5354409; fax: +886 3 2454499. E-mail address: [email protected] (H.M. Wee).
0895-7177/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2010.02.005
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refurbishment, repair and upgrading can also yield substantial cost benefits. Schrady [12] was the earliest author to propose a deterministic model with instantaneous production rate for manufacturing and remanufacturing. Schrady [12] argued that optimal lot sizes for manufacturers and remanufacturers can be determined by the classical EOQ formula. Koh et al. [13] analyzed finite manufacturing/remanufacturing rates. Savaskan et al. [14] determined the optimal collection channel configuration of a monopolistic manufacturer. Bautista and Pereira [15] proposed a method of identifying these collection channel configuration problems. Jaber and Saadany [16] developed a manufacturing and remanufacturing inventory system under the condition of lost sale. Chung et al. [17] developed a model with single manufacturing and single remanufacturing cycles (i.e., P = 1, R = 1). Yuan and Gao [18] derived two separate models: one model with single manufacturing and multi-remanufacturing cycles (i.e., P = 1, R ≥ 1); the other model with multi-manufacturing and single remanufacturing cycles (i.e., P ≥ 1, R = 1). The above researches didn’t consider the factor of deterioration. This paper develops a sequential and global optimization model considering the effect of deterioration; and synthesizes the two studies by Chung et al. [17] and Yuan and Gao [18] for multi-manufacturing and multi-remanufacturing cycles (i.e., P ≥ 1, R ≥ 1). Assumptions and notations The mathematical models in the analysis have the following assumptions: (1) (2) (3) (4) (5) (6) (7) (8) (9)
An infinite planning horizon. Deterioration is considered. Constant lead-time, manufacturing and remanufacturing rates. The multi-echelon inventory system contains a single item. The product demand rate and return rate are constant, and the return rate is less than demand rate. The number of deliveries within the manufacturing cycle is an integer. The setup cost per run and the annual holding cost fraction are known and constant. The remanufactured products are comparable to newly manufactured products. Single supplier, single manufacture, single retailer and single collector in the closed-loop supply chain multi-echelon inventory system are considered. (10) Multiple manufacturing and multiple remanufacturing cycles are considered. The following denotes the retailer’s parameters: d Ar Fr Pe Pr Tr TCr (TPr )
annual demand rate ordering cost per time annual percentage holding cost per dollar retailer price to end consumer wholesale price to retailer ordering cycle time annual total cost (total profit).
The following denotes the manufacturer’s parameters: AM (AR ) setup cost per manufacturing (remanufacturing) run AM w (ARw ) ordering cost for the material (used product) warehouse M Fixed cost for processing buyer order of any size FM (FM w /FRw )annual percentage holding cost per dollar for finished product (material/used product) PM (PR ) unit purchase cost to supplier (collector) TR1 (TR2 ) reproduction (non production) period in each remanufacturing cycle TM1 (TM2 ) production (non production) period in each manufacturing cycle mp (mr ) production (reproduction) rate, where mp > d (mr > d) m(n) number of deliveries per manufacturing (remanufacturing) cycle to the retailer (decision variable) P (R) number of manufacturing (remanufacturing) cycles in this model cycle N number of deliveries to the retailer in this model cycle, where N is an integer and N = Rn + Pm TCM w (TCRw ) annual manufacturing (remanufacturing) total cost TCm (TPm ) annual total cost (profit). The following denotes the collector’s parameters: u Ac Fc Pc k Acm TCc (TPc )
annual return rate setup cost per run for the collector annual percentage holding cost per dollar unit return cost number of deliveries per Tr1 to the manufacturer (decision variable) fixed cost to process manufacturer orders of any size annual total cost (total profit).
The following denotes the supplier’s parameters: As fixed cost per order Fs material percentage holding cost per dollar per year
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163
Fig. 1. The integrated multiple manufacturing/remanufacturing cycles closed-loop supply chain inventory system (adapted from Chung et al. [17]).
Ps unit purchase cost l number of deliveries per TM1 to the manufacturer (decision variable) Asm fixed cost to process manufacturer orders of any size TCs (TPs ) total cost (total profit) per unit time. Other related parameters:
θ deterioration rate for each finished product θ1 deterioration rate for each material and used product. 2. Model development The integrated closed-loop supply chain inventory system with multiple manufacturing/remanufacturing cycles is depicted in Fig. 1. From Fig. 2, the inventory levels are depleted by demand and deterioration. The buyer inventory differential equation is dIr (t )
+ θ Ir (t ) = −d, 0 ≤ t ≤ Tr dt with boundary condition of Ir (t ) = 0, t = Tr . The product inventory differential equations during TR1 and TR2 in a remanufacturing cycle are dIR1 (t ) dt
+ θ IR1 (t ) = mr − d,
dIR2 (t )
+ θ IR2 (t ) = −d, dt with boundary conditions: (i) IR1 (t ) = α,
t = 0;
0 ≤ t ≤ TR1
(2)
0 ≤ t ≤ TR2
(ii) IR2 (t ) = α,
(1)
(3)
t = TR2
and
(iii) IR1 (TR1 ) = IR2 (0).
From Spiegel [19], the differential equations are solved as Ir (t ) =
d
θ
IR1 (t ) = IR2 (t ) =
[exp(θ (Tr − t )) − 1],
mr − d
θ d
θ
0 ≤ t ≤ Tr
(1 − exp(−θ t )) + α,
[exp(TR2 − t ) − 1] + α,
0 ≤ t ≤ TR1
0 ≤ t ≤ TR2
(4) (5) (6)
where α can be derived as follows (see Appendix A):
α=
d2 Tr (2 + θ Tr ) 2mr
.
(7)
With boundary condition IR1 (TR1 ) = IR2 (0), the relation can be expressed as
(mr − d)(1 − exp(−θ TR1 )) = −d(1 − exp(θ TR2 )).
(8)
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Fig. 2. The manufacturer, retailer, collector and supplier multi-delivery inventory model with remanufacturing and deterioration (case of P = 2, R = 2, N = 8, k = 4, l = 7).
For θ TR1 1, eθ TR1 in (8) is replaced by 1 + θ TR1 + 3 θ 3 TR1
1 2
(θ TR1 )2 +
1 3!
2 θ 2 TR1
(θ TR1 )3 (Taylor series approximation). The percentage θ 3T 3
error for the fourth term in Taylor series is ( 3! )/(1 + θ TR1 + 2 + 3!R1 ). When θ TR1 ≤ 0.00649, the percentage error is about 0.00000454%. It will be smaller for term higher than four. Therefore, the term four and higher are neglected. By Taylor series expansion and the assumption of θ TR1 1, it is
1 1 (mr − d)TR1 1 − θ TR1 = dTR2 1 + θ TR2 . 2
(9)
2
From Misra [20], it is derived as dTR2 (2 + θ TR2 )
TR1 ≈
2(mr − d)
.
(10)
Since TR = TR1 + TR2 , TR can be expressed as TR ≈
TR2 (2mr + dθ TR2 ) 2(mr − d)
.
(11)
Similarly, TM1 and TM are TM1 ≈ TM ≈
dTM2 (2 + θ TM2 )
(12)
2(mp − d) TM2 (2mp + dθ TM2 ) 2(mp − d)
.
(13)
The lot size delivered to the retailer is Ir (t ) when t = 0. One has Ir (0) =
d
θ
(eθ Tr − 1) ≈
d
θ
1 dTr (2 + θ Tr ) θ Tr + θ 2 Tr2 = . 2
2
(14)
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165
The retailer’s average inventory level (I r ) is Ir =
1 Tr
Tr
Z
Ir (t )dt =
0
d Tr θ 2
[−1 − θ Tr + exp(θ Tr )] ≈
dTr
1
+ dθ Tr2 .
2
(15)
6
The manufacturer’s average inventory level (I m ) is IR1 (t )dt +
R
(m + n)Tr
Z
TR2
IR2 (t )dt + P
IM1 (t )dt +
TM2
Z
IM2 (t )dt
− ¯Ir
0
0
0
0
TM1
Z
R d2 Tr (2 + θ Tr )(TR1 + TR2 ) 2 2 (mr − d)TR1 + dTR2 +
1
≈
TR1
Z
1
¯Im =
(m + n)Tr 2 mr P d2 Tr (2 + θ Tr )(TM1 + TM2 ) 2 2 + (mp − d)TM1 + dTM2 + − ¯Ir . 2
(16)
mp
The retailer’s total profit (TPr ) is
TPr = Pe d −
Ar
+
Tr
(Fr + θ )Pr dTr
+
2
Pr dTr (2 + θ Tr )
2Tr
.
(17)
In (17), the first term is the sales revenue; the second terms are ordering cost, holding and deteriorating cost, and purchase cost. The manufacturer’s total profit (TPm ) is Pr dTr (2 + θ Tr )
TPm (P , R, l, k, N ) =
2Tr
" −
T
−
PlAM w
+
" −
RkARw T
+
PR (FRw + θ1 )kRmr l2R 1 + 13 θ1 lR
2T
PM (FM w + θ1 )Plmp l2M 1 + 13 θ1 lM 2T
PAM + RAR + (nR + mP )M
(nR + mP )Tr
+
+
+
RkPR mr lR (2 + θ1 lR )
#
2T
mp PPM lM l(2 + θ1 lM )
#
2T
(FM + θ )(PR Rn + PM Pm) Im . (nR + mP )
(18)
In (18), the first term is the sales revenue; the second terms are the used products’ ordering cost, holding and deteriorating cost, and purchase cost; the third terms are the material’s ordering cost, holding and deteriorating cost, and the purchase cost; the last term is the finished products’ holding and deteriorating costs. From Appendix B, the collector’s average inventory level and total profit are Ic = −
+
kRmr l2R 2T
1+
1 3
2 2 u R(mr − u)TR1 + (R − 1)uTR2 θ1 lR + (TR2 + mPTr )(2lR + TR2 + mPTr ) + 2T
ulR (2 − θ1 lR )(RTR − TR2 ) 2T
+
2T
R(R − 1)TR 4T
[(mr − u)TR1 (2 + θ1 TR1 ) − uTR2 (2 − θ1 TR2 )]
(19)
and TPc =
RkPR mr lR (2 + θ1 lR ) 2T
− Pc (Fc + θ1 )I c −
Ac + RkAcm T
− uPc
(20)
respectively. In (20), the first term is the sales revenue; the second terms are the holding and deteriorating costs; the third term is the setup cost; the last term is the purchase cost. From Appendix C, the supplier’s average inventory level and total profit are 2 Pmp TM1 − ll2M 1 + 31 θ1 lM
¯Is =
(21)
2T
and TPs =
mp PM lM lP (2 + θ1 lM ) 2T
− Ps (Fs + θ1 )I s −
P (As + lAsm ) T
−
PPs mp TM1 (2 + θ1 TM1 ) 2T
(22)
respectively. In (22), the first term is the sales revenue; the second terms are the holding and deteriorating costs; the third term is the setup cost; the last term is the purchase cost. The total profit per unit time is TP (P , R, l, k, N , T r , T R2 , T M2 ) = TPm + TPr + TPc + TPs
(23)
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Subject to u= m=
nRd
dTR2 (2 + θ TR2 )
(27)
2(mr − d) TR1 k
TM =
=
dTR2 (2 + θ TR2 )
(28)
2k(mr − d)
dTM2 (2 + θ TM2 )
TM1 =
Tr =
(26)
2(mr − d) TR2 (2mr + dθ TR2 )
TR =
lM =
(25)
P
TR1 =
lR =
(24)
(Pm + Rn) (N − nR)
(29)
2(mp − d) TM2 (2mp + dθ TM2 )
(30)
2(mp − d) TM1 l TR n
=
=
T = NTr =
dTM2 (2 + θ TM2 )
(31)
2l(mp − d)
TR2 (2mr + dθ TR2 )
(32)
2n(mr − d) NTR2 (2mr + dθ TR2 )
(33)
2n(mr − d)
N , P , R, l and k are positive integers.
(34)
3. Solution procedure The solution procedures for two cases are illustrated as follows: The first case is by the sequential (or local) optimization where the retailer and the manufacturer derive their respective optimal solution independently. The second case is to derive the global optimization of the integrated supply chain system. Case 1. Sequential or local optimization Step 1: Take derivatives of retailer’s profit (17) with respective to Tr and equating the result to zero. Solve the value of Tr , then substitute the value of Tr into (18). Step 2.1: Let P = 1 and take derivatives of manufacturer’s setup and holding cost (i.e., last term of (18), { PR Rn+PM Pm) [ (FM +θ)( ]I m }) (nR+mP )
PAM +RAR +(nR+mP )M (nR+mP )Tr
+
with respective to R and N, and equating the results to zero. Solve them and round them up or down the values of R and N. Step 2.2: Take the derivatives of the manufacturer’s used-product cost (i.e., the second term of (18), [ PR (FRw +θ1 )kRmr l2R
RkARw T
+
RkPR mr lR (2+θ1 lR ) 2T
(1 + θ ) + ]) with respective to k, and equating the result to zero. Solve it and 2T round it up or down the value of k so that the constraint of k ≥ 1 is observed. 1 l 3 1 R
Step 2.3: Take the derivatives of the manufacturer’s new-material cost (i.e., the third term of (18), [ PM (FM w +θ1 )Plmp l2M
PlAM w T
+
mp PPM lM l(2+θ1 lM ) 2T
(1 + θ ) + ]) with respective to l, and equating the result to zero. Solve it and 2T round it up or down the value of l so that the constraint of l ≥ 1 is observed. Step 2.4: Explore the local optimization by satisfying the following conditions 1 l 3 1 M
TP (P ∗ , R∗ − 1, k∗ , l∗ , N ∗ ) 5 TP (P ∗ , R∗ , k∗ , l∗ , N ∗ ) = TP (P ∗ , R∗ + 1, k∗ , l∗ , N ∗ )
(35)
TP (P , R , k − 1, l , N ) 5 TP (P , R , k , l , N ) = TP (P , R , k + 1, l , N )
(36)
TP (P , R , k , l − 1, N ) 5 TP (P , R , k , l , N ) = TP (P , R , k , l + 1, N )
(37)
TP (P ∗ , R∗ , k∗ , l∗ , N ∗ − 1) 5 TP (P ∗ , R∗ , k∗ , l∗ , N ∗ ) = TP (P ∗ , R∗ , k∗ , l∗ , N ∗ + 1).
(38)
∗ ∗
∗ ∗
∗ ∗
∗
∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗
∗
∗
∗
and
P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
167
Table 1 Exploration of sequential optimal solution without integration. Variables
P
R
k
l
N
n
m
TP
P =1
1 1 1 1 1 1 1 1 1
3 2 4 3 3 2 4 3 3
1 1 1 1 1 1 1 2 1
1 1 1 1 1 1 1 1 2
18 18 18 17 16 17 17 17 17
4.500 6.750 3.375 4.250 4.000 6.375 3.188 4.250 4.250
4.500 4.500 4.500 4.250 4.000 4.250 4.250 4.250 4.250
171,192 170,090 171,014 171,235# 171,183 170,361 170,900 169,735 170,394
R=1
1 1 1 1 1 2 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 2 1
1 1 1 1 1 1 1 2
6 5 7 8 9 8 8 8
4.500 3.750 5.250 6.000 6.750 6.000 6.000 6.000
1.500 1.250 1.750 2.000 2.250 1.000 2.000 2.000
168,173 166,776 168,922 169,139 169,109 164,327 168,564 167,424
Step 3.1: Let R = 1 and take derivatives of manufacturer’s setup and holding cost (i.e., last term of (18), {
PAM +RAR +(nR+mP )M (nR+mP )Tr
Remark: # is the local optimal solution.
PR Rn+PM Pm) [ (FM +θ)( ]I m }) (nR+mP )
+
with respective to P and N, and equating the results to zero. Solve them and round them up or down the values of P and N. RkA Step 3.2: Take the derivatives of the manufacturer’s used-product cost (i.e., the second term of (18), [ TRw + PR (FRw +θ1 )kRmr l2R
RkP m l (2+θ l )
R 1 R (1 + 13 θ1 lR ) + R r 2T ]) with respective to k, and equate to zero, and solve it. Round up and 2T down the value of k and obey the constraint of k ≥ 1. PlA Step 3.3: Take the derivatives of the manufacturer’s new-material cost (i.e., the third term of (18), [ TM w +
PM (FM w +θ1 )Plmp l2M
(1 + 13 θ1 lM ) +
mp PPM lM l(2+θ1 lM ) 2T
]) with respective to l, and equating the result to zero. Solve it and round it up or down the value of l so that the constraint of l ≥ 1 is observed. Step 3.4: Explore the local optimization by satisfying (36)–(38) and the following condition 2T
TP (P ∗ − 1, R∗ , k∗ , l∗ , N ∗ ) 5 TP (P ∗ , R∗ , k∗ , l∗ , N ∗ ) = TP (P ∗ + 1, R∗ , k∗ , l∗ , N ∗ ).
(39)
Step 4: Find the better solution from Step 2.4 and Step 3.4. Case 2. Global optimization Step 1: Let the values of P, R, k, l and N solved from Case 1 be the initial values. Step 2: Substituting the initial value into joint profit (23), take the derivative of (23) with respective of Tr , and equating the result to zero. Solve it and substitute Tr into (23), the joint profit can be obtained. Step 3: Explore the proximity of P, R, k and l, and derive Tr by using the method of step 2. Substituting Tr into (23), the joint profit can be obtained. Step 4: Find the global optimization by satisfying conditions (35)–(39). 4. Numerical example The numerical data are set as follows: d = 2000, Ar = 100, Fr = 0.3, Pe = 175, Pr = 150, AM = 750, AR = 500, AM w = 350, ARw = 350, M = 350, FM = 0.3, FM w = 0.2, FRw = 0.2, PM = 115, PR = 110, mp = 5000, mr = 4000, u = 1500, Ac = 250, Fc = 0.1, Pc = 70, Acm = 150, As = 200, Fs = 0.3, Ps = 90, Asm = 150, θ = 0.08 and θ1 = 0.05. By using the solution procedure in Section 4, the local optimal solution is P = 1, R = 3, k = 1, l = 1, N = 17 and TP = 171,235.11 (Table 1) and the global optimal solution is P = 1, R = 3, k = 1, l = 1, N = 7 and TP = 175,190.68 (Table 2). The percentage of total profit increase (denoted by PTPI) is defined as PTPI =
Total profit from Case 2 − Total profit from Case 1 Total profit from Case 1
.
From Table 3, the percentage of joint profit increase is 2.31%, the percentage of profit increase for the manufacturer and collector is positive, and the percentage of profit increase for the retailer and supplier is negative. Table 3 shows that the value of PTPI is 2.31%, the manufacturer and the collector benefit from integration; but the retailer and the supplier suffer lose. For lasting partnership, profit sharing between all parties should be considered.
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Table 2 Exploitation of global optimal solution with integration. P
R
k
l
N
n
m
Tr0
TP (Tr0 )
1 1 1 1 1 1 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1
3 2 4 5 6 7 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 4 3 4 4 4 4 4 4 4 4 4 3 2 3 3 3 3 3 3 2
1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1
1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1
17 17 17 17 17 17 17 17 17 18 16 16 16 16 15 14 13 12 11 10 9 8 10 10 10 10 10 10 10 10 10 9 8 7 8 8 8 8 8 9 7 6 7 7 7 7
4.250 6.375 3.188 2.550 2.125 1.822 2.550 2.550 2.550 2.700 2.400 2.400 2.400 2.400 2.250 2.100 1.950 1.800 1.650 1.500 1.350 1.200 1.500 1.500 1.500 1.250 1.875 2.500 1.875 1.875 1.875 1.688 1.500 1.313 1.500 1.500 1.500 2.000 3.000 2.250 1.750 1.500 1.750 1.750 1.750 2.625
4.250 4.250 4.250 4.250 4.250 4.250 2.125 2.125 2.125 2.250 2.000 2.000 2.000 4.000 3.750 3.500 3.250 3.000 2.750 2.500 2.250 2.000 2.500 2.500 1.250 2.500 2.500 2.500 2.500 2.500 1.250 2.250 2.000 1.750 2.000 2.000 1.000 2.000 2.000 2.250 1.750 1.500 1.750 1.750 0.875 1.750
0.05341 0.04581 0.05935 0.06434 0.06872 0.07267 0.07095 0.07298 0.07914 0.06853 0.07361 0.07579 0.08222 0.06672 0.06935 0.07229 0.07559 0.07933 0.08361 0.08856 0.09439 0.10134 0.09018 0.10063 0.09808 0.09494 0.08139 0.07300 0.09186 0.08307 0.09036 0.08670 0.09304 0.10078 0.09520 0.10554 0.10357 0.08339 0.07140 0.07773 0.09031 0.09901 0.09289 0.10186 0.10124 0.07735
172,810 170,862 173,554 173,781 173,743 173,553 173,824 172,942 172,829 173,675 173,959 173,059 172,896 173,978 174,163 174,331 174,479 174,602 174,691 174,735 174,720 174,623 174,081 172,807 174,239 174,331 174,960 174,816 173,469 174,264 174,244 175,034 175,038 174,942 174,294 173,271 174,086 175,140 174,454 175,006 175,191∗ 175,117 174,343 173,807 173,787 174,714
Remark: ∗ is the global optimization. Table 3 PTPI.
Case 1 (I) Case 2 (II) (II) −1 (I)
TP r
TP m
TP c
TP s
TP
44,746.43 42,661.31 −4.66%
57,809.00 63,837.14 10.43%
56,700.25 56,726.12 0.05%
11,979.43 11,966.11 −0.11%
171,235.11 175,190.68 PTPI = 2.31%
5. Sensitivity analysis Sensitivity analysis of return rate, deterioration rate, holding cost and demand rate is analyzed when each parameter increases or decreases a fixed rate. The results are as follows: 5.1. Sensitivity analysis of return rate From Table 4, Figs. 3 and 4, it is seen that 1. The total profit after integration is better than that before integration. 2. When the return rate decreases, the value of PTPI increases. It means a a higher significance for less return rate.
P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
169
Table 4 Relation between u and (P , R, TP and PTPI). u
200
400
600
800
1000
1200
1400
1600
1800
Case 1 without integration
P R l k N TP
7 1 1 1 42 144,150
4 1 1 1 23 148,149
2 1 1 1 14 152,525
1 1 1 1 9 156,719
1 1 1 1 9 160,813
1 2 1 1 13 164,952
1 3 1 1 16 169,071
1 5 1 1 24 173,208
1 8 1 1 35 177,638
Case 2 with integration
P R l k N TP
5 1 1 1 11 148,747
3 1 1 1 8 152,640
2 1 1 1 6 156,678
1 1 1 1 4 160,825
1 1 1 1 4 164,879
1 2 1 1 5 168,961
1 3 1 1 7 173,050
1 4 1 1 9 177,266
1 7 1 1 14 18,1570
3.19
3.03
2.72
2.62
2.53
2.43
2.35
2.34
2.21
PTPI
Fig. 3. Relation between u and (P , R) with/without integration.
Fig. 4. Relation between u and (TP , PTPI ) with/without integration.
3. The P value increases when the return rate increases. 4. The values of R and P decrease after integration.
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Fig. 5. Relation between (θ, θ1 ) and (TP and PTPI) with/without integration. Table 5 Relation between (θ, θ1 ) and (P , R, TP and PTPI) with/without integration. 0.056 0.035
−20% 0.064 0.040
−10% 0.072 0.045
0% 0.080 0.050
10% 0.088 0.055
20% 0.096 0.060
30% 0.104 0.065
Case 1 without integration
Tr P R l k N TP
0.04023 1 3 1 1 16 172,296
0.03947 1 3 1 1 16 171,936
0.03875 1 3 1 1 16 171,580
0.03807 1 3 1 1 17 171,235
0.03742 1 3 1 1 17 170,893
0.03681 1 3 1 1 17 170,556
0.03623 1 3 1 1 17 170,222
Case 2 with integration
Tr P R l k N TP
0.09306 1 3 1 1 7 175,918
0.09212 1 3 1 1 7 175,673
0.09120 1 3 1 1 7 175,431
0.09031 1 3 1 1 7 175,191
0.08944 1 3 1 1 7 174,953
0.08860 1 3 1 1 7 174,717
0.08778 1 3 1 1 7 174,484
2.10
2.17
2.24
2.31
2.38
2.44
2.50
% Parameter
θ θ1
PTPI
−30%
5.2. Sensitivity analysis of deterioration rate From Table 5 and Fig. 5, it is seen that 1. The total profit after integration is better. 2. When the deterioration rate increases, the value of PTPI increases as well. It means that for higher deterioration rate, integration is more significant. 5.3. Sensitivity analysis of holding cost From Table 6 and Fig. 6, it is seen that 1. The total profit after integration is greater. 2. When the percentage holding cost increases, the value of PTPI increases as well. It means that for higher holding cost, integration is most critical. 5.4. Sensitivity analysis of demand rate From Table 7 and Fig. 7, it is seen that 1. The total profit after integration is better. 2. When the demand rate decreases, the value of PTPI increases. It means that for smaller demand rate, there is a greater need to integrate.
P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
171
Fr ,FM ,FMw ,FRw ,Fc ,Fs Fig. 6. Relation between (Fr , FM , FM w , FRw , Fc and Fs ) and (TP and PTPI) with/without integration. Table 6 Relation between (Fr , FM , FM w , FRw , Fc and Fs ) and (P , R, TP and PTPI) with/without integration. %
−30%
−20%
−10%
0%
10%
20%
30%
Without integration
Tr P R l k N TP
0.04245 1 3 1 1 17 174,731
0.04082 1 3 1 1 17 173,514
0.03937 1 3 1 1 17 172,351
0.03807 1 3 1 1 17 171,235
0.03689 1 3 1 1 17 170,160
0.03581 1 3 1 1 16 169,127
0.03482 1 3 1 1 16 168,128
With integration
Tr P R l k N TP
0.10375 1 3 1 1 7 178,408
0.09863 1 3 1 1 7 177,285
0.09419 1 3 1 1 7 176,215
0.09031 1 3 1 1 7 175,191
0.08687 1 3 1 1 7 174,207
0.08380 1 3 1 1 7 173,260
0.08103 1 3 1 1 7 172,345
2.10
2.17
2.24
2.31
2.38
2.44
2.51
PTPI
Table 7 Relation between d and (P , R, TP and PTPI). d
1600
1800
2000
2200
2400
2600
Without integration
P R l k N TP
1 11 1 1 46 140,785
1 6 1 1 27 155,747
1 3 1 1 17 171,235
1 3 1 1 17 186,578
1 2 1 1 13 202,297
1 2 1 1 13 217,819
With integration
P R l k N TP
1 10 1 1 17 144,675
1 5 1 1 10 159,776
1 3 1 1 7 175,191
1 3 1 1 7 190,547
1 2 1 1 6 206,259
1 2 1 1 6 221,763
2.76
2.59
2.31
2.13
1.96
1.81
PTPI
6. Conclusion A closed-loop supply chain deteriorating inventory system with multi-manufacturing cycles and multi-remanufacturing cycles is analyzed using sequential and global optimization. The result shows that the global optimization is better than the sequential optimization. From the sensitivity analysis, one can see that (i) the PTPI value is more significant for smaller return rate; (ii) the greater number of manufacturing cycles results due to a small return rate,; (iii) the larger return rate results in a greater the number of the remanufacturing cycles; (iv) after integration, the values of P and R are smaller; (v) the
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Fig. 7. Relation between d and (TP and PTPI).
PTPI value is more significant with higher deterioration rate and holding cost; (vi) the PTPI value is more significant with smaller demand rate; and (vii) the total profit after integration is greater than that before integration. In this study, a single retailer, constant lead time, constant demand and return rates are assumed. Further research can be done for the system with multiple retailers and stochastic demands. Appendix A. Solve the values of α and β in Fig. A.1 From Fig. A.1, subject to the remanufacturing and deterioration rates, the manufacturer differential equation and the derived inventory level during β period are dIR (t ) dt
+ θ IR (t ) = mr ,
0≤t ≤β
(A.1)
and mr
IR (t ) =
θ
(1 − exp(−θ t )),
0≤t ≤β
(A.2)
respectively. Since the remanufacturing quantity in period β is equal to the retailer lot size, the relation is mr
θ
(1 − exp(−θ β)) =
d
θ
[exp(θ Tr ) − 1].
(A.3)
During the remanufacturing period, the differential equation of manufacturer’s total product inventory is dIR1 (t )
+ θ IR1 (t ) = mr − d.
dt
(A.4)
Since the total product inventory is α when t = 0, the derived inventory level is IR1 (t ) =
mr − d
θ
(1 − exp(−θ t )) + α.
(A.5)
From (A.2) and (A.5), the remanufactured quantity during the period β is equal to the retailer’s lot size, that is mr − d
α=
(1 − exp(−θβ)) + α =
d
[exp(θ Tr ) − 1]. θ Solving (A.3) and (A.6) simultaneously, α and β are solved as θ
d2
θ mr
[exp(θ Tr ) − 1]
(A.6)
(A.7)
and 1
β = − ln θ
mr + d − d exp(θ Tr ) mr
(A.8)
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173
Fig. A.1. The manufacturer and retailer multi-delivery inventory model with remanufacturing and deterioration.
respectively. If θ Tr 1 and, by Taylor’s series expansion, exp(x) = 1 + x + 12 x2 for x 1, α is derived as
α=
d2 Tr (2 + θ Tr ) 2mr
.
(A.9)
Using Misra [20], β is derived from (A.8) as
β=
dTr (2 + θ Tr ) 2mr
.
(A.10)
Appendix B. The collector’s average inventory and the manufacturer’s average used product inventory As illustrated in Fig. B.1, the collector’s differential equation of the used product inventory level during the nonremanufacturing period, I3 (t ), affected by return and deterioration rates, can be expressed as follows: dI3 (t )
+ θ1 I3 (t ) = u. (B.1) dt The two boundary conditions are set as follows: (i) I3 (t ) = 0, when t = 0 and (ii) I3 (t ) = α1 , when t = lR . The differential equation and the value of α1 are I3 (t ) =
u
θ1
(1 − exp(−θ1 t )),
t ≥0
(B.2)
and u
α1 =
θ1
(1 − exp(−θ1 lR )) ≈
u
θ1
1 ulR (2 − θ1 lR ) θ1 lR − θ12 l2R = 2
(B.3)
2
respectively. The differential equation of the manufacturer total inventory during the time period TR1 is dI1 (t )
+ θ1 I1 (t ) = mr − u, 0 ≤ t ≤ TR1 . (B.4) dt The two boundary conditions are set as follows: (i) I1 (t ) = 0, when t = TR1 , and (ii) I1 (t ) = β1 , when t = 0. The differential equation of the manufacturer’s total inventory during TR1 is I1 (t ) =
mr − u
θ1
[exp(θ1 (TR1 − t )) − 1],
0 ≤ t ≤ TR1 .
(B.5)
The value of β1 derived as
β1 =
mr − u
θ1
(exp(θ1 TR1 ) − 1) ≈
mr − u
θ1
1 2 2 (mr − u)TR1 (2 + θ1 TR1 ) θ1 TR1 + θ1 TR1 = . 2
2
(B.6)
The differential equation of the manufacturer total inventory during TR2 is dI2 (t ) dt
+ θ1 I2 (t ) = u,
0 ≤ t ≤ TR2 .
(B.7)
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P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
Fig. B.1. The collector multi-delivery inventory model with remanufacturing and deterioration.
The boundary condition is I2 (t ) = 0, when t = 0. The differential equation is u (1 − exp(−θ1 t )), 0 ≤ t ≤ TR2 . I2 ( t ) =
θ1
(B.8)
The value of γ1 is β1 − I2 (TR2 ), that is u γ1 = β1 − (1 − exp(−θ1 TR2 ))
≈
θ1 (mr − u)TR1 (2 + θ1 TR1 )
=
(mr − u)TR1 (2 + θ1 TR1 ) − uTR2 (2 − θ1 TR2 )
2
uTR2 (2 − θ1 TR2 )
−
2
.
2 The differential equation of the manufacturer’s used product inventory level, I4 (t ) is dI4 (t )
+ θ1 I4 (t ) = −mr , 0 ≤ t ≤ lR . dt The boundary condition is I4 (t ) = 0 when t = lR . The differential equation and lot size, I4 (0), are mr I4 ( t ) = [exp(θ1 (lR − t )) − 1], 0 ≤ t ≤ lR θ1
(B.9)
(B.10)
(B.11)
and mr
I4 (0) =
θ1
[exp(θ1 lR ) − 1] ≈
mr
θ1
1
θ1 lR + θ12 l2R 2
=
mr lR (2 + θ1 lR ) 2
.
(B.12)
The average used product inventory in the manufacturer warehouse (I Rw ) is lR
Z I Rw = kR
I4 (t )dt /T =
0
kR
mr
θ12
T
(e
θ1 lR
kRmr l2R (1 + 13 θ1 lR ) − 1 − θ1 lR ) ≈ . 2T
(B.13)
The average used product inventory level of the collector warehouse (I c ) is
Z
lR +TR2 +PmTr
Ic =
I3 (t )dt + R
lR
TR1
Z
I1 (t )dt + (R − 1)
TR2
Z
0
I2 (t )dt
0
1
+ α1 [RTR1 + (R − 1)TR2 ] + R(R − 1)γ1 TR − I Rw
2
≈− + +
kRmr l2R
1+ θ
1 l 3 1 R
2T
+
u 2T
(TR2 + mPTr )(2lR + TR2 + mPTr )
2 2 R(mr − u)TR1 + (R − 1)uTR2
2T R(R − 1)TR 4T
+
ulR (2 − θ1 lR )(RTR − TR2 ) 2T
[(mr − u)TR1 (2 + θ1 TR1 ) − uTR2 (2 − θ1 TR2 )] .
(B.14)
The collector’s total profit per unit time is RkPR mr lR (2 + θ1 lR )
Ac + RkAcm − Pc (Fc + θ1 )I c − − uPc . (B.15) 2T T In (B.15), the first term is the sales revenue; the second terms are the holding and deteriorating costs; the third term is setup cost; the last term is the collecting cost. TPc =
P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
175
Fig. C.1. The supplier multi-delivery inventory model with remanufacturing and deterioration.
Appendix C. Supplier’s average inventory and the manufacturer’s average new material inventory As illustrated in Fig. C.1, the supplier’s total new material inventory level, I5 (t ), is dI5 (t )
+ θ1 I5 (t ) = −mp , 0 ≤ t ≤ TM1 . (C.1) dt The boundary condition is I5 (t ) = 0 when t = TM1 . The supplier’s total inventory level and the purchase lot size, I5 (0), are I5 (t ) =
mp
I5 (0) =
mp
[exp(θ1 (TM1 − t )) − 1],
θ1
0 ≤ t ≤ TM1
(C.2)
and
(exp(θ1 TM1 ) − 1) ≈
θ1
mp
θ1
1
θ1 TM1 + θ 2
2 2 1 TM1
=
mp TM1 (2 + θ1 TM1 ) 2
.
(C.3)
The manufacturer new material inventory level, I6 (t ), is dI6 (t )
+ θ1 I6 (t ) = −mp , 0 ≤ t ≤ lM . dt The boundary condition is I6 (t ) = 0 when t = lM . The inventory level and delivery lot size, I6 (0), are I6 (t ) =
mp
I6 (0) =
mp
[exp(θ1 (lM − t )) − 1],
θ1
0 ≤ t ≤ lM
(C.4)
(C.5)
and
θ1
[exp(θ1 lM ) − 1] ≈
mp
θ1
1
θ1 lM + θ12 l2M
=
2
mp lM (2 + θ1 lM )
(C.6)
2
respectively. The supplier’s average new material inventory level (I M w ) can be determined by subtracting the average material inventory level of the manufacturer from the average total inventory in the supplier-manufacturer inventory system. lM
Z I M w = Pl
I6 (t )dt /T = Pl
0
Plmp l2M 1 ( exp (θ l ) − 1 − θ l ) T ≈ 1 + θ l . 1 M 1 M 1 M 2
mp
θ1
2T
3
(C.7)
The average new material inventory in the manufacturer warehouse (I s ) is TM1
Z Is = P
I5 (t )dt
T − I Mw
0
≈
2 Pmp TM1 − ll2M 1 + 31 θ1 lM
.
(C.8)
2T The supplier’s total profit per unit time is TPs =
=
mp PM lM lP (2 + θ1 lM ) 2T mp PM lM lP (2 + θ1 lM )
− Ps (Fs + θ1 )I s − −
P (As + lAsm ) T
2 Ps (Fs + θ1 ) TM1 − ll2M
PPs
[I5 (0)] 1 + 13 θ1 lM P (As + lAsm ) PPs mp TM1 (2 + θ1 TM1 ) − − . −
T
(C.9) 2T 2T T 2T In (C.9), the first term is the sales revenue; the second terms are the holding and deteriorating costs; the third term is the setup cost; the last term is the purchase cost.
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References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]
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Contents lists available at ScienceDirect
Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Sequential and global optimization for a closed-loop deteriorating inventory supply chain P.C. Yang a , H.M. Wee b,∗ , S.L. Chung c , P.C. Ho a a
Industrial Engineering & Management Department, St. John’s University, Tamsui, Taipei 25135, Taiwan, ROC
b
Industrial Engineering Department, Chung Yuan Christian University, Chungli 32023, Taiwan, ROC
c
Information Management Department, St. John’s University, Tamsui, Taipei 25135, Taiwan, ROC
article
info
Article history: Received 19 September 2009 Received in revised form 1 February 2010 Accepted 9 February 2010 Keywords: Supply chain management Closed-loop supply chain Multi-echelon inventory Deteriorating items
abstract Products such as IC chips, computers and mobile phones can become out of date due to technological innovation. However, these outdated products can be remanufactured and resold to market. In this paper, a closed-loop supply chain inventory system with multimanufacturing cycles and multi-remanufacturing cycles is analyzed using sequential and global optimization. In the case of sequential optimization, the decision is made initially by the down-stream player, then by the up-stream player. In the case of global optimization, the decision is made jointly by all the players. The supply chain includes a manufacturer, a retailer, a collector and a material supplier. This paper considers out of date as a type of deterioration and generalizes the following three cases: (i) single manufacturing cycle and single remanufacturing cycle, (ii) single manufacturing cycle and multi-remanufacturing cycles, and (iii) multi-manufacturing cycles and single remanufacturing cycle. The analytical results of this study show that a significant increase in the joint profit will result when the integrated policy is adopted. © 2010 Elsevier Ltd. All rights reserved.
1. Introduction One of the keys to successful supply chain management is the vendor–buyer inventory system integration. Clark and Scarf [1] presented a concept of serial multi-echelon structures to determine the optimal policy. Banerjee [2] derived a joint economic lot size model for a single vendor, single buyer system when the vendor has a finite production rate. Goyal [3] extended Banerjee’s model by relaxing the lot-for-lot production assumption. Ha and Kim [4] analyzed the integrated mathematical model between the buyer and the producer. Wee and Jong [5] studied the integration between parts and finished product with multi-lot size and deterioration. Yang et al. [6] derived a collaborative vendor–buyer inventory system with declining market. Wee considers product outdate or deterioration as decay, damage, spoilage, evaporation, obsolescence, pilferage, loss of utility or loss of marginal value of a commodity that results in decreasing usefulness from the original one [7] Ghare and Schrader [8] were the first authors to consider the on-going deterioration of inventory. Other authors such as Kang and Kim [9] and Raafat et al. [10] assumed either instantaneous or finite production with different assumptions on the patterns of deterioration. Yang and Wee [11] used a heuristic approach to develop an integrated vendor–buyer inventory model for deteriorating items. Recently many enterprises have focused their attention on reverse supply chain to meet environmental concerns/ regulations and social liability. Product remanufacturing such as transforming used items into marketable products through
∗
Corresponding author. Tel.: +886 3 5354409; fax: +886 3 2454499. E-mail address: [email protected] (H.M. Wee).
0895-7177/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2010.02.005
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refurbishment, repair and upgrading can also yield substantial cost benefits. Schrady [12] was the earliest author to propose a deterministic model with instantaneous production rate for manufacturing and remanufacturing. Schrady [12] argued that optimal lot sizes for manufacturers and remanufacturers can be determined by the classical EOQ formula. Koh et al. [13] analyzed finite manufacturing/remanufacturing rates. Savaskan et al. [14] determined the optimal collection channel configuration of a monopolistic manufacturer. Bautista and Pereira [15] proposed a method of identifying these collection channel configuration problems. Jaber and Saadany [16] developed a manufacturing and remanufacturing inventory system under the condition of lost sale. Chung et al. [17] developed a model with single manufacturing and single remanufacturing cycles (i.e., P = 1, R = 1). Yuan and Gao [18] derived two separate models: one model with single manufacturing and multi-remanufacturing cycles (i.e., P = 1, R ≥ 1); the other model with multi-manufacturing and single remanufacturing cycles (i.e., P ≥ 1, R = 1). The above researches didn’t consider the factor of deterioration. This paper develops a sequential and global optimization model considering the effect of deterioration; and synthesizes the two studies by Chung et al. [17] and Yuan and Gao [18] for multi-manufacturing and multi-remanufacturing cycles (i.e., P ≥ 1, R ≥ 1). Assumptions and notations The mathematical models in the analysis have the following assumptions: (1) (2) (3) (4) (5) (6) (7) (8) (9)
An infinite planning horizon. Deterioration is considered. Constant lead-time, manufacturing and remanufacturing rates. The multi-echelon inventory system contains a single item. The product demand rate and return rate are constant, and the return rate is less than demand rate. The number of deliveries within the manufacturing cycle is an integer. The setup cost per run and the annual holding cost fraction are known and constant. The remanufactured products are comparable to newly manufactured products. Single supplier, single manufacture, single retailer and single collector in the closed-loop supply chain multi-echelon inventory system are considered. (10) Multiple manufacturing and multiple remanufacturing cycles are considered. The following denotes the retailer’s parameters: d Ar Fr Pe Pr Tr TCr (TPr )
annual demand rate ordering cost per time annual percentage holding cost per dollar retailer price to end consumer wholesale price to retailer ordering cycle time annual total cost (total profit).
The following denotes the manufacturer’s parameters: AM (AR ) setup cost per manufacturing (remanufacturing) run AM w (ARw ) ordering cost for the material (used product) warehouse M Fixed cost for processing buyer order of any size FM (FM w /FRw )annual percentage holding cost per dollar for finished product (material/used product) PM (PR ) unit purchase cost to supplier (collector) TR1 (TR2 ) reproduction (non production) period in each remanufacturing cycle TM1 (TM2 ) production (non production) period in each manufacturing cycle mp (mr ) production (reproduction) rate, where mp > d (mr > d) m(n) number of deliveries per manufacturing (remanufacturing) cycle to the retailer (decision variable) P (R) number of manufacturing (remanufacturing) cycles in this model cycle N number of deliveries to the retailer in this model cycle, where N is an integer and N = Rn + Pm TCM w (TCRw ) annual manufacturing (remanufacturing) total cost TCm (TPm ) annual total cost (profit). The following denotes the collector’s parameters: u Ac Fc Pc k Acm TCc (TPc )
annual return rate setup cost per run for the collector annual percentage holding cost per dollar unit return cost number of deliveries per Tr1 to the manufacturer (decision variable) fixed cost to process manufacturer orders of any size annual total cost (total profit).
The following denotes the supplier’s parameters: As fixed cost per order Fs material percentage holding cost per dollar per year
P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
163
Fig. 1. The integrated multiple manufacturing/remanufacturing cycles closed-loop supply chain inventory system (adapted from Chung et al. [17]).
Ps unit purchase cost l number of deliveries per TM1 to the manufacturer (decision variable) Asm fixed cost to process manufacturer orders of any size TCs (TPs ) total cost (total profit) per unit time. Other related parameters:
θ deterioration rate for each finished product θ1 deterioration rate for each material and used product. 2. Model development The integrated closed-loop supply chain inventory system with multiple manufacturing/remanufacturing cycles is depicted in Fig. 1. From Fig. 2, the inventory levels are depleted by demand and deterioration. The buyer inventory differential equation is dIr (t )
+ θ Ir (t ) = −d, 0 ≤ t ≤ Tr dt with boundary condition of Ir (t ) = 0, t = Tr . The product inventory differential equations during TR1 and TR2 in a remanufacturing cycle are dIR1 (t ) dt
+ θ IR1 (t ) = mr − d,
dIR2 (t )
+ θ IR2 (t ) = −d, dt with boundary conditions: (i) IR1 (t ) = α,
t = 0;
0 ≤ t ≤ TR1
(2)
0 ≤ t ≤ TR2
(ii) IR2 (t ) = α,
(1)
(3)
t = TR2
and
(iii) IR1 (TR1 ) = IR2 (0).
From Spiegel [19], the differential equations are solved as Ir (t ) =
d
θ
IR1 (t ) = IR2 (t ) =
[exp(θ (Tr − t )) − 1],
mr − d
θ d
θ
0 ≤ t ≤ Tr
(1 − exp(−θ t )) + α,
[exp(TR2 − t ) − 1] + α,
0 ≤ t ≤ TR1
0 ≤ t ≤ TR2
(4) (5) (6)
where α can be derived as follows (see Appendix A):
α=
d2 Tr (2 + θ Tr ) 2mr
.
(7)
With boundary condition IR1 (TR1 ) = IR2 (0), the relation can be expressed as
(mr − d)(1 − exp(−θ TR1 )) = −d(1 − exp(θ TR2 )).
(8)
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Fig. 2. The manufacturer, retailer, collector and supplier multi-delivery inventory model with remanufacturing and deterioration (case of P = 2, R = 2, N = 8, k = 4, l = 7).
For θ TR1 1, eθ TR1 in (8) is replaced by 1 + θ TR1 + 3 θ 3 TR1
1 2
(θ TR1 )2 +
1 3!
2 θ 2 TR1
(θ TR1 )3 (Taylor series approximation). The percentage θ 3T 3
error for the fourth term in Taylor series is ( 3! )/(1 + θ TR1 + 2 + 3!R1 ). When θ TR1 ≤ 0.00649, the percentage error is about 0.00000454%. It will be smaller for term higher than four. Therefore, the term four and higher are neglected. By Taylor series expansion and the assumption of θ TR1 1, it is
1 1 (mr − d)TR1 1 − θ TR1 = dTR2 1 + θ TR2 . 2
(9)
2
From Misra [20], it is derived as dTR2 (2 + θ TR2 )
TR1 ≈
2(mr − d)
.
(10)
Since TR = TR1 + TR2 , TR can be expressed as TR ≈
TR2 (2mr + dθ TR2 ) 2(mr − d)
.
(11)
Similarly, TM1 and TM are TM1 ≈ TM ≈
dTM2 (2 + θ TM2 )
(12)
2(mp − d) TM2 (2mp + dθ TM2 ) 2(mp − d)
.
(13)
The lot size delivered to the retailer is Ir (t ) when t = 0. One has Ir (0) =
d
θ
(eθ Tr − 1) ≈
d
θ
1 dTr (2 + θ Tr ) θ Tr + θ 2 Tr2 = . 2
2
(14)
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165
The retailer’s average inventory level (I r ) is Ir =
1 Tr
Tr
Z
Ir (t )dt =
0
d Tr θ 2
[−1 − θ Tr + exp(θ Tr )] ≈
dTr
1
+ dθ Tr2 .
2
(15)
6
The manufacturer’s average inventory level (I m ) is IR1 (t )dt +
R
(m + n)Tr
Z
TR2
IR2 (t )dt + P
IM1 (t )dt +
TM2
Z
IM2 (t )dt
− ¯Ir
0
0
0
0
TM1
Z
R d2 Tr (2 + θ Tr )(TR1 + TR2 ) 2 2 (mr − d)TR1 + dTR2 +
1
≈
TR1
Z
1
¯Im =
(m + n)Tr 2 mr P d2 Tr (2 + θ Tr )(TM1 + TM2 ) 2 2 + (mp − d)TM1 + dTM2 + − ¯Ir . 2
(16)
mp
The retailer’s total profit (TPr ) is
TPr = Pe d −
Ar
+
Tr
(Fr + θ )Pr dTr
+
2
Pr dTr (2 + θ Tr )
2Tr
.
(17)
In (17), the first term is the sales revenue; the second terms are ordering cost, holding and deteriorating cost, and purchase cost. The manufacturer’s total profit (TPm ) is Pr dTr (2 + θ Tr )
TPm (P , R, l, k, N ) =
2Tr
" −
T
−
PlAM w
+
" −
RkARw T
+
PR (FRw + θ1 )kRmr l2R 1 + 13 θ1 lR
2T
PM (FM w + θ1 )Plmp l2M 1 + 13 θ1 lM 2T
PAM + RAR + (nR + mP )M
(nR + mP )Tr
+
+
+
RkPR mr lR (2 + θ1 lR )
#
2T
mp PPM lM l(2 + θ1 lM )
#
2T
(FM + θ )(PR Rn + PM Pm) Im . (nR + mP )
(18)
In (18), the first term is the sales revenue; the second terms are the used products’ ordering cost, holding and deteriorating cost, and purchase cost; the third terms are the material’s ordering cost, holding and deteriorating cost, and the purchase cost; the last term is the finished products’ holding and deteriorating costs. From Appendix B, the collector’s average inventory level and total profit are Ic = −
+
kRmr l2R 2T
1+
1 3
2 2 u R(mr − u)TR1 + (R − 1)uTR2 θ1 lR + (TR2 + mPTr )(2lR + TR2 + mPTr ) + 2T
ulR (2 − θ1 lR )(RTR − TR2 ) 2T
+
2T
R(R − 1)TR 4T
[(mr − u)TR1 (2 + θ1 TR1 ) − uTR2 (2 − θ1 TR2 )]
(19)
and TPc =
RkPR mr lR (2 + θ1 lR ) 2T
− Pc (Fc + θ1 )I c −
Ac + RkAcm T
− uPc
(20)
respectively. In (20), the first term is the sales revenue; the second terms are the holding and deteriorating costs; the third term is the setup cost; the last term is the purchase cost. From Appendix C, the supplier’s average inventory level and total profit are 2 Pmp TM1 − ll2M 1 + 31 θ1 lM
¯Is =
(21)
2T
and TPs =
mp PM lM lP (2 + θ1 lM ) 2T
− Ps (Fs + θ1 )I s −
P (As + lAsm ) T
−
PPs mp TM1 (2 + θ1 TM1 ) 2T
(22)
respectively. In (22), the first term is the sales revenue; the second terms are the holding and deteriorating costs; the third term is the setup cost; the last term is the purchase cost. The total profit per unit time is TP (P , R, l, k, N , T r , T R2 , T M2 ) = TPm + TPr + TPc + TPs
(23)
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Subject to u= m=
nRd
dTR2 (2 + θ TR2 )
(27)
2(mr − d) TR1 k
TM =
=
dTR2 (2 + θ TR2 )
(28)
2k(mr − d)
dTM2 (2 + θ TM2 )
TM1 =
Tr =
(26)
2(mr − d) TR2 (2mr + dθ TR2 )
TR =
lM =
(25)
P
TR1 =
lR =
(24)
(Pm + Rn) (N − nR)
(29)
2(mp − d) TM2 (2mp + dθ TM2 )
(30)
2(mp − d) TM1 l TR n
=
=
T = NTr =
dTM2 (2 + θ TM2 )
(31)
2l(mp − d)
TR2 (2mr + dθ TR2 )
(32)
2n(mr − d) NTR2 (2mr + dθ TR2 )
(33)
2n(mr − d)
N , P , R, l and k are positive integers.
(34)
3. Solution procedure The solution procedures for two cases are illustrated as follows: The first case is by the sequential (or local) optimization where the retailer and the manufacturer derive their respective optimal solution independently. The second case is to derive the global optimization of the integrated supply chain system. Case 1. Sequential or local optimization Step 1: Take derivatives of retailer’s profit (17) with respective to Tr and equating the result to zero. Solve the value of Tr , then substitute the value of Tr into (18). Step 2.1: Let P = 1 and take derivatives of manufacturer’s setup and holding cost (i.e., last term of (18), { PR Rn+PM Pm) [ (FM +θ)( ]I m }) (nR+mP )
PAM +RAR +(nR+mP )M (nR+mP )Tr
+
with respective to R and N, and equating the results to zero. Solve them and round them up or down the values of R and N. Step 2.2: Take the derivatives of the manufacturer’s used-product cost (i.e., the second term of (18), [ PR (FRw +θ1 )kRmr l2R
RkARw T
+
RkPR mr lR (2+θ1 lR ) 2T
(1 + θ ) + ]) with respective to k, and equating the result to zero. Solve it and 2T round it up or down the value of k so that the constraint of k ≥ 1 is observed. 1 l 3 1 R
Step 2.3: Take the derivatives of the manufacturer’s new-material cost (i.e., the third term of (18), [ PM (FM w +θ1 )Plmp l2M
PlAM w T
+
mp PPM lM l(2+θ1 lM ) 2T
(1 + θ ) + ]) with respective to l, and equating the result to zero. Solve it and 2T round it up or down the value of l so that the constraint of l ≥ 1 is observed. Step 2.4: Explore the local optimization by satisfying the following conditions 1 l 3 1 M
TP (P ∗ , R∗ − 1, k∗ , l∗ , N ∗ ) 5 TP (P ∗ , R∗ , k∗ , l∗ , N ∗ ) = TP (P ∗ , R∗ + 1, k∗ , l∗ , N ∗ )
(35)
TP (P , R , k − 1, l , N ) 5 TP (P , R , k , l , N ) = TP (P , R , k + 1, l , N )
(36)
TP (P , R , k , l − 1, N ) 5 TP (P , R , k , l , N ) = TP (P , R , k , l + 1, N )
(37)
TP (P ∗ , R∗ , k∗ , l∗ , N ∗ − 1) 5 TP (P ∗ , R∗ , k∗ , l∗ , N ∗ ) = TP (P ∗ , R∗ , k∗ , l∗ , N ∗ + 1).
(38)
∗ ∗
∗ ∗
∗ ∗
∗
∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗ ∗
∗
∗
∗
∗
and
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167
Table 1 Exploration of sequential optimal solution without integration. Variables
P
R
k
l
N
n
m
TP
P =1
1 1 1 1 1 1 1 1 1
3 2 4 3 3 2 4 3 3
1 1 1 1 1 1 1 2 1
1 1 1 1 1 1 1 1 2
18 18 18 17 16 17 17 17 17
4.500 6.750 3.375 4.250 4.000 6.375 3.188 4.250 4.250
4.500 4.500 4.500 4.250 4.000 4.250 4.250 4.250 4.250
171,192 170,090 171,014 171,235# 171,183 170,361 170,900 169,735 170,394
R=1
1 1 1 1 1 2 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 2 1
1 1 1 1 1 1 1 2
6 5 7 8 9 8 8 8
4.500 3.750 5.250 6.000 6.750 6.000 6.000 6.000
1.500 1.250 1.750 2.000 2.250 1.000 2.000 2.000
168,173 166,776 168,922 169,139 169,109 164,327 168,564 167,424
Step 3.1: Let R = 1 and take derivatives of manufacturer’s setup and holding cost (i.e., last term of (18), {
PAM +RAR +(nR+mP )M (nR+mP )Tr
Remark: # is the local optimal solution.
PR Rn+PM Pm) [ (FM +θ)( ]I m }) (nR+mP )
+
with respective to P and N, and equating the results to zero. Solve them and round them up or down the values of P and N. RkA Step 3.2: Take the derivatives of the manufacturer’s used-product cost (i.e., the second term of (18), [ TRw + PR (FRw +θ1 )kRmr l2R
RkP m l (2+θ l )
R 1 R (1 + 13 θ1 lR ) + R r 2T ]) with respective to k, and equate to zero, and solve it. Round up and 2T down the value of k and obey the constraint of k ≥ 1. PlA Step 3.3: Take the derivatives of the manufacturer’s new-material cost (i.e., the third term of (18), [ TM w +
PM (FM w +θ1 )Plmp l2M
(1 + 13 θ1 lM ) +
mp PPM lM l(2+θ1 lM ) 2T
]) with respective to l, and equating the result to zero. Solve it and round it up or down the value of l so that the constraint of l ≥ 1 is observed. Step 3.4: Explore the local optimization by satisfying (36)–(38) and the following condition 2T
TP (P ∗ − 1, R∗ , k∗ , l∗ , N ∗ ) 5 TP (P ∗ , R∗ , k∗ , l∗ , N ∗ ) = TP (P ∗ + 1, R∗ , k∗ , l∗ , N ∗ ).
(39)
Step 4: Find the better solution from Step 2.4 and Step 3.4. Case 2. Global optimization Step 1: Let the values of P, R, k, l and N solved from Case 1 be the initial values. Step 2: Substituting the initial value into joint profit (23), take the derivative of (23) with respective of Tr , and equating the result to zero. Solve it and substitute Tr into (23), the joint profit can be obtained. Step 3: Explore the proximity of P, R, k and l, and derive Tr by using the method of step 2. Substituting Tr into (23), the joint profit can be obtained. Step 4: Find the global optimization by satisfying conditions (35)–(39). 4. Numerical example The numerical data are set as follows: d = 2000, Ar = 100, Fr = 0.3, Pe = 175, Pr = 150, AM = 750, AR = 500, AM w = 350, ARw = 350, M = 350, FM = 0.3, FM w = 0.2, FRw = 0.2, PM = 115, PR = 110, mp = 5000, mr = 4000, u = 1500, Ac = 250, Fc = 0.1, Pc = 70, Acm = 150, As = 200, Fs = 0.3, Ps = 90, Asm = 150, θ = 0.08 and θ1 = 0.05. By using the solution procedure in Section 4, the local optimal solution is P = 1, R = 3, k = 1, l = 1, N = 17 and TP = 171,235.11 (Table 1) and the global optimal solution is P = 1, R = 3, k = 1, l = 1, N = 7 and TP = 175,190.68 (Table 2). The percentage of total profit increase (denoted by PTPI) is defined as PTPI =
Total profit from Case 2 − Total profit from Case 1 Total profit from Case 1
.
From Table 3, the percentage of joint profit increase is 2.31%, the percentage of profit increase for the manufacturer and collector is positive, and the percentage of profit increase for the retailer and supplier is negative. Table 3 shows that the value of PTPI is 2.31%, the manufacturer and the collector benefit from integration; but the retailer and the supplier suffer lose. For lasting partnership, profit sharing between all parties should be considered.
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Table 2 Exploitation of global optimal solution with integration. P
R
k
l
N
n
m
Tr0
TP (Tr0 )
1 1 1 1 1 1 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1
3 2 4 5 6 7 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 4 3 4 4 4 4 4 4 4 4 4 3 2 3 3 3 3 3 3 2
1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1
1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1
17 17 17 17 17 17 17 17 17 18 16 16 16 16 15 14 13 12 11 10 9 8 10 10 10 10 10 10 10 10 10 9 8 7 8 8 8 8 8 9 7 6 7 7 7 7
4.250 6.375 3.188 2.550 2.125 1.822 2.550 2.550 2.550 2.700 2.400 2.400 2.400 2.400 2.250 2.100 1.950 1.800 1.650 1.500 1.350 1.200 1.500 1.500 1.500 1.250 1.875 2.500 1.875 1.875 1.875 1.688 1.500 1.313 1.500 1.500 1.500 2.000 3.000 2.250 1.750 1.500 1.750 1.750 1.750 2.625
4.250 4.250 4.250 4.250 4.250 4.250 2.125 2.125 2.125 2.250 2.000 2.000 2.000 4.000 3.750 3.500 3.250 3.000 2.750 2.500 2.250 2.000 2.500 2.500 1.250 2.500 2.500 2.500 2.500 2.500 1.250 2.250 2.000 1.750 2.000 2.000 1.000 2.000 2.000 2.250 1.750 1.500 1.750 1.750 0.875 1.750
0.05341 0.04581 0.05935 0.06434 0.06872 0.07267 0.07095 0.07298 0.07914 0.06853 0.07361 0.07579 0.08222 0.06672 0.06935 0.07229 0.07559 0.07933 0.08361 0.08856 0.09439 0.10134 0.09018 0.10063 0.09808 0.09494 0.08139 0.07300 0.09186 0.08307 0.09036 0.08670 0.09304 0.10078 0.09520 0.10554 0.10357 0.08339 0.07140 0.07773 0.09031 0.09901 0.09289 0.10186 0.10124 0.07735
172,810 170,862 173,554 173,781 173,743 173,553 173,824 172,942 172,829 173,675 173,959 173,059 172,896 173,978 174,163 174,331 174,479 174,602 174,691 174,735 174,720 174,623 174,081 172,807 174,239 174,331 174,960 174,816 173,469 174,264 174,244 175,034 175,038 174,942 174,294 173,271 174,086 175,140 174,454 175,006 175,191∗ 175,117 174,343 173,807 173,787 174,714
Remark: ∗ is the global optimization. Table 3 PTPI.
Case 1 (I) Case 2 (II) (II) −1 (I)
TP r
TP m
TP c
TP s
TP
44,746.43 42,661.31 −4.66%
57,809.00 63,837.14 10.43%
56,700.25 56,726.12 0.05%
11,979.43 11,966.11 −0.11%
171,235.11 175,190.68 PTPI = 2.31%
5. Sensitivity analysis Sensitivity analysis of return rate, deterioration rate, holding cost and demand rate is analyzed when each parameter increases or decreases a fixed rate. The results are as follows: 5.1. Sensitivity analysis of return rate From Table 4, Figs. 3 and 4, it is seen that 1. The total profit after integration is better than that before integration. 2. When the return rate decreases, the value of PTPI increases. It means a a higher significance for less return rate.
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169
Table 4 Relation between u and (P , R, TP and PTPI). u
200
400
600
800
1000
1200
1400
1600
1800
Case 1 without integration
P R l k N TP
7 1 1 1 42 144,150
4 1 1 1 23 148,149
2 1 1 1 14 152,525
1 1 1 1 9 156,719
1 1 1 1 9 160,813
1 2 1 1 13 164,952
1 3 1 1 16 169,071
1 5 1 1 24 173,208
1 8 1 1 35 177,638
Case 2 with integration
P R l k N TP
5 1 1 1 11 148,747
3 1 1 1 8 152,640
2 1 1 1 6 156,678
1 1 1 1 4 160,825
1 1 1 1 4 164,879
1 2 1 1 5 168,961
1 3 1 1 7 173,050
1 4 1 1 9 177,266
1 7 1 1 14 18,1570
3.19
3.03
2.72
2.62
2.53
2.43
2.35
2.34
2.21
PTPI
Fig. 3. Relation between u and (P , R) with/without integration.
Fig. 4. Relation between u and (TP , PTPI ) with/without integration.
3. The P value increases when the return rate increases. 4. The values of R and P decrease after integration.
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Fig. 5. Relation between (θ, θ1 ) and (TP and PTPI) with/without integration. Table 5 Relation between (θ, θ1 ) and (P , R, TP and PTPI) with/without integration. 0.056 0.035
−20% 0.064 0.040
−10% 0.072 0.045
0% 0.080 0.050
10% 0.088 0.055
20% 0.096 0.060
30% 0.104 0.065
Case 1 without integration
Tr P R l k N TP
0.04023 1 3 1 1 16 172,296
0.03947 1 3 1 1 16 171,936
0.03875 1 3 1 1 16 171,580
0.03807 1 3 1 1 17 171,235
0.03742 1 3 1 1 17 170,893
0.03681 1 3 1 1 17 170,556
0.03623 1 3 1 1 17 170,222
Case 2 with integration
Tr P R l k N TP
0.09306 1 3 1 1 7 175,918
0.09212 1 3 1 1 7 175,673
0.09120 1 3 1 1 7 175,431
0.09031 1 3 1 1 7 175,191
0.08944 1 3 1 1 7 174,953
0.08860 1 3 1 1 7 174,717
0.08778 1 3 1 1 7 174,484
2.10
2.17
2.24
2.31
2.38
2.44
2.50
% Parameter
θ θ1
PTPI
−30%
5.2. Sensitivity analysis of deterioration rate From Table 5 and Fig. 5, it is seen that 1. The total profit after integration is better. 2. When the deterioration rate increases, the value of PTPI increases as well. It means that for higher deterioration rate, integration is more significant. 5.3. Sensitivity analysis of holding cost From Table 6 and Fig. 6, it is seen that 1. The total profit after integration is greater. 2. When the percentage holding cost increases, the value of PTPI increases as well. It means that for higher holding cost, integration is most critical. 5.4. Sensitivity analysis of demand rate From Table 7 and Fig. 7, it is seen that 1. The total profit after integration is better. 2. When the demand rate decreases, the value of PTPI increases. It means that for smaller demand rate, there is a greater need to integrate.
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171
Fr ,FM ,FMw ,FRw ,Fc ,Fs Fig. 6. Relation between (Fr , FM , FM w , FRw , Fc and Fs ) and (TP and PTPI) with/without integration. Table 6 Relation between (Fr , FM , FM w , FRw , Fc and Fs ) and (P , R, TP and PTPI) with/without integration. %
−30%
−20%
−10%
0%
10%
20%
30%
Without integration
Tr P R l k N TP
0.04245 1 3 1 1 17 174,731
0.04082 1 3 1 1 17 173,514
0.03937 1 3 1 1 17 172,351
0.03807 1 3 1 1 17 171,235
0.03689 1 3 1 1 17 170,160
0.03581 1 3 1 1 16 169,127
0.03482 1 3 1 1 16 168,128
With integration
Tr P R l k N TP
0.10375 1 3 1 1 7 178,408
0.09863 1 3 1 1 7 177,285
0.09419 1 3 1 1 7 176,215
0.09031 1 3 1 1 7 175,191
0.08687 1 3 1 1 7 174,207
0.08380 1 3 1 1 7 173,260
0.08103 1 3 1 1 7 172,345
2.10
2.17
2.24
2.31
2.38
2.44
2.51
PTPI
Table 7 Relation between d and (P , R, TP and PTPI). d
1600
1800
2000
2200
2400
2600
Without integration
P R l k N TP
1 11 1 1 46 140,785
1 6 1 1 27 155,747
1 3 1 1 17 171,235
1 3 1 1 17 186,578
1 2 1 1 13 202,297
1 2 1 1 13 217,819
With integration
P R l k N TP
1 10 1 1 17 144,675
1 5 1 1 10 159,776
1 3 1 1 7 175,191
1 3 1 1 7 190,547
1 2 1 1 6 206,259
1 2 1 1 6 221,763
2.76
2.59
2.31
2.13
1.96
1.81
PTPI
6. Conclusion A closed-loop supply chain deteriorating inventory system with multi-manufacturing cycles and multi-remanufacturing cycles is analyzed using sequential and global optimization. The result shows that the global optimization is better than the sequential optimization. From the sensitivity analysis, one can see that (i) the PTPI value is more significant for smaller return rate; (ii) the greater number of manufacturing cycles results due to a small return rate,; (iii) the larger return rate results in a greater the number of the remanufacturing cycles; (iv) after integration, the values of P and R are smaller; (v) the
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Fig. 7. Relation between d and (TP and PTPI).
PTPI value is more significant with higher deterioration rate and holding cost; (vi) the PTPI value is more significant with smaller demand rate; and (vii) the total profit after integration is greater than that before integration. In this study, a single retailer, constant lead time, constant demand and return rates are assumed. Further research can be done for the system with multiple retailers and stochastic demands. Appendix A. Solve the values of α and β in Fig. A.1 From Fig. A.1, subject to the remanufacturing and deterioration rates, the manufacturer differential equation and the derived inventory level during β period are dIR (t ) dt
+ θ IR (t ) = mr ,
0≤t ≤β
(A.1)
and mr
IR (t ) =
θ
(1 − exp(−θ t )),
0≤t ≤β
(A.2)
respectively. Since the remanufacturing quantity in period β is equal to the retailer lot size, the relation is mr
θ
(1 − exp(−θ β)) =
d
θ
[exp(θ Tr ) − 1].
(A.3)
During the remanufacturing period, the differential equation of manufacturer’s total product inventory is dIR1 (t )
+ θ IR1 (t ) = mr − d.
dt
(A.4)
Since the total product inventory is α when t = 0, the derived inventory level is IR1 (t ) =
mr − d
θ
(1 − exp(−θ t )) + α.
(A.5)
From (A.2) and (A.5), the remanufactured quantity during the period β is equal to the retailer’s lot size, that is mr − d
α=
(1 − exp(−θβ)) + α =
d
[exp(θ Tr ) − 1]. θ Solving (A.3) and (A.6) simultaneously, α and β are solved as θ
d2
θ mr
[exp(θ Tr ) − 1]
(A.6)
(A.7)
and 1
β = − ln θ
mr + d − d exp(θ Tr ) mr
(A.8)
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173
Fig. A.1. The manufacturer and retailer multi-delivery inventory model with remanufacturing and deterioration.
respectively. If θ Tr 1 and, by Taylor’s series expansion, exp(x) = 1 + x + 12 x2 for x 1, α is derived as
α=
d2 Tr (2 + θ Tr ) 2mr
.
(A.9)
Using Misra [20], β is derived from (A.8) as
β=
dTr (2 + θ Tr ) 2mr
.
(A.10)
Appendix B. The collector’s average inventory and the manufacturer’s average used product inventory As illustrated in Fig. B.1, the collector’s differential equation of the used product inventory level during the nonremanufacturing period, I3 (t ), affected by return and deterioration rates, can be expressed as follows: dI3 (t )
+ θ1 I3 (t ) = u. (B.1) dt The two boundary conditions are set as follows: (i) I3 (t ) = 0, when t = 0 and (ii) I3 (t ) = α1 , when t = lR . The differential equation and the value of α1 are I3 (t ) =
u
θ1
(1 − exp(−θ1 t )),
t ≥0
(B.2)
and u
α1 =
θ1
(1 − exp(−θ1 lR )) ≈
u
θ1
1 ulR (2 − θ1 lR ) θ1 lR − θ12 l2R = 2
(B.3)
2
respectively. The differential equation of the manufacturer total inventory during the time period TR1 is dI1 (t )
+ θ1 I1 (t ) = mr − u, 0 ≤ t ≤ TR1 . (B.4) dt The two boundary conditions are set as follows: (i) I1 (t ) = 0, when t = TR1 , and (ii) I1 (t ) = β1 , when t = 0. The differential equation of the manufacturer’s total inventory during TR1 is I1 (t ) =
mr − u
θ1
[exp(θ1 (TR1 − t )) − 1],
0 ≤ t ≤ TR1 .
(B.5)
The value of β1 derived as
β1 =
mr − u
θ1
(exp(θ1 TR1 ) − 1) ≈
mr − u
θ1
1 2 2 (mr − u)TR1 (2 + θ1 TR1 ) θ1 TR1 + θ1 TR1 = . 2
2
(B.6)
The differential equation of the manufacturer total inventory during TR2 is dI2 (t ) dt
+ θ1 I2 (t ) = u,
0 ≤ t ≤ TR2 .
(B.7)
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P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
Fig. B.1. The collector multi-delivery inventory model with remanufacturing and deterioration.
The boundary condition is I2 (t ) = 0, when t = 0. The differential equation is u (1 − exp(−θ1 t )), 0 ≤ t ≤ TR2 . I2 ( t ) =
θ1
(B.8)
The value of γ1 is β1 − I2 (TR2 ), that is u γ1 = β1 − (1 − exp(−θ1 TR2 ))
≈
θ1 (mr − u)TR1 (2 + θ1 TR1 )
=
(mr − u)TR1 (2 + θ1 TR1 ) − uTR2 (2 − θ1 TR2 )
2
uTR2 (2 − θ1 TR2 )
−
2
.
2 The differential equation of the manufacturer’s used product inventory level, I4 (t ) is dI4 (t )
+ θ1 I4 (t ) = −mr , 0 ≤ t ≤ lR . dt The boundary condition is I4 (t ) = 0 when t = lR . The differential equation and lot size, I4 (0), are mr I4 ( t ) = [exp(θ1 (lR − t )) − 1], 0 ≤ t ≤ lR θ1
(B.9)
(B.10)
(B.11)
and mr
I4 (0) =
θ1
[exp(θ1 lR ) − 1] ≈
mr
θ1
1
θ1 lR + θ12 l2R 2
=
mr lR (2 + θ1 lR ) 2
.
(B.12)
The average used product inventory in the manufacturer warehouse (I Rw ) is lR
Z I Rw = kR
I4 (t )dt /T =
0
kR
mr
θ12
T
(e
θ1 lR
kRmr l2R (1 + 13 θ1 lR ) − 1 − θ1 lR ) ≈ . 2T
(B.13)
The average used product inventory level of the collector warehouse (I c ) is
Z
lR +TR2 +PmTr
Ic =
I3 (t )dt + R
lR
TR1
Z
I1 (t )dt + (R − 1)
TR2
Z
0
I2 (t )dt
0
1
+ α1 [RTR1 + (R − 1)TR2 ] + R(R − 1)γ1 TR − I Rw
2
≈− + +
kRmr l2R
1+ θ
1 l 3 1 R
2T
+
u 2T
(TR2 + mPTr )(2lR + TR2 + mPTr )
2 2 R(mr − u)TR1 + (R − 1)uTR2
2T R(R − 1)TR 4T
+
ulR (2 − θ1 lR )(RTR − TR2 ) 2T
[(mr − u)TR1 (2 + θ1 TR1 ) − uTR2 (2 − θ1 TR2 )] .
(B.14)
The collector’s total profit per unit time is RkPR mr lR (2 + θ1 lR )
Ac + RkAcm − Pc (Fc + θ1 )I c − − uPc . (B.15) 2T T In (B.15), the first term is the sales revenue; the second terms are the holding and deteriorating costs; the third term is setup cost; the last term is the collecting cost. TPc =
P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
175
Fig. C.1. The supplier multi-delivery inventory model with remanufacturing and deterioration.
Appendix C. Supplier’s average inventory and the manufacturer’s average new material inventory As illustrated in Fig. C.1, the supplier’s total new material inventory level, I5 (t ), is dI5 (t )
+ θ1 I5 (t ) = −mp , 0 ≤ t ≤ TM1 . (C.1) dt The boundary condition is I5 (t ) = 0 when t = TM1 . The supplier’s total inventory level and the purchase lot size, I5 (0), are I5 (t ) =
mp
I5 (0) =
mp
[exp(θ1 (TM1 − t )) − 1],
θ1
0 ≤ t ≤ TM1
(C.2)
and
(exp(θ1 TM1 ) − 1) ≈
θ1
mp
θ1
1
θ1 TM1 + θ 2
2 2 1 TM1
=
mp TM1 (2 + θ1 TM1 ) 2
.
(C.3)
The manufacturer new material inventory level, I6 (t ), is dI6 (t )
+ θ1 I6 (t ) = −mp , 0 ≤ t ≤ lM . dt The boundary condition is I6 (t ) = 0 when t = lM . The inventory level and delivery lot size, I6 (0), are I6 (t ) =
mp
I6 (0) =
mp
[exp(θ1 (lM − t )) − 1],
θ1
0 ≤ t ≤ lM
(C.4)
(C.5)
and
θ1
[exp(θ1 lM ) − 1] ≈
mp
θ1
1
θ1 lM + θ12 l2M
=
2
mp lM (2 + θ1 lM )
(C.6)
2
respectively. The supplier’s average new material inventory level (I M w ) can be determined by subtracting the average material inventory level of the manufacturer from the average total inventory in the supplier-manufacturer inventory system. lM
Z I M w = Pl
I6 (t )dt /T = Pl
0
Plmp l2M 1 ( exp (θ l ) − 1 − θ l ) T ≈ 1 + θ l . 1 M 1 M 1 M 2
mp
θ1
2T
3
(C.7)
The average new material inventory in the manufacturer warehouse (I s ) is TM1
Z Is = P
I5 (t )dt
T − I Mw
0
≈
2 Pmp TM1 − ll2M 1 + 31 θ1 lM
.
(C.8)
2T The supplier’s total profit per unit time is TPs =
=
mp PM lM lP (2 + θ1 lM ) 2T mp PM lM lP (2 + θ1 lM )
− Ps (Fs + θ1 )I s − −
P (As + lAsm ) T
2 Ps (Fs + θ1 ) TM1 − ll2M
PPs
[I5 (0)] 1 + 13 θ1 lM P (As + lAsm ) PPs mp TM1 (2 + θ1 TM1 ) − − . −
T
(C.9) 2T 2T T 2T In (C.9), the first term is the sales revenue; the second terms are the holding and deteriorating costs; the third term is the setup cost; the last term is the purchase cost.
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P.C. Yang et al. / Mathematical and Computer Modelling 52 (2010) 161–176
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