Expert Systems with Applications 38 (2011) 13482–13486
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Short communication
The simplified solution procedures for the optimal replenishment decisions under two levels of trade credit policy depending on the order quantity in a supply chain system Kun-Jen Chung ⇑ College of Business, Chung Yuan Christian University, Chung Li, Taiwan, ROC
a r t i c l e
i n f o
Keywords: Inventory Two levels of trade credit Delay in payments
a b s t r a c t Kreng and Tan [Kreng, V. B. & Tan, S. J. (2010). The optimal replenishment decisions under two levels of trade credit policy depending on the order quantity. Expert Systems with Applications 37, 5514–5522] explore the optimal replenishment decisions under two levels of trade credit policy depending on the order quantity. Their inventory model is new and interesting. However, this paper indicates that their solution procedures can be improved further. So, the main purpose of this paper is to incorporate and simplify them for Kreng and Tan (2010). Ó 2011 Elsevier Ltd. All rights reserved.
1. Introduction Huang (2003) was the first person to consider the inventory model under two levels of trade credit independent of the order quantity. Khouja and Mehrez (1996) discussed the optimal inventory policy under supplier credit policies depending on the order quantity. Kreng and Tan (2010) adopt the viewpoint of Khouja and Mehrez (1996) to generalize Huang (2003) such that the supplier credit policy depends on the order quantity. In the latter case, suppliers use favorable credit terms to encourage the retailer to order large quantities. In other words, the favorable credit terms apply only at large order quantities. Basically, the inventory model presented by Kreng and Tan (2010) is new and interesting. However, this paper indicates that their solution procedures can be improved further. So, the main purpose of this paper is to incorporate and simplify them for Kreng and Tan (2010). 2. Model formulation
(5) M P N. (6) If Q < W, the delays in payments offered by both the supplier and the retailer are not permitted, the payment must be made immediately after receiving the items, and the retailer starts paying for charges on the items in stocks with rate Ik. (7) If Q P W, both the fixed trade credit period M offered by the supplier and the trade credit N offered by the retailer are permitted. The retailer can accumulate revenue and earn interest during the period N to M with rate Ie. When T P M, the account is settled at T = M, keeps the profit for the use of the other activities, and starts paying for the interest charges on the items in stocks with rate Ik. When T 6 M, the account is settled at T = M and the retailer does not need to pay any interest charge. Kreng and Tan (2010) assume Ik P Ie. However, this paper assumes that neither s = c nor Ik P Ie. Based on the above notation and assumptions, there are three cases to occur [i.e. Case 1: 0
Assumptions: Case 1: 0 < W < N. D (1) (2) (3) (4)
Demand rate is known and constant. Shortages are not allowed. Time horizon is infinite. Replenishments are instantaneous with a known and constant lead time.
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Under this case, Kreng and Tan (2010) show that the annual total relevant cost TVC(T) can be expressed as follows:
8 TVC 1 ðTÞ > > > < TVC ðTÞ 2 TVCðTÞ ¼ > TVC 3 ðTÞ > > : TVC 4 ðTÞ
if T P M;
ðaÞ
if N 6 T 6 M; ðbÞ if W 6 T 6 N; D if 0 < T < W : D
ðcÞ ðdÞ
ð1Þ
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K.-J. Chung / Expert Systems with Applications 38 (2011) 13482–13486
Nomenclature Notation D demand rate per year Q the order quantity W the fixed quantity at which the delay in payments is permitted A ordering cost per order c unit purchasing cost s unit selling price, s P c h holding cost per unit per year excluding interest charges
where
A DTh cIk DðT MÞ2 sIe DðM2 N2 Þ þ ; TVC 1 ðTÞ ¼ þ T 2 2T 2T 2
A DTh þ sIe DðM NÞ; T 2
TVC 3 ðTÞ ¼
T3
TVC 02 ðTÞ ¼ ð3Þ TVC 002 ðTÞ ¼
2A þ sIe DN2
2T 2 2A þ sIe DN 2 T3
ð4Þ TVC 03 ðTÞ ¼
A DTh cIk DT þ þ : T 2 2
ð5Þ
For convenience, we let all TVCi(T)(i = 1, 2, 3, 4) be defined on T > 0. Since TVC 4 W > TVC 3 W ; TVC 2 ðNÞ ¼ TVC 3 ðNÞ, and TVC1(M) = D D TVC2(M). So, TVC(T) is continuous except T ¼ W . Furthermore, we D have
TVC 4 ðTÞ > TVC 3 ðTÞ if T > 0;
ð6Þ
TVC 4 ðTÞ > TVC 2 ðTÞ if N 6 T 6 M;
ð7Þ
and
;
ð12Þ
TVC 003 ðTÞ ¼
A T2
2A T3
TVC 04 ðTÞ ¼
Dðh þ sIe Þ ; 2
> 0;
Dh ; 2
> 0;
A T
þ
þ
2
þ
Dh cIk D þ ; 2 2
ð13Þ ð14Þ
ð15Þ
ð16Þ
ð17Þ
and
TVC 004 ðTÞ ¼
2A T3
> 0:
ð18Þ
Let
TVC 4 ðTÞ > TVC 1 ðTÞ if T P M:
ð8Þ
Under this case, Kreng and Tan (2010) show that the annual total relevant cost TVC(T) can be expressed as follows:
8 ðaÞ > < TVC 1 ðTÞ if M 6 T; 6 T 6 M; ðbÞ TVCðTÞ ¼ TVC 2 ðTÞ if W D > : TVC 4 ðTÞ if 0 < T < W : ðcÞ D Since TVC 4
W D
> TVC 2
a ¼ 2A þ cIk DM2 sIe DðM2 N2 Þ:
W D
Lemma 1 (A) TVC1(T) is convex on T > 0 if a > 0. Furthermore, TVC1(T) is increasing on T > 0 if a 6 0. (B) TVCi(T) is convex on T > 0 for all i = 2, 3, 4.
ð9Þ
Proof
. ; TVCðTÞ is continuous except T ¼ W D
(A) If a > 0, Eq. (12) implies that TVC1(T) is convex on T > 0. On the other hand, if a 6 0, Eq. (11) reveals TVC 01 ðTÞ > 0 for all T > 0. So, TVC1(T) is increasing on T > 0 if a 6 0. Therefore, Lemma 1(A) holds. (B) Eqs. (14), (16) and (18) imply that TVCi(T) is convex on T > 0 for all i = 2, 3, 4.
Case 3: M 6 W : D Under this case, Kreng and Tan (2010) show that the annual total relevant cost TVC(T) can be expressed as follows:
(
TVC 1 ðTÞ if W 6 T; ðaÞ D : ðbÞ TVC 4 ðTÞ if 0 < T < W D
Since W P M; TVC 4 D T¼W . D
W D
> TVC 1
ð10Þ
W : So, TVC(T) is continuous except D
3. The convexity of TVCi(T) (i = 1, 2, 3, 4)
Combining (A) and (B), we have completed the proof of Lemma 1. Let TVC 0i ðT i Þ ¼ 0 for all i = 1, 2, 3, 4. We can obtain
T 1
Eqs. (2)–(5) yield that:
TVC 01 ðTÞ ¼
ð19Þ
Then, we have the following results.
Case 2: N 6 W < M. D
TVCðTÞ ¼
2A þ cIk DM2 sIe DðM 2 N2 Þ
ð2Þ
and
TVC 4 ðTÞ ¼
N TVC(T) T⁄
interest earned per $ per year interest charged per $ in stocks per year by the supplier the cycle time supplier’s trade credit period offered by supplier in years retailer’s trade credit period offered by supplier in years annual total relevant cost, which is a function of T optimal cycle time of TVC(T)
TVC 001 ðTÞ ¼
2
A DTh sIe Dð2MT N T Þ ; TVC 2 ðTÞ ¼ þ T 2 2T
Ie Ik T M
2A þ cIk DM2 sIe DðM 2 N2 Þ 2T 2
Dðh þ cIk Þ ; þ 2
ð11Þ
T 2
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2A þ cIk DM 2 sIe DðM2 N2 Þ ¼ if a > 0; Dðh þ cIk Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2A þ sIe DN 2 ¼ ; Dðh þ sIe Þ
ð20Þ
ð21Þ
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K.-J. Chung / Expert Systems with Applications 38 (2011) 13482–13486
rffiffiffiffiffiffiffi 2A ; T3 ¼ Dh
ð22Þ
Then, we have the following results. Lemma 2
and
T 4
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2A : ¼ Dðh þ cIk Þ
ð23Þ
Furthermore, Eqs. (11), (13), (15) and (17) yield that
TVC 01 ðMÞ ¼ TVC 02 ðMÞ ¼
2A þ sIe DðM2 N 2 Þ þ DhM 2M
2
2
¼
D1 2M2
; ð24Þ
TVC 02 ðNÞ ¼ TVC 03 ðNÞ ¼
TVC 03
2A þ DhN 2N
2
¼
2
D2 2N2
;
2 2A þ Dh W W D3 D ¼ ¼ 2 ; W 2 D 2 D 2 W D
2 2A þ Dðh þ cIk Þ W W D4 D ¼ 2 ; ¼ TVC 04 2 W D 2 W 2 D D TVC 02
2 2A þ Dðh þ sIe Þ W sIe DN 2 W D5 D ¼ ¼ 2 ; W 2 D 2 D 2 W D
ð25Þ
ð26Þ
Proof (A) If D1 6 0, Eq. (30) implies 2
2A sIe DðM2 N2 Þ P DhM :
ð39Þ
Eqs. (19) and (39) reveal
ð27Þ
a ¼ 2A sIe DðM2 N2 Þ þ DcIk M2 P Dðh þ cIk ÞM2 > 0: ð40Þ ð28Þ
and
TVC 01
(A) If D1 6 0, then (a1) a > 0, (a2) T 1 exists, (a3) M 6 T 1 , (a4) TVC1(T) is convex on T > 0. (B) If D6 6 0, then (b1) a > 0, (b2) T 1 exists, (b3) W 6 T 1 , D (b4) TVC1(T) is convex on T > 0.
Hence, Eqs. (12) and (20) demonstrate that Lemma 2(A) holds. (B) If D6 6 0, Eq. (35) implies
W 2A cIk DM2 þ sIe DðM 2 N2 Þ þ W 2 Dðh þ cIk Þ D6 ¼ ¼ ; D 2W 2 2W 2
2A sIe DðM2 N2 Þ P W 2 Dðh þ cIk Þ DcIk M 2 :
ð41Þ
Eqs. (19) and (41) reveal
ð29Þ
a ¼ 2A sIe DðM2 N2 Þ þ DcIk M2 P W 2 ðh þ cIk Þ > 0:
ð42Þ
where 2
D1 ¼ 2A þ sIe DðM2 N2 Þ þ DhM ; 2
ð30Þ
D2 ¼ 2A þ DhN ;
ð31Þ
2 W D3 ¼ 2A þ Dh ; D
ð32Þ
2 W ; D
ð33Þ
2 W sIe DN 2 ; D
ð34Þ
D4 ¼ 2A þ Dðh þ cIk Þ
D5 ¼ 2A þ Dðh þ sIe Þ
Hence, Eqs. (12) and (20) demonstrate that Lemma 2(B) holds. Combining (A) and (B), we have completed the proof of Lemma 2. Lemmas 1 and 2 imply that if T i exists, then
8 > < < 0 if 0 < T < T i ; ðaÞ 0 TVC i ðTÞ ¼ 0 if T ¼ T i ; ðbÞ > : > 0 if T > T i : ðcÞ
ð43Þ
Eq. (43) reveals that TVCi(T) is decreasing on ð0; T i and increasing on ½T i ; 1Þ for all i = 1, 2, 3, 4. Furthermore, Eqs. (24)–(29) and (43) yield the following results. Lemma 3
and
D6 ¼ 2A cIk DM2 þ sIe DðM 2 N2 Þ þ W 2 Dðh þ cIk Þ:
ð35Þ
Case 1: If 0 < W < N, then D
D1 P D2 > D3 ;
ð36Þ
(A) (B) (C) (D) (E)
If If If If If
(F) If (G) If
and
(H) If
ð37Þ
D4 > D3 : Case 2: If N 6
W D
(I) If (J) If
< M; then
(K) If
D1 > D5 :
ð38Þ
(L) If
D1 > 0, then TVC1(T) and TVC2(T) are increasing on [M, 1). D1 6 0, then TVC1(T) and TVC2(T) are decreasing on (0, M]. D2 > 0, then TVC2(T) and TVC3(T) are increasing on [N, 1). D2 6 0, then TVC2(T) and TVC3(T) aredecreasing on (0, N]. D3 > 0, then TVC3(T) is increasing on W ;1 : D D3 6 0, then TVC3(T) is decreasing on 0; W : W D D4 > 0, then TVC4(T) is increasing on D ; 1 : D4 6 0, then TVC4(T) is decreasing on 0; W : W D D5 > 0, then TVC2(T) is increasing on D ; 1 : D5 6 0, then TVC2(T) is decreasing on 0; W : W D D6 > 0, then TVC1(T) is increasing on D ; 1 : D6 6 0, then TVC1(T) is decreasing on 0; W : D
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K.-J. Chung / Expert Systems with Applications 38 (2011) 13482–13486
Proof (A): If D1 > 0, then Eq. (24) implies TVC 01 ðMÞ ¼ TVC 02 ðMÞ > 0: By the convexity of TVC2(T) and Eq. 43(c), we see that TVC2(T) is increasing on [M, 1). However, about TVC1(T), there are two cases to occur: (a1) If a > 0, then TVC1(T) is convex on T > 0. By the convexity of TVC1(T) and Eq. 43(c), we see that TVC1(T) is increasing on [M, 1). (a2) If a 6 0, Lemma 1(A) implies that TVC1(T) is increasing on T > 0. Combining (a1) and (a2), we conclude that TVC1(T) is increasing on [M, 1). So, Lemma 3(A) holds. (B): If D1 6 0, then Eq. (24) implies TVC 01 ðMÞ ¼ TVC 02 ðMÞ 6 0:Furthermore, Lemma 2(A) explains that TVC1(T) is convex on T > 0. By the convexities of TVCi(T)(i = 1, 2), Eq. 43(c) demonstrates that TVC1(T) and TVC2(T) are decreasing on (0, M]. (C)-(L): Lemma 1(B) and Eqs. 43(a–c) imply that Lemma 3(C)-(L) hold. Based on the above arguments, we have completed the proof of Lemma 3. 4. Theorems for the optimal cycle time T⁄ of TVC(T) Theorem 1. Suppose 0 < W D < N: Hence,
(A) If D3 > 0, then TVCðT Þ ¼ min TVC 3 W and ; TVCðT 4 Þ D T ¼W or T 4 associated with the less cost. D (B) If D2 > 0 P D3 and D4 > 0, then TVCðT Þ ¼ TVC 3 ðT 3 Þ and T ¼ T 3 : (C) If D2 > 0 P D3 and D4 6 0, then TVCðT Þ ¼ TVC 3 ðT 3 Þ and T ¼ T 3 : (D) If D1 > 0 P D2 and D4 > 0, then TVCðT Þ ¼ min TVC 2 ðT 2 Þ; TVC 4 ðT 4 Þg and T ¼ T 2 or T 4 associated with the less cost. (E) If D1 > 0 P D2 and D4 6 0, then TVCðT Þ ¼ TVC 2 ðT 2 Þ and T ¼ T 2 : (F) If D1 6 0 and D4 > 0, then TVCðT Þ ¼ min TVC 1 ðT 1 Þ; TVC 4 ðT 4 Þg and T ¼ T 1 or T 4 associated with the less cost. (G) If D1 6 0 and D4 6 0, then TVCðT Þ ¼ TVC 1 ðT 1 Þ and T ¼ T 1 : Proof
(b4) TVC 4W(T) is decreasing on ð0; T 4 and increasing on T 4; D :
Combining (b1)–(b4), we conclude that
TVCðT Þ ¼ min TVC 3 ðT 3 Þ; TVC 4 ðT 4 Þ :
ð44Þ
Furthermore, from Eq. (6), we have
TVC 4 ðT 4 Þ > TVC 3 ðT 4 Þ P TVC 3 ðT 3 Þ:
ð45Þ
Eqs. (44) and (45) demonstrate TVCðT Þ ¼
TVC 3 ðT 3 Þ
and T ¼
T 3 .
(C): If D2 > 0 P D3 and D4 6 0, then Lemmas 1 and 3(A, C, H) reveal that: (c1) TVC1(T) is increasing on [M, 1). (c2) TVC2(T) is increasing on [N, M]. (c3) TVC3(T) is decreasing on W ; T 3 and increasing on D ½T 3 ; N: (c4) TVC4(T) is decreasing on 0; W : D Since TVC 4 W > TVC 3 W ; combining (c1)–(c4), we conclude that D D TVCðT Þ ¼ TVC 3 ðT 3 Þ and T ¼ T 3 : (D): If D1 > 0 P D2 and D4 > 0, then, Lemmas 1 and 3(A, D, G) reveal that: (d1) TVC1(T) is increasing on [M, 1). (d2) TVC2(T) is decreasing on ½N; T 2 and increasing on ½T 2 ; M: (d3) TVC3(T) is decreasing on W ;N : D (d4) TVC4(T) is decreasing on ð0; T 4 and increasing on W T 4; D : Combining (d1)–(d4), we conclude.
TVCðT Þ ¼ min TVC 2 ðT 2 Þ; TVC 4 ðT 4 Þ and T ¼ T 2 or T 4 associated with the less cost. (E): If D1 > 0 P D2 and D4 6 0, then, Lemmas 1 and 3(A, F, H) reveal that: (e1) TVC1(T) is increasing on [M, 1). (e2) TVC2(T) is decreasing on ½N; T 2 and increasing on ½T 2 ; M: (e3) TVC3(T) is decreasing on W ; N : D (e4) TVC4(T) is decreasing on 0; W : D W Since TVC 4 W D > TVC 3 D ; combining (e1)–(e4), we conclude that TVCðT Þ ¼ TVC 2 ðT 2 Þ and T ¼ T 2 .
(A): If D3 > 0, Eqs. (36) and (37) imply Di > 0 for all i = 1, 2, 3, 4. Then, Lemmas 1 and 3(A, C, E) reveal that: (a1) TVC1(T) is increasing on [M, 1). (a2) TVC2(T) is increasing on [N, M]. ;N : (a3) TVC3(T) is increasing on W D (a4) TVC4(T) is decreasing on ð0; T 4 and increasing on ½T 4 ; W Þ: D
(F): If D1 6 0 and D4 > 0, then, Lemmas 1 and 3(D, F) reveal that: (f1) TVC1(T) is decreasing on ½M; T 1 and increasing on ½T 1 ; 1: (f2) TVC2(T) is decreasing on [N, M]. (f3) TVC3(T) is decreasing on W ;N : D (f4) TVC4(T) is decreasing on ð0; T 4 and increasing on W T 4; D :
Since TVC 4 W > TVC 3 W ; combining (a1)–(a4), we conclude that D D
and T ¼ W or T 4 associated ; TVC TVCðT Þ ¼ min TVC 3 W 4 ðT 4 Þ D D with the less cost.
Combining (f1)–(f4), we conclude.
TVCðT Þ ¼ min TVC 1 ðT 1 Þ; TVC 4 ðT 4 Þ and T ¼ T 1 or T 4 associated with the less cost.
(B): If D2 > 0 P D3 and D4 > 0, then, Lemmas 1 and 3(A, C) reveal that: (b1) TVC1(T) is increasing on [M, 1). (b2) TVC2(T) is increasing on [N, M]. ; T 3 and increasing on (b3) TVC3(T) is decreasing on W D ½T 3 ; N:
(G): If D1 6 0 and D4 6 0, then, Lemmas 1 and 3(D, F, H) reveal that: (g1) TVC1(T) is decreasing on ½M; T 1 and increasing on ½T 1 ; 1: (g2) TVC2(T) is decreasing on [N, M]. (g3) TVC3(T) is decreasing on W ; N : D W (g4) TVC4(T) is decreasing on 0; D :
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K.-J. Chung / Expert Systems with Applications 38 (2011) 13482–13486
W Since TVC 4 W D > TVC 3 D ; combining (g1)–(g4), we conclude that TVCðT Þ ¼ TVC 1 ðT 1 Þ and T ¼ T 1 : Based on all arguments of (A)–(G), we have completed the proof of Theorem 1. Theorem 1 in this paper incorporates and simplifies Theorems 2 and 3 in Kreng and Tan (2010). Furthermore, if N 6 W < M, then D TVC 4 W > TVC 2 W . With Lemmas 1, 2(A), Lemma 3(A, B, G, H, I, D D J), Eqs. (6)–(8) and (38), the techniques of arguments of Theorem 1 in this paper can be used to prove the following results. Theorem 2. Suppose N 6 W D < M: Hence,
(A) if D5 > 0 and D4 > 0, then TVCðT Þ ¼ min TVC 4 ðT 4 Þ; TVC 2 W D and T ¼ T 4 or W associated with the less cost. D W (B) if D5 > 0 and D4 6 0, then TVCðT Þ ¼ TVC 2 D and T ¼ W : D (C) if D1 > 0 P D5 and D4 > 0, then TVCðT Þ ¼ min TVC 2 ðT 2 Þ; TVC 4 ðT 4 Þg and T ¼ T 2 or T 4 associated with the less cost. (D) if D1 > 0 P D5 and D4 6 0, then TVCðT Þ ¼ TVC 2 ðT 2 Þ and T ¼ T 2 .
(E) if D1 6 0 and D4 > 0, then TVCðT Þ ¼ min TVC 1 ðT 1 Þ; TVC 4 ðT 4 Þ and T ¼ T 1 or T 4 associated with the less cost. (F) if D1 6 0 and D4 6 0, then TVCðT Þ ¼ TVC 1 ðT 1 Þ and T ¼ T 1 . Theorem 2 in this paper incorporates and simplifies Theorems 4 W and 5 in Kreng W and Tan (2010). Similarly, if M 6 D , then TVC 4 W > TVC . With Lemmas 1, 2(B), 3(G, H, K, L) and Eqs. 1 D D (6)–(8), the techniques of arguments of Theorem 1 can be used to prove the following results. Theorem 3. Suppose M 6 W D . Hence,
(A) if D4 > 0 and D6 > 0, then, TVCðT Þ ¼ min TVC 1 W ; TVC 4 ðT 4 Þ D W and T ¼ D or T 4 associated with the less cost. (B) if D4 > 0 and D6 6 0, then, TVCðT Þ ¼ min TVC 1 ðT 1 Þ; TVC 4 ðT 4 Þg and T ¼ T 1 or T 4 associated with the less cost. (C) if D4 6 0 and D6 > 0, then, TVCðT Þ ¼ TVC 1 W : and T ¼ W D D (D) if D4 6 0 and D6 6 0, then, TVCðT Þ ¼ TVC 1 ðT 1 Þ and T ¼ T 1 : Theorem 3 in this paper incorporates and simplifies Theorems 6 and 7 in Kreng and Tan (2010). 5. Conclusions According to the above arguments, we have the following conclusions.
Case 1: 0 < W < N: D (1) Theorems 2 and 3 in Kreng and Tan (2010) are used to locate the optimal solutions of TVC(T). However, Theorem 1 in this paper incorporates and simplifies them. (2) Since TVC 4 W > TVC 3 W ; TVC 4 W is never the optimal D D D value of TVC(T). Eqs. (6)–(8) reveal that Theorem 1 [B, C, D, E, F, G] in this paper can simplify the corresponding situations of Theorems 2 and 3 in Kreng and Tan (2010). Case 2: N 6 W < M: D (1) Theorems 4 and 5 in Kreng and Tan (2010) are used to locate the optimal solutions of TVC(T). However, Theorem 2 in this paper incorporates and simplifies them. (2) Since TVC 4 W > TVC 2 W ; TVC 4 W is never the optimal D D D value of TVC(T). Eqs. (6)–(8) reveal that Theorem 2 [B, C, D, E, F] in this paper can simplify the corresponding situations of Theorems 4 and 5 in Kreng and Tan (2010). Case 3: M 6 W : D (1) Theorems 6 and 7 in Kreng and Tan (2010) are used to locate the optimal solutions of TVC(T). However, Theorem 3 in this paper incorporates and simplifies them. (2) Since TVC 4 W > TVC 1 W ; TVC 4 W is never the optimal D D D value of TVC(T). Eqs. (6)–(8) reveal that Theorem 3 [B, C, D] in this paper can simplify the corresponding situations of Theorems 6 and 7 in Kreng and Tan (2010). In sum, Kreng and Tan (2010) need six theorems to locate the optimal solutions for all three cases. However, this paper just needs three theorems to incorporate and simplify Kreng and Tan (2010). Kreng and Tan (2010) do not use Eqs. (6)–(8). However, Eqs. (6)–(8) play important roles for locating the optimal solutions such that this paper can simplify the corresponding situations of Kreng and Tan (2010). Consequently, this paper simplifies and improves Kreng and Tan (2010). References Huang, Y. F. (2003). Optimal retailer’s replenishment ordering policies in the EOQ model under trade credit financing. Journal of the Operational Research Society, 54, 1011–1015. Khouja, M., & Mehrez, A. (1996). Optimal inventory policy under different supplier credit policies. Journal of Manufacturing Systems, 15, 334–339. Kreng, V. B., & Tan, S. J. (2010). The optimal replenishment decisions under two levels of trade credit policy depending on the order quantity. Expert Systems with Applications, 37, 5514–5522.