The algorithm for the optimal cycle time and pricing decisions for an integrated inventory system with order-size dependent trade credit in supply chain management

Applied Mathematics and Computation 268 (2015) 322–333

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The algorithm for the optimal cycle time and pricing decisions for an integrated inventory system with order-size dependent trade credit in supply chain management Kun-Jen Chung a,b,c, Jui-Jung Liao d, Pin-Shou Ting c, Shy-Der Lin e, H.M. Srivastava f,∗ a

Chung Yuan Christian University, Chung-Li 32023, Taiwan, ROC National Taiwan University of Science and Technology, Taipei 10607, Taiwan, ROC c Department of International Business Management, Shih Chien University, Taipei 10462, Taiwan, ROC d Department of Business Administration, Chihlee Institute of Technology, Taipei 22050, Taiwan, ROC e Departments of Applied Mathematics and Business Administration, Chung Yuan Christian University, Chung-Li 32023, Taiwan, ROC f Department of Mathematics and Statistics, University of Victoria, Victoria, British Columbia V8W 3R4, Canada b

a r t i c l e

i n f o

MSC: 91B24 93C15 90B30 Keywords: Inventory problems and optimization Inventory control and integrated model Pricing and finance Supply chain management Order-size dependent trade credit Mathematical solution procedures

a b s t r a c t A given inventory problem consists of two parts: (1) the modeling part and (2) the solution procedure part. The modeling part can provide insight into the solution of the inventory problem and the solution procedure part involves the implementation of the inventory model. Both the modeling part and the solution procedure part of the inventory problem are equally important. Recently, Ouyang et al. [17] developed an integrated inventory model with a pricesensitive demand rate and determined both the economic lot size of the buyer’s ordering and the supplier’s production batch in order to maximize the total profit per unit time. Basically, their modeling is correct and interesting. They developed an algorithm based upon the firstorder condition and the second-order condition to locate the optimal solution. However, the fundamentals of mathematics and the numerical examples which are considered in this paper illustrate that their algorithm based upon the first-order condition and the second-order condition to locate the optimal solution has several shortcomings. These shortcomings are shown here to influence the accuracy of the implementation of the inventory model. Since there exist reasons and motivations to present the correct solution procedures to the targeted readers, the main purpose of this paper is to adopt the rigorous methods of mathematical analysis in order to develop the complete solution procedures to locate the optimal solution for removing shortcomings in the earlier investigation by Ouyang et al. [17]. © 2015 Elsevier Inc. All rights reserved.

1. Introduction and motivation A large number of articles dealing with inventory models focus only on determining the optimal policy for the retailer or the supplier. However, in practice, many purchasing situations reveal that different considerations from both the retailer and the supplier frequently conflict with one another. In this connection, some recent investigations by (for example) Chung et al. [5] to [9] may be cited. Goyal [12] was probably the first researcher to develop the seller–customer integrated inventory model. In ∗

Corresponding author. Tel.: +1 250 472 5313; fax: +1 250 721 8962. E-mail addresses: [email protected] (K.-J. Chung), [email protected] (J.-J. Liao), [email protected] (P.-S. Ting), [email protected] (S.-D. Lin), [email protected] (H.M. Srivastava). http://dx.doi.org/10.1016/j.amc.2015.06.039 0096-3003/© 2015 Elsevier Inc. All rights reserved.

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323

general, the integrated models usually have the advantage of reducing the total cost. In the modern global competitive market, the supplier and the retailer should be treated as strategic partners in the supply chain with a long-term cooperative relationship. Goyal [11] extended the lot-for-lot policy discussed in Goyal [13] and illustrated the fact that the inventory cost could be significantly reduced if the supplier’s economic production quantity was an integer multiple of the retailer’s purchase quantity. Lu [16] further generalized the work of Goyal [13] by relaxing the assumption that the supplier could supply the retailer only after completing the entire lot size. On the other hand, in the real world, suppliers usually provide a delay period in payment to encourage retailers to buy more of the ordered quantity. During the trade credit period, the retailers can obtain the interest from the nonpayment and sales income, while suppliers lose the interest income during the same time. Goyal [12] developed the EOQ (economic order quantity) model under conditions of permissible delay in payments. Chen and Kang [1] first incorporated the works by Goyal (see [11] and [12]) to establish some integrated vendor–buyer cooperative inventory models with permissible delay in payments. In addition, inventory models involving trade credit usually make the assumption that the credit period offered by the supplier is constant. However, in practice, the length of the credit period might be adjusted with the order quantities and much attention has been paid recently to this fact. Shinn and Hwang [20] assumed that an order-size dependent trade credit is offered by the supplier. In their model, once a purchasing volume breakpoint is exceeded, a certain credit period is offered to the buyer. A longer credit period is offered for a larger amount of purchase. Ouyang et al. [17] incorporated the aforementioned earlier works by Goyal [11], Lu [16] and Shinn and Hwang [20] to establish an integrated inventory model to determine how the order decisions (timing and size), the delivery decision (shipments from the supplier to the buyer in one production run) and the pricing decision should be made in order to maximize the joint total profit per unit time. Basically, their inventory problem is rather interesting meaningful. They developed an algorithm based upon the first-order condition and the second-order condition to locate the optimal solution. However, the fundamentals of mathematics and the numerical examples (which we propose to consider in this paper) illustrate the fact that their algorithm based upon the first-order condition and the second-order condition to locate the optimal solution has several shortcomings. These shortcomings are shown to influence the accuracy of the implementation of the inventory model. Furthermore, since there exist reasons and motivations to present the correct solutions to readers, the main purpose of this paper is to adopt the rigorous methods of mathematical analysis in order to develop the complete solution procedures with a view to locating the optimal solution by appropriately removing the shortcomings in the work of Ouyang et al. [17]. Several other interesting developments on the subject of our present investigation can be found in a number of related recent works (see, for example, [2], [3], [6–8], [10], [14], [15], [18], [19], [21] and [24]). 2. The mathematical model The following assumptions and notations are adopted to develop the proposed model: 1. The discussion and analysis in this paper is restricted to the case of a single-supplier and single-buyer of a specific product. 2. Shortages are not permitted and the time horizon over which the product is ordered by the buyer and supplied by the supplier is infinite. 3. The demand for the product is assumed to be retail-price sensitive and is given by D( p) = ap−δ , where a > 0 is a scaling factor, and δ > 1 is a price-elasticity coefficient. For notational simplicity, D(p) and D will be used interchangeably in this paper. 4. For the buyer, the inventory cycle length is T and order quantity is Q (=DT) per order. The buyer incurs an ordering cost, SB , which includes the administrative costs of placing, tracking, shipping, receiving and paying for each order. 5. During the production period, the supplier manufactures in batches of size nQ where n is an integer and incurs a batch setup cost SV . Once the first Q units are produced, the supplier delivers them to the buyer and then continuous making the delivery on average every T(=Q/D) units of time until the supplier’s inventory level falls to zero. 6. The capacity utilization, ρ , is the ratio of the demand rate, D, to the production rate, R, which is given and is less than 1, that is, ρ = D/R and ρ < 1. 7. For each unit of product, the supplier spends $c in production and receives $v from the buyer. After that, the buyer sells it by $p to his/her customers. The relationship among them is p > v > c. 8. The carrying cost rate, excluding interest charge for the supplier, is rV and rB for the buyer. 9. The credit period Mj ( j = 1, 2, . . . , k ), which is offered by the supplier, is related to the buyer’s order quantity where the higher order quantity the longer the credit period, and is given as follows:

⎧ M1 ⎪ ⎪ ⎨

M2 M= . . ⎪ ⎪ ⎩. Mk

q1  Q < q2 , q2  Q < q3 , qk  Q < qk+1 ,

where

1  q1 < q2 < · · · < qk+1 = ∞ is the sequence of quantities at which a specific credit period is offered. Mj denotes the trade credit applicable to lot size falls in the interval qj to q j+1 with

0 < M1 < M2 < · · · < Mk .

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10. During the credit period, the buyer sells the items and uses the sales revenue to earn interest at a rate of IBe . At the end of the permissible delay period, the buyer pays the purchasing cost to the supplier and incurs a capital opportunity cost at a rate of IBp , for the items in stock where IBp ࣛIBe . 11. In offering trade credit to the buyer, the supplier endures a capital opportunity cost at rate IVp with the time gap between product shipped and paid. Adopting the above notations and assumptions, Ouyang et al. [17] showed that the joint total profit per unit time for the supplier and buyer can be expressed as follows:



(n, p, T ) =

j

⎧ (n, p, T ) ⎪ ⎨

if T < M j

(2.1a)

⎪ ⎩

if T  M j

(2.1b)

1j



(n, p, T )

2j

where



(n, p, T ) = −

1j









1 SV T + SB + ap−δ p − c + ( pIBe − vIV p )M j − {vrB + pIBe + c (rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]} , T n 2 (2.2)

 2j



  ( pIBe − vIBp )M2j T 1 SV + SB + ap−δ p − c + v(IBp − IV p )M j + − {v(rB + IBp ) (n, p, T ) = − T n 2T 2

+c (rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]}

(2.3)

for j = 1, 2, . . . , k. If n and p are fixed, Eqs. (2.2) and (2.3) imply



(n, p, M j ) =

1j



(n, p, M j )

(2.4)

2j

    Therefore, j (n, p, T) is continuous on T > 0. Additionally, 1j (n, p, T), 2j (n, p, T) and j (n, p, T )( j = 1, 2, . . . , k ) are defined on ∗ ∗ ∗ T > 0. The object in this article is to determine the optimal values (n , p , T ) subject to Assumptions (9) and (10) such that





(n , p , T ) = max ∗

∗

∗



n∗j ,

p∗j , T j∗





| j = 1, 2, . . . , k

(2.5)

j

where







n∗j , p∗j , T j∗ =

j

max

n1,p>v,T >0



⎧ ⎫ (n, p, T )⎪ ⎪ ⎨ ⎬

 (n∗ , p∗ , T ) =

j

max

n1,p>v,T >0

1j



⎪ ⎩

(n, p, T )⎪ ⎭

(2.6)

2j

3. The concavities of



j (n,

p, T ) ( j = 1, 2, . . . , k )

Let n and p be fixed. Then, by taking the first and the second partial derivatives of the Eqs. (2.2) and (2.3) with respect to T, we get

∂ ∂

 1j

 2j

  1 (n, p, T ) 1 SV = 2 + SB − ap−δ {vrB + pIBe + c (rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]}, ∂T n 2 T   (n, p, T ) 1 SV 1 = 2 + SB − ap−δ ( pIBe − vIBp )M2j ∂T n T 2T 2   1 − ap−δ {v(rB + IBp ) + c (rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]} , 2

∂2 and

∂2



  (n, p, T ) 2 SV + S = − <0 B ∂T2 T3 n

1j





 (n, p, T ) 1 SV −δ 2 + SB + ap (vIBp − pIBe )M j , =− 3 2 n ∂T2 T

2j

(3.1)

(3.2)

(3.3)

(3.4)

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325



where j = 1, 2, . . . , k. Eq. (3.3) implies that 1 j (n, p, T ) ( j = 1, 2, . . . , k ) is concave on T > 0 when n and p are fixed and Eq. (3.4)  implies that 2 j (n, p, T )( j = 1, . . . , k ) is concave when n and p are fixed and

2

S

V

n



+ SB + ap−δ (vIBp − pIBe )M2j > 0.

Moreover, if

2

S

V

n



+ SB + ap−δ (vIBp − pIBe )M2j < 0,

  then 2j (n, p, T) is not concave. Consequently, (n, p, T) is not necessarily concave or convex. We now let

G j (n, p) = 2

S



+ SB + ap−δ (vIBp − pIBe )M2j

V

n

( j = 1, 2, . . . , k ).

(3.5)

Upon solving

∂ and

∂



1j

 2j

(n, p, T ) =0 ∂T

(3.6)

(n, p, T ) = 0, ∂T

(3.7)

we obtain



T1 j (n, p) = and

ap−δ



2

n

+ SB



vrB + pIBe + c(rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]

 T2 j (n, p) =

SV

2



SV n





+ SB + ap−δ [vIBp − pIBe ]M2j

v(rB + IBp ) + c(rV + IV p )[(n − 1 )(1 − ρ ) + ρ ] S  V + SB + ap−δ [vIBp − pIBe ]M2j > 0 if 2 ap−δ



n

as the respective solutions of Eqs. (3.6) and (3.7) for j = 1, 2, . . . , k. Furthermore, if Tij (n, p) exists, then j = 1, 2, . . . , k) is concave on T > 0 for any fixed n and p. So, Eqs. (3.1) and (3.2) will yield

∂

⎧ ⎨> 0 ( n, p, T ) ij =0 ∂T ⎩



<0

(3.8)

(3.9) 

ij (n,

p) (i = 1, 2 and

if 0 < T < Ti j (n, p)

( 3.10a)

if T = Ti j (n, p)

(3.10b)

if T > Ti j (n, p).

(3.10c)



Eqs. (3.10a)–(3.10c) reveal that ij (n, p, T) is increasing on (0, Tij (n, p)] and decreasing on [Tij (n, p), ∞) for i = 1, 2, j = 1, 2, . . . , k and for any fixed n and p. Based on the above arguments, the following results hold true. Lemma 1.  (A) For fixed n and p, 1j (n, p, T) is concave on T > 0 if j = 1, 2, . . . , k.   (B) For fixed n and p, 2j (n, p, T) is concave on T > 0 if Gj (n, p) > 0. Otherwise, if Gj (n, p)ࣚ0, then 2j (n, p, T) is decreasing on T > 0 if j = 1, 2, . . . , k. Eqs. (2.2) and (2.3) yield

∂



1j

(n, p, M j ) ∂ = ∂T

where

 j (n, p) = 2

S

V

n

 2j

(n, p, M j )  j (n, p) = , ∂T 2M2j

(3.11)



+ SB − ap−δ {vrB + pIBp + c (rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]}M2j ( j = 1, 2, . . . , k ).

(3.12)

Eq. (3.12) implies that

 j (n, p) >  j+1 (n, p),

(3.13)

 j (n, p) >  j (n + 1, p)

(3.14)

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and

if M j = 0, then  j (n, p) > 0 for all n  1 and j = 1, 2, . . . , k.

(3.15)

Then the following result holds true. Lemma 2. If

 j (n, p)  0

( j = 1, 2, . . . , k ),

then (a) Gj (n, p) > 0; (b) T2j (n, p) exists;  (c) 2j (n, p, T) is concave on T > 0 for fixed n and p. Proof. If j (n, p) ࣛ 0, then

2

S

V

n



+ SB  ap−δ {vrB + pIBp + c (rV + IV p [(n − 1 )(1 − ρ ) + ρ ] )}M2j .

(3.16)

Eqs. (3.5), (3.14) and the Assumption (10) reveal that





G j (n, p)  ap−δ (vrB + vIBp ) + p(IBp − IBe ) M2j > 0.

(3.17)

Eqs. (3.4), (3.9) and (3.17) imply that Lemma 2 holds true.  4. Theorems for the optimal solution (n∗j , p∗j , T j∗ ) of



1j (n,

p, T) or



2j (n,

p, T)

Combining Lemmas 1 and 2, we have the following results. Theorem 1. For any given n and p, the following assertions hold true: (A) If j (n, p) < 0, then (i) 0 < T1j (n, p) < Mj ,  (ii) T1j (n, p) is the optimal solution of j (n, p, T) (B) If j (n, p)ࣛ0, then (i) T2j (n, p)ࣛMj ,  (ii) T2j (n, p) is the optimal solution of j (n, p, T). Proof. (A) If j (n, p) < 0, then there are two situations that would occur:   (1) If Gj (n, j) > 0, then 2j (n, p, T) is concave; so, 2j (n, p, T) is decreasing on [Mj , ∞).  (2) If Gj (n, p)ࣚ0, then Lemma 1(B) implies that 2j (n, p, T) is decreasing on [Mj , ∞).  Combining (A1) and (A2), we conclude that 2j (n, p, T) is decreasing on [Mj , ∞). Therefore, if j (n, p) < 0, Lemma 1 (A, B) and Eq. (3.11) imply that  (3) 1j (n, p, T) is increasing on (0, T1j (n, p)] and decreasing on [T1j (n, p), Mj ].  (4) 2j (n, p, T) is decreasing on [Mj , ∞).  Incorporating (A3) and (A4), we conclude that T1j (n, p, T) is the optimal solution of j (n, p, T).   (B) If j (n, p)ࣛ0, Lemma 2 implies that 2j (n, p, T) is concave on T > 0. Eq. (3.11) reveals that j (n, p, T) is concave on T > 0. We also have  (1) 1j (n, p, T) is increasing on (0, Mj ].  (2) 2j (n, p, T) is increasing on [Mj , T2j (n, p)] and decreasing on [T2j (n, p), ∞).  Incorporating (B1) and (B1), we see that T2j (n, p, T) is the optimal solution of j (n, p, T). Combining (A) and (B), we have completed the proof of Theorem 1.  According to Theorem 1 and Assumption (9), Ouyang et al. [17] demonstrated the following results. Theorem 2. For any given n, p and j = 1, 2, . . . , k, let ∗

Q j = ap−δ T j , ∗

where T j = T1 j (n, p) or T2j (n, p). Then each of the following assertions holds true: (a) If q j  Q j < q j+1 , then



j (n,

∗

p, T) at point T = T j has a maximum value.

∗

(b) If Q j  q j+1 , the lot size ordered by the buyer is over the upper-bound under the credit period Mj , then T j is not a feasible solution.

K.-J. Chung et al. / Applied Mathematics and Computation 268 (2015) 322–333

327 ∗

(c) If Qj < qj and the lot size ordered by the buyer is less than the lower-bound under the credit period Mj , then T j is not a feasible   solution. Furthermore, j (n, p, T) is a strictly decreasing function in T ∈ [q j /D, q j+1 /D] and j (n, p, T) at point T = q j /D has a maximum value. ∗

Theorem 2 indicates that, when the buyer’s optimal replenishment cycle length exists, then it is either T j or qj /D. The remaining part of this subsection will discuss the following two possible conditions in detail. ∗

Situation 1. The buyer’s optimal replenishment cycle length is T j . Substituting Eq. (3.8) into Eq. (2.2), we let



(n, p) =



1j

(n, p, T1 j (n, p))

(4.1)

1j

for j = 1, 2, . . . , k. Eq. (4.1) reveals that

∂



(n, p) = ∂n



1j

− −



ap−δ 1 SV [vrB + pIBe + c (rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]] + + SB 2 (T1 j (n, p))2 n

ap−δ T1 j (n, p) SV c (rV + IV p )(1 − ρ ) + . 2 T1 j (n, p)n2

Eqs. (3.4) and (3.6) yield

−

 ∂ T (n, p) 1j ∂n (4.2)





ap−δ 1 SV + SB = 0 {vrB + pIBe + c(rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]} + 2 (T1 j (n, p))2 n

(4.3)

Substituting Eq. (4.3) into Eq. (4.1), we have

∂

  (n, p) ap−δ c (rV + IV p )(1 − ρ ) 1 SV 2 = − ( T ( n, p )) 1 j ∂n T1 j (n, p) n2 2



1j

(4.4)

for j = 1, 2, . . . , k. According to Eq. (3.8), after rearrangements, Eq. (4.4) can be rewritten as follows:

∂



(n, p) SV {vrB + pIBe + c(rV + IV p )(2ρ − 1 )} − c(rV + IV p )(1 − ρ )SB n2 = ∂n n2 T1 j (n, p){vrB + pIBe + c (rV + IV p )[(n − 1 )(1 − ρ ) + ρ ]}

1j

for j = 1, 2, . . . , k. The number ρ denotes the capacity utilization. In practice, ρ 

1 2.

(4.5)

So, we have

vrB + pIBe + c(rV + IV p )(2ρ − 1 ) > 0.

(4.6)

For simplification, we assume that Eq. (4.6) holds true. Let n1j (p) denote the solution of Eq. (4.7):

∂



(n, p) =0 ∂n

1j

(4.7)

for j = 1, 2, . . . , k. Eqs. (4.5) and (4.6) reveal that



n1 j ( p ) =

SV {vrB + pIBe + c (rV + IV p )(2ρ − 1 )} > 0, SB c (rV + IV p )(1 − ρ )

(4.8)

for j = 1, 2, . . . , k and a fixed p. Then Eq. (4.5) implies that

⎧ > 0 if 0 < n < n1 j ( p) ∂ 1 j (n, p) ⎨ = 0 if n = n1 j ( p) ∂n ⎩ < 0 if n > n1 j ( p). 

(4.9a) (4.9b) (4.9c)

Eqs. (4.9 a), (4.9 b) and (4.9 c) reveal that the following results hold true.  Lemma 3. 1 j (n, p) is increasing on (0, n1j (p)] and decreasing on [n1j (p), ∞) for j = 1, 2, . . . , k and a fixed p. Since Eq. (4.8) is independent of j, we let K ( p) = n1 j ( p) for j = 1, 2, . . . , k, where x denotes the greatest integer ࣚx. Then Lemma 3 demonstrates that

max





⎧ ⎫ (K ( p) + 1, p, T1 j (K ( p)) + 1, p)⎪ ⎪ ⎨ ⎬ 1j

(n, p, T1 j (n, p)) : n  1 = max  ⎪ ⎩ (K ( p), p, T1 j (K ( p), p)) 1j 1j

Situation 2. The buyer’s optimal replenishment cycle length is qj /D.

.

⎪ ⎭

(4.10)

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Substituting Ti j = q j /(ap−δ ) (i = 1, 2 ) into Eqs. (2.2) and (2.3), respectively, we can get the joint total profit function as follows:

  1j

and

  2j



(n, p) ≡

1j



(n, p) ≡



qj ap−δ S (n, p, T1 j ) = − + ap−δ p − c + ( pIBe − vIV p )M j − (ϕ + vrB + pIBe ) qj 2ap−δ

(4.11)

(n, p, T2 j )

2j

=−







( pIBe − vIV p )ap−δ M2j qj ap−δ S + ap−δ p − c + v(IBp − IV p )M j + − [ϕ + v(rB + IBp )] , qj 2q j 2ap−δ

(4.12)

where

S=

SV + SB n

and

ϕ = c(rV + IV p )[(n − 1 )(1 − ρ ) + ρ ].

Eqs. (4.11) and (4.12) yield



∂  1 j (n, p) ap−δ SV c(rV + IV p )(1 − ρ )q j = − ∂n 2 n2 q j and

(4.13)



∂  2 j (n, p) ap−δ SV c(rV + IV p )(1 − ρ )q j = . − ∂n 2 n2 q j Let

1  n j ( p) = qj



(4.14)

2ap−δ SV c (rV + IV p )(1 − ρ )

(4.15)

for j = 1, 2, . . . , k and a fixed p. Then Eqs. (4.13) and (4.14) imply that

⎧ >0  ∂  i j (n, p) ⎨ =0 ∂ ⎩

<0

if 0 < n <  n j ( p ), if n =  n j ( p ),

(4.16a)

if n >  n j ( p ).

(4.16c)

(4.16b)

Eqs. (4.16a), (4.16b) and (4.16c) reveal that the following results hold true.  n j ( p)] and decreasing on [ n j ( p), ∞ ) for i = 1, 2 and j = 1, 2, . . . , k and a fixed p. Lemma 4.  i j (n, p) is increasing on (0,   ( p) =  Let K n j ( p) for i = 1, 2, . . . , k. Then Lemma 4 demonstrates that j

⎫ ⎧  K j ( p) + 1, p, q j /(ap−δ ) ⎪ ⎪ ⎨ ⎬   ij max n, p, q j /(ap−δ ) : n  1 = max   j ( p), p, q j /(ap−δ ) ⎪ ⎪ K ⎩ ⎭ ij



(4.17)

ij

for i = 1,  2 and j = 1,  2, . . . , k. Let n∗j ( p), p, T j∗ n∗j ( p), p

represent the optimal solution of



j (n,

p, T) subject to Assumptions (9) and (10). Referring to

Lemmas 1, 2 and 3, Theorems 1 and 2, and Eqs. (3.13) and (3.14), we have the following results. Theorem 3. For given p and j, each of the following assertions holds true: (A) If j (1, p) < 0, then there are six cases to occur: (1) If

ap−δ T1 j (K ( p), p) ∈ [q j , q j+1 ) and

ap−δ T1 j (K ( p) + 1, p) ∈ [q j , q j+1 ), then

 j

⎧ (K ( p), p, T1 j (K ( p), p)) ⎪ ⎨ 1j

⎫ ⎪ ⎬

(n∗j , p, Tj∗ ) = max  . ⎪ ⎩ (K ( p) + 1, p, T1 j (K ( p) + 1, p))⎪ ⎭ 1j

(4.18)

K.-J. Chung et al. / Applied Mathematics and Computation 268 (2015) 322–333

329

(2) If

ap−δ T1 j (K ( p), p) < q j and

ap−δ T1 j (K ( p) + 1, p) ∈ [q j , q j+1 ), then

⎧ ⎫ (K ( p) + 1, p, T1 j (K ( p) + 1, p))⎪ ⎪ ⎪ ⎪ ⎪ 1j ⎪ ⎪ ⎪ ⎪ ⎪ 

 ⎨ ⎬  −δ  K ( p ) , p, q / ( ap ) ∗ ∗ j j (n j , p, Tj ) = max . 1j ⎪ ⎪ ⎪ ⎪ j 

 ⎪ ⎪ ⎪ ⎪ j ( p) + 1, p, q j /(ap−δ ) ⎪ ⎪ K ⎩ ⎭

(4.19)

1j

(3) If

ap−δ T1 j (K ( p), p)  q j+1 and

ap−δ T1 j (K ( p) + 1, p) ∈ [q j , q j+1 ), then



(n∗j , p, Tj∗ ) =

j



(K ( p) + 1, p, T1 j (K ( p) + 1, p)).

(4.20)

1j

(4) If

ap−δ T1 j (K ( p), p) ∈ [q j , q j+1 ) and

ap−δ T1 j (K ( p) + 1, p) < q j , then

⎧ ⎫ (K ( p), p, T1 j (K ( p), p)) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1j ⎪ ⎪ ⎪ ⎪  ⎨  ⎬  −δ K ( p ) , p, q / ( ap ) ∗ ∗ j j (n j , p, Tj ) = max . 1j ⎪ ⎪ ⎪ ⎪ j ⎪ ⎪  ⎪ ⎪ ⎪ K j ( p) + 1, p, q j /(ap−δ ) ⎪ ⎩ ⎭

(4.21)

1j

(5) If

ap−δ T1 j (K ( p), p) ∈ [q j , q j+1 ) and

ap−δ T1 j (K ( p) + 1, p)  q j+1 , then

 j

(n∗j , p, Tj∗ ) =



(K ( p), p, T1 j (K ( p), p)).

(4.22)

1j

(6) If

ap−δ T1 j (K ( p), p)  q j+1 and

ap−δ T1 j (K ( p) + 1, p)  q j+1 , then both T1j (K(p), p) and T1 j (K ( p) + 1, p) are not feasible. (B) If j (1, p)ࣛ0, let n0j (p) be the smallest positive integer such that j (n0j (p), p) < 0. Suppose also that

A1 = {n : ap−δ T1 j (n, p) ∈ [q j , q j+1 ) and A2 = {n : ap−δ T1 j (n, p) < q j

n = K ( p) or

j ( p) and n = K

A4 = {n : ap−δ T2 j (n, p) ∈ [q j , q j+1 ) and

K ( p ) + 1 },

j ( p) + 1}, or K

1  n  n0 j ( p ) − 1}

and

A5 = {n : ap−δ T2 j (n, p) < q j

j ( p) and n = K

j ( p) + 1}. or K

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K.-J. Chung et al. / Applied Mathematics and Computation 268 (2015) 322–333

(1) If n0j (p)ࣚK(p), then

 ⎫ ⎧  ⎪ ⎪ ⎪ ⎪ ⎪max (n, p, T1 j (n, p)) : n ∈ A1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1j ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ ⎪ − δ ⎪ ⎪ max ( n, p, q / ( ap )) : n ∈ A 2 j ⎪ ⎪ ⎬ ⎨  1j ∗ ∗  . (n j , p, Tj ) = max ⎪ ⎪ ⎪ j ⎪ ⎪max (n, p, T (n, p)) : n ∈ A4 ⎪ ⎪ ⎪ 2j ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2j ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ −δ ⎪ ⎪ (n, p, q j /(ap )) : n ∈ A5 ⎪ ⎭ ⎩max 



(4.23)

2j

where max i j (n, p, T ) : n ∈ A = 0 if A = φ (the empty set). (2) Suppose that n0j (p) > K(p). Then each of the following assertions holds true: (a) If ap−δ T1 j (n0 j ( p), p) ∈ [q j , q j+1 ), then

⎫ ⎧ (n0 j ( p), p, T1 j (n0 j ( p), p)) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1j ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ⎬ ⎨  max ( n, j, T2 j (n, p)) : n ∈ A4 ∗ ∗ (n j , p, Tj ) = max . 2j ⎪ ⎪ ⎪ ⎪ j  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ −δ ⎪ max n, j, q / ( ap ) : n ∈ A 5 ⎪ j ⎭ ⎩

(4.24)

2j

(b) If

ap−δ T1 j (n0 j ( p),

p) < q j , then

⎫ ⎧ (n0 j ( p), p, q j /(ap−δ )) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1j ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪  ⎬ ⎨  max ( n, j, T2 j (n, p)) : n ∈ A4 ∗ ∗ (n j , p, Tj ) = max . 2j ⎪ ⎪ ⎪ ⎪ j  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ ⎪ −δ ⎪ max n, j, q / ( ap ) : n ∈ A 5 ⎪ j ⎭ ⎩

(4.25)

2j

(c) If

ap−δ T1 j (n0 j ( p),

p)  q j+1 , then

 ⎫ ⎧  ⎪ ⎪ ⎪ ⎪ ⎪ max (n, j, T2 j (n, p)) : n ∈ A4 ⎪ ⎪ ⎪ ⎨ ⎬  2j ∗ ∗  . (n j , p, Tj ) = max ⎪ ⎪   ⎪ ⎪ j ⎪ ⎪ ⎪ n, j, q j /(ap−δ ) : n ∈ A5 ⎪ ⎩max ⎭

(4.26)

2j

5. The algorithm to locate the optimal solution (n∗ , p∗ , T∗ ) of

 (n, p, T)

Referring to [4] and Theorem 3, we have the following results. Step 1: Input values of all parameters to involve in the inventory model. Step 2: Set j = 1.   ∗ = T ∗ (n∗ , p∗ ), and ∗ ∗ ∗ Step 3: p = v + 0.01, pstop = 0, p∗opt = p, n∗opt = n∗1 j ( p∗opt ), Topt opt = 11 (nopt , popt , Topt ). opt 1 j opt Step 4: (A) If j (1, p) < 0, then there are six cases to occur. Otherwise, go to Step (4B). (1) If ap−δ T1 j (K ( p), p) ∈ [q j , q j+1 ) and ap−δ T1 j (K ( p) + 1, p) ∈ [q j , q j+1 ), then (n∗j , p, T j∗ (n∗j , p)) is determined by Eq. (4.18) and go to Step 5. (2) If ap−δ T1 j (K ( p), p) < q j and ap−δ T1 j (K ( p) + 1, p) ∈ [q j , q j+1 ), then (n∗j , p, T j∗ ) is determined by Eq. (4.19) and go to Step 5. (3) If ap−δ T1 j (K ( p), p)  q j+1 and ap−δ T1 j (K ( p) + 1, p) ∈ [q j , q j+1 ), then (n∗j , p, T j∗ ) is determined by Eq. (4.20) and go to Step 5.

K.-J. Chung et al. / Applied Mathematics and Computation 268 (2015) 322–333

331

(4) If ap−δ T1 j (K ( p), p) ∈ [q j , q j+1 ) and ap−δ T1 j (K ( p) + 1, p) < q j , then (n∗j , p, T j∗ ) is determined by Eq. (4.21) and go to Step 5. (5) If ap−δ T1 j (K ( p), p) ∈ [q j , q j+1 ) and ap−δ T1 j (K ( p) + 1, p)  q j+1 , then (n∗j , p, T j∗ ) is determined by Eq. (4.22) and go to Step 5. (6) If ap−δ T1 j (K ( p), p)  q j+1 and ap−δ T1 j (K ( p) + 1, p)  q j+1 , set p = p + 0.01 and go to Step 4. (B) If j (1, p)ࣛ0, let n0j (p) be the smallest positive integer such that j (n0j (p), p) < 0. (1) If n0j (p)ࣚK(p), then there are the following two cases that would occur: (a) If A1 ∪A2 ∪A4 ∪A5 = φ (the empty set), then (n∗j , p, T j∗ ) is determined by Eq. (4.23) and go to Step 5. (b) If A1 ∪ A2 ∪ A4 ∪ A5 = φ , set p = p + 0.01 and go to Step 4. (2) If n0j (p) > K(p), then there are two cases to occur: (a) If A4 ∪A5 = φ , then there are three cases to occur as follows: (i) If

ap−δ T1 j (n0 j ( p), p) ∈ [q j , q j+1 ), then (n∗j , p, T j∗ ) is determined by Eq. (4.24) and we go to Step 5. (ii) If

ap−δ T1 j (n0 j ( p), p) < q j , then (n∗j , p, T j∗ ) is determined by Eq. (4.25) and we go to Step 5. (iii) If

ap−δ T1 j (n0 j ( p), p)  q j+1 , then (n∗j , p, T j∗ ) is determined by Eq. (4.26) and we go to Step 5. (b) If A4 ∪ A5 = φ , then the following three cases occur: (i) If

ap−δ T1 j (n0 j ( p), p) ∈ [q j , q j+1 ), then



(n∗j , p, Tj∗ ) =

j



(n0 j ( p), p, T1 j (n0 j ( p), p))

1j

and we go to Step 5. (ii) If

ap−δ T1 j (n0 j ( p), p) < q j , then 

(n∗j , p, Tj∗ ) =

j



(n0 j ( p), p, q j /(ap−δ ))

1j

and we go to Step 5. (iii) If

ap−δ T1 j (n0 j ( p), p)  q j+1 , we set p = p + 0.01 and go to Step 4.    ∗  ∗ = T ∗, ∗ Step 5: (A) If opt  j (n∗j , p, T j∗ ), set p∗opt = p, n∗opt = n∗j , Topt opt = j (n j , p, T j ), pstop = 0, p = p + 0.01 and go to j Step 4.   (B) Suppose that opt > j (n∗j , p, T j∗ ). If pstop > 30, go to Step 6. Otherwise, pstop = pstop + 1, p = p + 0.01 and go to Step 4.   ∗ , p = p∗ and = opt . Set j = j + 1. If j > k, go to Step 7. Otherwise, go to Step 3. Step 6: Set n j = n∗opt , T j = Topt j  opt j  ∗ ∗ ∗ ∗ ∗ ∗ Step 7: Let (n j , p , T j ) = max j (n j , p j , T j ) and find (n , p , T ). Go to Step 8.  ∗ ∗ ∗ Step 8: Exit the optimal solution (n , p , T ) and the optimal value (n∗ , p∗ , T∗ ). 6. Illustrative numerical examples Given a = 105 , δ = 1.5, ρ = 0.95, c = $2/unit, v = $4.2/unit, SV = $300/setup, SB = $50/order, rV = 0.05, rB = 0.1, IBe = 0.06, IBp = 0.08, IV P = 0.04 and q1 = 1. The trade credit terms offered by the supplier are listed in Table 1 in this paper. Let (n∗oc , p∗oc , Toc∗ ) denote the optimal solution obtained by Ouyang et al. [17]. Following the algorithm to locate the optimal solution (n∗ , p∗ , T∗ ) of (n, p, T) described in this paper, we obtain the compu∗ tational results for various values of q2 and q3 as shown in Table 2 below. All of the optimal solutions (n∗ , p∗ , T∗ ) and (n∗oc , p∗oc , Toc ) of the corresponding examples except for q2 = 900 and q3 = 1200 approximate very well. Furthermore, all numbers

 ∗ ∗ ∗ (n , p , T )  ∗ ∗ (noc , poc , Toc∗ )

are 1.000. Basically, our algorithm is rather accurate.

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K.-J. Chung et al. / Applied Mathematics and Computation 268 (2015) 322–333 Table 1 Credit terms of the numerical example. Total amount of purchase (unit/order)

Credit period (days)

1ࣚQ < 500 500ࣚQ < 1000 1000ࣚQ

M1 = 30 M2 = 45 M3 = 60

Table 2 Numerical comparisons between Ouyang et al. [17] and the present paper. q2

q3

The optimal solutions of the present paper  ∗ ∗ ∗ (n , p , T ) n∗ p∗ T ∗ = Ti j Q∗

500

800 900 1000 1100 1200 1400 900 1000 1100 1200 1400 1000 1100 1200 1400 1100 1200 1400

25 23 21 19 25 25 23 21 19 25 23 21 19 23 23 19 17 21

800

900

1000

6.20 6.18 6.15 6.13 6.19 6.19 6.18 6.15 6.13 6.19 6.19 6.15 6.13 6.18 6.18 6.13 6.11 6.16

T13 T13 T13 T23 T22 T22 T13 T13 T23 T22 T22 T13 T23 T22 T22 T23 T23 T22

= 46.51 = 50.47 = 55.67 = 60.94 = 46.51 = 46.51 = 50.47 = 55.67 = 60.94 = 46.51 = 46.51 = 55.67 = 60.94 = 50.47 = 50.47 = 60.94 = 66.15 = 55.80

825 900 1000 1100 827 827 900 1000 1100 827 827 1000 1100 900 900 1100 1200 1000

26449.90 26446.76 26436.68 26418.46 26395.61 26395.61 26446.76 26436.68 26418.46 26395.61 26395.61 26436.68 26418.46 26392.88 26392.88 26418.46 26394.42 26382.97

The optimal solutions of Ouyang et al. [17]  ∗ ∗ ∗ n∗oc p∗oc Toc∗ = Ti j Qoc (noc , poc , Toc∗ ) 25 23 21 19 25 25 23 21 19 25 23 21 19 17 23 19 17 21

6.20 6.18 6.16 6.14 6.19 6.19 6.18 6.16 6.14 6.19 6.19 6.16 6.14 6.13 6.18 6.14 6.13 6.16

T13 = 46.51 T13 = 50.48 T13 = 55.78 T23 = 61.09 T22 = 46.53 T22 = 46.53 T13 = 50.48 T13 = 55.78 T23 = 61.09 T22 = 46.53 T22 = 46.53 T13 = 55.77 T23 = 61.09 T23 = 66.42 T23 = 50.44 T23 = 61.09 T23 = 66.42 T22 = 55.77

825 900 1000 1100 827 827 900 1000 1100 827 827 1000 1100 1200 900 1100 1200 1000

26449.90 26449.94 26435.59 26417.86 26395.62 26395.62 26449.94 26435.59 26417.86 26395.62 26395.62 26435.59 26417.86 26395.41 26392.88 26417.86 26395.41 26382.03



 ∗ ∗ ∗  (n∗ ,p∗ ,T ∗) (noc ,poc ,Toc )

∂

1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

−0.1854 0.4058 −0.8265 0.1954 2.2548 2.2548 0.4058 −0.8265 0.1954 2.2548 2.2548 −0.8265 0.1954 −1.7820 −1.4255 0.1954 −1.7820 −1.3570

ij

(n∗oc ,p∗oc ,Toc∗ ) ∂p

7. The shortcomings of Algorithm 1 of Ouyang et al. [17] Shortcoming 1. According to Steps 2-2 [(a), (b)] and 3(c) [(i), (ii)] in Algorithm 1 in Ouyang et al. [17], the optimal solution obtained by them should satisfy the first-order conditions as in the equations (24), (26), (31) or (33) in Ouyang et al. [17], respectively. Furthermore, the equations (24), (26), (31) or (33) and Table 2 in Ouyang et al. [17] yield Table 1 above. However, in fact, all absolute values of

  ∗ ∗ ∗   ∂ i j (noc , poc , Toc )      ∂p

in Table 2 above are between 0.1856 and 2.2548. So, all values

∂



ij

(n∗oc , p∗oc , Toc∗ ) ∂p

do not approach zero. Consequently, it is questionable that all the optimal solutions obtained by Algorithm 1 in Ouyang et al. [17] should satisfy the first-order conditions as in equations (24), (26), (31) or (33) of Ouyang et al. [17], respectively. Assumption (9) in Ouyang et al. [17] [or in the present paper] yields Step 3 of Algorithm 1 in Ouyang et al. [17]. However, Step 3(b) will cause T1j or T2j obtained by Step 2-2 (a, b) of Algorithm 1 in Ouyang et al. [17] to become unfeasible. It means that the number p1j or p2j satisfying the first-order condition as in the equation (24) and the second-order condition as in the equation (25) of Ouyang et al. [17] may be eliminated by Step 3(b) such that Algorithm 1 in Ouyang et al. [17] has drawbacks from the logical viewpoints of mathematical analysis. It influences the validity of Algorithm 1 in Ouyang et al. [17]. However, the algorithm to locate the optimal  solution (n∗ , p∗ , T∗ ) of (n, p, T) described in this paper is free of the usages of the first-order and the second-order conditions to avoid the logical drawbacks of mathematical analysis. Shortcoming 2. We first state the following classical result (see, for example, [23, p. 164]; see also [22]). Theorem B. (The Second Derivative Test). Let f (x) and f

(x) exist at every point in an open interval (a, b) containing the point x = c. Suppose that f (c ) = 0 and also that that both (i) and (ii) hold true. (i) The first-order condition f (c ) = 0. (ii) The second-order condition f

(c) < 0. Then f(c) is a local maximum value of f(x).

K.-J. Chung et al. / Applied Mathematics and Computation 268 (2015) 322–333

333

In general, any point c satisfying both of the conditions (i) and (ii) on f is not necessarily such that f(c) is the maximum value of f as Example 5 in Varberg [23, p. 165] (see also [22]) shows this phenomenon. Similarly, although any price p satisfies both  the first-order and the second-order conditions on ij (n, p) such as equations [24] and [25], [26] and [27], [31] and [32] or [33]  and [34] in Ouyang et al. [17], p is not necessarily the optimal retailer price of the corresponding ij (n, p) when n is fixed. Hence, clearly, the fundamentals of mathematical analysis about Algorithm 1 in Ouyang et al. [17] are not completely satisfied. Shortcoming 3. In order to make Properties 2 and 3 in Ouyang et al. [17] hold true, Ouyang et al. [17] need to assume that

pIBe  vIBp

and

IV p  IBp .

However, this paper does not need those assumptions. 8. Concluding remarks and observations Theorem B (see, for example, [23, p. 164]; see also [22]) illustrates that the first-order and the second-order conditions can  only assure the existence of the local maximum value of (n, p, T), but they are unable to draw a conclusion about the maxima without more information. In addition, Assumption (9) may cause that the number p1j or p2j satisfying the first-order and the second-order conditions described in Ouyang et al. [17] may be eliminated by Step 3(b) such that Algorithm 1 in Ouyang et al. [17] has shortcomings from the logical viewpoints of mathematical analysis as the last column of Table 2 shows. This paper adopts the rigorous methods of mathematical analysis to develop the complete solution algorithm free of using the first-order and the second-order conditions in order to remove all logical shortcomings of mathematics occurring in the work of Ouyang et al. [17]. 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