Research Article EOQ Model for Deteriorating Items with Stock-Level-Dependent Demand Rate and Order-Quantity-Dependent Trade Credit

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1 Mathematical Problems in Engineering, Article I , 14 pages Research Article EOQ Model for eteriorating Items with Stock-Level-ependent emand Rate and Order-Quantity-ependent Trade Credit Jie Min, 1 Jian Ou, 1 Yuan-Guang Zhong, 2 and Xin-Bao Liu 3 1 School of Mathematics and Physics, Anhui Jianzhu University, Hefei , China 2 School of Business Administration, South China University of Technology, Guangzhou , China 3 School of Management, Hefei University of Technology, Hefei , China Correspondence should be addressed to Jie Min; minjie@ahjzu.edu.cn Received 24 July 2014; Accepted 29 October 2014; Published 16 ecember 2014 Academic Editor: Zhen-Lai Han Copyright 2014 Jie Min et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. This paper develops a generalized inventory model for exponentially deteriorating items with current-stock-dependent demand rate and permissible delay in payments. In the model, the payment for the item must be made immediately if the order quantity is less than the predetermined quantity; otherwise, a fixed trade credit period is permitted. The maximization of the average profit per unit of time is taken as the inventory system s objective. The necessary and sufficient conditions and some properties of the optimal solution to the model are developed. Simple solution procedures are proposed to efficiently determine the optimal ordering policies of the considered problem. Numerical example is also presented to illustrate the solution procedures obtained. 1. Introduction In classical inventory models, the demand rate of items was often assumed to be either constant or time dependent. However, many marketing researchers and practitioners have recognized that the demand rate of many retail items is usually influenced by the amount of inventory displayed. For example, Whitin [1] stated, For retail stores the inventory control problem for style goods is further complicated by the fact that inventory and sales are not independent of one another. An increase in inventories may bring about increased sales of some items. Levin et al. [2] andsliver and Peterson [3] also observed that sales at the retail level tend to be proportional to the inventory displayed and that large piles of goods displayed in a supermarket will lead customers to buy more. The reason behind this phenomenon is a typical psychology of customers. They may have the feeling of obtaining a wide range for selection when a large amount is stored or displayed. Conversely, they may have doubt about the freshness or quality of the product when a small amount is stored. Based on the observed phenomenon, itisclearthatinreallifethedemandrateofitemsmaybe influenced by the stock levels. Many researchers have developed lots of inventory models to cover this phenomenon. Gupta and Vrat [4] were the first to develop an inventory model with the initial-stock-dependent consumption rate. Padmanabhan and Vrat [5] proposed an EOQ model for initial-stock-dependent demand and exponential decay items. Giri et al. [6] presented a model under inflation for initial-stock-dependent consumption rate and exponential decay. In addition,baker and Urban [7] investigated another kind of inventory model in which the demand of item is a polynomial function of the instantaneous stock level. Mandal and Phaujdar [8] developed an inventory model for deteriorating items with linearly current-stock-dependent demand. Pal et al. [9] extended the model of Baker and Urban [7] to the case of deteriorating items. Padmanabhan and Vrat [10] further developed three models for deteriorating items under current-stock-level-dependent demand rate: without backorder, with complete backorder, and with partial backorder. Chung et al. [11]modifiedPadmanabhan andvrat s[10] models by showing the necessary and sufficient conditions of the existing optimal solution for models with no backorder and complete backorder. Zhou and Yang [12] developed

2 2 Mathematical Problems in Engineering an inventory model for a general inventory-level-dependent demand with limited storage space. Moreover, many further extensive models were developed by researchers such as Balkhi and Benkherouf [13], ye and Ouyang [14], Zhou and Yang [15], Wu et al. [16], and Alfares [17]. In all the above inventory models with stock-leveldependent demand rate assumed that the retailer must pay fortheitemsassoonastheitemsarereceived.however, in the real life, the supplier sometimes will offer the retailer adelayperiod,thatis,atradecreditperiod,inpayingfor the amount of purchasing cost. Before the end of the trade credit period, the retailer can sell the goods and accumulate the revenue to earn interest. A higher interest is charged if the payment is not settled by the end of trade credit period. In real world, the supplier often makes use of this policy to promote their commodities. uring the recent two decades, the effect of a permissible delay in payments on the optimal inventory systems has received more attention from numerous researchers. Goyal [18] explored a single item EOQ model under permissible delay in payments. Aggarwal and Jaggi [19] extended Goyal s [18] model to the case with deteriorating items. Jamal et al. [20] further generalized the above inventory models to allow for shortages. Other related articles can be found in Sarker et al. [21], Teng [22], Huang [23], Chang et al. [24, 25], Chung and Liao [26], Teng et al. [27], Liao [28] and Teng and Goyal[29], and their references. More recently, Chen et al. [30] studied an economic order quantity model under conditionally permissible delay in payments, in which the supplier offers the retailer a fully permissible delay periods if the retailer orders more than or equal to a predetermined quantity, otherwise partial payments will be provided. Chang et al. [31] developedan appropriate inventory model for noninstantaneously deteriorating items where suppliers provide a permissible payment delay schedule linked to order quantity. However, all of the abovementioned papers with trade credit financing did not address inventory-level-dependent demand rate. The present paper establishes an EOQ model for deteriorating items with current-inventory-level-dependent demand rate and permissible delay in payments which is dependent on the retailer s order quantity. That is, if the order quantity is less than the predetermined quantity, the paymentfortheitemmustbemadeimmediately;otherwise, a fixed trade credit period is permitted. We then establish the proper mathematical model and study the necessary and sufficient conditions of the optimal solution to the model. Some properties of the optimal solution to the model are proposed for obtaining the optimal replenishment quantity andreplenishmentcycle,whichmaketheaverageprofitofthe system maximized. The remainder of the paper is organized as below. The next section introduces some notations and assumptions used in this paper. We then present in Section 3 the formulation of the inventory model with current-stock-leveldependent demand rate and order-quantity-dependent trade credit. Section 4 examines the existence and uniqueness of the optimal solution to the considered inventory system and shows some properties of the optimal ordering policies. In Section 5, numerical example is given to illustrate the solution procedure obtained in this paper. Some conclusions are included in Section Notations and Assumptions The following notations and assumptions are used in the paper. Notations Assumptions K: ordering cost one order in dollars, c: purchase cost per unit in dollars, s: selling price per unit in dollars (s >c), h: stockholdingcostperyearperunitindollars (excluding interest charges), θ: deteriorating rate of items (0 <θ<1), Q d : the qualified quantity in units at or beyond which the delay in payments is permitted, T d : the time interval in years in which Q d units are depleted to zero due to both demand and deterioration, I e : interest which can be earned per $ per year by the retailer, I p : interest charges per $ in stocks per year by the supplier, M: the retailer s trade credit period offered by supplier in years, T:inventorycyclelengthinyears(decisionvariable), Q: the retailer s order quantity in units per cycle, Π(T): the retailer s average profit function per unit time in dollars. (1) The demand rate R(t) is a known function of retailer s instantaneous stock level I(t),whichisgivenby R (t) =+αi(t), where and α are positive constants. (2) The time horizon of the inventory system is infinite. (3) The lead time is negligible. (4)Shortagesarenotallowedtooccur. (5) When the order quantity is less than the qualified quantity Q d,thepaymentfortheitemmustbemade immediately. Otherwise, the fixed trade credit period M is permitted. (6) If the trade credit period M is offered, the retailer would settle the account at t = M and pay for the interest charges on items in stock with rate I p over the interval [M, T] as T M,whereastheretailer settles the account at t=manddoes not need to pay any interest charge of items in stock during the whole cycle as T M. (1)

3 Mathematical Problems in Engineering 3 (7) The retailer can accumulate revenue and earn interest from the very beginning of inventory cycle until the end of the trade credit period offered by the supplier. Thatis,theretailercanaccumulaterevenueandearn interest during the period from t=0to t=mwith rate I e under the condition of trade credit. (8) s c, I p I e,whicharereasonableassumptionsin reality. 3. Mathematical Formulation of the Model Based on the above assumptions, the considered inventory system goes like this: at the beginning (say, the time t=0) of each replenishment cycle, items of Q units are held, and the items are depleted gradually in the interval [0, T] due to the combined effects of demand and deterioration. At time t=t, the inventory level reaches zero, and the whole process is repeated. The variation of inventory level, I(t), with respect to time can be described by the following differential equation: di (t) = αi(t) θi(t), dt 0 t T (2) with the boundary condition I(T) = 0. The solution to the above differential equation is I (t) = α+θ (e(α+θ)(t t) 1), 0 t T. (3) So the retailer s order size per cycle is Q=I(0) = α+θ (e(α+θ)t 1). (4) We observe that if the order quantity Q < Q d, then the paymentmustbemadeimmediately.otherwise,theretailer will get a certain credit period, M. Hence,from(4) one has thefactthattheinequalityq<q d holds if and only if T<T d, where T d is given as below: 1 T d = α+θ ln [1+(α+θ) Q d ]. (5) The elements comprising the profit function per cycle of the retailer are listed below. (a) The ordering cost = K. (b) The holding cost (excluding interest charges) = h T I(t)dt = (h/(α + 0 θ)2 )[e (α+θ)t (α+θ)t 1]. (c) The purchasing cost = cq = (c/(α + θ))[e (α+θ)t 1]. (d)thesalesrevenue=s T R(t)dt = s[θt/(α + θ) + 0 (α/(α + θ) 2 )(e (α+θ)t 1)]. (e) According to the above assumption, there are two casestooccurininterestpayableineachordercycle. Case 1 (T T d ). In this case, the delay in payments is not permitted. That is, all products in inventory have to be financed with annual rate I p. Therefore, interest payable in each order cycle T =ci p I (t) dt = 0 ci p (α+θ) 2 [e(α+θ)t (α+θ) T 1]. (6) Case 2 (T T d ). In this case, the delay credit period M in payments is permitted, which arouses two subcases: Subcase 2.1 where T Mand Subcase 2.2 where T M. Subcase 2.1 (T M). Since the cycle time T is shorter than the credit period M, there is no interest paid for financing the inventory in stock. So the interest payable in each order cycle = 0. Subcase 2.2 (T M). When the credit period M is shorter than or equal to the replenishment cycle time T, the retailer starts paying the interest for the items in stock after time M with rate I p. Hence, the interest payable in each order cycle T =ci p I (t) dt M = ci p (α+θ) 2 [e(α+θ)(t M) (α+θ)(t M) 1]. (7) (f) Similarly, there are also two cases to occur in interest earned in each order cycle. Case 1 (T T d ).Inthiscase,thedelayinpaymentsisnot permitted, so the interest earned in each order cycle = 0. Case 2 (T T d ). In this case, the delay in payments is permitted, which causes two subcases as below. Subcase 2.1 (T M). Since the cycle time T is shorter than the credit period M, fromtime0tot the retailer sells the goods and continues to accumulate the sales revenue to earn interest si e T 0 t R(u)du dt, and from time T to M the retailer 0 can use the sales revenue generated in [0, T] to earn interest si e T R(u)du(M T). Therefore, the interest earned from 0 time 0 to M per cycle T =si e [ 0 t 0 T R (u) du dt + R (u) du (M T)] 0 =si e { θt2 2 (α+θ) α (α+θ) 3 (e(α+θ)t 1) + αt (α+θ) 2 e(α+θ)t +[ θt α+θ + α (α+θ) 2 (e(α+θ)t 1)](M T)}. Subcase 2.2 (T M). In this case, the delay in payments is permitted, and during time 0 through M the retailer sells the goods and continues to accumulate sales revenue and earns (8)

4 4 Mathematical Problems in Engineering the interest with rate I e. Therefore, the interest earned from time 0 to M per cycle M t =si e R (u) du dt 0 0 =si e [ θm2 2 (α+θ) + αm (α+θ) 2 e(α+θ)t + α (α+θ) 3 e(α+θ)t (e (α+θ)m 1)]. From the above, the profit function for the retailer can be expressed as Π (T) = {Sales revenue + interest earned ordering cost holding cost purchasing cost interest payable} T 1. (10) On simplification and summation, we get the profit function as the following three forms: (A) if T T d,then Π (T) =Π 1 (T) = 1 T { E (α+θ) 2 (e(α+θ)t 1) (B) if T>T d but T M,then Π (T) =Π 2 (T) + T α+θ (h + θs + ci p) K}, = 1 T { F (α+θ) 2 (e(α+θ)t 1) + T α+θ (h + θs + θsi em+ αsi e α+θ ) θsi et 2 2 (α+θ) K}, (C) if T>T d and T>M,then Π (T) =Π 3 (T) = 1 T { F (α+θ) 2 (e(α+θ)t 1) + (α+θ) 2 (e(α+θ)(t M) 1)( αsi e α+θ ci p) (9) (11) (12) + T α+θ (h + θs + ci p)+ M α+θ ( αsi e α+θ ci p) + θsi em 2 2 (α+θ) K}, (13) where the meaning of notations E and F in the above expressions is E=αs (α+θ) c h ci p, F=αs (α+θ) c h+αsi e M αsi e α+θ. (14) For convenience, we will consider the problem through the two following situations: (1) M T d and (2) M<T d. (1) Suppose That M T d.inthesituationofm T d, Π(T) has three different expressions as follows: Π (T) =Π 1 (T), 0 < T T d, (15a) Π (T) =Π 2 (T), T d T M, (15b) Π (T) =Π 3 (T), T M. (15c) It is easy to verify that, in the case of M T d, Π(T) continues except at T=T d. In fact, if ci p si e then we could prove that Π 1 (T d )<Π 2 (T d ).(ProofofΠ 1 (T d )<Π 2 (T d ) is shown in Appendix A.) (2) Suppose That M < T d.ifm < T d,thenπ(t) has two different expressions as follows: Π (T) =Π 1 (T), 0<T T d, (16a) Π (T) =Π 3 (T), T T d. (16b) Here, Π 1 (T) and Π 3 (T) are given by (11) and (13), respectively. Π(T) continues except at T = T d.infact,we could prove that Π 1 (T d )<Π 3 (T d ) when M<T d.(proofof Π 1 (T d )<Π 3 (T d ) is shown in Appendix B.) In the next section, we will determine the retailer s optimal cycle time T for the above two situations using some algebraic method. 4. The Solution Procedures In order to decide the optimal solution T of function Π(T), we need first to study the properties of function Π i (T) (i = 1, 2, 3) on(0, + ), respectively. The first-order necessary condition for Π 1 (T) in (11) to be maximum is dπ 1 (T)/dT = 0, which implies that E (α+θ) 2 [(α+θ) Te(α+θ)T e (α+θ)t +1] K=0. (17) For convenience, we set the left-hand expression of (17) as f 1 (T). Taking the derivative of f 1 (T) with respect to T, we get df 1 (T) dt = ETe(α+θ)T. (18) From (18) we know that if E < 0,thendf 1 (T)/dT > 0; thatis,f 1 (T) is increasing on (0, + ). It is obvious that f 1 (0) = K < 0 and lim T + f 1 (T) = + ; therefore the Intermediate Value Theorem implies that f 1 (T) = 0, that

5 Mathematical Problems in Engineering 5 is, dπ 1 (T)/dT = 0, hasauniqueroot(sayt 0 1 )on(0, + ). Moreover, taking the second derivative of Π 1 (T) with respect to T gives d 2 Π 1 (T) dt 2 = 2K T 3 + E (α+θ) 2 T 3 [(α+θ) 2 T 2 e (α+θ)t 2(α+θ) Te (α+θ)t +2e (α+θ)t 2]. (19) It is easy to show that the inequality x 2 e x 2xe x +2e x 2 > 0 holds for all x>0,sod 2 Π 1 (T)/dT 2 will always be negative on (0, + ) ife < 0;thatis,Π 1 (T) is convex on (0, + ). Therefore, T 0 1 istheuniquemaximumpointofπ 1(T) on (0, + ). But if E 0, (18) means that f 1 (T) is decreasing on (0, + ); since f 1 (0) = K < 0,thenf 1 (T) is negative and dπ 1 (T)/dT is positive for all T>0;henceΠ 1 (T) is strictly increasing on (0, + ). Then we have the following lemma to describe the property of Π 1 (T) on (0, + ). Lemma 1. If E<0, Π 1 (T) is convex on (0, + ) and T 0 1 is the unique maximum point of Π 1 (T).Otherwise,Π 1 (T) is strictly increasing on (0, + ). Similarly, the first-order necessary condition for Π 2 (T) in (12) to be maximized is dπ 2 (T)/dT = 0, which leads to F (α+θ) 2 [(α+θ) Te(α+θ)T e (α+θ)t +1] + θsi et 2 2 (α+θ) K=0. (20) We set the left-hand expression of (20) as f 2 (T). Takingthe first derivative of f 2 (T) with respect to T gives df 2 (T) dt = (Fe(α+θ)T θsi e )T. (21) α+θ Consequently, from the sign of F there are two distinct cases for finding the property of Π 2 (T) as follows. (1) When F 0,wehavedf 2 (T)/dT > 0;thatis,f 2 (T) is strictly increasing on (0, + ).Sincef 2 (0) = K < 0 and lim T + f 2 (T) = +, f 2 (T) = 0 has a unique solution (say T 0 2 )on(0, + ). Moreover, from (13) we have d 2 Π 2 (T) dt 2 = 2K T 3 + F (α+θ) 2 T 3 [(α+θ) 2 T 2 e (α+θ)t 2(α+θ) Te (α+θ)t +2e (α+θ)t 2]. (22) Since F 0and (α + θ) 2 T 2 e (α+θ)t 2(α+θ)Te (α+θ)t + 2e (α+θ)t 2 > 0,wegetd 2 Π 2 (T)/dT 2 <0.Hence, Π 2 (T) is convex on (0, + ) and T 0 2 is the unique maximum point of Π 2 (T). (2) When F > 0, there will exist a unique root to the equation df 2 (T)/dT = 0.enotethisrootbyT # ;then T # = 1 α+θ [ln θsi e ln (α+θ) F]. (23) (A) If T # 0,thatis0<F θsi e /(α + θ), then df 2 (T)/dT > 0 for T (0,T # ) and df 2 (T)/dT < 0 for T (T #,+ );thatis,f 2 (T) is increasing on (0, T # ) and decreasing on (T #,+ ). (a) When f 2 (T # )>0, f 2 (T) = 0 has a unique root (say T 1 2 )on(0, T# ) due to f 2 (0) = K < 0. Hence,Π 2 (T) is increasing on (0, T 1 2 ) and decreasing on (T 1 2,T# ). On the other interval of (T #,+ ),however,becauseof f 2 (T # ) > 0 and lim T + f 2 (T) = for F > 0, f 2 (T) has a unique zero point (say T 2 2 )on(t#,+ ).Thus,Π 2 (T) is decreasing on (T #,T 2 2 ) and increasing on (T 2 2,+ ).Hence,Π 2(T) is increasing on (0, T 1 2 ),decreasingon(t1 2,T2 2 ), and increasing on (T 2 2,+ ). (b) When f 2 (T # ) 0, f 2 (T) is negative on (0, T # ) because f 2 (T) is increasing on (0, T # ),whereasf 2 (T) is also negative on (T #,+ ) since f 2 (T) is decreasing on (T #,+ ). Therefore, f 2 (T) is always nonpositive on (0, + ) and Π 2 (T) is increasing on (0, + ). (B) If T # <0,thatis,F>θsI e /(α+θ),thenforany T 0 we have df 2 (T)/dT < 0; thatis,f 2 (T) is strictly decreasing on (0, + ). Sincef 2 (0) = K < 0, f 2 (T) is negative on (0, + ). Hence Π 2 (T) is increasing on (0, + ). Then we have the following lemma to describe the property of Π 2 (T) on (0, + ). Lemma 2. (1) For F 0, Π 2 (T) is convex on (0, + ) and T 0 2 is the unique maximum point of Π 2 (T). (2) For 0<F θsi e /(α+θ), Π 2 (T) is increasing on (0, T 1 2 ), decreasing on (T 1 2,T2 2 ),andincreasingon(t2 2,+ )if f 2(T # )> 0,andΠ 2 (T) is increasing on (0, + ) otherwise. (3) For F>θsI e /(α + θ), Π 2 (T) is always increasing on (0, + ). Likewise, the first-order necessary condition for Π 3 (T) in (13) to be maximum is dπ 3 (T)/dT = 0, which implies that F (e (α+θ)t 1) (α+θ) 2 + [1 (α+θ) T] FT α+θ (α+θ) 2 (e(α+θ)(t M) 1)( αsi e α+θ ci p)

6 6 Mathematical Problems in Engineering [1 (α+θ) T] (T M) α+θ ( αsi e α+θ ci p) + θsi em 2 2 (α+θ) K=0. (24) We set the left-hand expression of (24) as f 3 (T). Takingthe first derivative of f 3 (T) with respect to T gives df 3 (T) dt = [F+(αsI e α+θ ci p)e (α+θ)m ]Te (α+θ)t. (25) We consider the following two cases. (1) If F (ci p αsi e /(α+θ))e (α+θ)m,thendf 3 (T)/dT 0. Therefore, f 3 (T) is increasing on (0, + ). It is easy to get f 3 (0) = (/(α + θ) 2 )(e (α+θ)m 1)(αsI e /(α + θ) ci p ) + θsi e M 2 /2(α + θ) + αsi e M/(α + θ) 2 ci p M/(α + θ) K and lim T + f 3 (T) = +. Thus, if f 3 (0) < 0 then f 3 (T) = 0 has a unique solution (say T 0 3 )on(0, + ). Hence, f 3(T) will be negative on (0, T 0 3 )andpositiveon(t0 3,+ ), which implies Π 3 (T) is increasing on (0, T 0 3 ) and decreasing on (T 0 3,+ ). That is, Π 3(T) is unimodal on (0, + ) and T 0 3 is the unique maximum point of Π 3(T) if f 3 (0) < 0. In contrast, if f 3 (0) 0 then f 3 (T) will always be nonnegative on (0, + ); hence Π 3 (T) is always decreasing on (0, + ). (2) If F > (ci p αsi e /(α+θ))e (α+θ)m,thenf 3 (T) is strictly decreasing on (0, + ). Noting that lim T + f 3 (T) = for F > (αsi e /(α+θ) ci p )e (α+θ)m,oneeasily obtains that if f 3 (0) > 0 there will exist a unique root (say T 1 3 )totheequationf 3(T) = 0; hencef 3 (T) is positive on (0, T 1 3 ) and negative on (T1 3,+ ), which implies Π 3 (T) is decreasing on (0, T 1 3 ) and increasing on (T 1 3,+ ). But if f 3(0) 0, f 3 (T) is always negative on (0, + ); then Π 3 (T) is increasing on (0, + ). Summarizing the above arguments, the following lemma to describe the property of Π 3 (T) on (0, + )willbeobtained. Lemma 3. (1) For F (ci p αsi e /(α+θ))e (α+θ)m, Π 3 (T) is increasing on (0, T 0 3 ) and decreasing on (T0 3,+ )if f 3(0) < 0, and Π 3 (T) is decreasing on (0, + ) otherwise. (2) For F > (ci p αsi e /(α+θ))e (α+θ)m, Π 3 (T) is decreasing on (0, T 1 3 ) and increasing on (T1 3,+ )if f 3(0) > 0,andΠ 3 (T) is increasing on (0, + ) otherwise. Based upon Lemmas 1 3, we now will decide the optimal solution T to Π(T) from the below two situations ecision Rule of the Optimal Cycle Time T When M T d. When M T d,thepiecewiseprofitfunctionπ(t) has three different expressions, that is, Π 1 (T), Π 2 (T),andΠ 3 (T), respectively. From (17), (20), and(24), we have f 1 (T d )= f 2 (T d )= f 2 (M) =f 3 (M) E (α+θ) 2 [(α+θ) T de (α+θ)t d +1] K, F (α+θ) 2 [(α+θ) T de (α+θ)t d +1] + θsi et 2 d 2 (α+θ) K, = F (α+θ) 2 [(α+θ) Me(α+θ)M e (α+θ)m +1] + θsi em 2 2 (α+θ) K. (26) (27) (28) As shown in Appendix C, wewillfindthatf 1 (T d )>f 2 (T d ). For convenience, we rewrite the notation E as E=F ci p + αsi e /(α+θ) αsi e M,andthefollowingtwotheoremswould be obtained to decide the optimal solution T when M T d. Theorem 4. If M T d and (θsi e /(α+θ))e (α+θ)t d (ci p αsi e /(α + θ))e (α+θ)m, then one has the following results. (1) When F (θsi e /(α+θ))e (α+θ)m,thefollowingapplies. (A) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) 0, then Π(T )=Π 3 (T 0 3 ). Hence T is T 0 3. (B) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) > 0, then Π(T )=Π 2 (T 3 2 ). Hence T is T 3 2,whereT3 2 (T d,m)and satisfies (20). (C) If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) > 0, then Π(T )=max{π 1 (T 0 1 ), Π 2(T 3 2 )}. Hence T is T 0 1 or T 3 2 () If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) 0, then Π(T )=max{π 1 (T 0 1 ), Π 3(T 0 3 )}. Hence T is T 0 1 or T 0 3 (E) If f 1 (T d )>0, f 2 (T d ) 0,andf 2 (M) > 0, then Π(T )=max{π 1 (T 0 1 ), Π 2(Td)}. Hence T is T 0 1 or T d (2) When (θsi e /(α + θ))e (α+θ)m < F (θsi e /(α + θ))e (α+θ)t d,thefollowingapplies. (A) While f 2 (T # )>0, one has the following. (a) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) 0, then Π(T )=max{π 2 (T 1 2 ), Π 3(T 0 3 )}. Hence T is T 1 2 or T0 3 (b) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) > 0, then Π(T )=Π 2 (T 1 2 ). Hence T is T 1 2.

7 Mathematical Problems in Engineering 7 (c) If f 1 (T d ) > 0, f 2 (T d ) < 0,andf 2 (M) 0, then Π(T ) = max{π 1 (T 0 1 ), Π 2(T 1 2 ), Π 3 (T 0 3 )}. Hence T is T 0 1, T1 2,orT0 3 associated with maximum (d) If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) > 0, then Π(T )=max{π 1 (T 0 1 ), Π 2(T 1 2 )}. Hence T is T 0 1 or T1 2 (e) If f 1 (T d ) > 0, f 2 (T d ) 0,andf 2 (M) 0, then Π(T ) = max{π 1 (T 0 1 ), Π 2(T d ), Π 3 (T 0 3 )}. Hence T is T 0 1, T d,ort 0 3 associated with maximum (f) If f 1 (T d )>0, f 2 (T d ) 0,andf 2 (M) > 0, then Π(T )=max{π 1 (T 0 1 ), Π 2(T d )}. Hence T is T 0 1 or T d (B) While f 2 (T # ) 0, one has the following. (a) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) < 0, then Π(T )=Π 3 (T 0 3 ). Hence T is T 0 3. (b) If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) < 0, then Π(T )=max{π 1 (T 0 1 ), Π 3(T 0 3 )}. Hence T is T 0 1 or T0 3 (3) When (θsi e /(α + θ))e (α+θ)t d <F (ci p αsi e /(α + θ))e (α+θ)m,thefollowingapplies. (A) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) < 0, then Π(T )=Π 3 (T 0 3 ). Hence T is T 0 3. (B) If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) < 0, then Π(T )=max{π 1 (T 0 1 ), Π 3(T 0 3 )}. Hence T is T 0 1 or T 0 3 (C) If f 1 (T d )>0, f 2 (T d ) 0,andf 2 (M) < 0, then Π(T ) = max{π 1 (T 0 1 ), Π 2(T d ), Π 3 (T 0 3 )}. Hence T is T 0 1, T d,ort 0 3 () If f 1 (T d )>0, f 2 (T d ) 0,andf 2 (M) 0, then Π(T )=max{π 1 (T 0 1 ), Π 2(T d )}. Hence T is T 0 1 or T d (4) When F > (ci p αsi e /(α + θ))e (α+θ)m, T is +. Proof. See Appendix E for detail. Theorem 5. If M T d and (θsi e /(α + θ))e (α+θ)t d >(ci p θsi e /(α + θ))e (α+θ)m, then one has the following results. (1) When F (θsi e /(α + θ))e (α+θ)m, the results are the same as (1) intheorem 4. (2) When (θsi e /(α + θ))e (α+θ)m < F (ci p αsi e /(α + θ))e (α+θ)m, the results are the same as (2) in Theorem 4. (3) When F > (ci p αsi e /(α + θ))e (α+θ)m, T is +. Proof. See Appendix E for detail ecision Rule of the Optimal Cycle Time T When M< T d. When M<T d,thepiecewiseprofitfunctionπ(t) has only two different expressions, that is, Π 1 (T) and Π 3 (T), respectively. From (24), we have f 3 (T d )= F (e(α+θ)t d 1) (α+θ) 2 + [1 (α+θ) T d ] FT d α+θ (α+θ) 2 (e(α+θ)(t d M) 1)( αsi e α+θ ci p) [1 (α+θ) T d ] (T d M) α+θ ( αsi e α+θ ci p) + θsi em 2 2 (α+θ) K. (29) As shown in Appendix, wewillfindthatf 1 (T d ) > f 3 (T d ).Thefollowingtheoremwouldbegiventodecidethe system s optimal cycle time T when M<T d. Theorem 6. If M<T d, then one has the following results. (1) When F (ci p αsi e /(α + θ))e (α+θ)m,thefollowing applies. (A) If f 1 (T d ) 0 and f 3 (T d ) < 0,thenΠ(T ) = Π 3 (T 0 3 ). Hence T is T 0 3. (B) If f 1 (T d ) > 0 and f 3 (T d ) 0,thenΠ(T ) = max{π 1 (T 0 1 ), Π 3(Td)}. Hence T is T 0 1 or T d associated with the maximum (C) If f 1 (T d ) > 0 and f 3 (T d ) < 0,thenΠ(T ) = max{π 1 (T 0 1 ), Π 3(T 0 3 )}. Hence T is T 0 1 or T0 3 associated with the maximum (2) When F > (ci p αsi e /(α + θ))e (α+θ)m, T is +. Proof. See Appendix F for detail. 5. Numerical Example In this section, we will provide the following numerical example to illustrate the present model. Example 1. Given Q d = 200 units, = 1500 units, α = 0.4, θ = 0.20, K =$50,h =$1/unit/year,c =$5per unit, s =$9perunit,I e = 0.13/$/year, I p = 0.19/$/year, and M = 0.3 year, it is easy to get T d = , (θsi e /(α + θ))e (α+θ)t d (ci p αsi e /(α+θ))e (α+θ)m,andf (θsi e /(α+ θ))e (α+θ)m. Consequently, by using Bisection method or Newton-Raphson method we could obtain T 0 1 = , T 0 2 = , andt 0 3 = Becauseoff 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) 0, according to Theorem 4(1) the retailer s optimal cycle time should be T = T 0 3 = ;

8 8 Mathematical Problems in Engineering Table 1: The impact of change of α, θ, Q d,andm on the optimal solutions. α θ Q d M Parameters T 0 1 T 0 2 T 0 3 T Q Π hence the optimal order quantity and the maximum average profit are Q = and Π = ,respectively. Next, we further study the effects of changes of parameters α, θ, Q d,andm on the optimal solutions. The values of other parameters keep the same as in the above example when each of parameters α, θ, Q d,andm varies. Table 1 presents the observed results with various parameters α, θ, Q d,andm. The following inferences can be made based on Table 1. (1) The retailer s optimal order quantity Q and the optimal average profit Π increase as the value of α increases. It implies that when the market demand is more sensitive to the inventory level, the retailer will increase his/her order quantity to make more profit under permissible delay in payments. (2) The optimal order quantity Q and the maximum average profit Π decrease as the value of θ increases. (3) When Q d increases, the optimal order quantity Q and the optimal cycle time T are increasing but the optimal average profit Π is decreasing. (4) The retailer s order quantity Q is increasing as M increases. This verifies the fact that the retailer could indeed increase the sales quantity by adopting the trade credit policy provided by his/her supplier. 6. Conclusions In this paper, we develop an EOQ model for deteriorating items under permissible delay in payments. The primary difference of this paper as compared to previous related studies is the following four aspects: (1) the demand rate of items is dependent on the retailer s current stock level, (2) many items deteriorate continuously such as fruits and vegetables, (3) the retailer who purchases the items enjoys a fixed credit period offered by the supplier if the order quantity is greater than or equal to the predetermined quantity, and (4) we here use maximizing profit as the objective to find the optimal replenishment policies, which could better describe the essence of the inventory system than a cost-minimization model. In addition, we have discussed in detail the conditions of the existence and uniqueness of the optimal solutions to themodel.threeeasy-to-usetheoremsaredevelopedtofind the optimal ordering policies for the considered problem, and these theoretical results are illustrated by numerical example. By studying the effects of α, θ, Q d,andm on the optimal order quantity Q and the optimal average profit Π,some managerial insights are derived. The presented model can be further extended to some more practical situations. For example, we could allow for shortages, quantity discounts, time value of money and inflation, price-sensitive and stock-dependent demand, and so forth.

9 Mathematical Problems in Engineering 9 Appendices A. Proof of Π 1 (T d )<Π 2 (T d ) If ci p si e,wehavef E=αsI e M+cI p αsi e /(α + θ) > 0. From (11) and (12),weget Π 2 (T d ) Π 1 (T d ) C. Proof of f 1 (T d )>f 2 (T d ) If ci p si e,from(26) and (27),wehave f 2 (T d ) f 1 (T d ) = (α+θ) 2 (αsi em+ci p αsi e α+θ ) = 1 (F E) { T d (α+θ) 2 (e (α+θ)t d (α+θ) T d 1) +si e MT d θsi et 2 d 2 (α+θ) }. (A.1) [(α+θ) T d e (α+θ)t d +1]+ θsi et 2 d 2 (α+θ) (α+θ) 2 (αsi em+ θsi e α+θ ) It is evident that M T d and (e (α+θ)t d (α+θ)t d 1)>0for T d >0;then Π 2 (T d ) Π 1 (T d ) > 1 T d { T2 d 2 (αsi em αsi e α+θ +ci p) +si e MT d θsi et 2 d 2 (α+θ) } = T d 2 (αsi em+ci p ) si et d +si 2 e M T d 2 (αsi em+ci p )+ si em >0. 2 As a result, we have Π 1 (T d )<Π 2 (T d ). This completes the proof of Appendix A. B. Proof of Π 1 (T d )<Π 3 (T d ) (A.2) If ci p si e,wehavef E=αsI e M+cI p αsi e /(α + θ) > 0. From (11) and (13),weget Π 3 (T d ) Π 1 (T d ) = 1 (F E) { T d (α+θ) 2 [e (α+θ)t d e (α+θ)(td M) (α+θ) M] + αsi em (α+θ) 2 (e(α+θ)(t d M) 1) + αsi em 2 (α+θ) + θsi em 2 2 (α+θ) }. (B.1) Because of T d >Mand e (α+θ)t d e (α+θ)(t d M) (α+θ)m > 0, we have Π 3 (T d ) Π 1 (T d )>0. This completes the proof of Appendix B. [(α+θ) T d e (α+θ)t d +1]+ θsi et 2 d 2 (α+θ) = si e (α+θ) 2 {(αm + θ α+θ ) [(α+θ) T d e (α+θ)t d +1] θ (α+θ) T2 d }. 2 (C.1) Let G(T d ) = (αm + θ/(α + θ))[(α + θ)t d e (α+θ)t d + 1] θ(α+θ)t 2 d /2;then G (T d )=(αm+ θ α+θ ) (α+θ)2 T d e (α+θ)t d θ(α+θ) T d αm (α+θ) =θ(α+θ) T d [( +1)e (α+θ)t d 1]>0. θ (C.2) Hence, G(T d )>G(0)=0;thatis,f 1 (T d )>f 2 (T d ). This completes the proof of Appendix C.. Proof of f 1 (T d )>f 3 (T d ) From (26) and (24),we have f 3 (T d ) f 1 (T d ) = = (α+θ) 2 [(α+θ) T de (α+θ)t d +1] ( αsi e α+θ ci p αsi e M) (α+θ) 2 [(α+θ) T d e (α+θ)(t d M) e (α+θ)(t d M) +1 (α+θ) M] ( αsi e α+θ ci p)+ θsi em 2 2 (α+θ) (α+θ) 2 [(α+θ) T de (α+θ)t d

10 10 Mathematical Problems in Engineering (α+θ) T d e (α+θ)(t d M) +e (α+θ)(t d M) + (α+θ) M] ( αsi e α+θ ci p)+ θsi em 2 2 (α+θ) αsi em (α+θ) 2 [(α+θ) T d e (α+θ)t d +1]. (.1) For ci p si e and (α + θ)t d e (α+θ)t d (α+ θ)t d e (α+θ)(td M) +e (α+θ)(td M) +(α+θ)m>0,then f 3 (T d ) f 1 (T d ) θsi e (α+θ) 3 {(e(α+θ)t d e (α+θ)(t d M) )[(α+θ) T d 1] + (α+θ) M (α+θ)2 M 2 } 2 αsi em (α+θ) 2 [(α+θ) T de (α+θ)t d +1]. (.2) Let H(T d )=(e (α+θ)t d e (α+θ)(t d M) )[(α + θ)t d 1] + (α + θ)m (α+θ) 2 M 2 /2;thenitiseasytoobtainH (T d ) = (α + θ) 2 T 2 d e(α+θ)t d (1 e (α+θ)m )>0for T d >0,soH(T d )>H(M)> 0;hencef 1 (T d )>f 3 (T d ). This completes the proof of Appendix. E. Proof of Theorems 4 and 5 Before the proof of Theorems 4 and 5, according to the anterior analysis in Section 4, we can get the following results. (a) If F<cI p αsi e /(α+θ)+αsi e M, f 1 (T) is increasing on (0, T d ). Otherwise, f 1 (T) is decreasing on (0, T d ). (b) If F 0, f 2 (T) is increasing on (T d,m). Further, if F> 0 and T # M,thatis,0<F (θsi e /(α + θ))e (α+θ)m, f 2 (T) is also increasing on (T d,m). As a result, if F (θsi e /(α + θ))e (α+θ)m then f 2 (T) is always increasing on (T d,m),andinthiscaseiff 2 (T d )<0 and f 2 (M) > 0 there must exist a unique root to the equation f 2 (T) = 0 on (T d,m), and we denote this root by T 3 2 to differentiate with T0 2 (0,+ ) and T 1 2 (0,T# ).ButifT d T # <M,thatis,(θsI e /(α + θ))e (α+θ)m < F (θsi e /(α+θ))e (α+θ)t d,then df 2 (T)/dT > 0 for T (T d,t # ) and df 2 (T)/dT < 0 for T (T #,M).However,whenT # < T d,thatis, F>(θsI e /(α+θ))e (α+θ)t d,wehavedf 2 (T)/dT < 0 on (T d,m),andinthiscaseiff 2 (T d )>0and f 2 (M) < 0 there must exist a unique root (say T 4 2 )totheequation f 2 (T) = 0 on (T d,m). (c) If F (ci p αsi e /(α+θ))e (α+θ)m, f 3 (T) is increasing on (M, + ). Otherwise, f 3 (T) is decreasing on (M, + ). Therefore, to solve the optimal order cycle time T over (0, + ), the possible interval which F belongs to needs be considered; hence we should compare the size relations among values of ci p αsi e /(α + θ) + αsi e M, (θsi e /(α + θ))e (α+θ)m,and(θsi e /(α + θ))e (α+θ)t d with (ci p αsi e /(α + θ))e (α+θ)m.sincem T d and ci p si e, one will derive (θsi e /(α+θ))e (α+θ)m (θsi e /(α+θ))e (α+θ)t d and (θsi e /(α+ θ))e (α+θ)m (ci p αsi e /(α + θ))e (α+θ)m ci p αsi e /(α + θ) + αsi e M. So, there are three cases to occur: (i) (θsi e /(α + θ))e (α+θ)t d (ci p αsi e /(α+θ))e (α+θ)m, (ii) (ci p αsi e /(α+ θ))e (α+θ)m <(θsi e /(α + θ))e (α+θ)t d ci p αsi e /(α+θ)+ αsi e M, and (iii) (θsi e /(α + θ))e (α+θ)t d >ci p αsi e /(α+θ)+ αsi e M.NoticethatonceF > (ci p αsi e /(α + θ))e (α+θ)m,the subfunction Π 3 (T) is always increasing over (M, + ), so the profit function Π(T) has no optimal solution over (0, + ). Hence, we need only to study the two cases as below: (i) (θsi e /(α+θ))e (α+θ)t d (ci p αsi e /(α + θ))e (α+θ)m and (ii) (θsi e /(α + θ))e (α+θ)t d > (ci p αsi e /(α + θ))e (α+θ)m. According to these two cases and from some analysis, the proof of Theorems 4 and 5 would be obtained. Proof of Theorem 4. If M T d and (θsi e /(α+θ))e (α+θ)t d (ci p αsi e /(α+θ))e (α+θ)m, then we could obtain that (θsi e /(α + θ))e (α+θ)m (θsi e /(α + θ))e (α+θ)t d (ci p αsi e /(α + θ))e (α+θ)m ci p αsi e /(α + θ) + αsi e M.Hence, we have the following. (1) When F (θsi e /(α + θ))e (α+θ)m,sincef 1 (T d ) > f 2 (T d ) and f 2 (T) is increasing on (T d,m)thereare thefollowingfivepossiblecasestooccur. (A) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) 0, then T 0 1 T d and T 0 3 M. These yield that (i) Π 1(T) is increasing on (0, T d ); (ii) Π 2 (T) is increasing on (T d,m); (iii) Π 3 (T) is increasing on (M, T 0 3 ) anddecreasingon(t 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconcludethat Π(T )=Π 3 (T 0 3 );hencet is T 0 3. (B) If f 1 (T d ) 0, f 2 (T d ) < 0,andf 2 (M) > 0, then T 0 1 T d, T d < T 3 2 < M,andT 0 3 < M. These yield that (i) Π 1 (T) is increasing on (0, T d ); (ii) Π 2 (T) is increasing on (T d,t 3 2 )and decreasing on (T 3 2,M); (iii) Π 3(T) is decreasing on (M, + ). Combining (i) (iii) and Π 1 (T d )< Π 2 (T d ),weconcludethatπ(t ) = Π 2 (T 3 2 ); hence T is T 3 2. (C) If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) > 0, then 0 < T 0 1 < T d, T d < T 3 2 < M,andT0 3 < M. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing (T 0 1,T d); (ii) Π 2 (T) is increasing on (T d,t 3 2 ) and decreasing on (T3 2,M); (iii) Π 3 (T) is decreasing on (M, + ). Combining (i) (iii) and Π 1 (T d )<Π 2 (T d ),weconcludethat Π(T )=max{π 1 (T 0 1 ), Π 2(T 3 2 )}. HenceT is T 0 1 or T 3 2

11 Mathematical Problems in Engineering 11 () If f 1 (T d ) > 0, f 2 (T d ) < 0,andf 2 (M) 0, then 0 < T 0 1 < T d and T 0 3 M.These yield that (i) Π 1 (T) is increasing on (0, T 0 1 )and decreasing (T 0 1,T d); (ii) Π 2 (T) is increasing on (T d,m); (iii) Π 3 (T) is increasing on (M, T 0 3 ) anddecreasingon(t 0 3,+ ). Combining (i) (iii) and Π 1 (T d )<Π 2 (T d ),wecouldconclude that Π(T )=max{π 1 (T 0 1 ), Π 3(T 0 3 )}. HenceT is T 0 1 or T0 3 (E) If f 1 (T d )>0, f 2 (T d ) 0,andf 2 (M) > 0, then 0<T 0 1 <T d and T 0 3 <M. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing on (T 0 1,T d); (ii) Π 2 (T) is decreasing on (T d,m); (iii) Π 3 (T) is decreasing on (M, + ). Combining (i) (iii) and Π 1 (T d )<Π 2 (T d ),weconcludethat Π(T )=max{π 1 (T 0 1 ), Π 2(T d )}. HenceT is T 0 1 or T d (2) When (θsi e /(α+θ))e (α+θ)m < F (θsi e /(α + θ))e (α+θ)t d,thendf 2 (T)/dT > 0 for T (T d,t # ) and df 2 (T)/dT < 0 for T (T #,M). (A) While f 2 (T # )>0,wehavethefollowing. (a) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) 0, then T 0 1 T d, T d < T 1 2 < T #, T # < T 2 2 M,andT 0 3 M.Theseyield that (i) Π 1 (T) is increasing on (0, T d ); (ii) Π 2 (T) is increasing on (T d,t 1 2 ), decreasing on (T 1 2,T2 2 ), and increasing on (T2 2,M); (iii) Π 3 (T) is increasing on (M, T 0 3 )and decreasing on (T 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconclude that Π(T )=max{π 2 (T 1 2 ), Π 3(T 0 3 )}.Hence T is T 1 2 or T0 3 (b) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) > 0, then T 0 1 T d, T d <T 1 2 <T#,andT 0 3 <M. These yield that (i) Π 1 (T) is increasing on (0, T d ); (ii) Π 2 (T) is increasing on (T d,t 1 2 ) anddecreasingon(t 1 2,M); (iii) Π 3(T) is decreasing on (M, + ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconclude that Π(T )=Π 2 (T 1 2 ).HenceT is T 1 2. (c) If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) 0, then 0 < T 0 1 < T d, T d < T 1 2 < T #, T # <T 2 2 M,andT0 3 M.Theseyield that (i) Π 1 (T) is increasing on (0, T 0 1 )and decreasing (T 0 1,T d); (ii) Π 2 (T) is increasing on (T d,t 1 2 ), decreasing on (T1 2,T2 2 ), and increasing on (T 2 2,M); (iii) Π 3(T) is increasing on (M, T 0 3 ) and decreasing on (T 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ), we conclude that Π(T ) = max{π 1 (T 0 1 ), Π 2(T 1 2 ), Π 3(T 0 3 )}. Hence T is T 0 1, T1 2,orT0 3 associated with maximum (d) If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) > 0, then 0 < T 0 1 < T d, T d < T 1 2 < T #, and T 0 3 < M. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing (T 0 1,T d); (ii) Π 2 (T) is increasing on (T d,t 1 2 ) and decreasing on (T 1 2,M); (iii) Π 3(T) is decreasing on (M, + ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconclude that Π(T )=max{π 1 (T 0 1 ), Π 2(T 1 2 )}.Hence T is T 0 1 or T1 2 (e) If f 1 (T d )>0, f 2 (T d ) 0,andf 2 (M) 0, then 0 < T 0 1 < T d, T # < T 2 2 M, and T 0 3 M. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing (T 0 1,T d); (ii) Π 2 (T) is decreasing on (T d,t 2 2 ) and increasing on (T 2 2,M); (iii) Π 3(T) is increasing on (M, T 0 3 ) and decreasing on (T 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ), we conclude that Π(T ) = max{π 1 (T 0 1 ), Π 2(T d ), Π 3 (T 0 3 )}. Hence T is T 0 1, T d,ort 0 3 associated with maximum (f) If f 1 (T d ) > 0, f 2 (T d ) 0,andf 2 (M) > 0, then0 < T 0 1 < T d,andt 0 3 < M. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing (T0 1,T d); (ii) Π 2 (T) is decreasing on (T d,m); (iii) Π 3 (T) is decreasing on (M, + ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconclude that Π(T )=max{π 1 (T 0 1 ), Π 2(T d )}.Hence T is T 0 1 or T d (B) While f 2 (T # ) 0,thenf 2 (T d )<0and f 2 (M) < 0. (a) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) < 0, then T 0 1 T d and T 0 3 >M.Theseyieldthat (i) Π 1 (T) is increasing on (0, T d ); (ii) Π 2 (T) is increasing on (T d,m); (iii) Π 3 (T) is increasing on (M, T 0 3 ) and decreasing on (T 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ), we conclude that Π(T )=Π 3 (T 0 3 ).HenceT is T 0 3. (b) If f 1 (T d )>0, f 2 (T d )<0,andf 2 (M) < 0, then 0 < T 0 1 < T d and T 0 3 > M. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing on (T0 1,T d); (ii) Π 2 (T) is increasing on (T d,m); (iii) Π 3 (T) is increasing on (M, T 0 3 ) and decreasing on (T 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconclude that Π(T )=max{π 1 (T 0 1 ), Π 3(T 0 3 )}.Hence T is T 0 1 or T0 3

12 12 Mathematical Problems in Engineering (3) When (θsi e /(α+θ))e (α+θ)t d <F (ci p αsi e /(α + θ))e (α+θ)m,sincef 1 (T d ) > f 2 (T d ) and f 2 (T) is decreasing on (T d,m) there are the following four possible cases to occur. (A) If f 1 (T d ) 0, f 2 (T d )<0,andf 2 (M) < 0, then T 0 1 T d and T 0 3 >M. These yield that (i) Π 1(T) is increasing on (0, T d ); (ii) Π 2 (T) is increasing on (T d,m); (iii) Π 3 (T) is increasing on (M, T 0 3 ) anddecreasingon(t 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconcludethat Π(T )=Π 3 (T 0 3 ).HenceT is T 0 3. (B) If f 1 (T d ) > 0, f 2 (T d ) < 0,andf 2 (M) < 0, then 0 < T 0 1 < T d and T 0 3 > M.These yield that (i) Π 1 (T) is increasing on (0, T 0 1 )and decreasing on (T 0 1,T d); (ii) Π 2 (T) is increasing on (T d,m); (iii) Π 3 (T) is increasing on (M, T 0 3 ) anddecreasingon(t 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconcludethat Π(T )=max{π 1 (T 0 1 ), Π 3(T 0 3 )}. HenceT is T 0 1 or T 0 3 (C) If f 1 (T d ) > 0, f 2 (T d ) 0,andf 2 (M) < 0, then 0 < T 0 1 < T d, T d T 4 2 < M and T 0 3 >M. These yield that (i) Π 1(T) is increasing on (0, T 0 1 ) and decreasing on (T0 1,T d); (ii) Π 2 (T) is decreasing on (T d,t 4 2 ) and increasing on (T 4 2,M); (iii) Π 3(T) is increasing on (M, T 0 3 ) anddecreasingon(t 0 3,+ ). Combining (i) (iii) and Π 1 (T d ) < Π 2 (T d ),weconcludethat Π(T )=max{π 1 (T 0 1 ), Π 2(T d ), Π 3 (T 0 3 )}. Hence T is T 0 1, T d,ort 0 3 () If f 1 (T d )>0, f 2 (T d ) 0,andf 2 (M) 0, then 0<T 0 1 <T d and T 0 3 M. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing on (T 0 1,T d); (ii) Π 2 (T) is decreasing on (T d,m); (iii) Π 3 (T) is decreasing on (M, + ). Combining (i) (iii) and Π 1 (T d )<Π 2 (T d ),weconcludethat Π(T )=max{π 1 (T 0 1 ), Π 2(T d )}. HenceT is T 0 1 or T d (4) When F > (ci p αsi e /(α+θ))e (α+θ)m,becauseonthe interval of (M, + )theaverageprofitfunctionπ(t) is always increasing, the optimal solution T is + evidently. This completes the proof of Theorem 4. Proof of Theorem 5. If (θsi e /(α+θ))e (α+θ)t d >(ci p αsi e /(α+ θ))e (α+θ)m,thenonehas(θsi e /(α + θ))e (α+θ)m (ci p αsi e /(α + θ))e (α+θ)m < (θsi e /(α + θ))e (α+θ)t d ci p αsi e /(α + θ) + αsi e M or (θsi e /(α + θ))e (α+θ)m (ci p αsi e /(α+θ))e (α+θ)m ci p αsi e /(α + θ) + αsi e M (θsi e /(α + θ))e (α+θ)t d.sinceπ(t) has no optimal solution when F > (ci p αsi e /(α + θ))e (α+θ)m,sotherearethree possible situations to occur as below. (1) When F (θsi e /(α+θ))e (α+θ)m, it is obvious that the results are the same as (1) in Theorem 4. (2) When (θsi e /(α + θ))e (α+θ)m < F (ci p αsi e /(α + θ))e (α+θ)m, the results are the same as (2) in Theorem 4. (3) When F > (ci p αsi e /(α+θ))e (α+θ)m,becauseonthe interval of (M, + )theaverageprofitfunctionπ(t) is always increasing, the optimal solution T is + evidently. This completes the proof of Theorem 5. F. Proof of Theorem 6 According to the anterior analysis in Section 4, we could get the below results. (a) If F<cI p αsi e /(α+θ)+αsi e M, f 1 (T) is increasing on (0, T d ). Otherwise, f 1 (T) is decreasing on (0, T d ). (b) If F (ci p αsi e /(α + θ))e (α+θ)m,thenf 3 (T) is increasing on (T d,+ ). Otherwise, f 3 (T) is strictly decreasing on (T d,+ ). Similarly, in order to find the system s optimal cycle time T, the possible interval which F belongs to needs be considered if M < T d ; hence we should compare the size relations between ci p αsi e /(α + θ) + αsi e M and (ci p αsi e /(α+θ))e (α+θ)m.ifci p si e,thenonehas(ci p αsi e /(α+ θ))e (α+θ)m ci p αsi e /(α + θ) + αsi e M. (1) When F (ci p αsi e /(α+θ))e (α+θ)m,wehavethe following. (A) If f 1 (T d ) 0and f 3 (T d )<0,thenT 0 1 T d and T 0 3 >T d. These yield that (i) Π 1 (T) is increasing on (0, T d ); (ii) Π 3 (T) is increasing on (T d,t 0 3 ) and decreasing on (T 0 3,+ ). Combining (i)- (ii) and Π 1 (T d ) < Π 3 (T d ),weconcludethat Π(T )=Π 3 (T 0 3 ).HenceT is T 0 3. (B) If f 1 (T d ) > 0 and f 3 (T d ) 0,then0 < T 0 1 < T d and T 0 3 T d. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing on (T 0 1,T d); (ii) Π 3 (T) is decreasing on (T d,+ ). Combining (i)-(ii) and Π 1 (T d ) < Π 3 (T d ),we conclude that Π(T ) = max{π 1 (T 0 1 ), Π 3(T d )}. Hence T is T 0 1 or T d associated with the maximum (C) If f 1 (T d ) > 0 and f 3 (T d ) < 0,then0 < T 0 1 < T d and T 0 3 > T d. These yield that (i) Π 1 (T) is increasing on (0, T 0 1 ) and decreasing on (T 0 1,T d); (ii) Π 3 (T) is increasing on (T d,t 0 3 ) and decreasing on (T 0 3,+ ). Combining (i)-(ii) and Π 1 (T d ) < Π 3 (T d ),wecanconcludethat Π(T )=max{π 1 (T 0 1 ), Π 3(T 0 3 )}. HenceT is T 0 1 or T 0 3 associated with the maximum

13 Mathematical Problems in Engineering 13 (2) When F > (ci p αsi e /(α+θ))e (α+θ)m,becauseonthe interval of (M, + ) the average profit function Π(T) is always increasing, the optimal solution T is + evidently. This completes the proof of Theorem 6. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. Acknowledgments This work was supported in part by the National Natural Science Foundation of China Project under project Grant nos , , , , and The authors have benefited substantially from the insightful comments of the editor and anonymous referees, which were instrumental for elevating the quality of this work. References [1] T. M. Whitin, The Theory of Inventory Management, Princeton University Press, Princeton, NJ, USA, [2] R. I. Levin, C. P. McLaughlin, R. P. Lamone, and J. F. 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