Inventory models for deteriorating items with maximum lifetime under downstream partial trade credits to credit-risk customers by discounted cash-flow analysis

Author’s Accepted Manuscript Inventory models for deteriorating items with maximum lifetime under downstream partial trade credits to credit-risk customers by discounted cashflow analysis Jiang Wu, Faisal B. Al-khateeb, Jinn-Tsair Teng, Leopoldo Eduardo Cárdenas-Barrón www.elsevier.com/locate/ijpe

PII: DOI: Reference:

S0925-5273(15)00401-6 http://dx.doi.org/10.1016/j.ijpe.2015.10.020 PROECO6270

To appear in: Intern. Journal of Production Economics Received date: 12 July 2015 Revised date: 21 October 2015 Accepted date: 22 October 2015 Cite this article as: Jiang Wu, Faisal B. Al-khateeb, Jinn-Tsair Teng and Leopoldo Eduardo Cárdenas-Barrón, Inventory models for deteriorating items with maximum lifetime under downstream partial trade credits to credit-risk customers by discounted cash-flow analysis, Intern. Journal of Production Economics, http://dx.doi.org/10.1016/j.ijpe.2015.10.020 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting galley proof before it is published in its final citable form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

Inventory models for deteriorating items with maximum lifetime under downstream partial trade credits to credit-risk customers by discounted cash-flow analysis

Inventory models for deteriorating items with maximum lifetime under downstream partial trade credits to credit-risk customers by discounted cash-flow analysis Jiang Wu School of Statistics Southwestern University of Finance & Economics, Chengdu 611130, China Faisal B. Al-khateeb

Department of Business Administration Southern University at New Orleans, LA 70126, U.S.A. Jinn-Tsair Teng Department of Marketing and Management Sciences William Paterson University of New Jersey, Wayne, New Jersey 07470-2103, U.S.A. Leopoldo Eduardo Cárdenas-Barrón* School of Engineering and Sciences, Tecnológico de Monterrey E. Garza Sada 2501 Sur, C.P. 64849, Monterrey, Nuevo León, México

Abstract Getting loans from banks are almost impossible after 2008 global financial crisis. As a result, about 80% of companies in United Kingdom and United States offer their products on various short-term, free-interest loans (i.e., trade credit) to customers. Numerous researchers and academicians apply discounted cash flow (DCF) analysis merely to compute the interest earned and charged during credit period but not to the revenue and other costs which are considerably larger than the interest earned and charged. For a rigorous analysis, the DCF on all relevant costs is applied. In addition, many products deteriorate continuously and cannot be sold after their maximum lifetimes or expiration dates. However, very few researchers and investigators have implemented the product lifetime expectance into their models. In this paper, a supplier-retailer-customer chain system is developed in which the retailer gets an upstream full trade credit from the supplier whereas offers a downstream partial trade credit to credit-risk 

*

E-mail:

Corresponding author. Tel. +52 81 83284235, Fax +52 81 83284153.

[email protected] (L.E. Cárdenas-Barrón)

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customers, the deterioration rate is non-decreasing over time and near 100% particularly close to its expiration date, and DCF analysis is applied to compute all relevant costs. This paper demonstrates that the retailer’s optimal replenishment cycle time not only exists but also is unique. Thus, the search of the global optimal reduces to finding a local solution. Finally, several numerical examples and sensitivity analysis are performed in order to illustrate the problem and obtain managerial insights.

Keywords: Supply Chain Management; Expiration dates; Trade credit; Discount cash flow. Acknowledgements: The authors would like to express sincere appreciation to Editor Peter Kelle, three anonymous referees, and Dr. Phillip Lu for their encouragement and insightful suggestions which helped us improve the quality of the paper significantly.

1. Introduction In traditional business transactions, it was implicitly assumed that the buyer must pay the procurement cost when products are received. However, in today’s competitive markets, most companies offer buyers various credit terms (e.g., permissible delay in payments, cash discounts, etc.) to stimulate sales and hence reduce inventory. Seifert et al. (2013) stipulated that “estimates suggest that more than 80% of business-to-business transactions in the United Kingdom are made on credit, while about 80% of United States firms offer their products on trade credit.” Conversely, it is known that trade credit decreases the buyer’s inventory holding cost, and therefore affects the buyer’s economic order quantity (thereafter, EOQ). In review of the literature for inventory models with trade credit financing, Goyal (1985) derived the retailer’s optimal EOQ when the supplier provides a permissible delay in payment (i.e., an upstream trade credit). Aggarwal and Jaggi (1995) extended the EOQ model for non-deteriorating items (i.e., products can be stored indefinitely) to deteriorating items (i.e., products deteriorate constantly). Jamal et al. (1997) further expanded the model to permit

2

shortages. Teng (2002) modified the previous models by using sales revenue (instead of purchase amount) to compute the interest earned from sales. Huang (2003) then explored the problem to a supply chain system in which the retailer receives an upstream trade credit from the supplier while offers a downstream trade credit to customers. Liao (2008) further generalized Huang’s model with an unlimited replenishment rate to an economic production quantity (thereafter, EPQ) model with a limited replenishment rate for deteriorating items. Min et al. (2010) proposed an EPQ model with both upstream and downstream trade credits when the demand depends on on-hand displayed stocks. Teng et al. (2012) extended the constant demand rate to the increasing demand pattern for the growth stage of product life cycle. Chern et al. (2013, 2014) discussed non-cooperative Stackelberg and Nash two-player equilibrium solutions between the supplier and the retailer. Chen et al. (2014a, 2014b) discussed the retailer’s optimal EOQ/EPQ under different credit terms. Liao et al. (2014) studied optimal strategy for deteriorating items with capacity constraints under upstream and downstream trade credits. Various products such as volatile liquids, blood banks, vegetables, fruits, fashion merchandises and high-tech products deteriorate continuously due to evaporation, spoilage, obsolescence, among other reasons. Ghare and Schrader (1963) built an inventory model with an exponentially decaying inventory. Covert and Philip (1973) extended the constant deterioration rate to Weibull failure rate. Dave and Patel (1981) proposed an EOQ model for deteriorating items with linearly increasing demand. Sachan (1984) further generalized the EOQ model to permit shortages. Hariga (1996) derived EOQ models for deteriorating items when the demand function is log-concave with time (e.g., a constant demand or an increasing demand pattern is a special case). Teng et al. (1999) further expanded EOQ models to allow for shortages and any continuous demand pattern (which includes a log-concave demand as a special case). Teng et al. (2002) further explored the model to permit partial backlogging. Dye

3

(2013) investigated the effect of technology investment on refrigeration to improve the profit for deteriorating products. It is important to mention that none of the above cited papers took the expiration date into consideration until studies such as in Chen and Teng (2014), Sarkar (2012), Wang et al. (2014), Wu et al (2014) and Sarkar et al (2015). There were some research works related to DCF analysis. For example, Hill and Pakkala (2005) applied DCF analysis to determine the optimal inventory policy that maximizes the expected net present value of the cash flows. Chung and Liao (2006) using DCF developed an inventory model for deteriorating items with permissible delay in payments. Dye et al. (2007) took pricing into consideration and derived the optimal inventory and pricing strategies to maximize the net present value of total profit. Chang et al. (2010) established the optimal ordering policies for deteriorating items using DCF analysis when the seller offers a trade credit if the buyer orders more than or equal to the predetermined amount. Mousavi et al. (2013) used meta-heuristic algorithms to solve the multi-item and multi-period inventory systems with DCF. Recently, Chen and Teng (2015) applied the DCF analysis to obtain the optimal lot size and credit period in a supply chain with upstream and downstream trade credit financing. In reality, a credit-worthy retailer generally obtains a permissible delay on the entire purchasing quantity without collateral deposits from the supplier (i.e., an upstream full trade credit). However, a retailer often asks for credit-risk customers to cover a fraction of the purchasing cost as a collateral deposit at the time of placing an order, and then provides a permissible delay on the remaining quantity (i.e., a downstream partial trade credit). The use of a downstream partial trade credit as a strategy to reduce default (or bad debt) risks with credit-risk customers has received relatively little attention by the researchers and investigators. Additionally, the majority of the recent studies considers merely the opportunity loss (i.e., time value of money) of trade credit but ignores to take the opportunity loss of the other costs into consideration. For a rigorous and sound analysis, the DCF analysis must be used in order to

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take the effect of inflation and time value of money on all relevant revenue and costs. As a result, in this paper, a supplier-retailer-customer supply chain system is proposed in which the supplier provides the retailer an upstream full credit period of S years, meanwhile the retailer in turn gives a downstream partial credit period of R years to the customer, the deteriorating rate is constant or increasing and closer to 100% especially near to its expiration date, and the DCF analysis is applied to incorporate the effects of inflation and time value of money. In fact, the most commonly used Weibull non-decreasing deterioration (or failure) rate is a special case of the proposed generalized deterioration rate. This paper formulates the retailer’s objective functions under different possible alternatives. By applying existing theorems in concave functions, it is proved that there exists a unique global optimal solution to each alternative. Finally, some numerical examples are solved in order to illustrate the problem and also some managerial insights are given. The remaining of the paper is designed as follows. Section 2 defines notation and makes necessary assumptions. Section 3 derives the present value of the retailer’s annual total profit under each alternative. Section 4 shows that the optimal cycle time exists uniquely which simplifies the search for the global optimal to a local one. Section 5 presents numerical examples in order to illustrate the model, and obtain managerial insights. Finally, Section 6 provides the conclusions and the future research directions.

2. Notation and assumptions The following notation and assumptions are used in the entire paper.

2.1. Notation: For simplicity, the symbols for parameters, decision variable, and functions are defined accordingly.

5

Parameters:



Fraction of the purchasing cost must be paid at the time of placing an order, 0    1.

c

Purchasing cost per unit in dollars.

D

Demand rate in units per year.

h

Holding cost per unit per year in dollars excluding interest charge.

i

Annual compound interest per dollar per year.

m

Maximum lifetime or expiration time in years.

o

Ordering cost per order in dollars.

p

Selling price per unit in dollars.

R

Downstream credit period in years by the retailer.

S

Upstream credit period in years by the supplier.

t

Time in years.

Ic

Interest charged per dollar per year.

Ie

Interest earned per dollar per year.

Decision variable: T

Replenishment cycle time in years.

Functions: Q

Oder quantity in units per replenishment cycle, which is a function of T.

 (t )

Deterioration rate at time t, which is a non-decreasing function in t.

I(t)

Inventory level at time t.

PTP(T )

Present value of annual total profit, which is a function of T.

For convenience, the asterisk symbol on a variable is denoted the optimal solution of the

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variable. For instance, T * is the optimal solution of T. Next, the necessary assumptions to build up the mathematical model are given below.

2.2. Assumptions: The following assumptions are made to build the mathematical inventory model. 1. In a supplier-retailer-customer supply chain system, the retailer obtains a full upstream credit period of S years from his/her supplier (i.e., if the retailer orders items at time 0, and pays off at time S, then there is no interest charges), and in turn gives a partial downstream trade credit to his/her credit-risk customers who must cover  portion of the purchasing cost at the time of placing an order (i.e., cash payment), and then get a credit period of R years on the outstanding quantity (i.e., credit payment). Note that to good-credit customers, the retailer may provide a full trade credit in which just set  = 0. Therefore, the proposed model includes the special case in which the retailer gives a downstream full trade credit to his/her customers. 2. If S  R , then the retailer deposits the sales revenue into an interest bearing account after time R. If S  T  R (i.e., the upstream credit period is longer than the time at which the retailer gets the last payment from his/her customers), then the retailer collects all sales revenue and pays off the entire purchasing cost at time S. Otherwise (if S  T  R ), the retailer pays the supplier both the credit payment for all units sold by S  R and the cash payment obtained from time 0 to S, retains the profit for the use of the other activities, and begins paying for the interest charges on the items sold after S  R . 3. If S  R , the retailer gets cash payments from customers, and immediately deposits them into an interest bearing account until time S. As to credit payments, the retailer must finance (1   )cDT at time S, and then pay off the loan from time R to time T + R.

4. A deteriorating item deteriorates continuously and cannot be sold after its maximum lifetime

7

or expiration date. Therefore, its deterioration rate is 100% near to its expiration date. As a result, it is assumed without loss of generality (WLOG) that the deterioration rate  (t ) at time t, 0  t  m , satisfies the following conditions: 0   (t )  1,  ' (t )  0 , and  (m)  1 .

(1)

Note that the following deterioration rate as in Sarkar (2012), Chen and Teng (2014), Wang et al. (2014), Wu and Chan (2014), Wu et al. (2014), Dye et al. (2014) and Sarkar et al. (2015):

 (t )  1 /(1  m  t ) , 0  t  m ,

(2)

is also a special case of (1). 5. Since the deterioration rate reaches 100% after expiration date, it is assumed WLOG that both upstream and downstream credit periods R and S, and the replenishment cycle time T are less than or equal to the expiration date m. R  m, S  m, and T  m.

(3)

6. Shortages are not permitted. 7. Replenishment rate is instantaneous and complete. The problem here is for the retailer to determine his/her optimal replenishment cycle time T * such that the present value of his/her annual total relevant profit is maximized. Given the above notation and assumptions, it is time to derive the present value of the retailer’s annual total profit as a function of cycle time T for a deteriorating item with expiration date into a mathematical model.

3. Mathematical model During the replenishment cycle [0, T], the inventory level is depleted by demand and deterioration, and hence governed by the following differential equation:

I ' (t )   D   (t ) I (t ), 0  t  T ,

(4)

8

with the boundary condition I(T) = 0. Note that the prime symbol and the double-prime symbol on a variable are denoted the first-order and the second-order derivatives with respect to the variable throughout the entire paper, respectively. Solving the differential equation (4), one gets T

I (t )  e  ( t )  e ( u ) Ddu , 0  t  T ,

(5)

t

where t

 (t )    (u )du is non-decreasing, 0  t  T .

(6)

0

The retailer provides his/her customers a downstream credit period of R years. Thus, the retailer receives the cash payment from time 0 to T, and the credit payment from time R to

T  R . Consequently, the present value of the retailer’s sales revenue per cycle time T is SR  pD   e it dt  (1   )   0 R

T R

T

e it dt  . 

(7)

At time 0, the retailer orders deteriorating items. Hence, the present value of the retailer’s ordering cost per cycle time T is OC = o.

(8)

Since the upstream trade credit is S years, the retailer must pay the supplier the whole purchasing cost cQ at time S. As a result, the present value of the retailer’s purchasing cost per cycle time T is T

PC  ce iS Q  ce iS I (0)  cD  e (t )iS dt .

(9)

0

T

As noted, the order quantity Q  I (0)  D  e (u ) du . Likewise, the present value of the 0

retailer’s holding cost per cycle time T is T

T

0

0

HC  h  e it I (t ) dt  hD 



T

t

e (u ) (t )it dudt.

(10)

From the values of R and S, there exist two potential cases: R  S , and R  S . Both cases are

9

discussed them separately.

3.1. The case of R  S Based on values of S, T, and T  R (i.e. the time at which the retailer receives the credit payment from the last customer), three sub-cases can occur: (i) S  T , (ii) T  S  T  R , and (iii) T  R  S . Notice that the case of S  T is applicable for both Cases (i) S  T , and (ii) T  S  T  R . Similarly, the case of S  T  R is applicable for both Cases (ii) T  S  T  R , and (iii) T  R  S . For these three cases, the interest earned and the interest

charged are derived accordingly.

Sub-case 1.1. S  T In this sub-case, the retailer accumulates revenue and receives interest from two sources: (i) the cash payment beginning from time 0 through S, and (ii) the credit payment beginning from time R through S. Hence, the present value of the interest earned per cycle is S S IE  pI e D   te it dt  (1   )  (t  R)e it dt  .  0  R

(11)

On the other hand, the retailer provides his/her customers a credit period of R years, and gets customers’ credit payments from time R through T+R. At time S, the retailer obtains

cDS dollars from cash payment and (1   )cD(S  R) dollars from credit payment, and therefore pays his/her supplier cDS  (1   )cD(S  R) dollars. Consequently, the retailer must finance all items sold after time S for the cash payment, and all items sold after time S  R for the credit payment at an interest charged I c per dollar per year. As a result, the

present value of the interest charged per cycle is given by

IC  cI c D   (T  t ) e it dt  (1   )   S S T

T R

(T  R  t ) e it dt  . 

As a result, the present value of the retailer’s annual total profit by using (7) - (12) is:

(12)

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PTP1 (T ) =

=

1 ( SR  PC  HC  OC  IC  IE ) T

T R pD  T it cD T  (t )iS hD T T  (u ) (t )it   e dt  (1   )  e it dt   e dt  e du dt   T 0 R T  0 T  0 t





T R o cI c D  T    (T  t ) e it dt  (1   )  (T  R  t ) e it dt   S T T  S S pI e D  S it   te dt  (1   ) (t  R)eit dt  .  R T  0

(13)

Next, we derive the present value of the retailer’s annual total profit for the sub-case in which S  T  R .

Sub-case 1.2. T  S  T  R Here, again, the retailer accumulates revenue and obtains interest from two sources: (i) the cash payment starting from time 0 through S, and (ii) the credit payment starting from time R through S. Thus, the present value of the interest earned per cycle is T S S IE  pI e D    te it dt   Teit dt   (1   )  (t  R)e it dt  .   0  T R 

(14)

Likewise, the retailer receives all cash payments by time T (  S ) so that there is no interest charged for the cash payment. However, the retailer must finance all items sold during the time interval [S  R, T ] . Therefore, the annual interest charged is IC  cI c D(1   ) 

T R S

(T  R  t ) e it dt .

(15)

From (7) - (10), (14) and (15), it is known that the present value of the retailer’s annual total relevant profit is PTP2 (T ) =

=

1 ( SR  PC  HC  OC  IC  IE ) T

T R pD  T it cD T  (t )iS hD T T  (u ) (t )it   e dt  (1   )  e it dt   e dt  e du dt    0 R 0  T T  T  0 t

11





T R o cI c D  (1   )  (T  R  t ) e it dt S T T

S pI e D   T it it   (1   ) S (t  R)e it dt  .  te dt  Te dt  T R   T    0

(16)

Now, the last sub-case in which T  R  S is discussed.

Sub-case 1.3. T  R  S In this sub-case, the retailer receives the all cash and credit payments before the supplier’s upstream credit period S, and therefore there is no interest charged. However, the present value of the interest earned per cycle is given by T S T R S IE  pI e D    te it dt   Teit dt   (1   )  (t  R)e it dt   Teit dt  .   0 T R T  R   

(17)

So, the present value of the retailer’s annual total profit is PTP3 (T ) =

=

1 ( SR  PC  HC  OC  IE ) T

T R pD  T it cD T  (t )iS hD T T  (u ) (t )it   e dt  (1   )  e it dt   e dt  e du dt  T  0 R T  0 T  0 t



S o pI e D   T it it   (1   ) T  R (t  R)eit dt  S Teit dt  .   te dt  Te dt  T T  R   R  T T    0

(18)

Combining (13), (16), and (18), the present value of the retailer’s annual total profit is given as

 PTP1 (T ), if S  T .  PTP(T )   PTP2 (T ), if S  R  T  S .  PTP (T ), if T  S  R. 3 

(19)

It is clear from (13), (16), and (18) that

PTP1 (S )  PTP2 (S ) , and PTP2 (S  R)  PTP3 (S  R) . Hence, PTP(T) is continuous in T  0. Then we proceed with the case of R  S.

3.2. The case of R  S

(20)

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Likewise, the following two sub-cases can occur based on values of S and T: S  T , and S  T . Let’s discuss them accordingly.

Sub-case 2.1. S  T Since R  S, there is no interest earned from the credit payment. However, the present value of the annual interest earned from the cash payment is S

IE  pI e D  te it dt . 0

(21)

In this sub-case, at time S, the retailer must finance cD(T  S ) for the cash payment, and (1   )cDT for the credit payment, respectively. Then the retailer pays off the loan for the cash

payment at time T, but pays off the loan for the credit payment at t = T + R. Therefore, the present value of the interest charged per cycle is T R T R IC  cI c D   (T  t ) e it dt  (1   )  Teit dt   (T  R  t ) e it dt  .  S R  S 

(22)

Consequently, the present value of the retailer’s annual total profit is PTP4 (T ) =

=

1 ( SR  PC  HC  OC  IC  IE ) T

T R pD  T it cD T  (t )iS hD T T  (u ) (t )it   e dt  (1   )  e it dt   e dt  e du dt   T 0 R T  0 T  0 t





o cI c D  T T

 T(T  t ) eit dt  (1   ) R Teit dt  T  R (T  R  t ) eit dt   S  R   S  

pI e D

te it dt .

T



S 0

We then discuss the last sub-case in which R  S  T .

Sub-case 2.2. S  T Similarly, the present value of the interest earned from the cash payment per cycle is

(23)

13 T S IE  pI e D  te it dt   Teit dt  T  0 

(24)

However, there is no interest charged for the cash payment. On the other hand, the retailer must finance (1   )cDT for the credit payment at time S, and start paying off the loan from time R to T + R. Therefore, the present value of the interest charged per cycle is R T R IC  (1   )cI c D   Teit dt   (T  R  t ) e it dt  . R  S 

(25)

Consequently, the present value of the retailer’s annual total profit is PTP5 (T ) =

=

1 ( SR  PC  HC  OC  IC  IE ) T

T R pD  T it cD T  (t )iS hD T T  (u ) (t )it   e dt  (1   )  e it dt   e dt  e du dt   T 0 R T  0 T  0 t





T R o (1   )cI c D  R it it    S Te dt   R (T  R  t ) e dt    T T

pI e D  T

T S it it   0 te dt  T Te dt  .  

(26)

Combining (23) and (26), we know that the present value of the retailer’s annual total relevant profit is

PTP4 (T ), if S  T . PTP(T )   PTP5 (T ), if S  T .

(27)

It is clear that PTP(T ) is continuous in T, and has the following properties

PTP4 (S )  PTP5 (S ) .

(28)

Next, we discuss theoretical results and the unique optimal solution.

4. Theoretical results and optimal solution To solve the problem, the following theoretical result in generalized concave functions is applied. If f(x) is non-negative, differentiable and (strictly) concave, and g(x) is positive,

14

differentiable and convex, then the real-value function z ( x)  f ( x ) / g ( x)

(29)

is (strictly) pseudo-concave. For detailed proof, please see Cambini and Martein (2009, p. 245). By applying (29), we can prove that the present value of the retailer’s annual total profit

PTPj (T ) for j = 1, 2, 3, 4, and 5 is strictly pseudo-concave in T. As a result, there exists a unique global optimal solution T j* such that PTPj (T ) is maximized. Next, we characterize the optimal solution in two cases: the case of R  S first, and then the case of R  S .

4.1. Optimal replenishment cycle time for the case of R  S We maximize PTPj (T ) for j = 1, 2, and 3 separately. By applying the above mentioned theoretical result in (29), we can obtain the following results.

Theorem 1. (a) PTP1 (T ) is a strictly pseudo-concave function in T, and hence exists a unique maximum solution T1 . (b) If S  T1 , then PTP1 (T ) subject to S  T is maximized at T1*  T1 . (c) If S  T1 , then PTP1 (T ) subject to S  T is maximized at T1*  S . Proof. See Appendix A. Taking the first-order derivative of PTP1 (T ) with respect to T, setting the result to zero, and re-arranging terms, we get the necessary condition of T1 as follow: T T R T pD   TeiT   e i t dt   (1   ) Tei (T  R )   e i t dt   cD Te (T )iS   e (t )iS dt      0 R 0   

T T T  hD T  e (T ) (t )i t dt    e (u ) (t )i t du dt   o 0 t  0 

15

 cI c D   te it dt  (1   )   S S

T R

T

(t  R)e it dt  

S S  pI e D   te it dt  (1   )  (t  R)e it dt  = 0.  0  R

(30)

From Theorem 1, we know that (30) has a unique solution T1 . If S  T1 , then PTP1 (T ) is maximized at T1 . Otherwise, PTP1 (T ) is maximized at S. By using the analogous argument, we have the following results.

Theorem 2. (a) PTP2 (T ) is a strictly pseudo-concave function in T, and hence exists a unique maximum solution T2 . (b) If T2  S  R , then PTP2 (T ) subject to T  S  T  R is maximized at T2*  S  R . (c) If S  R  T2  S , then PTP2 (T ) subject to T  S  T  R is maximized at T2*  T2 . (d) If T2  S , then PTP2 (T ) subject to T  S  T  R is maximized at T2*  S . Proof. See Appendix B. By taking the first-order derivative of PTP2 (T ) with respect to T, setting the result to zero, and re-arranging terms, we obtain the necessary condition of T2 as follow: T T R T pD   TeiT   e i t dt   (1   ) Tei (T  R )   e i t dt   cD Te (T )iS   e (t )iS dt      0 R 0   

T T T  hD T  e (T ) (t )i t dt    e (u ) (t )i t du dt   o 0 t  0 

 (1   ) cI c D 

T R S

T S (t  R)e it dt  pI e D   te it dt  (1   )  (t  R)e it dt  = 0.  0  R

(31)

It is clear from Theorem 2 that (31) has a unique solution T2 . If T2  S  R , then PTP2 (T ) is maximized at S  R . If S  R  T2  S , then PTP2 (T ) is maximized at T2 . If S  T2 , then

PTP2 (T ) is maximized at S. Finally, for the case of T  S  R , we have the following similar

16

results for PTP3 (T ) .

Theorem 3. (a) PTP3 (T ) is a strictly pseudo-concave function in T, and hence exists a unique maximum solution T3 . (b) If T3  S  R , then PTP3 (T ) subject to T  R  S is maximized at T3*  T3 . (c) If T3  S  R , then PTP3 (T ) subject to T  R  S is maximized at T3*  S  R . Proof. See Appendix C. Similarly, taking the first-order derivative of PTP3 (T ) with respect to T, setting the result to zero, and re-arranging terms, we have the necessary condition of T3 as follow: T T R T pD   TeiT   e i t dt   (1   ) Tei (T  R )   e i t dt   cD Te (T )iS   e (t )iS dt       0 R 0    

T T T  hD T  e (T ) (t )i t dt    e (u ) (t )i t du dt   o 0 0 t  

 pI e D   te it dt  (1   )   0 R T

T R

(t  R)e it dt  = 0. 

(32)

We know from Theorem 3 that (32) has a unique solution T3 . If T3  S  R , then PTP3 (T ) is maximized at T3 . If T3  S  R , then PTP3 (T ) is maximized at S  R .

4.2. Optimal replenishment cycle time for the case of S  R Similar to the case of S  R , we can easily obtain the following results.

Theorem 4. (a) PTP4 (T ) is a strictly pseudo-concave function in T, and hence exists a unique maximum solution T4 .

17

(b) If S  T4 , then PTP4 (T ) subject to S  T is maximized at T4*  T4 . (c) If S  T4 , then PTP4 (T ) subject to S  T is maximized at T4*  S . Proof. See Appendix D. To find T4 , taking the first-order derivative of PTP4 (T ) with respect to T, setting the result to zero, and re-arranging terms, we derive the necessary condition of T4 as follow: T T R T pD   TeiT   e i t dt   (1   ) Tei (T  R )   e i t dt   cD Te (T )iS   e (t )iS dt      0 R 0   

T T T  hD T  e (T ) (t )i t dt    e (u ) (t )i t du dt   o 0 t  0 

 cI c D   te it dt  (1   )   S R T

T R

(t  R)e it dt   pI e D  te it dt = 0.  0 S

(33)

It is clear from Theorem 4 that (33) has a unique solution T4 . If T4  S , then PTP4 (T ) is maximized at T4 . Otherwise, PTP4 (T ) is maximized at S. By using the analogous argument, we have the following results.

Theorem 5. (a) PTP5 (T ) is a strictly pseudo-concave function in T, and hence exists a unique maximum solution T5 . (b) If S  T5 , then PTP5 (T ) subject to S  T is maximized at T5*  T5 . (c) If S  T5 , then PTP5 (T ) subject to S  T is maximized at T5*  S . Proof. See Appendix E. To get T5 , taking the first-order derivative of PTP5 (T ) with respect to T, setting the result to zero, and re-arranging terms, we derive the necessary condition of T5 as follow: T T R T pD   TeiT   e i t dt   (1   ) Tei (T  R )   e i t dt   cD Te (T )iS   e (t )iS dt      0 R 0   

18 T T T  hD T  e (T ) (t )i t dt    e (u ) (t )i t du dt   o 0 t  0 

 (1   ) cI c D 

T R R

T

(t  R) e it dt  pI e D  te it dt = 0. 0

(34)

It is clear from Theorem 5 that (34) has a unique solution, T5 . If T5  S , then PTP5 (T ) is maximized at T5 . If T5  S , then PTP5 (T ) is maximized at S.

4.3. An algorithm to find the optimal solution Summarizing the results in Theorems 1 -5, we propose the following simple algorithm to find the optimal solution. Algorithm Step 1. Input values of all parameters. Step 2. Compare the values of R and S. If R  S , then go to Step 3. Otherwise, go to Step 5. Step 3. Compute all PTPj (T j* ) , for j = 1, 2, and 3. Step 3.1. Find the unique root T1 in (30). If S  T1 we set T1*  T1 else we set T1*  S . Calculate PTP1 (T1* ) by (13). Step 3.2. Find the unique root T2 in (31). If T2  S  R, we set T2*  S  R . If S  R  T2  S , we set T2*  T2 . If T2  S , we set T2*  S . Calculate PTP2 (T2* ) by (16). Step 3.3. Find the unique root T3 in (32). If T3  S  R, we set T3*  T3 . Otherwise, we set T3*  S  R . Calculate PTP3 (T3* ) by (18).

Step 4. Find the maximum among PTPj (T j* ) for j = 1, 2, and 3, set the optimal solution

T , PTP(T ) accordingly, and then stop. *

*

Step 5. Compute PTPj (T j* ) , for j = 4 and 5. Step 5.1. Find the unique root T4 in (33). If S  T4 we set T4*  T4 else we set T4*  S .

19

Calculate PTP4 (T4* ) by (23). Step 5.2. Find the unique root T5 in (34). If S  T5 we set T5*  T5 else we set T5*  S . Calculate PTP5 (T5* ) by (26). Step 6. Find the maximum among PTPj (T j* ) for j = 4 and 5, set the optimal solution

T , PTP(T ) accordingly, and then stop. *

*

In the next section, we provide some numerical examples to illustrate the theoretical results as well as to gain some managerial insights.

5. Numerical examples Recently, many researchers have adopted the deterioration rate as  (t )  1 /(1  m  t ) with t  m to incorporate the fact that the deterioration rate is 100% near to its expiration date such

as Sarkar (2012), Chen and Teng (2014), Wang et al. (2014), Wu and Chan (2014), Wu et al. (2014), Dye et al. (2014) and Sarkar et al. (2015). Hence, we use the newly adopted deterioration rate to run the first two numerical examples, and then obtain the third numerical example by using the most commonly used Weibull non-decreasing deterioration rate as

 (t )  k t  1 , 0  k  1 ,   1 , and t  0 . Notice that if   1 , then Weibull rate is reduced to a constant deterioration rate.

Example 1. Let’s assume that  (t )  1 /(1  m  t ) ,  = 0.20, c = $10 per unit, D = 1,000 units per year, h = $5 per unit per year, i = 0.04 per dollar per year, m = 1 year, o = $50 per order, p = $20 per unit, R = 0.25 years, S = 0.30 or 0.50 years, I c = 0.05 per dollar per year, and I e = 0.04 per dollar per year. By using the proposed algorithm, we obtain the optimal solutions to S = 0.30 and 0.50 for each case as shown in Table 1.

20

Table 1. Optimal solutions when  (t )  1 /(1  m  t ) and S  R Parameter

Case i

S = 0.30

1

325.04

0.3000

$8,046.44

2

94.14

0.0920

$8,957.47

3

50.64

0.0500

$8,748.20

1

575.36

0.5000

$6,948.97

2

267.06

0.2500

$8,557.28

3

93.64

0.0915

$9,193.27

S = 0.50

Qi*

PTPi*

Ti *

As a result, if S = 0.30, then the optimal solution to the problem is

Q * = 94.14 units, T * = 0.0920 years = 34 days, and PTP * = $8,957.47. Figure 1 demonstrates that PTP* (T ) is a strictly pseudo-concave function in T. Notice that the black point on the curve is the location of the optimal T * . However, if S = 0.50, then the optimal solution to the problem becomes

Q * = 93.64 units, T * = 0.0915 years = 33 days, and PTP * = $9,193.27. Likewise, Figure 2 demonstrates that PTP* (T ) is also a strictly pseudo-concave function in T. Again, the black point on the curve is the location of the optimal T * . In addition, Table 1 reveals that a higher value of the upstream trade credit S causes a higher value of the present value of the annual total profit PTP * but slightly lower values of the order quantity Q * and cycle time T * .

21

Figure 1. Optimal solution for Example 1 with S = 0.30

Figure 2. Optimal solution for Example 1 with S = 0.50

Example 2. Using the same data as those in Example 1 except R = 0.30 years, S = 0.25 years. Again, applying the proposed algorithm, we obtain the optimal solutions to m = 1 year and 2 years for each case as shown in Table 2.

22

Table 2. Optimal solutions when  (t )  1 /(1  m  t ) and S  R Parameter

Case i

m=1

4

267.06

0.2500

$8,233.45

5

94.39

0.0922

$8,854.97

4

261.03

0.2500

$8,482.41

5

102.23

0.1005

$8,940.11

m=2

Qi*

PTPi*

Ti *

Consequently, if m = 1, then the optimal solution to the problem is

Q * = 94.39 units, T * = 0.0922 years = 34 days, and PTP * = $8,854.97. However, if m = 2, then the optimal solution to the problem becomes

Q * = 102.23 units, T * = 0.1005 years = 37 days, and PTP * = $8,940.11. As a result, Table 2 reveals that the higher the maximum lifetime m, the higher the present value of the annual total profit PTP * as well as the order quantity Q * and the replenishment cycle time T * .

Example 3.

Using the same data as those in Example 1, we change the time-varying

deterioration rate to Weibull deterioration rate as  (t )  k t  1 , where k  0.2 ,   3 , and t  1.29 . Similarly, using the proposed algorithm we obtain the optimal solutions to S = 0.30

and 0.50 for each case as shown in Table 3.

Table 3. Optimal solutions when  (t )  k t  1 , where k  0.2 ,   3 , and t  1.29 Parameter

Case i

Qi*

Ti *

PTPi*

23

S = 0.30

S = 0.50

1

300.41

0.3000

$8,898.67

2

124.86

0.1249

$9,232.75

3

50.00

0.0500

$8,874.81

1

503.15

0.5000

$8,483.32

2

250.20

0.2500

$9,246.72

3

122.99

0.1230

$9,463.73

As a result, if S = 0.30, then the optimal solution to the problem is

Q * = 124.86 units, T * = 0.1249 years = 46 days, and PTP * = $9,232.75. We know from Figure 3 that PTP* (T ) is a strictly pseudo-concave function in T. However, if S = 0.50, then the optimal solution to the problem becomes

Q * = 122.99 units, T * = 0.1230 years = 45 days, and PTP * = $9,463.73. Similarly, Figure 4 shows that PTP* (T ) is also a strictly pseudo-concave function in T. Notice that the black point on the curve in Figure 3 or 4 is the location of the optimal T * . Additionally, Table 3 also reveals that a higher value of the upstream trade credit S causes a higher value of the present value of the annual total profit PTP * while slightly lower values of the order quantity Q * and cycle time T * . Based on the numerical results in Examples 1 and 3, we know that these two different time-varying deterioration rates lead to similar conclusions. Hence, we study the sensitivity analysis only based on Example 1.

24

Figure 3. Optimal solution for Example 3 with S = 0.30

Figure 4. Optimal solution for Example 3 with S = 0.50

Example 4. Using the same data as those in Example 1, we study the sensitivity analysis on the optimal solution with respect to each parameter in appropriate unit. The computational results are shown in Table 4.

25

Table 4. Sensitivity analysis on parameters Parameter   0.2   0.3   0.5 c = 10 c = 12 c = 14 D = 1000 D = 1250 D = 1500 h=5 h = 10 h = 15 i = 0.04 i = 0.06 i = 0.08 m=1 m=2 m=3 o = 50 o = 100 o = 150 p = 20 p = 25 p = 30 R = 0.15 R = 0.20 R = 0.25 S = 0.3 S = 0.4 S = 0.5

Ic Ic Ic Ie Ie Ie

 0.05  0.08  0.12  0.04  0.08  0.12

Q* 94.14 94.05 93.87 94.14 89.99 86.37 94.14 105.14 115.06 94.14 78.44 68.63 94.14 92.77 91.47 94.14 101.95 106.61 94.14 133.49 163.71 94.14 93.01 91.92 93.46 93.47 94.14 94.14 93.56 93.64 94.14 93.48 92.63 94.14 92.76 91.39

T* 0.0920 0.0919 0.0917 0.0920 0.0880 0.0845 0.0920 0.0824 0.0753 0.0920 0.0769 0.0675 0.0920 0.0906 0.0894 0.0920 0.1003 0.1052 0.0920 0.1291 0.1572 0.0920 0.0909 0.0899 0.0913 0.0913 0.0920 0.0920 0.0914 0.0915 0.0920 0.0913 0.0905 0.0920 0.0906 0.0893

PTP* $8,957.47 $8,996.91 $9,075.81 $8,957.47 $6,934.82 $4,914.21 $8,957.47 $11,340.20 $13,735.20 $8,957.47 $8,745.25 $8,563.50 $8,957.47 $8,921.08 $8,885.14 $8,957.47 $9,042.21 $9,086.98 $8,957.47 $8,505.50 $8,156.36 $8,957.47 $13,920.90 $18,884.40 $9,081.98 $9,018.54 $8,957.47 $8,957.47 $9,074.49 $9,193.27 $8,957.47 $8,955.23 $8,952.32 $8,957.47 $9,006.55 $9,055.87

26

The sensitivity analysis reveals the following managerial insights: A higher value of c, h, or i causes lower values of Q* , T * , and PTP* . By contrast, a higher value of m causes higher values of Q* , T * , and PTP* . A higher value of  or p causes lower values of Q* and T * , but a higher value of PTP* . The higher the ordering cost o, the higher the order quantity Q* as well as the cycle time T * , while the lower the present value of the total annual profit PTP* . However, the higher the demand rate D, the higher the order quantity Q* as well as the present value of the total annual profit PTP* , meanwhile the shorter the replenishment cycle time T * . It is obvious that the higher the interest charged I c as well as the downstream trade credit R, the lower the present value of the total annual profit PTP* . On the other hand, the higher the interest earned I e as well as the upstream trade credit S, the higher the present value of the total annual profit PTP* .

6. Conclusions and future research To incorporate the facts that the retailer usually uses a downstream partial trade credit as a strategy to reduce default risks with credit-risk customers, and the deterioration rate is time-varying and near 100% particularly close to its expiration date, we have built an EOQ model for deteriorating items in a general framework which includes Goyal (1985), Teng (2002), Huang (2003), Teng and Goyal (2007), Teng (2009), Chen and Teng (2014), and Wu and Chan (2014) as special cases. In addition, to reflect the effects of inflation and time value of money, a discounted cash-flow analysis has been adopted to obtain the present value of the total profit. By applying the existing theoretical results in concave functions, we have demonstrated that the optimal solution for each possible alternative exists uniquely, which simplifies the search for the global optimal to finding a local solution. Then we have used two most commonly used deterioration rate to run several numerical simulations. Finally, results of

27

sensitivity analysis have demonstrated many managerial insights. For instance, an increase in the purchasing cost, the holding cost, or the compounded interest rate reduces the order quantity, the cycle time and the present value of the annual total profit. In contrast, a longer product maximum lifespan elevates the order quantity, the cycle time, and the present value of the annual total profit. In addition, the total profit is very sensitive to the purchasing cost and the selling price. Consequently, to increase profit, retailers must negotiate with suppliers to obtain a low purchase cost and invest in preservation technology (such as refrigeration) to prolong product lifespan. It may be profitable to have a closeout sale at a markdown price as time approaches to the expiration date. For example, all unsold bakery and sushi are half price after 7 PM in many United States and Japanese grocery stores. As a result, one may extend our EOQ model from zero-ending inventory to non-zero-ending inventory. Furthermore, one might consider to allow for shortages and partial backlogging. Also, one could generalize the EOQ model to EPQ model and allow for failure, scrap, and rework. Finally, the proposed model with single player can be extended to an integrated cooperative model for both the retailer and the customer or a non-cooperative Nash or Stackelberg equilibrium solution.

Appendix A. Proof of Theorem 1 From (13), let’s define f1 (T )  pD   e it dt  (1   )   0 R

T R

T

e it dt   cD  e (t )iS dt  hD   0 0 T

 o  cI c D   (T  t ) e it dt  (1   )   S S T

T R



T

t

e (u ) (t )it du dt

(T  R  t ) eit dt  

S S  pI e D   te it dt  (1   )  (t  R)e it dt  .  0  R

and

T

(A1)

28

g1 (T )  T  0 .

(A2)

Taking the first-order and second-order derivatives of f1 (T ) with respect to T, we have: T

f1' (T )  pD[e iT  (1   )e i (T  R ) ]  cDe (T )iS  hD  e (T ) (t )i t dt 0

 cI c D   e it dt  (1   )   S S

T R

T

e it dt  , 

(A3)

and T f1" (T )  ipD[e iT  (1   )e i (T  R ) ]  cD (T )e (T )iS  hD e iT   (T )  e (T ) (t )i t dt    0

 cI c D[eiT  (1   )ei (T R) ]  0.

(A4)

Therefore, PTP1 (T )  f1 (T ) / g1 (T ) is a strictly pseudo-concave function in T, which completes the proof of Part (a) of Theorem 1. The proof of Parts (b) and (c) immediately follows from Part (a) of Theorem 1. This completes the proof of Theorem 1.

Appendix B. Proof of Theorem 2 From (16), we let f 2 (T )  pD   e it dt  (1   )   0 R

T R

T

 o  (1   ) cI c D 

T R S

e it dt   cD  e (t )iS dt  hD   0 0 T

T



T

t

e (u ) (t )it du dt

(T  R  t ) e it dt

T S S  pI e D    te it dt   Teit dt   (1   )  (t  R)e it dt  .   0  T R 

(B1)

and

g2 (T )  T  0 .

(B2)

By taking the first-order and second-order derivatives of f 2 (T ) with respect to T, we have: T

f 2' (T )  pD[e iT  (1   )e i (T  R ) ]  cDe (T )iS  hD  e (T ) (t )i t dt 0

29

 (1   ) cI c D 

T R S

S

e it dt  pI e D  e it dt ,

(B3)

T

and T f 2" (T )  ipD[e iT  (1   )e i (T  R ) ]  cD (T )e (T )iS  hD e iT   (T )  e (T ) (t )i t dt    0

 (1   )cI c Dei (T R )  pI e DeiT  0.

(B4)

Hence, PTP2 (T )  f 2 (T ) / g 2 (T ) is a strictly pseudo-concave function in T, which completes the proof of Part (a) of Theorem 2. The proof of Parts (b) - (d) immediately follows from Part (a) of Theorem 2. This completes the proof of Theorem 2.

Appendix C. Proof of Theorem 3 Again, by using (18), we define f 3 (T )  pD   e it dt  (1   )   0 R T

T R

e it dt   cD  e (t )iS dt  hD   0 0 T

T



T

t

e (u ) (t )it du dt

T S T R S  o  pI e D    te it dt   Teit dt   (1   )  (t  R)e it dt   Teit dt  .   0 T R T  R   

(C1)

and g3 (T )  T  0 .

(C2)

Taking the first-order and second-order derivatives of f 3 (T ) with respect to T, we have: T

f 3' (T )  pD[e iT  (1   )e i (T  R ) ]  cDe (T )iS  hD  e (T ) (t )i t dt 0

S S  pI e D   e it dt  (1   )  e it dt  ,  T  T R

(C3)

and T f 3" (T )  ipD[e iT  (1   )e i (T  R ) ]  cD (T )e (T )iS  hD e iT   (T )  e (T ) (t )i t dt    0

 pI e D[eiT  (1   )ei (T R ) ]  0.

(C4)

30

Therefore, PTP3 (T )  f 3 (T ) / g3 (T ) is a strictly pseudo-concave function in T, which completes the proof of Part (a) of Theorem 3. The proof of Parts (b) and (c) immediately follows from Part (a) of Theorem 3. This completes the proof of Theorem 3.

Appendix D. Proof of Theorem 4 By (23), we define f 4 (T )  pD   e it dt  (1   )   0 R

T R

T

e it dt   cD  e (t )iS dt  hD   0 0 T

T



T

t

e (u ) (t )it du dt

T R T R  o  cI c D   (T  t ) e it dt  (1   )  Teit dt   (T  R  t ) e it dt   S R  S 

S

 pI e D  te it dt .

(D1)

0

and

g4 (T )  T  0 .

(D2)

Taking the first-order and second-order derivatives of f 4 (T ) with respect to T, we have: T

f 4' (T )  pD[e iT  (1   )e i (T  R ) ]  cDe (T )iS  hD  e (T ) (t )i t dt 0

 cI c D   e it dt  (1   )   S S T

T R

e it dt  , 

(D3)

and T f 4" (T )  ipD[e iT  (1   )e i (T  R ) ]  cD (T )e (T )iS  hD e iT   (T )  e (T ) (t )i t dt    0

 cI c D[eiT  (1   )ei (T R) ]  0.

(D4)

As a result, PTP4 (T )  f 4 (T ) / g 4 (T ) is a strictly pseudo-concave function in T, which completes the proof of Part (a) of Theorem 4. The proof of Parts (b) and (c) immediately follows from Part (a) of Theorem 4. This completes the proof of Theorem 4.

31

Appendix E. Proof of Theorem 5 Similarly, using (26), we define f 5 (T )  pD   e it dt  (1   )   0 R

T R

T

e it dt   cD  e (t )iS dt  hD   0 0 T

T



T

t

e (u ) (t )it du dt

R T R  o  (1   )cI c D   Teit dt   (T  R  t ) e it dt  R  S  T S  pI e D   te it dt   Teit dt  . T  0 

(E1)

and g5 (T )  T  0 .

(E2)

By taking the first-order and second-order derivatives of f 5 (T ) with respect to T, we have: T

f 5' (T )  pD[e iT  (1   )e i (T  R ) ]  cDe (T )iS  hD  e (T ) (t )i t dt 0

 (1   ) cI c D 

T R S

S

e it dt  pI e D  e it dt , T

(E3)

and T f 5" (T )  ipD[e iT  (1   )e i (T  R ) ]  cD (T )e (T )iS  hD e iT   (T )  e (T ) (t )i t dt    0

 (1   ) cI c Dei (T  R )  pI e Dei T  0.

(E4)

Consequently, PTP5 (T )  f 5 (T ) / g5 (T ) is a strictly pseudo-concave function in T, which completes the proof of Part (a) of Theorem 5. The proof of Parts (b) and (c) immediately follows from Part (a) of Theorem 5. This completes the proof of Theorem 5.

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