An economic order quantity model with partial ...

Computers & Industrial Engineering 82 (2015) 21–32

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An economic order quantity model with partial backordering and incremental discount Ata Allah Taleizadeh a, Irena Stojkovska b, David W. Pentico c,⇑ a

School of Industrial Engineering, College of Engineering, University of Tehran, Tehran, Iran Department of Mathematics, Faculty of Natural Sciences and Mathematics, Ss. Cyril and Methodius University, Skopje, Macedonia c Palumbo-Donahue School of Business, Duquesne University, Pittsburgh, PA 15282, USA b

a r t i c l e

i n f o

Article history: Received 14 August 2014 Received in revised form 2 December 2014 Accepted 5 January 2015 Available online 14 January 2015

a b s t r a c t Determining an order quantity when quantity discounts are available is a major interest of material managers. A supplier offering quantity discounts is a common strategy to entice the buyers to purchase more. In this paper, EOQ models with incremental discounts and either full or partial backordering are developed for the first time. Numerical examples illustrate the proposed models and solution methods. Ó 2015 Elsevier Ltd. All rights reserved.

Keywords: EOQ Incremental discounts Full backordering Partial backordering

1. Introduction and literature review Since Harris (1913) first published the basic EOQ model, many variations and extensions have been developed. In this paper we combine two of those extensions: partial backordering and incremental quantity discounts. Montgomery, Bazaraa, and Keswani (1973) were the first to develop a model and solution procedure for the basic EOQ with partial backordering (EOQ–PBO) at a constant rate. Others taking somewhat different approaches have appeared since then, including Pentico and Drake (2009), which will be one of the two bases for our work here. In addition, many authors have developed models for the basis EOQ-PBO combined with other situational characteristics, such as Wee (1993) and Abad (2000), both of which included a finite production rate and product deterioration, Sharma and Sadiwala (1997), which included a finite production rate with yield losses and transportation and inspection costs, San José, Sicilia, and García-Laguna (2005), which included models with a non-constant backordering rate, and Taleizadeh, Wee, and Sadjadi (2010), which included production and repair of a number of items on a single machine. Descriptions of all of these models and others may be found in Pentico and Drake (2011). Enticing buyers to purchase more by offering either all-units or incremental quantity discounts is a common strategy. With the ⇑ Corresponding author. E-mail addresses: [email protected] (A.A. Taleizadeh), [email protected], [email protected] (I. Stojkovska), [email protected] (D.W. Pentico). http://dx.doi.org/10.1016/j.cie.2015.01.005 0360-8352/Ó 2015 Elsevier Ltd. All rights reserved.

all-units discount, purchasing a larger quantity results in a lower unit purchasing price for the entire lot, while incremental discounts only apply the lower unit price to units purchased above a specific quantity. So the all-units discount results in the same unit price for every item in the given lot, while the incremental discount can result in multiple unit prices for an item within the same lot (Tersine, 1994). In the following we focus on the research using only an incremental discount or both incremental and all-units discounts together. Since Benton and Park (1996) prepared an extensive survey of the quantity discount literature until 1993, we will describe newer research, along with a short history of incremental discounts and older research which is more related to this paper. The EOQ model with incremental discounts was first discussed by Hadley and Whitin (1963). Tersine and Toelle (1985) presented an algorithm and a numerical example for the incremental discount and examined the methods for determining an optimal order quantity under several types of discount schedules. Güder, Zydiak, and Chaudhry (1994) proposed a heuristic algorithm to determine the order quantities for a multi-product problem with resource limitations, given incremental discounts. Weng (1995) developed different models to determine both all-units and incremental discount policies and investigated the effects of those policies with increasing demand. Chung, Hum, and Kirca (1996) proposed two coordinated replenishment dynamic lot-sizing problems with both incremental and all-units discounts strategies. Lin and Kroll (1997) extended a newsboy problem with both all-units and incremental discounts to maximize the expected profit subject to a constraint that the probability of achieving a target profit level is no less than

22

A.A. Taleizadeh et al. / Computers & Industrial Engineering 82 (2015) 21–32

a predefined risk level. Hu and Munson (2002) investigated a dynamic demand lot-sizing problem when product price schedules offer incremental discounts. Hu, Munson, and Silver (2004) continued their previous work and modified the Silver-Meal heuristic algorithm for dynamic lot sizing under incremental discounts. Rubin and Benton (2003) considered the purchasing decisions facing a buying firm which receives incrementally discounted price schedules for a group of items in the presence of budgets and space limitations. Rieksts, Ventura, Herer, and Sun (2007) proposed a serial inventory system with a constant demand rate and incremental quantity discounts. They showed that an optimal solution is nested and follows a zero-inventory ordering policy. Haksever and Moussourakis (2008) proposed a model and solution method to determine the ordering quantities for multi-product multiconstraint inventory systems from suppliers who offer incremental quantity discounts. Mendoza and Ventura (2008) incorporated quantity discounts, both incremental and all-units, on the purchased units into an EOQ model with transportation costs. Taleizadeh, Niaki, and Hosseini (2009) developed a constrained multi-product bi-objective single-period problem with incremental discounts and fully lost-sale shortages. Ebrahim, Razm, and Haleh (2009) proposed a mathematical model for supplier selection and order lot sizing under a multiple-price discount environment in which different types of discounts including all-unit, incremental, and total business volume are considered. Taleizadeh, Niaki, Aryanezhad, and Fallah-Tafti (2010) developed a multi-products multi-constraints inventory control problem with stochastic period length in which incremental discounts and partial backordering situations are assumed. Munson and Hu (2010) proposed procedures to determine the optimal order quantities and total purchasing and inventory costs when products have either all-units or incremental quantity discount price schedules. Bai and Xu (2011) considered a multi-supplier economic lot-sizing problem in which the retailer replenishes his inventory from several suppliers who may offer either incremental or all-units quantity discounts. Chen and Ho (2011) developed an analysis method for the single-period (newsboy) inventory problem with fuzzy demands and incremental discount. Taleizadeh, Barzinpour, and Wee (2011) discussed a constrained newsboy problem with fuzzy demand, incremental discounts, and lost-sale shortages. Taleizadeh, Niaki, and Nikousokhan (2011) developed a multiconstraint joint-replenishment EOQ model with uncertain unit cost and incremental discounts when shortages are not permitted. Bera, Bhunia, and Maiti (2013) developed a two-storage inventory model for deteriorating items with variable demand and partial backordering. Lee, Kang, Lai, and Hong (2013) developed an integrated model for lot sizing and supplier selection and quantity discounts including both all units and incremental discounts. Archetti, Bertazzi, and Speranza (2014) studied the economic lot-sizing problem with a modified all-unit discount transportation cost function and with incremental discount costs. According to the above mentioned research, it is clear that no researchers have developed an EOQ model with partial backordering and incremental discounts. Taleizadeh and Pentico (2014) developed an EOQ model with partial backordering and all-units discounts. In this paper we develop EOQ models with fully and partially backordered shortages when the supplier offers incremental discounts to the buyer.

Parameters A Fixed cost to place and receive an order b The fraction of shortages that will be backordered Cj The purchasing unit cost at the jth break point D Demand quantity of product per period g The goodwill loss for a unit of lost sales i Holding cost rate per unit time n Number of price breaks qj Lower bound for the order quantity for price j P Selling price of an item p Backorder cost per unit per period p0j The lost sale cost per unit at the jth break point of unit purchasing cost, p0j ¼ P C j þ g > 0 Decision variables B The back ordered quantity F The fraction of demand that will be filled from stock Q The order quantity T The length of an inventory cycle Dependent variables ATC Annual total cost ATP Annual total profit CTC Cyclic total cost CTP Cyclic total profit

2.1. EOQ models with no discount In this section we briefly discus EOQ models with fully or partially backordered shortages when discounts are not available. For the first case, the EOQ models with fully backordered shortages (see Fig. 1), Pentico and Drake (2009) derived the optimal values of F and T as:

p þ iC p rffiffiffiffiffiffiffiffirffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2A p þ iC T ¼ iCD p

F ¼

ð1Þ ð2Þ

For the second case, the EOQ model with partial backordering, Pentico and Drake (2009) showed that the values of F and T that minimize annual total cost are

ð1 bÞp0 þ bpT ðiC þ bpÞT sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2A iC þ bp ½ð1 bÞp0 2 T ¼ iCD bp biC p

F ¼

ð3Þ ð4Þ

only if b is at least as large as a critical value b0 given by Eq. (5)

pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2AiCD b ¼1 Dp0 0

ð5Þ

I DFT D

(1 − F )T

2. Model development In this section we model the defined problem under two different conditions: full backordering and partial backordering. But first we briefly discuss the EOQ model with full or partial backordering when discounts are not assumed. We use the following notation.

FT

t D (1 − F )T

Fig. 1. EOQ model with fully backordered shortages.

23


2.2. EOQ model with incremental discount without shortages Consider an EOQ model in which the supplier offers the volume-based unit purchasing costs shown in Eq. (6) (Q ¼ DT).

8 C1 > > > > < C2 Cj ¼ . > .. > > > : Cn

q1 ¼ 0 6 Q < q2 ATCj ðT; FÞ ¼

q2 6 Q < q3 ...

ð6Þ

M j ¼ X j þ C j DT;

j ¼ 1; 2; . . . ; n;

ð7Þ

where j X qk ðC k1 C k Þ;

A þ X j iX j F 2 iC j DF 2 T pDð1 FÞ2 T þ þ þ þ CjD 2 2 T 2

ð14Þ

Thus, the cost function that has to be minimized has the form

qn 6 Q

where C 1 > C 2 > > C n and q1 ¼ 0 < q2 < < qn . The purchasing cost per order is:

Xj ¼

where C 0j is the purchasing cost per unit given by Formula (10). Substituting Formula (10) into (13) and dividing by T we get the annual total cost for ordering the quantity from the interval ½qj ; qjþ1 Þ:

j ¼ 2; 3; . . . ; n and X 1 ¼ 0;

ð8Þ

8 ATC1 ðT; FÞ ; 0 < DT < q2 > > > > < ATC2 ðT; FÞ ; q2 6 DT < q3 ATCðT; FÞ ¼ .. > > . > > : ATCn ðT; FÞ ; qn 6 DT

ð15Þ

The minimization is performed over the region T > 0; 0 6 F 6 1 (see Fig. 1).

k¼2

Proposition 1. The function ATCðT; FÞ, defined by (14) and (15), is continuous.

From the definitions of C j ; qj and X j , we have that:

X j P 0;

j ¼ 1; 2; . . . ; n:

ð9Þ Proof. See Appendix B. h

Then the purchasing cost per unit is (Tersine, 1994)

C 0j ¼

Mj Xj ¼ þ Cj: DT DT

ð10Þ

and the optimal cycle length for ordering from the quantity from the jth interval ½qj ; qjþ1 Þ is

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðA þ X j Þ : Tj ¼ iC j D

ð11Þ

The optimal order quantity is Q j ¼ DT j , with minimal annual total cost of

ATCj

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi iX j ¼ 2DiC j ðA þ X j Þ þ þ C j D: 2

then the

optimal

min

T>0;06F61

ATCðT; FÞ ¼ min

16j6n

min ATCj ðT; FÞ

ðT;FÞ2Xj

acceptable

order

quantity is

Q j ¼ qjþ1 . For the latter two cases, the corresponding annual total cost, calculated using Eq. (A1) in Appendix A, is the new optimal annual total cost ATCj . Finally, ATCj for j ¼ 1; 2; . . . ; n are compared to find the minimal value among ATCj ; j ¼ 1; 2; . . . ; n, which will be the optimal annual cost for the EOQ model with incremental discount, and the corresponding Q j will be the optimal order quantity for the EOQ model with incremental discount. This solution procedure is justified, because we can prove that if Q j P qjþ1 , then there is an order quantity which costs less to order than Q j does (see Appendix A). In the following sub-sections we model the defined problem under two different conditions: full backordering and partial backordering, which are developed in Sections 2.3 and 2.4 respectively. 2.3. EOQ model with full backordering and incremental discounts We will consider an EOQ model in which all shortages will be backordered and the supplier offers incremental volume-based unit purchasing cost discounts. Then, according to Fig. 1, the cyclic total cost for ordering the quantity from the interval ½qj ; qjþ1 Þ is Holding Cost

Backordering Cost

zfflfflfflfflfflffl}|fflfflfflfflfflffl{ zfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflffl{ Purchasing Cost Fixed Cost zffl}|ffl{ z}|{ iC 0j DF 2 T 2 pDð1 FÞ2 T 2 0 CTC j ðT; FÞ ¼ A þ C j DT þ þ 2 2

ð13Þ

ð16Þ

where

X1 ¼ fðT; FÞj 0 < T 6 q2 =D; 0 6 F 6 1g; Xj ¼ fðT; FÞj qj =D 6 T 6 qjþ1 =D; 0 6 F 6 1g; j ¼ 2; 3; . . . ; n 1; and Xn ¼ fðT; FÞj qn =D 6 T; 0 6 F 6 1g: ð16aÞ

ð12Þ

The optimal order quantity Q j is acceptable if qj 6 Q j < qjþ1 . If Q j < qj , then the optimal acceptable order quantity is Q j ¼ qj . If Q j P qjþ1 ,

As a consequence of Proposition 1, the minimization problem can be transformed into

Note that the sign < is changed into 6 in the upper bounds, which is allowed by the continuity of ATCðT; FÞ. In what follows we will use the notation T j and F j for T and F, respectively, when we are minimizing the annual total cost for ordering the quantity from the interval ½qj ; qjþ1 Þ defined by Eq. (14). To solve the jth subproblem in (16), i.e. the problem

min ATCj ðT j ; F j Þ;

ð17Þ

ðT;FÞ2Xj

we first find the first partial derivatives of ATCj ðT j ; F j Þ with respect to T j and F j . 2

@ATCj A þ X j iC j DF j pDð1 F j Þ2 ðT j ; F j Þ ¼ þ þ ; @T j 2 2 T 2j

ð18Þ

@ATCj ðT j ; F j Þ ¼ iX j F j þ iC j DF j T j pDð1 F j ÞT j : @F j

ð19Þ

Setting the first derivatives (18) and (19) equal to 0, and solving the corresponding system with respect to T j and F j , remembering that T j > 0, we get

Tj ¼

T j ðF j Þ

¼

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðA þ X j Þ

D½iC j F 2j þ pð1 F j Þ2 iX j F j þ D iC j F j pð1 F j Þ T j ¼ 0:

ð20Þ ð21Þ

To find the solution of the system (20) and (21), we substitute (20) in (21), and obtain an equation with respect to F j :

iX j F j þ D iC j F j pð1 F j Þ T j ðF j Þ ¼ 0;

ð22Þ

which can be solved numerically with a solver like MatLab, Mathematica, or Excel Solver. Let us denote the solution of (22)

24


by F j . If we denote the left side of Eq. (22) by wðF j Þ ¼ iX j F j þ D iC j F j pð1 F j Þ T j ðF j Þ, then from wð1Þ ¼ iX j þ DiC j T j ð1Þ > 0; wð0Þ ¼ DpT j ð0Þ < 0, and wðF j Þ being continuous, we have that there exists a solution F j of (22) in the interval [0, 1]. So, we can formulate the following proposition.

I

DFT D

Proposition 2. There exists a solution F j of Eq. (22), for which 0 6 F j 6 1. Then, from Eq. (20) we have:

T j ¼ T j ðF j Þ ¼

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðA þ X j Þ 2

D½iC j F 2 j þ pð1 F j Þ

(1 − F )T

ð23Þ

:

Proof. See Appendix C. h Note that wðF j Þ is a monotone nondecreasing function since @wðF j Þ=@F j ¼ @ 2 /ðF j Þ=@F 2j > 0 (see (C6) in Appendix C). Thus the solution F j of Eq. (22) is the unique solution in the interval [0, 1]. If ðT j ; F j Þ 2 Xj , then ðT j ; F j Þ is the unique global minimizer of the function ATCj ðT j ; F j Þ on the set Xj . From Proposition 2 we have that 0 6 F j 6 1, but for T j it might not be always true that qj =D 6 T j 6 qjþ1 =D. If T j < qj =D, then the global solution ðT j ; F j Þ of Subproblem (17) lies on the lower boundary of T j , i.e. T j ¼ qj =D, and F j ¼ pqj =ððiC j þ pÞqj þ iX j Þ. If T j > qjþ1 =D, then the global solution ðT j ; F j Þ of Subproblem (17) lies on

the

upper

boundary

of

Tj,

i.e.

T j ¼ qjþ1 =D,

and

F j

¼ pqjþ1 =ððiC j þ pÞqjþ1 þ iX j Þ. This is true because of the convexity of ATCj ðT j ; F j Þ with respect to T j (see (C1) in Appendix C), and minimizing ATCj ðqj =D; F j Þ and ATCj ðqjþ1 =D; F j Þ, respectively, in order to obtain the last two values for F j . From the above discussion we can conclude that the global optimal solution ðT ; F Þ that minimizes the annual total cost given in Eq. (15) is the pair ðT j ; F j Þ for which the corresponding ATCj ðT j ; F j Þ is minimal over all j = 1, 2, . . . , n. That is,

n o ðT ; F Þ ¼ arg min ATCj ðT j ; F j Þ :

ð24Þ

16j6n

t

D (1 − β ) (1 − F )T

βD

If ðT j ; F j Þ 2 Xj , then it is the optimal solution of the Subproblem (17). The following proposition stands. Proving the global optimality of ðT j ; F j Þ can be also done as in Stojkovska (2013). Proposition 3. Assume that ðT j ; F j Þ 2 Xj , where F j is the solution of (22), T j is defined by (23), and Xj is the feasible region of Subproblem (17). Then ðT j ; F j Þ is the global optimal solution of Subproblem (17).

β D (1 − F )T

FT

Fig. 2. EOQ model with partially backordered shortages.

2. Find the optimal solution as the pair ðT j ; F j Þ for which the corresponding ATCj ðT j ; F j Þ is minimal over all j = 1, 2, . . . , n. 3. Calculate Q ¼ DT and B ¼ Dð1 F ÞT . 2.4. The EOQ with incremental discounts and partial backordering Unlike the full backordering model in which we minimized the annual total cost to obtain the optimal solutions, in the partial backordering model, in order to facilitate reaching the optimal solution using the approach in Pentico and Drake (2009), we will first model the profit function and then by maximizing it we will get the optimal solutions. According to Fig. 2, in which it is clear that the order quantity will be Q ¼ DT½F þ bð1 FÞ, the unit purchasing cost becomes

C 0j ¼

Xj þ Cj D½F þ bð1 FÞT

ð25Þ

where X j is given by Eq. (8). Then the cyclic total profit for ordering the quantity from the interval ½qj ; qjþ1 Þ is Revenue

zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{ CTPj ðF;TÞ ¼ PD½F þ bð1 FÞT 8 9 Holding Cost > zfflfflfflfflfflffl}|fflfflfflfflfflffl{ > > > Purchasing Cost > > Fixed Cost > 0 2 2 zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{ iC DF T > > > z}|{ > > j 0 > > > > < A þC j D½F þ bð1 FÞT þ = 2 2 2 > > > > > > þ pbDð1 FÞ T þgð1 bÞð1 FÞDT > > > > > > > > |fflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl {zfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl } 2 > |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} > : ; Backordering Cost

Lost Sale Cost

ð26Þ

Substituting (25) in (26) and dividing by T gives the average annual profit for ordering the quantity from the interval ½qj ; qjþ1 Þ:

We have the following solution procedure for EOQ model with incremental discount and full backordering. Solution procedure for the EOQ model with incremental discounts and full backordering

ATPj ðF; TÞ ¼ PD½F þ bð1 FÞ 9 8 2 = < AþXj þ C j D½F þ bð1 FÞ þ iX j F T 2ðFþbð1FÞÞ 2 ; : iC j DF T pbDð1FÞ2 T þ 2 þ þ gð1 bÞð1 FÞD 2

1. For j = 1, 2, . . . , n: 1.1. Solve (22) using some numerical procedure, to obtain F j . Calculate T j from (23). 1.2. If qj =D 6 T j 6 qjþ1 =D (with q1 ¼ 0 and qnþ1 ¼ 1), then ðT j ; F j Þ is an acceptable solution (or Q j ¼ DT j is an acceptable order quantity). 1.3. If T j < qj =D and j 2 f2; . . . ; ng, then calculate the new ðT j ; F j Þ using T j ¼ qj =D and F j ¼ pqj =ððiC j þ pÞqj þ iX j Þ 1.4. If T j > qjþ1 =D and j 2 f1; . . . ; n 1g, then calculate the new ðT j ; F j Þ using T j ¼ qjþ1 =D and F j ¼ pqjþ1 =ððiC j þ pÞqjþ1 þ iX j Þ 1.5. Calculate ATCj ðT j ; F j Þ.

After some algebraic transformations and letting p0j ¼ P C j þ g, we have:

ATPj ðF; TÞ ¼ P C j D ( A þ Xj iX j F 2 iC j DF 2 T þ þ 2 T 2ð1 ð1 bÞð1 FÞÞ ) pbDð1 FÞ2 T þ þ p0j ð1 bÞð1 FÞD 2

ð27Þ

ð28Þ

The function to be maximized over the region T > 0; 0 6 F 6 1, has the form

25


8 ATP1 ðT; FÞ ; 0 < DTðF þ bð1 FÞÞ < q2 > > > > < ATP2 ðT; FÞ ; q2 6 DTðF þ bð1 FÞÞ < q3 ATPðT; FÞ ¼ .. > > . > > : ATPn ðT; FÞ ; qn 6 DTðF þ bð1 FÞÞ

which can be solved numerically with a solver like MatLab, Mathematica, or Excel Solver. Let us denote the solution of (37)

ð29Þ

The function ATPðT; FÞ defined by (28) and (29), is continuous (see Appendix D). Thus, the maximization problem can be written as

( max

T>0;06F61

ATPðT; FÞ ¼ max 16j6n

) max ATPj ðT; FÞ

iX F

j j by F j . If we denote the left side of Eq. (37) by nðF j Þ ¼ F þbð1F j jÞ 2

iX j F j ð1bÞ 0 then we 2 þ D iC j F j pbð1 F j Þ T j ðF j Þ pj ð1 bÞD, 2ðF j þbð1F j ÞÞ have nð0Þ ¼ DpbT j ð0Þ p0j ð1 bÞD < 0, since p0j ¼ P C j þ g > 0, qffiffiffiffiffiffiffiffiffiffiffiffi iX ð1bÞ 2ðAþX j Þ and nð1Þ ¼ iX j j 2 þ DiC j T j ð1Þ p0j ð1 bÞD ¼ iX j þ DiC j DiC j

iX j þ p0j D ð1 bÞ > 0, only if the condition; 2

ð30Þ

~ ðT;FÞ2X j

b>1

iX j þ

iX j 2

with

~ j ¼ fðT; FÞj q 6 DTðF þ bð1 FÞÞ 6 q ; T > 0; 0 6 F 6 1g; X j jþ1 j ¼ 2; 3; . . . ; n 1; ~ n ¼ fðT; FÞj q 6 DTðF þ bð1 FÞÞ; T > 0; 0 6 F 6 1g: and X n Note that the sign < is changed into 6 in the upper bounds for the order quantity, which is allowed by the continuity of ATPðT; FÞ (see Appendix D). Since maximizing ATPj ðT; FÞ is equivalent to minimizing the function

A þ Xj iX j F 2 iC j DF 2 T uj ðT; FÞ ¼ þ þ 2 T 2ðF þ bð1 FÞÞ 2

þ p0j ð1 bÞð1 FÞD;

ð31Þ

Problem (30) is transformed into

( max

T>0;06F61

)

ATPðT; FÞ ¼ max ðP C j ÞD min uj ðT; FÞ : ~ ðT;FÞ2X j

16j6n

ð32Þ

As in Section 2.3, we will use T j and F j for T and F respectively when we are minimizing the function uj ðT; FÞ defined by Eq. (31). In order to minimize the function uj ðT j ; F j Þ, we first take the first partial derivatives: 2 @ uj A þ X j iC j DF j pbDð1 F j Þ2 ðT j ; F j Þ ¼ þ þ ; @T j 2 2 T 2j

2iX j F j F j þ bð1 F j Þ iX j F 2j ð1 bÞ @ uj ðT j ; F j Þ ¼

2 @F j 2 F j þ bð1 F j Þ h i þ iC j F j T j pbð1 F j ÞT j p0j ð1 bÞ D:

ð33Þ

T j ¼ T j ðF j Þ ¼

D½iC j F 2j þ pbð1 F j Þ2

;

ð34Þ

ð35Þ

iX j F 2j ð1 bÞ iX j F j

þ D iC j F j pbð1 F j Þ T j F j þ bð1 F j Þ 2 F j þ bð1 F j Þ 2 p0j ð1 bÞD ¼ 0:

ð36Þ

Substituting (35) into (36), we obtain an equation with respect to Fj:

iX j F 2j ð1 bÞ iX j F j

þ D iC j F j pbð1 F j Þ T j ðF j Þ F j þ bð1 F j Þ 2 F j þ bð1 F j Þ 2 p0j ð1 bÞD ¼ 0;

ð38Þ

For F j given by the solution of Eq. (37), we define T j by

T j

¼

T j ðF j Þ

¼

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðA þ X j Þ 2

D½iC j F 2 j þ pbð1 F j Þ

ð39Þ

:

Thus, if Condition (38) is satisfied, ðT j ; F j Þ is the solution of the system (35) and (36) for which 0 6 F j 6 1. Note that, when b0j < 0, it is clear that b > b0j (since b P 0) and Condition (38) is satisfied. Also note that if b ¼ b0j , where b0j is defined by (38), then nð1Þ ¼ 0, and F j ¼ 1 is the solution of Eq. (37). For F j ¼ 1, from pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (39) we have that T j ¼ 2ðA þ X j Þ=ðDiC j Þ which is the optimal cycle length for the cost of C j in the EOQ model with incremental discount without shortages (see Section 2.2, Eq. (11)). We can also prove that if Condition (38) is satisfied and ~ j , then it is the global minimizer of the function ðT ; F Þ 2 X j

j

~ j . The following proposition stands. uj ðT j ; F j Þ over the domain X Proving the global optimality of ðT j ; F j Þ can be also done as in Stojkovska (2013). Proposition 5. Assume that Condition (38) is satisfied and ~ j , where F is the solution of (37), T is defined by (39), ðT j ; F j Þ 2 X j j ~ and Xj is the feasible region defined by (30a). Then ðT ; F Þ is the j

j

~ j. global minimizer of the function uj ðT j ; F j Þ over the domain X Proof. See Appendix E. h

Setting the first derivatives (33) and (34) equal to 0, and solving the corresponding system with respect to T j and F j , remembering that T j > 0, we have:

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðA þ X j Þ

¼ b0j

Proposition 4. If Condition (38) is satisfied, then there exists a solution F j of Eq. (37), for which 0 6 F j 6 1.

ð30aÞ

pbDð1 FÞ2 T

þ p0j D

is satisfied. Then, because of the continuity of the function nðF j Þ, we can formulate the following proposition.

~ 1 ¼ fðT; FÞj DTðF þ bð1 FÞÞ 6 q ; T > 0; 0 6 F 6 1g; X 2

þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðA þ X j ÞDiC j

ð37Þ

As in the full backordering case, note that nðF j Þ is a monotone non-decreasing function since @nðF j Þ=@F j ¼ @ 2 gðF j Þ=@F 2j > 0 (see (E6) in Appendix E). Thus, if Condition (38) is satisfied, the solution F j of Eq. (37) is the unique solution in the interval [0, 1], and if ~ j , then ðT ; F Þ is the unique global minimizer of the ðT ; F Þ 2 X j

j

j

j

~ j . If Condition (38) is not satisfied, function uj ðT j ; F j Þ on the set X then 0 6 b < b0j , which is equivalent to nð1Þ < 0, and from nðF j Þ being a monotonic function, we have that there is no solution of Eq. (37) in the interval [0, 1]; consequently, partial backordering cannot be optimal. So, in this case (0 6 b < b0j ), the optimal decision is either meeting all demand (EOQ model with incremental discount and no shortages, Section 2.2) with the optimal value of pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the cycle length T j ¼ 2ðA þ X j Þ=ðDiC j Þ and the optimal value of the fill rate F j ¼ 1, or losing all sales with T j ¼ þ1 and F j ¼ 0. From Proposition 4 and the above discussion about the values for T j and F j , we always have T j > 0 and 0 6 F j 6 1 when Condition (38) is met, but it might not be always true that qj 6 DT j ðF j þ bð1 F j ÞÞ 6 qjþ1 , in which case the pair ðT j ; F j Þ would

26


be infeasible and cannot be the minimizer of the function uj ðT j ; F j Þ ~ j . We have the following proposition. on the set X Proposition 6. Assume that Condition (38) is satisfied, but ~ j , where F is the solution of (37), T is defined by (39), ðT j ; F j Þ R X j j ~ j is the feasible region defined by (30a). Then the minimizer and X ~ j is defined by: ðT j ; F j Þ of the function uj ðT j ; F j Þ over the domain X (i) If DT j ðF j þ bð1 F j ÞÞ < qj and one of the following conditions are met 0 j

p

AþX j 0 A þ Xj 1 þ b 2D pj qj P 0 and > 1b qj iX j þ iC j qj

ð40aÞ

or

2D A þ Xj 1þb p0j < 0 and > bð1 bÞ qj

AþX j qj

p0j

ð40bÞ

pqj

then F j is the solution of

! ! ðA þ X j Þ iX j 0 pj ð1 bÞD þ F j þ iC j F j pbð1 F j Þ DT j ðF j Þ qj qj ! iX j 2 F j þ iC j F 2j þ pbð1 F j Þ2 qj

D2 ð1 bÞ 2 T j ðF j Þ ¼ 0 2qj

q

ð40cÞ

q

j j where T j ðF j Þ ¼ DðF j þbð1F , and T j ¼ DðF þbð1F . ÞÞ j ÞÞ

DT j ðF j

(ii) If þ bð1 tions are met

p0j

F j ÞÞ

j

AþX j

A þ Xj 1 þ b 2D pj qjþ1 P 0 and > iX j þ iC j qjþ1 1b qjþ1

or

p0j

j

> qjþ1 and one of the following condi-

2D A þ Xj 1þb < 0 and > bð1 bÞ qjþ1

0

AþX j qjþ1

ð41aÞ

p0j

ð41bÞ

then F j is the solution of

! ! ðA þ X j Þ iX j 0 pj ð1 bÞD þ F j þ iC j F j pbð1 F j Þ DT j ðF j Þ qjþ1 qjþ1 ! iX j 2 2 2 F þ iC j F j þ pbð1 F j Þ qjþ1 j

D2 ð1 bÞ 2 T j ðF j Þ ¼ 0 2qjþ1

q

ð41cÞ

q

jþ1 jþ1 where T j ðF j Þ ¼ DðF j þbð1F , and T j ¼ DðF þbð1F . ÞÞ j ÞÞ j

j

Proof. See Appendix F. h When b < b0j and b0j P 0, as we saw earlier, the optimal decision is meeting all demand from the EOQ model with incremental discount and no shortages, i.e., the minimizer of uj ðT j ; F j Þ lies on the pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi boundary F j ¼ 1, so T j ¼ 2ðA þ X j Þ=ðDiC j Þ and F j ¼ 1. From the convexity of

hðT j Þ ¼ ATCj ðT j ; 1Þ ¼

A þ X j iX j iC j DT j þ þ þ Cj D Tj 2 2

the corresponding optimal profit is PD hðT j Þ. Note that, in the second case, we can exclude the point ðT j ; F j Þ from the set of candidates for the optimal solution, since the corresponding order quantity is not the overall optimal order quantity (see Appendix A). ~ j , but When Condition (38) is met (b P b0j ) and ðT j ; F j Þ R X neither (40a) or (40b) nor (41a) or (41b) is satisfied, this means that partial backordering cannot be optimal, so the optimal decision is meeting all demand from the EOQ model with incremental discount and no shortages (see Section 2.2) or losing all sales. In this case we should search for the optimal decision as in the b < b0j and b0j P 0 case. We can conclude that the global optimal solution ðT ; F Þ that maximizes the annual total profit, Function (29), is as one of the points ðT j ; F j Þ for which the corresponding profit is maximal over all j = 1, 2, . . . , n. The following solution procedure for the EOQ model with incremental discounts and partial backordering summarizes the details of the preceding theoretical results and their implications for the optimal solution. Solution procedure for the EOQ model with incremental discounts and partial backordering 1. For j ¼ 1; 2; . . . ; n: 1.1. Calculate b0j according to Formula (38). 1.2. If b P b0j P 0 or b0j < 0, solve Eq. (37) to obtain F j and calculate T j according to Formula (39). 1.2.1. If qj 6 DT j ðF j þ bð1 F j ÞÞ 6 qjþ1 (with q1 ¼ 0 and qnþ1 ¼ 1), then ðT j ; F j Þ is an acceptable solution (or Q j ¼ DT j ðF j þ bð1 F j ÞÞ is acceptable). Calculate the profit ATPj ðT j ; F j Þ, using Formula (28). Compare the profit ATPj ðT j ; F j Þ with the profit from not stocking, p0j D, and take the higher profit. If the profit from not stocking is higher, set T j ¼ þ1 and F j ¼ 0. 1.2.2. If DT j ðF j þ bð1 F j ÞÞ < qj and j 2 f2; . . . ; ng, then

pqjþ1

then the minimizer lies on the upper boundary T j ¼ qj1 =D, and

ð42Þ

(see (C1) in Appendix C for F j ¼ 1), if DT j ðF j þ bð1 F j ÞÞ ¼ DT j ð1 þ bð1 1ÞÞ ¼ DT j < qj , then the minimizer lies on the lower boundary T j ¼ qj =D, and the corresponding optimal profit is PD hðT j Þ. If DT j ðF j þ bð1 F j ÞÞ ¼ DT j ð1 þ bð1 1ÞÞ ¼ DT j > qjþ1 ,

(1.2.2.i) If one of the Conditions (40a) or (40b) is satisfied, find F j as the solution of Eq. (40c), and set T j ¼ qj =ðDðF j þ bð1 F j ÞÞÞ. Calculate the profit ATPj ðT j ; F j Þ using Formula (28). pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (1.2.2.ii) Set F j ¼ 1, and calculate T j ¼ 2ðA þ X j Þ=ðDiC j Þ. If T j < qj =D, set T j ¼ qj =D, and if T j > qjþ1 =D, set T j ¼ qjþ1 =D. Calculate the profit PD hðT j Þ, where hðT j Þ is given by Formula (42). (1.2.2.iii) Calculate the profit from not stocking, p0j D, and set T j ¼ þ1 and F j ¼ 0. (1.2.2.iv) Compare the profits from (1.2.2i), (1.2.2.ii), (1.2.2.iii) to determine the optimal (highest) profit if DT j ðF j þ bð1 F j ÞÞ < qj , and set T j ¼ T j and F j ¼ F j for the optimal solution. 1.2.3. If DT j ðF j þ bð1 F j ÞÞ > qjþ1 and j 2 f1; . . . ; n 1g (1.2.3.i) If one of the Conditions (41a) or (41b) is satisfied, find F j as the solution of Eq. (41c), and set T j ¼ qjþ1 =ðDðF j þ bð1 F j ÞÞÞ. Calculate the profit ATPj ðT j ; F j Þ, using Formula (28). pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (1.2.3.ii) Set F j ¼ 1, and calculate T j ¼ 2ðA þ X j Þ=ðDiC j Þ. If T j < qj =D, set T j ¼ qj =D, and if T j > qjþ1 =D, set T j ¼ qjþ1 =D. Calculate the profit PD hðT j Þ, where hðT j Þ is given by Formula (42). (1.2.3.iii) Calculate the profit from not stocking, p0j D, and set T j ¼ þ1 and F j ¼ 0.

27


(1.2.3.iv) Compare the profits from (1.2.3i), (1.2.3.ii), (1.2.3.iii) to determine the optimal (highest) profit if DT j ðF j þ bð1 F j ÞÞ > qjþ1 , and set T j ¼ T j and F j ¼ F j for the optimal solution. b0j ,

F j

T j

1.3. If 06b< set ¼1 and calculate ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðA þ X j Þ=ðDiC j Þ. 1.3.1. If qj =D 6 T j 6 qjþ1 =D (with q1 ¼ 0 and qnþ1 ¼ 1), then ðT j ; F j Þ is acceptable (or Q j ¼ DT j is acceptable). Calculate the profit PD - hðT j Þ ¼ PD C j D pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

iX j =2 þ 2ðA þ X j ÞDiC j , where hðT j Þ is given by Formula (42). 1.3.2. If T j < qj =D and j 2 f2; . . . ; ng, then set T j ¼ qj =D and calculate the profit PD hðT j Þ, where hðT j Þ is given by Formula (42). Compare the profit with the profit from not stocking, p0j D, and take the higher profit, If the profit from not stocking is higher, set T j ¼ þ1 and F j ¼ 0. 1.3.3. If T j > qjþ1 =D and j 2 f1; . . . ; n 1g, set T j ¼ qjþ1 =D and calculate the profit PD hðT j Þ, where hðT j Þ is given by Formula (42). Compare the profit with the profit from not stocking, p0j D, and take the higher profit. If the profit from not stocking is higher, set T j ¼ þ1 and F j ¼ 0. 2. Identify the maximum profit; the point ðT j ; F j Þ at which it is attained is the global optimal solution ðT ; F Þ. 3. If the optimal policy is partial backordering, calculate Q ¼ DT ðF þ bð1 F ÞÞ and B ¼ bDð1 F ÞT . If the optimal policy is meeting all demand with incremental discount, calculate Q ¼ DT . If the optimal policy is losing all sales, then Q ¼ 0. 3. Numerical examples We give numerical examples for both the full and partial backordering models with incremental discounts proposed in the above sections. The solution procedures are coded in Wolfram Mathematica, using built-in functions to solve nonlinear equations. Example 1 (EOQ model with incremental discounts and full backordering). We will use the values of all common parameters from the numerical example for Taleizadeh and Pentico’s (2014) all-units discount model: P = $9/unit, D = 200 units/period, i = 0.3/period, p = $2/unit/period, C ¼ ðC 1 ; C 2 ; C 3 Þ = $(6, 5, 4)/unit, g = $2/unit, q ¼ ðq1 ; q2 ; q3 Þ = (0, 75, 150) units, p0 ¼ ðp01 ; p02 ; p03 Þ = $(5, 6, 7)/unit. We set the fixed order cost A to $30/order. Values for T j ; F j and ATCj ðT j ; F j Þ for each j ¼ 1; 2; 3, are displayed in Table 1. Rows that are noted as ‘‘correction (j)’’, display the values of T j and F j after correcting T j for not being in the interval qj =D 6 T j 6 qjþ1 =D. Then, ATCj ðT j ; F j Þ is calculated for those corrected values for T j and F j . According to Table 1, the annual total cost is minimized for j = 3, so the overall optimal solution is T ¼ T 3 ¼ 1:83941; F ¼ F 3 ¼ 0:591107, with the optimal cost ATCðT ; F Þ ¼ ATC3 ðT 3 ; F 3 Þ ¼ 1089:06. The optimal order quantity is Q ¼ DT ¼ 367:881, with the maximum backordered quantity B ¼ Dð1 F ÞT ¼ 150:424.

Example 2 (EOQ model with incremental discounts and partial backordering). We use the same values for the parameters as in Example 1, and we will vary the backordering parameter b ¼ 0:95; 0:80; 0:50. The results are displayed in Table 2. Rows that are noted as ‘‘PBO correction (j)’’, display the values of T j and F j after correcting Q j ¼ DT j ðF j þ bð1 F j ÞÞ for not being into the

Table 1 Results for EOQ model with full backordering and incremental discounts (Example 1). j

T j

F j

1 Correction (1) 2 Correction (2) 3

0.562731 > q2 /D = 0.375 0.375 1.10621 > q3 /D = 0.75 0.75 1.83941

0.526316 0.526316 0.555294 0.547945 0.591107

ATCj ðT j ; F j Þ 1315.53 1207.81 1089.06

interval ½qj ; qjþ1 Þ, and if the correction is possible, i.e., if Conditions (40a) or (40b) or Conditions (41a) or (41b) is satisfied. Then, ‘‘profit (j)’’ is calculated for those corrected values for T j and F j . If corrections of the PBO model are done, then the row indicated with ‘‘NBO model (j)’’ is filled, and if Q j ¼ DT j from NBO model is not in the interval ½qj ; qjþ1 Þ, then the ‘‘NBO correction (j)’’ is done, and ‘‘profit (j)’’ is calculated for those corrected values for T j and F j . For each j, the profit from not stocking is calculated and is displayed in the row ‘‘not stocking (j)’’. The highest profit is taken as the over-all profit. According to Table 2, when b ¼ 0:95, the annual profit is maximized for j = 3, under the partial backordering policy, with T ¼ T 3 ¼ 1:85987 and F ¼ F 3 ¼ 0:636287, with the optimal profit ATPðT ; F Þ ¼ ATP3 ðT 3 ; F 3 Þ ¼ 686:411. The optimal order quantity is Q ¼ DT ðF þ bð1 F ÞÞ ¼ 365:21 and the maximum backordered quantity is B ¼ bDð1 F ÞT ¼ 128:528.

For b ¼ 0:80, the annual profit is maximized for j = 3, under the partial backordering policy, with T ¼ T 3 ¼ 1:74378 and F ¼ F 3 ¼ 0:805105, with the optimal profit ATPðT ; F Þ ¼ ATP3 ðT 3 ; F 3 Þ ¼ 630:197. The optimal order quantity is Q ¼ DT ðF þ bð1 F ÞÞ ¼ 335:161, and the maximum backordered quantity is B ¼ bDð1 F ÞT ¼ 54:3765. For b ¼ 0:50, the annual profit is maximized for j = 3, under the policy of meeting all demand, with T ¼ T 3 ¼ 1:45774; F ¼ F 3 ¼ 1, with the optimal profit ATPðT ; F Þ ¼ 616:393. The optimal order quantity is Q ¼ DT ¼ 291:548. All examples showed that if the jth optimal quantity is not in the jth interval ½qj ; qjþ1 Þ, then it cannot be the overall optimal quantity, even if it is corrected to the relevant interval endpoint. This was proved for the EOQ model with incremental discount and no backordering (see Appendix A). It is left to be proven that this might be also true for the proposed EOQ models with incremental discount – full and partial backordering respectively. From the examples we can see that keeping all parameters fixed and by varying the backordering rate, the total profit decreases when the backordering rate is decreasing. 4. Sensitivity analysis There are at least two possible objectives for sensitivity analysis: 1. Assess the relative impact of mis-estimation of different model parameters on the model’s performance. 2. Assess the relative importance of the different model parameters in determining the values of the decision variables and the performance function. 4.1. Study plan Both objectives can be addressed by changing a single parameter’s value by given percentages, repeating the analysis for each parameter of interest, using the same percentage changes.

28


Table 2 Results for EOQ model with partial backordering and incremental discounts (Example 2). j b = 0.95 1 PBO correction (1) NBO model (1) NBO correction (1) Not stocking (1) 2 PBO correction (2) NBO model (2) NBO correction (2) Not stocking (2) 3 Not stocking (3) b = 0.80 1 NBO model (1) NBO correction (1) Not stocking (1) 2 PBO correction (2) NBO model (2) NBO correction (2) Not stocking (2) 3 Not stocking (3) b = 0.50 1 NBO model (1) NBO correction (1) Not stocking (1) 2 NBO model (2) NBO correction (2) Not stocking (2) 3 NBO model (3) Not stocking (3)

b0j

T j

F j

Q j

0.853031 6 b

0.553432 0.381025 0.408248 0.375

0.635602 0.68374 1 1

108.67 > q2 75 81.6496 > q2 75

1.11088 0.763765 0.83666 0.75

0.619732 0.639558 1 1

217.952 > q3 150 167.332 > q3 150

1.85987

0.636287

365.21

0.408248 0.375

1 1

81.6496 > q2 75

0.920152 0.754037 0.83666 0.75

0.903821 0.973234 1 1

180.49 > q3 150 167.332 > q3 150

1.74378

0.805105

335.161

0.408248 0.375

1 1

81.6496 > q2 75

0.83666 0.75

1 1

167.332 > q3 150

1.45774

1

291.548

0.774202 6 b

0.708905 6 b

Profit (j)

466.148 452.5 1000 570.496 536.25 1200 686.411 1400

0.853031 > b

0.774202 6 b

0.708905 6 b

452.5 1000 536.394 536.25 1200 630.197 1400

0.853031 > b 452.5 1000

0.774202 > b 536.25 1200

0.708905 > b

The parameters in our model can be divided into two groups: (1) Parameters that have known values. (2) Parameters that are estimated. The second group can again be divided into at least two groups: those for which the estimates are probably fairly accurate and those that are less certain. For this model the breakdown is: Known: selling price ðPÞ, purchase cost ({C j }), number of different unit costs ðnÞ, cost breakpoints ({qj }) Estimated: More confident: ordering cost ðAÞ, demand ðDÞ, holding cost rate ðiÞ. Less confident: backordering rate (b), goodwill loss for stockout ðgÞ, backordering cost (p) There is one other relevant parameter group, the lost sale cost per unit ðfp0j gÞ, but that is derived from P, {C j }, and g, so we do not need to consider it separately. We use the problem solved in Example 2 with b = 0.80 as the base case and then resolve it with changes of ±25%, ±20%, ±15%, ±10%, and ±5% in each of the estimated parameters, keeping all the other parameters constant. The performance measure is percent reduction in the average profit per period (ATP) for the variation relative to the optimal ATP from using the original parameter values. Base case parameters: P = $9/unit, D = 200 units/period, A = $30/ order, i = 0.3/period, p = $2/unit/period, b ¼ 0:80, C ¼ ðC 1 ; C 2 ; C 3 Þ = $(6,5,4)/unit, q ¼ ðq1 ; q2 ; q3 Þ = (0, 75, 150) units, g = $2/unit, p0 ¼ ðp01 ; p02 ; p03 Þ = $(5, 6, 7)/unit. Base case optimal values: T⁄ = 1.74378, F⁄ = 0.805105, ⁄ Q = 335.161, B⁄ = 54.3765, ATP⁄ = 630.197/period.

616.393 1400

4.2. Study results 4.2.1. Effects of parameter changes on ATP The details of the results of the changes in the estimated parameters are shown in Table 3. The percentage changes in ATP are shown graphically in Fig. 3. From these results we can draw the following conclusions about how the estimated parameter changes affected the ATP: 1. As would be expected, the further the changed parameter’s value is from the value in the base case, the greater the decrease in the value of the ATP. There is one exception to this conclusion, b, for which the percentage changes in ATP are identical for changes in b of 15%, 20%, and 25%. The reason for this is that changes in b by these percentages bring b below its critical value for which partial backordering is optimal. As can be seen in those rows of Table 3, the optimal values of T and F for those cases are 1.45774 and 1.0, giving Q = 291.548, and B = 0. That is, the optimal solution for those cases is to use the basic EOQ with no stockouts for these parameter sets. As shown in Example 2, the minimum value of b for which partial backordering is optimal when j = 3 is 0.708905, a reduction of 11.39 percent from the base case value of 0.80. Note also that an increase of 25% in the value of b increases its value to 1.0, which means that all shortages will be backordered. This solution, which is shown in the last row of the b section of Table 3, results in a decrease in ATP of over 3.5 percent. 2. For all parameters except b and g, decreases in the parameter value resulted in greater reductions from the base case value than did the same-sized increases. The reason for this difference

29

A.A. Taleizadeh et al. / Computers & Industrial Engineering 82 (2015) 21–32 Table 3 Sensitivity analysis for Example 2 problem with b = 0.80. Parameter

Change (%)

Values of variables

Changes in variables

T

F

Q

A

25 20 15 10 5 +5 +10 +15 +20 +25

1.71247 1.71878 1.72506 1.73132 1.73756 1.74997 1.75615 1.76230 1.76843 1.77455

0.809171 0.808341 0.807519 0.806706 0.805902 0.804317 0.803537 0.802764 0.802000 0.801243

329.422 330.578 331.730 332.878 334.021 336.297 337.429 338.557 339.681 340.801

52.286 52.707 53.127 53.545 53.961 54.790 55.203 55.614 56.024 56.433

630.157 630.172 630.183 630.191 630.195 630.195 630.191 630.183 630.173 630.160

1.80 1.43 1.07 0.71 0.36 +0.36 +0.71 +1.06 +1.41 +1.76

+0.51 +0.40 +0.30 +0.20 +0.10 0.10 0.19 0.29 0.39 0.48

1.71 1.37 1.02 0.68 0.34 +0.34 +0.68 +1.01 +1.35 +1.68

3.84 3.07 2.30 1.53 0.76 +0.76 +1.52 +2.28 +3.03 +3.78

0.0063 0.0040 0.0022 0.0010 0.0002 0.0002 0.0010 0.0022 0.0038 0.0059

D

25 20 15 10 5 +5 +10 +15 +20 +25

2.08822 2.00781 1.93400 1.86587 1.80267 1.68868 1.63694 1.58821 1.54215 1.49850

0.754638 0.765050 0.775271 0.785334 0.795270 0.814862 0.824560 0.834220 0.843858 0.853490

397.141 382.693 369.416 357.153 345.771 325.230 315.902 307.110 298.798 290.919

81.979 75.478 69.540 64.086 59.050 50.022 45.950 42.127 38.527 35.127

626.155 627.728 628.867 629.629 630.060 630.069 629.701 629.114 628.324 627.344

+19.75 +15.14 +10.91 +7.00 +3.38 3.16 6.13 8.92 11.56 14.07

6.27 4.98 3.71 2.46 1.22 +1.21 +2.42 +3.62 +4.81 +6.01

+18.49 +14.18 +10.22 +6.56 +3.17 2.96 5.75 8.37 10.85 13.20

+50.76 +38.81 +27.89 +17.86 +8.59 8.01 15.50 22.53 29.15 35.40

0.6413 0.3918 0.2111 0.0901 0.0217 0.0203 0.0786 0.1718 0.2972 0.4526

i

25 20 15 10 5 +5 +10 +15 +20 +25

1.85593 1.82890 1.80453 1.78242 1.76226 1.72676 1.71103 1.69645 1.68287 1.67020

0.896370 0.876837 0.857954 0.839714 0.822104 0.788700 0.772866 0.757584 0.742831 0.728587

363.493 356.770 350.653 345.057 339.912 330.758 326.662 322.839 319.263 315.907

30.773 36.041 41.012 45.712 50.160 58.379 62.181 65.799 69.245 72.530

623.762 626.307 628.127 629.325 629.990 630.010 629.483 628.664 627.593 626.305

+6.43 +4.88 +3.48 +2.22 +1.06 0.98 1.88 2.71 3.49 4.22

+11.34 +8.91 +6.56 +4.30 +2.11 2.04 4.00 5.90 7.73 9.50

+8.45 +6.45 +4.62 +2.95 +1.42 1.31 2.54 3.68 4.74 5.74

43.41 33.72 24.58 15.94 7.75 +7.36 +14.35 +21.01 +27.34 +33.39

1.0211 0.6172 0.3285 0.1384 0.0328 0.0297 0.1133 0.2433 0.4132 0.6175

g

25 20 15 10 5 +5 +10 +15 +20 +25

1.77433 1.76847 1.76248 1.75637 1.75014 1.73729 1.73067 1.72393 1.71705 1.71005

0.782088 0.786601 0.791158 0.795760 0.800408 0.809852 0.814651 0.819503 0.824410 0.829375

339.401 338.598 337.773 336.925 336.055 334.244 333.303 332.339 331.351 330.338

61.864 60.382 58.893 57.396 55.890 52.855 51.325 49.786 48.240 46.684

629.969 630.050 630.114 630.160 630.187 630.187 630.158 630.110 630.041 629.952

+1.75 +1.42 +1.07 +0.72 +0.36 0.37 0.75 1.14 1.53 1.93

2.86 2.30 1.73 1.16 0.58 +0.59 +1.19 +1.79 +2.40 +3.01

+1.26 +1.03 +0.78 +0.53 +0.27 0.27 0.55 0.84 1.14 1.44

+13.77 +11.04 +8.31 +5.55 +2.78 2.80 5.61 8.44 11.29 14.15

0.0362 0.0233 0.0132 0.0059 0.0015 0.0015 0.0061 0.0138 0.0247 0.0389

p

25 20 15 10 5 +5 +10 +15 +20 +25

1.82468 1.80495 1.78728 1.77135 1.75692 1.73176 1.72072 1.71055 1.70114 1.69242

0.762912 0.772790 0.781858 0.790214 0.797939 0.811771 0.817989 0.823803 0.829251 0.834368

347.631 344.586 341.861 339.406 337.184 333.313 331.616 330.054 328.610 327.272

69.218 65.616 62.381 59.457 56.801 52.155 50.110 48.223 46.475 44.851

629.389 629.728 629.975 630.099 630.174 630.178 630.126 630.048 629.950 627.979

+4.64 +3.51 +2.49 +1.58 +0.75 0.69 1.32 1.91 2.44 2.94

5.24 4.01 2.89 1.85 0.89 +0.83 +1.60 +2.32 +3.00 +3.63

+3.72 +2.81 +2.00 +1.27 +0.60 0.55 1.06 1.52 1.95 2.35

+27.29 +20.67 +14.72 +9.34 +4.46 4.09 7.85 11.32 14.53 17.52

0.1281 0.0743 0.0381 0.0155 0.0036 0.0030 0.0113 0.0236 0.0391 0.0572

b

25 20 15 10 5 +5 +10 +15 +20 +25

1.45774 1.45774 1.45774 1.50695 1.64754 1.80699 1.84420 1.86005 1.85772 1.83941

1.0 1.0 1.0 0.966650 0.873217 0.750996 0.705254 0.664648 0.627097 0.591107

291.548 291.548 291.548 299.381 321.152 343.400 347.098 347.060 343.934 337.796

0 0 0 8.0411 33.421 71.992 86.971 99.804 110.840 120.339

616.393 616.393 616.393 620.486 628.332 628.906 625.654 620.955 614.982 607.705

16.40 16.40 16.40 13.28 5.52 +3.63 +5.76 +6.67 +6.53 +5.48

+24.21 +24.21 +24.21 +20.07 +8.46 6.72 12.40 17.45 22.11 26.58

13.01 13.01 13.01 10.68 4.18 +2.46 +3.56 +3.55 +2.59 +0.79

100.00 100.00 100.00 85.21 38.54 +32.39 +59.94 +83.54 +103.84 +121.31

2.1904 2.1904 2.1904 1.5456 0.2960 0.2049 0.7208 1.4665 2.4143 3.5690

B

ATP

for b was just discussed. The reason for g is unclear, but we note that the reductions in ATP for the same-sized negative and positive changes are very close and less than 0.04 percent. 3. Changes in A result in the least reduction in ATP, followed by g, p, D; i, and b, in that order. Note, however, that, with the exception of b and negative changes in i, the reductions in ATP are less than one percent from the base case, even for 25 percent changes in the parameter value.

T (%)

F (%)

Q (%)

B (%)

ATP (%)

Since changes in the value of b in 5 percent decrements, which means changes of 4 percentage points, quickly resulted in solutions that did not use partial backordering, we looked at the effects of changes in 1 b, the complementary percentage of unfilled demands that will not be backordered. b = 0.80 for the base case, so the base case value of 1 b is 0.20. Five percent changes in 1 b are only one percentage point, which is much smaller than the changes in b, so we looked at the effect of 10 percent changes

30


4 3.5 3 2.5 2 1.5 1 0.5 0

A D i β g

Reduction in ATP when (1-β) changes 2.5

ATP reduction (%)

ATP Reduction (%)

Percent Reduction in ATP When Parameters Are Changed One at a Time 2 1.5 1 0.5 0

π

-25

-20

-15

-10

-5

0

5

10

15

20

-50

-40

-30

-20

-10

0

10

20

30

40

50

(1-β) change (%)

25

Parameter Change (%)

Fig. 4. Change in ATP based on the change in 1 b.

Fig. 3. Percent reduction in ATP when parameters are changed one at a time.

(2 percentage points each). As shown in Table 4 and Fig. 4, only a 50% increase in 1 b to 0.30 or b = 0.70, resulted in the solution to the problem with a changed value of 1 b not being partial backordering. Since b = 0.70 is less than the minimum value of b for which partial backordering is optimal, this large an increase in 1 b results in the optimal solution for the altered case to be the EOQ with no stockouts (F = 1.0). 4.2.2. Effects of parameter changes on decision variable values The percentage changes in the values of the four decision variables that resulted from changing the parameter values are also shown in Tables 3 and 4. As was the case with the percentage changes in ATP, there are similarities and differences among the variables. 1. For all the parameters except b, the changes in T; F; Q , and B were consistent as the parameter value increased from 25% to +25%. However, these changes were not necessarily in the same direction for all four variables. For A; D; g, and p, the changes for T; Q , and B were in the same direction with F in the opposite direction. For i the changes in T; F, and Q were in the same direction, with B in the opposite direction. This is summarized in Table 5, which shows the direction of the changes for the variables as each parameter increases in value. The inconsistent results for b are, as discussed above, due to the fact that large decreases in the value of b led to the basic EOQ without backordering being optimal and an increase in b of 25 percent to 1.0 led to full backordering being optimal. As can be seen in Table 4, these inconsistencies with respect to b disappear when looking at the effects of changes in the value of 1 b. 2. The columns of Tables 3 and 4 that give the percentage changes in the decision variables also make it possible to see which variables have the greatest impact on the values of ATP and the decision variables. Looking only the results for ±25%, although the same conclusions would be reached if the other sizes are considered, changes in A have the least effect on ATP, followed

Table 5 Direction of changes in decision variable values as a parameter increases. Increase in parameter

Change in

A D i g

p b

T

F

Q

B

Increase Decrease Decrease Decrease Decrease Nonmonotone

Decrease Increase Decrease Increase Increase Decrease

Increase Decrease Decrease Decrease Decrease Nonmonotone

Increase Decrease Increase Decrease Decrease Increase

by g, p, D, i, and b (or 1 b). The results for the changes in the values of the four decision variables are very similar, with A; g, and p in some order having the least impact and i; D, and b (or 1 b) in some order having the greatest impact. To illustrate the sizes and directions of the effects of a parameter change graphically, the relative changes in the four decision variables as D changes are shown in Fig. 5. 4.2.3. Implications Our analysis of the effects of changes in the six unknown parameters values on ATP and the four decision variables – T; F; Q , and B – leads to two basic conclusions: 1. As is shown for the basic EOQ model in many introductory texts on inventory control model, even relatively large changes in or mis-estimation of the value of a model parameter have relatively small effects on the value of the model’s performance measure. Our conclusion here is basically the same. The only model parameter that generated changes in ATP of more than approximately one percent for a parameter change of ±25% was b. Thus, if the user’s interest is primarily finding a solution that will give a value of ATP close to the optimal without worrying about whether the values of the decision variables are approximately correct, keeping the parameter estimates within about 25% of the true values should be sufficient.

Table 4 Sensitivity analysis when parameter b changes its value (through changes in 1 b with 10% increments). Change (%)

Value

Values of variables

1b

b

T

F

Q

B

ATP

T (%)

Changes in variables F (%)

Q (%)

B (%)

ATP (%)

50 40 30 20 10 +10 +20 +30 +40 +50

0.90 0.88 0.86 0.84 0.82 0.78 0.76 0.74 0.72 0.70

1.85457 1.84420 1.82852 1.80699 1.77900 1.70036 1.64754 1.58376 1.50695 1.45774

0.684458 0.705254 0.727318 0.750996 0.776732 0.836901 0.873217 0.915651 0.96665 1.0

347.507 347.098 345.759 343.400 339.912 328.978 321.152 311.408 299.381 291.548

93.6315 86.9714 79.7765 71.9916 63.5511 44.3722 33.4208 21.3741 8.0411 0

623.469 625.654 627.482 628.906 629.848 629.779 628.332 625.446 620.456 616.393

+6.35 +5.76 +4.86 +3.63 +2.02 2.49 5.52 9.18 13.58 16.40

14.99 12.40 9.66 6.72 3.52 +3.95 +8.46 +13.73 +20.07 +24.21

+3.68 +3.56 +3.16 +2.46 +1.42 1.84 4.18 7.09 10.68 13.01

+72.19 +59.94 +46.71 +32.39 +16.87 18.40 38.54 60.69 85.21 100.00

1.0676 0.7208 0.4307 0.2049 0.0553 0.0663 0.2960 0.7539 1.5456 2.1904


References

Changes (%)

Changes (%) in output variables, when parameter D changes 60 50 40 30 20 10 0 -10 -25 -20 -30 -40

T F -20

-15

-10

-5

0

5

10

15

20

25

31

Q B

D change (%) Fig. 5. Percent changes in T; F; Q , and B when D changes by a given percent.

2. If, on the other hand, the user is equally as interested in having the values of T; F; Q , and B be approximately correct, then less attention can be paid to estimating the values of A; g, and p and more attention needs to be paid to estimating the values of i; D, and b. One final comment on sensitivity analysis is relevant. Due to the relative complexity of the equations for T and Fand, as a result, for the ATP, we used, as is most frequently done in assessing the sensitivity of a model to changes in its inputs, a numerical approach in this study. As was pointed out by Chu and Chung (2004) in their discussion of sensitivity analysis of a basic EOQ with partial backordering, ‘‘the conclusions made by the analyses of sensitivities based on the computational results of a set of numerical examples are questionable since different conclusions may be made if different sets of numerical examples are analyzed.’’ While we are confident that our conclusions above are fairly general, any user of this or a similarly complex model needs to conduct his or her own study.

5. Conclusion We extended the basic EOQ model with incremental discounts by combining the basic solution procedure for that problem with repeated use of Pentico and Drake’s (2009) models for the EOQ with full or partial backordering at a constant rate b to determine the best order quantity for each possible cost. Minimum cost (or maximum profit) was then used to choose among the best full (or partial) backordering solution, meeting all demand and losing all sales. We developed a condition under which partial backordering is optimal and guarantees global optimal values of period length, fraction of demand that will be filled from stock, and order quantity. We illustrated the developed models and proposed solution procedures with examples. A numerical study based on one of the partial backordering example problems was used to evaluate the sensitivity of the model’s results to the changes or mis-estimation of the various parameters. Extending the proposed model to include different fixed ordering costs for different price intervals and also considering the pricing issue to determine the optimal selling price of the ordered quantity are some directions for future research.

Funding The research for the first author was supported by the Iran National Science Foundation (INSF), Fund No. [INSF-93027686].

Appendix A–F. Supplementary material Supplementary data associated with this article can be found, in the online version, at http://dx.doi.org/10.1016/j.cie.2015.01.005.

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