Computers & Industrial Engineering 82 (2015) 21–32

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Computers & Industrial Engineering

journal homepage: www.elsevier .com/ locate/caie

An economic order quantity model with partial backorderingand incremental discount

http://dx.doi.org/10.1016/j.cie.2015.01.0050360-8352/� 2015 Elsevier Ltd. All rights reserved.

⇑ Corresponding author.E-mail addresses: taleizadeh@ut.ac.ir (A.A. Taleizadeh), irenatra@pmf.ukim.mk,

irena.stojkovska@gmail.com (I. Stojkovska), pentico@duq.edu (D.W. Pentico).

Ata Allah Taleizadeh a, Irena Stojkovska b, David W. Pentico c,⇑a School of Industrial Engineering, College of Engineering, University of Tehran, Tehran, Iranb Department of Mathematics, Faculty of Natural Sciences and Mathematics, Ss. Cyril and Methodius University, Skopje, Macedoniac Palumbo-Donahue School of Business, Duquesne University, Pittsburgh, PA 15282, USA

a r t i c l e i n f o

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

Keywords:EOQIncremental discountsFull backorderingPartial backordering

a b s t r a c t

Determining an order quantity when quantity discounts are available is a major interest of materialmanagers. A supplier offering quantity discounts is a common strategy to entice the buyers to purchasemore. In this paper, EOQ models with incremental discounts and either full or partial backordering aredeveloped for the first time. Numerical examples illustrate the proposed models and solution methods.

� 2015 Elsevier Ltd. All rights reserved.

1. Introduction and literature review all-units discount, purchasing a larger quantity results in a lower

Since Harris (1913) first published the basic EOQ model, manyvariations and extensions have been developed. In this paper wecombine two of those extensions: partial backordering andincremental quantity discounts.

Montgomery, Bazaraa, and Keswani (1973) were the first todevelop a model and solution procedure for the basic EOQ withpartial backordering (EOQ–PBO) at a constant rate. Others takingsomewhat different approaches have appeared since then,including Pentico and Drake (2009), which will be one of the twobases for our work here. In addition, many authors have developedmodels for the basis EOQ-PBO combined with other situationalcharacteristics, such as Wee (1993) and Abad (2000), both of whichincluded a finite production rate and product deterioration,Sharma and Sadiwala (1997), which included a finite productionrate with yield losses and transportation and inspection costs,San José, Sicilia, and García-Laguna (2005), which included modelswith a non-constant backordering rate, and Taleizadeh, Wee, andSadjadi (2010), which included production and repair of a numberof items on a single machine. Descriptions of all of these modelsand others may be found in Pentico and Drake (2011).

Enticing buyers to purchase more by offering either all-units orincremental quantity discounts is a common strategy. With the

unit purchasing price for the entire lot, while incremental dis-counts only apply the lower unit price to units purchased abovea specific quantity. So the all-units discount results in the sameunit price for every item in the given lot, while the incremental dis-count can result in multiple unit prices for an item within the samelot (Tersine, 1994). In the following we focus on the research usingonly an incremental discount or both incremental and all-units dis-counts together. Since Benton and Park (1996) prepared an exten-sive survey of the quantity discount literature until 1993, we willdescribe newer research, along with a short history of incrementaldiscounts and older research which is more related to this paper.

The EOQ model with incremental discounts was first discussedby Hadley and Whitin (1963). Tersine and Toelle (1985) presentedan algorithm and a numerical example for the incremental dis-count and examined the methods for determining an optimal orderquantity under several types of discount schedules. Güder, Zydiak,and Chaudhry (1994) proposed a heuristic algorithm to determinethe order quantities for a multi-product problem with resourcelimitations, given incremental discounts. Weng (1995) developeddifferent models to determine both all-units and incremental dis-count policies and investigated the effects of those policies withincreasing demand. Chung, Hum, and Kirca (1996) proposed twocoordinated replenishment dynamic lot-sizing problems with bothincremental and all-units discounts strategies. Lin and Kroll (1997)extended a newsboy problem with both all-units and incrementaldiscounts to maximize the expected profit subject to a constraintthat the probability of achieving a target profit level is no less than

I

DFT

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

a predefined risk level. Hu and Munson (2002) investigated adynamic demand lot-sizing problem when product price schedulesoffer incremental discounts. Hu, Munson, and Silver (2004) contin-ued their previous work and modified the Silver-Meal heuristicalgorithm for dynamic lot sizing under incremental discounts.Rubin and Benton (2003) considered the purchasing decisionsfacing a buying firm which receives incrementally discounted priceschedules for a group of items in the presence of budgets and spacelimitations. Rieksts, Ventura, Herer, and Sun (2007) proposed aserial inventory system with a constant demand rate and incre-mental quantity discounts. They showed that an optimal solutionis nested and follows a zero-inventory ordering policy. Hakseverand Moussourakis (2008) proposed a model and solution methodto determine the ordering quantities for multi-product multi-constraint inventory systems from suppliers who offer incrementalquantity discounts. Mendoza and Ventura (2008) incorporatedquantity discounts, both incremental and all-units, on the pur-chased units into an EOQ model with transportation costs.Taleizadeh, Niaki, and Hosseini (2009) developed a constrainedmulti-product bi-objective single-period problem with incremen-tal discounts and fully lost-sale shortages. Ebrahim, Razm, andHaleh (2009) proposed a mathematical model for supplierselection and order lot sizing under a multiple-price discountenvironment in which different types of discounts includingall-unit, incremental, and total business volume are considered.Taleizadeh, Niaki, Aryanezhad, and Fallah-Tafti (2010) developeda multi-products multi-constraints inventory control problem withstochastic period length in which incremental discounts and par-tial backordering situations are assumed. Munson and Hu (2010)proposed procedures to determine the optimal order quantitiesand total purchasing and inventory costs when products haveeither all-units or incremental quantity discount price schedules.Bai and Xu (2011) considered a multi-supplier economic lot-sizingproblem in which the retailer replenishes his inventory fromseveral suppliers who may offer either incremental or all-unitsquantity discounts. Chen and Ho (2011) developed an analysismethod for the single-period (newsboy) inventory problem withfuzzy demands and incremental discount. Taleizadeh, Barzinpour,and Wee (2011) discussed a constrained newsboy problem withfuzzy demand, incremental discounts, and lost-sale shortages.Taleizadeh, Niaki, and Nikousokhan (2011) developed a multi-constraint joint-replenishment EOQ model with uncertain unitcost and incremental discounts when shortages are not permitted.Bera, Bhunia, and Maiti (2013) developed a two-storage inventorymodel for deteriorating items with variable demand and partialbackordering. Lee, Kang, Lai, and Hong (2013) developed anintegrated model for lot sizing and supplier selection and quantitydiscounts including both all units and incremental discounts.Archetti, Bertazzi, and Speranza (2014) studied the economiclot-sizing problem with a modified all-unit discount transportationcost function and with incremental discount costs.

According to the above mentioned research, it is clear that noresearchers have developed an EOQ model with partial backorder-ing and incremental discounts. Taleizadeh and Pentico (2014)developed an EOQ model with partial backordering and all-unitsdiscounts. In this paper we develop EOQ models with fully and par-tially backordered shortages when the supplier offers incrementaldiscounts to the buyer.

tFT

( )1 F T−

D

( )1 FD T−

Fig. 1. EOQ model with fully backordered shortages.

2. Model development

In this section we model the defined problem under two differ-ent conditions: full backordering and partial backordering. But firstwe briefly discuss the EOQ model with full or partial backorderingwhen discounts are not assumed. We use the following notation.

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 � Cj 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 par-tially 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 ofF and T as:

F� ¼ ppþ iC

ð1Þ

T� ¼ffiffiffiffiffiffiffiffi2AiCD

r ffiffiffiffiffiffiffiffiffiffiffiffiffiffipþ iC

p

rð2Þ

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

F� ¼ 1� bð Þp0 þ bpT�

iC þ bpð ÞT� ð3Þ

T� ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2AiCD

iC þ bpbp

� �� 1� bð Þp0½ �2

biCp

sð4Þ

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

b0 ¼ 1�ffiffiffiffiffiffiffiffiffiffiffiffiffiffi2AiCDp

Dp0ð5Þ

A.A. Taleizadeh et al. / Computers & Industrial Engineering 82 (2015) 21–32 23

2.2. EOQ model with incremental discount without shortages

Consider an EOQ model in which the supplier offers thevolume-based unit purchasing costs shown in Eq. (6) (Q ¼ DT).

Cj ¼

C1 q1 ¼ 0 6 Q < q2

C2 q2 6 Q < q3

..

. ...

Cn qn 6 Q

8>>>><>>>>:

ð6Þ

where C1 > C2 > � � � > Cn and q1 ¼ 0 < q2 < � � � < qn. The purchasingcost per order is:

Mj ¼ Xj þ CjDT; j ¼ 1;2; . . . ;n; ð7Þ

where

Xj ¼Xj

k¼2

qkðCk�1 � CkÞ; j ¼ 2;3; . . . ;n and X1 ¼ 0; ð8Þ

From the definitions of Cj; qj and Xj, we have that:

Xj P 0; j ¼ 1;2; . . . ;n: ð9Þ

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

C 0j ¼Mj

DT¼ Xj

DTþ Cj: ð10Þ

and the optimal cycle length for ordering from the quantity fromthe jth interval ½qj; qjþ1Þ is

T�j ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ

iCjD

s: ð11Þ

The optimal order quantity is Q �j ¼ DT�j , with minimal annual totalcost of

ATC�j ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2DiCjðAþ XjÞ

qþ iXj

2þ CjD: ð12Þ

The optimal order quantity Q �j is acceptable if qj 6 Q �j < qjþ1. IfQ �j < qj, then the optimal acceptable order quantity is Q �j ¼ qj. IfQ �j P qjþ1, then the optimal acceptable order quantity isQ �j ¼ qjþ1. For the latter two cases, the corresponding annual totalcost, calculated using Eq. (A1) in Appendix A, is the new optimalannual total cost ATC�j . Finally, ATC�j for j ¼ 1;2; . . . ; n are comparedto find the minimal value among ATC�j ; j ¼ 1;2; . . . ;n, which will bethe optimal annual cost for the EOQ model with incremental dis-count, and the corresponding Q �j will be the optimal order quantityfor the EOQ model with incremental discount. This solution proce-dure is justified, because we can prove that if Q �j P qjþ1, then thereis an order quantity which costs less to order than Q �j does (seeAppendix A).

In the following sub-sections we model the defined problemunder two different conditions: full backordering and partialbackordering, which are developed in Sections 2.3 and 2.4respectively.

2.3. EOQ model with full backordering and incremental discounts

We will consider an EOQ model in which all shortages will bebackordered and the supplier offers incremental volume-basedunit purchasing cost discounts. Then, according to Fig. 1, the cyclictotal cost for ordering the quantity from the interval ½qj; qjþ1Þ is

CTCjðT; FÞ ¼ Az}|{Fixed Cost

þ C 0jDTzffl}|ffl{Purchasing Cost

þiC0jDF2T2

2

zfflfflfflfflfflffl}|fflfflfflfflfflffl{Holding Cost

þpDð1� FÞ2T2

2

zfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflffl{Backordering Cost

ð13Þ

where C0j is the purchasing cost per unit given by Formula (10).Substituting Formula (10) into (13) and dividing by T we get theannual total cost for ordering the quantity from the interval½qj; qjþ1Þ:

ATCjðT; FÞ ¼Aþ Xj

Tþ iXjF

2

2þ iCjDF2T

2þ pDð1� FÞ2T

2þ CjD ð14Þ

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

ATCðT; FÞ ¼

ATC1ðT; FÞ ;0 < DT < q2

ATC2ðT; FÞ ; q2 6 DT < q3

..

.

ATCnðT; FÞ ; qn 6 DT

8>>>><>>>>:

ð15Þ

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

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

Proof. See Appendix B. h

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

minT>0;06F61

ATCðT; FÞ ¼ min16j6n

minðT;FÞ2Xj

ATCjðT; FÞ� �

ð16Þ

where

X1 ¼ fðT;FÞj 0< T 6 q2=D; 06 F 6 1g;Xj ¼ fðT;FÞj qj=D6 T 6 qjþ1=D; 06 F 6 1g; j¼ 2;3; . . . ;n�1; and

Xn ¼ fðT;FÞj qn=D6 T; 06 F 6 1g:ð16aÞ

Note that the sign < is changed into 6 in the upper bounds,which is allowed by the continuity of ATCðT; FÞ. In what followswe will use the notation Tj and Fj for T and F, respectively, whenwe are minimizing the annual total cost for ordering the quantityfrom the interval ½qj; qjþ1Þ defined by Eq. (14). To solve the jthsubproblem in (16), i.e. the problem

minðT;FÞ2Xj

ATCjðTj; FjÞ; ð17Þ

we first find the first partial derivatives of ATCjðTj; FjÞ with respectto Tj and Fj.

@ATCj

@TjðTj; FjÞ ¼ �

Aþ Xj

T2j

þiCjDF2

j

2þ pDð1� FjÞ2

2; ð18Þ

@ATCj

@FjðTj; FjÞ ¼ iXjFj þ iCjDFjTj � pDð1� FjÞTj: ð19Þ

Setting the first derivatives (18) and (19) equal to 0, and solvingthe corresponding system with respect to Tj and Fj, rememberingthat Tj > 0, we get

Tj ¼ T�j ðFjÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2ðAþ XjÞD½iCjF

2j þ pð1� FjÞ2�

sð20Þ

iXjFj þ D iCjFj � pð1� FjÞ�

Tj ¼ 0: ð21Þ

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

iXjFj þ D iCjFj � pð1� FjÞ�

T�j ðFjÞ ¼ 0; ð22Þ

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

t

I

FT

( )1 F T−

D

βD

( )1 FD Tβ −

( ) ( )1 1 FD Tβ− −

DFT

Fig. 2. EOQ model with partially backordered shortages.

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

by F�j . If we denote the left side of Eq. (22) bywðFjÞ ¼ iXjFj þ D iCjFj � pð1� FjÞ

� T�j ðFjÞ, then from wð1Þ ¼ iXjþ

DiCjT�j ð1Þ > 0;wð0Þ ¼ �DpT�j ð0Þ < 0, and wðFjÞ 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.

Proposition 2. There exists a solution F�j of Eq. (22), for which0 6 F�j 6 1.

Then, from Eq. (20) we have:

T�j ¼ T�j ðF�j Þ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ

D½iCjF�2j þ pð1� F�j Þ

2�

s: ð23Þ

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 optimal-ity 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 ofSubproblem (17). Then ðT�j ; F

�j Þ is the global optimal solution of

Subproblem (17).

Proof. See Appendix C. h

Note that wðFjÞ is a monotone nondecreasing function since@wðFjÞ=@Fj ¼ @2/ðFjÞ=@F2

j > 0 (see (C6) in Appendix C). Thus thesolution 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 ATCjðTj; FjÞ 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 theglobal solution ðT�j ; F

�j Þ of Subproblem (17) lies on the lower bound-

ary of Tj, i.e. T�j ¼ qj=D, and F�j ¼ pqj=ððiCj þ pÞqj þ iXjÞ. IfT�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, andF�j ¼ pqjþ1=ððiCj þ pÞqjþ1 þ iXjÞ. This is true because of the convex-ity of ATCjðTj; FjÞ with respect to Tj (see (C1) in Appendix C), andminimizing ATCjðqj=D; FjÞ and ATCjðqjþ1=D; FjÞ, respectively, inorder to obtain the last two values for F�j .

From the above discussion we can conclude that the globaloptimal solution ðT�; F�Þ that minimizes the annual total cost givenin Eq. (15) is the pair ðT�j ; F

�j Þ for which the corresponding

ATCjðT�j ; F�j Þ is minimal over all j = 1,2, . . . ,n. That is,

ðT�; F�Þ ¼ arg min16j6n

ATCjðT�j ; F�j Þ

n o: ð24Þ

We have the following solution procedure for EOQ model withincremental discount and full backordering.

Solution procedure for the EOQ model with incremental discountsand full backordering

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 accept-

able 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=ððiCj þ pÞqj þ iXjÞ

1.4. If T�j > qjþ1=D and j 2 f1; . . . ;n� 1g, then calculate the newðT�j ;F

�j Þ using T�j ¼ qjþ1=D and F�j ¼ pqjþ1=ððiCjþpÞqjþ1þ iXjÞ

1.5. Calculate ATCjðT�j ; F�j Þ.

2. Find the optimal solution as the pair ðT�j ; F�j Þ for which the

corresponding ATCjðT�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 theannual total cost to obtain the optimal solutions, in the partialbackordering model, in order to facilitate reaching the optimalsolution using the approach in Pentico and Drake (2009), we willfirst model the profit function and then by maximizing it we willget the optimal solutions. According to Fig. 2, in which it is clearthat the order quantity will be Q ¼ DT½F þ bð1� FÞ�, the unit pur-chasing cost becomes

C0j ¼Xj

D½F þ bð1� FÞ�T þ Cj ð25Þ

where Xj is given by Eq. (8). Then the cyclic total profit for orderingthe quantity from the interval ½qj; qjþ1Þ is

CTPjðF;TÞ¼PD½Fþbð1�FÞ�Tzfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflffl{Revenue

�Az}|{Fixed Cost

þC0jD½Fþbð1�FÞ�Tzfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{Purchasing Cost

þiC 0jDF2T2

2

zfflfflfflfflfflffl}|fflfflfflfflfflffl{Holding Cost

þpbDð1�FÞ2T2

2|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}Backordering Cost

þgð1�bÞð1�FÞDT|fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}Lost Sale Cost

8>>>>>>>><>>>>>>>>:

9>>>>>>>>=>>>>>>>>;

ð26Þ

Substituting (25) in (26) and dividing by T gives the averageannual profit for ordering the quantity from the interval ½qj; qjþ1Þ:

ATPjðF; TÞ ¼ PD½F þ bð1� FÞ�

�AþXj

T þ CjD½F þ bð1� FÞ� þ iXjF2

2ðFþbð1�FÞÞ

þ iCjDF2T2 þ pbDð1�FÞ2T

2 þ gð1� bÞð1� FÞD

8<:

9=; ð27Þ

After some algebraic transformations and letting p0j ¼ P � Cj þ g, wehave:

ATPjðF; TÞ ¼ P � Cj �

D

� Aþ Xj

Tþ iXjF

2

2 1� ð1� bÞð1� FÞð Þ þiCjDF2T

2

(

þ pbDð1� FÞ2T2

þ p0jð1� bÞð1� FÞD)

ð28Þ

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

A.A. Taleizadeh et al. / Computers & Industrial Engineering 82 (2015) 21–32 25

ATPðT; FÞ ¼

ATP1ðT; FÞ ; 0 < DTðF þ bð1� FÞÞ < q2

ATP2ðT; FÞ ; q2 6 DTðF þ bð1� FÞÞ < q3

..

.

ATPnðT; FÞ ; qn 6 DTðF þ bð1� FÞÞ

8>>>><>>>>:

ð29Þ

The function ATPðT; FÞ defined by (28) and (29), is continuous (seeAppendix D).

Thus, the maximization problem can be written as

maxT>0;06F61

ATPðT; FÞ ¼max16j6n

maxðT;FÞ2~Xj

ATPjðT; FÞ( )

ð30Þ

with

~X1 ¼ fðT; FÞj DTðF þ bð1� FÞÞ 6 q2; T > 0;0 6 F 6 1g;~Xj ¼ fðT; FÞj qj 6 DTðF þ bð1� FÞÞ 6 qjþ1; T > 0; 0 6 F 6 1g;

j ¼ 2;3; . . . ;n� 1;

and ~Xn ¼ fðT; FÞj qn 6 DTðF þ bð1� FÞÞ; T > 0; 0 6 F 6 1g:ð30aÞ

Note that the sign < is changed into 6 in the upper bounds forthe order quantity, which is allowed by the continuity of ATPðT; FÞ(see Appendix D).

Since maximizing ATPjðT; FÞ is equivalent to minimizing thefunction

ujðT; FÞ ¼Aþ Xj

Tþ iXjF

2

2 F þ bð1� FÞð Þ þiCjDF2T

2

þ pbDð1� FÞ2T2

þ p0jð1� bÞð1� FÞD; ð31Þ

Problem (30) is transformed into

maxT>0;06F61

ATPðT; FÞ ¼max16j6n

ðP � CjÞD� minðT;FÞ2~Xj

ujðT; FÞ( )

: ð32Þ

As in Section 2.3, we will use Tj and Fj for T and F respectivelywhen we are minimizing the function ujðT; FÞ defined by Eq. (31).

In order to minimize the function ujðTj; FjÞ, we first take the firstpartial derivatives:

@uj

@TjðTj; FjÞ ¼ �

Aþ Xj

T2j

þiCjDF2

j

2þ pbDð1� FjÞ2

2; ð33Þ

@uj

@FjðTj; FjÞ ¼

2iXjFj Fj þ bð1� FjÞ �

� iXjF2j ð1� bÞ

2 Fj þ bð1� FjÞ �2

þ iCjFjTj � pbð1� FjÞTj � p0jð1� bÞh i

D: ð34Þ

Setting the first derivatives (33) and (34) equal to 0, and solvingthe corresponding system with respect to Tj and Fj, rememberingthat Tj > 0, we have:

Tj ¼ T�j ðFjÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2ðAþ XjÞD½iCjF

2j þ pbð1� FjÞ2�

s; ð35Þ

iXjFj

Fj þ bð1� FjÞ�

iXjF2j ð1� bÞ

2 Fj þ bð1� FjÞ �2 þ D iCjFj � pbð1� FjÞ

�Tj

� p0jð1� bÞD ¼ 0: ð36Þ

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

iXjFj

Fj þ bð1� FjÞ�

iXjF2j ð1� bÞ

2 Fj þ bð1� FjÞ �2 þ D iCjFj � pbð1� FjÞ

�T�j ðFjÞ

� p0jð1� bÞD ¼ 0; ð37Þ

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

by F�j . If we denote the left side of Eq. (37) by nðFjÞ ¼iXjFj

Fjþbð1�FjÞ�

iXjF2j ð1�bÞ

2 Fjþbð1�FjÞð Þ2þ D iCjFj � pbð1� FjÞ

�T�j ðFjÞ � p0jð1� bÞD, then we

have nð0Þ ¼ �DpbT�j ð0Þ � p0jð1� bÞD < 0, since p0j ¼ P � Cj þ g > 0,

and nð1Þ ¼ iXj �iXjð1�bÞ

2 þDiCjT�j ð1Þ �p0jð1� bÞD¼ iXj þDiCj

ffiffiffiffiffiffiffiffiffiffiffiffi2ðAþXjÞ

DiCj

q�

iXj

2 þp0jD�

ð1� bÞ> 0, only if the condition;

b > 1� iXj þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞDiCj

piXj

2 þ p0jD¼ b0j ð38Þ

is satisfied. Then, because of the continuity of the function nðFjÞ, wecan formulate the following proposition.

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

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

T�j ¼ T�j ðF�j Þ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ

D½iCjF�2j þ pbð1� F�j Þ

2�

s: 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

(39) we have that T�j ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ=ðDiCjÞ

pwhich is the optimal

cycle length for the cost of Cj in the EOQ model with incrementaldiscount without shortages (see Section 2.2, Eq. (11)).

We can also prove that if Condition (38) is satisfied andðT�j ; F

�j Þ 2 ~Xj, then it is the global minimizer of the function

ujðTj; FjÞ over the domain ~Xj. The following proposition stands.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

ðT�j ; F�j Þ 2 ~Xj, where F�j is the solution of (37), T�j is defined by (39),

and ~Xj is the feasible region defined by (30a). Then ðT�j ; F�j Þ is the

global minimizer of the function ujðTj; FjÞ over the domain ~Xj.

Proof. See Appendix E. h

As in the full backordering case, note that nðFjÞ is a monotonenon-decreasing function since @nðFjÞ=@Fj ¼ @2gðFjÞ=@F2

j > 0 (see(E6) in Appendix E). Thus, if Condition (38) is satisfied, the solutionF�j of Eq. (37) is the unique solution in the interval [0,1], and if

ðT�j ; F�j Þ 2 ~Xj, then ðT�j ; F

�j Þ is the unique global minimizer of the

function ujðTj; FjÞ on the set ~Xj. If Condition (38) is not satisfied,then 0 6 b < b0j, which is equivalent to nð1Þ < 0, and from nðFjÞbeing a monotonic function, we have that there is no solution ofEq. (37) in the interval [0,1]; consequently, partial backorderingcannot be optimal. So, in this case (0 6 b < b0j), the optimal decisionis either meeting all demand (EOQ model with incrementaldiscount and no shortages, Section 2.2) with the optimal value ofthe cycle length T�j ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ=ðDiCjÞ

pand 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 Condi-tion (38) is met, but it might not be always true thatqj 6 DT�j ðF

�j þ bð1� F�j ÞÞ 6 qjþ1, in which case the pair ðT�j ; F

�j Þwould

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

be infeasible and cannot be the minimizer of the function ujðTj; FjÞon the set ~Xj. We have the following proposition.

Proposition 6. Assume that Condition (38) is satisfied, butðT�j ; F

�j Þ R ~Xj, where F�j is the solution of (37), T�j is defined by (39),

and ~Xj is the feasible region defined by (30a). Then the minimizerðT�j ; F

�j Þ of the function ujðTj; FjÞ over the domain ~Xj is defined by:

(i) If DT�j ðF�j þ bð1� F�j ÞÞ < qj and one of the following conditions

are met

p0j �Aþ Xj

qjP 0 and

1þ b1� b

>2D p0j �

AþXj

qj

� iXj þ iCjqj

ð40aÞ

or

p0j �Aþ Xj

qj< 0 and

1þ bbð1� bÞ >

2D AþXj

qj� p0j

� pqj

ð40bÞ

then F�j is the solution of

ðAþ XjÞqj

�p0j

!ð1� bÞDþ iXj

qjFj þ iCjFj �pbð1� FjÞ

!DTjðFjÞ

� iXj

qjF2

j þ iCjF2j þpbð1� FjÞ2

!

�D2ð1� bÞ

2qj� T2

j ðFjÞ ¼ 0 ð40cÞ

where TjðFjÞ ¼qj

DðFjþbð1�FjÞÞ, and T�j ¼

qj

DðF�j þbð1�F�j ÞÞ.

(ii) If DT�j ðF�j þ bð1� F�j ÞÞ > qjþ1 and one of the following condi-

tions are met

p0j �Aþ Xj

qjþ1P 0 and

1þ b1� b

>2D p0j �

AþXj

qjþ1

� iXj þ iCjqjþ1

ð41aÞ

or

p0j �Aþ Xj

qjþ1< 0 and

1þ bbð1� bÞ >

2D AþXj

qjþ1� p0j

� pqjþ1

ð41bÞ

then F�j is the solution of

ðAþXjÞqjþ1

�p0j

!ð1�bÞDþ iXj

qjþ1Fjþ iCjFj�pbð1� FjÞ

!DTjðFjÞ

� iXj

qjþ1F2

j þ iCjF2j þpbð1� FjÞ2

!

�D2ð1�bÞ2qjþ1

� T2j ðFjÞ ¼ 0 ð41cÞ

where TjðFjÞ ¼qjþ1

DðFjþbð1�FjÞÞ, and T�j ¼

qjþ1DðF�j þbð1�F�j ÞÞ

.

Proof. See Appendix F. h

When b < b0j and b0j P 0, as we saw earlier, the optimal decisionis meeting all demand from the EOQ model with incremental dis-count and no shortages, i.e., the minimizer of ujðTj; FjÞ lies on the

boundary Fj ¼ 1, so T�j ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ=ðDiCjÞ

pand F�j ¼ 1. From the

convexity of

hðTjÞ ¼ ATCjðTj;1Þ ¼Aþ Xj

Tjþ iXj

2þ iCjDTj

2þ CjD ð42Þ

(see (C1) in Appendix C for Fj ¼ 1), if DT�j ðF�j þ bð1� F�j ÞÞ ¼

DT�j ð1þ bð1� 1ÞÞ ¼ DT�j < qj, then the minimizer lies on the lowerboundary T�j ¼ qj=D, and the corresponding optimal profit isPD� hðT�j Þ. If DT�j ðF

�j þ bð1� F�j ÞÞ ¼ DT�j ð1þ bð1� 1ÞÞ ¼ DT�j > qjþ1,

then the minimizer lies on the upper boundary T�j ¼ qj1=D, andthe corresponding optimal profit is PD� hðT�j Þ. Note that, in thesecond case, we can exclude the point ðT�j ; F

�j Þ from the set of

candidates for the optimal solution, since the correspondingorder quantity is not the overall optimal order quantity (seeAppendix A).

When Condition (38) is met (b P b0j) and ðT�j ; F�j Þ R ~Xj, but

neither (40a) or (40b) nor (41a) or (41b) is satisfied, this meansthat partial backordering cannot be optimal, so the optimaldecision is meeting all demand from the EOQ model withincremental discount and no shortages (see Section 2.2) or losingall sales. In this case we should search for the optimal decisionas in the b < b0j and b0j P 0 case.

We can conclude that the global optimal solution ðT�; F�Þ thatmaximizes the annual total profit, Function (29), is as one of thepoints 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 incre-

mental discounts and partial backordering summarizes the detailsof the preceding theoretical results and their implications for theoptimal solution.

Solution procedure for the EOQ model with incremental discountsand 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 ATPjðT�j ; F�j Þ, using Formula (28). Compare

the profit ATPjðT�j ; F�j Þ with the profit from not stock-

ing, �p0jD, and take the higher profit. If the profitfrom 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

(1.2.2.i) If one of the Conditions (40a) or (40b) is satisfied,find F�j as the solution of Eq. (40c), and setT�j ¼ qj=ðDðF

�j þ bð1� F�j ÞÞÞ. Calculate the profit

ATPjðT�j ; F�j Þ using Formula (28).

(1.2.2.ii) Set F�j ¼ 1, and calculate T�j ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ=ðDiCjÞ

p. If

T�j < qj=D, set T�j ¼ qj=D, and if T�j > qjþ1=D, setT�j ¼ qjþ1=D. Calculate the profit PD� hðT�j Þ, wherehðTjÞ is given by Formula (42).

(1.2.2.iii) Calculate the profit from not stocking, �p0jD, and setT�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 ifDT�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 setT�j ¼ qjþ1=ðDðF

�j þ bð1� F�j ÞÞÞ. Calculate the profit

ATPjðT�j ; F�j Þ, using Formula (28).

(1.2.3.ii) Set F�j ¼ 1, and calculate T�j ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ=ðDiCjÞ

p. If

T�j < qj=D, set T�j ¼ qj=D, and if T�j > qjþ1=D, setT�j ¼ qjþ1=D. Calculate the profit PD� hðT�j Þ, wherehðTjÞ is given by Formula (42).

(1.2.3.iii) Calculate the profit from not stocking, �p0jD, and setT�j ¼ þ1 and F�j ¼ 0.

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

j T�j F�j ATCjðT�j ; F�j Þ

1 0.562731 > q2/D = 0.375 0.526316Correction (1) 0.375 0.526316 1315.532 1.10621 > q3/D = 0.75 0.555294Correction (2) 0.75 0.547945 1207.813 1.83941 0.591107 1089.06

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

1.3. If 0 6 b < b0j, set F�j ¼ 1 and calculate T�j ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞ=ðDiCjÞ

p.

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 accept-

able). Calculate the profit PD - hðT�j Þ ¼ PD� CjD

� iXj=2þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðAþ XjÞDiCj

p �, where hðTjÞ 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 hðTjÞ is givenby Formula (42). Compare the profit with the profitfrom not stocking, �p0jD, and take the higher profit,If the profit from not stocking is higher, setT�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=Dand calculate the profit PD� hðT�j Þ, where hðTjÞ isgiven by Formula (42). Compare the profit with theprofit from not stocking, �p0jD, and take the higherprofit. If the profit from not stocking is higher, setT�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 optimalpolicy is meeting all demand with incremental discount, calcu-late Q � ¼ DT�. If the optimal policy is losing all sales, thenQ � ¼ 0.

3. Numerical examples

We give numerical examples for both the full and partial back-ordering models with incremental discounts proposed in the abovesections. The solution procedures are coded in Wolfram Mathem-atica, using built-in functions to solve nonlinear equations.

Example 1 (EOQ model with incremental discounts and full backor-dering). We will use the values of all common parameters from thenumerical example for Taleizadeh and Pentico’s (2014) all-unitsdiscount model: P = $9/unit, D = 200 units/period, i = 0.3/period,p = $2/unit/period, C ¼ ðC1; C2;C3Þ = $(6,5,4)/unit, g = $2/unit,q ¼ ðq1; q2; q3Þ = (0,75,150) units, p0 ¼ ðp01;p02;p03Þ = $(5,6,7)/unit.We set the fixed order cost A to $30/order. Values for T�j ; F

�j and

ATCjðT�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 aftercorrecting T�j for not being in the interval qj=D 6 T�j 6 qjþ1=D. Then,ATCjðT�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� ¼F�3 ¼ 0:591107, with the optimal cost ATCðT�; F�Þ ¼ ATC3ðT�3; F

�3Þ ¼

1089:06. The optimal order quantity is Q� ¼ DT� ¼ 367:881, withthe maximum backordered quantity B� ¼ Dð1� F�ÞT� ¼ 150:424.

Example 2 (EOQ model with incremental discounts and partial back-ordering). We use the same values for the parameters as in Exam-ple 1, and we will vary the backordering parameterb ¼ 0:95;0:80;0:50. The results are displayed in Table 2. Rows thatare noted as ‘‘PBO correction (j)’’, display the values of T�j and F�jafter correcting Q �j ¼ DT�j ðF

�j þ bð1� F�j ÞÞ for not being into the

interval ½qj; qjþ1Þ, 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 correc-tions of the PBO model are done, then the row indicated with ‘‘NBOmodel (j)’’ is filled, and if Q �j ¼ DT�j from NBO model is not in theinterval ½qj; qjþ1Þ, 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 therow ‘‘not stocking (j)’’. The highest profit is taken as the over-allprofit.

According to Table 2, when b ¼ 0:95, the annual profit ismaximized for j = 3, under the partial backordering policy, withT� ¼ T�3 ¼ 1:85987 and F� ¼ F�3 ¼ 0:636287, with the optimal profitATPðT�; F�Þ ¼ ATP3ðT�3; F

�3Þ ¼ 686:411. The optimal order quantity is

Q� ¼ DT�ðF� þ bð1� F�ÞÞ ¼ 365:21 and the maximum backorderedquantity is B� ¼ bDð1� F�ÞT� ¼ 128:528.

For b ¼ 0:80, the annual profit is maximized for j = 3, under thepartial backordering policy, with T� ¼ T�3 ¼ 1:74378 and F� ¼ F�3 ¼0:805105, with the optimal profit ATPðT�; F�Þ ¼ ATP3ðT�3; F

�3Þ ¼

630:197. The optimal order quantity is Q � ¼ DT�ðF� þ bð1� F�ÞÞ ¼335:161, and the maximum backordered quantity isB� ¼ bDð1� F�ÞT� ¼ 54:3765.

For b ¼ 0:50, the annual profit is maximized for j = 3, under thepolicy 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 orderquantity is Q � ¼ DT� ¼ 291:548.

All examples showed that if the jth optimal quantity is not inthe jth interval ½qj; qjþ1Þ, then it cannot be the overall optimal quan-tity, even if it is corrected to the relevant interval endpoint. Thiswas proved for the EOQ model with incremental discount and nobackordering (see Appendix A). It is left to be proven that thismight be also true for the proposed EOQ models with incrementaldiscount – full and partial backordering respectively. From theexamples we can see that keeping all parameters fixed and byvarying the backordering rate, the total profit decreases whenthe backordering rate is decreasing.

4. Sensitivity analysis

There are at least two possible objectives for sensitivityanalysis:

1. Assess the relative impact of mis-estimation of different modelparameters on the model’s performance.

2. Assess the relative importance of the different model parame-ters in determining the values of the decision variables andthe performance function.

4.1. Study plan

Both objectives can be addressed by changing a single parame-ter’s value by given percentages, repeating the analysis for eachparameter of interest, using the same percentage changes.

Table 2Results for EOQ model with partial backordering and incremental discounts (Example 2).

j b0j T�j F�j Q�j Profit (j)

b = 0.951 0.853031 6 b 0.553432 0.635602 108.67 > q2

PBO correction (1) 0.381025 0.68374 75 466.148NBO model (1) 0.408248 1 81.6496 > q2

NBO correction (1) 0.375 1 75 452.5Not stocking (1) �10002 0.774202 6 b 1.11088 0.619732 217.952 > q3

PBO correction (2) 0.763765 0.639558 150 570.496NBO model (2) 0.83666 1 167.332 > q3

NBO correction (2) 0.75 1 150 536.25Not stocking (2) �12003 0.708905 6 b 1.85987 0.636287 365.21 686.411Not stocking (3) �1400

b = 0.801 0.853031 > bNBO model (1) 0.408248 1 81.6496 > q2

NBO correction (1) 0.375 1 75 452.5Not stocking (1) �10002 0.774202 6 b 0.920152 0.903821 180.49 > q3

PBO correction (2) 0.754037 0.973234 150 536.394NBO model (2) 0.83666 1 167.332 > q3

NBO correction (2) 0.75 1 150 536.25Not stocking (2) �12003 0.708905 6 b 1.74378 0.805105 335.161 630.197Not stocking (3) �1400

b = 0.501 0.853031 > bNBO model (1) 0.408248 1 81.6496 > q2

NBO correction (1) 0.375 1 75 452.5Not stocking (1) �10002 0.774202 > bNBO model (2) 0.83666 1 167.332 > q3

NBO correction (2) 0.75 1 150 536.25Not stocking (2) �12003 0.708905 > bNBO model (3) 1.45774 1 291.548 616.393Not stocking (3) �1400

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

The parameters in our model can be divided into two groups:(1) Parameters that have known values. (2) Parameters that areestimated. The second group can again be divided into at leasttwo groups: those for which the estimates are probably fairly accu-rate and those that are less certain. For this model the breakdownis:

Known: selling price ðPÞ, purchase cost ({Cj}), number of differ-ent unit costs ðnÞ, cost breakpoints ({qj})Estimated:

More confident: ordering cost ðAÞ, demand ðDÞ, holding costrate ðiÞ.Less confident: backordering rate (b), goodwill loss forstockout ðgÞ, backordering cost (p)

There is one other relevant parameter group, the lost sale costper unit ðfp0jgÞ, but that is derived from P, {Cj}, and g, so we donot need to consider it separately.

We use the problem solved in Example 2 with b = 0.80 as thebase case and then resolve it with changes of ±25%, ±20%, ±15%,±10%, and ±5% in each of the estimated parameters, keeping allthe other parameters constant. The performance measure ispercent reduction in the average profit per period (ATP) for thevariation relative to the optimal ATP from using the originalparameter values.

Base case parameters: P = $9/unit, D = 200 units/period, A = $30/order, i = 0.3/period, p = $2/unit/period, b ¼ 0:80, C ¼ ðC1;C2;C3Þ =$(6,5,4)/unit, q ¼ ðq1; q2; q3Þ = (0,75,150) units, g = $2/unit, p0 ¼ðp01;p02;p03Þ = $(5,6,7)/unit.

Base case optimal values: T⁄ = 1.74378, F⁄ = 0.805105,Q⁄ = 335.161, B⁄ = 54.3765, ATP⁄ = 630.197/period.

4.2. Study results

4.2.1. Effects of parameter changes on ATPThe details of the results of the changes in the estimated param-

eters are shown in Table 3. The percentage changes in ATP areshown graphically in Fig. 3. From these results we can draw thefollowing conclusions about how the estimated parameter changesaffected the ATP:

1. As would be expected, the further the changed parameter’svalue is from the value in the base case, the greater the decreasein the value of the ATP. There is one exception to this conclu-sion, b, for which the percentage changes in ATP are identicalfor changes in b of �15%, �20%, and �25%. The reason for thisis that changes in b by these percentages bring b below its crit-ical value for which partial backordering is optimal. As can beseen in those rows of Table 3, the optimal values of T and Ffor those cases are 1.45774 and 1.0, giving Q = 291.548, andB = 0. That is, the optimal solution for those cases is to use thebasic EOQ with no stockouts for these parameter sets. As shownin Example 2, the minimum value of b for which partial backor-dering is optimal when j = 3 is 0.708905, a reduction of 11.39percent from the base case value of 0.80. Note also that anincrease of 25% in the value of b increases its value to 1.0, whichmeans that all shortages will be backordered. This solution,which is shown in the last row of the b section of Table 3, resultsin a decrease in ATP of over 3.5 percent.

2. For all parameters except b and g, decreases in the parametervalue resulted in greater reductions from the base case valuethan did the same-sized increases. The reason for this difference

Table 3Sensitivity analysis for Example 2 problem with b = 0.80.

Parameter Change (%) Values of variables Changes in variables

T F Q B ATP T (%) F (%) Q (%) B (%) ATP (%)

A �25 1.71247 0.809171 329.422 52.286 630.157 �1.80 +0.51 �1.71 �3.84 �0.0063�20 1.71878 0.808341 330.578 52.707 630.172 �1.43 +0.40 �1.37 �3.07 �0.0040�15 1.72506 0.807519 331.730 53.127 630.183 �1.07 +0.30 �1.02 �2.30 �0.0022�10 1.73132 0.806706 332.878 53.545 630.191 �0.71 +0.20 �0.68 �1.53 �0.0010�5 1.73756 0.805902 334.021 53.961 630.195 �0.36 +0.10 �0.34 �0.76 �0.0002+5 1.74997 0.804317 336.297 54.790 630.195 +0.36 �0.10 +0.34 +0.76 �0.0002

+10 1.75615 0.803537 337.429 55.203 630.191 +0.71 �0.19 +0.68 +1.52 �0.0010+15 1.76230 0.802764 338.557 55.614 630.183 +1.06 �0.29 +1.01 +2.28 �0.0022+20 1.76843 0.802000 339.681 56.024 630.173 +1.41 �0.39 +1.35 +3.03 �0.0038+25 1.77455 0.801243 340.801 56.433 630.160 +1.76 �0.48 +1.68 +3.78 �0.0059

D �25 2.08822 0.754638 397.141 81.979 626.155 +19.75 �6.27 +18.49 +50.76 �0.6413�20 2.00781 0.765050 382.693 75.478 627.728 +15.14 �4.98 +14.18 +38.81 �0.3918�15 1.93400 0.775271 369.416 69.540 628.867 +10.91 �3.71 +10.22 +27.89 �0.2111�10 1.86587 0.785334 357.153 64.086 629.629 +7.00 �2.46 +6.56 +17.86 �0.0901�5 1.80267 0.795270 345.771 59.050 630.060 +3.38 �1.22 +3.17 +8.59 �0.0217+5 1.68868 0.814862 325.230 50.022 630.069 �3.16 +1.21 �2.96 �8.01 �0.0203

+10 1.63694 0.824560 315.902 45.950 629.701 �6.13 +2.42 �5.75 �15.50 �0.0786+15 1.58821 0.834220 307.110 42.127 629.114 �8.92 +3.62 �8.37 �22.53 �0.1718+20 1.54215 0.843858 298.798 38.527 628.324 �11.56 +4.81 �10.85 �29.15 �0.2972+25 1.49850 0.853490 290.919 35.127 627.344 �14.07 +6.01 �13.20 �35.40 �0.4526

i �25 1.85593 0.896370 363.493 30.773 623.762 +6.43 +11.34 +8.45 �43.41 �1.0211�20 1.82890 0.876837 356.770 36.041 626.307 +4.88 +8.91 +6.45 �33.72 �0.6172�15 1.80453 0.857954 350.653 41.012 628.127 +3.48 +6.56 +4.62 �24.58 �0.3285�10 1.78242 0.839714 345.057 45.712 629.325 +2.22 +4.30 +2.95 �15.94 �0.1384�5 1.76226 0.822104 339.912 50.160 629.990 +1.06 +2.11 +1.42 �7.75 �0.0328+5 1.72676 0.788700 330.758 58.379 630.010 �0.98 �2.04 �1.31 +7.36 �0.0297

+10 1.71103 0.772866 326.662 62.181 629.483 �1.88 �4.00 �2.54 +14.35 �0.1133+15 1.69645 0.757584 322.839 65.799 628.664 �2.71 �5.90 �3.68 +21.01 �0.2433+20 1.68287 0.742831 319.263 69.245 627.593 �3.49 �7.73 �4.74 +27.34 �0.4132+25 1.67020 0.728587 315.907 72.530 626.305 �4.22 �9.50 �5.74 +33.39 �0.6175

g �25 1.77433 0.782088 339.401 61.864 629.969 +1.75 �2.86 +1.26 +13.77 �0.0362�20 1.76847 0.786601 338.598 60.382 630.050 +1.42 �2.30 +1.03 +11.04 �0.0233�15 1.76248 0.791158 337.773 58.893 630.114 +1.07 �1.73 +0.78 +8.31 �0.0132�10 1.75637 0.795760 336.925 57.396 630.160 +0.72 �1.16 +0.53 +5.55 �0.0059�5 1.75014 0.800408 336.055 55.890 630.187 +0.36 �0.58 +0.27 +2.78 �0.0015+5 1.73729 0.809852 334.244 52.855 630.187 �0.37 +0.59 �0.27 �2.80 �0.0015

+10 1.73067 0.814651 333.303 51.325 630.158 �0.75 +1.19 �0.55 �5.61 �0.0061+15 1.72393 0.819503 332.339 49.786 630.110 �1.14 +1.79 �0.84 �8.44 �0.0138+20 1.71705 0.824410 331.351 48.240 630.041 �1.53 +2.40 �1.14 �11.29 �0.0247+25 1.71005 0.829375 330.338 46.684 629.952 �1.93 +3.01 �1.44 �14.15 �0.0389

p �25 1.82468 0.762912 347.631 69.218 629.389 +4.64 �5.24 +3.72 +27.29 �0.1281�20 1.80495 0.772790 344.586 65.616 629.728 +3.51 �4.01 +2.81 +20.67 �0.0743�15 1.78728 0.781858 341.861 62.381 629.975 +2.49 �2.89 +2.00 +14.72 �0.0381�10 1.77135 0.790214 339.406 59.457 630.099 +1.58 �1.85 +1.27 +9.34 �0.0155�5 1.75692 0.797939 337.184 56.801 630.174 +0.75 �0.89 +0.60 +4.46 �0.0036+5 1.73176 0.811771 333.313 52.155 630.178 �0.69 +0.83 �0.55 �4.09 �0.0030

+10 1.72072 0.817989 331.616 50.110 630.126 �1.32 +1.60 �1.06 �7.85 �0.0113+15 1.71055 0.823803 330.054 48.223 630.048 �1.91 +2.32 �1.52 �11.32 �0.0236+20 1.70114 0.829251 328.610 46.475 629.950 �2.44 +3.00 �1.95 �14.53 �0.0391+25 1.69242 0.834368 327.272 44.851 627.979 �2.94 +3.63 �2.35 �17.52 �0.0572

b �25 1.45774 1.0 291.548 0 616.393 �16.40 +24.21 �13.01 �100.00 �2.1904�20 1.45774 1.0 291.548 0 616.393 �16.40 +24.21 �13.01 �100.00 �2.1904�15 1.45774 1.0 291.548 0 616.393 �16.40 +24.21 �13.01 �100.00 �2.1904�10 1.50695 0.966650 299.381 8.0411 620.486 �13.28 +20.07 �10.68 �85.21 �1.5456�5 1.64754 0.873217 321.152 33.421 628.332 �5.52 +8.46 �4.18 �38.54 �0.2960+5 1.80699 0.750996 343.400 71.992 628.906 +3.63 �6.72 +2.46 +32.39 �0.2049

+10 1.84420 0.705254 347.098 86.971 625.654 +5.76 �12.40 +3.56 +59.94 �0.7208+15 1.86005 0.664648 347.060 99.804 620.955 +6.67 �17.45 +3.55 +83.54 �1.4665+20 1.85772 0.627097 343.934 110.840 614.982 +6.53 �22.11 +2.59 +103.84 �2.4143+25 1.83941 0.591107 337.796 120.339 607.705 +5.48 �26.58 +0.79 +121.31 �3.5690

A.A. Taleizadeh et al. / Computers & Industrial Engineering 82 (2015) 21–32 29

for b was just discussed. The reason for g is unclear, but we notethat the reductions in ATP for the same-sized negative and posi-tive 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 excep-tion of b and negative changes in i, the reductions in ATP areless than one percent from the base case, even for 25 percentchanges in the parameter value.

Since changes in the value of b in 5 percent decrements, whichmeans changes of 4 percentage points, quickly resulted in solutionsthat did not use partial backordering, we looked at the effects ofchanges in 1 � b, the complementary percentage of unfilleddemands 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 in1 � b are only one percentage point, which is much smaller thanthe changes in b, so we looked at the effect of 10 percent changes

00.51

1.52

2.53

3.54

-25 -20 -15 -10 -5 0 5 10 15 20 25

ATP

Red

uctio

n (%

)

Parameter Change (%)

Percent Reduction in ATP When Parameters Are Changed One at a Time

A

D

i

β

g

π

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

0

0.5

1

1.5

2

2.5

-50 -40 -30 -20 -10 0 10 20 30 40 50

ATP

redu

ctio

n (%

)

(1-β) change (%)

Reduction in ATP when (1-β) changes

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

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

Increase inparameter

Change in

T F Q B

A Increase Decrease Increase IncreaseD Decrease Increase Decrease Decreasei Decrease Decrease Decrease Increaseg Decrease Increase Decrease Decreasep Decrease Increase Decrease Decreaseb Nonmonotone Decrease Nonmonotone Increase

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

(2 percentage points each). As shown in Table 4 and Fig. 4, only a50% increase in 1 � b to 0.30 or b = 0.70, resulted in the solutionto the problem with a changed value of 1 � b not being partialbackordering. Since b = 0.70 is less than the minimum value of bfor which partial backordering is optimal, this large an increasein 1 � b results in the optimal solution for the altered case to bethe EOQ with no stockouts (F = 1.0).

4.2.2. Effects of parameter changes on decision variable valuesThe percentage changes in the values of the four decision vari-

ables that resulted from changing the parameter values are alsoshown in Tables 3 and 4. As was the case with the percentagechanges in ATP, there are similarities and differences among thevariables.

1. For all the parameters except b, the changes in T; F;Q , and Bwere consistent as the parameter value increased from �25%to +25%. However, these changes were not necessarily in thesame direction for all four variables. For A;D; g, and p, thechanges for T;Q , and B were in the same direction with F inthe opposite direction. For i the changes in T; F, and Q were inthe same direction, with B in the opposite direction. This issummarized in Table 5, which shows the direction of thechanges for the variables as each parameter increases in value.The inconsistent results for b are, as discussed above, due to thefact that large decreases in the value of b led to the basic EOQwithout backordering being optimal and an increase in b of25 percent to 1.0 led to full backordering being optimal. Ascan be seen in Table 4, these inconsistencies with respect to bdisappear when looking at the effects of changes in the valueof 1 � b.

2. The columns of Tables 3 and 4 that give the percentage changesin the decision variables also make it possible to see which vari-ables have the greatest impact on the values of ATP and thedecision variables. Looking only the results for ±25%, althoughthe same conclusions would be reached if the other sizes areconsidered, changes in A have the least effect on ATP, followed

Table 4Sensitivity analysis when parameter b changes its value (through changes in 1 � b with 1

Change (%) Value Values of variables

1 � b b T F Q B

�50 0.90 1.85457 0.684458 347.507 93.6315�40 0.88 1.84420 0.705254 347.098 86.9714�30 0.86 1.82852 0.727318 345.759 79.7765�20 0.84 1.80699 0.750996 343.400 71.9916�10 0.82 1.77900 0.776732 339.912 63.5511+10 0.78 1.70036 0.836901 328.978 44.3722+20 0.76 1.64754 0.873217 321.152 33.4208+30 0.74 1.58376 0.915651 311.408 21.3741+40 0.72 1.50695 0.96665 299.381 8.0411+50 0.70 1.45774 1.0 291.548 0

by g, p, D, i, and b (or 1 � b). The results for the changes in thevalues 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 (or1 � b) in some order having the greatest impact. To illustratethe sizes and directions of the effects of a parameter changegraphically, the relative changes in the four decision variablesas D changes are shown in Fig. 5.

4.2.3. ImplicationsOur 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 textson inventory control model, even relatively large changes in ormis-estimation of the value of a model parameter have rela-tively small effects on the value of the model’s performancemeasure. Our conclusion here is basically the same. The onlymodel parameter that generated changes in ATP of more thanapproximately one percent for a parameter change of ±25%was b. Thus, if the user’s interest is primarily finding a solutionthat will give a value of ATP close to the optimal without wor-rying about whether the values of the decision variables areapproximately correct, keeping the parameter estimates withinabout 25% of the true values should be sufficient.

0% increments).

Changes in variables

ATP T (%) F (%) Q (%) B (%) ATP (%)

623.469 +6.35 �14.99 +3.68 +72.19 �1.0676625.654 +5.76 �12.40 +3.56 +59.94 �0.7208627.482 +4.86 �9.66 +3.16 +46.71 �0.4307628.906 +3.63 �6.72 +2.46 +32.39 �0.2049629.848 +2.02 �3.52 +1.42 +16.87 �0.0553629.779 �2.49 +3.95 �1.84 �18.40 �0.0663628.332 �5.52 +8.46 �4.18 �38.54 �0.2960625.446 �9.18 +13.73 �7.09 �60.69 �0.7539620.456 �13.58 +20.07 �10.68 �85.21 �1.5456616.393 �16.40 +24.21 �13.01 �100.00 �2.1904

-40-30-20-100102030405060

-25 -20 -15 -10 -5 0 5 10 15 20 25

Cha

nges

(%)

D change (%)

Changes (%) in output variables, when parameter D changes

T

F

Q

B

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

A.A. Taleizadeh et al. / Computers & Industrial Engineering 82 (2015) 21–32 31

2. If, on the other hand, the user is equally as interested in havingthe values of T; F;Q , and B be approximately correct, then lessattention can be paid to estimating the values of A; g, and pand more attention needs to be paid to estimating the valuesof i;D, and b.

One final comment on sensitivity analysis is relevant. Due to therelative complexity of the equations for T and Fand, as a result, forthe ATP, we used, as is most frequently done in assessing the sen-sitivity of a model to changes in its inputs, a numerical approach inthis study. As was pointed out by Chu and Chung (2004) in theirdiscussion of sensitivity analysis of a basic EOQ with partial back-ordering, ‘‘the conclusions made by the analyses of sensitivitiesbased on the computational results of a set of numerical examplesare questionable since different conclusions may be made if differ-ent sets of numerical examples are analyzed.’’ While we are confi-dent that our conclusions above are fairly general, any user of thisor a similarly complex model needs to conduct his or her ownstudy.

5. Conclusion

We extended the basic EOQ model with incremental discountsby combining the basic solution procedure for that problem withrepeated use of Pentico and Drake’s (2009) models for the EOQwith full or partial backordering at a constant rate b to determinethe best order quantity for each possible cost. Minimum cost (ormaximum profit) was then used to choose among the best full(or partial) backordering solution, meeting all demand and losingall sales. We developed a condition under which partial backorder-ing is optimal and guarantees global optimal values of periodlength, fraction of demand that will be filled from stock, and orderquantity. We illustrated the developed models and proposed solu-tion procedures with examples. A numerical study based on one ofthe partial backordering example problems was used to evaluatethe sensitivity of the model’s results to the changes or mis-estima-tion of the various parameters. Extending the proposed model toinclude different fixed ordering costs for different price intervalsand also considering the pricing issue to determine the optimalselling price of the ordered quantity are some directions for futureresearch.

Funding

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

Appendix A–F. Supplementary material

Supplementary data associated with this article can be found, inthe online version, at http://dx.doi.org/10.1016/j.cie.2015.01.005.

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