research article inventory model with partial backordering...

Research ArticleInventory Model with Partial Backordering When BackorderedCustomers Delay Purchase after Stockout-Restoration

Ren-Qian Zhang1 Yan-Liang Wu1 Wei-Guo Fang1 and Wen-Hui Zhou2

1School of Economics and Management Beihang University Beijing 100191 China2School of Business Administration South China University of Technology Guangzhou 510640 China

Correspondence should be addressed to Wen-Hui Zhou whzhouscuteducn

Received 20 October 2015 Revised 11 January 2016 Accepted 14 January 2016

Academic Editor Young Hae Lee

Copyright copy 2016 Ren-Qian Zhang et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Many inventory models with partial backordering assume that the backordered demand must be filled instantly after stockoutrestoration In practice however the backordered customers may successively revisit the store because of the purchase delaybehavior producing a limited backorder demand rate and resulting in an extra inventory holding cost Hence in this paperwe formulate the inventory model with partial backordering considering the purchase delay of the backordered customers andassuming that the backorder demand rate is proportional to the remaining backordered demand Particularly wemodel the problemby introducing a new inventory cost component of holding the backordered items which has not been considered in the existingmodels We propose an algorithmwith a two-layer structure based on Lipschitz Optimization (LO) to minimize the total inventorycost Numerical experiments show that the proposed algorithm outperforms two benchmarks in both optimality and efficiencyWealso observe that the earlier the backordered customer revisits the store the smaller the inventory cost and the fill rate are but thelonger the order cycle is In addition if the backordered customers revisit the store without too much delay the basic EOQ withpartial backordering approximates our model very well

1 Introduction

It is observed usually in daily life that customers who haveincurred a stockout might not be avid to pay and fetch thebackordered item definitely at the next replenishment pointwhen the stocked out item is available again Consider thata customer visits an electronics store for a computer butincurs a stockout If the customer is willing to backorderthe demand the manager will inform herhim of the datewhen the computer is available again In an ideal situationas assumed in the existing model of economic order quantitywith partial backordering (EOQ-PBO) the customer willinstantly revisit the store and fetch the backordered computerexactly at the next replenishment point [1 2] It implies thatthe backordered customers generate an unlimited demandrate and the store need not keep any inventory for thebackordered demand

However in some practical situations it need take time tomeet the backordered customer and the customer who is not

in urgency is more possible to delay the purchase or decreasethe order frequency to the store after incurring a stockout[3ndash5] Thus among the customers who have backorderedthe computer because of the stockout some of them mayrepurchase the item immediately as soon as the inventoryis replenished again but some of them will revisit the storerelatively later after the stockout-restoration although theyhave incurred a stockout Similarly some other backorderedcustomers may revisit the store more lately and so on

Because of the purchase delay behavior the backorderedcustomers will repurchase the item generating a limited notan unlimited demand rate Consequently the vendor has tokeep inventory for the backordered customers resulting in anew cost component in the inventory model which thereforeleads to a different inventory policy for the vendor as a resultof the different cost structureThis paperwill reinvestigate theretailer-consumer inventory problem considering that thebackordered customers do not repurchase the item exactlyat the next replenishment point but rather revisit the store

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 6425403 16 pageshttpdxdoiorg10115520166425403

2 Mathematical Problems in Engineering

successively during the next in-stock period because of thepurchase delay behavior

2 Literature Review

Assuming that a fraction of unsatisfied customers incurringstockouts is willing to backorder their demand and besatisfied by the next replenishment led to the problem ofEOQ with partial backordering (EOQ-PBO) [1] The basicEOQ-PBO model assumes that a constant percentage 120573 ofthe demand during the stockout period will be backorderedand the remaining fraction 1 minus 120573 of the unsatisfied demandis lost [6ndash10] Among different variants of the model [2] oneof the most frequent extensions is to introduce a changingpattern for the backordering rate rather than assumingthat it is a constant percentage [11ndash15] These models haveconsidered the psychology regarding how the backorderedcustomers will wait the stockout-restoration before the nextreplenishment point which is often varying with the waitingtime

Some researches considered new demand patterns ratherthan using the constant demand rate The often adoptedpattern is that demand is a time-varying function [16ndash22]Another demand pattern adopted in EOQ-PBO is inventory-level-dependent demand which is also used in inventorymodel of perishable items [23ndash27] Besides some researchesassumed that the demand rate was dependent on price andtime [28 29]We refer the reader to the survey of Pentico andDrake [2] for a more extensive survey Recently the problemwas extended to consider prepayment delayed paymentlinear and fixed backordering costs discounts and so forth[30ndash33]

The above literature shows thatmost variants of the EOQ-PBO model are focused on different patterns of demand rateor backordering rate The backordering rate describes howthe unsatisfied customers will wait for the next replenishmentHowever it does not consider the ldquosequelrdquo after the customercompletes the wait that is how the backordered customersrevisit the store and fetch their backordered items after thenext replenishment arrives and the item is available again

Duran et al [34] pointed out that unsatisfied customersmay delay their purchase in the next replenishment periodbecause they wish to obtain greater customer utility (eitherincreased service or decreased price) More generally Green-leaf and Lehmann [35] and G-R Chen and M-L Chen[36] reported that customers may delay their purchase justbecause they do not have enough time feel unpleasant forshopping the product experience perceived risk rely onadvice from others wait for a lower price and so forth Itis also possible that one may delay occasionally a purchasebecause of unexpected events [37] for example unexpectedbusiness meeting obstructs the backordered customer revis-iting the store in time or even the backordered customerdelays the purchase due to avoidance and indecision [38]Moore and Fitzsimons [39] further showed that consumersmay respond negatively to the vendor after the stockout-restoration For online purchase situations although retailerscan promptly inform consumers of product availabilityconsumers do not immediately recover the negative effects of

the stockout [40] We can reasonably speculate that all thesefactors will considerably delay the repurchasing action of thebackordered customers

On the other hand in some practical situations theretailer may also delay delivering goods to the customers whohave lower priorities [41] Even if the backordered customersrequire that the vendor sends out the item as soon as it isavailable they need still wait a number of days until theproduct is delivered [42] In case of cash on delivery it meansthat the repurchase is delayed and the retailer has to bear theinventory cost for keeping the backordered item on passage

In sum despite the possible delay of vendor or customeran extra inventory of backordered demand should be heldwith longer time compared with the rigid assumption inmost of the existing EOQ-PBO models that the backordereddemandmust be fulfilled completely as soon as the new orderarrives We have checked the above-mentioned literaturesthat are related to the customerrsquos purchase delay behavior butnone of themhad introduced the factor of purchase delay intoinventory model [3ndash5 34ndash41]

This paper extends the EOQ-PBO problem consideringthe purchase delay of the backordered customer that isthe backordered demand will be fulfilled gradually duringthe subsequent replenishment period rather than being metoutright at the very beginning of the contiguous inventoryperiod The key of the problem is to model the endogenousdemand rate derived from the backordered customers and toformulate the inventory cost function considering new coststructure

The rest of the paper is organized as follows In Section 3we present the problem definition and further list the newcontributions of our study Section 4 formulates themodel byderiving the backorder demand rate and the total inventorycost function Section 5first briefly introduces LipschitzOpti-mization (LO) used in the proposed algorithm Subsequentlythe internal-layer algorithm for obtaining the optimal ordercycle with given fill rate is presented based on which theexternal-layer algorithm for finding the globally optimalsolution is proposed Section 6 evaluates the performance ofthe proposed algorithm and presents managerial insights bynumerical computations Conclusions are given in Section 7

3 Problem Definition

The traditional EOQ-PBO model assumes one constantdemand rate but the vendor in fact must handle two kinds ofdemandduring each inventory period (1) a stable demand forordinary customers who have not incurred any stockouts (wecall the corresponding demand per unit time routine demandrate) (2) an endogenous demand derived from the previous-period backordered customers which is characterized by thebackorder demand rate

Specifically we assume that the endogenous demand isdependent on the amount of the backordered customerswho have not been satisfied which thereby generates adecreasing and convex backorder demand rate Differentfrom the existing inventory models the inventory holdingcost in this case will include two parts the inventory holdingcost of the item for meeting the routine demand and the

Mathematical Problems in Engineering 3

Table 1 Cost components in related literature and our model

Inventory cost componentsfactors Papers reviewed Our modelStream I Stream II

Ordering cost radic times radic

Inventory holding cost of in-stock items radic times radic

Backorder cost of stockouts radic times radic

Opportunity cost of lost sales radic times radic

Repurchase delay of backordered customers times radic radic

Describeanalyze the behavior of purchase delay times radic radic

Endogenous demand rate of backordered customers times times radic

Inventory holding cost of backordered demand times times radic

Stream I

Montgomery et al [1] Rosenberg [6] Park [7 8] Wee [9] Padmanabhan and Vrat[23] Wee [16] Abad [28 29] Abboud and Sfairy [11] Skouri and Papachristos [17]Zhou et al [18] Ghosh and Chaudhuri [19] San Jose et al [12ndash14] Hou [24] Jain etal [25] Uthayakumar and Geetha [26] Pentico and Drake [10] Shah [20] Yang etal [27] Dye and Hsieh [15] Taleizadeh et al [30ndash32] and Wee et al [33]

Stream IIAjzen [37] Greenleaf and Lehmann [35] Darpy [38] Zinn and Liu [3] Duran et al[34] Liberopoulos and Tsikis [4] Li and Chen [41] G-R Chen and M-L Chen[36] Rao et al [5] Pizzi and Scarpi [40] and Moore and Fitzsimons [39]

inventory holding cost of the item for satisfying the endoge-nous backordered demand Moreover considering orderingcost and the opportunity cost of lost sales we will determinethe optimal fill rate and the optimal replenishment interval soas to minimize the total inventory cost

Our study is based on the literature regarding twostreams One is focused on the optimal inventory lot-sizingdetermined by minimizing the total inventory cost (StreamI in Table 1) The other is focused on customer purchasingbehavior referring mainly to the reason-explanation fordelaying purchase and the related statistical analysis (StreamII in Table 1) Because of the endogenous backordereddemand the components of inventory cost are differentfrom those of the existing inventory models We compareour model with the existing models in Table 1 for clearlyhighlighting the new contribution of this paper

To our knowledge this is the first time that the back-ordered customerrsquos purchase delay is introduced into theinventory model to minimize the total inventory cost Modelwith this assumption is more complicate on account thatwe must handle with a new inventory cost componentthat is the inventory holding cost of the item kept for thebackordered customers What is more a two-layer globaloptimization algorithm is needed for the solution because ofthe mathematical intractability

4 Formulation of the Model

41 Notations and Assumptions We keep all assumptions ofthe basic EOQ-PBO model [2 10] except assuming that thebackordered demand is filled instantly at the beginning of thevery next order cycle with an infinitely large demand rateIf the backordered customers delay their repurchasing andtherefore do not fetch the backordered items immediately

at the beginning of the next order cycle the correspondingdemand rate is limited Therefore the store will incur aninventory holding cost of the backordered demand because ofkeeping inventory for the backordered customers as in rightside of Figure 1

The notations in Notations section are used to model theproblem

Obviously the total order quantity is equal to 119863119865119879 + 119861It can be virtually divided into two parts the first part 119863119865119879is used for satisfying the routine demand of the first timevisiting (first-visiting) customers by the shelf stock duringin-stock period The corresponding demand rate is 119863 Thesecond part 119861 = 120573119863(1 minus 119865)119879 is ordered for satisfying thedemand that is backordered during the previous stockoutperiod and will be consumed according to the backorderdemand rate 119863119887(119905) We can see from the left side of Figure 1that if the backordered demand is satisfied instantly thenthe inventory of the order quantity 119861 for the backorderedcustomers becomes 0 immediately as soon as the inventoryis replenished without causing any inventory holding costotherwise the corresponding inventory level will graduallydecrease to 0 and the store must bear an extra inventoryholding cost (see the right side of Figure 1)

Different repurchasing behavior of backordered cus-tomers generates different backorder demand rate 119863119887(119905)Because of the complexity of the purchase delay factorswe consider an ldquoaverage effectrdquo of the purchase delay forfacilitating modeling the inventory problem That is all thebackordered customers are regarded as homogeneous so thatat any time during the next in-stock period each backorderedcustomer has the same probability of revisiting the storeand buying the backordered item It means that during anyunit time the amount of the backordered customers whoimplement the repurchasing action should be proportional


Backordered demand is satisfied instantly with Backordered demand is satisfied gradually

DFT DD DFT

TT TT

unlimited demand rate Db = infin (basic EOQ-PBO) with a limited demand rate Db(t)

Db(t) lt infin

Ib(t)Ib(t)B = 120573D(1 minus F)T B = 120573D(1 minus F)T

Db = infin

Figure 1 Two different cases of satisfying the backordered demand

to the remaining backordered customers which is equallyexpressed in Assumption 1 as follows

Assumption 1 The backorder demand rate is proportionalto the remained unsatisfied backordered customers that is119863119887(119905) = 120572B(119905) and 120572 is a positive real number

Moreover for the convenience in management a vendoroften commits one-time backorder for a backordered cus-tomer It is possible that a so-called ldquobackorderedrdquo customermight miss the entire very next in-stock period due toexcessive purchase delay (the delay time is longer than 119865119879)In this case the customer will suffer a second stockout andit is highly possible that the corresponding demand willbe lost forever due to customerrsquos impatience [43 44] Butin this case we should not fault the vendor again for theldquosecond-timerdquo stockout because it is completely caused bythe customerrsquos excessive purchase delay Thus this kind ofldquobackorderedrdquo customers does not lead to any inventory costand therefore does not impact the inventory model In factif a ldquobackorderedrdquo customer misses the next replenishmentagain and fades forever the store should not replenish forthe corresponding demand This is equal to the case thatthe customer does not backorder the item at all when theyincur the first time stockout Thus we can calibrate thebackordering rate 120573 to eliminate the ldquobackorderedrdquo customerwho incurs a second stockout because of the excessivepurchase delay time longer than 119865119879 Hence to facilitatemodeling the inventory problem we have Assumption 2

Assumption 2 All the backordered customers must be com-pletely satisfied during all the next in-stock period of [0 119865119879]

It is also possible that all backordered customers aresatisfied during a period of time that is shorter than 119865119879 If ithappens our formulation will not precisely model the inven-tory problem But we can approximate this case by adopting avery large attenuation coefficient which ensures that almostall the backordered customers are satisfied before 119865119879 Fromthis point we approximately assume in Assumption 2 thatthe backordered customers are satisfied during all the in-stock period of [0 119865119879]

42 The Backorder Demand Rate Based on the backorderdemand rate assumption (Assumption 1) we have a differen-tial equation as

119863119887 (119905) = minus119889B (119905)

119889119905= 120572B (119905) 120572 gt 0 (1)

The solution to the differential equation is B(119905) = B0119890minus120572119905

where B0 is a constant that is dependent on boundaryconditions Thus the backorder demand rate at time 119905 is

119863119887 (119905) = minus119889B (119905)

119889119905= 120572B0119890

minus120572119905 (2)

Thebackorder demand rate119863119887(119905) is an exponentially decreas-ing function in time 119905 where 120572 in (2) is the attenuationcoefficient of the backorder demand rate 120572B0 represents theinitial demand rate at time 119905 = 0 which is dependent onhow many demand has been backordered We can see thata larger 120572 corresponds to a larger backorder demand rate atthe beginning of order cycle but with a faster attenuation

Since all the backordered customers are satisfied duringthe next in-stock period (Assumption 2) the total backo-rdered demand 119861meets

119861 = int119865119879

119905=0

119863119887 (119905) 119889119905 = int119865119879

119905=0

120572B0119890minus120572119905119889119905

= B0 (1 minus 119890minus120572119865119879)

(3)

The total quantity of backordered demand during each ordercycle is 119861 = 120573119878 = 120573119863(1 minus 119865)119879 [10] Therefore B0 isdetermined by the following equation

B0 =120573119863 (1 minus 119865) 119879

1 minus 119890minus120572119865119879 (4)

43 The Inventory Cost Function

431 The Inventory Holding Cost of the Backordered DemandFigure 2 presents the inventory level curves where the upperpart of the figure indicates how the backordered demands aresatisfied in the coming order cycle and the lower part presentshow the routine demand is satisfied


D

T

FT

DFT

Sb

Ib(t)Ib(t)

B

BDb(t)

Figure 2 Inventory levels over time

The backorder demand rate 119863119887(119905) exponentially atten-uates and gradually reaches 0 so that the backorderedcustomers successively revisit the store during the in-stockperiod [0 119865119879] Obviously the inventory level correspondingto the backorder demand changes as

119868119887 (119905) = 119861 minus int119905

120591=0

119863119887 (120591) 119889120591 = 119861 minus int119905

120591=0

120572B0119890minus120572120591119889120591

= B0 (119890minus120572119905 minus 119890minus120572119865119879)

(5)

Regarding the area under the inventory level curve of thebackorder demand we have

119878119887 = int119865119879

119905=0

119868119887 (119905) 119889119905 = B0 [1

120572(1 minus 119890minus120572119865119879) minus 119890minus120572119865119879119865119879] (6)

Considering that B0 = 120573119863(1 minus 119865)119879(1 minus 119890minus120572119865119879) (see (4))the inventory holding cost per unit time of the backordereddemand is

119862ℎ119878119887119879

=119862ℎ119879B0 [

1

120572(1 minus 119890minus120572119865119879) minus 119890minus120572119865119879119865119879]

=120573119863119862ℎ (1 minus 119865)

120572minus

120573119863119862ℎ (1 minus 119865) 119865119879

119890120572119865119879 minus 1

(7)

432 The Total Inventory Cost Function From the existingmodels it is easy to know that the ordering cost of oneunit time is 119860119879 the backordering cost per unit timeis 120573119863119862119887(1 minus 119865)21198792 the opportunity cost of lost sales is119862119900119863(1 minus 120573)(1 minus 119865) and the holding cost of the inventory forsatisfying the first-visiting customers is119863119862ℎ119865

21198792 Summingup all the cost terms yields the total inventory cost per unittime as in the following

Γ (119879 119865) =119860

119879+

119863119862ℎ1198652 + 120573119863119862119887 (1 minus 119865)2

2119879

minus120573119863119862ℎ (1 minus 119865)

120572

120572119865119879

119890120572119865119879 minus 1+

120573119863119862ℎ (1 minus 119865)

120572

+ 119862119900119863(1 minus 120573) (1 minus 119865)

(8)

Introduce auxiliary functions as follows

119906 (119865) =[119863119862ℎ119865

2 + 120573119863119862119887 (1 minus 119865)2]

2

V (119865) =120573119863119862ℎ (1 minus 119865)

120572

119908 (119865) = V (119865) + 119862119900119863(1 minus 120573) (1 minus 119865)

120579 (119909) = 119909 (119890119909 minus 1)minus1

(9)

Then the total cost function is recast as

Γ (119879 119865) =119860

119879+ 119906 (119865) 119879 minus V (119865) 120579 (119909) + 119908 (119865) (10)

where 119909 = 120572119865119879 For the convenience we denote 1198911015840(119909) =119889119891(119909)119889119909 in this paper

Comparedwith the existing EOQ-PBOmodel the inven-tory total cost function of (10) has a new term of 120573119863119862ℎ(1 minus

119865)120572 minus (120573119863119862ℎ(1 minus 119865)120572)(120572119865119879(119890120572119865119879 minus 1)) It leads to theintractability in finding the closed-form solution whichforces us to develop an algorithm based on Lipschitz Opti-mization (LO)

Note that the backordered customermay prepay a depositfor the backordered item In this case the store will incurless financial cost for the capital of inventory where the unitinventory holding cost per unit time of the backordered itemshould be smaller than 119862ℎ say 119888ℎ lt 119862ℎ To adapt the modelfor this case we need only to substitute 119888ℎ for 119862ℎ in theauxiliary function V(119865) in the model To be concise we donot differentiate 119888ℎ and 119862ℎ in this paper

5 Solution

51 Solution Scheme and Lipschitz Optimization (LO)

511 Solution Scheme The convexity of the cost functionΓ(119879 119865) cannot be obtained and to derive a closed-formsolution for the problem of min Γ(119879 119865) is very difficult sowe consider a two-layer strategy to find the optimal solutionSpecifically in the internal layer we minimize functionΓ(119879 119865) supposing that 119865 is given say 119865 = This yieldsthe solution to the problem of min119879 Γ(119879 ) for any given119865 = isin [0 1] Define a new function

Γ () = min119879

Γ (119879 ) (11)

representing the minimal value of Γ(119879 119865) for 119865 = In the external layer of the algorithm we search the

optimal value of that minimizes function Γ() whichfinally yields the globally optimal solution to the problemof min Γ(119879 119865) However it is still very difficult to derivea closed-form solution even optimizing the single variableproblem min119879 Γ(119879 ) for 119865 = so we use LipschitzOptimization (LO) to solve the problem of min119879 Γ(119879 ) andminΓ() respectively

Function 119891(119909) is Lipschitzian on interval [119886 119887] if forany two points 119909 119910 isin [119886 119887] the absolute difference of the


a b a b a b

Lower bound Lower bound Lower bound

minusL minusL minusL +L+L+L

x1 = X(a b f L) x2 = X(x1 b f L)

x1x1 x2

x3 = X(a x1 f L)

Figure 3 The basic idea of Shubertrsquos algorithm

a b a

Lower bounds

b

+K

+K

+K

+K

+K

minusK

minusK

minusK

minusK

minusK

plusmnK998400

c1 a2 c2 b2 c3 ai ci bi a2 c2 b2 c3

Figure 4 Interval partition of the DIRECT algorithm

function is bounded by a constant multiplying the distancebetween 119909 and 119910 That is |119891(119909) minus 119891(119910)| le 119871|119909 minus 119910|where 119871 is the Lipschitz constant We have found that it iseasy to determine the Lipschitz constant for function Γ(119879 )

given 119865 = and therefore the LO algorithm with knownLipschitz constant can be applied to optimizing functionΓ(119879 ) However the Lipschitz constant of function Γ() isvery difficult to be obtained so we have to optimize functionΓ() without knowing Lipschitz constant Below we brieflyintroduce the basic idea of LO algorithms and the reader isreferred to the related literature for the details [45 46]

512 Shubertrsquos Algorithm with a Known Lipschitz ConstantThere are several LO algorithms for optimizing a Lipschitzfunction if the Lipschitz constant 119871 is known The passivealgorithm is the simplest one that evaluates the function onpoints 119886 + 120576119871 119886 + 3120576119871 119886 + 5120576119871 isin [119886 119887] where thefunction value on the best point does not differ more than120576 from the global optimal value

A more efficient LO method is Shubertrsquos algorithm Shu-bertrsquos algorithm uses piecewise linear functions with slopes of119871 and minus119871 to approximate the objective function from belowon interval [119886 119887] (see Figure 3)

Based on the piecewise linear function approximationthe algorithm continuously selects the interval with theminimal lower bound and partitions it into subintervals untiloptima are obtained under a given preciseness We refer thereader to Shubert [45] for the details In this paper we willadopt Shubertrsquos algorithm to minimize Γ(119879 ) obtaining theoptimal value of 119879 with given 119865

513 The DIRECT Algorithm without Lipschitz ConstantJones et al [46] proposed a new variant of LO algorithmnamed DIRECT which does not need to know the Lipschitzconstant It is different from Shubertrsquos algorithm that theDIRECT algorithm approximates the objective function fromthe middle point of the interval and partitions the selectedinterval (the interval with minimal lower bound) into threeequal parts as shown in Figure 4

Figure 4 shows that we first partition the interval intothree equal parts whose centers are 1198881 1198882 and 1198883 respectivelyThe two lines crossing the middle points 1198881 1198882 and 1198883 withslopesplusmn119870 provide lower bounds for the objective function onintervals [119886 1198862] [1198862 1198872] and [1198872 119887] respectively Then selectthe interval with minimal lower bound that is [119886 1198862] for thefurther trisection while updating 119870 with 1198701015840

Generally let 119888119894 = (119886119894 + 119887119894)2 denote the center ofinterval [119886119894 119887119894] The two lines crossing the middle point119888119894 with slopes plusmn119870 provide lower bound for the objectivefunction on interval [119886119894 119887119894] Denote 119888119894 = 119887119894 minus 119886119894 the widthof interval [119886119894 119887119894] It is noticeable that there are multipleintervals corresponding to each value of 119888119894 Each pair of 119888119894and 119891(119888119894) corresponds to a point with coordinates (119888119894 119891(119888119894))on a two-dimensional plane As long as DIRECT selects theinterval for the further trisection the value of 119870 can beupdated for the new iteration by finding the lower right ofthe convex hull of the set of points (119888119894 119891(119888119894)) which can berealized by the conhull algorithm proposed in Bjorkman andHolmstrom [47] The above procedure is repeated until theconvergence basin of the optimum is found Obviously theDIRECT algorithm continuously renews 119870 to determine theslope of the crossing lines for approximating the objective


function so the Lipschitz constant is not needed To savespace we refer the reader to Jones et al [46] for the detailsof the DIRECT algorithm

52 The Optimal Value of 119879 When 119865 Is Given

521 The First-Order Condition of Optimality If 119865 is givensay 119865 = we introduce two functions 119903(119879) and 119902(119879) asfollows

119903 (119879) =119860

119879+ 119906 () 119879

119902 (119879) = V () 120579 (120572119879)

(12)

Then the cost function with given 119865 is recast as

Γ (119879) = Γ (119879 ) = 119903 (119879) minus 119902 (119879) + 119908 () (13)

The optimal value of Γ(119879 ) defined as Γ() = min119879 Γ(119879 )in (11) is

Γ () = min119879

Γ (119879 ) = min119879

Γ (119879) (14)

Proposition 3 gives the first-order necessary condition forthe minima of function Γ(119879)

Proposition 3 The minima of function Γ(119879) solve the follow-ing equation

Γ1015840

(119879) = 1199031015840 (119879) minus 1199021015840 (119879) = 0 (15)

where

1199031015840 (119879) = minus119860

1198792+ 119906 ()

1199021015840 (119879) = 120572 sdot V () sdot 1205791015840 (120572119879)

(16)

Proof When 119879 rarr 0 119903(119879) rarr infin and 119902(119879) gt 0 so Γ(119879) rarr

infin On the other hand if 119879 rarr infin we also have Γ(119879) rarrinfin Moreover Γ(119879) is continuous and differentiable and fora given 119879 Γ(119879) has a finite value Hence the minima offunction Γ(119879) meet the first-order condition This completesthe proof

522 Bounds of the Optimal Value of 119879with Given 119865 Denote119865 the optimal value of119879with given119865 But even if119865 is knownthe convexity of function Γ(119879) cannot be obtained and it isalso difficult to solve the first-order necessary condition toobtain 119865 for minimizing Γ(119879) To use LO to find the optimalsolution for a given 119865 we first determine the bounds of 119865

Proposition 4 Let 119879119865 denote the upper bound of 119865 anddefine a sequence 119879

(119896)

119865 as

119879(119896)

119865= radic

119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909(119896minus1)) (17)

where 119909(119896minus1) = 120572119879(119896minus1)

119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906() Then 119865 le 119879(119896)

119865such that for any 119896 119879

(119896)

119865provides

an upper bound for 119865 and 119879119865 = lim119896rarrinfin 119879(119896)

119865is the minimal

upper bound of 119865

Proof See Appendix B

Proposition 5 Let 119879119865denote the lower bound of 119865 and

define a sequence 119879(119896)119865

as

119879(119896)119865

= radic119860


where 119909(119896minus1) = 120572119879(119896minus1)119865

(119896 = 1 infin) and 119879(0)119865

=

radic119860(119906() + 120572 sdot V()2) Then 119865 ge 119879(119896)119865

such that for any119896 119879(119896)119865

provides a lower bound for 119865 and 119879119865

= lim119896rarrinfin 119879(119896)119865

is the maximal lower bound of 119865


Propositions 4 and 5 give the bounds of 119865 For thelimits of upper and lower bounds when 119896 rarr infin we haveProposition 6

Proposition 6 The limits of upper and lower bounds bothsolve the first-order necessary condition given by (15)


Corollary 7 If 120572 rarr infin that is the backorder demand rateis infinitely large the upper and lower bounds are identicaland both solve the first-order condition of the basic EOQ-PBOmodel

Proof It is easy to verify that lim119909rarrinfin1205791015840(119909) = 0 (see

Lemma A2 in Appendix A) Thus when 120572 rarr infin we have119909 = 120572119865119879 rarr infin and so lim119896rarrinfin 119879

(119896)

119865= lim119896rarrinfin 119879(119896)

119865=

radic2119860119906(119865) = radic2119860119863[119862ℎ1198652 + 120573119862119887(1 minus 119865)2] This is equation

(9) of Pentico and Drake [10] which solves the first-ordercondition of the basic EOQ-PBO model The proof is com-pleted

Corollary 7 shows that the basic EOQ-PBO is a specialcase of our model If 120572 is sufficiently large we can use thebasic EOQ-PBO model as an approximation The numericalexperiment in Section 6 suggests that if 120572 gt 30 the basicEOQ-PBO can approximate our model very well

Corollary 8 For 120572 lt infin if the limits of upper and lowerbounds are identical that is 119879119865 = 119879

119865 then they are the

unique optimal solution to the problem of min119879 Γ(119879) that is119865 = 119879119865 = 119879

119865

Proof It is straightforward from Propositions 3 and 6

Corollary 8 allows us to improve the efficiency of findingthe optimal solution 119865 For a given 119865 we first check if the


limits of the upper and lower bounds are identical Onlywhen the two limits are not equal should we further usesome optimization algorithms such as the LO algorithm tominimize function Γ(119879) otherwise the optimal value 119865 =

119879119865 = 119879119865is attained Proposition 9 presents a sufficient

condition under which 119879119865 and 119879119865are equal to each other

Proposition 9 Theoptimal value of order cycle119879with given119865

can be determined by 119865 = 119879119865 = 119879119865if the following inequality

holds

12057222

V ()radic119860 lt 12 [119906 () minus 120572V () 1205791015840 (120572119879119865)]32

(19)


523 Lipschitz Optimization on 119879 with Given 119865 Based onthe bounds of 119879 determined by Propositions 4 and 5 wecan use LO to minimize Γ(119879) when 119865 is given for whichProposition 10 presents the Lipschitz constant

Proposition 10 For 1198791 1198792 isin [119879119865 119879119865] one has |Γ(1198791) minus

Γ(1198792)| le 119871sdot |1198791 minus 1198792| where 119871 is the Lipschitz constant that

is determined by

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(20)


Additionally Proposition 9 shows that if the limits of theupper and lower bounds are identical then the limit directlygives the optimal value of 119879 as 119865 = 119879119865 = 119879

119865 In this case

we do not need to use the LO method The procedure of thealgorithm optimizing 119879 with given 119865 is detailed as followsWe name this algorithm Opt Local for the convenience

Algorithm Opt Local

Step 1 For given119865 = calculate the limit of the upper bound119879119865 from (17) by iterations

Step 2 Check whether (19) holds If yes the optimal valueof 119879 with given 119865 is obtained by 119865 = 119879119865 and go to Step 6otherwise go to Step 3

Step 3 Calculate the limit of the lower bound 119879119865from (18)

by iterations

Step 4 If 119879119865= 119879119865 then the optimal value of 119879 with given 119865

is obtained by 119865 = 119879119865= 119879119865 and go to Step 6 otherwise go

to Step 5

Step 5 Determine the Lipschitz constant from (20) ApplyLipschitz Optimization (Shubertrsquos algorithm) to the problemof min119879 Γ(119879) and the optimal value 119865 is obtained

Step 6 Output the optimal solution 119865 and Γ() =

min119879 Γ(119879) = Γ(119865)

In Step 5 we set the stop criterion based on relativeimprovements Specifically if the relative improvement onthe objective function Γ(119879) between two successive iterationskeeps smaller than a very small positive number 120576 for 119901

consecutive iterations then stop In this paper we set 120576 = 10minus6

and 119901 = 10 Thus if the relative improvement keeps smallerthan 00001 for 10 consecutive iterations stop

53 Algorithm for Finding the Globally Optimal SolutionTo obtain the global minima we need minimize functionΓ() = min119879 Γ(119879 ) to find the optimal fill rate 119865 Theconvexity of Γ() cannot be obtained so the LO algorithmis employed again Unfortunately it is very difficult to deriveclosed-form expression for the Lipschitz constant of Γ()so we adopt the DIRECT algorithm that does not need theLipschitz constant The condition of using DIRECT is thatthe objective function should at least be continuous withinthe neighborhood of global optima [46]

Consider two values of 119865 say 119865 = and 119865 = 1015840 As

mentioned above with given 119865 the inventory cost functionΓ(119879 119865) is optimized by algorithm Opt Local yielding Γ() =

Γ( ) and Γ(

1015840

) = Γ(1015840 1015840

) respectively Since functionΓ(119879 119865) is differentiable and continuous in (119879 119865) if thedifference between

1015840 and is small enough Γ(1015840) can besufficiently close to Γ() that is lim

1015840

rarrΓ(1015840

) = Γ()Therefore Γ() is continuous in

The one-dimensional DIRECT algorithm optimizingfunction Γ() named Algorithm Opt Global is depicted asfollows

Algorithm Opt Global

Step 1 Divide the selected intervals within [0 1] by theDIRECT algorithm into three equal sections and yield can-didate points of isin [0 1]

Step 2 Determine the value of Γ() on each candidate pointof by using Algorithm Opt Local and obtain the currentbest solution

Step 3 Check whether the stop criterion is met If yes go toStep 5 otherwise go to Step 4

Step 4 Update 119870 the slope of the crossing linear functionsby calling the algorithm of conhull [47] and approximatethe lower bound of function Γ() for each interval by usingthe crossing linear function with slope 119870 Intervals with thesmallest lower bound of function Γ() are selected for thefurther trisection Go to Step 1

Step 5 Output the globally optimal solution

Note that in the first iteration after partitioning theoriginal interval [0 1] into three equal subintervals we select


the subinterval for the next trisection directly based on thefunction value of the middle point of each subinterval with-out approximating the objective function by the crossing lineswith slope 119870 119870 will be obtained and updated starting fromthe second iteration Moreover if the relative improvementon the objective function between two successive iterationskeeps smaller than 120576 = 10minus6 for 119901 = 10 consecutive iterationsthen stop In addition as Zhang [48] showed that we mustfurther compare Γlowast the optimal value of Γ(119879 119865) with 119862119900119863the opportunity cost of not stocking at all to determinethe optimal inventory policy The smaller one is the optimaldecision

6 Computational Study

61 Benchmark Algorithms To evaluate the performance ofAlgorithm Opt Global we use two algorithms as bench-marks The first one is a grid search (GS) algorithm forevaluating the optimality of Algorithm Opt Global and thesecond one is a two-dimensional DIRECT algorithm [46] forevaluating the computational efficiency

611 Grid Search on We use grid search (GS) to approxi-mately optimize function Γ() and compare the solutionwiththat obtained by Algorithm Opt Global For implementingthe GS algorithm we divide the interval isin [0 1] by stepsof 10minus4 For each problem instance we need to optimizefunction Γ(119879) = Γ(119879 ) for ten thousand times by callingAlgorithm Opt Local and keep the value of with minimalinventory cost as the optimal solution

612 Two-Dimensional DIRECT Algorithm The problemcan also be solved by using the two-dimensional DIRECTalgorithm (TD) proposed by Jones et al [46] Howeverthe two-dimensional split of the space can be very time-consuming if requiring a high precision We here comparethe performance of the Opt Global algorithm with that ofthe TD algorithm to show howmuch Opt Global will reducethe computational effort The stop condition of the TDalgorithm is the same as that of Algorithm Opt Global thatis the relative improvement keeps smaller than 10minus6 for 10consecutive iterations

62 Outline of Numerical Examples We investigate theperformance of our model and algorithm under differentscenarios by solving numerical examples The parameter ofthe numerical examples should cover different cases as manyas possibleThuswe enumerate different values of parametersto construct different examples referring to the parameterscheme of themajor item of the numerical examples in Zhanget al [49] as follows

(i) Ordering Cost We consider four values 119860 = $100order$1000order $2500order and $5000order respectively rep-resenting the difference in lower and higher ordering cost

(ii) Inventory Holding Cost and Backordering Cost Set 119862ℎ =$5unityear $10unityear $25unityear and $50unityear

representing the inventory holding cost changes from low tohigh Similarly we set backorder cost as 119862119887 = $5unityear$10unityear $25unityear or $50unityear respectively

(iii) OpportunityCost Set119862119900 =$5unit $10unit $25unit and$50unit respectively

(iv) Backordering Rate Backordering rate 120573 is changed from01 to 09 with a step of 02

(v) Demand Set the routine demand rate 119863 = 100 unitsyear1000 unitsyear 5000 unitsyear and 10000 unitsyearrespectively

(vi) Attenuation Coefficient of the BackorderDemandRate Set120572 = 01 05 1 5 10 50 100 500 representing the backorderdemand rate attenuates from slow to fast

By combination of 119860 times 119862ℎ times 119862119887 times 119862119900 times 120573 times 119863 times 120572 = 4 times4 times 4 times 4 times 5 times 4 times 8 we generate 40960 different numericalexamples for computational experiments

63 Performance of the Algorithm We apply respectively thethree algorithms Algorithm Opt Global grid search (GS)and the two-dimensional DIRECT (TD) algorithm to solvethe generated 40960 numerical examples Let ΓOpt ΓGS andΓTD denote the best value of the objective function obtainedby Opt Global GS and TD respectively Let (ΓGSminusΓOpt)ΓOptindicate the relative deviation of the solution found by GSfrom that obtained by the Opt Global algorithm and (ΓTD minusΓOpt)ΓOpt indicate the relative deviation of TD We group allthe numerical examples by the attenuation coefficient 120572 andthe computational results are presented in Table 2

It can be observed from Table 2 that AlgorithmOpt Global performs better than grid search (GS) and two-dimensional DIRECT (TD) This verifies the optimality ofOpt Global Moreover if we naively apply TD to solvethe problem it yields the worst solution among the threealgorithms

We also evaluate the computational efficiency through thecomputational time (on a PC with 34GHz Intel CPU and4GB memory) which is presented in Table 3

Table 3 shows that the Opt Global algorithm is muchmore efficient than GS and TD in terms of computationaltime of CPU In particular TD may consume more than onehour to get the solution for some cases but obtain inferiorresults compared with that of Opt Global (Table 2) Thus itis reasonable to conclude that the two-layer structure in theproposed algorithm significantly improves the computationalefficiency

Note that we have tried searching the optimal value of 119865by smaller steps of 10minus6 to improve the preciseness of the GSalgorithm but it cannot improve the solution significantlyWhat is more under the search step of 10minus6 it costs morethan 4000 seconds for the computer to solve one example byGS and we totally needmore than three years to complete thecomputation of all the numerical examples on our computerThus a search step of 10minus4 for 119865 is used with which all thenumerical examples were solved within 20 days by the GSalgorithm


Table 2 Relative deviations of the two benchmarks

120572Dev of ΓGS () Dev of ΓTD ()

Min Avg Max Min Avg Max01 lt10minus6 803 times 10minus6 206 times 10minus2 lt10minus6 243 times 10minus5 616 times 10minus1

05 lt10minus6 357 times 10minus4 388 times 10minus2 lt10minus6 325 times 10minus4 215 times 10minus1







Table 3 Computational times of the three algorithms

120572CPU time of Opt Global (s) CPU time of GS (s) CPU time of TD (s)

Min Avg Max Min Avg Max Min Avg Max01 041 190 285 4114 4441 5301 2902 1792 476905 040 168 274 4276 4426 5313 3058 1856 49571 041 192 314 4238 4419 5327 3125 1866 50235 041 192 375 4215 4412 5342 3117 1922 512910 039 187 400 4219 4250 5385 3211 1784 488350 039 159 426 4270 4292 4724 4259 1956 5212100 025 183 419 4223 4289 5230 3568 1829 5019500 042 195 400 4201 4329 5313 3689 1865 5217

64 Optimal Solution Changing with Attenuation Coefficient120572 We have shown that when 120572 rarr infin the optimal solutionof our model converges to the optima of the basic EOQ-PBO model (Corollary 7) Let 119879infin 119865infin and Γinfin representthe optimal order cycle the optimal fill rate and the cor-responding optimal inventory cost when 120572 rarr infin whichcan be obtained by solving the basic EOQ-PBO model [10]Correspondingly let119879120572119865120572 and Γ120572 indicate the optimal ordercycle the optimal fill rate and the optimal inventory costfor a limited value of 120572 lt infin Densify 120572 by respectivelysetting 120572 = 01 025 05 1 3 5 10 25 50 75 100 300 500Implementing the scheme in Section 62 to generate numer-ical examples each value of 120572 corresponds to 119860 times 119862ℎ times 119862119887 times119862119900 times 120573 times119863 = 4 times 4 times 4 times 4 times 5 times 4 = 5120 problem instancesSolving all the numerical examples by Opt Global for each 120572the average minimum and maximum values of the ratio of119865120572119865infin and 119879120572119879infin are presented in Figure 5

It can be observed that 119865120572 is decreasing in 120572 untilreaching 119865infin when 120572 rarr infin (119865120572119865infin rarr 1) which has anintuitive interpretation A smaller 120572 corresponds to a smallerbackorder demand rate so that the unsatisfied customer ofthe previous order cycle will fetch the backordered item laterIn this case the store must keep the backordered item for alonger time and incur a larger inventory holding cost Thusif 120572 is smaller the vendor should backorder less resulting in alarger fill rate to avoid toomuch inventory holding cost of thebackordered demand Conversely if 120572 increases the optimalfill rate 119865120572 decreases

Moreover another method for avoiding a high inventoryholding cost caused by the backordered demand in case of

F120572F

infin

35

30

25

20

15

10

Max(F120572Finfin)Avg(F120572Finfin)Min(F120572Finfin)

Max(T120572Tinfin)Avg(T120572Tinfin)Min(T120572Tinfin)

00

02

04

06

08

10

12

T120572T

infin540 1 2 3 6 7minus2 minus1minus3

ln 120572

Figure 5 The ratio of the optimal solution 119865120572119865infin and 119879120572119879infinunder different 120572

a small 120572 is to diminish the maximum backorder level bypromising a shorter stockout period Intuitively this can berealized by replenishing inventorymore frequently based on ashorter order cycle It follows that under a smaller120572 a shorterorder cycle ismore preferable In otherwords119879120572 is increasingin 120572 which can also be observed in Figure 5


minus3 minus2 minus1 0 1 2 3 4 5 6 7

000

005

010

015

020

025

(Γ120572minusΓinfin

)Γinfin

ln 120572

Max((Γ120572 minus Γinfin)Γinfin)Avg((Γ120572 minus Γinfin)Γinfin)Min((Γ120572 minus Γinfin)Γinfin)

Figure 6 The optimal inventory cost decreases in 120572 until reachingthat of the basic EOQ-PBO

In addition Figure 5 shows that for some numericalexamples the optimal order cycle and the fill rate are alwaysequal to their counterparts of the basic EOQ-PBO (119879120572119879infinand 119865120572119865infin are equal to 1) We have checked these examplesand found that if the basic EOB-PBOrsquos optimal policy is 119865infin =1 that is to meet all demands without any lost sales then 119865120572is also equal to 1 no matter what value of 120572 is In fact since119865120572 is decreasing in 120572 and 119865infin le 119865120572 le 1 it is an inevitableoutcome that 119865120572 must be equal to 1 in case of 119865infin = 1

Figure 6 presents the relative deviations of the optimalinventory cost Γ120572 from Γinfin where Γinfin represents the optimalinventory cost of the basic EOQ-PBO model

We can see that the inventory cost is decreasing in 120572and the optimal solution converges quickly to the case of120572 rarr infin as 120572 increases In particular when ln120572 gt 34 thatis 120572 gt 30 the optimal solution will be sufficiently close tothat of the basic EOQ-PBO with a relative deviation less than5That means if the backordered customers revisit the storewithout too much purchase delay the basic EOQ-PBO canapproximate our model very well

7 Conclusions

It is frequent that if demands are not filled and the unsatisfiedcustomers are willing to accept the backorder shehe willfetch the demanded item in the next order cycle The basicEOQ with partial backordering assumes that all backorderedcustomers will fetch their products simultaneously at thebeginning of the very next order cycle and the store will notincur inventory holding cost for the backordered demandHowever this assumption may be violated in some caseswhere the backordered customer may revisit the store suc-cessively because of the purchase delay and therefore thebackordered demand is not filled instantly

This paper proposed a new inventory model with partialbackordering by assuming that the backordered customersrepurchase the backordered products successively and thearriving rate of the backordered customer is proportionalto the backordered customers who have not been filledThis leads to an exponentially decreasing demand rate intime regarding the backordered demand based on whichwe formulated the total inventory cost function To solvethe model we developed an efficient algorithm based onLipschitz Optimization and conduct numerical experimentsproving the efficiency and effectiveness of the algorithm

From the theoretical results and numerical computationswe mainly find that (1) the larger the attenuation coefficientof the backorder demand rate the lower the fill rate withwhich the store will satisfy the customer to avoid holdingbackordered items for a longer time resulting in higherinventory holding cost (2) the more slowly the backorderedcustomer revisits the store the shorter the order cycle thatthe store should set is to diminish the backorder level andtherefore decrease the inventory holding cost of the backo-rdered demand (3) a larger backorder demand attenuationcoefficient leads to a smaller total inventory cost so the storeshould deliver the item to the backordered customers as soonas possible

Compared with the traditional EOQ-PBO model thenew parameter needed when implementing our model inthe real-world is the attenuation coefficient 120572 120572 reflects theproportion of the backordered customer who has revisitedthe store and fetched the backordered item To estimatethis parameter the vendor need record the number of thebackordered customers who revisit the store per unit timewhich is compared with the total number of the backorderedcustomers of the previous stockout Thereby the attenuationcoefficient 120572 can be calibrated

The study can be extended by combining the limitedbackorder demand rate considered in this paper with otherinventory problems for example deteriorating items time-varying demand rate and the vendor-retailer supply chainto further enrich the study of EOQ-PBO models Theconsumer behavior model describing how the backorderedcustomers repurchase their demanded items also deservesmore researches In addition this paper mainly concentrateson the theoretical analysis of the new inventory problem Toimplement a theoretical inventory model based on practicaldata is very important for the research on operations man-agement which will be carried out as the empirical study inthe future work

Appendices

A Lemmas

For the convenience of proof we first present some propertiesof function 120579(119909) = 119909(119890119909 minus 1)

minus1 which are straightforward byLrsquoHopitalrsquos rule (Bernoullirsquos rule) as Lemmas A1 and A2

LemmaA1 120579(119909) is decreasing and convex in 119909 isin (0infin) andlim119909rarr0 120579(119909) = 1 lim119909rarrinfin 120579(119909) = 0


TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)

Figure 7 The curves of functions 119903(119879) and 119902(119879)

Lemma A2 For 119909 isin (0infin) regarding the derivatives of 120579(119909)with respect to 119909 one has

lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)

Moreover regarding function 119902(119879) = V()120579(120572119879)Lemma A3 can be easily proved based on Lemma A2 asshown below

Lemma A3 119902(119879) is decreasing and convex in 119879 with given119865 In particular lim119879rarr0 119902(119879) = V() and lim119879rarrinfin 119902(119879) = 0lim119879rarr0 119902

1015840(119879) = minus120572 sdot V()2 and lim119879rarrinfin 1199021015840(119879) = 0

B Proofs

Proof of Proposition 4 From Lemma A3 the curves offunction 119903(119879) and function 119902(119879) are as in Figure 7

The right side of Figure 7 shows that the optimal value of119879 should not be larger than the optimum of function 119903(119879) =

119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)

Note that 119902(119879) is decreasing and convex in 119879 Thus if119865 le 119879

(0)

119865 then on the optimal point 119879 = 119865 the absolute

value of the first-order derivative of 119902(119879)must not be smallerthan the first-order derivative on 119879 = 119879

(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)

119865)| Proposition 3 shows that on the optimal point we

have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)

119865 Because 1199031015840(119879) le 0 that is minus119860

2

119865+

119906() le 0 if 119879 le 119879(0)

119865 and 1205791015840(119909) lt 0 it follows that

119860

2

119865

minus 119906 () ge minus120572 sdot V () sdot 1205791015840 (119909(0)) (B3)

Hence we have

119865 le radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909(0)) (B4)

Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)

119865and the upper bound of 119865

is diminishedSimilarly the absolute value of the first derivative of 119902(119879)

on the optimal point 119879 = 119865 must not be smaller thanthe derivative on 119879 = 119879

(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|

By Proposition 3 that 1199031015840(119865) = 1199021015840(119865) we have |1199031015840(119865)| =

|1199021015840(119865)| ge |1199021015840(119879(1)

119865)| which yields

119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865

From Lemma A2 1205791015840(119909) is strictly increasing in 119909 lt infinbecause 12057910158401015840(119909) gt 0 for a limited value of 119909 As mentionedabove 119879

(1)

119865lt 119879(0)

119865implying that 119909(1) lt 119909(0) so 1205791015840(119909(1)) lt

1205791015840(119909(0)) As a result we have 119906() minus 120572 sdot V() sdot 1205791015840(119909(1)) gt

119906() minus 120572 sdot V() sdot 1205791015840(119909(0)) and it follows that 119879(2)

119865lt 119879(1)

119865


The upper bound of 119865 is again diminished Repeating theabove derivation we can obtain a decreasing sequence 119879

(119896)

119865

that bounds the maximum value of 119865 as follows

119865 le 119879(119896)

119865= radic

119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909(119896minus1)) (B7)


119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)

119865of any 119896 gives an upper bound for 119865 and

119879119865 = lim119896rarrinfin 119879(119896)

119865is the best upper bound of 119865 The proof

is completed

Proof of Proposition 5 Because function 119902(119879) is convex anddecreasing in 119879 from the left side of Figure 7 we can seethat the absolute value of 1199021015840(119879) reaches its maximum value if119879 rarr 0 Thus |1199021015840(119879)| le |lim119879rarr0 119902

1015840(119879)| Proposition 3 showsthat on the optimal point 119879 = 119865 we have 1199031015840(119865) = 1199021015840(119865)Thus |1199031015840(119865)| = |1199021015840(119865)| le |lim119879rarr0 119902

1015840(119879)| In addition fromLemma A3 we have lim119879rarr0 119902

1015840(119879) = minus120572 sdot V()2 so that

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus120572 sdotV ()

2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)

Note that minus1198602

119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865

is a lower bound of 119865 Furthermore because119902(119879) is decreasing and convex in119879 then on the optimal point119879 = 119865 the absolute value of the first derivative of 119902(119879)

must not be larger than that on 119879 = 119879(0)119865 that is |1199021015840(119865)| le

|1199021015840(119879(0)119865

)| By Proposition 3 that 1199031015840(119865) = 1199021015840(119865) we have|1199031015840(119865)| = |1199021015840(119865)| le |1199021015840(119879(0)

119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)

where 119909(0) = 120572119879(0)119865 The above inequality yields

119860

2

119865

minus 119906 () le minus120572 sdot V () sdot 1205791015840 (119909(0)) (B11)

Hence we have

119865 ge 119879(1)119865

= radic119860


Lemma A2 shows that minus12 lt 1205791015840(119909) lt 0 It follows that119906()minus120572sdotV()sdot1205791015840(119909(0)) lt 119906()+120572sdotV()2 so we have119879(1)

119865gt

119879(0)119865 The lower bound of 119865 is increased by 119879(1)

119865 Repeating

the above procedure (18) yields an increasing sequence 119879(119896)119865

that bounds the minimum value of 119865 as

119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865

of any 119896 can be a lowerbound of 119865 and 119879

119865= lim119896rarrinfin 119879(119896)

119865is the best lower bound

The proof is completed

Proof of Proposition 6 For the sequence of the upper bound119879(119896)

119865= radic119860(119906() minus 120572 sdot V() sdot 1205791015840(119909(119896minus1))) if the limit exists

we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)

Thus the limit of the upper bound solves the first-ordernecessary condition Similarly it can be proved the limitof the lower bound also solves the first-order necessarycondition This completes the proof

Proof of Proposition 9 Suppose that function 119891(119909) and itsfirst-order derivative 1198911015840(119909) are continuous on interval [119886 119887]that is 119891(119909) 1198911015840(119909) isin C[119886 119887] According to Theorem 22 inMathews and Fink [50] if |1198911015840(119909)| le L lt 1 (L is a realnumber) and 119891(119909) isin [119886 119887] for all 119909 isin [119886 119887] then the fixed-point iteration 119901119899 = 119891(119901119899minus1) converges to the unique fixedpoint 119875 isin [119886 119887]

From Propositions 4 and 5 the iterative formula of theupper and lower bounds is

119879(119896)119865

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879(119896minus1)119865

)

(119896 = 1 infin)

(B16)

Define a function as120601(119879) = radic119860(119906() minus 120572 sdot V() sdot 1205791015840(120572119879))Obviously the limit of119879(119896)

119865is the fixed point of function 120601(119879)

Note that from Propositions 4 and 5 we have 119879 isin [119879119865 119879119865]

Propositions 4 and 5 show that119879(119896+1)119865

ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865

so 119879119865le 120601(119879) le 119879119865 for all 119879 isin [119879

119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus

120572V()1205791015840(120572119879)]minus3212057910158401015840(120572119879) Lemma A2 shows that1205791015840(120572119879) le 0 and 0 le 12057910158401015840(120572119879) le 16 so

0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)

According to Lemma A2 1205791015840(120572119879) is increasing in 119879 so

1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)

As mentioned above for function 120601(119879) if |1206011015840(119879)| le L lt1 then its fixed point is unique Note that 1206011015840(119879) ge 0 so asufficient condition ensuring 1206011015840(119879) le L lt 1 is

12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)

which yields 12057222V()radic119860 lt 12[119906() minus 120572V()1205791015840(120572119879119865)]32

This is (19) The proof is completed

Proof of Proposition 10 With given 119865 the first-order deriva-tive of the inventory cost function with respect to 119879 is

119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)

Considering that minus12 le 1205791015840(119909) le 0 if 119889Γ(119879)119889119879 ge 0 themaximum value of 119889Γ(119879)119889119879 is 119906() + 120572 sdot V()2 minus 119860119879

2

119865

If 119889Γ(119879)119889119879 le 0 the minimum value of 119889Γ(119879)119889119879 is 119906() minus1198601198792119865 so that the absolute value of 119889Γ(119879)119889119879 meets

100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)

Hence the Lipschitz constant for the model with given 119865 canbe taken as

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)

This completes the proof

Notations

119863 The routine demand rate that is the demandrate of the first time visiting customer who hasnot incurred any stockouts in unitsunit time

119860 The ordering cost for placing and receiving anorder in $order

119862119900 The opportunity cost of unit lost sale includingthe lost profit and goodwill loss in $unit

119862ℎ The cost to hold one unit item for one unittime in $unitunit time

119862119887 The cost to keep one unit backordered for oneunit time in $unitunit time

120572 Attenuation coefficient of the backorderdemand rate

120573 The backordering rate that is the fraction ofstockouts that will be backordered

119879 The order cycle that is the time intervalbetween two replenishments

119865 The fill rate that is the percentage of theroutine demand that is filled from the shelfstock

119861 The maximum backorder level during onestockout period in units

119905 The time elapses from the beginning of anorder cycle

B(119905) The unsatisfied backordered demand remainedat time 119905 during an in-stock period in units

119868119861(119905) The inventory level corresponding toB(119905) inunits

119863119887(119905) The backorder demand rate at time 119905 which isgenerated by the backordered customers whohave not been satisfied in unitsunit time

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The research of Ren-Qian Zhang is supported byNSFC under71271010 and 71571006 and the research of Wen-Hui Zhou issupported by NSFC under 71271089


References

[1] D C Montgomery M S Bazaraa and A K Keswani ldquoInven-tory models with a mixture of backorders and lost salesrdquo NavalResearch Logistics Quarterly vol 20 no 2 pp 255ndash263 1973

[2] D W Pentico and M J Drake ldquoA survey of deterministicmodels for the EOQ and EPQ with partial backorderingrdquoEuropean Journal of Operational Research vol 214 no 2 pp179ndash198 2011

[3] W Zinn and P C Liu ldquoConsumer response to retail stockoutsrdquoJournal of Business Logistics vol 22 no 1 pp 49ndash71 2001

[4] G Liberopoulos and I Tsikis ldquoDo stockouts undermine imme-diate and future sales Statistical evidence from a wholesalerrsquoshistorical datardquo Working Paper Department of MechanicalEngineering University of Thessaly Volos Greece 2008

[5] S Rao S EGriffis andT J Goldsby ldquoFailure to deliver Linkingonline order fulfillment glitches with future purchase behaviorrdquoJournal of OperationsManagement vol 29 no 7-8 pp 692ndash7032011

[6] D Rosenberg ldquoA new analysis of a lot-size model with partialbackorderingrdquoNaval Research Logistics Quarterly vol 26 no 2pp 349ndash353 1979

[7] K S Park ldquoInventory model with partial backordersrdquo Interna-tional Journal of Systems Science vol 13 no 12 pp 1313ndash13171982

[8] K S Park ldquoAnother inventory model with a mixture ofbackorders and lost salesrdquo Naval Research Logistics Quarterlyvol 30 no 3 pp 397ndash400 1983

[9] H M M Wee ldquoOptimal inventory policy with partial backo-rderingrdquoOptimal Control Applications andMethods vol 10 no2 pp 181ndash187 1989

[10] D W Pentico and M J Drake ldquoThe deterministic EOQ withpartial backordering a new approachrdquo European Journal ofOperational Research vol 194 no 1 pp 102ndash113 2009

[11] N E Abboud and R G Sfairy ldquoTime-limited free back-ordersEOQmodelrdquo Applied Mathematical Modelling vol 21 no 1 pp21ndash25 1997

[12] L A San Jose J Sicilia and J Garcıa-Laguna ldquoThe lot size-reorder level inventory system with customers impatiencefunctionsrdquo Computers amp Industrial Engineering vol 49 no 3pp 349ndash362 2005

[13] L A San Jose J Sicilia and J Garcıa-Laguna ldquoAnalysis ofan inventory system with exponential partial backorderingrdquoInternational Journal of Production Economics vol 100 no 1pp 76ndash86 2006

[14] L A San-Jose J Sicilia and J Garcıa-Laguna ldquoAn economiclot-sizemodel with partial backlogging hinging onwaiting timeand shortage periodrdquo Applied Mathematical Modelling vol 31no 10 pp 2149ndash2159 2007

[15] C-Y Dye and T-P Hsieh ldquoAn optimal replenishment policyfor deteriorating itemswith effective investment in preservationtechnologyrdquo European Journal of Operational Research vol 218no 1 pp 106ndash112 2012

[16] H-M Wee ldquoA deterministic lot-size inventory model fordeteriorating items with shortages and a declining marketrdquoComputers amp Operations Research vol 22 no 3 pp 345ndash3561995

[17] K Skouri and S Papachristos ldquoA continuous review inven-tory model with deteriorating items time-varying demandlinear replenishment cost partially time-varying backloggingrdquoApplied Mathematical Modelling vol 26 no 5 pp 603ndash6172002

[18] Y-W Zhou H-S Lau and S-L Yang ldquoA finite horizonlot-sizing problem with time-varying deterministic demandandwaiting-time-dependent partial backloggingrdquo InternationalJournal of Production Economics vol 91 no 2 pp 109ndash119 2004

[19] S K Ghosh and K S Chaudhuri ldquoAn EOQmodel for a deterio-rating itemwith trended demand and variable backloggingwithshortages in all cyclesrdquo Advanced Modeling and Optimizationvol 7 no 1 pp 57ndash68 2005

[20] N H Shah ldquoStorersquos optimal policy for time-dependent deteri-orating inventory with exponentially decreasing demand andpartial backloggingrdquo ASOR Bulletin vol 28 no 2 pp 9ndash142009

[21] J-T Teng J Min and Q Pan ldquoEconomic order quantitymodel with trade credit financing for non-decreasing demandrdquoOmega vol 40 no 3 pp 328ndash335 2012

[22] B Sarkar ldquoAn EOQ model with delay in payments and timevarying deterioration raterdquo Mathematical and Computer Mod-elling vol 55 no 3-4 pp 367ndash377 2012

[23] G Padmanabhan and P Vrat ldquoEOQ models for perishableitems under stock dependent selling raterdquo European Journal ofOperational Research vol 86 no 2 pp 281ndash292 1995

[24] K-L Hou ldquoAn inventory model for deteriorating items withstock-dependent consumption rate and shortages under infla-tion and time discountingrdquo European Journal of OperationalResearch vol 168 no 2 pp 463ndash474 2006

[25] S Jain M Kumar and P Advani ldquoAn inventory model withinventory level dependent demand rate deterioration partialbacklogging and decrease in demandrdquo International Journal ofOperations Research vol 5 no 3 pp 154ndash159 2008

[26] R Uthayakumar and K V Geetha ldquoA replenishment policy fornon-instantaneous deteriorating inventory system with partialbackloggingrdquo Tamsui Oxford Journal of Mathematical Sciencesvol 25 no 3 pp 313ndash332 2009

[27] H-L Yang J-T Teng and M-S Chern ldquoAn inventory modelunder inflation for deteriorating items with stock-dependentconsumption rate and partial backlogging shortagesrdquo Interna-tional Journal of Production Economics vol 123 no 1 pp 8ndash192010

[28] P L Abad ldquoOptimal pricing and lot-sizing under conditionsof perishability and partial backorderingrdquoManagement Sciencevol 42 no 8 pp 1093ndash1104 1996

[29] P L Abad ldquoOptimal pricing and lot-sizing under conditionsof perishability finite production and partial backordering andlost salerdquo European Journal of Operational Research vol 144 no3 pp 677ndash685 2003

[30] A A Taleizadeh D W Pentico M S Jabalameli and MAryanezhad ldquoAn economic order quantity model with multiplepartial prepayments and partial backorderingrdquo Mathematicaland Computer Modelling vol 57 no 3-4 pp 311ndash323 2013

[31] A A Taleizadeh D W Pentico M Saeed Jabalameli and MAryanezhad ldquoAn EOQmodel with partial delayed payment andpartial backorderingrdquo Omega vol 41 no 2 pp 354ndash368 2013

[32] A A Taleizadeh andDW Pentico ldquoAnEconomicOrderQuan-tity model with partial backordering and all-units discountrdquoInternational Journal of Production Economics vol 155 pp 172ndash184 2014

[33] H-M Wee Y-D Huang W-T Wang and Y-L Cheng ldquoAnEPQmodel with partial backorders considering two backorder-ing costsrdquo Applied Mathematics and Computation vol 232 pp898ndash907 2014


[34] S Duran T Liu D Simchi-Levi and J L Swann ldquoPolicies uti-lizing tactical inventory for service-differentiated customersrdquoOperations Research Letters vol 36 no 2 pp 259ndash264 2008

[35] E A Greenleaf and D R Lehmann ldquoReasons for substantialdelay in consumer decision makingrdquo The Journal of ConsumerResearch vol 22 no 2 pp 186ndash199 1995

[36] G-R Chen and M-L Chen ldquoThe link between purchase delayand resale price maintenance using the real options approachrdquoInternational Journal of Economics and Finance vol 3 no 1 pp171ndash177 2011

[37] I Ajzen ldquoFrom intentions to actions a theory of plannedbehaviorrdquo in Action Control From Cognition to Behavior JKuhl and J Beckmann Eds SSSP Springer Series in SocialPsychology pp 11ndash39 Springer Berlin Germany 1985

[38] D Darpy ldquoConsumer procrastination and purchase delayrdquo inProceedings of the 29th Annual Conference of the EuropeanMarketing Academy (EMAC rsquo00) Rotterdam The NetherlandsMay 2000

[39] S G Moore and G J Fitzsimons ldquoYes we have no bananasconsumer responses to restoration of freedomrdquo Journal ofConsumer Psychology vol 24 no 4 pp 541ndash548 2014

[40] G Pizzi and D Scarpi ldquoWhen out-of-stock products DObackfire managing disclosure time and justification wordingrdquoJournal of Retailing vol 89 no 3 pp 352ndash359 2013

[41] X P Li and Y R Chen ldquoImpacts of supply disruptions andcustomer differentiation on a partial-backordering inventorysystemrdquo Simulation Modelling Practice and Theory vol 18 no5 pp 547ndash557 2010

[42] S M Nowlis N Mandel and D B McCabe ldquoThe effect ofa delay between choice and consumption on consumptionenjoymentrdquo Journal of Consumer Research vol 31 no 3 pp502ndash510 2004

[43] S Netessine and N Rudi ldquoCentralized and competitive inven-tory models with demand substitutionrdquo Operations Researchvol 51 no 2 pp 329ndash335 2003

[44] D Huang H Zhou and Q-H Zhao ldquoA competitive multiple-product newsboy problem with partial product substitutionrdquoOmega vol 39 no 3 pp 302ndash312 2011

[45] B O Shubert ldquoA sequential method seeking the global maxi-mum of a functionrdquo SIAM Journal on Numerical Analysis vol9 no 3 pp 379ndash388 1972

[46] D R Jones C D Perttunen and B E Stuckman ldquoLipschitzianoptimization without the Lipschitz constantrdquo Journal of Opti-mizationTheory andApplications vol 79 no 1 pp 157ndash181 1993

[47] M Bjorkman and K Holmstrom ldquoGlobal optimization usingthe DIRECT algorithm in Matlabrdquo Applied Optimization andModeling vol 1 no 2 pp 17ndash37 1999

[48] R-Q Zhang ldquoA note on the deterministic EPQ with partialbackorderingrdquo Omega vol 37 no 5 pp 1036ndash1038 2009

[49] R-Q Zhang I Kaku and Y-Y Xiao ldquoDeterministic EOQ withpartial backordering and correlated demand caused by cross-sellingrdquo European Journal of Operational Research vol 210 no3 pp 537ndash551 2011

[50] J H Mathews and K D Fink Numerical Methods UsingMATLAB Prentice Hall Upper Saddle River NJ USA 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


successively during the next in-stock period because of thepurchase delay behavior

2 Literature Review

Assuming that a fraction of unsatisfied customers incurringstockouts is willing to backorder their demand and besatisfied by the next replenishment led to the problem ofEOQ with partial backordering (EOQ-PBO) [1] The basicEOQ-PBO model assumes that a constant percentage 120573 ofthe demand during the stockout period will be backorderedand the remaining fraction 1 minus 120573 of the unsatisfied demandis lost [6ndash10] Among different variants of the model [2] oneof the most frequent extensions is to introduce a changingpattern for the backordering rate rather than assumingthat it is a constant percentage [11ndash15] These models haveconsidered the psychology regarding how the backorderedcustomers will wait the stockout-restoration before the nextreplenishment point which is often varying with the waitingtime

Some researches considered new demand patterns ratherthan using the constant demand rate The often adoptedpattern is that demand is a time-varying function [16ndash22]Another demand pattern adopted in EOQ-PBO is inventory-level-dependent demand which is also used in inventorymodel of perishable items [23ndash27] Besides some researchesassumed that the demand rate was dependent on price andtime [28 29]We refer the reader to the survey of Pentico andDrake [2] for a more extensive survey Recently the problemwas extended to consider prepayment delayed paymentlinear and fixed backordering costs discounts and so forth[30ndash33]

The above literature shows thatmost variants of the EOQ-PBO model are focused on different patterns of demand rateor backordering rate The backordering rate describes howthe unsatisfied customers will wait for the next replenishmentHowever it does not consider the ldquosequelrdquo after the customercompletes the wait that is how the backordered customersrevisit the store and fetch their backordered items after thenext replenishment arrives and the item is available again

Duran et al [34] pointed out that unsatisfied customersmay delay their purchase in the next replenishment periodbecause they wish to obtain greater customer utility (eitherincreased service or decreased price) More generally Green-leaf and Lehmann [35] and G-R Chen and M-L Chen[36] reported that customers may delay their purchase justbecause they do not have enough time feel unpleasant forshopping the product experience perceived risk rely onadvice from others wait for a lower price and so forth Itis also possible that one may delay occasionally a purchasebecause of unexpected events [37] for example unexpectedbusiness meeting obstructs the backordered customer revis-iting the store in time or even the backordered customerdelays the purchase due to avoidance and indecision [38]Moore and Fitzsimons [39] further showed that consumersmay respond negatively to the vendor after the stockout-restoration For online purchase situations although retailerscan promptly inform consumers of product availabilityconsumers do not immediately recover the negative effects of

the stockout [40] We can reasonably speculate that all thesefactors will considerably delay the repurchasing action of thebackordered customers

On the other hand in some practical situations theretailer may also delay delivering goods to the customers whohave lower priorities [41] Even if the backordered customersrequire that the vendor sends out the item as soon as it isavailable they need still wait a number of days until theproduct is delivered [42] In case of cash on delivery it meansthat the repurchase is delayed and the retailer has to bear theinventory cost for keeping the backordered item on passage

In sum despite the possible delay of vendor or customeran extra inventory of backordered demand should be heldwith longer time compared with the rigid assumption inmost of the existing EOQ-PBO models that the backordereddemandmust be fulfilled completely as soon as the new orderarrives We have checked the above-mentioned literaturesthat are related to the customerrsquos purchase delay behavior butnone of themhad introduced the factor of purchase delay intoinventory model [3ndash5 34ndash41]

This paper extends the EOQ-PBO problem consideringthe purchase delay of the backordered customer that isthe backordered demand will be fulfilled gradually duringthe subsequent replenishment period rather than being metoutright at the very beginning of the contiguous inventoryperiod The key of the problem is to model the endogenousdemand rate derived from the backordered customers and toformulate the inventory cost function considering new coststructure

The rest of the paper is organized as follows In Section 3we present the problem definition and further list the newcontributions of our study Section 4 formulates themodel byderiving the backorder demand rate and the total inventorycost function Section 5first briefly introduces LipschitzOpti-mization (LO) used in the proposed algorithm Subsequentlythe internal-layer algorithm for obtaining the optimal ordercycle with given fill rate is presented based on which theexternal-layer algorithm for finding the globally optimalsolution is proposed Section 6 evaluates the performance ofthe proposed algorithm and presents managerial insights bynumerical computations Conclusions are given in Section 7

3 Problem Definition

The traditional EOQ-PBO model assumes one constantdemand rate but the vendor in fact must handle two kinds ofdemandduring each inventory period (1) a stable demand forordinary customers who have not incurred any stockouts (wecall the corresponding demand per unit time routine demandrate) (2) an endogenous demand derived from the previous-period backordered customers which is characterized by thebackorder demand rate

Specifically we assume that the endogenous demand isdependent on the amount of the backordered customerswho have not been satisfied which thereby generates adecreasing and convex backorder demand rate Differentfrom the existing inventory models the inventory holdingcost in this case will include two parts the inventory holdingcost of the item for meeting the routine demand and the












Stream I














DFT DD DFT

TT TT


Db(t) lt infin


Db = infin








119863119887 (119905) = minus119889B (119905)

119889119905= 120572B (119905) 120572 gt 0 (1)



119863119887 (119905) = minus119889B (119905)

119889119905= 120572B0119890

minus120572119905 (2)



119861 = int119865119879

119905=0

119863119887 (119905) 119889119905 = int119865119879

119905=0

120572B0119890minus120572119905119889119905

= B0 (1 minus 119890minus120572119865119879)

(3)


B0 =120573119863 (1 minus 119865) 119879

1 minus 119890minus120572119865119879 (4)




D

T

FT

DFT

Sb

Ib(t)Ib(t)

B

BDb(t)



119868119887 (119905) = 119861 minus int119905

120591=0

119863119887 (120591) 119889120591 = 119861 minus int119905

120591=0

120572B0119890minus120572120591119889120591


(5)


119878119887 = int119865119879

119905=0

119868119887 (119905) 119889119905 = B0 [1



119862ℎ119878119887119879

=119862ℎ119879B0 [

1


=120573119863119862ℎ (1 minus 119865)

120572minus

120573119863119862ℎ (1 minus 119865) 119865119879

119890120572119865119879 minus 1

(7)



Γ (119879 119865) =119860

119879+

119863119862ℎ1198652 + 120573119863119862119887 (1 minus 119865)2

2119879

minus120573119863119862ℎ (1 minus 119865)

120572

120572119865119879

119890120572119865119879 minus 1+

120573119863119862ℎ (1 minus 119865)

120572

+ 119862119900119863(1 minus 120573) (1 minus 119865)

(8)


119906 (119865) =[119863119862ℎ119865

2 + 120573119863119862119887 (1 minus 119865)2]

2

V (119865) =120573119863119862ℎ (1 minus 119865)

120572

119908 (119865) = V (119865) + 119862119900119863(1 minus 120573) (1 minus 119865)

120579 (119909) = 119909 (119890119909 minus 1)minus1

(9)


Γ (119879 119865) =119860

119879+ 119906 (119865) 119879 minus V (119865) 120579 (119909) + 119908 (119865) (10)





5 Solution



Γ () = min119879

Γ (119879 ) (11)





a b a b a b



x1 = X(a b f L) x2 = X(x1 b f L)

x1x1 x2

x3 = X(a x1 f L)


a b a

Lower bounds

b

+K

+K

+K

+K

+K

minusK

minusK

minusK

minusK

minusK

plusmnK998400















119903 (119879) =119860

119879+ 119906 () 119879

119902 (119879) = V () 120579 (120572119879)

(12)


Γ (119879) = Γ (119879 ) = 119903 (119879) minus 119902 (119879) + 119908 () (13)


Γ () = min119879

Γ (119879 ) = min119879

Γ (119879) (14)



Γ1015840

(119879) = 1199031015840 (119879) minus 1199021015840 (119879) = 0 (15)

where

1199031015840 (119879) = minus119860

1198792+ 119906 ()

1199021015840 (119879) = 120572 sdot V () sdot 1205791015840 (120572119879)

(16)





(119896)

119865 as

119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906() Then 119865 le 119879(119896)

119865such that for any 119896 119879

(119896)

119865provides







as

119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=


such that for any119896 119879(119896)119865


= lim119896rarrinfin 119879(119896)119865









(119896)

119865= lim119896rarrinfin 119879(119896)

119865=







119865









holds

12057222

V ()radic119860 lt 12 [119906 () minus 120572V () 1205791015840 (120572119879119865)]32

(19)





is determined by

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(20)



119865 In this case


Algorithm Opt Local




by iterations



to Step 5



min119879 Γ(119879) = Γ(119865)







Γ( ) and Γ(

1015840

) = Γ(1015840 1015840



1015840

rarrΓ(1015840















































F120572F

infin

35

30

25

20

15

10



00

02

04

06

08

10

12

T120572T


ln 120572





000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra




















Stream I














DFT DD DFT

TT TT


Db(t) lt infin


Db = infin








119863119887 (119905) = minus119889B (119905)

119889119905= 120572B (119905) 120572 gt 0 (1)



119863119887 (119905) = minus119889B (119905)

119889119905= 120572B0119890

minus120572119905 (2)



119861 = int119865119879

119905=0

119863119887 (119905) 119889119905 = int119865119879

119905=0

120572B0119890minus120572119905119889119905

= B0 (1 minus 119890minus120572119865119879)

(3)


B0 =120573119863 (1 minus 119865) 119879

1 minus 119890minus120572119865119879 (4)




D

T

FT

DFT

Sb

Ib(t)Ib(t)

B

BDb(t)



119868119887 (119905) = 119861 minus int119905

120591=0

119863119887 (120591) 119889120591 = 119861 minus int119905

120591=0

120572B0119890minus120572120591119889120591


(5)


119878119887 = int119865119879

119905=0

119868119887 (119905) 119889119905 = B0 [1



119862ℎ119878119887119879

=119862ℎ119879B0 [

1


=120573119863119862ℎ (1 minus 119865)

120572minus

120573119863119862ℎ (1 minus 119865) 119865119879

119890120572119865119879 minus 1

(7)



Γ (119879 119865) =119860

119879+

119863119862ℎ1198652 + 120573119863119862119887 (1 minus 119865)2

2119879

minus120573119863119862ℎ (1 minus 119865)

120572

120572119865119879

119890120572119865119879 minus 1+

120573119863119862ℎ (1 minus 119865)

120572

+ 119862119900119863(1 minus 120573) (1 minus 119865)

(8)


119906 (119865) =[119863119862ℎ119865

2 + 120573119863119862119887 (1 minus 119865)2]

2

V (119865) =120573119863119862ℎ (1 minus 119865)

120572

119908 (119865) = V (119865) + 119862119900119863(1 minus 120573) (1 minus 119865)

120579 (119909) = 119909 (119890119909 minus 1)minus1

(9)


Γ (119879 119865) =119860

119879+ 119906 (119865) 119879 minus V (119865) 120579 (119909) + 119908 (119865) (10)





5 Solution



Γ () = min119879

Γ (119879 ) (11)





a b a b a b



x1 = X(a b f L) x2 = X(x1 b f L)

x1x1 x2

x3 = X(a x1 f L)


a b a

Lower bounds

b

+K

+K

+K

+K

+K

minusK

minusK

minusK

minusK

minusK

plusmnK998400















119903 (119879) =119860

119879+ 119906 () 119879

119902 (119879) = V () 120579 (120572119879)

(12)


Γ (119879) = Γ (119879 ) = 119903 (119879) minus 119902 (119879) + 119908 () (13)


Γ () = min119879

Γ (119879 ) = min119879

Γ (119879) (14)



Γ1015840

(119879) = 1199031015840 (119879) minus 1199021015840 (119879) = 0 (15)

where

1199031015840 (119879) = minus119860

1198792+ 119906 ()

1199021015840 (119879) = 120572 sdot V () sdot 1205791015840 (120572119879)

(16)





(119896)

119865 as

119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906() Then 119865 le 119879(119896)

119865such that for any 119896 119879

(119896)

119865provides







as

119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=


such that for any119896 119879(119896)119865


= lim119896rarrinfin 119879(119896)119865









(119896)

119865= lim119896rarrinfin 119879(119896)

119865=







119865









holds

12057222

V ()radic119860 lt 12 [119906 () minus 120572V () 1205791015840 (120572119879119865)]32

(19)





is determined by

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(20)



119865 In this case


Algorithm Opt Local




by iterations



to Step 5



min119879 Γ(119879) = Γ(119865)







Γ( ) and Γ(

1015840

) = Γ(1015840 1015840



1015840

rarrΓ(1015840















































F120572F

infin

35

30

25

20

15

10



00

02

04

06

08

10

12

T120572T


ln 120572





000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











DFT DD DFT

TT TT


Db(t) lt infin


Db = infin








119863119887 (119905) = minus119889B (119905)

119889119905= 120572B (119905) 120572 gt 0 (1)



119863119887 (119905) = minus119889B (119905)

119889119905= 120572B0119890

minus120572119905 (2)



119861 = int119865119879

119905=0

119863119887 (119905) 119889119905 = int119865119879

119905=0

120572B0119890minus120572119905119889119905

= B0 (1 minus 119890minus120572119865119879)

(3)


B0 =120573119863 (1 minus 119865) 119879

1 minus 119890minus120572119865119879 (4)




D

T

FT

DFT

Sb

Ib(t)Ib(t)

B

BDb(t)



119868119887 (119905) = 119861 minus int119905

120591=0

119863119887 (120591) 119889120591 = 119861 minus int119905

120591=0

120572B0119890minus120572120591119889120591


(5)


119878119887 = int119865119879

119905=0

119868119887 (119905) 119889119905 = B0 [1



119862ℎ119878119887119879

=119862ℎ119879B0 [

1


=120573119863119862ℎ (1 minus 119865)

120572minus

120573119863119862ℎ (1 minus 119865) 119865119879

119890120572119865119879 minus 1

(7)



Γ (119879 119865) =119860

119879+

119863119862ℎ1198652 + 120573119863119862119887 (1 minus 119865)2

2119879

minus120573119863119862ℎ (1 minus 119865)

120572

120572119865119879

119890120572119865119879 minus 1+

120573119863119862ℎ (1 minus 119865)

120572

+ 119862119900119863(1 minus 120573) (1 minus 119865)

(8)


119906 (119865) =[119863119862ℎ119865

2 + 120573119863119862119887 (1 minus 119865)2]

2

V (119865) =120573119863119862ℎ (1 minus 119865)

120572

119908 (119865) = V (119865) + 119862119900119863(1 minus 120573) (1 minus 119865)

120579 (119909) = 119909 (119890119909 minus 1)minus1

(9)


Γ (119879 119865) =119860

119879+ 119906 (119865) 119879 minus V (119865) 120579 (119909) + 119908 (119865) (10)





5 Solution



Γ () = min119879

Γ (119879 ) (11)





a b a b a b



x1 = X(a b f L) x2 = X(x1 b f L)

x1x1 x2

x3 = X(a x1 f L)


a b a

Lower bounds

b

+K

+K

+K

+K

+K

minusK

minusK

minusK

minusK

minusK

plusmnK998400















119903 (119879) =119860

119879+ 119906 () 119879

119902 (119879) = V () 120579 (120572119879)

(12)


Γ (119879) = Γ (119879 ) = 119903 (119879) minus 119902 (119879) + 119908 () (13)


Γ () = min119879

Γ (119879 ) = min119879

Γ (119879) (14)



Γ1015840

(119879) = 1199031015840 (119879) minus 1199021015840 (119879) = 0 (15)

where

1199031015840 (119879) = minus119860

1198792+ 119906 ()

1199021015840 (119879) = 120572 sdot V () sdot 1205791015840 (120572119879)

(16)





(119896)

119865 as

119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906() Then 119865 le 119879(119896)

119865such that for any 119896 119879

(119896)

119865provides







as

119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=


such that for any119896 119879(119896)119865


= lim119896rarrinfin 119879(119896)119865









(119896)

119865= lim119896rarrinfin 119879(119896)

119865=







119865









holds

12057222

V ()radic119860 lt 12 [119906 () minus 120572V () 1205791015840 (120572119879119865)]32

(19)





is determined by

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(20)



119865 In this case


Algorithm Opt Local




by iterations



to Step 5



min119879 Γ(119879) = Γ(119865)







Γ( ) and Γ(

1015840

) = Γ(1015840 1015840



1015840

rarrΓ(1015840















































F120572F

infin

35

30

25

20

15

10



00

02

04

06

08

10

12

T120572T


ln 120572





000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










D

T

FT

DFT

Sb

Ib(t)Ib(t)

B

BDb(t)



119868119887 (119905) = 119861 minus int119905

120591=0

119863119887 (120591) 119889120591 = 119861 minus int119905

120591=0

120572B0119890minus120572120591119889120591


(5)


119878119887 = int119865119879

119905=0

119868119887 (119905) 119889119905 = B0 [1



119862ℎ119878119887119879

=119862ℎ119879B0 [

1


=120573119863119862ℎ (1 minus 119865)

120572minus

120573119863119862ℎ (1 minus 119865) 119865119879

119890120572119865119879 minus 1

(7)



Γ (119879 119865) =119860

119879+

119863119862ℎ1198652 + 120573119863119862119887 (1 minus 119865)2

2119879

minus120573119863119862ℎ (1 minus 119865)

120572

120572119865119879

119890120572119865119879 minus 1+

120573119863119862ℎ (1 minus 119865)

120572

+ 119862119900119863(1 minus 120573) (1 minus 119865)

(8)


119906 (119865) =[119863119862ℎ119865

2 + 120573119863119862119887 (1 minus 119865)2]

2

V (119865) =120573119863119862ℎ (1 minus 119865)

120572

119908 (119865) = V (119865) + 119862119900119863(1 minus 120573) (1 minus 119865)

120579 (119909) = 119909 (119890119909 minus 1)minus1

(9)


Γ (119879 119865) =119860

119879+ 119906 (119865) 119879 minus V (119865) 120579 (119909) + 119908 (119865) (10)





5 Solution



Γ () = min119879

Γ (119879 ) (11)





a b a b a b



x1 = X(a b f L) x2 = X(x1 b f L)

x1x1 x2

x3 = X(a x1 f L)


a b a

Lower bounds

b

+K

+K

+K

+K

+K

minusK

minusK

minusK

minusK

minusK

plusmnK998400















119903 (119879) =119860

119879+ 119906 () 119879

119902 (119879) = V () 120579 (120572119879)

(12)


Γ (119879) = Γ (119879 ) = 119903 (119879) minus 119902 (119879) + 119908 () (13)


Γ () = min119879

Γ (119879 ) = min119879

Γ (119879) (14)



Γ1015840

(119879) = 1199031015840 (119879) minus 1199021015840 (119879) = 0 (15)

where

1199031015840 (119879) = minus119860

1198792+ 119906 ()

1199021015840 (119879) = 120572 sdot V () sdot 1205791015840 (120572119879)

(16)





(119896)

119865 as

119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906() Then 119865 le 119879(119896)

119865such that for any 119896 119879

(119896)

119865provides







as

119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=


such that for any119896 119879(119896)119865


= lim119896rarrinfin 119879(119896)119865









(119896)

119865= lim119896rarrinfin 119879(119896)

119865=







119865









holds

12057222

V ()radic119860 lt 12 [119906 () minus 120572V () 1205791015840 (120572119879119865)]32

(19)





is determined by

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(20)



119865 In this case


Algorithm Opt Local




by iterations



to Step 5



min119879 Γ(119879) = Γ(119865)







Γ( ) and Γ(

1015840

) = Γ(1015840 1015840



1015840

rarrΓ(1015840















































F120572F

infin

35

30

25

20

15

10



00

02

04

06

08

10

12

T120572T


ln 120572





000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










a b a b a b



x1 = X(a b f L) x2 = X(x1 b f L)

x1x1 x2

x3 = X(a x1 f L)


a b a

Lower bounds

b

+K

+K

+K

+K

+K

minusK

minusK

minusK

minusK

minusK

plusmnK998400















119903 (119879) =119860

119879+ 119906 () 119879

119902 (119879) = V () 120579 (120572119879)

(12)


Γ (119879) = Γ (119879 ) = 119903 (119879) minus 119902 (119879) + 119908 () (13)


Γ () = min119879

Γ (119879 ) = min119879

Γ (119879) (14)



Γ1015840

(119879) = 1199031015840 (119879) minus 1199021015840 (119879) = 0 (15)

where

1199031015840 (119879) = minus119860

1198792+ 119906 ()

1199021015840 (119879) = 120572 sdot V () sdot 1205791015840 (120572119879)

(16)





(119896)

119865 as

119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906() Then 119865 le 119879(119896)

119865such that for any 119896 119879

(119896)

119865provides







as

119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=


such that for any119896 119879(119896)119865


= lim119896rarrinfin 119879(119896)119865









(119896)

119865= lim119896rarrinfin 119879(119896)

119865=







119865









holds

12057222

V ()radic119860 lt 12 [119906 () minus 120572V () 1205791015840 (120572119879119865)]32

(19)





is determined by

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(20)



119865 In this case


Algorithm Opt Local




by iterations



to Step 5



min119879 Γ(119879) = Γ(119865)







Γ( ) and Γ(

1015840

) = Γ(1015840 1015840



1015840

rarrΓ(1015840















































F120572F

infin

35

30

25

20

15

10



00

02

04

06

08

10

12

T120572T


ln 120572





000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119903 (119879) =119860

119879+ 119906 () 119879

119902 (119879) = V () 120579 (120572119879)

(12)


Γ (119879) = Γ (119879 ) = 119903 (119879) minus 119902 (119879) + 119908 () (13)


Γ () = min119879

Γ (119879 ) = min119879

Γ (119879) (14)



Γ1015840

(119879) = 1199031015840 (119879) minus 1199021015840 (119879) = 0 (15)

where

1199031015840 (119879) = minus119860

1198792+ 119906 ()

1199021015840 (119879) = 120572 sdot V () sdot 1205791015840 (120572119879)

(16)





(119896)

119865 as

119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906() Then 119865 le 119879(119896)

119865such that for any 119896 119879

(119896)

119865provides







as

119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=


such that for any119896 119879(119896)119865


= lim119896rarrinfin 119879(119896)119865









(119896)

119865= lim119896rarrinfin 119879(119896)

119865=







119865









holds

12057222

V ()radic119860 lt 12 [119906 () minus 120572V () 1205791015840 (120572119879119865)]32

(19)





is determined by

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(20)



119865 In this case


Algorithm Opt Local




by iterations



to Step 5



min119879 Γ(119879) = Γ(119865)







Γ( ) and Γ(

1015840

) = Γ(1015840 1015840



1015840

rarrΓ(1015840















































F120572F

infin

35

30

25

20

15

10



00

02

04

06

08

10

12

T120572T


ln 120572





000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra















holds

12057222

V ()radic119860 lt 12 [119906 () minus 120572V () 1205791015840 (120572119879119865)]32

(19)





is determined by

119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(20)



119865 In this case


Algorithm Opt Local




by iterations



to Step 5



min119879 Γ(119879) = Γ(119865)







Γ( ) and Γ(

1015840

) = Γ(1015840 1015840



1015840

rarrΓ(1015840















































F120572F

infin

35

30

25

20

15

10



00

02

04

06

08

10

12

T120572T


ln 120572





000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra














































F120572F

infin

35

30

25

20

15

10



00

02

04

06

08

10

12

T120572T


ln 120572





000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











000

005

010

015

020

025


)Γinfin

ln 120572






7 Conclusions






Appendices

A Lemmas





TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










TT

r(T)

r(T)

q(T)

q(T)

T(0)F T(1)

F T(2)F

T(2)

F T(1)

F T(0)

F = radicAu(F)



lim119909rarr0

1205791015840 (119909) = minus1

2

lim119909rarrinfin

1205791015840 (119909) = 0

1205791015840 (119909) le 0

lim119909rarr0

12057910158401015840 (119909) =1

6

lim119909rarrinfin

12057910158401015840 (119909) = 0

12057910158401015840 (119909) ge 0

(A1)




B Proofs



119860119879 + 119906()119879 that is

119865 le 119879(0)

119865= radic

119860

119906 () (B1)


(0)



(0)

119865 that is |1199021015840(119865)| ge

|1199021015840(119879(0)


have 1199031015840(119865) = 1199021015840(119865) Thus |1199031015840(119865)| = |1199021015840(119865)| ge |1199021015840(119879(0)

119865)|

that is10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

ge10038161003816100381610038161003816120572 sdot V () sdot 1205791015840 (119909(0))

10038161003816100381610038161003816 (B2)

where 119909(0) = 120572119879(0)


2

119865+

119906() le 0 if 119879 le 119879(0)


119860

2

119865


Hence we have

119865 le radic119860


Denote

119879(1)

119865= radic

119860


Note that 1205791015840(119909) lt 0 so 119879(1)

119865lt 119879(0)




(1)

119865 that is |1199021015840(119865)| ge |1199021015840(119879

(1)

119865)|


|1199021015840(119865)| ge |1199021015840(119879(1)


119865 le 119879(2)

119865= radic

119860


where 119909(1) = 120572119879(1)

119865


(1)

119865lt 119879(0)




119865lt 119879(1)

119865



(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











(119896)

119865


119865 le 119879(119896)

119865= radic

119860



119865 (119896 = 1 infin) and 119879

(0)

119865=

radic119860119906()

Obviously 119879(119896)


119879119865 = lim119896rarrinfin 119879(119896)


is completed





10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816100381610038161003816


2

10038161003816100381610038161003816100381610038161003816100381610038161003816

(B8)


119865+ 119906() lt 0 which yields

119865 ge 119879(0)119865

= radic2119860

2119906 () + 120572 sdot V () (B9)

Thus 119879(0)119865



|1199021015840(119879(0)119865


119865)| that is

10038161003816100381610038161003816100381610038161003816100381610038161003816

minus119860

2

119865

+ 119906 ()

10038161003816100381610038161003816100381610038161003816100381610038161003816

le10038161003816100381610038161003816120572 sdot V () 1205791015840 (119909(0))

10038161003816100381610038161003816 (B10)


119860

2

119865


Hence we have

119865 ge 119879(1)119865

= radic119860



119865gt


119865 Repeating



119865 ge 119879(119896)119865

= radic119860



(119896 = 1 infin) and 119879(0)119865

=

radic2119860(2119906() + 120572 sdot V()) 119879119865

= 119879(119896)119865


119865= lim119896rarrinfin 119879(119896)





we have

119879119865 = radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (119909)

= radic119860

119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

(B14)

which is recast as

minus119860

1198792

119865

+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879119865)

= 1199031015840 (119879119865) minus 1199021015840 (119879119865) = 0

(B15)




119879(119896)119865

= radic119860


)

(119896 = 1 infin)

(B16)





ge 119879(119896)119865

and119879(119896+1)

119865le 119879(119896)

119865


119865 119879119865]


Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Moreover 1206011015840(119879) = (12057222

V()radic1198602)[119906() minus


0 le 1206011015840 (119879)

le12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879)]

minus32

(B17)


1206011015840 (119879) le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879)]minus32

le12057222

V ()radic119860

12[119906 ()

minus 120572V () 1205791015840 (120572119879119865)]minus32

(B18)


12057222

V ()radic119860

12[119906 () minus 120572V () 1205791015840 (120572119879119865)]

minus32

lt 1

(B19)




119889Γ (119879)

119889119879= 1199031015840 (119879) minus 1199021015840 (119879)

= minus119860

1198792+ 119906 () minus 120572 sdot V () sdot 1205791015840 (120572119879)

(B20)


2

119865


100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

or100381610038161003816100381610038161003816100381610038161003816

119889Γ (119879)

119889119879

100381610038161003816100381610038161003816100381610038161003816le

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B21)


119871= max

10038161003816100381610038161003816100381610038161003816100381610038161003816

119906 () +120572 sdot V ()

2minus

119860

1198792

119865

10038161003816100381610038161003816100381610038161003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816119906 () minus

119860

1198792119865

100381610038161003816100381610038161003816100381610038161003816

(B22)


Notations

















Acknowledgments



References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










References



























































Volume 2014




Journal of











Journal of


Function Spaces






Algebra


































Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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