Research Article Mathematical Programming...

Hindawi Publishing CorporationAdvances in Operations ResearchVolume 2013 Article ID 272648 7 pageshttpdxdoiorg1011552013272648

Research ArticleMathematical Programming Approach to theOptimality of the Solution for Deterministic InventoryModels with Partial Backordering

Irena Stojkovska

Department of Mathematics Faculty of Natural Sciences and Mathematics Saints Cyril and Methodius University Gazi Baba bb1000 Skopje Macedonia

Correspondence should be addressed to Irena Stojkovska irenastojkovskagmailcom

Received 26 March 2013 Accepted 5 June 2013

Academic Editor Michael N Katehakis

Copyright copy 2013 Irena Stojkovska This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We give an alternative proof of the optimality of the solution for the deterministic EPQ with partial backordering (EPQ-PBO)[Omega vol 37 no 3 pp 624ndash636 2009] Our proof is based on the mathematical programming theory We also demonstrate thedetermination of the optimal decision policy through solving the correspondingmathematical programming problemWe indicatethat the same approach can be used within other inventory models with partial backordering and we consider additional models

1 Introduction

A basic inventory control model is the economic order quan-tity (EOQ) model known by its simplicity and its restrictivemodeling assumptions Managing the inventory of a singleitem the purpose of the EOQ model is to determine howmuch to purchase (order quantity) and when to place theorder (the reorder point) By relaxing the assumption thatshortages are not allowed and allowing stockouts with partialbackordering the EOQwill extend to EOQwith partial back-ordering (EOQ-PBO) But when the assumption of instanta-neous replenishment is replaced with the assumption that thereplenishment order is received at a constant finite rate overtime EOQ is extended to the economic production quantity(EPQ) model and by allowing stockouts with partial backo-rdering it will extend to EPQwith partial backordering (EPQ-PBO) There are many papers that propose an inventorymodel with partial backordering a very few of them are [1ndash7]For a survey of the deterministic models for the EOQ andEPQ with partial backordering see Penticiorsquos and Drakersquospaper [8]

Pentico et al [5] proposed EPQ-PBO model whosemain characteristics are that the derived equations are morelike those for the classical EPQ model and it gives theoptimal solution in a closed form that helps in understanding

the behavior of the inventory system Let us briefly give thenotations from [5] that are used in the total cost function andin the solution of the Pentico et alrsquos EPQ-PBO as well Theyare

119863 demand per year119875 production rate per year if constantly producing1198620 the fixed cost of placing and receiving an order

119862ℎ the cost to hold a unit in inventory for a year

119862119887 the cost to keep a unit backordered for a year

1198621 the cost for a lost sale

120573 the backordering rate119879 the length of an order cycle (a decision variable)119865 the fill rate (a decision variable)

The derived total averaged cost per year for the Penticoet alrsquos EPQ-PBO model according to [5] that has to beminimized is

Γ (119879 119865) =1198620

119879+1198621015840

ℎ1198631198791198652

2+1205731198621015840

119887119863119879(1 minus 119865)

2

2+ 1198621015840

1119863(1 minus 119865)

(1)

2 Advances in Operations Research

where 1198621015840ℎ= 119862ℎ(1 minus 119863119875) 1198621015840

119887= 119862119887(1 minus 120573119863119875) and 119862

1015840

1=

1198621(1 minus 120573) Setting the first partial derivatives of Γ(119879 119865) equal

to zero Pentico et al [5] find the following solutions

119879lowast=radic

21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

(2)

119865lowast= 119865 (119879

lowast) =

1198621015840

1+ 1205731198621015840

119887119879lowast

119879lowast (1198621015840

ℎ+ 1205731198621015840

119887) (3)

Comparing the optimal time length of inventory cycle 119879lowast forthe partial backordering model with the one for the standardEPQ model if backordering is optimal Pentico et al [5]derived condition under which (119879

lowast 119865lowast) gives an optimal

solution that is

120573 ge 120573lowast= 1 minus radic

211986201198621015840

ℎ

1198631198622

1

(4)

where 120573lowast denotes the critical value of the backordering rateThen Pentico et al [5] determine the optimal inventory pol-icy in a way that if condition (4) is satisfied then the optimalpolicy is to allow stockouts with partial backordering andmeet the fractional demand otherwise the optimal policy iseither to meet all demand or to lose all sales When condition(4) is satisfied Pentico et al [5] proved the optimality of thesolution for the EPQ-PBO given with (2) and (3) by exam-ining the characteristics of the partial derivatives and theboundary conditions In this paper we are going to prove theoptimality of the solution by using the mathematical pro-gramming approach

Recognition of the (global) optimal solution for thegeneral nonlinear programming problem using the nonlinearprogramming theory is not an easy task There are sufficientconditions for the local optimality but there are very few ortoo restrictive for the global optimality Among the inventorymodels there are many with a convex objective function butmost problems arise when the objective function is noncon-vex because otherwise the local optimal solution is globalIn this case with the nonconvex objective proving the globaloptimality of the solution for the model is usually done byexamining the partial derivatives and the boundary condi-tions some of them are [2 5 9 10] There are some globaloptimality theories for general nonlinear programming prob-lems that are quite effective in practice for certain classes ofproblems One of them is the canonical duality and trivialtheory proposed by Gao and Sherali which has good per-formance for quadratic programming problems inmechanics[11] Algebraical approach is another approach that is usedwith inventory models see [12 13] Proving the optimalityof the solution becomes more complex when there is someuncertainty in the inventory system see [14 15]

The paper is organized as follows In Section 2 Pentico etalrsquos EPQ-PBO is formulated as a mathematical programmingproblem then mathematical programming approach forproving the optimality of the solution is presented and thesame approach is used for two other inventory models withpartial backordering an extension of Pentico et alrsquos EPQ-PBO [6] and Hu et alrsquos inventory model [2] There are

comments on howmathematical programming approach canbe used for the models presented in Taleizadeh et al [16]Taleizadeh et al [10] Taleizadeh and Pentico [9] andCardenas-Barron [12]

2 Mathematical Programming Approach

21 Formulation as a Mathematical Programming ProblemFor the decision variables the length of an order cycle 119879

and the fill rate 119865 there are natural restrictions that is thefollowing conditions should be met

119879 gt 0 0 le 119865 le 1 (5)

So the corresponding mathematical programming problemfor Pentico et alrsquos EPQ-PBO [5] is a box-constrained nonlin-ear programming problem and is formulated as follows

min(119879119865)isinR2

Γ (119879 119865)

subject to 119879 gt 0 0 le 119865 le 1

(6)

where Γ(119879 119865) is given with (1) Denote the feasible region forthe problem (6) by F = (119879 119865) isin R2 | 119879 gt 0 0 le 119865 le 1We should note that in their later work on the extension ofthe EPQ-PBO [1] the authors impose these conditions on thedecision variables119879 and 119865 and formulated the correspondingmathematical programming problem and for the proof of theoptimality of the solution they still have the same approach asthat for the basic EPQ-PBO [5]

In the next subsection using the nonlinear programmingtheory we are going to prove that the solution (119879

lowast 119865lowast) given

by (2) and (3) is a global minimizer of the problem (6) ifthe condition on the backordering rate that is 120573 ge 120573

lowast issatisfied We will also make some comments on choosing theoptimal policy among all three policies to meet fractionaldemand to meet all demand or to lose all sales as aconsequence of solving the problem (6) via nonlinear pro-gramming theory

22 Proving the Optimality of the Solution First let us notethat the equivalent form of the condition 120573 ge 120573

lowast where 120573lowastis the critical value of the backordering rate from (4) is thefollowing inequality

21198620

1198631198621015840

ℎ

ge

(1198621015840

1)2

(1198621015840

ℎ)2 (7)

With the following two lemmas we will prove that underthe condition 120573 ge 120573

lowast the solution (119879lowast 119865lowast) given by (2) and

(3) is well defined (Lemma 1) and is a feasible solution for theproblem (6) (Lemma 2)

Lemma 1 If 120573 ge 120573lowast then (119879

lowast 119865lowast) is well defined

Advances in Operations Research 3

Proof Assume that120573 ge 120573lowastThen using inequality (7) for the

expression under the square root in the formula (2) we have

21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

ge

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

=

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

minus1198621015840

ℎ

1205731198621015840

119887

) =

(1198621015840

1)2

(1198621015840

ℎ)2ge 0

(8)

so 119879lowast is well defined which implies that 119865lowast is also well

defined

Lemma 2 120573 ge 120573lowast if and only if 119865lowast le 1

Proof Using (2) (3) and (7) we have

119865lowastle 1 lArrrArr

1198621015840

1+ 1205731198621015840

119887119879lowast

119879lowast (1198621015840

ℎ+ 1205731198621015840

119887)le 1

lArrrArr 1198621015840

1+ 1205731198621015840

119887119879lowastle 119879lowast(1198621015840

ℎ+ 1205731198621015840

119887)

lArrrArr 119879lowastge1198621015840

1

1198621015840

ℎ

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

ge

(1198621015840

1)2

(1198621015840

ℎ)2

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) ge

(1198621015840

1)2

(1198621015840

ℎ)2(

1198621015840

ℎ

1205731198621015840

119887

+ 1)

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) ge

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

)

lArrrArr21198620

1198631198621015840

ℎ

ge

(1198621015840

1)2

(1198621015840

ℎ)2lArrrArr 120573 ge 120573

lowast

(9)

which completes the proof

So if 120573 ge 120573lowast we have that (119879lowast 119865lowast) is well defined

(Lemma 1) and because it is obvious that 119879lowast gt 0 and con-sequently 119865lowast gt 0 and we have that 119865lowast le 1 (Lemma 2) we canfinally conclude that (119879lowast 119865lowast) isin F that is (119879lowast 119865lowast) is a feasi-ble point for the problem (6) Since the statement in Lemma 2is true in both directions the value 120573lowast can also be consideredas a critical value for the feasibility of the solutionThis is whywhen 120573 lt 120573

lowast due to the fact that in that case the solution(119879lowast 119865lowast) is not feasible the policy of meeting the fractional

demand cannot be optimal and we should decide betweenmeeting all demand or losing all sales as Pentico et al [5]proposed But when 120573 lt 120573

lowast the optimal policy is to meet alldemand for sure as it is shown by Stojkovska [17]

In the remaining part of this subsection we will provethe optimality of the solution (119879

lowast 119865lowast) by examining the

optimality conditions for the problem (6) We will use theoptimality conditions that deal with the linear independence-constrained qualification (LICQ) that is LICQ holds at apoint if the set of active constrained gradients in this pointis linearly independent and the KKT point that is the pointfor which there are Lagrangianmultipliers such that the KKTsystem is satisfied which are not always easily verified seefor example [18] But the problem under our consideration(6) has the suitable form for checking these conditions as wewill demonstrate

To prove the optimality of the solution (119879lowast 119865lowast) we will

first find all KKT points for the problem (6) (Proposition 3)Then we will prove that (119879

lowast 119865lowast) is a local minimizer

using the second-order sufficient conditions and the globaloptimality of (119879lowast 119865lowast) will be a consequence of the previousfindings (Proposition 4)

Proposition 3 Assume that 120573 ge 120573lowast where 120573lowast is defined by

(4) Then (119879lowast 119865lowast) defined by (2) and (3) is the only KKT pointfor the problem (6)

The KKT point for the problem (6) lies on the boundary119865 = 1 if and only if 120573 = 120573

lowast

Proof The problem (6) has three constraints 119879 gt 0 119865 ge 0and 1 minus 119865 ge 0 So the corresponding KKT system for theproblem (6) is

120597Γ

120597119879minus 1205821= 0

120597Γ

120597119865minus 1205822+ 1205823= 0 (10)

119879 gt 0 119865 ge 0 1 minus 119865 ge 0 (11)

1205821119879 = 0 120582

2119865 = 0 120582

3(1 minus 119865) = 0 (12)

1205821ge 0 120582

2ge 0 120582

3ge 0 (13)

where 1205821 1205822 and 120582

3are the Lagrangian multipliers and the

first partial derivatives are

120597Γ

120597119879= minus

1198620

1198792+1198621015840

ℎ1198631198652

2+1205731198621015840

119887119863(1 minus 119865)

2

2

120597Γ

120597119865= 1198621015840

ℎ119863119879119865 minus 120573119862

1015840

119887119863119879 (1 minus 119865) minus 119862

1015840

1119863

(14)

To obtain all KKT points we need to solve the KKT system(10)ndash(13) with respect to 119879 119865 120582

1 1205822 and 120582

3 Note that one

easy approach for solving the KKT system is to start withcomplementarity conditions (12) Because 119879 gt 0 from thefirst complementarity condition in (12) we have 120582

1= 0 and

from the first equation in (10) we have 120597Γ120597119879 = 0Assume that119865 = 0Then from the third complementarity

condition in (12) we have 1205823

= 0 and from the secondequation in (10) we have 120597Γ120597119865minus120582

2= 0 Substituting119865 = 0 in

the last equation we have 1205822= 120597Γ120597119865 = minus120573119862

1015840

119887119863119879minus119862

1015840

1119863 lt 0

with its strict inequality sign since 119879 gt 0 This contradictswith the second inequality in (13) So it must be that 119865 gt 0

Now from 119865 gt 0 and the second complementaritycondition in (12) we have 120582

2= 0


Assume that 1205823= 0 Then from the second equation in

(10) we have 120597Γ120597119865 = 0 Recall that (119879lowast 119865lowast) is the solutionobtained by setting the first partial derivatives to zero andtaking the positive value for 119879 Now from Lemma 2 when120573 ge 120573

lowast (119879lowast 119865lowast) satisfies the feasibility conditions (11) Weobtained that for 120582

1= 1205822= 1205823= 0 (119879lowast 119865lowast) satisfies the

KKT system so (119879lowast 119865lowast) is a KKT pointWhat is left to completely solve the KKT system (10)ndash

(13) is to assume that 119865 = 1 Then substituting 119865 = 1 into120597Γ120597119879 = 0 we have 1198792 = 2119862

01198621015840

ℎ119863 and 119879 = radic2119862

01198621015840

ℎ119863 gt 0

is the only positive solution Now from the second equationin (10) and 120582

2= 0 we have 120582

3= minus120597Γ120597119865 = minus119862

1015840

ℎ119863119879 + 119862

1015840

1119863 =

minus1198621015840

ℎ119863radic2119862

01198621015840

ℎ119863+1198621015840

1119863 = 119862

1015840

ℎ119863(1198621015840

11198621015840

ℎminusradic2119862

01198621015840

ℎ119863) Having

in mind that (7) is equivalent to 120573 ge 120573lowast we have that 120582

3le 0

if and only if 120573 ge 120573lowast But from the third inequality in (13)

we have that 1205823= 0 and this is true if and only if 120573 = 120573

lowastSince all three Lagrangian multipliers in this case are zero1205821= 1205822= 1205823= 0 the newly found solution is actually the

solution (119879lowast 119865lowast) when 120573 = 120573

lowast in which case we proved thatit lies on the boundary 119865 = 1

So if 120573 ge 120573lowast then (119879

lowast 119865lowast) is the unique KKT point that

lies on the boundary 119865 = 1 if and only if 120573 = 120573lowast which we

wanted to prove


(4) Then (119879lowast 119865lowast) defined by (2) and (3) is a local minimizer

of the problem (6)Moreover if 120573 ge 120573

lowast then (119879lowast 119865lowast) is the global minimizer

of the problem (6)

Proof Assume that 120573 ge 120573lowast From Proposition 3 we have that

(119879lowast 119865lowast) is a KKT point for the problem (6) We are going to

prove that the Hessian matrix of the Lagrangian calculated atthe point (119879lowast 119865lowast) is positive definite

For the problem (6) theHessianmatrix of the Lagrangiancoincides with the Hessian matrix of the objective functionthat is

119867(119879 119865)=[

[

21198620

11987931198621015840

ℎ119863119865minus120573119862

1015840

119887119863 (1minus119865)

1198621015840

ℎ119863119865minus120573119862

1015840

119887119863 (1minus119865) 119862

1015840

ℎ119863119879+120573119862

1015840

119887119863119879

]

]

(15)

Now let us examine the signs of the principals of the aboveHessian calculated at (119879lowast 119865lowast) But first let us make someobservations We know that the first partial derivative withrespect to 119865 calculated at the solution (119879lowast 119865lowast) is zero that is

1198621015840

ℎ119863119879lowast119865lowastminus 1205731198621015840

119887119863119879lowast(1 minus 119865

lowast) minus 1198621015840

1119863 = 0 (16)

From the last equality we obtain the following one

119879lowast(1198621015840

ℎ119865lowastminus 1205731198621015840

119887(1 minus 119865

lowast)) = 119862

1015840

1 (17)

Now for the first principal of the above Hessian we have

119867(119879lowast 119865lowast)11

=21198620

(119879lowast)3gt 0 (18)

And for the second principal by using (7) and (17) we have

119867(119879lowast 119865lowast)22

=

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

21198620

(119879lowast)3

1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast)

1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast) 119862

1015840

ℎ119863119879lowast+ 1205731198621015840

119887119863119879lowast

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=21198620

(119879lowast)3(1198621015840

ℎ119863119879lowast+ 1205731198621015840

119887119863119879lowast)

minus (1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast))2

=119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887)

minus119863(119879lowast)2(1198621015840

ℎ119865lowastminus 1205731198621015840

119887(1 minus 119865

lowast))2

)

=119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887) minus 119863(119862

1015840

1)2

)

ge119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887) minus 119863

211986201198621015840

ℎ

119863)

=119863

(119879lowast)2211986201205731198621015840

119887

gt 0

(19)

Both principals of the Hessian calculated at (119879lowast 119865lowast) are

positive so the Hessian calculated at (119879lowast 119865lowast) is a positivedefinite matrix This together with the fact that (119879lowast 119865lowast) isa KKT point for the problem (6) when 120573 ge 120573

lowast implies that(119879lowast 119865lowast) is a localminimizer of the problem (6) when120573 ge 120573

lowastLet us note that for the box-constrained problem (6)

LICQ holds at every feasible point So if there is another localminimizer ( 119865) for the problem (6) different from (119879

lowast 119865lowast)

then ( 119865) is also a KKT point for the problem (6) whichcontradicts with the uniqueness of the KKT point (119879lowast 119865lowast)(Proposition 3) This implies that (119879lowast 119865lowast) is the global mini-mizer for the problem (6) which completes the proof

It can be seen from the proof of Proposition 3 that whenthe solution (119879

lowast 119865lowast) lies on the boundary 119865 = 1 then it

coincides with the basic EPQ solution which is always on theboundary 119865 = 1 Since (119879lowast 119865lowast) minimizes the cost functionΓ(119879 119865) when 120573 ge 120573

lowast (Proposition 4) the partial backorder-ing policy is optimal compared to the policy of meeting alldemand when 120573 ge 120573

lowast But as Zhang [19] noticed when120573 ge 120573

lowast it is not always preferable to meet the fractionaldemand losing all sales can actually be a better decisionLater the proposed condition by Zhang under which meet-ing the fractional demand is optimal compared to losing allsales when 120573 ge 120573

lowast is corrected by Stojkovska [17]If the production rate is infinitely large the Pentico et alrsquos

EPQ-PBO [5] will degenerate into Pentico and Drakersquos EOQ-PBO [4] When we substitute 119862

ℎfor 1198621015840ℎ 119862119887 for 1198621015840

119887the above

propositions can be used for Pentico and Drakersquos EOQ-PBO[4]


23 Implementation to Other Inventory Models with PartialBackordering It is difficult to use the above mathematicalprogramming approach as a general solution procedurebecause its implementation depends on finding all KKTpoints for the corresponding mathematical programmingproblem and examining the Hessian in the candidate solu-tion But many inventory models are suitable for thisapproach Here we consider two more models

231 Implementation to an Extension of Pentico et alrsquos EPQ-PBO [6] One of the extensions to the Pentico et alrsquos EPQ-PBO is the EPQ-PBO and phase-dependent backorderingrate [6] Relaxing the assumption on a constant backorderingrate 120573 they considered two phases of constant backorderingrate During the first phase before the start of the productionthe backordering rate is 120573 and during the second phase afterthe production starts the backordering rate is 120588120573 where 1 le

120588 le 1120573 For this extension by using a different methodologythe arithmetic-geometric-mean-inequality approach Hsiehand Dye [13] derived the solution and proved its optimalityWe will prove the optimality of the solution for this extensionwith the mathematical programming approach

For thismodel the total cost function Γ(119879 119865) the optimallength of an order cycle 119879

lowast and the optimal fill rate 119865lowast

are of the same form as the ones for the EPQ-PBO givenwith the formulas (1) (2) and (3) respectively where 1198621015840

ℎ=

119862ℎ(1 minus 119863119875) 1198621015840

119887= (119862119887(1 minus 120588120573119863119875))(1 minus (120588 minus 1)120573119863119875) and

1198621015840

1= 1198621(1 minus 120573(1 minus (120588 minus 1)120573119863119875)) The critical value of the

backordering rate 120573 developed by Pentico et al [6] which wewill denote by 120573lowastnew is

120573lowast

new =120573lowast

1 + (120588 minus 1) (119863119875) 120573lowast (20)

where 120573lowast is defined by (4) Pentico et al [6] proved that if120573 ge 120573

lowast

new then (119879lowast 119865lowast) is the optimal solution They based

the proof of the optimality on the same approach presentedfor the basic EPQ-PBO [5] We will use the mathematicalprogramming approach for proving the optimality of thesolution for this extension

First let us notice that the equivalent form of the condi-tion 120573 ge 120573

lowast

new where 120573lowast

new is defined by (20) is the conditionof the form (7) where 1198621015840

ℎ= 119862ℎ(1 minus 119863119875) and 119862

1015840

1= 1198621(1 minus

120573(1 minus (120588 minus 1)120573119863119875)) are notations that correspond to theextension Therefore the proof that the solution (119879

lowast 119865lowast) is

well defined and feasible under condition 120573 ge 120573lowast

new imitatesthe proofs of Lemmas 1 and 2 for the solution of the basicEPQ-PBO from the previous subsection For the EPQ-PBOand phase-dependent backordering rate it can be also provenin a same way as in Propositions 3 and 4 that if 120573 ge 120573

lowast

new(119879lowast 119865lowast) is the only KKT point which lies on the boundary

119865 = 1 if and only if 120573 = 120573lowast

new and when 120573 ge 120573lowast

new then(119879lowast 119865lowast) is the global minimizer of the objective Γ(119879 119865)

The same approach can be used for other extensions ofPentico and Drakersquos basic EOQ-PBO [4] or Pentico et alrsquosbasic EPQ-PBO [5] One of the most recent extensions ofPentico and Drakersquos basic EOQ-PBO [4] is models presentedby Taleizadeh et al [16] Taleizadeh et al [10] and Taleizadehand Pentico [9] In these papers the proof of the optimality

of the solution is made using the same methodology as in[4] since the objective function has the similar form Sothemathematical programming approach presented here canalso be implemented for these inventory models in a similarway It can be shown that when some condition on thebackordering rate (and a condition on the price increase forthe model presented in [9]) is met then the correspondingmathematical programming problem has a unique KKTpoint which is the global optimizer of the objective functionWhen condition on the backordering rate is satisfied with itsequality part then the optimal point lies on the boundary

232 Implementation toHu et alrsquos InventoryModel [2] Hu etal [2] proposed an EOQ-PBO inventorymodel with constantbackordering rate and unit backorder cost increasing linearlywith the duration of the shortage Their decision variablesare the cycle time 119879 and the time to which the inventory ispositive 119905

1 In their notation the total cost per unit time that

has to be minimized is

TRC (119879 1199051) =

1

119879(119860 +

1198621198941199052

1120572

2

+ 119903120572(119887

6(119879 minus 119905

1)3+1198620

2(119879 minus 119905

1)2)

+119877 (1 minus 119903) 120572 (119879 minus 1199051) )

(21)

and the minimization is performed under the constrains 119879 gt

0 0 le 1199051le 119879 Then Hu et al determine the optimal solution

by approximately solving by a numerical search an equationwith respect to 119905

1and the obtained value they substitute in an

equation for 119879 They established a critical value for the hold-ing cost per unit and by examining the partial derivatives andboundary conditions they proved that if the condition

119862119894 ge1198772(1 minus 119903)

2120572

2119860

(22)

is satisfied then partial backordering is optimal otherwisethe optimal policy is not to allow backordering We are goingto find the solution and prove its optimality by themathemat-ical programming approach

If we put 1199051= 119865119879 where 0 le 119865 le 1 is the fill rate then we

can transform the total cost function into

TRC (119879 119865) =119860

119879+1198621198941198652119879120572

2+119903120572119887

61198792(1 minus 119865)

3

+1199031205721198620

2119879(1 minus 119865)

2+ 119877 (1 minus 119903) 120572 (1 minus 119865)

(23)

The corresponding mathematical programming problem is

min(119879119865)isinR2

TRC (119879 119865)


(24)

where TRC(119879 119865) is given with (23) We should note that theintroduction of new decision variables was not needed to


demonstrate the mathematical programming approach andprove the optimality of the solution This transformation isdone mainly to shorten the proofs related to this model andto be able to lean on the ones for Pentico et alrsquos EPQ-PBOpresented in Section 22

First partial derivatives of TRC(119879 119865) with respect to 119879

and 119865 respectively are

120597TRC (119879 119865)

120597119879= minus

119860

1198792+1198621198941198652120572

2+119903120572119887

3119879(1 minus 119865)

3

+1199031205721198620

2(1 minus 119865)

2

120597TRC (119879 119865)

120597119865= 119862119894119865119879120572 minus

119903120572119887

21198792(1 minus 119865)

2

minus 1199031205721198620119879 (1 minus 119865) minus 119877 (1 minus 119903) 120572

(25)

If we solve (120597TRC(119879 119865))120597119865 = 0with respect to 119865 0 le 119865 le 1we will have

119865=119865 (119879)

= 1minus

minus (1199031198620+119862119894)+radic(119903119862

0+119862119894)2minus4119903119887 (119877 (1minus119903)minus119862119894119879)

119903119887119879

(26)

If we substitute the expressions for (1 minus 119865)119879 and 119865119879 that arederived form (26) in (120597TRC(119879 119865))120597119879 = 0 we will have

119862119894120572

2(119879 minus 120593 (119879))

2+119903120572119887

3120593(119879)3+1199031205721198620

2120593(119879)2= 119860 (27)

where 120593(119879) = (minus(1199031198620

+ 119862119894) +

radic(1199031198620+ 119862119894)2minus 4119903119887(119877(1 minus 119903) minus 119862119894119879))(119903119887) Let us denote by

119879lowast the solution of (27) and by 119865lowast = 119865(119879

lowast) = 1 minus 120593(119879

lowast)119879lowast

Under the condition (22) (27) has the unique solution119879lowaston119879 gt 119877(1minus119903)(119862119894) since the function on the left side of (27)is nondecreasing with respect to 119879 with a value less orequal to 119860 at 119877(1 minus 119903)(119862119894) (from where the condition (22)is derived) and infinity when 119879 rarr infin And since theexpression under the square root in (26) is positive on 119879 gt

119877(1minus119903)(119862119894) we have thatwhen the condition (22) is satisfiedthe solution (119879lowast 119865lowast) is well defined and unique on 119879 gt 119877(1minus

119903)(119862119894)From the equivalence 119865lowast le 1 hArr 119879

lowastge 119877(1 minus 119903)(119862119894) we

have that if the condition (22) is satisfied then the solution(119879lowast 119865lowast) is well defined unique and feasible for the problem

(24) on 119879 gt 119877(1 minus 119903)(119862119894) Otherwise if condition (22) isnot satisfied then (27) does not have the solution on 119879 gt

119877(1minus119903)(119862119894) which implies that in that case the partial backo-rdering solution (119879lowast 119865lowast) is not feasible (since 119865lowast gt 1) and inthat case the optimal policy is to meet all demand accordingto the EOQmodel with no shortage period with optimal timecycle 119879lowastEOQ = radic(2119860)(119862119894120572)

Similarly to Proposition 3 it can be shown that when thecondition (22) is satisfied then (119879lowast 119865lowast) is the only KKTpointfor the problem (24) which lies on the boundary 119865 = 1whenthe condition (22) is satisfied with equality sign and in that

case the solution (119879lowast 119865lowast) coincides with the solution from

the EOQ model with no shortages And finally similarly toProposition 4 it can be shown that when the condition (22)is satisfied then the Hessian calculated at (119879lowast 119865lowast) is positivedefinite so (119879

lowast 119865lowast) is the local minimizer and due to its

uniqueness as a KKT point it is the global minimizer for theproblem (24)

This approach of transforming the mathematical pro-gramming problem into decision variables 119879 and 119865 can beused within other models for example the EPQ model withfull backordering considered by Cardenas-Barron [12] Aftertransformation it can be proven that there is a unique KKTpoint and it lies in the interior of the feasible region andat this point the Hessian matrix is positive definite whichimplies the optimality of the solution

3 Conclusions

In this paper we give an alternative way of deriving the solu-tion for the Pentico et alrsquos EPQ-PBO and proving its optimal-ity by using the mathematical programming approach Wealso determine the optimal decision policy among meetingfractional demand meeting all demand and losing all saleswhile solving the correspondingmathematical programmingproblem We use the same approach for two other inventorymodels with partial backordering If we can find all KKTpoints for themathematical programming problem then thisapproach can be used for deriving and proving optimality ofthe solution for the corresponding inventory model

References

[1] M J Drake D W Pentico and C Toews ldquoUsing the EPQ forcoordinated planning of a product with partial backorderingand its componentsrdquo Mathematical and Computer Modellingvol 53 no 1-2 pp 359ndash375 2011

[2] W Hu S L Kim and A Banerjee ldquoAn inventory model withpartial backordering and unit backorder cost linearly increas-ing with the waiting timerdquo European Journal of OperationalResearch vol 197 no 2 pp 581ndash587 2009

[3] Z Pang ldquoOptimal dynamic pricing and inventory controlwith stock deterioration and partial backorderingrdquo OperationsResearch Letters vol 39 no 5 pp 375ndash379 2011

[4] 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

[5] D W Pentico M J Drake and C Toews ldquoThe deterministicEPQ with partial backordering a new approachrdquo Omega vol37 no 3 pp 624ndash636 2009

[6] DW PenticoM J Drake and C Toews ldquoThe EPQwith partialbackordering and phase-dependent backordering raterdquoOmegavol 39 no 5 pp 574ndash577 2011

[7] H-L Yang ldquoA partial backlogging inventory model for deterio-rating items with fluctuating selling price and purchasing costrdquoAdvances in Operations Research vol 2012 Article ID 385371 15pages 2012

[8] 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


[9] A A Taleizadeh and D W Pentico ldquoAn economic orderquantity model with a known price increase and partial backo-rderingrdquo European Journal of Operational Research vol 228 no3 pp 516ndash525 2013

[10] 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

[11] D Y Gao and H D Sherali ldquoCanonical duality theory connec-tion between nonconvex mechanics and global optimizationrdquoin Advances in Mechanics and Mathematics vol 3 pp 249ndash316Springer 2007

[12] L E Cardenas-Barron ldquoThe economic production quantity(EPQ) with shortage derived algebraicallyrdquo International Jour-nal of Production Economics vol 70 no 3 pp 289ndash292 2001

[13] T P Hsieh and C Y Dye ldquoA note on lsquoThe EPQ with partialbackordering and phase-dependent backordering ratersquordquoOmegavol 40 no 1 pp 131ndash133 2012

[14] Y Zhao and M N Katehakis ldquoOn the structure of opti-mal ordering policies for stochastic inventory systems withminimum order quantityrdquo Probability in the Engineering andInformational Sciences vol 20 no 2 pp 257ndash270 2006

[15] B Zhou Y Zhao and M N Katehakis ldquoEffective controlpolicies for stochastic inventory systems with aminimum orderquantity and linear costsrdquo International Journal of ProductionEconomics vol 106 no 2 pp 523ndash531 2007

[16] A A Taleizadeh D W Pentico M Aryanezhad and S MGhoreyshi ldquoAn economic order quantity model with partialbackordering and a special sale pricerdquo European Journal ofOperational Research vol 221 no 3 pp 571ndash583 2012

[17] I Stojkovska ldquoOn the optimality of the optimal policies for thedeterministic EPQ with partial backorderingrdquo Omega vol 41no 5 pp 919ndash923 2013

[18] J Nocedal and S J Wright Numerical Optimization SpringerNew York NY USA 1999

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

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


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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


where 1198621015840ℎ= 119862ℎ(1 minus 119863119875) 1198621015840

119887= 119862119887(1 minus 120573119863119875) and 119862

1015840

1=

1198621(1 minus 120573) Setting the first partial derivatives of Γ(119879 119865) equal

to zero Pentico et al [5] find the following solutions

119879lowast=radic

21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

(2)

119865lowast= 119865 (119879

lowast) =

1198621015840

1+ 1205731198621015840

119887119879lowast

119879lowast (1198621015840

ℎ+ 1205731198621015840

119887) (3)

Comparing the optimal time length of inventory cycle 119879lowast forthe partial backordering model with the one for the standardEPQ model if backordering is optimal Pentico et al [5]derived condition under which (119879

lowast 119865lowast) gives an optimal

solution that is

120573 ge 120573lowast= 1 minus radic

211986201198621015840

ℎ

1198631198622

1

(4)

where 120573lowast denotes the critical value of the backordering rateThen Pentico et al [5] determine the optimal inventory pol-icy in a way that if condition (4) is satisfied then the optimalpolicy is to allow stockouts with partial backordering andmeet the fractional demand otherwise the optimal policy iseither to meet all demand or to lose all sales When condition(4) is satisfied Pentico et al [5] proved the optimality of thesolution for the EPQ-PBO given with (2) and (3) by exam-ining the characteristics of the partial derivatives and theboundary conditions In this paper we are going to prove theoptimality of the solution by using the mathematical pro-gramming approach

Recognition of the (global) optimal solution for thegeneral nonlinear programming problem using the nonlinearprogramming theory is not an easy task There are sufficientconditions for the local optimality but there are very few ortoo restrictive for the global optimality Among the inventorymodels there are many with a convex objective function butmost problems arise when the objective function is noncon-vex because otherwise the local optimal solution is globalIn this case with the nonconvex objective proving the globaloptimality of the solution for the model is usually done byexamining the partial derivatives and the boundary condi-tions some of them are [2 5 9 10] There are some globaloptimality theories for general nonlinear programming prob-lems that are quite effective in practice for certain classes ofproblems One of them is the canonical duality and trivialtheory proposed by Gao and Sherali which has good per-formance for quadratic programming problems inmechanics[11] Algebraical approach is another approach that is usedwith inventory models see [12 13] Proving the optimalityof the solution becomes more complex when there is someuncertainty in the inventory system see [14 15]

The paper is organized as follows In Section 2 Pentico etalrsquos EPQ-PBO is formulated as a mathematical programmingproblem then mathematical programming approach forproving the optimality of the solution is presented and thesame approach is used for two other inventory models withpartial backordering an extension of Pentico et alrsquos EPQ-PBO [6] and Hu et alrsquos inventory model [2] There are

comments on howmathematical programming approach canbe used for the models presented in Taleizadeh et al [16]Taleizadeh et al [10] Taleizadeh and Pentico [9] andCardenas-Barron [12]

2 Mathematical Programming Approach

21 Formulation as a Mathematical Programming ProblemFor the decision variables the length of an order cycle 119879

and the fill rate 119865 there are natural restrictions that is thefollowing conditions should be met

119879 gt 0 0 le 119865 le 1 (5)

So the corresponding mathematical programming problemfor Pentico et alrsquos EPQ-PBO [5] is a box-constrained nonlin-ear programming problem and is formulated as follows

min(119879119865)isinR2

Γ (119879 119865)


(6)

where Γ(119879 119865) is given with (1) Denote the feasible region forthe problem (6) by F = (119879 119865) isin R2 | 119879 gt 0 0 le 119865 le 1We should note that in their later work on the extension ofthe EPQ-PBO [1] the authors impose these conditions on thedecision variables119879 and 119865 and formulated the correspondingmathematical programming problem and for the proof of theoptimality of the solution they still have the same approach asthat for the basic EPQ-PBO [5]

In the next subsection using the nonlinear programmingtheory we are going to prove that the solution (119879

lowast 119865lowast) given

by (2) and (3) is a global minimizer of the problem (6) ifthe condition on the backordering rate that is 120573 ge 120573

lowast issatisfied We will also make some comments on choosing theoptimal policy among all three policies to meet fractionaldemand to meet all demand or to lose all sales as aconsequence of solving the problem (6) via nonlinear pro-gramming theory

22 Proving the Optimality of the Solution First let us notethat the equivalent form of the condition 120573 ge 120573

lowast where 120573lowastis the critical value of the backordering rate from (4) is thefollowing inequality

21198620

1198631198621015840

ℎ

ge

(1198621015840

1)2

(1198621015840

ℎ)2 (7)

With the following two lemmas we will prove that underthe condition 120573 ge 120573

lowast the solution (119879lowast 119865lowast) given by (2) and

(3) is well defined (Lemma 1) and is a feasible solution for theproblem (6) (Lemma 2)

Lemma 1 If 120573 ge 120573lowast then (119879

lowast 119865lowast) is well defined




21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

ge

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

=

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

minus1198621015840

ℎ

1205731198621015840

119887

) =

(1198621015840

1)2

(1198621015840

ℎ)2ge 0

(8)


defined




1198621015840

1+ 1205731198621015840

119887119879lowast

119879lowast (1198621015840

ℎ+ 1205731198621015840

119887)le 1

lArrrArr 1198621015840

1+ 1205731198621015840

119887119879lowastle 119879lowast(1198621015840

ℎ+ 1205731198621015840

119887)


1

1198621015840

ℎ

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

ge

(1198621015840

1)2

(1198621015840

ℎ)2

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) ge

(1198621015840

1)2

(1198621015840

ℎ)2(

1198621015840

ℎ

1205731198621015840

119887

+ 1)

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) ge

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

)

lArrrArr21198620

1198631198621015840

ℎ

ge

(1198621015840

1)2

(1198621015840

ℎ)2lArrrArr 120573 ge 120573

lowast

(9)

















lowast


120597Γ

120597119879minus 1205821= 0

120597Γ

120597119865minus 1205822+ 1205823= 0 (10)

119879 gt 0 119865 ge 0 1 minus 119865 ge 0 (11)

1205821119879 = 0 120582

2119865 = 0 120582

3(1 minus 119865) = 0 (12)

1205821ge 0 120582

2ge 0 120582

3ge 0 (13)

where 1205821 1205822 and 120582



120597Γ

120597119879= minus

1198620

1198792+1198621015840

ℎ1198631198652

2+1205731198621015840

119887119863(1 minus 119865)

2

2

120597Γ

120597119865= 1198621015840

ℎ119863119879119865 minus 120573119862

1015840

119887119863119879 (1 minus 119865) minus 119862

1015840

1119863

(14)


1 1205822 and 120582

3 Note that one


1= 0 and






1015840

119887119863119879minus119862

1015840

1119863 lt 0



2= 0








01198621015840

ℎ119863 and 119879 = radic2119862

01198621015840

ℎ119863 gt 0


2= 0 we have 120582

3= minus120597Γ120597119865 = minus119862

1015840

ℎ119863119879 + 119862

1015840

1119863 =

minus1198621015840

ℎ119863radic2119862

01198621015840

ℎ119863+1198621015840

1119863 = 119862

1015840

ℎ119863(1198621015840

11198621015840


01198621015840

ℎ119863) Having


3le 0









wanted to prove





of the problem (6)





119867(119879 119865)=[

[

21198620

11987931198621015840

ℎ119863119865minus120573119862

1015840

119887119863 (1minus119865)

1198621015840

ℎ119863119865minus120573119862

1015840

119887119863 (1minus119865) 119862

1015840

ℎ119863119879+120573119862

1015840

119887119863119879

]

]

(15)


1198621015840


119887119863119879lowast(1 minus 119865


1119863 = 0 (16)


119879lowast(1198621015840

ℎ119865lowastminus 1205731198621015840

119887(1 minus 119865

lowast)) = 119862

1015840

1 (17)


119867(119879lowast 119865lowast)11

=21198620

(119879lowast)3gt 0 (18)


119867(119879lowast 119865lowast)22

=

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

21198620

(119879lowast)3

1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast)

1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast) 119862

1015840

ℎ119863119879lowast+ 1205731198621015840

119887119863119879lowast

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=21198620

(119879lowast)3(1198621015840

ℎ119863119879lowast+ 1205731198621015840

119887119863119879lowast)

minus (1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast))2

=119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887)

minus119863(119879lowast)2(1198621015840

ℎ119865lowastminus 1205731198621015840

119887(1 minus 119865

lowast))2

)

=119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887) minus 119863(119862

1015840

1)2

)

ge119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887) minus 119863

211986201198621015840

ℎ

119863)

=119863

(119879lowast)2211986201205731198621015840

119887

gt 0

(19)
















ℎfor 1198621015840ℎ 119862119887 for 1198621015840

119887the above









ℎ=

119862ℎ(1 minus 119863119875) 1198621015840


1198621015840



120573lowast

new =120573lowast

1 + (120588 minus 1) (119863119875) 120573lowast (20)


lowast




lowast



ℎ= 119862ℎ(1 minus 119863119875) and 119862

1015840

1= 1198621(1 minus





lowast










TRC (119879 1199051) =

1

119879(119860 +

1198621198941199052

1120572

2

+ 119903120572(119887

6(119879 minus 119905

1)3+1198620

2(119879 minus 119905

1)2)

+119877 (1 minus 119903) 120572 (119879 minus 1199051) )

(21)






119862119894 ge1198772(1 minus 119903)

2120572

2119860

(22)




TRC (119879 119865) =119860

119879+1198621198941198652119879120572

2+119903120572119887

61198792(1 minus 119865)

3

+1199031205721198620

2119879(1 minus 119865)

2+ 119877 (1 minus 119903) 120572 (1 minus 119865)

(23)


min(119879119865)isinR2

TRC (119879 119865)


(24)






120597TRC (119879 119865)

120597119879= minus

119860

1198792+1198621198941198652120572

2+119903120572119887

3119879(1 minus 119865)

3

+1199031205721198620

2(1 minus 119865)

2

120597TRC (119879 119865)

120597119865= 119862119894119865119879120572 minus

119903120572119887

21198792(1 minus 119865)

2


(25)


119865=119865 (119879)

= 1minus

minus (1199031198620+119862119894)+radic(119903119862


119903119887119879

(26)


119862119894120572

2(119879 minus 120593 (119879))

2+119903120572119887

3120593(119879)3+1199031205721198620

2120593(119879)2= 119860 (27)

where 120593(119879) = (minus(1199031198620

+ 119862119894) +



lowast) = 1 minus 120593(119879

lowast)119879lowast














3 Conclusions


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

ge

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

=

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

minus1198621015840

ℎ

1205731198621015840

119887

) =

(1198621015840

1)2

(1198621015840

ℎ)2ge 0

(8)


defined




1198621015840

1+ 1205731198621015840

119887119879lowast

119879lowast (1198621015840

ℎ+ 1205731198621015840

119887)le 1

lArrrArr 1198621015840

1+ 1205731198621015840

119887119879lowastle 119879lowast(1198621015840

ℎ+ 1205731198621015840

119887)


1

1198621015840

ℎ

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) minus

(1198621015840

1)2

1205731198621015840

ℎ1198621015840

119887

ge

(1198621015840

1)2

(1198621015840

ℎ)2

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) ge

(1198621015840

1)2

(1198621015840

ℎ)2(

1198621015840

ℎ

1205731198621015840

119887

+ 1)

lArrrArr21198620

1198631198621015840

ℎ

(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

) ge

(1198621015840

1)2

(1198621015840

ℎ)2(1198621015840

ℎ+ 1205731198621015840

119887

1205731198621015840

119887

)

lArrrArr21198620

1198631198621015840

ℎ

ge

(1198621015840

1)2

(1198621015840

ℎ)2lArrrArr 120573 ge 120573

lowast

(9)

















lowast


120597Γ

120597119879minus 1205821= 0

120597Γ

120597119865minus 1205822+ 1205823= 0 (10)

119879 gt 0 119865 ge 0 1 minus 119865 ge 0 (11)

1205821119879 = 0 120582

2119865 = 0 120582

3(1 minus 119865) = 0 (12)

1205821ge 0 120582

2ge 0 120582

3ge 0 (13)

where 1205821 1205822 and 120582



120597Γ

120597119879= minus

1198620

1198792+1198621015840

ℎ1198631198652

2+1205731198621015840

119887119863(1 minus 119865)

2

2

120597Γ

120597119865= 1198621015840

ℎ119863119879119865 minus 120573119862

1015840

119887119863119879 (1 minus 119865) minus 119862

1015840

1119863

(14)


1 1205822 and 120582

3 Note that one


1= 0 and






1015840

119887119863119879minus119862

1015840

1119863 lt 0



2= 0








01198621015840

ℎ119863 and 119879 = radic2119862

01198621015840

ℎ119863 gt 0


2= 0 we have 120582

3= minus120597Γ120597119865 = minus119862

1015840

ℎ119863119879 + 119862

1015840

1119863 =

minus1198621015840

ℎ119863radic2119862

01198621015840

ℎ119863+1198621015840

1119863 = 119862

1015840

ℎ119863(1198621015840

11198621015840


01198621015840

ℎ119863) Having


3le 0









wanted to prove





of the problem (6)





119867(119879 119865)=[

[

21198620

11987931198621015840

ℎ119863119865minus120573119862

1015840

119887119863 (1minus119865)

1198621015840

ℎ119863119865minus120573119862

1015840

119887119863 (1minus119865) 119862

1015840

ℎ119863119879+120573119862

1015840

119887119863119879

]

]

(15)


1198621015840


119887119863119879lowast(1 minus 119865


1119863 = 0 (16)


119879lowast(1198621015840

ℎ119865lowastminus 1205731198621015840

119887(1 minus 119865

lowast)) = 119862

1015840

1 (17)


119867(119879lowast 119865lowast)11

=21198620

(119879lowast)3gt 0 (18)


119867(119879lowast 119865lowast)22

=

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

21198620

(119879lowast)3

1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast)

1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast) 119862

1015840

ℎ119863119879lowast+ 1205731198621015840

119887119863119879lowast

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=21198620

(119879lowast)3(1198621015840

ℎ119863119879lowast+ 1205731198621015840

119887119863119879lowast)

minus (1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast))2

=119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887)

minus119863(119879lowast)2(1198621015840

ℎ119865lowastminus 1205731198621015840

119887(1 minus 119865

lowast))2

)

=119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887) minus 119863(119862

1015840

1)2

)

ge119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887) minus 119863

211986201198621015840

ℎ

119863)

=119863

(119879lowast)2211986201205731198621015840

119887

gt 0

(19)
















ℎfor 1198621015840ℎ 119862119887 for 1198621015840

119887the above









ℎ=

119862ℎ(1 minus 119863119875) 1198621015840


1198621015840



120573lowast

new =120573lowast

1 + (120588 minus 1) (119863119875) 120573lowast (20)


lowast




lowast



ℎ= 119862ℎ(1 minus 119863119875) and 119862

1015840

1= 1198621(1 minus





lowast










TRC (119879 1199051) =

1

119879(119860 +

1198621198941199052

1120572

2

+ 119903120572(119887

6(119879 minus 119905

1)3+1198620

2(119879 minus 119905

1)2)

+119877 (1 minus 119903) 120572 (119879 minus 1199051) )

(21)






119862119894 ge1198772(1 minus 119903)

2120572

2119860

(22)




TRC (119879 119865) =119860

119879+1198621198941198652119879120572

2+119903120572119887

61198792(1 minus 119865)

3

+1199031205721198620

2119879(1 minus 119865)

2+ 119877 (1 minus 119903) 120572 (1 minus 119865)

(23)


min(119879119865)isinR2

TRC (119879 119865)


(24)






120597TRC (119879 119865)

120597119879= minus

119860

1198792+1198621198941198652120572

2+119903120572119887

3119879(1 minus 119865)

3

+1199031205721198620

2(1 minus 119865)

2

120597TRC (119879 119865)

120597119865= 119862119894119865119879120572 minus

119903120572119887

21198792(1 minus 119865)

2


(25)


119865=119865 (119879)

= 1minus

minus (1199031198620+119862119894)+radic(119903119862


119903119887119879

(26)


119862119894120572

2(119879 minus 120593 (119879))

2+119903120572119887

3120593(119879)3+1199031205721198620

2120593(119879)2= 119860 (27)

where 120593(119879) = (minus(1199031198620

+ 119862119894) +



lowast) = 1 minus 120593(119879

lowast)119879lowast














3 Conclusions


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















01198621015840

ℎ119863 and 119879 = radic2119862

01198621015840

ℎ119863 gt 0


2= 0 we have 120582

3= minus120597Γ120597119865 = minus119862

1015840

ℎ119863119879 + 119862

1015840

1119863 =

minus1198621015840

ℎ119863radic2119862

01198621015840

ℎ119863+1198621015840

1119863 = 119862

1015840

ℎ119863(1198621015840

11198621015840


01198621015840

ℎ119863) Having


3le 0









wanted to prove





of the problem (6)





119867(119879 119865)=[

[

21198620

11987931198621015840

ℎ119863119865minus120573119862

1015840

119887119863 (1minus119865)

1198621015840

ℎ119863119865minus120573119862

1015840

119887119863 (1minus119865) 119862

1015840

ℎ119863119879+120573119862

1015840

119887119863119879

]

]

(15)


1198621015840


119887119863119879lowast(1 minus 119865


1119863 = 0 (16)


119879lowast(1198621015840

ℎ119865lowastminus 1205731198621015840

119887(1 minus 119865

lowast)) = 119862

1015840

1 (17)


119867(119879lowast 119865lowast)11

=21198620

(119879lowast)3gt 0 (18)


119867(119879lowast 119865lowast)22

=

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

21198620

(119879lowast)3

1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast)

1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast) 119862

1015840

ℎ119863119879lowast+ 1205731198621015840

119887119863119879lowast

10038161003816100381610038161003816100381610038161003816100381610038161003816100381610038161003816

=21198620

(119879lowast)3(1198621015840

ℎ119863119879lowast+ 1205731198621015840

119887119863119879lowast)

minus (1198621015840

ℎ119863119865lowastminus 1205731198621015840

119887119863(1 minus 119865

lowast))2

=119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887)

minus119863(119879lowast)2(1198621015840

ℎ119865lowastminus 1205731198621015840

119887(1 minus 119865

lowast))2

)

=119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887) minus 119863(119862

1015840

1)2

)

ge119863

(119879lowast)2(21198620(1198621015840

ℎ+ 1205731198621015840

119887) minus 119863

211986201198621015840

ℎ

119863)

=119863

(119879lowast)2211986201205731198621015840

119887

gt 0

(19)
















ℎfor 1198621015840ℎ 119862119887 for 1198621015840

119887the above









ℎ=

119862ℎ(1 minus 119863119875) 1198621015840


1198621015840



120573lowast

new =120573lowast

1 + (120588 minus 1) (119863119875) 120573lowast (20)


lowast




lowast



ℎ= 119862ℎ(1 minus 119863119875) and 119862

1015840

1= 1198621(1 minus





lowast










TRC (119879 1199051) =

1

119879(119860 +

1198621198941199052

1120572

2

+ 119903120572(119887

6(119879 minus 119905

1)3+1198620

2(119879 minus 119905

1)2)

+119877 (1 minus 119903) 120572 (119879 minus 1199051) )

(21)






119862119894 ge1198772(1 minus 119903)

2120572

2119860

(22)




TRC (119879 119865) =119860

119879+1198621198941198652119879120572

2+119903120572119887

61198792(1 minus 119865)

3

+1199031205721198620

2119879(1 minus 119865)

2+ 119877 (1 minus 119903) 120572 (1 minus 119865)

(23)


min(119879119865)isinR2

TRC (119879 119865)


(24)






120597TRC (119879 119865)

120597119879= minus

119860

1198792+1198621198941198652120572

2+119903120572119887

3119879(1 minus 119865)

3

+1199031205721198620

2(1 minus 119865)

2

120597TRC (119879 119865)

120597119865= 119862119894119865119879120572 minus

119903120572119887

21198792(1 minus 119865)

2


(25)


119865=119865 (119879)

= 1minus

minus (1199031198620+119862119894)+radic(119903119862


119903119887119879

(26)


119862119894120572

2(119879 minus 120593 (119879))

2+119903120572119887

3120593(119879)3+1199031205721198620

2120593(119879)2= 119860 (27)

where 120593(119879) = (minus(1199031198620

+ 119862119894) +



lowast) = 1 minus 120593(119879

lowast)119879lowast














3 Conclusions


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















ℎ=

119862ℎ(1 minus 119863119875) 1198621015840


1198621015840



120573lowast

new =120573lowast

1 + (120588 minus 1) (119863119875) 120573lowast (20)


lowast




lowast



ℎ= 119862ℎ(1 minus 119863119875) and 119862

1015840

1= 1198621(1 minus





lowast










TRC (119879 1199051) =

1

119879(119860 +

1198621198941199052

1120572

2

+ 119903120572(119887

6(119879 minus 119905

1)3+1198620

2(119879 minus 119905

1)2)

+119877 (1 minus 119903) 120572 (119879 minus 1199051) )

(21)






119862119894 ge1198772(1 minus 119903)

2120572

2119860

(22)




TRC (119879 119865) =119860

119879+1198621198941198652119879120572

2+119903120572119887

61198792(1 minus 119865)

3

+1199031205721198620

2119879(1 minus 119865)

2+ 119877 (1 minus 119903) 120572 (1 minus 119865)

(23)


min(119879119865)isinR2

TRC (119879 119865)


(24)






120597TRC (119879 119865)

120597119879= minus

119860

1198792+1198621198941198652120572

2+119903120572119887

3119879(1 minus 119865)

3

+1199031205721198620

2(1 minus 119865)

2

120597TRC (119879 119865)

120597119865= 119862119894119865119879120572 minus

119903120572119887

21198792(1 minus 119865)

2


(25)


119865=119865 (119879)

= 1minus

minus (1199031198620+119862119894)+radic(119903119862


119903119887119879

(26)


119862119894120572

2(119879 minus 120593 (119879))

2+119903120572119887

3120593(119879)3+1199031205721198620

2120593(119879)2= 119860 (27)

where 120593(119879) = (minus(1199031198620

+ 119862119894) +



lowast) = 1 minus 120593(119879

lowast)119879lowast














3 Conclusions


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













120597TRC (119879 119865)

120597119879= minus

119860

1198792+1198621198941198652120572

2+119903120572119887

3119879(1 minus 119865)

3

+1199031205721198620

2(1 minus 119865)

2

120597TRC (119879 119865)

120597119865= 119862119894119865119879120572 minus

119903120572119887

21198792(1 minus 119865)

2


(25)


119865=119865 (119879)

= 1minus

minus (1199031198620+119862119894)+radic(119903119862


119903119887119879

(26)


119862119894120572

2(119879 minus 120593 (119879))

2+119903120572119887

3120593(119879)3+1199031205721198620

2120593(119879)2= 119860 (27)

where 120593(119879) = (minus(1199031198620

+ 119862119894) +



lowast) = 1 minus 120593(119879

lowast)119879lowast














3 Conclusions


References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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