research article an inventory model with price and...

Hindawi Publishing CorporationAdvances in Operations ResearchVolume 2013 Article ID 795078 8 pageshttpdxdoiorg1011552013795078

Research ArticleAn Inventory Model with Price and Quality DependentDemand Where Some Items Produced Are Defective

Tapan Kumar Datta

BITS PilanimdashDubai Campus Dubai International Academic City PO Box 345055 Dubai UAE

Correspondence should be addressed to Tapan Kumar Datta dattap12rediffmailcom

Received 24 January 2013 Revised 15 May 2013 Accepted 17 May 2013

Academic Editor Ching-Jong Liao

Copyright copy 2013 Tapan Kumar DattaThis 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

This paper analyzes an inventory system for joint determination of product quality and selling price where a fraction of itemsproduced are defective It is assumed that only a fraction of defective items can be repairedreworked The demand rate dependsupon both the quality and the selling price of the product The production rate unit price and carrying cost depend upon thequality of the items produced Quality index is used to determine the quality of the product An algorithm is provided to solve themodel with given values of model parameters Sensitivity analysis has also been performed

1 Introduction

In every business product quality is an important factorthat attracts customers For durable goods quality dependsupon several factors Some of such factors are typequalityof the raw materials used typequality of the machines usedin the production process skills of workers engaged in theproduction system and so forth It is obvious that the unitcost for a high quality product will be high In general unitcost increases with quality Quality measure is an importantissue in all production systems There is no well-definedmethod for measuring quality In fact quality characteristicsare not the same for all types of items It varies from one typeof items to another type There are several research articleson quality measure Maynes [1] described the concept ofevaluating quality index as a measure of quality for durablegoods He suggested to combine characteristics of varietyand the characteristics of seller to evaluate quality indexJiang [2] defined quality index as a ratio of two different lifemeasures based on fractile life one represents life utilizationextent and the other represents the quality improvementpotential He derived quality index formulae for severalknown lifetime distributions Some authors proposed qualityindexmethod tomeasure quality of sea foods Huidobro et al[3] proposed quality determination method for raw Gilthead

seabream (Sparusaurata) based on the quality parametersmdashflesh elasticity odor clarity shape of fish Barbosa and Vaz-Pires [4] proposed the development of a sensorial scheme tomeasure quality of common octopus

Though customers have the tendency to buy a high qual-ity product sometimes due to high price they compromisewith the quality Thus a challenging task for a productionmanager is to produce units in suitable quality and setting areasonable selling price for these units Normally customersrsquodemand decreases when selling price increases

In some production systems all items manufactured arenot goodperfectThismay be seen in failure-pronemanufac-turing systemwhere the produced items are amixture of goodas well as defective items This situation can be found in theindustries where units are produced in large numbers Someof the research articles on defective products are authored byRosenblatt and Lee [5] Kim andHong [6] Salameh and Jaber[7] Chung and Hou [8] Chiu [9] Sana [10] Datta [11] andMhada et al [12] Rosenblatt and Lee [5] studied the effectof the imperfect production process on optimal productioncycle They assumed the system deteriorates during the pro-duction process and produces some proportion of defectiveitems Kim andHong [6] analyzed a production systemwhichdeteriorates randomly and shifts from in-control state to out-of-control state They determined the optimal production

2 Advances in Operations Research

run length Salameh and Jaber [7] developed an EOQ modelwhere all items produced are not perfect They assumedthat the imperfect items would be sold in a single batchby the end of the screening process Chung and Hou [8]analyzed a system with deteriorating production process andallowable shortages Chiu [9] developed a finite productionrate model by assuming that a fraction of defective productscan be reworked and the rest will be disposed Sana [10]analyzed an inventory model to determine optimal productreliability optimal production rate in a faulty productionsystem In his article the demand rate is assumed to be timedependent Datta [11] developed an inventory model withadjustable production rate and selling price sensitive demandrate In his article he assumed that all items produced are notperfect This model jointly determines optimal productionrate production period and selling price Mhada et al [12]proposed a model with perfectly mixed good and defectiveparts But these research articles did not consider productquality as a decision parameter Some researchers realized theimportance of product quality and incorporated the qualityfactor in their models Some of such articles are authoredby Chen [13] Mahapatra and Maiti [14] Chen and Liu [15]and Chen [16] Chen [13] developed a model to find optimalquality level purchase price and selling quantity for theimmediate firms but his model is not valid in a productionsystemwhere all units produced are not goodMahapatra andMaiti [14] developed a multiobjective multi-item inventorymodel with quality and stock-dependent demand rate Theirpaper did not focus neither on the influence of selling pricein demand nor on the defective product Chen and Liu [15]proposed an optimal consignment policy considering a fixedfee and a per unit commission Their model determines ahigher manufacturerrsquos profit than the traditional productionsystem and coordinates the retailer to obtain a large supplychain profit Chen [16] modified the model of Chen and Liu[15] by incorporating the influence of retailerrsquos order quantityon the manufacturerrsquos product quality He considered thequality of the product as normally distributed None of theabove articles explained the joint determination of best sellingprice product quality and product quantity under quality-dependent production rate and costs

In the present paper the author has attempted to developan inventory model to integrate the above mentioned factorsThe salient features of this developed model are as follows

(i) demand rate depends upon quality and selling price(selling markup rate)

(ii) a fraction of items produced are defective and only afraction of defective items are repairable

(iii) unit cost and carrying cost are variables dependentupon quality

(iv) production rate is quality dependent

This model maximizes the average net profit per unit timeand determines the best suitable quality and the best possiblemarkup for selling price It also determines the optimumproduction quantity in each cycle This model is suitablefor a manufacturing system where the manufacturer wantsto jointly determine the quality (grade) of the item that

should be produced and the selling price of the items tomaximize the average net profit per unit time The model isillustrated by numerical examples A sensitivity analysis hasbeen performed

2 Assumptions and Notation

The following assumptions and notations are used in thedeveloped model

Production ProcessThe production process is not completelyperfect A fraction of the items produced are defective Afraction of the defective items can be repaired to make itperfect The rest cannot be repaired and will be disposed

Quality of the ProductThe item can be produced in differentqualities but the manufacturer wants to market a particularquality which will be most profitable Actually types ofraw materials used skills of the workers working in theproduction line and quality of machineries used in theproduction system are responsible for the qualityThe qualityis assumed to be under manufacturerrsquos control Quality isrepresented by a quality index 119903 0 lt 119903

1le 119903 le 1 Here 119903

1is

the minimum quality that is required to market the product119903 = 1 indicates the top quality

Production Rate The production rate depends upon thequality The production rate 119875(119903) is a decreasing function of119903 It means the rate decreases when the quality improvesThisrate is taken in the following form

119875 (119903) = 1198751+

1198752

119903

0 lt 1199031le 119903 le 1 (1)

where 1198751represents the constant part of the production rate

which does not depend upon the quality and the second part1198752119903 decreases with quality It can be noticed that 119875min = 1198751 +1198752and 119875max = 1198751 + (11987521199031) This type of production rate can

be seen in Datta [11]Unit Cost The unit cost 119862

119906(119903) of the item depends on the

quality The cost increases with quality The following linearform is taken for unit cost 119862

119906(119903) 119862

119906(119903) = 119886 + 119887119903 where

119886 and 119887 are two positive constants It can be noticed that119886 + 119887119903

1le 119862119906le 119886 + 119887

Carrying Cost Carryingholding cost is 119862ℎ(119903) per unit time

This cost increases with quality 119903 and is taken in the quadraticform 119862

ℎ(119903) = 119901 + 119902119903

2 where 119901 and 119902 are positive constantsConsider the following120572 fraction of items produced that are defective120573 fraction of defective items that can be repaired tomake it perfect119862119904 setup cost per production run

119862119889 disposal cost per unit for disposing nonrepairable

items119862119903 cost of repairing one unit

119896 markup rate for selling (119896 gt 0) Selling price is 119904 =119896119862119906(119903) for each unit of the product of quality 119903 But to

earn profit 119896 should be greater than 1

Advances in Operations Research 3

Stoc

k le

vel

Time

S1 S2

t = t1 t = t2 t = Tt = 0

Figure 1 Pictorial representation of the system

Time Horizon Time horizon is infinite

Shortages Shortages are not allowed

Demand Rate Demand rate 119863(119896 119903) depends upon bothmarkup rate 119896 and the quality 119903 119863(119896 119903) decreases with 119896Normally 119863(119896 119903) is an increasing function of 119903 but in somesituations customers judge the quality by its price If price istoo low they will doubt the quality and refuse to buy Thisimplies that for a given 119896 demand will increase with 119903 up toa certain level and then it will decrease

Further119863(119896 119903) should be concave in 119896 Thus the follow-ing conditions should be satisfied by119863(119896 119903)119863

119896lt 0119863

119896119896lt 0

where suffix (lowast) indicates the partial derivative with respect toldquolowastrdquo

Following condition must be satisfied by119863(119896 119903)

(1 minus 120572) 119875 (119903) minus 119863 (119896 119903) gt 0 (2)

3 The Proposed System

Production starts just after time 119905 = 0 Each unit will beinspected by an automated inspection system immediatelyafter its production Defective units will be immediatelyshifted to the repairing shop repairing shop will separate therepairable and nonrepairable defective units An amount 119862

119889

(disposal cost) will be charged for disposing each unit 119862119903

is the cost of repairing each unit to make it perfect Theserepaired units will be brought to the selling area immediatelywhen the stock level at the selling area is zero A typicalgraph of the system is shown in Figure 1 Production andconsumption will jointly continue during [0 119905

1] The level of

inventory at time 119905 = 1199051is 1198781 Production stops at time 119905 = 119905

1

There will be only consumption during [1199051 1199052] Stock level

becomes zero at time 119905 = 1199052 Immediately repaired units will

be brought which will raise the inventory level to 1198782at time

119905 = 1199052

This initial stock level 1198782will gradually decrease due to

market demand and becomes zero at time 119905 = 119879 The timeperiod [0 119879] defines a complete replenishment cycle

4 Revenue Calculation

The following costs are involved in the proposed inventorysystem setup cost unit cost carrying cost repairing costand disposal costThe policy adopted here is to maximize the

average net revenue (ANR) per unit time over a replenish-ment cycleThe decision variables are production period (119905

1)

markup rate (119896) and the quality level (119903)Revenue calculation details

Total amount produced during [0 1199051] = 119875(119903)119905

1

Total amount of defective items produced = 120572119875(119903)1199051

Total amount of repairable defective items =120572120573119875(119903)119905

1

Total amount of nonrepairable defective items thatwill be disposed = 120572(1 minus 120573)119875(119903)119905

1

Total amount of perfect items produced = (1 minus120572)119875(119903)119905

1

The inventory level 1198781at time 119905 = 119905

1can be expressed

as 1198781= [(1 minus 120572)119875(119903) minus 119863(119896 119903)]119905

1

Fresh level of inventory at time 119905 = 1199052is 1198782= 120572120573119875(119903)119905

1

The following expressions of 1199052and 119879 are obtained

1199052= 1199051+ (1198781119863(119896 119903)) = (1 minus 120572)119875(119903)119905

1119863(119896 119903) and

119879 = 1199052+ (1198782119863(119896 119903)) = (1 minus 120572 + 120572120573)119875(119903)119905

1119863(119896 119903)

Total production cost in a cycle = 119862119906(119903)119875(119903)119905

1

Setup cost = 119862119904

Total disposal cost = 119862119889120572(1 minus 120573)119875(119903)119905

1

Total repairing cost = 1198621199031205721205731199051119875(119903) (119862

119903lt 119862119906)

Total carrying cost = (119862ℎ(119903)2)[119878

11199052+ 1198782(119879 minus 119905

2)] =

(119862ℎ(119903)2119863(119896 119903))[120572

21205732+(1 minus 120572)

2119875(119903)

2minus119863(119896 119903)(1minus

120572)119875(119903)]1199052

1

Gross revenue = 119896119862119906(119903)(1 minus 120572 + 120572120573)119875(119903)119905

1

Hence the average net revenue (ANR) per unit time is

ANR (1199051 119896 119903)

=

1

119879

times[

[

gross revenue minus setup costminusproduction cost minus repairing costminusdisposal cost minus holding cost

]

]

= 119896119862119906(119903)119863 (119896 119903) minus

119863 (119896 119903)

1 minus 120572 + 120572120573

times [

119862119904

119875 (119903) 1199051

+ 119862119906(119903) + 119862

119889120572 (1 minus 120573) + 119862

119903120572120573]

minus

119862ℎ(119903) 1199051

2 (1 minus 120572 + 120572120573)

times [12057221205732+ (1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)]

(3)


5 Solution of the Model

The necessary conditions for the existence of a maximumvalue of ANR for a given value of 119903 are

120597ANR1205971199051

= 0

120597ANR120597119896

= 0

Now 120597ANR1205971199051

= 0 997904rArr

(4)

1199051

= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903)[120572

21205732+ (1 minus 120572)

2119875(119903) minus (1 minus 120572)119863 (119896 119903)]

(5)

Also

1205972ANR1205971199051

2= minus

2119862119904119863 (119896 119903)

(1 minus 120572 + 120572120573) 119875 (119903) 1199051

3lt 0 (6)

This implies that for given values of 119896 and 119903 ANR ismaximumat 1199051defined by (5)

Property 1 The expression under the square root is positive

Proof The expression under the square root will be positiveif 12057221205732 + (1 minus 120572)2119875(119903) minus (1 minus 120572)119863(119896 119903) is positive

Now

12057221205732+ (1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)

= 12057221205732119875 (119903) + (1 minus 120572) [(1 minus 120572) 119875 (119903) minus 119863 (119896 119903)]

(7)

But by (2) (1 minus 120572)119875(119903) minus 119863(119896 119903) gt 0Hence the expression under the square root is positive

(proved)

Moreover

120597ANR120597119896

= 0 997904rArr

119862119906(119903)119863 (119896 119903) + [119870119862

119906(119903) minus 119883 (119905

1) + 119884 (119905

1)]119863119896= 0

(8)

where

119883(1199051) =

1

1 minus 120572 + 120572120573

times [

119862119904

119875 (119903) 1199051

+ 119862119906(119903) + 120572 (1 minus 120573)119862

119889+ 120572120573119862

119903]

119884 (1199051) =

119862ℎ(1 minus 120572) 119905

1

2 (1 minus 120572 + 120572120573)

119883 (1199051) 119884 (119905

1) gt 0

(9)

The sufficient conditions for maximum are

(a) 1205972ANR1205971199051

2lt 0

(b) 1205972ANR1205971198962lt 0

(c) 1205972ANR1205971199051

2sdot

1205972ANR1205971198962minus

1205972ANR1205971199051120597119896

2

gt 0

(10)

The first condition (a) has already been proved If we lookat (3) we will see that 119863(119896 119903) sdot 119883(119905

1) is the sum of ordering

cost unit cost disposal cost and repairing cost per unittime whereas 119896119862

119906119863(119896 119903) is the selling price per unit For any

business 119896119862119906sdot 119863(119896 119903) gt 119883(119905

1)119863(119896 119903) or 119896119862

119906gt 119883(119905

1) This

leads to the result 119896119862119906minus 119883(119905

1) + 119884(119905

1) gt 0

Hence 1205972ANR1205971198962 = 2119862119906119863119896(119896 119903) + [119896119862

119906minus 119883(119905

1) +

119884(1199051)]119863119896119896lt 0 This proved (b)

It is very difficult to give an analytical proof of (c)However for given values of the model parameters this canbe proved For a given value of 119903 the most economic values119905lowast

1and 119896lowastof 119905

1and 119896 can be obtained by jointly solving (5) and

(8) These values of 1199051and 119896 will give the maximum value of

ANR for a given quality 119903The optimum production quantity119876lowast in each cycle will satisfy the following equation

119876lowast= 119875 (119903) 119905

1

= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

21205732+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(11)

An algorithm is given below for solving the problem

Algorithm

Step 1 Select a demand pattern Enter the values of themodelparameters

Step 2 Select the increment of 119903 say 119894 The value of ldquo119894rdquo can betaken as 01 or 001 Use a counter 119895 Take 119895 = 1

Step 3 Take 119903 = 1199031 119896 = 1

Step 4 Find 1199051by using (5)

Step 5 Find 119896 by (8)

Step 6 Repeat Steps 3 4 and 5 until the values of 119896 and1199051become stable

Step 7 Calculate ANR using (3) Let this value be ANR(119895)

Step 8 119903 = 1199031+ 119894

Step 9 If 119903 gt 1 move to the next step Else 119895 = 119895 + 1 and goto Step 3


Table 1 Values of the model parameters

Parameter 119898 119899 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 120 25 12 20 300 100 02 6 8 04 05 5 15 200

Table 2 Optimum results and hessian value

119903lowast

119896lowast

119905lowast

1 ANRlowast ANR11990511199051

ANR119896119896

ANR1199051119896

119867

054 233788 031657 28955 minus204203 minus908223 minus54801 155430979Suffix ldquolowastrdquo in ANR indicates partial derivative with respect to ldquolowastrdquo ldquo119867rdquo stands for the hessian at the optimal values of the decision variables


Parameter 119906 V 119908 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 200 30 80 12 20 300 50 02 6 8 02 05 5 15 1000

Step 10 Compare the values of ANR(119895) and find 119895 for whichit is maximum The corresponding values of 119903 119896 119905

1will give

us the solution of the model

Step 11 Stop

6 Some Special Cases

Case 1 When 120572 = 0

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [1 minus (119863 (119896 119903) 119875 (119903))]

(12)

This is optimal production quantity (OPQ) when all itemsproduced are perfect It can be noted that this expressionof OPQ is similar to the basic inventory model with finiteproduction rate and without shortages Further if we takelimit as 119875(119903) rarr infin then 119876lowast = radic2119862

119904119863(119896 119903)119862

ℎ(119903) which

is classical EOQ formula

Case 2 When 120573 = 0

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) (1 minus 120572) [(1 minus 120572) minus (119863 (119896 119903) 119875 (119903))]

(13)

This is OPQ when all defective items are nonrepairableIf we further take limit as 119875(119903) rarr infin then 119876lowast =

radic2119862119904119863(119896 119903)119862

ℎ(119903)(1 minus 120572)

2 which is similar to (10) ofSalameh and Jaber [7] when defective rate 119901 is constant andscreening rate 119909 rarr infin

Case 3 When 120573 = 1

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

2+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(14)

This is OPQ when all defective items are repairable

Case 4 If we substitute 120573 = 0 in (5) it will give us theoptimum production period in each cycle as

1199051= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903) [(1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)]

(15)

which is similar to (2) of Datta [11] except the last term in thedenominator Dattarsquos [11] article assumed that the defectiveitems would be separated at the end of the production periodDue to this reason the last term is different

7 Numerical Examples

Example 1 Demand pattern119863(119896 119903) = 119898119903 minus 11989911990321198962

The demand rate is taken in a nonlinear form 119863(119896 119903) =119898119903 minus 119899119903

21198962 where119898 and 119899 are positive constants (119898 gt 119899) It

can be observed that119863119896lt 0119863

119896119896lt 0 For a positive demand

119896 lt radic119898119899119903 = 1198961 Thus the value of 119896 lies in the interval

(0 1198961) Parameter values are given in Table 1

Solution Results are shown in Table 2

Example 2 Linear demand pattern119863(119896 119903) = 119906 + V119903 minus 119908119896

Let us analyze the model for a linear demand patternin the form 119863(119896 119903) = 119906 + V119903 minus 119908119896 where 119906 V 119908 gt 0 Itcan be observed that 119863 satisfies the condition 119863

119896lt 0 One

advantage of this linear form is that one can easily estimatethe constants 119906 V 119908 by using multiple linear regression Thedemand must be positive that is119863 gt 0 This condition gives119896 lt (119906 + V119903)119908 = 119896

1 say Thus the value of 119896 lies in the open

interval (0 1198961) Parameter values are given in Table 3




119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867

1 21263 03689 78198 minus75717 minus5120 minus191952 35119890 + 07Suffix ldquolowastrdquo in ANR indicates partial derivative with respect to ldquolowastrdquo ldquo119867rdquo stands for the hessian at the optimal values of the decision variables

minus50minus40minus30minus20minus10

0102030405060

minus12 minus8 minus4 0 4 8 12

Change in ANR with respect to m ()Change in ANR with respect to k ()Change in ANR with respect to n ()Change in ANR with respect to t1 ()

Figure 2 change in ANR (Example 1)

8 Sensitivity Analysis

In this section a sensitivity analysis has been performedto analyze the effects of demand parameters and decisionvariables 119896 119905

1on the average net profit ANR The previ-

ous numerical problems are considered for this sensitivityanalysis The changes of ANR for the changes of theseparametersvariables are shown in Table 5

81 Analysis of Nonlinear Demand Pattern (Example 1) Itcan be observed that demand parameters 119898 and 119899 are verysensitive parameters ANR increases rapidly with the increaseof the value of 119898 But rate of increase is more for highervalues of 119898 The parameter 119899 is relatively less sensitive Theparameter 119899 is relatively more sensitive for lower values thanhigher valuesThedecision parameter 119896 ismore sensitive thanthe other decision parameter 119905

1 A change in the value of 119896 by

12 causes more than 20 decrease in the value of ANR

82 Analysis of Linear Demand Pattern (Example 2) Thedemand parameters119906 and119908 are very sensitiveTheparameter119906 is more sensitive for higher values but 119908 is more sensitivefor lower values The demand parameter V is relatively lesssensitive The decision variable 119905

1is very less sensitive The

sensitivity of 119896 is moderate To get a better idea about thesensitiveness previous results are graphically presented inFigures 2 and 3 The following are the managerial insights ofthe above analysis

minus70minus60minus50minus40minus30minus20minus10

0102030405060708090


Change in ANR with respect to u ()Change in ANR with respect to w ()Change in ANR with respect to t1 ()Change in ANR with respect to ()Change in ANR with respect to k ()


(i) proper care should be taken while estimating thedemand parameters A small error in these parame-ters may mislead the production manager

(ii) if the production manager is forced to change theoptimal production time by a small amount of timeit will not make a significant difference in the optimalprofit

9 Concluding Remarks

This paper described how to solve a managerial problemassociated with a faulty production system which producesa combination of perfect and defective units It is assumedthat only a fraction of defective products are repairable Thismodel jointly determines the quality of the product andselling price which will maximize the average net revenueper unit time A general model is developed For illustratingthe model two numerical problems are presented with twodemand patterns Sensitivity analysis is performed for bothexamples An algorithm is provided for solving the modelThis algorithm helps in generating computer codes How-ever standard optimizing software like MATLAB can beused for solving this model In MATLAB fmincon functioncan be used The model considers quality as a continuousvariable described by the quality index 119903 but a situationmay arise where manufacturerbusiness organization mayface the problem to select the appropriate quality out of afinite number of quality options It means that the qualityhas a discrete set of options The previous model can beused to tackle this situation with a minor modification In


Table 5 Results of sensitivity analysis

Parameterdecision variable

change ANR change in ANR

Example 1

119898

minus12 17864 minus383minus8 21165 minus269minus4 24852 minus142+4 33503 157+8 38531 331+12 44082 522

119899

minus12 38838 341minus8 35062 211minus4 31796 98+4 26467 minus86+8 24275 minus162+12 22342 minus228

119896

minus12 22702 minus216minus8 26137 minus97minus4 28242 minus25+4 28226 minus25+8 26014 minus102+12 22310 minus229

1199051


Example 2

119906


V


119908


Table 5 Continued

Parameterdecision variable change ANR change in ANR

119896


1199051


this situation find the best ANR for each of these possiblequalities and then compare the values to find the mostappropriate quality

There is enough scope to extend thismodel by incorporat-ing shortages inflation and stock-dependent demand rate

Conflict of Interests

The author declares no conflict of interests with the softwarepackage MATLAB

Acknowledgments

The author deeply appreciates anonymous referees for theirvaluable commentssuggestions

References

[1] E SMaynesHousehold Production and Consumption NationalBureau of Economic Research Cambridge Mass USA 1976

[2] R Jiang ldquoA product quality index derived from product lifetimedistributionrdquo in Proceedings of the International Conference onMeasuring Technology andMechatronics Automation (ICMTMArsquo09) pp 573ndash577 April 2009

[3] A Huidobro A Pastor and M Tejada ldquoQuality index methoddeveloped for raw gilthead seabream (Sparus aurata)rdquo Journalof Food Science vol 65 no 7 pp 1202ndash1205 2000

[4] A Barbosa and P Vaz-Pires ldquoQuality index method (QIM)development of a sensorial scheme for common octopus (Octo-pus vulgaris)rdquo Food Control vol 15 no 3 pp 161ndash168 2004

[5] M J Rosenblatt and H L Lee ldquoEconomic production cycleswith imperfect production processesrdquo IIE Transactions vol 18no 1 pp 48ndash55 1986

[6] C H Kim and Y Hong ldquoAn optimal production run lengthin deteriorating production processesrdquo International Journal ofProduction Economics vol 58 no 2 pp 183ndash189 1999

[7] M K Salameh andM Y Jaber ldquoEconomic production quantitymodel for items with imperfect qualityrdquo International Journal ofProduction Economics vol 64 no 1 pp 59ndash64 2000


[8] K Chung and K Hou ldquoAn optimal production run timewith imperfect production processes and allowable shortagesrdquoComputers and Operations Research vol 30 no 4 pp 483ndash4902003

[9] Y P Chiu ldquoDetermining the optimal lot size for the finiteproduction model with random defective rate the reworkprocess and backloggingrdquo Engineering Optimization vol 35no 4 pp 427ndash437 2003

[10] S S Sana ldquoA production-inventory model in an imperfectproduction processrdquo European Journal of Operational Researchvol 200 no 2 pp 451ndash464 2010

[11] T K Datta ldquoProduction rate and selling price determinationin an inventory system with partially defective productsrdquo ISTTransactions of Applied MathematicsmdashModeling and Simula-tion vol 1 no 2 pp 15ndash19 2010

[12] F Mhada A Hajji R Malhame A Gharbi and R PellerinldquoProduction control of unreliable manufacturing systems pro-ducing defective itemsrdquo Journal of Quality in MaintenanceEngineering vol 17 no 3 pp 238ndash253 2011

[13] C Chen ldquoOptimal determination of quality level selling quan-tity and purchasing price for intermediate firmsrdquo ProductionPlanning amp Control vol 11 no 7 pp 706ndash712 2000

[14] NKMahapatra andMMaiti ldquoMulti-objective inventorymod-els of multi-items with quality and stock-dependent demandand stochastic deteriorationrdquo Advanced Modeling and Opti-mization vol 7 no 1 pp 69ndash84 2005

[15] S L Chen and C L Liu ldquoThe optimal consignment policy forthe manufacturer under supply chain co-ordinationrdquo Interna-tional Journal of Production Research vol 46 no 18 pp 5121ndash5143 2008

[16] C Chen ldquoThe joint determination of optimum process meanand economic order quantityrdquo Tamkang Journal of Science andEngineering vol 14 no 4 pp 303ndash312 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

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Applied MathematicsJournal of


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Journal of


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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


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Algebra

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Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


run length Salameh and Jaber [7] developed an EOQ modelwhere all items produced are not perfect They assumedthat the imperfect items would be sold in a single batchby the end of the screening process Chung and Hou [8]analyzed a system with deteriorating production process andallowable shortages Chiu [9] developed a finite productionrate model by assuming that a fraction of defective productscan be reworked and the rest will be disposed Sana [10]analyzed an inventory model to determine optimal productreliability optimal production rate in a faulty productionsystem In his article the demand rate is assumed to be timedependent Datta [11] developed an inventory model withadjustable production rate and selling price sensitive demandrate In his article he assumed that all items produced are notperfect This model jointly determines optimal productionrate production period and selling price Mhada et al [12]proposed a model with perfectly mixed good and defectiveparts But these research articles did not consider productquality as a decision parameter Some researchers realized theimportance of product quality and incorporated the qualityfactor in their models Some of such articles are authoredby Chen [13] Mahapatra and Maiti [14] Chen and Liu [15]and Chen [16] Chen [13] developed a model to find optimalquality level purchase price and selling quantity for theimmediate firms but his model is not valid in a productionsystemwhere all units produced are not goodMahapatra andMaiti [14] developed a multiobjective multi-item inventorymodel with quality and stock-dependent demand rate Theirpaper did not focus neither on the influence of selling pricein demand nor on the defective product Chen and Liu [15]proposed an optimal consignment policy considering a fixedfee and a per unit commission Their model determines ahigher manufacturerrsquos profit than the traditional productionsystem and coordinates the retailer to obtain a large supplychain profit Chen [16] modified the model of Chen and Liu[15] by incorporating the influence of retailerrsquos order quantityon the manufacturerrsquos product quality He considered thequality of the product as normally distributed None of theabove articles explained the joint determination of best sellingprice product quality and product quantity under quality-dependent production rate and costs

In the present paper the author has attempted to developan inventory model to integrate the above mentioned factorsThe salient features of this developed model are as follows

(i) demand rate depends upon quality and selling price(selling markup rate)

(ii) a fraction of items produced are defective and only afraction of defective items are repairable

(iii) unit cost and carrying cost are variables dependentupon quality

(iv) production rate is quality dependent

This model maximizes the average net profit per unit timeand determines the best suitable quality and the best possiblemarkup for selling price It also determines the optimumproduction quantity in each cycle This model is suitablefor a manufacturing system where the manufacturer wantsto jointly determine the quality (grade) of the item that

should be produced and the selling price of the items tomaximize the average net profit per unit time The model isillustrated by numerical examples A sensitivity analysis hasbeen performed

2 Assumptions and Notation

The following assumptions and notations are used in thedeveloped model

Production ProcessThe production process is not completelyperfect A fraction of the items produced are defective Afraction of the defective items can be repaired to make itperfect The rest cannot be repaired and will be disposed

Quality of the ProductThe item can be produced in differentqualities but the manufacturer wants to market a particularquality which will be most profitable Actually types ofraw materials used skills of the workers working in theproduction line and quality of machineries used in theproduction system are responsible for the qualityThe qualityis assumed to be under manufacturerrsquos control Quality isrepresented by a quality index 119903 0 lt 119903

1le 119903 le 1 Here 119903

1is

the minimum quality that is required to market the product119903 = 1 indicates the top quality

Production Rate The production rate depends upon thequality The production rate 119875(119903) is a decreasing function of119903 It means the rate decreases when the quality improvesThisrate is taken in the following form

119875 (119903) = 1198751+

1198752

119903

0 lt 1199031le 119903 le 1 (1)

where 1198751represents the constant part of the production rate

which does not depend upon the quality and the second part1198752119903 decreases with quality It can be noticed that 119875min = 1198751 +1198752and 119875max = 1198751 + (11987521199031) This type of production rate can

be seen in Datta [11]Unit Cost The unit cost 119862

119906(119903) of the item depends on the

quality The cost increases with quality The following linearform is taken for unit cost 119862

119906(119903) 119862

119906(119903) = 119886 + 119887119903 where

119886 and 119887 are two positive constants It can be noticed that119886 + 119887119903

1le 119862119906le 119886 + 119887

Carrying Cost Carryingholding cost is 119862ℎ(119903) per unit time

This cost increases with quality 119903 and is taken in the quadraticform 119862

ℎ(119903) = 119901 + 119902119903

2 where 119901 and 119902 are positive constantsConsider the following120572 fraction of items produced that are defective120573 fraction of defective items that can be repaired tomake it perfect119862119904 setup cost per production run

119862119889 disposal cost per unit for disposing nonrepairable

items119862119903 cost of repairing one unit

119896 markup rate for selling (119896 gt 0) Selling price is 119904 =119896119862119906(119903) for each unit of the product of quality 119903 But to

earn profit 119896 should be greater than 1


Stoc

k le

vel

Time

S1 S2

t = t1 t = t2 t = Tt = 0






119896lt 0119863

119896119896lt 0



(1 minus 120572) 119875 (119903) minus 119863 (119896 119903) gt 0 (2)



119889



1] The level of


1




119905 = 1199052






1)



1



1


1


1


1can be expressed

as 1198781= [(1 minus 120572)119875(119903) minus 119863(119896 119903)]119905

1


1


1199052= 1199051+ (1198781119863(119896 119903)) = (1 minus 120572)119875(119903)119905

1119863(119896 119903) and

119879 = 1199052+ (1198782119863(119896 119903)) = (1 minus 120572 + 120572120573)119875(119903)119905

1119863(119896 119903)


1

Setup cost = 119862119904


1


119903lt 119862119906)


11199052+ 1198782(119879 minus 119905

2)] =

(119862ℎ(119903)2119863(119896 119903))[120572

21205732+(1 minus 120572)

2119875(119903)

2minus119863(119896 119903)(1minus

120572)119875(119903)]1199052

1


1


ANR (1199051 119896 119903)

=

1

119879

times[

[


]

]

= 119896119862119906(119903)119863 (119896 119903) minus

119863 (119896 119903)

1 minus 120572 + 120572120573

times [

119862119904

119875 (119903) 1199051

+ 119862119906(119903) + 119862

119889120572 (1 minus 120573) + 119862

119903120572120573]

minus

119862ℎ(119903) 1199051

2 (1 minus 120572 + 120572120573)

times [12057221205732+ (1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)]

(3)




120597ANR1205971199051

= 0

120597ANR120597119896

= 0

Now 120597ANR1205971199051

= 0 997904rArr

(4)

1199051

= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903)[120572

21205732+ (1 minus 120572)

2119875(119903) minus (1 minus 120572)119863 (119896 119903)]

(5)

Also

1205972ANR1205971199051

2= minus

2119862119904119863 (119896 119903)

(1 minus 120572 + 120572120573) 119875 (119903) 1199051

3lt 0 (6)




Now

12057221205732+ (1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)


(7)


(proved)

Moreover

120597ANR120597119896

= 0 997904rArr

119862119906(119903)119863 (119896 119903) + [119870119862

119906(119903) minus 119883 (119905

1) + 119884 (119905

1)]119863119896= 0

(8)

where

119883(1199051) =

1

1 minus 120572 + 120572120573

times [

119862119904

119875 (119903) 1199051

+ 119862119906(119903) + 120572 (1 minus 120573)119862

119889+ 120572120573119862

119903]

119884 (1199051) =

119862ℎ(1 minus 120572) 119905

1

2 (1 minus 120572 + 120572120573)

119883 (1199051) 119884 (119905

1) gt 0

(9)


(a) 1205972ANR1205971199051

2lt 0

(b) 1205972ANR1205971198962lt 0

(c) 1205972ANR1205971199051

2sdot

1205972ANR1205971198962minus

1205972ANR1205971199051120597119896

2

gt 0

(10)





business 119896119862119906sdot 119863(119896 119903) gt 119883(119905

1)119863(119896 119903) or 119896119862

119906gt 119883(119905

1) This


1) + 119884(119905

1) gt 0

Hence 1205972ANR1205971198962 = 2119862119906119863119896(119896 119903) + [119896119862

119906minus 119883(119905

1) +

119884(1199051)]119863119896119896lt 0 This proved (b)






119876lowast= 119875 (119903) 119905

1

= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

21205732+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(11)


Algorithm



Step 3 Take 119903 = 1199031 119896 = 1





Step 8 119903 = 1199031+ 119894




Parameter 119898 119899 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 120 25 12 20 300 100 02 6 8 04 05 5 15 200


119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



Parameter 119906 V 119908 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 200 30 80 12 20 300 50 02 6 8 02 05 5 15 1000


1will give


Step 11 Stop


Case 1 When 120572 = 0

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [1 minus (119863 (119896 119903) 119875 (119903))]

(12)


119904119863(119896 119903)119862

ℎ(119903) which


Case 2 When 120573 = 0

119876lowast= radic

2119862119904119863 (119896 119903)


(13)


radic2119862119904119863(119896 119903)119862

ℎ(119903)(1 minus 120572)


Case 3 When 120573 = 1

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

2+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(14)



1199051= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903) [(1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)]

(15)













119896lt 0 One







119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



0102030405060















0102030405060708090












Example 1

119898


119899


119896


1199051


Example 2

119906


V


119908


Table 5 Continued


119896


1199051






Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Stoc

k le

vel

Time

S1 S2

t = t1 t = t2 t = Tt = 0






119896lt 0119863

119896119896lt 0



(1 minus 120572) 119875 (119903) minus 119863 (119896 119903) gt 0 (2)



119889



1] The level of


1




119905 = 1199052






1)



1



1


1


1


1can be expressed

as 1198781= [(1 minus 120572)119875(119903) minus 119863(119896 119903)]119905

1


1


1199052= 1199051+ (1198781119863(119896 119903)) = (1 minus 120572)119875(119903)119905

1119863(119896 119903) and

119879 = 1199052+ (1198782119863(119896 119903)) = (1 minus 120572 + 120572120573)119875(119903)119905

1119863(119896 119903)


1

Setup cost = 119862119904


1


119903lt 119862119906)


11199052+ 1198782(119879 minus 119905

2)] =

(119862ℎ(119903)2119863(119896 119903))[120572

21205732+(1 minus 120572)

2119875(119903)

2minus119863(119896 119903)(1minus

120572)119875(119903)]1199052

1


1


ANR (1199051 119896 119903)

=

1

119879

times[

[


]

]

= 119896119862119906(119903)119863 (119896 119903) minus

119863 (119896 119903)

1 minus 120572 + 120572120573

times [

119862119904

119875 (119903) 1199051

+ 119862119906(119903) + 119862

119889120572 (1 minus 120573) + 119862

119903120572120573]

minus

119862ℎ(119903) 1199051

2 (1 minus 120572 + 120572120573)

times [12057221205732+ (1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)]

(3)




120597ANR1205971199051

= 0

120597ANR120597119896

= 0

Now 120597ANR1205971199051

= 0 997904rArr

(4)

1199051

= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903)[120572

21205732+ (1 minus 120572)

2119875(119903) minus (1 minus 120572)119863 (119896 119903)]

(5)

Also

1205972ANR1205971199051

2= minus

2119862119904119863 (119896 119903)

(1 minus 120572 + 120572120573) 119875 (119903) 1199051

3lt 0 (6)




Now

12057221205732+ (1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)


(7)


(proved)

Moreover

120597ANR120597119896

= 0 997904rArr

119862119906(119903)119863 (119896 119903) + [119870119862

119906(119903) minus 119883 (119905

1) + 119884 (119905

1)]119863119896= 0

(8)

where

119883(1199051) =

1

1 minus 120572 + 120572120573

times [

119862119904

119875 (119903) 1199051

+ 119862119906(119903) + 120572 (1 minus 120573)119862

119889+ 120572120573119862

119903]

119884 (1199051) =

119862ℎ(1 minus 120572) 119905

1

2 (1 minus 120572 + 120572120573)

119883 (1199051) 119884 (119905

1) gt 0

(9)


(a) 1205972ANR1205971199051

2lt 0

(b) 1205972ANR1205971198962lt 0

(c) 1205972ANR1205971199051

2sdot

1205972ANR1205971198962minus

1205972ANR1205971199051120597119896

2

gt 0

(10)





business 119896119862119906sdot 119863(119896 119903) gt 119883(119905

1)119863(119896 119903) or 119896119862

119906gt 119883(119905

1) This


1) + 119884(119905

1) gt 0

Hence 1205972ANR1205971198962 = 2119862119906119863119896(119896 119903) + [119896119862

119906minus 119883(119905

1) +

119884(1199051)]119863119896119896lt 0 This proved (b)






119876lowast= 119875 (119903) 119905

1

= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

21205732+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(11)


Algorithm



Step 3 Take 119903 = 1199031 119896 = 1





Step 8 119903 = 1199031+ 119894




Parameter 119898 119899 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 120 25 12 20 300 100 02 6 8 04 05 5 15 200


119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



Parameter 119906 V 119908 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 200 30 80 12 20 300 50 02 6 8 02 05 5 15 1000


1will give


Step 11 Stop


Case 1 When 120572 = 0

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [1 minus (119863 (119896 119903) 119875 (119903))]

(12)


119904119863(119896 119903)119862

ℎ(119903) which


Case 2 When 120573 = 0

119876lowast= radic

2119862119904119863 (119896 119903)


(13)


radic2119862119904119863(119896 119903)119862

ℎ(119903)(1 minus 120572)


Case 3 When 120573 = 1

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

2+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(14)



1199051= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903) [(1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)]

(15)













119896lt 0 One







119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



0102030405060















0102030405060708090












Example 1

119898


119899


119896


1199051


Example 2

119906


V


119908


Table 5 Continued


119896


1199051






Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












120597ANR1205971199051

= 0

120597ANR120597119896

= 0

Now 120597ANR1205971199051

= 0 997904rArr

(4)

1199051

= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903)[120572

21205732+ (1 minus 120572)

2119875(119903) minus (1 minus 120572)119863 (119896 119903)]

(5)

Also

1205972ANR1205971199051

2= minus

2119862119904119863 (119896 119903)

(1 minus 120572 + 120572120573) 119875 (119903) 1199051

3lt 0 (6)




Now

12057221205732+ (1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)


(7)


(proved)

Moreover

120597ANR120597119896

= 0 997904rArr

119862119906(119903)119863 (119896 119903) + [119870119862

119906(119903) minus 119883 (119905

1) + 119884 (119905

1)]119863119896= 0

(8)

where

119883(1199051) =

1

1 minus 120572 + 120572120573

times [

119862119904

119875 (119903) 1199051

+ 119862119906(119903) + 120572 (1 minus 120573)119862

119889+ 120572120573119862

119903]

119884 (1199051) =

119862ℎ(1 minus 120572) 119905

1

2 (1 minus 120572 + 120572120573)

119883 (1199051) 119884 (119905

1) gt 0

(9)


(a) 1205972ANR1205971199051

2lt 0

(b) 1205972ANR1205971198962lt 0

(c) 1205972ANR1205971199051

2sdot

1205972ANR1205971198962minus

1205972ANR1205971199051120597119896

2

gt 0

(10)





business 119896119862119906sdot 119863(119896 119903) gt 119883(119905

1)119863(119896 119903) or 119896119862

119906gt 119883(119905

1) This


1) + 119884(119905

1) gt 0

Hence 1205972ANR1205971198962 = 2119862119906119863119896(119896 119903) + [119896119862

119906minus 119883(119905

1) +

119884(1199051)]119863119896119896lt 0 This proved (b)






119876lowast= 119875 (119903) 119905

1

= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

21205732+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(11)


Algorithm



Step 3 Take 119903 = 1199031 119896 = 1





Step 8 119903 = 1199031+ 119894




Parameter 119898 119899 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 120 25 12 20 300 100 02 6 8 04 05 5 15 200


119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



Parameter 119906 V 119908 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 200 30 80 12 20 300 50 02 6 8 02 05 5 15 1000


1will give


Step 11 Stop


Case 1 When 120572 = 0

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [1 minus (119863 (119896 119903) 119875 (119903))]

(12)


119904119863(119896 119903)119862

ℎ(119903) which


Case 2 When 120573 = 0

119876lowast= radic

2119862119904119863 (119896 119903)


(13)


radic2119862119904119863(119896 119903)119862

ℎ(119903)(1 minus 120572)


Case 3 When 120573 = 1

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

2+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(14)



1199051= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903) [(1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)]

(15)













119896lt 0 One







119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



0102030405060















0102030405060708090












Example 1

119898


119899


119896


1199051


Example 2

119906


V


119908


Table 5 Continued


119896


1199051






Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











Parameter 119898 119899 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 120 25 12 20 300 100 02 6 8 04 05 5 15 200


119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



Parameter 119906 V 119908 119886 119887 1198751 1198752 1199031 119901 119902 120572 120573 119862119889

119862119903

119862119904

Value 200 30 80 12 20 300 50 02 6 8 02 05 5 15 1000


1will give


Step 11 Stop


Case 1 When 120572 = 0

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [1 minus (119863 (119896 119903) 119875 (119903))]

(12)


119904119863(119896 119903)119862

ℎ(119903) which


Case 2 When 120573 = 0

119876lowast= radic

2119862119904119863 (119896 119903)


(13)


radic2119862119904119863(119896 119903)119862

ℎ(119903)(1 minus 120572)


Case 3 When 120573 = 1

119876lowast= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) [120572

2+ (1 minus 120572)

2minus (1 minus 120572) (119863 (119896 119903) 119875 (119903))]

(14)



1199051= radic

2119862119904119863 (119896 119903)

119862ℎ(119903) 119875 (119903) [(1 minus 120572)

2 119875 (119903) minus (1 minus 120572)119863 (119896 119903)]

(15)













119896lt 0 One







119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



0102030405060















0102030405060708090












Example 1

119898


119899


119896


1199051


Example 2

119906


V


119908


Table 5 Continued


119896


1199051






Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119903lowast

119896lowast

119905lowast


ANR119896119896

ANR1199051119896

119867



0102030405060















0102030405060708090












Example 1

119898


119899


119896


1199051


Example 2

119906


V


119908


Table 5 Continued


119896


1199051






Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













Example 1

119898


119899


119896


1199051


Example 2

119906


V


119908


Table 5 Continued


119896


1199051






Acknowledgments


References

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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