A note on fuzzy inventory model with storage space and budget constraints

A note on fuzzy inventory model with storage space and budget constraints

Applied Mathematical Modelling 33 (2009) 4069–4077 Contents lists available at ScienceDirect Applied Mathematical Modelling journal homepage: www.el...

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Applied Mathematical Modelling 33 (2009) 4069–4077

Contents lists available at ScienceDirect

Applied Mathematical Modelling journal homepage: www.elsevier.com/locate/apm

A note on fuzzy inventory model with storage space and budget constraints Shuo-Yan Chou a,*, Peterson C. Julian b, Kuo-Chen Hung c a b c

Department of Industrial Management, National Taiwan University of Science and Technology, 43 Keelung Road, Section 4, Taipei 106, Taiwan Department of Traffic Science, Central Police University, Taiwan Department of Logistics Management, National Defense University, Taiwan

a r t i c l e

i n f o

Article history: Received 19 June 2007 Received in revised form 31 January 2009 Accepted 16 February 2009 Available online 21 February 2009

Keywords: Inventory management Fuzzy inventory model Budget constraint Storage space constraint

a b s t r a c t In 1997, Roy and Maiti developed a fuzzy EOQ model with fuzzy budget and storage capacity constraints where demand is influenced by the unit price and the setup cost varies with the quantity purchased [T.K. Roy, M. Maiti, A fuzzy EOQ model with demand-dependent unit cost under limited storage capacity, Eur. J. Oper. Res. 99 (1997) 425–432]. However, their procedure has some questionable points and their numerical examples contain rather peculiar results. The purpose of this paper is threefold. First, for the same inventory model with fuzzy constraints, based on the max–min operator, we proposed an improved solution procedure. Second, we review the solution procedure by Roy and Maiti that is based on Kuhn–Tucker approach to point out their questionable results. Third, we compare Roy and Maiti’s approach with ours to explain why our approach can solve the problem and theirs cannot. Numerical examples provided by them also support our findings. Ó 2009 Elsevier Inc. All rights reserved.

1. Introduction Fuzzy methodologies provide a useful way to model vagueness in human recognition and judgment. Uncertainties characterized by imprecise expression can be represented by fuzzy sets. However, some fuzzy procedures are too complex to examine, which may even lead to false interpretation and subsequently erroneous decisions. In this paper, we examine a paper by Roy and Maiti [1], which investigated the fuzzy EOQ model with demand-dependent unit cost under limited storage capacity. Fuzziness is introduced in both storage capacity and invested capital budget. They used both fuzzy nonlinear and geometric programming techniques to solve the optimization problem. However, there are some questionable results in their work, which will be illustrated and corrected in this paper. Although there were some incorrect mathematical formulation on Roy and Maiti [1], the original work done by Roy and Maiti [1] by using the fuzzy nonlinear/geometric programming method to solve the fuzzy inventory control system was a great contribution to the academic. After their initial work on the fuzzy EOQ model, Roy and Maiti continued to co-author four related papers. They considered a multi-period inventory model with fuzzy inventory costs and demand and then solved the problem with a shortest path algorithm and a fuzzy dynamic programming method [2]. They also developed multi-item fuzzy inventory models of deteriorating items with stock-dependent demand [3]. In Das et al. [4], they formulated a multi-item inventory model with budgetary and floor-space constraints in a fuzzy environment such that fuzziness is introduced in both the objective function and constraint goals. In Kar et al. [5], they developed an inventory model under the budgetary constraint for deteriorating multi-item with cost-price dependent demand in a fuzzy environment. The fuzzy inventory models have also been explored by many other researchers. Gen et al. [6] solved an inventory control problem such that input data are described by triangular fuzzy numbers. Petrovic et al. [7] discussed fuzzy modelling and simulation of a supply chain in an uncertain environment to provide a dynamic view and to assess the impact of decisions * Corresponding author. Tel.: +886 227376327; fax: +886 227376344. E-mail address: [email protected] (S.-Y. Chou). 0307-904X/$ - see front matter Ó 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.apm.2009.02.001

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recommended by the fuzzy models. Buckley et al. [8] generated good approximate solutions to single-item, N-period, fuzzy inventory control problems. Yao et al. [9] examined the inventory problem without backorder such that both order and the total demand quantities are triangular fuzzy numbers. Petrovic [10] considered the supply chain behavior and performance in the presence of uncertainty. Samanta and Al-Araimi [11] constructed a periodic review model to apply fuzzy logic for inventory control with variable order quantity. Kao and Hsu, [12] constructed a single-period inventory model for cases of fuzzy demand to find the optimal order quantity in terms of the cost. Katagiri and Ishii [13] developed an inventory control model for a single perishable product with a fuzzy shortage cost and a fuzzy outdating cost to investigate effects of the fuzziness on the obtained solutions. Ouyang and Chang [14] applied a minimax distribution free procedure for mixed inventory models with variable lead-time and fuzzy lost sales. There have been more than 30 papers that have referred to Roy and Maiti [1]. These papers can be classified by models with fuzzy backorders [14–17], with imperfect quality items [18–22], with multi-item inventory model [23–27], with fuzzy demand [12,28,29], with signed distance of fuzzy sets [30,31], with fuzzy lead-time [32,33], for two-echelon inventory model [34,35], for newsvendor problems [36,37], for multi-objective inventory models [3,38], for a displayed inventory model [17], with a genetic algorithm [39], with fuzzy order quantity [40], with fuzzy capacity [41], with polynomial geometric programming [42], for single-period inventory problem [43] and so on. The high citation number of Roy and Maiti [1] warrants the study and improvement of their work. Despite that a mass amount of new approaches have been developed based on the paper by Roy and Maiti [1], the solution approach of their paper is in fact constrained and not being able to find the optimal solution beyond a certain boundary value. The main issue of their procedure is in its complexity and lack of proper interpretation for the seemingly unexpected results from varying the value of the storage space constraint. In this paper, we first investigate their procedure to shed the light of what causes their questionable results. Then, we consider the same fuzzy inventory model to defuzzy it by linear membership function proposed by Roy and Maiti [1], and study the max–min operator accordingly. In order for the solution approach to be manageable, we propose a deterministic objective functional algorithm to derive a formulated solution to prepare a framework for the crisp inventory model that is useful for this fuzzy inventory model. This paper is divided into three parts. First, for the same EOQ model without fuzzy constraints, we directly solve the model and derive an explicit expression for the minimum solution that will be useful to find the optimal solution under fuzzy environment. Using the convex property, we establish an improved mathematical formulation, the theorem on the existence and uniqueness of the solution, and the optimal policy under the fuzzy environment. Second, followed Kuhn–Tucker’s method applied by them for the inventory model with such constraints, we derive the criterion to explain why their procedure will imply peculiar results and show that their numerical examples support our finding. More, we explore existing problems in their solution procedure for Kuhn–Tucker’s method of Roy and Maiti [1]. We discover that only to consider results of intersection of two membership functions is not complete for max–min operator to this kind of fuzzy inventory model. Third, we compare the Kuhn–Tucker approach with ours to demonstrate that there are several difficult unsolved problems contained in Kuhn–Tucker’s approach proposed by Roy and Maiti [1]. It may explain why they cannot finish their solution procedure. Our results are compared with those obtained by Roy and Maiti [1] for the same numerical examples to prepare a detailed explanation for unexpected phenomena in their examples. 2. Notation and assumptions

q D C1 C3 C0 B p A

We use the same notation and assumptions as in Roy and Maiti [1]. number of order quantity demand per unit time holding cost per item per unit time setup cost ¼ C 03 qc with 0 < c < 1 fuzzy objective goal with tolerance P0 fuzzy space constraint goal with tolerance P unit production cost, KDb , with b > 1 the storage space for one item Assumptions include the lead-time being zero, no back order being permitted and replenishment rate being infinite.

3. The fuzzy inventory model of Roy and Maiti While Roy and Maiti [1] dealt with the same problem with the fuzzy nonlinear programming and the fuzzy geometric programming approaches individually, we only focus on the fuzzy nonlinear programming. In the following, we will review their inventory model with fuzzy storage space and budget constraints. Roy and Maiti [1] introduced the fuzziness in both the budget and storage space constraint, and proposed the following fuzzy model:

S.-Y. Chou et al. / Applied Mathematical Modelling 33 (2009) 4069–4077

q M~inCðD; qÞ ¼ C 03 qr1 D þ KD1b þ C 1 ; 2

4071

ð1Þ

~ 0 with D; q > 0. Roy and Maiti used the linear fuzzy membership functions for the capital ~ and CðD; qÞ 6 C such that Aq 6 B constraint, l1 , and the storage space constraint, l2 , so that the membership functions for the objective and resources were assumed to be

li ðg i ðxÞÞ ¼

8 > < > :

1 if i 1  gi ðxÞb if Pi

g i ðxÞ < bi bi 6 g i ðxÞ 6 bi þ Pi

for i ¼ 0; 1; 2; . . . ; m:

ð2Þ

g i ðxÞ > bi þ P i

if

0

Using the max–min operator, the membership function of the decision set was given by

lD ðxÞ ¼ minfl0 ðxÞ; l1 ðxÞ; l2 ðxÞ; . . . ; lm ðxÞg for all x 2 X; and; lD ðxmax Þ ¼ max½minfl0 ðxÞ; l1 ðxÞ; l2 ðxÞ; . . . ; lm ðxÞg: xP0

With the above settings, the fuzzy problem was transformed to a crisp nonlinear programming problem

Max a; s:t: l0 ðxÞ P a; li ðxÞ P a; i ¼ 1; 2; . . . ; m; x P 0; 0 6 a 6 1:

ð3Þ

We study the single-item inventory model with demand-dependent unit price and variable setup cost under storage space constraint, formulated as

q CðD; qÞ ¼ C 03 qc1 D þ KD1b þ C 1 ; 2

ð4Þ

such that according to Eq. (2), the budget and storage space constraints are represented by linear fuzzy membership functions for the capital constraint, l1 , and the storage space constraint, l2 ,

l1 ðCÞ ¼

8 > < > :

1 if 0 1  CC P0 if 0

if

C 6 C0 C 0 6 C 6 C 0 þ P0 ;

ð5Þ

C P C 0 þ P0

and

l2 ðqÞ ¼

8 > <

1 if 1  AqB if P > : 0 if

Aq 6 B B 6 Aq 6 B þ P ;

ð6Þ

Aq P B þ P

respectively, for the fuzzy inventory model. As in Eq. (5), C ¼ CðD; qÞ, l1 ðCÞ is thus in variable C. Later when we try to find minfl1 ðCÞ; l2 ðqÞg, we need to use the same variable, q. We solve the minimization problem in two steps: first without the storage space and budget constraints (crisp model) and then with the storage space and budget constraints. 4. Proposed procedure under crisp case We begin by solving the problem without constraints. By taking the derivative of CðD; qÞ, we have

o CðD; qÞ ¼ C 03 qc1  Kðb  1ÞDb ; oD and

o2

CðD; qÞ ¼ Kbðb  1ÞDb1 ;

oD2

for b > 1:

ð7Þ

2

o Since oD 2 CðD; qÞ > 0, we know that CðD; qÞ is a convex function in D when q is fixed. Hence the minimization of CðD; qÞ is o o CðD; qÞ ¼ 0. From Eq. (7), when oD CðD; qÞ ¼ 0, we have equivalent to solving for the root of oD

 D¼

1 Kðb  1Þ b : C 03 qc1

Therefore, by plugging Eq. (8) into Eq. (4), we can then consider the following problem

ð8Þ

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f ðqÞ ¼ aqð1cÞðb

1

1Þ

q þ C1 2

with a ¼

 1 b Kðb  1Þ b : C 03 b1 C 03

ð9Þ

We first derive 1

f 0 ðqÞ ¼ að1  cÞðb1  1Þqð1cÞðb

1Þ1

þ

C1 ; 2

ð10Þ

and

f 00 ðqÞ ¼ að1  cÞðb1  1Þ½ð1  cÞðb1  1Þ  1qð1cÞðb

1

1Þ2

:

The inequalities 0 < c < 1 and b > 1 can be rearranged to obtain b1  1 < 0 and ð1  cÞðb1  1Þ < 0, which yields f ðqÞ > 0. This means that f ðqÞ is a convex function with the absolute minimum point at f 0 ðqÞ ¼ 0. For the convenience of later discussion, we denote the solution of f 0 ðqÞ ¼ 0 as q# . By Eq. (10), we can obtain 00

1   1þð1cÞð1b1 Þ a q# ¼ 2 ð1  cÞð1  b1 Þ : C1

ð11Þ

Following Eq. (8), similarly we assume that

" #

D ¼

Kðb  1Þ

C 03 ðq# Þc1

#1b ð12Þ

:

From the above discussion, we can establish the following theorem. Theorem 1. Without the storage space and the budget constraints, q# is the absolute minimum solution for f ðqÞ and ðD# ; q# Þ is the absolute minimum solution for CðD; qÞ under the crisp environment.

5. Proposed procedure under fuzzy constraints We then examine the minimization problem under the consideration of the storage space and the budget constraints. CðDðqÞ; qÞ is a convex function with minimum value at ðDðq# Þ; q# Þ, where DðqÞ satisfies Eq. (8) such that l1 ðq# Þ attains its maximum. We divide it into three cases: Case 1: q# 6 AB, Case 2: AB 6 q# 6 qI , and Case 3: AB 6 qI 6 q# , where U 1 ðqÞ and U 2 ðqÞ 0 and U 2 ðqÞ ¼ 1  AqB are the linear parts of the membership functions l1 and l2 , intersect at qI and U 1 ðCðD; qÞÞ ¼ 1  CðD;qÞC P0 P respectively. For Case 1, l2 ðq# Þ ¼ 1 and l1 ðq# Þ attains its maximum, which implies that the optimal solution is ðDðq# Þ; q# Þ. For Case 2, as we know that

( minfl1 ðqÞ; l2 ðqÞg ¼

l1 ðqÞ; l2 ðqÞ;

B A

6 q 6 qI qI 6 q

;

ð13Þ

the maximum value then occurs at q# and the optimal solution is ðDðq# Þ; q# Þ. For Case 3, it yields that

( minfl1 ðqÞ; l2 ðqÞg ¼

l1 ðqÞ; l2 ðqÞ;

B A

6 q 6 qI qI 6 q

;

ð14Þ

so that the maximum value occurs at qI and then the optimal solution is ðDðqI Þ; qI Þ. We summarize our discussion for the three cases to indicate that

q ¼ minfq# ; qI g

ð15Þ

is the optimal solution for the order quantity. Theorem 2. With the storage space and the budget constraints, ðDðq Þ; q Þ is the optimal solution for the fuzzy inventory model. 6. Discussion of the solution procedure by Roy and Maiti i P a, which means that g i ðxÞ 6 bi þ ð1  aÞPi . Thus, the proFrom Eq. (2) and the constraint li ðxÞ P a, we have 1  gi ðxÞb Pi posed model in Eq. (3) can be reduced to

S.-Y. Chou et al. / Applied Mathematical Modelling 33 (2009) 4069–4077

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Max a; q s:t: C 03 qc1 D þ KD1b þ C 1 6 C 0 þ ð1  aÞP0 ; 2 Aq 6 B þ ð1  aÞP;

ð16Þ

D; q > 0; a 2 ð0; 1Þ; By the Kuhn–Tucker’s necessary conditions, the following system was solved:

q C 03 qc1 D þ KD1b þ C 1  C 0  ð1  aÞP0 6 0; 2 Aq  B  ð1  aÞP 6 0; h i q k1 C 03 qc1 D þ KD1b þ C 1  C 0  ð1  aÞP0 ¼ 0; 2 k2 ½Aq  B  ð1  aÞP ¼ 0; 1  k1 P0  k2 P ¼ 0;   C1 k1 C 03 ðc  1Þqc2 D þ þ k2 A ¼ 0; 2

ð17Þ ð18Þ ð19Þ ð20Þ ð21Þ ð22Þ

and

k1 ½C 03 qc1 þ Kð1  bÞDb  ¼ 0:

ð23Þ

By solving these simultaneous equations, Roy and Maiti obtained the optimal values

q ¼

B þ ð1  a ÞP ; A

ð24Þ

D ¼

 1b  1c Kðb  1Þ B þ ð1  a ÞP b ; C A

ð25Þ

and

where a is the root of the following equation cÞ ð1bÞð1  1b  b Kðb  1Þ b B þ ð1  aÞP H½B þ ð1  aÞP Kb þ  ð1  aÞP0 ¼ 0: C 03 A 2A

ð26Þ

To simplify the expression, we define

gðaÞ ¼ gðDðaÞ; qðaÞÞ ¼ C 03 ðc  1Þqc2 D þ

C1 : 2

From Eqs. (21) and (22), we know that k1 and k2 are the two solutions of the simultaneous equations

k1 P0 þ k2 P ¼ 1; and

k1 gðD; qÞ þ k2 A ¼ 0; which yields that

k1 ¼

1 ; P0  AP gðD; qÞ

k2 ¼

gðD; qÞ k1 : A

and

ð27Þ

With A > 0, we can then have k1 – 0. If we obtain k2 – 0, then Eqs. (19), (20), and (23) can be reduced to

q C 03 qc1 D þ KD1b þ C 1  C 0  ð1  aÞP0 ¼ 0; 2 C 03 qc1 þ Kð1  bÞDb ¼ 0;

ð28Þ ð29Þ

and

Aq  B  ð1  aÞP ¼ 0;

ð30Þ

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Eq. (30) shows that Eq. (12) in Roy and Maiti [1] is correct. However, by Eq. (29), the expression of Eq. (13) in Roy and Maiti [1] should be changed to

D ¼

1c  1  Kðb  1Þ b B þ ð1  a ÞP b : C 03 A

ð31Þ

By plugging Eqs. (24) and (31) into Eq. (30), the correct expression for Eq. (26) of Roy and Maiti [1] can be shown to be cÞ ð1bÞð1  1b  b Kðb  1Þ b B þ ð1  aÞP C 1 ½B þ ð1  aÞP Kb þ  C 0  ð1  aÞP 0 ¼ 0: C 03 A 2A

ð32Þ

With Eq. (32) we can then point out a questionable result in the procedure of Roy and Maiti [1]. To simplify the expression, we assume that cÞ ð1bÞð1  1b  b Kðb  1Þ b B þ ð1  aÞP C 1 ½B þ ð1  aÞP  C 0  ð1  aÞP0 : hðaÞ ¼ Kb þ C 03 A 2A

ð33Þ

Roy and Maiti did not discuss whether or not hðaÞ ¼ 0 has a solution for a 2 ð0; 1Þ. The most likely way to show that hðaÞ has a unique root for 0 < a < 1 is to prove that 0

hð0Þ < 0; hð1Þ > 0 and h ðaÞ > 0:

ð34Þ

We found that cÞ  1b  ð1bÞð1 b Kðb  1Þ b B C1B  C0; hð1Þ ¼ Kb þ C 03 A 2A cÞ ð1bÞð1  1b  b Kðb  1Þ b B þ P C 1 ½B þ P hð0Þ ¼ Kb þ  C 0  P0 ; C 03 A 2A

and cÞ ð1bÞð1  1b  b KP Kðb  1Þ b B þ ð1  aÞP C1P h ðaÞ ¼  ðb  1Þð1  cÞ þ P0 : A C 03 A 2A

0

Let us consider whether or not that

P0 

C1P > 0: 2A

ð35Þ

From the examples in Roy and Maiti [1], that is, varying P 0 from 20 to 5  105 and P from 15 to 162, we verify that Eq. (35) 0 is always satisfied. With the satisfaction of Eq. (35), we can then have h ðaÞ > 0. Therefore, the condition hð0Þ < 0 is essential for hðaÞ ¼ 0 to have a solution for 0 < a < 1. However, even for Roy and Maiti’s own example, the condition hð0Þ < 0 is sometimes violated and was observed by Roy and Maiti [1] and recorded in Table 3 of their paper for researchers to investigate. We can illustrate this phenomenon with numerical examples.

7. Numerical examples We now consider the numerical examples of Roy and Maiti [1] for showing the differences between our results and their results. For an EOQ problem, they have the first example with C 03 ¼ $4, K ¼ 100, C 1 ¼ $2, c ¼ 0:5, b ¼ 1:5, A ¼ 10 units, B ¼ 50 units, C 0 ¼ $40, P 0 ¼ $20, P ¼ 15 units. They performed sensitivity analysis by varying P from 15 to 162. We quote their results (Tables 3 of Roy and Maiti [1]) in Table 1. Using Eq. (33), we can obtain that if P ¼ 161, hð0Þ ¼ 0:02 < 0; and if P ¼ 162; hð0Þ ¼ 0:05 > 0. Hence, we know that hð0Þ > 0 if P P 162; and hðaÞ ¼ 0 does not have any solution for 0 < a < 1. Moreover, from the numerical results of Roy and Maiti [1] in our Table 1 (their Table 3), when P ¼ 162, they could not obtain a unique root for hðaÞ ¼ 0 for 0 < a < 1. It is a surprising result. With more potential storage space, there should be benefit to derive an optimal solution that satisfies the looser constraint. However, in Roy and Maiti [1], when the extra storage space increases to P ¼ 162, an unexpected result a < 0 was derived by Roy and Maiti [1]. Our detailed analysis is useful to provide a reasonable explanation for why their approach failed. Their numerical example therefore also supports our findings. In their paper, Roy and Maiti mentioned that the values of C  ðD ; q Þ in our Table 1 indicate that the objective value was at first decreasing but then turned increasing when P became greater than 41. However, this claim is contradictory to the basic principle for the inventory policy; since increasing the storage space did not increase the total cost. This first decreasing then increasing behavior shown by Roy and Maiti [1] is therefore false and their numerical approach erroneous. The cost should be decreased until it attained its minimum then fixed at the minimum as P increases. Our findings in Table 2 will support this prediction.

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S.-Y. Chou et al. / Applied Mathematical Modelling 33 (2009) 4069–4077 Table 1 Effect of variation of P from Roy and Maiti [1]. P

a

q

D

T 0 ¼ ð1  aÞP 0

T ¼ ð1  aÞP

C  ðD ; q Þ

15 16 20 23 36 38 40 41 42 44 156 160 161 162

0.30 0.30 0.30 0.31 0.31 0.31 0.31 0.31 0.31 0.31 0.30 00 00 0.005

6.04 6.11 6.38 6.59 7.46 7.60 7.73 7.80 7.87 8.01 19.99 20.84 21.06 21.29

9.81 9.85 9.99 10.10 10.53 10.59 10.65 10.68 10.72 10.78 14.61 14.82 14.87 14.93

13.93 13.91 13.84 13.79 13.70 13.69 13.69 13.69 13.70 13.70 19.22 19.80 19.96 20.116

10.45 11.13 13.84 15.86 24.65 26.02 27.39 28.08 28.76 30.14 152.02 158.43 160.66 162.94

53.93 53.91 53.84 53.79 53.70 53.69 53.69 53.69 53.70 53.70 59.23 59.80 59.95 60.116

Table 2 The results of our proposed method. P

q ¼ qI

a ¼ l1 ðqI Þ ¼ l2 ðqI Þ

CðDðq Þ; q Þ

15 16 20 23 36 38 39

6.044936 6.112904 6.383801 6.586328 7.465591 7.602017 7.670421

0.303376 0.304435 0.308100 0.310292 0.315114 0.315259 0.315277

53.932 53.911 53.838 53.794 53.698 53.695 53.694

P

q ¼ q#

a ¼ l1 ðq# Þ

l2 ðq# Þ

CðDðq Þ; q Þ

qI

l1 ðqI Þ ¼ l2 ðqI Þ

40 160 162

7.670637 7.670637 7.670637

0.315277 0.315277 0.315277

0.3223 0.833085 0.835146

53.694 53.694 53.694

7.738965 20.843 21.29396

0.315259 0.00981 0

8. A detailed analysis of Roy and Maiti’s result In this section, we will provide a detailed analysis for why Roy and Maiti [1] could not obtain the optimal solution by taking the Kuhn–Tucker approach. In particular, we will show why the seemingly infeasible result of a ¼ 0:005 at P ¼ 162 happened while under the condition of 0 6 a 6 1. Given a value of P, there is a unique q, denoted by qðPÞ, that satisfies U 2 ðqÞ ¼ U 1 ðqÞ, that is

U 2 ðqÞ ¼ 1 

Aq  B ¼ U 1 ðqÞ: P

When P increases, qðPÞ also increases. Then, P such that qðPÞ ¼ q# is to be found. With q# ¼ 7:67 and yields that



Aq#  B ¼ 39:003: 1  l1 ðq# Þ

ð36Þ

l1 ðq# Þ ¼ 0:315, it ð37Þ

We know that if P < 39:003, the optimal ordering quantity qI depends on P such that qI ðPÞ increases as P increases. The same results as that of Roy and Maiti’s can be found by our approach. However, if P P 39:003, the optimal ordering quantity, fixed as q# , can be improved by our approach. When P ¼ 40, the intersection qI ¼ 7:738965, as obtained by Roy and Maiti, is larger than the true optimum q# ¼ 7:670637. The maximum value therefore occurs at a ¼ l1 ðq# Þ ¼ 0:315277 rather than at the intersection point l1 ðqI Þ ¼ l2 ðqI Þ ¼ 0:315259. From another perspective, it can be observed that when P ¼ 160, the intersection qI ¼ 20:843 is larger than q# ¼ 7:671; so the optimal value at a ¼ l1 ðq# Þ ¼ 0:315 is again better than that from the intersection. When P ¼ 162, by our approach, the optimal ordering quantity q# can be derived. Roy and Maiti derived the solution a ¼ 0:00580, which is supposed to be the intersection of U 1 and U 2 . However, in the case, the two membership functions, l1 and l2 , have no intersection within the feasible range ð0; 1Þ. In detail, when q ¼ 21:12698, we have CðDðqÞ; qÞ ¼ 60 and l1 ðCðDðqÞ; qÞÞ ¼ 0, which implies that l1 ðCðDðqÞ; qÞÞ ¼ 0 when q P 21:12698. On the other hand, when q ¼ 21:2, then l2 ðqÞ ¼ 0, which implies that l2 ðqÞ ¼ 0 when q P 21:2. Hence, for P ¼ 162, the intersection of l1 and l2 for the range ð0; 1Þ does not happen such that

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l1 ð21:29396Þ ¼ 0 ¼ l2 ð21:29396Þ:

ð38Þ

Roy and Maiti [1] would derive the intersection of U 1 and U 2 such that

U 1 ð21:29396Þ ¼ 0:00580 ¼ U 2 ð21:29396Þ:

ð39Þ

In addition to the above problem, there exists another issue in their solution procedure with the Kuhn–Tucker’s method. From Eq. (27), for k2 –0, Roy and Maiti [1] required the condition that

gða Þ ¼ gðD ; q Þ ¼ C 03 ðc  1Þðq Þc2 D þ

C1 –0; 2

ð40Þ

where q ; D and a satisfy Eqs. (24), (31), and (32), respectively. However, Roy and Maiti [1] missed the point that Eq. (40) needs to be verified. Since a is the implicit solution of Eq. (32) where q and a are related by Eq. (24), the verification of Eq. (40) is in fact an unsolved problem in the Roy and Maiti’s [1] solution procedure. 9. Conclusion For the fuzzy inventory model with storage space constraints and total cost constraints, we have shown that the analytical procedure by Roy and Maiti [1] suffers from the lack of ability to obtain the minimum solution and the difficulty in interpretation of their sensitive analysis with varying storage spaces. Our proposed procedure can resolve these problems by employing formulated solutions based on the max–min operator. Our results provide an analytical approach to solve the deterministic inventory model converted from the fuzzy inventory model. References [1] T.K. Roy, M. 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