Research note on the criteria for the optimal solution
of the inventory model with a mixture of partial
backordering and lost sales
Gino K. Yang
a,*, Shuo-Yan Chou
b, Chih-Young Hung
c,
Jennifer Shu-Jen Lin
d, Peter Chu
eaDepartment of Logistics Management, National Defense University, Taiwan, ROC
bDepartment of Industrial Management, National Taiwan University of Science and Technology, Taiwan, ROC cGraduate Institute of Technology Management, National Chiao Tung University, Taiwan, ROC dInstitute of Technology and Vocational Education, National Taipei University of Technology, Taiwan, ROC
e
Department of Traffic Science, Central Police University, Taiwan, ROC Received 1 November 2005; received in revised form 1 May 2007; accepted 4 June 2007
Available online 24 June 2007
Abstract
Perishable products are commonly seen in inventory management. By allowing shortages and backlogging, the impact on the cost from the decay of the products can be balanced out. In a recent paper published in Computers and Industrial Engineering [P.L. Abad, Optimal lot size for a perishable good under conditions of finite production and partial backor-dering and lost sale, Comput. Ind. Eng. 38 (2000) 457–465] considered a problem in such context. However, his algorithm was incomplete due to flaws in his solution procedure. The purpose of this note is to explore the same production inventory models with a mixture of partial backordering and lost sales for deteriorated items. We find the criteria for the optimal solution for different cases and derive a formulated minimum value. By theoretical analysis, we develop a few lemmas to reveal parameter effects and optimal solution procedure. The solutions are illustrated by solving the same examples from Abad’s paper to illustrate the accuracy and completeness of our procedure.
2007 Published by Elsevier Inc.
Keywords: Inventory; Perishable items; Backorder; Lost sales
1. Introduction
Permitting limited planned shortages can reduce the pressure on high production capacity and hence result in a smoother production schedule. Firms are able to maintain a backlog of orders to certain loyal customers without losing their business. However, the costs of shortages or lost sales should not be exorbitant to
0307-904X/$ - see front matter 2007 Published by Elsevier Inc. doi:10.1016/j.apm.2007.06.022
* Corresponding author.
E-mail address:yangklung@yahoo.com.tw(G.K. Yang).
Available online at www.sciencedirect.com
Applied Mathematical Modelling 32 (2008) 1758–1768
facilitate the feasibility of the strategy. If the cost of holding inventory is significantly higher than the shortage cost, permitting occasional brief shortages to lower the average inventory level may be a sound business prac-tice to reduce the total cost.
There has been a great amount of research considering partial backordering in the inventory model.
Mont-gomery et al.[1]established continuous review and periodic review inventory models that considered a
mix-ture of backorders and lost sales. Kim and Park [2] considered a continuous review system with constant
lead-time where a fraction of the unfilled demand was backordered and the backorder cost was assumed to
be proportional to the length of the shortage period. Padmanabhan and Vrat [3] developed an inventory
model with a mixture of backorders and lost sales such that the backlogged demand rate was dependent upon the negative inventory level during the stock out period. Raafat et al.[4]also derived an alternative method for
finding the optimal replenishment schedule for Mak’s [5]model in which an inventory model with Weibull
distributed deterioration and backlogging is considered. Wee [6]developed an economic production lot size
model for deteriorating items with partial backordering and obtained the time intervals and cycle times that
minimize the total cost function. Padmanabhan and Vrat [7] presented inventory models for deteriorating
items with stock-dependent selling rates and derived the profit functions with and without backlogging and
complete backlogging cases. DeCroix and Arreola-Risa[8]explored the potential benefits of offering economic
incentives to backorder as a strategy for inventory management when the system involves an unreliable
sup-ply. Chung et al.[9]considered the Padmanabhan and Vrat[7]problem and developed the necessary and
suf-ficient conditions for the optimal profit per unit time function solutions. Abad[10]considered the problem of
determining the lot size for perishable goods under finite production with exponential decay, partial
backor-dering and lost sales. Zeng[11]studied the effects of using a partial backordering approach to control
inven-tory under deterministic and stochastic demands, respectively. Wu and Ouyang[12]investigated the lot size,
reorder point inventory model, including variable lead-time with partial backorders and an imperfect produc-tion process.
The model studied by this note is identical to that of Wee [6]and Abad[10]where they showed that the
inventory model is a constrained, non-linear problem with convexity characteristics. Abad used the Solver in MS/Excel to solve for the solutions. However, there are critical flaws in his analytical process, rendering the resulting solutions incorrect. We will point out the questionable proofs in his model and establish the nec-essary and sufficient conditions for the minimum solution to occur inside the interior section of the cost func-tion. A theorem to determine the criterion for the existence and uniqueness of the minimum solution is subsequently developed. We will derive simple formulated optimal solutions for each case, respectively. Numerical examples are given to illustrate all results obtained in this note.
2. Notations and assumptions
Notations and assumptions from Wee[6]and Abad[10]are adopted except for a few minor modifications
from Abad and simplification of some expressions. We outline these notations in the following for the sake of completeness and easy reference.
Notations
I(t) net stock (on hand – backorders) level at time t
p production rate for the item (units/period)
d demand rate (units/period)
h wastage coefficient, assumed to be constant (i.e., exponential decay)
hI(t) wastage rate at time t
C duration of inventory cycle when there is positive inventory
k duration of inventory cycle for which there exists stock out
b, w interim time-spans
C unit product cost
c1 setup cost
c3 backordering cost/unit/period
c4 penalty cost of lost sale including lost profit ($/unit)
B the fraction of demand backordered
(C*, k*) the pair of minimum solution of the inventory model
B(C) c3
2Bdmðk
2ðCÞ þ 2CkðCÞÞ þ c
4ð1 BÞdmC c1 ðc þch2ÞðpbðCÞ dCÞ
m pdþBdpd A(C) pþdðexpdðexpChCh1Þ1Þ
D ðc þc2 hÞðp dÞ c4ð1 BÞdm C0 1hln 1þ ðpdÞðchþc2ÞðpdÞcc4ð1BÞdmh4ð1BÞdmh h i , when D > 0 k(C) the solution ofðc þc2 hÞ ðpdÞdðexpCh1Þ
pþdðexpCh1Þ ¼ c3Bdmkþ c4ð1 BÞdm, when D > 0, for C06C<1
B(C0) c4ð1 BÞdmC0 c1 ðc þch2Þ½pbðC0Þ dC0 B(1) phðc þc2 hÞ ln p d c1þ ½ðchþc2ÞðpdÞc4ð1BÞdmh2 2c3Bdmh2 F(C, k) c1þ ðc þch2ÞðpbðCÞ dCÞ þc23Bdmk2þ c4ð1 BÞdmk C(C, k) ðC þ kÞðc þc2 hÞ ðpdÞdðexpCh1Þ pþdðexpCh1Þ F ðC; kÞ e C ifðc þc2 hÞ p hln p
d c1>0, then eC satisfies that Cð eC;0Þ ¼ 0
Assumptions
1. The planning horizon is infinite.
2. Demand occurs at a known steady constant rate d.
3. The production rate is a constant p and is strictly greater than the demand rate, i.e., p > d. 4. The goods decays at an exponential rate h.
5. When stockout occurs, demand is partially backlogged. The fraction of demand backordered B is assumed to be between zero and one, i.e., 0 6 B 6 1.
6. Production quantity in each cycle is kept constant.
7. The cost of a deteriorated item (taking into account the salvage value) is known.
8. There are no space or budget limitations. Nor are there any limitations in terms of production lot size or number of setups per year.
3. Mathematical model
The problem that Wee[6]and Abad[10]tried to solve is the minimization of the average total cost during
the inventory and shortage cycle of time-span C + k. That is
Min PðC; kÞ ¼FðC; kÞ
Cþ k ð1Þ
subject to C P 0, k P 0 and C + k, where FðC; kÞ ¼ c1þ c þ c2 h ðpbðCÞ dCÞ þc3 2 Bdmk 2þ c 4ð1 BÞdmk ð2Þ
is the total cost during the cycle of time-span C + k with bðCÞ ¼1 hln½1 þ
d pðexp
Ch 1Þ. Abad[10]constructed
Assumption 1and provedProposition 1 as follows.
Assumption 1. The set G = {(C, k)jC + k > 0, F(C, k) > 0} is not a null set.
InAppendix Aof this note, we prove thatAssumption 1of Abad[10] is unnecessary and can be removed. Proposition 1. P(C, k) is a strictly pseudoconvex function on G.
InProposition 1of Abad[10], he proved that the objective function P(C, k) is a strictly pseudoconvex func-tion on G. In Bazarra et al.[13], after showing that P(C, k) being strictly pseudoconvex, if (C1, k1) is a solution
for the first partial derivative system, then (C1, k1) will be the global minimum. Abad[10]predicted that owing
to the boundaries C P 0 and k P 0 being linear, the function P(C, k) will have a unique global minimum. In the following, we establish the necessary and sufficient conditions for which the minimum solution occurs in the interior section. We will also establish the criteria for which the minimum solution may degen-erate to infinity on the boundary when the system of first partial derivatives has no solution. In other words, Abad’s paper contains questionable results. Moreover, we derive a formulated minimum value. From the
same numerical examples of Abad [10], we demonstrate that our method improves the minimum value with
an average saving of 27.28%. 4. Improved mathematical model
Taking the first partial derivatives for P(C, k) yields oP oC ¼ 1 ðC þ k2Þ ðC þ kÞ c þ c2 h ðp dÞdðexpCh 1Þ pþ dðexpCh 1Þ F ðC; kÞ ð3Þ and oP oC ¼ 1 ðC þ kÞ2½ðC þ kÞ½c3Bdmkþ c4ð1 BÞdm F ðC; kÞ: ð4Þ
Solving the system ofoP
oC¼ 0 and oP oC¼ 0 results in cþc2 h ðp dÞdðexpCh 1Þ pþ dðexpCh 1Þ ¼ c3Bdmkþ c4ð1 BÞdm: ð5Þ
Assume that for AðCÞ ¼pþdðexpdðexpChCh1Þ1Þ for C P 0, which yields dAðCÞ
dC ¼
pdhexpCh
½pþdðexpCh1Þ2>0, A(0) = 0 and
limC!1A(C) = 1. On the other hand, assume that C0satisfies the equation
cþc2 h
ðp dÞdðexpC0h 1Þ
pþ dðexpC0h 1Þ ¼ c4ð1 BÞdm ð6Þ
and by solving Eq.(6), we have C0¼1hln 1þ ðdpÞðchþc2ÞðpdÞcc4ð1BÞdmh4ð1BÞdmh
h i
. The necessary and sufficient condition with respect to a given k for C can thus be established in the following lemma.
Lemma 1. With a given k, Eq.(5)has a solution for C if and only ifðc þc2
hÞðp dÞ > c3Bdmkþ c4ð1 BÞdm.
To simplify the expression, we assume that D¼ ðc þc2
hÞðp dÞ c4ð1 BÞdm. From Lemma 1, we know
that when D > 0, for each k satisfying D
c3Bdm>k P0, there exists a unique C such that the pair (k, C) satisfies Eq.(5). On the other hand, when D > 0, it means that for each C satisfying1 > C P C0, there exists a k, say
k(C), such that (k(C), C) satisfies Eq.(5). By Eq.(6), it yields that k(C0) = 0. Consequently, there exists a
one-to-one and onto relationship between C and k.
Next, we try to solveoP
ok¼ 0, that is, solving
½C þ k½c3Bdmkþ c4ð1 BÞdm ¼ c1þ c þ c2 h ðpbðCÞ þ dCÞ þc3 2Bdmk 2 þ c4ð1 BÞdmk: ð7Þ
Motivated by Eq.(7) and k(C) satisfying Eq.(5)for C06C<1, we define B(C) as
BðCÞ ¼c3 2 Bdmðk 2 ðCÞ þ 2CkðCÞÞ þ c4ð1 BÞdmC c1 c þ c2 h ðpbðCÞ dCÞ: ð8Þ
Using Eq.(5), we rewrite Eq.(8)as
BðCÞ ¼ cþc2 h ðpdÞdðexpCh1Þ pþdðexpCh1Þ c4ð1 BÞdm h i2 2c3Bdm þ ðp dÞdðexp Ch 1ÞC pþ dðexpCh 1Þ ðpbðCÞ dCÞ cþc2 h c1: ð9Þ
As given inAppendix B, we can show that B(C) is a strictly increasing function of C. Moreover, we know that BðC0Þ ¼ c4ð1 BÞdmC0 c1 ðc þch2Þ½pbðC0Þ dC0. We now consider limC!1B(C) and define B(1) =
limC!1B(C). Then as given inAppendix C, we can show that
Bð1Þ ¼p h cþ c2 h lnp d c1þ ½ðch þ c2Þðp dÞ c4ð1 BÞdmh 2 2c3Bdmh2 : ð10Þ
In the above discussion, we have found the criteria that ensure the existence of the solution for the first
partial derivative system of P(C, k), which leads toLemma 2as established in the following.
Lemma 2. There exists a solution for the first partial derivative system of P(C, k) if and only if D > 0, B(C0) 6 0
and B(1) > 0.
Next, if we assume that C(C, k) satisfyingoP oC¼
CðC;kÞ
ðCþkÞ2and limC!1C(C, k) = C(1, k), we have from Eq.(3)
oCðC; kÞ oC ¼ ðC þ kÞ c þ c2 h ðp dÞpdhexpCh ½p þ dðexpCh 1Þ2>0: ð11Þ
It can be easily verified that
Cð1; 0Þ ¼ c þc2 h p h ln p d c1: ð12Þ From CðC ¼ 0; kÞ ¼ c1c23Bdmk2 c4ð1 BÞdmk < 0 to Cð1; kÞ ¼ ðc þch2Þðkðp dÞ þphlnðpdÞÞ þ CðC ¼ 0; kÞ,
depending on the value of C(1, k), we have two cases:
(a) If C(1, k) > 0, the minimum value of {P(C, k) : C P 0} occurs at C(k), where C(C(k), k) = 0; and (b) if C(1, k) 6 0, the minimum value of {P(C, k) : C P 0} occurs as C approaches to infinity with its
min-imum value limC!1PðC; kÞ ¼ ðc þch2Þðp dÞ.
In the following, we consider the minimization problem of P(C, k). We have shown thatAssumption 1of
Abad[10]always holds, and also the minimization problem always has an optimal solution, since it is bounded
below by zero. If the interior points of {(C, k) : C P 0, k P 0 and C + k > 0} do not have a solution for this
minimization problem, the minimum must occur on the boundary. Hence, as shown inAppendix D, we
con-sider the following four cases for the boundary points: (a) C = 0, (b) k = 0, (c) k! 1, and (d) C ! 1. We list
the key results in the next lemma.
Lemma 3. The minimum value for P(C, k) along the boundary k = 0 satisfies that
(1) if C(1, 0) > 0, then it occurs at eC where Cð eC;0Þ ¼ 0 with oP
oC ¼ CðC;kÞ
ðCþkÞ2 and the minimum value Pð eC;0Þ ¼ ðc þc2
hÞ
ðpdÞdðexpeC h1Þ pþdðexpeC h1Þ ; and
(2) if C(1, 0) 6 0, then the minimum can be obtained as C ! 1 such that the minimum value satisfies limC!1PðC; 0Þ ¼ ðc þch2Þðp dÞ.
By combining the above results,Theorem 1can be established as follows.
Theorem 1. The minimum solution (C*, k*) of this inventory model can be divided into the following cases:
(1) If the conditions D > 0, B(C0) 6 0 and B(1) > 0 hold, then (C*, k*) = (C#, k#(C#)) where C#satisfies Eq.(8)
as B(C#) = 0 and k#(C#) satisfies Eq.(5). (2) otherwise,
(a) if C(1, 0) > 0, then ðC;kÞ ¼ ð e
C;0Þ where Cð eC;0Þ ¼ 0, and (b) if C(1, 0) 6 0, then C*=1.
Proof. It is sufficient to show that PðC#; k#ðC#ÞÞ < Pð e C;0Þ. UsingoP oC¼ 0 at (C # , k#(C#)), it can be derived that PðC#;k# ðC#ÞÞ ¼FðC #;k# ðC#ÞÞ C#þ k# ðC#Þ ¼ c þ c2 h ðp dÞdðexpC#h 1Þ pþ dðexpC#h 1Þ : ð13Þ
Since eCsatisfies that Cð eC;0Þ ¼ 0, we have Pð eC;0Þ ¼c1þ c þ c2 h ðpbð eCÞ d eCÞ e C ¼ c þ c2 h ðp dÞdðexpeCh 1Þ pþ dðexpeCh 1Þ : ð14Þ
From Eq.(9), we know that
CðC#;0Þ ¼ cþ c2 h ðpdÞdðexpC# h1Þ pþdðexpC# h1Þ c4ð1 BÞdm h i2 2c3Bdm <0: ð15Þ
Moreover, from Cð eC;0Þ ¼ 0, Eq.(15)and C(C, 0) being an increasing function of C, it follows that C#< eC.
Observing that A(C) is an increasing function of C and then by Eqs. (13) and (14), we establish that
PðC#;
k#ðC#ÞÞ < Pð e
C;0Þ. h
Next, we consider Case (2) with C(1, 0) 6 0. We know that by Appendix D, Case (d), when C! 1,
limC!1PðC; kÞ ¼ ðc þch2Þðp dÞ is the minimum value. However, owing to some operational constraints,
for examples, capacity of storage spaces and limited budget, we may only extend the inventory period to, say C1. Hence, we turn to minimizing P(C1, k) for 0 6 k <1. According to Eqs. (4), (6) and (7), solving
d dkPðC1;kÞ ¼ 0 is equivalent to solving k2þ 2C1kþ a0¼ 0; ð16Þ where a0¼c 2 3Bdm c4ð1 BÞdmC1 c1 c þ c2 h ðpbðC1Þ dC1Þ
. Therefore, the complex solutions for Eq.(16)
are k¼ C1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C2
1 a0
q
. To simplify the expression, we assume k1¼ C1þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C2
1 a0
q
, and then divide the minimization problem for P(C1, k) into the following two cases: (i) a060, and (ii) a0> 0.
For Case (i), it yields that k1P0. Moreover, as dkdPðC1;kÞ < 0, for 0 6 k < k1 and dkdPðC1;kÞ > 0, for
k1< k <1, k1is therefore the minimum solution for P(C1, k).
For Case (ii), as it implies that there is no nonnegative solution for Eq. (16) and d
dkPðC1;kÞ > 0, for
0 6 k <1, k = 0 is therefore the minimum solution for P(C1, k).
These findings are summarized byLemma 4.
Lemma 4. When Case (2) with C(1, 0) 6 0 happens and the largest possible inventory period is represented as C1, the minimum solution for P(C1, k) can be divided into the following two cases:
(i) If a060, then k1¼ C1þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C21 a0
q
is the minimum solution. (ii) If a0> 0, then k = 0 is the minimum solution.
Finally, we consider the instantaneous case where p! 1. From m ! 1, Eq.(5)can be revised as
ðch þ c2ÞðexpCh 1Þ ¼ c3Bhkþ c4ð1 BÞh: ð17Þ
Moreover, since limp!1pbðCÞ ¼dhðexp
Ch1Þ, we obtain the special case of B(C) as
BðCÞ ¼ dh 2 2c3B cþc2 h ðexpCh 1Þ c 4ð1 BÞ h i2 þ c þc2 h d hðChexp Chþ 1 expChÞ c 1: ð18Þ
On the other hand, it implies that
CðC; 0Þ ¼ c þc2 h d hðChexp Chþ 1 expChÞ c 1: ð19Þ
From D¼ ðc þc2
hÞðp dÞ c4ð1 BÞdm and Eqs. (10) and (12), it yields that when p! 1, then D > 0,
B(1) > 0, and C(1, 0) > 0. InAppendix E, we show the procedure for simplifying the expression of B(C0).
It implies that when p! 1
BðC0Þ ¼ d h cþ c2 h ½ð1 þ zÞ lnð1 þ zÞ z c1; ð20Þ where z¼c4ð1BÞh
chþc2 . Consequently, the results for the infinite production rate can be established inLemma 5.
Lemma 5. The minimum solution (C*, k*) of the inventory model for the instantaneous replenishment case can be
divided into the following two cases:
(1) If the conditions of Eq.(20), B(C0) 6 0, holds, then (C*, k*) = (C #
, k#(C#)), where C# satisfies Eq.(18)as B(C#) = 0 and k#(C#) satisfies Eq.(17).
(2) Otherwise,ðC;kÞ ¼ ð eC;0Þ, where Cð eC;0Þ ¼ 0 from Eq.(19).
5. Numerical examples
To demonstrate the advantage of our method, we consider the same numerical example as Abad[10]with
the following data: p = 750 units/week; d = 400 units/week; c = $20/units; c1= $1000/production run;
c2= $2/unit/week; h = 0.1; c3= $4/unit/week backordered; c4= $10/unit lost sale and B = 0.7. Abad [10]
examined the sensitivity analyses with respect to various relevant parameters. We quote his results inTable 1.
Using our method, for example, B = 0.7, we have D = 13,300 > 0, C0= 0.896, B(C0) =701 < 0 and
B(1) = 334,000 > 0, fromTheorem 1case (1), so the minimum solution is the solution for the first partial deriv-ative system, (C#, k#(C#)). When B = 0.3, we have D = 11,910 > 0, C0= 2.838, B(C0) = 1939 > 0,ðc þch2Þphlnpd¼
188; 582 and c1= 1000 from Case (2) ofTheorem 1, then the minimum solution is the minimum solution for the
boundary along k = 0ð eC;0Þ. We compute the same examples and list the results inTable 2.
Comparison ofTables 1 and 2indicates that our method can find the optimal solution. The range of our
saving for these five examples from 40.85% to 15.44% and its average is 27.28%.
Finally, we illustrate the extreme case where setup costs become very big, for example, c1= 4· 10 5
. We list
some possibilities to demonstrate that the optimal solution will be attained when C! 1. When C is chosen,
the value of k is derived according to Eq.(7).
Table 1
Sensitivity analysis with respect to B (reproduced from Abad[10, Table 1])
B 0.1 0.3 0.7 0.85 1
C 1.175 1.175 0.921 0.791 0.674
k 0 0 1.105 1.251 1.335
P(C, k) 1717.163 1717.163 1354.421 1167.58 996.75
Table 2
Sensitivity analysis with respect to B by our method
B 0.1 0.3 0.7 0.85 1
C 1.645 1.645 1.505 1.334 1.162
k 0 0 0.722 0.994 1.156
P(C, k) 1219.115 1219.115 1116.081 990.223 863.421
Table 3
The extreme case for B = 0.7 and c1= 4· 10 5
C C= 1 C= 10 C= 100 C= 1000 C= 5000 C= 7000
k 34.444 25.480 16.433 15.157 15.032 15.023
When C is chosen, the value of k is derived according to Lemma 4.
From Table 3, we observe that when C approaches infinity then the value of P(C, k) will decrease to its
minimum valueðc þc2
hÞðp dÞ ¼ 14; 000. When having a high setup cost, it implies inventory holding duration
should be extended as long as possible to lessen average total cost. From a practical view, the product types must be simplified to decrease the effect on setup cost for rearranging production procedures, including equip-ment preparation and adjustequip-ment. Stock must be held continuously and the machine should operate uninter-ruptedly to reduce shutdown loss (because of setup and shortage) in regular procedure, to gain minimum average total cost.
6. Conclusion
We had pointed out in this note the questionable results in Abad’s paper and provided a new and correct solution. From our theorem, the decision maker can decide where to search for the optimal solution. Review-ing the sensitivity analysis in Abad’s paper, he examined 25 examples. However, he was not aware that some-times the optimal solution approaches infinity. Therefore, our detailed analytical work patches the leak in Abad’s paper, with a variety of proposed examples explaining the background and strategy that we meet in real cases. Finally, we deduced the optimal values via complete procedures that are mathematically sound. Acknowledgements
The authors are grateful to the two anonymous referees for constructive comments that led to important improvement in this paper. All their suggestions were incorporated directly into the text. In addition, the authors wish to express appreciation to Miss Bonnie Shuan Wang for her assistance of English improvement.
Appendix A. The proof forAssumption 1of Abad[10]
Here we prove thatAssumption 1of Abad[10]is always valid. The verification is divided into the following
cases: (1) C > 0, and (2) C = 0.
Under the condition of C > 0, to show that F(C, k) is positive for any combination of constants c1, cþch2, c3
and c4, it is sufficient to prove that pb(C) dC > 0 for C > 0, when h > 0 and limh!0pbðCÞdCh exists and is
posi-tive. We will divide it into two cases: (a): h > 0 and (b): h = 0. For Case (a), since pb(C) dC > 0 is equivalent to phlnð1 þd
pexp
Ch 1Þ > dC, that is, equivalent to d
pðexp
Ch 1Þ > expdpCh 1, we assume that d
p¼ x and hC = y with 0 < x < 1 and y > 0. By fixing x with
0 < x < 1 and letting f(y) = x(expy 1) (expxy 1) for y P 0, with f(0) = 0 anddf
dy ¼ xðexp
y expxyÞ > 0,
it yields that f(y) > 0 for y > 0. Consequently, for h > 0, we derive that d pðexp
Ch 1Þ > expdpCh 1. For case (b), we compute
lim h!0 pbðCÞ dC h ¼ limh!0 pln 1þd pðexp Ch 1Þ h i dC h2 ¼ lim h!0 p 1þd pðexp ChÞ h i1 d pCexp Ch dC 2h ¼ lim h!0dC expCh 1 d pðexp Ch 1Þ 2h 1þd pðexp Ch 1Þ h i ¼ lim h!0 dCð1 dp1Þp expCh 2 1þd pðexpCh 1Þ h i þ 2hd pCexpCh ¼d 2C 2 1d p :
On the other hand, we findc2 2ðp dÞb 2 þc2 2dðC bÞ 2 with bðC; 0Þ ¼ lim h!0bðC; hÞ ¼ limh!0 ln 1þd pðexp Ch 1Þ h i h ¼ limh!0 dp1CexpCh 1þd pðexp Ch 1Þ¼ d pC: Then, we havec2 2ðp dÞð d pCÞ 2 þc2 2dð1 d pÞ 2 C2¼c2 2 d pðp dÞC 2
, which means when h = 0 FðC; kÞ ¼ c1þ c2 2 d pðp dÞC 2þc3 2 Bdmk 2þ c 4ð1 BÞdmk;
which is the total cost per cycle in the model without deterioration.
For the case of C = 0, from b(0) = 0, it yields that Fð0; kÞ ¼ c1þc23Bdmk2þ c4ð1 BÞdmk such that F(0, k)
has the desired property.
Therefore, we finish the proof forAssumption 1of Abad[10]that F(C, k) > 0 is always valid. Consequently,
Assumption 1of Abad’s[10]should thus be dropped from the discussion. Appendix B. The proof for B(C) being a strictly increasing function of C
From d dCBðCÞ ¼ c3 2Bdm½2kðCÞ dkðCÞ dC þ 2kðCÞ þ 2k dkðCÞ dC þ c4ð1 BÞdm ðc þ c2 hÞ þ ðpdÞdðexpCh1Þ
pþdðexpCh1Þ , since k(C) sat-isfies Eq.(5), we may simplifydBðCÞdC ¼ c3BdmðC þ kðCÞÞdkðCÞdC . Using Eq.(5), we obtain
dkðCÞ dC ¼ ðch þ c2Þðp dÞ c3Bdmh dAðCÞ dC ¼ ðch þ c2Þðp dÞ c3Bdmh pdhexpCh ½p þ dðexpCh 1Þ2:
Therefore, we derive that dBðCÞ dC ¼ ðC þ kðCÞÞ c þ c2 h ðp dÞpdhexpCh ½p þ dðexpCh 1Þ2>0:
We therefore have the condition that B(C) is a strictly increasing function of C. Appendix C. The value of B(‘)
We know that limC!1 exp
Ch exphbðCÞ¼ expCh 1þd pðexpCh1Þ¼ p dand limC!1kðCÞ ¼ ðcþc2hÞðpdÞc4ð1BÞdm c3Bdm . Since BðCÞ ¼c3 2Bdmk 2 ðCÞ þ C½c3BdmkðCÞ þ c4ð1 BÞdm c1 c þ c2 h ðpbðCÞ dCÞ ¼c3 2Bdmk 2 ðCÞ þ Cðp dÞdðexp Ch 1Þ pþ dðexpCh 1Þ ðpbðCÞ dCÞ cþc2 h c1; we consider lim C!1 Cðp dÞdðexpCh 1Þ pþ dðexpCh 1Þ ðpbðCÞ dCÞ ¼ limC!1 ðp dÞdðexpCh 1ÞC ðpbðCÞ dCÞ½p þ dðexpCh 1Þ pþ dðexpCh 1Þ ¼ lim C!1 ðp dÞdhexpChC ðpbðCÞ dCÞdhexpCh dhexpCh ¼ lim C!1ðp dÞC ðpbðCÞ dCÞ ¼ lim C!1 p hðhC hbðCÞÞ ¼p hln limC!1 expCh expbðCÞh ¼p hln p d:
Consequently, we derive that Bð1Þ ¼p h cþ c2 h lnp d c1þ ½ðch þ c2Þðp dÞ c4ð1 BÞdmh 2 2c3Bdmh2 :
Appendix D. For the boundary cases
For Case (a), along the boundary C = 0, recall Eq.(11). We have shown that when k is fixed, the minimum
of P(C, k) will at (a1) C(k), where C(k) satisfies the equation C(C(k), k) = 0, or (a2) C! 1, when
C(C(k), k) = 0 does not have solution. Therefore, the minimum solution will not occur along the boundary
C= 0.
For case (b), k = 0, it is the special situation for case (a) with k = 0. Then we know that (b1) ifðc þc2
hÞ p hln
p
d c1>0, there is a unique point, say eC, such that Cð eC;0Þ ¼ 0 and eCis the optimal
solu-tion for P(C, 0) and Pð eC;0Þ ¼ ðc þc2 hÞ
ðpdÞdðexpeC h1Þ pþdðexpeC h1Þ ; and
(b2) on the other hand, if ðc þc2
hÞ p hln
p
d c160, the minimum will occur when C! 1 such that
limC!1PðC; 0Þ ¼ ðc þch2Þðp dÞ.
For Case (c), k! 1, since we can rewrite P(C, k) as
PðC; kÞ ¼c3
2 Bdmðk CÞ þ c4ð1 BÞdm þ
c1þ ðc þch2ÞðpbðCÞ dCÞ c4ð1 BÞdmC þc23BdmC 2
Cþ k :
When C is fixed, if we take k! 1, the value of P(C, k) ! 1. Then, we do not need to consider the case
k! 1 for this minimum problem.
For Case (d), C! 1, we have
lim C!1PðC; kÞ ¼ limC!1 ðc þc2 hÞ½p 1 hlnð1 þ d pðexp Ch 1ÞÞ dC Cþ k ¼ lim C!1 cþ c2 h p h dp1hexpCh 1þ dp1ðexpCh 1Þ d ¼ c þc2 h ðp dÞ:
Since Pð eC;0Þ < limC!1PðC; 0Þ ¼ ðc þch2Þðp dÞ, we do not need to consider Case (d) on the boundary as
C! 1.
Appendix E. For the instantaneous replenishment case, the value of B(C0)
From C0¼1hln 1þ ðdpÞðchþc c4ð1BÞdmh 2ÞðpdÞc4ð1BÞdmh
h i
and z¼c4ð1BÞh
chþc2 , when p! 1, we derive that limp!1C0¼
1
hlnð1 þ zÞ and limp!1pbðC0Þ ¼dzh.
Hence, when p! 1, it follows that
lim p!1BðC0Þ ¼ c4ð1 BÞdC0 c1 c þ c2 h d hðexp C0h 1Þ dC 0 ¼ðch þ c2Þz h d hlnð1 þ zÞ c1 ðch þ c2Þ h dz h þ ðch þ c2Þ h d h lnð1 þ zÞ ¼dðch þ c2Þ h2 ½ð1 þ zÞ lnð1 þ zÞ z c1:
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