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A generalization of Chu and Chung’s (2004) sensitivity of the inventory model with partial backorders
European Journal of Operational Research 196 (2009) 554–562
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Production, Manufacturing and Logistics
A generalization of Chu and Chung’s (2004) sensitivity of the inventory model with partial backorders Kit-Nam Francis Leung * Department of Management Sciences, City University of Hong Kong, 83 Tat Chee Avenue, Kowloon Tong, Hong Kong
a r t i c l e
i n f o
Article history: Received 2 January 2007 Accepted 1 April 2008 Available online 8 April 2008 Keywords: Inventory model Partial backorders Sensitivity analysis Monotonic properties
a b s t r a c t In this paper, we use the elementary techniques of differential calculus to investigate the sensitivity analysis of Montgomery et al.’s [Montgomery, D.C., Bazaraa, M.S., Keswani, A.K., 1973. Inventory models with a mixture of backorders and lost sales. Naval Research Logistics Quarterly 20, 225–263] inventory model with a mixture of backorders and lost sales and generalize Chu and Chung’s [Chu, P., Chung, K.J., 2004. The sensitivity of the inventory model with partial backorders. European Journal of Operational Research 152, 289–295] sensitivity analysis. We provide three numerical examples to demonstrate our findings, and remark the interpretation of the global minimum of the average annual cost at which the complete backordering occurs. Ó 2008 Elsevier B.V. All rights reserved.
1. Introduction Sensitivity analyses of many inventory systems (such as Chung and Lin (2001) and Hwang and Hahn (2000)) are based on the computational results of a set of numerical examples, from which some conclusions can be made. However, from the standpoint of mathematics, such conclusions are questionable because they highly depend on the values of the parameters used in the numerical examples. For instance, Park (1982) had discussed the sensitivity analysis through numerical examples, but Chu and Chung (2004) mathematically showed that some of Park’s results are not always true. Hence researchers should be careful of using the conclusions of sensitivity analyses drawn from numerical examples. Montgomery et al. (1973) presented a deterministic inventory model for a situation in which a fraction b of the demand is backordered during the stockout period, while the remaining fraction (1 b) is lost. The sensitivity analysis (here equivalently monotonic properties) of Montgomery et al.’s model lead to conclusions (a)–(d) listed in Section 6. The main contribution of this paper is twofold: (1) We use the elementary techniques of differential calculus to investigate the monotonic properties of the four optimal decision varito b. ables R*(b), S*(b), Q*(b) and M*(b) and the resulting minimum objective function K b , each with respect pffiffi B 2ADh < b 6 1, where B ¼ ðDp (2) We derive the necessary and sufficient condition to assure that S*(b) strictly increases as b (0 6 1 2 and 1h l0 Þ pb h ¼ p ) increases in a simple, direct and natural manner. In addition, a heuristic is proposed to readily locate a b-value of demarcation l0 when the condition is not met.
2. Assumptions and notation Montgomery et al.’s (1973) model is developed under the following assumptions: (1) A single item is considered and the annual demand is deterministic and constant. (2) The unit cost is independent of the quantity purchased. (3) The demand, order quantity, inventory level and shortage can be treated as continuous variables.
* Tel.: +852 27888589; fax: +852 27888560. E-mail address: [email protected] 0377-2217/$ - see front matter Ó 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2008.04.003
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(4) The order quantity will not vary from one cycle to another. (5) The lead time is zero. (6) Only a fraction of the demand during the stockout period is made up in the beginning of the next order cycle. The remaining fraction is lost. (7) The planning horizon is infinite. The following symbols are used in the expression of the average annual cost. A D h p pb pl p0 pl0 = pl + p0 b R S Q ¼ M þ bS ¼ R bS M = Q bS = R S Kb(R, S)
fixed ordering cost per inventory cycle (dollars per order) demand per year (units per year) inventory holding cost per unit per year (dollars per unit per year) shortage cost per backorder per year (dollars per unit per year) fixed penalty cost per unit backorder (dollars per unit) fixed penalty cost per unit lost (dollars per unit); this is greater than pb when the cost of goodwill is included profit per unit (dollars per unit) fixed penalty cost per unit lost plus profit per unit (dollars per unit) is lost fraction of the demand is backordered during the stockout period, while the remaining fraction ð1 bÞ b total demand per cycle (units per cycle, a decision variable) total demand per cycle during the stockout period (units per cycle, a decision variable) order quantity (units per cycle, a decision variable) positive inventory level (units per cycle, a decision variable) average annual cost as a function of R and S with a parameter b (dollars per year, the objective function)
We recognize that the four optimal decision variables R*, S*, Q* and M* are dependent on the parameter b, i.e., R* R*(b), S* S*(b), Q* Q*(b) and M* M*(b). In some circumstances, we can simply write the resulting minimum objective function Kb(R*, S*) as K b . 2ADh p S , k ¼ hp, T ¼ Dp and h ¼ ppb . Note that To simplify the presentation of the subsequent mathematical expressions, we designate B ¼ ðDp Þ2 l0
l0
l0
h < 1 because pb < pl, where pl includes both fixed penalty cost per unit lost and the cost of goodwill. 3. Solution and feasibility of the EOQ model with partial backorders By considering the inventory geometry for this system as shown in Fig. 1, we can obtain the average annual cost which is the sum of ordering, holding, and stockout costs, and which includes lost profit, as bS2 þ 2Dðpb b þ pl0 bÞS 2AD þ hðR SÞ2 þ p ; 2R ¼ 1 b. where pl0 = pl + p0 and b Partially differentiating Eq. (1) with respect to R and S and then setting the two partial derivatives K b ðR; SÞ ¼
2
2
hR
bÞS ¼ 2AD þ ðh þ p
S ¼
hR Dðpb b þ pl0 bÞ ; b hþp
; þ 2Dðpb b þ pl0 bÞS
ð1Þ oK b ðR;SÞ oR
and
oK b ðR;SÞ oS
to be zero yield ð2Þ
and ð3Þ
M R
Q
ts βS
S (1–β )S
t Fig. 1. Behavior of the inventory system. t ¼ DR = cycle length (measured in a fraction of a year) and t s ¼ DS = time interval when stockout occurs (in a fraction of a year).
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which can be written as R ¼
bÞS þ Dðpb b þ pl0 bÞ ðh þ p : h
ð4Þ
Substituting Eq. (3) into (2), we obtain the explicit expression of R* given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2ADðh þ p bÞ ½Dðpb b þ pl0 bÞ R ¼ : h pb
ð5Þ
Substituting Eq. (5) into (3), we obtain the explicit expression of S* given by 9 8sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi = 2g bÞ ½Dðpb b þ pl0 bÞ 1 < hf2ADðh þ p S ¼ Dðpb b þ pl0 bÞ : ; b : b hþp p
ð6Þ
Substituting Eq. (4) into (2) yields the minimum condition of Eq. (1), namely 2 þ h ½ pbS þ Dðpb b þ pl0 bÞ pbS2 ¼ 2ADh:
ð7Þ
We notice from Eq. (3) that R* P S*, and hence the EOQ with partial backorders and the maximum positive inventory level are ; Q ¼ R bS
ð8Þ
M ¼ R S :
ð9Þ
and Substituting Eq. (2) into (1), we obtain the expression of K b given by
K b ¼ hM :
ð10Þ *
*
Eqs. (5) and (6) reveal that R > 0 and S > 0 provided 2; bÞ > ½Dðpb b þ pl0 bÞ 2ADðh þ p or taking the square root becomes qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ADðh þ p bÞ > Dðpb b þ pl0 bÞ;
ð11Þ
and sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2g bÞ ½Dðpb b þ pl0 bÞ hf2ADðh þ p > Dðpb b þ pl0 bÞ; b p which can easily be simplified to pffiffiffiffiffiffiffiffiffiffiffiffi 2ADh > Dðpb b þ pl0 bÞ:
ð12Þ *
*
Clearly, condition (12) is satisfied and so is condition (11). This implies that real R exists whenever S > 0. pffiffiffiffiffiffiffiffiffiffiffiffi Condition (12) together with the condition 0 6 B 6 1 (i.e., 0 6 2ADh 6 Dpl0 Þ can be combined and written as 06
pffiffiffi 1 B < b 6 1; 1h
ð13Þ
pb 2ADh where B ¼ ðDp 2 and h ¼ p . l0 l0 Þ p ffiffiffiffiffiffiffiffiffiffiffiffi is the sum of annual penalty cost and Note that 2ADh is the minimum average annual cost in the classical EOQ model and Dðpb b þ pl0 bÞ pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi of demand is lost. Based on the condition that 0 6 2ADh 6 Dpl0 , on the relative magnitudes of 2ADh and annual lost profit if a fraction b and on whether p >p0ffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0, there are three analytic cases: pffiffi or p Dðpb b þ pl0 bÞ, P 0 (i.e., when 0 6 1 B < b 6 1 or condition (13) is satisfied) and p > 0, R* and S* Analytic case (i): When Dpl0 P 2ADh > Dðpb b þ pl0 bÞ 1h are determined using Eqs. (5) and p (3). pffiffi ffiffiffiffiffiffiffiffiffiffiffiffi P 0 (i.e., when 0 6 1 B < b 6 1 or condition (13) is satisfied) and p ¼ 0, we have Analytic case (ii): When Dpl0 P 2ADh > Dðpb b þ pl0 bÞ 1h R* = 1 from Eq. (5) and hence S* = 1 from Eq. (3), which are meaningless. What is going on? In this case, we have using Eqs. (3), (9) and (10), i.e., the optimal policy is to partially backorder everything and the resulting minK b ¼ hðR S Þ ¼ Dðpb b þ pl0 bÞ in which we note that pl0 is regarded as an opportunity cost relating to negative inventory (i.e., imum average annual cost is Dðpb b þ pl0 bÞ, shortage), similar to regarding h as an opportunity cost relating to positive inventory, and pl0 equals the sum of the cost of goodwill and unit profit (not including thepfixed (2006, p. 63). pffiffi ffiffiffiffiffiffiffiffiffiffiffiffi penalty cost). This is a modified result in Sphicas 6 Dpl0 (i.e., when 0 6 b 6 1 BÞ (no matter whether p > 0 or p ¼ 0Þ, we have S* = 0, i.e., Analytic case (iii): When 2ADh 6 Dðpb b þ pl0 bÞ 1h the optimal solution is q toffiffiffiffiffiffi allow no backorders or lost sales to occur. As a result, Eq. (1) becomes EOQ model, and the optimal pffiffiffiffiffiffiffiffiffiffiffiffi the classical and the resulting minimum average annual cost is KðQ Þ ¼ 2ADh ¼ hQ . order quantity is Q ¼ 2AD h Two remarks for this section are as follows:
(1) If condition (12) is satisfied, then the EQQ model given by Eq. (1) is feasible. If condition (11) is met, then the cost function given by the same equation is convex and hence the global minimum is guaranteed. Condition (12) is satisfied and so is condition (11) and thus we can conclude that model feasibility implies function convexity for the EOQ model under study. We have confirmed this conclusion in the Appendix.
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(2) To solve or analyze the EOQ model given by Eq. (1), we can use the complete squares method proposed in Leung (2008a,b). It is an algebraic method which requires no working knowledge of multi-variable calculus and is usually taught when introducing the notion of a general quadratic equation in Form 2 at high school. Hence the method can be understood by ordinary readers. On the other hand, to investigate the monotonic property of an optimal expression in Section 4, we must use differential calculus.
4. Sensitivity analysis when 0 < B < 1 and
pffiffi 1 B 1h
pffiffi B < b <1 (i.e., when 0 < 1 < b <1 or condition (13) is satisfied) 1h
pffiffi B
Differentiating Eq. (5) with respect to b, we have 2 2D2 bðpb b þ pl0 bÞðp b pl0 Þ 2ADh dR ½Dðpb b þ pl0 bÞ ¼ : 2 db 2h pb R The expression for R* strictly increases with b iff dR >0 db
2 2D2 bðpb b þ pl0 bÞðp b pl0 Þ 2ADh > 0 iff ½Dðpb b þ pl0 bÞ
2 2bðhb þ bÞðh 1Þ B > 0 iff ðhb þ bÞ pffiffiffiffiffiffiffiffiffiffiffiffi 1B iff b < : 1h pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi It is easy to show that 1 B 6 1 B when 0 6 B 6 1. As a result, we need consider two cases: (1) If (2) If
pffiffiffiffiffiffiffi 1B 1h pffiffiffiffiffiffiffi 1B 1h
ð14Þ
pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi pffiffi B 1B 1B increases, but R* strictly decreases as b 1h < 1, then R* strictly increases as b 0 6 1 < b 6 1h 6 b 6 1 increases. 1h pffiffi B P 1, then R* strictly increases as b 0 6 1 < b 6 1 increases. 1h
* In particular, when hp=ffiffiffiffiffiffiffiffiffiffiffi 0 (i.e., ffi pb = 0 in Park’s (1982) model), then R strictly increases as b (0 6 1 strictly decreases as b ð 1 B 6 b 6 1Þ increases.
pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi B < b 6 1 BÞ increases, but R*
pffiffi B
To facilitate the investigation, we are better off writing Eq. (7) as þ hbÞ2 þ kbT 2 ¼ B; ðbT þ b
ð15Þ 2
which is obtained by first dividing by the factor (Dpl0) throughout Eq. (7) and next using the designations of B ¼ h ¼ ppb . l0 Implicitly differentiating Eq. (15) with respect to b, we have
2ADh , ðDpl0 Þ2
þ hbÞ kT 2 dT 2ð1 h T ÞðbT þ b ¼ ; þ hb db 2b½ðk þ bÞT þ b
k¼
h , p
T ¼
p S Dpl0
and
ð16Þ
and dT >0 db
þ hbÞ kT 2 > 0 iff 2ð1 h T ÞðbT þ b iff b <
2ð1 h T Þ kT 2
2ð1 h T Þ
2
ð17Þ
:
pffiffi B We consider such a scenario that T* strictly increases as b 0 6 1 < b 6 1 increases. This scenario happens iff 1h 1<
2½1 h T ð1Þ k½T ð1Þ2
2½1 h T ð1Þ2 k iff 1 þ ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ < 0; 2
ð18Þ
where
T ð1Þ ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Bð1 þ kÞ kh2 h 1þk
ð19Þ
;
S ð1Þ 2ADh which can be obtained by first setting b = 1 in Eq. (6) and next using T ð1Þ ¼ pDp , B ¼ ðDp , k ¼ hp and h ¼ ppb . Þ2 l0
l0
l0
2 In particular, when h = 0 (i.e., pb = 0 in Park’s (1982) model), Eqs. (18)q and ffiffiffiffiffiffi (19) become T ð1Þ < 2þk, where T ð1Þ ¼ dT ðbÞ B 2 the necessary and sufficient condition for db > 0 when h = 0 becomes 1þk < 2þk, which can be simply written as
B<
1þk 1 þ k þ ð2k Þ2
:
qffiffiffiffiffiffi B . Consequently, 1þk
ð20Þ
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K.-N.F. Leung / European Journal of Operational Research 196 (2009) 554–562 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 1Bð1þ 1BÞ , B
2 1þk In the Appendix, we show that B < 1þkþð k Þ2 is the same as k < 2
which was ingeniously deduced by Chu and Chung (2004). How-
ever, condition (20) is deduced (or derived) in a simple, direct and natural manner, as well as appearing concisely. If condition (18) (or (20) in particular) is not satisfied, we cannot explicitly obtain a b-value of demarcation because the right-hand side 2½1hT ðb0 Þk½T ðb0 Þ2 * b0 ¼ 0. Then T* strictly increases of inequality (17) involves T (b), i.e., to deduce an explicit expression of b0 from solving pffiffi 2½1hT ðb0 Þ2 * 1 B as b 0 6 1h < b 6 b0 increases, but T strictly decreases as b (b0 6 b 6 1) increases. In the Appendix, we also show that there exists a robust initial value, namely b1 ¼
2½1 h T ð1Þ k½T ð1Þ2 2½1 h T ð1Þ2
ð21Þ
;
whereby we can readily determine a b-value that is extremely close to the exact value of b0, provided that
kT ðb1 Þ > 1
T ðb1 Þ : 1h
ð22Þ
To facilitate the determination, a three-step heuristic procedure is proposed in the Appendix. This heuristic is illustrated by Examples 2 and 3 in Section 5. pffiffi B 4.3. The monotonic property of Q* with respect to b when 0 6 1
Substituting Eq. (4) into (8) and using the symbol designations, we obtain the expression of Q* given by Q ¼
Dpl0 þ hb: ½ð1 þ kÞbT þ b h
ð23Þ
Differentiating Eq. (23) with respect to b, we have dQ Dpl0 dT ¼ ð1 þ kÞb þ ð1 þ kÞT 1 þ h ; db h db
ð24Þ
and then substituting Eq. (16) into (24), we obtain dQ >0 db
þ 2hð1 hÞb > 0 iff ð1 þ kÞT 2 þ 2ð1 T Þð1 hÞb
2 2 ð1 hÞb ½ð1 hÞb þ 2hð1 hÞb > 0 þ 2ð1 hÞb iff ð1 þ kÞ T 1þk 1þk 2 1 þ 2k þ b þ hb ð1 hÞb þ 2hð1 hÞb > 0: iff ð1 þ kÞ T þ ð1 hÞb 1þk 1þk pffiffi B The last inequality is always true because h < 1. Thus Q* strictly increases as b 0 6 1 < b 6 1 increases. 1h
ð25Þ
pffiffi B 4.4. The monotonic properties of M* and K b with respect to b when 0 6 1
Substituting Eq. (4) into (9) and using the symbol designations, we obtain the expression of M* given by M ¼
Dpl0 þ hbÞ: ðbT þ b h
ð26Þ
Differentiating Eq. (26) with respect to b, we have dM Dpl0 dT ¼ b þ T 1 þ h ; db h db
ð27Þ
and then substituting Eq. (16) into (27), we obtain dM <0 db
iff T < 2ð1 hÞ:
The last inequality is always true because pffiffiffi 1 B < b 6 1: T < 1 h for 0 6 1h
ð28Þ
ð29Þ pffiffi B ð1 ; 1 1h
The validity of result (29) is proved by contradiction as follows: Assume that there exists a b-value denoted by b lying in such that pffiffi B ; 1 and the assumption of T*(b) P 1 h, the left-hand side of Eq. (15) indicates that T*(b) P 1 h. With T* > 0 for any b-value in ð1 1h þ hb 2 þ kb ½T ðb Þ2 > ½b ð1 hÞ þ b þ hb 2 ¼ 1; ½b T ðb Þ þ b
pffiffi B and is greater than B. Because 0 6 B 6 1, the assumption contradicts equality (15) which is valid for any b-value in ð1 ; 1. In other words, 1h pffiffi * 1 B we have shown that T < 1 h for any b-value in ð 1h ; 1. pffiffi B Thus M* strictly decreases as b 0 6 1 < b 6 1 increases, and so does K b because K b ¼ hM , using Eq. (10). Moreover, we have 1h pffiffi pffiffiffiffiffiffiffi dM ðbÞ B 1B ¼ dRdbðbÞ dSdbðbÞ < 0 or dRdbðbÞ < dSdbðbÞ. This implies that dSdbðbÞ > 0 while dRdbðbÞ > 0 for 0 6 1 < b 6 1h . That is, db 1h p ffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi ffi dR ðbÞ dS ðbÞ 1 B 1B0) > 0 for 0 6 1h 1h db db
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5. Numerical examples Example 1. Suppose that an item has the characteristics as those in Montgomery et al. (1973, p. 259): A = $50 per order, D = 250 units per ¼ $0:1 per unit per year, pb = pl = $0.5 per unit, p0 = $2 per unit and pl0 = $2.5 per unit. year, h = $2 per unit per year, p pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 pffiffiffiffiffiffiffi pffiffi Bð1þkÞkh h B 1B First we compute B ¼ 2ADh2 ¼ 0:128, k ¼ hp ¼ 20, h ¼ ppl0b ¼ 0:2, 1 ¼ 0:0559. 1h ¼ 0:8028, 1h ¼ 1:1673 and T ð1Þ ¼ 1þk ðDpl0 Þ Next we draw the following conclusions: (1) (2) (3) (4)
pffiffiffiffiffiffiffi 1B ¼ 1:1673 > 1,pffiffiffiffiffiffi R*(b) Because 1h ffi strictly increases as b (0.8028 < b 6 1) increases. 1B ¼ 1:1673 > 1 indicates that S*(b) strictly increases as b (0.8028 < b 6 1) increases. Result (30) together with 1h * Q (b) strictly increases as b (0.8028 < b 6 1) increases. M*(b) or K b strictly decreases as b (0.8028 < b 6 1) increases.
Example 2. Further suppose that the values of h and pb become 4 and 0.05 (rather than 2 and 0.5), respectively, and the other parameter values remain the same. pffiffiffiffiffiffiffi pffiffi * B 1B First we obtain B = 0.256, k = 40, h = 0.02, 1 1h ¼ 0:5041, 1h ¼ 0:8802 and T (1) = 0.0785. Next we draw the following conclusions (see Table 1): pffiffiffiffiffiffiffi 1B ¼ 0:8802 < 1, R*(b) strictly increases as b (0.5041 < b 6 0.8802) increases, but R*(b) strictly decreases as b (1) Because 1h (0.8802 6 b 6 1) increases (see column 2 in Table 1). (2) Because ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ ¼ 0:0344 > 0, i.e., because condition (18) is not satisfied, S*(b) is not an increasing function of b for 0.5041 < b 6 1. To determine b0, first we compute b1 = 0.9576 using Eq. (21); second we know that b0 < b1 because condition (22) holds; third letting b = 0.9569, 0.9570 or 0.9571, we have the associated values of S*(b) (see column 3 in Table 1). Thus S* strictly increases as b (0.5041 < b 6 0.9570) increases, but S* strictly decreases as b (0.9570 6 b 6 1) increases. (3) Q*(b) strictly increases as b (0.5041 < b 6 1) increases. (4) M*(b) or K b strictly decreases as b (0.5041 < b 6 1) increases.
Example 3. Suppose that another item has the characteristics as those in Park (1982, p. 1316) as follows: A = $50 per order, D = 200 units ¼ $0:3 (instead of $1) per unit per year. = $2 per unit, with theffi exception that p per year, h = $3 per unit per year, pb = $0 per unit and pp l0ffiffiffi pffiffiffiffiffiffiffiffiffiffiffi First we compute B ¼ 2ADh2 ¼ 0:375, k ¼ hp ¼ 10, 1 B ¼ 0:3876 and 1 B ¼ 0:7906. ðDpl0 Þ Next we draw the following conclusions (see Table 2): (1) R*(b) strictly increases as b (0.3876 < b 6 0.7906) increases, but R*(b) strictly decreases as b (0.7906 6 b 6 1) increases (see column 2 in Table 2). * 1þk (2) Because 0:375 ¼ B P 1þkþð k Þ2 ¼ 0:3056 i.e., because condition (20) is not satisfied, S (b) is not an increasing function of b for 2
0.3876 < b 6 1. To determine b0, first we compute b1 = 0.9701 using Eq. (21); second we know that b0 < b1 because condition (22) holds; third letting b = 0.9697, 0.9698 or 0.9699, we have the associated values of S*(b) (see column 3 in Table 2). Thus S* strictly increases as b (0.5041 < b 6 0.9698) increases, but S* strictly decreases as b (0.9698 6 b 6 1) increases. (3) Q*(b) strictly increases as b (0.3876 < b 6 1) increases. (4) M*(b) or K b strictly decreases as b (0.3876 < b 6 1) increases.
Table 1 ¼ 0:1, pb = 0.05, pl0 = 2.5) Sensitivity analysis for ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ P 0 in Example 2 (A = 50, D = 250, h = 4, p b
R*(b) (units)
S*(b) (units)
T*(b) < 0.98
Q*(b) (units)
M*(b) (units)
K b ($)
0 to 0.5041 0.51 0.60 0.70 0.80 0.8801 0.8802 0.8803 0.90 0.95 0.9569 b0 = 0.9570 0.9571 0.9576 0.96 0.97 0.98 0.99 1.00b
79.06 131.74 382.94 475.23 511.66 518.9697182 518.9697197a 518.9697007 518.57 514.31 513.3781808 513.3641085 513.3500197 513.2793275 512.93 511.40 509.70 507.84 505.83
0 52.91 313.86 418.84 468.54 486.7747688 486.7885624 486.8023358 489.13 491.85 491.8863423 491.8863525a 491.8863466 491.8860749 491.88 491.76 491.47 491.03 490.44
0 0.0084656 0.0502176 0.0670144 0.0749664 0.077883963 0.077886169 0.077888373 0.0782608 0.0786960 0.078701814 0.078701816a 0.078701815 0.078701772 0.0787008 0.0786816 0.0786352 0.0785648 0.0784704
79.06 105.81 257.40 349.58 417.95 460.61 460.65 460.70 469.66 489.42 492.18 492.21 492.25 492.42 493.25 496.65 499.87 502.93 505.83
79.06 78.83 69.08 56.39 43.12 32.19 32.18 32.16 29.44 22.46 21.49 21.48 21.46 21.39 21.05 19.64 18.23 16.81 15.39b
316.24 315.32 276.32 225.56 172.48 128.76 128.72 128.64 117.76 89.84 85.96 85.92 85.84 85.56 84.20 78.56 72.92 67.24 61.56b
b = fraction of the demand backordered during the stockout period, R*(b) = optimal total demand per cycle, S*(b) = maximum total demand during the stockout period per pffiffi S ðbÞ * 1 B , Q*(b) = optimal cycle, T ðbÞ ¼ pDp qffiffiffiffiffiffiffi order quantity, M (b) = maximum positive inventory level, and K b = minimum average annual cost. When b is in ½0; 1h , we use the l0 ¼ R ¼ M . classical EOQ formula: Q ¼ 2AD h a Global maximum occurs. b Global minimum occurs.
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Table 2 1þk ¼ 0:3, pb = 0, pl0 = 2) Sensitivity analysis for B P 1þkþð k Þ2 in Example 3 (A = 50, D = 200, h = 3, p 2
b
R*(b) (units)
S*(b) (units)
T*(b) < 1
Q*(b) (units)
M*(b) (units)
K b ($)
0 to 0.3876 0.39 0.40 0.50 0.60 0.70 0.7905 0.7906 0.7907 0.80 0.90 0.95 0.96 0.9697 b0 = 0.9698 0.9699 0.9700 0.9701 0.98 0.99 1.00b
81.65 89.38 115.47 226.08 265.27 281.15 284.8348180 284.8348196a 284.8348132 284.80 280.65 276.36 275.34 274.3139367 274.3030681 274.2921947 274.2813164 274.2704333 273.17 272.01 270.80
0 7.74 34.11 151.82 199.94 225.37 238.0811683 238.0913198 238.1014638 239.01 245.24 246.30 246.36 246.3822499 246.3822508a 246.3822472 246.3822392 246.3822268 246.36 246.29 246.18
0 0.0058050 0.0255825 0.1138650 0.1499550 0.1690275 0.178560876 0.178568489 0.178576097 0.1792575 0.1839300 0.1847250 0.1847700 0.184786687 0.184786688a 0.184786685 0.184786679 0.184786670 0.1847700 0.1847175 0.1846350
81.65 84.66 95.00 150.17 185.29 213.54 234.96 234.98 235.00 237.00 256.13 264.05 265.49 266.85 266.86 266.88 266.89 266.90 268.24 269.55 270.80
81.65 81.64 81.36 74.26 65.33 55.78 46.75 46.74 46.73 45.79 35.41 30.06 28.98 27.93 27.92 27.91 27.90 27.89 26.81 25.72 24.62b
244.95 244.92 244.08 222.78 195.99 167.34 140.25 140.22 140.19 137.37 106.23 90.18 86.94 83.79 83.76 83.73 83.70 83.67 80.43 77.16 73.86b
b = fraction of the demand backordered during the stockout period, R*(b) = optimal total demand per cycle, S*(b) = maximum total demand during the stockout period per pffiffi S ðbÞ * 1 B , Q*(b) = optimal cycle, T ðbÞ ¼ pDp qffiffiffiffiffiffiffi order quantity, M (b) = maximum positive inventory level, and K b =minimum average annual cost. When b is in ½0; 1h , we use the l0 ¼ R ¼ M . classical EOQ formula: Q ¼ 2AD h a Global maximum occurs. b Global minimum occurs.
S ðbÞ Using Eqs. (5), (3) and (8)–(10) and the designation of T ðbÞ ¼ pDp , we obtain the results of Examples 2 and 3 shown in Tables 1 and 2, l0 respectively. We remark that although the global minimum average annual cost occurs when b = 1 (see the last columns in Tables 1 and 2), we must use the partial backordering model for a situation where b is actually equal to 0.7, say. On the one hand, if we assume that all demand during the stockout period can be backordered even though this is not strictly true, then substituting the optimal solution obtained by setting b = 1 in Eqs. (5) and (3) (as deduced from the so-called ‘‘EOQ model with complete backorders”) into Eq. (1) yields a non-minimum average annual cost. In Examples 2 and 3, we have (505.83, 490.44, $232.57) and (270.80, 246.18, $172.87). (Note that if b = 1 were true, then the resulting minimum average annual cost would be $61.56 and $73.86.) On the other hand, substituting the optimal solution obtained by setting b = 0.7 in Eqs. (5) and (3) (as deduced from the so-called ‘‘EOQ model with partial backorders”) into Eq. (1) or equivalently Eqs. (9) and (10) yields the minimum average annual cost. In Examples 2 and 3, we have (475.23, 418.84, $225.56) and (281.15, 225.37, $167.34). Thus failure to use the appropriate model has cost management $232.57 $225.56 = $7.01 and $172.87 $167.34 = $5.53 per year for these two items. The effects of this in a multi-item inventory may be quite substantial.
6. Conclusions Summarizing the monotonic properties when 0 6 B 6 1 and
pffiffi 1 B 1h
pffiffi B < b 6 1 (i.e., when 0 6 1 < b 6 1), we have 1h
pffiffiffiffiffiffiffi pffiffi pffiffiffiffiffiffiffi B 1B 1B < 1, then R*(b) strictly increases as b 0 6 1 < b 6 1h 6 b 6 1 increases. increases, but R*(b) strictly decreases as b 1h 1h pffiffiffiffiffiffiffi pffiffi 1B B However, if 1h P 1, then both R*(b) and S*(b) strictly increase as b 0 6 1 6 b 6 1 increases. In particular, when h = 0 (i.e., pb = 0 1h pffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffi in Park’s (1982) model), R*(b) strictly increases as b 0 6 1 B < b 6 1 B increases, but R*(b) strictly decreases as pffiffiffiffiffiffiffiffiffiffiffiffi bð 1 B 6 b 6 1Þ increases. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ffi Bð1þkÞkh h * 1þk (b) On the one hand, if ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ < 0 where T ð1Þ ¼ (or B < 1þkþð kÞ2 when h = 0), then S (b) 1þk 2 pffiffi 2 1 B k strictly increases as b 0 6 1h < b 6 1 increases. On the other hand, if ð1 þ 2Þ½T ð1Þ ð1 2hÞT ð1Þ hð1 hÞ P 0, then S*(b) pffiffi B strictly increases as b 0 6 1 < b 6 b0 increases, but S*(b) strictly decreases as b (b0 6 b 6 1) increases, where b0 satisfies 1h (a) If
pffiffiffiffiffiffiffi 1B 1h
2½1hT ðb0 Þk½T ðb0 Þ2 2½1hT ðb0 Þ2 *
b0 ¼ 0.
pffiffi B (c) Q (b) strictly increases as b 0 6 1 < b 6 1 increases. 1h pffiffi B < b 6 1 increases. (d) Both M*(b) and K b strictly decrease as b 0 6 1 1h In addition, we have (1) (2) (3) (4)
pffiffi B < b 6 1 is true, T*(b) < 1 h. When 0 6 1 1h pffiffi 1 B When 0 6 1h 1 (i.e., 2ADh > Dpl0 Þ and h = 0 are true, the conclusions of the monotonic properties of the optimal expressions and the optimal inventory polices can be found in Yang (2007).
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In the Appendix, we tabulate some symbols used in this paper and compare them to those used in Montgomery et al. (1973), Park (1982) and Chu and Chung (2004) so that readers can easily reconcile the materials. Finally, we would like to indicate a minor mistake pffiffiffi at the start of Section 3 in Chu and Chung (2004, p. 291). This should read that Kb (R, S) is convex with respect to R and S for all 0 6 1 B < b 6 1 (not 0 6 b 6 1). This can be confirmed in the Appendix in Park (1982, p. 1317). Acknowledgements The author is grateful to the two anonymous referees for constructive comments that led to essential improvement in this paper. Their appropriate suggestions were included in the text. In addition, the author would like to thank Referee 1 for arousing his awareness of the paper written by Yang (2007) and to thank Referee 2 for insisting him on rewriting the paper according to Montgomery et al.’s (1973) model. Appendix A1. Partially differentiating the function K Kb(R, S) with respect to the two decision variables R and S twice and computing the determinant of the Hessian matrix, we have o2 K 2
oR
¼
bÞS2 þ 2Dðpb b þ pl0 bÞS 2AD þ ðh þ p 3
R
;
o2 K oS
2
¼
b hþp ; R
bÞS þ Dðpb b þ pl0 bÞ o2 K ½ðh þ p ; ¼ oRoS R2
and o2 K o2 K j H j¼ 2 2 oR oS
!2 2 bÞ ½Dðpb b þ pl0 bÞ o2 K 2ADðh þ p ¼ ; oRoS R4 2
2
;S Þ ;S Þ which is greater than zero provided condition (11) is satisfied. Clearly, o KðR > 0 and o KðR > 0; thus K has a global minimum point at oR2 oS2 (R*, S*). A2. The equivalence of the two necessary and sufficient conditions is shown as follows: 2
B<
1þk Bk 2 < 1 þ k () 4B þ 4Bk þ Bk < 4 þ 4k 2 () B þ Bk þ 4 1 þ k þ 2k
k<
pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi 2 1 Bð1 þ 1 BÞ () Bk < 2 1 B þ 2ð1 BÞ () Bðk þ 2Þ 2 < 2 1 B B () B2 ðk þ 2Þ2 4Bðk þ 2Þ þ 4 < 4ð1 BÞ () Bðk þ 2Þ2 4k 4 < 0
and
2
2
() Bðk þ 4k þ 4Þ 4k 4 < 0 () 4B þ 4Bk þ Bk < 4 þ 4k: A3. Denote the expressions on the left and right-hand sides of inequality (17) by f(b) and g(b), i.e., let f ðbÞ ¼ b; and gðbÞ ¼
2ð1 h T Þ kT 2ð1 h T Þ2
2
:
g(β ) f (β ) =β
1
(1, g(1))
β1
0
β 1− B 1−θ
β0 β1
1
1þk Fig. 2. Behavior of the function g(b) when ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ P 0 (or B P 1þkþð k Þ2 when h = 0). 2
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Differentiating g(b) with respect to b yields
dT dgðbÞ ð1 h T Þ½kð1 hÞT ð1 h T Þ db : ¼ 4 db ð1 h T Þ
Since
dT db
¼ 0 at b = b0,
kT ðb1 Þ > 1
dT db
< 0 for b0 < b 6 1, and result (29), we have
dgðbÞ db
¼ 0 at b = b0 and
dgðbÞ db
> 0 for b0 < b 6 1 provided that
ðb1 Þ
T : 1h
ð22Þ
Define b1 ¼ f ðb1 Þ ¼ gð1Þ ¼
2½1 h T ð1Þ k½T ð1Þ2 2½1 h T ð1Þ2
ð21Þ
:
If condition (22) is satisfied at b = b1, then we have dgðbÞ > 0 at b = b1 and deduce that b0 < b1. This means that we can regard b1 as an initial db value to determine b0 efficiently. Fig. 2 shows the behavior of the function g(b) when ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ P 0 (or 1þk B P 1þkþð kÞ2 when h = 0). 2
We propose a heuristic procedure with three steps to determine the value of b0: (1) Compute b1 using Eq. (21). ðb1 Þ holds. If so, then we know that b0 < b1. (2) Check whether the inequality kT ðb1 Þ > 1 T1h (3) Regard b1 as an initial value and use trial and error to determine the value of b0. A4. Some symbols used in this paper, Montgomery et al. (1973), Park (1982) and Chu and Chung (2004) are tabulated as follows: Symbol description
Leung (this paper)
Montgomery et al. (1973)
Park (1982)
Chu and Chung (2004)
Ordering cost per cycle Demand per year Holding cost per unit per year Backorder cost per unit per year Backorder cost per unit
A D h
A D I C = IC
A d h
A d h
p
p
p
p
pb
p
Assign zero
Lost-sale cost per unit
pl
p
Profit per unit
p0
p0
Lost-sale cost plus profit per unit Fraction of backordered demand Fraction of lost demand Demand per cycle Shortage per cycle Order quantity Positive inventory level
pl0 = pl + p0
p + p0
Assign zero Not available Not available p
b
b
b
b
¼1b b R S Q M
1b U S Q V
1b R S Q Not available
Average annual cost Designations
Kb(R, S) S ðbÞ ; h ¼ ppl0b . B ¼ 2ADh2 ; k ¼ hp ; T ðbÞ ¼ pDp l0
K(Q, S) or K(U, V) Not applicable
1b R S Q Not available K(R, S;b) Not applicable
ðDpl0 Þ
Not available Not available p
Kb(R, S) B ¼ 2Adh2 ; k ¼ hp ; T ðbÞ ¼ pSdpðbÞ ; h ¼ 0. ðdpÞ
References Chu, P., Chung, K.J., 2004. The sensitivity of the inventory model with partial backorders. European Journal of Operational Research 152, 289–295. Chung, K.J., Lin, C.N., 2001. Optimal inventory replenishment models for deteriorating items taking account of time discounting. Computers and Operations Research 28, 67–83. Hwang, H., Hahn, K.H., 2000. An optimal procurement policy for items with an inventory level-dependent demand rate and fixed lifetime. European Journal of Operational Research 127, 537–545. Leung, K.N.F., 2008a. Technical note: A use of the complete squares method to solve and analyze a quadratic objective function with two decision variables exemplified via a deterministic inventory model with a mixture of backorders and lost sales. International Journal of Production Economics 113, 275–281. Leung, K.N.F., 2008b. Using the complete squares method to analyze a lot size model when the quantity backordered and the quantity received are both uncertain. European Journal of Operational Research 187, 19–30. Montgomery, D.C., Bazaraa, M.S., Keswani, A.K., 1973. Inventory models with a mixture of backorders and lost sales. Naval Research Logistics Quarterly 20, 255–263. Park, K.S., 1982. Inventory model with partial backorders. International Journal of Systems Science 13, 1313–1317. Sphicas, G.P., 2006. EOQ and EPQ with linear and fixed backorder costs: Two cases identified and models analyzed without calculus. International Journal of Production Economics 100, 59–64. Yang, G.K., 2007. Note on sensitivity analysis of inventory model with partial backorders. European Journal of Operational Research 177, 865–871.
Contents lists available at ScienceDirect
European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor
Production, Manufacturing and Logistics
A generalization of Chu and Chung’s (2004) sensitivity of the inventory model with partial backorders Kit-Nam Francis Leung * Department of Management Sciences, City University of Hong Kong, 83 Tat Chee Avenue, Kowloon Tong, Hong Kong
a r t i c l e
i n f o
Article history: Received 2 January 2007 Accepted 1 April 2008 Available online 8 April 2008 Keywords: Inventory model Partial backorders Sensitivity analysis Monotonic properties
a b s t r a c t In this paper, we use the elementary techniques of differential calculus to investigate the sensitivity analysis of Montgomery et al.’s [Montgomery, D.C., Bazaraa, M.S., Keswani, A.K., 1973. Inventory models with a mixture of backorders and lost sales. Naval Research Logistics Quarterly 20, 225–263] inventory model with a mixture of backorders and lost sales and generalize Chu and Chung’s [Chu, P., Chung, K.J., 2004. The sensitivity of the inventory model with partial backorders. European Journal of Operational Research 152, 289–295] sensitivity analysis. We provide three numerical examples to demonstrate our findings, and remark the interpretation of the global minimum of the average annual cost at which the complete backordering occurs. Ó 2008 Elsevier B.V. All rights reserved.
1. Introduction Sensitivity analyses of many inventory systems (such as Chung and Lin (2001) and Hwang and Hahn (2000)) are based on the computational results of a set of numerical examples, from which some conclusions can be made. However, from the standpoint of mathematics, such conclusions are questionable because they highly depend on the values of the parameters used in the numerical examples. For instance, Park (1982) had discussed the sensitivity analysis through numerical examples, but Chu and Chung (2004) mathematically showed that some of Park’s results are not always true. Hence researchers should be careful of using the conclusions of sensitivity analyses drawn from numerical examples. Montgomery et al. (1973) presented a deterministic inventory model for a situation in which a fraction b of the demand is backordered during the stockout period, while the remaining fraction (1 b) is lost. The sensitivity analysis (here equivalently monotonic properties) of Montgomery et al.’s model lead to conclusions (a)–(d) listed in Section 6. The main contribution of this paper is twofold: (1) We use the elementary techniques of differential calculus to investigate the monotonic properties of the four optimal decision varito b. ables R*(b), S*(b), Q*(b) and M*(b) and the resulting minimum objective function K b , each with respect pffiffi B 2ADh < b 6 1, where B ¼ ðDp (2) We derive the necessary and sufficient condition to assure that S*(b) strictly increases as b (0 6 1 2 and 1h l0 Þ pb h ¼ p ) increases in a simple, direct and natural manner. In addition, a heuristic is proposed to readily locate a b-value of demarcation l0 when the condition is not met.
2. Assumptions and notation Montgomery et al.’s (1973) model is developed under the following assumptions: (1) A single item is considered and the annual demand is deterministic and constant. (2) The unit cost is independent of the quantity purchased. (3) The demand, order quantity, inventory level and shortage can be treated as continuous variables.
* Tel.: +852 27888589; fax: +852 27888560. E-mail address: [email protected] 0377-2217/$ - see front matter Ó 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2008.04.003
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(4) The order quantity will not vary from one cycle to another. (5) The lead time is zero. (6) Only a fraction of the demand during the stockout period is made up in the beginning of the next order cycle. The remaining fraction is lost. (7) The planning horizon is infinite. The following symbols are used in the expression of the average annual cost. A D h p pb pl p0 pl0 = pl + p0 b R S Q ¼ M þ bS ¼ R bS M = Q bS = R S Kb(R, S)
fixed ordering cost per inventory cycle (dollars per order) demand per year (units per year) inventory holding cost per unit per year (dollars per unit per year) shortage cost per backorder per year (dollars per unit per year) fixed penalty cost per unit backorder (dollars per unit) fixed penalty cost per unit lost (dollars per unit); this is greater than pb when the cost of goodwill is included profit per unit (dollars per unit) fixed penalty cost per unit lost plus profit per unit (dollars per unit) is lost fraction of the demand is backordered during the stockout period, while the remaining fraction ð1 bÞ b total demand per cycle (units per cycle, a decision variable) total demand per cycle during the stockout period (units per cycle, a decision variable) order quantity (units per cycle, a decision variable) positive inventory level (units per cycle, a decision variable) average annual cost as a function of R and S with a parameter b (dollars per year, the objective function)
We recognize that the four optimal decision variables R*, S*, Q* and M* are dependent on the parameter b, i.e., R* R*(b), S* S*(b), Q* Q*(b) and M* M*(b). In some circumstances, we can simply write the resulting minimum objective function Kb(R*, S*) as K b . 2ADh p S , k ¼ hp, T ¼ Dp and h ¼ ppb . Note that To simplify the presentation of the subsequent mathematical expressions, we designate B ¼ ðDp Þ2 l0
l0
l0
h < 1 because pb < pl, where pl includes both fixed penalty cost per unit lost and the cost of goodwill. 3. Solution and feasibility of the EOQ model with partial backorders By considering the inventory geometry for this system as shown in Fig. 1, we can obtain the average annual cost which is the sum of ordering, holding, and stockout costs, and which includes lost profit, as bS2 þ 2Dðpb b þ pl0 bÞS 2AD þ hðR SÞ2 þ p ; 2R ¼ 1 b. where pl0 = pl + p0 and b Partially differentiating Eq. (1) with respect to R and S and then setting the two partial derivatives K b ðR; SÞ ¼
2
2
hR
bÞS ¼ 2AD þ ðh þ p
S ¼
hR Dðpb b þ pl0 bÞ ; b hþp
; þ 2Dðpb b þ pl0 bÞS
ð1Þ oK b ðR;SÞ oR
and
oK b ðR;SÞ oS
to be zero yield ð2Þ
and ð3Þ
M R
Q
ts βS
S (1–β )S
t Fig. 1. Behavior of the inventory system. t ¼ DR = cycle length (measured in a fraction of a year) and t s ¼ DS = time interval when stockout occurs (in a fraction of a year).
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K.-N.F. Leung / European Journal of Operational Research 196 (2009) 554–562
which can be written as R ¼
bÞS þ Dðpb b þ pl0 bÞ ðh þ p : h
ð4Þ
Substituting Eq. (3) into (2), we obtain the explicit expression of R* given by sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2ADðh þ p bÞ ½Dðpb b þ pl0 bÞ R ¼ : h pb
ð5Þ
Substituting Eq. (5) into (3), we obtain the explicit expression of S* given by 9 8sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi = 2g bÞ ½Dðpb b þ pl0 bÞ 1 < hf2ADðh þ p S ¼ Dðpb b þ pl0 bÞ : ; b : b hþp p
ð6Þ
Substituting Eq. (4) into (2) yields the minimum condition of Eq. (1), namely 2 þ h ½ pbS þ Dðpb b þ pl0 bÞ pbS2 ¼ 2ADh:
ð7Þ
We notice from Eq. (3) that R* P S*, and hence the EOQ with partial backorders and the maximum positive inventory level are ; Q ¼ R bS
ð8Þ
M ¼ R S :
ð9Þ
and Substituting Eq. (2) into (1), we obtain the expression of K b given by
K b ¼ hM :
ð10Þ *
*
Eqs. (5) and (6) reveal that R > 0 and S > 0 provided 2; bÞ > ½Dðpb b þ pl0 bÞ 2ADðh þ p or taking the square root becomes qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ADðh þ p bÞ > Dðpb b þ pl0 bÞ;
ð11Þ
and sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2g bÞ ½Dðpb b þ pl0 bÞ hf2ADðh þ p > Dðpb b þ pl0 bÞ; b p which can easily be simplified to pffiffiffiffiffiffiffiffiffiffiffiffi 2ADh > Dðpb b þ pl0 bÞ:
ð12Þ *
*
Clearly, condition (12) is satisfied and so is condition (11). This implies that real R exists whenever S > 0. pffiffiffiffiffiffiffiffiffiffiffiffi Condition (12) together with the condition 0 6 B 6 1 (i.e., 0 6 2ADh 6 Dpl0 Þ can be combined and written as 06
pffiffiffi 1 B < b 6 1; 1h
ð13Þ
pb 2ADh where B ¼ ðDp 2 and h ¼ p . l0 l0 Þ p ffiffiffiffiffiffiffiffiffiffiffiffi is the sum of annual penalty cost and Note that 2ADh is the minimum average annual cost in the classical EOQ model and Dðpb b þ pl0 bÞ pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi of demand is lost. Based on the condition that 0 6 2ADh 6 Dpl0 , on the relative magnitudes of 2ADh and annual lost profit if a fraction b and on whether p >p0ffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0, there are three analytic cases: pffiffi or p Dðpb b þ pl0 bÞ, P 0 (i.e., when 0 6 1 B < b 6 1 or condition (13) is satisfied) and p > 0, R* and S* Analytic case (i): When Dpl0 P 2ADh > Dðpb b þ pl0 bÞ 1h are determined using Eqs. (5) and p (3). pffiffi ffiffiffiffiffiffiffiffiffiffiffiffi P 0 (i.e., when 0 6 1 B < b 6 1 or condition (13) is satisfied) and p ¼ 0, we have Analytic case (ii): When Dpl0 P 2ADh > Dðpb b þ pl0 bÞ 1h R* = 1 from Eq. (5) and hence S* = 1 from Eq. (3), which are meaningless. What is going on? In this case, we have using Eqs. (3), (9) and (10), i.e., the optimal policy is to partially backorder everything and the resulting minK b ¼ hðR S Þ ¼ Dðpb b þ pl0 bÞ in which we note that pl0 is regarded as an opportunity cost relating to negative inventory (i.e., imum average annual cost is Dðpb b þ pl0 bÞ, shortage), similar to regarding h as an opportunity cost relating to positive inventory, and pl0 equals the sum of the cost of goodwill and unit profit (not including thepfixed (2006, p. 63). pffiffi ffiffiffiffiffiffiffiffiffiffiffiffi penalty cost). This is a modified result in Sphicas 6 Dpl0 (i.e., when 0 6 b 6 1 BÞ (no matter whether p > 0 or p ¼ 0Þ, we have S* = 0, i.e., Analytic case (iii): When 2ADh 6 Dðpb b þ pl0 bÞ 1h the optimal solution is q toffiffiffiffiffiffi allow no backorders or lost sales to occur. As a result, Eq. (1) becomes EOQ model, and the optimal pffiffiffiffiffiffiffiffiffiffiffiffi the classical and the resulting minimum average annual cost is KðQ Þ ¼ 2ADh ¼ hQ . order quantity is Q ¼ 2AD h Two remarks for this section are as follows:
(1) If condition (12) is satisfied, then the EQQ model given by Eq. (1) is feasible. If condition (11) is met, then the cost function given by the same equation is convex and hence the global minimum is guaranteed. Condition (12) is satisfied and so is condition (11) and thus we can conclude that model feasibility implies function convexity for the EOQ model under study. We have confirmed this conclusion in the Appendix.
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(2) To solve or analyze the EOQ model given by Eq. (1), we can use the complete squares method proposed in Leung (2008a,b). It is an algebraic method which requires no working knowledge of multi-variable calculus and is usually taught when introducing the notion of a general quadratic equation in Form 2 at high school. Hence the method can be understood by ordinary readers. On the other hand, to investigate the monotonic property of an optimal expression in Section 4, we must use differential calculus.
4. Sensitivity analysis when 0 < B < 1 and
pffiffi 1 B 1h
pffiffi B < b <1 (i.e., when 0 < 1 < b <1 or condition (13) is satisfied) 1h
pffiffi B
Differentiating Eq. (5) with respect to b, we have 2 2D2 bðpb b þ pl0 bÞðp b pl0 Þ 2ADh dR ½Dðpb b þ pl0 bÞ ¼ : 2 db 2h pb R The expression for R* strictly increases with b iff dR >0 db
2 2D2 bðpb b þ pl0 bÞðp b pl0 Þ 2ADh > 0 iff ½Dðpb b þ pl0 bÞ
2 2bðhb þ bÞðh 1Þ B > 0 iff ðhb þ bÞ pffiffiffiffiffiffiffiffiffiffiffiffi 1B iff b < : 1h pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi It is easy to show that 1 B 6 1 B when 0 6 B 6 1. As a result, we need consider two cases: (1) If (2) If
pffiffiffiffiffiffiffi 1B 1h pffiffiffiffiffiffiffi 1B 1h
ð14Þ
pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi pffiffi B 1B 1B increases, but R* strictly decreases as b 1h < 1, then R* strictly increases as b 0 6 1 < b 6 1h 6 b 6 1 increases. 1h pffiffi B P 1, then R* strictly increases as b 0 6 1 < b 6 1 increases. 1h
* In particular, when hp=ffiffiffiffiffiffiffiffiffiffiffi 0 (i.e., ffi pb = 0 in Park’s (1982) model), then R strictly increases as b (0 6 1 strictly decreases as b ð 1 B 6 b 6 1Þ increases.
pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi B < b 6 1 BÞ increases, but R*
pffiffi B
To facilitate the investigation, we are better off writing Eq. (7) as þ hbÞ2 þ kbT 2 ¼ B; ðbT þ b
ð15Þ 2
which is obtained by first dividing by the factor (Dpl0) throughout Eq. (7) and next using the designations of B ¼ h ¼ ppb . l0 Implicitly differentiating Eq. (15) with respect to b, we have
2ADh , ðDpl0 Þ2
þ hbÞ kT 2 dT 2ð1 h T ÞðbT þ b ¼ ; þ hb db 2b½ðk þ bÞT þ b
k¼
h , p
T ¼
p S Dpl0
and
ð16Þ
and dT >0 db
þ hbÞ kT 2 > 0 iff 2ð1 h T ÞðbT þ b iff b <
2ð1 h T Þ kT 2
2ð1 h T Þ
2
ð17Þ
:
pffiffi B We consider such a scenario that T* strictly increases as b 0 6 1 < b 6 1 increases. This scenario happens iff 1h 1<
2½1 h T ð1Þ k½T ð1Þ2
2½1 h T ð1Þ2 k iff 1 þ ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ < 0; 2
ð18Þ
where
T ð1Þ ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Bð1 þ kÞ kh2 h 1þk
ð19Þ
;
S ð1Þ 2ADh which can be obtained by first setting b = 1 in Eq. (6) and next using T ð1Þ ¼ pDp , B ¼ ðDp , k ¼ hp and h ¼ ppb . Þ2 l0
l0
l0
2 In particular, when h = 0 (i.e., pb = 0 in Park’s (1982) model), Eqs. (18)q and ffiffiffiffiffiffi (19) become T ð1Þ < 2þk, where T ð1Þ ¼ dT ðbÞ B 2 the necessary and sufficient condition for db > 0 when h = 0 becomes 1þk < 2þk, which can be simply written as
B<
1þk 1 þ k þ ð2k Þ2
:
qffiffiffiffiffiffi B . Consequently, 1þk
ð20Þ
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K.-N.F. Leung / European Journal of Operational Research 196 (2009) 554–562 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 1Bð1þ 1BÞ , B
2 1þk In the Appendix, we show that B < 1þkþð k Þ2 is the same as k < 2
which was ingeniously deduced by Chu and Chung (2004). How-
ever, condition (20) is deduced (or derived) in a simple, direct and natural manner, as well as appearing concisely. If condition (18) (or (20) in particular) is not satisfied, we cannot explicitly obtain a b-value of demarcation because the right-hand side 2½1hT ðb0 Þk½T ðb0 Þ2 * b0 ¼ 0. Then T* strictly increases of inequality (17) involves T (b), i.e., to deduce an explicit expression of b0 from solving pffiffi 2½1hT ðb0 Þ2 * 1 B as b 0 6 1h < b 6 b0 increases, but T strictly decreases as b (b0 6 b 6 1) increases. In the Appendix, we also show that there exists a robust initial value, namely b1 ¼
2½1 h T ð1Þ k½T ð1Þ2 2½1 h T ð1Þ2
ð21Þ
;
whereby we can readily determine a b-value that is extremely close to the exact value of b0, provided that
kT ðb1 Þ > 1
T ðb1 Þ : 1h
ð22Þ
To facilitate the determination, a three-step heuristic procedure is proposed in the Appendix. This heuristic is illustrated by Examples 2 and 3 in Section 5. pffiffi B 4.3. The monotonic property of Q* with respect to b when 0 6 1
Substituting Eq. (4) into (8) and using the symbol designations, we obtain the expression of Q* given by Q ¼
Dpl0 þ hb: ½ð1 þ kÞbT þ b h
ð23Þ
Differentiating Eq. (23) with respect to b, we have dQ Dpl0 dT ¼ ð1 þ kÞb þ ð1 þ kÞT 1 þ h ; db h db
ð24Þ
and then substituting Eq. (16) into (24), we obtain dQ >0 db
þ 2hð1 hÞb > 0 iff ð1 þ kÞT 2 þ 2ð1 T Þð1 hÞb
2 2 ð1 hÞb ½ð1 hÞb þ 2hð1 hÞb > 0 þ 2ð1 hÞb iff ð1 þ kÞ T 1þk 1þk 2 1 þ 2k þ b þ hb ð1 hÞb þ 2hð1 hÞb > 0: iff ð1 þ kÞ T þ ð1 hÞb 1þk 1þk pffiffi B The last inequality is always true because h < 1. Thus Q* strictly increases as b 0 6 1 < b 6 1 increases. 1h
ð25Þ
pffiffi B 4.4. The monotonic properties of M* and K b with respect to b when 0 6 1
Substituting Eq. (4) into (9) and using the symbol designations, we obtain the expression of M* given by M ¼
Dpl0 þ hbÞ: ðbT þ b h
ð26Þ
Differentiating Eq. (26) with respect to b, we have dM Dpl0 dT ¼ b þ T 1 þ h ; db h db
ð27Þ
and then substituting Eq. (16) into (27), we obtain dM <0 db
iff T < 2ð1 hÞ:
The last inequality is always true because pffiffiffi 1 B < b 6 1: T < 1 h for 0 6 1h
ð28Þ
ð29Þ pffiffi B ð1 ; 1 1h
The validity of result (29) is proved by contradiction as follows: Assume that there exists a b-value denoted by b lying in such that pffiffi B ; 1 and the assumption of T*(b) P 1 h, the left-hand side of Eq. (15) indicates that T*(b) P 1 h. With T* > 0 for any b-value in ð1 1h þ hb 2 þ kb ½T ðb Þ2 > ½b ð1 hÞ þ b þ hb 2 ¼ 1; ½b T ðb Þ þ b
pffiffi B and is greater than B. Because 0 6 B 6 1, the assumption contradicts equality (15) which is valid for any b-value in ð1 ; 1. In other words, 1h pffiffi * 1 B we have shown that T < 1 h for any b-value in ð 1h ; 1. pffiffi B Thus M* strictly decreases as b 0 6 1 < b 6 1 increases, and so does K b because K b ¼ hM , using Eq. (10). Moreover, we have 1h pffiffi pffiffiffiffiffiffiffi dM ðbÞ B 1B ¼ dRdbðbÞ dSdbðbÞ < 0 or dRdbðbÞ < dSdbðbÞ. This implies that dSdbðbÞ > 0 while dRdbðbÞ > 0 for 0 6 1 < b 6 1h . That is, db 1h p ffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi ffi dR ðbÞ dS ðbÞ 1 B 1B
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K.-N.F. Leung / European Journal of Operational Research 196 (2009) 554–562
5. Numerical examples Example 1. Suppose that an item has the characteristics as those in Montgomery et al. (1973, p. 259): A = $50 per order, D = 250 units per ¼ $0:1 per unit per year, pb = pl = $0.5 per unit, p0 = $2 per unit and pl0 = $2.5 per unit. year, h = $2 per unit per year, p pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 pffiffiffiffiffiffiffi pffiffi Bð1þkÞkh h B 1B First we compute B ¼ 2ADh2 ¼ 0:128, k ¼ hp ¼ 20, h ¼ ppl0b ¼ 0:2, 1 ¼ 0:0559. 1h ¼ 0:8028, 1h ¼ 1:1673 and T ð1Þ ¼ 1þk ðDpl0 Þ Next we draw the following conclusions: (1) (2) (3) (4)
pffiffiffiffiffiffiffi 1B ¼ 1:1673 > 1,pffiffiffiffiffiffi R*(b) Because 1h ffi strictly increases as b (0.8028 < b 6 1) increases. 1B ¼ 1:1673 > 1 indicates that S*(b) strictly increases as b (0.8028 < b 6 1) increases. Result (30) together with 1h * Q (b) strictly increases as b (0.8028 < b 6 1) increases. M*(b) or K b strictly decreases as b (0.8028 < b 6 1) increases.
Example 2. Further suppose that the values of h and pb become 4 and 0.05 (rather than 2 and 0.5), respectively, and the other parameter values remain the same. pffiffiffiffiffiffiffi pffiffi * B 1B First we obtain B = 0.256, k = 40, h = 0.02, 1 1h ¼ 0:5041, 1h ¼ 0:8802 and T (1) = 0.0785. Next we draw the following conclusions (see Table 1): pffiffiffiffiffiffiffi 1B ¼ 0:8802 < 1, R*(b) strictly increases as b (0.5041 < b 6 0.8802) increases, but R*(b) strictly decreases as b (1) Because 1h (0.8802 6 b 6 1) increases (see column 2 in Table 1). (2) Because ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ ¼ 0:0344 > 0, i.e., because condition (18) is not satisfied, S*(b) is not an increasing function of b for 0.5041 < b 6 1. To determine b0, first we compute b1 = 0.9576 using Eq. (21); second we know that b0 < b1 because condition (22) holds; third letting b = 0.9569, 0.9570 or 0.9571, we have the associated values of S*(b) (see column 3 in Table 1). Thus S* strictly increases as b (0.5041 < b 6 0.9570) increases, but S* strictly decreases as b (0.9570 6 b 6 1) increases. (3) Q*(b) strictly increases as b (0.5041 < b 6 1) increases. (4) M*(b) or K b strictly decreases as b (0.5041 < b 6 1) increases.
Example 3. Suppose that another item has the characteristics as those in Park (1982, p. 1316) as follows: A = $50 per order, D = 200 units ¼ $0:3 (instead of $1) per unit per year. = $2 per unit, with theffi exception that p per year, h = $3 per unit per year, pb = $0 per unit and pp l0ffiffiffi pffiffiffiffiffiffiffiffiffiffiffi First we compute B ¼ 2ADh2 ¼ 0:375, k ¼ hp ¼ 10, 1 B ¼ 0:3876 and 1 B ¼ 0:7906. ðDpl0 Þ Next we draw the following conclusions (see Table 2): (1) R*(b) strictly increases as b (0.3876 < b 6 0.7906) increases, but R*(b) strictly decreases as b (0.7906 6 b 6 1) increases (see column 2 in Table 2). * 1þk (2) Because 0:375 ¼ B P 1þkþð k Þ2 ¼ 0:3056 i.e., because condition (20) is not satisfied, S (b) is not an increasing function of b for 2
0.3876 < b 6 1. To determine b0, first we compute b1 = 0.9701 using Eq. (21); second we know that b0 < b1 because condition (22) holds; third letting b = 0.9697, 0.9698 or 0.9699, we have the associated values of S*(b) (see column 3 in Table 2). Thus S* strictly increases as b (0.5041 < b 6 0.9698) increases, but S* strictly decreases as b (0.9698 6 b 6 1) increases. (3) Q*(b) strictly increases as b (0.3876 < b 6 1) increases. (4) M*(b) or K b strictly decreases as b (0.3876 < b 6 1) increases.
Table 1 ¼ 0:1, pb = 0.05, pl0 = 2.5) Sensitivity analysis for ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ P 0 in Example 2 (A = 50, D = 250, h = 4, p b
R*(b) (units)
S*(b) (units)
T*(b) < 0.98
Q*(b) (units)
M*(b) (units)
K b ($)
0 to 0.5041 0.51 0.60 0.70 0.80 0.8801 0.8802 0.8803 0.90 0.95 0.9569 b0 = 0.9570 0.9571 0.9576 0.96 0.97 0.98 0.99 1.00b
79.06 131.74 382.94 475.23 511.66 518.9697182 518.9697197a 518.9697007 518.57 514.31 513.3781808 513.3641085 513.3500197 513.2793275 512.93 511.40 509.70 507.84 505.83
0 52.91 313.86 418.84 468.54 486.7747688 486.7885624 486.8023358 489.13 491.85 491.8863423 491.8863525a 491.8863466 491.8860749 491.88 491.76 491.47 491.03 490.44
0 0.0084656 0.0502176 0.0670144 0.0749664 0.077883963 0.077886169 0.077888373 0.0782608 0.0786960 0.078701814 0.078701816a 0.078701815 0.078701772 0.0787008 0.0786816 0.0786352 0.0785648 0.0784704
79.06 105.81 257.40 349.58 417.95 460.61 460.65 460.70 469.66 489.42 492.18 492.21 492.25 492.42 493.25 496.65 499.87 502.93 505.83
79.06 78.83 69.08 56.39 43.12 32.19 32.18 32.16 29.44 22.46 21.49 21.48 21.46 21.39 21.05 19.64 18.23 16.81 15.39b
316.24 315.32 276.32 225.56 172.48 128.76 128.72 128.64 117.76 89.84 85.96 85.92 85.84 85.56 84.20 78.56 72.92 67.24 61.56b
b = fraction of the demand backordered during the stockout period, R*(b) = optimal total demand per cycle, S*(b) = maximum total demand during the stockout period per pffiffi S ðbÞ * 1 B , Q*(b) = optimal cycle, T ðbÞ ¼ pDp qffiffiffiffiffiffiffi order quantity, M (b) = maximum positive inventory level, and K b = minimum average annual cost. When b is in ½0; 1h , we use the l0 ¼ R ¼ M . classical EOQ formula: Q ¼ 2AD h a Global maximum occurs. b Global minimum occurs.
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K.-N.F. Leung / European Journal of Operational Research 196 (2009) 554–562
Table 2 1þk ¼ 0:3, pb = 0, pl0 = 2) Sensitivity analysis for B P 1þkþð k Þ2 in Example 3 (A = 50, D = 200, h = 3, p 2
b
R*(b) (units)
S*(b) (units)
T*(b) < 1
Q*(b) (units)
M*(b) (units)
K b ($)
0 to 0.3876 0.39 0.40 0.50 0.60 0.70 0.7905 0.7906 0.7907 0.80 0.90 0.95 0.96 0.9697 b0 = 0.9698 0.9699 0.9700 0.9701 0.98 0.99 1.00b
81.65 89.38 115.47 226.08 265.27 281.15 284.8348180 284.8348196a 284.8348132 284.80 280.65 276.36 275.34 274.3139367 274.3030681 274.2921947 274.2813164 274.2704333 273.17 272.01 270.80
0 7.74 34.11 151.82 199.94 225.37 238.0811683 238.0913198 238.1014638 239.01 245.24 246.30 246.36 246.3822499 246.3822508a 246.3822472 246.3822392 246.3822268 246.36 246.29 246.18
0 0.0058050 0.0255825 0.1138650 0.1499550 0.1690275 0.178560876 0.178568489 0.178576097 0.1792575 0.1839300 0.1847250 0.1847700 0.184786687 0.184786688a 0.184786685 0.184786679 0.184786670 0.1847700 0.1847175 0.1846350
81.65 84.66 95.00 150.17 185.29 213.54 234.96 234.98 235.00 237.00 256.13 264.05 265.49 266.85 266.86 266.88 266.89 266.90 268.24 269.55 270.80
81.65 81.64 81.36 74.26 65.33 55.78 46.75 46.74 46.73 45.79 35.41 30.06 28.98 27.93 27.92 27.91 27.90 27.89 26.81 25.72 24.62b
244.95 244.92 244.08 222.78 195.99 167.34 140.25 140.22 140.19 137.37 106.23 90.18 86.94 83.79 83.76 83.73 83.70 83.67 80.43 77.16 73.86b
b = fraction of the demand backordered during the stockout period, R*(b) = optimal total demand per cycle, S*(b) = maximum total demand during the stockout period per pffiffi S ðbÞ * 1 B , Q*(b) = optimal cycle, T ðbÞ ¼ pDp qffiffiffiffiffiffiffi order quantity, M (b) = maximum positive inventory level, and K b =minimum average annual cost. When b is in ½0; 1h , we use the l0 ¼ R ¼ M . classical EOQ formula: Q ¼ 2AD h a Global maximum occurs. b Global minimum occurs.
S ðbÞ Using Eqs. (5), (3) and (8)–(10) and the designation of T ðbÞ ¼ pDp , we obtain the results of Examples 2 and 3 shown in Tables 1 and 2, l0 respectively. We remark that although the global minimum average annual cost occurs when b = 1 (see the last columns in Tables 1 and 2), we must use the partial backordering model for a situation where b is actually equal to 0.7, say. On the one hand, if we assume that all demand during the stockout period can be backordered even though this is not strictly true, then substituting the optimal solution obtained by setting b = 1 in Eqs. (5) and (3) (as deduced from the so-called ‘‘EOQ model with complete backorders”) into Eq. (1) yields a non-minimum average annual cost. In Examples 2 and 3, we have (505.83, 490.44, $232.57) and (270.80, 246.18, $172.87). (Note that if b = 1 were true, then the resulting minimum average annual cost would be $61.56 and $73.86.) On the other hand, substituting the optimal solution obtained by setting b = 0.7 in Eqs. (5) and (3) (as deduced from the so-called ‘‘EOQ model with partial backorders”) into Eq. (1) or equivalently Eqs. (9) and (10) yields the minimum average annual cost. In Examples 2 and 3, we have (475.23, 418.84, $225.56) and (281.15, 225.37, $167.34). Thus failure to use the appropriate model has cost management $232.57 $225.56 = $7.01 and $172.87 $167.34 = $5.53 per year for these two items. The effects of this in a multi-item inventory may be quite substantial.
6. Conclusions Summarizing the monotonic properties when 0 6 B 6 1 and
pffiffi 1 B 1h
pffiffi B < b 6 1 (i.e., when 0 6 1 < b 6 1), we have 1h
pffiffiffiffiffiffiffi pffiffi pffiffiffiffiffiffiffi B 1B 1B < 1, then R*(b) strictly increases as b 0 6 1 < b 6 1h 6 b 6 1 increases. increases, but R*(b) strictly decreases as b 1h 1h pffiffiffiffiffiffiffi pffiffi 1B B However, if 1h P 1, then both R*(b) and S*(b) strictly increase as b 0 6 1 6 b 6 1 increases. In particular, when h = 0 (i.e., pb = 0 1h pffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffi in Park’s (1982) model), R*(b) strictly increases as b 0 6 1 B < b 6 1 B increases, but R*(b) strictly decreases as pffiffiffiffiffiffiffiffiffiffiffiffi bð 1 B 6 b 6 1Þ increases. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ffi Bð1þkÞkh h * 1þk (b) On the one hand, if ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ < 0 where T ð1Þ ¼ (or B < 1þkþð kÞ2 when h = 0), then S (b) 1þk 2 pffiffi 2 1 B k strictly increases as b 0 6 1h < b 6 1 increases. On the other hand, if ð1 þ 2Þ½T ð1Þ ð1 2hÞT ð1Þ hð1 hÞ P 0, then S*(b) pffiffi B strictly increases as b 0 6 1 < b 6 b0 increases, but S*(b) strictly decreases as b (b0 6 b 6 1) increases, where b0 satisfies 1h (a) If
pffiffiffiffiffiffiffi 1B 1h
2½1hT ðb0 Þk½T ðb0 Þ2 2½1hT ðb0 Þ2 *
b0 ¼ 0.
pffiffi B (c) Q (b) strictly increases as b 0 6 1 < b 6 1 increases. 1h pffiffi B < b 6 1 increases. (d) Both M*(b) and K b strictly decrease as b 0 6 1 1h In addition, we have (1) (2) (3) (4)
pffiffi B < b 6 1 is true, T*(b) < 1 h. When 0 6 1 1h pffiffi 1 B When 0 6 1h
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K.-N.F. Leung / European Journal of Operational Research 196 (2009) 554–562
In the Appendix, we tabulate some symbols used in this paper and compare them to those used in Montgomery et al. (1973), Park (1982) and Chu and Chung (2004) so that readers can easily reconcile the materials. Finally, we would like to indicate a minor mistake pffiffiffi at the start of Section 3 in Chu and Chung (2004, p. 291). This should read that Kb (R, S) is convex with respect to R and S for all 0 6 1 B < b 6 1 (not 0 6 b 6 1). This can be confirmed in the Appendix in Park (1982, p. 1317). Acknowledgements The author is grateful to the two anonymous referees for constructive comments that led to essential improvement in this paper. Their appropriate suggestions were included in the text. In addition, the author would like to thank Referee 1 for arousing his awareness of the paper written by Yang (2007) and to thank Referee 2 for insisting him on rewriting the paper according to Montgomery et al.’s (1973) model. Appendix A1. Partially differentiating the function K Kb(R, S) with respect to the two decision variables R and S twice and computing the determinant of the Hessian matrix, we have o2 K 2
oR
¼
bÞS2 þ 2Dðpb b þ pl0 bÞS 2AD þ ðh þ p 3
R
;
o2 K oS
2
¼
b hþp ; R
bÞS þ Dðpb b þ pl0 bÞ o2 K ½ðh þ p ; ¼ oRoS R2
and o2 K o2 K j H j¼ 2 2 oR oS
!2 2 bÞ ½Dðpb b þ pl0 bÞ o2 K 2ADðh þ p ¼ ; oRoS R4 2
2
;S Þ ;S Þ which is greater than zero provided condition (11) is satisfied. Clearly, o KðR > 0 and o KðR > 0; thus K has a global minimum point at oR2 oS2 (R*, S*). A2. The equivalence of the two necessary and sufficient conditions is shown as follows: 2
B<
1þk Bk 2 < 1 þ k () 4B þ 4Bk þ Bk < 4 þ 4k 2 () B þ Bk þ 4 1 þ k þ 2k
k<
pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi 2 1 Bð1 þ 1 BÞ () Bk < 2 1 B þ 2ð1 BÞ () Bðk þ 2Þ 2 < 2 1 B B () B2 ðk þ 2Þ2 4Bðk þ 2Þ þ 4 < 4ð1 BÞ () Bðk þ 2Þ2 4k 4 < 0
and
2
2
() Bðk þ 4k þ 4Þ 4k 4 < 0 () 4B þ 4Bk þ Bk < 4 þ 4k: A3. Denote the expressions on the left and right-hand sides of inequality (17) by f(b) and g(b), i.e., let f ðbÞ ¼ b; and gðbÞ ¼
2ð1 h T Þ kT 2ð1 h T Þ2
2
:
g(β ) f (β ) =β
1
(1, g(1))
β1
0
β 1− B 1−θ
β0 β1
1
1þk Fig. 2. Behavior of the function g(b) when ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ P 0 (or B P 1þkþð k Þ2 when h = 0). 2
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K.-N.F. Leung / European Journal of Operational Research 196 (2009) 554–562
Differentiating g(b) with respect to b yields
dT dgðbÞ ð1 h T Þ½kð1 hÞT ð1 h T Þ db : ¼ 4 db ð1 h T Þ
Since
dT db
¼ 0 at b = b0,
kT ðb1 Þ > 1
dT db
< 0 for b0 < b 6 1, and result (29), we have
dgðbÞ db
¼ 0 at b = b0 and
dgðbÞ db
> 0 for b0 < b 6 1 provided that
ðb1 Þ
T : 1h
ð22Þ
Define b1 ¼ f ðb1 Þ ¼ gð1Þ ¼
2½1 h T ð1Þ k½T ð1Þ2 2½1 h T ð1Þ2
ð21Þ
:
If condition (22) is satisfied at b = b1, then we have dgðbÞ > 0 at b = b1 and deduce that b0 < b1. This means that we can regard b1 as an initial db value to determine b0 efficiently. Fig. 2 shows the behavior of the function g(b) when ð1 þ 2kÞ½T ð1Þ2 ð1 2hÞT ð1Þ hð1 hÞ P 0 (or 1þk B P 1þkþð kÞ2 when h = 0). 2
We propose a heuristic procedure with three steps to determine the value of b0: (1) Compute b1 using Eq. (21). ðb1 Þ holds. If so, then we know that b0 < b1. (2) Check whether the inequality kT ðb1 Þ > 1 T1h (3) Regard b1 as an initial value and use trial and error to determine the value of b0. A4. Some symbols used in this paper, Montgomery et al. (1973), Park (1982) and Chu and Chung (2004) are tabulated as follows: Symbol description
Leung (this paper)
Montgomery et al. (1973)
Park (1982)
Chu and Chung (2004)
Ordering cost per cycle Demand per year Holding cost per unit per year Backorder cost per unit per year Backorder cost per unit
A D h
A D I C = IC
A d h
A d h
p
p
p
p
pb
p
Assign zero
Lost-sale cost per unit
pl
p
Profit per unit
p0
p0
Lost-sale cost plus profit per unit Fraction of backordered demand Fraction of lost demand Demand per cycle Shortage per cycle Order quantity Positive inventory level
pl0 = pl + p0
p + p0
Assign zero Not available Not available p
b
b
b
b
¼1b b R S Q M
1b U S Q V
1b R S Q Not available
Average annual cost Designations
Kb(R, S) S ðbÞ ; h ¼ ppl0b . B ¼ 2ADh2 ; k ¼ hp ; T ðbÞ ¼ pDp l0
K(Q, S) or K(U, V) Not applicable
1b R S Q Not available K(R, S;b) Not applicable
ðDpl0 Þ
Not available Not available p
Kb(R, S) B ¼ 2Adh2 ; k ¼ hp ; T ðbÞ ¼ pSdpðbÞ ; h ¼ 0. ðdpÞ
References Chu, P., Chung, K.J., 2004. The sensitivity of the inventory model with partial backorders. European Journal of Operational Research 152, 289–295. Chung, K.J., Lin, C.N., 2001. Optimal inventory replenishment models for deteriorating items taking account of time discounting. Computers and Operations Research 28, 67–83. Hwang, H., Hahn, K.H., 2000. An optimal procurement policy for items with an inventory level-dependent demand rate and fixed lifetime. European Journal of Operational Research 127, 537–545. Leung, K.N.F., 2008a. Technical note: A use of the complete squares method to solve and analyze a quadratic objective function with two decision variables exemplified via a deterministic inventory model with a mixture of backorders and lost sales. International Journal of Production Economics 113, 275–281. Leung, K.N.F., 2008b. Using the complete squares method to analyze a lot size model when the quantity backordered and the quantity received are both uncertain. European Journal of Operational Research 187, 19–30. Montgomery, D.C., Bazaraa, M.S., Keswani, A.K., 1973. Inventory models with a mixture of backorders and lost sales. Naval Research Logistics Quarterly 20, 255–263. Park, K.S., 1982. Inventory model with partial backorders. International Journal of Systems Science 13, 1313–1317. Sphicas, G.P., 2006. EOQ and EPQ with linear and fixed backorder costs: Two cases identified and models analyzed without calculus. International Journal of Production Economics 100, 59–64. Yang, G.K., 2007. Note on sensitivity analysis of inventory model with partial backorders. European Journal of Operational Research 177, 865–871.