Improved inventory models with service level and lead time

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Improved inventory models with service level and lead time

Peter Chu

a;∗

_{, Kuo-Lung Yang}

b

_{, Patrick S. Chen}

c

a_{Department of Trac Science, Central Police University, Taoyuan, Taiwan}

b_{Department of Industrial Engineering and Management, National Chiao Tung University, Hsinchu, Taiwan} c_{Department of Information Management, Central Police University, Taiwan}

Abstract

This paper explores the mixed inventory backorder and lost sales problem in which both the lead time and order quantity are treated as decision variables. In a recent paper on Computers and Operations Research, Ouyang and Wu considered this problem. However, their algorithms might not 3nd the optimal solution due to 4aws in their solution procedure. We develop some lemmas to reveal the parameter e5ects and then present two complete procedures for 3nding the optimal solution for the models. The savings are illustrated by solving the same examples from Ouyang and Wu’s paper to demonstrate the superiority of our revised algorithms. ? 2003 Elsevier Ltd. All rights reserved.

Keywords: Inventory; Lead time; Backorder; Lost sales; Distribution-free approach

1. Introduction

Today, the just-in-time (JIT) production system is in vogue. It emphasizes high quality, low stock and short lead time. Many business scholars have focused on “high quality, low stock and short lead time as competitive business goals. The ultimate goal of JIT is a smooth, rapid 4ow of materials through the system. The idea is to make the process time as short as possible using resources in the best-possible way. Sometimes, the lead time is too long prolonging the process. In this situation lead time reduction is important and a serial improvement objective. Continual improvement can reduce internal production time and satisfy the exterior market. The business service level and competitiveness can then be raised. The issue of lead time reduction in inventory management has thus become a matter of great interest.

Recent studies by scholars on how to control lead time are reviewed in this paper. In the traditional

inventory model, as described by Silver and Peterson [1], lead time is considered as a predetermined

∗_{Corresponding author.}

E-mail address: [email protected](P. Chu).

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constant or a stochastic parameter. Liao and Shyu [2] stated that lead time is negotiable and can be decomposed into several components, each having a di5erent piecewise linear crash cost function for lead time reduction. Ben-Daya and Raouf [3] extended Liao and Shyu’s [2] work to consider both lead time and order quantity as decision variables. Moon and Gallego [4] assumed unfavorable lead time demand distribution and solved both the continuous review and periodic review models with a mixture of backorders and lost sales using the minmax distribution-free approach. Ouyang et al. [5] generalized Ben-Daya and Raouf’s [3] assumption that shortages were allowed and constructed variable lead time from a mixed inventory model with backorders and lost sales. Moon and Choi [6] and Lan et al. [7] pointed out the problem in Ouyang et al’s. [5] method. They found indi-vidual optimal order quantities and optimal lead time for a mixed inventory model, and developed a simpli3ed solution procedure. Ouyang and Wu [8] extended the Ouyang et al. [5] article. They relaxed the assumption about the cumulative lead time distribution demand and applied the minimax distribution-free procedure to determine the optimal order quantity and optimal lead time. Wu and Tsai [9] considered that the lead time demands from di5erent customers are not identical. They developed a mixed inventory model with backorders and lost sales for variable lead time demand with a mixed normal distribution. Pan and Hsiao [10] presented inventory models with backorder discount and variable lead time to ensure that customers would be willing to wait for backorders.

This article will study the same inventory model as Ouyang and Wu [11] who considered both lead time and order quantity as decision variables for a mixed inventory model. Ouyang and Wu [11] thought that it is often diKcult to determine the stock-out cost value in inventory systems. Therefore, they replaced the stock-out cost with a service level condition. In their paper, 3rst they assumed that the lead time demand follows a normal distribution. They then relaxed the assumption about the lead time distribution demand function and applied the minmax distribution-free procedure to solve the problem. However, there are critical 4aws in their solution algorithms under di5erent assumptions. The model studied in this article is the same used by Ouyang and Wu [11]. We will point out the questionable algorithms in their model. Their algorithms are complicated and cannot obtain the optimal solution, as demonstrated by their examples. We will construct correct and eKcient algorithms to 3nd the optimum order quantity and reorder point simultaneously when the lead time probability distribution is normal or free. We developed lemmas to reveal the parameter e5ects and illustrate our improvement by solving the same examples.

2. Notation and assumptions

We use the same notations and assumptions as Ouyang and Wu [11]. Notations:

A 3xed ordering cost per order

D average demand per year

h inventory holding cost per item per year

L length of lead time

Q order quantity

X the lead time demand which has a distribution function F with 3nite mean L and standard derivation √L (¿ 0)

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x+ _{maximum value of x and 0, i.e. x}∗_{= max{x; 0}}

proportion of demands that are not met from stock so 1 − is the service level

fraction of the demand during the stock-out period that will be backordered

C(L) Lead time crashing cost

EAC(Q; L) total expected annual cost

Assumptions. (1) The reorder point r = expected demand during lead time + safety stock (SS), and

SS = k√L, that is, r = L + k√L, where k is the safety factor and satis3es P(X ¿ r) = q, q

representing the allowable stock-out probability during L.

(2) B(r) = E[X − r]+ _{is the expected demand shortage at the end of cycle. Hence, B(r) are the}

backordered quantities and (1 − )B(r) are the lost sales. Therefore, the total demand during lead time period equals L−(1−)B(r) and the expected net inventory level just before the order arrives is r − ( L − (1 − )B(r)). Moreover, the expected net inventory level at the beginning of the cycle is Q + r − ( L − (1 − )B(r)), so the expected holding cost per cycle is

Q D Q 2 + k √ L + (1 − )B(r) :

(3) If X has a normal distribution function F(x), according to Ravindran et al. [12], then B(r) =

√L (k), where (k) = (k) − k[1 − (k)] ¿ 0, and , are the standard normal probability density function and distribution function, respectively.

(4) Inventory is continuously reviewed. Replenishments are made whenever the inventory level falls to the reorder point r.

(5) The lead time L has n mutually independent components. The ith component has a minimum duration ai, and normal duration bi, and a crash cost per unit time ci. Further, we assume that c16 c26 · · · 6 cn.

(6) The lead time components are crashed one at a time starting with the least ci component and so on.

(7) If we let L0=nj=1 bj and Li be the length of lead time with components 1; 2; : : : ; i crash to their minimum durations, then Li=n_j=i+1 bj+i_j=1 aj. The lead time crash cost C(L) per cycle for a given L ∈ [Li; Li−1], is given by C(L) = ci(Li−1− L) +i_j=1−1 cj(bj− aj).

3. Normal distribution model

The total expected annual cost is the sum of the ordering cost, holding cost and lead time crash cost, subject to a service level constraint. Hence, the problem is

Min EAC(Q; L) = AD_Q + h Q 2 + k √ L + (1 − )B(r) +D_QC(L) (1) s:t: B(r)_Q 6 ; (2)

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3.1. Review of normal distribution model of Ouyang and Wu

Ouyang and Wu considered that the lead time demand X has a normal distribution function F(x). They then solved the following problem:

Min EAC(Q; L) = AD_Q + h Q 2 + k √ L + (1 − )√L (k) + D_QC(L) (3) s:t: √ L (k) Q 6 ; (4)

where 0 ¡ Q ¡ ∞ and L ∈ [Li; Li−1] for i = 1; 2; : : : ; n.

They ignored the service level constraint and took the partial derivatives of EAC(Q; L) with respect to Q and L in each time interval (Li; Li−1), and they found

@ @QEAC(Q; L) = h 2 − AD Q2 − D Q2C(L); (5) @ @LEAC(Q; L) = hk 2√L + h(1 − ) (k) 2√L − ci D Q; (6) @2 @Q2EAC(Q; L) = 2AD Q3 + 2D Q3C(L) ¿ 0 (7) and @2 @L2EAC(Q; L) = − hk 4√L3+ h(1 − ) (k) 4√L3 ¡ 0: (8)

Ouyang and Wu derived that for 3xed L ∈ (Li; Li−1), EAC(Q; L) is convex in Q and for 3xed Q,

EAC(Q; L) is concave in L ∈ (Li; Li−1). Setting Eq. (5) to zero and solving for Q, they obtained

Q = 2D h [A + C(L)] ₁₌₂ : (9)

Ouyang and Wu [11] established an iterative algorithm to 3nd the optimal lead time and optimal

order quantity. To save space, we do not quote their algorithm, but directly o5er the improved algorithm.

3.2. Modi7cation for normal distribution model

First, we review the normal distribution model and then point out the error in the Ouyang and Wu algorithm [11]. From Eq. (8), we know that for 3xed Q, EAC(Q; L) is concave in L ∈ [Li; Li−1].

Hence, the problem is reduced to consider Min EAC(Q; Li) = AD_Q + h Q 2 + k Li+ (1 − ) Li (k) +D_QC(Li); (10)

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Table 1 Lead time data

Lead time Normal duration, bi Minimum duration, bi− ai Unit crashing cost,

component, i (days) ai (days) (weeks) ci ($/week)

1 20 6 2 2.8 2 20 6 2 8.4 3 16 9 1 35 where Li (k) 6 Q and i = 0; 1; : : : ; n: (11)

Using Eq. (7), for a given i = 0; 1; : : : ; n, EAC(Q; Li) is convex in [(= )√Li (k); ∞). From

Eq. (9), we know that Qi= 2D h [A + C(Li)] ₁₌₂ (12)

is the minimum point without the constraint in Eq. (4). Hence, we have that (a) if (= )√Li (k)

6 Qi, then Qi is the local minimum of EAC(Q; Li) and (b) if (= )√Li (k) ¿ Qi, then (= )√Li (k) is the local minimum of EAC(Q; Li).

Therefore, the optimal order quantity of EAC(Q; Li) for Q ∈ [√Li (k); ∞) is maxQi;

Li (k)

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3.3. Modi7ed algorithm for normal distribution model Now, we o5er our improved algorithm.

(1) Step 1: For each Li, i = 0; 1; : : : ; n, compute Qi, using Eq. (12).

(2) Step 2: Let xi= max{Qi; (= )√Li (k)} for i = 0; 1; : : : ; n.

(3) Step 3: If EAC(xs; Ls) = mini=0;1;:::;nEAC(xi; Li), then the optimal solution is (xs; Ls).

From our algorithm, this inventory model always has feasible solutions. Conversely, in Step 5 of

Algorithm 1 by Ouyang and Wu [11], they predicted that this inventory model sometimes did not

have feasible solutions. Hence, their algorithm involves errors. 3.4. Numerical example

We use Example 1 from Ouyang and Wu [11] with the following data: D=600 unit=year, A=$200 per order, h = $20, = 7 units=week, = 1, the service level 1 − = 0:985, q = 0:2 (in this situation, the safety factor k = 0:845) and the lead time has three components with the data shown in Table 1. The results from our algorithm are listed in Table 3 and we have the optimal quantity Q∗_=125:9663

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Table 2

Summary of the optimal solution from Ouyang and Wu [11] (Li in week)

i Li C(Li) B(r) Qi EAC(Qi; Li) B(r)=Qi

0 8 0 2.1818 110 2525.21 0.0198

1 6 5.6 1.8895 111 2511.13 0.0170

2 4 22.4 1.5428 116 2546.94 0.0133

3 3 57.4 1.3361 124 2690.39 0.0108

Table 3

Summary of the optimal solution using the proposed method (Li in week)

i B(r) = √ Li (k) Qi xi EAC(xi; Li) 0 145.4533 109.5445 145.4533 2614.14 1 125.9663 111.0675 125.9663 2528.75 2 102.8510 115.5162 115.5162 2546.94 3 89.0716 124.2739 124.2739 2690.39

We reproduced Table 2 from Ouyang and Wu [11] as our Table 3. Even though EAC(Q1; L1)

is the minimum among EAC(Qi; Li) for i = 0; 1; 2; 3, however, B(r)=Q1 = 0:017 ¿ = 0:015, so Q1 and L1 are not feasible solutions under the service level constraint. They obtained the optimal quantity Q∗ _{= 116 units; optimal lead time L}∗ _{= 4 and the minimum total expected annual cost}

EAC(Q∗_{; L}∗_{) = $2546:94. Therefore, using our algorithm, we have saved $18:19.}

4. Distribution-free model

The information about the form of the probability distribution of lead time is often limited in practice. This makes the probability distribution of lead time not easy to know. We can derive mean and standard derivation from some related information. For facing this kind of problem, many scholars have used minmax distribution-free approach to get optimal order quantity and optimal lead time in some references and had good outcomes.

4.1. Review of Distribution-free model from Ouyang and Wu

Ouyang and Wu [11] used a minimax distribution-free procedure to solve the inventory model with distribution function F(x) of lead time demand that has mean L, standard derivation √L and unknown probability distribution. According to Gallego and Moon [13] and assumption (1) r − L = k√L, they asserted that

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Ouyang and Wu [11] changed the problem in Eq. (3) to consider Min EAC(Q; L) = AD_Q + h Q 2 + k √ L + (1 − )₂ √L(1 + k2_{− k)} +_QDC(L) (15) s:t: _2Q1 √L(1 + k2_{− k) 6
;} ₍₁₆₎

where 0 ¡ Q ¡ ∞ and L ∈ [Li; Li−1] for i = 1; 2; : : : ; n.

They ignored the service level constraint and used the 3rst and second partial derivative of

EAC(Q; L) with respect to Q and L. Comparing Eqs. (3) and (15), the only di5erence is (k)

and (√1 + k2_{− k)=2. Hence, the monotonic and concave properties of EAC(Q; L) in Eq. (}₃_{) still} hold for EAC(Q; L) in Eq. (15).

Using Proposition 2 from Ouyang and Wu [11], they derived that q ≡ P(X ¿ r) 6 1

1 + k2: (17)

Hence, they obtained the range for k as 0 6 k 6q−1_{− 1. Moreover, they selected a number,}

say N, that is large enough to partition the interval [0;q−1_{− 1] into N equal subintervals and let}

kj= (j=N)q−1− 1 for j = 0; 1; : : : ; N, and they then established an algorithm to obtain a suitable

kj and the optimal Q and L. To save space, their algorithm is not repeated here.

4.2. Modi7cation for Distribution-free model

Ouyang and Wu [11] could not manage the problem for 0 6 k 6q−1_{− 1. They used the discrete}

numbers kj=(j=N)

q−1_{− 1 for j =0; 1; : : : ; N. However, this is not trivial work to choose a proper}

number for N. In Ouyang and Wu [11], they took N = 200; however, this number is not big enough to insure the existence of a minimum solution.

Now, we consider the distribution-free model as follows: Min EAC(Q; L; k) = AD_Q + h Q 2 + k √ L + (1 − )₂ √L(1 + k2_{− k)} + D_QC(L) (18) s:t: ₂1 √L(1 + k2_{− k) 6 Q;} ₍₁₉₎ and 0 6 k 6q−1_{− 1} ₍₂₀₎ with L ∈ [Li; Li−1] for i = 1; 2; : : : ; n. Similar to Eq. (8), we 3nd @2 @L2EAC(Q; L; k) = −h 4√L3 k +(1 − ) 2 ( 1 + k2_{− k)} ¡ 0: We reduce the problem to the following:

Min EAC(Q; Li; k) = A_QD+ h Q 2 + k Li+(1 − )₂ Li( 1 + k2_{− k)} +D_QC(Li) (21)

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s:t: ₂1 Li(

1 + k2_{− k) 6 Q;} ₍₂₂₎

and 0 6 k 6q−1_{− 1} ₍₂₃₎

with i = 0; 1; : : : ; n.

For a given i, we obtain the partial derivatives with respect to Q and k as follows: @ @k EAC(Q; Li; k) = h Li 1 +1 −  2 k √ 1 + k2 − 1 ; (24) @2 @k2EAC(Q; Li; k) = (1 − )h√Li 2(1 + k2₎3 ¿ 0; (25) @ @QEAC(Q; Li; k) = h 2 − D[A + C(Li)] Q2 (26) and @2 @Q2EAC(Q; Li; k) = 2D[A + C(Li)] Q3 ¿ 0: (27)

Since is the backordered fraction, we have 0 6 6 1. Moreover, using the inequality −1 6 k=√1 + k2_{− 1 ¡ 0, it shows} −1=2 61 − ₂ k √ 1 + k2 − 1 6 0: Hence, @ @k EAC(Q; Li; k) ¿ 0 (28)

First, we solve Eq. (26), and then we know that (@=@Q)EAC(Q; Li; k) = 0 occurs at

(2D=h)[A + C(Li)]. To simplify the notation, we assume ai=

2D

h [A + C(Li)]: (29)

Here, we consider constraint (22). Let

g(k) = √ Li 2 ( 1 + k2_{− k):} ₍₃₀₎

Furthermore, we assume that for a 3xed k,

Q∗_{(k) = min point of Q for EAC(Q; L}

i; k) subject to g(k) 6 Q: (31) From g(k) = √ Li 2 k √ 1 + k2 − 1 ¡ 0 and g(k) = √ Li 2 (1 + k2₎3¿ 0;

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we get that g(k) decreases and is concave up for k ∈ [0;q−1_{− 1]. Now, we derive several lemmas}

to develop our theorem. Based on Eq. (28), EAC(Q; Li; k) is convex in Q. Therefore, we have the

following Lemma 1.

Lemma 1. For a number k with k ∈ [0;q−1_{− 1], we have Q}∗_{(k) = max{g(k); a}_i_}.

We compare g(k) with ai. Solving g(k) = ai, we obtain k = √Li=4ai − ai =√Li. To simplify

the expression, we assume

bi= √ Li 4ai − ai √Li; (32)

that is, we have g(bi) = ai.

In the sequel, we distinguish among three cases as follows:

Case 1: 0 6 bi6q−1− 1. Case 2: bi¡ 0. Case 3: bi¿q−1− 1.

Lemma 2. For Case 1, we show that the minimum value of EAC(Q; Li; k), subject to g(k) 6 Q and

bi6 k 6q−1− 1, occurs at k = bi and Q = ai.

Proof. Since g(k) decreases for k ∈ [bi;

q−1_{− 1] then max{g(k); a}_i_{} = a}_i_{. Using Lemma} ₁_{, to}

solve min EAC(Q; Li; k), subject to g(k) 6 Q and bi6 k 6

q−1_{− 1, is equivalent to solving}

min EAC(Q = ai; Li; k), subject to bi6 k 6q−1− 1. Using Eq. (28), EAC(Q = ai; Li; k) increases on k, so minimum value occurs at k = bi.

Lemma 3. For Case 1, we prove that the optimal point (Q∗_{; k}∗_{) attaining the minimum value}

of EAC(Q; Li; k), subject to g(k) 6 Q and 0 6 k 6 bi, can be divided into the following three

scenarios:

(1) If di¿ 0 then k∗= 0 and Q∗= g(0).

(2) If di¡ 0 and ci6 − di then k∗= bi and Q∗= g(bi) = ai.

(3) If di¡ 0 and ci¿ − di then k∗= min{−di=c2i − d2i; bi} and Q∗= g(k∗), with

ci=2D √_L i[A + C(Li)] + h Li 1 4 + 1 −  2 ; and di=2D √_L i [A + C(Li)] + h Li 1 − 1 4 − 1 −  2 :

Proof. When k ∈ [0; bi], g(k) decreases, then max{g(k); ai} = g(k). Using Lemma 1, to solve

min EAC(Q; Li; k) under g(k) 6 Q and 0 6 k 6 bi is equivalent to solving min EAC(Q = g(k); Li; k), subject to 0 6 k 6 bi. If we assume ci=2D √_L i [A + C(Li)] + h Li 1 4 + 1 −  2 and di=2D √_L i [A + C(Li)] + h Li 1 −₄1 −1 − ₂

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then we can rewrite EAC(Q = g(k); Li; k) as

ci

1 + k2_{+ d}_i_k: ₍₃₃₎

Motivated by Eq. (33), we assume

h(k) = ci

1 + k2_{+ d}_i_k: ₍₃₄₎

We have h_{(k) = c}_i_k=√_{1 + k}2_{+ d}_i _{and h}_{(k) = −c}_i₌_{(1 + k}2₎3_{¿ 0. Consequently, we divide this}

into three scenarios:

(1) If di¿ 0, it shows h(k) ¿ 0 and h(k) increases so k∗= 0 and Q∗= g(0).

(2) If di¡ 0 and ci6 − di, then h(k) ¡ 0 and h(k) decreases so k∗= bi and Q∗= g(bi) = ai.

(3) If di¡ 0 and ci¿ − di, solving h(k) = 0 then k = −di=

c2

i − d2i. We have h(k) ¿ 0, so we

know k∗_{= min{−d}_i₌_c2

i − d2i; bi} and Q∗= g(k∗).

Combining Lemmas 2 and 3, we get the following:

Lemma 4. For Case 1 with 0 6 bi6

q−1_{− 1, we show that}

(1) If di¿ 0, then k∗= 0 and Q∗= g(0).

(2) If di¡ 0 and ci6 − di, then k∗= bi and Q∗= g(bi) = ai.

(3) If di¡ 0 and ci¿ − di, then k∗= min{−di=

c2

i − d2i; bi} and Q∗= g(k∗).

Lemma 5. For Case 2, when bi¡ 0, then k∗= 0 and Q∗= g(bi) = ai.

Proof. Eq. (32) shows that 1 ¡ 2ai =√Li.

Since √1 + k2 _{− k = 1=(}√_{1 + k}2 _{+ k) ¡ 1 we get that g(k) ¡ a}_i _{and max{g(k); a}_i_{} = a}_i _for k ∈ [0;q−1_{− 1]. Based on Lemma} ₁_{, Q}∗_{= a}_i _{for k ∈ [0;}_q−1_{− 1). Using Eq. (}₂₈_{), we derive}

k∗_{= 0.}

Lemma 6. For Case 3, when bi¿q−1− 1, we have the same results as Lemma 3.

Proof. Using Eq. (30), g(k) decreases. Recalling that g(bi) = ai, for k ∈ [0;q−1− 1], we

ob-tain g(k) ¿ ai. Based on Lemma 1, with k ∈ [0;

q−1_{− 1], than Q}∗_{(k) = g(k). Therefore, to solve}

min EAC(Q; Li; k) subject to g(k) 6 Q and 0 6 k 6

q−1_{− 1 is equivalent to solving}

min EAC(Q = g(k); Li; k), subject to 0 6 k 6

q−1_{− 1. Thus, for Case 3, we have the same results}

as Lemma 3.

Finally, putting Lemmas 4–6 together, we derive the main theorem. Theorem. We prove that

(1) If bi¡ 0, then k∗= 0 and Q∗= g(bi) = ai.

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(3) If 0 6 bi, di¡ 0 and ci6 − di, then k∗= bi and Q∗= g(bi) = ai.

(4) If 0 6 bi, di¡ 0 and ci¿ − di, then k∗= min{−di=

c2

i − d2i; bi} and Q∗= g(k∗).

4.3. Modi7ed algorithm for Distribution-free model

Based on our Theorem, a revised and simpli3ed algorithm is produced. (1) Step 1: Let ai= 2D h [A + C(Li)]; bi= √Li 4ai − ai √Li; g(k) = √Li 2 ( 1 + k2_{− k);} yi=2D √_L i[A + C(Li)]; zi= h Li 1 4 + 1 −  2 ; ci= yi+ zi and di= yi+ h Li− zi:

(2) Step 2: If bi¡ 0, then k∗= 0 and Q∗= ai.

(3) Step 3: If 0 6 bi and di¿ 0, then k∗= 0 and Q∗= g(0).

(4) Step 4: If 0 6 bi, di¡ 0 and ci6 − di, then k∗= bi and Q∗= ai.

(5) Step 5: If 0 6 bi, di¡ 0 and ci¿ − di, then k∗= min{−di=

c2

i − d2i; bi} and Q∗= g(k∗).

4.4. Numerical example

Example 2 from Ouyang and Wu [11] is used as our second example. The data are shown in

Example 1 except that the probability distribution for the lead time demand is free. They solved the problem, when q = 0:2 and took N = 200. Using their algorithm, constraint (16) does not hold for i = 0; 1. Hence, they did not have minimum solutions for i = 0; 1. We quote their results for i = 2; 3 in Table 4. Ouyang and Wu [11] accepted that the optimal order quantity Q∗ _{= 116, optimal lead}

time L∗_{= 4 weeks, the suitable safety factor k}∗_{= 1:89 and the minimum total expected annual cost}

EAC(Q∗_{; L}∗_{) = $2839:06.}

Based on our algorithm, the results are listed in Table 5. We have Q∗ _{= 143:1506, L}∗ _{= 4,}

k∗_{= 1:4766 and EAC(Q}∗_{; L}∗_{) = $2777:12.}

Comparing Tables 4 and 5, we know that for i = 0; 1, the algorithm by Ouyang and Wu [11] cannot 3nd any solutions. Likewise, for i = 2; 3, the algorithm by Ouyang and Wu [11] cannot 3nd the minimum solutions. Therefore, we may conclude that the Algorithm by Ouyang and Wu [11] cannot attain the optimal value.

Table 4

Summary of the optimal solution from Ouyang and Wu [11]

i Li C(Li) Qi ks(i) EAC(Qi; Li) √Li(1 + k2

s(i)− ks(i))=2Qi

2 4 22.4 116 1.89 2839.06 0.015

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Table 5

Summary of the optimal solution from the proposed method

i ai bi ci di −di=c2 i − d2i k∗ Q∗ EAC(Q∗; k∗) 0 109.5445 2.9293 6781 −6022 1.9309 1.9309 160.7542 3118.63 1 111.0675 2.4758 5931 −5157 1.7596 1.7596 151.0649 2930.66 2 115.5162 1.8962 4953 −4101 1.4766 1.4766 143.1506 2777.12 3 124.2739 1.4723 4424 −3417 1.2162 1.2162 144.8213 2809.53 5. Conclusion

Good inventory management is often the mark of a well-run organization. Inventory levels must be planned carefully to balance the costs and reasonable levels for good customer service. Controlling lead time properly and taking the optimal order quantity are very important in attaining the minimum total expected annual cost. In the above discussions we pointed out two questionable procedures in

the paper by Ouyang and Wu [11]. We o5ered two simpli3ed algorithms to replace the algorithms by

Ouyang and Wu. Their methods are too complicated and lose control of the computation procedure. Our re3ned algorithms are easy to use and mathematically sound.

References

[1] Silver EA, Peterson R. Decision systems for inventory management and production planning. New York: Wiley; 1985.

[2] Liao CJ, Shyu CH. An analytical determination of lead time with normal demand. International Journal of Operations Production Management 1991;11:72–8.

[3] Ben-Daya M, Raouf A. Inventory models involving lead time as decision variable. Journal of the Operational Research Society 1994;45:579–82.

[4] Moon I, Gallego G. Distribution free procedures for some inventory models. Journal of the Operational research Society 1994;45:651–8.

[5] Ouyang LY, Yeh NC, Wu KS. Mixture inventory models with backorders and lost sales for variable lead time. Journal of the Operational Research Society 1996;47:829–32.

[6] Moon I, Choi S. A note on lead time and distribution assumptions in continuous reviews inventory models. Computer and Operations Research 1998;25:1007–12.

[7] Lan SP, Chu P, Chung KJ, Wan WJ. A simple method to locate the optimal solution of inventory model with variable lead time. Computer and Operations Research 1999;26:599–605.

[8] Ouyang LY, Wu KS. A mixture distribution free procedure for mixed inventory model with variable lead time. International Journal of Production Economics 1998;56:511–6.

[9] Wu JK, Tsai HY. Mixture inventory model with back orders and lost sales for variable lead time demand with the mixture of normal distribution. International Journal of Systems Science 2001;32:259–68.

[10] Pan CH, Hsiao YC. Inventory models with back-order discounts and variable lead time. International Journal of Systems Science 2001;32:925–9.

[11] Ouyang LY, Wu KS. Mixture inventory model involving variable lead time with a service level constraint. Computers and Operations Research 1997;24:875–82.

[12] Ravindran A, Phillips DT, Solberg JJ. Operations research: principles and practice. New York: Wiley; 1987. [13] Gallego G, Moon I. The distribution free newsboy problem: review and extensions. Journal of the Operational