A Study on Iterative Algorithm for Stochastic Distribution Free Inventory Models

Abstract

We studied the iterative algorithm in Tung et al. (2010) to find out that their assertion is questionable. We derived two new relations between safe factor and order quantity so that we can execute the iterative algorithm proposed by Wu and Ouyang (2001). We have proved that three generated sequences indeed converge to provide a theoretical validation for their iterative procedure.

1. Introduction

Paknejad et al. [1] developed inventory models where lead time and defective rate are constant. Wu and Ouyang [2] generalized their results to include crashable lead time, and defective items are random variables that follow a probabilistic distribution with known mean and derivation. For the distribution free inventory model, Wu and Ouyang [2] used the minimax approach of Moon and Gallego [3] to study the minimum problem for an upper bound for the expected average cost. Wu and Ouyang [2] mentioned that the optimal order quantity and safety factor can be derived by iterative algorithm. Tung et al. [4] developed an analytical approach to prove that the optimal solutions for the order quantity and safety factor exist and are unique. Moreover, they claimed that iterative algorithm cannot be operated by the formulas in Wu and Ouyang [2].

Tung et al. [4] offered an analytical proof to prove the existence and uniqueness of the optimal solution for the inventory model of Wu and Ouyang [2]. During their derivations, they found two upper bounds and one lower bound, and then they used numerical examination to compare these two upper bounds to decide the minimum upper bound. They only studied the first derivative system such that they only considered the interior minimum. Moreover, they examined the iterative algorithm in Wu and Ouyang [2] to claim that the formulas in Wu and Ouyang [2] cannot be used to locate the optimal order quantity and safety factor. In this paper, we will show that the formulas in Wu and Ouyang [2] after two modifications are workable for the iterative algorithm.

2. Review of Previous Results

To be compatible with the results of Wu and Ouyang [2] and Tung et al. [4], we use the same notation and assumptions. We study the paper of Tung et al. [4]. They considered the stochastic inventory model of Wu and Ouyang [2] with crashable lead time, defective items, and minimax approach for distribution free demand with the following objective function: 𝐸𝐴𝐶^𝑢(𝑄,𝑘,𝐿)=𝐴𝐷𝑄𝑝+ℎ1−𝐸2𝑄𝑝𝐸𝑝1−𝐸+𝑄²−𝐸²𝑝𝑝+𝐸𝑝1−𝐸1−𝑝𝑝√1−𝐸+ℎ𝑘𝜎𝐿+1−𝛽2𝜎√𝐿1+𝑘²+𝐷−𝑘𝜋+𝜋₀1−𝛽𝑝𝜎√2𝑄1−𝐸𝐿1+𝑘²+()ℎ−𝑘𝑄−1^′𝐸𝑝1−𝑝𝑝+1−𝐸𝐷𝑣𝑝+𝐷1−𝐸𝑄𝑝×𝑐1−𝐸_𝑖𝐿_𝑖−1−𝐿+^𝑖−1_𝑗=1𝑐_𝑗𝑏_𝑗−𝑎_𝑗, (1) for 𝐿∈[𝐿_𝑖,𝐿_𝑖−1], where 𝐸𝐴𝐶^𝑢(𝑄,𝑘,𝐿) is a least upper bound of 𝐸𝐴𝐶(𝑄,𝑘,𝐿). Wu and Ouyang [2] derived that 𝐸𝐴𝐶^𝑢(𝑄,𝑘,𝐿) is a concave function of 𝐿 with 𝐿∈[𝐿_𝑖,𝐿_𝑖−1]; so, the minimum must occur at boundary point 𝐿_𝑖 or 𝐿_𝑖−1. To simplify the expression, we will use 𝐿 instead of 𝐿_𝑖 or 𝐿_𝑖−1 as Tung et al. [4]. For 𝐸𝐴𝐶^𝑢(𝑄,𝑘,𝐿), Wu and Ouyang [2] computed the first partial derivatives with respect to 𝑄 and 𝑘, separately.

Using (𝜕/𝜕𝑄)𝐸𝐴𝐶^𝑢(𝑄,𝑘,𝐿)=0 and (𝜕/𝜕𝑘)𝐸𝐴𝐶^𝑢(𝑄,𝑘,𝐿)=0, Wu and Ouyang [2] derived that 𝑄=2𝐷ℎ𝛿𝐴+𝑐_𝑖𝐿_𝑖−1−𝐿+^𝑖−1_𝑗=1𝑐_𝑗𝑏_𝑗−𝑎_𝑗+𝜋+𝜋₀1−𝛽2𝜎√𝐿1+𝑘²−𝑘^1/2, (2) where 𝛿=1−2𝐸(𝑝)+𝐸(𝑝²)+2(ℎ^′/ℎ)𝐸(𝑝(1−𝑝)), and 21+𝑘²1+𝑘²𝐷−𝑘=1−𝛽+𝜋+𝜋₀1−𝛽𝑝.ℎ𝑄1−𝐸 (3) Wu and Ouyang [2] claimed that the optimal solution can be obtained by the iterative method. In Tung et al. [4], they assumed that 𝑘₀=0 to plug it into (2) to derive 𝑄₁=315.62, and then they plugged it into (3) to derive a relation for 𝑘₁ as 1+𝑘²₁1+𝑘²₁−𝑘₁=11.9811. (4) Tung et al. [4] mentioned that, in (4), they only derived the value for 1+𝑘²₁/(1+𝑘²₁−𝑘₁) such that they can not use (3) to execute the iterative process. In this paper, we will show that (3) can be improved to operate the iterative process.

3. Our Revision for Tung et al. [4]

When running the iterative process, after we derive 𝑄_𝑛,  we then plug it into (3) to find a relation of 𝑘_𝑛 as 1+𝑘²_𝑛1+𝑘²_𝑛−𝑘_𝑛=12𝐷1−𝛽+𝜋+𝜋₀1−𝛽ℎ𝑄_𝑛1−𝐸𝑝. (5) To abstractly handle this problem, we assume that 12𝐷1−𝛽+𝜋+𝜋₀1−𝛽ℎ𝑄_𝑛1−𝐸𝑝=𝑑_𝑛. (6) Under the assumption of Tung et al. [4], it yields that 𝑄<√𝛼₁+𝛼₂<𝛼₄1+𝛽𝛼₃, (7) with 𝛼₃=ℎ(1−𝐸(𝑝)), and 𝛼₄=𝐷(𝜋+𝜋₀(1−𝛽)) to imply that 𝑑_𝑛=12𝛼1−𝛽+⁴𝛼₃𝑄_𝑛>121−𝛽+1+𝛽=1. (8) From (5) and (6), and with 𝑑_𝑛>1, we obtain that 1+𝑘²_𝑛𝑑_𝑛−1=𝑑_𝑛𝑘_𝑛, (9) and then squaring both sides, we find that 𝑘_𝑛=𝑑_𝑛−12𝑑_𝑛−1 (10) to show that there is a unique 𝑘_𝑛 that can be derived by (3). The assertion of Tung et al. [4] that (3) can not be used to execute the iterative process can be improved.

4. The Proof for the Convergence of the Proposed Three Iterated Sequences

In this section, we will prove that the two iterative sequences proposed for (2) and (10) indeed converge.

We combine (6) and (10) to derive that 𝑘_𝑛=𝐷𝜋+𝜋₀1−𝛽/ℎ𝑄_𝑛1−𝐸𝑝−1+𝛽2𝐷𝜋+𝜋₀1−𝛽/ℎ𝑄_𝑛1−𝐸𝑝−𝛽^1/2. (11) For later purpose, we rewrite the iterative process based on (2) as follows: 𝑄_𝑛+1=𝐵₁+𝐵₂1+𝑘²_𝑛−𝑘_𝑛^1/2, (12) where 𝐵₁=[(2𝐷/ℎ𝛿){𝐴+𝑐_𝑖(𝐿_𝑖−1−𝐿)+∑^𝑖−1_𝑗=1𝑐_𝑗(𝑏_𝑗−𝑎_𝑗)}] and 𝐵₂=[(2𝐷/ℎ𝛿){((𝜋+𝜋₀(1−𝛽))/2)𝜎√𝐿}] with 𝐵₁>0 and 𝐵₂>0.

Based on (12), we derive that 𝑄²_𝑛+2−𝑄²_𝑛+1=𝐵₂𝑘_𝑛−𝑘_𝑛+1+1+𝑘²_𝑛+1−1+𝑘²_𝑛=𝐵₂𝑘_𝑛−𝑘_𝑛+1+𝑘²_𝑛+1−𝑘²_𝑛1+𝑘²_𝑛+1+1+𝑘²_𝑛=𝐵₂𝑘_𝑛+1−𝑘_𝑛𝑘_𝑛+1+𝑘_𝑛1+𝑘²_𝑛+1+1+𝑘²_𝑛−1. (13) We know that √1+𝑥²>𝑥 such that (𝑘_𝑛+1+𝑘_𝑛)/(1+𝑘²_𝑛+1+1+𝑘²_𝑛)−1<0. Hence, we obtain the following lemma.

Lemma 1. If 𝑘_𝑛+1>𝑘_𝑛, then 𝑄_𝑛+2<𝑄_𝑛+1.
One recalls (10); then one evaluates that 𝑘²_𝑛+1−𝑘²_𝑛=𝑑_𝑛+1−1²2𝑑_𝑛+1−𝑑−1^𝑛−1²2𝑑_𝑛=𝑑−1_2𝑛+1−2𝑑_𝑛+1+12𝑑_𝑛+1−𝑑−1^2𝑛−2𝑑_𝑛+12𝑑_𝑛=𝑑−1_2𝑛+12𝑑_𝑛+1−𝑑−12𝑑^2𝑛_𝑛.−1 (14)

We need the next lemma for our future proof.

Lemma 2. If 𝑑_𝑛+1>𝑑_𝑛>1, then 𝑘_𝑛+1>𝑘_𝑛.
After cross-multiplication of (14), one finds that 𝑘²_𝑛+1>𝑘²_𝑛 if and only if 𝑑_𝑛2𝑑²_𝑛+1+𝑑_𝑛>𝑑_𝑛+1𝑑_𝑛+1+2𝑑²_𝑛; (15) that is, 2𝑑²_𝑛+1+𝑑_𝑛𝑑_𝑛+1>2𝑑²_𝑛+𝑑_𝑛+1𝑑_𝑛. (16) One rewrites (16) as follows: 2𝑑_𝑛+1+𝑑_𝑛𝑑_𝑛+1>2𝑑_𝑛+𝑑_𝑛+1𝑑_𝑛 (17) that is equivalent to 2𝑑_𝑛+1−𝑑_𝑛>𝑑_𝑛+1𝑑_𝑛−𝑑_𝑛𝑑_𝑛+1=𝑑²_𝑛+1−𝑑²_𝑛𝑑_𝑛𝑑_𝑛+1. (18) One can cancel out the common factor 𝑑_𝑛+1−𝑑_𝑛>0 from the previous inequality and still preserve the same direction of the inequality sign.
Consequently, one tries to show that 2𝑑_𝑛𝑑_𝑛+1>𝑑_𝑛+𝑑_𝑛+1. (19) One knows that 2𝑑_𝑛𝑑_𝑛+1−𝑑_𝑛−𝑑_𝑛+1=𝑑_𝑛(𝑑_𝑛+1−1)+𝑑_𝑛+1(𝑑_𝑛−1)>0, owing to the condition of 𝑑_𝑛+1>𝑑_𝑛>1.

From the iterative procedure, 𝑘₀=0, which is plugged it into (2) to derive 𝑄₁>0, and then we plug 𝑄₁ into (11) to obtain that 𝑘₁>0.

Owing to 𝑘₁>𝑘₀, we apply Lemma 1 to obtain 𝑄₂<𝑄₁.

We may simplify (6) as follows: 𝑑_𝑛=𝐵₃+𝐵₄𝑄_𝑛, (20) where 𝐵₃=(1−𝛽)/2 and 𝐵₄=(𝐷(𝜋+𝜋₀(1−𝛽))/2ℎ(1−𝐸(𝑝))) with 𝐵₃≥0 and 𝐵₄>0.

Using 𝑄₂<𝑄₁, by (20), and we have 𝑑₂>𝑑₁, applying Lemma 2, we know that 𝑘₂>𝑘₁.

Repeating the previous argument, we derived that (𝑘_𝑛) is an increasing sequence and (𝑄_𝑛) is a decreasing sequence and bounded below by zero such that (𝑄_𝑛) must converge, and then sequence (𝑑_𝑛) converges. Finally, the sequence (𝑘_𝑛) converges too.

5. Numerical Example to Support Our Proof

We will use the same numerical examples in Wu and Ouyang [2] and Tung et al. [4] to illustrate our findings in Section 5. For the precious space of this journal, please refer to Wu and Ouyang [2] for the detailed data, and we refer to the findings of Tung et al. [4] to consider the representative case of 𝛽=0.5 and 𝐿₃=3 to demonstrate the convergence of the proposed two sequences (𝑘_𝑖)_𝑖≥0 and (𝑄_𝑖)_𝑖≥1 with our auxiliary sequence (𝑑_𝑖)_𝑖≥1. We list the numerical results in Table 1.

Table 1 

For 𝛽=0.5 and 𝐿₃=3, the convergence of proposed three sequences.

If we observe Table 1, then the sequence (𝑘_𝑖)_𝑖≥0 increases to its least upper bound and the sequence (𝑄_𝑖)_𝑖≥1 decreases to its greatest lower bound as described by our proposed analysis. Our finding is consistent with Tung et al. [4] in which they mentioned that for 𝛽=0.5 and 𝐿₃=3, 𝑄^∗=171.14.