# A Study on Iterative Algorithm for Stochastic Distribution Free Inventory Models

Authors: Jennifer Lin
Journal: Abstract and Applied Analysis
 (2013)

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## 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.


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## Body

## 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:
(1)EACu(Q,k,L)=ADQ(1-E(p))+h2{E(p2)-E2(p)1-E(p)Q(1-E(p))+QE(p2)-E2(p)1-E(p)+E(p(1-p))1-E(p)}+h{kσL+1-β2σL(1+k2-k)}+D(π+π0(1-β))2Q(1-E(p))σL(1+k2-k)+(Q-1)h′E(p(1-p))1-E(p)+Dv1-E(p)+DQ(1-E(p))×(ci(Li-1-L)+∑j=1i-1cj(bj-aj)),
for L∈[Li,Li-1], where EACu(Q,k,L) is a least upper bound of EAC(Q,k,L). Wu and Ouyang [2] derived that EACu(Q,k,L) is a concave function of L with L∈[Li,Li-1]; so, the minimum must occur at boundary point Li or Li-1. To simplify the expression, we will use L instead of Li or Li-1 as Tung et al. [4]. For EACu(Q,k,L), Wu and Ouyang [2] computed the first partial derivatives with respect to Q and k, separately.Using(∂/∂Q)EACu(Q,k,L)=0 and (∂/∂k)EACu(Q,k,L)=0, Wu and Ouyang [2] derived that
(2)Q=[2Dhδ{A+ci(Li-1-L)+∑j=1i-1cj(bj-aj)+π+π0(1-β)2σL(1+k2-k)∑j=1i-1cj(bj-aj)}]1/2,
where δ=1-2E(p)+E(p2)+2(h′/h)E(p(1-p)), and
(3)21+k21+k2-k=1-β+D(π+π0(1-β))hQ(1-E(p)).
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
(4)1+k121+k12-k1=11.9811.
Tung et al. [4] mentioned that, in (4), they only derived the value for 1+k12/(1+k12-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 deriveQn,  we then plug it into (3) to find a relation of kn as
(5)1+kn21+kn2-kn=12(1-β+D(π+π0(1-β))hQn(1-E(p))).
To abstractly handle this problem, we assume that
(6)12(1-β+D(π+π0(1-β))hQn(1-E(p)))=dn.
Under the assumption of Tung et al. [4], it yields that
(7)Q<α1+α2<α4(1+β)α3,
with α3=h(1-E(p)), and α4=D(π+π0(1-β)) to imply that
(8)dn=12(1-β+α4α3Qn)>12(1-β+1+β)=1.
From (5) and (6), and with dn>1, we obtain that
(9)1+kn2(dn-1)=dnkn,
and then squaring both sides, we find that
(10)kn=dn-12dn-1
to show that there is a unique kn 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
(11)kn=D(π+π0(1-β))/hQn(1-E(p))-(1+β)2(D(π+π0(1-β))/hQn(1-E(p))-β)1/2.
For later purpose, we rewrite the iterative process based on (2) as follows:
(12)Qn+1=(B1+B2(1+kn2-kn))1/2,
where B1=[(2D/hδ){A+ci(Li-1-L)+∑j=1i-1cj(bj-aj)}] and B2=[(2D/hδ){((π+π0(1-β))/2)σL}] with B1>0 and B2>0.Based on (12), we derive that
(13)Qn+22-Qn+12=B2[(kn-kn+1)+1+kn+12-1+kn2]=B2[(kn-kn+1)+kn+12-kn21+kn+12+1+kn2]=B2(kn+1-kn)[kn+1+kn1+kn+12+1+kn2-1].
We know that 1+x2>x such that (kn+1+kn)/(1+kn+12+1+kn2)-1<0. Hence, we obtain the following lemma.Lemma 1.
Ifkn+1>kn, then Qn+2<Qn+1.
One recalls (10); then one evaluates that
(14)kn+12-kn2=(dn+1-1)22dn+1-1-(dn-1)22dn-1=dn+12-2dn+1+12dn+1-1-dn2-2dn+12dn-1=dn+122dn+1-1-dn22dn-1.We need the next lemma for our future proof.Lemma 2.
Ifdn+1>dn>1, then kn+1>kn.
After cross-multiplication of (14), one finds that kn+12>kn2 if and only if
(15)dn(2dn+12+dn)>dn+1(dn+1+2dn2);
that is,
(16)2dn+12+dndn+1>2dn2+dn+1dn.
One rewrites (16) as follows:
(17)2dn+1+dndn+1>2dn+dn+1dn
that is equivalent to
(18)2(dn+1-dn)>dn+1dn-dndn+1=dn+12-dn2dndn+1.
One can cancel out the common factor dn+1-dn>0 from the previous inequality and still preserve the same direction of the inequality sign.
Consequently, one tries to show that(19)2dndn+1>dn+dn+1.
One knows that 2dndn+1-dn-dn+1=dn(dn+1-1)+dn+1(dn-1)>0, owing to the condition of dn+1>dn>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 tok1>k0, we apply Lemma 1 to obtain Q2<Q1.We may simplify (6) as follows:
(20)dn=B3+B4Qn,
where B3=(1-β)/2 and B4=(D(π+π0(1-β))/2h(1-E(p))) with B3≥0 and B4>0.UsingQ2<Q1, by (20), and we have d2>d1, applying Lemma 2, we know that k2>k1.Repeating the previous argument, we derived that(kn) is an increasing sequence and (Qn) is a decreasing sequence and bounded below by zero such that (Qn) must converge, and then sequence (dn) converges. Finally, the sequence (kn) 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 (ki)i≥0 and (Qi)i≥1 with our auxiliary sequence (di)i≥1. We list the numerical results in Table 1.Table 1
Forβ=0.5 and L3=3, the convergence of proposed three sequences.
i
k
i
Q
i
d
i
0
0
1
1.925382
271.444019
8.884377
2
2.433960
179.201796
13.328831
3
2.490503
171.872873
13.886532
4
2.495817
171.206597
13.939601
5
2.496308
171.145237
13.944509
6
2.496353
171.139579
13.944962
7
2.496357
171.139060
13.945003
8
2.496357
171.139014
13.945007
9
2.496357
171.139014
13.945007If we observe Table1, then the sequence (ki)i≥0 increases to its least upper bound and the sequence (Qi)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.



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*Source: 251694-2013-03-19.xml*