# Novel Solution Method for Inventory Models with Stochastic Demand and Defective Units

Authors: Xu-Ren Luo; Chun-Hsiao Chu; Henry C. J. Chao
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

This paper is a response to two papers. We improve the lengthy proof for the first paper by an elegant verification. For the second paper, we point out the three-sequence approach will result in different convergent rates such that when the other two sequences are converged, the ordering quantity sequence may still not converge to the optimal solution. We construct a novel iterative method to simplify the previous approach proposed by the three-sequence approach for the optimal solution. By the same numerical examples of three published papers, we demonstrate that we can control our findings to converge more accurately than previous results. Moreover, we show that there are three distinct features of our proposed approach. (i) It converges to the desired solution within the preassigned threshold value. (ii) We estimate the convergent ratio. (iii) We find the dominant factors for our proposed convergent sequence.


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## Body

## 1. Introduction

For the past several decades, many important and interesting inventory models had been developed by researchers. For example, Rosenblatt and Lee [1] examined the relationship between imperfect quality of goods and lot size. Paknejad et al. [2] studied the connection between defective items and lead time for inventory models. Huang [3], Ouyang et al. [4], and Wu et al. [5] worked on inventory systems with the imperfect quality of goods or stochastic property of lead time. Wee et al. [6] focused on the interaction between the imperfect quality and shortages or nonshortages in his model. Chung [7] derived the necessary and sufficient constraints for the existence of the optimal solution for a single-vendor single-buyer integrated production-inventory problem under the condition of an unreliable process. Chung and Wee [8] constructed an integrated inventory model with warranty policy, tactical price, imperfect manufacture process, and inspection procedure. There are several researchers to explore the shortages during the lead time, for example, Das [9], Deng et al. [10], Ben-Daya and Raouf [11], Ouyang and Wu [12], and Wadhwa et al. [13]. Recently, there is a trend to improve previously published papers, for example, Deng [14], Deng et al. [15, 16], Lan et al. [17], Chang et al. [18], Jung et al. [19], and Tang et al. [20], that had provided useful analytical works to revise some questionable results in previous papers.We will follow this trend to examine inventory models with stochastic demand, crashable lead time, and defective items that were developed by Wu and Ouyang [21], but they did not show the uniqueness of the optimal solution for their minimum cost problem. There are thirty-five papers that had cited Wu and Ouyang [21] in their references. We list them in the following: Mostard et al. [22], Wu et al. [5], Ho [23], Annadurai and Uthayakumar [24], Tung et al. [25], Çapar et al. [26], Drezner et al. [27], Lin et al. [28], Wang [29], Zheng and Liu [30], Gholami-Qadikolaei et al. [31], Ma and Qiu [32], Ouyang et al. [33], Lin [34–37], Parvathi et al. [37], Lin [38], Jindal and Solanki [39], Lin [40], Lin et al. [41], Vishkaei et al. [42], Wang et al. [43], Madhusoodhanan et al. [44, 45], Chang et al. [46], Jindal and Solanki [47], Madhusoodhanan et al. [48], Chang et al. [49], Sarkar and Mahapatra [50], Chuang et al. [51], Hu et al. [52], Maleki Vishkaei et al. [53], Noh et al. [54], and Singha et al. [55]. Most of them only mentioned Wu and Ouyang [21] in their introduction but did not offer further study for Wu and Ouyang [21].On the contrary, Tung et al. [25] and Lin [38] provided a detailed examination for Wu and Ouyang [21] concerning the solution procedure of the iterative sequence whether or not they will converge to the optimal solution. Three conditions were obtained by Tung et al. [25] which are two upper bounds and one lower bound. They used a numerical examination to compare two upper bounds to decide the smaller one. Under their conditions, they merged the system of two first partial derivatives into a function of order quantity and verify the uniqueness of the optimal solution. Moreover, Tung et al. [25] criticized the iterative approach in Wu and Ouyang [21] to point out that the original development suggested by Wu and Ouyang [21] cannot be executed. And then Tung et al. [25] provided an example to illustrate that the results presented in Wu and Ouyang [21] did not match the optimal solution derived by Tung et al. [25] through a revised iterative approach proposed by Tung et al. [25]. Three papers had cited Tung et al. [25] in their references: Yang [56], Lin [38], and Hu et al. [50]. Two of them, Yang [56] and Hu et al. [50], mentioned Tung et al. [25] in their introduction, but did not provide any discussion for the solution approach proposed by Tung et al. [25]. Only one paper, Lin [38], constructed a new system that contains three convergent sequences to generate the desired sequence that converges to the optimal ordering quantity. Lin [38] further considered the iterative approach of Tung et al. [25] to point out that they did not verify the convergence of the iterative approach in their paper. She constructed three sequences to prove the convergence of them. Only one paper, Hu et al. [50], had cited Lin [38] in their reference. However, Hu et al. [50] did not provide any comments for the three-sequence approach proposed by Lin [38]. In this paper, firstly we will provide a simplified verification to replace the complicated proof proposed by Tung et al. [25]. Secondly, we develop a novel approach to generate a new sequence and then prove its convergence. Moreover, by the same numerical example of Tung et al. [25] and Lin [38], we show that our sequence converges better than that of Lin [38] so our findings are simple and yield better convergent result.Several related papers concerning inventory models with defective items are worthy to mention: Khanna et al. [57], Gautam et al. [58], Gautam and Khanna [59], and Khanna et al. [60]. The organization of this paper is explained as follows. Section 2 provides notation and assumptions for the examined inventory model. Section 3 describes the results of Wu and Ouyang [21]. Section 4 reviews Tung et al. [25]. Section 5 discusses the findings of Lin [38]. Section 6 provides our new proof to verify that the solution for the first partial system is unique and is the optimal solution. Section 7 presents a numerical example to illustrate that our approach can obtain an optimal solution. Section 8 shows the three distinct features of our approach which is the key contribution of this paper. Section 9 concludes our paper.


## 2. Notation and Assumptions

We adopt the same notation and assumptions as Wu and Ouyang [21], Tung et al. [25], and Lin [38] and list them as follows:D: expected demand per yearA: setup cost per setuph: nondefective holding cost per unit per yearh′: defective holding cost per unit per year, h′<hπ: shortage cost per unit shortπ0: marginal profit per unitv: unit inspection costβ: the fraction of the demand during the stock out period will be backordered, 0≤β≤1p: the defective rate in an order lot (independent of lot size), 0≤p<1, and it is a random variablegp: the probability density function (p.d.f.) of pf: the proportion of order quantity inspectedQ: lot size (order quantity), a decision variableL: length of the lead time, a decision variableX: the lead time demand with finite mean μL, and standard deviation σL for lead time Lr: reorder point, r=μL+kσL, where k is the safety factor that is a decision variableAn arrival order may contain some defective items. We assume that the number of defective items in an arriving order of sizeQ is a binomial random variable with parameters Q and p, where p0≤p≤1 represents the defective rate in an order lot. Upon the arrival of an order, all the items are inspected and defective items in each lot will be returned to the vendor at the time of delivery of the next lot.The inventory is continuously reviewed. Replenishments are made whenever the inventory level (based on the number of nondefective items) falls to the reorder pointr.The lead timeL has n mutually independent components. The ith component has a minimum duration a1 and normal duration bi, and a crashing cost per unit time ci. Furthermore, for convenience, we rearrange ci such that c1≤c2≤⋯≤cn.The components of lead time are crashed one at a time starting with component 1 (because it has the minimum unit crashing cost) and then component 2, etc.If we letL0=∑j=1nbj and Li=∑i+1nbj+∑j=1iaj, the lead time crashing cost RL per cycle for a given L∈Li,Li−1 is given by RL=ciLi−1−L+∑j=1i−1cjbj−aj.


## 3. Recap of Wu and Ouyang [21]

Wu and Ouyang [21] constructed inventory models with stochastic demand, crashable lead time, and defective items. They applied the minimax distribution-free approach of Moon and Gallego [61] to develop the next minimum problem for variables: order quantity, Q, safety factor, k, and lead time L:(1)EACuQ,k,L=ADQ1−Ep+h2Q1−Ep+QEp2−E2p1−Ep+Ep1−p1−Ep+hkσL+1−β2σL1+k2−k+Dπ+π01−β2Q1−EpσL1+k2−k+Q−1h′Ep1−p1−Ep+Dv1−Ep+DQ1−EpciLi−1−L+∑j=1i−1cjbj−aj,for L∈Li,Li−1, with i=1,…,n.Wu and Ouyang [21] proved that EACuQ,k,L concaves down for L∈Li,Li−1 such that the minimum will happen on the two boundary points Li or Li−1. Consequently, under the restriction of L=Li or L=Li−1, they derived the first partial derivatives for Q and k.To simplify the expression, Wu and Ouyang [21] used L to represent L=Li or L=Li−1 in the following derivations.Using∂/∂QEACuQ,k,L=0 and ∂/∂kEACuQ,k,L=0, Wu and Ouyang [21] obtained that(2)Q=2DhδA+ciLi−1−L+∑j=1i−1cjbj−aj+π+π01−β2σL1+k2−k1/2,(3)21+k21+k2−k=1−β+Dπ+π01−βhQ1−Ep,where(4)δ=1−2Ep+Ep2+2h′hEp1−p.Nevertheless, the expressions in equations (2) and (3) and the variables Q and k are mixed. Hence, Wu and Ouyang [21] cannot verify the uniqueness of the optimal solution. On the contrary, Wu and Ouyang [21] mentioned that they will apply equations (2) and (3) to generate two sequences which will converge to the optimal solution as follows:(5)Qn+1=2DhδA+ciLi−1−L+∑j=1i−1cjbj−aj+π+π01−β2σL1+kn2−kn1/2,(6)21+kn+121+kn+12−kn+1=1−β+Dπ+π01−βhQn1−Ep.Whenβ=0 and L3=3, Wu and Ouyang [21] informed us that Q∗=183 without offering a detailed description for their solution procedure.However, Tung et al. [25] pointed out that the results of Wu and Ouyang [21] did not match the reproduction examined by Tung et al. [25] which will be explained after Tables 1 and 2 of this paper.Table 1
Reproduction Table2 of Tung et al. [25]. Comparisons between Wu and Ouyang [21] and Tung et al. [25].
Wu and Ouyang [21]Tung et al. [25]βQEACuQ,k,LQEACuQ,k,L01835697.95179.745697.41Table 2
Reproduction Table4 of Tung et al. [25]. The revised iterative method of Wu and Ouyang [21] and executed by Tung et al. [25].
n=1n=2n=3n=4n=5n=6Qn315.62194.45181.25179.90179.75179.74kn2.29163.02483.13863.15103.15233.1524


## 4. Review of Tung et al. [25]

Based on the results of equations (2) and (3), Tung et al. [25] obtained that(7)Q2=α1+α21+k2−k,(8)1+k2−k1+k2=2α3Q1−βα3Q+α4,where(9)α1=2DhδA+ciLi−1−L+∑j=1i−1cjbj−aj,(10)α2=Dhδπ+π01−βσL,(11)α3=h1−Ep,(12)α4=Dπ+π01−β,that satisfy αj>0, for j=1,…,4.Based on equation (8), Tung et al. [25] obtained that(13)1−1+k2−k1+k2=1−2α3Q1−βα3Q+α4,and then Tung et al. [25] simplified equation (13) to find that(14)k1+k2=α4−1+βα3Q1−βα3Q+α4.Tung et al. [25] observed equation (14), under the condition that k is the safety factor, with the condition k≥0. The solution for the first partial derivative system is an interior solution and then the restriction is modified from k≥0 to k>0.Hence, Tung et al. [25] derived an upper bound for the ordering quantity, Q, as(15)α4>1+βα3Q.We follow the solution approach of Tung et al. [25] to flip over equation (14) to derive that(16)1+k2k=1−βα3Q+α4α4−1+βα3Q,and then we take square on both sides of equation (16) to yield that(17)1+k2k2=1−βα3Q+α42α4−1+βα3Q2.We take “minutes one” from both sides of equation (17) to imply that(18)1+k2k2−1=1−βα3Q+α42α4−1+βα3Q2−1,and then we simplify equation (18) to find that(19)1k2=4α3Qα4−βα3Qα4−1+βα3Q2.After lengthy derivation, Tung et al. [25] obtained a relation to express k as a function in the variable Q as(20)k=α4−1+βα3Q2α3Qα4−βα3Q.Based on equation (20), Tung et al. [25] further derived that(21)1+k2=1+α4−1+βα3Q24α3Qα4−βα3Q1/2=α4+1−βα3Q2α3Qα4−βα3Q.Plugging equations (20) and (21) into equation (7), then(22)Q2=α1+α2α3Qα4−βα3Q.Based on equation (22), Tung et al. [25] obtained a lower bound of Q as(23)Q>α1.If we recall the inequality of equation (15), then(24)α4−βα3Q>α3Q.Then, it follows that(25)1>α3Qα4−βα3Q,that is,(26)1>α3Qα4−βα3Q.Based on equation (26), Tung et al. [25] found another upper bound of Q as(27)Q<α1+α2.Tung et al. [25] applied a numerical method to use data from Wu and Ouyang [21] to compare two upper bounds: α4/1+βα3 by equation (15) and α1+α2 by equation (27). We cite Table 3 of Tung et al. [25] in the following.Table 3
Reproduction Table 3 of Tung et al. [25]. The ratio of α4/1+βα3α1+α2.
L=L0L=L1L=L2L=L3β=018.85520.09021.85622.962β=0.59.67510.26411.07311.513β=0.86.2116.5526.9947.180β=14.1744.3694.6004.648Based on the results of Table3, Tung et al. [25] observed that(28)α41+βα3>α1+α2.Tung et al. [25] rewrote the rational function of equation (22) as follows:(29)Q2−α1α4−βα3Q−α2α3Q=0,under the conditions(30)α41+βα3>α1+α2>Q>α1,with conditions of equations (9)–(12).Based on equation (29), Tung et al. [25] assumed an auxiliary function:(31)fQ=Q2−α1α4−βα3Q−α2α3Q.Tung et al. [25] found that(32)f′Q=2Qα4−βα3Q+Q2−α1−βα32α4−βα3Q−α2α32Q,f″Q=2α4−βα3Qα4−2βα3Q+α2α3α4−βα33−Q3βα32Q2−α14α4−βα3QQ3.Tung et al. [25] obtained excellent derivations to claim that(33)α4≥1+βα3Q≥2βα3Q,(34)α3α4−βα3Q3≥α3α3Q3=Q3α32≥Q3βα32.Applying equations (33) and (34) with the inequality of equation (27), Tung et al. [25] proved the convexity property of fQ, that is,(35)f″Q>0.They further checked that(36)fα1=−α2α1α3<0,fα1+α2=α2α4−1+βα3α1+α2α4−βα3α1+α2+α3α1+α21/4.Based on the results of Table3, Tung et al. [25] obtained that(37)fα1+α2>0.Hence,fQ has a unique root for α1+α2>Q>α1 which satisfies the first-order partial derivatives of ∂/∂QEACuQ,k,L=0 and ∂/∂kEACuQ,k,L=0, so it is the optimal solution of EACuQ,k,L. On the contrary, Tung et al. [25] pointed out that the iterative approach proposed by Wu and Ouyang [21] was not consistent with their findings in the numerical examples.Tung et al. [25] mentioned that given k0=0 in equation (5), then Q1 is obtained by the following formula:(38)Q1=2DhδA+ciLi−1−L+∑j=1i−1cjbj−aj+π+π01−β2σL1+k02−k01/2.However, pluggingQ1 into equation (8), researchers only find that(39)21+k121+k12−k1=1−β+Dπ+π01−βhQ11−Ep,which is a relation containing k1. However, researchers cannot directly derive k1.Therefore, Tung et al. [25] applied their finding of equation (20), together with equation (7) to generate the following approach:(40)Qn+1=α1+α21+kn2−kn,kn+1=α4−1+βα3Qn2α3Qnα4−βα3Qn.For completeness, we reproduce Tables2 and 4 of Tung et al. [25] to show that the findings of Wu and Ouyang [21] did not satisfy their claim.From Table2, Tung et al. [25] mentioned that if researchers execute the iterative method proposed by Wu and Ouyang [21], then the optimal ordering quantity, Q∗=179.74. However, Wu and Ouyang [21] claimed that Q∗=183. Hence, Tung et al. [25] asserted that the iterative method proposed by Wu and Ouyang [21], which is too complicated, cannot execute their method.In Section6, we will provide one simplification and another alternative approach to replace the lengthy solution procedure proposed by Tung et al. [25].


## 5. Discussion of Lin [38]

Lin [38] further examined the sequence convergent problem discussed by Wu and Ouyang [21] and Tung et al. [25]. In the following, we provide our comments for Lin [38]. Lin [38] admitted that Tung et al. [25] and improved the iterative approach of Wu and Ouyang [21], but Tung et al. [25] did not prove the convergence of their new approach. Hence, Lin [38] developed a new derivation that consists of three sequences as follows:(41)Qn+1=B1+B21+kn2−kn,with(42)B1=2DhδA+ciLi−1−L+∑j=1i−1cjbj−aj,B2=2Dhδπ+π01−β2σL,dn+1=121−β+Dπ+π01−βhQn+11−Ep,kn+1=dn+1−12dn+1−1.Moreover, Lin [38] proved the convergence of her approach. For interested readers, please refer to Lin [38] for her detailed proof. In the next section, we will develop our method which will be shown superior to that of Lin [38] which will be demonstrated by the example in Section 7. Therefore, to save the precious space of this journal, we did not further discuss the detailed proof proposed by Lin [38] concerning the convergence of her three convergent sequence approach.For later discussion, we cite results derived by Lin [38] in the following.Table 4
Reproduction from Lin [38]. For β=0.5 and L3=3, the three sequences generated by Lin [38].
ikiQidi00——11.925382271.4440198.88437722.433960179.20179613.32883132.490503171.87287313.88653242.495817171.20659713.93960152.496308171.14523713.94450962.496353171.13957913.94496272.496357171.13906013.94500382.496357171.13901413.94500792.496357171.13901413.945007


## 6. Our Improved Approach

Now, we begin to develop our verification for the uniqueness of the optimal ordering quantity to simplify the lengthy approach proposed by Tung et al. [25]. We observe (8) to imply that(43)1+k2−k21+k2=α3Q1−βα3Q+α4.IfA/B=C/D, then A/−A+B=C/−C+D. We apply this rule to equation (43) to derive that(44)1+k2−k1+k2+k=α3Qα4−βα3Q.We know that(45)1+k2−k=11+k2+k.Using equation (54), we find that(46)1+k2−k2=α3Qα4−βα3Q.If we plug our findings of equation (55) into equation (7), then we derive the results of equation (22) as proposed by Tung et al. [25]. Therefore, we demonstrate a simplified approach to find the same result as Tung et al. [25] without deriving 1+k2 of equation (21) or finding k of equation (20).After simplifying the approach of Tung et al. [25], we will show our new solution method. Based on equation (8), we express Q as a function in the variable k; then,(47)Q=1+k2−kα4α31+β1+k2+1−βk.We plug the findings of equation (47) into (7) to derive that(48)1+k2−kα4α31+β1+k2+1−βk2=α1+α21+k2−k.We rewrite equation (48) as(49)1+k2−kα4α3β1+k2−k+1/1+k2−k2=α1+α21+k2−k.Therefore, our goal becomes to prove that there is a unique point that satisfies equation (49).We observe equation (49) to find if we assume a new variable, denoted as s, where(50)s=1+k2−k.With the new variable, we rewrite equation (49) as follows:(51)α32βs2+12α1+α2s=α42s4.Owing to equation (51), we assume an auxiliary function, denoted as gs, where(52)gs=α32βs2+12α1+α2s−α42s4.We derive that(53)g′s=4βα32sβs2+1α1+α2s+α2α32βs2+12−4α42s3,(54)g″s=8β2α32s2α1+α2s+4βα32βs2+1α1+3α2s−12α42s2,(55)g′″s=24β2α32sα1+2α2s+12βα2α32βs2+1−24α42s,(56)g4s=24β2α32α1+5α2s−24α42.For the later discussion, we will begin to show that(57)3α1>α2,(58)α1>0,(59)α2>0.We recallα1=2D/hδA+ciLi−1−L+∑j=1i−1cjbj−aj of equation (9), and from equation (10), we know that α2=D/hδπ+π01−βσL such that two inequalities of equations (58) and (59) are valid.We will recall examples from Wu and Ouyang [23], Tung et al. [20], and Lin [13]. Hence, we adopt the same data as them with D=600 units/year, A=200 per order, h=20, h′=12, v=1.6, π=50, π0=150, and σ=7; the lead time has three components such that L0=8, RL0=0, L1=6, RL1=5.6, L2=4, RL2=22.4, L3=3, and RL2=57.4; the defective rate p has a Beta distribution, with the probability density function gp=41−p3 to imply that Ep=1/5; and Ep2=1/15, the fraction of the backordered demand, β, has four values, 0,0.5,0.8, and 1. From the above data, we know that the lower bound α1=136.683 and the upper bound α1+α2=271.444.Based onα1=18.68×103 and α2=54.99×103, we derive that(60)3α1−α2=1.05×103>0.Hence, the inequality of equation (57) is valid under data from Wu and Ouyang [21], Tung et al. [25], and Lin [38].We recall thats=1+k2−k of equation (50), with k>0, and then we derive that(61)0<s<1.Based on equations (58), (59), and (61), we find that(62)α2s<α1+α2.Based on equation (28), we imply that(63)α42>1+β2α32α1+α2,and then owing to 0≤β≤1, we know that(64)2β≤1+β.We derive that(65)g0=α1α32>0,(66)g1=α321+β2α1+α2−α42<0,owing to equation (28). Next, we find that(67)g′0=α2α32>0,(68)g′1=4βα321+βα1+α2+α2α321+β2−4α42=22β1+βα32α1+α2−α42+α2α321+β2−α42−α42<0.And then, we compute that(69)g″0=4βα1α32>0,(70)g′″1=8β2α32α1+α2+4βα321+βα1+3α2−12α42=22β2α32α1+α2−α42+22β1+βα32α1+3α2−3α42−4α42<0.Finally, we find that(71)g′″0=12βα2α32>0,(72)g‴1=24β2α32α1+2α2+12βα2α321+β−24α42=62β2α32α1+2α2−2α42+62β1+βα2α32−α42−6α42<0.From the inequality3α1>α2 of equation (57), we know that(73)4α1+α2>α1+5α2.Hence, we derive that(74)α42>4β2α32α1+α2>β2α32α1+5α2.Based on equation (74), we rewrite g4s of equation (56) to show that(75)g4s=24β2α32α1+5α2s−α42<0.From equation (75), we know that g4s<0, for 0<s<1 and then g3s is a strictly increasing function. We recall that g30>0 of equation (71) and g31<0 of equation (72) such that we know that there is a point, denoted as s#, that satisfying(76)g′″s#=0,with(77)g′″s>0,for 0<s<s#, and(78)g′″s<0,for s#<s<1.Based on our findings of equations (77) and (78), we derive that g″s is an increasing function for 0<s<s#, and g″s is a decreasing function for s#<s<1.Together withg″0>0 of equation (69) and g″1<0 of equation (70), we know that there is a point, denoted as sΔ that satisfies g″sΔ=0 such that(79)g″s>0,for 0<s<sΔ, and(80)g″s<0,for sΔ<s<1.Based on our findings of equations (79) and (80), we derive that g′s is an increasing function for 0<s<sΔ, and g′s is a decreasing function for sΔ<s<1.Together withg′0>0 of equation (67) and g′1<0 of equation (68), we know that there is a point, denoted as sΩ that satisfies g′sΩ=0 such that(81)g′s>0,for 0<s<sΩ, and(82)g′s<0,for sΩ<s<1.Based on our findings of equations (81) and (82), we derive that gs is an increasing function for 0<s<sΩ, and gs is a decreasing function for sΩ<s<1.Together withg0>0 of equation (65) and g1<0 of equation (66), we know that there is a point, denoted as s∗ that satisfies gs∗=0 such that(83)gs>0,for 0<s<s∗, and(84)gs<0,for s∗<s<1.Based on the above discussion, we prove that there is a unique point that satisfies0<s∗<1, and(85)gs∗=0.Based on equation (50), we find that(86)k∗=1−s∗22s∗satisfies equation (48).We recall equation (47), and we find that(87)Q∗=1+k∗2−k∗α4α31+β1+k∗2+1−βk∗.From the above discussion, we provide an alternative solution approach to show the uniqueness of the optimal solution.Our proof needs an extra condition:3α1>α2, which was supported by numerical examples in Wu and Ouyang [21], Tung et al. [25], and Lin [38]. Hence, we provide a simplified proof to replace the complicated verification in Tung et al. [25] with an extra condition of α4>1+βα3α1+α2 to find an upper bound and a lower bound of the order quantity.From equation (31), we rewrite fQ=0 to yield the following fifth-degree polynomial:(88)βα3Q5−α4Q4−2βα1α3Q3+2α1α4Q2+βα12+α22α3Q−α12α4=0.Based on equation (31), we derive that(89)Q3=2α1α4Q2+βα12+α22α3Q−α12α4Qα4−βα3Q+2βα1α3.Owing to the upper bound in equation (30), we know that α4−βα3Q>0, and the lower bound in equation (30) yields that α1α4Q2−α12α4>0 such that the numerator and the denominator of equation (89) are both positive.Motivated by equation (89), we will use the following relation:(90)Qn+1=2α1α4Qn2+βα12+α22α3Qn−α12α4Qnα4−βα3Qn+2βα1α33,to construct a sequence, Qn, with an initial point Q1. We will show that Qn is a convergent sequence. We evaluate Qn+23−Qn+13 and express the result as(91)Qn+23−Qn+13=Qn+1−QnNM,where(92)N=y1Qn+1+Qn+y3QnQn+1+y4>0,(93)M=ΔnΔn+1>0,with(94)y1=3βα12α3α4,y2=βα32βα12+α22,y3=2α1α42+y2,y4=α12α42+2α1y2,Δn=Qnα4−βα3Qn+2βα1α3.Depending on the relation ofQ2−Q1, our solution results are divided into three cases. Case (a) Q2−Q1<0; Case (b) Q2−Q1=0; Case (c) Q2−Q1>0.For Case (a), owing toQ2−Q1<0 and equations (91)–(93), we know that Q3−Q2<0, and then Qn+1−Qn<0 for n=1,2,…, such that Qn is an increasing sequence which is bounded above by α1+α2, and then the increasing sequence will converge to its least upper bound.For Case (b), fromQ2=Q1, it yields that Q3=Q2 and Qn+1=Qn for n=1,2,…, such that Qn is a constant sequence that converges to Q1.For Case (c), fromQ2−Q1>0 and equations (91)–(93), it yields that Q3−Q2>0, and then Qn+1−Qn>0 for n=1,2,…, such that Qn is a decreasing sequence which is bounded below by α1, and then the decreasing sequence will converge to its greatest lower bound.Now, we summarize our results for three different cases to derive that our proposed sequenceQn will converge.From our findings, we can claim that the optimal solution is between the limit of Case (a) and the limit of Case (c) such that depending on the accuracy of the optimal solution, we can obtain the optimal solution as accuracy as desired which will be demonstrated in the next section.


## 7. Numerical Examples

We will compare our findings with that of Wu and Ouyang [21], Tung et al. [25], and Lin [38]. Hence, we adopt the same data as them with D=600 units/year, A=200 per order, h=20, h′=12, v=1.6, π=50, π0=150, and σ=7; the lead time has three components such that L0=8, RL0=0, L1=6, RL1=5.6, L2=4, RL2=22.4, L3=3, and RL3=57.4; the defective rate p has a Beta distribution, with probability density function gp=41−p3 to imply that Ep=1/5; and Ep2=1/15, the fraction of the backordered demand, β, has four values, 0,0.5,0.8, and 1. From the above data, we know that the lower bound α1=136.683 and the upper bound α1+α2=271.444. Therefore, for Case (a), we select Q1=α1+α2 with L=L3=3 and β=0.5 to derive our first numerical example. To save the precious space of the journal, we use different decimal places without influencing the tendency of the limit for our proposed decreasing sequence. We derive that Q1=271.4, Q2=231.1, Q3=191.6, Q4=181.9, Q5=177.0, Q6=174.4, Q7=173.0, Q8=172.2, Q9=171.7, Q10=171.5, Q11=171.33, Q12=171.25, Q13=171.20, Q14=171.18, Q15=171.16, Q16=171.151, Q17=171.146, Q18=171.143, Q19=171.1412, Q20=171.1403, Q21=171.1397, Q22=171.1394, Q23=171.1392, Q24=171.1391, Q25=171.139085, Q26=171.139052, Q27=171.139033, Q28=171.139022, Q29=171.139016, Q30=171.139013, Q31=171.139010, Q32=171.1390093, Q33=171.1390087, Q34=171.1390083, Q35=171.1390081, Q36=171.1390080, Q37=171.13900793, Q38=171.13900789, Q39=171.13900786, Q40=171.13900785, Q41=171.139007844, and Q42=171.139007840.From the above sequence, if we decide to stop up to the eighth decimal place, we can claim that the optimal solution(95)Q∗≤171.13900784.For our second numerical example, we selectQ1=α1 with L=L3=3 and β=0.5 and then list our results for the increasing sequence as follows: Q1=136.7, Q2=147.2, Q3=155.6, Q4=161.6, Q5=165.4, Q6=167.8, Q7=169.2, Q8=170.0, Q9=170.5, Q10=170.8, Q11=170.9, Q12=171.02, Q13=171.07, Q14=171.10, Q15=171.116, Q16=171.126, Q17=171.132, Q18=171.135, Q19=171.1366, Q20=171.1376, Q21=171.1382, Q22=171.13855, Q23=171.138746, Q24=171.138858, Q25=171.138922, Q26=171.138959, Q27=171.138980, Q28=171.138992, Q29=171.138999, Q30=171.1390026, Q31=171.1390048, Q32=171.1390061, Q33=171.1390069, Q34=171.1390073, Q35=171.13900751, Q36=171.13900765, Q37=171.13900773, Q38=171.13900777, Q39=171.1390077998, and Q40=171.1390078145.From the above sequence, if we decide to stop up to the eighth decimal place, we can state that the optimal solution(96)Q∗≥171.13900781.We have computed to the tenth decimal place to obtain that(97)171.1390078339≤Q∗≤171.1390078341.Because we have constructed a decreasing sequence withQ1=α1+α2 and an increasing sequence with Q1=α1, we can accurately estimate the optimal solution.If we compare our findings with Tung et al. [25], then they derived that Q∗=171.139009 which was obtained by a decreasing sequence proposed by Tung et al. [25]. The result of Tung et al. [25] is consistent with our results when we check our decreasing sequence with Q32=171.1390093 and Q33=171.1390087 to express the finding to the sixth decimal place; then, we will stop at Q∗≤171.139009.Furthermore, we compare our results with Lin [38]. She derived that Q∗=171.139014 which is found by a decreasing sequence proposed by Lin [38]. It indicates that the decreasing sequence Qi of Lin [38] did not converge to the desired result. The implicit reason for her unsuccessful converge may result from that there are three sequences in the iterative process of Lin [38]. We can predict that ki or di converge faster than Qi so that the equal value to the sixth decimal place of ki or di mislead the convergence for Qi. The above observation supports that our derivation is better than the three-sequence approach of Lin [38]. Hence, our approach implies one-sequence is a good method. In the next section, we will explore three distinct characters proposed by our approach.


## 8. Three Distinct Features of Our New Approach

In this section, we show that there are three distinct features of our new approach.The first feature is that we can derive the optimal solution within the preassigned threshold value. If we obtained a decreasing sequence, sayxn, with x1=271.4 and an increasing sequence, say zn, with z1=136.7, the exact optimal solution is denoted as Q∗.We know that(98)z1<z2<⋯<zn<⋯<Q∗<⋯<xn<⋯<x2<x1.For a given threshold value, sayε, after we find a number, say m that satisfies(99)xm−zm<ε,then we assume our optimal solution, denoted as QΔ with(100)QΔ=xm+zm2,such that we obtain(101)zm<Q∗<xm.We combine our findings of equations (99)–(101) to derive that(102)QΔ−Q∗<ε,to indicate our selected solution, QΔ, is as close to the exact optimal solution, Q∗, within the preassigned threshold value.The second feature is that we can estimate the converge ratio of our increasing sequence. Based on equation (91), we find that(103)Qn+2−Qn+1Qn+1−Qn≈2y1Q∗+y3Q∗2+y43Q∗2Q∗α4−βα3Q∗+2βα1α32.From our results, we obtain that(104)2y1Q∗+y3Q∗2+y43Q∗2Q∗α4−βα3Q∗+2βα1α32=0.571840729.We compute several related difference ratios and then list them in the following:(105)Q29−Q28Q28−Q27=0.571840853,(106)Q30−Q29Q29−Q28=0.571840779,(107)Q31−Q30Q30−Q29=0.571840770.We compare results from equations (105)–(107) with our estimation of equation (104) to show that our estimation of equation (103) is very accurate when n is big enough.The third feature is that we can estimate the dominant factors to influence the convergence ratio proposed by our approach.To estimate which factor influences the convergent ratio, we treat(108)α4−βα3Q∗≈α4,such that we rewrite equation (104) as follows:(109)y3Q∗2+2y1Q∗+y43Q∗2α4Q∗+2βα1α32.Based on(110)y3Q∗2=6.16773×1018,(111)2y1Q∗=2.15036×1017,(112)y4=1.97858×1018,(113)α4Q∗=12835425.59,(114)2βα1α3=298916.129,we compare magnitudes of equations (110)–(114) to decide the dominant factors are (i) Q∗, (ii) α4, (iii) y3, and (iv) y4. Hence, we improve our findings of equation (103) to claim that(115)Qn+2−Qn+1Qn+1−Qn≈y3Q∗2+y43α42Q∗4.We compute that(116)y3Q∗2+y4/3α42Q∗42y1Q∗+y3Q∗2+y4/3Q∗2Q∗α4−βα3Q∗+2βα1α32=98.412%.According to equation (116), we claim that our dominant factors can explain 98% of our convergent ratio.We may provide a hypothetical example as follows:Q1=5.327, Q2=2.049, Q3=2.043, and Q4=2.041 with limn⟶∞Qn=1.The researcher assumed that he will adopt the second decimal place to imply thatQ1=5.33, Q2=2.05, Q3=2.04, and Q4=2.04.Because the decimal expressions ofQ3 and Q4 are the same, the researcher will accept that Q∗=2.04.However, the convergent result at the second decimal place is not the optimal solution. The above example points out that depending on the convergence of the truncated decimal place without considering the convergent accuracy and ratio may not derive the optimal solution.Our proposed approach can control the convergent accuracy and ratio that will be useful for future researchers to develop their convergent algorithms.


## 9. Conclusion

We follow the research trend to revise previously published papers of Wu and Ouyang [21], Tung et al. [25], and Lin [38]. Our approach simplifies the complex derivation in Tung et al. [25], and our generated sequence converges better than that of Lin [38]. At last, we show that our proposed method contained three distinct features. Our findings will provide a complete method for future papers in their proof for the existence and uniqueness of the optimal solution and explanation for their convergent environment.



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*Source: 3528706-2020-10-10.xml*