A Nonpolynomial Optimal Algorithm for Sequencing Inspectors in a Repeat Inspection System with Rework

Abstract

Assuming that two types of inspection errors are nonidentical and that only the items rejected by an inspector are reworked and sent to the next inspection cycle, we formulate a combinatorial optimization problem for simultaneously determining both the minimum frequency of inspection-rework cycles and the optimal sequence of inspectors selected from a set of available inspectors, in order to meet the constraints of the outgoing quality level. Based on the inherent properties from our mathematical model, we provide a nonpolynomial optimal algorithm with a time complexity of O ( [superscript] 2 m [/superscript] ) .

Academic Editor:Francisco Chicano

Department of Industrial Engineering, Dankook University, Cheonan, Chungnam 330-714, Republic of Korea

Received 26 November 2014; Accepted 27 April 2015; 3 September 2015

This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Criticisms of the "Zero Defects (ZD)" concept frequently center around allegations of extreme cost in meeting the standard. Proponents say that it is an entirely reachable ideal and that claims of extreme cost result from misapplication of the principles. In order to reach ZD or to meet the constraints of the outgoing quality with the ppm (parts per million) level demanded by consumers, many authors have proposed repeated inspection systems (RIS), such that the few nonconforming items that escape detection at the first inspection operation would then be caught during the second or third inspection operations. Especially for critical items that could result in catastrophic system failure, any type of RIS designed for critical item can turn out to be cost effective in terms of the expected total cost, since the cost resulting from a falsely accepted item is much higher than that of repeated inspections. In designing or redesigning a cost-effective RIS, many factors besides the number of repeated inspection cycles should be considered, such as objective functions, the number of quality characteristics (single or multiple), sequences of multiple quality characteristics, the type and assumption of inspection errors (type I and/or type II, identical or nonidentical), a series of fractions for inspection (partial/full or incomplete/complete), and sequences of inspectors if their errors are not identical.

Several models of RISs subject to inspection errors under different conditions have appeared in the literature. We chronologically describe some papers partially related to our model. In the 1980s, Raouf et al. [1] were the first to develop a model for determining the optimal number of repeat inspections for multicharacteristic components to minimize the total expected cost. In the same year, Raz and Thomas [2] proposed a series of multiple inspections where only the items perceived to be conforming by all inspectors would be finally accepted, while the others would be discarded along the inspection line. Assuming that a group of inspectors operate at both different cost levels and nonidentical inspection errors in each stage in the sequence, a branch-and-bound technique was presented for determining an optimum sequencing inspection plan for obtaining a target quality level, minimal unit cost of production and inspection, and a series of optimal fractions for inspection. Based on dynamic programming, Drury et al. [3] examined different ways of combining the results of two different sequential inspections. Garcia-Diaz et al. [4] determined the number of repeated inspections at a certain cycle where the minimum expected total cost is achieved. Jaraiedi et al. [5] proposed a multiple-inspection system similar to Raz and Thomas's model, but it differed in that an inspector must examine all the characteristics of a unit one by one. This model can be used for determining the minimum number of inspection stages to meet a target average outgoing quality. Tang [6] provided a rule for determining the optimal sequence of multiple quality characteristics for minimizing the cost of inspection within each inspection stage. Lee [7] dealt with the model discussed by Raouf et al. and developed a stop rule for seeking the optimal number of inspection stages.

In the 1990s, Raz and Bricker [8] proposed three basic types of inspection sequences (complete, fixed, and variable) and studied the problem of sequencing inspection operations subject to errors in order to minimize the expected sum of inspection and penalty costs. Based on a branch-and-bound approach and recursive calls to a sequence of evaluation functions, the optimal algorithm for each type was provided, as well as a family of heuristics for the variable sequence problem. By modifying Raz and Thomas's cost model slightly and under some regular conditions, Liou et al. [9] derived an analytical solution to determine the optimal number of inspectors and the optimal sequence of inspectors. Chen and Labbrecht [10] used marginal analysis and gave an efficient algorithm to optimize the sequence and frequency of inspections of multicharacteristic components. Chiou [11] proposed a multiple-inspection system, where the inspection error is identical, and rejected items are repairable. With or without the assumption of AOQL constraint (average outgoing quality limit) and reinspection policy, four types of mathematical models were suggested, and algorithms to solve the optimal number of inspection cycles as well as its optimal inspection fraction for each cycle were provided.

In the 2000s, Yang [12] proposed an RIS with rework assuming identical inspection errors, repairable items, and a percent-defective target and provided an analytical formula for determining the minimum number of inspection-rework cycles that gave less than a target AOQ (average outgoing quality). Duffuaa and Khan [13] developed a general repeat inspection plan for dependent multicharacteristic critical components by extending the model given by Duffuaa and Nadeem [14]. Assuming six types of inspection errors, a procedure was provided for determining a local optimal number of cycles that minimized the total expected cost. Up to now, the effects of selecting and sequencing inspectors on AOQ or total inspection cost have rarely been studied, due to its inherent mathematical complexity. Only a few, such as Raouf et al. [1] and Raz and Thomas [2], have dealt with similar topics assuming that rejected items are discarded.

In this paper, we assume that (1) all the items accepted by an inspector are not reinspected but stored in a specific storage area, (2) all the items rejected by an inspector are reworked and sent to the next inspection cycle, (3) an inspector performs a single characteristic inspection, and (4) the types I and II errors of inspectors are nonidentical, fixed, and known. In Section 2, our problem is described in detail. In Section 3, we derive AOQ as a function of a sequence of inspectors and prove some fundamental properties. Based on the properties, we provide a practical algorithm for simultaneously determining an optimal frequency and an optimal sequence of inspectors, which gives a target AOQ.

2. Problem Statement

Suppose that the expected initial defective rate of items produced in production lines is given as a constant [figure omitted; refer to PDF] and that we need repeated inspections in order to meet the constraints of the target AOQ level ( [figure omitted; refer to PDF] ). Let [figure omitted; refer to PDF] be a set of [figure omitted; refer to PDF] available inspectors, represented as [figure omitted; refer to PDF] , and let [figure omitted; refer to PDF] ( [figure omitted; refer to PDF] ) be the inspector, selected from the set [figure omitted; refer to PDF] and assigned to the [figure omitted; refer to PDF] th inspection cycle. The following assumptions are made:

(1) [figure omitted; refer to PDF] examines a single characteristic of an item at a time and examines all the items given during the [figure omitted; refer to PDF] th inspection cycle, and each inspector must be assigned only once to one cycle.

(2) [figure omitted; refer to PDF] makes two kinds of errors: rejecting a conforming item (type I error) or accepting a nonconforming item (type II error), and type I and type II error probabilities are not variables but constants measured as [figure omitted; refer to PDF] . The error probabilities of inspectors are not identical. That is, [figure omitted; refer to PDF] for integers [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , with [figure omitted; refer to PDF] .

(3) The item inspected by [figure omitted; refer to PDF] is accepted as conforming only if the characteristic is correctly or falsely classified as conforming by [figure omitted; refer to PDF] . Otherwise, it is rejected as nonconforming. All the items accepted by [figure omitted; refer to PDF] are not reinspected anymore but accumulated in a specific storage area, and all the items rejected by [figure omitted; refer to PDF] are reworked by a repairman at a constant defective rate ( [figure omitted; refer to PDF] ).

(4) The reworked items are sent to either the next inspection-rework cycle or the storage area only when the AOQ of all the items accumulated in the area is less than or equal to [figure omitted; refer to PDF] . The above procedure is continued until we obtain an AOQ less than or equal to [figure omitted; refer to PDF] .

Let [figure omitted; refer to PDF] , represented by [figure omitted; refer to PDF] , be a sequence of [figure omitted; refer to PDF] inspectors selected from [figure omitted; refer to PDF] and assigned to [figure omitted; refer to PDF] inspection cycles. Let [figure omitted; refer to PDF] be the sequence with size [figure omitted; refer to PDF] from [figure omitted; refer to PDF] such that [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] is an AOQ resulting from a sequence [figure omitted; refer to PDF] . Now, under the assumption that [figure omitted; refer to PDF] , our problem can be described as follows:

SP:: given [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] as input variables where [figure omitted; refer to PDF] , determine both [figure omitted; refer to PDF] and [figure omitted; refer to PDF] simultaneously, where [figure omitted; refer to PDF] is the minimum number of inspection-rework cycles such that [figure omitted; refer to PDF] .

Note that the total number of possible sequences will be as many as [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] , since the number of combinations of [figure omitted; refer to PDF] inspectors from [figure omitted; refer to PDF] is [figure omitted; refer to PDF] , and the number of permutations (or sequences) of [figure omitted; refer to PDF] inspectors from a combination of [figure omitted; refer to PDF] inspectors is [figure omitted; refer to PDF] .

3. Analysis and Optimization of Our Inspection Problem

3.1. Derivation of AOQ and Properties

Suppose that a sequence with size [figure omitted; refer to PDF] is given as [figure omitted; refer to PDF] for [figure omitted; refer to PDF] . For an integer [figure omitted; refer to PDF] with [figure omitted; refer to PDF] , let

: [figure omitted; refer to PDF] = expected number of conforming items entering the ( [figure omitted; refer to PDF] )-st inspection cycle,

: [figure omitted; refer to PDF] = expected number of nonconforming items entering the ( [figure omitted; refer to PDF] )-st inspection cycle,

: [figure omitted; refer to PDF] = expected number of items correctly accepted by [figure omitted; refer to PDF] ,

: [figure omitted; refer to PDF] = expected number of items falsely rejected by [figure omitted; refer to PDF] ,

: [figure omitted; refer to PDF] = expected number of items falsely accepted by [figure omitted; refer to PDF] ,

: [figure omitted; refer to PDF] = expected number of items correctly rejected by [figure omitted; refer to PDF] ,

: [figure omitted; refer to PDF] = expected number of items entering the [figure omitted; refer to PDF] th rework operation,

: [figure omitted; refer to PDF] = expected number of items entering the first cycle,

: [figure omitted; refer to PDF] = expected number of items correctly or falsely accepted by [figure omitted; refer to PDF] .

Then, for an integer [figure omitted; refer to PDF] with [figure omitted; refer to PDF] , we have [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] and [figure omitted; refer to PDF] . Since the items classified as conforming by [figure omitted; refer to PDF] are accumulated in a specific storage area, we have [figure omitted; refer to PDF] and [figure omitted; refer to PDF] . Hence, we have [figure omitted; refer to PDF]

After reworking at the [figure omitted; refer to PDF] th rework operation, we have [figure omitted; refer to PDF]

After completing the [figure omitted; refer to PDF] th inspection-rework cycle, since the expected number of nonconforming items in the storage area will be the sum of the expected number of items falsely accepted and the expected number of nonconforming items in the [figure omitted; refer to PDF] th rework operation, [figure omitted; refer to PDF] can be derived as [figure omitted; refer to PDF]

For convenience, let [figure omitted; refer to PDF] and [figure omitted; refer to PDF] for [figure omitted; refer to PDF] . Note that [figure omitted; refer to PDF] can be interpreted as the probability with which [figure omitted; refer to PDF] classifies an item as nonconforming and that [figure omitted; refer to PDF] . From the four recurrence relations equations (1) through (4), the mathematical expressions for [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] can be derived as closed-form solutions in the following lemma.

Lemma 1.

For an integer [figure omitted; refer to PDF] with [figure omitted; refer to PDF] ,

(1) [figure omitted; refer to PDF] ,

(2) [figure omitted; refer to PDF] ,

(3) [figure omitted; refer to PDF] ,

(4) [figure omitted; refer to PDF]

(5) [figure omitted; refer to PDF]

Proof.

Using (3) and (4), [figure omitted; refer to PDF] can be reduced to [figure omitted; refer to PDF] Using (8), we have [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] , from (1), we have [figure omitted; refer to PDF] Using (9), since [figure omitted; refer to PDF] , we have [figure omitted; refer to PDF] Using (5), (8), and (9), we have [figure omitted; refer to PDF]

Example 2.

When [figure omitted; refer to PDF] = (10,000 units, 5%, 4) and [figure omitted; refer to PDF] are given as shown in Table 1, compute [figure omitted; refer to PDF] for all possible sequences of [figure omitted; refer to PDF] and find the minimum of [figure omitted; refer to PDF] .

Table 1: Input data of [figure omitted; refer to PDF] and [figure omitted; refer to PDF] .

[figure omitted; refer to PDF]	[figure omitted; refer to PDF] (%)	[figure omitted; refer to PDF] (%)	[figure omitted; refer to PDF] (%)	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF] (%)	[figure omitted; refer to PDF] (%)	[figure omitted; refer to PDF] (%)	[figure omitted; refer to PDF]
1	0.92	4.50	5.42	0.0454	3	0.72	4.60	5.32	0.0463
2	0.97	4.51	5.48	0.0455	4	0.85	5.18	6.03	0.0522

Suppose that [figure omitted; refer to PDF] is scheduled; Inspector 3 is assigned to the first inspection cycle and Inspector 1 to the second. Since [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , Inspector 3 has [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] after completion of the first inspection cycle as shown in Table 2. It follows that the number of items sent to the storage area is [figure omitted; refer to PDF] since these items have been classified as conforming by Inspector 3 even though some of them have been falsely accepted. The number of items correctly or falsely rejected by Inspector 3 is [figure omitted; refer to PDF] and these items are sent to the rework shop, where they are repaired with [figure omitted; refer to PDF] and [figure omitted; refer to PDF] and sent to the second inspection cycle assigned by Inspector 1. Now, since [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , Inspector 1 has [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] after completion of the second inspection cycle. It follows that the number of items sent to the storage area is [figure omitted; refer to PDF] since those items have been classified as conforming by Inspector 1. The number of items rejected by Inspector 1 is [figure omitted; refer to PDF] and these items are sent to the rework shop, where they are repaired with [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , and they are sent to the storage area instead of the third inspection cycle since two inspection cycles are given.

Table 2: All possible sequences with size 2 and AOQs.

[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	Repair	[figure omitted; refer to PDF]	Repair	[figure omitted; refer to PDF] (ppm)
[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]
[figure omitted; refer to PDF]	9,413	87	23	478	9,435	565	537	28	531	5	1	27	533	32	31	2	2,538
[figure omitted; refer to PDF]	9,408	92	23	477	9,430	570	541	28	536	5	1	27	537	32	31	2	2,544
[figure omitted; refer to PDF]	9,413	87	23	478	9,435	565	537	28	533	4	1	27	534	31	29	2	2,534
[figure omitted; refer to PDF]	9,432	68	23	477	9,455	545	518	27	513	5	1	26	515	31	29	2	2,577
[figure omitted; refer to PDF]	9,413	87	23	478	9,435	565	537	28	532	5	1	27	534	31	30	2	2,553
[figure omitted; refer to PDF]	9,419	81	26	474	9,445	555	527	28	522	5	1	26	524	31	30	2	2,872
[figure omitted; refer to PDF]	9,408	92	23	477	9,430	570	541	28	537	4	1	27	539	31	30	2	2,541
[figure omitted; refer to PDF]	9,432	68	23	477	9,455	545	518	27	513	5	1	26	514	31	30	2	2,578
[figure omitted; refer to PDF]	9,408	92	23	477	9,430	570	541	28	537	5	1	27	538	32	30	2	2,561
[figure omitted; refer to PDF]	9,419	81	26	474	9,445	555	527	28	522	5	1	26	523	32	30	2	2,873
[figure omitted; refer to PDF]	9,432	68	23	477	9,455	545	518	27	514	4	1	26	515	30	29	2	2,593
[figure omitted; refer to PDF]	9,419	81	26	474	9,445	555	527	28	523	4	1	26	525	30	29	2	2,869

Since the total number of nonconforming items in the storage area is [figure omitted; refer to PDF] , the value of [figure omitted; refer to PDF] can be computed as 2,577 ppm, which can be also computed by using Lemma 1-(5). In the similar method above, the value of AOQ for each possible sequence with size 2 can be computed and summarized in Table 2. Note that the total number of possible sequences with size 2 is [figure omitted; refer to PDF] and that the sequence [figure omitted; refer to PDF] gives the smallest AOQ (=2,534 ppm) of all possible sequences with size 2.

In fact, the input data given in the example are partially extracted from the original data sets of a Korean back-light unit supplier, and, from the data sets, the averages of the type I and type II errors were estimated as 0.8641% and 4.5031%, respectively, by Yang and Cho [15].

Consider the minimization problem of [figure omitted; refer to PDF] . Suppose that [figure omitted; refer to PDF] with [figure omitted; refer to PDF] from a set [figure omitted; refer to PDF] of [figure omitted; refer to PDF] available inspectors is assigned to the single inspection cycle. Then, from Lemma 1-(5), [figure omitted; refer to PDF] can be expressed as [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] . Since [figure omitted; refer to PDF] is constant, [figure omitted; refer to PDF] is minimized if and only if the inspector with [figure omitted; refer to PDF] is assigned to the cycle. In the above example, since [figure omitted; refer to PDF] , Inspector 3 must be allocated in order to attain the minimum AOQ = 5,027 ppm.

Consider the minimization problem of [figure omitted; refer to PDF] . This minimization problem is not as easy as the minimization problem of [figure omitted; refer to PDF] . However, if the first inspector has been optimally assigned, that is, [figure omitted; refer to PDF] , the minimization problem of [figure omitted; refer to PDF] given [figure omitted; refer to PDF] can be easily solved. When [figure omitted; refer to PDF] with [figure omitted; refer to PDF] from a set of ( [figure omitted; refer to PDF] ) available inspectors, expressed as [figure omitted; refer to PDF] , is assigned to the second inspection cycle, [figure omitted; refer to PDF] given [figure omitted; refer to PDF] can be expressed as [figure omitted; refer to PDF] from Lemma 1-(5). Since [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] are constants, [figure omitted; refer to PDF] given [figure omitted; refer to PDF] can be minimized if and only if the inspector with the smallest value of [figure omitted; refer to PDF] among [figure omitted; refer to PDF] is assigned to the second inspection cycle. Generally, when [figure omitted; refer to PDF] with [figure omitted; refer to PDF] from [figure omitted; refer to PDF] is assigned to the [figure omitted; refer to PDF] th inspection cycle for [figure omitted; refer to PDF] , [figure omitted; refer to PDF] given [figure omitted; refer to PDF] can be derived as [figure omitted; refer to PDF] where [figure omitted; refer to PDF] . Since [figure omitted; refer to PDF] and [figure omitted; refer to PDF] are constants, [figure omitted; refer to PDF] given [figure omitted; refer to PDF] can be minimized if and only if the inspector with [figure omitted; refer to PDF] among [figure omitted; refer to PDF] is assigned to the [figure omitted; refer to PDF] th inspection cycle. For this reason, it may be likely to conjecture that the optimal solution of the minimization of [figure omitted; refer to PDF] is one of [figure omitted; refer to PDF] -increasing sequences. This conjecture turned out to be wrong. The counterexample can be constructed from the above example. Consider the minimization problem of [figure omitted; refer to PDF] . Since [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] , the optimal sequence of the minimization of [figure omitted; refer to PDF] given [figure omitted; refer to PDF] is [figure omitted; refer to PDF] . However, the optimal sequence of the minimization of [figure omitted; refer to PDF] turned out to be [figure omitted; refer to PDF] instead of [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] = 2,577 ppm > [figure omitted; refer to PDF] = 2,534 ppm as shown in Table 2.

Basically the minimization of [figure omitted; refer to PDF] is different from the minimization of [figure omitted; refer to PDF] given [figure omitted; refer to PDF] . It turns out that when [figure omitted; refer to PDF] , the optimal sequence [figure omitted; refer to PDF] of the minimization of [figure omitted; refer to PDF] is one of [figure omitted; refer to PDF] -increasing sequences with size [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] is the ratio of the type II error to one minus the type I error. In the above example, since the [figure omitted; refer to PDF] -increasing sequence with size 4 is [figure omitted; refer to PDF] , the [figure omitted; refer to PDF] -increasing sequences with size 2 are [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] , and the sequence with the smallest value of AOQ of them is optimal. Since [figure omitted; refer to PDF] gives the minimum value of AOQ as shown in Table 2, [figure omitted; refer to PDF] becomes [figure omitted; refer to PDF] . Similarly, [figure omitted; refer to PDF] is the sequence with the smallest value of AOQ of the candidate solutions: [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] . It turns out that [figure omitted; refer to PDF] with AOQ = 2,394 ppm.

In order to prove generally that [figure omitted; refer to PDF] is one of [figure omitted; refer to PDF] -increasing sequences with size [figure omitted; refer to PDF] when [figure omitted; refer to PDF] , let

: [figure omitted; refer to PDF] = a subset of [figure omitted; refer to PDF] inspectors selected from [figure omitted; refer to PDF] ,

: [figure omitted; refer to PDF] = a collection of subsets with size [figure omitted; refer to PDF] of [figure omitted; refer to PDF] ; that is, [figure omitted; refer to PDF] ,

: [figure omitted; refer to PDF] = the [figure omitted; refer to PDF] th sequence or permutation generated from [figure omitted; refer to PDF] ,

: [figure omitted; refer to PDF] = a collection of sequences with size [figure omitted; refer to PDF] generated from [figure omitted; refer to PDF] ; that is, [figure omitted; refer to PDF] .

In the above example, since [figure omitted; refer to PDF] , the collection [figure omitted; refer to PDF] of subsets with size 2 of [figure omitted; refer to PDF] becomes [figure omitted; refer to PDF] . Suppose that [figure omitted; refer to PDF] . Then, all possible permutations generated from [figure omitted; refer to PDF] are [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , and the collection [figure omitted; refer to PDF] of sequences with size 2 generated from [figure omitted; refer to PDF] becomes [figure omitted; refer to PDF] . Using the notations above, [figure omitted; refer to PDF] can be redefined as the sequence with size [figure omitted; refer to PDF] from [figure omitted; refer to PDF] such that [figure omitted; refer to PDF]

Suppose that the number of completed inspection-rework cycles is [figure omitted; refer to PDF] for [figure omitted; refer to PDF] and that the sequence with size [figure omitted; refer to PDF] is represented as [figure omitted; refer to PDF] . By choosing the [figure omitted; refer to PDF] th and ( [figure omitted; refer to PDF] )-st adjacent inspectors from [figure omitted; refer to PDF] and by swapping them, we can construct the swapped sequence; [figure omitted; refer to PDF] . The AOQ of the swapped sequence [figure omitted; refer to PDF] turned out to be smaller than [figure omitted; refer to PDF] , if and only if the following necessary and sufficient condition holds.

Lemma 3.

[figure omitted; refer to PDF] if and only if [figure omitted; refer to PDF] for [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] .

Proof.

Consider two cases as follows: Case 1 for [figure omitted; refer to PDF] and Case 2 for [figure omitted; refer to PDF] .

Case 1 . When [figure omitted; refer to PDF] , by swapping [figure omitted; refer to PDF] and [figure omitted; refer to PDF] and using Lemma 1-(5), [figure omitted; refer to PDF] and [figure omitted; refer to PDF] can be derived, respectively, as [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] , we have [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] , Lemma 3 holds when [figure omitted; refer to PDF] .

Case 2 . When [figure omitted; refer to PDF] , two subcases are considered as follows: Subcase 1 for [figure omitted; refer to PDF] and Subcase 2 for [figure omitted; refer to PDF] .

Subcase 1. For [figure omitted; refer to PDF] , from Lemma 1-(5), [figure omitted; refer to PDF] can be derived as [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] , Lemma 3 holds when [figure omitted; refer to PDF] and [figure omitted; refer to PDF] .

Subcase 2. When [figure omitted; refer to PDF] , from Lemma 1-(5), [figure omitted; refer to PDF] can be derived as [figure omitted; refer to PDF]

Since [figure omitted; refer to PDF] , Lemma 3 holds when [figure omitted; refer to PDF] and [figure omitted; refer to PDF] . Therefore, Lemma 3 holds.

Lemma 4.

For [figure omitted; refer to PDF] , [figure omitted; refer to PDF] is one of [figure omitted; refer to PDF] -increasing sequences with size [figure omitted; refer to PDF] .

Proof.

It is enough to prove that if [figure omitted; refer to PDF] is not one of [figure omitted; refer to PDF] -increasing sequences, then [figure omitted; refer to PDF] does not give a minimum AOQ among a set of [figure omitted; refer to PDF] . If [figure omitted; refer to PDF] does not satisfy [figure omitted; refer to PDF] -increasing order, then there must exist at least two adjacent inspectors violating [figure omitted; refer to PDF] -increasing order, and, using Lemma 3, we can further reduce AOQ by swapping the adjacent inspectors. Thus, [figure omitted; refer to PDF] does not give a minimum AOQ among a set of [figure omitted; refer to PDF] , and Lemma 4 holds.

Suppose that we take [figure omitted; refer to PDF] inspectors from [figure omitted; refer to PDF] in [figure omitted; refer to PDF] -increasing order. Let this ordered set be [figure omitted; refer to PDF] , expressed as [figure omitted; refer to PDF] . Let [figure omitted; refer to PDF] be the sequence with size [figure omitted; refer to PDF] , constructed by taking the first [figure omitted; refer to PDF] inspectors from [figure omitted; refer to PDF] . That is, [figure omitted; refer to PDF] . Then, it may be conjectured that [figure omitted; refer to PDF] will give the minimum AOQ if the optimal number of inspection-rework cycles is known as [figure omitted; refer to PDF] . However, unfortunately, this is not always true. For a counterexample, when the input data for [figure omitted; refer to PDF] are given as in Table 1 and when [figure omitted; refer to PDF] is given as (3,000 ppm, 5%), it turns out that [figure omitted; refer to PDF] and we have [figure omitted; refer to PDF] and [figure omitted; refer to PDF] . However, [figure omitted; refer to PDF] is not [figure omitted; refer to PDF] but [figure omitted; refer to PDF] since [figure omitted; refer to PDF] > [figure omitted; refer to PDF] = 2,534 ppm as shown in Table 2.

Note that [figure omitted; refer to PDF] is not [figure omitted; refer to PDF] but one of [figure omitted; refer to PDF] -increasing sequences with size [figure omitted; refer to PDF] out of [figure omitted; refer to PDF] available inspectors. Also note that if all inspection error probabilities are zero, that is, [figure omitted; refer to PDF] for all [figure omitted; refer to PDF] , then [figure omitted; refer to PDF] converges to zero, since [figure omitted; refer to PDF] can be reduced to [figure omitted; refer to PDF] . If [figure omitted; refer to PDF] for all [figure omitted; refer to PDF] , then [figure omitted; refer to PDF] converges to one, since [figure omitted; refer to PDF] can be reduced to [figure omitted; refer to PDF] . From the results, it can be said that our RIS does not always guarantee that [figure omitted; refer to PDF] is a decreasing function of [figure omitted; refer to PDF] . However, as proved in the following two lemmas, if [figure omitted; refer to PDF] is given, we can construct [figure omitted; refer to PDF] such that [figure omitted; refer to PDF] by selecting the [figure omitted; refer to PDF] th inspector with [figure omitted; refer to PDF] . Similarly, if [figure omitted; refer to PDF] is given, we can construct [figure omitted; refer to PDF] such that [figure omitted; refer to PDF] by selecting the [figure omitted; refer to PDF] th inspector with [figure omitted; refer to PDF] . Hence, [figure omitted; refer to PDF] can be a strictly decreasing function of [figure omitted; refer to PDF] .

Lemma 5.

For [figure omitted; refer to PDF] , [figure omitted; refer to PDF] if and only if [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] .

Proof.

Since [figure omitted; refer to PDF] , we have [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] , Lemma 5 holds when [figure omitted; refer to PDF] . When [figure omitted; refer to PDF] , by using Lemma 1-(5), we have [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] , Lemma 5 holds when [figure omitted; refer to PDF] . Therefore Lemma 5 holds.

Lemma 6.

For [figure omitted; refer to PDF] , [figure omitted; refer to PDF] if and only if [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] .

Proof.

If [figure omitted; refer to PDF] is given for an integer with [figure omitted; refer to PDF] and if [figure omitted; refer to PDF] , we can construct [figure omitted; refer to PDF] using Lemma 5 such that [figure omitted; refer to PDF] by assigning the inspector [figure omitted; refer to PDF] with [figure omitted; refer to PDF] . Since [figure omitted; refer to PDF] by definition, it follows that [figure omitted; refer to PDF] . Reversely, if [figure omitted; refer to PDF] , then, from Lemma 5, there exists [figure omitted; refer to PDF] with [figure omitted; refer to PDF] .

It remains whether or not an inspector with [figure omitted; refer to PDF] satisfying [figure omitted; refer to PDF] exists among unassigned inspectors of the set [figure omitted; refer to PDF] . However, since [figure omitted; refer to PDF] do not exceed 100% in most practical cases, we assume hereafter that [figure omitted; refer to PDF] for [figure omitted; refer to PDF] .

3.2. An Optimal Algorithm

Using the previously proven properties, we can construct an optimal algorithm, ALGSP, for determining both [figure omitted; refer to PDF] and [figure omitted; refer to PDF] simultaneously. Our algorithm consists of three phases: an initialization phase, a local optimization phase, and a global optimization phase. In the initialization phase, in order to reduce the computational execution time in the local optimization phase, we construct the [figure omitted; refer to PDF] -increasing set [figure omitted; refer to PDF] from [figure omitted; refer to PDF] . Without loss of generality, let [figure omitted; refer to PDF] be [figure omitted; refer to PDF] .

In the local optimization phase, we find [figure omitted; refer to PDF] for [figure omitted; refer to PDF] . Let [figure omitted; refer to PDF] be the [figure omitted; refer to PDF] -increasing sequence with size [figure omitted; refer to PDF] generated from a given ordered combination, [figure omitted; refer to PDF] , such that [figure omitted; refer to PDF] . For example, if [figure omitted; refer to PDF] is given as [figure omitted; refer to PDF] , then [figure omitted; refer to PDF] can be immediately obtained as [figure omitted; refer to PDF] , where [figure omitted; refer to PDF] for [figure omitted; refer to PDF] . In the local optimization phase, given [figure omitted; refer to PDF] , we construct [figure omitted; refer to PDF] from [figure omitted; refer to PDF] using [figure omitted; refer to PDF] . Next, from [figure omitted; refer to PDF] , we find the sequence [figure omitted; refer to PDF] such that [figure omitted; refer to PDF] . In the outer loop, changing [figure omitted; refer to PDF] from one to [figure omitted; refer to PDF] , we store a set of the local optimal sequences, [figure omitted; refer to PDF] . In the global optimization phase, we search for [figure omitted; refer to PDF] satisfying [figure omitted; refer to PDF] . If [figure omitted; refer to PDF] is not found, we terminate our algorithm with a message " [figure omitted; refer to PDF] ." See Algorithm 1.

Algorithm 1: ALGSP.

Step 1 (Initialization Phase):

Take inspectors in [figure omitted; refer to PDF] -increasing order using [figure omitted; refer to PDF] and let this ordered set be [figure omitted; refer to PDF] .

Step 2 (Local Optimization Phase)

For [figure omitted; refer to PDF] to [figure omitted; refer to PDF] do

Begin

For [figure omitted; refer to PDF] to [figure omitted; refer to PDF]

Begin

[figure omitted; refer to PDF]

Begin

Construct [figure omitted; refer to PDF] and [figure omitted; refer to PDF]

Compute [figure omitted; refer to PDF] using Lemma 1( [figure omitted; refer to PDF] ).

If [figure omitted; refer to PDF] min, then [figure omitted; refer to PDF] ,

[figure omitted; refer to PDF] , [figure omitted; refer to PDF]

End

End

End

Step 3 (Global Optimization Phase)

For [figure omitted; refer to PDF] to [figure omitted; refer to PDF] do

Begin

If [figure omitted; refer to PDF] , then [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , [figure omitted; refer to PDF]

If [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , then print " [figure omitted; refer to PDF] greater than [figure omitted; refer to PDF] " and stop

End

Theorem 7.

[figure omitted; refer to PDF] determined by ALGSP is an optimal sequence to SP.

Proof.

[figure omitted; refer to PDF] is a decreasing function of [figure omitted; refer to PDF] from Lemma 6, and [figure omitted; refer to PDF] determined by ALGSP is the smallest number of inspection-rework cycles such that [figure omitted; refer to PDF] . Hence, [figure omitted; refer to PDF] determined by ALGSP is an optimal sequence to SP.

Theorem 8.

ALGSP solves SP in [figure omitted; refer to PDF] .

Proof.

From Theorem 7, [figure omitted; refer to PDF] gives an optimal solution to SP. Steps 1 and 3 require [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , respectively. Step 2 requires [figure omitted; refer to PDF] , since [figure omitted; refer to PDF] can be obtained by enumerating not all the permutations but all the combinations of [figure omitted; refer to PDF] inspectors from [figure omitted; refer to PDF] ; that is, [figure omitted; refer to PDF] . Hence, ALGSP solves SP in [figure omitted; refer to PDF] .

Even though the time complexity of ALGSP is [figure omitted; refer to PDF] , our algorithm is practically efficient since the optimal number of inspection-rework cycles is usually less than five cycles in most practical cases.

3.3. A Numerical Example

Suppose that [figure omitted; refer to PDF] = (2,500 ppm, 5%, 4) and that [figure omitted; refer to PDF] are given as shown in Table 1. Using ALGSP, the optimal solution to our example will be obtained as follows.

Step 1.

The [figure omitted; refer to PDF] -increasing sequence is [figure omitted; refer to PDF] and [figure omitted; refer to PDF] .

Step 2.

As shown in Table 3, when [figure omitted; refer to PDF] , we generate four sequences [figure omitted; refer to PDF] for [figure omitted; refer to PDF] , and we have [figure omitted; refer to PDF] (or [figure omitted; refer to PDF] ) with [figure omitted; refer to PDF] ppm. When [figure omitted; refer to PDF] , we generate six sequences [figure omitted; refer to PDF] where [figure omitted; refer to PDF] is obtained by finding the combinations of two inspectors from [figure omitted; refer to PDF] , and we have [figure omitted; refer to PDF] with [figure omitted; refer to PDF] ppm. When [figure omitted; refer to PDF] , we generate four sequences [figure omitted; refer to PDF] obtained by finding the combinations of three inspectors from [figure omitted; refer to PDF] , and we have [figure omitted; refer to PDF] with [figure omitted; refer to PDF] ppm. When [figure omitted; refer to PDF] , we generate one sequence [figure omitted; refer to PDF] , which is the only [figure omitted; refer to PDF] -increasing sequence, and we have [figure omitted; refer to PDF] with [figure omitted; refer to PDF] ppm. After Step 2, [figure omitted; refer to PDF] and [figure omitted; refer to PDF] for [figure omitted; refer to PDF] can be obtained as summarized in Table 4.

Table 3: The changing values in Step 2.

[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]
1	1	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	5075	[figure omitted; refer to PDF]	5075
2	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	5103	[figure omitted; refer to PDF]	5075
3	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	5027	[figure omitted; refer to PDF]	5027
4	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	5364	[figure omitted; refer to PDF]	5027
2	1	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2538	[figure omitted; refer to PDF]	2538
2	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2534	[figure omitted; refer to PDF]	2534
3	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2553	[figure omitted; refer to PDF]	2534
4	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2541	[figure omitted; refer to PDF]	2534
5	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2561	[figure omitted; refer to PDF]	2534
6	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2593	[figure omitted; refer to PDF]	2534
3	1	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2394	[figure omitted; refer to PDF]	2394
2	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2395	[figure omitted; refer to PDF]	2394
3	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2396	[figure omitted; refer to PDF]	2394
4	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2403	[figure omitted; refer to PDF]	2394
4	1	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2386	[figure omitted; refer to PDF]	2386

Table 4: [figure omitted; refer to PDF] and [figure omitted; refer to PDF] , which resulted from Step 2.

[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]
1	[figure omitted; refer to PDF]	5027
2	[figure omitted; refer to PDF]	2534
3	[figure omitted; refer to PDF]	2394
4	[figure omitted; refer to PDF]	2386

Step 3.

Since [figure omitted; refer to PDF] ppm > [figure omitted; refer to PDF] ppm > [figure omitted; refer to PDF] ppm, ALGSP terminates with [figure omitted; refer to PDF] and [figure omitted; refer to PDF] .

For the reader's convenience, we have included the computational results of all the possible sequences generated from the set [figure omitted; refer to PDF] in Tables 5, 6, and 7.

Table 5: [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] when [figure omitted; refer to PDF] .

[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]
[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2538	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2541
[figure omitted; refer to PDF]	2544	[figure omitted; refer to PDF]	2578
[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2534	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2561
[figure omitted; refer to PDF]	2577	[figure omitted; refer to PDF]	2873
[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2553	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2593
[figure omitted; refer to PDF]	2872	[figure omitted; refer to PDF]	2869

Table 6: [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] when [figure omitted; refer to PDF] .

[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]
[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2394	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2396
[figure omitted; refer to PDF]	2396	[figure omitted; refer to PDF]	2412
[figure omitted; refer to PDF]	2399	[figure omitted; refer to PDF]	2439
[figure omitted; refer to PDF]	2402	[figure omitted; refer to PDF]	2457
[figure omitted; refer to PDF]	2438	[figure omitted; refer to PDF]	2731
[figure omitted; refer to PDF]	2439	[figure omitted; refer to PDF]	2733
[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2395	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2403
[figure omitted; refer to PDF]	2412	[figure omitted; refer to PDF]	2418
[figure omitted; refer to PDF]	2400	[figure omitted; refer to PDF]	2440
[figure omitted; refer to PDF]	2419	[figure omitted; refer to PDF]	2457
[figure omitted; refer to PDF]	2731	[figure omitted; refer to PDF]	2731
[figure omitted; refer to PDF]	2731	[figure omitted; refer to PDF]	2733

Table 7: [figure omitted; refer to PDF] , [figure omitted; refer to PDF] , and [figure omitted; refer to PDF] when [figure omitted; refer to PDF] .

[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]
[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2386	[figure omitted; refer to PDF]	[figure omitted; refer to PDF]	2431
[figure omitted; refer to PDF]	2387	[figure omitted; refer to PDF]	2432
[figure omitted; refer to PDF]	2388	[figure omitted; refer to PDF]	2431
[figure omitted; refer to PDF]	2389	[figure omitted; refer to PDF]	2432
[figure omitted; refer to PDF]	2404	[figure omitted; refer to PDF]	2449
[figure omitted; refer to PDF]	2404	[figure omitted; refer to PDF]	2449
[figure omitted; refer to PDF]	2392	[figure omitted; refer to PDF]	2723
[figure omitted; refer to PDF]	2392	[figure omitted; refer to PDF]	2723
[figure omitted; refer to PDF]	2394	[figure omitted; refer to PDF]	2723
[figure omitted; refer to PDF]	2395	[figure omitted; refer to PDF]	2723
[figure omitted; refer to PDF]	2411	[figure omitted; refer to PDF]	2725
[figure omitted; refer to PDF]	2411	[figure omitted; refer to PDF]	2725

4. Conclusion

Assuming that all the items rejected by an inspector are reworked in a constant defective rate and are sent to the next inspection-rework cycle, we have addressed the sequencing problem of inspectors from a set of available inspectors in order to meet a constraint on the outgoing quality level, subject to nonidentical errors and several assumptions. We provided a practical nonpolynomial time-optimal algorithm, which simultaneously determines the minimum number of inspection-rework cycles and an optimal sequence in [figure omitted; refer to PDF] .

Our strong conjecture is that our problem is NP-complete. Not by our conjecture but by our computational experience, we would like to suggest two heuristic algorithms with the time complexity of [figure omitted; refer to PDF] based on [figure omitted; refer to PDF] -increasing or [figure omitted; refer to PDF] -increasing order. The near-optimal sequence with size [figure omitted; refer to PDF] can be determined just by taking the first [figure omitted; refer to PDF] inspectors from the set of available inspectors listed in [figure omitted; refer to PDF] -increasing or [figure omitted; refer to PDF] -increasing order. Further research may be concentrated on either proving that our problem is NP-complete or finding a polynomial-time algorithm by discovering more properties in our problem, while the combinatorial optimization problem for determining an optimal sequence in terms of total cost or the number of inspections may be studied in depth. Our basic methods used in proofs could be utilized for solving other similar sequencing problems.

Acknowledgments

The authors would like to thank the reviewers very much for their comments. The present research was conducted with the support of the research fund of Dankook University, in 2013.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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