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

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(2).


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 costeffective 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.

Problem Statement
Suppose that the expected initial defective rate of items produced in production lines is given as a constant  and that we need repeated inspections in order to meet the constraints of the target AOQ level (  ).Let  be a set of  available inspectors, represented as { 1 ,  2 , . . .,   }, and let   (∈ ) be the inspector, selected from the set  and assigned to the th inspection cycle.The following assumptions are made: (1)   examines a single characteristic of an item at a time and examines all the items given during the th inspection cycle, and each inspector must be assigned only once to one cycle.
( (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   .The above procedure is continued until we obtain an AOQ less than or equal to   .
Note that the total number of possible sequences will be as many as ∑  =1 (   ) !, where (   ) = !/(−)!!, since the number of combinations of  inspectors from  is (   ), and the number of permutations (or sequences) of  inspectors from a combination of  inspectors is !.

Analysis and Optimization of Our
Inspection Problem Then, for an integer  with 1 ≤  ≤ , we have , where  0 = (1 − ) 0 and  0 =  0 .Since the items classified as conforming by   are accumulated in a specific storage area, we have   =   +   and   =   +   .Hence, we have After reworking at the th rework operation, we have After completing the 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 th rework operation, AOQ(Ψ()) can be derived as For convenience, let   = (1 − )  + (1 −   ) and Γ  = ∏  =1   for 1 ≤  ≤ .Note that   can be interpreted as the probability with which   classifies an item as nonconforming and that 0 <   < 1.From the four recurrence relations equations ( 1) through ( 4), the mathematical expressions for   ,   ,   ,   , and AOQ(Ψ()) can be derived as closed-form solutions in the following lemma.

Lemma 1. For an integer
(5) Proof.Using ( 3) and ( 4),   can be reduced to Using ( 8), we have , we have Using ( 9), since Using ( 5), (8), and ( 9), we have Example 2. When ( 0 , , ) = (10,000 units, 5%, 4) and {(  ,   ),  = 1, 2, 3, 4} are given as shown in Table 1, compute AOQ(Ψ( 2)) for all possible sequences of Ψ( 2) and find the minimum of AOQ(Ψ(2)).Suppose that Ψ(2) = ⟨3, 1⟩ is scheduled; Inspector 3 is assigned to the first inspection cycle and Inspector 1 to the second.Since  0 = 9,500 and  0 = 500, Inspector 3 has  1 = 9,432,  1 = 68,  1 = 23, and  1 = 477 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  1 =  1 + 1 = 9,455 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  1 = 545 and these items are sent to the rework shop, where they are repaired with  1 = 518 and  1 = 27 and sent to the second inspection cycle assigned by Inspector 1. Now, since  1 = 518 and  1 = 27, Inspector 1 has  2 = 513,  2 = 5,  2 = 1, and  2 = 26 after completion of the second inspection cycle.It follows that the number of items sent to the storage area is  2 =  2 +  2 = 515 since those items have been classified as conforming by Inspector 1.The number of items rejected by Inspector 1 is  2 = 31 and these items are sent to the rework shop, where they are repaired with  2 = 29 and  2 = 2, and they are sent to the storage area instead of the third inspection cycle since two inspection cycles are given.
Since the total number of nonconforming items in the storage area is  1 +  2 +  2 = 25.77, the value of AOQ(⟨3, 1⟩) 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 2 ) 2! = 12 and that the sequence ⟨1, 3⟩ 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 AOQ(Ψ(1)).Suppose that   with (  ,   ) from a set  of  available inspectors is assigned to the single inspection cycle.Then, from Lemma 1-(5), AOQ(Ψ(1)) can be expressed as (1 − )V  +  2 , where V  =   +   .Since  is constant, AOQ(Ψ(1)) is minimized if and only if the inspector with V min = min 1≤≤ [V  ] is assigned to the cycle.In the above example, since V min = V 3 = 5.32%, Inspector 3 must be allocated in order to attain the minimum AOQ = 5,027 ppm.
Proof.It is enough to prove that if Ψ * () is not one of increasing sequences, then Ψ * () does not give a minimum AOQ among a set of Ψ().If Ψ * () does not satisfy increasing order, then there must exist at least two adjacent inspectors violating -increasing order, and, using Lemma 3, we can further reduce AOQ by swapping the adjacent inspectors.Thus, Ψ * () does not give a minimum AOQ among a set of Ψ(), and Lemma 4 holds.
It remains whether or not an inspector with (  ,   ) satisfying V  < 1 exists among unassigned inspectors of the set .However, since V  do not exceed 100% in most practical cases, we assume hereafter that V  < 1 for 1 ≤  ≤ .

An Optimal Algorithm.
Using the previously proven properties, we can construct an optimal algorithm, ALGSP, for determining both  * and Ψ * ( * ) 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 -increasing set   from .Without loss of generality, let   be {1, 2, . . ., }.
Even though the time complexity of ALGSP is (2  ), our algorithm is practically efficient since the optimal number of inspection-rework cycles is usually less than five cycles in most practical cases.

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 Table 6:   (3), Ψ  (  (3)), and AOQ(Ψ  (  (3))) when  = 3. 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 (2  ).
Our strong conjecture is that our problem is NPcomplete.Not by our conjecture but by our computational experience, we would like to suggest two heuristic algorithms with the time complexity of (⌈log ⌉) based on -increasing or V-increasing order.The near-optimal sequence with size  can be determined just by taking the first  inspectors from the set of available inspectors listed in -increasing or V-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.

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