Mathematical Problems in Engineering

Mathematical Problems in Engineering / 2015 / Article

Research Article | Open Access

Volume 2015 |Article ID 601029 | https://doi.org/10.1155/2015/601029

Moon Hee Yang, Jae Hyung Cho, "A Nonpolynomial Optimal Algorithm for Sequencing Inspectors in a Repeat Inspection System with Rework", Mathematical Problems in Engineering, vol. 2015, Article ID 601029, 9 pages, 2015. https://doi.org/10.1155/2015/601029

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

Academic Editor: Francisco Chicano
Received26 Nov 2014
Accepted27 Apr 2015
Published03 Sep 2015

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 .

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 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 , 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.(2) 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 . The error probabilities of inspectors are not identical. That is, for integers and , with .(3)The item inspected by is accepted as conforming only if the characteristic is correctly or falsely classified as conforming by . Otherwise, it is rejected as nonconforming. All the items accepted by are not reinspected anymore but accumulated in a specific storage area, and all the items rejected by are reworked by a repairman at a constant defective rate ().(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 .

Let , represented by , be a sequence of inspectors selected from and assigned to inspection cycles. Let be the sequence with size from such that , where is an AOQ resulting from a sequence . Now, under the assumption that , our problem can be described as follows:SP:given , , , and as input variables where , determine both and simultaneously, where is the minimum number of inspection-rework cycles such that .

Note that the total number of possible sequences will be as many as , 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 .

3. Analysis and Optimization of Our Inspection Problem

3.1. Derivation of AOQ and Properties

Suppose that a sequence with size is given as for . For an integer with , let = expected number of conforming items entering the ()-st inspection cycle, = expected number of nonconforming items entering the ()-st inspection cycle, = expected number of items correctly accepted by , = expected number of items falsely rejected by , = expected number of items falsely accepted by , = expected number of items correctly rejected by , = expected number of items entering the th rework operation, = expected number of items entering the first cycle, = expected number of items correctly or falsely accepted by .Then, for an integer with , we have , , , and , where and . 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, can be derived as

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

Lemma 1. For an integer with , (1),(2),(3),(4)(5)

Proof. Using (3) and (4), can be reduced toUsing (8), we haveSince , from (1), we have Using (9), since , we have Using (5), (8), and (9), we have

Example 2. When = (10,000 units, 5%, 4) and are given as shown in Table 1, compute for all possible sequences of and find the minimum of .


(%) (%) (%) (%) (%) (%)

10.924.505.420.045430.724.605.320.0463
20.974.515.480.045540.855.186.030.0522

Suppose that is scheduled; Inspector 3 is assigned to the first inspection cycle and Inspector 1 to the second. Since and , Inspector 3 has , , , and 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 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 and these items are sent to the rework shop, where they are repaired with and and sent to the second inspection cycle assigned by Inspector 1. Now, since and , Inspector 1 has , , , and after completion of the second inspection cycle. It follows that the number of items sent to the storage area is since those items have been classified as conforming by Inspector 1. The number of items rejected by Inspector 1 is and these items are sent to the rework shop, where they are repaired with and , and they are sent to the storage area instead of the third inspection cycle since two inspection cycles are given.


RepairRepair (ppm)

9,41387234789,435565537285315127533323122,538
9,40892234779,430570541285365127537323122,544
9,41387234789,435565537285334127534312922,534
9,43268234779,455545518275135126515312922,577
9,41387234789,435565537285325127534313022,553
9,41981264749,445555527285225126524313022,872
9,40892234779,430570541285374127539313022,541
9,43268234779,455545518275135126514313022,578
9,40892234779,430570541285375127538323022,561
9,41981264749,445555527285225126523323022,873
9,43268234779,455545518275144126515302922,593
9,41981264749,445555527285234126525302922,869

Since the total number of nonconforming items in the storage area is , the value of 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 and that the sequence 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 . Suppose that with from a set of available inspectors is assigned to the single inspection cycle. Then, from Lemma 1-(5), can be expressed as , where . Since is constant, is minimized if and only if the inspector with is assigned to the cycle. In the above example, since , Inspector 3 must be allocated in order to attain the minimum AOQ = 5,027 ppm.

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

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

In order to prove generally that is one of -increasing sequences with size when , let = a subset of inspectors selected from , = a collection of subsets with size of; that is, , = the th sequence or permutation generated from , = a collection of sequences with size generated from ; that is, .

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

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

Lemma 3. if and only if for and , where .

Proof. Consider two cases as follows: Case 1 for and Case 2 for .
Case 1. When , by swapping and and using Lemma 1-(5), and can be derived, respectively, asSince , we haveSince , Lemma 3 holds when .
Case 2. When , two subcases are considered as follows: Subcase 1 for and Subcase 2 for .
Subcase 1. For , from Lemma 1-(5), can be derived asSince , Lemma 3 holds when and .
Subcase 2. When , from Lemma 1-(5), can be derived as Since , Lemma 3 holds when and . Therefore, Lemma 3 holds.

Lemma 4. For , is one of -increasing sequences with size .

Proof. It is enough to prove that if is not one of -increasing sequences, then does not give a minimum AOQ among a set of . If 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, does not give a minimum AOQ among a set of , and Lemma 4 holds.

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

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

Lemma 5. For , if and only if , where .

Proof. Since , we haveSince , Lemma 5 holds when . When , by using Lemma 1-(5), we have Since , Lemma 5 holds when . Therefore Lemma 5 holds.

Lemma 6. For , if and only if , where .

Proof. If is given for an integer with and if , we can construct using Lemma 5 such that by assigning the inspector with . Since by definition, it follows that . Reversely, if , then, from Lemma 5, there exists with .

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

3.2. An Optimal Algorithm

Using the previously proven properties, we can construct an optimal algorithm, ALGSP, for determining both and 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 .

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

Step  1 (Initialization Phase):
 Take inspectors in -increasing order using and let this ordered set be .
Step  2 (Local Optimization Phase)
 For to do
 Begin
  For to
   Begin
    
    Begin
     Construct and
     Compute using Lemma 1().
     If min, then ,
       ,
    End
   End
 End
Step  3 (Global Optimization Phase)
 For to do
  Begin
   If , then , ,
   If and , then print “ greater than ” and stop
  End

Theorem 7. determined by ALGSP is an optimal sequence to SP.

Proof. is a decreasing function of from Lemma 6, and determined by ALGSP is the smallest number of inspection-rework cycles such that . Hence, determined by ALGSP is an optimal sequence to SP.

Theorem 8. ALGSP solves SP in .

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

Even though the time complexity of ALGSP is , 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 = (2,500 ppm, 5%, 4) and that are given as shown in Table 1. Using ALGSP, the optimal solution to our example will be obtained as follows.

Step 1. The -increasing sequence is and .

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



1150755075
251035075
350275027
453645027

2125382538
225342534
325532534
425412534
525612534
625932534

3123942394
223952394
323962394
424032394

412386