ch10 acceptance sampling

8/17/2019 Ch10 Acceptance Sampling

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10

SUPPLEMENT OUTLINE

Introduction and Sampling Plans, 2

Operating Characteristic Curve, 3

Determining Single SamplingPlans, 9

Average Quality of Inspected Lotsand a Related Sampling Plan, 12

Key Terms, 14

Solved Problems, 14

Discussion and ReviewQuestions, 16

Internet Exercises, 16

Problems, 16

Mini-case: CRYSTAL S.A., 19

Mini-case: Gustave RoussyInstitute, 20

Acceptance Sampling

SUPPLEMENT TO

CHAPTER

LEARNING OBJECTIVES

Af ter completing th is supplement,

you should be able to:

LO1 Explain acceptance sampling,and contrast single andmultiple sampling plans.

LO2 Construct and use anoperating characteristic curve.

LO3 Determine single samplingplans.

LO4 Determine the averagequality of inspected lots, anddetermine a related sampling

plan.


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PART FOUR Quality2

LO1 INTRODUCTION AND SAMPLING PLANS Acceptance sampling is a form of inspection that is applied to lots or batches of itemseither before or after a process. In the majority of cases, the lots represent incoming pur-chased items or final products awaiting shipment to warehouses or customers. The pur-pose of acceptance sampling is to decide whether a lot satisfies predetermined standards(specifications) for important characteristics of the item. Lots that satisfy these standards

arepassed or accepted ; those that do not are rejected. Rejected lots may be subjected to100 percent inspection or, if purchased, returned to the supplier for credit or replacement(especially if destructive testing is involved).

The alternatives to acceptance sampling are (a) 100% inspection and (b) no inspection. The decision of which one to choose is mainly based on the costs. A measure used is thebreakeven point BEP1:

BEP= cost of inspection per item / cost of later repair due to a defective item

LetP = estimated proportion of defectives in the lot. The decision is:

If P ≈ BEP, use acceptance sampling

If P > BEP, use 100% inspection

If P


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SUPPLEMENT TO CHAPTER 10 Acceptance Sampling 3

or equal to a specified number of defectives,c , the lot is accepted; otherwise (if it containsmore thanc defectives) the lot is rejected.

Double-Sampling PlansA double-sampling plan allows taking a second sample if the results of the initial sample isinconclusive. Specifically, if the quality of the initial sample is high, the lot can be acceptedwithout the need for a second sample. If the quality in the initial sample is poor, the lotis rejected (and there is also no need for a second sample). For results between those twocases, a second sample is then taken and the items inspected, after which the lot is eitheraccepted or rejected on the basis of the evidence obtained from both samples. A double-sampling plan specifies the size of the initial sample, the accept/reject criteria for the initialsample, the size of the second sample, and a single overall acceptance number.

With a double-sampling plan, two values are specified for the number of defective itemsin the first sample, a lower level, c 1, and an upper level, r 1. For instance, the lower levelmight be two defectives and the upper level might be five defectives. If the number ofdefective items in the first sample is less than or equal to the lower value (i.e., c 1), the lotis judged to be good and sampling is terminated. Conversely, if the number of defectivesin the first sample equals or exceeds the upper value (i.e., r 1), the lot is rejected. If thenumber of defectives in the first sample falls somewhere in betweenc 1 and r 1, a second

sample is taken and the total number of defectives in both samples is compared to a thirdvalue, c 2. For example, c 2 might be six. If the combined number of defectives does notexceedc 2, the lot is accepted; otherwise, the lot is rejected.

Multiple-Sampling PlansA multiple-sampling plan is similar to a double-sampling plan except that more than twosamples may be required. A multiple sampling plan will specify each sample size andtwo limits for each sample. If, for any sample, the cumulative number of defectives found(i.e., those in the present sample plus those found in all previous samples) is greater thanor equal to the upper limit specified for that sample, sampling is terminated and the lotis rejected. If the cumulative number of defectives is less than or equal to the lower limit,sampling is terminated and the lot is accepted. If the cumulative number of defectivesis between the two limits, another sample is taken. The process continues until the lot iseither accepted or rejected.

Choosing a Sampling Plan The cost and time required for inspection often dictate the type of sampling plan used. The two primary considerations are the number of samples needed and the total numberof observations required. Single-sampling plans involve only a single sample, but thesample size is larger than the expected number of observations taken under double- ormultiple-sampling plans.This stems from the fact that a very good or very poor quality lotwill often be accepted or rejected early in a multiple-sampling plan, and sampling can beterminated. Where the cost to obtain a sample is relatively high compared with the costto analyze the observations, a single-sampling plan is more desirable. Conversely, whereitem inspection costs are relatively high, such as destructive testing, it may be better to use

double or multiple sampling because the average number of items inspected per lot willbe lower. Another advantage of a single sampling plan is that it is easy to understand anduse. For this reason, we will focus on single sampling in this supplement.

LO2 OPERATING CHARACTERISTIC CURVEAn important feature of a sampling plan is how it discriminates between lots of high andlow quality. The ability of a sampling plan to discriminate between lots of high and lowquality is described by its operating characteristic (OC) curve. A typical OC curve fora single-sampling plan is shown in Figure 10S-1. The curve shows the probabilities ofaccepting lots with various qualities (propor tion defectives) . For example, it shows that alot with 3 percent defectives (a proportion of defectives of .03) would have a probability

operating characteristic(OC) curve Curve that showsthe probabilities of accepting

lots with various quality

(proportion defective).


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PART FOUR Quality4

of .80 of being accepted (or a probability of 1.00 − .80= .20 of being rejected). Note thedownward relationship: as lot quality decreases, the probability of lot acceptance decreases,although the relationship is not linear.

A sampling plan cannot provide perfect discrimination between good and bad lots; somelow-quality lots will inevitably be accepted, and some high-quality lots will inevitably be rejected.Even lots containing more than 20 percent defectives still have some probability of being accepted,whereas lots with as few as 3 percent defectives have some chance of being rejected.

The degree to which a sampling plan discriminates between good and bad lots is a func-tion of the steepness of its OC curve: the steeper the curve, the more discriminating thesampling plan (see Figure 10S-2.) Principles of sampling imply that the larger the samplesizen , the steeper the curve. Note the curve for an ideal plan (i.e., one that can discriminateperfectly between good and bad lots). To achieve that, you need to inspect 100 percent ofeach lot. Obviously, if you are going to do that, theoretically all of the defectives can beeliminated (although errors due to boredom might result in a few defectives remaining).However, the cost of additional discrimination may be larger than the cost of additionalinspection (i.e., it may not be cost-effective).

Lot quality (proportion defective)0 .05 .10 .15 .20 .25

1.00

.90

.80

.70

.60

.50

.40

.30

.20

.10

.00

P r o

a

i i t y

o

a c c e p

t i n g

t

e

o t

3%

FIGURE 10S-1

A typical OC curve

0“Good”

P r o

a

i i t y

o

a c c e p t i n g

t

e

o t

“Bad”

Ideal

Better

1.00

Not very discriminating

Lot quality (proportion defective)

FIGURE 10S-2

The steeper the OC cur ve,

the more discriminating the

sampli ng plan


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Buyers (or consumers) are generally willing to accept lots that contain small percent-ages of defectives as “good,” especially if the cost related to a few defects is low. Oftenthis percentage is in the range of .01 to 2 percent. This figure is known as theacceptablequality level (AQL). AQL should be set based on the criticality of the characteristic thatis being inspected—the more critical the characteristic, the smaller the AQL should be.For example, for spoons, the defect of being cracked may have AQL of .5% whereas thedefect of being scratched may have AQL of 2%. Also, AQL should be set so that the good

incoming lots have quality equal to AQL. Otherwise the supplier will be overwhelmedwith rejected lots and there may not be enough accepted lots for the buyer to continueproduction. For example, if the incoming lots generally have 1 to 4% defectives, then AQLshould be set to 1%.

Because of the inability of random sampling to identify all lots that contain more thanAQL percentage of defectives, consumers (buyers) recognize that some lots that actuallycontain more defectives than AQL will be accepted. However, there is usually an upperlimit on the percentage of defective items that a consumer is willing to tolerate in acceptedlots. The percentage just larger than this is known as thelot tolerance percent defective(LTPD). Thus, consumers want quality equal to or better than the AQL, and are willing tolive with poorer quality, but they prefer not to accept any lots with a defective percent-age greater or equal to the LTPD. LTPD should also be set based on the criticality of the

characteristic that is being inspected—the more critical the characteristic, the smaller theLTPD. Also, LTPD should be set so that the bad incoming lots have quality equal to LTPD.Otherwise the seller may receive more rejected lots than expected or the buyer may losethe opportunity to demand higher quality for bad lots. For example, if the incoming lotsgenerally have 1 to 4% defectives, then LTPD should be set to 4%.

As mentioned above, sampling plans are not perfect in discriminating between good andbad lots, i.e., mistakes will be made in acceptance of bad lots and rejection of good lots. The probability that a bad lot containing defectives equal to the LTPD will be accepted isknown as theconsumer’s risk , orbeta (β ), or the probability of making aType II error. Onthe other hand, the probability that a good lot containing defectives equal to the AQL willbe rejected is known as theproducer’s risk , or alpha (α ), or the probability of making aType I error. Many sampling plans are designed to have a producer’s risk of 5 percent anda consumer’s risk of 10 percent, although other combinations are also used. Figure 10S-3

illustrates an OC curve with the AQL, LTPD, producer’s risk, and consumer’s risk. Notethat the probability of accepting a lot with AQL quality is 1 – α.

A certain amount of insight is gained by constructing an OC curve. The probability ofobserving up to and includingc defectives in a sample of sizen from a lot with proportionof defectivesP is given by cumulative hyper-geometric formula:

P x c

NP

x

N NP

n x

N

n

NP

x NP x

N

x

c

( )

!

!( )!

(

≤ =

−−

= −−

=∑

0

N NP

n x N NP n x

N

n N n x

c )!

( )!( )!!

!( )!

− − − +

−=

∑0

wherex =number of defectives in the sample, and N n represents the number of combina-

tions (different ways) of choosing a sample ofn from the lotN , etc. Note that the expectednumber of defectives in the lot isNP . Intuitively, probability of observingx defectives ina sample of size n from a lot of sizeN equals the number of ways to choosex defectivesfrom all the defectives in the lot (NP ) times numbers of ways to pick n – x non-defectivesfrom all the non-defectives in the lot (N – NP) over the number of ways to choose a sampleof n from the population N . For example, suppose a short quiz will contain two ques-tions. Each question is from a different topic. There are four topics. You know two out ofthe four topics. There will be six different pairs of topics in the quiz. One pair you knowboth questions, and another you don’t know either question. The remaining pairs contain

acceptable quality level(AQL) The percentage ofdefects at which a consumer

(buyer) is willing to accept lots

as “good.”

lot tolerance percentdefective (LTPD) Thepercentage just larger than the

upper limit of the percentage

of defectives of a lot that aconsumer is willing to accept.

consumer’s risk Theprobability that a bad lot

containing defects equal to the

LTPD will be accepted on the

basis of sample data.

producer’s risk Theprobability that a good lot

containing defects equal to the

AQL will be rejected on the

basis of sample data.

(10S-1)


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PART FOUR Quality6

one question you know and one you don’t. There are four of these because there are twoways to get the question you know multiplied by two ways to get the question you don’tknow. The probably that you will know both questions is 1/6, the probably that you willnot know either question is 1/6, and the probably that you will know exactly 1 of the twoquestions is 4/6.

Because there are no hyper-geometric tables, if the lot sizeN

is large relative to sample sizen (so that

n

N ≤ 0. ) we can approximate the hyper-geometric probabilities by binomial

probabilities. The difference is that binomial probabilities assume that a sampled item isput back in a lot after being inspected before the next item is selected from the lot. Theprobability of observing up to and including c defectives in a sample (with replacement)of size n from a lot with proportion defectiveP is given by cumulative binomial formula:

P x c n

x n x P P x n x

x

c

( )!

!( )!( )≤ =

− − −

=∑ 1

0

(10S-2)

wherex = number of defectives in the sample.

Draw the OC curve for a situation in which a sample of n = 10 items is drawn from a lotcontainingN = 2,000 items, and the lot is accepted if no more thanc = 1 defect is foundand rejected if 2 or more defects are found in the sample.

Example S-2

Solution Because the sample size is small relative to the lot size (10/2000 = .005


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.10

P r o

b a

b i l i t y

o f a c c e p t a n c e

1.00

.90

.80

.70

.60

.50

.40

.30

.20

.10

.00

Proportion defective in lot

.20 .30 .40 .50 .600

.9139

.7361

.5443

.3758

.2440

.1493

.0464.0860

.0233.0107

.0045.0017

FIGURE 10S-4

QC curve for single sampli ng

plan n = 10, c = 1

To use the table, select various lot qualities (values of p listed across the top of the table),beginning with .05, and find the probability that a lot with that percentage of defects wouldbe accepted (i.e., the probability of finding zero or one defect in this case). For p = .05,the probability of one or no defects is .9139. For a lot with 10 percent defective (i.e., aproportion defective of .10), the probability of one or fewer defects drops to .7361, andfor 15 percent defective, the probability of acceptance is .5443. In effect, you simply readthe probabilities across the row for c = 1. By plotting these points (e.g., .05 and .9139,.10 and .7361) on a graph and connecting them, you obtain the OC curve illustrated inFigure 10S-4.

If the lot sizeN is large relative to sample size n (so thatn /N ≤ 0.1) andnp


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PART FOUR Quality8

The probabilities of acceptance, P ac , are drawn against lot proportion defectives P toconstruct the OC curve:

Use the cumulative Poisson table at the end of the textbook to construct an OC curve forthe following single sampling plan:

N = 5,000,n = 80, c = 2

Example S-3

Solution Note that 80/5000= .016


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Effect of sample size n on the shape of the OC curve

Proportion defective

.01.00

.10

.20

.30

.40

.50

.60

.70

.80

.90

1.00

.02 .03 .04 .05 .06 .07 .08 .09 .1 .11 .12 .13

P r o

a

i i t y

o

a c c e p t a n c e

o

t

e

o t

c = 2

n = 80

n = 50

n

= 125n = 200

On the other hand, for fixed sample size (n = 80), the OC curve becomes steeper (willhave more discriminatory power) as acceptance number c decreases (see the graph below).

Perhaps it is easier to understand this by considering very largec values: in this case, mostlots will be accepted, i.e., no discrimination.

Effect of acceptance number c on the shape of the OC curve

Proportion defective.01

.00

.10.20

.30

.40

.50

.60

.70

.80

.901.00

.02 .03 .04 .05 .06 .07 .08 .09 .1 .11 .12 .13

P r o

a

i i t y

o

a c c e p t a n c e

n = 80

c = 0c = 1c = 2c = 3

LO3 DETERMINING SINGLE SAMPLING PLANSA sampling plan and its operating characteristic (OC) curve have a one-to-one relation-ship. Therefore, determining the sample size n and acceptance numberc is equivalent todetermining the sampling plan’s OC curve. One way to determine an OC curve is to specifytwo points on it, for example, the two points (AQL, 1 – α) and (LTPD,β). There are otherapproaches to determine the sampling plan as well. Two of the important methods are

Dodge-Romig and MIL-STD-105E. Dodge and Romig in AT&T used the point (LTPD,β) and expected lot proportion defectiveP to minimize the expected total items inspected(assuming that a rejected lot is completely inspected). Another approach, MIL-STD-105E,created by the U.S. Armed Forces, uses the point (AQL, 1 – α), lot size N , and a choseninspection level to determine the sampling plan. We will illustrate these methods below.

Using (AQL, 1 – α ) and (LTPD, β )Substituting (AQL, 1 – α ) in the cumulative binomial formula 10S-2:

1 10

− =−

− −=

∑α n x n x AQL AQLx n x x

c !

!( )!( ) ( ) (10S-4)


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PART FOUR Quality10

Using (LTPD, β) Lot size N , and Expected Lot ProportionDefective P

– to Minimize Expected Total Items Inspected

(The Dodge-Romig Approach)Let I = Expected total items inspected. If a lot is rejected, we assume that it is 100%inspected. In this case, I = n + Probability of rejecting the lot× (N – n ). Dodge and Romigused the Poisson approximation to binomial to calculate the probability of accepting a lot. Therefore, we have:

I n e nP

x N n

nP x

x

c

= + −

× −

− −

=∑1

0

( )

!( ) (10S-6)

LetP t = LTPD. We also use Poisson probabilities to ensure that the probability of accept-ing a lot with proportion of defectives= P t isβ. Dodge-Romig used β = .10. Therefore,we have:

e nP

x

nP t

x

x

c t −

=∑ =( )! .

0

1 (10S-7)

The Dodge-Romig approach is to determinen andc so that I, given in (10S-6), is mini-mized subject to satisfying the equation (10S-7). Again, this problem is difficult to solvebecause of the non-linear functions involved. Fortunately, Dodge and Romig have providedsome curves to do this: Figures DR1 to DR3 at the end of this chapter supplement.

Substituting (LTPD,β) in the cumulative binomial formula 10S-2:

β =−

− −=

∑ n x n x LTPD LTPD x n x x

c !

!( )!( ) ( )1

0

(10S-5)

We can try to solve equations (10S-4) and (10S-5) simultaneously to determine the twounknown quantities n and c . However, this is not easy because these equations are non-linear. Fortunately, Larson has determined a nomograph (a graphical calculating device)that will provide the solution (see page 24 of this supplement).

Suppose AQL = .02 withα = 5% and LTPD = .08 withβ = 10%. Use Larson’s nomographat the end of this supplement to determinen andc .

Example S-4

Solution Larson’s nomograph on page 24 can be used as follows: the vertical line on the left-handside is for lot percentage defectives such as AQL and LTPD. The vertical line on the right-hand side is for the probability of lot acceptance such as (1 –α) andβ. Connect AQL with(1 – α) and LTPD with β with straight lines. The intersection of these two lines gives thesample sizen and acceptance numberc . In this case, n = 90 andc = 3 (see page 24).

Suppose that the expected lot proportion defectiveP = .02, LTPD = .08 withβ = 10%, andlot sizeN = 200. Determine n and c using Dodge-Romig approach.

Example S-5

Solution Fi rst, we need to find the ratioP

LTPD = =

.

..

02

082 . Next, we need to find

LTPD ×N =.08(200) =16. Then, in Figure DR1 on page 25 at the end of this supplement,we draw a vertical line at x -axis value .25 and a horizontal line at y -axis value 16. Theyintersect in an area associated with c =2. Next, in Figure DR2 on page 26, we draw avertical line at x -axis value 16 and see at what horizontal line it intersects the c =2 curve.


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They intersect at approximately 5. This isn ×LTPD. Therefore, n =5 / .08 =62.5 or 63. To determine the minimum expected total items inspected Imin, in Figure DR3 on page 27,we draw a vertical line at x -axis value .25 and a horizontal line at y -axis value 16. Theyintersect close to curve numbered 6. This is LTPD ×Imin. Therefore, Imin =6 / .08 =75.

Using MIL-STD-105E The MIL-STD-105E uses the point (AQL, .95), the lot sizeN , the level of inspectiondesired, and the type of inspection required (depending on the past history of the supplier)to determinen andc .

The inspection level determines the relationship between the lot size and sample size. There are three general inspection levels (I, II, and III) and four special inspection levels(S-1 to S-4). Level II is designated as normal. Level I requires about half the amount ofinspection as level II, and is used when reduced sampling is required and a lower level ofdiscrimination can be tolerated. Level III requires about twice the amount of inspection aslevel II, and is used when more discrimination is needed. The four special inspection levelsS-1,S-2,S-3,S-4 use very small samples, and should be used when small sample sizes arenecessary, and when large sampling risks can be tolerated.

There are three types of inspection (other than Discontinue inspection): Normal inspec-tion, Reduced inspection, and Tightened inspection. Reduced inspection results in smallern (less discrimination after a good history) and Tightened inspection results in larger n (more discrimination after a bad history). The Discontinue inspection requires correctiveaction by the supplier before any new lots are purchased. The switching rules between the4 inspection types are displayed below:2

5 consecutive batches not rejected

1 batch not accepted

Reducedinspection

Normalinspection

Tightenedinspection

Discontinueinspection

2 of 5 consecutive batches rejected

10 consecutive batches remainon tightened inspection

5 consecutive batches not rejected

The following Web site will providen and c according to the MIL-STD-105E: http://www.sqconline.com/mil-std-105.html. Alternatively, a copy of the MIL-STD-105E docu-ment, which includes results tables, can be found online, e.g., at http://www.dianyuan.com/bbs/u/39/1142140688.pdf.

Suppose lot sizeN is 2,000 and AQL is 1 percent. We would like to use Normal type of

inspection at general inspection level 2. Determine the sample size n and the acceptancenumberc .

Example S-6

SolutionIn www.sqconline.com/mil-std-105.html we choose the right range forN and check to seethat default values for AQL, inspection level, and type of inspection are right. We clickon Submit.

2 http://www.sqconline.com/switching_rules_enter.php4


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PART FOUR Quality12

http://www.sqconline.com

The values for n and c for single sampling appear: n = 125 and c = 3 (see the top leftbox below). As an added bonus, the results for double sampling plan, which has a similarOC curve, and the OC curve are also provided:


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Construct the AOQ curve forN = 500,n = 10, andc = 1 using formula 10S-9: Example S-7

Solution

Incoming proportion defective

.10 .20 .30 .40

.08

.06

.04

.02

.00

A O Q

( O u t g o i n g p r o p o r t i o

n d e

f e c t i v e )

Approximate AOQL = .082

LO4 AVERAGE QUALITY OF INSPECTED LOTS AND A RELATED SAMPLING PLAN

It is fair to expect that acceptance sampling would reduce the proportion of defective itemsaccepted. Indeed, this is the case, provided that the rejected lots are not re-submitted foracceptance sampling without improvement in quality. If they are, they will eventually be

accepted and the whole point of acceptance sampling is lost (i.e., the quality of accepteditems will be the same as the quality of incoming or rejected items). This fact can be seen byconsidering the following example: suppose that the probability of a bad lot being acceptedis .1. If this lot is rejected and is returned to the supplier but the supplier sends it back to thecustomer unchanged, and this process is repeated 40 times, the probability of acceptancewithin 40 tries is .1+ .9(.1)+ .92(.1)+ ... + .939(.1)= .985. Therefore, the buyer has to eitherrequire the supplier to perform 100% inspection of rejected lots or do it himself.

Theaverage outgoing quality (AOQ) of the inspected lots is average percentage defec-tive of accepted lots assuming that rejected lots are 100 percent inspected and defectiveitems in those lots are replaced with good items. AOQ can be calculated using the fol-lowing formula:

AOQ = ×

−

P p N n

N ac (10S-8)

where

P ac = Probability of accepting the lot

p = Lot proportion defective

N = Lot size

n = Sample size

In practice, the last term in (10S-8) is often omitted because it is usually close to 1.0and therefore has little effect on the resulting values. The formula then becomes

AOQ = ×P p ac (10S-9)

average outgoing quality(AOQ) Average percentagedefective of accepted lots

assuming that rejected lots

are 100 percent inspected and

defective items in those lots are

replaced with good items.

Let values of p vary from .05 to .40 in steps of .05. You can read the probabilities ofacceptance,P ac , from the binomial table at the end of this supplement.

AOQ = ×P p ac

P P ac AOQ

.05 .9139 .046

.10 .7361 .074

.15 .5443 .082

.20 .3758 .075

.25 .2440 .061

.30 .1493 .045

.35 .0860 .030

.40 .0464 .019


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PART FOUR Quality14

Note that the average outgoing quality is best for either very good lots or very bad lots(i.e., the outgoing proportion defective is least for lots with either very low or very highincoming proportion defective). The reason very bad lots also will have high outgoingquality is that they will likely be rejected and then 100% inspected and rectified.

A quantity of interest to the buyer is the maximum outgoing proportion defective, alsocalled Average Outgoing Quality Limit (AOQL). AOQL is the worst quality of outgoing(i.e., accepted) items. In Example 7, The AOQL is approximately 8 percent, associatedwith incoming lots of approximately 15% defective.

An approximate value for AOQL can be obtained using the formula.

( . )4

125 1n

c + .3 For

Example 7, AOQL or≈ × + =.

( . ) . , %4

10125 1 1 09 9 .

An AOQL based Sampling PlanDodge and Romig have provided a sampling plan involving AOQL similar to that givenby them in the previous section, but instead of using the (LTPD, β), a desired value for theAOQL is assumed. Recall thatP = expected lot proportion defective and I = Expected totalitems inspected. As for the case of (LTPD, β), I is determined from equation (10S-6).

LetP L = the desired value of AOQL, i.e., a limit on the worst outgoing quality possible,andP the proportion defective of an incoming lot. Using equation (10S-8), we get:

Max PN n

N

e nP

x P

nP x

x

c

L

−

=

− −

=∑ ( )!

0

(10S-10)

Taking the derivative of the term inside the curly brackets in (10S-10) with respect toP and setting it equal to 0, after some work we get

e nP

x

e nP

c

nP x

x

c nP c − −

=

− +

∑ =( )!( )

!0

1

(10S-11)

LetP m be the solution to equation (10S-11) for P , given values for n and c. P m is thex -axis value of lot proportion defective that is associated with P L. Substituting P m for P

in (10S-10) and omitting the Max [becauseP m maximizes the left term in formula 10S-10],we get:

P N n

N

e nP

x P m

nP m

x

x

c

L

m −

=

− −

=∑ ( )!

0

(10S-12)

The Dodge-Romig approach is to determinen andc so that I in (10S-6) is minimizedsubject to determiningP m from (10S-11) and satisfying equation (10S-12) for AOQL. Again,this problem is difficult to solve because of the non-linear functions involved. Fortunately,Dodge and Romig have provided Figure DR4 (on page 28) and Table DR1 (on page 29)at the end of this supplement that can be used to determinen andc .

average outgoing qualitylimit (AOQL) The worstquality of outgoing (i.e.,

accepted) items.

3

J. M. Juran and F. M. Gryna,Quali ty Planning and Analysis , 3rd ed., 1993, New York: McGraw-Hill, p. 25.13.

Suppose that the expected lot proportion defective P = .02, AOQL = .05, and lot sizeN = 200. Determinen andc .

First, we need to find the ratioP

AOQL = =

.

..

02

05. Next, we need to findP N × = =. ( ) .02 200 4

Then, in Figure DR4 on page 28 at the end of this supplement we draw a vertical lineat x -axis value .4 and a horizontal line at y -axis value 4. These lines intersect in an areaassociated with c = 1. Next, in Table DR1 on page 29, for c = 1, we pick up x = 1.62 andy = .84. Now we can calculaten = yN/(AOQL × N + y ) = .84(200)/(.05× 200+ .84)= 15.5,round to 16. AOQL occurs atP m = x /n = 1.62 / 16= .10.

Example S-8

Solution


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Key Termsacceptable quality level (AQL), 5acceptance sampling, 2

average outgoing quality (AOQ), 12

average outgoing quality limit (AOQL), 13

consumer’s risk, 5

lot tolerance percent defective (LTPD), 5

operating characteristic (OC) curve, 3

producer’s risk, 5

sampling plans, 2

Problem 2Shipments of 300 boxes of glassware are received at a warehouse. Random samples of fiveboxes are checked, and the lot is rejected if more than one box contains a breakage. Construct

the OC curve for this sampling plan.

Solved ProblemsA process for manufacturing shock absorbers for light trucks produces .5 percent defectives. In-spection cost per shock is $.40. Currently 100 percent inspection is performed, which is assumed

to catch all the defectives. If a defective shock absorber were to be installed on a truck, it must

eventually be replaced at a cost of $120 per shock. Is 100 percent inspection justified?

SolutionBecause n/N =5/300= .0167


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PART FOUR Quality16

Develop the AOQ curve for the previous problem using formula 10S-9.

What is acceptance sampling and what is its purpose? (LO1)1.

How does acceptance sampling differ from process control using control charts? (LO1)2.

When should a buyer use sampling inspection vs. 100% inspection vs. no inspection?3.

(LO1)

What general factors govern the choice between single-sampling and multiple-sampling plans?4.

(LO1)

What is an operating characteristic curve, and how is it useful in acceptance sampling?5.

(LO2)

Briefly explain or define each of these terms. (LO2)6.

AQL.a.

LTPD.b.

Producer’s risk.c.

Consumer’s risk.d.

When can each of the following distributions be used in calculating the probability of accep-7.

tance of a lot? (LO2)

Hyper-geometric.a.

Binomial.b.

Poisson.c.

Discussion andReview Questions

Problem 3

Solution AOQ = P

ac × p

(Values of probability of acceptance P ac , can be taken from the top portion of the binomial table

shown on the previous page)

Incoming proportion defective, p

.16

.12

.08

.04

0 .1 .2 .3 .4 .5 .6 .7 .8

Max = .158

A O Q

( O u t g o i n g p r o p o r t i o n

d e

f e c t i v e )

p P ac AOQ p P ac AOQ

.05 .9974 .050 .45 .2562 .115

.10 .9185 .092 .50 .1875 .094

.15 .8352 .125 .55 .1312 .072

.20 .7373 .147 .60 .0870 .052

.25 .6328 .158 .65 .0540 .035

.30 .5258 .158 .70 .0380 .027

.35 .4284 .150 .75 .0156 .012

.40 .3370 .135 .80 .0067 .005


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Visit http://www.itl.nist.gov/div898/handbook/pmc/section2/pmc22.htm, read about (a)1.

sequential, and (b) skip lot sampling plans, and define or explain them. (LO1)

Visit http://www.astm.org/SNEWS/JF_2010/datapoints_jf10.html, and summarize why we2.

still need Acceptance Sampling (given the existence of control charts). (LO1)

Internet Exercises

An assembly operation for the trigger mechanism of a semiautomatic spray gun produces a1.

small percentage of defective mechanisms. Management must decide whether to continue the

current practice of 100 percent inspection, perform acceptance sampling, or replace defective

mechanisms after final assembly when all guns are inspected. Replacement at final assembly

costs $30 each; inspection during trigger assembly costs $12 per hour for labour and overhead.

The inspection rate is one trigger per minute. (LO1)

Would 100 percent inspection during trigger assembly be justified if there are (1) 4 percenta.

defective? (2) 1 percent defective?

At what point would management prefer acceptance sampling?b.

Random samples of2. n = 20 circuit breakers are tested for damage caused by shipment in

each lot of 4,000 received. Lots with more than one defective are pulled and subjected to 100

percent inspection. (LO2 & 4)

Construct the OC curve for this sampling plan.a.

Construct the AOQ curve for this plan using formula 10S-9, assuming defectives foundb.

during 100 percent inspection are replaced with good parts. What is the approximate

AOQL?

Auditors use a technique called3. discovery sampli ng in which a random sample of items is

inspected. If any defects are found, the entire lot is subjected to 100 percent inspection. (LO2

& 4)

Draw an OC curve for the case where a sample of 15 credit accounts will be inspected outa.

of a total of 8,000 accounts.

Draw an OC curve for the case where 150 accounts out of 8,000 accounts will be examined.b.

(Hint: Use p = .001, .002, .003, . ..)

Draw the AOQ curve for the preceding case (part b), and determine the approximatec.

AOQL.

Random samples of lots of textbooks are inspected for defective books just prior to shipment4.

to the publisher’s warehouse. Each lot contains 3,000 books. (LO2)

On a single graph, construct OC curves fora. n = 100 and (1) c = 0, (2) c = 1, and (3)c = 2.(Hint: Use p = .001, .002, .003, . ..)

On a single graph, construct OC curves forb. c = 2 and (1) n = 5, (2) n = 20, and (3)

n = 120.

A manufacturer receives shipments of several thousand parts from a supplier every week. The5.

manufacturer has the option of inspection before accepting the parts. Inspection cost is $1 per

unit. If parts are not inspected, defectives become apparent during a later assembly operation,

at which time replacement cost is $6.25 per unit. (LO1 & 2)

At what proportion defective would the manufacturer prefer acceptance sampling?a.

For the sample sizeb. n = 15, what acceptance numberc would result in probability of accep-

tance close to .95 for AQL = 2%?

Problems

When would you use each of the four methods given in this supplement for determining single8.

sampling plans? (LO3)

Explain or define each of the following: (LO4)9.

AOQ.a.

AOQL.b.


18/29

PART FOUR Quality18

If the shipment actually contains 1 percent defective items and AQLc. =2 percent:

What is the correct decision?i.

What is the probability that the lot would be accepted if acceptable numberii. c = 0?

What is the probability that it would be rejected ifiii. c = 0?

Answer the questions in partiv. c for a shipment that contains 3 percent defective

items.

Suppose there are two defective units in a sample. (LO2 & 4)6.

If the acceptance number isa. c = 1, what decision should be made? What type of error is

possible?

If the acceptance number isb. c = 3, what decision should be made? What type of error is

possible?

Use formula 10S-9 to determine the average outgoing quality for each of the followingc.

percent defectives if c = 1 andn =15.

5 percent.i.

10 percent.ii.

15 percent.iii.

20 percent.iv.Suppose lot size7. N is 432 and acceptable quality level AQL is .65 percent. We would like to use

Normal inspection at general inspection level II. Determine the sample sizen and acceptance

numberc using MIL-STD-105E.4 (LO3)

A manufacturer of colour TV picture tubes is wondering if its current sampling procedure can8.

be improved.5 Currently, the defects are classified into critical (C : e.g., contaminated anode),

major (B: e.g., bent pins) and minor (A: e.g., wrong label). The company has also grouped its

customers into three groups and has one plan for each group: (LO3)

Existing Sampli ng Plans

Plan1 Plan2 Plan3

Nonconformity class C&B A C&B A C&B ALot size (N) 48 48 48 48 48 48

Documented AQL 1.0 1.0 2.5 2.5 2.5 2.5

Documented inspection severity Normal Reduced Reduced

Sample size (n) 8 8 5 5 3 3

Acceptance number (c) 0 1 1 1 0 0

For each plan and defect category, determine the sample sizen and acceptance numberc

using MIL-STD-105E. Compare your results with the current plans above.

A single sampling plan uses sample size9. n = 100 and the inspector accepts the lot if there are

2 or fewer defectives in the sample.6 You may use the Poisson approximation to answer the

following questions. (LO2 & 4)

What is the probability of accepting a lot with proportion defectivea. p = .01?

What protection does the buyer have against accepting lots with proportion defectiveb.

p = .05?

What is the average outgoing quality AOQ forc. p = .01? Use formula 10S-9.

What is the average outgoing quality AOQ ford. p = .05? Use formula 10S-9.

4 E. F. Bauer, “A Move from Attribute to Variables Acceptance Sampling in an ISO-Certified ManufacturingPlant,” M.S. thesis, California State University, Dominguez Hills, 2000.

5 E. Gamino, “Improvement to the Acceptance Control System of a Manufacturer of Color Picture Tubes,” M.S.thesis, California State University, Dominguez Hills, 2005.

6 P. W. M. John, Stati stical Methods in Engineer ing and Quali ty Assurance , New York: Wiley, 1990, p. 188.


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What is the average outgoing quality limit AOQL for this plan? You may use the approxi-e.

mation.

( . )4

125 1n

c + . *10. A manufacturer wishes to sample a purchased component used in its assembly operation.7

The company wishes to reject lots that are 5% defective. The components are received in

lots of 1,000 units, which average 2.2% defective. The supplier has agreed to perform 100%

inspection of all rejected lots. Find a sampling plan method that meets these conditions and

determinen andc . (LO3) *11. You are the quality manager for a company receiving large quantities of material from a sup-

plier in lots of 1,000 units.8 The cost of inspecting the items is $.76 per unit. The cost of repair

if bad material is introduced into your product is $15.20 per unit. A single sampling plan of

75 units with acceptance number of 2 has been suggested by one of your quality inspectors.

In the past, lots submitted by this supplier have averaged 3.4% defective. (LO1–4)

a. Is acceptance sampling economically justified?

b. If you want to accept only lots of 4% defective or better, what do you think of the sampling

plan of the inspector?

c. Suppose that rejected lots are 100% inspected. If a supplier submits many 4% defective

lots, what will be the average outgoing quality of these lots? Use formula 10S-8.

12. a. Determine a sampling plan that will have AQL = 1% with producer risk α = .05 and

LTPD = 5% with consumer riskβ = .10.9 (LO3)

b. Suppose a sample is taken according to the sampling plan derived in parta and two non-

conforming units are found. What action should be taken?

13. A housing development company buys heavy-duty nails in lots of 10,000 nails. A destruc-

tive test is performed to determine the strength of the nails.10 AQL is 1% and LTPD is 10%.

A single sample of n = 100 andc = 2 is used. Determineα andβ. (LO3)

14. A manufacturer inspects all of its shipments to its customers prior to delivery using a double

sampling MIL-STD-105E standard plan, Inspection level II, Normal type of inspection, and

an AQL of 1%.11 The lot is 500 units. (LO3)

a. Find the sampling plan used and explain it in words.

b. Upon delivery of the product, the customer also inspects the lot using a single sampling

MIL-STD-105E standard plan, AQL = 1.5%, inspection level II, and Normal type of inspec-tion. Find the customer’s sampling plan.

c. If a lot that is 10% defective is produced, calculate the probability that it will pass the

manufacturer’s inspection.

d. Calculate the probability that a 10% defective lot will pass the customer’s inspection.

e. What is the probability that a 10% defective lot will pass both inspections?

15. Find the Dodge-Romig single sampling plan for AOQL = 4%, lot size of 125, and average lot

percentage defective= 1%.12 (LO4)

16. Binder clips are packaged 12 to a box and 12 boxes to a carton.13 You have received a lot

consisting of four cartons of binder clips. (LO3)

a. Use MIL-STD-105E to determine a single sampling plan to decide whether to accept or

reject the lot. Use Inspection Level II, Normal type of inspection, and an AQL of 2.5%. b. If in inspecting your sample you find three defective binder clips, what would you do?

7 J. M. Juran and F. M. Gryna,Quali ty Planning and Analysis , 3rd ed., 1993, New York: McGraw-Hill, p. 486.8 J. M. Juran and F. M. Gryna,Quali ty Planning and Analysis , 3rd ed, 1993, New York: McGraw-Hill, p. 487.9 J. M. Juran and F. M. Gryna, Juran’s Quali ty Control Handbook , 4th ed., 1988, New York: McGraw-Hill,pp. 25.26–25.27.

10 S. Nahmias,Producti on and Operations Analysis , 3rd ed., 1997, Chicago: Irwin.11 E. I. Grant and R. S. Leavenworth, Statistical Quali ty Control , 4th ed., 1972, New York: McGraw-Hill,

p. 445.12 E. G. Schilling, Acceptance Sampling in Quali ty Control , 1982, New York: Marcel Dekker, p. 398.13 http://www.shsu.edu/ ~mgt_ves/mgt481/lesson9/lesson9.htm.


20/29

PART FOUR Quality20

MINI-CASE

Gustave Roussy InstituteAcceptance sampling is used in the Department of Clinical

Pharmacy of Gustave Roussy Institute in Villejuif, France.15

Most of the over 10,000 custom-made units of 40 or so che-

motherapy drugs produced during a semester (half year) for

5 semesters starting in 2001 were tested (at the cost of $1.50

each) to make sure that their content, dosage, and concentration

were acceptable (i.e., within±10% of the specification). Causes

of defects were investigated and fixed so that percentage defec-

tive was reduced from an average of 8.9% to 2.2%. Based on

50,000 test results, the test administrators classified the drugs

into three classes:

Category of Drugs Drugs Concerned

Group 1 (17 drugs)


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22/29


23/29


24/29

PART FOUR Quality24

Larson’s Binomial Nomograph

.01

0 1000

N u m b e r o f t r i a l s o r s a m

p l e s i z e ( n )

700

500400

300

200

140

100

70

50

40

30

20

20

1

2

3

45

O c c u r r e n c e s ( c )

7

9102030405070100140

10

5

5

10

20

N u m

b e r o

f o c c u r r

e n c e s ( c

)

40

50

70

100

140

200

.02

.03

.04

.05

.06

.07

.08

.09

.10

P r o

b a

b i l i t y

o f o c c u r r e n c e

i n

a

s i n g

l e

t r i a l ( p )

.15

.20

.25

.30

.35

.40

.45

.50

.001.10

.02

.08.95

n = 90

n

N

c = 3

.005

.01

.02

.05

.10

.20

.30

.40

P r o

b a

b i l i t y o

f c

o r

f e w e r o c c u r r e n c e s i n n

t r i a l s ( P )

.50

.60

.70

.80

.90

.95

.98

.99

.995

.999

p

P

P = P {m � c } = p m(1 – p )n – mc

m = 0

n!

m!(n – m)!

� 0.1


25/29


Ratio of process average to tolerance

[P/Pt = (Process average fraction defective)/(Tolerance fraction defective)]

A c c e p t a

n c e n u

m b e

r

1000

800

600

500

400

300

200

100

80

60

50

40

30

20

10 0

1

2

3

5

6

7

8910

1112

13

14

15

16

17

18

1920

25

To

To

To

30

21

26

35

31

4

8

6

5

4

3

2

10 0.1 0.2 0.3 0.4 0.5 0.6

Single samplingconsumer’s risk, 0.10

T o

l e r a n c e n u m

b e r o

f d

e f e c t i v e s

[ P t N =

( T o

l e r a n c e

f r a c t i o n

d e f e

c t i v e ) ( L o t s i z e ) ]

Figure DR1 Dodge-Romig Cur ves for Finding the Acceptance Number

Source:H. F. Dodge and H. G. Romig, Sampling Inspection Tables , 2nd ed, 1959,New York: John Wiley, p. 14.


26/29

PART FOUR Quality26

24

22

20

18

16

T o e r a n c e t i m e s s a m p e s i z e

[ P t n = ( T o

l e r a n c e

f r a c t i o n

d e

f e c t i v e ) ( S a m p l e

s i z e ) ]

14

12

10

8

6

4

2

01 2 3 4 5 7 10 20

Tolerance number of defectives[Pt N = (Tolerance fraction defective) (Lot size)]

30 50 70 100 200 300 500 1000

01 2

34

56

7 8910

1112

1314151617

1819

Simple sampling

consumers risk, O.IO

A c c e p t a

n c e

n u m b

e r

Figure DR2 Dodge-Romig Curves for Finding the Size of Sample

Source: H. F. Dodge and H. G. Romig, Sampli ng I nspecti on Tables , 2nd ed, 1959, New York: John Wiley, p. 15.


27/29


1000

800

600

500

400

300

200

100

80

60

50

40

30

20

10

8

6

5

4

3

2

0 0.1 0.2 0.3 0.4 0.5 0.6

1.5

2

3

4

5

6

7 8

91011

1315

20

25

30

35

40

1

Single samplingconsumer’s risk, 0.10

Ratio of process average to tolerance[P/Pt = (Process average fraction defective)/(Tolerance fraction defective)]

T o e r a n c e n u m

e r o

e e c t i v

e s

[ P t N

= ( T o

l e r a n c e

f r a c t i o n

d e

f e c t i v e )

( L o t s i z e ) ]

T o l e r a n c e t i m e s m i n i m u m a v e r a g e i n s p e c t e d

[ P t I m i n = ( T

o l e r a n c e

f r a c t i o n

d e

f e c t i v e ) ( M i n i m u m

a v e r a g

e n u m

b e r i n s p e c t e d p e r

l o t ) ]

Figure DR3 Dodge-Romig Curves for Finding the Minimum Amount of I nspection per Lot

Source: H. F. Dodge and H. G. Romig, Sampli ng I nspection Tables , 2nd ed, 1959,New York: John Wiley, p. 16.


28/29

PART FOUR Quality28

100008000

600050004000

3000

2000

1000

800

600500

400

300

200

100

80

6050

40

30

20

108

65

4

3

2

C = 0

1

2

3

4

10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5

k = (ratio of process average fraction defective to AOQL)p

pL

M =

p N

( p r o c e s s a v e r a g e

f r a c t i o n

d e

f e c t i v e t i m e s

l o t s i z e )

5

6

7 8910

12

1416

18

20

21 to

25

15

C

17

19

Figure DR4 Dodge-Romig Curves for Determining the Acceptance Number c; AOQL

Protection

Source: H. F. Dodge and H. G. Romig,Sampling Inspection Tables , 1944, New York: John Wiley, p. 50.


29/29


Given c x y Given c x y Given c x y Given c x y

0 1.00 0.3679 10 8.05 6.528 20 15.92 13.89 30 24.11 21.70

1 1.62 0.8400 11 8.82 7.233 21 16.73 14.66 31 24.95 22.50

2 2.27 1.371 12 9.59 7.948 22 17.54 15.43 32 25.78 23.30

3 2.95 1.942 13 10.37 8.670 23 18.35 16.20 33 26.62 24.10

4 3.64 2.544 14 11.15 9.398 24 19.17 16.98 34 27.45 24.90

5 4.35 3.168 15 11.93 10.13 25 19.99 17.76 35 28.29 25.71

6 5.07 3.812 16 12.72 10.88 26 20.81 18.54 36 29.13 26.52

7 5.80 4.472 17 13.52 11.62 27 21.63 19.33 37 29.97 27.33

8 6.55 5.146 18 14.31 12.37 28 22.46 20.12 38 30.82 28.14

9 7.30 5.831 19 15.12 13.13 29 23.29 20.91 39 31.66 28.96

10 8.05 6.528 20 15.92 13.89 30 24.11 21.70 40 32.51 29.77

Table DR1 Dodge-Romig values of x and y for Determining AOQL

Source: H. F. Dodge and H. G. Romig,Sampling Inspection Tables , 1944, New York: John Wiley, p. 49.

ch10 acceptance sampling

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