Chapter 15Dr. Shokri Selim, KFUPM1 15 Lot-by-Lot Acceptance Sampling for Attributes

Chapter 15 Dr. Shokri Selim, KFUPM 1

15 Lot-by-Lot Acceptance Sampling for Attributes

The acceptance sampling problem


• An old method used in the 1930’s and 40’s. • The purpose of AS is to inspect received lots of products

and decide whether to accept or reject the lot; lot disposition, or lot sentencing.• Accepted lots are put into production• Rejected lots may be returned to supplier or

subjected to other lot-disposition action• Sampling methods may also be used during various

stages of production


Note that:1.The purpose of AS is sentencing lots and not estimate lot quality.2. It could happen that, lots of same quality be sentenced differently based on the sample.3. AS is an audit tool4. Control charts are used to signal departures from quality.

Approaches to lot sentencing

1. Accept without any inspectionSupplier process is very good and defectives are rare or

there is no economic justification to look for defectives

2. 100% inspectionUse if defective components can cause high failure cost or

supplier process does not meet specifications

3. Acceptance sampling



When is AS used?

Advantages of sampling


Disadvantages of sampling


1. There is a risk of accepting “bad” lots and rejecting “good” lots.2. Less information is generated about the product or about the

process that manufactured it.3. Acceptance sampling requires planning and documentation where

as 100% inspection does not


• One major classification is by data type, variables and attributes

• Another is based on the number of samples required for a decision.

Types of sampling plans

• Single-sampling plansSelect a sample of size n. If the number of defectives ≤ c, accept lot, else, reject the lot.

• Double-sampling plansSelect a sample of size n, then depending on the number of defective

accept the lot reject the lot

take a second sample and decide on both samples• Multiple-sampling plans

similar to double but with more than 2 samples

• Sequential-sampling plansunits are selected one at a time and decision to accept or reject or continue is madeChapter 15 Dr. Shokri Selim, KFUPM 9

Types based on number of samples


Single-, double-, multiple-, and sequential sampling plans can be designed to produce equivalent results.

A lot of the some quality level has the same probability of being accepted by these plans.

Factors to consider include:• Administrative efficiency• Type of information produced by the plan• Average amount of inspection required by plan• Impact of the procedure on manufacturing flow


There are a number of important considerations in forming lots for inspection, including:

1. Lots should be homogeneous.Same machine, same raw material, same operator

2. Larger lots are preferred over smaller ones.More economic

3. Lots should be conformable to materials-handling systems used in both supplier and consumer facilities.

Lot packaging minimizes risk of damageSelection of sample is easier

Lot formation

1. Assign a number to each item and select n random numbers to form a sample or

2. Randomly select the length, depth and width in the container or

3. Stratify the lot into layers, then cubes.


Random sampling

• The selection of a sampling plan depends on the objective and the history of the supplier.

• Non-static nature of AS plansIf supplier is known for quality, start with sampling plan for attributes. If quality is proven, may use skip-a-lot policy. If capability is high may stop sampling.

If supplier quality is not known, use an attribute plan. If quality is good may use a variable plan, and help them in SPC

• Companies start with AS and shift to SPC as the quality improves.


Guidelines for using acceptance sampling


The plan is defined by the sample size n and the acceptance number c.

If the number of defectives, d ≤ c accept the lot,else reject the lot.

Single sampling plans for attributes

Definition of a single sampling plan


Type A: Lot size is finiteType B: Lot size in infinite

Types of Lot Size


The OC curve gives the probability of accepting the lot given the lot fraction defective

Main assumption: lot size is very largeLet p = probability a unit is defective d = number of defectives in a sample of size n

)defectives (dP

)c (dPPa

Type B OC curves

dnd ppd

n

)1(

c

od

dnd ppd

n)1(

Probability of accepting a lot:

Sample OC curve



Reject lot if p > 0.025

Can we choose n and c that give similar OCC?

The ideal OC curve


Effect of n


Effect of C

Chapter 15 21

Effect of n and c on OC curves


• If the lot size is vey large, we use the binomial distribution to model the P(d ≤ c)

• If the lot size is finite we use the hypergeometric distribution.

Type-A and Type-B OC curves


Dd

n

N

dn

N-D

d

D

dP

for sample) theis defective (

Dc

n

N

dn

N-D

d

D

Pcd

da

for

0 DEMO

Constructing Type A OCCLetN be the lot sizeD be the number of defectives in the lotN be the sample sizeC be the maximum number of defectives allowed


If N is large (N ≥ 10n ) both graphs are close.

Relation between Types A and B

DEMO

Type A and Type B OC curves


Behavior of Type B OC curve for c = 0

OC curve far from the ideal OCC.

Pa falls sharply as p increases

Plans with c = 0.

na pP )1(


Behavior of Type A OC curve for c = 0

Plans with c = 0 and N = 10n

OC curve far from the ideal OCC.

Pa falls sharply as p increases

N = 10n but the OCCs behave differently.

/

n

N

n

N-DPa

p = D/N

Chapter 15 27

• Acceptable quality level, AQL = the least quality level for the supplier’s process that a consumer would consider to be acceptable. The consumer assigns high acceptance probability to it. • Lot tolerance percent defective, LTPD = Rejectable quality level, RQL = the least quality level, that the consumer is willing to accept with small acceptance probability.

• We can design a plan that almost satisfies both conditions.

Specific points on the OC curve

Dr. Shokri Selim, KFUPM


1 10

2 20

1 1

1

c n dd

d

c n dd

d

np p

d

np p

d

Designing a single-sampling plan with a specified OC curve

DEMO

p1 = AQLp2 = RQL


Binomial monograph

N ≥ 10 n

0( ) (1 )

c i n i

i

nP d c p p

i


N ≥ 10 n

0( ) (1 )

c i n i

i

nP d c p p

i

Binomial monograph

Chapter 15 31

Incoming lotsFraction Defectives

P0

Outgoing lotsFraction Defectives

P1 < P0Accepted lot has P0 fraction

defectives

No of defectives = P0(N-n)


Rectifying inspection

Inspectionactivity

Rejected lot has 0 defectives

Average number of defective units =

pa (N – n)p0 + ( 1 – pa)x0

Average outgoing quality, AOQ = pa(N – n)p0/N

N = 1000, n =100, c = 3, p = 0.01

pa = 0.9816

Average outgoing quality =

AOQ = pa(N – n)p/N = 0.009


Example


AOQ limit = worst AQLFor the example; AOQL = 0.019417

0.00

5000

...

0.01

0000

...

0.01

5

0.02

0000

...

0.02

5000

...

0.03

0000

...

0.03

5000

...

0.04

0000

...

0.04

5

0.05

0000

...

0.05

5000

...

0.06

0000

...

0.06

5000

...

0.07

0.07

5000

...

0.08

0000

...

0.08

5000

...

0.09

0000

...

0.09

5000

... 0.1

0

0.005

0.01

0.015

0.02

0.025

AOQ for N very large

AOQ = pa(N – n)p/N = pa(1 – n/N)p ≈ pa p

AOQL is the maximum point on

the curve

Number of inspected items:

If d ≤ c, n units will be inspected.

If d > c, N units will be inspected.

ATI = pa n + ( 1 – pa )*N = n + ( 1 – pa )*( N – n)

If N= 1000, n= 100, c = 3, p= 0.01

pa = 0.9816 and ATI = 116.56


Average total inspection


ATI = n + ( 1- pa )*( N – n)

ATI for sampling plan: n = 89, c = 2 for lot sizes of 1000, 5000, 10,000

Does the behavior of the graph makes

sense?

• If AOQL is specified, the solution is not unique• We find n and c that minimize ATI given some AOQL

value


Selection of n and c based on AOQL and ATI


• n1 = sample size on the first sample

• c1 = acceptance number of the first sample

• n2 = sample size on the second sample

• c2 = acceptance number of the second sample

• If d1 in the first sample is ≤ c1 accept the lot

• If d1 in the first sample is > c2 reject the lot

• Otherwise take 2nd sample.• If d1 + d2 ≤ c2 accept the lot

• Otherwise, reject the lot.

Double Sampling Plans

Double, multiple, and sequential sampling




HW: read the advantages and disadvantages

Advantage of double sampling over single sampling plans


0,

)2(,2

)1(,1

1

sample second on the acceptance of probabilty be

samplefirst on the acceptance of probabilty be

samples combind theof acceptance of probabilty be

221

12211

12211

01

1 111

1

dcdP

ccdcdP

ccdcdPP

ppd

nP

PPP

P

P

PLet

IIa

dndc

d

Ia

IIa

Iaa

IIa

Ia

a

Constructing the OC curve for double sampling plans


Primary OC curve

Supplementary OC curve

Supplementary OC curve

Example


n1 = 50, C1 = 3, n2 = 150, C2 = 6

0,61,52,4 212121 ddPddPddPP IIa

21

2

2221

2

2221

116

0,6

115

1,5

114

2,4

66121

1

02

255121

2

02

244121

nn

d

dndn

d

dndn

pppn

ddP

ppd

npp

nddP

ppd

npp

nddP


It is the average number of inspected units

ASN = n1 + n2 Pr( c1 < d1 ≤ c2 )

What is the assumption here ?

The assumption

We complete the inspection of the second sample even after the total number of

defectives has exceeded c2

The average sample number


It is stoppage of sampling when the rejection condition is satisfied

Do not do it with the first sample

Why ?

To be able to estimate the fraction defective.

However, can do on the second

sample.

Curtailment

Rectifying Inspection with double sampling


N

pnnNPnNPAOQ II

aI

a 211

1 1 2

1 2 1 2

1

1 1

I IIa a a

Ia a

ATI n P n n P N P

n n P N n n P

If all defective items are discovered, either in sampling or 100% inspection, and are replaced with good ones:

Multiple Sampling Plans


Cumulative sample size

Acceptance number

Rejection number

20 0 340 1 460 3 580 5 7

100 8 9

Example:



Acceptance number

Rejection number

20 0 340 1 460 3 580 5 7

100 8 9

At the completion of stage i:

If d1 + d2 + … + di ≤ acceptance number → Accept lot

If d1 + d2 + … + di ≥ rejection number → Reject lot

Otherwise take the next sample

Usually, the first sample is inspected 100%.

Usually, subsequent samples are subject to curtailment



Acceptance number

Rejection number

20 0 340 1 460 3 580 5 7

100 8 9Suppose d1 = 1

Take 2nd sample, what value of d2 will result in accepting lot?

what value of d2 will result in rejecting lot?

suppose d2 = 1

Take 3nd sample, what value of d3 will result in accepting lot?

what value of d3 will result in rejecting lot?

The values of the d’s that lead to lot acceptance


0 3 0 1 2 d1

1 4 0 1 2 0 1 d2

3 5 0 1 2 0 1 0 1 2 0 1 0 1 d3

5 7 0 1 2 0 1 2 0 1 2 0 1 0 1 2 d4

8 9 0 1 2 0 1 2 0 1 2 0 1 2 d5


Acceptance number

Rejection number

20 0 340 1 460 3 580 5 7

100 8 9

Item by Item Sequential Sampling Plans

The x-axis shows cumulative number of items inspected.

The y-axis shows cumulative number of defectives

If the point is between the two lines, take one more item

If the point falls on or above the top line, reject lot

If the point falls on or below the bottom line, accept lot



Given p1 and 1-α, p2 and β.

)1()1(

log

11

log1

log

linerejection The

)1()1(

log

11

log1

log

:line acceptance The

21

12

2

1

21

12

2

1

pppp

pp

n

y

pppp

pp

n

y

Suppose p1 = 0.01, α = 0.05,

p2 = 0.06, β = 0.1

ny

ny

028.057.1

linerejection The

0.028n-1.22

)06.01(01.0)01.01(06.0

log

06.0101.01

log05.01

1.0log

:line acceptance The

The acceptance and rejection lines


Sequential sampling could be truncated if the number of units inspected reaches three times the sample size of the equivalent single sample plan

For the previous example, the equivalent single sample plan has n = 89. If sentencing does not takes place after 267 units, stop sampling and accept lot

Truncation of sampling

Chapter 15 Dr. Shokri Selim, KFUPM 55Chapter 15 Dr. Shokri Selim, KFUPM 55

Homework

Project 1:Suppose the fraction defective is 0.05, find the ASN for double sampling plan defined by (n1 = 50, c1 = 1, n2 = 50, c2 = 2), in case of curtailment at the second sample

Project 2:A single sample plan has n = 50; find c that will result in least ATI and AOQL ≤ 0.01

Project 3:Consider a double sampling plan with n1=50, c1=1, c2 =2. Find the smallest n2 that will result in AQL = 0.01 with α ≤ 0.05, and RQL = 0.1 with β ≤ 0.2.

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Chapter 15Dr. Shokri Selim, KFUPM1 15 Lot-by-Lot Acceptance Sampling for Attributes