md. imrul kaes - acceptance sampling 2013-4-25

7/30/2019 Md. Imrul Kaes - Acceptance Sampling 2013-4-25

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INTRODUCTION TO ACCEPTANCE

SAMPLING PLAN


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HISTORY

Acceptance sampling plan was applied by the US

military to the testing of bullet during World War ll. If

every bullet was tested in advance, no bullets

would be left to ship, since testing was required for

firing. If on the other hand, none were tested,malfunction might occur in the battle field with

potential disastrous result.

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WHY ACCEPTANC SAMPLING

The are several situations when 100% inspection is not

practical:

When testing is destructive, otherwise all the products will be

lost. When inspection cost is very high.

When many similar products are

to be tested.

When efforts required for testing

is very high.

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When time and technology limitations are very high.

When lot size is very large.

Suppliers quality history is good enough to justify less than

100% inspection.

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ADVANTAGES

It is economical.

It requires less time and less effort.

It requires less personnel, etc.

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DISADVANTAGES

There is always high risk for both the producer and

the customer.

Requires expertise in statistical aspects.

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SOME DEFINITIONS RELATED TO

SAMPLING PLAN

Acceptance Quality Level (AQL): this is the

poorest quality level of the suppliers process that

the customer would consider to be acceptable as a

process average.

Lot Tolerance Percentage Defect (LTPD): This is

the poorest quality that a consumer is willing to

tolerate in an individual lot.

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Customers Risk: this is the probability of

accepting a bad lot.

Producers Risk: this is the probability of rejecting

a good lot.

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SINGLE SAMPLING PLAN

A single sampling plan is the simplest and shortest

plan. Here decision is based on single trial. A

sample size (n) is drawn from a batch or lot (N). If

the number of non conforming units is less or equal

to a predetermined number(c),then the lot is accepted, otherwise rejected.

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Let lot size, N=1000, sample size=30, acceptance number,

c=3.

This means that the batch to be inspected contains 1000pieces of garments. 30 pieces garments are randomly

drawn from the batch and inspected. If the number of

nonconforming units if found less than or equal to 3 pieces,

then entire batch is accepted, if the number of

nonconforming units is more than 3 pieces, then the entirebatch is rejected.

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OPERATING CHARACTERISTIC CURVEOC curve


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PERMUTATIONAND COMBINATION

A permutationis any arrangement

ofr objects selected from n

possible objects. The order of

arrangement is important in

permutations.

EXAMPLE

Suppose that in addition toselecting the group, he must also

rank each of the players in that

starting lineup according to their

ability.

A combination is the number of ways to

choose r objects from a group ofn

objects without regard to order.

EXAMPLE

There are 12 players on the Carolina

Forest High School basketball team.Coach Thompson must pick five players

among the twelve on the team to

comprise the starting lineup. How many

different groups are possible?

792)!512(!5

!12512

C040,95)!512(

!12512

P

5-12


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PROBABILITYOFACCEPTANCE

SINGLESAMPLINGPLAN

Suppose,

N = 1000

n = 30

c = 2

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PROBABILITYOFACCEPTANCEWHENACTUAL

NONCONFORMANCE 1% OR 0.01

P(0) = 30C0x (0.01)0x (1-0.01)30-0

=0.7397

P(1) = 30C1x (0.01)1x (1-0.01)30-1

=0.22415

P(2) = 30C2x (0.01)2x (1-0.01)30-2

=.03283

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Total probability of acceptance when actual non

conformance 1% or 0.01

Pa (0.01) = P(0) + P(1) + P(2)

= 0.7397 + 0.22415 + 0.03283

= 0.99668= 99.668 %

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Total probability of acceptance when actual non

conformance 15% or 0.15

Pa (0.15) = P(0) + P(1) + P(2)

= 0.00763 + 0.04039 + 0.10337

= 0.15139= 15.139%

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OC CURVE SINGLE SAMPLING PLAN4

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This curve plots the probability of accepting the lot

(Y-axis) versus the lot fraction nonconforming or

percent defectives (X-axis).

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DOUBLE SAMPLING PLAN

Lot Size = N

Sample size in the 1st trial = n1

Sample size in the 2nd trial = n2

Acceptance no. for 1st

trial = c12nd Acceptance no.= c2

No. of defects for 1st sample = D1

No. of defects for 2nd sample = D2

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In the first trial, sample n1 is taken from the lot N. If the

defects D1 is less or equal to acceptance no c1, then

the entire lot is accepted, without requiring the second

trial. If D1 is greater than acceptance no c2, then the

entire lot is rejected without requiring the 2nd trial. Incase D1 is greater than c1 but less or equal to c2,

then the 2nd sample size n2 is taken. If D1+D2 is less

than or equal to c2, then the lot is accepted otherwise

rejected.

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Suppose, N=3000, n1=40, n2 = 80, c1=1, c2=4

A lot contains 3000 units of garments, in the 1st trial 40units are randomly drawn. If the defective unit is less or

equal to 1, lot is accepted, without requiring the 2nd trial.

If the no defective units are greater than 4 then the

entire lot is rejected without requiring the 2nd

trial. If thedefective units are 2 or 3 or 4 then 2nd

sample 80 pieces are taken. If the total defective units

(1st + 2nd ) is less or equal to

4 then the lot is accepted, otherwise rejected.

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Sample n1 is taken

Accept lotIs D1 c 2? Reject lot

Sample n2 is taken

Is D1+D2 > c2 ?

Accept lot

Reject lot

N

N

N

Y

Y

Y


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PROBABILITYOFACCEPTANCE

DOUBLESAMPLINGPLAN

Suppose,

N = 3000

n1 = 40, c1 = 2

n2 = 80, c2 = 4D1 = nonconforming from n1= 40 sample

D2 = nonconforming from n2= 80 sample

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PROBABILITYOFACCEPTANCEIN 1STSAMPLE

WHENACTUALNONCONFORMANCE 5% OR 0.05

P(0) = 40C0x (0.05)0x (1-0.05)40-0

=0.128512

P(1) = 40C1x (0.05)1x (1-0.05)40-1

=0.27055

P(2) = 40C2x (0.05)2x (1-0.04)40-2

=0.27767

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Total probability of acceptance in 1st sample when

actual non conformance 5% or 0.05

P1 = P(0) + P(1) + P(2)

= 0.128512 + 0.27055 + 0.27767

= 0.67673= 67.673%

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PROBABILITYOFACCEPTANCEIN 2ND SAMPLE

WHENACTUALNONCONFORMANCE 5% OR 0.05

P2 (D1=3 & D2=0)

= { 40C3x (0.05)3x (1-0.05)40-3 } X { 80C0x (0.05)

0x (1-0.05)80-0 }

=0.003057

P2 (D1=3 & D2=1) = 40C1x (0.05)1x (1-0.05)40-1

= { 40C3x (0.05)3x (1-0.05)40-3 } X { 80C1x (0.05)

1x (1-0.05)80-1 }

=0.01287

P2 (D1=4 & D2=0) = 40C2x (0.05)2x (1-0.04)40-2

= { 40C4x (0.05)4x (1-0.05)40-4 } X { 80C0x (0.05)

0x (1-0.05)80-0 }

=0.001488

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TOTALPROBABILITYOFACCEPTANCEIN 2ND

SAMPLEWHENACTUALNONCONFORMANCE

5% OR 0.05

P2

= P2 (D1=3 & D2=0) + P2 (D1=3 & D2=1) + P2 (D1=4 & D2=0)

= 0.003057 + 0.01287 + 0.001488

= 0.017415

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TOTALCOMBINEDPROBABILITYOFACCEPTANCE

INCLUDING 1ST & 2ND SAMPLEWHENACTUALNON

CONFORMANCEIS 5% OR 0.05

Pa = P1 + P2

= 0.67673 + 0.017415

= 0.694145

= 69.4145%

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OC CURVE DOUBLE SAMPLING PLAN4

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MULTIPLE SAMPLING PLAN(extension of double sapling plan)


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In double sampling plan, samples are taken

maximum twice, where in multiple sampling plan,

samples are taken many times before a decision is

taken, regarding acceptance or rejection.

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Similar to double sampling plan, an acceptance

number and a rejection number is specified. At any

sample, if cumulative number of defectives is less

than or equal to the acceptance number, the entire

lot is accepted and no more samples are taken or inthe other way to say, the process of sampling is

stopped. At any sample, if the cumulative number of

defectives is more than or equal to rejection

number, the entire lot is rejected and no moresample is taken. The process of taking samples

continues only if the number of defectives is more

then the acceptance number but less than the

rejection number.

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Sample

Number

Sample Size Cumulative

Sample Size

Cumulative

Acceptance

Number

Cumulative

Rejection

Number

Take another sample if

number of cumulative

defectives

1 50 50 1 3 22 50 100

2 4 33 50 150

3 5 4

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SEQUENTIALSAMPLINGPLAN(extension of double sampling plan)


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The acceptance number or rejection number are

not by fixed numbers, rather in terms acceptance

and rejection region.

There are two types of sequecial sampling with

respect to nature of drawing the sample

1. Item by item

2. Group

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For example, if

= 0.05 = probability of type 1 error

= 0.10 = probability of type 2 error

P1 = 0.01 = fraction nonconforming for whichprobability of acceptance is high

P2 = 0.10 = fraction nonconforming for whichprobability of acceptance is low

Then,

X1 = -a1 + b*n = -0.393 + 0.04 * n

X2 = a2 + b*n = 1.205 + 0.04 * n

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Both acceptance number and rejection numbers

must be integers. The acceptance number is the

next integer less than or equal to X1 and the

rejection number is the next number greater than or

equal to X2.

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For, n =1

X1 = -0.899 ~ -1

X2 = 1.245 ~ 2

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Units Inspected Accept the lot ifcumulative

defectives are

Reject the lot ifcumulative

defectives are

1

2 2

3 24 2

5 2

6 2

7 2

8 2

9 2

10 2


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Units Inspected Accept the lot ifcumulative

defectives are

Reject the lot ifcumulative

defectives are

11 2

12 2

13 214 2

15 2

16 2

17 2

18 2

19 2

20 3


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Units Inspected Accept the lot if

cumulative

defectives are

Reject the lot if

cumulative

defectives are

21 3

22 3

23 3

24 3

25 3

26 348 3

49 3

58 4

74 4

83 5

100 5

109 6

md. imrul kaes - acceptance sampling 2013-4-25

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