Applied Reliability Applied Reliability

Copyright David C. Trindade, Ph. D.

S

TAT-

T

ECH ®

Spring 2010

Applied

Reliability

Techniques for Reliability

Analysis

with

Applied Reliability Tools (ART) (an EXCEL Add-In)

and

JMP® Software

AM216 Class 6 Notes

Santa Clara University

Reference

Applied Reliability – Quality Control

Material Based on Chapter 9 in

Text:

Applied Reliability, 2nd _ed.

by Paul A. Tobias and David C. Trindade

Published 1995 by Chapman & Hall, New York

Table of Contents

Combinatorics

Multiplication Rule

Permutations (Spreadsheet Function) Combinations (Spreadsheet Function)

Binomial Distribution

Binomial Experiment Probability Calculations

Parameter Estimation and Application Confidence Intervals

Poisson Distribution

Approximation to Binomial Probability Calculations

Parameter Estimation and Application Confidence Intervals Hypergeometric Distribution Probability Calculations Application Acceptance Sampling Risks

Operating Characteristic Curves Sampling Plan

LTPD Plans

Adjusting Sampling Plans Minimum Sample Size Plans AOQ Curve and AOQL

Control Charts for Reliability

Multiplication Rule for Outcomes

If the first experiment can result in n₁ possible outcomes, and for each outcome, the second experiment can result in n₂ possible outcomes, then there are a total of n₁ n₂ possible outcomes for the two experiments together.

The multiplication rule is extendable to any number of experiments in a sequence.

Example:

On a restaurant menu, there are five appetizers, seven entrees, and six desserts. How many different three course meals are possible?

Combinatorics

Permutations

The number of ways to arrange n objects in order:

If all n objects are used: n(n-1)(n-2)···1 = n!

If only r objects of n

are used:

|--- r terms ---|

Examples:

Your child receives five birthday presents. How many different ways can your child open them?

Answer: 5! = 5x4x3x2x1 = 120.

Two musical notes are needed to complete the song. In one octave, thirteen possible notes exist. Without repeating the same note, how many different ways can the song be ended?

Answer: 13x12 = 156.







_{  }

! 1 ( 2) 1 ! n n n n n r n r      

Combinations

The number of ways of combining r objects taken from n available in which order is not important:

Note: 0! = 1

Examples:

Eight players offer to play for the company golf team in a tournament. How many different four person teams can I make up?

Answer: 8!/(4!4!) = (8x7x6x5)/(4x3x2x1) = 70.

There are seven blade servers and five open positions in the rack. How many ways can any five be chosen for installing in any order on the five rack positions? (Two will be left out.)

Answer: 7!/(5!2!) = (7x6x5x4x3)/(5x4x3x2x1) = 21.

Combinatorics



 







1 1 _! ! ! n n n r _n r! r n r      

 

nr

 

₀n



1

 

₁n



n

Class Project #1

Combinatorics:

Consider the three objects: A, B, C.

1. Taking all three objects, how many permutations are there?

List them:

2. Taking two objects at a time, how many permutations are there?

List them:

3. Taking two objects at a time, how many combinations are there?

Class Project #1

Combinatorics:

Consider the three objects: A, B, C.

1. Taking all three objects, how many permutations are there?

3! = 3x2x1 = 6. List them:

ABC ACB BAC BCA CAB CBA

2. Taking two objects at a time, how many permutations are there?

3!/(3-2)! = (3x2x1)/(1) = 6. List them:

AB AC BA BC CA CB 3. Taking two objects at a time, how many combinations are there?

3!/(2!1!) = 3. List them:

Class Project #2

Combinatorics

1. How many different eleven letter words

can be formed from the letters in the word

MISSISSIPPI?

2. Let S denote a survival and F, a failure.

Given two failures among five items on stress, how many different ways can the two failures occur on the five objects? (Hint: One way is

SSSFF.) List all words.

3.How many different ways can r failures occur among n units on stress?

This problem is analogous to asking how many different words of n letters can be formed from r F's and (n-r) S's.

Class Project #2

Combinatorics

1. How many different eleven letter words can be formed from the letters in the word MISSISSIPPI?

2. Let S denote a survival and F, a failure. Given two failures among five items on stress, how many different ways can the two failures occur on the five objects? (Hint: One way is SSSFF.) List all words.

SSSFF SSFSF SSFFS SFFSS SFSFS SFSSF FFSSS FSFSS FSSFS FSSSF

3.How many different ways can r failures occur among n units on stress?

 

5 2 5! 10 2!3!   11! 34,650 4!4!2! 



_n

_

_r







_



n

_r



_



r

n

!

!

!

Permutations and

Combinations in EXCEL

Binomial Experiment

Necessary Conditions

1. Only two possible outcomes (success, failure).

2. Fixed number of trials, n.

3. Constant probability of success, p, across trials.

Binomial Distribution

Derivation

 

 

_{n x}





n x x

p

1

p

x

P







 

n

x





n x x

p

1

p





Let p be the probability of failure, given

by the CDF at time t; then 1 - p is the

probability of survival. Consider n units

on test for time t. Designate a failure by F

and a survival by S. For the specific

sequence with x failures and (n-x) survivals:

F F ... F S S ... S

The probability of this sequence is:

Since there are

ways of x failures occurring among n units,

the probability of x failures among n units

on stress is given by the binomial distribution probability mass function (PMF)

Binomial Distribution

Probability Function

20 units are placed on stress for 100 hours. The CDF at 100 hours is 20%. The graph below shows the binomial distribution

probabilities for x failures at 100 hours,

x = 0,1,2,… 0.0% 5.0% 10.0% 15.0% 20.0% 25.0% 0 1 2 3 4 5 6 7 8 9 10 11 x = Number of Failures P ro b a b il it y o f x F a il u re s

n = 20, p = 0.2

:

Binomial Distribution

Cumulative Function

 

0

1

)

one

least

at

(

1

)

(

)

(

)

(

)

1

(

)

(

Note

)

(

)

(

)

1

(

)

0

(

)

(

1 0

P

P

x

X

P

x

X

P

k

P

x

X

P

x

X

P

k

P

x

P

P

P

x

X

P

n x k x k







































  



To get the cumulative probability of x or less events, sum the individual binomial

Binomial Distribution

Cumulative Function

20 units are placed on stress for 100

hours. The CDF at 100 hours is 20%. The graph below shows the binomial

distribution cumulative probabilities of x or

less failures at 100 hours, x = 0,1,2,…

n = 20, p = 0.2

:

Binomial CDF 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 1 2 3 4 5 6 7 8 9 10 11 12 x = Number of Failures P ro b a b il it y o f x o r le s s F a il u re s

Binomial Distribution

Properties

Mean or expected number:

Expected number of successes (or failures) in

n trials with probability of success (or failure) p

per trial equals

Variance: Standard Deviation:

np









2

1

np

p









1



np

p







Binomial Distribution

Calculations in EXCEL

=BINOMDIST(x, n, p, 0 or 1)

Fourth argument is:

0 or false = individual probability 1 or true = cumulative probability

Class Project #3

Binomial Distribution:

A fair coin is tossed one hundred times.

1. What is the expected number of heads? 2. What is the standard deviation?

3. What is the probability of exactly fifty heads? 4. What is the probability of at most fifty heads? 5. What is the probability of 60 or more heads? 6. What is the probability of 40 or less heads?

7. What is the probability of greater than 40 but less than 60 heads?

Class Project #3

Binomial Distribution:

A fair coin is tossed one hundred times.

1. What is the expected number of heads? np = 100(0.5) = 50

2. What is the standard deviation?

3. What is the probability of exactly fifty heads? 4. What is the probability of at most fifty heads? 5. What is the probability of 60 or more heads? 6. What is the probability of 40 or less heads? 7. What is the probability of greater than 40 but less than 60 heads?

8. What is the probability of 100 heads?

      np 1 p 100(0.5)(0.5) 25 5   _    _{   }         P X 50 100 50 100 1 1 50 7.96% 50 2 2







 P X 60 2.844%    P X 50 53.98%







 P X 40 2.844%



 



   P 41 X 59 1 2(2.844) 94.31%  _ _      _             P X x 100 100 100 31 100 1 1 100 7.89 10 100 2 2

Binomial Distribution PMF

n = 100, p = 0.5

Cumulative Binomial Distribution

n = 100, p = 0.5

Binomial Distribution

Parameter Estimation

Estimation of

p

:

In a binomial experiment, the probability of failure p at a given time is estimated by the

proportion of failures observed out of n units.

Thus,

where x is the number of failures:

x = 0,1,2,...,n.

Note this is the same statistic for estimating the CDF F(t) at time t, that is,

ˆ

_/

p



x n

 

t

F

p

ˆ



ˆ

Confidence Intervals on

p

:

Instead of just a point estimate for p, we may specify a confidence interval that captures the true (population) p with a certain level of confidence.

Binomial Distribution

95% Confidence Intervals for Proportions

Binomial Distribution

Exact Confidence Limits Using

EXCEL:

Exact confidence limits for the population parameter

p may be found using the Inverse Beta function in

EXCEL.

Lower limit: 0 for x = 0; =BETAINV(a/2, x, n-x+1) for

x between 1 and n.

Upper limit: =BETAINV(1-a/2, x+1, n-x) for x between 1 and n-1; 1 for x = n.

Example: n =100, x = 2,

The estimate of p = 2/100 = 0.02 = 2%.

Thus, a 90% 2-sided confidence interval on the population p is:

LCL: = BETAINV(.05, 2, 99) = 0.36% UCL: = BETAINV(.95, 3, 98) = 6.16%

Class Project #4

Binomial Distribution:

Fifty devices are stressed for 168 hrs. The

probability of a device failing by 168 hrs is 0.05 or 5%.

Determine:

A. The probability all devices survive 168 hours.

B. The expected number of failures.

C. The probability of at least one failure.

D. At the end of the test there are 10 failures. Provide a 95% confidence interval for the

population failure proportion. Does the interval include the assumed 5% failure probability?

Class Project #4

Binomial Distribution:

Fifty devices are stressed for 168 hrs. The

probability of a device failing 168 hrs is 0.05 or 5%.

Determine:

A. The probability all devices survive 168 hours.

B. The expected number of failures.

np = 50x0.05 = 2.5

C. The probability of at least one failure.

D. At the end of the test there are 10 failures.

Provide a 95% confidence interval for the population failure proportion. Does the interval include the

assumed 5% failure probability?

From the Clopper Pearson chart, we get the interval (10% to 34%),

which does not include the 5% failure probability.





50

( 0) 1 .05 0.077

P X    

( 1) 1 ( 0) 1 0.077 0.923

The Poisson Distribution

Approximation to Binomial

Distribution

For n large and p small, the Poisson

distribution provides an excellent

approximation to the binomial distribution.

Let l = n x p, the expected number. The probability of exactly x occurrences is:

The mean or expected number is l. The

variance is also l. The standard deviation is Approximation is good for large n and small p,

such that l= n x p is less than, roughly, 7.

 

!

x

e

P x

x

l

l





.

l

The Poisson Distribution

Comparison of Binomial and Poisson Probabilitiess n = 90, p = 0.01, l = np = 0.9 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0 1 2 Number of Failures P rob a bi li ty Fu nc ti on Binomial Poisson

     

0 1 2 ) 2 (X P P P P    



X  2



 0.40470.36790.16540.9381 P

Approximation to Binomial:

There are ninety units placed on stress for 100 hours. If the probability of failure is 0.01 for an individual unit, what is the probability of 2 or less units failing? We want Binomial Solution: Poisson Solution:



X 2



0.40660.36590.1647 0.9337 P

Poisson Distribution

Applications of Poisson Distribution

The Poisson is also a distribution on its own to model situations where the probability of a single event over a period of time or

space is constant.

For example, for calculations involving

density, such as the number of defects per

wafer, the Poisson distribution (with l = the

mean number of defects for a given area)

can be used to model the defect distribution.

The Poisson has been applied to the

number of phone calls in a given period, the number of repairs in time, the number of bugs in software, the number of raisins in a box of cereal, the number of people in a queue, the number of flaws per yard of

insulated wire, the number of misprints per page, and so on.

Poisson Distribution

Calculations in EXCEL

Class Project #5

Poisson Distribution:

The average defect density is 0.01 defects per square inch. The wafer area is 200 square inches. Ten wafers are randomly selected from a lot of fifty.

What is:

1. The expected number of defects per wafer?

2. The expected number of defects for the ten wafers? 3. The probability any wafer is defect free?

4. The probability any wafer has at least one defect? 5. The probability any wafer has exactly five defects? 6. The probability all ten wafers are defect free?

7. The probability exactly two of the ten wafers are defect free?

Class Project #5

Poisson Distribution:

The average defect density is 0.01 defects per square inch. The wafer area is 200 square inches. Ten wafers are randomly selected from a lot of fifty.

1. The expected number of defects per wafer? np = 200x0.01 = 2.0

2. The expected number of defects for the ten wafers? 10x2 = 20

3. The probability any wafer is defect free?

4. The probability any wafer has at least one defect? 5. The probability any wafer has exactly five defects? 6. The probability all ten wafers are defect free?

7. The probability exactly two of the ten wafers are defect free? Binomial: Poisson: l = np = 10(0.135) = 1.35 2 1P(0)  1 e  0.865 036 . 0 ! 5 25e2 _

 

e2 10  e20  0

   

2



8 10 2 2 2 0.258 (2) 1 P  e e  2 1.35 1.35 (2) 0.236 2 e P   

 

_

2

_

0

0.135

P

e

Poisson Distribution

Estimation of

l

The estimate of the population

parameter l is just the average observed number of occurrences

over the collection of time periods, samples, objects, lines, etc.

Confidence Intervals for

l

Tables for the 90%, 95%, and 99%

confidence levels for various observed

counts are shown next. Note the confidence limits depend only on the

confidence level and the observed

count. Conversion to a CI for l may be

done using these tables.

Exact confidence limits can also be calculated directly using functions in EXCEL.

Poisson Distribution

Table of Confidence Limits Based

on Observed Count

Confidence Level alpha Observed Count

X lower limit upper limit lower limit upper limit lower limit upper limit 0 0.00 3.00 0.00 3.69 0.00 5.30 1 0.05 4.74 0.03 5.57 0.01 7.43 2 0.36 6.30 0.24 7.22 0.10 9.27 3 0.82 7.75 0.62 8.77 0.34 10.98 4 1.37 9.15 1.09 10.24 0.67 12.59 5 1.97 10.51 1.62 11.67 1.08 14.15 6 2.61 11.84 2.20 13.06 1.54 15.66 7 3.29 13.15 2.81 14.42 2.04 17.13 8 3.98 14.43 3.45 15.76 2.57 18.58 9 4.70 15.71 4.12 17.08 3.13 20.00 10 5.43 16.96 4.80 18.39 3.72 21.40 11 6.17 18.21 5.49 19.68 4.32 22.78 12 6.92 19.44 6.20 20.96 4.94 24.14 13 7.69 20.67 6.92 22.23 5.58 25.50 14 8.46 21.89 7.65 23.49 6.23 26.84 15 9.25 23.10 8.40 24.74 6.89 28.16 16 10.04 24.30 9.15 25.98 7.57 29.48 17 10.83 25.50 9.90 27.22 8.25 30.79 18 11.63 26.69 10.67 28.45 8.94 32.09 19 12.44 27.88 11.44 29.67 9.64 33.38 20 13.25 29.06 12.22 30.89 10.35 34.67 21 14.07 30.24 13.00 32.10 11.07 35.95 22 14.89 31.41 13.79 33.31 11.79 37.22 23 15.72 32.59 14.58 34.51 12.52 38.48 24 16.55 33.75 15.38 35.71 13.26 39.74 25 17.38 34.92 16.18 36.90 14.00 41.00 26 18.22 36.08 16.98 38.10 14.74 42.25 27 19.06 37.23 17.79 39.28 15.49 43.50 28 19.90 38.39 18.61 40.47 16.25 44.74 29 20.75 39.54 19.42 41.65 17.00 45.98 30 21.59 40.69 20.24 42.83 17.77 47.21 90% 0.1 99% 95% 0.05 0.01

Poisson Distribution

Exact Confidence Limits Using

EXCEL:

Exact confidence limits for the population parameter

l may be found using the inverse chi-squared

distribution (CHIINV(probability; degrees of freedom)) in EXCEL.

Lower limit: 0 for x = 0; =0.5* CHIINV(1-a/2, 2x) for x > 0.

Upper limit: =0.5* CHIINV(a/2, 2*(x+1)) for x > 0.

Example: Observe 2 bugs in 100 lines of code, The estimate of l: x = 2 bugs/100 LOC Thus, a 90% 2-sided confidence interval on the

population l = bugs/100 lines of code is: LCL: = 0.5* CHIINV(.95, 4) = 0.36

Poisson Confidence Limits

Distinguish between Poisson confidence limits on: • the total count (as in the previous table)

• the average count per unit of measure (as a % in ART).

EXAMPLE:

Suppose we open three boxes of cereal and find a total

of 9 raisins. The average estimate lis 3 raisins per box. The 90% confidence interval on:

• the total count for three boxes is 4.70 to 15.71 • the average per box l is 1.57 to 5.24.

(CI width is 3.67.)

Suppose we open ten boxes and find a total of 30

raisins, The average number of raisins per box is 3, as before. The 90% confidence interval on:

• the total count for ten boxes is 21.59 to 40.69 • the average per box l is 2.159 to 4.069.

(CI width for l is now 1.91 since the sample is larger.)

The width of the confidence interval depends on the total count.

Class Project #6

Poisson Distribution:

1000 randomly selected lines of code are inspected. Twenty three bugs are found. Assuming a Poisson distribution:

A. What's the estimate of the expected number of bugs per 1000 lines of code (KLOC) in the population?

B. Provide a 90% confidence interval on the expected number of bugs per KLOC in the population.

Class Project #6

Poisson Distribution:

1000 randomly selected lines of code are inspected. Twenty three bugs are found. Assuming a Poisson distribution:

A. What's the estimate of the expected number of bugs per 1000 lines of code (KLOC) in the population?

Answer: 23

B. Provide a 90% confidence interval on the expected number of bugs per KLOC in the population.

Answer:

From Table of CL for Poisson, CI is 15.72 to 32.59

Using EXCEL:

LCL: = 0.5* CHIINV(0.95, 46) = 15.72 UCL: = 0.5* CHIINV(0.05, 48) = 32.59

Poisson Distribution

No Limit on Defect Count

The Poisson distribution has no limit (like n

for the binomial distribution) on the number of defects per unit of measure that can be

counted.

Defective Unit versus Defects

A unit is defective if it has one or more defects (a defect is defined as a nonconformance to specifications). Thus, defective units are

treated separately from the number of defects per unit or area.

A defective unit is handled by the binomial or hypergeometric distributions. For calculations involving density (e.g., defects per wafer,

failures per period, etc.), the Poisson distribution is often used.

Hypergeometric Distribution

Application:

Fixed number of trials.

Two possible outcomes: Success or failure.

Sampling from finite population. Hence,

probability of success with each trial changes and events are not independent.

Example: What’s the probability of drawing two aces in a row from a well shuffled deck of

cards?

Example: 20 lines of code in 100 are defective (contain a bug). Randomly select 10 lines. What’s the probability of finding 0,1,2,… defective lines?

Hypergeometric Distribution

Formula:

The probability of getting exactly x rejects in a sample of size n drawn from a finite lot of size N

that contains a total of m rejects is

where x = 0,1,2,3,...

For n<0.1N, binomial distribution is a good approximation for P(X) where p = m/N.

 

  

_{ }

_N n m N x n m x

x

P

 



Class Project #7

Hypergeometric Distribution:

Your dresser drawer contains twenty socks, half red and half black. It’s early morning and still dark. You don’t want to turn on the light and disturb your spouse. So in darkness you randomly select two socks.

1. What is the probability that you

choose a matched pair of black socks?

2. What is the probability that you

Class Project #7

Hypergeometric Distribution:

1. What is the probability that you

choose a matched pair of black socks?

2. What is the probability that you

choose a matched pair of either color?

P(matched pair, either color) = 2xP(2) = 0.474

 

  

_{ }

  

102

_{ }

100 20 2 45 1 2 0.237 190 m N m x n x N n x

P

 









Hypergeometric Distribution

Calculations in EXCEL

Fisher’s Exact Test

In two groups we observe failure proportions 1/6 and 7/9. The small sample sizes are small, but we want to test whether Group 1 is

significantly better than Group 2. The results are displayed in a 2x2 contingency table with margin totals:

Is the population from which the first group was drawn significantly better than the second

group population or is the difference likely to occur by chance alone?

For small sample sizes, Fisher’s Exact Test,

based on the hypergeometric distribution, calculates the probability of observing

differences equal to or greater than the results observed.

Group 1 Group 2 Totals

Pass 5 2 7

Fail 1 7 8

Fisher’s Exact Test

For both groups, note the total sample size is N

= 15 and the total failures is m = 8. Assume a randomly drawn sample size of n = 6 for Group 1. The probability of getting exactly 1 failure out of the 8 possible is given by the hypergeometric distribution

Group 1 Group 2 Totals

Pass 5 2 7 Fail 1 7 8 Totals 6 9 15

 

  

_{ }







P

8 15 8 1 6 1 15 6

1

0.03357

Fisher’s Exact Test

(Continued)

An even more extreme event (zero failures in Group 1) with the same marginal totals is

shown below.

This probability is

So the one-sided probability of observing the results or stronger is the sum

0.03357 + 0.001399 = 0.034965 3.5%

At ~96.5% confidence, the test results are significantly different.

 

  

_{ }

   P 8 15 8 0 6 0 15 6 0 0.001399

Group 1 Group 2 Totals

Pass 6 1 7

Fail 0 8 8

Totals 6 9 15

Class Project #8

Fisher’s Exact Test

In an experiment to compare different

treatments, the old method produced four rejects out of twenty. In the second

experiment, the new procedure resulted in zero rejects out of fifteen. How statistically significant is the improvement between the new and old?

Class Project #8

Fisher’s Exact Test

In an experiment to compare different

treatments, the old method produced four rejects out of twenty. In the second

experiment, the new procedure resulted in zero rejects out of fifteen. How statistically significant is the improvement between the new and old?

New Old Totals

Pass 15 16 31 Fail 0 4 4 Totals 15 20 35

 

0

  

_{ }

₃₅

0

.

0925

15 4 35 0 15 4 0







P

=HYPGEOMDIST(0,4,35,15)=0.092532

Acceptance Sampling

< Y % > Y % Accept Correct Type II error

Reject Type I error Correct

Decision on Lot

Population Value (% Defective)

Matrix of Possible Choices

Risks

Accept or reject a lot based on the

results of looking at a sample of size n randomly drawn from a lot of size N. Several possible risks involved in deciding whether or not a lot (the population) has a maximum allowed percent defective level, say Y%:

-Reject a good lot with percent

defective below Y% and commit a Type 1, a, or producer's error.

-Accept a bad lot with percent

defective above Y% and commit a Type 2, b, or consumer's error.

Operating Characteristic

(O.C.) Curve

Plot of P

A

versus

p

:

n = 50 c = 3 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0% 2% 4% 6% 8% 10% 12% 14% 16% 18% 20%

Lot Percent Defective

P ro b a b il ity o f A c c e p ta n c e OC Curve

For a given sampling plan (sample size n and acceptance number c), the probability of

acceptance will depend on the incoming lot

percent defective p. A plot of the probability of

acceptance P_A versus the lot percent defective

Calculations for an O.C. Curve

Use binomial distribution with n, varying p, to

calculate probability of acceptance, that is, P( X < c ):

Example:

Sketch the O.C. curve for a sample size n = 100 and

acceptance number c = 2.

The probability of acceptance is:

With n = 100, now try different p values, using ART. For p = 0.01, the probability of acceptance P is

0.3660+0.3697+0.1849 = 0.9206.

For p = 0.05, the probability of acceptance P is 0.0059+0.0312+0.0812=0.1183.

For p = 0.10, the probability of acceptance P is 0.0000+0.0003+0.0016=0.0019.





 

 

 

 





 





 









1 2 2 2 0 1 2 0 1 1 1 1 2 1 2 n n n P X P P P P p P np p n n P p p              Method:

OC Curves

Fixed c, Varying Sample Sizes n:

Fixed n, Varying Acceptance Numbers c:

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0% 5% 10% 15% 20% 25%

Lot Percent Defective

P ro b a b il it y o f A c c e p ta n c e c = 3 n = 500 n = 100 n = 300 n = 50 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0% 5% 10% 15% 20% 25%

Lot Percent Defective

P ro b a b il it y o f A c c e p ta n c e n = 50 c = 0 c = 1 c = 2 c = 3

O.C. Curve Calculations

Using ART

Sampling Plan

A sampling plan is a specified sample size n and

acceptance number c. A sampling plan generates a specific O.C. curve.

The sampling plan is determined by specifying four

values, equivalent to two points on the O.C. curve:

1. The p value for an incoming lot for which the

probability of acceptance is very high (e.g., 95% or

producer’s risk a = 5%).

2. The p value for an incoming lot for which the

probability of rejection is very high ( e.g., 90% or

consumer’s risk b = 10%).

These four numbers (two p values and associated

risks) uniquely determine the sampling plan.

We assume the lot size N is large relative to n, or

equivalently, that we are sampling from a process, rather than from lots. This assumption allows using simpler binomial distribution for calculations rather than the more complex hypergeometric distribution.

Several procedures exist to generate sampling plans: published tables, nomograph, computer programs, iterative calculations, etc.

Generating a Sampling Plan

Using ART

2%, 5%; 8%, 10%

LTPD Sampling Plans

What if our primary concern was viewing sampling plans that would assure rejection with high probability, say 90%, of a defect level that was the maximum value a

consumer could tolerate.

Such plans are called LTPD (lot tolerance percent defective) plans. The next table is based on a 10% consumer’s risk at various rejectable quality levels.

LTPD Table

Adjusting Sampling Plans

Reducing the Sample Size:

Suppose the qualification sampling plan is to accept on 3 or less rejects out of 300 units. Because of the cost or lack of

availability of 300 units, we wish to reduce the sample size while holding the

consumer’s risk constant. What is the

sample size if 2 or less rejects are allowed, or 1 or less, or even zero?

Adjusting Sampling Plans

Procedure:

Refer to figures following.

1. Find the graph corresponding to the present acceptance number, say 3.

2. Find the intersection of the sample size on the horizontal x-axis with the diagonal line in the graph labeled with the probability of acceptance value “10,” corresponding to a 10% consumer’s risk.

3. Read, on the vertical y-axis, the percent defective value associated with the 10% consumer’s risk.

4. Find the graph with a lower acceptance number and reverse the above procedure.

5. Using the percent defective value previously found for the consumer’s risk on the y-axis of the new graph, find its intersection with the labeled “10” line.

6. Dropping vertically at this intersection point to the x - axis, read off the new sample size for this lower acceptance number.

Graphs for Adjusting

Sampling Plans

Graphs for Adjusting

Sampling Plans

Class Project #9

Adjusting a Qualification Sampling Plan:

The qualification plan calls for allowing two rejects out of two hundred units. The engineer needs to reduce the

sample size of the study. If he wants to keep the same defect level at which the consumer’s risk is 10% for lot

acceptance, what is the necessary sample size, allowing only one reject. What is the LTPD?

Class Project #9

Adjusting a Qualification Sampling Plan:

The qualification plan calls for allowing two rejects out of two hundred units. The engineer needs to reduce the

sample size of the study. If he wants to keep the same defect level at which the consumer’s risk is 10% for lot

acceptance, what is the necessary sample size, allowing only one reject. What is the LTPD?

From Figure 9.15 (c = 3), 10% consumer’s

risk is at 2.2% defective.

From Figure 9.13 (c = 1) , at 2.2% defective, the 10% consumer’s risk corresponds to sample size 175 units.

Minimum Sample Size Plans

The smallest sample size for a plan having a specified rejectable quality level occurs when the acceptance number is zero.

Minimum Sample Size Plans

Zero Failures

Protection Against Consumer's Risk:

For zero failures and rejection

probability (1 - b), the minimum

sample size to reject a lot with fraction

defective value p is given by:





ln

ln 1

n

p

b





Using ART “Minimum Size

Sampling Plans”

Class Project #10

Sampling Plans:

We're willing to accept at most a 10% chance of permitting a failure fallout as high as 1.0%.

A. If we allow up to three rejects, what sample size do we need ? See Table 9.3 or ART Sampling Plans.

B. It turns out, we have only 500 pieces available. What should our acceptance number be now?

C. What is the minimum sample size we need, that is, allowing zero failures?

Class Project #10

Sampling Plans:

We're willing to accept at most a 10% chance of permitting a failure fallout as high as 1.0%. A. Allowing up to three rejects, what sample size do we need?

Table 9.3: 668

B. It turns out, we have only 500 pieces available. What should our acceptance number be now?

Table 9.3: c = 1 for 390 units

c = 2 for 533 units

C. What is the minimum sample size we need, that is, allowing zero failures?.

Table 9.3: 231 ART (Min. SS): 230 Formula: ln0.1 229 ln(1 .01) n   

Sampling Plans

The Meaning of AQL and LTPD

AQL: Acceptable quality level

The incoming lot percent defective with typically a 95% chance of acceptance. If a lot is rejected (that is, the number of rejects is above the acceptance number), we state with 95% confidence that the lot defect level is above the AQL.

LTPD: Lot tolerance percent defective

The incoming lot percent defective with a 10% chance of acceptance. If a lot is

accepted (that is, the number of rejects is

equal to or below the acceptance number),

we can state with 90% confidence that the lot defect level is below the LTPD.

Sampling Plans

Outgoing Quality

AOQ: Average outgoing quality

Average level of defects shipped. Found by multiplying lot percent defective by probability of acceptance, that, is pxP. Generates AOQ curve. Assumes rejected lots are screened to make lot perfect.

AOQL: Average outgoing quality limit

Maximum value of AOQ curve. With 100% inspection of rejected lots, AOQ to customer is never any worse than AOQL.

Sampling Plans

AOQ Curve and AOQL

Statistical Process Control

Control Charts for Reliability

If p is the historical process average, the upper control limit is given by

Three-Sigma Control Chart for Binomial Proportions





1 1 UCL p z p p n a    

Cumulative Count Control Chart

Control Based on Good Units

CCC chart concentrates on number of

good units produced instead of number of defective items.

Accumulated number of good units is

plotted in time and compared to control limits such that a cumulative count of good units will fall inside the limits when a

Cumulative Count Control Chart

Cumulative Count Control Chart

Points Outside Control Limits

Interpretation

If the first failure occurs before the cumulative count exceeds the lower

limit, the process is not capable of

meeting the PPM level specified for the control limits.

If the first failure occurs after the cumulative count exceeds the upper

limit, the process is demonstrating even

better performance than the PPM level

Cumulative Count Control

Charts for Low PPM

p

.

Median



0

6931







p



ln

/

ln

n

_LCL



a





1

2

1







p



ln

/

ln

n

_UCL



a



1

2

Centerline Set at Median:

Lower Control Limit:

Upper Control Limit:

Centerline and 90% LCL and UCL can be

obtained from the following table by reading in the row for a given PPM under the columns labeled 0.5, 0.95, and 0.05, respectively.

Minimum Sample Sizes for Zero

Rejects at Various Probabilities

Probability of Zero Rejects

PPM 0.999 0.995 0.99 0.975 0.95 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.025 0.01 0.005 0.001 1 1001 5013 10051 25318 51294 105361 223144 356675 510826 693147 916291 1203973 1609438 2302584 2995731 3688878 4605168 5298315 6907752 6 sigma: 3.4 295 1475 2956 7447 15087 30989 65631 104905 150243 203867 269497 354110 473364 677230 881097 1084963 1354460 1558326 2031690 5 201 1003 2011 5064 10259 21073 44629 71335 102165 138630 183258 240794 321887 460516 599145 737775 921032 1059661 1381548 10 101 502 1006 2532 5130 10536 22315 35668 51083 69315 91629 120397 160943 230258 299572 368887 460515 529830 690773 20 51 251 503 1266 2565 5268 11158 17834 25542 34658 45815 60199 80472 115129 149786 184443 230257 264914 345385 50 21 101 202 507 1026 2108 4463 7134 10217 13863 18326 24079 32188 46051 59914 73776 92102 105964 138152 75 14 67 134 338 684 1405 2976 4756 6811 9242 12217 16053 21459 30700 39942 49184 61400 70642 92100 100 11 51 101 254 513 1054 2232 3567 5109 6932 9163 12040 16094 23025 29956 36887 46050 52981 69075 200 6 26 51 127 257 527 1116 1784 2554 3466 4581 6020 8047 11512 14978 18443 23024 26489 34536 5 sigma: 233 5 22 44 109 221 453 958 1531 2193 2975 3933 5167 6907 9882 12856 15831 19763 22737 29644 500 3 11 21 51 103 211 447 714 1022 1386 1833 2408 3219 4605 5990 7376 9209 10594 13813 1000 1 6 11 26 52 106 224 357 511 693 916 1204 1609 2302 2995 3688 4603 5296 6905 5000 1 3 6 11 22 45 72 102 139 183 241 322 460 598 736 919 1058 1379 4 sigma: 6210 1 2 5 9 17 36 58 83 112 148 194 259 370 481 593 740 851 1109 10000 1 3 6 11 23 36 51 69 92 120 161 230 299 368 459 528 688 3 sigma: 66807 1 2 4 6 8 11 14 18 24 34 44 54 67 77 100

Class Project #11

Cumulative Count Control Chart:

A process is assumed to run at 500 PPM. A cumulative count control chart is

desired for monitoring purposes.

Find the centerline and upper and lower non-defective cumulative count values for a 95% control limit band.

Class Project #11

Cumulative Count Control Chart:

A process is assumed to run at 500 PPM.

A cumulative count control chart is desired for monitoring purposes.

Find the centerline and upper and lower non-defective cumulative count values for a 95% control limit band.

From Table: CL = 1386, LCL = 51, UCL = 7376

Formulas:

Compare to the tabulated values for “Minimum Sample Sizes for Zero Rejects at Various

Probabilities.” Matches! 6 0.6931 1386 500 10 ln(1 0.05 / 2) 51 ln(1 0.0005) ln(0.05 / 2) 7376 ln(1 0.0005) CL LCL UCL           

Using ART for “Cumulative Count

Control Charts”