Chap008tn SPC

7/29/2019 Chap008tn SPC

1/46

1

Technical Note 8

Process Capability and

Statistical Quality Control

7/29/2019 Chap008tn SPC

2/46

2

Process Variation

Process Capability

Process Control Procedures

Variable data

Attribute data

Acceptance Sampling

Operating Characteristic Curve

OBJECTIVES

7/29/2019 Chap008tn SPC

3/46

3

Basic Forms of VariationAssignable variation

is caused byfactors that can beclearly identified

and possiblymanaged

Common variationisinherent in theproduction process

Example: A poorly trained

employee that creates

variation in finished

product output.

Example: A moldingprocess that always leaves

burrs or flaws on a

molded item.

7/29/2019 Chap008tn SPC

4/46

4Taguchis View of Variation

Incremental

Cost of

Variability

High

Zero

Lower

Spec

Target

Spec

Upper

Spec

Traditional View

Incremental

Cost of

Variability

High

Zero

Lower

Spec

Target

Spec

Upper

Spec

Taguchis View

Exhibits

TN8.1 &

TN8.2

Traditional view is that quality within the LS and US is good

and that the cost of quality outside this range is constant, where

Taguchi views costs as increasing as variability increases, so seek

to achieve zero defects and that will truly minimize quality costs.

7/29/2019 Chap008tn SPC

5/46

5Process CapabilityProcess limits

Specification limits

How do the limits relate to one another?

7/29/2019 Chap008tn SPC

6/46

6

Process Capability Index, Cpk

3

X-UTLor

3

LTLXmin=Cpk

Shifts in Process Mean

Capability Index shows

how well parts being

produced fit into design

limit specifications.

As a production process

produces items small

shifts in equipment or

systems can cause

differences inproduction

performance from

differing samples.

7/29/2019 Chap008tn SPC

7/46

7

A simple ratio:

Specification Width_________________________________________________________

Actual Process Width

Generally, the bigger the better.

Process Capability A Standard

Measure of How Good a Process Is.

7/29/2019 Chap008tn SPC

8/46

8

Process Capability

This is a one-sided Capability Index

Concentration on the side which is closest to

the specification - closest to being bad

3

;3

XUTLLTLXMinCpk

7/29/2019 Chap008tn SPC

9/46

9The Cereal Box Example

We are the maker of this cereal. Consumer reportshas just published an article that shows that we

frequently have less than 15 ounces of cereal in a

box.

Lets assume that the government says that wemust be within 5 percent of the weight advertised

on the box.

Upper Tolerance Limit = 16 + .05(16) = 16.8 ounces

Lower Tolerance Limit = 16 .05(16) = 15.2 ounces

We go out and buy 1,000 boxes of cereal and find

that they weight an average of 15.875 ounces with a

standard deviation of .529 ounces.

7/29/2019 Chap008tn SPC

10/46

10

Cereal Box Process Capability

Specification orTolerance Limits Upper Spec = 16.8 oz

Lower Spec = 15.2 oz

Observed Weight Mean = 15.875 oz

Std Dev = .529 oz

3

;3

XUTLLTLXMinCpk

)529(.3

875.158.16;

)529(.3

2.15875.15MinCpk

5829.;4253.MinCpk

4253.pkC

7/29/2019 Chap008tn SPC

11/46

11

What does a Cpk of .4253 mean?

An index that shows how well the units

being produced fit within the specification

limits.This is a process that will produce a

relatively high number of defects.

Many companies look for a Cpk of 1.3 orbetter 6-Sigma company wants 2.0!

12

7/29/2019 Chap008tn SPC

12/46

12

Types of Statistical Sampling

Attribute (Go or no-go information) Defectives refers to the acceptability of product

across a range of characteristics.

Defects refers to the number of defects per unit

which may be higher than the number ofdefectives.

p-chart application

Variable (Continuous) Usually measured by the mean and the

standard deviation.

X-bar and R chart applications

13

7/29/2019 Chap008tn SPC

13/46

13

Statistical

Process

Control(SPC) Charts

UCL

LCL

Samples

over time

1 2 3 4 5 6

UCL

LCL

Samples

over time

1 2 3 4 5 6

UCL

LCL

Samples

over time

1 2 3 4 5 6

Normal Behavior

Possible problem, investigate

Possible problem, investigate

14

7/29/2019 Chap008tn SPC

14/46

14Control Limits are based on theNormal Curve

x

0 1 2 3-3 -2 -1z

m

Standarddeviation

units or z

units.

15

7/29/2019 Chap008tn SPC

15/46

15Control LimitsWe establish the Upper Control Limits (UCL) and

the Lower Control Limits (LCL) with plus or minus

3 standard deviations from some x-bar or mean

value. Based on this we can expect 99.7% of our

sample observations to fall within these limits.

xLCL UCL

99.7%

16

7/29/2019 Chap008tn SPC

16/46

16

Example of Constructing a p-Chart:Required Data

1 100 4

2 100 2

3 100 5

4 100 3

5 100 6

6 100 4

7 100 3

8 100 7

9 100 1

10 100 2

11 100 3

12 100 2

13 100 2

14 100 8

15 100 3

Sample

No.

No. of

Samples

Number of

defects found

in each sample

17

7/29/2019 Chap008tn SPC

17/46

17

Statistical Process Control Formulas:Attribute Measurements (p-Chart)p =

T o t al N u m b e r o f D e fe c tiv e s

T o ta l N u m b e r o f O b s e rv a tio n s

n

s

)p-(1p=p

p

p

z-p=LCL

z+p=UCL

s

s

Given:

Compute control limits:

18

7/29/2019 Chap008tn SPC

18/46

18

1. Calculate the

sample proportions,

p (these are what

can be plotted on thep-chart) for eachsample

Sample n Defectives p1 100 4 0.04

2 100 2 0.02

3 100 5 0.05

4 100 3 0.03

5 100 6 0.066 100 4 0.04

7 100 3 0.03

8 100 7 0.07

9 100 1 0.01

10 100 2 0.02

11 100 3 0.03

12 100 2 0.02

13 100 2 0.02

14 100 8 0.08

15 100 3 0.03

Example of Constructing a p-chart: Step 1

19

7/29/2019 Chap008tn SPC

19/46

19

2. Calculate the average of the sample proportions

0.036=

1500

55=p

3. Calculate the standard deviation of the

sample proportion

.0188=100

.036)-.036(1=

)p-(1p=p

n

s

Example of Constructing a p-chart: Steps 2&3

20

7/29/2019 Chap008tn SPC

20/46

20

4. Calculate the control limits

3(.0188).036

UCL = 0.0924LCL = -0.0204 (or 0)

p

p

z-p=LCL

z+p=UCL

s

s

Example of Constructing a p-chart: Step 4

21

7/29/2019 Chap008tn SPC

21/46

21

Example of Constructing a p-Chart: Step 5

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Observation

p

UCL

LCL

5. Plot the individual sample proportions, the averageof the proportions, and the control limits

22

7/29/2019 Chap008tn SPC

22/46

22

Example of x-bar and R Charts:Required DataSample Obs 1 Obs 2 Obs 3 Obs 4 Obs 5

1 10.68 10.689 10.776 10.798 10.714

2 10.79 10.86 10.601 10.746 10.779

3 10.78 10.667 10.838 10.785 10.723

4 10.59 10.727 10.812 10.775 10.73

5 10.69 10.708 10.79 10.758 10.6716 10.75 10.714 10.738 10.719 10.606

7 10.79 10.713 10.689 10.877 10.603

8 10.74 10.779 10.11 10.737 10.75

9 10.77 10.773 10.641 10.644 10.725

10 10.72 10.671 10.708 10.85 10.71211 10.79 10.821 10.764 10.658 10.708

12 10.62 10.802 10.818 10.872 10.727

13 10.66 10.822 10.893 10.544 10.75

14 10.81 10.749 10.859 10.801 10.701

15 10.66 10.681 10.644 10.747 10.728

23

7/29/2019 Chap008tn SPC

23/46

23

Example of x-bar and R charts: Step 1.Calculate sample means, sample ranges,mean of means, and mean of ranges.Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 5 Avg Range

1 10.68 10.689 10.776 10.798 10.714 10.732 0.116

2 10.79 10.86 10.601 10.746 10.779 10.755 0.259

3 10.78 10.667 10.838 10.785 10.723 10.759 0.171

4 10.59 10.727 10.812 10.775 10.73 10.727 0.221

5 10.69 10.708 10.79 10.758 10.671 10.724 0.1196 10.75 10.714 10.738 10.719 10.606 10.705 0.143

7 10.79 10.713 10.689 10.877 10.603 10.735 0.274

8 10.74 10.779 10.11 10.737 10.75 10.624 0.669

9 10.77 10.773 10.641 10.644 10.725 10.710 0.132

10 10.72 10.671 10.708 10.85 10.712 10.732 0.179

11 10.79 10.821 10.764 10.658 10.708 10.748 0.16312 10.62 10.802 10.818 10.872 10.727 10.768 0.250

13 10.66 10.822 10.893 10.544 10.75 10.733 0.349

14 10.81 10.749 10.859 10.801 10.701 10.783 0.158

15 10.66 10.681 10.644 10.747 10.728 10.692 0.103

Averages 10.728 0.220400

24

7/29/2019 Chap008tn SPC

24/46

24

Example of x-bar and R charts: Step 2. DetermineControl Limit Formulas and Necessary TabledValuesx Chart Control Limits

UCL = x + A R

LCL = x - A R

2

2

R Chart Control Limits

UCL = D R

LCL = D R

4

3

From Exhibit TN8.7

n A2 D3 D4

2 1.88 0 3.27

3 1.02 0 2.57

4 0.73 0 2.28

5 0.58 0 2.11

6 0.48 0 2.00

7 0.42 0.08 1.92

8 0.37 0.14 1.869 0.34 0.18 1.82

10 0.31 0.22 1.78

11 0.29 0.26 1.74

25

7/29/2019 Chap008tn SPC

25/46

25Example of x-bar and R charts: Steps 3&4.Calculate x-bar Chart and Plot Values

10.601

10.856

=).58(0.2204-10.728RA-x=LCL

=).58(0.2204-10.728RA+x=UCL

2

2

10.550

10.600

10.650

10.700

10.750

10.800

10.850

10.900

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Sample

Means

UCL

LCL

26

7/29/2019 Chap008tn SPC

26/46

Example of x-bar and R charts: Steps 5&6.Calculate R-chart and Plot Values

00.46504

)2204.0)(0(RD=LCL)2204.0)(11.2(RD=UCL

3

4

0 . 0 0 0

0 . 1 0 0

0 . 2 0 0

0 . 3 0 0

0 . 4 0 0

0 . 5 0 0

0 . 6 0 0

0 . 7 0 0

0 . 8 0 0

1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5

S a m p l e

RUCL

LCL

27

7/29/2019 Chap008tn SPC

27/46

Basic Forms of Statistical Samplingfor Quality Control

Acceptance Sampling is sampling

to accept or reject the immediate lot

ofproduct at hand

Statistical Process Control is

sampling to determine if the

process is within acceptable limits

28

7/29/2019 Chap008tn SPC

28/46

Acceptance Sampling

Purposes Determine quality level

Ensure quality is within predetermined level

Advantages

Economy Less handling damage

Fewer inspectors

Upgrading of the inspection job

Applicability to destructive testing Entire lot rejection (motivation for

improvement)

29

7/29/2019 Chap008tn SPC

29/46

Acceptance Sampling (Continued)Disadvantages

Risks of accepting bad lots and

rejecting good lots Added planning and documentation

Sample provides less information

than 100-percent inspection

30

7/29/2019 Chap008tn SPC

30/46

Acceptance Sampling:Single Sampling PlanA simple goal

Determine (1) how many units, n,to sample from a lot, and (2) themaximum number of defective

items, c, that can be found in thesample before the lot is rejected

31

7/29/2019 Chap008tn SPC

31/46

Risk Acceptable Quality Level (AQL)

Max. acceptable percentage of defectives

defined by producer

The a (Producers risk)

The probability of rejecting a good lot Lot Tolerance Percent Defective (LTPD)

Percentage of defectives that defines

consumers rejection point The (Consumers risk)

The probability of accepting a bad lot

32

7/29/2019 Chap008tn SPC

32/46

Operating Characteristic Curve

n = 99

c = 4

AQL LTPD

00.1

0.2

0.3

0.4

0.50.6

0.7

0.8

0.9

1

1 2 3 4 5 6 7 8 9 10 11 12

Percent defective

Probability

ofac

ceptance

=.10(consumers risk)

a = .05 (producers risk)

The OCC brings the concepts of producers risk, consumers

risk, sample size, and maximum defects allowed together

The shape

or slope of

the curve is

dependent

on a

particular

combination

of the four

parameters

33

7/29/2019 Chap008tn SPC

33/46

Example: Acceptance Sampling ProblemZypercom, a manufacturer of video interfaces,purchases printed wiring boards from an outside

vender, Procard. Procard has set an acceptable

quality level of 1% and accepts a 5% risk of rejecting

lots at or below this level. Zypercom considers lotswith 3% defectives to be unacceptable and will assume

a 10% risk of accepting a defective lot.

Develop a sampling plan for Zypercom and determinea rule to be followed by the receiving inspection

personnel.

34

7/29/2019 Chap008tn SPC

34/46

Example:Step 1. What is given and what is not?

In this problem, AQL is given to be 0.01 and LTPD

is given to be 0.03. We are also given an alpha of

0.05 and a beta of 0.10.

What you need to determine is your sampling

plan is c and n.

35

7/29/2019 Chap008tn SPC

35/46

Example: Step 2. Determine c

First divide LTPD by AQL.

LTPD

AQL

=.03

.01

= 3

Then find the value for c by selecting the value in the

TN7.10 n(AQL)column that is equal to or just greater than

the ratio above.

Exhibit TN 8.10

c LTPD/AQL n AQL c LTPD/AQL n AQL

0 44.890 0.052 5 3.549 2.6131 10.946 0.355 6 3.206 3.286

2 6.509 0.818 7 2.957 3.981

3 4.890 1.366 8 2.768 4.695

4 4.057 1.970 9 2.618 5.426

So, c = 6.

36

7/29/2019 Chap008tn SPC

36/46

Example: Step 3. Determine Sample Size

c = 6, from Table

n (AQL) = 3.286, from Table

AQL = .01, given in problem

Sampling Plan:Take a random sample of 329 units from a lot.

Reject the lot ifmore than 6 units are defective.

Now given the information below, compute the sample

size in units to generate your sampling plan

n(AQL/AQL) = 3.286/.01 = 328.6, or 329 (always round up)

37

7/29/2019 Chap008tn SPC

37/46

Computers in Quality Control

Records about quality testing and results limit

a firms exposure in the event of a product

liability suit.

Recall programs require that manufacturers

Know the lot number of the parts that are

responsible for the potential defects

Have an information storage system that can tie

the lot numbers of the suspected parts to the finalproduct model numbers

Have an information system that can track the

model numbers of final products to customers

38

7/29/2019 Chap008tn SPC

38/46

Computers in Quality Control

With automation, inspection and testing canbe so inexpensive and quick that

companies may be able to increase sample

sizes and the frequency of samples, thusattaining more precision in both control

charts and acceptance plans

39

7/29/2019 Chap008tn SPC

39/46

Question BowlA methodology that is used to show how well parts

being produced fit into a range specified by

design limits is which of the following?

a. Capability index

b. Producers riskc. Consumers risk

d. AQL

e. None of the above

Answer: a. Capability index

40

7/29/2019 Chap008tn SPC

40/46

Question BowlOn a quality control chart if one of the values plotted falls outside

a boundary it should signal to the production manager to dowhich of the following?

a. System is out of control, should be stopped and fixed

b. System is out of control, but can still be operated without any

concern

c. System is only out of control if the number of observations

falling outside the boundary exceeds statistical expectations

d. System is OK as is

e. None of the aboveAnswer: c. System is only out of

control if the number of observationsfalling outside the boundary exceeds

statistical expectations

(We expect with Six Sigma that 3 out of 1,000 observations will fall outside

the boundaries normally and those deviations should not lead managers

to conclude the system is out of control.)

41

7/29/2019 Chap008tn SPC

41/46

Question BowlYou want to prepare a p chart and you

observe 200 samples with 10 in each,

and find 5 defective units. What is the

resulting fraction defective?

a. 25b. 2.5

c. 0.0025

d. 0.00025e. Can not be computed on data above

Answer: c. 0.0025 (5/(2000x10)=0.0025)

42

7/29/2019 Chap008tn SPC

42/46

Question BowlYou want to prepare an x-bar chart. If the number of

observations in a subgroup is 10, what is the

appropriate factor used in the computation of

the UCL and LCL?

a. 1.88

b. 0.31

c. 0.22

d. 1.78


Answer: b. 0.31 (from Exhibit TN8.7)

43

7/29/2019 Chap008tn SPC

43/46

Question BowlYou want to prepare an R chart. If the number of

observations in a subgroup is 5, what is the

appropriate factor used in the computation

of the LCL?

a. 0b. 0.88

c. 1.88

d. 2.11


Answer: a. 0 (from Exhibit TN8.7)

44

7/29/2019 Chap008tn SPC

44/46

Question BowlYou want to prepare an R chart. If the

number of observations in a subgroup

is 3, what is the appropriate factor

used in the computation of the UCL?

a. 0.87b. 1.00

c. 1.88

d. 2.11


Answer: e. None of the above (from Exhibit

TN8.7 the correct value is 2.57)

45

7/29/2019 Chap008tn SPC

45/46

Question BowlThe maximum number of defectives that can be

found in a sample before the lot is rejected is

denoted in acceptance sampling as which of

the following?

a. Alphab. Beta

c. AQL

d. c


Answer: d. c

46

7/29/2019 Chap008tn SPC

46/46

Chap008tn SPC

Documents