1 © the mcgraw-hill companies, inc., 2004. 2 technical note 7 process capability and statistical...

1

©The McGraw-Hill Companies, Inc., 2004

2


Technical Note 7

Process Capability and Statistical Quality Control

3


• Process Variation

• Process Capability

• Process Control Procedures– Variable data– Attribute data

• Acceptance Sampling– Operating Characteristic Curve

OBJECTIVES

4


Basic Forms of Variation

Assignable variation is caused by factors that can be clearly identified and possibly managed

Common variation is inherent in the production process

Example: A poorly trained employee that creates variation in finished product output.

Example: A poorly trained employee that creates variation in finished product output.

Example: A molding process that always leaves “burrs” or flaws on a molded item.

Example: A molding process that always leaves “burrs” or flaws on a molded item.

5


Taguchi’s View of Variation

IncrementalCost of Variability

High

Zero

LowerSpec

TargetSpec

UpperSpec

Traditional View

IncrementalCost of Variability

High

Zero

LowerSpec

TargetSpec

UpperSpec

Taguchi’s View

Exhibits TN7.1 & TN7.2

Exhibits TN7.1 & TN7.2

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.

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.

6


Process Capability

• Process limits

• Tolerance limits

• How do the limits relate to one another?

7


Process Capability Index, Cpk

3

X-UTLor

3

LTLXmin=C pk

Shifts in Process Mean

Capability Index shows how well parts being produced fit into design limit specifications.

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 in production performance from differing samples.

As a production process produces items small shifts in equipment or systems can cause differences in production performance from differing samples.

8


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 of defectives.

– p-chart application

• Variable (Continuous)– Usually measured by the mean and the standard

deviation.– X-bar and R chart applications

9


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 BehaviorNormal Behavior

Possible problem, investigatePossible problem, investigate

Possible problem, investigatePossible problem, investigate

Statistical Process Control (SPC) Charts

10


Control Limits are based on the Normal Curve

x

0 1 2 3-3 -2 -1z

Standard deviation units or “z” units.

Standard deviation units or “z” units.

11


Control Limits

We 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%

12


Example of Constructing a p-Chart: Required Data

1 100 42 100 23 100 54 100 35 100 66 100 47 100 38 100 79 100 1

10 100 211 100 312 100 213 100 214 100 815 100 3

Sample

No.

No. of

Samples

Number of defects found in each sample

13


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

p =Total Number of Defectives

Total Number of Observationsp =

Total Number of Defectives

Total Number of Observations

ns

)p-(1 p = p n

s)p-(1 p

= p

p

p

z - p = LCL

z + p = UCL

s

s

p

p

z - p = LCL

z + p = UCL

s

s

Given:

Compute control limits:

14


1. Calculate the sample proportions, p (these are what can be plotted on the p-chart) for each sample

1. Calculate the sample proportions, p (these are what can be plotted on the p-chart) for each sample

Sample n Defectives p1 100 4 0.042 100 2 0.023 100 5 0.054 100 3 0.035 100 6 0.066 100 4 0.047 100 3 0.038 100 7 0.079 100 1 0.01

10 100 2 0.0211 100 3 0.0312 100 2 0.0213 100 2 0.0214 100 8 0.0815 100 3 0.03

Example of Constructing a p-chart: Step 1

15


2. Calculate the average of the sample proportions2. Calculate the average of the sample proportions

0.036=1500

55 = p 0.036=1500

55 = p

3. Calculate the standard deviation of the sample proportion 3. Calculate the standard deviation of the sample proportion

.0188= 100

.036)-.036(1=

)p-(1 p = p n

s .0188= 100

.036)-.036(1=

)p-(1 p = p n

s

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

16


4. Calculate the control limits4. Calculate the control limits

3(.0188) .036 3(.0188) .036

UCL = 0.0924LCL = -0.0204 (or 0)UCL = 0.0924LCL = -0.0204 (or 0)

p

p

z - p = LCL

z + p = UCL

s

s

p

p

z - p = LCL

z + p = UCL

s

s

Example of Constructing a p-chart: Step 4

17


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 average of the proportions, and the control limits

5. Plot the individual sample proportions, the average of the proportions, and the control limits

18


Example of x-bar and R Charts: Required Data

Sample Obs 1 Obs 2 Obs 3 Obs 4 Obs 51 10.68 10.689 10.776 10.798 10.7142 10.79 10.86 10.601 10.746 10.7793 10.78 10.667 10.838 10.785 10.7234 10.59 10.727 10.812 10.775 10.735 10.69 10.708 10.79 10.758 10.6716 10.75 10.714 10.738 10.719 10.6067 10.79 10.713 10.689 10.877 10.6038 10.74 10.779 10.11 10.737 10.759 10.77 10.773 10.641 10.644 10.72510 10.72 10.671 10.708 10.85 10.71211 10.79 10.821 10.764 10.658 10.70812 10.62 10.802 10.818 10.872 10.72713 10.66 10.822 10.893 10.544 10.7514 10.81 10.749 10.859 10.801 10.70115 10.66 10.681 10.644 10.747 10.728

19


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 0.1162 10.79 10.86 10.601 10.746 10.779 10.755 0.2593 10.78 10.667 10.838 10.785 10.723 10.759 0.1714 10.59 10.727 10.812 10.775 10.73 10.727 0.2215 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 0.1437 10.79 10.713 10.689 10.877 10.603 10.735 0.2748 10.74 10.779 10.11 10.737 10.75 10.624 0.6699 10.77 10.773 10.641 10.644 10.725 10.710 0.13210 10.72 10.671 10.708 10.85 10.712 10.732 0.17911 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 0.25013 10.66 10.822 10.893 10.544 10.75 10.733 0.34914 10.81 10.749 10.859 10.801 10.701 10.783 0.15815 10.66 10.681 10.644 10.747 10.728 10.692 0.103

Averages 10.728 0.220400

20


Example of x-bar and R charts: Step 2. Determine Control Limit Formulas and

Necessary Tabled Values

x Chart Control Limits

UCL = x + A R

LCL = x - A R

2

2

x 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

R Chart Control Limits

UCL = D R

LCL = D R

4

3

From Exhibit TN7.7From Exhibit TN7.7

n A2 D3 D42 1.88 0 3.273 1.02 0 2.574 0.73 0 2.285 0.58 0 2.116 0.48 0 2.007 0.42 0.08 1.928 0.37 0.14 1.869 0.34 0.18 1.82

10 0.31 0.22 1.7811 0.29 0.26 1.74

21


Example 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

10.601

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

Mea

ns

UCL

LCL

22


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

0

0.46504

)2204.0)(0(R D= LCL

)2204.0)(11.2(R D= UCL

3

4

0

0.46504

)2204.0)(0(R D= LCL

)2204.0)(11.2(R D= UCL

3

4

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

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

Sample

RUCL

LCL

23


Basic Forms of Statistical Sampling for Quality Control

• Acceptance Sampling is sampling to accept or reject the immediate lot of product at hand

• Statistical Process Control is sampling to determine if the process is within acceptable limits

24


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)

25


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

26


Acceptance Sampling: Single Sampling Plan

A simple goal

Determine (1) how many units, n, to sample from a lot, and (2) the maximum number of defective items, c, that can be found in the sample before the lot is rejected

27


Risk• Acceptable Quality Level (AQL)

– Max. acceptable percentage of defectives defined by producer

• The(Producer’s risk)– The probability of rejecting a good lot

• Lot Tolerance Percent Defective (LTPD)– Percentage of defectives that defines consumer’s

rejection point

• The (Consumer’s risk)– The probability of accepting a bad lot

28


Operating Characteristic Curve

n = 99c = 4

AQL LTPD

00.10.20.30.40.50.60.70.80.9

1

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

Percent defective

Pro

bab

ilit

y of

acc

epta

nce

=.10(consumer’s risk)

= .05 (producer’s risk)

The OCC brings the concepts of producer’s risk, consumer’s risk, sample size, and maximum defects allowed together

The OCC brings the concepts of producer’s risk, consumer’s 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

The shape or slope of the curve is dependent on a particular combination of the four parameters

29


Example: Acceptance Sampling Problem

Zypercom, 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 lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot.

Develop a sampling plan for Zypercom and determine a rule to be followed by the receiving inspection personnel.

Zypercom, 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 lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot.

Develop a sampling plan for Zypercom and determine a rule to be followed by the receiving inspection personnel.

30


Example: Step 1. What is given and what

is not?

In this problem, AQL is given to be 0.01 and LTDP is given to be 0.03. We are also given an alpha of 0.05 and a beta of 0.10.

In this problem, AQL is given to be 0.01 and LTDP 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.”

What you need to determine is your sampling plan is “c” and “n.”

31


Example: Step 2. Determine “c”

First divide LTPD by AQL.First divide LTPD by AQL.LTPD

AQL =

.03

.01 = 3

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.

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 7.10Exhibit TN 7.10

c LTPD/AQL n AQL c LTPD/AQL n AQL0 44.890 0.052 5 3.549 2.6131 10.946 0.355 6 3.206 3.2862 6.509 0.818 7 2.957 3.9813 4.890 1.366 8 2.768 4.6954 4.057 1.970 9 2.618 5.426

So, c = 6.So, c = 6.

32


Example: Step 3. Determine Sample Size

c = 6, from Tablen (AQL) = 3.286, from TableAQL = .01, given in problem

c = 6, from Tablen (AQL) = 3.286, from TableAQL = .01, given in problem

Sampling Plan:Take a random sample of 329 units from a lot. Reject the lot if more than 6 units are defective.

Sampling Plan:Take a random sample of 329 units from a lot. Reject the lot if more than 6 units are defective.

Now given the information below, compute the sample size in units to generate your sampling plan

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

33


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 indexb. Producer’s riskc. Consumer’s riskd. AQLe. None of the aboveAnswer: a. Capability index

34


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

outside a boundary it should signal to the production manager to do which of the following?

a. System is out of control, should be stopped and fixedb. System is out of control, but can still be operated without

any concernc. System is only out of control if the number of observations

falling outside the boundary exceeds statistical expectationsd. System is OK as ise. None of the above

Answer: c. System is only out of control if the number of observations falling 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.)

35


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. 2.5c. 0.0025d. 0.00025e. Can not be computed on data above

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

36


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

e. None of the above

Answer: b. 0.31 (from Exhibit TN7.7)

37


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. 0.88c. 1.88d. 2.11e. None of the above

Answer: a. 0 (from Exhibit TN7.7)

38


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. 1.00c. 1.88d. 2.11e. None of the aboveAnswer: e. None of the above (from Exhibit TN7.7 the correct value is 2.57)

39


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. Betac. AQLd. ce. None of the above

Answer: d. c

40


End of Technical Note 7

1 © the mcgraw-hill companies, inc., 2004. 2 technical note 7 process capability and statistical...

Documents