Chapter 9A

Process Capability and SPC

• Process Variation

• Process Capability

• Process Control Procedures– Variable data– Attribute data

OBJECTIVES

9A-2

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.

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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

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.

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Process Capability

• Process limits

• Specification limits

• How do the limits relate to one another?

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.

9A-6

A simple ratio:

Specification Width_________________________________________________________

Actual “Process Width”

Generally, the bigger the better.

Process Capability – A Standard Measure of How Process Capability – A Standard Measure of How Good a Process Is.Good a Process Is.

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Process CapabilityProcess Capability

This is a “one-sided” Capability Index

Concentration on the side which is closest to the specification - closest to being “bad”

3

;3

XUTLLTLXMinC pk

9A-8

The Cereal Box Example

• We are the maker of this cereal. Consumer reports has just published an article that shows that we frequently have less than 16 ounces of cereal in a box.

• Let’s assume that the government says that we must 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.

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Cereal Box Process Capability

• Specification or Tolerance Limits– Upper Spec = 16.8 oz– Lower Spec = 15.2 oz

• Observed Weight– Mean = 15.875 oz– Std Dev = .529 oz

3

;3

XUTLLTLXMinC pk

)529(.3

875.158.16;

)529(.3

2.15875.15MinC pk

5829.;4253.MinC pk

4253.pkC

9A-10

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 or better… 6-Sigma company wants 2.0!

9A-11

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

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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 BehaviorNormal Behavior

Possible problem, investigatePossible problem, investigate

Possible problem, investigatePossible problem, investigate

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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.

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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%

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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

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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:

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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

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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

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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

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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

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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

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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 Range1 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

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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 9A.6From Exhibit 9A.6

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

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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

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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

9A-27

Question Bowl

A 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 above

Answer: a. Capability index

9A-29

Question Bowl

You 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)

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Question Bowl

You 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

9A-31

Question Bowl

You 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

9A-32

Question Bowl

You 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 above

Answer: e. None of the above

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End of Chapter 9A

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