process capability training

8/18/2019 Process Capability Training

1/54

1

Using control charts,distribution analysis, rational

sampling and Ppk to betterunderstand manufacturing

process capability

Process Capability


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2

Process Capability

StandardPPAP Ppk > 1.67

OngoingProduction

Ppk > 1.33


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3

What is the problem that needs to be addressed?

First we will examine the components of

variation that are measured by Cpk and Ppk


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4

Subgroup Number

20

90

Total Variation

Components of Variation


5/54

5

Subgroup Number

20

90

Between Subgroup Variation



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6

Subgroup Number

20

90

Within Subgroup Variation



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7

Subgroup Number

20

90

Within

Within Between

Between Total

Total

+ =



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8

Total VariationWithin Subgroup

Variation

Between Subgroup

Variation+ =

Cpk measures within

Subgroup variation Ppk measuresTotal variation

Ppk and Cpk

σ2within σ2between σ2total


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9

Ppk and Cpk

Now for a look at the formulas


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10

)ˆ3

)X(or

ˆ3min(

22 / / d Rd R

LSL) X (USLCpk

σ σ

−−

=

Cpk (Process Potential Capability)

Ppk (Process Performance)

)ˆ3

)(orˆ3

min(S S

LSL X ) X (USLPpk

σ σ

−−

=

What is the difference above?

Ppk vs. Cpk Formulas


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11

Standard Deviation

2 / / ˆ:Cpk 2 d Rd R =σ

∑=

−

−

=

n

i

i

S

n

X X

1

2

1

)(ˆ :Ppk σ

Ppk vs. Cpk Formulas

Uses all of the data

to estimate sigma!

Uses min and max of each

subgroup to estimate sigma.For subgroup size of 5, this amounts

to only 40% of the data.


12/54

12S u b g r o u p Nu m b e r54321

9 0

8 0

7 0

6 0

5 0

4 0

3 0

2 0

S u b g r o u p Nu m b e r

54321

9 0

8 0

7 0

6 0

5 0

4 0

3 0

2 0

R1

R2

R3

R4

R5

X

Ppk standard deviation uses the

difference between each reading and

the mean

Ppk

Cpk

Cpk vs. Ppk Training

Total Variation

Within Subgroup VariationCpk standard deviation uses the

average range divided by D2.

Rave. = (R1+R2+R3+R4+R5)/5

D2 is a constant.

X


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13

Risks of using only Cpk

• Cpk ignores between subgroup variation(It is possible to have out of specification parts and have a Cpk > 1.67)

• Note: OK to use Pp, Cpk, Cp, etc. with Ppk forinvestigation


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14

Results – Cpk vs. Ppk

Now consider the following example


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15

Sample

D y n a m i c R a t e

24222018161412108642

100

80

60

40

20

0

Run Chart with Less Between Variation

Sample

D y n a m i c R a t

e

24222018161412108642

100

80

60

40

20

0

Run Chart with More Between Variation

Ppk vs. Cpk ResultsThe subgroups have the same ranges

Cpk = 2.53

Ppk = 0.52

Cpk = 2.53

Ppk = 1.65

In the bottomexample Cpk = 2.53

even though parts

are out of

specification,

because Cpk

measures only

within subgroupvariation.

USL = 100, LSL = 10


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Rational Sampling

How do I sample parts so the controlcharts show process shifts?

16


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Rational Subgroup

• Includes only short term variation(No shifts in the process)

• Consecutive parts

• Typically 3-7 consecutive parts

• Typical 300 piece PPAP rational sampling plan (Measure 5,skip 7)

• Measure parts 1- 5,

• Skip parts 6-12,

• Measure parts 13 – 17,

• Skip parts 18-24)………

• Options for 1000 piece run when measuring 125 samples

1. subgroups of 5 evenly spaced: Measure 5 parts, skip 36 parts…2. subgroups of 3 evenly spaced: Measure 3 parts, skip 21 parts…..

17

Why would you use a subgroup of 5 vs. 3 or 3 vs. 5?


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1501401301201101009080706050403020101

0.007

0.006

0.005

0.004

0.003

0.002

Order of Production

Run Chart of Profile

Rational Subgroups Example

18

Good Rational Sampling

Subgroups are parts builtconsecutively and most include

only short term variation.

1501401301201101009080706050403020101

0.007

0.006

0.005

0.004

0.003

0.002

Order of Production


Subgroup 1 Subgroup 4

Subgroup 3

Subgroup 2

Subgroup 6

Subgroup 5

Incorrect Sampling

Single subgroup includes

multiple shifts in the process.

Is this an acceptable rational sampling plan?


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1501401301201101009080706050403020101

0.007

0.006

0.005

0.004

0.003

0.002

Order of Production


20

Subgroup Subgroup

Subgroup

Subgroup

Subgroup

Subgroup

What happens when consecutive parts are chosen for subgroups?

28252219161310741

0.006

0.005

0.004

0.003

Sample

S a m p l e M e a n

_ _ X=0.004453

UCL=0.005093

LCL=0.003813

28252219161310741

0.003

0.002

0.001

0.000

Sample

S a m p l e R a n g e

_ R=0.001109

UCL=0.002346

LCL=0

11111

1111

1

11111111

11

111

1

11

1111

1

Xbar-R Chart of Subgroup - Consecutive Parts

Process ShiftsARE Detected!


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21

Subgroups – Charts below are from same process

Subgroups that include

Process Shifts

Rational Subgroups

(Consecutive Samples)

Process Shifts Detected NO Yes

X bar Control Limits 0.0027 - 0.006 (Inflated) 0.0038 - 0.005

Range Upper ControlLimit

0.006 (Inflated) 0.002

28252219161310741

0.006

0.005

0.004

0.003

Sample

S a m p l e M e a n

_ _ X=0.004425

UCL=0.006095

LCL=0.002756

28252219161310741

0.0060

0.0045

0.0030

0.0015

0.0000

Sample


_ R=0.002894

UCL=0.006120

LCL=0

Xbar-R Chart (Subgroups include Process Shifts)

28252219161310741

0.006

0.005

0.004

0.003

Sample

S a m p l e M e a n

_ _ X=0.004453

UCL=0.005093

LCL=0.003813

28252219161310741

0.003

0.002

0.001

0.000

Sample


_ R=0.001109

UCL=0.002346

LCL=0

11111

111

1

1

11111111

11

111

1

11

1111

1

Xbar-R Chart (Subgroups with Consecutive Parts)


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Rational Subgroup Take Away

• If parts in a subgroup span process shifts then

control limits are inflated and process shifts areNOT detected.

• Use 3-7 consecutive parts to minimize risk ofincluding process shifts inside a given subgroup

22


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Sampling Plan – Initial Capability Study

What parts should I measure and what variablesdo I need to keep track of?

23


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Sampling Plan - Exhaust Manifold

• 500 manifolds are being produced for phase 0 PPAP.

• MFG Equipment: One milling machine with two spindles.One broaching bar is used to make the spline.

• Time constraints and resources only allow for themeasurement of roughly 150 samples.

How should samples be collected for the capability study of theSCs?

Increase number of subgroups (Use 3 parts instead of 5 parts)

25 subgroups of 3 from spindle 1, 25 subgroups of 3 from spindle 2

Evenly space subgroups (e.g. spindle 1 – measure parts 1-3, skip 4- 10,

measure 11-13, skip 14-20…. Same process for spindle 2.


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Sampling Plan - Rear Differential Side Gear

• Supplier can measure all SCs on 250 parts• 1040 gears are being manufactured

1. Two CNC lathes.2. Two spindles per CNC lathe3. After the CNC operation all parts enter the process to machine

the spline. This process uses two broaching bars.4. After broaching the parts are heat treated in one of the two furnaces

each of which hold 520 parts.5. Each furnace is known to have a risk for variation between top and

bottom and left and right.6. Parts cannot be etched and any mark put on parts prior to heat treat

will be burned off during the heat treatment.

Put together a sampling plan for the PPAP run.

21 subgroups of 3 for each of the 4 spindles

Keep track of which broach is used on each part and order of broaching

Divide parts between two furnaces and area of furnaces (top, bottom, left, right)


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Sampling Plan – Cast Aluminum Engine Block

• Process: Molten aluminum forms around 14 sand cores to make anengine block. Sand is then removed leaving a cast block withpassage ways. Block is then cubed (Sides are machined)

• Seven molds make a total of the 14 sand cores

• There are 6 sets of the seven molds each of which makes a set ofthe 14 needed sand cores.• Sets 1-3 are run on production line 1 and sets 4-6 are run on

production line 2.• One machining cell machines all of the parts.

• Part has one SC (cast locator to machined surface), and eight HICs(as cast bore locations to cast datum)

Put together a sampling plan for the PPAP run.

Run 300 pieces for each of the 6 core sets (1800 pieces total)

Rational sampling of 25 subgroups of 5 for each 6 core set (Total of 750 measured parts).

Measure parts 1-5, skip 6-12, measure parts 13-17…..


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Sampling Plan Summary

• Number samples and divide subgroups evenly overproduction run (e.g. measure 5, skip 7………)

• Allocate subgroups across important variables(e.g. machines, spindles, mold cavities, furnace area etc.)

• Keep track of important variables for each measuredpart so their affect can be analyzed

• If possible divide subgroups so control charts can bedone separately for key variables(e.g. 25 subgroups of 3 for spindle 1 and 25 subgroups of 3 for spindle 2.

• Think of the initial PPAP runs as passive DOEs that canbe used to help quantify affect of critical variables

27

Good sampling strategy can lead to quicker improvements!


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28

Normality, Control, & Stability

• Capability numbers (Ppk, Pp, Cpk, Cp) must beignored pending a review of the Control, Stability

and the Distribution.


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29

Control and Stability

Sample

S a m p l e M e a

n

464136312621161161

180

175

170

165

160

_ _ X=171.59

UC L=177.19

LCL=166.00

Sample


464136312621161161

24

18

12

6

0

_ R=7.68

UC L=17.51

LCL=0

111

1111

1

11

1111

1

11

11

1

1

1

1

11

Xbar-R Chart of C15

Not Stable

Not In ControlC p 2.68

CPL 2.82

CPU 2.54

Cpk 2.54

Pp 1.59

PPL 1.67

PPU 1.50

Ppk 1.50

C pm *

O v erall C apability

Potential (Within) C apability

Within

Overall

XX

Out of control points = Special Cause

(Do

NOTuse Ppk, Pp, Cpk, Cp to predict future performance)

3 sigma

limits


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30

Is this process in control?

UCL

LCL

BoreDiameter

Time

Does the answer change

if the source of variation

is from known Non-Random

variation such as tool wear?

Demonstrating e idence of stabilit and


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31

Demonstrating evidence of stability andcontrol with Non-Random variation

UCL One solution is to use

Modified Control Limits

BoreDiameter

Time

LCL

Demonstrating evidence of stability and


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32

UCL

LCL

Modified control charts are sometimes difficult to

manage. An alternative is endpoint control.

Insuring the tool begins and ends where it should.

Any variation from the expected outcome would be

analogous to a “trend” failure for random systems.

First piece inspection to get in the ball park

add 4 more pieces to form a sub-group

Variation from “best fit line” is common cause.

Dotted line is the desired

“Saw Tooth Pattern”

A five piece sub-group to ensuring weended the tool in the right spot

Control limitscentered on

nominal

A second solution is

to use endpoint control

Demonstrating evidence of stability andcontrol with Non-Random variation

Control and Stability with Non Random


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33

Control and Stability with Non-RandomVariation

• Some processes have known causes of variation (Non-

Random) that can cause a process to shift or drift (e.g. batch

to batch variation, machining – tool wear)

• Non-Random variation may be acceptable if there are controls

in place for the shifts and drifts

Reference the Ford STA SPC training and see your site STA or

Master Black Belt for further guidance


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34

Control and Stability Take Away

1. The chances of a subgroup falling outside the 3 sigma limits is

extremely small (1 subgroup out of 370) thus any out of control

points indicate special causes.

Reference the Ford STA SPC training and see your

Master Black Belt for help.

2. For future production the special causes may come back and triggerthe process to make wild swings and out of specification parts.

3. When special causes exist Ppk does not predict the future

performance

5. If special causes exist they must be fixed or controlledprior to using any capability calculations to predict the

future or for approval of PPAP

4. Non-Random variation may be acceptable if there are controls in

place for the shifts and drifts

P k & C k ti


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35

Distribution and Normality

8 . 88 . 07 . 26 . 45 . 64 . 8

Pilot OD

P e r c e n t

0.68450.68400.68350.68300.68250.6820

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Mean 0.6830

StDev 0.0004121

N 125

AD 3.361

P-Value .05

Skewed – Non-

Normal

P value < .05

Ppk & Cpk equations assume

Normal Distribution (Bell Curve).

Calculations are often extremelyinaccurate if data is not normal.


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36

Distribution and Normality

If process experts or a review of historical data

indicates that the process output is not normally

distributed (and should not be), then the data can be• Fitted to the correct distribution (you should

be working with a Master Black Belt)

This will ensure a more accurate estimate of Ppk.

See example on next page


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Is this process capable?

37

Ppk

Normal Distribution Model 1.73

Weibull Distribution Model 0.77

11.29.68.06.44.83.21.60.0

LB USL

LB 0

Target *

USL 7

Sample Mean 0.901049

S ample N 5000

S hape 0.769151

Scale 0.77244

Process Data

Pp *

PPL *

PPU 0.77

Ppk 0.77

Overall Capability

Process Capability of Dimension

Calculations Based on Weibull Distribution Model

121086420-2

LB USL

LB 0

Target *

USL 7

Sample Mean 0.901049

Sample N 5000

StDev(Within) 1.17469

StDev(Overall) 1.17466

Process Data

CPU 1.73

Cpk 1.73

PPU 1.73

Ppk 1.73


Potential (Within) C apability

Process Capability of Dimension

Calculations Based on Normal Distribution Model

Should the data shown below be fit to the lognormal or the


38/54

Should the data shown below be fit to the lognormal or thenormal distribution?

38

252321191715131197531

6.0

5.5

5.0

Sample

S a m p l e M e a n

_ _ X=5.261

UCL=5.588

LCL=4.934

252321191715131197531

1.2

0.9

0.6

0.3

0.0

Sample


_ R=0.567

UCL=1.199

LCL=0

1

111

111

1

111

Xbar-R Chart of Hole Diameter

Based on the control chart does

your answer change?

Looking at the control chart what

is a potential cause for the data

being non-normal?

7.6.56.05.55.04.54.0

99.9

99

95

80

50

20

5

1

0.1

Hole Diameter

P e r c

e n t

Goodness of Fit Test

Lognormal

AD = 0.604

P-Value = 0.114

Probability Plot for Hole DiameterLognormal - 95% CI

7.06.56.05.55.04.54.0

99.9

99

95

80

50

20

5

1

0.1

Hole Diameter

P e r c e

n t

Goodness of Fit Test

Normal

AD = 0.820

P-Value = 0.033

Probability Plot for Hole Diameter

Normal - 95% CI

P value > 0.05 P value < 0.05

Wh t if th d t i d b / t hift?


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6.46.26.05.85.65.45.2

99

95

90

80

70

60

50

40

30

20

10

5

1

Hole Diameter(Post-tool change)

P e

r c e n t Mean 5.726

StDev 0.2296

N 40 AD 0.391

P-Value 0.366

Post Tool Change - Probability Plot of Hole Diameter

Normal

6.05.55.04.54.0

99.9

99

95

90

8070605040

3020

10

5

1

0.1

Hole Diameter (Pre-tool change)

P e

r c e n t

Mean 5.042

StDev 0.2607N 85

AD 0.128

P-Value 0.984

Pre-tool Change - Probability Plot of Hole Diameter

Normal

What if the data is grouped by pre/post process shift?

39

Analyzing the pre & post toolchange data separately shows

the process is normally

distributed.

252321191715131197531

6.0

5.5

5.0

Subgroup

S a m p l e M e a n

Pre-tool change Post-tool change

252321191715131197531

1.2

0.9

0.6

0.3

0.0

Subgroup




Always check the control chartto see if a process shift or out of

control point is causing the data

to be non-normal.

If the process is expected to have a normal distribution and robust controls


40/54

If the process is expected to have a normal distribution and robust controlsare in place for tool changes is this data acceptable for PPAP?

40

252321191715131197531

6.0

5.5

5.0

Subgroup

S a m

p l e M e a n


252321191715131197531

1.2

0.9

0.6

0.3

0.0

Subgroup




6.05.55.04.54.0

99.9

99

95

90

807060504030

20

10

5

1

0.1

Hole Diameter (Pre-tool change)

P e r c e n t

Mean 5.042

StDev 0.2607N 85

AD 0.128

P-Value 0.984

Pre-tool Change - Probability Plot of Hole Diameter

Normal

6.46.26.05.85.65.45.2

99

95

90

80

70

60

5040

30

20

10

5

1

Hole Diameter(Post-tool change)

P e r c e n t Mean 5.726

StDev 0.2296N 40

AD 0.391

P-Value 0.366

Post Tool Change - Probability Plot of Hole Diameter

Normal

Pre-tool change data is stable

Post-tool change

data is stable

Pre-tool change data is on control

Post-tool change

data is in control

Pre-tool change

data is normal

Post-tool change

data is normal

7.26.66.05.44.84.23.63.0

LSL USL

Cp 3.08

CPL 3.09

CPU 3.06

Cpk 3.06

Pp 1.85

PPL 1.85

PPU 1.84Ppk 1.84

Cpm *

Overall Capability

Potential (Within) Capability

Within

Overall

Process Capability of Hole Diameter

Ppk > 1.67

Yes

Resolution and Normal Distributions


41/54

5.45.35.25.15.04.94.84.74.64.5

99.99

99

95

80

50

20

5

1

0.01

Length

P e r c e n t

Mean 4.950

StDev 0.1042

N 5000

AD 188.900

P-Value


42/54

42

Distribution and Normality Take Away

1. Use process experts and historical data to determine

expected distribution. This is a physics discussion.

(Do NOT blindly fit data to a distribution)

Reference the Ford STA SPC training and see your

Master Black Belt for help.

2. If the data does not match the expected distribution determine the

special causes and fix the process.

3. If the data matches the expected distribution but is not normal

(Anderson-Darling P value < .05) fit the data to the correctdistribution and calculate Ppk. (If process is stable and in

control)

4. If data matches the expected distribution and is normal or

symmetrical calculate Ppk. (If process is stable and in control)See following slide for list of non-normal distributions

N l di t ib ti


43/54

43

Non-normal distributions

Manufacturing ProcessDescription

Expected Process Distribute type

Bilateral Process Normal

Grinding process with auto correction Uniform

Tapered roller bearing Fatigue Weibull

Output signal Voltage Log Normal

Flatness/Roundness Log Normal

Plating Thickness Family of Weibull Distributions

FTT - Non Normal - piled up near 100% Log Normal

Sensor output transient voltage Exponential

PPAP O i P d i


44/54

44

PPAP vs. Ongoing Production

When does the process change from a PPAP (Ppk> 1.67) to ongoing Production (Ppk > 1.33) ?

The ongoing production requirement of 1.33 is used afterPPAP once the process and data contains the expectedsources of variation that occur over time.

With many commodities the initial Phase 2 PPAP run will notexperience all of the sources of variation (e.g.: different shifts,lots of components, machine wear, etc.)

The PPAP Ppk requirement of 1.67 is higher than the ongoing

production requirement of 1.33 because it is recognized thatthere will be a degradation as all of the expected sources ofvariation are experienced in ongoing production.


45/54

45

Summary

Process must be in control and stable before looking at

capability index values

The data must be fit to the expected

distribution

(Do NOT blindly fit the data to a distribution.

Exercise 1


46/54

46

Exercise 1

42363024181260

USL

LSL *

Target *

USL 10

Sample Mean 1.64497

S ample N 10000

StDev (O v erall) 2.11492

Process Data

Pp *

PPL *

PPU 1.32

Ppk 1.32

C pm *


PPM < LSL *

PPM > USL 9700.00

PPM Total 9700.00

O bserv ed Performance

PPM < LSL *

PPM > USL 38.99

PPM Total 38.99

Exp. Ov erall Performance

Process Capability of C1

Ppk = 1.32 for ongoing production. Assuming the Engineer is

willing to modify the specification so the Ppk is greater than

1.33 is this process capable?

Exercise 1 Answer

No


47/54

47

Exercise 1 Answer

Ppk

Original data 1.32

Lognormal DistributionModel

0.46

We don’t know if process is stable and in control. We also need historicaldata or process expert to help determine the expected distribution.

42363024181260

USL

Pp *

PPL *

P PU 0.46

P pk 0.46

O verall C apability

PPM < LSL *

PPM > USL 10940.34

PPM Total 10940.34


Process Capability of C1Calculations Based on Lognormal Distribution Model

Answers are drastically

different depending upon

the distribution that is used.

No

Exercise 2


48/54

48

Exercise 2

Sample

S a m p l e

M e a n

554943373125191371

32

28

24

20

_ _ X=22.17UC L=22.87

LCL=21.47

Sample


554943373125191371

2.4

1.8

1.2

0.6

0.0

_ R=1.210

UC L=2.559

LCL=0

1

Xbar-R Chart of Rule 1 Data

Given that the Ppk = 1.87 can the PPAP be signed?

No: There is an out of control point which is likely from aspecial cause. Do NOT look at capability numbers until the process

is stable and in control.

Exercise 3


49/54

49

Exercise 3

With a Ppk = 1.66 can the PPAP be signed?

Pilot OD

P e r c e n t

0.68450.68400.68350.68300.68250.6820

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Mean 0.6830

StDev 0.0004121

N 125

AD 3.361

P-Value USL 0.00

PPM Tota l 0.00

Observ ed Performance

PPM < LSL 0.34

PPM > USL 0.12

PPM Tota l 0.46

Exp. Within Performance

PPM < LSL 1.04

PPM > USL 0.39

PPM Tota l 1.43


Within

Overall

Process Capability of Pilot OD

Sample

S a m p l e M e a n

252321191715131197531

0.68350

0.68325

0.68300

0.68275

0.68250

_ _ X=0.68296

UC L=0.683489

LCL=0.682431

Sample


252321191715131197531

0.0020

0.0015

0.0010

0.0005

0.0000

_ R=0.000918

UC L=0.001941

LCL=0

1

Xbar-R Chart of Pilot OD

Out of control condition

Rule # 1

X

X

Answer: No

Data is notnormal

P value < .05

Exercise 4


50/54

50

I n d i v

i d u a l V a l u e

10987654321

0.0

-0.8

-1.6

_ X=-0.596

UCL=0.388

LCL=-1.580

M o

v i n g R a n g e

10987654321

1.0

0.5

0.0

__ MR=0.37

UCL=1.209

LCL=0

Observation

V a l u e s

108642

0.0

-0.5

-1.0

1.81.20.60.0-0.6-1.2-1.8

LSL USL

LSL -2

USL 2

Specifications

0.0-0.5-1.0-1.5

Within

Overall

Specs

StDev 0.328014

C p 2.03

C pk 1.43

Within

StDev 0.26604

Pp 2.51

Ppk 1.76

C pm *

O v erall

Process Capability Sixpack of 10 Parts

I Chart

Moving Range Chart

Last 10 Observations

Capability Histogram

Normal Prob Plot

A D: 0.234, P: 0.723

Capability Plot

With a Ppk = 1.76 can this acceptable for PPAP?

NoOnly 10 pieces used in study are not enough toallow a robust assessment of normality, stability or control.

The PPAP requirement is 25 subgroups with typical subgroup size of 5

Exercise 5


51/54

51

With a process that has an expected normal distribution are these

results acceptable for PPAP?

25 subgroups with 5 parts each were used per Ford

Specific PPAP

S a m p

l e M e a n

252321191715131197531

-0.4

-0.6

-0.8

_ _ X=-0.5726

UCL=-0.2586

LCL=-0.8865

S a m

p l e R a n g e

252321191715131197531

1.0

0.5

0.0

_ R=0.544

UCL=1.151

LCL=0

Sample

V a l u e s

252015105

0.0

-0.5

-1.0

1.81.20.60.0-0.6-1.2-1.8

LSL USL

LSL -2

USL 2

Specifications

0.0-0.5-1.0-1.5

Within

O verall

Specs

StDev 0.234026

C p 2.85

Cpk 2.03

Within

StDev 0.238404

Pp 2.8

Ppk 2

C pm *

O v erall

Process Capability Sixpack of 125 Parts

Xbar Char t

R Chart

Last 25 Subgroups

Capability Histogram

Normal Prob Plot A D: 0.665, P: 0.081

Capability Plot

Process is stable

Process is normal

P value > .05

Process is in control

Ppk > 1.67

Yes

Exercise 6


52/54

1501401301201101009080706050403020101

0.007

0.006

0.005

0.004

0.003

0.002

Order of Production


Exercise 6

52

Is this a rational sampling plan?

SubgroupNo – subgroup

includes process shifts

Exercise 7b


53/54

1501401301201101009080706050403020101

0.007

0.006

0.005

0.004

0.003

0.002

Order of Production


Exercise 7

53

Is this a rational sampling plan?

Yes – subgroups are

consecutive pieces and are lesslikely to include process shifts

Subgroup 1Subgroup 4

Subgroup 3

Subgroup 2

Subgroup 6

Subgroup 5

Exercise 8


54/54

1501401301201101009080706050403020101

0.007

0.006

0.005

0.004

0.003

0.002

Order of Production


54

SubgroupUsing the subgroup sampling

plan shown on the left results

in the control chart shownbelow

What is wrong with thissampling plan and what is the

effect on the X bar R chart?

28252219161310741

0.006

0.005

0.004

0.003

Sample

S a m p l e M e a n

_ _ X=0.004425

UCL=0.006095

LCL=0.002756

28252219161310741

0.0060

0.0045

0.0030

0.0015

0.0000

Sample


_ R=0.002894

UCL=0.006120

LCL=0

Xbar-R Chart of Subgroup with Process Shifts

1. Subgroups are not consecutive partsand include process shifts

2. Process shifts will NOT be detected

Control limits are inflated (too large)

process capability training

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