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1 Process Capability Process Capability Assessment Assessment

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Process Capability AssessmentProcess Capability Assessment

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Process Capability vs. Process ControlProcess Capability vs. Process Control

Evaluating Process Performance– Ability of process to produce parts that

conform to engineering specifications

(CONFORMANCE)(CONFORMANCE)– Ability of process to maintain a state of

statistical control; i.e., be within control limits

(CONTROL)(CONTROL)

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– Process Capability…dealing with individual measurements, e.g., X

(LSL, USL)(LSL, USL)

Linkages Between Process Control & Linkages Between Process Control & Process CapabilityProcess Capability

Process must be in statistical control before assessing process capability. Why?

Statistical aspects– Process Control…use summary statistics from a

sample (subgroup); dealing with sampling distributions, e.g., and R

(LCL, UCL)(LCL, UCL)X

4

Process may be in statistical control, but not capable (of meeting specifications)

– Process is off-center from nominal (bias)

– Process variability is too large relative to specifications (variation)

– Process is both off-center and has large variation.

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Relationship Between Process Relationship Between Process Variability and Product SpecificationVariability and Product Specification

(a) Process variation is small relative to the specifications so that the process mean can shift about without causing the process to degrade its capability. This will reduce the defects per million (DPM), reduce the cost of quality (COQ), and hence increase profitability.

Upper Specification

Lower Specification

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Relationship Between Process Relationship Between Process Variability and Product SpecificationVariability and Product Specification

(b) Process variation is large relative to the specifications such that the process must remain well centered for the process capability to be maintained at a tolerable level. Variation, however, must be reduced. This will increase process capability, reduce DPMs, reduce COQ, and increase profitability.

Upper Specification

Lower Specification

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Relationship Between Process Relationship Between Process Variability and Product SpecificationVariability and Product Specification

(c) Process variation is large relative to the specifications so that the process cannot be considered capable regardless of the process centering. Hence we have a severe and urgent problem. Process variation must be reduced drastically.

Upper Specification

Lower Specification

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Statistical Assessment of Process Statistical Assessment of Process CapabilityCapability

Get Process in Statistical Control

Statistical Assessment (Minitab or Excel)– Construct histogram of individual measurements

– Compute probability of exceeding specifications P(•)• Empirically (observed)

• Mathematically: … assume N(,) and compute (expected)

• Convert to defects per million (DPM) and sigma capability

• Compute process capability indices … Cp, Cpk

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Alternatives for Improving Process Alternatives for Improving Process CapabilityCapability

If bias– Recenter and recompute P(•), dpm, and sigma

capability If too much variation

– Sort by 100% inspection – Widen tolerance – Use a more precise process (e.g., better or new

technology) to reduce variation – Use statistical methods to identify variation reduction

opportunities for existing process

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Summary: Comparison of Specification Summary: Comparison of Specification Limits and Control LimitsLimits and Control Limits

Spec limits or tolerances for product quality

characteristics are:– Characteristic of the part/item (product) in question

– Based on functional design considerations

– Related to/compared with an individual part

measurement

– Used to establish a part’s conformance to design intent

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Control limitsControl limits on a control chart on a control chart are:are:

Characteristic of the process in question

Based on the process mean and variation

Dependent on sampling parameters, viz., sample

size and -risk (Type I error)

Used to identify presence/absence of special-cause

variation in the process

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Some Common Indices of Process CapabilitySome Common Indices of Process Capability

Cp Formula

x

p 6

LSLUSLC

Specification Range

Variation of Distribution of Individual Product

USL ~ LSLUSL ~ LSL

N(N(XX, , XX))

rejectrejectrejectreject

-3X(1) +3X(1)

(1)

TTXX

-3X(2)+3X(2)

(2)Cp(1) < Cp(2)

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Relationship Between CRelationship Between Cpp, DPM, and , DPM, and

Sigma of Process (Assumes No Bias)Sigma of Process (Assumes No Bias)Cp = (USL-LSL) / 6 *Sigma of

Process = 3Cp

DPM2-sided spec limits

DPM1-sided spec limits

Remarks

0 0 Very high Very High Worst Case: Itemsproduced haveenormous variation

0.5 1.50 133,615 66,8080.75 2.25 24,449 12,2251.00 3.00 2,700 1,350 Minimally

Acceptable Case1.25 3.75 177 891.33 4.00 63 31.51.50 4.50 7 3.51.75 5.25 0.2 0.12.00 6.00 0.0 0.0 Motorola 0.0 0.0 Ideal Case: Each item

produced right ontarget (cloned)

*DPM = (Probability of Exceeding Specs) * 106

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CCpkpk

Purpose: To promote adherence of process mean to target (nominal) value of spec.

Formulas:

,USL

ZX

XUSL

X

XLSL

LSLZ

]Z,Z[MINZ LSLUSLmin

3/ZC minpk

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ExampleExampleUSLUSLLSLLSL

100100 190 x190 xN(130, 10)N(130, 10) T = 145T = 145

((XX-T) = bias-T) = bias

50.1)10(6

100190Cp

00.13/ZC

3)]3(,6[MINZ

310

130100Z

610

130190Z

)Bias(:C

minpk

min

LSL

USL

pk

50.13/ZC

5.4)]5.4(,5.4[MINZ

5.410

145100Z

5.410

145190Z

)Centering:Unbias(:C

minpk

min

LSL

USL

pk

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Questions?Questions?

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PCB Exercise (USL = +8, LSL = -8)PCB Exercise (USL = +8, LSL = -8)

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Microeconomics of Quality: Loss Microeconomics of Quality: Loss Due to Variation (Taguchi)Due to Variation (Taguchi)

Linking Cost of Quality Due to Bias and Variation to DPM and Process Capability in PCB Manufacture

Variation is Related to Functional Form (Distribution) of Process Output

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Loss Function Representation of Loss Function Representation of Quality for PCBsQuality for PCBs

LSLLSL(-8 microns)(-8 microns)

Target (T)Target (T)(Normal)(Normal)

USLUSL(+8 microns)(+8 microns)

Quality LossQuality Loss

Reject Reject (Scrap)(Scrap)

Reject Reject (Scrap)(Scrap)

•Failure Cost @ Failure Cost @ USL = $2.40/unitUSL = $2.40/unit•Failure Rate Failure Rate (Probability) = (Probability) = 100%100%

Probability Distribution of Quality Characteristic Produced by Process

Quadratic Loss Function

Poor

FairGood

BestGood

Fair

Poor

X (Microns)X (Microns)

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Loss Function ApproachLoss Function Approach Measured loss Function: LM(X)

LM(X) = k (x - T)2

where k is an unknown constantx is a value of the quality characteristicT is the target

Determining the Constant kLM(x) @ USL = k (USL - T)2

where LM(x) @ USL is a known measured cost of scrap(=Cost of a failure @ USL * failure rate (probability))

USL & T are knownk = (2.40)(1.0)/(8)2

=0.0375

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Total Loss Function (Measured + Hidden): LT(x)

LT(x) = ak (x - T)2

where a is the hidden “cost of quality” multiplier

(6 < a < 50)

If we assume a = 28, then ak = 28 * 0.0375 = 1.05

Evaluation of Expected Total Process loss: ET{L(x)}

ET{L(x)} = ak {x2 + (x - T2)}

Where x2 is process variance and

(x - T) is process bias (mean from target)

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Effect of Bias & Variation (Variance) of Process Variable on Cost of Quality in

Millions of Dollars

(ak = 1.05; produce 10 million units/year)

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Questions?Questions?

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(x -T) = 0

x = 6

ET {L(x)} = 1.05 {62 + 02} * 107 = $378 million

Figure 2. Evaluation of Quality Loss Function (N(0,62))

TT

Loss ($)Loss ($)

LSLLSL USLUSL x (microns)

Probability Probability DistributionDistribution

Loss Loss FunctionFunction

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Analysis (Figure 2.) Analysis (Figure 2.)

Process Capability

No. Standard Deviation from Mean (z)

z = 3Cp = 3 (0.444) = 1.332

This is a 1.332 Process.

444.0)6(6

16

6

LSLUSLCp

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Defects Per Million (dpm): ExcelDefects Per Million (dpm): Excel

dpm = (1-NORMSDIST(1.332)) * 2 * 10 ^ 6 = 182,423 dpm

MMLSLLSL-1.332-1.332

RR RR

USLUSL+1.332+1.332

AA

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(x -T) = 0

x = 2

ET {L(x)} = 1.05 {22 + 02} * 107 = $42 million

Figure 3. Evaluation of Quality Loss Function (N(0,22))

TT

Loss ($)Loss ($)

LSLLSL USLUSL x (microns)

Probability Probability DistributionDistribution

Loss Loss FunctionFunction

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Analysis (Figure 3)Analysis (Figure 3)

Process Capability

No. Standard Deviation from Mean (z)

z = 3Cp = 3 (1.333) = 4.000

This is a 4 Process.

333.1)2(6

16

6

LSLUSLCp

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Defects Per Million (dmp): Excel

dpm = (1-NORMSDIST(4.000)) * 2 * 106

= 63 dpm

Expected Cost Change (ECC)

9

1

36

4

378

42

)}7({

)}2({

LE

LEECC

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ET {L(x)} = 1.05 {22 + 22} * 107 = $84 million

Figure 4. Quality Loss Consequences of Shifting the Process Mean Toward the Upper Specification

TTLSLLSL USLUSLx (microns)(x -T) = 2

x = 2

(x -T) = 2

(x -T) = 2

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Analysis (Figure 4)Analysis (Figure 4)

Process Capability

333.1)2(6

16

6

LSLUSLCp

1

}67.1;1{Value Abs Min

)2(3

)2(8;

)2(3

28Value Abs Min

3;

3Value Abs Min

LSLUSLCpk

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Defects Per Million (dpm): Excel

dpm = (1-NORMSDIST(3))+ NORMSDIST (-5.01)) * 10^6

= 1349.97 + .27 = 1350.24

Expected Cost Change (ECC)

9

2

36

8

378

84

)}1({

)}3({

LE

LEECC

No. Standard Deviation from Mean (z)

ZUSL = 3 (1) = 3

ZLSL = 3(-1.67) = -5.01

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Questions?Questions?

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Normal Distribution: N(0, 2.672)

x (microns)

Los

s ($)

LSL USL(x -T) = 0x = 2.67

(Note: 3x = 8, thus x = 8/3 = 2.67)

ET{L(x)} = 1.05 {2.672 + 02} * 107 = $74.85 million

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Uniform Distribution: U(0, 4.622)

x (microns)

Los

s ($)

LSL USL(x -T) = 0x = 4.62

(Note: x2 = (b-a)2/12 = 21.33; x = 4.62)

ET{L(x)} = 1.05 {4.622 + 02} * 107 = $224.12 million

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Analysis (Figure 5)Analysis (Figure 5)

Process Capability

Normal Distribution

000.1)67.2(6

16

6

LSLUSLCp

Uniform Distribution (Inspection; Adjustment)

577.0)62.4(6

16

6

LSLUSLCp

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Defects Per Million (dmp): Excel

Normal Distribution

dpm = (1-NORMSDIST(3.000)* 2 * 106 = 2700

Uniform Distribution

Theoretically, there are no units exceeding the specification limits; however, there are many more units further away from the target value (T) than with a normal distribution. This accounts for the higher variance (4.622 vs 2.672) and, as we shall see, higher cost of quality.

No. Standard Deviation from Mean (z)

Normal Distribution Z = 3 Cp= 3 (1.000) = 3.000 This is a 3 Process.

Uniform Distribution Z = 3 Cp= 3 (0.577) = 1.731 Process

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Cost of QualityCost of QualityNormal (N) Distribution

E{L(x)} = ak {2 + ( - T)2} = ak * 107 {2.672 + 02} = ak * 107 * 7.13

Uniform (U) DistributionE{L(x)} = ak * 107 { 4.622 + 02}

= ak * 107 * 21.34

Expected Cost Change (ECC): Normal vs UniformExpected Cost Change (ECC): Normal vs Uniform

334.034.21

13.7

)}U(L{E

)}N(L{EECC

39

SummarySummary

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Comparison of Process (1 vs 2)Comparison of Process (1 vs 2)

(. , .) ( - T), bias, variation

AA(0, 6)(0, 6)

BB(0, 2)(0, 2)Process

T = 0

Max

Min

Max

Min

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Comparison of Process (1 vs 3)Comparison of Process (1 vs 3)

AA(0, 6)(0, 6)

CC(2, 2)(2, 2)Process

T = 0

Max

Min

Max

Min

(. , .) ( - T), bias, variation

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Questions?Questions?

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Understanding the DifferencesUnderstanding the Differences

3 Capability Historical Standard

4 Capability Current Standard

6 Capability New Standard

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Understanding the DifferencesUnderstanding the Differences

3Floor space of a small hardware store

1.5 misspelled words per page in a book

$2.7 million indebtedness per $1 billion in assets

3 1/2 months per century

Coast-to-coast trip

4Floor space of a typical living room

1 misspelled word per 30 pages in a book

$63,000 indebtedness per $1 billion in assets

2 1/2 days per century

45 minutes of freeway driving (in any direction)

5Size of the bottom of your telephone

1 misspelled word in a set of encyclopedias

$570 indebtedness per $1 billion in assets

30 minutes per century

A trip to the local gas station

Sigma Area Spelling Money Time Distance

6Size of a typical diamond

1 misspelled word in all of the books contained in a small library

$2 indebtedness per $1 billion assets

6 seconds per century

4 steps in any direction

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Understanding the DifferenceUnderstanding the Difference

Suppose a process produced 294,118 units of product. If the process capability was 4then the defects produced could be represented by the matrix of dots given below. If the capability was 6, only one dot would appear in the entire matrix.

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Understanding the DifferenceUnderstanding the Difference• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

4 Capability: Defect Dots = 1849

6 Capability: Defect Dots = 1

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The Inspection ExerciseThe Inspection ExerciseTask: Count the number of times the 6th letter of the alphabet appears in the following text.

The Necessity of Training Farm Hands for First Class Farms in the Fatherly Handling of Farm Live Stock is Foremost in the Eyes of Farm Owners. Since the Forefathers of the Farm Owners Trained the Farm Hands for First Class Farms in the Fatherly Handling of Farms Live Stock, the Farm Owners Feel they should Carry on with the Family Tradition of Training Farm Hands of First Class Farmers in the Fatherly Handling of Farm Live Stock Because They Believe it is the basis of Good Fundamental Farm Management.

Questions?Questions?