Transcript Process Capability Assessment

Process Capability Assessment
1
Process Capability vs. Process Control

Evaluating Process Performance
–
–
Ability of process to produce parts that conform
to engineering specifications
(CONFORMANCE)
Ability of process to maintain a state of
statistical control; i.e., be within control limits
(CONTROL)
2
Linkages Between Process Control &
Process 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., X and R
(LCL, UCL)
–
Process Capability…dealing with individual
measurements, e.g., X
(LSL, USL)
3

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.
4
Relationship Between Process
Variability and Product Specification
Upper Specification
Lower 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.
5
Relationship Between Process
Variability and Product Specification
Upper Specification
Lower 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.
6
Relationship Between Process
Variability and Product Specification
Upper Specification
Lower 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.
7
Statistical Assessment of Process
Capability

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(m,s) and compute (expected)
•
Convert to defects per million (DPM) and sigma capability
•
Compute process capability indices … Cp, Cpk
8
Alternatives for Improving Process
Capability

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 
9
Summary: Comparison of Specification
Limits 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
10
Control limits on a control chart
are:

Characteristic of the process in question

Based on the process mean and variation

Dependent on sampling parameters, viz., sample
size and a-risk (Type I error)

Used to identify presence/absence of special-cause
variation in the process
11
Some Common Indices of Process Capability
Cp Formula
Cp 
Specification Range
USL  LSL
6s x
reject
Variation of Distribution
of Individual Product
reject
USL ~ LSL
Cp(1) < Cp(2)
(2)
(1)
-3sX(1)
T
-3sX(2)
N(mX, sX)
+3sX(1)
+3sX(2)
X
12
Relationship Between Cp, DPM, and
Sigma of Process (Assumes No Bias)
0
*Sigma of
Process = 3Cp
0
DPM
2-sided spec limits
Very high
DPM
1-sided spec limits
Very High
0.5
0.75
1.00
1.50
2.25
3.00
133,615
24,449
2,700
66,808
12,225
1,350
1.25
1.33
1.50
1.75
2.00

3.75
4.00
4.50
5.25
6.00

177
63
7
0.2
0.0
0.0
89
31.5
3.5
0.1
0.0
0.0
Cp = (USL-LSL) / 6s
Remarks
Worst Case: Items
produced have
enormous variation
Minimally
Acceptable Case
Motorola
Ideal Case: Each item
produced right on
target (cloned)
*DPM = (Probability of Exceeding Specs) * 106
13
Cpk
Purpose: To promote adherence of process
mean to target (nominal) value of spec.
 Formulas:

ZUSL
USL  m X

,
sX
ZLSL
LSL  m X

sX
Zmin  MIN[ZUSL ,ZLSL ]
Cpk  Zmin / 3
14

Example
LSL
Cp 
USL
(mX-T) = bias
190 100
 1.50
6(10)
C pk : (Bias)
190 130
6
10
100 130
Z LSL 
 3
10
Z min  MIN[6,(3)]  3
Z USL 
C pk  Z min / 3  1.00
C pk : ( Unbias : Centering)
190 145
 4.5
10
100 145
Z LSL 
 4.5
10
Z min  MIN[4.5,(4.5)]  4.5
Z USL 
100
N(130, 10) T = 145
190 x
C pk  Z min / 3  1.50
15
Questions?
16
PCB Exercise (USL = +8, LSL = -8)
Exercise
No.
1
2
N(mX, sX)
N(-.27, 6.44)
N(0.06, 4.07)
Cp
_______
_______
Cpk
_______
_______
Sigma of
Process
(Centered)
_______
_______
DPM
(Centered)
_______
_______
17
Microeconomics of Quality: Loss
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

18
Loss Function Representation of
Quality for PCBs
Reject
(Scrap)
Probability Distribution
of Quality Characteristic
Produced by Process
Quadratic Loss
Function
Reject
(Scrap)
Quality Loss
Poor
Poor
Fair
Fair
Good
LSL
(-8 microns)
Best
Target (T)
(Normal)
•Failure Cost @
USL = $2.40/unit
•Failure Rate
(Probability) =
100%
Good
X (Microns)
USL
(+8 microns)
19
Loss Function Approach

Measured loss Function: LM(X)
LM(X) = k (x - T)2
where k is an unknown constant
x is a value of the quality characteristic
T is the target
Determining the Constant k
LM(x) @ USL = k (USL - T)2

where LM(x) @ USL is a known measured cost of scrap
(=Cost of a failure @ USL * failure rate (probability))
USL & T are known
k = (2.40)(1.0)/(8)2
=0.0375
20

Total Loss Function (Measured + Hidden): LT(x)
LT(x) = ak (x - T)2
where a is the hidden “cost of quality” multiplier
(6 < a < 50)
If we assume a = 28, then ak = 28 * 0.0375 = 1.05

Evaluation of Expected Total Process loss: ET{L(x)}
ET{L(x)} = ak {sx2 + (mx - T2)}
Where sx2 is process variance and
(mx - T) is process bias (mean from target)
21
Effect of Bias & Variation (Variance) of
Process Variable on Cost of Quality in Millions
of Dollars
(ak = 1.05; produce 10 million units/year)
Variation sx: (microns)
6
2
420*
84.00

15.00**
(3.00)
0
42.00
(1.5)
388.50
(13.88)
52.50
(1.88)
10.50
(0.38)
378.00
 (13.50)
* Total Cost (Measured + Hidden)
** Measured Cost
42.00
 (1.50)
Bias (mx-T): (microns)
2
1
0
0
(0)
22
Questions?
23
Loss ($)
Loss
Function
LSL
Probability
Distribution
T
USL
x (microns)
(mx -T) = 0
sx = 6
ET {L(x)} = 1.05 {62 + 02} * 107 = $378 million
Figure 2. Evaluation of Quality Loss Function (N(0,62))
24
Analysis (Figure 2.)
Process Capability
USL  LSL 16
Cp 

 0.444
6s
6(6)
No. Standard Deviation from Mean (z)
z = 3Cp = 3 (0.444) = 1.332
This is a 1.332 s Process.
25
Defects Per Million (dpm): Excel
R
A
R
LSL
-1.332
M
USL
+1.332
dpm = (1-NORMSDIST(1.332)) * 2 * 10 ^ 6
= 182,423 dpm
26
Loss ($)
Loss
Function
LSL
Probability
Distribution
T
USL
x (microns)
(mx -T) = 0
sx = 2
ET {L(x)} = 1.05 {22 + 02} * 107 = $42 million
Figure 3. Evaluation of Quality Loss Function (N(0,22))
27
Analysis (Figure 3)
Process Capability
USL  LSL 16
Cp 

 1.333
6s
6(2)
No. Standard Deviation from Mean (z)
z = 3Cp = 3 (1.333) = 4.000
This is a 4 s Process.
28
Defects Per Million (dmp): Excel
dpm = (1-NORMSDIST(4.000)) * 2 * 106
= 63 dpm
Expected Cost Change (ECC)
E{L(2)}
42
4
1
ECC 



E{L(7)} 378 36 9
29
(mx -T) = 2
(mx -T) = 2
LSL
T
(mx -T) = 2
sx = 2
USL
x (microns)
ET {L(x)} = 1.05 {22 + 22} * 107 = $84 million
Figure 4. Quality Loss Consequences of Shifting the
Process Mean Toward the Upper Specification
30
Analysis (Figure 4)
Process Capability
USL  LSL 16
Cp 

 1.333
6s
6(2)
USL  m LSL  m 
Cpk  Min Abs Value
;

3s
 3s

 8  2  8  ( 2) 
 Min Abs Value
;

3( 2) 
 3( 2)
 Min Abs Value{1;1.67}
1
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No. Standard Deviation from Mean (z)
ZUSL = 3 (1) = 3
ZLSL = 3(-1.67) = -5.01
Defects Per Million (dpm): Excel
dpm = (1-NORMSDIST(3))+ NORMSDIST (-5.01)) * 10^6
= 1349.97 + .27 = 1350.24
Expected Cost Change (ECC)
E{L(3)}
84
8
2
ECC 



E{L(1)} 378 36 9
32
Questions?
33
Normal Distribution: N(0, 2.672)
Loss ($)
(Note: 3sx = 8, thus sx = 8/3 = 2.67)
LSL
(mx -T) = 0
sx = 2.67
USL
x (microns)
ET{L(x)} = 1.05 {2.672 + 02} * 107 = $74.85 million
34
Uniform Distribution: U(0, 4.622)
Loss ($)
(Note: sx2 = (b-a)2/12 = 21.33; sx = 4.62)
LSL
(mx -T) = 0
sx = 4.62
USL
x (microns)
ET{L(x)} = 1.05 {4.622 + 02} * 107 = $224.12 million
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Analysis (Figure 5)
Process Capability
Normal Distribution
USL  LSL
16
Cp 

 1.000
6s
6(2.67)
Uniform Distribution (Inspection; Adjustment)
USL  LSL
16
Cp 

 0.577
6s
6(4.62)
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No. Standard Deviation from Mean (z)
Normal Distribution
Z = 3 Cp= 3 (1.000) = 3.000 This is a 3s Process.
Uniform Distribution
Z = 3 Cp= 3 (0.577) = 1.731s Process
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.
37
Cost of Quality
Normal (N) Distribution
E{L(x)} = ak {s2 + (m - T)2}
= ak * 107 {2.672 + 02}
= ak * 107 * 7.13
Uniform (U) Distribution
E{L(x)} = ak * 107 { 4.622 + 02}
= ak * 107 * 21.34
Expected Cost Change (ECC): Normal vs Uniform
E{L( N)} 7.13
ECC 

 0.334
E{L( U)} 21.34
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Summary
Process Output
(m-T, s)
Process
Capability
(Cpk)
No. Std. Dev.
From m
z
Defects per
Million
(dpm)
Cost of
Quality
E{L}
(0, 6)
0.444
1.332
182,423
36
(0, 2)
1.333
4.000
63
4
(2, 2)
1.000
3.000
-5.010
1,349.7
0.27
8
Normal(0, 2.67)
1.000
3.000
2,700
7.13
Uniform(0, 4.62)
0.577
1.731
0
21.34
USL
LSL
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Comparison of Process (1 vs 2)
Max
Max
T=0
Min
Min
A
(0, 6)
Process
B
(0, 2)
(. , .)  (m - T), s
 bias, variation
40
Comparison of Process (1 vs 3)
Max
Max
Min
T=0
Min
A
(0, 6)
Process
C
(2, 2)
(. , .)  (m - T), s
 bias, variation
41
Questions?
42
Understanding the Differences
3s Capability
Historical Standard
4s Capability
Current Standard
6s Capability
New Standard
43
Understanding the Differences
Sigma
3s
4s
5s
6s
Area
Spelling
Money
Time
Distance
Floor space of a 1.5 misspelled $2.7 million
3 1/2 months Coast-tosmall hardware words per page indebtedness per $1
per century
coast trip
store
in a book
billion in assets
45 minutes
Floor space of a 1 misspelled
$63,000 indebtedness
of freeway
2 1/2 days per
typical living word per 30
per $1 billion in
driving (in
century
room
pages in a book assets
any
direction)
Size of the
1 misspelled
$570 indebtedness
A trip to
30 minutes per
bottom of your word in a set of per $1 billion in
the local
century
telephone
encyclopedias assets
gas station
1 misspelled
4 steps in
Size of a typical word in all of the $2 indebtedness per 6 seconds per
any
diamond
books contained $1 billion assets
century
direction
in a small library
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Understanding the Difference
Suppose a process produced 294,118 units of
product. If the process capability was 4s,then
the defects produced could be represented by
the matrix of dots given below. If the capability
was 6s, only one dot would appear in the entire
matrix.
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Understanding the Difference
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4s Capability: Defect Dots = 1849
6s Capability: Defect Dots = 1
46
The Inspection Exercise
Task: 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.
47
Questions?