Process Capability Process
capability compares the width of the process parameter or output characteristic to the specification width. Process vs Specification - Capable
f(x)
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
9
10
11
12
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14
Specification is 10.00mm + 2.00 mm
Diam eter (.01m m )
G. Baker, Department of Statitsics University of South Carolina: Slide 1
There’s Capable & Then There’s Capable Process vs Specification
f(x)
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
9
10
11
12
13
14
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
Diam eter (.01m m )
7
8
9
10
11
10
11
12
13
14
12
13
14
Process vs Specification
f(x)
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
9
Diam eter (.01m m )
12
13
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
14
6
7
8
Diam eter (.01m m )
9
10
11
Diam eter (.01m m )
G. Baker, Department of Statitsics University of South Carolina: Slide 2
…and then of course there’s Not Capable Process vs Specification
f(x)
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
9
10
11
12
13
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
14
6
7
8
Diam e ter (.01m m )
7
8
9
10
11
Diam eter (.01m m )
10
11
12
13
14
12
13
14
Process vs Specification
f(x)
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
9
Diam ete r (.01m m )
12
13
14
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
9
10
11
Diam e ter (.01m m )
G. Baker, Department of Statitsics University of South Carolina: Slide 3
1
Consider Process Width CP index compares the width of the
specification to that of the process.
Width of the process is called the Natural Tolerance.
c = P
USL − LSL 6σˆ
G. Baker, Department of Statitsics University of South Carolina: Slide 4
Location! Location! Location! Process capability depends not only on the width of the process, but also on its location.
C = PU
USL − μˆ 3σˆ
C = PL
μˆ − LSL 3σˆ
C = min{C , C } PK
PU
PL
G. Baker, Department of Statitsics University of South Carolina: Slide 5
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
9
10
11
12
13
14
12
13
14
Diam eter (.01m m )
Cp = 1 Cpu = 1 Cpu = 1 Cpk = 1
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
9
10
11
Diam eter (.01m m )
Cp = 2 Cpl = 2 Cpu = 2 Cpk = 2
G. Baker, Department of Statitsics University of South Carolina: Slide 6
2
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
9
10
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Diam e ter (.01m m )
Cp = 1.5 Cpl = 1 Cpu = 2 Cpk = 1
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6
7
8
9
10
11
12
13
14
Diam e te r (.01m m )
G. Baker, Department of Statitsics University of South Carolina: Slide 7
Cp = 1 Cpl = 1.5 Cpu = 0.5 Cpk = 0.5
Process Capability Study (1)
Determine process stability.
(2)
Estimate process location and standard deviation.
(3)
Is normal assumption reasonable for individual measurements?
(4)
Calculate appropriate indexes and ppm.
G. Baker, Department of Statitsics University of South Carolina: Slide 8
Ball Bearing Data Sample Mean
Ball Bearing M aximum Diameter (.01mm) UCL=10.56
10.5 10.0
Mean=9.898
9.5 LCL=9.236 9.0
Subgroup
Sample Range
Sample
0
5
10
15
20
25
10:30
13:30
16:30
19:00
22:30
UCL=2.424 2 R =1.146
1
0
LCL=0
After eliminating points at 8:00,12:00,16:00,19:30,20:00
G. Baker, Department of Statitsics University of South Carolina: Slide 9
3
Minitab Normal Capability Analysis Ball Bearing Data LS L
Targ et
USL W ith in O v erall
P rocess D ata LS L 7.00000 T arget 10.00000 USL 13.00000 S am ple M ean 9.89768 S am ple N 125 S tD ev (Within) 0.49286 S tD ev (O v erall) 0.50280
P otential (Within) C apa bility Cp 2.03 C PL 1.96 C PU 2.10 C pk 1.96 C C pk 2.03 O v erall C apability Pp PPL PPU P pk C pm
7.2 O bserv ed P erform ance P P M < LS L 0.00 P P M > U S L 0.00 P P M T otal 0.00
8 .0
8.8
9 .6
E xp. Within P erform ance P P M < LS L 0.00 P P M > U S L 0.00 P P M T otal 0.00
10 .4 11.2 12.0
1.99 1.92 2.06 1.92 1.95
12 .8
E xp. O v erall P erform ance P P M < LS L 0.00 P P M > U S L 0.00 P P M T otal 0.00
G. Baker, Department of Statitsics University of South Carolina: Slide 10
Minitab Normal Sixpack Quality Display Ball Be ar ing D ata Sample Mean
X ba r C ha r t
C a p a b i lity H i s to g r a m
10.5
UC L = 10.559
10.0
_ _ X = 9.898
9.5 LC L= 9.236 2
4
6
8
10
12
14
16
18
20
22
24
8.8
9.2
R C har t Sample Range
10.0
10.4
10.8
2 _ R = 1.146
1 0
LC L= 0 2
4
6
8
10
12
14
16
18
20
22
24
8
9
L a st 2 5 Subgr oups
10
11
C a p a b ili ty P lo t
11 Values
9.6
No r m a l P r o b P lo t A D : 0 .2 4 4 , P : 0 .7 5 9
UC L = 2.424
W ithin S tD ev 0 . 4 92 8 6 Cp 2 .0 3 C pk 1 .9 6 C C pk 2 .0 3
10 9
Within
Overall
O v e ra ll S tD e v 0 .5 0 2 8 0 Pp 1.9 9 P pk 1.9 2 C pm 1.9 5
Spec s
5
10
15
20
25
Sa m ple
G. Baker, Department of Statitsics University of South Carolina: Slide 11
What is a Six Sigma Process?
μˆ = 10.00 σˆ = 0.67
f(x)
Process vs Specification 0.9 0.8 0.7 0.6 LSL 0.5 0.4 0.3 0.2 0.1 0
= 6.00
USL = 14.00
6σ
6
7
6σ
8
9
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12
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14
Diam eter (.01m m ) G. Baker, Department of Statitsics University of South Carolina: Slide 12
4
Six Sigma PPM is Based on a 1.5 σ Shift Process vs Specification
LSL
6
USL
7
8
9
10
11
12
13
14
Diameter (.01 mm)
G. Baker, Department of Statitsics University of South Carolina: Slide 13
How are Specifications Determined? By
– Customers – Engineering Based
on – History – Theoretical Calculations – Devine Inspiration G. Baker, Department of Statitsics University of South Carolina: Slide 14
Tolerancing Consider
the following assembly:
1
2
3
1.500+0.010
1.250+0.005 1.000+0.002 Process producing components has + 4 σ capability. What is the tolerance for the assembly based on the tolerances for the components?
G. Baker, Department of Statitsics University of South Carolina: Slide 15
5
Approach 1: Additivity Assembly tolerance = sum of component tolerances Assembly tolerance =
0.010 + 0.002 + 0.005 = 0.017 Assembly specification is 3.750 + 0.017 This tolerance turns out to be too loose. G. Baker, Department of Statitsics University of South Carolina: Slide 16
Approach 2: Statistical Assignment of Tolerances Based
on –Statistical distribution of component dimensions –Random assembly
1
2
0.020
3
0.004
0.010 G. Baker, Department of Statitsics University of South Carolina: Slide 17
Component Distributions 4 σ capability implies: Part 1: + 0.010
σ = 0.0025
Part 2: + 0.002
σ = 0.0005
Part 3: + 0.005
σ = 0.00125
1
2
3
G. Baker, Department of Statitsics University of South Carolina: Slide 18
6
Additive Law of Variances 2 σ assembly = σ12 + σ 22 + σ 32
σ assembly = .00252 + .00052 + .001252
= .00284 G. Baker, Department of Statitsics University of South Carolina: Slide 19
Comparison of Approaches Statistical
Assignment of Tolerances:
Virtually all assemblies will fall within + 4(.00284) = + .01136 Additivity
Assignment of Tolerances:
+ .01700 G. Baker, Department of Statitsics University of South Carolina: Slide 20
Consider Starting with Assembly Tolerance Consider
the following assembly: 1
2
3
0.900 + 0.009 Process
producing components has + 4 σ capability. What tolerance is needed for the component parts? Assume equal tolerances on all parts.
G. Baker, Department of Statitsics University of South Carolina: Slide 21
7
Additivity of Tolerances
± .009 = ±.003 + ±.003 + ±.003 4σ part = 0.003 → σ part = 0.00075 But this is a tighter than needed. G. Baker, Department of Statitsics University of South Carolina: Slide 22
Statistical Tolerancing
4σ assembly = 0.009
σ assembly = 0.009 / 4 = 0.00225
σ assembly = σ 12 + σ 22 + σ 32 2 .00225 = 3σ part
σ part = .0013
Recall Adding Tolerances:
σ part = 0.00075 G. Baker, Department of Statitsics University of South Carolina: Slide 23
Statistical Tolerancing To have a 4σ process for parts:
σ part = .0013 Parts will need to be +(4)(0.0013) = + 0.0052
G. Baker, Department of Statitsics University of South Carolina: Slide 24
8
Comparison of Approaches Setting part tolerances at + 4 Ď&#x192;: Using statistical tolerancing, virtually all assemblies will be within + 0.009 when part tolerance is + 0.0052 (as compared to + 0.0030 using additivity of tolerances) G. Baker, Department of Statitsics University of South Carolina: Slide 25
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