Transcript Statistical Analysis & Specification

SMU
EMIS 7364
NTU
TO-570-N
Statistical Quality Control
Dr. Jerrell T. Stracener,
SAE Fellow
Tolerance Limits
Statistical Analysis & Specification
Updated: 2/14/02
1
Product Specification
Lower
Specification
Limit
Nominal
Specification
Upper
Specification
Limit
x
Target
(Ideal level for use in product)
Tolerance
(Product
characteristic)
(Maximum range of variation of the product
characteristic that will still work in the product.)
2
Traditional US Approach to Quality
(Make it to specifications)
No-Good
Loss ($)
No-Good
good
LSL
T
USL
x
3
4
Setting Specification Limits on Discrete Components
5
Variability Reduction
Variability reduction is a modern concept of design
and manufacturing excellence
• Reducing variability around the target value leads
to better performing, more uniform, defect-free
product
• Virtually eliminates rework and waste
• Consistent with continuous improvement concept
Don’t just conform to specifications
reject
accept
Reduce variability
around the target
reject
target
6
True Impact of Product Variability
• Sources of loss
- scrap
- rework
- warranty obligations
- decline of reputation
- forfeiture of market share
• Loss function - dollar loss due to deviation of
product from ideal characteristic
• Loss characteristic is continuous - not a step
function.
7
Representative Loss Function Characteristics
Loss
$
Loss
$
T
Loss
$
x
x
X nominal is best X smaller is better
L = k (x - T)2
L = k (x2)
x
X larger is better
L = k (1/x2)
8
Variability-Loss Relationship
LSL
USL
Target
Loss
Maximum
$ loss
per item
$ savings
due to
reduced
variability
9
Loss Computation for Total Product Population
X nominal is best
L = k (x - T)2
1
f(x) 
e
σ 2π
 (x  T) 2
2σ 2
Loss
$
Loss
$
T
x
T
x
10
Statistical Tolerancing - Convention
Normal Probability
Distribution
0.00135
LTL
-3
0.9973
Nominal

0.00135
UTL
+3
11
Statistical Tolerancing - Concept
LTL
Nominal
UTL
x
12
Caution
For a normal distribution, the natural tolerance
limits include 99.73% of the variable, or put
another way, only 0.27% of the process output will
fall outside the natural tolerance limits. Two points
should be remembered:
1. 0.27% outside the natural tolerances sounds
small, but this corresponds to 2700 nonconforming
parts per million.
2. If the distribution of process output is non
normal, then the percentage of output falling
outside   3 may differ considerably from 0.27%.
13
Normal Distribution
Probability Density Function:
1
f (x) 
e
 2
where

1
2
2

x



2
<x<
 = 3.14159...
e = 2.7183...
14
Normal Distribution
• Mean or expected value of X
Mean = E(X) = 
• Median value of X
X0.5 = 
• Standard deviation
Var(X )  
15
Normal Distribution
Standard Normal Distribution
X 
Z
If X ~ N(, ) and if
 , then Z ~ N(0, 1).
A normal distribution with  = 0 and  = 1, is called
the standard normal distribution.
16
Normal Distribution - example
The diameter of a metal shaft used in a disk-drive unit
is normally distributed with mean 0.2508 inches and
standard deviation 0.0005 inches. The specifications
on the shaft have been established as 0.2500 
0.0015 inches. We wish to determine what fraction of
the shafts produced conform to specifications.
17
Normal Distribution - example solution
Pmeeting spec   P0.2485  x  0.2515
 Px  0.2515  P0.2485  x 
 0.2515 - 0.2508 
 0.2485 - 0.2508 
 
  

0.0005
0.0005




 1.40   4.60
 0.91924  0.0000
 0.91924
f(x)
0.2500
0.2485
LSL
nominal
0.2508
0.2515
USL
x
18
Normal Distribution - example solution
Thus, we would expect the process yield to be
approximately 91.92%; that is, about 91.92% of
the shafts produced conform to specifications. Note
that almost all of the nonconforming shafts are too
large, because the process mean is located very
near to the upper specification limit. Suppose we
can recenter the manufacturing process, perhaps
by adjusting the machine, so that the process mean
is exactly equal to the nominal value of 0.2500.
Then we have
19
Normal Distribution - example solution
P0.2485  x  0.2515  Px  0.2515  P0.2485  x 
 0.2515 - 0.2500 
 0.2485 - 0.2500 
 
  

0.0005
0.0005




 3.00   3.00
 0.99865  0.00135
 0.9973
f(x)
0.2485
LSL
nominal
0.2500
x
0.2515
USL
20
What is the magnitude of the difference between sigma levels?
Sigma
One
Area
Floor of
Astrodome
Two
Spelling
170 typos/page
in a book
Time
31 years/century
Distance
earth to moon
Large supermarket 25 typos/page
in a book
4 years/century
1.5 times around
the earth
Three
small hardware
store
1.5 typos/page
in a book
3 months/century
CA to NY
Four
Typical living
room
1 typo/30 pages
~(1 chapter)
2 days/century
Dallas to Fort Worth
Five
Size of the bottom 1 typo in a set of
of your telephone encyclopedias
30 minutes/century SMU to 75 Central
Six
Size of a typical
diamond
1 typo in a
small library
6 seconds/century
Seven
Point of a sewing
needle
1 typo in several
large libraries
1 eye-blink/century 1/8 inch
four steps
21
Linear Combination of Tolerances
Xi = part characteristic for ith part, i = 1, 2, ... , n
Xi ~ N(i, i)
X1, X2, ..., Xn are independent
22
Linear Combination of Tolerances
Y = assembly characteristic
If
n
Y   a i Xi
i 1
then
, where the a1, ..., an are
constants,
Y ~ N(Y, Y),
where
n
 Y   a ii
i 1
and
Y 
n
2 2
a
 i i
i 1
23
Concept
x1
x2
.
.
.
xn
n
y   xi
i 1
y
24
Statistical Tolerancing - Concept
f(x)
0.2485
LSL
nominal
0.2500
0.2515
USL
x
25
Tolerance Analysis - example
The mean external diameter of a shaft is S = 1.048
inches and the standard deviation is S = 0.0020
inches. The mean inside diameter of the mating
bearing is b = 1.059 inches and the standard
deviation is b = 0.0030 inches. Assume that both
diameters are normally and independently
distributed.
(a) What is the required clearance, C, such that the
probability of an assembly having a clearance
less than C is 1/1000?
(b) What is the probability of interference?
26
Tolerance Analysis - example solution
Bearing
Xb
Shaft
XS
250.00
200.00
150.00
fb(x)
fs(x)
100.00
f(xb)
f(xs)
diameter
1.0325
1.0350
1.0375
1.0400
1.0425
1.0450
1.0475
1.0500
1.0525
1.0550
1.0575
1.0600
1.0625
1.0650
1.0675
1.0700
1.0725
1.0750
1.0775
1.0800
f(x b )
0.00
0.00
0.00
0.00
0.00
0.00
0.09
1.48
12.72
54.67
117.36
125.79
67.33
18.00
2.40
0.16
0.01
0.00
0.00
0.00
f(x s)
0.00
0.00
0.00
0.07
4.55
64.76
193.33
120.99
15.87
0.44
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
50.00
0.00
1.0300
1.0400
1.0500
1.0600
1.0700
1.0800
27
Tolerance Analysis - example solution
150.00
125.00
100.00
Intersection
Region
fb(x)
fs(x)
75.00
50.00
25.00
0.00
1.045
1.047
1.049
1.051
1.053
1.055
1.057
28
The Normal Model - example solution
D = xb-xs = Clearance of bearing inside diameter minus shaft
outside diameter
D = b - S = 0.011
D = (b2 + S2)1/2 = 0.0036
so D~N(0.011,0.0036)
Clearance Probability Density Function
100.00
fD(x)
50.00
0.00
0.000
0.005
0.010
0.015
0.020
0.025
0.030
d=xb-xs
29
The Normal Model - example solution
1
(a) Find c such that P(D < c) =
1000
c  0.011 

so that P Z 
  0.001
0.0036 

From the normal table (found in the resource section of the
website), the Z = -3.09
PZ  3.09  0.001
 c  D 

so that  3.09  
 D 
 c  0.011 
 3.09  

 0.0036 
and
c  0.000124
30
The Normal Model - example solution
Since c < 0, there is no value of c for which the
probability is equal to 0.001
(b) Find the probability of interference, i.e.,
Pinterferen ce  PD  0
0  0.011 

 P D 

0.0036 

 PZ  -3.06
 0.0011
From the normal table (found in the resource section of the
website), the Z of -3.1 = 0.0011
31
Tolerance Analysis - example
Using Monte Carlo Simulation (n=1000):
(a) What is the required clearance, C, such that the
probability of an assembly having a clearance less
than C is 1/1000?
(b) What is the probability of interference?
32
Tolerance Analysis - example
Using Monte Carlo Simulation
N( b ,b )
First generate random samples from
(I used n=1000)
1.056233
1.059985
1.065796
1.055505
1.059354
1.062841
1.055726
1.058989
1.061587
1.058047
1.060637
1.05933
1.056185
1.058666
Xbi~N(b, b) = N(1.059, 0.0030)
and
Xsi~N(s, s) = N(1.048, 0.0020)
N( s,s)
1.058537 1.045935
1.047846
1.05248
1.048849
1.047922
1.047164
1.049267
1.047815
1.048584
1.047335
1.047325
1.047536
1.046401
1.047694
1.045457
33
Tolerance Analysis - example
Then calculate the differences
Estimate
xdi  xbi  xsi for i  1,..., n
Estimate s by taking the mean.
(You can use the AVERAGE() function.)
̂  xd
Estimate s by calculating the standard deviation.
(You can use the STDEV() function.)
ˆ  sd
1000  1
1000
34
Tolerance Analysis - example
̂ D= 0.01093
and ˆ D = 0.00371
(a)
P̂(clearance  C)  0.001
 c  ˆ D 

 3.09  
 ˆ D 
 c  0.01093 


 0.00371 
c  0.000534
This is close to c = -0.000124.
35
Tolerance Analysis - example
(b) P̂(interfere nce)  no. for which d  0
n
2

1000
 0.002
This can be compared to P(I) = 0.000968.
36
Statistical Tolerance Analysis Process
Assembly consists of K components
• Specifications Assembly:
x AN  t A
• Specifications Component:
i, for i  1,..., K, xin  ti
• Assembly Nominal
K
x An   ai xin
i 1
where ai = 1 or -1 as appropriate
37
Statistical Tolerance Analysis Process
• Assembly tolerance
K
2
t
i
tA 
i 1
• If dimension
X i is normally Distribute d
with parameters  i and  i , then
X i ~ N( A ,  A )
where
K
 A   ai xi  x A
n
i 1
and
A 
K
2

 i,
i 1
n
t
3
i  i .
38
Statistical Tolerance Analysis Process
•
x AN  t A is specified
•
xin  ti is determined during design

• P x AN  t A  X A  x AN  t A
Case 1:

is calculated
if probability is too small, then
(1) component tolerance(s) must be reduced
or
(2) tA must be increased
39
Statistical Tolerance Analysis Process
Case 2:
if probability is too large, then some or
all components tolerances must be
increased.
Note: Do not perform a worst-case tolerance
analysis
40
Estimating the Natural Tolerance
Limits of a Process
41
Tolerance Limits Based on the Normal Distribution
Suppose a random variable x is distributed with
mean  and variance 2, both unknown. From a
random sample of n observations, the sample
mean x and sample variance S2 may be computed.
A logical procedure for estimating the natural
tolerance limits  ± Za/2  is to replace  by x and 
by S, yielding.
x  Zα/2S
42
Tolerance Intervals - Two-Sided
Since x and S are only estimates and not the true
parameters values, we cannot say that the above
interval always contains 100(1 - a)% of the
distribution. However, one may determine a
constant K, such that in a large number of samples
a faction
x  Zα/2S
43
Tolerance Limits
Based on the Normal Distribution
44
Tolerance Intervals - Two-Sided
If X1, X2, …, Xn is a random sample of size n from
a normal distribution with unknown mean  and
unknown standard deviation , then a two-sided
tolerance interval is (LTL,UTL), i.e., an interval that
contains at least the proportion P of the population,
with g 100% confidence is:
LTL  X  K 2S
and
UTL  X  K 2S
K 2 is a function of n, P, and g and may be obtained
from the table Factors for Two-Sided Tolerance Limits
for Normal Distributions (Located in the resource section on the website).
45
Tolerance Intervals - One-Sided
If X1, X2, …, Xn is a random sample of size n from
a normal distribution with unknown mean  and
unknown standard deviation , then a one-sided
lower (upper) tolerance interval is defined by the
lower tolerance limit LTL (upper tolerance limit UTL),
the value for which at least the proportion P of the
population lies above (below) LTL (UTL) with g100%
confidence where
LTL  X  K1S
UTL  X  K S.
1
K1 is a function of n, P, and g and may be obtained
from the table Factors for Once-Sided Tolerance Limits
for Normal Distributions (Located in the resource section on the website). 46
Tolerance Intervals - Two-Sided Example
Ten washers are selected at random from a
population that can be described by a normal
distribution. The measured thicknesses, in inches,
are:
.123
.132
.124
.123
.126
.126
.129
.129
.120
.128
Establish an interval that contains at least 90% of
the population of washer thicknesses with 95%
confidence.
47
Tolerance Intervals - Two-Sided Example Solution
From the sample data
X  .1260
and
S  0.00359
The K value can be found on Tolerance Limits TableTwo-Sided with gamma 95 and 99 and n=2 to 27 (Located in the resource
section on the website).
K 2  K10,0.90,0.95
 2.829
48
Tolerance Intervals - Two-Sided Example Solution
so that LTL  X  K 2 S
 0.1260  2.8290.00359 
 0.116
UTL X  K 2 S
 0.1260  2.8390.00359
 0.136
Therefore, with 95% confidence at least 90% of the
population of washer thicknesses, in inches, will be
contained in the interval (0.116,0.136).
49