Setting Specification Limits on Discrete Components

1
(No Transcript)
2
(No Transcript)
3
(No Transcript)
4
(No Transcript)
5
Setting Specification Limits on Discrete
Components
6

• 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

Dont just conform to specifications
Reduce variability around the target
accept
reject
reject
target
7

• 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.

8
(No Transcript)
9
Variability-Loss Relationship
LSL
USL
Target
Loss
savings due to reduced variability
Maximum loss per item
10
(No Transcript)
11
(No Transcript)
12
(No Transcript)
13
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.
14
(No Transcript)
15
(No Transcript)
16
Normal Distribution Standard Normal
Distribution If X N(?, ?) and if , then Z
N(0, 1). A normal distribution with ? 0 and
? 1, is called the standard normal distribution.
17
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.
18
Normal Distribution - example solution
19
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
20
(No Transcript)
21
What is the magnitude of the difference between
sigma levels? Sigma Area Spelling Time Distanc
e One Floor of 170 typos/page 31 years/century
earth to moon Astrodome in a
book Two Large supermarket 25 typos/page 4
years/century 1.5 times around in a
book the earth Three small hardware 1.5
typos/page 3 months/century CA to NY store in
a book Four Typical living 1 typo/30 pages 2
days/century Dallas to Fort Worth room (1
chapter) Five Size of the bottom 1 typo in a
set of 30 minutes/century SMU to 75 Central of
your telephone encyclopedias Six Size of a
typical 1 typo in a 6 seconds/century four
steps diamond small library Seven Point of a
sewing 1 typo in several 1 eye-blink/century 1/8
inch needle large libraries
22
Linear Combination of Tolerances Xi part
characteristic for ith part, i 1, 2, ... ,
n Xi N(?i, ?i) X1, X2, ..., Xn are
independent
23
(No Transcript)
24
Concept
x1
x2 . . .
xn
y
25
(No Transcript)
26
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?
27
Tolerance Analysis - example solution
fb(x)
fs(x)
28
Tolerance Analysis - example solution
Intersection Region
fb(x)
fs(x)
29
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 DN(0.011,0.0036)
fD(x)
dxb-xs
30
The Normal Model - example solution (a) Find c
such that P(D lt c) so that From the normal
table (found in the resource section of the
website), the Z -3.09 so that and
31
(No Transcript)
32
Tolerance Analysis - example Using Monte Carlo
Simulation (n1000) (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?
33
Tolerance Analysis - example Using Monte Carlo
Simulation First generate random samples from
(I used n1000) XbiN(mb, sb) N(1.059,
0.0030) and XsiN(ms, ss) N(1.048, 0.0020)
34
Tolerance Analysis - example Then calculate the
differences Estimate Estimate ms by taking the
mean. (You can use the AVERAGE()
function.) Estimate ss by calculating the
standard deviation.(You can use the STDEV()
function.)
35
(No Transcript)
36
Tolerance Analysis - example (b) This
can be compared to P(I) 0.000968.
37

• Statistical Tolerance Analysis Process
• Assembly consists of K components
• Specifications Assembly
• Specifications Component
• Assembly Nominal
• where ai 1 or -1 as appropriate

38

• Statistical Tolerance Analysis Process
• Assembly tolerance
• If dimension
• with parameters and , then
• where
• and

39

• Statistical Tolerance Analysis Process
• is specified
• is determined during design
• is calculated
• Case 1 if probability is too small, then
• (1) component tolerance(s) must be reduced
• or (2) tA must be increased

40
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
41
Estimating the Natural ToleranceLimits of a
Process
42
Tolerance Limits Based on the Normal
Distribution Suppose a random variable x is
distributed with mean m and variance s2, both
unknown. From a random sample of n observations,
the sample mean and sample variance S2 may be
computed. A logical procedure for estimating the
natural tolerance limits m Za/2 s is to replace
m by and s by S, yielding.
43
Tolerance Intervals - Two-Sided Since 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
44
Tolerance LimitsBased on the Normal Distribution
45
(No Transcript)
46
(No Transcript)
47
(No Transcript)
48
Tolerance Intervals - Two-Sided Example
Solution From the sample data and The K value
can be found on Tolerance Limits Table- Two-Sided
with gamma 95 and 99 and n2 to 27 (Located in
the resource section on the website).
49
Tolerance Intervals - Two-Sided Example
Solution so that 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).