lecture # 28 prof. john w. sutherland oct. 31, 2005 · dept. of mechanical engineering -...
TRANSCRIPT
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Lecture # 28
Prof. John W. Sutherland
Oct. 31, 2005
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Another Example
Three identical parts
Manufacturing process -- Cp = 1.5 (at least)
321
0.5± 0.5± 0.5±
1.5 0.050±
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
What will the individual process σx be, if tolerancesare obtained by simple division?
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Find (i) standard deviation for the processes(ii) tolerances for individual parts
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Another Example
For now, let’s assume hole & pin producingprocesses are centered at the nominal values andthat processes have Cp values of 1.0.
What does clearance dist. look like??
1.00 ± 0.015 0.965 ± 0.015
PinHole
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Clearance Distribution
C = H - P = 0.035
---- σc = 0.007
µC µH µP–=
σC2
σH2
σP2+=
.014 .035 .056
Clearance
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
What can go Wrong?
• We have already seen that if our individual processes(in this case pin and hole) do not remain centered -the results can be disastrous.
• What if our processes are not maintained in a state ofstatistical control?
0.985 1.000 1.015
Hole
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Histogram - Hole Dimension
020406080
100120
0.9865 0.9895 0.9925 0.9955 0.9985 1.0015 1.0045 1.0075 1.0105 1.0135
Hole Diameter
Freq
uenc
y
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Histogram - Pin Dimension
020406080
100120
0.9515 0.9545 0.9575 0.9605 0.9635 0.9665 0.9695 0.9725 0.9755 0.9785
Pin Diameter
Freq
uenc
y
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Histogram - Clearance Dimension
0
50
100
150
200
250
0.004 0.012 0.02 0.028 0.036 0.044 0.052 0.06 0.068 0.076
Clearance
Freq
uenc
y About91.4%Capable
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Capability Assessmentvia a Loss Function
Quadratic Loss Function
Nominal
Qua
lity
loss
Quality characteristic
x
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Defining the Loss Function
• for an x value of 0, the loss, L, is $0.This means that m = 0 and L0 = 0
• when x= 1.5, the loss, L, is $2.25L/(x-m)2 = k k = $2.25/(1.5 - 0)2 = 1
The $2.25 loss is obtained from the fact that whenx=1.5, the probability of a $45 complaint is 5%, and$45 * (0.05) = $2.25
L L0 k x m–( )2+=
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Loss Function Interpretationof Engineering Specifications
LS m US
Qua
lity
loss
x
Cost to Rework/Replace
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
More on Loss / Specifications
So, we want to set our specifications at the point ofindifference: where the loss = cost of replacement
If cost of replacement = $2 for our previous example,
L = 1 (x - 0)2 = 2 ----- x = ±1.414
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
More on the Loss Function
m 1.5
Qua
lity
loss
x
$2.25
Can we come upwith a loss assoc.with this distrib.
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Expected Process Loss
where,
and f(t) is the pdf
after simplification,
E L X( )[ ] L t( )f t( ) td
all X∫=
L x( ) k x m–( )2=
E L X( )[ ] k µX m–( )2
σX2+
=
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Loss Function: Example #1
m=0, k=$1.0, LSL=-1, USL=1, X normal w/ mean 0 and σx = 0.33
E[L] = 1 [0.332 + (0-0)2] = $0.11 How to interpret??
-1.5 -1.0 -0.5 0 0.5 1.0 1.5
Qua
lity
loss
x
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Loss Function: Example #2
σx = 0.58
E[L] = 1 [0.582 + (0-0)2] = $0.33
-1.5 -1.0 -0.5 0 0.5 1.0 1.5
Qua
lity
loss
x
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Filling in a Past Blank
Should know from prior Statistics ClassFor a Uniform Distribution between a & b
f x( ) 1 b a–( )⁄=
µx xf x( ) xd
a
b∫
xb a–( )
----------------- xd
a
b∫
x2
2 b a–( )--------------------
a
bb a+( )
2-----------------= = = =
σx2 x µx–( )
2f x( ) xd
a
b∫
b a–( )2
12--------------------= =
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Loss Function: Example #3
Say we reduce σx = 0.33 to σx = 0.10
E[L] = 1 [0.102 + (0-0)2] = $0.01
-1.5 -1.0 -0.5 0 0.5 1.0 1.5
Qua
lity
loss
x
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Quality Engineering (MEEM 4650 / 5650)Dept. of Mechanical Engineering - Engineering MechanicsMichigan Technological University © John W. Sutherland
Loss Function: Example #4
Can save $0.10 by reducing process mean from 0 to -0.7
E[L] = 1 [0.102 + (-0.7 - 0)2] = $0.50 WOW!!
Taguchi’s comment:
-1.5 -1.0 -0.5 0 0.5 1.0 1.5
Qua
lity
loss
x
σx = 0.10