Six Sigma, Statistics expert needed
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Robustness.pdf
Robust Design for Products & Processes
Variation influence on the Optimum
2
Input
O u tp
u t
Upper Specification Limit
Save SideShow Bob’s life
2000 feet
Pool is 200 feet long Copyright the Simpsons
Engineering
• r = range
• V = velocity
• q = angle (radians)
• g = acceleration due to gravity (32.2 ft/s2)
( ) g
2sinV r
2 θ
= Objectives
• Target range (r) is 2000 feet
• Minimise variation of range (less than 100 feet)
Response Sensitivity
0 10 20 30 40 50 60 70 80 90
Angle
D is
ta n
ce
Smaller spread
Spread in distance
Distribution Moments
− == dxxxfxE )()(
2222 )()()()( −=−=
− dxxfxxExExV
Transformations
( ) dy
dw ywfyg =)( (3.1)
where: f (x) is the probability density function for x,
y = u(x) is the transformation function, and
x = w(y) is the inverse of the transformation function, and has only one root.
Example y x= +4 12
x
y =
−12
4
dx
dy =
1
4
The exponential probability density function is
f x e x
( ) = −
, x > 0
Substituting into Equation 3.Error! Bookmark not defined. gives
( )
=
−−
4
1 )(
4/)12( y eyg
, y > 12
The exponential probability density function is
f x e x
( ) = −
, x > 0
Substituting into Equation 3.Error! Bookmark not defined. gives
( )
=
−−
4
1 )(
4/)12( y eyg
, y > 12
Transformations
( ) 2111
, xxuy = and ( ) 2122
, xxuy = . If the inverse of the transformation functions,
x w y y 1 1 1 2 = ( , ) and x w y y
2 2 1 2 = ( , ) have single roots, the joint probability density
function for y1 and y2 is
( ) ( ) Jyywyywfyyg 21221121
,,,),( = (3.1)
where J is the Jacobian and is defined as the determinant of the partial derivatives;
J x y x y
x y x y =
1 1 1 2
2 1 2 2
/ /
/ /
System characteristics • Properties of means and variances of functions of
several variables, where y = f(x1, x2, x3, ..., xn).
• An approximation for the mean of a function of several variables is:
• An approximation for the variance of a function of several variables is:
2
x
2 n
1i i
2
y ix
y V(y) σσ
=
=
=
+=
n
1i
2
x2
i
2
y ix
y
2
1 yE(y) σμ
Circuit Example
P = V2/R
• Voltage (V) is Normally distributed • mean = 12
• standard deviation = 0.1
• Resistance (R) is Normally distributed • mean = 2
• standard deviation = 0.2
Where does input data come from?
• Current production • Supplier • Discuss how to obtain mean and standard deviation for
voltage and resistance
Circuit Example 1. What does the power distribution look like?
• Specifications
• 72 ± 10
• Will this circuit design meet specifications?
2. Change standard deviation of voltage to 0.3 and standard deviation of resistance to 0.4
• What is the mean and standard deviation of power?
3. Change standard deviation of voltage to 0.5 and standard deviation of resistance to 0.6
Statistical Bias
0
100
200
300
400
500
600
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10
Resistance
P o
w e
r Resistance decreased from 1.0 to 0.5
Power increased from 144 to 288
Power change = 144
Resistance increased from 1.0 to 1.5
Power decreased from 144 to 96
Power change = 48
Statistical Bias
0
20
40
60
80
100
120
0 2 4 6 8 10 12 14 16 18 20
Length
W e
ig h
t
Length increases from 9 to 10
Weight increases from 58.7 to 64.57
Weight change = 5.87
Length decreases from 9 to 8
Weight decreases from 58.7 to 52.83
Weight change = 5.87
Circuit Example
• Use robustness approximations to compute • Statistical bias
• Power standard deviation
• How do these results compare to the simulation?
Example Y = 25 + 200P - 2P2 + 14T
1. Determine bias of Y & STD of Y using robustness equations & verify with simulation
• P = 0 • Y = 7000
2. Determine values of T & P that minimize the STD of Y using robustness equations & verify with simulation
• Y = 7000
T = Normal (std = 10)
P = Normal (std = 5)
Example – Optimise using derivatives • Y = 25 + 200P - 2P2 + 14T
• Determine T and P
• Y = 7000
• Minimise variation of Y
( ) ( ) 22222
2
T
2
2
P
2
140101454P200
T
Y
P
Y V(y)
=+−=
+
= σσ
( ) ( )
( ) 5054 2
1
T
Y
P
Y
2
1 Bias
2
2
T2
2 2
P2
2
−=−=
+
= σσ
4P200 P
Y −=
( ) 4
P
Y 2
2
−=
14 T
Y =
( ) 0
T
Y 2
2
=
1st Order Der. P =
1st Order Der. T =
2nd Order Der. P =
2nd Order Der. T =
Example – Optimise using derivatives
Excel > Tools > Solver
T & P optimised by Solver with minimal StD(Y)
• Friction factor (f) 0.01 < f < 0.05
• Pipe diameter (d) 0.5 < d < 10.0
• Flow rate (v) 7 < v < 50
• Pipe length (L) 2000
• Discharge level below reservoir surface (D) 25 < D < 500
• Specification 5 < HP < 45
• Friction factor (f) StDev = 0.002
• Pipe diameter (d) StDev = 0.1
• Flow rate (v) StDev = 2
• Pipe length (L) StDev = 1
• Discharge level below reservoir surface (D) StDev = 3
• HorsePower within specifications and minimise variance of HorsePower
Robustness example – Reservoir flow
D−+= 2323
0006188.0000115.000000961.0 vddfLvdvHP
Class exercise – Salt tank mixing
• Q = grams of salt after t minutes
• Tank volume (V): 500 litres max
• Salt density (d): 260 grams/litre max
• Water flow (f) litres/min: (V/5) maximum, 0.1 minimum
• Fresh water in, uniform stirring, mixed water out
• Required
• More than 90 g of salt after 2 minutes
• Between 70 g and 80 g of salt after 5 minutes
• Inputs are Normally distributed with standard deviation below:
• StD(V) = 2 StD(d) = 0.04 StD(f) = 0.3
V
tf
dVeQ −
=
Better Estimate of System Variance
22
2 2
2
2
1
)()( 2
1 )(
jii xx
n
i
n
j ji
x
n
i i
Y σσ
xx
Y σ
x
Y σ
+
=
=
Example
Y = 25 + 200P - 2P2 + 14T
• Determine values of T & P that minimize the STD of Y using robustness equations & verify with simulation
– Y = 7000
T = Normal (std = 10)
P = Normal (std = 5)
Large Scale Systems
End
