Robust Design forProducts & Processes

Variation influence on the Optimum

2

Input

Outp

ut

Upper Specification Limit

Save SideShow Bob’s life

2000 feet

Pool is 200 feet longCopyright the Simpsons

Engineering

• r = range

• V = velocity

• q = angle (radians)

• g = acceleration dueto gravity (32.2 ft/s2)

( ) g

2sinVr

2θ

=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

Dis

tan

ce

Smaller spread

Spread in distance

Distribution Moments

−== dxxxfxE )()(

2222 )()()()( −=−=

−dxxfxxExExV

Transformations

( ) dy

dwywfyg =)( (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 ex

( ) =−

, x > 0

Substituting into Equation 3.Error! Bookmark not defined. gives

( )

=

−−

4

1)(

4/)12( yeyg

, y > 12

The exponential probability density function is

f x ex

( ) =−

, x > 0

Substituting into Equation 3.Error! Bookmark not defined. gives

( )

=

−−

4

1)(

4/)12( yeyg

, y > 12

Transformations

( )2111

, xxuy = and ( )2122

, xxuy = . If the inverse of the transformation functions,

x w y y1 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

( ) ( ) Jyywyywfyyg21221121

,,,),( = (3.1)

where J is the Jacobian and is defined as the determinant of the partial derivatives;

Jx 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

2n

1i i

2

yix

yV(y) σσ

=

=

=

+=

n

1i

2

x2

i

2

yix

y

2

1yE(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 Example1. 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

Po

we

rResistance 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

We

igh

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?

ExampleY = 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

YV(y)

=+−=

+

= σσ

( ) ( )

( ) 50542

1

T

Y

P

Y

2

1Bias

2

2

T2

22

P2

2

−=−=

+

= σσ

4P200P

Y−=

( )4

P

Y2

2

−=

14T

Y=

( )0

T

Y2

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 Solverwith 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) = 2StD(d) = 0.04StD(f) = 0.3

V

tf

dVeQ−

=

Better Estimate of System Variance

22

22

2

2

1

)()(2

1 )(

jii xx

n

i

n

j ji

x

n

i i

Yσσ

xx

Yσ

x

Yσ

+

=

=

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