[PPT] - Part1: Full factorial designs for 2-level factors Prepared by: PowerPoint Presentation

SLIDE 1

Part1: Full factorial designs for 2-level factors

Prepared by: Paul Funkenbusch, Department of Mechanical Engineering, University of Rochester

SLIDE 2

 Terminology  Experiment  Model relationship  Full factorial experiments  Interactions and sparsity of effects

2 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 3

Terms

Example

(measure the volume of a balloon as a function of temperature and pressure)

 Factors  variables whose

influence you want to study.

 Levels specific values given

to a factor during experiments (initially limit ourselves to 2-levels)

 Treatment condition  one

running of the experiment

 Response  result measured

for a treatment condition

 Temperature,

Pressure

 50C, 100C

1Pa, 2Pa

 Set T = 50C, P = 1Pa

and measure volume

 measured volume

3 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 4

Leve vel Factor

1

+1

X1. Temperature (C)

50 100

X2. Pressure (Pa)

1 2 TC TC X1 X1 X2 X2 y 1

1
1

y1 2 +1

1

y2 3

1

+1 y3 4 +1 +1 y4 etc. Use X1, X2, etc. to designate factors. Use -1, +1 to designate levels X1 at level -1 means T = 50 C Use a table to show factor levels and response (a) for each treatment condition. For example, during TC2, set T = 100C and P = 1Pa, measure the balloon volume = y2

y  response y = volume of balloon

4 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 5

 “Effects” calculated from

responses (y1, y2)

 m*

=(y1 + y2)/2
= overal

all l average age

 DX1

= (avg. at +1) – (avg. at -1)
= m+1 – m-1 = y2 – y1
= “effect of X1”

Leve vel Factor

1

+1

X1. Temperature (C)

50 100 TC TC X1 X1 y 1

1

y1 2 +1 y2

1

1 y

X1 X1

5 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 6

 Can fit a straight line  ypred = ao + a1X1  ao

intercept
related to m*
(= m* for X1on ±1 scale)

 a1

slope
related to DX1
(= DX1/2 for X1on ±1 scale)

TC TC X1 X1 y 1

1

y1 2 +1 y2

1

1 y

X1 X1

D

m*

6 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 7

 DOF or DF

Counter for information

 Experimental data

2 data points (y1 & y2)
= 2 DOF

 Analysis

2 results (m*, DX1  ao, a1)
= 2 DOF

 Information is conserved

2 DOF  2 DOF

TC TC X1 X1 y 1

1

y1 2 +1 y2

1

1 y

X1 X1

7 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 8

 One At a Time experiment

“OAT”
what most people are taught
pick a baseline (X1=-1,

X2=-1)

Change one factor at a time

 Results

baseline = y1
DX1 = y2 – y1
Dx2 = y3 – y1

Leve vel Factor

1

+1

X1. Temperature (C)

50 100

X2. Pressure (Pa)

1 2 TC TC X1 X1 X2 X2 y 1

1
1

y1 2 +1

1

y2 3

1

+1 y3

8 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 9

 Choice of baseline is

critical (next slide)

 Inefficient use of data

baseline = y1
DX1 = y2 – y1
Dx2 = y3 – y1
Only use part of data to

calculate each effect

Leve vel Factor

1

+1

X1. Temp (C)

50 100

X2. Pressure (Pa)

1 2 TC TC X1 X1 X2 X2 y 1

1
1

y1 2 +1

1

y2 3

1

+1 y3

9 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 10

 Study plant growth as a

function of watering and sunlight

 Results

baseline = y1 = 0
DX1 = y2 – y1 = 0
Dx2 = y3 – y1 = 0

 Conclusion

Watering and sunlight do

not affect plant growth

Leve vel Factor

1

+1

X1. watering

never daily

X2. daily sunlight

none 10 hrs TC TC X1 X1 X2 X2 growth th 1

1
1

2 +1

1

3

1

+1

10 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 11

 Test all combinations  Results

m* = (y1+y2+y3+y4)/4
DX1 = (y3+y4)/2 - (y1+y2)/2
Dx2 = (y2+y4)/2 - (y1+y3)/2

 Effect of X1

Averaged over both X2 levels
No baseline

 All results  all effects

Use all data in each calculation
Efficient use of data

Leve vel Factor

1

+1

X1. Temp (C)

50 100

X2. Pressure (Pa)

1 2 TC TC X1 X1 X2 X2 y 1

1
1

y1 2

1

+1 y2 3 +1

1

y3 4 +1 +1 y4

11 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 12

 a12  D12

= (avg. for X1X2 =+1) – (avg. for X1X2 =-1)
= (y1+y4)/2 - (y2+y3)/2 = “effect of X1*X2 interaction”

 Can show interaction as a column to help calculations

(Note: This simple algebra works only for 2-level factors, with

levels set to ±1, but the underlying ideas also apply to other designs.)

 Four degrees of freedom  Model can include an extra

(interaction) term

 ypred = ao+a1X1+a2X2+a12 12X1X2

TC TC X1 X1 X2 X2 X1X2 X2 y 1

1
1

+1 y1 2

1

+1

1

y2 3 +1

1
1

y3 4 +1 +1 +1 y4

12 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 13

Leve vel Factor

1

+1

X1. applied load (kg)

2 3

X2. previous cuts

(new) 20

 Data on the removal

rate of osteotomy drills is collected as a function of the load applied and the number of previous cuts made.

 Find the effects of the

two factors and the

interaction. Which

is/are most important?

 Build a simple model to

predict the removal rate.

TC TC X1 X1 X2 X2 X1X2 X2 Remov

val

al rate (mm3/s) /s) 1

1
1

+1 3 2

1

+1

1

2 3 +1

1
1

5 4 +1 +1 +1 2

13 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 14

TC TC X1 X1 X2 X2 X1X2 X2 Remov

val

al rate (mm3/s) /s) 1

1
1

+1 3 2

1

+1

1

2 3 +1

1
1

5 4 +1 +1 +1 2

 Effects

m* = ?
DX1 = ?
Dx2 =?
Dx1x2 = ?

14 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 15

TC TC X1 X1 X2 X2 X1X2 X2 Remov

val

al rate (mm3/s) /s) 1

1
1

+1 3 2

1

+1

1

2 3 +1

1
1

5 4 +1 +1 +1 2

 Effects

m* = (y1+y2+y3+y4)/4 =(3+2+5+2)/4 =

3

DX1 = (y3+y4)/2 - (y1+y2)/2 = (5+2)/2 – (3+2)/2 =

1

Dx2 = (y2+y4)/2 - (y1+y3)/2 = (2+2)/2 – (3+5)/2 =
2
Dx1x2 = (y1+y4)/2 - (y2+y3)/2 = (3+2)/2 – (2+5)/2 = -1

 Number of previous

cuts (i.e. X2) is the more important factor (given the range of values tested).

15 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 16



Effects

m* =

3

DX1 =

1

Dx2 =
2
Dx1x2 =
1

 Use the normalized (±1) scale for the levels.  Then the midpoint of the design corresponds to the

intercept (0) for both factors  a0 = m* = 3.0

 And the slope (a) for each effect is given by D/2

a1 = DX1 /2 = 0.5; a2 = DX2 /2 = -1.0; a12 = DX12 /2 = -0.5

 ypred = 3.0 + (0.5)X1 - (1.0)X2 - (0.5)X1X2

(y in units of mm3/s)

1

1 y

X1 X1

m* a0

1

1 y

X1 X1

D 2

16 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 17

Leve vel Factor

1

+1

X1. applied load (kg)

2 3

X2. previous cuts

20  L = load (kg)  C = # of cuts  X1 = -5 + 2L  X2 = -1 + 0.1C  ypred = 3.0 + (0.5)X1 - (1.0)X2 - (0.5)X1X2

(y in mm3/s, for L in kg)

17 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 18

 4 data points, 4 constants

 will hit all of the original data exactly

 New trials under same conditions?

 How good was the original data? (error)

 Other conditions (interpolation, extrapolation)?

 How good is the linear fit to the real dependence?

18 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 19

 require more TC to include all combinations

# of TC = 2n , where n= # of factors

 model includes more terms  # of interaction terms increases rapidly  includes “higher-order” interactions

“order” = number of factors in the interaction term
X1X2  second order; X1X2 X3 third order, etc.

19 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 20

 23 = 8  8 responses (8 y’s)  m* , Dx1 , Dx2 , Dx3 ,

Dx1x2 , Dx1x3 , Dx2x3 , Dx1x2x3

 ypred = ao+a1X1+a2X2+ a2X3

+ a12X1X2+a13X1X3+a23X2X3 +a123X1X2X3

TC TC X1 X1 X2 X2 X3 X3 y 1

1
1
1

y1 2

1
1

+1 y2 3

1

+1

1

y3 4

1

+1 +1 y4 5 +1

1
1

y5 6 +1

1

+1 y6 7 +1 +1

1

y7 8 +1 +1 +1 y8

 8 DOF

m*

 1 DOF

factors

 3 DOF

2-factor inter.

 3 DOF

3-factor inter.

 1 DOF

20 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 21

More factors  much more effort to measure interactions

 4 factors at 2 levels  24 = 16  16 DOF

m*

 1 DOF

factors

 4 DOF

2-factor inter.

 6 DOF

3-factor inter.

 4 DOF

4-factor inter.

 1 DOF

 5 factors at 2 levels  25 = 32  32 DOF

m*

 1 DOF

factors

 5 DOF

2-factor inter.

 10 DOF

3-factor inter.

 10 DOF

4-factor inter.

 5 DOF

5-factor inter.

 1 DOF

21 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 22

 System is usually dominated by the effects of

factors and lower-order interactions.

 Usually don’t really need 3-factor, 4-factor,

etc. interactions ~ 0.

 But have just seen, that we may spend a lot

f effort (i.e. lots of DOF) on them.

 Can we make better use of these DOF?

 will discuss in Part II.

22 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 23

 Experiments can be influenced by time related

changes

Temperature changes during the course of a day
Drift of measurement apparatus
Operator fatigue
etc.

 Randomize the order in which TCs are run

Reduces the chance that time related changes will be

misattributed to factor effects

Exceptions should be justified on a case-by-case

basis

23 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 24

 Continuous Factor

Involves something that can be quantified on a continuous scale
Temperature, pressure, time, voltage, length, etc.
Algebraic model  ypred = ao+a1X1+a2X2+a12X1X2…

 Discrete (or nominal, categorical, etc.) Factor

Cannot be put on a continuous scale
Supplier, country, gender, “style”, etc.
e.g. Toyota vs. Ford as suppliers no “halfway” point
Can still write a model equation , but the corresponding X can have
nly the discrete (tested) levels

 Can use Full Factorial designs for either or both factor types

Assume continuous for now  not important for 2-level designs
Does affect designs with 3 or more levels  will discuss in Part III

24 DOE mini-course, part 1, Paul Funkenbusch, 2015

SLIDE 25

 Full factorial design = all combinations

“effect” = difference in average value at the two levels

 Advantages of full factorial designs

Not dependent on choice of a baseline
All of the data is used to calculate each effect (“efficient”)
Can measure interactions between factors
Convert easily to a multi-factor model

 Disadvantages of full factorial designs

Work best with only 2 (or maybe 3) levels for factor
Many DOF used to measure higher-order interactions

 But may be able to take advantage of this  Part II

25 DOE mini-course, part 1, Paul Funkenbusch, 2015