Design of Experiments (DOE) ME 470 Fall 2013 – Day 2.
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Transcript of Design of Experiments (DOE) ME 470 Fall 2013 – Day 2.
Design of Experiments (DOE)
ME 470
Fall 2013 – Day 2
We will use statistics to make good design decisions!
We may be forced to run experiments to characterize our system. We will use valid statistical tools such as Linear Regression, DOE, and Robust Design methods to help us make those characterizations.
DOE is a powerful tool for analyzing and predicting system behavior.
At the end of the DOE module, students should be able to perform the following actions:• Explain the advantages of designed experiments (DOE) over
one-factor-at-a-time• Define key terminology used in experimental design• Analyze experimental data using the DOE techniques
introduced• Make good design decisions!!!
Be prepared to define the terms below.
Factor - A controllable experimental variable thought to influence response (in the case of the Frisbee thrower: angle, motor speed, tire pressure)
Response - The outcome or result; what you are measuring (distance Frisbee goes)
Levels - Specific value of the factor (15 degrees vs. 30 degrees) Interaction - Factors may not be independent, therefore combinations
of factors may be important. Note that these interactions can easily be missed in a straight “hold all other variables constant” scientific approach. If you have interaction effects you can NOT find the global optimum using the “OFAT” (one factor at a time) approach!
Replicate – performance of the basic experiment
There are six suggested steps in DOE.
1. Statement of the Problem2. Selection of Response Variable 3. Choice of Factors and Levels
Factors are the potential design parameters, such as angle or tire pressure
Levels are the range of values for the factors, 15 degrees or 30 degrees
4. Choice of Design screening tests response prediction factor interaction
5. Perform Experiment6. Data Analysis
23 Factorial Design Example
#1. Problem Statement: A soft drink bottler is interested in obtaining more uniform heights in the bottles produced by his manufacturing process. The filling machine theoretically fills each bottle to the correct target height, but in practice, there is variation around this target, and the bottler would like to understand better the sources of this variability and eventually reduce it.#2. Selection of Response Variable: Variation of height of liquid from target#3. Choice of Factors: The process engineer can control three variables during the filling process:
(A) Percent Carbonation(B) Operating Pressure(C) Line Speed
Pressure and speed are easy to control, but the percent carbonation is more difficult to control during actual manufacturing because it varies with product temperature. It can be controlled in a lab setting.
23 Factorial Design Example
#3. Choice of Levels – Each test will be performed for both high and low levels
#4. Choice of Design – Interaction effects
#5. Perform Experiment
Determine what tests are required using tabular data or Minitab
Determine the order in which the tests should be performed
Stat>DOE>Factorial>Create Factorial Design
3 factors
Full Factorial
Number of replicates
Enter Information
Ask for random
runs
Term
Standardized Effect
AC
BC
ABC
AB
C
B
A
876543210
2.306Factor NameA % CarbonationB Pressure
C Line Speed
Pareto Chart of the Standardized Effects(response is Deviation from Target, Alpha = .05)
The Pareto Chart shows the significant effects. Anything to the right of the red line is significant at a (1-a) level. In our case =0.05, a so we are looking for significant effects at the 0.95 or 95% confidence level. So what is significant here?
Residual
Perc
ent
10-1
99
90
50
10
1
Fitted Value
Resi
dual
6420-2
1.0
0.5
0.0
-0.5
-1.0
Residual
Fre
quency
1.00.50.0-0.5-1.0
6.0
4.5
3.0
1.5
0.0
Observation Order
Resi
dual
16151413121110987654321
1.0
0.5
0.0
-0.5
-1.0
Normal Probability Plot of the Residuals Residuals Versus the Fitted Values
Histogram of the Residuals Residuals Versus the Order of the Data
Residual Plots for Deviation from Target
Estimated Effects and Coefficients for Deviation from Target (coded units)
Term Effect Coef SE Coef T PConstant 1.0000 0.1976 5.06 0.001%Carbonation 3.0000 1.5000 0.1976 7.59 0.000Pressure 2.2500 1.1250 0.1976 5.69 0.000Line Speed 1.7500 0.8750 0.1976 4.43 0.002%Carbonation*Pressure 0.7500 0.3750 0.1976 1.90 0.094%Carbonation*Line Speed 0.2500 0.1250 0.1976 0.63 0.545Pressure*Line Speed 0.5000 0.2500 0.1976 1.26 0.242%Carb*Press*Line Speed 0.5000 0.2500 0.1976 1.26 0.242
S = 0.790569 PRESS = 20R-Sq = 93.59% R-Sq(pred) = 74.36% R-Sq(adj) = 87.98%
We could construct an equation from this to predict Deviation from Target. Deviation = 1.00 + 1.50*(%Carbonation) +1.125*(Pressure) + 0.875*(Line Speed) + 0.375*(%Carbonation*Pressure) + 0.125*(%Carbonation*Line Speed) + 0.250*(Pressure*Line Speed) + 0.250*(%Carbonation*Pressure*Line Speed)We can actually get a better model, which we will discuss in a few slides.
Mean o
f Devia
tion fro
m T
arg
et
1210
2
1
0
3025
250200
2
1
0
%Carbonation Pressure
Line Speed
Main Effects Plot (data means) for Deviation from Target
Estimated Effects and Coefficients for Deviation from Target (coded units)
Term Effect Coef SE Coef T PConstant 1.0000 0.1976 5.06 0.001%Carbonation 3.0000 1.5000 0.1976 7.59 0.000Pressure 2.2500 1.1250 0.1976 5.69 0.000Line Speed 1.7500 0.8750 0.1976 4.43 0.002%Carbonation*Pressure 0.7500 0.3750 0.1976 1.90 0.094%Carbonation*Line Speed 0.2500 0.1250 0.1976 0.63 0.545Pressure*Line Speed 0.5000 0.2500 0.1976 1.26 0.242%Carb*Press*Line Speed 0.5000 0.2500 0.1976 1.26 0.242
S = 0.790569 PRESS = 20R-Sq = 93.59% R-Sq(pred) = 74.36% R-Sq(adj) = 87.98%
The statisticians at Cummins suggest that you remove all terms that have a p value greater than 0.2. This allows you to have more data to estimate the values of the coefficients.
Here is the final model from Minitab with the appropriate terms.
Estimated Effects and Coefficients for Deviation from Target (coded units)
Term Effect Coef SE Coef T PConstant 1.0000 0.2030 4.93 0.000%Carbonation 3.0000 1.5000 0.2030 7.39 0.000Pressure 2.2500 1.1250 0.2030 5.54 0.000Line Speed 1.7500 0.8750 0.2030 4.31 0.001%Carbon*Press 0.7500 0.3750 0.2030 1.85 0.092
S = 0.811844 PRESS = 15.3388R-Sq = 90.71% R-Sq(pred) = 80.33% R-Sq(adj) = 87.33%
Deviation from Target = 1.000 + 1.5*(%Carbonation) + 1.125*(Pressure) + 0.875*(Line Speed) + 0.375*(%Carbonation*Pressure)
Estimated Effects and Coefficients for Deviation from Target (coded units).The term coded units means that the equation uses a -1 for the low value and a +1 for the high value of the data.
Deviation from Target = 1.000 + 1.5*(%Carbonation) + 1.125*(Pressure) + 0.875*(Line Speed) + 0.375*(%Carbonation*Pressure)
Let’s check this for %Carbonation = 10, Pressure = 30 psi, and Line Speed = 200 BPM%Carbonation is at its low value, so it gets a -1. Pressure is at its high value, so it gets +1, Line Speed is at its low value, so it gets a -1.
Deviation from Target = 1.000 + 1.5*(-1) + 1.125*(1)+ 0.875*(-1)+ 0.375*(-1*-1)
Deviation from Target = -0.625 tenths of an inch
How does this compare with the actual runs at those settings?
>Stat>DOE>Factorial>Response Optimizer
Now that we have our model, we can play with it to find items of interest.
Select C8 Deviation (tenths of an inch)
CurHigh
Low1.0000D
Optimal
d = 1.0000
Targ: 0.0Deviatio
y = 0.0
1.0000DesirabilityComposite
200.0
250.0
25.0
30.0
10.0
12.0Pressure Line Spe%Carbona
[10.0] [29.3713] [223.2464]
The engineer wants the higher line speed and decides to put the
target slightly negative. Why??
NEVER GIVE THIS SETTING TO PRODUCTION UNTIL YOU HAVE VERIFIED THE MODEL!!!
Not too Noisy
Noise Level < 75 db
VOC
System Spec
Lawn Mower
Example
Engine Noise Blade Assy Noise
Combustion Noise
Muffler Noise
Muffler Volume
Hole Area
Diameter
Blade Speed
Blade Area
Blade Width
Blade Length
Grass Height
Blade to Hsg Clearance
Let’s revisit the Frisbee Thrower and see what the data shows us.
BC
ABC
B
AC
AB
A
C
9876543210
Term
Standardized Effect
2.306
A Speed %B Tire Pressure (psi)C Angle (Degrees)
Factor Name
Pareto Chart of the Standardized Effects(response is Distance (ft), Alpha = 0.05)
5.02.50.0-2.5-5.0
99
90
50
10
1
Residual
Perc
ent
70605040
5.0
2.5
0.0
-2.5
-5.0
Fitted Value
Resi
dual
420-2-4
2.0
1.5
1.0
0.5
0.0
Residual
Fre
quency
16151413121110987654321
5.0
2.5
0.0
-2.5
-5.0
Observation Order
Resi
dual
Normal Probability Plot Versus Fits
Histogram Versus Order
Residual Plots for Distance (ft)
Factorial Fit: Distance (ft versus Speed %, Tire Pressure, Angle (Degrees)
Estimated Effects and Coefficients for Distance (ft) (coded units)
Term Effect Coef SE Coef T PConstant 51.775 0.9604 53.91 0.000Speed % 6.000 3.000 0.9604 3.12 0.014Tire Pressure (psi) -3.125 -1.562 0.9604 -1.63 0.142Angle (Degrees) -15.650 -7.825 0.9604 -8.15 0.000Speed %*Tire Pressure (psi) -5.225 -2.612 0.9604 -2.72 0.026Speed %*Angle (Degrees) -4.400 -2.200 0.9604 -2.29 0.051Tire Pressure (psi)*Angle (Degrees) 0.075 0.037 0.9604 0.04 0.970Speed %*Tire Pressure (psi)* 1.775 0.887 0.9604 0.92 0.382 Angle (Degrees)
S = 3.84171 PRESS = 472.28R-Sq = 92.02% R-Sq(pred) = 68.09% R-Sq(adj) = 85.04%
Following recommended procedures, we achieved a reduced model.
B
AC
AB
A
C
9876543210
Term
Standardized Effect
2.228
A Speed %B Tire Pressure (psi)
C Angle (Degrees)
Factor Name
Pareto Chart of the Standardized Effects(response is Distance (ft), Alpha = 0.05)
840-4-8
99
90
50
10
1
Residual
Perc
ent
70605040
5.0
2.5
0.0
-2.5
-5.0
Fitted Value
Resi
dual
6420-2-4
4
3
2
1
0
Residual
Fre
quency
16151413121110987654321
5.0
2.5
0.0
-2.5
-5.0
Observation Order
Resi
dual
Normal Probability Plot Versus Fits
Histogram Versus Order
Residual Plots for Distance (ft)
Let’s look at both main effects and interaction effects.
STAT >DOE>FACTORIAL>FACTORIAL PLOTS
10050
60
55
50
45
4530
3015
60
55
50
45
Speed %
Mean
Tire Pressure (psi)
Angle (Degrees)
Main Effects Plot for Distance (ft)Data Means
4530 3015
60
50
40
60
50
40
Speed %
Tire Pressure (psi)
Angle (Degrees)
50100
Speed %
3045
(psi)Pressure
Tire
Interaction Plot for Distance (ft)Data Means