Augmenting full factorial design to CCF

"Ali"

Registered
I am using Minitab for modeling a continuous response as a function of two continuous factors. Initially I created a full factorial DOE with two replicates and three center points. The results showed that the curvature is actually significant. Now I want to augment the DOE with axial points to have a face-centered central composite design.

  1. Would it be necessary to also replicate the axial points? If not, what are the disadvantages to expect though?
  2. Would it make sense to increase the number of center points?
 

Miner

Forum Moderator
Leader
Admin
  1. Replicating the axial points would add some precision to the estimate of the coefficients for the quadratic terms but is not strictly necessary.
  2. You definitely want to add some additional center points. Since you are adding the axial points and running these experiments on a different day than you ran the original full factorial experiment, you will want to block on the different days. Having a set of center points in each block provides an easy way to check for a block effect and remove it from the analysis.
 

"Ali"

Registered
Thank you Miner for your reply.

I know that adding the axial points and center points is actually the right way to model the curvature and to see which factor(s) contributing to it. However in the mean time, I created a custom response surface design solely based on the original full factorial design which had already three center points and I got the results below. The model could only consider a quadratic term of one factor. Based on this model I could also find an optimum which maximises the response. But if I choose to have the quadratic term of the second factor, then I will end up getting a different optimum with a lower response. In this case, the model which delivered a higher response is more relevant for us.

Considering the good ANOVA results, is my argument valid? In other words, would it be statistically reasonable to rely on the full factorial model with center points and thus avoid the cost of running more experiments? If not, what is the added value of augmenting the original design to ccf? Would I get completely new optimum with a higher response?

Analysis of Variance

Source
DF​
Seq SS​
Contribution​
Adj SS​
Adj MS​
F-Value​
Model
3​
85902,4​
96,59%​
85902,4​
28634,1​
66,06​
Linear
2​
75714,5​
85,13%​
75714,5​
37857,3​
87,34​
A
1​
19602,0​
22,04%​
19602,0​
19602,0​
45,22​
B
1​
56112,5​
63,09%​
56112,5​
56112,5​
129,45​
Square
1​
10187,9​
11,46%​
10187,9​
10187,9​
23,50​
A*A
1​
10187,9​
11,46%​
10187,9​
10187,9​
23,50​
Error
7​
3034,2​
3,41%​
3034,2​
433,5​
Lack-of-Fit
1​
12,5​
0,01%​
12,5​
12,5​
0,02​
Pure Error
6​
3021,7​
3,40%​
3021,7​
503,6​
Total
10​
88936,5​
100,00%​


Source
P-Value
Model
0,000
Linear
0,000
A
0,000
B
0,000
Square
0,002
A*A
0,002
Error
Lack-of-Fit
0,880
Pure Error
Total

Total
Model Summary

S​
R-sq​
R-sq(adj)
PRESS​
R-sq(pred)
20,8195​
96,59%​
95,13%
7032,56​
92,09%
Coded Coefficients

Term
Coef​
SE Coef​
95% CI​
T-Value​
P-Value​
VIF​
Constant
352,3​
12,0​
(323,9; 380,8)​
29,31​
0,000​
A
49,50​
7,36​
(32,09; 66,91)​
6,72​
0,000​
1,00​
B
-83,75​
7,36​
(-101,16; -66,34)​
-11,38​
0,000​
1,00​
A*A
-68,3​
14,1​
(-101,7; -35,0)​
-4,85​
0,002​
1,00​
Regression Equation in Uncoded Units

C=-206 + 45,95 A - 3,045 B - 0,683 A*A
 

Miner

Forum Moderator
Leader
Admin
I know that adding the axial points and center points is actually the right way to model the curvature and to see which factor(s) contributing to it. However in the mean time, I created a custom response surface design solely based on the original full factorial design which had already three center points and I got the results below. The model could only consider a quadratic term of one factor. Based on this model I could also find an optimum which maximises the response. But if I choose to have the quadratic term of the second factor, then I will end up getting a different optimum with a lower response. In this case, the model which delivered a higher response is more relevant for us.

Considering the good ANOVA results, is my argument valid? In other words, would it be statistically reasonable to rely on the full factorial model with center points and thus avoid the cost of running more experiments? If not, what is the added value of augmenting the original design to ccf? Would I get completely new optimum with a higher response?
I was not able to follow what you actually did during this process, but the easiest way to determine whether it was legitimate is to verify whether the predicted optimum works. Run an experiment at the levels that your model says are optimum and see whether you actually achieve the predicted results. If it works, the rest of the discussion is moot.
 

"Ali"

Registered
DOE_1: Full Factorial, 2 replicates, 3 center points --> Significant terms: A, B, and curvature --> Used to build two response surface model with only one quadratic term either A*A or B*B:
  • RSM_1 with A*A: optimum with response r_opt_1
  • RSM_2 with B*B: optimum with response r_opt_2 < r_opt_1
--> Choose the response surface model RSM_1

DOE_2: DOE_1 + Axial points x1 + additional center points x2 --> response surface model RSM_3 --> optimum with response r_opt_3
  1. Given that the ANOVA results of DOE_1 were actually pretty good, are there other statistical measures/tests which could be used to argue that the DOE_2 is actually better? One point I could think about it is the statistical power, but how could the statistical power of the DOE_2 be measured in Minitab?
  2. Would it be possible that using the RSM_1, one misses the absolute optimum? In other words, using RSM_3 one could get r_opt_3 > r_opt_1. If this is the case, then verifying the RSM_1 would not solve the issue.
Thinking more general, I would like to figure out why a RSM based on a CCF design would still be better than a RSM based on full factorial design including center points eventhough the latter delivers good ANOVA results i.e. p-values, R-sq(adj), R-sq(pred), Lack of fit...
 

Miner

Forum Moderator
Leader
Admin
How did you build an RSM model using center points alone (DOE_1)? Center points can only identify the presence of curvature. It cannot isolate it to a specific factor, nor model it.
 

"Ali"

Registered
How did you build an RSM model using center points alone (DOE_1)? Center points can only identify the presence of curvature. It cannot isolate it to a specific factor, nor model it.
I created a custom response surface design in Minitab using the DOE_1. I could then build the model RSM_1/RSM_2 using the response surface design created. However the model could only consider one quadratic term (A*A or B*B) in addition to the significant terms A & B identified before. It is as if the curvature captured by the full factorial design was represented by the A*A or B*B term. Another way to get the same models RSM_1/RSM_2 is to fit a regression model to the data manually or to let the assistent do it using the muliple regression.
 

Miner

Forum Moderator
Leader
Admin
If I understand correctly, you took a standard full factorial design with center points and defined it as an RSM design without adding any axial points? You cannot do that! I know Minitab will allow you to do it, but you should not do it. Defining it as an RSM does not make it an RSM design.

See DOE-center-points-what-they-are-why-theyre-useful
1686168834172.png


And How-minitab-handles-center-points-in-a-2-level-factorial-design/
1686169213414.png


Note the requirement for axial points.

What you actually did when you defined it as an RSM design and when you fit a regression model, was to arbitrarily decide which term had the curvature then forced a model to fit your assumption. You did not KNOW which factor actually had curvature or whether both factors had curvature. This is a correlation does not imply causation situation. That is the difference between an experiment (DOE) and regression (observational study). A properly designed experiment firmly establishes causation. Regression only establishes correlation.
 

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"Ali"

Registered
Yes Miner you understood it correctly. I will execute the DOE_2 and see what results come out. Thank you again for your answers.
 
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