Full Factorial Design leads to experiments where at least one trial is included for all possible combinations of factors and levels. This exhaustive approach makes it impossible to miss any interactions, as all factor interactions are accounted for. However, the thoroughness of this approach makes it quite expensive and time-consuming for experiments with multiple factors – this increases exponentially with the number of factors and levels.
| Factor 1 | Factor 2 | Factor 3 | ||||
|---|---|---|---|---|---|---|
| Trials | L1 | L2 | L1 | L2 | L1 | L2 |
| 1 | ✔ | | ✔ | | ✔ | |
| 2 | ✔ | | ✔ | | | ✔ |
| 3 | ✔ | | | ✔ | ✔ | |
| 4 | ✔ | | | ✔ | | ✔ |
| 5 | | ✔ | ✔ | | ✔ | |
| 6 | | ✔ | ✔ | | | ✔ |
| 7 | | ✔ | | ✔ | ✔ | |
| 8 | | ✔ | | ✔ | | ✔ |
Sample factorial design table for a three-factor experiment with two levels per factor
Calculating the Number of Trials
The number of trials required for a full factorial experimental run is the product of the levels of each factor:
No. of trials = F1 level count x F2 level count x … x Fn level count
How Many trials in a Full Factorial Design?
Found by taking the number of levels as the base and the number of factors as the exponent:
Ex1. a design of 4 factors with 3 levels each would be: 3 x 3 x 3 x 3 = 3^4 = 81
Ex 2. 4 factors (A = 3, B = 2, C = 5, D = 4 levels). 3 x 2 x 5 x 4 = 120 observations.
Example
Let’s look at an experiment with four factors:
- The first factor has two possible levels.
- The second factor has five possible levels.
- The third factor has three possible levels.
- The fourth factor has six possible levels.
To cover all of the potential combinations, the experiment will need:
No. of trials = 2 x 5 x 3 x 6 = 180 trials
Full Factorial Design Notes
- Full factorial designs include all possible combinations of every level of every factor.
- Full factorial designs can require a lot of trials, which can take a lot of time.
- Full factorial designs can require a lot of trials, which can cost a lot of money.
- Requires at least one observation for every combination of factors and levels.
- Allows for the measurement of all possible interactions.
- Expensive and time-consuming.
Analyzing Full Factorial Designs
Factorial ANOVA
You can use an Analysis of Variation – ANOVA to determine the results of full factorial design experiments.
Yates Analysis
Yates analysis is used in experiments with multiple factors, all having two levels. In some circumstances, the two levels can be ‘high’ and ‘low’ data points. It can be used in both full and fractional factorial design experiments. Read more about Yates analysis.
Why You Would Use Partial or Fractional Factorial Design Instead
One of the big drawbacks of fractional factorial design is the potential to miss important interactions.
Fractional factorials (like Latin and Graeco-Latin Squares) will not allow the analysis of interactions. The interactions are confounded with other effects.
Moving from Full Factorial to Partial Factorial
- There will be fewer trials
- There will be confounding
- Resolution will decrease
Design of Experiments Factorial Design Video
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Comments (15)
How many factor at least can be used in factorial design
In factorial design, you can start with as few as two factors. In fact, the simplest factorial experiment is a 2² factorial design, which examines two factors, each at two levels.
Factorial designs are highly scalable, so you can include as many factors as are practical and meaningful for your experiment; however, the number of required runs increases exponentially with the number of factors and levels. For example:
This scalability is why factorial design is so powerful because it enables you to study the interaction between multiple variables efficiently.
For Usha, you can use at least two factors in a factorial design.
How can i fit and diagnose my model?
Lana – I’m going to need a bit more context than that to be able to help. Can you elaborate?
Hi Ted,
Thank you for the detailed explanation.
Can you please include some details about Linear & Quadratic Mathematical Models. Would be grateful for it.
Thanks
VJ
Linear Model
A linear model assumes that the response variable is a straight-line function of the input factors. In a two-factor system, it typically looks like:
Here:
Yis the response variableX₁andX₂are the input variables (factors)β₀is the interceptβ₁andβ₂are the coefficients representing the effect of each factorLinear models are useful when the relationship between the factor and response is expected to be constant and additive.
Quadratic Model
A quadratic model includes squared terms to account for curvature in the response surface. A typical model would be:
This model:
β₁₂X₁X₂)β₁₁X₁²andβ₂₂X₂²)Quadratic models are often used in >Response Surface Methodology (RSM), such as in Central Composite Designs and Box-Behnken designs.
Best ,
Ted
How many experimental runs exist in a Full Factorial and fully randomized design for 4 factors with 2 replicates for the Corner Points and no Center Points? The factors in the experiment are only at 2-levels *
To calculate the number of experimental runs in a Full Factorial and fully randomized design with 4 factors at 2 levels, and with 2 replicates for the corner points and no center points, follow this logic:
Step-by-Step Calculation
1. Determine the number of runs for a 2-level factorial design:
For 4 factors at 2 levels:
2^4 = 16corner points (unique treatment combinations).2. Account for the number of replicates:
Each corner point is replicated 2 times:
16 × 2 = 32total runs.3. No center points:
Since there are no center points, we do not add any additional runs.
Answer:
Total experimental runs = 32
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When the search space is large, genetic algorithms can be tailored to find the optimal parameters. Better than a Partial design.
That’s really interesting. Where could we learn more?
Hi,
i want to ask a question,i have been working on minitab lately.
i saw a question and i cant this solve question.I see error=0 in my anova table
Would you help me please?
0 (Low)1 (High)
Factors Repeats
A B C 1 2 3 4
0 0 0 46 50 50 58
1 0 0 70 59 52 61
0 1 0 51 39 59 53
1 1 0 51 55 52 40
0 0 1 48 48 44 66
1 0 1 36 40 37 33
0 1 1 48 47 65 47
1 1 1 39 66 62 47
Thank you for your question! When you encounter an error = 0 in your ANOVA table in Minitab, it typically indicates that your model is saturated. This means that all the degrees of freedom (DF) have been used to estimate the model terms, leaving none to estimate the error term. As a result, Minitab cannot compute the Mean Square Error (MSE), F-values, or p-values, and these may appear as asterisks (*) in the output.
Understanding the Saturated Model
In your case, you have a 2-level, 3-factor full factorial design with 4 replicates per treatment combination, totaling 32 observations. If you include all main effects and interactions (A, B, C, AB, AC, BC, ABC) in your model, the degrees of freedom are allocated as follows:
With 32 observations, the total DF is 31 (n – 1). Subtracting the model DF (7), you should have 24 DF for error. However, if your model includes all possible terms and interactions, and the data perfectly fits the model (i.e., no variability left unexplained), the residual error becomes zero, leading to an error DF of zero.
How to Address This Issue
To resolve this, consider the following steps:
By simplifying the model or increasing the data, you can obtain a more reliable estimate of the error term, enabling Minitab to compute the necessary statistics for your ANOVA.
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Full factorial designs do not adequately account for any factor interactions that may exist. Pls explain about this?
Hi Tharanga,
I’m unsure where you are seeing that. My text above states that full factor design is exhaustive. “This exhaustive approach makes it impossible for any interactions to be missed as all factor interactions are accounted for.”
Best, Ted.