# Fractional Factorial Design when some factors have more than 2 levels

Let's say I was conducting an experiment with 4 factors.

Three of them have qualitative settings, so I can simply designate one as +1 and one as -1. But one of the factors is numeric, and I would like to investigate three levels. If the other three were numeric as well I could treat this as a 2^4 with center points, but that's not the case.

The full general factorial is thus 24 design points per replicate. Is there a fractional approach in this case? Any references for how to proceed?

• Can you give some more details? What is your intended sample size? How many replicates do you need? Do you need blocking, and if so, what is the block size? ... Commented Nov 17, 2022 at 16:46
• This is for a school assignment. We've only covered 2^k designs so I just wasn't sure about the one factor having three levels. I would like to try to get through three replicates for better error estimation. I would probably do one replicate per day and block by day.
– jerH
Commented Nov 17, 2022 at 18:43
• If you can do one block of 24 there is no need of fractions. For smaller fractions maybe look into algorithmic design, such as D-optimal (search this site). But since it is an assignment, you should add the tag self-study, and read its wiki. I will try an answer! Commented Nov 17, 2022 at 19:10

There is fractional approaches also in the mixed levels case, but less straightforward than for the $$2^{n-k}$$-designs. One idea is to use orthogonal arrays (OA), see wikipedia.

In your case $$2^3\cdot 3$$ a full factorial have 24 runs, so we can try for a half-fraction, with 12 runs. We use symbols 0,1 for the 2-level factors, 0,1,2 for the 3-level factor. We start writing down columns as we would for a $$2^3$$-design. What is important is that all possible symbol combinations, at least taken two columns a time, will appear with the same frequency (that will give an OA of strength 2).

0 0 0 0
0 0 1 1
0 0 0 2
0 1 1 0
0 1 0 1
0 1 1 2
1 0 1 0
1 0 0 1
1 0 1 2
1 1 0 0
1 1 1 1
1 1 0 2


Note that filling in the third column, reaching halfway down we had to "mirror" the pattern. Note that taking one two-level and the 3-level column, there are $$6=2\cdot 3$$ possible combinations, so the maximum possible strength is two, which is achieved. For two 2-level columns, there are 4 possible combinations, so 3 copies of each is possible (and obtained). So the 3 first columns by itself is an OA of strength 3.