Basic Question - Central Composite Design

[#1]

Hello, I wondered if I could ask a question about orthogonality of the squared term columns in a Central Composite Design (CCD) with two factors A and B. The dot product of any pair of columns in the model matrix is always zero - I always understood this to mean that the column (vectors) of say A with B or B with AB or B with B^2 are orthogonal (at right angles to) each other so their beta's can be calculated independently. The one exception is column A^2 with B^2 - their dot product can never be zero as you have -1 and/or +1 in 4 rows and they all become +1 when you square them which makes me think that their effects (the betas) would be aliased (so wouldn't be able to determine the coefficients for the A^2 and B^2 term independently from each other). I suspect I'm misunderstanding the meaning of orthogonality and that you can estimate these parameters independently of each other and of the other terms in the model otherwise what would be the point of the CCD design. Can anyone help with my lack of understanding please? Thankyou 

[#2]

Read this:

https://www.jmp.com/support/help/en/18.1/?os=mac&source=application#page/jmp/choose-a-design.shtml%2...

[#3]

Hi Statman, Thanks for the link. It was an interesting read but I couldn't find an answer to my question. I thought I'd draw a 2 factor CCD to explain. As you can see, any pair of columns are orthogonal to each other (dot product is zero, i always thought that was the criterion for orthogonality but maybe it's not - see below). The only exception is that for the A^2 and B^2 columns the dot product can never be zero and so I'm interpreting this (probably incorrectly) that those factors are not orthogonal. However, if I plot A^2 and B^2 as a pair of points, they all sit at the 4 corners of a square and if you try to fit a trend line you get an R value of zero. Maybe I've answered my question - Is it the case that it's the R=0 which means A^2 and B^2 are orthogonal to each other and so their betas can be independently estimated. So a non zero dot product of two column vectors doesn't mean non-orthogonality

[#4]

For this to happen, the axial points have to be at ±Sqrt(2) in a 2 factor RSM. Ignore my earlier post, sorry for the confusion.

[#5]

Reading my response I realize I didn't answer your question. Rather than muddling things even more by editing it, let me add another comment. Orthogonality is a function of the entire design and not just any subset of columns. If X is the design matrix then the design is orthogonal if all off-diagonal elements of Inverse(Transpose(X)*X) are zero except those associated with the intercepts. In this case * corresponds to matrix multiplication.

[#6]

Hi Don,

            Thanks for your reply. This is really helpful. I decided to watch a few more videos on CCD's and read through what I could find online and in my DoE books over the weekend. I thought I'd look at the CCD case (for 2 factors) where the design is both rotatable and orthogonal. This is a special case I believe in that there are two conditions: 1. alpha = sqrt(2) = 1.414 for obvious reason (all not centre points on circle / equidistant from centre) and 2. There needs to be a certain number of centre points which I calculated to be nc=8. However, when I do the matrix operations in Excel (see below) there are two elements that aren't zero. Maybe by nc calculation is wrong and nc isn't 8, maybe Excel isn't so great at matrix arithmetic (but have no reason to suspect that), maybe I've got it wrong about conditions 1 & 2 above being those that give a design with both orthogonality and rotatability? Any idea why I can't get all off-diagonal elements (excepting the first row that's associated with the intercept) to be zero? Many Thanks

[#7 · Accepted solution]

The first column/row is the intercept and is ignored when considering orthogonality. I am fairly certain that for common designs it will always be correlated with one or more effects in the model.

[#8]

Thanks Don. You're right and the conditions alpha = sqrt(2) and nc=8 (nc = sqrt(nf) + 4 - k where nf = number of factorial (corner) points = 4 and k = number of factors = 2) do give a rotatable and orthogonal CCD in 2 factors.
I tried nc = 7, 8 and 9 and nc=8 was the only one that gave all off-diagonal elements of zero (not counting the first row and column associated with the intercept). Having only spent time on full and fractional factorial designs at 2 levels up to this point I thought that a design was orthogonal if the dot product of all pairs of columns was zero then that meant it was an orthogonal designs as those column vectors were at right angles to each other. I think that's to do with aliasing and not orthogonality maybe.

[#9]

I should add that I'd like to figure out if it's possible to have a rotatable and orthogonal ((XTX)^-1) with all off-diagonal elements zero) for a single block case (as if i'd carried out all the runs at one time but random order, rather than the more likely case where you'd augment the factorial points (subset of larger screening design) with start and centre points (so you'd have two blocks). This discussion is more about orthogonality / aliasing / dot product between terms in the design matrix than about CCD's. Thanks