Full additive model

[JK2020]

Hello again,

After my data collection I'll be estimating individual utility functions for the alternatives and an overall utility function that is made up of the alternatives.

For example, if I have 3 alternatives (A, B, and C) I'd have four models. For purposes of simplicity, let's assume they all only have 3 levels. The alternative specific utility models are:

U(A) = A.effect[0 | 0]*A[1,2,3] /
U(B) = B.effect[0 | 0]*B[1,2,3] /
U(C) = C.effect[0 |0 ]*C[1,2,3] $

However, shouldn't I be able to also include the full additive model. I can't find anything in the manual about doing so, I tried adding:

U(A, B, C) = A.effect[0 | 0]*A[1,2,3] + B.effect[0 | 0]*B[1,2,3] + C.effect[0 |0 ]*C[1,2,3]

However, it then states that I have a model that wasn't specified in the ';alts' property. I then considered adding an extra alternative called 'Overall' and using the function:

U(Overall) = A.effect[0 | 0]*A[1,2,3] + B.effect[0 | 0]*B[1,2,3] + C.effect[0 |0 ]*C[1,2,3]

The program then runs successfully, however, then instead of having three columns for my three alternatives (A, B, and C) I have six columns (A, B, C, Overall.A, Overall.B, Overall.C).

So, I'm left wondering if there is a way to include the full additive utility model? Or is it unnecessary?

Many thanks and happy holidays to all!

[Michiel Bliemer]

I am not sure what the purpose is of adding an overall utility? I have never seen this before. You only need to include utility functions for each alternative. Perhaps you are mixing up attributes and alternatives? An alternative consists of multiple attributes. An alternative does not have levels but an attribute does.

With U(A) you define the utility of an ALTERNATIVE, which is a linear function of ATTRIBUTES (similar to what you wrote in the "overall" model formulation), which have LEVELS.

How many options do you let respondents choose between? For example, if you have two options/alternatives, then you have formulate something like:

U(alt1) = A.effect[0 | 0]*A[1,2,3] + B.effect[0 | 0]*B[1,2,3] + C.effect[0 |0 ]*C[1,2,3] /
U(alt2) = A.effect[0 | 0]*A[1,2,3] + B.effect[0 | 0]*B[1,2,3] + C.effect[0 |0 ]*C[1,2,3]

Michiel