by johnr » Thu Jan 15, 2015 9:19 am
Hi Deniz
It will depend on how you specify your syntax, as to what output the program generates. For example
Design
;alts = alt1, alt2
;rows = 12
;orth = sim
;model:
U(alt1) = b1 + b2 * A[0,1] + b3 * B[0,1] /
U(alt2) = b2 * A + b4 * C[2,4,6,8] $
Does not specify any efficiency measure to optimise on, so it will simply give you the first orthogonal design it can find.
Design
;alts = alt1, alt2
;rows = 12
;eff=(mnl,d)
;orth = sim
;model:
U(alt1) = b1 + b2 * A[0,1] + b3 * B[0,1] /
U(alt2) = b2 * A + b4 * C[2,4,6,8] $
Specifies an efficiency measure (d-error), but no priors (so they are set to zero). It will attempt to find the most efficient orthogonal design possible, however it is probable that given the zero priors, the logit probabilities become constants = 1/J and the model approximates a linear model, all orthogonal designs are equally optimal. In this case, again, the program will not be able to locate a better orthogonal design than the first reported.
In the following, non-zero priors are specified as well as an efficiency measure, and the design will likely be able to locate a more efficient orthogonal design. In this case, the program will report each new improved (in terms of efficiency) orthogonal design.
Design
;alts = alt1, alt2
;rows = 12
;eff=(mnl,d)
;orth = sim
;model:
U(alt1) = b1[0.6] + b2[-0.2] * A[0,1] + b3[-0.4] * B[0,1] /
U(alt2) = b2 * A + b4[-0.1] * C[2,4,6,8] $
So to specifically answer your question, there are likely to be multiple orthogonal designs possible, however which design the program reports/locates, will depend on the priors and efficiency measure you specify.
John