Getting the same attribute level pair during my design

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Getting the same attribute level pair during my design

Postby neeraj85 » Tue Feb 03, 2015 6:37 pm

Hi,

I am trying to build an efficient design for an unlabelled experiment having 2 alternatives. Following is the script:

Design
;alts = alt1*, alt2*
;rows = 15
;eff = (mnl,d)
;model:
U(alt1) = b2[-0.001] * tt[0.8, 0.9, 1.0, 1.1, 1.2] + b3[-0.002] * tts[0.3, 0.4, 0.5, 0.6, 0.7] + b4[-0.003] * sn[0.8, 1.0, 1.2, 1.4, 1.6] + b5[-0.004] * vr[0.75, 0.875, 1.0, 1.125, 1.25] /
U(alt2) = b2 * tt + b3 * tts + b4 * sn + b5 * vr $


Following is one of the output that I got:

Choice situation alt1.tt alt1.tts alt1.sn alt1.vr alt2.tt alt2.tts alt2.sn alt2.vr
1 1.2 0.7 1 1 0.8 0.3 1.4 1
2 0.8 0.5 1.2 1.125 1.2 0.5 1.2 0.875
3 0.9 0.6 0.8 0.875 1.1 0.4 1.6 1.125
4 0.9 0.6 1.2 1.25 1.1 0.4 1.2 0.75
5 1.1 0.4 1.2 1.25 0.9 0.6 1.2 0.75
6 1.2 0.4 1.6 1.25 0.8 0.7 0.8 0.75
7 1 0.5 1.6 0.75 1 0.5 0.8 1.25
8 1 0.7 1.4 0.875 1 0.3 1 1.125
9 1 0.6 1.4 0.875 1 0.4 1 1.125
10 0.8 0.4 1 1 1.2 0.6 1.4 0.875
11 0.9 0.5 1.6 0.75 1.1 0.5 0.8 1.25
12 1.1 0.3 0.8 1 0.9 0.6 1.6 1
13 1.2 0.3 0.8 0.75 0.8 0.7 1.6 1.25
14 1.1 0.7 1 1.125 0.9 0.3 1.4 0.875
15 0.8 0.3 1.4 1.125 1.2 0.7 1 1

MY OBSERVATION: For the attribute "tt" the levels are always occurring in pairs (1.2 , 0.8) (1.1 , 0.9) (1 , 1) or the other way around. Similar thing can e observed with the attriute "tts" "sn" and "vr"

I looked into other outputs as well but the same issue persists in each one of them.


DOUBTS:
1. Why Ngene didn't generate, say (1.2 , 0.9) for the "tt" attribute?
2. Did I specifically code this condition accidentally in my script?


I'll be grateful if anyone can help me clarify the doubts.

Thanks
Neeraj
neeraj85
 
Posts: 31
Joined: Wed Jan 28, 2015 4:58 pm

Re: Getting the same attribute level pair during my design

Postby Michiel Bliemer » Tue Feb 03, 2015 6:52 pm

This is a question that has been asked many times on the forum.
The elements in the variance-covariance matrix are minimised if the Fisher information is maximised. The information from a choice task is largest if the trade-offs made between attributes across alternatives are largest. So comparing the highest level of tt with the lowest value of tt provides the largest trade-off, thereby minimising the standard errors. So this is actually an optimal way, which automatically comes out of the optimisation when maximising the Fisher information matrix in Ngene.

Note that also optimal orthogonal designs (OOD) as proposed by Street and Burgess have this property; their design generators are such that the same attribute levels are always traded off across alternatives.
Michiel Bliemer
 
Posts: 1885
Joined: Tue Mar 31, 2009 4:13 pm


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