unbalanced number of alternatives

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unbalanced number of alternatives

Postby mburton » Wed Aug 17, 2011 5:30 pm

I am trying to construct a model where I have a heterogeneous design, with different numbers of alternatives in each model. Page 133 of the manual (version 1.1) suggests its possible, but does not show an explicit example. The syntax below generates the error

Error: The model 'bad' that belongs to the 'fish' ;fisher specification is inconsistent with the first model. The number of alternatives does not match.

where am I going wrong?

Michael

Design
;alts(good) =alt1, alt2, alt3, alt4
;alts(bad) = alt1, alt2, alt3
;rows = 8

;eff = fish(mnl,d)
;fisher(fish) =des1(good[0.5]) + des2(bad[0.5])


;model(good):
U(alt1) = b1[-0.2]* price[10,20,30,40] + b2[0.2]*a1[0,1]+b3[0.2]*a2[0,1]+b4[0.2]*a3[0,1]/
U(alt2) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt3) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt4) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]


;model(bad):
U(alt1) = b1[-0.2]* price[10,20,30,40] + b2[0.2]*a1[0,1]+b3[0.2]*a2[0,1]+b4[0.2]*a3[0,1]/
U(alt2) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt3) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]$
mburton
 
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Joined: Wed May 27, 2009 7:21 pm

Re: unbalanced number of alternatives

Postby johnr » Wed Aug 31, 2011 6:04 pm

We are currently working on the theory for designs assuming different numbers of alternatives. In the meantime, the syntax you want should look like this where the seperation occurs via the eff command, not the fisher:

Design
;alts(good) =alt1, alt2, alt3, alt4
;alts(bad) = alt1, alt2, alt3
;rows = 8

;eff = good(mnl,d) + bad(mnl,d)
?;fisher(fish) =des1(good[0.5]) + des2(bad[0.5])

;model(good):
U(alt1) = b1[-0.2]* price[10,20,30,40] + b2[0.2]*a1[0,1]+b3[0.2]*a2[0,1]+b4[0.2]*a3[0,1]/
U(alt2) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt3) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt4) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]

;model(bad):
U(alt1) = b1[-0.2]* price[10,20,30,40] + b2[0.2]*a1[0,1]+b3[0.2]*a2[0,1]+b4[0.2]*a3[0,1]/
U(alt2) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt3) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]$
johnr
 
Posts: 168
Joined: Fri Mar 13, 2009 7:15 am

Re: unbalanced number of alternatives

Postby tibor » Wed Dec 12, 2012 9:49 pm

I have a similar problem to the one described below - unbalanced numbers of alternatives. seperating the models via the eff command works fine in the situation below. However, if I want to have different attribute levels in each model I believe that this procedure will not work anymore. Is this correct?

johnr wrote:We are currently working on the theory for designs assuming different numbers of alternatives. In the meantime, the syntax you want should look like this where the seperation occurs via the eff command, not the fisher:

Design
;alts(good) =alt1, alt2, alt3, alt4
;alts(bad) = alt1, alt2, alt3
;rows = 8

;eff = good(mnl,d) + bad(mnl,d)
?;fisher(fish) =des1(good[0.5]) + des2(bad[0.5])

;model(good):
U(alt1) = b1[-0.2]* price[10,20,30,40] + b2[0.2]*a1[0,1]+b3[0.2]*a2[0,1]+b4[0.2]*a3[0,1]/
U(alt2) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt3) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt4) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]

;model(bad):
U(alt1) = b1[-0.2]* price[10,20,30,40] + b2[0.2]*a1[0,1]+b3[0.2]*a2[0,1]+b4[0.2]*a3[0,1]/
U(alt2) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]/
U(alt3) = b1* price[10,20,30,40] + b2*a1[0,1]+b3*a2[0,1]+b4*a3[0,1]$
tibor
 
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Joined: Thu Jun 17, 2010 10:50 pm

Re: unbalanced number of alternatives

Postby Michiel Bliemer » Thu Dec 13, 2012 8:45 am

In case all attributes have different levels, I believe you could just generate two separate designs, there is no reason to put them together in a single design I think? If some attributes have the same levels, while others are different, then you can use the above approach, but you have to name the attributes with different levels differently.
Michiel Bliemer
 
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Joined: Tue Mar 31, 2009 4:13 pm

Re: unbalanced number of alternatives

Postby tibor » Fri Dec 14, 2012 12:53 am

Just to sum up this discussion: I can create heterogeneous designs (vie the eff-command) which have different numbers of alternatives for different respondents. All I got to know is the proportion of each group of respondents (to be used in the eff-command line) and the different model specifications. When implementing the survey the CE is conditioned such that each respondent gets the corresponding model, i.e. some will for instance have choice sets with 3 alternatives and others with 4. If some alternatives have different attribute levels I can accommodate this by defining new variables.

Correct?

Thanks for your support, it's really appreciated!
Cheers!
Tibor
tibor
 
Posts: 7
Joined: Thu Jun 17, 2010 10:50 pm

Re: unbalanced number of alternatives

Postby Michiel Bliemer » Mon Dec 17, 2012 3:44 pm

Yes that is correct.
Michiel Bliemer
 
Posts: 1733
Joined: Tue Mar 31, 2009 4:13 pm


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