unbalanced number of alternatives

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mburton
Posts: 3
Joined: Wed May 27, 2009 7:21 pm

unbalanced number of alternatives

Post by mburton »

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]$
johnr
Posts: 172
Joined: Fri Mar 13, 2009 7:15 am

Re: unbalanced number of alternatives

Post by johnr »

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
Posts: 7
Joined: Thu Jun 17, 2010 10:50 pm

Re: unbalanced number of alternatives

Post by tibor »

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]$
Michiel Bliemer
Posts: 2057
Joined: Tue Mar 31, 2009 4:13 pm

Re: unbalanced number of alternatives

Post by Michiel Bliemer »

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.
tibor
Posts: 7
Joined: Thu Jun 17, 2010 10:50 pm

Re: unbalanced number of alternatives

Post by tibor »

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
Michiel Bliemer
Posts: 2057
Joined: Tue Mar 31, 2009 4:13 pm

Re: unbalanced number of alternatives

Post by Michiel Bliemer »

Yes that is correct.
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