choice-metrics.com

Hello,

Two questions for one case. Let's take the transport example
Design
;alts(model1) = car, bus, train
;rows = 12
;eff = model1(d)
;alg = swap(reset=10000, resetinc=5000)
;model(model1):
U(car) = /
U(bus) = /
U(train) =
1- Is it possible to request for different number of alternatives?
for example..I can have 1 or 2 trains (night/day) to propose,
2 Is it possible to build a DoE proposing two levels of confort
U(train)=train1*price[100:200] + train2*confort[0,1]+ train3*additionalpriceconfort[20:30]

Thanks a lot for your help
Best
Naji

1. You can have a different number of alternatives in different models. You can make a heterogeneous design or use a modelling average approach. You cannot change the number of alternatives within a design/model.

2. You can specify the utility function as you like, Ngene does not have many restrictions. Note that you cannot specify multiple continuous attribute levels as it does not make sense (in the end only the end point will be used). A single continuous attribute can be used which is sufficient to optimise probabilities.

Thanks Bliemier

1- Sorry, I forgot I can specify different models within one DoE... It is described in the manual page 130-33..

2- For the second question, one will choose the Train / departure time (day/night) but then he can choose 1st, 2nd class .. How can I build such DoE
To be more precise, the respondent will be facing 3 alternatives, but he can make up to 6 choices... Should I specify in Ngene, that I have 6 alternatives and define the conditions that all attributes (except Comfort) should be the same between alt1&2, 3&4, 5&6?

Best
Naji

I do not understand what you mean with 3 alternatives and 6 choices.
Can you write out the utility functions that you wish to estimate?

tried to reply several times.. nothing appeared, sorry for the multiple posting

Bliemer


I made a confusion between alternatives and choices.
I want to build a choice situation with 4 alternatives that can be structured into 2 pairs (alt1&2, 3&4)
The attributes are brand, price, comfort, and additional price for comfort
Alt1 & Alt3 has no comfort option, 2 & 4 are same alternatives (1 & 3) with comfort at higher price..

Design
;alts = altl,alt2,alt3,alt4,none
;rows = 600
;block= 25
;eff = (mnl,d)
;cond:
if (alt1.brand=1 , alt2.brand=1),
if (alt1.brand=2 , alt2.brand=2),
if (alt1.brand=3 , alt2.brand=3),
if (alt1.price=100, alt2.price=100),
if (alt1.price=120, alt2.price=120),
if (alt1.price=145, alt2.price=145),
if (alt1.price=175, alt2.price=175),
if (alt1.price=210, alt2.price=210),
if (alt1.comfort > 0, alt1.comfort = 0),
if (alt2.comfort = 0, alt2.comfort > 0),
if (alt3.brand=1 , alt4.brand=1),
if (alt3.brand=2 , alt4.brand=2),
if (alt3.brand=3 , alt4.brand=3),
if (alt3.price=100, alt4.price=100),
if (alt3.price=120, alt4.price=120),
if (alt3.price=145, alt4.price=145),
if (alt3.price=175, alt4.price=175),
if (alt3.price=210, alt4.price=210),
if (alt3.comfort > 0, alt3.comfort = 0),
if (alt4.comfort = 0, alt4.comfort > 0),
;model:
U(alt1)=a.dummy[0|0]*brand[1,2,3] + b.dummy[0|0|0]*comfort[0,1,2,3] + c*price[100,120,145,175,200] /
U(alt2)=a*brand + b*comfort + c*price + d*additional[20:50] /
U(alt3)=a*brand + b*comfort + c*price /
U(alt4)=a*brand + b*comfort + c*price + d*additional
$

Hope it's clearer
Naji

It is perhaps easier to use scenarios, for example you can use

U(train1) = b1*comfort1[0,1,2] + ... /
U(train2) = b1*comfort2[comfort1] + ...

This is the same as putting in a constraint.

thanks a lot

I came out with the following syntax that works quite fine.

I would like that every respondent faces situations from both design
The ficher rejected that models show different number of alternatives...

Design
;alts(m3) = alt1,alt2,alt3,alt4,alt5,alt6,none
;alts(m2) = alt1,alt2,alt3,alt4,none
;rows = 500
;block= 50
;eff = m3(mnl,d) + m2(mnl,d)
;model(m3):
U(alt1)=a * brand1[1,2,3] + c * price1[100,120,145,175,200] /
U(alt2)=a * brand2[brand1] + c * price2[price1] + b * comfort[1,2,3] + d * additional[20:50] /
U(alt3)=a * brand3[1,2,3] + c * price3[100,120,145,175,200] /
U(alt4)=a * brand4[brand3] + c * price4[price3] + b * comfort[1,2,3] + d * additional[20:50] /
U(alt5)=a * brand5[1,2,3] + c * price5[100,120,145,175,200] /
U(alt6)=a * brand6[brand5] + c * price6[price5] + b * comfort[1,2,3] + d * additional[20:50]
;model(m2):
U(alt1)=a * brand1[1,2,3] + c * price1[100,120,145,175,200] /
U(alt2)=a * brand2[brand1] + c * price2[price1] + b * comfort[1,2,3] + d * additional[20:50] /
U(alt3)=a * brand3[1,2,3] + c * price3[100,120,145,175,200] /
U(alt4)=a * brand4[brand3] + c * price4[price3] + b * comfort[1,2,3] + d * additional[20:50]