how to decide the number of situation

[fengjie]

If alternative is 1, how to decide the number of situation
If alternative is 1, how to decide the number of situation

Dear all,

Thank goodness, I have received your confirmation timely.
The Ngene manual explains that "the number of choice situation is bounded from (J-1)S=K(p60, section 5.2.2)." I assume this means, in general designs, we can use this formula to obtain the number of situation. But, if there is only 1 alternative in the design, how could I get the number of the situation, and how to design the experiment ?
My design:
The only one alternative is flight, there are seven general attributes(A,B,C,D,E,F,G)
I would like to use the Eff design method for an unlabeled choice experiment design, well, I did not know all the priors, but I have expectations about the positive and negative of some parameter values.
The function of the flight utility is
U(flight)=b1+b2*A[0,1,2,3,4]+b3*B[0,1,2,3,4]+b4*C[0,1,2,3,4]+b5[+]*D[0,1,2,3,4]+b6[+]*E[0,1,2,3,4]+b7*F[0,1,2,3,4]+b8*G[0,1,2]
In the beginning, I did not know how to calculate the number of situation for just one alternative, so,I take out the attribute F(airline) as the alternatives,so the unlabeled choice experiment became a labeled choice experiment ,and the other attributes remained.So, the design of the labeled experiment is :
Design
;alts= alt1, alt2, alt3, alt4, alt5
;rows=15
;eff=(mnl, d)
;model:
U(alt1)=b1+b2*A[0,1,2,3,4]+b3*B[0,1,2,3,4]+b4*C[0,1,2,3,4]+b5[+]*D[0,1,2,3,4]+b6[+]*E[0,1,2,3,4]+b8*G[0,1,2]/
U(alt2)=b2*A+b3*B+b4*C+b5*D+b6*E+b8*G/
U(alt3)=b2*A+b3*B+b4*C+b5*D+b6*E+b8*G/
U(alt4)=b2*A+b3*B+b4*C+b5*D+b6*E+b8*G/
U(alt5)=b2*A+b3*B+b4*C+b5*D+b6*E+b8*G$
Questions:
1.If I still want to get the unlabeled design, and the alternative is just flight, how to decide the design?
2.if I take out the “F” attribute(airline) as the alternatives, then, the design changed to the labeled one. According to the Ngene manual, the minimum number of the choice situation is S=K/J-1=7/4, which is not a integer, so, how to decide the S?
3.I have browsed all the posts about "the prior" for the Efficient Design, and you have gave the solution for their design, but I still have some question on the priors and I hope you can give me some suggestion on my design. I have expectations about the positive and negative of some parameter values.(b5 and b6 are positive parameter).
I am sorry, questions might look a little messy! but I 'm also good in confusing myself.I am looking forward foe your early reply.
Thanks a lot.
Feng

[Michiel Bliemer]

The number of choice tasks S should be such that (J-1)*S >= K, where J is the number of alternatives, and K is the number of parameters.

If J = 5, and K = 7, then S needs to be at minimum 2 choice tasks.

Note that you cannot have just 1 alternative in a choice experiment, because then there is no longer a choice. You always need at least 2 alternatives. So if you have an unlabelled experiment with 2 alternatives, flightA and flight B (in which airline is a dummy-coded attribute), you will get something like:

;model:
U(flightA) = b1.dummy * airline[1,2,3,4,5] + b2 * A[0,1,2,3,4] + .... /
U(flightB) = b1.dummy * airline[1,2,3,4,5] + b2 * A{0,1,2,3,4] + ....

In this case, J = 2, K = 11 (7 attributes plus 4 dummy coded variables), such that S needs to be a minimum of 6.

If you use 3 alternatives, then the minimum number of choice tasks is 4.

Priors you typically determine by doing a pilot study. If you know the signs but not the actual values, you can simply put in a small positive or negative value, i.e. b2[-0.0001]. This will not effect the efficiency much, since it is still effectively zero, but you can use the signs to remove dominant alternatives by using ;alts = flightA*, flightB*

Michiel

[fengjie]

Dear Michiel,
Thanks for your prompt reply, I still have some questions.
If I have decided to conduct a labeled experiment using Efficient Design, which includes 5 alternatives,6 general alternative attributes, and the attributes levels of the 6 attributes were 5,5,5,5,5,3.
1.In the Efficient Design, whether the level balanced design is better than the unbalanced?
2.If I want to keep the design balance,then the number of choice situation is the least common multiple of five and three,is this correct?
3.If the question 1 is correct, then the Ngene will show a design with 15 choice situation, and in each situation,there are five alternatives, I do believe that it is a little difficult to require respondents to make 15 five select one in one time,so I am eager to know, whether I can assign the 15 situations randomly? Because I have not read the block function in the Efficient Design.Whether it is necessary to block the 15 situations?
4.If I would like to add the constant in utility function, I know the number of the constants is J-1, but how to arrange the constant?
Thanks in advance for your suggestion.
Feng

[Michiel Bliemer]

1. Level balance is a nice property, but completely level balance is not necessary. As long as you cover the attribute space sufficiently then strict level balance is not required.
2. Correct, so S = 15 is the smallest design satisfying strict attribute level balance.
3. Having 15 choice tasks with 5 alternatives is likely too much for your respondents. It would be best to block the design in blocks of 5 choice tasks, so each respondent only answers 5 of them. This can be done using ;block = 3. Blocking ensures that each respondent sees more or less a mix of attribute levels (i.e. a high level of attribute level balance within each block), to ensure that a single respondent does not only see high or low levels.
4. For a labelled experiment, it does not matter where you put the 4 constants. The alternative without a constant is the base alternative and all other constants are interpreted relative to this alternative. If you have an unlabelled experiment (in which airline is an attribute), then you do not include any constants.

Michiel

[fengjie]

Hi, Michiel,
Thanks for your patient guidance, I was still confused about some problems, I hope you can give me some advice later.
If J=5,K=7, and attributes levels were three levels and five levels mixed.
1.According to the Ngene manual, such that S needs to be a minimum of 2, so,in the unbalanced design, I set ";rows=2" in the syntax, and the output showed "D-error is 885724.56", then I set";rows=6",and the output showed "D-error is 0.048852.
My question is, if the design is unbalanced, how to decide the "rows" in the syntax, is it a multiple of the minimum of S?
2.The Ngene manual explains that "It is possible that one would like to have a no-choice alternative, this alternative should be added in the alts property, but not have a utility function in the model property. Ngene automatically recognizes the alternative without a utility function as a no-choice alternative.(P107, section 7.2.1)." I understood that if I want to add "no-choice" in the design, then,I should change the syntax from ";alts=alt1, alt2,alt3,alt4,alt5" to ";alts=alt1, alt2,alt3,alt4,alt5,alt6",then, the J=6,and I would use "J=6" to calculate the S and set the property "rows" in the syntax?
3.When using the ‘wtp’ property, the "D-error"of the Efficient Design changed into 6.821231,well, the "D-error" for the same design without "wtp" property was 0.04,is this change of "D-error" normal?(;eff(mnl,d) changed to ;eff(mnl,wtp(wtp1)) ;wtp=wtp1(b5,b6/b2)
4. The Ngene manual explains that" the design for the panel MMNL model is requires sampling of respondents. The number of sampled (simulated) respondents is set by the rep property,the higher this value, the more accurate the computations"(P112 section 7.2.3). What I have been worried about that, if I had set ";rep=500" for the accuracy of the design, but the number of questionnaires did not reach to 500, wether it had an impact on the parameter calibration of the model ?
Thank you in advance, looking forward for your reply.
Fengjie

[Michiel Bliemer]

1. If the design does not need to be balanced, you can set ;rows to any number larger than or equal to the minimum number S according to S*(J-1)>=K. It does not need to be a multiple of anything. If you require balancedness, S needs to be divisible by 3 and 5.
2. If you add a no-choice, then you indeed add another alternative. Note that this alternative may have a constant or may be set to zero. This depends on the model you would like to estimate. See other posts on this topic.
3. D-error values cannot be compared between different studies, designs, different models, and different efficiency measures (like wtp). The only rule is the lower the better, but sometimes 0.5 is very low, in other cases 0.01 is still very high. We have mentioned this is several other posts on this forum.
4. ;rep only internally simulates respondents, but the number of choice tasks of course remains what you have set in ;rows. The simulation of respondents is only needed to calculate the efficiency of the design, it does not make the design larger.

[sukunta]

Dear all,
I am master degree student faculty of pharmacy Khon-Kaen university Thailand. I don't know How to decide the number of situation about my model that

Design
;alts=cer, ba
;rows=12
;eff = (MNL, D)
;model:
U(cer) =b1.dummy[0 I 0]*Inf1 [1,2,3] +b2.dummy[0I0I0]*Logis1 [1,2,3,4] +b3[0]*salarycer[8985,13800] /
U(ba) = b0+b4.dummy[0I0I0]*Info2 [4,5,6,7] +b5.dummy[0 I0I0]*Logis2 [5,6,7,8] +b6[0]*salaryba[11000,15000,18000] $
info1= informatic competency of certificated PhT
Logis1= logistic competency of certificated PhT
info2= informatic competency of bachelor PhT
Logis1= logistic competency of bachelor PhT
salary= salary of PhT each alternative

when I ran Ngene it shown me to change the rows to 14 because I have 6 attributes and 2 labeled + degree of freedom=9 and want to level balance then =12 why it shown must 14 choice set ? and could I reduce no.of choice set and How many minimize choice to cover maximize parameter? and How to?
thank you for your kindness
Beat Wishes,
Sukunta

[Michiel Bliemer]

Dear Sukunta,

Please note that it is best to create a new topic for each new question, even though your question is related.

In order to determine the minimum number of choice tasks, the rule of S*(J-1)>= K needs to be applied. K is the number of coefficients (which is NOT necessarily the same as the number of attributes!), J is the number of alternatives, and S is the number of choice tasks.

In your case, J = 2, and K = 2 + 3 + 1 + 1 + 3 + 3 + 1 = 14 coefficients, of which there is one constant, two coefficients for continuous attributes, and eleven dummy variables. Then find S such that S*(2-1)>=14, means that S needs to be greater or equal to 14.

Since your attributes have 2, 3, and 4 levels in order to maintain attribute level balance you would need to go to S = 24, although it is not mandatory to impose attribute level balance.

If you add ;block = 2, then Ngene will split the design in two parts, e.g. 2 times 7 choice tasks. You can increase the number of blocks to decrease the number of choice tasks presented to a single respondent.

Michiel

[sukunta]

Dear Micheil Bliemer,

thank you very much for your kind and quick reply my question.

Best Wishes,
Sukunta