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  • #46
    Originally posted by Arne Risa Hole View Post
    Dear Sarah,

    As demonstrated in the dcreate help file, in many applications a sensible way of creating an opt-out alternative is to set the attribute levels for the fixed alternative to the lowest level for the other alternatives and include an ASC for the opt out alternative. As Ric Scarpa points out above this is not always appropriate - sometimes you want the opt-out to have other attribute levels - but you cannot give it attribute levels that do not appear in the other alternatives, as such a design is not identified. If attribute level "1" only appears in the opt-out you cannot identify the difference between attribute level "1" and "2" (for example) - this difference will be captured by the ASC. How you present the opt-out to the respondents in your survey is a different matter - the utility of the opt-out is different from the utility of another alternative with all of the attributes set to the lowest level due to the ASC, so you do not need to present the opt-out alternative in those terms (it could for example simply be a tick-box saying "neither").

    I can't help more than this I'm afraid - good luck.

    Arne
    Thanks for your response Arne. If I understand you correctly, my opt-out would be fixed at the lowest levels of alt1 and alt2, but when I present it to respondents, I just need to include choice sets where Alt 1 and Alt 2 do not have the lowest level?

    Comment


    • #47
      Dear Arne,

      First, thank you very much for the very helpful dcreate module!

      I encouter a curious situation where two choice sets come out identical when I attempt to generate a specific design (similar to Example 3 in the help file, with the additional opt-out alternative with associated ASC).

      Here is my code:

      Code:
      matrix levmat = 3,2,2
      genfact, levels(levmat)
      matrix optout = J(1,3,1)
      matrix b = J(1,9,0)
      dcreate i.x1##i.x2 i.x1##i.x3, nalt(2) nset(30) fixedalt(optout) asc(3) bmat(b)

      When the algorithm converges (after 3 iterations), choice sets #5 and #7 are exactly the same. Is this normal/to be expected?

      Thanks again,

      Claudio

      Comment


      • #48
        Claudio

        Yes, with simple designs (few attributes/levels) with many choice sets this can happen - there is nothing in the dcreate code that explicitly prevents it. Note that in your case you can only construct 144 different choice sets, and some of them will not convey any information (for example a choice set with two identical alternatives), so will not be selected by dcreate.

        Arne

        Comment


        • #49
          Originally posted by Arne Risa Hole View Post
          Claudio

          Yes, with simple designs (few attributes/levels) with many choice sets this can happen - there is nothing in the dcreate code that explicitly prevents it. Note that in your case you can only construct 144 different choice sets, and some of them will not convey any information (for example a choice set with two identical alternatives), so will not be selected by dcreate.

          Arne
          Thank you Arne for your quick response!

          Too bad there is nothing in the dcreate code to prevent generating duplicate choice sets.

          I know that 66 unique 2-alternative choice sets can be constructed in my case (12*11/2). I admit that I haven't investigated how the modified Fedorov algorithm works, but I didn't expect to have duplicates using the dcreate module with fewer choice sets than this number.

          For your information, after playing a bit with the command by changing the number in nset to 24 and 36, and examining the results explicitly for choice sets containing the same two alternatives (either as alt=1 or alt=2 in the choice set), I obtained these results:

          - With nset(24): 4 sets were duplicated twice ; 2 sets were duplicated thrice.
          - With nset(36): 1 set was duplicated thrice ; 3 sets were duplicated 4 times ; and 2 sets were duplicated 5 times.

          If I understand correctly, this would mean that 8 (4*1 + 2*2) and 19 (1*2 + 3*3 + 2*4) choice sets, respectively, do not add any new information over the remaining unique sets in these generated designs, translating into 33% (8/24) and 53% (19/36) wasted choice sets (again, respectively).

          In such cases, would you recommend removing the duplicate choice sets or trying to generate an efficient design with fewer choice sets, or something else entirely?

          Thank you again,

          Claudio


          Comment


          • #50
            Claudio

            I personally don't really see the issue here - I would just generate a design with fewer choice sets. Both 24 and 36 are way more than you need, and the repeated choice sets reflect that.

            Arne

            Comment


            • #51
              Originally posted by Arne Risa Hole View Post
              Claudio

              I personally don't really see the issue here - I would just generate a design with fewer choice sets. Both 24 and 36 are way more than you need, and the repeated choice sets reflect that.

              Arne
              Thank you Arne for your helpful insights.

              Claudio

              Comment


              • #52
                Originally posted by Arne Risa Hole View Post
                Dear Mariya,

                I am glad to hear that you have found dcreate useful. A design with a higher D-efficiency is more efficient than a design for the same experiment with a lower D-efficiency, but the numbers themselves don't mean much as they depend on things like the number of attributes, levels etc. in your choice experiment.

                Arne
                Thany you for this explanation. Therefore how could I produce several designs for the same experiment to compare the D-efficiencies, would it be by changing the seed? How many should I compare?

                Comment


                • #53
                  Originally posted by Arne Risa Hole View Post
                  Thanks Eleanya. To answer your question, if all your attributes are categorical your matrix should be "matrix b = J(1, 9, 0)", which is the same as "matrix b = 0,0,0,0,0,0,0,0,0" (see help matrix functions).

                  See also post #9 in this thread.

                  Arne
                  Dear Arne, Can you please explain the numbers in matrix b = J(? ? ?)? for beginners

                  Comment


                  • #54
                    Dear Arne,
                    Many thanks for the dcreate command. It has been very helpful. I tried the example below and I got the message below (There are 11 effects in the design and 9 coefficients in b)

                    . matrix levmat = 4,3,3,3,2,2
                    . genfact, levels(levmat)
                    . matrix b = J(1, 9, 0)
                    . dcreate i.x1 i.x2 i.x3 i.x4 i.x5 i.x6, nalt(2) nset(16) bmat(b)
                    There are 11 effects in the design and 9 coefficients in b

                    Secondly, the example below also gave a similar message
                    . clear
                    . matrix levmat = 4,3,3,3,2,2
                    . genfact, levels(levmat)
                    . matrix b = J(1,10,0)
                    . dcreate i.x1 i.x2 i.x3 i.x4 i.x5 i.x6, nalt(2) nset(16) bmat(b)
                    There are 11 effects in the design and 10 coefficients in b

                    How should I solve this?
                    Thanks

                    Comment


                    • #55
                      Kofi

                      See posts #9 and #28 in this thread. You need as many values in your b matrix as the number of effects to be estimated.

                      Arne

                      Comment


                      • #56
                        Dear Arne

                        I am using DCREATE to design a DCE.
                        I have 6 attributes and used the following syntax

                        // x1-installcost (Shillings-continuous), x2-reliability (%-continuous), x3-durability (Years-continuous), x4-maintenancecost (categorical), x5-movability (categorical), x6-defectID (categorical)

                        Code:
                        clear
                        matrix levmat = 3,3,3,3,2,2
                        genfact, levels(levmat)
                        matrix optout = J(1,6,1)
                        matrix b = -0.0000135, 1.6, 0.05, -0.07, -0.2, 0.83, 0.6
                        dcreate c.x1 c.x2 c.x3 i.x4 i.x5 i.x6 , nalt(3) nset(16) fixedalt(optout) bmat(b)
                        blockdes block, nblock(2)
                        evaldes c.x1 c.x2 c.x3 i.x4 i.x5 i.x6, bmat(b)

                        In the resulting experimental design, only 2 levels are represented (rather than the 3) for the attributes installcost & durability. Does this mean that I should just go ahead and use the extremes (the 2 levels) and leave out one level (the middle level) for these two attributes in my choice sets?

                        Thank you.
                        Last edited by Stephen Mailu; 07 Aug 2019, 04:14.

                        Comment


                        • #57
                          Stephen

                          See post #33 in this thread. It often makes sense to treat the attributes as categorical at the design stage even if you plan to model them as continuous. This also ensures that all the levels of the attribute are present in the design.

                          Arne

                          Comment


                          • #58
                            Originally posted by Arne Risa Hole View Post
                            Stephen

                            See post #33 in this thread. It often makes sense to treat the attributes as categorical at the design stage even if you plan to model them as continuous. This also ensures that all the levels of the attribute are present in the design.

                            Arne
                            Dear Arne

                            Thank you for the response. It does work and the levels that had been dropped are all included now when I specify all as categorical variables . I do hope that it matters not whether I intend to run a mixed logit or a conditional logit or even a latent class model etc for data analysis?

                            Comment


                            • #59
                              Originally posted by Stephen Mailu View Post
                              I do hope that it matters not whether I intend to run a mixed logit or a conditional logit or even a latent class model etc for data analysis?
                              Well, yes and no. The design is optimised for the conditional logit model, but in practice it is a common design strategy to generate a design for the conditional logit and then go on to estimate more flexible models after collecting the data.

                              Comment


                              • #60
                                Originally posted by Dimitris Damigos View Post
                                Dear Arne,

                                First of all thank you so much for providing this useful module. I want to design a DCE and I would mostly appreciate your help in clarifying a couple of issues.

                                The number of choice sets in the design has to be specified by the user. The minimum number of choice sets should be equal or greater than the degrees of freedom of the model. My model has 17 degrees of freedom and thus specifying 18 sets seems to be o.k. (I have some three-level attributes). Nevertheless, the minimum runs required for my model (based on the number of attributes and their levels) when searching for an OMEP to represent the profiles in the first option in the choice sets, is 36.

                                Do I have to try a number of different candidate sets (starting from the minimum number of choice sets required according to the df of the model) and use the one with the highest efficiency or is it more appropriate to use the minimum number of sets determined by the orthogonal array?

                                Thank you in advance for your time.

                                Kind regards,

                                Dimitris
                                Hallo Arne,
                                Can you explain about minimum number of nset, how many choice set should be there.

                                Comment

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