Dear Michiel

I also have another question about the none option in the survey. Could I write your own current situation in the none option or I only can put I prefer none of these in the none option?

Thanks again.

Best

Steve

**Moderators:** Andrew Collins, Michiel Bliemer, johnr

17 posts
• Page **2** of **2** • 1, **2**

Dear Michiel

I also have another question about the none option in the survey. Could I write your own current situation in the none option or I only can put I prefer none of these in the none option?

Thanks again.

Best

Steve

I also have another question about the none option in the survey. Could I write your own current situation in the none option or I only can put I prefer none of these in the none option?

Thanks again.

Best

Steve

- Steven Guu
**Posts:**10**Joined:**Wed Jun 15, 2022 12:54 am

Ngene reports a UTILITY balance of 99%, which has nothing to do with ATTRIBUTE LEVEL balance. The design is essentially 100% utility balanced because you are using zero priors such that the choice probabilities become 33-33-33% across the three alternatives.

You do not need utility balance in the design, nor do you need attribute level balance in the design, to be able to estimate your model. If your D-errors are not infinite, then you can estimate the model. It should be fine for the pilot study.

Michiel

You do not need utility balance in the design, nor do you need attribute level balance in the design, to be able to estimate your model. If your D-errors are not infinite, then you can estimate the model. It should be fine for the pilot study.

Michiel

- Michiel Bliemer
**Posts:**1433**Joined:**Tue Mar 31, 2009 4:13 pm

Dear Michiel

Thank you very much for advice. Sorry for keep asking these simple questions

Following your suggestions, to achieve attributes balance, I consider to use 12 rows, and block 3, in this case, people can keep facing 4 choice cards?

Do you think 16 rows is sufficient for the design?

In addition, if I also change blocks from 3 to 4, people will be shown 3 choice cards, which design do you think is better?

This is my codes

Design

;alts = alt1*,alt2*,none

;rows = 12

;block = 3,minsum

;eff = (mnl,d)

;con

;alg = mfederov(candidates=1000)

;model:

U(alt1) =b0[0]+ b1.dummy[0.001|0.002|0.003] * sorting[2,4,7,1]

+ b2.dummy[0.001|0.002] * collected[2,3,1]

+ b3.dummy[0.001|0.003] * point[2,3,1]

+ b4[-.001] * cost[20,40,60,80,100,200](1-2,1-2,1-2,1-2,1-2,1-2)

/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

$

Thank you very much for your time and help.

Best wishes

Steve

Thank you very much for advice. Sorry for keep asking these simple questions

Following your suggestions, to achieve attributes balance, I consider to use 12 rows, and block 3, in this case, people can keep facing 4 choice cards?

Do you think 16 rows is sufficient for the design?

In addition, if I also change blocks from 3 to 4, people will be shown 3 choice cards, which design do you think is better?

This is my codes

Design

;alts = alt1*,alt2*,none

;rows = 12

;block = 3,minsum

;eff = (mnl,d)

;con

;alg = mfederov(candidates=1000)

;model:

U(alt1) =b0[0]+ b1.dummy[0.001|0.002|0.003] * sorting[2,4,7,1]

+ b2.dummy[0.001|0.002] * collected[2,3,1]

+ b3.dummy[0.001|0.003] * point[2,3,1]

+ b4[-.001] * cost[20,40,60,80,100,200](1-2,1-2,1-2,1-2,1-2,1-2)

/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

$

Thank you very much for your time and help.

Best wishes

Steve

- Steven Guu
**Posts:**10**Joined:**Wed Jun 15, 2022 12:54 am

If you ask me, I would use the syntax below. It uses the default swapping algorithm so it has full attribute level balance. It uses 24 rows so it has sufficient variation, and it is blocked in 4 because people can easily answer 6 choice tasks. But you could block in more parts if you like, but if you only show 3 choice tasks to a respondent you obviously will need double the sample size to capture the same amount of information.

Michiel

- Code: Select all
`Design`

;alts = alt1*,alt2*,none

;rows = 24

;block = 4,minsum

;eff = (mnl,d)

;con

;model:

U(alt1) =b0[0]+ b1.dummy[0.001|0.002|0.003] * sorting[2,4,7,1]

+ b2.dummy[0.001|0.002] * collected[2,3,1]

+ b3.dummy[0.001|0.003] * point[2,3,1]

+ b4[-.001] * cost[20,40,60,80,100,200]

/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

$

Michiel

- Michiel Bliemer
**Posts:**1433**Joined:**Tue Mar 31, 2009 4:13 pm

Dear Michiel

Thank you very much for your prompt and thorough explanation. Clear now. I really appreciate your time and help.

Best regards

Steven

Thank you very much for your prompt and thorough explanation. Clear now. I really appreciate your time and help.

Best regards

Steven

- Steven Guu
**Posts:**10**Joined:**Wed Jun 15, 2022 12:54 am

Dear Michiel

Some months ago I asked you for advice regarding the pilot design for my research. I really appreciate your time and help. My data collection (150 respondents) went very well and all significant, I am very thankful for having been able to count on your expertise!

Now I’m going to design my main survey choice cards.

Since I have an unlabeled experiment experiment, with 2 alternatives + opt out choice.

- Three alternatives (two plus one none)

- 4 attribute, with 4, 3, 3, 6 levels respectively, which are method of sorting, collection, point and cost.

- none with no utility function

I tried to use the priors to construct a Bayesian efficient design.

Therefore, I create my design based on the following:

- Estimate an MNL model with ln (1) and without ln (1) (dummy code and effects code respectively) and obtain the parameter estimates and standard errors by using stata. Call this model1

- Specify a syntax for model_mnl in Ngene for generating a Bayesian efficient design for this MNL model using normal distributions with a mean set to the parameter estimate and a standard deviation set to the standard error

- Estimate an MMNL model with ln (1) and without ln (1) (dummy code and effects respectively) and obtain the parameter estimates (means and standard deviations and I estimate all normally distributed parameters). Call this model2

- Specify a syntax for model_rppanel in Ngene for evaluating an efficient design (with fixed priors) for this RPPANEL model using the parameter estimates.

- Optimise on model_mnl, and evaluate in model_rppanel

Then, I read one of the forum post and I follow your suggestion to run a mnl Bayesian model but to evaluate an efficient rppanel and I run the following syntax:

And I found I can get using effects code priors, I can get result, but using dummy code priors I get undefined results

The syntax will look something like:

1.Model 1: MNLln(1) (effects code) and Model 2: MMNLln (1) (effects code)

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b0 [2.1776]+b1.effects[(n,0.4817, 0.0982) |(n, 0.6718, 0.0911)|(n,0.0766,0.0801)]*sorting[2,4,7,1]+b2.effects[(n,0.2491, 0.1283)|(n,0.0986,0.2066)]*collected[2,3,1]+b3.effects[(n,0.1382, 0.0551)|(n,0.1964,0.0659)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;model(model2):

U(alt1)=b0 [2.6594]+b1.effects[n,0.9518, 1.1526 |n, 1.1794, 1.1526|n,0.1787,0.3335]*sorting[2,4,7,1]+b2.effects[n,0.3696, 0.5529|n,0.0850,0.9935]*collected[2,3,1]+b3.effects[n,0.1528, 0.2692|n,0.4018,0.4141]*point[2,3,1]+b4[n,-0.0048, 0.0192]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;alg=swap(stop=total(10mins))

$

2.Model 1: MNL(effects) and Model 2: MMNL(effects )

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b0 [2.1776]+b1.effects[(n,0.4817, 0.0982) |(n, 0.6718, 0.0911)|(n,0.0766,0.0801)]*sorting[2,4,7,1]+b2.effects[(n,0.2491, 0.1283)|(n,0.0986,0.2066)]*collected[2,3,1]+b3.effects[(n,0.1382, 0.0551)|(n,0.1964,0.0659)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;model(model2):

U(alt1)=b0 [2.5666]+b1.effects[n,1.0219, 1.2793 |n, 1.2695, 1.1102|n,0.2339,0.5641]*sorting[2,4,7,1]+b2.effects[n,0.4472, 0.4141|n,0.2192,1.1468]*collected[2,3,1]+b3.effects[n,0.1577, 0.1206|n,0.3696,0.4963]*point[2,3,1]+b4[n,-0.0022, 0.0122]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;alg=swap(stop=total(10mins))

$

3.Model 1: MNL(dummy) and Model 2: MMNL(dummy)

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b0 [0.6128]+b1.dummy[(n,1.7119, 0.1953) |(n, 1.9020, 0.1963)|(n,1.3068,0.1579)]*sorting[2,4,7,1]+b2.dummy[(n,0.2491, 0.1283)|(n,0.1505,0.1123)]*collected[2,3,1]+b3.dummy[(n,0.4728, 0.1089)|(n,0.5309,0.1257)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;model(model2):

U(alt1)=b0 [0.5318]+b1.dummy[n,2.2813, 1.1474 |n, 2.5509, 0.1700|n,1.7945,0.4958]*sorting[2,4,7,1]+b2.dummy[n,0.2876, 0.5755|n,-0.2078,0.8792]*collected[2,3,1]+b3.dummy[n,0.5745, 0.1680|n,0.6267,0.0042]*point[2,3,1]+b4[n,-0.0022, 0.0090]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;alg=swap(stop=total(10mins))

$

4.Model 1: MNL ln(1)(dummy) and Model 2: MMNLln(1)(dummy)

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b0 [0.6128]+b1.dummy[(n,1.7119, 0.1953) |(n, 1.9020, 0.1963)|(n,1.3068,0.1579)]*sorting[2,4,7,1]+b2.dummy[(n,0.2491, 0.1283)|(n,0.1505,0.1123)]*collected[2,3,1]+b3.dummy[(n,0.4728, 0.1089)|(n,0.5309,0.1257)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;model(model2):

U(alt1)=b0 [0.7429]+b1.dummy[n,2.1945, 1.0682 |n, 2.4467, 0.1767|n,1.7065,0.4894]*sorting[2,4,7,1]+b2.dummy[n,0.2059, 0.5370|n,0.2604,0.8903]*collected[2,3,1]+b3.dummy[n,0.5844, 0.2577|n,0.6827,0.0183]*point[2,3,1]+b4[n,-0.0042, 0.0154]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;alg=swap(stop=total(10mins))

$

However, since my alternative3 is none (opt out choice), it seems illogical to have a ASC in one of the alternatives 1,2. It should not have a constant there, the constant should be in alt3 to make sense.

Therefore, I tried to change the constants in b0 in alt12 to alt3 by using same mean but changed to negative.

5.Model 1: MNLln(1)(effects) and Model 2: MMNLln(1) (effects)

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b1.effects[(n,0.4817, 0.0982) |(n, 0.6718, 0.0911)|(n,0.0766,0.0801)]*sorting[2,4,7,1]+b2.effects[(n,0.2491, 0.1283)|(n,0.0986,0.2066)]*collected[2,3,1]+b3.effects[(n,0.1382, 0.0551)|(n,0.1964,0.0659)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b1*sorting+b2*collected+b3*point+b4*cost/

U(none)=b0 [-2.1776]

;model(model2):

U(alt1)=b1.effects[n,0.9518, 1.1526 |n, 1.1794, 1.1526|n,0.1787,0.3335]*sorting[2,4,7,1]+b2.effects[n,0.3696, 0.5529|n,0.0850,0.9935]*collected[2,3,1]+b3.effects[n,0.1528, 0.2692|n,0.4018,0.4141]*point[2,3,1]+b4[n,-0.0048, 0.0192]*cost[20,40,60,80,100,200]/

U(alt2) =b1*sorting+b2*collected+b3*point+b4*cost/

U(none)=b0 [-2.6594]

;alg=swap(stop=total(10mins))

$

6.Model 1: MNL(effects) and Model 2: MMNL(effects )

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b1.effects[(n,0.4817, 0.0982) |(n, 0.6718, 0.0911)|(n,0.0766,0.0801)]*sorting[2,4,7,1]+b2.effects[(n,0.2491, 0.1283)|(n,0.0986,0.2066)]*collected[2,3,1]+b3.effects[(n,0.1382, 0.0551)|(n,0.1964,0.0659)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b1*sorting+b2*collected+b3*point+b4*cost/

U(alt2) =b0[-2.1776]

;model(model2):

U(alt1)=b1.effects[n,1.0219, 1.2793 |n, 1.2695, 1.1102|n,0.2339,0.5641]*sorting[2,4,7,1]+b2.effects[n,0.4472, 0.4141|n,0.2192,1.1468]*collected[2,3,1]+b3.effects[n,0.1577, 0.1206|n,0.3696,0.4963]*point[2,3,1]+b4[n,-0.0022, 0.0122]*cost[20,40,60,80,100,200]/

U(alt2) =b1*sorting+b2*collected+b3*point+b4*cost/

U(alt2) =b0[-2.5666]

;alg=swap(stop=total(10mins))

$

If possible, could I ask some questions about my designs? the design 1 and design 5 are which I want pick one to use for my main survey, because the priors are significant and most reasonable. Other just for comparison.

First, does the way I changed the constant above was correct? If not, How can I delete the ASC in alt12 and add constant in alt3?

Secondly, How do I compare the efficiency between those models, and How do I compare MNL AND rp-panel results in one design? I haven’t found any from the manual?

Thirdly, do I need to add ;alg = mfederov(candidates = 1000) and delete ;alg=swap(stop=total(10mins)) to keep design balance?

Lastly, I read one of the forum post, and add ;alg=swap(stop=total(10mins)), I figured this just for quickly check the model which work or not, and I don’t know whether or not this is good for the design and why?

Best regards

Steven

Some months ago I asked you for advice regarding the pilot design for my research. I really appreciate your time and help. My data collection (150 respondents) went very well and all significant, I am very thankful for having been able to count on your expertise!

Now I’m going to design my main survey choice cards.

Since I have an unlabeled experiment experiment, with 2 alternatives + opt out choice.

- Three alternatives (two plus one none)

- 4 attribute, with 4, 3, 3, 6 levels respectively, which are method of sorting, collection, point and cost.

- none with no utility function

I tried to use the priors to construct a Bayesian efficient design.

Therefore, I create my design based on the following:

- Estimate an MNL model with ln (1) and without ln (1) (dummy code and effects code respectively) and obtain the parameter estimates and standard errors by using stata. Call this model1

- Specify a syntax for model_mnl in Ngene for generating a Bayesian efficient design for this MNL model using normal distributions with a mean set to the parameter estimate and a standard deviation set to the standard error

- Estimate an MMNL model with ln (1) and without ln (1) (dummy code and effects respectively) and obtain the parameter estimates (means and standard deviations and I estimate all normally distributed parameters). Call this model2

- Specify a syntax for model_rppanel in Ngene for evaluating an efficient design (with fixed priors) for this RPPANEL model using the parameter estimates.

- Optimise on model_mnl, and evaluate in model_rppanel

Then, I read one of the forum post and I follow your suggestion to run a mnl Bayesian model but to evaluate an efficient rppanel and I run the following syntax:

And I found I can get using effects code priors, I can get result, but using dummy code priors I get undefined results

The syntax will look something like:

1.Model 1: MNLln(1) (effects code) and Model 2: MMNLln (1) (effects code)

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b0 [2.1776]+b1.effects[(n,0.4817, 0.0982) |(n, 0.6718, 0.0911)|(n,0.0766,0.0801)]*sorting[2,4,7,1]+b2.effects[(n,0.2491, 0.1283)|(n,0.0986,0.2066)]*collected[2,3,1]+b3.effects[(n,0.1382, 0.0551)|(n,0.1964,0.0659)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;model(model2):

U(alt1)=b0 [2.6594]+b1.effects[n,0.9518, 1.1526 |n, 1.1794, 1.1526|n,0.1787,0.3335]*sorting[2,4,7,1]+b2.effects[n,0.3696, 0.5529|n,0.0850,0.9935]*collected[2,3,1]+b3.effects[n,0.1528, 0.2692|n,0.4018,0.4141]*point[2,3,1]+b4[n,-0.0048, 0.0192]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;alg=swap(stop=total(10mins))

$

2.Model 1: MNL(effects) and Model 2: MMNL(effects )

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b0 [2.1776]+b1.effects[(n,0.4817, 0.0982) |(n, 0.6718, 0.0911)|(n,0.0766,0.0801)]*sorting[2,4,7,1]+b2.effects[(n,0.2491, 0.1283)|(n,0.0986,0.2066)]*collected[2,3,1]+b3.effects[(n,0.1382, 0.0551)|(n,0.1964,0.0659)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;model(model2):

U(alt1)=b0 [2.5666]+b1.effects[n,1.0219, 1.2793 |n, 1.2695, 1.1102|n,0.2339,0.5641]*sorting[2,4,7,1]+b2.effects[n,0.4472, 0.4141|n,0.2192,1.1468]*collected[2,3,1]+b3.effects[n,0.1577, 0.1206|n,0.3696,0.4963]*point[2,3,1]+b4[n,-0.0022, 0.0122]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;alg=swap(stop=total(10mins))

$

3.Model 1: MNL(dummy) and Model 2: MMNL(dummy)

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b0 [0.6128]+b1.dummy[(n,1.7119, 0.1953) |(n, 1.9020, 0.1963)|(n,1.3068,0.1579)]*sorting[2,4,7,1]+b2.dummy[(n,0.2491, 0.1283)|(n,0.1505,0.1123)]*collected[2,3,1]+b3.dummy[(n,0.4728, 0.1089)|(n,0.5309,0.1257)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;model(model2):

U(alt1)=b0 [0.5318]+b1.dummy[n,2.2813, 1.1474 |n, 2.5509, 0.1700|n,1.7945,0.4958]*sorting[2,4,7,1]+b2.dummy[n,0.2876, 0.5755|n,-0.2078,0.8792]*collected[2,3,1]+b3.dummy[n,0.5745, 0.1680|n,0.6267,0.0042]*point[2,3,1]+b4[n,-0.0022, 0.0090]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;alg=swap(stop=total(10mins))

$

4.Model 1: MNL ln(1)(dummy) and Model 2: MMNLln(1)(dummy)

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b0 [0.6128]+b1.dummy[(n,1.7119, 0.1953) |(n, 1.9020, 0.1963)|(n,1.3068,0.1579)]*sorting[2,4,7,1]+b2.dummy[(n,0.2491, 0.1283)|(n,0.1505,0.1123)]*collected[2,3,1]+b3.dummy[(n,0.4728, 0.1089)|(n,0.5309,0.1257)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;model(model2):

U(alt1)=b0 [0.7429]+b1.dummy[n,2.1945, 1.0682 |n, 2.4467, 0.1767|n,1.7065,0.4894]*sorting[2,4,7,1]+b2.dummy[n,0.2059, 0.5370|n,0.2604,0.8903]*collected[2,3,1]+b3.dummy[n,0.5844, 0.2577|n,0.6827,0.0183]*point[2,3,1]+b4[n,-0.0042, 0.0154]*cost[20,40,60,80,100,200]/

U(alt2) =b0+ b1*sorting+b2*collected+b3*point+b4*cost

;alg=swap(stop=total(10mins))

$

However, since my alternative3 is none (opt out choice), it seems illogical to have a ASC in one of the alternatives 1,2. It should not have a constant there, the constant should be in alt3 to make sense.

Therefore, I tried to change the constants in b0 in alt12 to alt3 by using same mean but changed to negative.

5.Model 1: MNLln(1)(effects) and Model 2: MMNLln(1) (effects)

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b1.effects[(n,0.4817, 0.0982) |(n, 0.6718, 0.0911)|(n,0.0766,0.0801)]*sorting[2,4,7,1]+b2.effects[(n,0.2491, 0.1283)|(n,0.0986,0.2066)]*collected[2,3,1]+b3.effects[(n,0.1382, 0.0551)|(n,0.1964,0.0659)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b1*sorting+b2*collected+b3*point+b4*cost/

U(none)=b0 [-2.1776]

;model(model2):

U(alt1)=b1.effects[n,0.9518, 1.1526 |n, 1.1794, 1.1526|n,0.1787,0.3335]*sorting[2,4,7,1]+b2.effects[n,0.3696, 0.5529|n,0.0850,0.9935]*collected[2,3,1]+b3.effects[n,0.1528, 0.2692|n,0.4018,0.4141]*point[2,3,1]+b4[n,-0.0048, 0.0192]*cost[20,40,60,80,100,200]/

U(alt2) =b1*sorting+b2*collected+b3*point+b4*cost/

U(none)=b0 [-2.6594]

;alg=swap(stop=total(10mins))

$

6.Model 1: MNL(effects) and Model 2: MMNL(effects )

Design

;alts (model1)= alt1*, alt2*, none

;alts (model2)= alt1*, alt2*, none

;rows=24

;block=4,minsum

;eff = model1(mnl,d,mean)

;rdraws= gauss(3)

;bdraws= gauss(3)

;rep= 1000

;model(model1):

U(alt1)=b1.effects[(n,0.4817, 0.0982) |(n, 0.6718, 0.0911)|(n,0.0766,0.0801)]*sorting[2,4,7,1]+b2.effects[(n,0.2491, 0.1283)|(n,0.0986,0.2066)]*collected[2,3,1]+b3.effects[(n,0.1382, 0.0551)|(n,0.1964,0.0659)]*point[2,3,1]+b4[(n,-0.0011, 0.0009)]*cost[20,40,60,80,100,200]/

U(alt2) =b1*sorting+b2*collected+b3*point+b4*cost/

U(alt2) =b0[-2.1776]

;model(model2):

U(alt1)=b1.effects[n,1.0219, 1.2793 |n, 1.2695, 1.1102|n,0.2339,0.5641]*sorting[2,4,7,1]+b2.effects[n,0.4472, 0.4141|n,0.2192,1.1468]*collected[2,3,1]+b3.effects[n,0.1577, 0.1206|n,0.3696,0.4963]*point[2,3,1]+b4[n,-0.0022, 0.0122]*cost[20,40,60,80,100,200]/

U(alt2) =b1*sorting+b2*collected+b3*point+b4*cost/

U(alt2) =b0[-2.5666]

;alg=swap(stop=total(10mins))

$

If possible, could I ask some questions about my designs? the design 1 and design 5 are which I want pick one to use for my main survey, because the priors are significant and most reasonable. Other just for comparison.

First, does the way I changed the constant above was correct? If not, How can I delete the ASC in alt12 and add constant in alt3?

Secondly, How do I compare the efficiency between those models, and How do I compare MNL AND rp-panel results in one design? I haven’t found any from the manual?

Thirdly, do I need to add ;alg = mfederov(candidates = 1000) and delete ;alg=swap(stop=total(10mins)) to keep design balance?

Lastly, I read one of the forum post, and add ;alg=swap(stop=total(10mins)), I figured this just for quickly check the model which work or not, and I don’t know whether or not this is good for the design and why?

Best regards

Steven

- Steven Guu
**Posts:**10**Joined:**Wed Jun 15, 2022 12:54 am

Putting the same constant in alt1 and alt2 or putting a constant in none with inverse sign is exactly the same model, so it does not matter where you put the constant.

You can only compare efficiency across designs for the same model with the same priors (e.g. an orthogonal design and an efficient design). You cannot compare efficiency across models with different specification, e.g. dummy versus effects coding or mnl versus rppanel.

The default swapping algorithm will try to maintain attribute level balance, mfederov will not.

I never use a stopping criterion and 10 minutes is usually not enough; I usually run scripts overnight when I have Bayesian priors and observe if the D-error has stabilised.

Note that you are using 3^8 = 6,561 draws for the random parameters, and for rppanel model the default is also a sample of 500 (via ;rep = 500), so the total number of computations Ngene is doing for evaluating the mmnl model is more than 3 mnl! It will take a long time for Ngene to evaluate the resulting design. I generally omit such evaluations but you can try.

Michiel

PS: try to keep the questions short as otherwise I may not be able to answer them

You can only compare efficiency across designs for the same model with the same priors (e.g. an orthogonal design and an efficient design). You cannot compare efficiency across models with different specification, e.g. dummy versus effects coding or mnl versus rppanel.

The default swapping algorithm will try to maintain attribute level balance, mfederov will not.

I never use a stopping criterion and 10 minutes is usually not enough; I usually run scripts overnight when I have Bayesian priors and observe if the D-error has stabilised.

Note that you are using 3^8 = 6,561 draws for the random parameters, and for rppanel model the default is also a sample of 500 (via ;rep = 500), so the total number of computations Ngene is doing for evaluating the mmnl model is more than 3 mnl! It will take a long time for Ngene to evaluate the resulting design. I generally omit such evaluations but you can try.

Michiel

PS: try to keep the questions short as otherwise I may not be able to answer them

- Michiel Bliemer
**Posts:**1433**Joined:**Tue Mar 31, 2009 4:13 pm

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