Query about efficient design

[Michiel Bliemer]

Please refer to Section 8.8 of the Ngene manual for a discussion on unlabelled alternatives and dominance.

Michiel

[bpaudel]

Hi Michael,
Thank you so much for your every insightful informations.

Further, I have a query on prohibited pairs. Is there any provision in Ngene that i can get rid of prohibited pairs? I do not want to include prohibited pairs in my choice sets.

To clarify more to my context,
I got one of the option of choice set like as follow,

Price : $2
Use of Antibiotics : Antibiotics free
Production Method: Traditional
Use of Synthetic Growth Promoter: Yes

Here my concern is not to have Antibiotic free with price combination of $2 in any choice options as such combination doesn't make sense for my research question and might affect my result.

I kindly request you to provide insight in this matter.

Hoping to hearing from you soon.

With regards,
bpaudel

[Michiel Bliemer]

Yes you can easily include conditions on pairs of attribute levels or reject/require certain combinations. This is done through the ;cond (which uses the default algorithm) or ;reject/require commands (which uses the mfederov algorithm), see Section 8.2 of the Ngene manual.

For example, you would have something like:

;alg =mfederov
;reject:
alt1.antibiotics = 1 and alt1.price = 2
;model:
U(alt1) = b1.dummy * antibiotics[1,0] + b2 * price[1,2,3] /
...

Michiel

[bpaudel]

Hi Michael,
Thank you so much for your suggestion. But I am having a problem in running the design.

I have four attributes with their level as follows

Price ($/lb) : 2, 4, 6, 8
Use of antibiotics: Antibiotic free (dummy 2), Minimal (dummy 1) and Conventional (dummy 0)
Production Method: Traditional (dummy 1) and Standard (dummy 0)
Use of Synthetic growth promoter: No (dummy 1) and Yes (dummy 0)

I have following restrictions to put on my model,

Antibiotic free (dummy 2) should not occur with price of $2 in both the alternatives.
Traditional (dummy 1) should not occur with price of $2 in both the alternatives.
No use of Synthetic growth promoter (dummy 1) should not occur with price of $2 in both the alternatives.

I redesigned the model as follow:
Design
;alts= optA*, optB*, Neither
;rows=24
;eff=(mnl,d, mean)
;block=2
;alg= mfederov
;reject:
optA.Anti= 2 and optA.price= 2,
optB.anti= 2 and optB.price= 2,
optA.production= 1 and optA.price= 2,
optB.production= 1 and optB.price= 2,
optA.SGP= 1 and optA.price= 2,
optB.SGP= 1 and optB.price= 2
;model:
U(optA) = b1 [(n,-0.4, 0.03)] * price[2, 4, 6, 8]
+ b2.dummy[(n, 1.3, .70)| (n, 0.4, 0.04)] * Anti[2, 1, 0]
+ b3.dummy[(n, 0.8, .07)] * production[1,0]
+ b4.dummy[(n, 0.4, 0.01)] * SGP[1,0] /
U(optB) = b1 * price
+ b2 * Anti
+ b3 * production
+ b4 * SGP /
U(Neither) = b0 [-3.0]
$

But, when I am running this model the error occurs. Can you please tell me where I am going wrong?

Your suggestions regarding this matter will be very helpful to me.

Hoping to hearing from you soon.

With regards,
bpaudel

[Michiel Bliemer]

Given your constraints there only exist 810 feasible choice tasks, so please change this in your syntax and it will run:

;alg = mfederov(candidates = 810)

You can see that there are 810 feasible choice tasks by running this syntax and opening the design:

Design
;alts= optA*, optB*, Neither
;rows=all
;fact
;reject:
optA.Anti= 2 and optA.price= 2,
optB.anti= 2 and optB.price= 2,
optA.production= 1 and optA.price= 2,
optB.production= 1 and optB.price= 2,
optA.SGP= 1 and optA.price= 2,
optB.SGP= 1 and optB.price= 2
;model:
U(optA) = b1 [(n,-0.4, 0.03)] * price[2, 4, 6, 8]
+ b2.dummy[(n, 1.3, .70)| (n, 0.4, 0.04)] * Anti[2, 1, 0]
+ b3.dummy[(n, 0.8, .07)] * production[1,0]
+ b4.dummy[(n, 0.4, 0.01)] * SGP[1,0] /
U(optB) = b1 * price
+ b2 * Anti
+ b3 * production
+ b4 * SGP /
U(Neither) = b0 [-3.0]
$

Michiel

[bpaudel]

Hi Michael,
Thank you so much. It is really helpful for me.

With regards,
bpaudel