Minimum number of choice situations and SQ

[tpham]

Thank you very much for your response. In effect, I looked into Cooper et al. (2012), who proposed a hybrid coding with 2 base levels (one base level for attribute level only appears in the sq, coded as 0 and one other -1), such that:
if the original codes for attribute levels 3, 2, 1 are 2; 1; 0, then the hybrid code for the attribute level 3 would be equal 1 if level =3, -1 if level =2 and 0 if level = 1 (level 1 only appears in statu quo).

However, I don't know how exactly we should code in NGENE. The following syntax works, but I am not sure that it is what I am looking for:

Code: Select all

Design
;alts = alt1*, alt2*, sq*
;rows = 10
;block = 2
;eff = (mnl,d) + 0.5*(imbalance)
;alg = mfederov
;require:
sq.Risque_appt=1,
sq.Risque_maison=2,
sq.Risque_eco=3,
sq.visibility=0
;model:
U(alt1) = b0_alt1
+ b1.effects[0] * Diag1[1,2]  ? the coefficient of b1 is associated with level 1, the coeff of level 2 will be minus b1
+ b2.effects[0] * Diag2[1,2]
+ b3.dummy[0] * Risque_appt[1,2]
+ b4.dummy[0|0] * Risque_maison[1,2,3]
+ b5.dummy[0|0] * Risque_eco[1,2,3]
+ b6.dummy[0] * visibility[1,0]
+ b7.effects[0] * financement[1,2]
+ b8[0] * cout[25, 50, 75, 100, 125, 150]
/
U(alt2) = b0_alt2
+ b1 * Diag1
+ b2 * Diag2
+ b3 * Risque_appt
+ b4 * Risque_maison
+ b5 * Risque_eco
+ b6 * visibility
+ b7 * financement
+ b8 * cout
/
U(sq) = 0 ? The coeff of level 0 for Diag1, Diag2 and financement = 0
+ b3 * Risque_appt
+ b4 * Risque_maison
+ b5 * Risque_eco
+ b6 * visibility
+ b8 * coutsq[0]
$


It generates, no surprise, choice situations in which the sq alternative has only 5 attributes (without Diag1, Diag2 and financement)...

[Michiel Bliemer]

I am not familiar with the method Cooper et al. (2012) myself, this was just a suggestion mentioned by a colleague, but you could try to formulate reject constraints to impose that coding scheme.

Michiel