I’m preparing a design for an unlabelled choice experiment with 2 alternatives and a statu-quo. The status-quo levels are fixed and present in the other two alternatives for all attributes except the monetary attribute which level is 0 in the status quo and superieur to 0 in the alternatives. My code is the following:

Design

;alts=alt1, alt2, sq

;rows=?

;block=?

;eff=(mnl,d)

;alg = mfederov(candidates = 500)

;require:

sq.INF = 1,

sq.HEAT = 1

sq.CAR = 1,

sq.DISTANCE = 1

;model:

U(alt1)=b1.dummy[0|0|0|0]*INF[1,2,3,4,5] ? INF : Water substitution level INF=1 <10% INF=2 25% INF=3 50% INF=4 75% INF=5 100%

+b2.dummy[0|0|0|0]*HEAT[1,2,3,4,5] ? HEAT : Heat island effect surface reduction HEAT=1 0% HEAT=2 15% HEAT=3 30% HEAT=4 45% HEAT=5 60%

+b3.dummy[0]*CAR[1,2] ? CAR Reduction of road and parking space CAR=1 No Reduction CAR=2 Reduction

+b4.dummy[0|0|0|0]*DISTANCE[1,2,3,4,5] ? DISTANCE Distance of closest solution to home DISTANCE=1 No new green solution DISTANCE=2 100m DISTANCE=3 200m DISTANCE=4 500m DISTANCE=5 1KM

+b5[0]*BILL [10,20,30,40, 50] ? Increase of the water bill in €

U(alt2)=b1*INF+b2*HEAT+b3*CAR+b4*DISTANCE+b5*BILL

U(SQ)= b6[0]+ b1*INF+b2*HEAT+b3*CAR+b4*DISTANCE

$

I have 3 questions:

- Is my coding of the status-quo accurate?

- I'm struggling with the estimation of the minimum of choice situation. In the formula used to estimate the minimum number of choice situation, (J-1)S=K I wonder whether the status quo is considered an alternative. In other words, if J=2 or 3. If the SQ is not an alternative then K being equal to 15 (including constant). I have to include a minimum of 15 choice situations, 20 if I want to have attribute level balance. If the SQ counts then the minimum number is 7.5 and we can live with 10 choice situations.

- If I end up with 20 choice situations. Is it preferable to have 2 blocks of 10 or 4 blocks of 5.

Thanks a lot for your help