Hi,
I have an experiment where we want to have a fixed number of attributes differing in each choice set (so nine dimensions with either four or five levels, and we want four to differ in each choice set). We also need to be able to estimate a large number of two factor interactions between each of the levels of each of the coefficients with the willingness to pay numeraire (which is A below).
I have run it successfully without this constraint:
Design
;alts = alt1, alt2
;rows = 100
;eff = (mnl,wtp(ref1))
;wtp = ref1(i1,i2,i3,i4,i5,i6,i7,i8,i9,i10,i11,i12,i13,i14,i15,i16,i17,i18,i19,i20,i21,i22,i23,i24,i25,i26,i27,i28,i29,i30,i31,i32/b2)
;alg = swap(stop=noimprov(120 secs))
;model:
U(alt1) = b1[0] + b2[0.1] * A[10,5,2,1] + b3.dummy[0|0|0|0] * B[4,3,2,1,0] + b4.dummy[0|0|0|0] * C[4,3,2,1,0] + b5.dummy[0|0|0|0] * D[4,3,2,1,0] + b6.dummy[0|0|0|0] * E[4,3,2,1,0] + b7.dummy[0|0|0|0] * F[4,3,2,1,0] + b8.dummy[0|0|0|0] * G[4,3,2,1,0] + b9.dummy[0|0|0|0] * H[4,3,2,1,0] + b10.dummy[0|0|0|0] * I[4,3,2,1,0] + i1[0] * A * B.dummy[1] + i2[0] * A * B.dummy[2] + i3[0] * A * B.dummy[3] + i4[0] * A * B.dummy[4] + i5[0] * A * C.dummy[1] + i6[0] * A * C.dummy[2] + i7[0] * A * C.dummy[3] + i8[0] * A * C.dummy[4] + i9[0] * A * D.dummy[1] + i10[0] * A * D.dummy[2] + i11[0] * A * D.dummy[3] + i12[0] * A * D.dummy[4] + i13[0] * A * E.dummy[1] + i14[0] * A * E.dummy[2] + i15[0] * A * E.dummy[3] + i16[0] * A * E.dummy[4] + i17[0] * A * F.dummy[1] + i18[0] * A * F.dummy[2] + i19[0] * A * F.dummy[3] + i20[0] * A * F.dummy[4] + i21[0] * A * G.dummy[1]
+ i22[0] * A * G.dummy[2] + i23[0] * A * G.dummy[3] + i24[0] * A * G.dummy[4] + i25[0] * A * H.dummy[1] + i26[0] * A * H.dummy[2] + i27[0] * A * H.dummy[3] + i28[0] * A * H.dummy[4] + i29[0] * A * I.dummy[1] + i30[0] * A * I.dummy[2] + i31[0] * A * I.dummy[3] + i32[0] * A * I.dummy[4] /
U(alt2) = b2 * A + b3.dummy * B + b4.dummy * C + b5.dummy * D + b6.dummy * E + b7.dummy * F + b8.dummy * G + b9.dummy * H + b10.dummy * I + i1 * A * B.dummy[1] + i2 * A * B.dummy[2] + i3 * A * B.dummy[3] + i4 * A * B.dummy[4] + i5 * A * C.dummy[1] + i6 * A * C.dummy[2] + i7 * A * C.dummy[3] + i8 * A * C.dummy[4] + i9 * A * D.dummy[1] + i10 * A * D.dummy[2] + i11 * A * D.dummy[3] + i12 * A * D.dummy[4] + i13 * A * E.dummy[1] + i14 * A * E.dummy[2] + i15 * A * E.dummy[3] + i16 * A * E.dummy[4] + i17 * A * F.dummy[1] + i18 * A * F.dummy[2] + i19 * A * F.dummy[3] + i20 * A * F.dummy[4] + i21 * A * G.dummy[1]+ i22 * A * G.dummy[2] + i23 * A * G.dummy[3] + i24 * A * G.dummy[4] + i25 * A * H.dummy[1] + i26 * A * H.dummy[2] + i27 * A * H.dummy[3] + i28 * A * H.dummy[4] + i29 * A * I.dummy[1] + i30 * A * I.dummy[2] + i31 * A * I.dummy[3] + i32 * A * I.dummy[4]
$
I asked a colleague of mine about how to enforce this overlap condition, and she suggested using a large number of constraints like this one:
;cond:
if (alt1.A <> alt2.A and alt1.B <> alt2.B and alt1.C <> alt2.C and alt1.D <> alt2.D, alt1.E=alt2.E and alt1.F=alt2.F and alt1.G=alt2.G and alt1.H=alt2.H and alt1.I=alt2.I), etc etc
Thus, we would specify each combination of attributes that differ, and force the others to be the same. This gives a lot of constraints (112 I think), and causes Ngene to fall over. I tried fractional but couldn’t find a fraction which would prevent this.
I also tried defining the condition in terms of reject (i.e reject any profile where 5 are the same, or 6 are different), but that ended up with even more constraints, and the same problem of not being able to find an answer.
;reject:
Alt1.A = Alt2.A and Alt1.B=Alt2.B and Alt1.C=Alt2.C and Alt1.D = Alt2.D and Alt1.E=Alt2.E and Alt1.F=Alt2.F, etc etc
Is there a more sensible way of trying to do this?
Thanks in advance for any suggestions on this,
Richard