Page 253 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

Table 13.3. Solutions to Example 13.2 using Eqn 13.17.
Iteration number
Linear
Trait a Factor 0 1 4 8 13 model
BW Heifer origin
1 41.6633 41.5471 41.6262 41.6182 41.6195 41.6175
2 42.2530 42.1409 42.2178 42.2099 42.2112 42.2022
Calving season
1 −1.2350 −1.2345 −1.2346 −1.2343 −1.2344 −1.2387
2 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
Sex of calf
Male 3.1589 3.1890 3.1687 3.1690 3.1690 3.1845
Female 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
Sire
1 −0.4155 −0.2633 −0.3671 −0.3580 −0.3595 −0.3268
2 0.1048 0.1687 0.1246 0.1311 0.1300 0.1171
3 −0.3315 −0.2280 −0.3007 −0.2939 −0.2950 −0.2641
4 0.1364 0.3365 0.2035 0.2139 0.2122 0.1886
5 0.2730 0.3261 0.2893 0.2979 0.2965 0.2688
6 0.1545 0.2270 0.1770 0.1821 0.1813 0.1690
CD Heifer origin
1 0.1873 −1.0189 −1.4072 −1.3915 −1.3943 0.1349
2 0.1484 −1.2813 −1.7342 −1.7472 −1.7452 0.0876
Calving season
1 −0.0874 0.1871 0.1327 0.1415 0.1401 −0.0311
2 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
Sex of calf
Male 0.2756 0.3218 0.8621 0.8369 0.8411 0.2410
Female 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
Sire
1 −0.1180 0.0471 −0.0656 −0.0561 −0.0577 −0.0527
2 0.0144 0.0705 0.0319 0.0379 0.0369 0.0285
3 −0.0850 0.0185 −0.0546 −0.0477 −0.0488 −0.0427
4 −0.0380 0.1698 0.0319 0.0424 0.0407 0.0350
5 −0.0048 0.0362 0.0075 0.0163 0.0148 0.0195
6 0.0079 0.0702 0.0270 0.0315 0.0308 0.0323
a BW, birth weight; CD, calving difficulty

Using Eqn 13.18, the elements of q for animals 1, 2, 46 and 47 are:

q(1) = −{0(−1)f(0.0242)/F(0.0242) + (1 − 0)f(0.0242)/(1 − F(0.0242))}
= −{0(−1)0.3988/0.5097 + 1(0.3988/0.4903)} = −0.8134
q(2) = −{0(−1)f(−0.3800)/F(−0.3800) + (1 − 0)f(−0.3800)/(1 − F(−0.3800))}
= −{0(−1)0.3712/0.3520 + 1(0.3712/0.6480)} = −0.5727
q(46) = −{0(−1)f(0.1244)/F(0.1244) + (1 − 0)f(0.1244)/(1 − F(0.1244))}

= −{0(−1)0.3959/0.5495 + 1(0.3959/0.4505)} = −0.8787

Analysis of Ordered Categorical Traits 237