Page 105 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

k ∑
n = 11/ n jkl
−
i
where n is the size of the CG in which the daughter k of sire i made her lth lactation.
jkl j

An example of MACE for two countries

Example 5.5
The data set below consists of bull breeding values (kg) and DRP for fat yield for six bulls
from two countries. Two of the bulls have evaluations in both countries and in addition
each country had two other bulls, which were the only progeny tested in that country.
2
A MACE is implemented using the data set. Assume residual variances of 206.5 kg and
2
148.5 kg for countries 1 and 2, respectively, with corresponding sire additive genetic
2
2
variances of 20.5 kg and 9.5 kg . The sire genetic covariance between fat yield in both
countries was assumed to be 12.839 kg, giving a genetic correlation of 0.92.
Country 1 Country 2
Sire EDC BV DRP EDC BV DRP
1 58 9.0 9.7229 90 13.5 14.5088
2 150 10.1 9.9717 65 7.6 7.7594
3 20 15.8 19.2651 – – –
4 25 –4.7 –8.5711 – – –
5 – – – 30 19.6 23.9672
6 – – – 55 –5.3 –9.6226
EDC = effective daughter contribution; BV = breeding value; DRP = deregressed proof.

Assume that the sires in the data set have the following pedigree structure, with
unknown sires, MGS and MGD assigned to group G , with i = 1,... 5.
i

Bull Sire MGS MGD
1 7 G3 G5
2 8 9 G5
3 7 2 G5
4 1 G2 G5
5 8 G3 G4
6 1 9 G4
7 G1 G2 G4
8 G1 G2 G4
9 G1 G3 G4

Computing sire breeding values

The matrix G for Example 5.5 is:
−1
⎛ 0.31762 − 0.42925 ⎞
G −1 = ⎜ ⎝ −0.42925 0.68539 ⎟ ⎠

Multivariate Animal Models 89