Page 120 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition
P. 120

Assuming that G and R, respectively, are:

                ⎛20 18    4    9  ⎞            ⎛ 40 11 16    9⎞
                ⎜ 18  40  9   20  ⎟            ⎜ 11 30 112 14  ⎟
            G = ⎜                 ⎟   and  R  = ⎜              ⎟
                ⎜ 4   9  25    4.  ⎟ 5         ⎜ 16 12  70 10⎟
                ⎜                 ⎟            ⎜               ⎟
                           .
                ⎝ 920     4 5 32  ⎠            ⎝  91410     55⎠
         Applying Eqn 6.8 to G using the function factanal in the R package (The R Development
         Core Team, 2010) gives:
            F′ = (2.8532 6.3056 1.4250 3.1678) and  S′ = (11.860 0.200 22.975 21.952)
         This implies that the number of common factors, m, is equal to 1 for the example G
         above. Thus the column vector z* for animal i in the matrix Z *  in Eqn 6.11 equals F.
                                     i
                                             −1
         Therefore, for animal i with a record, z*′r z* is 1.361. However, for animal i, W  is
                                           i
                                               i
                                                                               i
                                         −1
         a diagonal matrix and therefore W ′R W  is computed as described for the MBLUP
                                       i     i
                                                −1
         model in Section 5.2. Thus for animal i, W ′R W  is:
                                             i     i
                       ⎛  0 0297                           ⎞
                          .
                       ⎜ −0 0079.  0 0419   symmetric      ⎟
                                    .
              ′
                 −1
            WR W    i  = ⎜                                 ⎟
              i
                       ⎜ −0 0052.  − 0 0041.  0 0163       ⎟
                                              .
                       ⎜ ⎝ −0 0019.  −  0 0086  − 0 0011 0 0209⎠ ⎟
                         0
                                              .
                                    .
                                                      .
            Although there were 48 equations in the MME defined in Eqn 6.11 for this
         example compared with 40 in the usual MBLUP, there were only 502 non-zero ele-
         ments in the XFA compared with 620 in MBLUP, illustrating the increased sparsity
         of the MME with the XFA model. Solving the MME gave the following solutions.
         The results from the usual MBLUP gave exactly the same solutions and these have
         not been presented.
         Solutions for sex of calf effects
             WWG     PWG     MSC     BFAT
         M   4.352   6.795   9.412   0.231
         F   3.487   5.959   7.095   0.535
         Animal and specific solutions
                         Specific effects solutions       Transformed solutions b
             COM a   WWG     PWG     BFAT    MSC    WWG     PWG      MSC    BFAT
         1    0.036  −0.008   0.095   0.000  0.005   0.095   0.227   0.340   0.010
         2   −0.012  −0.001  −0.073   0.021  0.000  −0.089  −0.073   0.313  −0.050
         3   −0.027   0.068   0.031   0.208  0.000  −0.086  −0.169   0.031  −0.032
         4    0.021   0.046   0.113  −0.021  0.000   0.168   0.136  −0.855   0.113
         5   −0.064  −0.191   0.000   0.005  0.000  −0.191  −0.407  −0.539  −0.029
         6    0.046   0.290   0.021   0.000  0.000   0.017   0.290   1.350  −0.082
         7   −0.063  −0.813   0.208   0.000  0.029  −0.208  −0.399  −0.813  −0.015
         8    0.028  −0.101  −0.021   0.000  0.000  −0.017   0.178   1.431  −0.101
                                  b
         a COM, solutions for common factor.  Transformed solutions from Eqn 6.12.

          104                                                             Chapter 6
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