The farmer chooses to feed his herd with the “grass-based” diet. He has around 43 cows and therefore 13 heifers and 26 calves. He grows 26 ha of wheat, 4 ha of maize and 19 ha of grassland. All his wheat production gets sold.
The optimum number of animas is as follows :
| Dairy cow | 42.857 |
| Heifers (alive) | 12.857 |
| Calves (sold) | 25.714 |
Optimum cropping pattern is as follows :
| Wheat | 26.104 ha |
| Maize | 3.896 ha |
| Grassland | 19.286 ha |
Income is as follows : 56 629.87€.
Checking the number of animals present and the renewal and culling rates :
0.3 * 42.857 = 12.857
0.6 * 42.857 = 25.714
Checking whether the food needs are covered by the arable land production :
With 19ha of grassland he produces :
Grazing: 5*19.286 = 96.43 T
Hay : 4.5*19.286 = 87.787 T
Wrapped bales : 2.2*19.286 = 42.42 T
Maize : 11*3.896 = 42.86 T
To feed his animals, he needs :
Grazing : 1.5*42.857 + 12.857*2.5 = 96.43 T
Hay : 2*12.857=25.714 T
Wrapped bales : 2.6*12.857=33.4282 T
Maize : 1*42.857=42.857 T
Production is therefore much more significant than needs.
The solution model is as follows : modelEco_dairyFarm_1.gms
