Length : around 40 minutes
This activity follows the livestock activity in lesson 16. You have to download the following model : modelEco_elevageLait_2.gms. This model is based on the model of lesson 16 but two technologies have been introduced which correspond to two dairy cattle feeding systems : one is more intensive, the other is grass-based.
In the “grass-based” diet, the share of grazing is more significant than maize and soya shares ; and in the “intensive” diet a significant part of the food is soya and maize-based. The first diet leads to an average production of 7000 litres (L) per dairy cow and per year, and the second diet to a slightly higher production of 7140 L.
« Dairy quotas were abolished by the European Union in April 2015. European production significantly increased whereas the optimistic projected increase in global demand did not become a reality. Milk prices in Europe and in France therefore crashed. First they fell to 0,3 €/L and then to 0,27 €/L ».
Compare the three following scenarios and analyse the production choices of the farmer :
- Before 2015 : Price of milk 0.35€/L and a quota of 300000L/farm
- Before 2015 : Price of milk 0.3€/L without quota
- After 2015 : Price of milk 0.27€/L without quota
Other prices and costs remain unchanged
Reminder of the problem statement :
« M. Fourhooves has a dairy farm in the west of France. It is a 50 ha farm, 30 ha of which is arable land and 20 ha is permanent grassland. The farmer can grow wheat and/or maize (for selling purposes) on the arable land. Maize is exclusively used as silage to feed the dairy cows. The permanent grassland cannot be tilled (CAP greening regulations). Every grassland hectare is grazed then mowed to produce hay and wrapped bales.
The cows graze for 6 months (beginning of April – end of September). During the winter period, they are fed maize silage (produced on the farm) and a supplement of soya-bean cakes (bought). The heifers are fed by grazing, hay and wrapped bales.
The average production per dairy cow is 7000 litres (L) of milk per year.
Plant production
| Yields (t /ha) | Operational costs (€ /ha) | ||||
| Wheat | 6 | 600 |
| Fodder | Fodder yields (t /ha) | Operational costs (€ /ha) | |||
| Maize | Silage maize | 11.0 | 620 | ||
| Grassland | Grazing | 5 | 300 |
||
| Hay | 4.5 | ||||
| Wrapped bales | 2.2 | ||||
Maize is only used for the animals, wheat is sold. Every hectare is grazed then mowed in order to produce hay and wrapped bales.
Animal activities
Certain technico-economic indicators concerning the organization of the herd are available : herd management parameters and herd food needs.
Herd management, breeding, culling
Every cow gives birth to 0,9 calves per year which are either sold (CALF), or kept for herd renewal (HEIFER). Every year, 0,3 cows are sold as cull cows and replaced by selected heifers. The culling rate is equal to the renewal rate.
Food needs
The animals are fed using both food produced on the farm and purchased food. Food needs were estimated for cows and heifers.
“intens” intensive feeding diet
| Fodder products | Cows | Heifers (what heifers need until they give birth) |
| Hay | 2 t /year | |
| Wrapped bales | 2,6 t /year | |
| Grazing | 1 t/year | 2.5 t/year |
| Maize | 2 t / year | |
| Supplements bought | Cows | Heifers |
| Soya | 0,57 t /year |
“Grass-based” feeding diet
| Fodder products | Cows | Heifers (what heifers need until they give birth) |
| Hay | 2 t /year | |
| Wrapped bales | 2,6 t /year | |
| Grazing | 1.5 t/year | 2.5 t/year |
| Maize | 1 t / year | |
| Supplements bought | Cows | Heifers |
| Soya | 0,45 t /year |
Technico-economic data
The price of milk is 0,35 €/L. The price of wheat is 150€ per ton (T). The livestock farmer is subject to milk quotas which limit annual milk production to 300 00 L.
The price of purchased soya is 360 €/t
Price of meat products :
Cows : 800€/cow
Calf : 170 €/ calf
Other expenses including structural costs (depreciation of buildings, …) are estimated at 1300 € per cow and per year regardless of the diet.
The objective of the farmer is to maximize his income.
Then check the solution : 
