Suboptimal (incomplete) diet
If cattle receive less feed than they need to achieve the optimal diet, the diet is considered suboptimal. This leads to the following consequences:
- The rate of weight gain decreases.
- Manure production is decreasing.
- Milk production decreases.
- The period of preparation for mating increases.
- The pregnancy period is extended.
To avoid negative consequences, it is important to provide complete, balanced feeding.
Table: Amount of feed required for the optimal diet depending on age group
| Age, days | Types of feed |
|---|---|
| 1–60 | 7 |
| 61+ | 6 |
Calculation of weight gain, milk production, and manure production with a suboptimal diet
For cattle aged 1–60 days:
N = Nₒ × A / 7
For cattle aged 61 days and older:
N = Nₒ × A / 6
Variables:
| Variable | Explanation |
|---|---|
| Nₒ | parameter value for optimal diet |
| A | number of feed types actually received |
Calculating the pregnancy length and preparing for mating with a suboptimal diet
For cattle aged 1–60 days:
N = Nₒ × 7 / A
For cattle over 60 days old:
N = Nₒ × 6 / A
The variables are similar to the previous ones.
Examples
Situation 1
On the 60th day, a young bull (1–60 days) received only 5 types of feed out of 7, and 2 types were missing.
- Weight gain with the optimal diet: 16 kg → with non-optimal:
X = 16 × 5 / 7 ≈ 11.42 kg
- Manure production with the optimal diet: 28 kg → with suboptimal:
X = 28 × 5 / 7 ≈ 20.00 kg
Situation 2
An adult cow (241–420 days, weight 350 kg) receives only 3 types of feed out of 6, half of one type → a total of 3.5 types.
Initial parameters with the optimal diet: weight gain 0.275 kg, manure production 21 kg, milk 5.5 l, preparation for mating 6 days, gestation 26 days.
- Weight gain:
X = 0.275 × 3.5 / 6 ≈ 0.16 kg
- Manure production:
X = 21 × 3.5 / 6 ≈ 12.25 kg
- Milk production:
X = 5.5 × 3.5 / 6 ≈ 3.20 l
- Preparing for mating:
X = 6 × 6 / 3.5 ≈ 10.28 days
- Pregnancy:
X = 26 × 6 / 3.5 ≈ 44.57 days
Situation 3
An adult cow (61–240 days) was not given any type of feed.
- Weight gain with optimal diet: 1.083 kg → weight loss:
X = 1.083 × 1 / 3 ≈ 0.36 kg
- Products (milk and manure) are not produced.
- The processes of pregnancy and preparation for mating stop for a day.