Suboptimal (incomplete) diet#

Suboptimal feed ration

If cattle receive less feed than they need to achieve an 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 period of pregnancy is extended.

To avoid negative consequences, it is important to provide complete, balanced feeding.

Table: Amount of feed required for an optimal diet depending on age group

Age, daysTypes of feed
1–607
61+6

Calculation of weight gain, milk production, and manure production with a suboptimal diet#

For cattle aged 1–60 days:

\( N = N_о * A / 7 \)

For cattle aged 61 days and older:

\( N = N_о * A / 6 \)

Variables:

VariableExplanation
\( Nо \)parameter value for optimal diet
\( A \)number of feed types actually received
\( B \)the number of feed types required for an optimal diet

Calculating the timing of pregnancy and preparation for mating with a suboptimal diet#

For cattle aged 1–60 days:

\( N = N_о * 7 / A \)

For cattle over 60 days old:

\( N = N_о * 6 / A \)

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 optimal diet: 16 kg → with non-optimal:
\( X = 16 * 5 / 7 \approx 11.42 \text{ kg} \)
  • Manure production with optimal diet: 28 kg → with non-optimal:
\( X = 28 * 5 / 7 \approx 20.00 \text{ kg} \)

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 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:
\( X = 0.275 * 3.5 / 6 \approx 0.16 \text{ kg} \)
  • Manure production:
\( X = 21 * 3.5 / 6 \approx 12.25 \text{ kg} \)
  • Milk production:
\( X = 5.5 * 3.5 / 6 \approx 3.20 \text{ l} \)
  • Preparing for mating:
\( X = 6 * 6 / 3.5 \approx 10.28 \text{ days} \)
  • Pregnancy:
\( X = 26 * 6 / 3.5 \approx 44.57 \text{ days} \)

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:
\( X = 1.083 * 1 / 3 \approx 0.36 \text{ kg} \)
  • Products (milk and manure) are not produced.
  • The processes of pregnancy and preparation for mating stop for a day.