Suboptimal (incomplete) diet

If on any given day an animal has received less feed than is necessary for an optimal ration, that feed ration is considered sub-optimal (that is, incomplete).

This will have the following consequences:

The rate of weight gain will decrease.
Manure production will decrease.
Milk production will decrease.
The preparation time for mating will increase.
The period of pregnancy will increase.

How do I know what the deviations from normal values will be? It's pretty simple.

Calculation of weight gain, milk and manure production at suboptimal rations

Calculation of the corresponding parameters in case of suboptimal diet is carried out according to the formula:

N = No - A/B,

Where:

No - value of the parameter at the optimal ration;
A - the number of actually received types of feed;
B - the number of necessary types of feed for an optimal diet (7 - for the age group of cattle 1-60 days; 6 - for other age groups).

Calculation of the timing of pregnancy and preparation for mating in case of suboptimal diet

Calculation of the corresponding parameters in case of suboptimal diet is carried out according to the formula:

N = No - B/A,

Where:

No - value of the parameter at the optimal ration;
A - the number of actually received types of feed;
B - the number of necessary types of feed for an optimal diet (7 - for the age group of cattle 1-60 days; 6 - for other age groups).

Examples

Below we will look at some examples of situations that will help you understand how to calculate the amount of production, as well as the timing of pregnancy and preparation for mating in cattle with sub-optimal (incomplete) feed rations.

The consequences of any intermediate deviations from the optimal feed ration are calculated in the same way.

Situation 1

A young steer (age group 1-60 days) on the 60th day received only 5 types of food out of 7, and 2 types did not receive at all. Take into account that the weight gain at the optimal diet on that day should have been 16 kg, and the manure production - 28 kg.

With the above non-optimal (incomplete) diet the weight gain will be:

X = 16 - 5/7 = ~11.42 kg.

At the specified non-optimal (incomplete) ration, manure production would be:

X = 28 - 5/7 = ~20.00 kg.

Situation 2

A mature cow (age group 241-420 days, initial weight 350 kg) continuously receives only 3 types of feed out of 6, and one type receives exactly half. Thus, the animal receives a total of 3.5 types of food. Consider that the weight gain at the optimal diet was to be 0.275 kg, manure production - 21 kg, milk production - 5.5 liters, the period of preparation for mating - 6 days, the period of pregnancy - 26 days.

With the above non-optimal (incomplete) diet the weight gain will be:

X = 0.275 - 3.5/6 = ~0.16 kg.

At the specified non-optimal (incomplete) ration, manure production would be:

X = 21 - 3.5/6 = ~12.25 kg.

At the specified suboptimal (incomplete) ration, milk production would be:

X = 5.5 - 3.5/6 = ~3.20 liters.

At the specified suboptimal (incomplete) ration, the period of preparation for mating will be:

X = 6 - 6/3.5 = ~10.28 days.

With the above non-optimal (incomplete) ration, the period of pregnancy will be:

X = 26 - 6/3.5 = ~44.57 days.

Situation 3

The adult cow (age group 61-240 days) did not receive any type of feed (remained hungry). Consider that the weight gain at the optimal diet should have been 1.083 kg.

In the absence of all types of food, the weight loss will be:

X = 1.083 - 1/3 = ~0.36 kg.

No products (milk and manure) will be produced. The processes of pregnancy and preparation for mating will stop on this day.

Optimal feed ration Types of products