# Sub-optimal (partial) rations for pigs

If it happens that on any day the animal receives less feed than necessary for an optimal ration, such a feed ration is considered sub-optimal (partial). This will have the following consequences:

• The rate of weight gain will decrease.
• Manure production will decrease.
• The results of sporting events will worsen.
• The preparation time for mating will increase.
• The gestation period will increase.

How do I know how strong these deviations from normal values will be? It's pretty simple.

Calculation of weight gain, manure production, and consideration of athletic performance in suboptimal diets

## Calculation of the corresponding parameters in case of a 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 - number of feed types needed for an optimal diet (7 - for the age group of pigs 1-30 days; 6 - for other age groups).

Calculation of gestation period 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 - number of feed types needed for an optimal diet (7 - for the age group of pigs 1-30 days; 6 - for other age groups).

## Examples

Below we will look at a few examples to help you understand how to calculate production, performance, and gestation and mating preparation times in pigs 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

The young boar (age group 1-30 days) on the 30th day received only 5 types of feed 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 6 kg, and the manure production - 8 kg. The boar also sports and has 5,000 Dexterity units.

With the above suboptimal diet, the weight gain will be:

X = 6 - 5/7 = ~4.28 kg.

At the specified suboptimal ration, manure production would be:

X = 8 - 5/7 = ~5.71 kg.

With the specified suboptimal diet, the animal's effective Dexterity score during the run will be:

X = 5,000 - 5/7 = ~3,571 units.

### Situation 2

An adult sow (age group 31-350 days, initial weight 100 kg) consistently 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. We take into account that the weight gain at the optimal diet should have been 0.5 kg, the production of manure - 8 kg, the period of preparation for mating - 10 days, the period of pregnancy - 35 days.

With the above suboptimal diet, the weight gain will be:

X = 0.5 - 3.5/6 = ~0.29 kg.

At the specified suboptimal ration, manure production would be:

X = 8 - 3.5/6 = ~4.66 kg.

With the above suboptimal diet, the preparation time for mating will be:

X = 10 - 6/3.5 = ~17.14 days.

With this sub-optimal diet, the gestation period will be:

X = 35 - 6/3.5 = ~60 days.

### Situation 3

The adult sow (age group 31-350 days) did not receive any type of feed (remained hungry). Take into account that the weight gain at the optimal diet should have been 0.5 kg.

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

X = 0.5 - 1/3 = ~0.16 kg.

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