Food is required by an animal for the following reasons:

For Maintenance:

  • To keep up the body temperature;
    • To supply energy for breathing;
    • To maintain circulation of the blood; and
    • To repair and to renew worn and damaged tissues.

For Production:

  • To provide for the growth of the young animal;
    • To provide for the growth of the foetus inside the mother;
    • To provide energy for work – e. g. horses or draught oxen; and
    • To provide for the production of milk or eggs or meat.


  • To satisfy the animal’s requirements for a particular purpose; i.e. maintenance or both maintenance and production.
  • A dairy cow that is dry will require only enough food to satisfy her maintenance requirements. A cow that is milking will require enough food for her maintenance and her milk production.
  • To use the available foods to the best advantage. All farms produce crop, and crop residues which can be fed to cattle, sheep, pigs or poultry. A knowledge of rationing helps a farmer to make the best possible use of these foods.
  • To feed animals as economically as possible while at the same time keeping up the best possible production. Cattle foods are expensive and a farmer cannot afford to overfeed his stock or waste food in any other way.


The requirements of animals must be satisfied in all these respects:

  • Appetite: The animal’s appetite must be satisfied, because once a cow, steer or pig is full it will lie down, rest and digest its food. It is while they are lying down resting that the cow is converting food into milk and the steer or pig is converting food into meat. The daily intake of an animal is measured in dry matter, and a general rule is that cattle and sheep will require up to 3% of their bodyweight in dry matter a day. The figure used for dairy cattle is 3% and for beef cattle 2.5%.
  • For example, a cow weighing 600kg will require 600 x 3/100 = 18kg
  • of dry matter and this is the MOST she can eat in a day and it will satisfy her appetite and she will not eat any more. Suppose this animal is being fed on hay only. Hay has a dry matter (or D.M.) of 90%, so that in one day it would eat 18 x 100 /30 = 20kg of Hay.
  • If the animal is being fed on maize silage, a succulent feed with a D.M. of 30%, it would eat 18 x 100/30 = 60kg of silage daily.

A cow weighing 400 kg would consume 400 x 3/100 = 12.0kg.

of D.M. a day and if it were fed on hay, it would eat 12 x 100/90 = 13.3kg of hay a day.

Energy: The energy requirement of the animal is expressed as kilograms of Total Digestible Nutrients, or T.D.N. This energy is supplied from the fat, carbohydrate and some of the protein in the ration. The T.D.N. requirements for various types and sizes of animals are obtained from table, and the table for dairy cattle is given below:

Table: daily maintenance requirements for cattle

  • Protein: The protein requirements are expressed as kilograms of Crude Protein (C.P.) or Digestible Crude Protein (D.C.P.), and those for dairy cattle are shown above.
  • Minerals: The mineral requirements are expressed in grams for each mineral, the important ones being calcium (Ca), and Phosphorus (P).
  • Vitamins: These requirements are expressed in International Units (I.U.) for each vitamin. It is not necessary to add vitamins to cattle rations unless the animals are closely confined.


The table above gives the amounts of nutrients required by cows of differing weights for maintenance each day.

A cow weighing 500kg would need 0.638kg of Crude Protein or 0.30 of Digestible Crude Protein, 3.7kg of T.D.N., 20g of Calcium, and 15g of Phosphorus to maintain her in normal condition each  day. These nutrients would have to be supplied within a dry matter limit of 500 x 3 = 15kg a day.


The first step in the calculation of any ration is to find out the nutrient requirements of the animal. In this case we are talking about the daily maintenance requirements of dairy cows.

Example 1:

Calculate the daily maintenance ration for a dairy cow weighing 650kg. The following food is available on the farm:

 DM %T.D.N. %D.C.P. %Ca %P %
Good Quality Hay90503.50.50.2

The first step is to refer to the table for the daily maintenance requirements for this animal. These are:

Body MassDM kgT.D.N. kgD.C.P. kgCa kgP kg

Looking at the analysis of the hay you can see immediately that 10 kg will provide the following nutrients:

Dry MatterT.D.N.D.C.P.CaP
  90/100 x 10 = 9kg   50/100 x 10 = 5kg  3.5/100 x 10 = 0.35  0.5/100 x 10 = 0.05  0.2/100 x 10 = 0.02

Now you can compare the requirements of the cow with the nutrients provided by the hay.

Requirements Provided by 10kg of Hay
DM19.5kg9kgWell within the maximum intake of the cow
T.D.N.4.5kg5kgSlightly high
D.C.P.0.365kg0.35kgSlightly low
P0.018kg0.02kgSlightly high
      Lignified: make rigid and woody by the deposition of lignin in cell walls.

This ration of hay would be well within the appetite of the cow. It would provide slightly too much energy (T.D.N.), and not quite enough protein (D.C.P. It is high in calcium and phosphorus. However, this type of good quality hay made from grass that has not matured and become too lignified is quite well balanced for the maintenance of dairy cows. How many kgs of this hay would be needed to supply the maintenance requirements of a cow weighing 500 kg?

Example 2:

 DM %T.D.N. %D.C.P. %Ca %P %
Maize Silage (average quality)

We will start with 25kg of silage which is a reasonable amount for a large dairy cow to eat each day. This would supply the following nutrients:

25kg Silage Requirements
Dry Matter27/100 x 25 = 6.75kg 19.5kg
T.D.N.17/100 x 25 = 4.25kg 4.5kg
D.C.P.1.0/100 x 25 = 0.25kg 0.365kg
Ca0.08/100 x 25 = 0.02kg 0.024kg
P0.06/100 x 25 = 0.01kg 0.018kg

From these results we see that the Dry Matter is well within the appetite of the cow. The T.D.N. is slightly low, the D.C.P. is very low and the calcium and phosphorus are slightly low. We need to add something to increase the protein content of the ration. Looking at the analysis of lucerne you can see it is high in protein and rather low in energy (T.D.N.). Adding 3 kg of freshly-cut lucerne will increase the nutrient value of the ration as follows:

3kg Lucerne 
Dry Matter25/100 x 3 = 0.75kg
T.D.N.14/100 x 3 = 0.42kg
D.C.P.3.5/100 x 3 = 0.10kg
Ca0.4/100 x 3 = 0.01
P0.8/100 x 3 = 0.02kg

The complete maintenance ration of 25kg of Maize Silage and 3kg of lucerne will supply:

Dry Matter6.75 +0.75 =7.50kg19.5kg
T.D.N.4.25 +0.42 =4.67kg4.5kg
D.C.P.0.25 +0.10 =0.35kg0.365kg
Ca0.02 +0.01 =0.03kg0.023kg
P0.01 +0.02 =0.03kg0.018kg

This ration is well within the appetite of the animal. It is slightly high in T.D.N., slightly low in D.C.P., and high in Ca and P. However, it would make a satisfactory maintenance ration for a cow of 650kg.

In practice, a maintenance ration would be worked out for a whole herd, using the average weight of all the cattle. This would be done for both dairy and beef herd.

If the above ration had been calculated for a dairy herd of 50 cows averaging 650kg, and this could be the case with Friesland or Holstein cows, the daily maintenance ration for the herd would be:

  • 25 x 50 = 1250kg of Maize Silage
  • 3 x 50 =   150kg of Fresh Lucerne

This ration would be fed in two feeds a day, half the ration in the morning and half in the evening. In the case of a beef herd the ration could be fed in two feeds or in a single feed given in the morning.

The important point to remember in rationing cattle is that, although it seems to be a matter of exact calculations, the nutrient content of foodstuffs does vary, as do the nutrient requirements of cattle.

Even when foodstuffs on the farm have been analysed silage and hay can vary within a silage pit or hay stack. In the case of cattle, one is working with average weights, and the nutrient requirements of individuals can vary.

The main objective in rationing is to make sure cattle are not being over fed or under fed, particularly with protein, and that food and money are not being wasted. Under feeding is as wasteful as over feeding.


In the case of cattle grazing on the veld or on pastures you have to look at feeding from a different viewpoint. The animals are not being rationed in the sense that their food is being restricted. They will fill themselves up with grass and satisfy their appetite, and the question is “Have they obtained sufficient nutrients from the grass?”. Consider the cow weighing 650kg; her daily capacity will be 19,5kg of dry matter. What nutrients will she get from this amount of grass?

An earlier lecture gave a table showing the nutrient value of grass in different stages of growth. In December, when the veld is at its most nutritious, the figures are:

  • T.D.N. 62.5%
    • D.C.P. 6.03%

These figures are percentages relating to Dry Matter so that the cow would get these nutrients from her 19,5kg of dry matter:

 Requirements of Cow
T.D.N.62.5/100 x 19.5 = 12.18kg 4.5kg
D.C.P.6.03/100 x 19.5 = 1.17kg 0.365kg

One can see that the cow is getting well above her maintenance requirements from the grass. The extra nutrients will be used for production, either of milk or fat.

Let us look at the situation in March when the grass is longer, lignified and less nutritious. The analysis is:

  • T.D.N. 47.2%
    • D.C.P. 0.58%

The cow eating 19,5kg of dry matter per day will get the following nutrients:

 Requirements of Cow
T.D.N.47.2/100 x 19.6 = 9.20kg4.5kg
D.C.P.0.58/100 x 19.5 = 0.11kg 0.365g

The situation has changed, and although the cow is getting more energy or T.D.N. than she needs for her maintenance, she is not getting enough protein; her deficiency is 0,255kg each day.

This is the problem with veld grass in winter. A deficiency of protein and with such a deficiency for maintenance, there can be no nutrients left over at that season for production.