In the last lecture we looked at the standards required for the maintenance of a dairy cow. In this lecture, we will consider the standards required for the cow’s production, which is milk. Nutrient requirements for milk vary according to the butterfat content, and the table below gives the nutrient requirements for milk from 2.5% to 6% butterfat.

Table 1: Milk production: nutrients required per kilogramme of milk

Percent FatCP kgD.C.P. kgT.D.N. kgCa kgP kg

Using this table we can calculate the nutrients required for any given amount of milk. Take the cow giving 20kg a day of milk with a butterfat of 4%; her production requirements would be:

  • T.D.N.         0,33 x 20    =   6.60kg a day
    • D.C.P.         0,051 x 20  =   1.02kg a day

A cow giving 5kg of milk a day with a butterfat of 3,5% would require:

  • T.D.N.         0,305 x 5 = 1.525 or 1,53kg a day
    • D.C.P.         0,048 x 5 = 0.240 or 0,24kg a day.

Remember that these are production requirements. In order to calculate the total requirements of a milking cow, we have to add together the maintenance and the production requirements.

As an example, consider the case of a cow weighing 600kg, and giving 20kg a day of milk with a butterfat of 3,5%. Her total daily requirements of Dry Matter, T.D.N and D.C.P. would be calculated as follows:


  • 3% of her bodyweight 600 x 3/100 = 18kg of D.M.


  • Requirement for Maintenance = 4.20kg per day
    • Requirement for production    = 0.305 x 20 = 6.10kg per day
    • Total Requirement = 10.30kg per day


  • Requirement for maintenance = 0.345kg per day
    • Requirement for-production    = 0.048 x 20= 0.960kg per day
    • Total Requirement = 1.305kg per day

The daily requirement for calcium and phosphorus can be calculated in exactly the same way, using the table for maintenance in the last lecture, and the table for production requirements given in this lecture.


  • Requirement for maintenance = 0.022kg per day
    • Requirement for production    = 0.026 x 20= 0.520kg per day
    • Total Requirement = 0.542kg per day


  • Requirement for maintenance = 0.017kg per day
    • Requirement for production    = 0.019x 20=0.380kg per day
    • Total Requirement = 0.39 kg per day

Now we can state the daily requirement of this cow as follows: (rounding the figures off to two decimal places):

Dry MatterT.D.N.D.C.P.CaP

Let us look at this same animal after she has been in-milk for some time and her daily production has dropped to 5kg of milk with a buttermilk of 3,5%. Her daily requirements will be:

Dry Matter 600 x 3/100 = 18kg of D.M


  • Requirement for maintenance = 4.20kg
    • Requirement for production    = 0.305 x 5 = 1.52kg
    • Total Requirement = 5.72kg


  • Requirement for maintenance = 0.345kg
    • Requirement for production    = 0.048 x 5 = 0.240kg
    • Total Requirement = 0.585kg


  • Requirement for maintenance = 0.022kg
    • Requirement for production    =    0.026 x5   =    0.130kg
    • Total Requirement = 0.152kg


  • Requirement for maintenance = 0.017kg
    • Requirement for production    = 0.019 x 5 = 0.090kg
    • Total Requirement = 0.107kg

Now we can compare the daily requirements of this cow when she is giving 20 kg of milk a day and when she is giving 5 kg.

Daily ProductionD.M.T.D.N.D.C.P.CaP

One can notice that her daily requirements have dropped so that she can be fed a ration containing fewer nutrients when she is giving 5kg of milk a day. Her total capacity for dry matter, or her appetite, has remained the same.


Having calculated the maintenance and production requirements for a milking cow, how do we feed a ration to meet these requirements? If you look at the requirements of the cow giving 20 kg of  milk, you can see that you could calculate a ration to meet these requirements and the same applies to the cow when she is giving 5kg. However, this would mean feeding all sorts of different rations to a herd of cows, because they would be producing very different amounts of milk. In practice, dairy cows are fed as a herd, using roughages or succulent foods to meet the maintenance requirements of the cows, and a concentrate ration for production.


These are bulky foods and they are divided into dry roughages and succulent roughages. Dry roughages are foods like hay, straw and stover. Succulent roughages are silage, green oats or lucerne and grass.


These are dry foods which are high in nutrients. Concentrate rations are based on cereals. They are mainly maize in Southern Africa and America and barley in Europe, with the protein content of the ration being increased by adding high protein foods like soybean meal, groundnut meal, cottonseed meal or urea.

In the dairy herd, a maintenance ration is worked out and fed to meet the maintenance requirements of the cows, using the average weight of the cows in the herd. The production ration is worked out to meet the requirements of the milk production, based on the butterfat average of the herd. This ration is fed according to the daily milk yield of each cow in the herd.


The normal practice with a dairy concentrate ration is to feed at the rate of 0.5kg of the concentrate for every 1kg of milk produced by the cow. A cow giving 1 kg of milk a day would be fed 0.5kg of concentrate a day. A cow giving 6kg of milk a day would be fed 0.5 x 6 = 3.0 kg of concentrate and a cow giving 25kg of milk a day would be fed 0.5 x 25 = 12.5kg of concentrate a day. It follows that, if we work out a ration in which 0.5kg will meet the requirements for 1kg of milk, this ration can be fed to all the cows in the dairy by varying the amount according to the amount of milk given by each cow.

Take the case of a dairy herd with a butterfat average of 4%. The requirements for 1kg of milk taken from the table on Page 2 are:

  • T.D.N.     0.33kg
    • D.C.P.     0.051kg

Now look at the nutrient content of maize, which has a T.D.N. of 80% and a D.C.P. of 6.5%. If maize was fed as a concentrate ration, 0.5kg would supply:

  • T.D.N. 80/100 x 0.5 = 0,40kg
  • D.C.P. 6.5/100 x 0.5 = 0.0325kg

This is supplying too much energy (T.D.N.) and too little protein – so that maize by itself is not balanced for milk production. We must add something to the maize before it can be called a concentrate ration for milk production. Suppose we use Groundnut Cake which has a T.D.N. of 75%, and a D.C.P. of 37%, and make a mixture of 60kg of Maize and 10kg of Groundnut Cake. This would supply the following nutrients:

60 kg of Maize would supply80/100 x 60 = 48kg and6.5/100 x 60 = 3.9kg
10kg of G.N.C. would supply75/100 x 10 = 7.5kg 37/100 x 10 = 3.7kg
70kg of mixture would supply55.5kg and7.6kg
1kg of mixture would supply55.5/70 = 0.79kg and 7.6 /70          0.109kg
0.5kg of mixture would supply0.79 x 0.5 = 0.39kg and 0.109 x 0.5 = 0.054

This percentage T.D.N. and D.C.P. in this mixture would be 79% T.D.N. and 10% D.C.P.

This ration would satisfy the requirements for milk production of 4% butterfat if fed at 0.5kg for each 1kg milk.

We can check this quite easily by considering the cow giving 20kg of milk a day and the cow giving 5kg of milk a day.


20kg a day
5kg a day
0.33 x 20 = 6.60kg
0.33 x   5 = 1.65kg
0.051 x 20 = 1.02
0.051 x   5 = 0.26

For 20kg a day, 20 x 0.5 = 10kg of mixture would be fed and this would supply:

  • 0.79 x 10 = 7.9kg of T.D.N.
  • 0.109 x 10 = 1.09kg of D.C.P.

For 5kg a day, 5 x 0.5 = 2.5kg of mixture would be fed and this would supply:

  • 0.79 x 2.5 = 1.98 of T.D.N.
  • 0.109 x 2.5 = 0.27 of D.C.P

In fact, this mixture is providing a bit too much energy. Try working out the ration using 50kg of maize and 10kg of G.N.C. and see if this reduces the energy in the final mixture.

This is a very simple ration containing only 2 ingredients, maize and groundnut cake. In practice, more ingredients would be used so that the protein would come from several sources because of the different Biological Value of different proteins.