RATION FORMULATION
Ration formulation is a process by which different feed ingredients are combined in a proportion necessary to provide the animal with proper amount of nutrients needed at a particular stage of
production.
• It requires the knowledge about nutrients, feedstuffs and animal in the development of nutritionally adequate rations that will be eaten in sufficient amounts to provide the level of production at a reasonable cost. The ration formulated should be palatable and will not cause any serious digestive disturbance or toxic effects to the animal.
o The nutrient requirements can be arrived using feeding standards.
o The list of commonly available feeds in that region is prepared.
o The nutritional value of the feeds is obtained from any standard source such as NRC.
• Using the above information rations can be prepared by several methods that include
o Square Method
o Simultaneous Equation Method
o Two-by-two Matrix method
o Trial and Error Method and
o Linear Programming (LP)
Factors to be considered in ration formulations
• Acceptability to the animal – The ration formulated has to be palatable.
• Digestibility – The nutrients in the feed have to be digested and released into the gastrointestinal tract to be utilized by the animal. Rations with high fiber content cannot be tolerated by poultry and swine.
• Cost – The requirement of the animal can be met through several combinations of feed ingredients. However, when the costs of these ingredients are considered, there can only be one least-cost formulation. The least-cost ration should ensure that tile requirements of the animal are met and the desired objectives are achieved.
• Presence of anti-nutritional factors and toxins. The presence of anti-nutritional factors in the feed affects the digestion of some nutrients and makes them unavailable to the
animal. The inclusion of these feed ingredients should therefore be limited in the formulation.
• Other factors that should be considered are texture, moisture and the processing the feed has to undergo.
• This is relatively simple and easy to follow. It satisfies only one nutrient requirement and uses only two feed ingredients. Another limitation is that the level of nutrient being computed should be intermediate between the nutrient concentration of the two feed ingredients being used.
The Pearson square or box method
• The Pearson square or box method of balancing rations is a simple procedure that has been used for many years. It is of greatest value when only two ingredients are to be mixed. The nutrient requirement is noted in the middle of the square this value in the middle of the square must be intermediate between the two values that are used on the left side of the square which are actually the nutrient content of the two ingredients that are to be used. For example, the 14 percent crude protein requirement has to be intermediate between the soybean meal that has 45 percent crude protein or the corn that has 10 percent crude protein. Subtract the nutrient value from the nutritional requirement on the diagonal and arrive at a numerical value and note it down on the right side of the square. Two sets of values will be got. By summing those parts and dividing by the total, you can determine the percent of the ration that each ingredient should represent in order to provide a specific nutrient level.
SIMULTANEOUS EQUATION METHOD
• This is an alternative method for the square method using a simple algebraic equation.
• Here, a particular nutrient requirement is satisfied using a combination of two feed ingredients.
Ration formulation poultry algebraic method
Fix the nutrient requirements for the feed to be formulated
• Slack space for additives 0.5%
• Slack space for lysine – 0.05
• Slack space for methionine – 0.15
• Slack space for common salt if fish meal not included – 0.4
• Slack space for calcium and phosphorus. Chicks 3, growers 2, layers 10 -12
• Fix level of animal origin protein source if it is going to be included max 10%
• Fix level of cereal milling byproducts to be included (refer maximum inclusion level) • Calculate the total of the ingredients so far added.
• Assuming that the following is the list of ingredients selected
Ingredients Inclusion level CP ME Kcal/Kg
Maize
Soya bean meal
Fish meal 8 4 224
Rice polish 5 0.55 150
Lysine 0.05
Methionine 0.15
Additives 0.5
Mineral mixture (Calcium, P etc) 2
4.55 324
• Total of ingredients other than Maize and Soya bean meal = 15.70 • Balance from maize and Soya bean meal = 100 – 15.70 = 84.30 • Total Protein requirement = 20 %
• Protein already supplied = 4.55
• Balance protein required = 15.45
Let X = Maize and Y = Soya bean meal
• X + Y = 84.3 (Equation 1)
• 0.09 X + 0.50 Y = 15.45 (Equation 2)
Multiplying Equation 1 with 0.50
• 0.50 X + 0.50 Y = 42.15 (Equation 3)
Subtracting equation 2 from equation 3
• 0.41 X = 26.7
• X = 65.12 = 65
• Y = 84.3 – 65.12 = 19.18 = 19.2
Ingredients Inclusion level CP ME Kcal/Kg
Maize 65 5.85 2145
Soya bean meal 19.2 9.6 518
Fish meal 8 4 224
Rice bran 5 0.55 135
Lysine 0.05
Methionine 0.15
Additives 0.5
Mineral mixture (Calcium, P etc) 2
Extra ricebran 0.1 0.011 2.7
Total 100 20.07 3024.7
TWO-BY-TWO MATRIX METHOD
This method solves two nutrient requirements using two different feed ingredients. A 2 x 2 matrix is set and a series of equations are done to come up with the solution to the problem.
TRIAL AND ERROR METHOD
• This is the most popular method of formulating rations for swine and poultry.
• As the name implies, the formulation is manipulated until the nutrient requirements of the
animal are met.
• This method makes possible the formulation of a ration that meets all the nutrient requirements of the animal.
• Greater control can be had on implementing restrictions and judging inclusion levels
• It is a time consuming method involving a lot of calculations and meeting out specifications may not be very precise.
LINEAR PROGRAMMING (LP)
• This is a method of determining the least-cost combination of ingredients using a series of mathematical equations. There are many possible solutions to each series of equations, but when the factor of cost is applied, there can only be one least cost combination.
• An electronic computer is capable of making thousands of calculations in a very short time. However, the machine is incapable of correcting errors resulting from incorrect data
and errors in setting up of the program. Therefore, the resultant rations obtained from linear programming will be no better than the information and values which are entered
FORMULATION OF RATIONS – CATTLE
CALCULATION OF NUTRIENT REQUIREMENTS:
The nutrient requirements can be calculated by using a few factors without referring to the table. The protein and energy requirement is related to metabolic body weight.
CALCULATION OF BODY WEIGHT:
Where it is not possible to weigh the animals the body weight can be calculated by using shaeffers formula.
FOR CATTLE:
LG2
W =
300
W = Body weight in pounds
L = Length of the animal in inches. It is from the point of the shoulder to the
point of the buttock
G = Girth in inches. It is the circumference measured just behind the point of
elbow.
The body weight can be converted to kgs.
1 kg. = 2.2 pounds.
FOR BUFFALOES:
GL
W =
Y
W = Weight of the animal in seers (1 seer is equal to 0.93 kgs.)
L = Length of the animal in inches.
G = Girth of animal in inches.
Y = Constant
When G is below 65 Y is 9.0
When G is between 65 and 80 Y is 8.5 When G is above 80 Y is 8.0
METABOLIC BODY WEIGHT (W0.75):
Body weight to the power of 0.75. This is calculated by using Logarithm tables of a calculator.
(1) Using Logarithm tables:
Logarithm value = 2.6021 x 0.75 = 1.9516
Antilogarithm }
value for metabolic = 89.44
boy weight
(2) Using a calculator:
Body weight to the power of 3 and press the square root key twice, you will get the metabolic body weight.
Eg. 400 x 400 x 400 = 64000000
8000 89.44
Dry matter intake
The total quantity of dry matter the animal can consume per day should be known, so that we can compute the ration in such a manner that the entire quantity of the nutrient requirement is present in the dry mater which the animal is able to consume. Cattle will generally consume 2.0 to 2.5 kg of dry matter per 100 kg. body weight. Buffaloes, crossbred animals and heavy yielders consume 2.5 to 3.0 kg dry matter per 100 kg body weight.
TDN REQUIREMENT FOR MAINTENANCE:
34g TDN/kg W0.75
W0.75 = Metabolic body weight.
DCP REQUIREMENT FOR MAINTENANCE:
2.84 g DCP/kg W0.75
W0.75 = Metabolic body weight.
NUTRIENT REQUIREMENT FOR LACTATION:
TDN requirement
330 g TDN/kg fat corrected milk.
Formula for fat correction of milk is 0.4 + 0.15F, where F is % of fat in milk.
DCP requirement
The protein content of milk can be calculated by using the following formula. Percentage of protein = 1.9 + 0.4F, where F is percentage of fat.
Then the DCP requirement is calculated by assuming that the Biological value of microbial protein in Cattle as 70.
Eg. The DCP requirement of 1 kg of milk with 5% fat
Percentage of protein = 1.9 + (0.4 x 5) = 3.9% or 39 g in 1 kg.
Quantity of DCP required is (39/70) x 100 = 55.7 or 56 g of DCP for 1 kg of milk with 5% fat.
Calculation of nutrient requirements for a cow weighing 400 kg and yielding 10 kg of milk with
5% fat. The metabolic body weight is 89.44.
CALCULATION OF DRY MATTER REQUIREMENT:
The dry matter requirement for crossbred heavy yielders is 2.5 to 3.0% of its body weight. For an animal weighing 400 kg. It is 400 x 2.5/100 = 10 kg or 400 x 3/100 = 12 kgs.
The dry matter requirement is 10 to 12 kg.
PARTITIONING OF DRY MATTER:
1/3 of dry matter from concentrate.
10 x 1/3 = 3.3 kg. }
12 x 1/3 = 4.0 kg
7 to 8 kg from roughage.
3 to 4 kg dry matter from concentrate
1/3 from green roughage = 2 to 3 kg.
2/3 from dry roughage = 4 to 5 kg.
TDN REQUIREMENT:
For maintenance: 34 x 89.44 = 3040g
For milk production: 330g TDN/kg. FCM (Fat corrected milk)
FCM = 0.4 + 0.15F
0.4 + (0.15 x 5) = 1.15
1 kg of milk with 5% fat = 1.15 kg of milk with 4% fat.
For 10 kg = 11.5 x 300 = 3795 g
Total TDN required = for maintenance = 3040 g
for Lactation = 3795 g
6835 g
DCP REQUIREMENT:
For maintenance 2.84 x 89.44 = 254 g.
For milk production:
10 kg of milk with 5% fat.
Percentage of protein = 1.9 + 0.4 F
= 1.9 + (0.4 x 5) = 3.9
DCP required = 3.9 x 100 = 5.6g/100g.
70
56g for 1 kg.milk.
560g DCP/10 kg.milk.
Total DCP required: Maintenance 254g
Lactation 560g
814g
COMPUTATION OF RATION:
D.M DCP TDN
Requirement 10-12 814g 6835g
Ingredients Quantity as fed – – –
Groundnut cake 1.1 kg. 1.0 kg 460g 790g
Cotton seed 2.2 kg. 2.0 kg 250g 1780g
Tapica thippi 1.1 kg. 1.0 kg 15g 830g
Bengal gram husk 1.1 kg. 1.0 kg – 600g
Green grass 10 kg. 2.5 kg 100g 1500g
Paddy straw 3.3 kg. 3.0 kg – 1350g
10.5 kg 825g 6850g
To the concentrate part add 2% mineral mixture and 1% salt. If green grass is not included in the ration Vitamin A should be supplemented 1 mg. of carotene = 400IU of Vitamin A.
Nutritive value of feeds and fodder
DM
% CP
% DCP
% TDN
% ME
kcal/kg
Cereal grains 90 8-10 4-7 70-85 2500-3000
Oil cake 90 40-50 38-45 55-65 2000-2500
Bran 90 6-12 5-10 50-60 1500-1800
Non legume fodder 85 6-8 2-3 50-60
Legume hay 85 15-18 11-12 50-60 –
Paddy straw 90 3 0 45 –
