Computer Feed Formulation
Introduction
Computer programs that formulate animal rations use terms such as dry matter percentage, dry matter price, nutrient concentration, least-cost balanced ration, linear programming, etc. This paper defines these terms and shows relationships between them. There is also an explanation of how linear programming restricts feeds and nutrients and optimizes properties of rations other than cost. What is not included are topics about the health, growth, and production of animals.
Basic Terminology
The first seven topics in this paper, from “Weights and dry matter” to “Linear programming”, cover the basic terminology and ideas that are common to all feed formulation programs. The remaining topics are mostly independent of one another and may be read in any order. Many of the topics in this paper are well-known; some -- like “Restricting dry matter in a ration”, “Requirement feeding” and “Optimizing properties of a ration” -- are less well-known. The topic “Feeding animals with fixed intake” is believed to be new.
Special Note
In this paper, water is a nutrient, but not a feed. There are two reasons why water will not be considered a feed. Firstly, it is unusual in the literature of feed formulation to see water listed as one of the feeds making up an animal ration. The second and more important reason is that some calculations made by computer feed formulation programs involve division by the dry matter weight of a feed, and the dry matter weight of water is zero.
Basic Terminology
Weights and dry matter
A feed is an ingredient or material (other than water, as explained earlier) fed to animals for the purpose of sustaining them. The term feed will include basic ingredients, such as alfalfa hay and limestone, as well as mixtures of basic ingredients, which are called rations, such as vitamin and mineral supplements and complete rations. Feeds are customarily weighed in pounds, kilograms or (American or metric) tons; and the pound, kilogram or ton is the unit of weight of the feed.
Suppose a container holds AF# pounds of a feed and after all moisture has been removed it holds DM# pounds of the feed. Then AF# is the as fed weight of the feed (in pounds), and DM# is the dry matter weight of the feed (in pounds).
Clearly, for a given amount AF# of feed, 0 < DM# < AF#. Then multiplication by 100 / AF# shows
0 < 100 • DM# / AF# < 100
provided AF# > 0.
The expression DM# / AF# is the dry matter portion of the feed and is written POR = DM# / AF#. The expression 100 • DM# / AF# is the dry matter percentage of the feed and is written DRY = 100 • DM# / AF# = 100 • POR, or
(1) DM# = AF# • (DRY / 100) = POR • AF#
Equation(1) is independent of the unit of weight. That is, it is valid if weight is measured in pounds (lb), kilograms (kg), or tons (ton).
The number 100 - DRY is the moisture percentage of the feed and is written MOIST = 100 - DRY. The numbers POR, DRY , and MOIST are properties of the feed and do not depend on the amount of feed in the container.
Example
Suppose a sack holds 100 pounds of corn (AF# = 100), and corn has a dry matter percentage of 90% (DRY = 90). The dry matter percentage of corn and the moisture percentage of corn (MOIST = 100 - DRY = 10) are both properties of corn and do not depend on the amount of corn in the sack. The dry matter percentage of corn (DRY = 90), however, does affect the dry matter weight of the sack, which is DM# = AF# • (DRY / 100) = 90 lb.
Suppose a container holds AF# > 0 pounds of feed and, after all of the moisture has been removed, it holds DM# pounds of feed. Then DM# > 0, because water is the only material that has 100% moisture, and we have agreed that water is not a feed. Therefore, if a container holds AF# > 0 pounds of feed, then
0 < DM# < AF#
0 < POR < 1
0 < DRY < 100
0 < MOIST < 100
Feed cost and feed prices
Example
Suppose a 100-pound sack of corn can be purchased for $8.00. Then the "as fed weight" of the sack is 100 lb, and the "as fed price" of corn is $8.00 / 100 lb = 0.08 $/lb. If corn is 90% dry matter, then the "dry matter weight" of the sack is 90 lb, and the "dry matter price" of corn is $8.00 / 90 = 0.09 $/lb. The amount $8.00 is known as the "cost" of the sack of corn.
Suppose a container holds AF# > 0 pounds (lb) of feed and, after all moisture has been removed from the feed, it holds DM# > 0 pounds of feed. If the feed in the container can be purchased for C dollars, then C is the feed cost (in $), C / AF# is the as fed price of the feed (in $/lb), and C / DM# is the dry matter price of the feed (in $/lb). These prices will be written AF$ and DM$, respectively. That is,
(2) AF$ = C / AF#
DM$ = C / DM#
Since AF$ • AF# = C = DM$ • DM#, the cost of the feed is
(3) AF$ • AF# = DM$ • DM#
Equation (3) shows that the cost of the feed is price times weight, and it does not matter if price and weight are on an as fed basis or on a dry matter basis. Equation (3) is valid if the unit of cost is the dollar ($), the euro (€) or the peso (P).
Equations (1) and (3) show that
(4) AF$ = DM$ • (DRY / 100)
Equation (4) is valid if price is measured in $/lb, €/kg or P/ton.
If Feed i has as fed weight AFi# > 0, dry matter weight DMi# > 0 and costs Ci dollars, for i = 1, 2, ..., n, then a mixture that consists of AFi# pounds of Feed i, for i = 1, 2, ..., n, has as fed price
(C1 + C2 + ... + Cn) / (AF1# + AF2# + ... + AFn#)
and dry matter price
(C1 + C2 + ... + Cn) / (DM1# + DM2# + ... + DMn#)
Example
Suppose a sack holds 100 pounds of corn (AF# = 100), corn has a dry matter percentage of 90% (DRY = 90) and the sack of corn can be purchased for $8.00 (C = 8). Then the as fed price of corn is AF$ = C / AF# = 0.08 $/lb. Since the dry matter weight of corn is DM# = AF# • (DRY / 100) = 90 lb, the dry matter price of corn is DM$ = C / DM# = 0.09 $/lb.
The as fed weight of a ration is always greater than or equal to its dry matter weight (AF# > DM#), and the as fed price of a ration is always less than or equal to its dry matter price (AF$ < DM$). These weights and prices are equal (AF# = DM# and AF$ = DM$) if and only if the ration has 100% dry matter (DRY = 100).
Nutrients and units
On page 70 of Feeds & Nutrition - Complete, first edition 1978, Ensminger and Olentine say “By definition, nutrients are the chemical substances found in feed materials that can be used and are necessary for the maintenance, production, and health of animals. The chief classes of nutrients are carbohydrates and fats (energy), proteins, minerals, vitamins, and water.”
Nutrients can be numerically measured. The unit of a nutrient is a name that suggests the method by which the nutrient is measured, and the quantity of a nutrient in a container of feed is the number that results from the measurement of the nutrient.
It is assumed that nutrient quantities can be added: If X pounds of Feed A with quantity P of nutrient N is mixed with Y pounds of Feed B with quantity Q of nutrient N, the resulting X + Y pounds of the mixture will have quantity P + Q of nutrient N. Also, nutrient quantities can be multiplied: If a > 0 and X pounds of Feed A has quantity P of nutrient N, then a • X pounds of Feed A will have quantity a • P of nutrient N. Finally, nutrient quantities are consistent: if X kg of Feed A has quantity Q of P, then any other sample of X kg of Feed A has quantity Q of P.
(These properties of nutrients, which are often assumed automatically by computer feed formulation programs, are ideals or approximations. For practical objections to the additive and multiplicative rules, see the later topic “Variable nutrient concentrations”, while the topic “Stochastic feed formulation” questions the consistency rule.)
Example
The following are some nutrient units: A unit of energy is the kilocalorie (kcal) or megacalorie (Mcal), and a unit of protein is the gram (g) or kilogram (kg). Minerals are measured in milligrams (mg), and vitamins are measured in micrograms (µg), milligrams (mg) or international units (IU). Water is measured in pounds (lb) or kilograms (kg).
The above list shows that many nutrient units are names of weights like the gram, kilogram or pound. For an example of nutrient quantities, a container of feed could have 5000 kcal of energy, 3 kg of protein, 20 g of magnesium, 400 IU of Vitamin A, and 8 kg of water.
Nutrient concentrations
Suppose a container holds AF# > 0 pounds of feed and, after all moisture has been removed from the feed, it holds DM# > 0 pounds of feed. If A is the quantity of a nutrient in the container, then A / AF# is the as fed concentration of the nutrient in the feed, and A / DM# is the dry matter concentration of the nutrient in the feed. These concentrations will be written AN* and DN*, respectively. That is
(5) AN* = A / AF#
DN* = A / DM#
If the unit of a nutrient is the same as the unit of weight of the feed containing the nutrient, then the as fed percentage of the nutrient is 100 • AN* and the dry matter percentage of the nutrient is 100 • DN*.
There are two special nutrient concentrations, percent (%) and parts per million (ppm), that don't appear to fit this pattern of something / unit of weight. But they can be made to conform by writing
x % = x kg/100kg = (x/100) kg/kg
x % = x g/100g = 10x g/1000g = 10x g/kg
x ppm = x mg/1000000mg = x mg/kg
Example
Suppose a container of feed has 5000 kcal of energy, 3 kg of protein, 20 g of magnesium, 400 IU of vitamin A, and 8 kg of water. If the as fed weight of the feed is 100 kg, then the as fed concentrations are 50 kcal/kg energy, 0.03 kg/kg (= 3%) protein, 0.2 g/kg (= 200 mg/kg or 200 ppm) magnesium, 4 IU/kg vitamin A, and 0.08 kg/kg (= 8%) water.
If the quantity of nutrient N is A = x lb and the as fed weight of the feed is AF# = 50 lb, then the as fed percentage of N is 100 • AN# = 100 • ( A / AF# ) = 100 • ( x lb / 50 lb ) = 100 • x / 50.
Equation (5) can be written
(6) A = AN* • AF# = DN* • DM#
Equation (6) shows that the quantity of a nutrient is concentration times weight, whether weight and concentration are on an as fed basis or dry matter basis.
Equations (1) and (6) show that
(7) AN* = DN* • (DRY/100)
100 • AN* = 100 • DN* • (DRY / 100)
Many tables give the nutrient concentrations or nutrient percentages of feeds. Some tables show them on an as fed basis, and some show them on a dry matter basis. Equation (7) is used to convert nutrient concentrations and nutrient percentages from one basis to the other.
If Feed i has concentration Ci of nutrient N, for i = 1, 2, ..., k, then a mixture that consists of Xi pounds of Feed i, for i = 1, 2, ..., k, will have the following concentration of nutrient N. (All ci's and Xi's are on an as fed basis, or all ci's and Xi's are on a dry matter basis.)
(8) (C1 • X1 + C2 • X2 + ... + Ck • Xk)/(X1 + X2 + ... + Xk)
This formula is established by the following argument in which all weights and concentrations are on an as fed basis, or all weights and concentrations are on a dry matter basis.
By Equation (6), Ci • Xi is the quantity of nutrient N in Xi pounds of Feed i, for i = 1, 2, ..., k. Since nutrient quantities can be added, C1 • X1 + C2 • X2 + ... + Ck • Xk is the quantity of nutrient N in X1 + X2 + ... + Xk pounds of the mixture. By Equation (5), the concentration of nutrient N in the mixture is given by Equation (8).
Nutrient prices
When formulating animal rations it is important to find inexpensive sources of nutrients. For example, needing more protein in a ration, it would be helpful to compare feeds high in protein and tell which of them is the most economical source of protein. That is, which feed has the smallest “price of protein” in the sense of the following definition.
If AF$ is the as fed price of a feed, and AN* > 0 is the as fed concentration of nutrient N in the feed, then the price of nutrient N in the feed is AF$ / AN*. In the case AN* = 0, the “price of nutrient N” is not defined (or is infinite).
Example
If corn has 8.5% protein (AN* = 8.5), and the price of corn is 155.00 $/kg (AF$ = 155), then the "price of protein in corn" is AF$ / AN* = 18.24 $/kg. If soybean has 37% protein (AF$ = 37) and the price of soybean is 300.00 $/kg (AF$ = 300), then the "price of protein in soybean" is AF$ / AN* = 8.11 $/kg, and soybean is cheaper than corn as a source of protein. The "price of protein" is not a fixed number, and depends on the choice of price units: $/kg, $/lb, €/ton, etc. But as long as the price units are the same for both ingredients, the ingredient with the smaller AF$ / AN* is the less expensive source for the nutrient.
To justify this definition, suppose k1 > 0 and k2 > 0 are numbers such that k1 • AF$ is the as fed price of a feed in $/kg and k2 • AN* > 0 is the as fed concentration of nutrient N in the feed in units of N/kg.
Then the cost of 1 kg of the feed is
k1 • AF$ • (1 kg)
but 1 kg of the feed contains k2 • AN* • (1 kg) units of N. Therefore, the cost of k2 • AN* • (1 kg) units of N is
k1 • AF$ • (1 kg)
Dividing by k2 • AN* • (1 kg), the cost of 1 unit of N is
k1 • AF$ • (1 kg) / k2 • AN* • (1 kg)
or, the cost of 1 unit of N is
(k1 / k2) • (AF$ / AN*)
Since k1 / k2 is a constant, the feed with the smallest nutrient price (AF$ / AN*) will also have the smallest cost of one unit of N, and this feed will be the most economical source of nutrient N.
When AN* > 0, dividing Equation (4) by Equation (7) gives
(9) AF$ / AN* = DM$ / DN*
showing that the price of a nutrient does not change if the quantities are on a dry matter basis.
NOTE: To see all nutrient prices, click By Nutrient in the MixitWin window. Then select a nutrient in the drop down box, and scroll to the rightmost column of the ingredient table.
Intake and daily Amounts
Dry matter intake (DMI) is the dry matter amount of feed that an animal consumes in a day. If an animal consumes AF# kg of feed, then Equation (1) shows that DMI = DM# kg. Dry matter intake depends on many factors including animal type, feed type, and feeding environment.
The daily amount (DA) of a nutrient is the number of grams, kilograms or pounds of the nutrient that an animal requires or consumes in one day. The daily amount that an animal consumes depends on the nutrient content of the ration and the animal’s dry matter intake. The daily unit is how the daily amount of a nutrient is measured. Some examples of daily units are grams per day (g/day), mega calories per day (mcal/day), and international units per day (IU/day).
Suppose a container holds AF# > 0 kg of feed and, after all moisture has been removed from the feed, it holds DM# > 0 kg of feed. If A is the quantity of a nutrient in the container, then Equation (6) shows that A = DN* • DM#, where DN* is the dry matter concentration of the nutrient in the feed. If AF# is the total amount of feed that the animal consumes in a day, then DM# = DMI and A = DA.
The daily amount of a nutrient in a feed is therefore
(10) DA = DN* • DMI
where DN* is the dry matter concentration of the nutrient in the feed and DMI is the dry matter intake of the animal.
Example
Care must be taken with the units of DN* and DMI. For example, if DN* = 20% and DMI = 3 kg, then
DA = (20%) • (3 kg/day) = (200 g/kg) • (3 kg/day) = 600 g/day
Example
Assume that each day an animal needs 0.8 g of Protein for each kg of body weight. If BW is the body weight of the animal in kg, then the animal needs
0.8 • BW g/day of Protein
This is the required intake (DA) of the animal. If you know the animal's dry matter intake (DMI), then Equation (10)shows
DN* = DA / DMI
where DN* is the dry matter concentration of Protein in the ration. Now DA / DMI is a constant, and the value of DN* in a ration can be made equal to DA / DMI by a process called linear programming.
Linear programming
Given N feeds, each containing M nutrients, the feed-mix problem is to find the as fed weights X1, X2, ..., XN of these feeds that satisfy certain feed and nutrient restrictions at minimum cost. For example, suppose three feeds have the protein concentrations and feed prices shown below.
Feed Protein Price
(as fed) (as fed)
Feed A 5% $3.00/lb
Feed B 8% $5.00/lb
Feed C 9% $6.00/lb
A feed-mix problem is to find a 100-pound mixture of these feeds that has between 7% and 8% protein, at least 50% of Feed B, and is not more expensive than any other 100-pound mixture of these three feeds that satisfy these feed and nutrient restrictions. A feed mixture is a balanced ration if it satisfies the feed and nutrient restrictions, and is a least-cost balanced ration if no other balanced ration has a smaller feed cost. This problem, whose solution (33.33 lb of Feed A and 66.67 lb of Feed B) is difficult to guess, is a simple feed-mix problem. Typical feed-mix problems have ten to twenty feeds and ten to twenty nutrients with minimum or maximum restrictions. Practical feed-mix problems are so large that only computers can solve them.
Feed mix problems are solved by a mathematical technique called linear programming, which can be used to restrict feeds, nutrients, ratios of feeds and ratios of nutrients on an as fed or dry matter basis. Linear programming can also be used to solve multiple feed-mix problems when feed inventory is limited, and to optimize feed properties other than cost. In the present case, the linear programming method assumes that the solution is Xa pounds of Feed A, Xb pounds of Feed B, and Xc pounds of Feed C, where initially Xa, Xb and Xc are unknown as fed weights, and sets up the following conditions.
Minimize: Z = 3.00Xa + 5.00Xb + 6.00Xc
Subject to: Xa + Xb + Xc = 100
Xb > 50
.05Xa + .08Xb + .09Xc > 7
.05Xa + .08Xb + .09Xc < 8
The function Z, called the objective function, is the feed cost that is to be minimized. The equality says that the ration must have exactly 100 pounds of as fed weight, followed by an inequality, called a feed restriction or feed constraint that says at least 50 of these 100 pounds must come from Feed B. The last two inequalities are nutrient restrictions or nutrient constraints and say that the ration must have at least 7% protein and at most 8% protein on an as fed basis. The coefficients of the X’s are then placed in a matrix and a solution is found by a procedure called the simplex algorithm. (See books on business mathematics for details.)
The weight of the above ration is AF# = 33.33 lb + 66.67 lb = 100 lb, and the cost is C = (3.00 $/lb) • (33.33 lb) + (5.00 $/lb) • (66.67 lb) = $433.34. By Equation (2), the as fed price of the ration is AF$ = C / AF# = 4.3334 $/lb.
Restricting dry matter in a ration
Some linear programming procedures restrict the as fed concentrations of nutrients and others restrict the dry matter concentrations of nutrients. Regardless of which linear programming procedure you are using, it is possible to restrict the dry matter percentage of a least-cost balanced ration to a number between MIN and MAX
(A) MIN < DMP < MAX
where DMP > 0 is the dry matter percentage of the balanced ration and 0 < MIN < MAX < 100. Restricting the dry matter percentage of a ration is the same as restricting its moisture percentage since moisture % = 100 - dry matter %.
As fed concentrations of nutrients are restricted
Define a nutrient with the name Dry Weight and unit %. For each feed Fi, set the as fed concentration of Dry Weight to Di, where Di is the dry matter percentage of Fi. Then, for each feed Fi, the as fed concentration of Dry Weight is AN* = Di, and the dry matter percentage of Fi is DRY = Di. By Equation (7), the dry matter concentration of Dry Weight is
DN* = AN* • (100 / DRY) = Di • (100 / Di) = 100
That is, for each feed, the as fed concentration of Dry Weight is the dry matter percentage of the feed, and the dry matter concentration of Dry Weight is 100.
If the linear programming procedure restricts the as fed concentrations of nutrients, the following inequality can be used to restrict the as fed concentration of Dry Weight to a number between MIN and MAX.
MIN < Dry Weight < MAX
Since the as fed concentration of Dry Weight in the ration (Dry Weight) is the dry matter percentage of the ration (DMP), the last inequality is the same as Equation (A).
Use Dry Weight when the as fed concentrations of nutrients are restricted. Use Wet Weight, explained below, when the dry matter concentrations of nutrients are restricted.
NOTE: In MixitWin, it is simplest to set the dry matter concentration of Dry Weight to 100 for each ingredient, and to set the as fed concentration of Wet Weight to 1 for each ingredient.
Dry matter concentrations of nutrients are restricted
Define a nutrient with the name Wet Weight and unit %. For each feed Fi, set the dry matter concentration of Wet Weight to 100/Di, where Di is the dry matter percentage of Fi. (Equation (1) says AF# = DM# • (100 / DRY). Letting DM# = 1 and DRY = Di shows that 100 / Di is the number of pounds of Fi that is required to produce one pound of dry matter.) Then, for each feed Fi, the dry matter concentration of Wet Weight is DN* = 100 / Di, and the dry matter percentage of Fi is DRY = Di. By Equation (7), the as fed concentration of Wet Weight is
AN* = DN* • (DRY / 100) = (100 / Di) • (Di / 100) = 1
That is, for each feed, the dry matter concentration of Wet Weight is 100 divided by the dry matter percentage of the feed, and the as fed concentration of Wet Weight is 1.
If the linear programming procedure restricts the dry matter concentrations of nutrients, the following inequality can be used to restrict the dry matter concentration of Wet Weight to a number between 100 / MAX and 100 / MIN.
100 / MAX < Wet Weight < 100 / MIN
Since the dry matter concentration of Wet Weight in the balanced ration is 100 / DMP, the last inequality is the same as
(B) 100 / MAX < 100 / DMP < 100 / MIN
The following calculation shows that Inequalities (A) and (B) are equivalent, and therefore the dry matter percentage of a balanced ration can always be restricted to a number between MIN and MAX, as long as a balanced ration is possible.
100 / MAX < 100 / DMP < 100 / MIN
1 / MAX < 1 / DMP < 1 / MIN
MIN < DMP < MAX
NOTE: An alternate approach is to restrict the moisture percentage of a ration by defining two nutrients, Wet Percent and Dry Percent, whose as fed and dry matter percentages are MOIST = 100 - DRY and MOIST • (100 / DRY), respectively, and using the equivalent inequalities
MIN < Wet Percent < MAX
100 • MIN < Dry Percent < 100 • MAX
100 - MIN 100 - MAX
Wet weight is consistent
The previous sections showed how Dry Weight and Wet Weight are used to restrict the percentage of dry matter in a ration when the nutrient constraints of a ration are on an as fed or dry matter basis, respectively. This section shows that defining the dry matter concentration of Wet Weight in Feed i as 100 / Di causes the dry matter concentration of Wet Weight of a mixture of these feeds to be 100 / DMP.
Suppose Feed i has dry matter percentage Di for i = 1,2. Then a mixture of X1 kg of Feed 1 and X2 kg of Feed 2 has dry matter percentage
DMP = D1 • X1 + D2 • X2
X1 + X2
If Ci is the dry matter concentration of nutrient N in Feed i, then Equation (8) shows that the dry matter concentration of N in a mixture of X1 kg of Feed 1 and X2 kg of Feed 2 has dry matter concentration
CON = C1 • D1 • X1 + C2 • D2 • X2
D1 • X1 + D2 • X2
If, for i = 1,2, we set Ci = 100 / Di, which is the dry matter concentration of Wet Weight in Feed i, then
CON = 100 • (X1 + X2) = 100
D1 • X1 + D2 • X2 DMP
which is the dry matter concentration of Wet Weight in the mixture.
Similarly, defining the as fed concentration of Dry Weight in Feed i as Di causes the as fed concentration of Dry Weight of a mixture of these feeds to be DMP, as shown next.
If Ci is the as fed concentration of a nutrient N in Feed i, then Equation (8) shows that the as fed concentration of N in a mixture of X1 kg of Feed 1 and X2 kg of Feed 2 is
CON = C1 • X1 + C2 • X2
X1 + X2
and if we set Ci = Di, which is the as fed concentration of Dry Weight in Feed i, then
CON = D1 • X1 + D2 • X2 = DMP
X1 + X2
which is the as fed concentration of Dry Weight in the mixture.
Restricting Feeds
In addition to the usual meaning of "nutrient", there are other ways a computer feed formulation program can use nutrients that extend its usefulness in formulating rations. For example, nutrients can be used to record numerical properties of feeds, such as “solubility” or “weight per cubic foot.” Assuming these properties carry over to mixtures of feeds, you can see and restrict the solubility or weight per cubic foot of rations.
Restricting classes of feeds
Nutrients can also be used to restrict classes of feeds. The following three examples show how to create least-cost balanced rations that restrict the amount of wheat products. Suppose you are balancing a ration with many feeds, including “Wheat bran”, “Wheat flour shorts”, and “Wheat grain hard”, and you want no more than 12% of the feeds to be wheat products.
MixitWin Examples
Example 1 Feeds restricted as fed, nutrients restricted as fed
Define a nutrient with the name “Wheat as fed” and the unit %. For each feed, enter 100% as fed if the feed is a wheat product and 0% as fed if it is not a wheat product. When you balance a ration, use the maximum value 12% for the nutrient “Wheat as fed”. No more than 12% (as fed) of the balanced ration will be wheat products.
Example 2 Feeds restricted dry matter, nutrients restricted dry matter
Define a nutrient with the name “Wheat dry matter” and the unit %. For each feed, enter 100% dry matter if the feed is a wheat product and 0% dry matter if it is not a wheat product. When you balance a ration, use the maximum value 12% for the nutrient “Wheat dry matter”. No more than 12% (dry matter) of the balanced ration will be wheat products.
Example 3 Feeds restricted as fed, nutrients restricted dry matter
A common situation occurs when you restrict the feeds of a ration on an as fed basis, the nutrients on a dry matter basis, and you want no more than 12% (as fed) of the balanced ration to be wheat products. You begin by guessing DRY, the dry matter percentage of the final ration.
Step 1. Define a nutrient “Wheat as fed” as in Example 1.
Step 2. Balance the ration without restricting “Wheat as fed” and let DRY be the dry matter percentage of this balanced ration. DRY is the first approximation.
Step 3. Enter the maximum value of (12)(100/DRY) for “Wheat as fed”, balance the ration and let DRY be the dry matter percentage of this new balanced ration. DRY is the next approximation.
Step 4. Repeat Step 3 as many times as necessary until no more than 12% (as fed) of the balanced ration is wheat products.
Restricting by ratios of nutrients
Computer feed formulation programs that restrict ratios of nutrients can also be used to restrict other properties of a ration. The present section shows how restricting ratios of nutrients can restrict ratios of feeds in a ration. The next section shows how to use a ratio of nutrients to limit the amount of a nutrient that comes from a particular feed.
MixitWin Examples
The following examples show how to create least-cost balanced rations that restrict the ratio of wheat products to corn products. If the ration has only one wheat product and only one corn product, the procedure will restrict the ratio of the single wheat product to the single corn product; otherwise, the ratio of all wheat products to all corn products is restricted.
Example 4 Feeds restricted on an as fed basis
Define a nutrient with the name “Wheat as fed” and the unit %. For each feed, enter 100% as fed if the feed is a wheat product and 0% as fed if it is not a wheat product. Define a second nutrient with the name “Corn as fed” and the unit %. For each feed, enter 100% as fed if the feed is a corn product and 0% as fed if it is not a corn product. When you balance a ration, restrict the ratio “Wheat as fed” / “Corn as fed” between minimum and maximum values.
Example 5 Feeds restricted on a dry matter basis
Define a nutrient with the name “Wheat dry matter” and the unit %. For each feed, enter 100% dry matter if the feed is a wheat product and 0% dry matter if it is not a wheat product. Define a second nutrient with the name “Corn dry matter” and the unit %. For each feed, enter 100% dry matter if the feed is a corn product and 0% dry matter if it is not a corn product. When you balance a ration, restrict the ratio “Wheat dry matter” / “Corn dry matter” between minimum and maximum values.
Restricting by limiting nutrients
The previous section showed how to use a ratio of nutrients to limit the ratio of wheat products to corn products in a ration; this section shows how to use a ratio of nutrients to limit the amount of a nutrient that comes from a particular feed.
Example 6 Limiting nutrients from feeds
Suppose you are mixing a liquid supplement with the following feeds on an as fed basis and you want a ration that is 32% protein and 0.75% phosphorus on an as fed basis. In addition, you want 20% of the phosphorus to come from ammonium polyphosphate and 80% from other sources
Feed Protein %Phosphorus
(as fed) (as fed)
Ammonium poly 62.5% 14.5%
Phosporic Acid 0% 23.7%
Urea 140% 0%
Molasses 4.4% 0%
To solve this problem, define two nutrients “Phosphorus from ammonium polyphosphate (Pammon)” and “Phosphorus from other sources (Pother)” with units %. Give “Ammonium polyphosphate (Ammonium poly)” 14.5% of “Pammon”, “Phosphoric acid” 23.7% of “Pother”, and “Molasses” 0.08% of “Pother”, as shown below.
Feed Protein %Phosphorus %Pammon %Pother
(as fed) (as fed) (as fed) (as fed)
Ammonium poly 62.5% 14.5% 14.5% 0%
Phosporic Acid 0% 23.7% 0% 23.7%
Urea 140% 0% 0% 0%
Molasses 4.4% 0% 0% 0.08%
Balance a ration with the following nutrient restrictions. The ratio of “Pammon” to “Pother” will be 20% / 80% = 0.25.
Nutrient Minimum Maximum
(as fed) (as fed)
Protein 32% 32%
Phosporus 0.75% 0.75%
Pammon 0% none
Pother 0% none
Pammon/Pother 0.25 0.25
Variable nutrient concentrations
It is known that the nutrient concentrations of a feed can change according to the percentage of the feed in a ration. What changes is the animal’s ability to digest the nutrients, not the nutrient concentrations themselves, and this digestibility decreases with increasing amounts of the feed in a ration. It will simplify the following discussion, however, to pretend that the nutrient concentrations themselves change, and use phrases like “The feed contains 7.15% Protein when it makes up 4% of the ration” instead of more accurately saying “Due to digestibility, the feed apparently contains 7.15% Protein when it makes up 4% of the ration.
Example
The feed Corn may contain
Table 1 - Available Values
7.15% Protein when it makes up between 0 and 4% of the ration
6.57% Protein when it makes up between 4 and 8% of the ration
6.09% Protein when it makes up between 8 and 12% of the ration
Define three feeds. “Corn 0-4%”, “Corn 4-8%” and “Corn 8-12%”, with the same nutrient concentrations as Corn, and then give these feeds the following concentrations of Protein.
Table 2 - Adjusted Values
FEED PROTEIN
Corn 0-4% 7.15%
Corn 4-8% 5.99%
Corn 8-12% 5.13%
To balance a ration with feeds that include Corn, insert the following feed constraints.
FEED MINIMUM MAXIMUM
Corn 0-4% 0% 4%
Corn 4-8% 0% 4%
Corn 8-12% 0% 4%
If the balanced ration contains 4% or less of Corn, it will use Corn 0-4%. If the balanced ration contains 5.5% of Corn, it will use 4% of Corn 0-4% and 1.5% of Corn 4-8%. And if the balanced ration contains 9% of Corn, it will use 4% of Corn 0-4%, 4% of Corn 4-8%, and 1% of Corn 812%. (If necessary, increase the prices of Corn 4-8% and Corn 8-12%.)
If a ration contains exactly 12% of Corn, its protein content from Corn should be 6.09%, according to Table 1. This is the same number that is found by adding the protein concentrations in Table 2 using Equation (8). That is, (4 • 7.15% + 4 • 5.99% + 4 • 5.13%) / 12 = 6.09%.
The way to calculate the nutrient concentrations of Table 2, based on the nutrient concentrations of Table 1, will be illustrated with a feed Meat, a nutrient N, and the percentages and nutrient concentrations that are shown in Table 3. The symbols C1 > C2 > C3 represent variable nutrient concentrations of N in Meat, and 0 < P1 < P2 < P3 < 100.
Assume the feed Meat contains
Table 3 - Available Values
C1 of N when it makes up between 0% and P1% of the ration
C2 of N when it makes up between P1% and P2% of the ration
C3 of N when it makes up between P2% and P3% of the ration
All percentages and concentrations in Table 3 should be on an as fed basis, or all should be on a dry matter basis. If the concentrations are on a dry matter basis and the percentages are on an as fed basis, multiply each Pi with D/100, where D is the dry matter percentage of Meat.
Define three feeds with the same nutrient concentrations as Meat, and then give these feeds the following concentrations of N.
Table 4 - Adjusted Values
FEED Concentration nutrient N
Meat 0% to P1% C1
Meat P1% to P2% (P2 • C2-P1C1) / (P2-P1)
Meat P2% to P3% (P3 • C3-P2C2) / (P3-P2)
To balance a ration with feeds that include Meat, use the following feed constraints.
FEED MINIMUM MAXIMUM
Meat 0% to P1% 0% P1%
Meat P1% to P2% 0% (P2-P1)%
Meat P2% to P3% 0% (P3-P2)%
The adjusted values of nutrient N in Table 4 above can be explained in the following way. By Equation (8), the nutrient concentration of N in a mixture consisting of P1, P2-P1, and P3-P2 pounds of each of the three feeds in Table 4, respectively, is
[P1 • C1 + (P2 - P1) • (P2 • C2 - P1 • C1) / (P2 - P1) + (P3 - P2) • (P3 • C3 - P2 • C2) / (P3 - P2)] / P3 = C3
This is the concentration of nutrient N that Table 3 shows for a ration with P3 pounds of Meat.
Multiple nutrient prices
The earlier section, Nutrient prices, defined the price of a nutrient N in a feed F to be AF$ / AN*, where AF$ is the as fed price of F and AN* > 0 is the as fed concentration of nutrient N in F. It showed that the feed with the smallest nutrient price also has the smallest cost of one unit of N, and therefore is the most economical source of nutrient N.
In particular,
The cost of 1 unit of N is (k1 / k2) • (AF$ / AN*)
The cost of U units of N is U • (k1 / k2) • (AF$ / AN*)
where k1 > 0 and k2 > 0 are numbers such that k1 • AF$ is the as fed price of feed F in $/kg and k2 • AN* > 0 is the as fed concentration of nutrient N in the feed in units of N/kg. The table below shows values of k1 and k2 for different as fed prices and concentrations, where 1 pound equals c = 0.45359237 kilograms..
AF$ k1 AN* k2
$/kg 1.00 mcal/kg 1.00
$/ton 1/1000 (metric) kcal/lb 1/c
$/lb 1/c IU/lb 1/c
$/ton 1/2000c (American) ppm=mg/kg 1.00
$/cwt 1/20c %=kg/100kg 0.01
We often want several nutrients, say N1 and N2, in a single feed and would like to know which feed is the most economical source of both nutrients. As stated, this problem has no obvious answer because the amounts of N1 and N2 are unknown. But suppose we are mixing feeds for an animal that needs U1 units of N1 and U2 units of N2.
In this case we define (for this animal) the price of nutrients N1 and N2 in a feed F to be
U1 • (k1 / k21) • (AF$ / AN1*) + U2 • (k1 / k22) • (AF$ / AN2*)
where
k1 • AF$ > 0 is the as fed price of feed F in $/kg
k21 • AN1* > 0 is the as fed concentration of nutrient N1
in feed F in units of N1/kg
k22 • AN2* > 0 is the as fed concentration of nutrient N2
in feed F in units of N2/kg
The price of nutrients N1 and N2 is the cost of U1 units of N1 and U2 units of N2. This definition can be easily extended to the price of n nutrients, N1, N2, ..., Nn, in a single feed F.
Requirement feeding
Theoretically, all animals have minimum daily requirements of nutrients for purposes of health, growth, production, etc. and these requirements depend on the age, weight and other characteristics of the animal. This suggests defining a “requirement feed” as a mixture of feed ingredients in amounts sufficient to meet the daily nutrient requirements of a particular animal at a certain stage of its development.
Since any mixture of feeds is a ration, the above definition can be stated in the following way: Given N feeds, a requirement feed is a ration R and an amount W so that an animal’s daily nutrient requirements are met by consuming W pounds (grams or kilograms) of R. Given N feeds, a least-cost requirement feed is a ration R and an amount W so that R and W are a requirement feed for the animal; and if R* and W* are also a requirement feed for the animal, then the cost of W does not exceed the cost of W*.
A least-cost requirement feed is not the same as a least-cost balanced ration. A least-cost balanced ration meets nutrient concentration restrictions such as “at least 6% protein” at minimum cost. A least-cost requirement feed meets nutrient requirement restrictions such as “at least 60 grams of protein” at minimum cost. The amount of a least-cost balanced ration, 100 or 2000 pounds or kilograms, has no effect on the concentrations of nutrients in the ration. The amount of a least-cost requirement feed is a number W that varies with the ration and the animal.
The topic Linear programming shows how linear programming finds a 100-pound, least-cost mixture of feeds that has between 7% protein and 8% protein. Linear programming can also be used to find a least-cost requirement feed that has between 2 and 3 pounds of protein.
Requirement feeding gives an economic advantage over a least-cost balanced ration when feeding animals without a fixed daily intake. Another way a least-cost requirement feed may be more economical than a least-cost balanced ration is in mixing vitamin and mineral supplements by specifying nutrient quantities rather than nutrient concentrations.
NOTE: MixitWin allows requirement feeding. After starting MixitWin, right-click the Balance This Formula button to open the Requirement window. Then press the F1 function key and read about requirement feeding.
Example
Assume that an animal needs at least 2 pounds of protein and 300 megacalories of energy every day, and we are looking for a ration R and an amount W of R that has at least 2 pounds of protein and 300 megacalories of energy.
From Equation(5), the as fed concentration of energy is
AN* = A / AF# = 300 / W Mcal/lb
and the sentence and Example below Equation (5) show that the as fed percentage of protein is
100 • AN* = 100 • A / AF# = 200 / W %
A computer feed formulation program that finds least-cost balanced rations and least-cost requirement feeds could solve this problem in one of the following ways.
Method I
Letting W = 100 pounds, find a least-cost balanced ration R with at least 2% protein and 3 Mcal/lb of energy, on an as fed basis, and feed the animal 100 pounds of R every day.
Method II
If W is approximately the daily as fed intake of the animal, find a least-cost balanced ration R with at least 200 / W % protein and 300 / W Mcal/lb of energy, on an as fed basis, and feed the animal W pounds of R every day.
Method III
Find a least-cost requirement feed R and an amount W of R that contains at least 2 pounds of protein and 300 megacalories of energy. Then feed the animal W pounds of R every day.
The cost of feeding by Method III is less than or equal to the cost of feeding by Method I or II, since the R and W of Method I or II is a requirement feed for the animal and the R and W of Method III is a requirement feed of minimum cost.
The previous paragraphs describe three methods of feeding animals so that they receive their minimum daily requirements of nutrients. The next example shows the economic advantage that can result from feeding by Method III instead of by Method I.
Example
Suppose an animal (or group of animals) needs at least 2 pounds of protein and 300 Mcal of energy every day, and these nutrients will be supplied by a mixture of the following two grains
Protein Energy Price
(as fed) (as fed) (as fed)
Grain A 2% 3 Mcal/lb 200 $/ton
Grain B 3% 2 Mcal/lb 100 $/ton
A least-cost requirement feed of Method III consists of 150 pounds of Grain B at a cost of $7.50 per day. This meets the animal’s minimum daily requirements since 150 pounds of Grain B contains (0.03)(150) = 4.5 pounds of protein and (2)(150) = 300 Mcal of energy.
To meet the animal’s minimum daily requirements by Method I calls for 100 pounds of a feed that contains at least 2 pounds of protein and 300 Mcal of energy. This feed must therefore have concentrations of at least 2% protein and 3 Mcal/lb of energy. A least-cost balanced ration of Method I consists of 100 pounds of Grain A at a cost of $10.00 per day. This meets the animal’s minimum daily requirements since 100 pounds of Grain A contains (0.02)(100) = 2 pounds of protein and (3)(100) = 300 Mcal of energy.
Both Method I and Method III give requirement feeds for the animal. Method III, however, gives a least-cost requirement feed, and therefore Method III always results in a daily cost as low as or lower than the cost of Method I. But what if R and W are a least-cost requirement feed and the animal does not consume W pounds of R? In feeding dairy cattle it is assumed that the dry matter intake of a cow is constant. (See the section Feeding animals with fixed intake.) A requirement feed for a dairy cow must satisfy another condition: the dry matter weight of W must equal the dry matter intake of the animal. This suggests another method, which linear programming can be used to calculate. The following section is an example of this method.
Method IV
Find a least-cost requirement feed R and an amount W of R that contains at least 2 pounds of protein and 300 megacalories of energy and such that the dry matter weight of W is 40 pounds. Then feed the animal W pounds of R every day.
Example
The following example comes from Table 14-7 in the NRC’s “Nutrient Requirements of Dairy Cattle”, Seventh Revised Edition, 2001 for Holstein cows with dry matter intake of 20.3 kg.
Nutrients Concentrations Daily Amounts
(% dry matter) (g in 20.3 kg)
TDN 65.0 13195.0
Crude protein 14.1 2862.3
Calcium 0.62 125.86
Ingredients DM TDN CP CA $/kg
% (%DM) (%DM) (%DM) (AF)
Bermudagrass coastal 87.1 52.9 10.4 0.49 0.070
Cotton seeds 37 whole 90.1 77.2 23.5 0.17 0.160
Corn 28 Grain steam flk 88.1 91.7 9.4 0.04 0.175
Limestone ground 100.0 0.0 0.0 34.00 0.300
Sorghum Sudan silage 28.8 54.4 10.8 0.64 0.100
Using these ingredients a least-cost balanced ration was compared with a least-cost requirement feed with the following results. Feeding 20.3 kg DM per cow daily of the requirement feed costs $2.59 and supplies the same minimum daily amounts (shown above) as feeding 20.3 kg DM per cow daily of the balanced ration, which costs $2.79.
The following requirement feed balanced with fixed dry matter weight has 88.21% dry matter.
Problem 1
Problem 1 Amounts $/ton
(fixed DM weight) (kg as fed) (as fed)
Bermudagrass coastal 12.957 70.00
Cotton seeds 37 whole 6.751 160.00 The cost
Corn 28 Grain steam flk 3.131 175.00 per cow
Limestone ground 0.174 300.00 daily is
Sorghum Sudan silage 0.000 30.00 $2.59
Total 23.013 112.43
The following ration balanced with fixed as fed weight has 39.75% dry matter.
Problem 2
Problem 2 Amounts $/ton
(fixed AF weight) (kg as fed) (as fed)
Sorghum Sudan silage 41.881 30.00
Cotton seeds 37 whole 6.264 160.00 The cost
Corn 28 Grain steam flk 2.818 175.00 per cow
Limestone ground 0.112 300.00 daily is
Bermudagrass coastal 0.000 70.00 $2.79
Total 51.075 54.53
Price ranges
Changing ingredient prices can obviously change the composition of a least-cost balanced ration. What is not so obvious is that changing ingredient prices may not always change the composition of a least-cost balanced ration. The reason for this lies in a branch of linear programming called sensitivity analysis, and, in particular, in the price ranges of ingredients.
After balancing a ration, many computer programs show the “low price” and the “high price” of each ingredient in the ration. If the current price of an ingredient is changed to a price that is below the low price or above the high price and the ration is rebalanced, a different ration will result. If the price of an ingredient is changed to a price that is above the low price and below the high price and the ration is rebalanced, there will be no change in the composition of the ration, although its cost may change.
The “price range” of an ingredient in the ration is the set of prices between the low price and the high price of the ingredient. The price range of an ingredient shows how much the price of a single ingredient can vary without changing the balanced ration. (This statement is correct when a single price is changed, but may not be true if two or more prices are changed simultaneously.) The price range of an ingredient depends on the other ingredients in the ration, the ration constraints and the ingredient prices and is usually different for different rations or ingredient prices.
Example
In the earlier topic Requirement feeding, Problem 2 (balanced with fixed as fed weight) results in the least-cost balanced ration and price ranges shown below
Ingredients Amounts (Problem 2
Bermudagrass coastal 0.000 with
Cotton seeds 37 whole 12.264 weights
Corn 28 Grain steam flk 5.517 adjusted
Limestone ground 0.219 to total
Sorghum Sudan silage 82.000 100)
Daily Cost 2.785
Low Price High Price
Bermudagrass coastal 0.00 infinite
Cotton seeds 37 whole 104.02 infinite
Corn 28 Grain steam flk 0.00 282.97
Limestone ground 0.00 infinite
Sorghum Sudan silage 0.00 100.60
Both Bermudagrass and Limestone have price ranges from zero to infinity. This means that the ration will not change no matter how the prices of Bermudagrass and Limestone are changed (although the ration cost and price ranges of the ingredients may change).
If the price of Cotton is changed to a number less than 104.02, the ration will change: probably more Cotton and a smaller cost. If the price of Sorghum is change to a number greater than 100.60, the ration will change: probably less Sorghum and a larger cost.
Rations based on energy
The dietary concentration of a nutrient is the amount of the nutrient in a diet. Dietary concentrations can be expressed by weight:
20 % protein (equivalent to 200 g / kg)
or in terms of the energy in the diet:
50 grams of protein per 1 mcal ME
Feed formulation programs usually work with weight-based dietary concentrations of nutrients, but it is easy to work with energy-based concentrations of nutrients. This section shows how to find the energy-based concentrations of a nutrient and how to restrict it in a ration.
Finding energy-based concentrations
Equation (10) says that the daily amount (DA) of a nutrient is
DA = DN* • DMI
where DN* is the dry matter concentration of the nutrient in the feed and DMI is the dry matter intake of the animal. The energy-based concentration of a nutrient is the ratio
DA1 = DN1* • DMI = DN1* = AN1*
DA2 DN2* • DMI DN2* AN2*
where DA1 and DN1* are the daily amount and dry matter concentration of the nutrient, and DA2 and DN2* are the daily amount and dry matter concentration of energy. The final ratio of as fed concentrations comes from Equation (7).
A linear programming procedure that can restrict ratios of concentrations can also restrict the energy-based concentration of a nutrient:
Minimum < DN1* / DN2* < Maximum
Example
If one of the nutrients is a protein like Crude Protein with DA = 156.263 g/day, and another nutrient is an energy like Metabolizable Energy with DA = 2.268 mcal/day, then the ratio of the daily amounts (156.263 g/day / 2.268 mcal/day = 68.9 g/mcal) gives the dietary concentration of protein based on energy, which can be written
68.9 grams of Crude Protein per 1 mcal of Metabolizable Energy
or simply
protein: 68.9 grams per mcal ME
Finally, there is nothing special about energy in the above ratios. In general, we can speak of the nutrient 2-based concentration of nutrient 1.
Rations of minimal price
The section Weights and dry matter defines feed and ration. It then defines, and gives an example of the as fed weight (AF#) and dry matter weight (DM#) of a quantity of feed, and the dry matter percentage (DRY) and dry matter portion (POR) of a feed. The section, Feed cost and feed prices, defines and gives an example of the cost (C), as fed price (AF$), and dry matter price (DM$) of a feed.
The section, Linear programming, defines and gives an example of a balanced ration and a least-cost balanced ration, and introduces the concept of "a ration of minimal (or least) cost". The present section introduces the related concepts of "a ration of minimal (or least) as fed price" and "a ration of minimal (or least) dry matter price". It makes a clear distinction between these three concepts by showing examples of balanced rations
of minimal cost, but not minimal as fed or dry matter price,
of minimal as fed price, but not minimal cost or dry matter price,
of minimal dry matter price, but not minimal cost or as fed price.
Least costing is the process of producing balanced rations with the smallest possible cost. Least pricing is the process of producing balanced rations with the smallest possible as fed or dry matter price.
The subject of minimal as fed and dry matter prices is continued in the next section, Constant weight linear programming, and plays a central role in the later section, Feeding animals with fixed intake.
As fed and dry matter prices
Given a particular animal and n feeds, F1, F2, ..., Fn, a balanced ration is a mixture of X1 kg of F1, X2 kg of F2, ..., Xn kg of Fn so that the animal’s nutritional requirements are met by consuming
AF# = X1 + X2 + ... + Xn
kg of the mixture every day.
If ai is the as fed price of feed Fi, for i = 1, 2, ..., n, the feed cost (in $) of this mixture is
C = a1 • X1 + a2 • X2 + ... + an • Xn
Equation(2) then shows that the as fed price (in $/kg) of this mixture is
AF$ = C / AF# = a1 • X1 + a2 • X2 + ... + an • Xn
X1 + X2 + ... + Xn
Equation (1) shows that the dry matter weight of this mixture is
DM# = p1 • X1 + p2• X2 + ... + pn • Xn
where pi is the dry matter portion of feed Fi for i = 1, 2, ..., n.
The dry matter price (in $/kg) of this mixture is
DM$ = C / DM# = a1 • X1 + a2 • X2 + ... + an • Xn
p1 • X1 + p2 • X2 + ... + pn • Xn
Example
This example will show that an animal’s daily requirements can be met by a balanced ration with minimum as fed price, a balanced ration with minimum dry matter price, or a balanced ration with minimum cost, and these three rations are different. Thus, least pricing is not the same as least costing.
Suppose an animal needs at least 1 kg of Protein and at least 50 Mcal of Energy every day and these minimum amounts will be supplied by a mixture of the following two grains.
DM AF Price DM Price Protein Energy
% $/kg $/kg % AF Mcal/kg AF
Wet Grain 40 0.14 0.35 3 3
Dry Grain 80 0.16 0.20 4 1
Since 100/3 kg of Wet Grain has (0.03)(100/3 kg) = 1 kg Protein and (3 Mcal/kg)(100/3 kg) = 100 Mcal Energy, a mixture of these two feeds will satisfy the minimum nutrient requirements and have the smallest as fed price if it consists of 100/3 kg of Wet Grain. (There is no way the as fed price of a mixture of these two feeds could be smaller than 0.14 $/kg.)
Since 50 kg of Dry Grain has (0.04)(50 kg) = 2 kg Protein and (1 Mcal/kg)(50 kg) = 50 Mcal Energy, a mixture of these two feeds will satisfy the minimum nutrient requirements and have the smallest dry matter price if it consists of 50 kg of Dry Grain. (There is no way the dry matter price of a mixture of these two feeds could be smaller than 0.20 $/kg.)
It can be shown that a least-cost balanced ration is a mixture of 100/9 kg of Wet Grain and 50/3 kg of Dry Grain. For the purpose of this example, however, it is not necessary for this to be a least-cost ration. It is only important that it is a balanced ration with a cost less than the cost of either of the other balanced rations.
Since 100/9 kg of Wet Grain has (0.03)(100/9 kg) = 1/3 kg Protein and (3 Mcal/kg)(100/9 kg) = 100/3 kg Energy, and 50/3 kg of Dry Grain has (0.04)(50/3 kg) = 2/3 kg Protein and (1 Mcal/kg)(50/3 kg) = 50/3 kg Energy, a mixture of 100/9 kg of Wet Grain and 50/3 kg of Dry Grain will satisfy the minimum nutrient requirements.
These three rations are shown below. Calculations of prices, weights and costs are left as an exercise for the reader.
Ration 1 Ration 2 Ration 3
Wet Grain (kg) 100/3 0 100/9
Dry Grain (kg) 0 50 50/3
Protein (kg) 1 2 1
Energy (Mcal) 100 50 50
AF Price ($/kg) 0.14 0.16 0.152
DM Price ($/kg) 0.35 0.20 0.2375
AF Weight (kg) 33.33 50.00 83.33
Cost ($) 4.667 8.00 4.222
The table shows that the three rations are balanced; that is, they satisfy the minimum requirements of Protein and Energy. Moreover
Ration 1 has the smallest as fed price
Ration 2 has the smallest dry matter price
Ration 3 has the smallest cost
It is significant that these three rations have different as fed weights. They are also requirement feeds, as defined in the earlier section Requirement feeding.
Constant weight linear programming
Computer feed formulation programs find least-cost balanced rations through a mathematical technique called linear programming. Linear programming uses a mathematical model of the feed formulation problem.
A constant weight linear program (CWLP) has the form:
Minimize: Z = a1 • X1 + a2 • X2 + ... + an • Xn
Subject to:
(weight equality) c1 • X1 + c2 • X2 + ... + cn • Xn = W
(k inequalities) ej1 • X1 + ej2 • X2 + ... + ejn • Xn < ej0
where Z is the objective function, W > 0 is the constant weight, Xi > 0 for 1 < i < n, and ai, ci, eji and ej0 are constants for 1 < j < k and 1 < i < n.
The n-tuple (X1, X2, ... , Xn) is a solution of the CWLP if (X1, X2, ... , Xn) satisfies the weight constraint and the k inequalities, and if (Y1, Y2, ... , Yn) satisfies the weight constraints and the k inequalities, then a1 • X1 + a2 • X2 + ... + an • Xn < a1 • Y1 + a2 • Y2 + ... + an • Yn. That is, z is minimal.
In feed formulation problems, for 1 < i < n, Xi is the (initially unknown) as fed weight of Feed i (in kilograms), ai is the as fed price of Feed i (in $/kg), and Z is the cost (in $) of purchasing a mixture of feeds that contains Xi kilograms of Feed i, for 1 < i < n. A solution is also called a least-cost balanced ration. The problem has constant as fed weight if ci = 1, for 1 < i < n, and constant dry matter weight if ci = pi, the dry matter portion of Feed i, for 1 < i < n.
The as fed price (in $/kg) of a solution (X1, X2, ... , Xn) is
a1 • X1 + a2 • X2 + ... + an • Xn
X1 + X2 + ... + Xn
If the CWLP has constant as fed weight, then W = X1 + X2 + ... + Xn is a constant and the as fed price (in $/kg) of a solution (X1, X2, ... , Xn) is
a1 • X1 + a2 • X2 + ... + an • Xn
W
Since W is a constant and linear programming guarantees that the cost, which is a1 • X1 + a2 • X2 + ... + an • Xn, is minimal, it follows that the as fed price, which is (a1 • X1 + a2 • X2 + ... + an • Xn) / W, is minimal.
This proves
Proposition 1
If a CWLP has constant as fed weight, the solution is a balanced ration of minimum cost and minimum as fed price.
The dry matter price (in $/kg) of a solution (X1, X2, ... , Xn) is
a1 • X1 + a2 • X2 + ... + an • Xn
p1 • X1 + p2 • X2 + ... + pn • Xn
where pi is the dry matter portion of Feed i, for 1 < i < n. If the CWLP has constant dry matter weight, then W = p1 • X1 + p2 • X2 + ... + pn • Xn is a constant and the dry matter price (in $/kg) of a solution (X1, X2, ... , Xn) is
a1 • X1 + a2 • X2 + ... + an • Xn
W
Since W is a constant and linear programming guarantees that the cost, which is a1 • X1 + a2 • X2 + ... + an • Xn, is minimal, it follows that the dry matter price, which is (a1 • X1 + a2 • X2 + ... + an • Xn) / W, is minimal.
This proves
Proposition 2
If a CWLP has constant dry matter weight, the solution is a balanced ration of minimum cost and minimum dry matter price.
A CWLP is simple if ej0 = 0 for 1 < j < k. A simple CWLP will also be called a simple linear program. The next proposition shows that a simple linear program results in a ration whose price does not depend on the weight W. These three Propositions, all concerning rations with minimal price, will be important in the next section, Feeding animals with fixed intake.
Proposition 3
Two simple CWLPs that are identical, except for their weights, have solutions with identical prices.
NOTE: A proof of Proposition 3 is found here: Topics in Linear Programmi
Feeding animals with fixed intake
Computer feed formulation programs use linear programming to find balanced rations of minimum cost for many types of animals. Some linear programming procedures hold the as fed weight of the ration constant and some hold the dry matter weight of the ration constant. When feeding animals with fixed intake, we will see what what is needed are balanced rations of minimum price. Although the section Rations of minimal price shows that rations of minimum cost may not have minimum price, and vice versa, a constant weight linear program (CWLP) produces balanced rations of constant weight and minimum price. This will turn out to be the most economical way to formulate rations for animals with fixed intake.
Animals with fixed intake
In feeding dairy cattle it is assumed that the dry matter intake of a cow is constant. The following calculation shows that the cost of feeding one cow for one day is the product of the cow’s dry matter intake in kg/day and the dry matter price of the ration in $/kg.
To see this, let
AF$ = as fed price of the ration ($/kg)
AFI = as fed intake of the animal (kg/day)
DM$ = dry matter price of the ration ($/kg)
DMI = dry matter intake of the animal (kg/day)
DRY = dry matter percentage of the ration
The equations relating as fed to dry matter price, and as fed to dry matter intake, are
AF$ = DM$ • DRY / 100
AFI = DMI • 100 / DRY
Multiplying these two equations shows the cost of feeding one cow for one day is
AF$ • AFI = DM$ • DMI
Assuming DMI is constant, the cost of feeding one cow for one day will be minimum when DM$, the dry matter price of the ration, is minimum. The above equality also shows that if AFI is constant, as in some poultry feeding, the daily cost will be minimum when AF$, the as fed price of the ration, is minimum.
The previous calculation shows that the most economical way to feed an animal with fixed dry matter intake D > 0 is to use a balanced ration with dry matter weight D and minimum dry matter price; and the most economical way to feed an animal with fixed as fed intake A > 0 is to use a balanced ration with as fed weight A and minimum as fed price. What is needed is a way to produce balanced rations of fixed weight and minimum price.
Proposition 1 and Proposition 2 of the section Constant weight linear programming show that the most economical way to feed an animal with fixed dry matter intake D > 0 is to use linear programming to find a least-cost balanced ration with constant dry matter weight D; and the most economical way to feed an animal with fixed as fed intake A > 0 is to use linear programming to find a least-cost balanced ration with constant as fed weight A.
This situation is not satisfactory since it ties the linear program to particular weights, D and A. What is wanted is a constant weight linear program that results in a ration whose (minimum) price does not depend on a particular weight W. It turns out that a “simple” CWLP is such a program. (See Proposition 3 in the section Constant weight linear programming.)
Conclusion
For an animal with fixed dry matter intake, use a simple linear program with fixed dry matter weight and get a balanced ration of minimum dry matter price. For an animal with fixed as fed intake, use a simple linear program with fixed as fed weight and get a balanced ration of minimum as fed price. The constant weight is arbitrary; for example, W = 100 can be used for all animals and all intakes.
The following is the most economical way to feed an animal with fixed dry matter intake D.
1. Use D to find the minimum and maximum dry matter concentrations of nutrients in the ration.
2. Enter these into a simple linear program with constant dry matter weight 100.
3. Feed the animal D / 100 kg of the solution of the linear program.
The following is the most economical way to feed animals with fixed as fed intake A.
1. Use A to find the minimum and maximum as fed concentrations of nutrients in the ration.
2. Enter these into a simple linear program with constant as fed weight 100.
3. Feed the animal A / 100 kg of the solution of the linear program.
The separate paper Topics in Linear Programming shows the economic disadvantage that may result from using a simple linear program with constant as fed weight to feed an animal with fixed dry matter intake, and using a simple linear program with constant dry matter weight to feed an animal with fixed as fed intake.
NOTE: MixitWin uses a simple linear program with constant dry matter weight 100 if nutrients are restricted on a dry matter basis, and it uses a simple linear program with constant as fed weight 100 if nutrients are restricted on an as fed basis. To override these settings, right-click the Balance button in the Balance This Formula window.
The paradox of as fed prices
Computer feed formulation programs prominently display the as fed price of rations and seem to suggest that rations with low as fed prices are the most economical. Why mix expensive rations, the argument goes, when less expensive rations are possible? If you think about it, however, we shouldn’t be looking for rations with low as fed prices. For example, if you mix 1 ton of a ration with as fed price $160/ton with 1 ton of water, you will get a ration with as fed price $80/ton. This is not a better ration, only wetter, and you will have to use twice as much of it to get the same amounts of the nutrients. What is significant is the dry matter
price: adding 1 ton of water to a ration does not change the dry matter price of the ration.
Puzzling results occur when minimizing the cost of a ration balanced with constant weight. In some cases you may actually be able to lower the as fed price of a ration by forcing a feed into the ration! The answer, of course, is that by forcing in the feed you also make the ration wet enough to lower the as fed price. Moreover, the ration with the smaller as fed price can be more expensive when feeding animals with fixed dry matter intake.
What can happen when balancing rations with as fed prices and constant dry matter weight will be illustrated by a simple linear program with two variables and no constraints other than constant dry matter weight.
Example
We assume the following feed values:
Dry matter As fed price As fed amount
Feed 1 90 % 100 $/ton X ton
Feed 2 30 % 40 $/ton Y ton
Problem D
Find numbers X > 0 and Y > 0 that
Minimize: Z = 100 • X + 40 • Y
Subject to: 0.9 • X + 0.3 • Y = 100
Problem D is solved by noting that
z = 100 • X + 40 • Y
= 100 • (100 - 0.3 • Y) / 0.9 + 40 • Y
= (10000 + 6 • Y) / 0.9
is minimum when Y = 0 tons and X = 100 / 0.9 tons. The minimum cost is Z = 100 • X + 40 • Y = 10000 / 0.9 $, the as fed price is (100 • X + 40 • Y) / (X + Y) = 100 $/ton, and the dry matter percentage is DRY = 100 • DM# / AF# = 100 • (0.9 • X + 0.3 • Y) / (X + Y) = 90 %. The solution has 0 tons of Feed 2, even though Feed 2 has the smaller as fed price!
Now watch what happens when we force 10 tons of Feed 2 into Problem D.
Problem D^
Find numbers X > 0 and Y > 0 that
Minimize: Z = 100 • X + 40 • Y
Subject to: 0.9 • X + 0.3 • Y = 100
Y > 10
Problem D^ is solved by noting that
z = 100 • X + 40 • Y
= 100 • (100 - 0.3 • Y) / 0.9 + 40 • Y
= (10000 + 6 • Y) / 0.9
is minimum when Y = 10 tons and X = 97 / 0.9 tons. The minimum cost is Z = 100 • X + 40 • Y = 10060 / 0.9 $, the as fed price is (100 • X + 40 • Y)/(X + Y) = 10060 / 106 $/ton, and the dry matter percentage is 100 • (0.9 • X + 0.3 • Y)/(X + Y) = 9000 / 106 %.
These results are summarized below, where the DM price is DM$ = AF$ • 100 / DRY. The last column is the daily cost of feeding a 500 kg cow with a fixed dry matter intake of 20 kg = 0.02 ton, which is the product of the DM price and the DM intake (see Feeding animals with fixed intake.)
Cost Z AF price DM price DM Cost
($) ($/ton) ($/ton) (%) ($/day)
Problem D 11111.11 100.00 111.11 90.00 2.22
Problem D^ 11177.78 94.91 111.78 84.91 2.24
When 10 tons of Feed 2 is forced into the ration, the cost Z of the ration increases, as it should. But the as fed price of the ration decreases (because the dry matter percentage decreased), the dry matter price increases and it becomes a more expensive ration to feed an animal with fixed dry matter intake.
The as fed price of a ration has its uses, however, since rations with low as fed prices are best for feeding animals with fixed as fed intake (see Feeding animals with fixed intake).
Optimizing properties of a ration
A property is any quality or attribute of a feed that can be numerically measured. The unit of a property is a name that suggests the method by which the property is measured, and the quantity of a property is the number that results from the measurement of the property. A property is independent, like nutrient concentrations (mg/kg, IU/kg, %) and feed prices ($/ton, €/kg), if its quantity does not depend on the weight of the feed. A property is dependent, like nutrient quantities (kg, IU) and feed costs ($, €) if its quantity depends on the weight of the feed. This section shows how to optimize dependent properties.
Dependent properties may be additive, multiplicative and consistent. A property P is additive if whenever X kg of Feed A with quantity Q units of P is mixed with Y kg of Feed B with quantity R units of P, the resulting X + Y kg of the mixture has quantity Q + R units of P. A property P is multiplicative if whenever a > 0 and X kg of Feed A has quantity Q units of P, then aX kg of Feed A has quantity aQ units of P. A property P is consistent if whenever X kg of Feed A has quantity Q units of P, then any other sample of X kg of Feed A has quantity Q units of P. A feed property is defined as a property that is additive, multiplicative, consistent, and dependent.
Example
Some feed properties are nutrients with units like IU, g, kg, kcal, etc. (See Nutrients and units)
Some feed properties are feed costs ($, €, etc.) that are routinely minimized by linear programming (See Feed cost and feed prices and Linear programming.)
Others are unusual feed properties such as density (kg/V) and electric charge (mEq). Readers may think of other feed properties to optimize.
Regardless of the type, any feed property can be optimized − that is, minimized or maximized − in a balanced ration by using the procedures of linear programming. (See Constant weight linear programming.)
A constant weight linear program (CWLP) has the general form:
Minimize: Z = a1 • X1 + a2 • 2 + ... + an • Xn
Subject to:
(weight equality) c1 • X1 + c2 • X2 + ... + cn • Xn = W
(k inequalities) ej1 • X1 + ej2 • X2 + ... + ejn • Xn < ej0
where W > 0 is the constant weight, Xi > 0 for 1 < i < n, and ai, ci, eji and ej0 are constants for 1 < j < k and 1 < i < n.
In feed formulation problems, Xi is the as fed weight of Feed i (in kg), ai is the as fed price of Feed i (in $/kg), and Z is the cost (in $) of purchasing a mixture of feeds that contains Xi kilograms of Feed i, for 1 < i < n. The problem has constant as fed weight if ci = 1, and constant dry matter weight if 100 • ci is the dry matter percentage of Feed i.
Just as a CWLP minimizes the feed property "cost" (with units $) by setting the coefficients ai to the feed prices (with units $/kg), a CWLP can minimize a feed property other than feed cost by setting the coefficients ai to an independent property of the form Unit/kg, where Unit is the unit of the feed property.
Example
By setting as fed feed prices, ai, to electric charges per kilogram (with units mEq/kg), a CWLP will minimize the electric charge of the ration. The electric change of a ration is a (dependent) feed property with unit mEq. Thus, if ai = 3 mEq/kg and Xi is a weight in kg, then ai • Xi = 3 mEq; and z = a1 • X1 + a2 • X2 + … + an • Xn is the total electric charge of the ration.
A CWLP can also maximize a feed property by setting the coefficients, ai, to the negatives of the property.
Example
A final example is the property "density" with units kg/V (kilograms per unit volume). Up to now independent properties have all been in the form Unit/weight, like price $/kg, so that when the CWLP minimizes z = a1 • X1 + … + an • Xn, it minimizes a feed property, like cost $.
Since the unit kg/V of density is upside down, it is convenient to work with the "reciprocal density" with units V/kg. For 1 < i < n, let 1/ai be the density of Feed i. Then z = a1 • X1 + a2 • X2 + … + an • Xn is a volume that can be minimized by a CWLP. Notice that minimizing the volume V is the same as maximizing the density kg/V.
Of course, feed properties, like electric charge, can simply be restricted in a ration to lie between minimum and maximum limits. (See Dietary cation anion balance.)
As a final note, some nutritionists use linear programming with dry matter prices replacing as fed prices as the coefficients of the objective function z. It is a mystery what this transformation accomplishes; that is, what property of the ration is minimized.
Constant weight linear programming
A constant weight linear program (CWLP) has the general form:
Minimize: Z = a1 • X1 + a2 • 2 + ... + an • Xn
Subject to:
(weight equality) c1 • X1 + c2 • X2 + ... + cn • Xn = W
(k inequalities) ej1 • X1 + ej2 • X2 + ... + ejn • Xn < ej0
where W > 0 is the constant weight, Xi > 0 for 1 < i < n, and ai, ci, eji and ej0 are constants for 1 < j < k and 1 < i < n.
In feed formulation problems, Xi is the as fed weight of Feed i (in kg), ai is the as fed price of Feed i (in $/kg), and Z is the cost (in $) of purchasing a mixture of feeds that contains Xi kilograms of Feed i, for 1 < i < n. The problem has constant as fed weight if ci = 1, and constant dry matter weight if 100 • ci is the dry matter percentage of Feed i.
A CWLP can minimize a feed property other than feed cost by setting the feed prices to the feed property concentrations; for example, by setting as fed feed prices to electric charges per kilogram. A CWLP can also maximize a feed property by setting the feed prices to the negatives of the feed property concentrations. Moreover, a CWLP can minimize or maximize the quantity of one feed property while holding the quantity of a second feed property constant. If the ai are the concentrations of feed property P1, and the ci are the concentrations of feed property P2, then Equation (II) shows that Z is the (minimum) quantity of feed property P1, and W is the (constant) quantity of feed property P2 in the mixture.
By setting the ai to the concentrations (or their negatives) of one feed property, and setting the ci to the concentrations of another feed property, a CWLP can perform the following tasks. (Select one feed property from the first list and a different feed property from the second list.)
minimize or maximize the while holding constant the
P1) feed cost P1) feed cost
P2) feed weight P2) feed weight
P3) quantity of a nutrient P3) quantity of a nutrient
P4) electric charge P4) electric charge
P5) another feed property P5) another feed property
Note: In MixitWin, it is possible to formulate while holding the as fed or dry matter weight of the ration constant. To hold other feed properties constant requires a dedicated program. MixitWin also formulates without a constant weight (see “Requirement feeding”
Dietary cation anion balance
Dietary cation-anion balance (DCAB) is used to balance dairy, swine and poultry rations more precisely. All animals need to be electrically neutral; that is, the sum of the positive charges must equal the sum of the negative charges. This is the principal of electrical neutrality, and it is the underlying principal of DCAB.
Two formulas are commonly used to show whether a diet will cause an acidic or basic response when fed to an animal. These formulas involve the Dietary Cation Anion Difference (DCAD).
DCAD = mEq(Na + K) - (Cl ) / 100 g dry matter
DCAD = mEq(Na + K) - (Cl + S) / 100 g dry matter
The unit of measure is the equivalent (Eq) or the milliequivalent (mEq). An Eq is the atomic weight adjusted for ionic charge; a mEq is one thousandth of an Eq. The mEq/100 g DM for the strong ions are shown below. The number 0.023 is 1/1000 of the atomic weight of Na divided by the valence of Na. The other three numbers are calculated in the same way.
% Na / 0.023 % K / 0.039 % Cl / 0.0355 % S / 0.016
Enter some of the common anionic minerals (aluminum sulfate, magnesium chloride, magnesium sulfate, ammonium chloride, ammonium sulfate, calcium chloride, calcium sulfate) and cationic salts (sodium bicarbonate, S-Carb(TM), Alkaten(TM) in the ingredient list. The following table may be used.
N Na Ca Mg S Cl Water Cost
Mineral or salt % % % % % % % $/ton
Aluminum sulfate 7.62 1700
Magnesium chloride 11.84 34.96 1850
Magnesium sulfate 9.76 13.03 500
Ammonium chloride 26.2 66.4 750
Calcium chloride 27.2 48.3 450
Ammonium sulfate 21.2 24.2 350
Calcium sulfate 23.2 18.6 200
Sodium bicarbonate 27.4
S-Carb(TM) 30.4 16.0
Alkaten(TM) 28.5 14.9
To see and control the Dietary Cation Anion Difference (DCAD) of a ration, create a new nutrient with the name DCAD and units mEq/100g, and add the following values.
Add (% Na) / 0.023 (multiply Sodium by 43.478 and add to DCAD)
Add (% K) / 0.039 (multiply Potassium by 25.641 and add to DCAD)
Subtract (% Cl) / 0.0355 (multiply Chlorine by -28.169 and add to DCAD)
Subtract (% S) / 0.016 (multiply Sulfur by -62.500 and add to DCAD)
The factor 43.478 is the reciprocal of 0.023, and similarly for the other three numbers.
You control the Dietary Cation Anion Difference of a ration by placing minimum or maximum constraints on DCAD. The only limitation may be that the minimum or maximum value of a constraint cannot be a negative number. There is, however, a way to place minimum and maximum constraints on the negative values of DCAD and use these constraints in calculating balanced rations. The way to do this is to create another nutrient with the name NegDCAD and units mEq/100g that has the same values of DCAD but with the positive and negative signs reversed.
You can now restrict negative values of DCAD in a ration by restricting corresponding positive values of NegDACD. For example, to restrict DCAD between a minimum of -3 and a maximum of -1, give NegDCAD a minimum of 1 and a maximum of 3. For another example, to restrict DCAD to the range -2 to + 3, give DCAD a minimum of 0 and a maximum of 3, and give NegDCAD a minimum of 0 and a maximum of 2.
Note: See step by step instructions, including screen captures for controlling DCAD in a ration in MixitWin.
Stochastic feed formulation
Stochastic feed formulation is a method of calculating least cost rations while controlling the probability that some of the nutrients meet minimum or maximum constraints. To use stochastic feed formulation you must know the average (mean) and variability (standard deviation) of selected nutrients and you must know these numbers for each ingredient in the ration. Many nutritionists will find that this information is not available and, for them, stochastic feed formulation becomes more a theoretical exercise than a practical approach to ration formulation. Nevertheless, by understanding the ideas of stochastic feed formulation a nutritionist is better equipped to set realistic nutrient constraints for least cost rations.
Suppose a bin holds 5 tons of Yellow Corn that has 9.44% Protein. What does the number 9.44% really mean? If the Yellow Corn was properly sampled and tested, then 9.44% is the average amount of Protein in each pound of corn in the bin. Of course, 9.44% could also be a number you got from a supplier or found in a book. If you remove a sack of corn from the bin, and the average amount of Protein in the bin is 9.44%, then 50% of the time the sack of corn you remove will have less than 9.44% Protein, and 50% of the time it will have more than 9.44% Protein. How much less or more depends on the variability of Protein in the bin of Yellow Corn.
The statement that something occurs “50% of the time” is a statement of the probability that something will occur, not a guarantee that it will occur. For example, the statement that “a coin will come up heads 50% of the time” does not mean that if you flip a coin twice you will get exactly one head and one tail. It only means that each time you flip a coin, a head is just as likely to appear as a tail. Similarly, the statement that something occurs “80% of the time” means it should be expected to occur, in the long run, 8 times out of 10. The phrases “80% of the time”, “8 times out of 10” and “an 80% probability” will be used interchangeability.
This section introduces the methods of stochastic feed formulation by calculating the probability that Protein meets a specific minimum constraint. Although this section refers to the nutrient Protein, similar comments apply to other nutrients.
The example in this section is based on an example in “Optimization for Profit” by Bender, Kahan and Mylander, The Haworth Press, New York, 1992, pages 458-467.
Example
Assume the following information for the three ingredients Yellow Corn, Soybean Meal, and Meat & Bone, where all values are on an as fed basis.
Protein Protein
average standard Fat Fiber Price
Ingredient (mean) deviation (mean) (mean) $/ton
Yellow Corn 9.44% 0.53 4.04% 2.58% 39
Soybean Meal 50.08% 0.72 1.26% 3.02% 77
Meat & Bone 47.66% 1.11 9.77% 0.00% 72
When these ingredients are used to calculate a least cost ration with the specifications:
Minimum Protein 24% as fed
Minimum Fat 2% as fed
Minimum Fiber 1% as fed
The following ration is the result:
Yellow Corn 61.9 lbs as fed
Meat & Bone 38.1 lbs as fed Price/ton 51.57
Due to the variability of Protein in the ingredients, when this ration is actually mixed you will find that half of the time your ration will have more than 24% Protein, and half of the time it will have less. That is, there is a 50% probability of meeting the minimum protein constraint. Instead of a 50% probability of success, you may prefer an 80% probability of success (8 times out of 10) in meeting the minimum Protein constraint. The following sections show how to formulate a ration so that it meets the minimum Protein requirement 80% of the time. Stochastic formulation does this by increasing the minimum Protein constraint by a small amount.
To guarantee that the 24% minimum amount of Protein is exceeded more than 50% of the time, you could raise the minimum of Protein by an arbitrary amount; for example, from 24% to 24.65%. (Some nutritionists routinely raise minimum amounts by arbitrary increments for this very reason.) If you then know the average (mean) and variability (standard deviation) of Protein in each ingredient, you can calculate the probability that the minimum amount of Protein in the ration meets or exceeds 24%. The following steps show how to perform this calculation
A ration that is produced by stochastic feed formulation is usually more expensive than an ordinary least cost ration. The requirement that the minimum of Protein is met 80% of the time increases the minimum Protein constraint, which in turn increases the price of the ration.
There is a straightforward way to calculate the probability that an actual ration will meet its minimum and maximums constraints. The method for a single nutrient involves the following four steps.
Step 1.
Increase the minimum of Protein from 24% to 24.65% (the amount of increase, 0.65, is only a guess at this time.)
Step 2.
Find a least cost ration using this new nutrient specification. You will get the following ration with a higher price.
Yellow Corn 60.2 lbs as fed
Meat & Bone 39.8 lbs as fed Price/ton 52.13
Step 3.
Calculate the standard deviation of this nutrient in the least cost ration. The standard deviation, S, is the square root of:
V = (X1 • S1) • (X1 • S1) + (X2 • S2) • (X2 • S2) + (X3 • S3) • (X3 • S3)
where
X1 = 0.602 (proportion of Yellow Corn in mixture)
X2 = 0.000 (proportion of Soybean Meal in mixture)
X3 = 0.398 (proportion of Meat & Bone in mixture)
S1 = 0.53 (standard deviation of Protein in Yellow Corn)
S2 = 0.72 (standard deviation of Protein in Soybean Meal)
S3 = 1.11 (standard deviation of Protein in Meat & Bone)
This calculation shows the standard deviation of Protein in the mixture to be S = 0.5449
Step 4.
Look up the probability in a table of areas under the normal curve.
First calculate:
Z = (New % - Old %) / S = (24.65 - 24.00) / 0.5449 = 1.19
A table of areas under the standard normal curve, available in most statistics books, shows that the area corresponding to Z = 1.19 is approximately 0.88, which corresponds to a probability of 88%. (Instead of using a table of areas under the standard normal curve, it is easier to use the Excel function NORMSDIST(z). The probability is the number NORMSDIST(1.19) = 0.882977.)
The previous steps showed that a ration that is 60.2% Yellow Corn and 39.8% Meat & Bone has an 88% probability of meeting the 24% Protein minimum. Instead of an 88% probability of success, what is really wanted is a ration with an 80% probability of success in meeting the 24% minimum Protein constraint. The following steps do this.
Step 1.
Increase the minimum of Protein from 24% to 24.46%.
The amount of increase is no longer a guess, but is estimated by the product (0.5449)(0.84) = 0.46 where 0.5449 is the standard deviation of the previous ration, and 0.84 is the value of Z in the table of areas under the normal curve that corresponds to 0.80 or 80%. (You can also use the Excel inverse function NORMSINV(0.80) = 0.841621 to find this number.)
Step 2.
Find a least cost ration using this new nutrient specification. You will get the following ration with a lower price because of the lower minimum of Protein.
Yellow Corn 60.7 lbs as fed
Meat & Bone 39.3 lbs as fed Price/ton 51.96
Step 3.
Calculate the standard deviation of this nutrient in the least cost ration, as shown in the previous section. The standard deviation turns out to be S = 0.5420.
Step 4.
Look up the probability in a table of areas under the normal curve.
First calculate:
Z = (New % - Old %) / S = (24.46 - 24.00) / 0.5420 = 0.85
A table of areas under the normal curve shows that the area corresponding to Z = 0.85 is approximately 0.80. (The function NORMSDIST(0.85) = 0.802338 also shows this.)
When you mix a ration with 60.7% Yellow Corn and 39.3% Meat & Bone, 8 times out of 10 the ration will contain at least 24% Protein.
Least cost feed formulation software since 1979.