Computer Feed FormulationIntroductionComputer 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.
Contents
Nutrient prices
The as fed weight of a ration is always greater than or equal to its dry matter weight (AF#
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). Top
or, the cost of 1 unit of N is (k1 / k2) • (AF$ / AN*)
showing that the price of a nutrient does not change if the quantities are on a dry matter basis.
Intake and daily Amounts
where DN* is the dry matter concentration of the nutrient in the feed and DMI is the dry matter intake of the animal.
Care must be taken with the units of DN* and DMI. For example, if DN* = 20% and DMI = 3 kg, then
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.
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 Minimize: Z = 3.00Xa + 5.00Xb + 6.00Xc The function Z, called the 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.
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 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).
MIN
CON =
Top
The feed Corn may contain
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.
To balance a ration with feeds that include Corn, insert the following feed constraints.
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%.
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.
The following example comes from Table 14-7 in the NRC’s
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.
The following ration balanced with fixed as fed weight has 39.75% dry matter.
Ingredients Amounts (Problem 2
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). Top The dietary concentration of a nutrient is the amount of the nutrient in a diet. Dietary concentrations can be expressed by weight:
If ai is the as fed price of feed Fi, for i = 1, 2, ..., n, the feed cost (in $) of this mixture is
Equation(2) then shows that the as fed price (in $/kg) of this mixture is
Equation (1) shows that the dry matter weight of this mixture is
Constant weight linear programming
If the CWLP has constant as fed weight, then W = X1 + X2 + ... + Xn is a constant and the as fed price (in $/kg) of a solution (X1, X2, ... , Xn) is
To see this, let
The equations relating as fed to dry matter price, and as fed to dry matter intake, are
Multiplying these two equations shows the cost of feeding one cow for one day is
1. Use D to find the minimum and maximum dry matter concentrations of nutrients in the ration.2. Enter these into a simple linear program with constant dry matter weight 100.3. Feed the animal D / 100 kg of the solution of the linear program.
1. Use A to find the minimum and maximum as fed concentrations of nutrients in the ration.2. Enter these into a simple linear program with constant as fed weight 100.3. Feed the animal A / 100 kg of the solution of the linear program.
Find numbers X
These results are summarized below, where the DM price is DM$ = 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
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.
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.
Assume the following information for the three ingredients Yellow Corn, Soybean Meal, and Meat & Bone, where all values are on an as fed basis.
When these ingredients are used to calculate a least cost ration with the specifications:
The following ration is the result:
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. 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 Look up the probability in a table of areas under the normal curve.
Z = (New % - Old %) / S = (24.65 - 24.00) / 0.5449 = 1.19A 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.) 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.) 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 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.
Z = (New % - Old %) / S = (24.46 - 24.00) / 0.5420 = 0.85A 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. |