A manufacturer produces a feeding mix for pet animals.
The feed mix contains two nutritive ingredients and one ingredient (filler)
to provide bulk.
One kg of feed mix must contain a minimum quantity of each of four nutrients
as below:
Nutrient
A
B
C
D
gram
80
50
25
5
The ingredients have the following nutrient values and cost
(gram/kg)
A
B
C
D
Cost/kg
Ingredient 1
100
50
40
10
40
Ingredient 2
200
150
10
-
60
Filler
-
-
-
-
0
The problem is to find the composition of the feeding mix that minimises
the production costs subject to the constraints above.
Stated otherwise: what is the optimal amount of ingredients in one kg of
feeding mix?
Mathematically this can be estimated by solving a linear programming problem:
min(∑ {Cost_i*x_i})
subject to
x_i>=0
Ex=f
Gx>=h
Where the Cost (to be minimised) is given by:
x_1*40+x_2*60
The equality ensures that the sum of the three fractions equals 1:
1 = x_1+x_2+x_3
And the inequalities enforce the nutritional constraints:
100*x_1+200*x_2>80
50*x_1+150*x_2>50
and so on
The solution is Ingredient1 (x1) = 0.5909, Ingredient2 (x2)=0.1364 and
Filler (x3)=0.2727.
Usage
Blending
Format
A list with matrix G and vector H that contain the inequality
conditions and with vector Cost, defining the cost function.
Columnnames of G or names of Cost are the names of the
ingredients, rownames of G and names of H are the nutrients.
Author(s)
Karline Soetaert <karline.soetaert@nioz.nl>.
See Also
linp to solve a linear programming problem.
Examples
# Generate the equality condition (sum of ingredients = 1)
E <- rep(1, 3)
F <- 1
G <- Blending$G
H <- Blending$H
# add positivity requirement
G <- rbind(G, diag(3))
H <- c(H, rep(0, 3))
# 1. Solve the model with linear programming
res <- linp(E = t(E), F = F, G = G, H = H, Cost = Blending$Cost)
# show results
print(c(res$X, Cost = res$solutionNorm))
dotchart(x = as.vector(res$X), labels = colnames(G),
main = "Optimal blending with ranges",
sub = "using linp and xranges", pch = 16,
xlim = c(0, 1))
# 2. Possible ranges of the three ingredients
(xr <- xranges(E, F, G, H))
segments(xr[,1], 1:ncol(G), xr[,2], 1:ncol(G))
legend ("topright", pch = c(16, NA), lty = c(NA, 1),
legend = c("Minimal cost", "range"))
# 3. Random sample of the three ingredients
# The inequality that all x > 0 has to be added!
xs <- xsample(E = E, F = F, G = G, H = H)$X
pairs(xs, main = "Blending, 3000 solutions with xsample")
# Cost associated to these random samples
Costs <- as.vector(varsample(xs, EqA = Blending$Cost))
hist(Costs)
legend("topright", c("Optimal solution",
format(res$solutionNorm, digits = 3)))