A linear inverse blending problem

Description

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:

The ingredients have the following nutrient values and cost

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)))