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:

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

1

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

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# 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)))