Blending {limSolve}R Documentation

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(sum {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 <k.soetaert@nioo.knaw.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

# 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=rbind(G,diag(3)),
              H=c(H,rep(0,3)))$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)))

[Package limSolve version 1.3 Index]