| Blending {limSolve} | R Documentation |
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.
Blending
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.
Karline Soetaert <k.soetaert@nioo.knaw.nl>.
linp to solve a linear programming problem.
# 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)))