limSolve (version 1.4)

Blending: 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:

rlllll{ Nutrient A B C D gram 80 50 25 5 }

The ingredients have the following nutrient values and cost

rllllll{ (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.

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)))