LIMBlending: A blending problem specification

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 linear inverse models LIMBlending and LIMinputBlending are generated from the file Blending.input which can be found in subdirectory /examples/LinearProg of the package directory

LIMBlending is generated by function Setup

LIMinputBlending is generated by function Read

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

LIMBlending
LIMinputBlending

Arguments

Format

LIMBlending is of type lim, which is a list of matrices, vectors, names and values that specify the linear inverse model problem.

see the return value of Setup for more information about this list

LIMinputBlending is of type liminput, see the return value of Read for more information.

A more complete description of these structures is in vignette("LIM")

Author

Karline Soetaert <karline.soetaert@nioz.nl>

Examples

Run this code


# 1. Solve the model with linear programming
res <- Linp(LIMBlending, ispos = TRUE)

# show results
print(c(res$X, Cost = res$solutionNorm))


# 2. Possible ranges of the three ingredients
(xr <- Xranges(LIMBlending, ispos = TRUE))
Nx <- LIMBlending$NUnknowns

# plot
dotchart(x = as.vector(res$X), xlim = range(xr),
         labels = LIMBlending$Unknowns,
         main = "Optimal blending with ranges",
         sub = "using linp and xranges", pch = 16)
segments(xr[ ,1], 1:Nx, xr[ ,2], 1:Nx)
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!
blend   <- LIMBlending
blend$G <- rbind(blend$G, diag(3))
blend$H <- c(blend$H, rep(0, 3))

xs <- Xsample(blend)

pairs(xs, main = "Blending, 3000 solutions with xsample")