Description Usage Format Author(s) See Also Examples

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.

1 |

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 <karline.soetaert@nioz.nl>.

`linp`

to solve a linear programming problem.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 | ```
# 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)))
``` |

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