FOLP.posib: Solves a fuzzy objective linear programming problem using Representation Theorem.

Description

The goal is to solve a linear programming problem having Trapezoidal Fuzzy Numbers as coefficients in the objective function ($f(x)=c_{1}^{f} x_1+\ldots+c_{n}^{f} x_n$). $$Max\, f(x)\ or\ Min\ f(x)$$ $$s.t.:\quad Ax<=b$$

FOLP.posib uses a possibilistic approach. This approach is based on Trapezoidal Fuzzy Numbers arithmetic, so the whole objective may be considered as a Fuzzy Number itself. Defining a notion of maximum for this kind of numbers (a weighted average of the minimum and maximum of the support of the Trapezoidal number).

Usage

FOLP.posib(objective, A, dir, b, maximum = TRUE, w1 = 0.5)

Value

FOLP.posib returns the solution for the given weights if the solver has found it or NULL if not.

Arguments

objective: A vector $(c_{1}^{f}, c_{2}^{f}, ..., c_{n}^{f})$ of Trapezoidal Fuzzy Numbers with the objective function coefficients $f(x)=c_{1}^{f} x_1+\ldots+c_{n}^{f} x_n$. Note that any of the coefficients may also be Real Numbers.
A: Technological matrix of Real Numbers.
dir: Vector of strings with the direction of the inequalities, of the same length as b. Each element of the vector must be one of "=", ">=", "<=", "<" or ">".
b: Vector with the right hand side of the constraints.
maximum: TRUE to maximize the objective function, FALSE to minimize the objective function.
w1: Weight to be used, w2 is calculated as w2=1-w1. w1 must be in the interval [0,1].

References

Dubois, D. and Prade, H. Operations in fuzzy numbers. International Journal of Systems Science, 9:613-626, 1978.

Examples

Run this code

## maximize:   [0,2,3]*x1 + [1,3,4,5]*x2
## s.t.:         x1 + 3*x2 <= 6
##               x1 +   x2 <= 4
##               x1, x2 are non-negative real numbers

obj <- c(FuzzyNumbers::TrapezoidalFuzzyNumber(0,2,2,3), 
         FuzzyNumbers::TrapezoidalFuzzyNumber(1,3,4,5))
A<-matrix(c(1, 1, 3, 1), nrow = 2)
dir <- c("<=", "<=")
b <- c(6, 4)
max <- TRUE

FOLP.posib(obj, A, dir, b, maximum = max, w1=0.2)