Calculator.LR.FNs (version 1.3)

support: Support of LR fuzzy number

Description

To determining the support of a LR fuzzy number one can use from this function. In other words, the support function is able to compute the smallest and biggest values \(x\) for which \(\mu(x)>0\).

Usage

support(M, Left.fun = NULL, Right.fun = NULL)

Arguments

M

A LR, RL or L fuzzy number

Left.fun

The left-shape function which usually defined before using LRFN.plot (see examples in bellow)

Right.fun

The right-shape function which usually defined before using LRFN.plot (see examples in bellow)

Value

The "support" function return a interval-valued vector in which the membership function value of LR fuzzy number is bigger than zero.

Examples

# NOT RUN {
Left.fun  = function(x)  { (1-x)*(x>=0)}
Right.fun = function(x)  { (exp(-x))*(x>=0)}
T = LR(1, 0.6, 0.2)
support(T)
LRFN.plot( T, xlim=c(-5,20), lwd=2, lty=3, col=4)

N = RL(3, 0.5, 2)
support(N)

Left.fun  = function(x)  { (1-x)*(x>=0)}
M = L(2,4,3)
support(M)

Left.fun  = function(x)  { (1-x^2)*(x>=0)}
Right.fun = function(x)  { (exp(-x))*(x>=0)}
support( LR(17,5,3))


## The function is currently defined as
function (M, Left.fun = NULL, Right.fun = NULL) 
{
    range1 = M[1] - M[2] - M[3] - 100
    range2 = M[1] + M[2] + M[3] + 100
    x = seq(range1, range2, len = 2e+05)
    if (M[4] == 0) {
        y = Left.fun((M[1] - x)/M[2]) * (x <= M[1]) + Right.fun((x - 
            M[1])/M[3]) * (M[1] < x)
    }
    else if (M[4] == 1) {
        y = Right.fun((M[1] - x)/M[2]) * (x <= M[1]) + Left.fun((x - 
            M[1])/M[3]) * (M[1] < x)
    }
    else if (M[4] == 0.5) {
        y = Left.fun((M[1] - x)/M[2]) * (x <= M[1]) + Left.fun((x - 
            M[1])/M[3]) * (M[1] < x)
    }
    supp = c()
    supp[1] = min(x[0 < y & y < 1])
    supp[2] = max(x[0 < y & y < 1])
    if (supp[1] == min(x)) {
        supp[1] = -Inf
    }
    if (supp[2] == max(x)) {
        supp[2] = +Inf
    }
    return(supp)

  }
# }