Novel multi-objective, multi-item and four-dimensional transportation problem with vehicle speed in LR-type intuitionistic fuzzy environment

Abstract

In this paper, we present some novel multi-objective, multi-item and four-dimensional transportation problems in LR-type intuitionistic fuzzy environment. Here, for the first time, the speed of different vehicles and rate of disturbance of speed due to the road condition of different routes for the time minimization objective are introduced. Furthermore, three models are presented under three different conditions. The reduced deterministic models are obtained on implementation of a defuzzification approach by using the accuracy function. Moreover, a new method for converting multi-objective problem into single-objective one is proposed and also we use convex combination method. The models are illustrated by some numerical examples and optimal results are presented.

Appendix

1.1 Preliminaries

In this section, we recall some basic definitions and results which will be used in the next sections of this paper.

Definition 1

An intuitionistic fuzzy set (IFS) is a generalization of the ordinary fuzzy sets, which is characterized by a membership function and a nonmembership function. Let $X=\{ x_1 , x_2 ,\ldots , x_n \}$ be a collection of some objects, then an intuitionistic fuzzy set $A^{I}$ in X is defined as the form of an ordered triplet $A^{I} = \{\langle x_i,\mu _{A^{I}}(x_i), \nu _{A^{I}}(x_i)\rangle / x_i \in X \}$, where $\mu _{A^{I}}(x_i): X \rightarrow [0, 1]$ is called the membership function or grade of membership of $x_i$ in $A^{I}$ and $\nu _{A^{I}}(x_i): X [0, 1]$ is called the nonmembership function or grade of nonmembership of $x_i$ in $A^{I}$ satisfying the condition $0 \le \mu _{A^{I}}(x_i) + \nu _{A^{I}}(x_i) \le 1$. $\pi _{A^{I}}(x)=1-\mu _{A^{I}}(x) - \nu _{A^{I}}(x)$ represents the degree of hesitation or the degree of indeterminacy of $x_i$ being in $A^{I}$ in X and $0\le \pi _{A^{I}}\le 1$.

Definition 2

The $\alpha$-cut of an IFS $A^{I}$ is denoted as $A^{I}_{\alpha }$ and is given by:

$$\begin{aligned} A^{I}_{\alpha } = \{x\in X:\mu _{A^{I}}(x)\ge \alpha \}, \quad \forall \alpha \in [0,1] \end{aligned}$$

Definition 3

The $\beta$-cut of an IFS $A^{I}$ is denoted as $A^{I}_{\beta }$ and is given by:

$$\begin{aligned} A^{I}_{\beta } = \{x\in X:\nu _{A^{I}}(x)\le \beta \}, \quad \forall \beta \in [0,1] \end{aligned}$$

Definition 4

The $(\alpha ,\beta )$-cut of an IFS $A^{I}$ is denoted as $A^{I}_{\alpha , \beta }$ and is given by:

$$\begin{aligned} A^{I}_{\alpha , \beta } = \{x\in X:\mu _{A^{I}}(x)\ge \alpha , \nu _{A^{I}}(x)\le \beta ; \alpha +\beta \le 1\}, \quad \forall \alpha , \beta \in [0,1] \end{aligned}$$

Definition 5

An IFS $A^{I} = \{\langle x,\mu _{A^{I}}(x), \nu _{A^{I}}(x)\rangle / x \in {\mathfrak {R}}\}$ is called an intuitionistic fuzzy number (IFN) if the following hold

• (i) There exists $m \in {\mathfrak {R}}$ such that $\mu _{A^{I}}(m)=1$ and $\nu _{A^{I}}(m)=0$ (m is called the mean value of $A^{I}$)

• (ii) The membership function $\mu _{A^{I}}$ and nonmembership function $\nu _{A^{I}}$ are piecewise continuous functions from R to the closed interval [0, 1] and $0 \le \mu _{A^{I}}(x) + \nu _{A^{I}}(x)\le 1,$ ,$\forall x\in {\mathfrak {R}}$. $\mu _{A^{I}}$, $\nu _{A^{I}}$ are of the following forms:

  $$\begin{aligned} \mu _{A^{I}}(x)= \left\{ \begin{array}{ll} f_{1}(x), &{} \quad \text{ for } m-l< x< m\\ 1,&{} \quad \text{ for } x=m \\ f_{2}(x),&{} \quad \text{ for } m< x <m+r\\ 0, &{} \quad \text{ otherwise } \end{array} \right. \end{aligned}$$

  and

  $$\begin{aligned} \nu _{A^{I}}(x)= \left\{ \begin{array}{ll} g_{1}(x), &{} \quad \text{ for } m-l'< x< m; 0\le f_{1}(x)+g_{1}(x)\le 1\\ 0,&{}\quad \text{ for } x=m \\ g_{2}(x),&{} \quad \text{ for } m< x <m+r'; 0\le f_{2}(x)+g_{2}(x)\le 1\\ 1, &{} \quad \text{ otherwise } \end{array} \right. \end{aligned}$$

Here, $f_{1}$ and $f_{2}$ are piecewise continuous, strictly increasing and strictly decreasing functions in $(m-l, m)$ and $(m, m+r),$ respectively. Again $g_{1}$ and $g_{2}$ are piecewise continuous, strictly decreasing and strictly increasing functions in $(m-l', m)$ and $(m, m+r'),$ respectively. l and r are the left and right spreads of membership function $\mu _{A^{I}},$ respectively. Again $l'$ and $r'$ are the left and right spreads of nonmembership function $\nu _{A^{I}},$ respectively. The IFN $A^{I}$ is represented by $(m; l, r; l', r')$.

Definition 6

An IFN $A^{I}$ is called trapezoidal intuitionistic fuzzy number (TIFN) if its membership function $\mu _{A^{I}}$ and nonmembership function $\nu _{A^{I}}$ are as follows:

$$\begin{aligned} \mu _{A^{I}}(x)= \left\{ \begin{array}{ll} \frac{x-r_{1}}{r_{2}-r_{1}}, &{} \quad \text{ for } r_{1}\le x\le r_{2} \\ 1,&{} \quad \text{ for } r_{2}\le x\le r_{3} \\ \frac{r_{4}-x}{r_{4}-r_{3}},&{} \quad \text{ for } r_{3}\le x \le r_{4}\\ 0, &{} \quad \text{ otherwise } \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \nu _{A^{I}}(x)= \left\{ \begin{array}{ll} \frac{r_{2}-x}{r_{2}-r_{1}'}, &{} \quad \text{ for } r_{1}'\le x\le r_{2} \\ 0,&{} \quad \text{ for } r_{2}\le x\le r_{3} \\ \frac{x-r_{3}}{r_{4}'-r_{3}},&{} \quad \text{ for } r_{3}\le x \le r_{4}'\\ 1, &{} \quad \text{ otherwise } \end{array} \right. \end{aligned}$$

where $r_{1}'\le r_{1}\le r_{2}\le r_{3}\le r_{4}\le r_{4}'$. The TIFN $A^{I}$ in ${\mathfrak {R}}$ is represented as $(r_{1}, r_{2},r_{3}, r_{4};r_{1}',r_{2}, r_{3}, r_{4}')$ with its membership function $\mu _{A^{I}}$ and nonmembership function $\nu _{A^{I}}$.

Definition 7

A function$f:[0,\infty )\rightarrow [0,1]$ is called a shape function if the following conditions hold:

• (a) $f(0)=1$

• (b) f is continuous function on $[0,\infty )$

• (c) f is decreasing on $[0,\infty )$ and

• (d) $\lim \nolimits _{x\rightarrow \infty }f(x)=0.$

Definition 8

An IFN $A^{I}$ is called LR-type IFN, so that for membership function $\mu _{A^{I}}(x)$ and nonmembership function $\nu _{A^{I}}(x)$, $0 \le \mu _{A^{I}}(x) + \nu _{A^{I}}(x)\le 1$ holds and is defined as follows:

$$\begin{aligned} \mu _{A^{I}}(x)= \left\{ \begin{array}{ll} L\left(\frac{m-x}{l}\right), &{} \quad \text{ for } x\le m \\ R\left(\frac{x-m}{r}\right),&{} \quad \text{ for } x \ge m \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \nu _{A^{I}}(x)= \left\{ \begin{array}{ll} 1-L\left(\frac{m-x}{l'}\right), &{}\quad \text{ for } x\le m \\ 1-R\left(\frac{x-m}{r'}\right),&{} \quad \text{ for } x \ge m \end{array} \right. \end{aligned}$$

Definition 9

The LR-type representation of a TIFN $A^{I}=(r_{1}, r_{2},r_{3}, r_{4};r_{1}',r_{2}, r_{3}, r_{4}')$ is given by $A^{I}=(r_{2}, r_{3};r_{2}-r_{1}, r_{4}-r_{3};r_{2}-r_{1}', r_{4}'-r_{3}),$ and its membership function $\mu _{A^{I}}$ and nonmembership function $\nu _{A^{I}}$ are defined by

$$\begin{aligned} \mu _{A^{I}}(x)= \left\{ \begin{array}{ll} L\left(\frac{r_{2}-x}{r_{2}-r_{1}}\right), &{} \quad \text{ for } x\le r_{2} \\ 1,&{} \quad \text{ for } r_{2}\le x\le r_{3} \\ R\left(\frac{x-r_{3}}{r_{4}-r_{3}}\right),&{} \quad \text{ for } x \ge r_{3} \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \nu _{A^{I}}(x)= \left\{ \begin{array}{ll} 1-L\left(\frac{r_{2}-x}{r_{2}-r_{1}'}\right), &{} \quad \text{ for } x\le r_{2} \\ 0,&{} \quad \text{ for } r_{2}\le x\le r_{3} \\ 1-R\left(\frac{x-r_{3}}{r_{4}'-r_{3}}\right),&{} \quad \text{ for } x \ge r_{3} \end{array} \right. \end{aligned}$$

where $L(x)=R(x)=\max \{0,1-x\}$; $r_{2}-r_{1}$ and $r_{4}-r_{3}$ are the left spread and right spread of the membership function $\mu _{A^{I}},$ respectively, whereas $r_{2}-r_{1}'$ and $r_{4}'-r_{3}$ are the left spread and right spread of the nonmembership function $\nu _{A^{I}} ,$ respectively.

Theorem 1

The$\alpha$-cut and$\beta$-cut of a TIFN$A^{I}=(r_{1}, r_{2},r_{3}, r_{4};r_{1}',r_{2}, r_{3}, r_{4}')$are given by$A^{I}_{\alpha }=[r_1+(r_2-r_1)\alpha , r_4-(r_4-r_3)\alpha ]$and$A^{I}_{\beta }=[r_2-(r_{2}-r_{1}')\beta , r_3+(r_4'-r_3)\beta ]$, $\forall \alpha ,\beta \in (0,1]]$

Proof

For $\alpha \in (0,1]$,

$$\begin{aligned}\ \mu _{A^{I}}(x)\ge \alpha\Rightarrow & {} \frac{x-r_1}{r_2-r_1}\ge \alpha , \frac{r_4-x}{r_4-r_3}\ge \alpha \\\Rightarrow & {} x\ge r_1+(r_2-r_1)\alpha , x\le r_4-(r_4-r_3)\alpha \\\Rightarrow & {} r_1+(r_2-r_1)\alpha \le x\le r_4-(r_4-r_3)\alpha \\\Rightarrow & {} A^{I}_{\alpha }=[r_1+(r_2-r_1)\alpha , r_4-(r_4-r_3)\alpha ] \end{aligned}$$

Now for $\beta \in (0,1]$,

$$\begin{aligned} \nu _{A^{I}}(x)\le \beta\Rightarrow & {} \frac{r_{2}-x}{r_{2}-r_{1}'}\le \beta , \frac{x-r_{3}}{r_{4}'-r_{3}}\le \beta \\\Rightarrow & {} x\ge r_2-(r_{2}-r_{1}')\beta , x\le r_3+(r_4'-r_3)\beta \\\Rightarrow & {} r_2-(r_{2}-r_{1}')\beta \le x\le r_3+(r_4'-r_3)\beta \\\Rightarrow & {} A^{I}_{\beta }=[r_2-(r_{2}-r_{1}')\beta , r_3+(r_4'-r_3)\beta ] \end{aligned}$$

Hence, it is proved. $\square$

Theorem 2

The$(\alpha , \beta )$-cut of a TIFN$A^{I}=(r_{1}, r_{2},r_{3}, r_{4};r_{1}',r_{2}, r_{3}, r_{4}')$is given by$A^{I}_{\alpha , \beta }=[r_1+(r_2-r_1)\alpha , r_4-(r_4-r_3)\alpha ]\cap [r_2-(r_{2}-r_{1}')\beta , r_3+(r_4'-r_3)\beta ]$, $\forall \alpha , \beta \in (0,1]]$

Proof

For $\alpha \in (0,1]$, the $\alpha$-cut of a TIFN $A^{I}=(r_{1}, r_{2},r_{3}, r_{4};r_{1}',r_{2}, r_{3}, r_{4}')$ is given by $A^{I}_{\alpha }=[r_1+(r_2-r_1)\alpha , r_4-(r_4-r_3)\alpha ]$.

For $\beta \in (0,1]$, the $\beta$-cut of the TIFN $A^{I}=(r_{1}, r_{2},r_{3}, r_{4};r_{1}',r_{2}, r_{3}, r_{4}')$ is given by $A^{I}_{\beta }=[r_2-(r_{2}-r_{1}')\beta , r_3+(r_4'-r_3)\beta ]$, $\forall \beta \in (0,1]]$.

So by Definition 4, $A^{I}_{\alpha , \beta }=[r_1+(r_2-r_1)\alpha , r_4-(r_4-r_3)\alpha ]\cap [r_2-(r_{2}-r_{1}')\beta , r_3+(r_4'-r_3)\beta ]$, $\forall \alpha \in (0,1], \alpha +\beta \le 1]$.

Hence, it is proved. $\square$

Theorem 3

Let$A^{I}=(m, n;l, r;l', r')$be a LR-type TIFN, wherelandrare the left spread and right spread of the membership function$\mu _{A^{I}}$, $l'$and$r'$are the left spread and right spread of the nonmembership function$\nu _{A^{I}}$. Then, its$\alpha$-cut and$\beta$-cut are given by$A^{I}_{\alpha }=[m-lL^{-1}(\alpha ), n+rR^{-1}(\alpha )]$and$A^{I}_{\beta }=[m-l'L^{-1}(1-\beta ), n+rR^{-1}(1-\beta )]$, $\forall \alpha ,\beta \in (0,1]$

Proof

For $\alpha \in (0,1]$,

$$\begin{aligned}\ \mu _{A^{I}}(x)\ge \alpha\Rightarrow & {} L\bigg (\frac{m-x}{l}\bigg )\ge \alpha , R\bigg (\frac{x-n}{r}\bigg )\ge \alpha \\\Rightarrow & {} x\ge m-lL^{-1}(\alpha ), x\le n+rR^{-1}(\alpha )\\\Rightarrow & {} m-lL^{-1}(\alpha ) \le x\le n+rR^{-1}(\alpha )\\\Rightarrow & {} A^{I}_{\alpha }=[m-lL^{-1}(\alpha ), n+rR^{-1}(\alpha )] \end{aligned}$$

Now for $\beta \in (0,1]$,

$$\begin{aligned} \nu _{A^{I}}(x)\le \beta\Rightarrow & {} 1-L\bigg (\frac{m-x}{l'}\bigg )\le \beta , 1-R\bigg (\frac{x-n}{r'}\bigg )\le \beta \\\Rightarrow & {} x\ge m-l'L^{-1}(1-\beta ), x\le n+r'R^{-1}(1-\beta )\\\Rightarrow & {} m-l'L^{-1}(1-\beta ) \le x\le n+r'R^{-1}(1-\beta )\\\Rightarrow & {} A^{I}_{\beta }=[m-l'L^{-1}(1-\beta ), n+r'R^{-1}(1-\beta )] \end{aligned}$$

Hence, it is proved. $\square$

Theorem 4

Let$A^{I}=(m, n;l, r;l', r')$be a LR-type TIFN. Then, its$(\alpha , \beta )$-cut is given by$A^{I}_{\alpha , \beta }=[m-lL^{-1}(\alpha ), n+rR^{-1}(\alpha )]\cap [m-l'L^{-1}(1-\beta ), n+r'R^{-1}(1-\beta )]$, $\forall \alpha , \beta \in (0,1]$and$\alpha +\beta \le 1$.

Proof

For $\alpha \in (0,1]$, The $\alpha$-cut of a LR-type TIFN $A^{I}=(m, n;l, r;l', r')$ is given by $A^{I}_{\alpha }=[m-lL^{-1}(\alpha ), n+rR^{-1}(\alpha )]$ For $\beta \in (0,1]$, The $\beta$-cut of the LR-type TIFN $A^{I}=(m, n;l, r;l', r')$ is given by $A^{I}_{\beta }=[m-l'L^{-1}(1-\beta ), n+rR^{-1}(1-\beta )]$, $\forall \beta \in (0,1]]$.

So by definition, $A^{I}_{\alpha , \beta }=[m-lL^{-1}(\alpha ), n+rR^{-1}(\alpha )]\cap [m-l'L^{-1}(1-\beta ), n+r'R^{-1}(1-\beta )]$, $\forall \alpha , \beta \in (0,1], \alpha +\beta \le 1$. Hence, it is proved. $\square$

Arithmetic Operations on LR -type TIFN

Proposition 1

(Addition) If${A_{1}}^{I}=(m_{1}, n_{1};l_{1}, r_{1};l_{1}', r_{1}')$and${A_{2}}^{I}=(m_{2}, n_{2};l_{2}, r_{2};l_{2}', r_{2}')$are two LR-type TIFNs, then${A_{1}}^{I} +{A_{2}}^{I}= (m_{1}+m_{2}, n_{1}+n_{2};l_{1}+l_{2}, r_{1}+r_{2};l_{1}'+l_{2}', r_{1}'+r_{2}')$.

Proof

For $\alpha \in (0,1]$ and $\beta \in (0,1]$, the $\alpha$-cut and $\beta$-cut of the LR-type TIFNs ${A_{1}}^{I}=(m_{1}, n_{1};l_{1}, r_{1};l_{1}', r_{1}')$ and ${A_{2}}^{I}=(m_{2}, n_{2};l_{2}, r_{2};l_{2}', r_{2}')$ are given by ${A_{1}}^{I}_{\alpha }=[m_{1}-l_{1}L^{-1}(\alpha ), n_{1}+r_{1}R^{-1}(\alpha )]$, ${A_{2}}^{I}_{\alpha }=[m_{2}-l_{2}L^{-1}(\alpha ), n_{2}+r_{2}R^{-1}(\alpha )]$ ; ${A_{1}}^{I}_{\beta }=[m_{1}-l_{1}'L^{-1}(1-\beta ), n_{1}+r_{1}'R^{-1}(1-\beta )]$, ${A_{2}}^{I}_{\alpha }=[m_{2}-l_{2}'L^{-1}(1-\beta ), n_{2}+r_{2}'R^{-1}(1-\beta )],$ respectively Now,

$$\begin{aligned} ({A_{1}}^{I}+{A_{2}}^{I})_{\alpha }&= {A_{1}}^{I}_{\alpha }+{A_{2}}^{I}_{\alpha }\nonumber \\&= [m_{1}-l_{1}L^{-1}(\alpha ), n_{1}+r_{1}R^{-1}(\alpha )]+[m_{2}-l_{2}L^{-1}(\alpha ), n_{2}+r_{2}R^{-1}(\alpha )]\nonumber \\&= [(m_{1}+m_{2})-(l_{1}+l_{2})L^{-1}(\alpha ), (n_{1}+n_{2})+(r_{1}+r_{2})R^{-1}(\alpha )] \end{aligned}$$

(35)

Since L and R are decreasing functions on $[0,\infty ]$ with $L(0)=R(0)=1$, $\exists$ some $\alpha =\alpha _0\in (0,1]$ such that $L^{-1}(\alpha _0)=R^{-1}(\alpha _0)=1$.

Therefore,

$$\begin{aligned} ({A_{1}}^{I}+{A_{2}}^{I})_{\alpha _0}&= [(m_{1}+m_{2})-(l_{1}+l_{2}), (n_{1}+n_{2})+(r_{1}+r_{2})] \end{aligned}$$

(36)

Now, by putting $\alpha =1$ in Eq. (35), we get the model point of ${A_{1}}^{I}+{A_{2}}^{I}$, which is given by

$$\begin{aligned} ({A_{1}}^{I}+{A_{2}}^{I})_{1}&= [m_{1}+m_{2}, n_{1}+n_{2}] \end{aligned}$$

(37)

Again

$$\begin{aligned}&({A_{1}}^{I}+{A_{2}}^{I})_{\beta }={A_{1}}^{I}_{\beta }+{A_{2}}^{I}_{\beta }\nonumber \\&\quad =[m_{1}-l_{1}'L^{-1}(1-\beta ), n_{1}+r_{1}'R^{-1}(1-\beta )]\nonumber \\&\qquad +\,[m_{2}-l_{2}'L^{-1}(1-\beta ), n_{2}+r_{2}'R^{-1}(1-\beta )]\nonumber \\&\quad =[(m_{1}+m_{2})-(l_{1}'+l_{2}')L^{-1}(1-\beta ), (n_{1}+n_{2})+(r_{1}'+r_{2}')R^{-1}(1-\beta )] \end{aligned}$$

(38)

Since L and R are decreasing functions on $[0,\infty ]$ with $L(0)=R(0)=1$, $\exists$ some $\beta =\beta _0\in (0,1]$ such that $L^{-1}(1-\beta _0)=R^{-1}(1-\beta _0)=1$.

Therefore,

$$\begin{aligned} ({A_{1}}^{I}+{A_{2}}^{I})_{\beta _0}&= [(m_{1}+m_{2})-(l_{1}'+l_{2}'), (n_{1}+n_{2})+(r_{1}'+r_{2}')] \end{aligned}$$

(39)

Since ${A_{1}}^{I}, {A_{2}}^{I}$ are two LR-type TIFNs, $0 \le l_{1} \le l_{1}'$, $0\le l_{2}\le l_{2}'$, $0 \le r_{1} \le r_{1}'$ and $0\le r_{2}\le r_{2}'$, so $0 \le l_{1} + l_{2}\le l_{1}' + l_{2}'$ and $0 \le r_{1} + r_{2}\le r_{1}' + r_{2}'$

So, from Eqs. (36), (37) and (39), we have ${A_{1}}^{I} +{A_{2}}^{I}= (m_{1}+m_{2}, n_{1}+n_{2};l_{1}+l_{2}, r_{1}+r_{2};l_{1}'+l_{2}', r_{1}'+r_{2}')$.

Hence, it is proved. $\square$

Proposition 2

(Subtraction) If${A_{1}}^{I}=(m_{1}, n_{1};l_{1}, r_{1};l_{1}', r_{1}')$and${A_{2}}^{I}=(m_{2}, n_{2};l_{2}, r_{2};l_{2}', r_{2}')$are two LR-type TIFNs, then${A_{1}}^{I} -{A_{2}}^{I}= (m_{1}-n_{2}, n_{1}-m_{2};l_{1}+r_{2}, r_{1}+l_{2};l_{1}'+r_{2}', r_{1}'+l_{2}')$.

Proof

The $\alpha$-cut and $\beta$-cut of the LR-type TIFNs ${A_{1}}^{I}=(m_{1}, n_{1};l_{1}, r_{1};l_{1}', r_{1}')$ and ${A_{2}}^{I}=(m_{2}, n_{2};l_{2}, r_{2};l_{2}', r_{2}')$ are given by

$$\begin{aligned} {A_{1}}^{I}_{\alpha }&= [m_{1}-l_{1}L^{-1}(\alpha ), n_{1}+r_{1}R^{-1}(\alpha )],\\ {A_{2}}^{I}_{\alpha }&= [m_{2}-l_{2}L^{-1}(\alpha ), n_{2}+r_{2}R^{-1}(\alpha )] ;\\ {A_{1}}^{I}_{\beta }&= [m_{1}-l_{1}'L^{-1}(1-\beta ), n_{1}+r_{1}^{'}R^{-1}(1-\beta )],\\ {A_{2}}^{I}_{\alpha }&= [m_{2}-l_{2}'L^{-1}(1-\beta ), n_{2}+r_{2}'R^{-1}(1-\beta )], \end{aligned}$$

respectively. Now,

$$\begin{aligned} ({A_{1}}^{I}-{A_{2}}^{I})_{\alpha }&= {A_{1}}^{I}_{\alpha }-{A_{2}}^{I}_{\alpha }\nonumber \\&= [m_{1}-l_{1}L^{-1}(\alpha ), n_{1}+r_{1}R^{-1}(\alpha )]-[m_{2}-l_{2}L^{-1}(\alpha ), n_{2}+r_{2}R^{-1}(\alpha )]\nonumber \\&= [m_{1}-n_{2}-(l_{1}L^{-1}(\alpha )+r_{2}R^{-1}(\alpha )), n_{1}-m_{2}+(r_{1}R^{-1}(\alpha )+l_{2}L^{-1}(\alpha ))] \end{aligned}$$

(40)

Since L and R are decreasing functions on $[0,\infty ]$ with $L(0)=R(0)=1$, $\exists$ some $\alpha =\alpha _0\in (0,1]$ such that $L^{-1}(\alpha _0)=R^{-1}(\alpha _0)=1$.

Therefore,

$$\begin{aligned} ({A_{1}}^{I}-{A_{2}}^{I})_{\alpha _0}&= [(m_{1}-n_{2})-(l_{1}+r_{2}), (n_{1}-m_{2})+(r_{1}+l_{2})] \end{aligned}$$

(41)

Now, by putting $\alpha =1$ in Eq. (40), we get the model point of ${A_{1}}^{I}-{A_{2}}^{I}$, which is given by

$$\begin{aligned} ({A_{1}}^{I}-{A_{2}}^{I})_{1}&= [m_{1}-n_{2}, n_{1}-m_{2}] \end{aligned}$$

(42)

Again

$$\begin{aligned} ({A_{1}}^{I}-{A_{2}}^{I})_{\beta }&= {A_{1}}^{I}_{\beta }-{A_{2}}^{I}_{\beta }\nonumber \\&= [m_{1}-l_{1}'L^{-1}(1-\beta ), n_{1}+r_{1}'R^{-1}(1-\beta )]\nonumber \\&-[m_{2}-l_{2}'L^{-1}(1-\beta ), n_{2}+r_{2}'R^{-1}(1-\beta )]\nonumber \\&= [m_{1}-n_{2}-(l_{1}'L^{-1}(1-\beta )+r_{2}'R^{-1}(1-\beta )), n_{1}\nonumber \\&-m_{2}+(r_{1}'R^{-1}(1-\beta )+l_{2}'L^{-1}(1-\beta ))] \end{aligned}$$

(43)

Since L and R are decreasing functions on $[0,\infty ]$ with $L(0)=R(0)=1$, $\exists$ some $\beta =\beta _0\in (0,1]$ such that $L^{-1}(1-\beta _0)=R^{-1}(1-\beta _0)=1$.

Therefore,

$$\begin{aligned} ({A_{1}}^{I}-{A_{2}}^{I})_{\beta _0}&= [(m_{1}-n_{2})-(l_{1}'+r_{2}'), (n_{1}-m_{2})+(r_{1}'+l_{2}')] \end{aligned}$$

(44)

Since ${A_{1}}^{I}, {A_{2}}^{I}$ are two LR-type TIFNs, $0 \le l_{1} \le l_{1}'$, $0\le l_{2}\le l_{2}'$, $0 \le r_{1} \le r_{1}'$ and $0\le r_{2}\le r_{2}'$, so $0 \le l_{1} + r_{2}\le l_{1}' + r_{2}'$ and $0 \le r_{1} + l_{2}\le r_{1}' + l_{2}'$.

Therefore, from Eqs. (41), (42) and (44), we have

$$\begin{aligned} {A_{1}}^{I} - {A_{2}}^{I}= (m_{1}-n_{2}, n_{1}-m_{2};l_{1}+r_{2}, r_{1}+l_{2};l_{1}'+r_{2}', r_{1}'+l_{2}'). \end{aligned}$$

Hence, it is proved.

Proposition 3

(Scalar multiplication) If${A_{1}}^{I}=(m_{1}, n_{1};l_{1}, r_{1};l_{1}', r_{1}')$is a LR-type TIFN and$\lambda$is any real number, then$\lambda {A_{1}}^{I} = (\lambda m_{1}, \lambda n_{1};\lambda l_{1}, \lambda r_{1};\lambda l_{1}', \lambda r_{1}')$, when$\lambda \ge 0$.

$\lambda {A_{1}}^{I} = (\lambda n_{1}, \lambda m_{1};-\lambda r_{1}, -\lambda l_{1};-\lambda r_{1}', -\lambda l_{1}')$, when$\lambda < 0$.

Proof

The $\alpha$-cut and $\beta$-cut of the LR-type TIFN ${A_{1}}^{I}=(m_{1}, n_{1};l_{1}, r_{1};l_{1}', r_{1}')$ are given by ${A_{1}}^{I}_{\alpha }=[m_{1}-l_{1}L^{-1}(\alpha ), n_{1}+r_{1}R^{-1}(\alpha )]$, ${A_{1}}^{I}_{\beta }=[m_{1}-l_{1}'L^{-1}(1-\beta ), n_{1}+r_{1}'R^{-1}(1-\beta )]$, respectively

Case I$\lambda \ge 0$

$$\begin{aligned} (\lambda {A_{1}}^{I})_{\alpha }&= \lambda {A_{1}}^{I}_{\alpha }\nonumber \\&= \lambda [m_{1}-l_{1}L^{-1}(\alpha ), n_{1}+r_{1}R^{-1}(\alpha )]\nonumber \\&= [\lambda m_{1}-\lambda l_{1}L^{-1}(\alpha ), \lambda n_{1}+\lambda r_{1}R^{-1}(\alpha )] \end{aligned}$$

(45)

Since L and R are decreasing functions on $[0,\infty ]$ with $L(0)=R(0)=1$, $\exists$ some $\alpha =\alpha _0\in (0,1]$ such that $L^{-1}(\alpha _0)=R^{-1}(\alpha _0)=1$.

Therefore,

$$\begin{aligned} (\lambda {A_{1}}^{I})&= [\lambda m_{1}-\lambda l_{1}, \lambda n_{1}+\lambda r_{1}] \end{aligned}$$

(46)

Now, by putting $\alpha =1$ in Eq. (45), we get the model point of $\lambda {A_{1}}^{I}$, which is given by

$$\begin{aligned} (\lambda {A_{1}}^{I})_{1}&= [\lambda m_{1}, \lambda n_{1}] \end{aligned}$$

(47)

Again

$$\begin{aligned} (\lambda {A_{1}}^{I})_{\beta }&= \lambda {A_{1}}^{I}_{\beta }\nonumber \\&= [\lambda m_{1}-\lambda l_{1}'L^{-1}(1-\beta ), \lambda n_{1}+ \lambda r_{1}'R^{-1}(1-\beta )] \end{aligned}$$

(48)

Since L and R are decreasing functions on $[0,\infty ]$ with $L(0)=R(0)=1$, $\exists$ some $\beta =\beta _0\in (0,1]$ such that $L^{-1}(1-\beta _0)=R^{-1}(1-\beta _0)=1$.

Therefore,

$$\begin{aligned} (\lambda {A_{1}}^{I})_{\beta _0}&= [\lambda m_{1}-\lambda l_{1}', \lambda n_{1} + \lambda r_{1}'] \end{aligned}$$

(49)

Since ${A_{1}}^{I}$ are LR-type TIFN and $\lambda > 0$, $0 \le l_{1} \le l_{1}'$ and $0 \le r_{1} \le r_{1}'$. So $0 \le \lambda l_{1}\le \lambda l_{1}'$ and $0 \le \lambda r_{1} \le \lambda r_{1}'$

So, from Eqs. (46), (47) and (49), we have

$$\begin{aligned} \lambda {A_{1}}^{I} = (\lambda m_{1}, \lambda n_{1};\lambda l_{1}, \lambda r_{1};\lambda l_{1}', \lambda r_{1}') \end{aligned}$$

Hence, it is proved.

Case II$\lambda < 0$

$$\begin{aligned} (\lambda {A_{1}}^{I})_{\alpha }&= \lambda {A_{1}}^{I}_{\alpha }\nonumber \\&= \lambda [m_{1}-l_{1}L^{-1}(\alpha ), n_{1}+r_{1}R^{-1}(\alpha )]\nonumber \\&= [\lambda n_{1}+\lambda r_{1}R^{-1}(\alpha ), \lambda m_{1}-\lambda l_{1}L^{-1}(\alpha )] \end{aligned}$$

(50)

Since L and R are decreasing functions on $[0,\infty ]$ with $L(0)=R(0)=1$, $\exists$ some $\alpha =\alpha _0\in (0,1]$ such that $L^{-1}(\alpha _0)=R^{-1}(\alpha _0)=1$.

Therefore,

$$\begin{aligned} (\lambda {A_{1}}^{I})&= [\lambda n_{1}+\lambda r_{1}, \lambda m_{1}-\lambda l_{1}] \end{aligned}$$

(51)

Now, by putting $\alpha =1$ in Eq. 50, we get the model point of $\lambda {A_{1}}^{I}$, which is given by

$$\begin{aligned} (\lambda {A_{1}}^{I})_{1}&= [\lambda n_{1}, \lambda m_{1}] \end{aligned}$$

(52)

Again

$$\begin{aligned} (\lambda {A_{1}}^{I})_{\beta }& = \lambda {A_{1}}^{I}_{\beta }\nonumber \\&= [\lambda n_{1}+ \lambda r_{1}'R^{-1}(1-\beta ), \lambda m_{1}-\lambda l_{1}'L^{-1}(1-\beta )] \end{aligned}$$

(53)

Since L and R are decreasing functions on $[0,\infty ]$ with $L(0)=R(0)=1$, $\exists$ some $\beta =\beta _0\in (0,1]$ such that $L^{-1}(1-\beta _0)=R^{-1}(1-\beta _0)=1$.

Therefore

$$\begin{aligned} (\lambda {A_{1}}^{I})_{\beta _0}&= [\lambda n_{1} + \lambda r_{1}', \lambda m_{1}-\lambda l_{1}'] \end{aligned}$$

(54)

Since ${A_{1}}^{I}$ are LR-type TIFN and $\lambda < 0$, $0 \le l_{1} \le l_{1}'$ and $0 \le r_{1} \le r_{1}'$. So $0 \le -\lambda l_{1}\le -\lambda l_{1}'$ and $0 \le -\lambda r_{1} \le -\lambda r_{1}'$

So, from Eqs. (51), (52) and (54), we have

$$\begin{aligned} \lambda {A_{1}}^{I} = (\lambda n_{1}, \lambda m_{1};-\lambda r_{1}, -\lambda l_{1};-\lambda r_{1}', -\lambda l_{1}'). \end{aligned}$$

Hence, it is proved. $\square$

Definition 10

[33] Let the $(\alpha ,\beta )$-cut of a LR-type TIFN be given by

$$\begin{aligned} A^{I}_{\alpha , \beta }=[A_{1}(\alpha ),A_{2}(\alpha )]\cap [{A'}_{1}(\beta ),A'_{2}(\beta )]; \alpha +\beta \le 1, \forall \alpha , \beta \in [0,1]; \end{aligned}$$

where $A_{1}(\alpha )=m-lL^{-1}(\alpha )$, $A_{2}(\alpha )=n+rR^{-1}(\alpha )$, ${A'}_{1}(\beta )=m-l'L^{-1}(1-\beta )$ and ${A'}_{1}(\beta )=n+r'R^{-1}(1-\beta )$. Then, by mean of $(\alpha , \beta )$-cut method, the representation of membership function is

$$\begin{aligned} R_{\mu }(A^{I})=\frac{1}{2}\int _{0}^{1}\bigg [A_{1}(\alpha )+A_{2}(\alpha )\bigg ]{\mathrm{d}}\alpha . \end{aligned}$$

Again by mean of $(\alpha , \beta )$-cut method, the representation of nonmembership function is

$$\begin{aligned} R_{\nu }(A^{I})=\frac{1}{2}\int _{0}^{1}\bigg [{A'}_{1}(\beta )+A'_{2}(\beta )\bigg ]{\mathrm{d}}\beta . \end{aligned}$$

The accuracy function of $A^{I}$ is denoted by $f(A^{I})$ and defined by

$$\begin{aligned} f(A^{I})=\frac{R_{\mu }(A^{I})+R_{\nu }(A^{I})}{2}, \end{aligned}$$

to defuzzify the given numbers as deterministic one.

Theorem 5

Let$A^{I}=(m, n;l, r;l', r')$be a LR-type TIFN. Then, its accuracy function is given by$f(A^{I})=\frac{2m+2n-\frac{l}{2}+\frac{r}{2}-\frac{l'}{2}+\frac{r'}{2}}{4}$.

Proof

By definition, the accuracy function of $A^{I}$ is defined by

$$\begin{aligned} f(A^{I})&= \frac{R_{\mu }(A^{I})+R_{\nu }(A^{I})}{2}\\&= \frac{\frac{1}{2}\int _{0}^{1}\bigg [A_{1}(\alpha )+A_{2}(\alpha )\bigg ] {\mathrm{d}}\alpha +\frac{1}{2}\int _{0}^{1}\bigg [{A'}_{1}(\beta )+A'_{2}(\beta )\bigg ]{\mathrm{d}}\beta }{2}\\&= \frac{1}{4}\bigg [\int _{0}^{1}\bigg \{m-lL^{-1}(\alpha )+n+rR^{-1}(\alpha ) \bigg \}{\mathrm{d}}\alpha \\ &\quad+\,&\int _{0}^{1}\bigg \{m-l'L^{-1}(1-\beta )+n+r'R^{-1}(1-\beta ) \bigg \}{\mathrm{d}}\beta \bigg ] \end{aligned}$$

Now, by Definition 9 for a TIFN, $L(x)=R(x)=max\{0,1-x\}, \forall x\ge 0$;

Since $\alpha \in (0,1]$, $L(1-\alpha )=\alpha \Rightarrow L^{-1}(\alpha )=\alpha$ and $R(1-\alpha )=\alpha \Rightarrow R^{-1}(\alpha )=\alpha$ Similarly, $L(\beta )=1-\beta \Rightarrow L^{-1}(1-\beta )=\beta$ and $R(\beta )=1-\beta \Rightarrow R^{-1}(1-\beta )=\beta$

$$\begin{aligned} f(A^{I})&= \frac{1}{4}\bigg [\int _{0}^{1}\bigg \{m-l(1-\alpha )+n+r(1-\alpha )\bigg \}{\mathrm{d}}\alpha +\int _{0}^{1}\bigg \{m-l'\beta +n+r'\beta \bigg \}{\mathrm{d}}\beta \bigg ]\nonumber \\ \Rightarrow f(A^{I})&= \frac{2m+2n-\frac{l}{2}+\frac{r}{2}-\frac{l'}{2}+\frac{r'}{2}}{4} \end{aligned}$$

(55)

Hence, it is proved. $\square$

Theorem 6

Let${A_{1}}^{I}=(m_{1}, n_{1};l_{1}, r_{1};l_{1}', r_{1}')$, ${A_{2}}^{I}=(m_{2}, n_{2};l_{2}, r_{2};l_{2}', r_{2}')$be any two LR-type TIFNs and$\lambda _{1}$and$\lambda _{2}$be any two real numbers, then$f(\lambda _{1}{A_{1}}^{I}+\lambda _{2}{A_{2}}^{I})=\lambda _{1} f({A_{1}}^{I})+\lambda _{2} f({A_{2}}^{I})$

Proof

Let $\lambda \ge 0$, now by using propositions 2 and 3, $\lambda {A_{1}}^{I}+\lambda {A_{2}}^{I}=(\lambda _{1}m_{1}+\lambda _{2}m_{2}, \lambda _{1}n_{1}+\lambda _{2}n_{2}; \lambda _{1}l_{1}+\lambda _{2}l_{2}, \lambda _{1}r_{1}+\lambda _{2}r_{2}; \lambda _{1}l_{1}'+\lambda _{2}l_{2}', \lambda _{1}r_{1}'+\lambda _{2}r_{2}')$ Thus, by using Eq. (55),

$$\begin{aligned}&f(\lambda {A_{1}}^{I}+\lambda {A_{2}}^{I})\nonumber \\&\quad =\frac{2(\lambda _{1}m_{1}+\lambda _{2}m_{2})+2(\lambda _{1}n_{1}+\lambda _{2}n_{2})- \frac{\lambda _{1}l_{1}+\lambda _{2}l_{2}}{2}+\frac{\lambda _{1}r_{1}+\lambda _{2}r_{2}}{2}- \frac{\lambda _{1}l_{1}'+\lambda _{2}l_{2}'}{2}+\frac{\lambda _{1}r_{1}'+\lambda _{2}r_{2}'}{2}}{4} \end{aligned}$$

(56)

Again $f({A_{1}}^{I})= \frac{2m_{1}+2n_{1}-\frac{l_{1}}{2}+\frac{r_{1}}{2}-\frac{l_{1}'}{2}+\frac{r_{1}'}{2}}{4}$ and $f({A_{2}}^{I})= \frac{2m_{2}+2n_{2}-\frac{l_{2}}{2}+\frac{r_{2}}{2}-\frac{l_{2}'}{2}+\frac{r_{2}'}{2}}{4}$ So,

$$\begin{aligned}&\lambda _{1} f({A_{1}}^{I})+\lambda _{2} f({A_{2}}^{I})\\&\quad =\lambda _{1}\bigg (\frac{2m_{1}+2n_{1}-\frac{l_{1}}{2}+\frac{r_{1}}{2}-\frac{l_{1}'}{2}+\frac{r_{1}'}{2}}{4}\bigg )+ \lambda _{2}\bigg (\frac{2m_{2}+2n_{2}-\frac{l_{2}}{2}+\frac{r_{2}}{2}-\frac{l_{2}'}{2}+\frac{r_{2}'}{2}}{4}\bigg )\\&\quad \Rightarrow \lambda _{1} f({A_{1}}^{I})+\lambda _{2} f({A_{2}}^{I})\\&\quad =\frac{2(\lambda _{1}m_{1}+\lambda _{2}m_{2})+2(\lambda _{1}n_{1}+\lambda _{2}n_{2})- \frac{\lambda _{1}l_{1}+\lambda _{2}l_{2}}{2}+\frac{\lambda _{1}r_{1}+\lambda _{2}r_{2}}{2}- \frac{\lambda _{1}l_{1}'+\lambda _{2}l_{2}'}{2}+\frac{\lambda _{1}r_{1}'+\lambda _{2}r_{2}'}{2}}{4} \end{aligned}$$

Thus, from Eqs. (56) and (57), for $\lambda _{1}\ge 0$ and $\lambda _{2}\ge 0$$f(\lambda _{1}{A_{1}}^{I}+\lambda _{2}{A_{2}}^{I})=\lambda _{1} f({A_{1}}^{I})+\lambda _{2} f({A_{2}}^{I}).$ In the same way, it can be proved for any value of $\lambda _{1}$ and $\lambda _{2}$.

Hence, it is proved. $\square$

1.2 Conversion technique for multi-objective into single objective

In this segment, we will present a discussion about conversion techniques for multi-objective into single objective. Now we discuss about convex combination method

1.3 Convex combination method (CCM)

We consider the multi-objective optimization problem together with some constraints as follows:

$$\begin{aligned} \left\{ \begin{array}{ll} \max & \quad [f_{i}(x),\ i=1,2, \ldots , M] \\ \hbox {s.t} &{} \quad g_{j}\ge 0; \ j=1,2,\ldots , N\\ &{} \quad x\in X \end{array} \right. \end{aligned}$$

(57)

Subsequently, by the convex combination method (Tanino et al. [34]), we shift the above problem into the following form as:

$$\begin{aligned} \left\{ \begin{array}{ll} \max &{} \quad {\sum \limits _{i=1}^{M}w_{i}f_{i}(x)}, \text{ where } \sum \limits _{i=1}^{M}w_{i}=1, 0<w_{i}<1\\ \hbox {s.t} &{} \quad g_{j}\ge 0; j=1,2,\ldots , N \\ &{} \quad x\in X \end{array} \right. \end{aligned}$$

(58)

where $w_{i}$ is the weight function of ith objective. Then, the corresponding x and $f_{i}(x)$ are the solutions of the problem in Eq. (57).