Commentary on “A reply to a Note on the paper “A simplified novel technique for solving fully fuzzy linear programming problems””

Abstract

Bhardwaj and Kumar (J. Optim. Theory Appl. 163(2), 685-696 (2014)) pointed out that Khan et al. (J. Optim. Theory Appl. 159(2), 536-546 (2013)) have assumed some mathematical incorrect assumptions in their proposed method for solving fully fuzzy linear programming problems (FFLPPs). However, Khan et al. (J. Optim. Theory Appl. 173(1), 353-356 (2017)) replied that their proposed method is valid and no mathematical incorrect assumptions have been considered. The aim of this commentary is to make the researchers aware that Khan et al. (J. Optim. Theory Appl. 173(1), 353-356 (2017)) are misleading the others. In actual case, Khan et al. (J. Optim. Theory Appl. 159(2), 536-546 (2013)) have considered mathematical incorrect assumptions in their proposed method due to which some of the elements of the optimal simplex table, obtained by Khan et al. (J. Optim. Theory Appl. 159(2), 536-546 (2013)), are not triangular fuzzy numbers.

Keywords

FFLPPs simplex method ranking function

1 Introduction

Khan et al. [4] proposed a method for solving FFLPPs. All the steps of the method, proposed by Khan et al. [4], are same as well-known simplex method for solving crisp linear programming problems (LPPs). The only difference between the well-known simplex method for solving crisp LPPs and the method, proposed by Khan et al. [4], are as follows.

Khan et al. [4] have represented the elements and variables in the simplex table astriangular fuzzy numbers (TFNs) instead of real numbers.

To check the optimality as well as to find the entering variable, there is need to check that a triangular fuzzy number (TFN) is a non-negative TFN or a non-positive TFN. In the literature, different methods have been proposed to check that a TFN is a non-negative TFN or a non-positive TFN.

Khan et al. [4] have adopted the following method to check the nature of a TFN, $\tilde{T} = (u, v, w)$ .

Find $(\tilde{T}) = u + w - σ$ where, $σ = \frac{w - u}{6}$ .

Case (i): If $R (\tilde{T}) ⩾ 0$ then $\tilde{T}$ is a non-negative TFN.

Case (ii): If $R (\tilde{T}) ⩽ 0$ then $\tilde{T}$ is a non-positive TFN.

Furthermore, in the literature, different methods have been proposed to find the minimum of TFNs. Khan et al. [4] have adopted the following method to find the minimum of two TFNs, ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1})$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2})$

Check that $R ({\tilde{T}}_{1}) > R ({\tilde{T}}_{2})$ or $R ({\tilde{T}}_{1}) < R ({\tilde{T}}_{2})$ or $R ({\tilde{T}}_{1}) = R ({\tilde{T}}_{2})$ where, $R (\tilde{T}) = u_{i} + w_{i} - σ_{i}$ , $σ_{i} = \frac{w_{i} - u_{i}}{6}, i = 1$ .

Case (i): If $R ({\tilde{T}}_{1}) > R ({\tilde{T}}_{2})$ then ${\tilde{T}}_{1} > {\tilde{T}}_{2}$ i.e., $minimum {{\tilde{T}}_{1}, {\tilde{T}}_{2}} = {\tilde{T}}_{2}$ .

Case (ii): If $R ({\tilde{T}}_{1}) < R ({\tilde{T}}_{2})$ then ${\tilde{T}}_{1} < {\tilde{T}}_{2}$ i.e., $minimum {{\tilde{T}}_{1}, {\tilde{T}}_{2}} = {\tilde{T}}_{1}$ .

Case (iii): If $R ({\tilde{T}}_{1}) = R ({\tilde{T}}_{2})$ then ${\tilde{T}}_{1} = {\tilde{T}}_{2}$ i.e., $minimum {{\tilde{T}}_{1}, {\tilde{T}}_{2}} = {\tilde{T}}_{1}, {\tilde{T}}_{2}$ .

Bhardwaj and Kumar [1] claimed that Khan et al. [4] have used the relations

$R (\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{2}}) = \frac{R ({\tilde{T}}_{1})}{R ({\tilde{T}}_{2})}$ .

${\tilde{T}}_{1} - {\tilde{T}}_{1} = (0, 0, 0)$ .

$\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{1}} = (1, 1, 1)$ .

in their proposed method. Whereas, in actual case,

$R (\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{2}}) \neq \frac{R ({\tilde{T}}_{1})}{R ({\tilde{T}}_{2})}$ .

${\tilde{T}}_{1} - {\tilde{T}}_{1} \neq (0, 0, 0)$ .

$\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{1}} \neq (1, 1, 1)$ .

Therefore, the method, proposed by Khan et al. [4], is not valid.

Khan et al. [5] replied that they have not used any of these relations in their proposed method [4]. Hence, Khan et al. [4] method is valid.

The aim of this commentary is to make the researchers aware that Khan et al. [5] are misleading the others. In actual case, Khan et al. [4] have considered the mathematical incorrect assumptions, pointed out by Bhardwaj and Kumar [1], in their proposed method.

2 Validity of the claim

The following clearly indicates that Khan et al. [4] have considered mathematical incorrect assumptions in their proposed method.

Khan et al. [4] have defined as well as used the following multiplication operation of two TFNs.

If ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1})$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2})$ are two TFNs then, $\begin{matrix} {\tilde{T}}_{1} \cdot {\tilde{T}}_{2} & = & (u_{1}, v_{1}, w_{1}) \cdot (u_{2}, v_{2}, w_{2}) \\ = & (u_{1} u_{2}, v_{1} v_{2}, w_{1} w_{2}) \end{matrix}$

Khan et al. [4] have assumed that this multiplication operation is valid for all TFNs. However, in actual case, this multiplication is valid only if u₁ ⩾ 0 and u₂ ⩾ 0.

For example, if ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1}) = (- 2, - 1, 3)$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2}) = (1, 4, 5)$ i.e., u₁ ⩾ 0 and u₂ ⩾ 0 then according to this multiplication

${\tilde{T}}_{1} \cdot {\tilde{T}}_{2} = (u_{1} u_{2}, v_{1} v_{2}, w_{1} w_{2}) = (- 2, - 4, 15)$

It is pertinent to mention that the obtained number (-2, - 4, 15) is not a TFN as for a TFN (u, v, w), the condition u ⩽ v ⩽ w should always be satisfied.

Khan et al. [4] have defined as well as used the following division operation of two TFNs. If ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1})$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2})$ are two TFNs then,

\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{2}} = \frac{(u_{1}, v_{1}, w_{1})}{(u_{2}, v_{2}, w_{2})} = (\frac{u_{1}}{w_{1}}, \frac{v_{1}}{v_{2}}, \frac{w_{1}}{u_{2}}) .

Khan et al. [4] have assumed that this division operation is valid for all TFNs. However, in actual case, this division is valid only if u₁ ⩾ 0 and.u₂ > 0.

For example, if ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1}) = (- 3, - 2, - 1)$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2}) = (1, 4, 5)$ i.e., u₁ ≱ 0 and u₂ > 0 then according to this division, $\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{2}} = (\frac{u_{1}}{w_{1}}, \frac{v_{1}}{v_{2}}, \frac{w_{1}}{u_{2}}) = (- \frac{3}{5}, - \frac{1}{2}, - 1)$ .

It is pertinent to mention that the obtained number $(- \frac{3}{5}, - \frac{1}{2}, - 1)$ is not a TFN as for a TFN (u, v, w), the condition u ⩽ v ⩽ w should always be satisfied.

The following clearly indicates that Khan et al. [4] have used the mathematical incorrect assumptions (i) and (ii) in their proposed method.

Khan et al. [4] solved the fully fuzzy linear programming problem (FFLPP) (P1) by their proposed method and claimed that Table 1 represents the optimal table of FFLPP (P1).

Table 1
Optimal table of FFLPP (P1)

${\tilde{x}}_{1}$ ${\tilde{x}}_{2}$ ${\tilde{x}}_{3}$ ${\tilde{x}}_{4}$ ${\tilde{x}}_{5}$ ${\tilde{x}}_{6}$ $\tilde{b}$

${\tilde{x}}_{4} (R_{1}^{'})$ $(- \frac{8}{5}, - \frac{1}{6}, 1)$ $(- 3, - \frac{281}{36}, - 10)$ (0, 0, 0) (1, 1, 1) (0, 0, 0) $(- \frac{6}{5}, - \frac{31}{30}, - 1)$ $(- 12, - \frac{299}{36}, 0)$

${\tilde{x}}_{5} (R_{2}^{'})$ $(- 2, \frac{23}{12}, \frac{11}{5})$ $(- 5, - \frac{101}{8}, - 8)$ (0, 0, 0) (0, 0, 0) (1, 1, 1) $(- 2, - \frac{105}{60}, - \frac{7}{5})$ $(- 20, - \frac{885}{72}, 8)$

${\tilde{x}}_{3} (R_{3}^{'})$ $(\frac{1}{5}, \frac{1}{2}, \frac{7}{5})$ $(\frac{1}{3}, \frac{17}{12}, 4)$ (1, 1, 1) (0, 0, 0) (0, 0, 0) $(\frac{1}{15}, \frac{1}{10}, \frac{1}{5})$ $(1, \frac{29}{12}, 6)$

$\tilde{Z} (R_{4}^{'})$ $(- 7, \frac{2}{3}, 19)$ $(- \frac{25}{3}, \frac{4}{9}, 57)$ (0, 0, 0) (0, 0, 0) (0, 0, 0) $(\frac{1}{3}, \frac{34}{30}, 3)$ $(5, \frac{986}{36}, 90)$

	${\tilde{x}}_{1}$	${\tilde{x}}_{2}$	${\tilde{x}}_{3}$	${\tilde{x}}_{4}$	${\tilde{x}}_{5}$	${\tilde{x}}_{6}$	$\tilde{b}$
${\tilde{x}}_{4} (R_{1}^{'})$	$(- \frac{8}{5}, - \frac{1}{6}, 1)$	$(- 3, - \frac{281}{36}, - 10)$	(0, 0, 0)	(1, 1, 1)	(0, 0, 0)	$(- \frac{6}{5}, - \frac{31}{30}, - 1)$	$(- 12, - \frac{299}{36}, 0)$
${\tilde{x}}_{5} (R_{2}^{'})$	$(- 2, \frac{23}{12}, \frac{11}{5})$	$(- 5, - \frac{101}{8}, - 8)$	(0, 0, 0)	(0, 0, 0)	(1, 1, 1)	$(- 2, - \frac{105}{60}, - \frac{7}{5})$	$(- 20, - \frac{885}{72}, 8)$
${\tilde{x}}_{3} (R_{3}^{'})$	$(\frac{1}{5}, \frac{1}{2}, \frac{7}{5})$	$(\frac{1}{3}, \frac{17}{12}, 4)$	(1, 1, 1)	(0, 0, 0)	(0, 0, 0)	$(\frac{1}{15}, \frac{1}{10}, \frac{1}{5})$	$(1, \frac{29}{12}, 6)$
$\tilde{Z} (R_{4}^{'})$	$(- 7, \frac{2}{3}, 19)$	$(- \frac{25}{3}, \frac{4}{9}, 57)$	(0, 0, 0)	(0, 0, 0)	(0, 0, 0)	$(\frac{1}{3}, \frac{34}{30}, 3)$	$(5, \frac{986}{36}, 90)$

$max \tilde{z} = (2, 5, 8) {\tilde{x}}_{1} + (3, \frac{37}{6}, 10) {\tilde{x}}_{2} + (5, \frac{34}{3}, 15) {\tilde{x}}_{3}$

Subject to ${\begin{matrix} (2, 5, 8) {\tilde{x}}_{1} + (3, \frac{41}{6}, 10) {\tilde{x}}_{2} + (5, \frac{31}{3}, 18) {\tilde{x}}_{3} ⩽ (6, \frac{50}{3}, 30) \\ (4, \frac{32}{3}, 12) {\tilde{x}}_{1} + (5, \frac{73}{6}, 20) {\tilde{x}}_{2} + (7, \frac{105}{6}, 30) {\tilde{x}}_{3} ⩽ (10, 30, 50) \\ (3, 5, 7) {\tilde{x}}_{1} + (5, 15, 20) {\tilde{x}}_{2} + (5, 10, 15) {\tilde{x}}_{3} ⩽ (2, \frac{145}{6}, 30) {\tilde{x}}_{1}, {\tilde{x}}_{2}, {\tilde{x}}_{3} ⩾ 0 \end{matrix} (P 1)$

It is well known fact that for a TFN, $\tilde{T} = (u, v, w)$ , the condition u ⩽ v ⩽ w should always be satisfied. While, it is obvious that

For the second element of first row i.e., $(- 3, \frac{281}{36}, - 10)$ , this condition is not satisfying i.e., $(- 3, \frac{281}{36}, - 10)$ is not a TFN.

For the second element of second row i.e., $(- 5, \frac{101}{8}, - 8)$ , this condition is not satisfying i.e., $(- 5, \frac{101}{8}, - 8)$ is not a TFN.

3 Exact multiplication and division operations of TFNs

In this section, the exact multiplication and division operations of TFNs are presented [2, 3].

If ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1})$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2})$ are two TFNs such that u₁ and u₂ are any real numbers. Then,

T_{1} \cdot {\tilde{T}}_{2} = (\begin{matrix} minimum {u_{1} u_{2}, u_{1} w_{2}, w_{1} u_{2}, w_{1} w_{2}}, v_{1} v_{2}, \\ max imium {u_{1} u_{2}, u_{1} w_{2}, w_{1} u_{2}, w_{1} w_{2} \end{matrix})

If ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1})$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2})$ are two TFNs such that u₁ is any realnumber and u₂ is any non-zero real number. Then,

\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{2}} = (\begin{matrix} minimum {\frac{u_{1}}{u_{2}}, \frac{u_{1}}{w_{2}}, \frac{w_{1}}{u_{2}}, \frac{w_{1}}{w_{2}}}, \frac{v_{1}}{v_{2}}, \\ max imium {\frac{u_{1}}{u_{2}}, \frac{u_{1}}{w_{2}}, \frac{w_{1}}{u_{2}}, \frac{w_{1}}{w_{2}}} \end{matrix})

4 Fuzzy optimal solution of a FFLPP by Khan et al. method

To point out the mathematical incorrect assumptions, considered by Khan et al. [4], there is need to solve a FFLPP (P1). Therefore, in this section, the FFLPP (P1), considered by Khan et al. [4] to illustrate their proposed method, is solved by Khan et al. method [4] in a very detailed manner.

Khan et al. [4] have used the following method to find the optimal table, represented by Table 1, of the FFLPP (P1).

Step 1: Table 2 [4] represents the initial simplex table of the FFLPP (P1).

Table 2
Initial simplex table of the FFLPP (P1) [4]

${\tilde{x}}_{1}$ ${\tilde{x}}_{2}$ ${\tilde{x}}_{3}$ ${\tilde{x}}_{4}$ ${\tilde{x}}_{5}$ ${\tilde{x}}_{6}$ $\tilde{b}$

${\tilde{x}}_{4} (R_{1})$ (2, 5, 8) $(3, \frac{41}{6}, 10)$ $(5, \frac{31}{3}, 18)$ (1, 1, 1) (0, 0, 0) (0, 0, 0) $(6, \frac{50}{3}, 30)$

${\tilde{x}}_{5} (R_{2})$ $(4, \frac{32}{3}, 12)$ $(5, \frac{73}{6}, 20)$ $(7, \frac{105}{3}, 30)$ (0, 0, 0) (1, 1, 1) (0, 0, 0) (10, 30, 50)

${\tilde{x}}_{6} (R_{3})$ (3, 5, 7) (5, 15, 20) (5, 10, 15) (0, 0, 0) (0, 0, 0) (1, 1, 1) $(2, \frac{145}{6}, 30)$

$\tilde{Z} (R_{4})$ (-8, - 5, - 2) $(- 10, - \frac{37}{6}, - 3)$ $(- 15, - \frac{34}{3}, - 5)$ (0, 0, 0) (0, 0, 0) (0, 0, 0) (0, 0, 0)

	${\tilde{x}}_{1}$	${\tilde{x}}_{2}$	${\tilde{x}}_{3}$	${\tilde{x}}_{4}$	${\tilde{x}}_{5}$	${\tilde{x}}_{6}$	$\tilde{b}$
${\tilde{x}}_{4} (R_{1})$	(2, 5, 8)	$(3, \frac{41}{6}, 10)$	$(5, \frac{31}{3}, 18)$	(1, 1, 1)	(0, 0, 0)	(0, 0, 0)	$(6, \frac{50}{3}, 30)$
${\tilde{x}}_{5} (R_{2})$	$(4, \frac{32}{3}, 12)$	$(5, \frac{73}{6}, 20)$	$(7, \frac{105}{3}, 30)$	(0, 0, 0)	(1, 1, 1)	(0, 0, 0)	(10, 30, 50)
${\tilde{x}}_{6} (R_{3})$	(3, 5, 7)	(5, 15, 20)	(5, 10, 15)	(0, 0, 0)	(0, 0, 0)	(1, 1, 1)	$(2, \frac{145}{6}, 30)$
$\tilde{Z} (R_{4})$	(-8, - 5, - 2)	$(- 10, - \frac{37}{6}, - 3)$	$(- 15, - \frac{34}{3}, - 5)$	(0, 0, 0)	(0, 0, 0)	(0, 0, 0)	(0, 0, 0)

Step 2: Since the considered problem is of maximization, so the initial fuzzy basic feasible solution, presented in Table 2, will be a fuzzy optimal solution if all the values of $\tilde{z}$ , corresponding to non-basic variables, will be non-negative TFNs i.e., the TFNs (-8, - 5, - 2), $(- 10, - \frac{37}{6}, - 3)$ and $(- 15, - \frac{34}{3}, - 5)$ , representing the values of corresponding to non-basic fuzzy variables ${\tilde{x}}_{1}$ , ${\tilde{x}}_{2}$ and ${\tilde{x}}_{3}$ respectively, will be a non-negative TFNs. As discussed in Section 1, Khan et al. [4] have assumed that a TFN ${\tilde{T}}_{1} = (u, v, w)$ will be non-negative TFN if $R (\tilde{T}) ⩾ 0$ , where $R (\tilde{T}) = u + w - σ$ , σ = $\frac{w - u}{6}$ .

Since, R (- 8, - 5, - 2) = -11, $R (- 10, - \frac{37}{6}, - 3) = - 14.16$ and $R (- 15, - \frac{34}{3}, - 5) = - 21.66$ i.e. (-8, - 5, - 2), $(- 10, - \frac{37}{6}, - 3)$ and $(- 15, - \frac{34}{3}, - 5)$ are not non-negative TFNs and hence, the optimality condition is not satisfying. So, the initial fuzzy basic feasible solution, presented in Table 2, is not a fuzzy optimal solution.

Step 3: Since, the initial fuzzy basic feasible solution, presented in Table 2, is not a fuzzy optimal solution, so the fuzzy non-basic variable ${\tilde{x}}_{3}$ , corresponding to which the value of $R (\tilde{z})$ is most negative, will enter into the basis.

Step 4: Since, all the elements of the column corresponding to the entering variable ${\tilde{x}}_{3}$ are non-negative TFNs i.e., $R (5, \frac{31}{3}, 18) > 0$ , $R (7, \frac{105}{6}, 30) > 0$ , R (5, 10, 15) >0. So, any of the fuzzy basic variables ${\tilde{x}}_{4}$ , ${\tilde{x}}_{5}$ and ${\tilde{x}}_{6}$ can leave from the basis. However, according to the minimum ratio rule, out of the fuzzy basic variables ${\tilde{x}}_{1} (i = 4, 5, 6)$ that fuzzy basic variable will leave from the basis corresponding to which $minimum {\frac{b_{i}}{a_{i 3}}, i = 1, 2, 3}$ will exist i.e., corresponding to which $minimum {\frac{(6, \frac{50}{3}, 30)}{(5, \frac{31}{3}, 18)}, \frac{(10, 30, 50)}{(7, \frac{105}{6}, 30)}, \frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)}}$ will exist.

Khan et al. [4] have used the following steps to find. $minimum {\frac{(6, \frac{50}{3}, 30)}{(5, \frac{31}{3}, 18)}, \frac{(10, 30, 50)}{(7, \frac{105}{6}, 30)}, \frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)}}$

Step 4(a): $R (6, \frac{50}{3}, 30) = 32$ , $R (5, \frac{31}{3}, 18) = 20.83$ , R (10, 30, 50) = 53.33, $R (7, \frac{105}{6}, 30) = 33.16$ , $R (2, \frac{145}{6}, 30) = 27.33$ and R (5, 100, 15) = 18.33. $\frac{R (6, \frac{50}{3}, 30)}{R (5, \frac{31}{3}, 18)} = \frac{32}{20.83} = 1.5362$ , $\frac{R (10, 30, 50)}{R (7, \frac{105}{6}, 30)} = \frac{53.33}{33.16} = 1.6082$ and $\frac{R (2, \frac{145}{6}, 30)}{R (5, 10, 15)} = \frac{27.33}{18.33} = 1.4909$ .

Step 4(b): Since,minimum {1.5362, 1.6082, 1.4909} = 1.4909, which is corresponding to $\frac{R (2, \frac{145}{6}, 30)}{R (5, 10, 15)}$ . Therefore, $minimum {\frac{(6, \frac{50}{3}, 30)}{(5, \frac{31}{3}, 18)}, \frac{(10, 30, 50)}{(7, \frac{105}{6}, 30)}, \frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)}} = \frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)}$ . This indicates that the fuzzy basic variable ${\tilde{x}}_{6}$ will leave from the basis.

Step 5: Since, the fuzzy non-basic variable ${\tilde{x}}_{3}$ is entering variable (third column of Table 2) and the fuzzy basic variable ${\tilde{x}}_{6}$ is the leaving variable (third row of Table 2). So, in the next simplex table the third element of the third column should be (1, 1, 1). To obtain the same, Khan et al. [4] have applied the arithmetic operation $\frac{R_{3}}{(5, 10, 15)}$ .

Khan et al. [4] claimed that after applying this operation the simplex Table 2 will be transformed into the simplex Table 3.

Table 3

First simplex table of the FFLPP (P1)

	${\tilde{x}}_{1}$	${\tilde{x}}_{2}$	${\tilde{x}}_{3}$	${\tilde{x}}_{4}$	${\tilde{x}}_{5}$	${\tilde{x}}_{6}$	$\tilde{b}$
${\tilde{x}}_{4} (R_{1})$	(2, 5, 8)	$(3, \frac{41}{6}, 10)$	$(5, \frac{31}{3}, 18)$	(1, 1, 1)	(0, 0, 0)	(0, 0, 0)	$(6, \frac{50}{3}, 30)$
${\tilde{x}}_{5} (R_{2})$	$(4, \frac{32}{3}, 12)$	$(5, \frac{73}{6}, 20)$	$(7, \frac{105}{3}, 30)$	(0, 0, 0)	(1, 1, 1)	(0, 0, 0)	(10, 30, 50)
${\tilde{x}}_{6} (R_{3})$	$(\frac{1}{5}, \frac{1}{2}, \frac{7}{5})$	$(\frac{1}{3}, \frac{17}{12}, 4)$	(1, 1, 1)	(0, 0, 0)	(0, 0, 0)	$(\frac{1}{15}, \frac{1}{10}, \frac{1}{5})$	$(1, \frac{29}{12}, 6)$
$\tilde{Z} (R_{4})$	(-8, - 5, - 2)	$(- 10, - \frac{37}{6}, - 3)$	$(- 15, - \frac{34}{3}, - 5)$	(0, 0, 0)	(0, 0, 0)	(0, 0, 0)	(0, 0, 0)

Step 6: Since, the remaining elements of the column corresponding to ${\tilde{x}}_{3}$ in the next simplex tableshould be (0, 0, 0). So, to obtain the same, Khan et al. [4] have used the arithmetic operations $R_{1} + (- 18, - \frac{31}{3}, - 5) R_{3}^{'},$ $R_{2} + (- 30, - \frac{105}{6}, - 7) R_{3}^{'},$

$R_{4} + (5, \frac{34}{3}, 15) R_{3}^{'} .$

Khan et al. [4] have claimed that after applying theses operations, the simplex Table 3 will be transformed into the simplex Table 4 (same as the simplex Table 1).

Step 7: It can be easily verified that in the simplex Table 4, $R (\tilde{Z})$ corresponding to each non-basic variable is a non-negative real number. So, the fuzzy basic feasible solution, obtained in Table 4, is a fuzzy optimal solution and hence, Table 4 is optimal simplex table of the FFLPP (P1).

Table 4

Optimal simplex table of the FFLPP (P1)

	${\tilde{x}}_{1}$	${\tilde{x}}_{2}$	${\tilde{x}}_{3}$	${\tilde{x}}_{4}$	${\tilde{x}}_{5}$	${\tilde{x}}_{6}$	$\tilde{b}$
${\tilde{x}}_{4} (R_{1})$	$(- \frac{8}{5}, - \frac{1}{6}, 1)$	$(- 3, - \frac{281}{36}, - 10)$	(0, 0, 0)	(1, 1, 1)	(0, 0, 0)	$(- \frac{6}{5}, - \frac{31}{30}, - 1)$	$(- 12, - \frac{299}{36}, 0)$
${\tilde{x}}_{5} (R_{2})$	$(- 2, \frac{23}{12}, \frac{11}{5})$	$(- 5, - \frac{101}{8}, - 8)$	(0, 0, 0)	(0, 0, 0)	(1, 1, 1)	$(- 2, - \frac{105}{60}, - \frac{7}{5})$	$(- 20, - \frac{885}{72}, 8)$
${\tilde{x}}_{3} (R_{3})$	$(\frac{1}{5}, \frac{1}{2}, \frac{7}{5})$	$(\frac{1}{3}, \frac{17}{12}, 4)$	(1, 1, 1)	(0, 0, 0)	(0, 0, 0)	$(\frac{1}{15}, \frac{1}{10}, \frac{1}{5})$	$(1, \frac{29}{12}, 6)$
$\tilde{Z} (R_{3})$	$(- 7, \frac{2}{3}, 19)$	$(- \frac{25}{3}, \frac{4}{9}, 57)$	(0, 0, 0)	(0, 0, 0)	(0, 0, 0)	$(\frac{1}{3}, \frac{34}{30}, 3)$	$(5, \frac{986}{36}, 90)$

5 Mathematical incorrect assumptions considered by Khan et al.

Bhardwaj and Kumar [1] pointed out that the following mathematical incorrect assumptions have been considered in Khan et al. method [4].

Khan et al. [4] have assumed that if $\tilde{A}$ and $\tilde{B}$ aretwo TFNs then $R (\frac{\tilde{A}}{\tilde{B}}) = \frac{R (\tilde{A})}{R (\tilde{B})}$ . While, inactualcase $R (\frac{\tilde{A}}{\tilde{B}}) \neq \frac{R (\tilde{A})}{R (\tilde{B})}$ .

Khan et al. [4] have assumed that if $\tilde{A}$ is a TFN then $\frac{\tilde{A}}{\tilde{A}} - (1, 1, 1)$ and $\tilde{A} - \tilde{A} = (0, 0, 0)$ . While, in actual case neither $\frac{\tilde{A}}{\tilde{A}} \neq (1, 1, 1)$ nor $\tilde{A} - \tilde{A} \neq (0, 0, 0)$ .

Khan et al. [5] replied that this claim of Bhardwaj and Kumar [1] is irrelevant as no such mathematical incorrect assumptions have been used in the existing method [4].

However, the following clearly indicates that the claim of Bhardwaj and Kumar [1] is relevant i.e., the mathematical incorrect assumptions, pointed out by Bhardwaj and Kumar [1], have been used in Khan et al. method [4].

It is obvious from Step 4 of Section 4 that Khan et al. [4] have assumed that to find $minimum {\frac{(6, \frac{50}{3}, 30)}{(5, \frac{31}{3}, 18)}, \frac{(10, 30, 50)}{(7, \frac{105}{6}, 30)}, \frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)}}$ is equivalent to find $minimum {\frac{R (6, \frac{50}{3}, 30)}{R (5, \frac{31}{3}, 18)},$ $\frac{R (10, 30, 50)}{R (7, \frac{105}{6}, 30)}, \frac{R (2, \frac{145}{6}, 30)}{R (5, 10, 15)}}$ .

While, in actual case to find $minimum {\frac{(6, \frac{50}{3}, 30)}{(5, \frac{31}{3}, 18)}, \frac{(10, 30, 50)}{(7, \frac{105}{6}, 30)}, \frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)}}$ is equivalent to find $minimum {R (\frac{(6, \frac{50}{3}, 30)}{(5, \frac{31}{3}, 18)})$ , $R (\frac{(10, 30, 50)}{(7, \frac{105}{6}, 30)}), R (\frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)})}$ .

This clearly indicates that Khan et al. [4] have assumed the property $R (\frac{\tilde{A}}{\tilde{B}}) = \frac{R (\tilde{A})}{R (\tilde{B})}$ . Whereas, Bhardwaj and Kumar [1] have shown that $R (\frac{\tilde{A}}{\tilde{B}}) \neq \frac{R (\tilde{A})}{R (\tilde{B})}$ .

It is obvious from Step 5 of Section 4 that Khan et al. [4] assumed that on applying theoperation $\frac{R_{3}}{(5, 10, 15)}$ on the elements of third row of simplex Table 2, the elements of thirdrow of the simplex Table 3 will be obtained i.e., Khan et al. [4] have assumed that

$\frac{(3, 5, 7)}{(5, 10, 15)} = (\frac{1}{5}, \frac{1}{2}, \frac{7}{5})$ .

$\frac{(5, 10, 15)}{(5, 10, 15)} = (\frac{1}{3}, \frac{17}{12}, 4)$ .

$\frac{(5, 10, 15)}{(5, 10, 15)} = (1, 1, 1)$ .

$\frac{(0, 0, 0)}{(5, 10, 15)} = (0, 0, 0)$ .

$\frac{(1, 1, 1)}{(5, 10, 15)} = (\frac{1}{15}, \frac{1}{10}, \frac{1}{5})$ .

$\frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)} = (1, \frac{\frac{145}{6}}{10}, \frac{30}{5})$ .

It is obvious that to find these values there is need to use the division operation of TFNs. As pointed out in (ii) of Section 2, Khan et al. [4] have used the following division operation to obtain these elements.

If ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1})$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2})$ are two TFNs. Then, $\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{2}} = (\frac{u_{1}}{w_{2}}, \frac{v_{1}}{v_{2}}, \frac{w_{1}}{u_{2}})$ .

On using this division operation $\frac{(3, 5, 7)}{(5, 10, 15)} = (\frac{1}{5}, \frac{1}{2}, \frac{7}{5})$ , $\frac{(0, 0, 0)}{(5, 10, 15)} = (0, 0, 0)$ , $\frac{(0, 0, 0)}{(5, 10, 15)} = (0, 0, 0)$ and $\frac{(1, 1, 1)}{(5, 10, 15)} = (\frac{1}{15}, \frac{1}{10}, \frac{1}{5})$ . But, $\frac{(5, 10, 15)}{(5, 10, 15)} = (\frac{5}{15}, \frac{10}{10}, \frac{15}{5}) = (\frac{1}{3}, 1, 3) \neq (1, 1, 1)$

Whereas, Khan et al. [4] have assumed that $\frac{(5, 10, 15)}{(5, 10, 15)} = (1, 1, 1)$ i.e., to obtain this element, Khan et al. [4] have used the operation $\frac{(u_{1}, v_{1}, w_{1})}{(u_{2}, v_{2}, w_{2})} = (\frac{u_{1}}{u_{2}}, \frac{v_{1}}{v_{2}}, \frac{w_{1}}{w_{2}})$ . However, the following example clearly indicates that this division operation is not valid in general.

If ${\tilde{T}}_{1} = (2, 3, 4)$ and ${\tilde{T}}_{2} = (5, 6, 7)$ are two TFNs then according to this division operation $\frac{{\tilde{T}}_{1}}{{\tilde{T}}_{2}} = (\frac{2}{5}, \frac{3}{6}, \frac{4}{7})$ , which is not a TFN as the necessary condition of a TFN, $\frac{2}{5} ⩽ \frac{3}{6} ⩽ \frac{4}{7}$ is not satisfying.

Furthermore, it is pertinent to mention that although on applying the operation $\frac{(u_{1}, v_{1}, w_{1})}{(u_{2}, v_{2}, w_{2})} = (\frac{u_{1}}{w_{2}}, \frac{v_{1}}{v_{2}}, \frac{w_{1}}{u_{2}})$ , the values of $\frac{(5, 15, 20)}{(5, 10, 15)}$ , $\frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)}$ should be $(\frac{1}{3}, \frac{3}{2}, 4)$ , $(\frac{2}{15}, \frac{29}{12}, 6)$ .

While, inadvertently, Khan et al. [4] have mentioned that the values of $\frac{(5, 15, 20)}{(5, 10, 15)}$ , $\frac{(2, \frac{145}{6}, 30)}{(5, 10, 15)}$ are $(\frac{1}{3}, \frac{17}{12}, 4)$ and $(1, \frac{29}{12}, 6)$ respectively.

It is obvious from Step 6 of Section 4 that Khan et al. [4] have assumed that on applying the operation $R_{1} + (- 18, - \frac{31}{3}, - 5) R_{3}^{'}$ on the elements of first row of Table 3, the elements of first row of Table 4 will be obtained i.e., Khan et al. [4] have assumed that

$(2, 5, 8) + (- 18, - \frac{31}{3}, - 5) (\frac{1}{5}, \frac{1}{2}, \frac{7}{5}) = (- \frac{8}{5}, - \frac{1}{6}, 1)$ .

$(3, \frac{41}{6}, 10) + (- 18, - \frac{31}{3}, - 5) (\frac{1}{3}, \frac{17}{12}, 4) = (- 3, - \frac{281}{36}, - 10)$ .

$(5, \frac{31}{3}, 18) + (- 18, - \frac{31}{3}, - 5) (1, 1, 1) = (0, 0, 0)$ .

$(1, 1, 1) + (- 18, - \frac{31}{3}, - 5) (0, 0, 0) = (1, 1, 1)$ .

$(0, 0, 0) + (- 18, - \frac{31}{3}, - 5) (0, 0, 0) = (0, 0, 0)$ .

$(0, 0, 0) + (- 18, - \frac{31}{3}, - 5) (\frac{1}{15}, \frac{1}{10}, \frac{1}{5}) = (- \frac{6}{5}, - \frac{31}{30}, - 1)$ .

$(6, \frac{50}{3}, 30) + (- 18, - \frac{31}{3} - 5) (1, \frac{29}{12}, 6) = (- 12, - \frac{299}{36}, 0)$ .

It is obvious that to find these values there is need to use the multiplication operation as well as the addition operation of TFNs. As pointed out in (i) of Section 2, that Khan et al [4] have used the following multiplication operation to obtain these elements.

If ${\tilde{T}}_{1} = (u_{1}, v_{1}, w_{1})$ and ${\tilde{T}}_{2} = (u_{2}, v_{2}, w_{2})$ are two TFNs. Then, ${\tilde{T}}_{1} \cdot {\tilde{T}}_{2} = (u_{1}, v_{1}, w_{1}) \cdot (u_{2}, v_{2}, w_{2}) = (u_{1} u_{2}, v_{1} v_{2}, w_{1} w_{2})$

On using this multiplication operation $(- 18, - \frac{31}{3}, - 5) (\frac{1}{5}, \frac{1}{2}, \frac{7}{5}) = (- \frac{18}{5}, - \frac{31}{6}, - 7)$ .

Furthermore, using the addition operation (u₁, v₁, w₁)+ (u₂, v₂, w₂) = (u₁ + u₂, v₁ + v₂, w₁ + w₂), $(2, 5, 8) + (- 18, - \frac{31}{3}, - 5) (\frac{1}{5}, \frac{1}{2},$ $\frac{7}{5}) = (2, 5, 8) + (- \frac{18}{5}, - \frac{31}{6}, - 7) = (- \frac{8}{5}, - \frac{1}{6}, 1)$ .

It is obvious that the element $(- \frac{18}{5}, - \frac{31}{6}, - 7)$ is not a TFN as for a TFN $\tilde{T} = (u, v, w)$ , the condition u ⩽ v ⩽ w should always be satisfied. Therefore, the element $(- \frac{8}{5}, - \frac{1}{6}, 1)$ is not correct.

Similarly, it can be easily verified that the remaining elements, obtained by Khan et al. [4], are not correct.

8 Conclusion

On the basis of the present study, it can be easily concluded that Khan et al. [4] have used mathematical incorrect assumptions in their proposed method. Furthermore, it can be concluded that as for a TFN $\tilde{T} = (u, v, w)$ the conditions $\tilde{T} - \tilde{T} = (0, 0, 0)$ , $\frac{\tilde{T}}{\tilde{T}} = (1, 1, 1)$ will never be satisfied. While, to solve a FFLPP by the simplex method, these conditions are required to make the identity column corresponding to the entering variable. Therefore, it is not possible to resolve the flaws of Khan et al. method [4].

Footnotes

Acknowledgments

The authors would like to thank Associate Editor “Professor Yucheng Dong” and the anonymous Reviewers for their constructive suggestions which have led to an improvement in both the quality and clarity of the paper.

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