A note on “Solving intuitionistic fuzzy linear programming problems by ranking function”

Abstract

Suresh et al. (Solving intuitionistic fuzzy linear programming problems by ranking function, Journal of Intelligent and Fuzzy Systems, 27, 3081–3087, 2014) proposed the ranking function for comparing triangular intuitionistic fuzzy numbers (TIFNs) and applied this ranking function to solve different types of intuitionistic fuzzy linear programming problems (IFLPPs). In this note, it is pointed out that the ranking function proposed by authors is not valid. Hence, the results of intuitionistic fuzzy linear programming problems, obtained by using this ranking function, are also not valid. Further, the exact ranking function is obtained by modifying existing ranking function and using the exact ranking function, the exact results of intuitionistic fuzzy linear programming problems considered by Suresh et al., are obtained.

Keywords

Triangular intuitionistic fuzzy number ranking function

1 Introduction

Linear programming is one of the most successively applied operation research techniques. Real world situations are represented by using any linear programming model which involves a lot of parameters, whose values are assigned by experts. However, both experts and decision maker frequently do not precisely know the value of those parameters. Therefore, it is useful to consider the knowledge of experts about the parameters as fuzzy data/intuitionistic fuzzy data [3, 4].

To find the fuzzy optimal solution of fuzzy linear programming problems (FLPPs)/intuitionistic fuzzy linear programming problems (IFLPPs), there is need to compare fuzzy numbers. Several methods [1, 2] have been proposed in the literature for comparing fuzzy numbers. However, very few methods has been proposed in literature for comparing intuitionistic fuzzy numbers.

Suresh et al. [5, Definition 4.2, pp. 3084] proposed a ranking function (named as Mag) for comparing triangular intuitionistic fuzzy numbers and used it to find the solution of the following types of IFLPPs.

(i) IFLPP (P₁) in which the coefficients of the variables in objective function are represented by triangular intuitionistic fuzzy numbers whereas all other variables and parameters are represented by real numbers. $\begin{matrix} Maximize / Minimize [\tilde{z} \approx \sum_{j = 1}^{n} {\tilde{c}}_{j}^{I} x_{j}] \\ Subject to (P_{1}) \\ \sum_{j = 1}^{n} a_{ij} x_{j} \leq, =, \geq b_{i}; i = 1, 2, . . ., m; \\ x_{j} \geq 0; j = 1, 2, . . ., n . \end{matrix}$ where ${\tilde{c}}_{j}^{I} = {({\underline{c}}_{j}^{μ}, c_{j}, {\bar{c}}_{j}^{μ}; w_{{\tilde{c}}_{j}}), ({\underline{c}}_{j}^{ν}, c_{j}, {\bar{c}}_{j}^{ν}; u_{{\tilde{c}}_{j}})}$ is atriangular intuitionistic fuzzy number.

(ii) IFLPP (P₂) in which the variables and coefficients of the variables in objective function are represented by real numbers whereas coefficients of the variables in constraints and right hand side vector is represented by triangular intuitionistic fuzzy numbers. $\begin{matrix} Maximize / Minimize [z = \sum_{j = 1}^{n} c_{j} x_{j}] \\ Subject to (P_{2}) \\ \sum_{j = 1}^{n} {\tilde{a}}_{ij}^{I} x_{j} ⪯, \approx, ⪰ \tilde{b_{i}^{I}}; i = 1, 2, . . ., m; \\ x_{j} \geq 0; j = 1, 2, . . ., n . \end{matrix}$ where ${\tilde{a}}_{ij}^{I} = {({\underline{a}}_{ij}^{μ}, a_{ij}, {\bar{a}}_{ij}^{μ}; w_{{\tilde{a}}_{ij}}), ({\underline{a}}_{ij}^{ν}, a_{ij}, {\bar{a}}_{ij}^{ν}; u_{{\tilde{a}}_{ij}})},$

$\tilde{b_{i}^{I}} = {({\underline{b}}_{i}^{μ}, b_{i}, {\bar{b}}_{i}^{μ}; w_{{\tilde{b}}_{i}}), ({\underline{b}}_{i}^{ν}, b_{i}, {\bar{b}}_{i}^{ν}; u_{{\tilde{b}}_{i}})}$ are triangular intuitionistic fuzzy numbers.

(iii) IFLPP (P₃) in which only variables are represented by real numbers whereas all other parameters are represented by triangular intuitionistic fuzzynumbers. $\begin{matrix} Maximize / Minimize [z \approx \sum_{j = 1}^{n} {\tilde{c}}_{j}^{I} x_{j}] \\ Subject to (P_{3}) \\ \sum_{j = 1}^{n} {\tilde{a}}_{ij}^{I} x_{j} ⪯, \approx, ⪰ \tilde{b_{i}^{I}}; i = 1, 2, . . ., m; \\ x_{j} \geq 0; j = 1, 2, . . ., n . \end{matrix}$ where ${\tilde{c}}_{j}^{I} = {({\underline{c}}_{j}^{μ}, c_{j}, {\bar{c}}_{j}^{μ}; w_{{\tilde{c}}_{j}}), ({\underline{c}}_{j}^{ν}, c_{j}, {\bar{c}}_{j}^{ν}; u_{{\tilde{c}}_{j}})},$

${\tilde{a}}_{ij}^{I} = {({\underline{a}}_{ij}^{μ}, a_{ij}, {\bar{a}}_{ij}^{μ}; w_{{\tilde{a}}_{ij}}), ({\underline{a}}_{ij}^{ν}, a_{ij}, {\bar{a}}_{ij}^{ν}; u_{{\tilde{a}}_{ij}})}$ and

2 Existing ranking function

Suresh et al. [5] proposed a ranking function $(Mag ({\tilde{A}}^{I}) = \frac{1}{12} {\frac{4 a - 2 ({\bar{a}}^{μ} + {\underline{a}}^{μ}) + 3 w_{\tilde{a}} ({\bar{a}}^{μ} + {\underline{a}}^{μ})}{w_{\tilde{a}}} +$ $\frac{2 (a + {\bar{a}}^{ν} + {\underline{a}}^{ν}) - 3 u_{\tilde{a}} ({\bar{a}}^{ν} + {\underline{a}}^{ν})}{1 - u_{\tilde{a}}}})$ to obtain a real number corresponding to a triangular intuitionistic fuzzy number ${\tilde{A}}^{I} = {({\underline{a}}^{μ}, a, {\bar{a}}^{μ}; w_{\tilde{a}}), ({\underline{a}}^{ν}, a, {\bar{a}}^{ν}; u_{\tilde{a}})}$ .

Suresh et al. [5] have used the following method to obtain the ranking function $(Mag ({\tilde{A}}^{I}) =$ $\frac{1}{12} {\frac{4 a - 2 ({\bar{a}}^{μ} + {\underline{a}}^{μ}) + 3 w_{\tilde{a}} ({\bar{a}}^{μ} + {\underline{a}}^{μ})}{w_{\tilde{a}}} + \frac{2 (a + {\bar{a}}^{ν} + {\underline{a}}^{ν}) - 3 u_{\tilde{a}} ({\bar{a}}^{ν} + {\underline{a}}^{ν})}{1 - u_{\tilde{a}}}})$ .

Step 1. Find the α- cut $[{\underline{a}}^{μ} + \frac{α (a - {\underline{a}}^{μ})}{w_{\tilde{a}}}, {\bar{a}}^{μ} - \frac{α ({\bar{a}}^{μ} - a)}{w_{\tilde{a}}}]$ corresponding to membership function of TIFN ${\tilde{A}}^{I} = {({\underline{a}}^{μ}, a, {\bar{a}}^{μ}; w_{\tilde{a}}), ({\underline{a}}^{ν}, a, {\bar{a}}^{ν}; u_{\tilde{a}})}$ .

Step 2. Find α- cut $[\frac{a - u_{\tilde{a}} {\underline{a}}^{ν} - α (a - {\underline{a}}^{ν})}{1 - u_{\tilde{a}}}, \frac{a - u_{\tilde{a}} {\bar{a}}^{ν} + α ({\bar{a}}^{ν} - a)}{1 - u_{\tilde{a}}}]$ corresponding to non-membership function of TIFN ${\tilde{A}}^{I} = {({\underline{a}}^{μ}, a, {\bar{a}}^{μ}; w_{\tilde{a}}), ({\underline{a}}^{ν}, a, {\bar{a}}^{ν}; u_{\tilde{a}})}$ .

Step 3. Find $Mag ({\tilde{A}}^{I}) =$ $\frac{1}{2} {\int_{0}^{1} ({\underline{a}}^{μ} + \frac{α (a - {\underline{a}}^{μ})}{w_{\tilde{a}}} + {\bar{a}}^{μ} - \frac{α ({\bar{a}}^{μ} - a)}{w_{\tilde{a}}}) f (α) d α}$ $+ \frac{1}{2} {\int_{0}^{1} (\frac{a - u_{\tilde{a}} {\underline{a}}^{ν} - α (a - {\underline{a}}^{ν})}{1 - u_{\tilde{a}}} + \frac{a - u_{\tilde{a}} {\bar{a}}^{ν} + α ({\bar{a}}^{ν} - a)}{1 - u_{\tilde{a}}}) f (α) d α}$ where f (α) is non-negative and increasing weighting function on [0, 1] with f (0) =0, f (1) =1 and $\int_{0}^{1} f (α) d α = \frac{1}{2} .$

3 Invalidity of the existing ranking function

The ranking function $(Mag ({\tilde{A}}^{I}) = \frac{1}{12} {\frac{4 a - 2 ({\bar{a}}^{μ} + {\underline{a}}^{μ}) + 3 w_{\tilde{a}} ({\bar{a}}^{μ} + {\underline{a}}^{μ})}{w_{\tilde{a}}}$ $+ \frac{2 (a + {\bar{a}}^{ν} + {\underline{a}}^{ν}) - 3 u_{\tilde{a}} ({\bar{a}}^{ν} + {\underline{a}}^{ν})}{1 - u_{\tilde{a}}}})$ , proposed by Suresh et al. [5], is not valid due to the following reasons.

1. It is obvious from Step 3 of Section 2 that Suresh et al. [5] have assumed that the upper limit of integration as 1. While, as in Step 1, the membership function is considered for finding the α- cut and the maximum membership value is $w_{\tilde{a}}$ . So, in Step 3 of the method, discussed in Section 2, the upper limit of integration $\frac{1}{2} {\int ({\underline{a}}^{μ} + \frac{α (a - {\underline{a}}^{μ})}{w_{\tilde{a}}} + {\bar{a}}^{μ} - \frac{α ({\bar{a}}^{μ} - a)}{w_{\tilde{a}}}) f (α) d α}$ should be $w_{\tilde{a}}$ . Similarly, in Step 2, non-membership is considered for finding α- cut and maximum value of non-membership from X- axis is $1 - u_{\tilde{a}}$ . So, in Step 3 of method, discussed in Section 2, the upper limit of integration $\frac{1}{2} {\int (\frac{a - u_{\tilde{a}} {\underline{a}}^{ν} - α (a - {\underline{a}}^{ν})}{1 - u_{\tilde{a}}} + \frac{a - u_{\tilde{a}} {\bar{a}}^{ν} + α ({\bar{a}}^{ν} - a)}{1 - u_{\tilde{a}}}) f (α) d α}$ should be $1 - u_{\tilde{a}}$ .

2. The values of α- cut, mentioned in Step 2 of Section 2, is obtained by putting $\frac{a - x + u_{\tilde{a}} (x - {\underline{a}}^{ν})}{a - {\underline{a}}^{ν}} = α$ , $\frac{x - a + u_{\tilde{a}} ({\bar{a}}^{ν} - x)}{{\bar{a}}^{ν} - a} = α,$ with the assumption that α will be distance from X- axis. While in case of non-membership function distance from X- axis will be 1 - α instead of α i.e., the exact value of α- cuts, obtained by putting $\frac{a - x + u_{\tilde{a}} (x - {\underline{a}}^{ν})}{a - {\underline{a}}^{ν}} = 1 - α$ , $\frac{x - a + u_{\tilde{a}} ({\bar{a}}^{ν} - x)}{{\bar{a}}^{ν} - a} = 1 - α$ is $[{\underline{a}}^{ν} + \frac{α (a - {\underline{a}}^{ν})}{1 - u_{\tilde{a}}}, {\bar{a}}^{ν} - \frac{α ({\bar{a}}^{ν} - a)}{1 - u_{\tilde{a}}}]$ .

4 Exact ranking function

It is obvious from Section 3 that the exact ranking function $(Mag ({\tilde{A}}^{I}))$ for TIFN ${\tilde{A}}^{I}$ will be $\begin{matrix} Mag ({\tilde{A}}^{I}) \\ = \frac{1}{2} {\int_{0}^{w_{\tilde{a}}} ({\underline{a}}^{μ} + \frac{α (a - {\underline{a}}^{μ})}{w_{\tilde{a}}} + {\bar{a}}^{μ} - \frac{α ({\bar{a}}^{μ} - a)}{w_{\tilde{a}}}) f (α) d α \\ + \int_{0}^{1 - u_{\tilde{a}}} ({\underline{a}}^{ν} + \frac{α (a - {\underline{a}}^{ν})}{1 - u_{\tilde{a}}} + {\bar{a}}^{ν} - \frac{α ({\bar{a}}^{ν} - a)}{1 - u_{\tilde{a}}}) f (α) d α} . \end{matrix}$

Assuming f (α) = α, we have $\begin{matrix} Mag ({\tilde{A}}^{I}) \\ = \frac{1}{2} {\int_{0}^{w_{\tilde{a}}} ({\underline{a}}^{μ} + \frac{α (a - {\underline{a}}^{μ})}{w_{\tilde{a}}} + {\bar{a}}^{μ} - \frac{α ({\bar{a}}^{μ} - a)}{w_{\tilde{a}}}) α d α \\ + \int_{0}^{1 - u_{\tilde{a}}} ({\underline{a}}^{ν} + \frac{α (a - {\underline{a}}^{ν})}{1 - u_{\tilde{a}}} + {\bar{a}}^{ν} - \frac{α ({\bar{a}}^{ν} - a)}{1 - u_{\tilde{a}}}) α d α} . \end{matrix}$

After simplification, $Mag ({\tilde{A}}^{I}) = \frac{1}{12} {w_{\tilde{a}}^{2} (4 a + {\bar{a}}^{μ} + {\underline{a}}^{μ}) + (1 - u_{\tilde{a}})^{2} (4 a + {\bar{a}}^{ν} + {\underline{a}}^{ν})} .$

5 Exact solution of existing problems

Suresh et al. [5] used the rankingfunction $Mag ({\tilde{A}}^{I}) = (\frac{1}{12} {\frac{4 a - 2 ({\bar{a}}^{μ} + {\underline{a}}^{μ}) + 3 w_{\tilde{a}} ({\bar{a}}^{μ} + {\underline{a}}^{μ})}{w_{\tilde{a}}} +$ $\frac{2 (a + {\bar{a}}^{ν} + {\underline{a}}^{ν}) - 3 u_{\tilde{a}} ({\bar{a}}^{ν} + {\underline{a}}^{ν})}{1 - u_{\tilde{a}}}})$ for finding the optimal solution of IFLPPs (P₄) [5, Example 1, pp. 3085], (P₅) [5, Example 2, pp. 3085], (P₆) [5, Example 3, pp. 3086] and (P₇) [5, Example 4, pp. 3086]. $\begin{matrix} Maximize {\tilde{z}}^{I} = \\ {(19, 25, 33; 0.9), (18, 25, 34; 0)} x_{1} + \\ {(44, 48, 54; 0.9), (43, 48, 56; 0)} x_{2} \\ Subject to \\ 15 x_{1} + 30 x_{2} \leq 45000, \\ 24 x_{1} + 6 x_{2} \leq 24000, \\ 21 x_{1} + 14 x_{2} \leq 28000, \\ x_{1}, x_{2} \geq 0 . \end{matrix}} (P_{4})$ $\begin{matrix} Maximize z = 25 x_{1} + 48 x_{2} \\ Subject to \\ {(14, 15, 17; 0.9), (10, 15, 18; 0)} x_{1} + \\ {(25, 30, 34; 0.9), (23, 30, 38; 0)} x_{2} ⪯ \\ {(44980, 45000, 45030; 0.9), \\ (44970, 45000, 45070; 0)}, \\ {(21, 24, 26; 0.9), (20, 24, 33; 0)} x_{1} + \\ {(4, 6, 8; 0.9), (2, 6, 11; 0)} x_{2} ⪯ \\ {(23980, 24000, 24050; 0.9), \\ (23940, 24000, 24060; 0)}, \\ {(17, 21, 22; 0.9), (16, 21, 26; 0)} x_{1} + \\ {(12, 14, 19; 0.9), (8, 14, 22; 0)} x_{2} ⪯ \\ {(27990, 28000, 28030; 0.9), \\ (27950, 28000, 28040; 0)}, \\ x_{1}, x_{2} \geq 0 . \end{matrix}} (P_{5})$

$\begin{matrix} Maximize \tilde{z} = {(19, 25, 33; 0.9), \\ (18, 25, 34; 0)} x_{1} + \\ {(44, 48, 54; 0.9), (43, 48, 56; 0)} x_{2} \\ Subject to \\ {(14, 15, 17; 0.9), (10, 15, 18; 0)} x_{1} + \\ {(25, 30, 34; 0.9), (23, 30, 38; 0)} x_{2} ⪯ \\ {(44980, 45000, 45030; 0.9), \\ (44970, 45000, 45070; 0)}, \\ {(21, 24, 26; 0.9), (20, 24, 33; 0)} x_{1} + \\ {(4, 6, 8; 0.9), (2, 6, 11; 0)} x_{2} ⪯ \\ {(23980, 24000, 24050; 0.9), \\ (23940, 24000, 24060; 0)}, \\ {(17, 21, 22; 0.9), (16, 21, 26; 0)} x_{1} + \\ {(12, 14, 19; 0.9), (8, 14, 22; 0)} x_{2} ⪯ \\ {(27990, 28000, 28030; 0.9), \\ (27950, 28000, 28040; 0)}, \\ x_{1}, x_{2} \geq 0 . \end{matrix}} (P_{6})$ $\begin{matrix} Maximize z = 3 x_{1} + 4 x_{2} \\ Subject to \\ {(0, 1, 3; 0.9), (0, 1, 4; 0)} x_{1} + \\ {(1, 2, 6; 0.9), (0, 2, 8; 0)} x_{2} ⪯ \\ {(18, 20, 32; 0.9), (16, 20, 34; 0)}, \\ {(2, 3, 8; 0.9), (1, 3, 10; 0)} x_{1} + \\ {(3, 5, 6; 0.9), (1, 5, 9; 0)} x_{2} ⪯ \\ {(56, 60, 66; 0.9), (55, 60, 68; 0)}, \\ x_{1}, x_{2} \geq 0 . \end{matrix}} (P_{7})$

However, as discussed in Section 2 that the ranking function, proposed by Suresh et al. [5], is not valid. Hence, the optimal solutions of problems (P₄) , (P₅) , (P₆) and (P₇) , obtained by Suresh et al. [5], are also not exact optimal solutions of these problems.

The exact optimal solutions of these problems, obtained by using the existing algorithm [5, Section 4, pp. 3084], with exact ranking function $Mag ({\tilde{A}}^{I}) = \frac{1}{12} {w_{\tilde{a}}^{2} (4 a + {\bar{a}}^{μ} + {\underline{a}}^{μ}) + (1 - u_{\tilde{a}})^{2} (4 a + {\bar{a}}^{ν} + {\underline{a}}^{ν})}$ instead of using the invalid ranking function $Mag ({\tilde{A}}^{I}) = \frac{1}{12} {\frac{4 a - 2 ({\bar{a}}^{μ} + {\underline{a}}^{μ}) + 3 w_{\tilde{a}} ({\bar{a}}^{μ} + {\underline{a}}^{μ})}{w_{\tilde{a}}} + \frac{2 (a + {\bar{a}}^{ν} + {\underline{a}}^{ν}) - 3 u_{\tilde{a}} ({\bar{a}}^{ν} + {\underline{a}}^{ν})}{1 - u_{\tilde{a}}}}$ , are shown in Table 1.

6 Conclusion

It is shown that the ranking function, proposed by Suresh et al. [5], is not valid and the exact ranking function is proposed. Also, the exact optimal solutions of the problems, solved by Suresh et al. [5] to illustrate their proposed algorithm, are obtained.

Footnotes

Acknowledgments

The authors would like to thank anonymous reviewers and the editor for their valuable comments and suggestions. Dr. Amit Kumar would like to acknowledge the adolescent blessings of Mehar (lovely daughter of his cousin sister Dr. Parmpreet Kaur). He believes that Mata Vaishno Devi has appeared on the earth in the form of Mehar and without Mehar’s blessings it was not possible to think the ideas presented in this manuscript.

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