Modified approaches to solve matrix games with payoffs of single-valued trapezoidal neutrosophic numbers

Abstract

Seikh and Dutta (Soft Comput 26: 921–936, 2022) claimed that there does not exist any approach to solve single-valued trapezoidal neutrosophic (SVTrN) matrix games (matrix games in which each payoff is represented by a SVTrN number). To fill this gap, Seikh and Dutta, firstly, proposed SVTrN non-linear programming problems (NLPPs) corresponding to Player-I and Player-II. Then, Seikh and Dutta proposed two different approaches to transform the proposed SVTrN NLPPs into crisp linear programming problems (CLPPs). Finally, Seikh and Dutta claimed that an optimal solution of the transformed CLPPs also represents an optimal solution of SVTrN NLPPs. Brikaa (Soft Comput 26: 9137–9139, 2022) pointed out that a mathematically incorrect result is considered in Seikh and Dutta’s first approach to transform SVTrN NLPPs into CLPPs. Therefore, the transformed CLPPs are not equivalent to SVTrN NLPPs. Hence, it is mathematically incorrect to assume that an optimal solution of the transformed CLPPs also represents an optimal solution of SVTrN NLPPs. Brikaa also proposed an approach to transform the SVTrN NLPPs into CLPPs. In this paper, it is pointed out that on solving the CLPPs, obtained by Brikaa’s approach corresponding to SVTrN NLPPs of Player-I and Player-II, different optimal value is obtained. Also, it is pointed out that on solving the CLPPs, obtained by Seikh and Dutta’s second approach corresponding to SVTrN NLPPs of Player-I and Player-II, a different optimal value is obtained. However, in the actual case, the obtained optimal value should be the same as in the literature; it is proved that the CLPPs corresponding to Player-I and Player-II represent a primal–dual pair. This indicates that neither the CLPPs, obtained by Brikaa’s approach nor the CLPPs, obtained by Seikh and Dutta’s second approach, are equivalent to the SVTrN NLPPs of Player-I and Player-II. Hence, it is inappropriate to use the CLPPs, obtained by Brikaa’s approach as well as Seikh and Dutta’s second approach to find an optimal solution for the SVTrN NLPPs of Player-I and Player-II. Also, Brikaa’s approach as well as Seikh and Dutta’s second approach is modified to transform SVTrN NLPPs into their equivalent CLPPs. Furthermore, it is proved that the CLPPs corresponding to SVTrN NLPPs of Player-I and Player-II, obtained by the proposed modified approaches, represent a primal–dual pair. Finally, the correct result of a SVTrN matrix game, considered by Seikh and Dutta to illustrate their approaches, is obtained by the proposed modified approaches.

Appendices

Appendix A

It is obvious from Sect. 6.1 that the base of the first proposed modified approach is the CLPPs (P41) and (P42). These CLPPs are obtained as follows:

Step 1: Using the expressions (11) and (12) (named as expressions (E.1) and (E.2), respectively), proved in Sect. 4.2.1, to find an optimal solution of the interval-valued NLPPs (P15) and (P16) is equivalent to find an optimal solution of the interval-valued NLPPs (A-1) and (A-2) respectively.

$${\left(\sum\limits_{j=1}^{n}\sum\limits_{i=1}^{m}\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{l}_{ij},{m}_{ij},{n}_{ij}\Bigg\rangle {u}_{i}{v}_{j}\right)}_{k}=\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}{\left({\widetilde{b}}_{ij}\right)}_{k}{u}_{i}\right){v}_{j}; k=\rho ,\sigma ,\tau $$

(E.1)

$${\left(\sum\limits_{j=1}^{n}\sum\limits_{i=1}^{m}\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{l}_{ij},{m}_{ij},{n}_{ij}\Bigg\rangle {u}_{i}{v}_{j}\right)}_{k}=\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}{\left({\widetilde{b}}_{ij}\right)}_{k}{v}_{j}\right){u}_{i}; k=\rho ,\sigma ,\tau $$

(E.2)

where,

$${\widetilde{b}}_{ij}=\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{minimum}_{\begin{array}{c}1\le i\le m\\ 1\le j\le n\end{array}}\left\{{l}_{ij}\right\},{maximum}_{\begin{array}{c}1\le i\le m\\ 1\le j\le n\end{array}}\left\{{m}_{ij}\right\},{maximum}_{\begin{array}{c}1\le i\le m\\ 1\le j\le n\end{array}}\left\{{n}_{ij}\right\}\Bigg\rangle $$

Problem (A-1)

$$Maximize \left\{\begin{array}{c}Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}{\left({\widetilde{b}}_{ij}\right)}_{\rho }{u}_{i}\right){v}_{j}\right\}\\ Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}{\left({\widetilde{b}}_{ij}\right)}_{\sigma }{u}_{i}\right){v}_{j}\right\}\\ Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}{\left({\widetilde{b}}_{ij}\right)}_{\tau }{u}_{i}\right){v}_{j}\right\} \\ {\text{Subject to}} \\ \sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n\end{array}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (A-2)

$$Minimize \left\{\begin{array}{c}Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}{\left({\widetilde{b}}_{ij}\right)}_{\rho }{v}_{j}\right){u}_{i}\right\}\\ Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}{\left({\widetilde{b}}_{ij}\right)}_{\sigma }{v}_{j}\right){u}_{i}\right\}\\ Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}{\left({\widetilde{b}}_{ij}\right)}_{\tau }{v}_{j}\right){u}_{i}\right\} \\ {\text{Subject to}} \\ \sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m\end{array}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 2: Using the expression (E.3) to find an optimal solution of the interval-valued NLPPs (A-1) and (A-2) is equivalent to find an optimal solution of the interval-valued NLPPs (A-3) and (A-4) respectively.

$${\left({\widetilde{b}}_{ij}\right)}_{k}=\left[{\left({b}_{ij}\right)}_{lk},{\left({b}_{ij}\right)}_{rk}\right];k=\rho ,\sigma ,\tau $$

(E.3)

Problem (A-3)

$$Maximize \left\{\begin{array}{c}Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\left[{\left({b}_{ij}\right)}_{l\rho },{\left({b}_{ij}\right)}_{r\rho }\right]{u}_{i}\right){v}_{j}\right\}\\ Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\left[{\left({b}_{ij}\right)}_{l\sigma },{\left({b}_{ij}\right)}_{r\sigma }\right]{u}_{i}\right){v}_{j}\right\}\\ Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\left[{\left({b}_{ij}\right)}_{l\tau },{\left({b}_{ij}\right)}_{r\tau }\right]{u}_{i}\right){v}_{j}\right\}\\ {\text{Subject to}} \\ \sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n\end{array}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (A-4)

$$Minimize \left\{\begin{array}{c}Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\left[{\left({b}_{ij}\right)}_{l\rho },{\left({b}_{ij}\right)}_{r\rho }\right]{v}_{j}\right){u}_{i}\right\}\\ Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\left[{\left({b}_{ij}\right)}_{l\sigma },{\left({b}_{ij}\right)}_{r\sigma }\right]{v}_{j}\right){u}_{i}\right\}\\ Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\left[{\left({b}_{ij}\right)}_{l\tau },{\left({b}_{ij}\right)}_{r\tau }\right]{v}_{j}\right){u}_{i}\right\}\\ {\text{Subject to}} \\ \sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m\end{array}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 3: Aggregating the objective function of the interval-valued NLPPs (A-3) and (A-4), to find an optimal solution of the interval-valued NLPPs (A-3) and (A-4) is equivalent to find an optimal solution of the interval-valued NLPPs (A-5) and (A-6) respectively.

Problem (A-5)

$$Maximize \left\{\begin{array}{c}Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\left[\frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }}{3},\frac{{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{3}\right]{u}_{i}\right){v}_{j}\right\}\\ {\text{Subject to}}\\ \sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n\end{array}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (A-6)

$$Minimize \left\{\begin{array}{c}Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\left[\frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }}{3},\frac{{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{3}\right]{v}_{j}\right){u}_{i}\right\}\\ {\text{Subject to}}\\ \sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m\end{array}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 4: To find an optimal solution of the interval-valued NLPPs (A-5) and (A-6) is equivalent to find an optimal solution of the crisp bi-objective NLPPs (A-7) and (A-8) respectively.

Problem (A-7)

$$Maximize \left\{\begin{array}{c}Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\left\{\begin{array}{c}\frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }}{3},\\ \frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{l\tau }+{\left({b}_{ij}\right)}_{r\tau }}{6}\end{array}\right\}{u}_{i}\right){v}_{j}\right\}\\ {\text{Subject to}}\\ \sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n\end{array}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (A-8)

$$Minimize \left\{\begin{array}{c}Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\left\{\begin{array}{c}\frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }}{3},\\ \frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{l\tau }+{\left({b}_{ij}\right)}_{r\tau }}{6}\end{array}\right\}{v}_{j}\right){u}_{i}\right\}\\ {\text{Subject to}}\\ \sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m\end{array}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 5: Using the weighted average method, to find an optimal solution of the crisp bi-objective NLPPs (A-7) and (A-8) is equivalent to find an optimal solution of the crisp NLPPs (A-9) and (A-11) or the equivalent crisp NLPPs (A-10) and (A-12) respectively.

Problem (A-9)

$$Maximize \left\{\begin{array}{c}Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\frac{1}{2}\left(\begin{array}{c}\frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }}{3}+\\ \frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{l\tau }+{\left({b}_{ij}\right)}_{r\tau }}{6}\end{array}\right){u}_{i}\right){v}_{j}\right\}\\ {\text{Subject to}}\\ \sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n\end{array}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (A-10)

$$Maximize \left\{\begin{array}{c}Minimize\left\{\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){u}_{i}\right){v}_{j}\right\}\\ {\text{Subject to}}\\ \sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n\end{array}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (A-11)

$$Minimize \left\{\begin{array}{c}Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\frac{1}{2}\left(\begin{array}{c}\frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }}{3}+\\ \frac{{\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{l\tau }+{\left({b}_{ij}\right)}_{r\tau }}{6}\end{array}\right){v}_{j}\right){u}_{i}\right\}\\ {\text{Subject to}}\\ \sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m\end{array}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Problem (A-12)

$$Minimize \left\{\begin{array}{c}Maximize\left\{\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){v}_{j}\right){u}_{i}\right\}\\ {\text{Subject to}}\\ \sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m\end{array}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 6: Since

$$\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){u}_{i}\right){v}_{j}$$

is a convex linear combination of

$$\sum\limits_{i=1}^{m}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){u}_{i}, j=\mathrm{1,2},\dots ,n$$

and

$$\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){v}_{j}\right){u}_{i}$$

is a convex linear combination of

$$\sum\limits_{j=1}^{n}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){v}_{j}, i=\mathrm{1,2},\dots ,m$$

. So, to find an optimal solution of the crisp NLPPs (A-10) and (A-12) is equivalent to find an optimal solution of the CLPPs (A-13) and (A-14).

Problem (A-13)

$$Maximize \left\{{Minimum}_{1\le j\le n}\left\{\sum\limits_{i=1}^{m}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){u}_{i}\right\}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0.$$

Problem (A-14)

$$Minimize \left\{{Maximum}_{1\le i\le m}\left\{\sum\limits_{j=1}^{n}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){v}_{j}\right\}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 7: Assuming

$${Minimum}_{1\le j\le n}\left\{\sum\limits_{i=1}^{m}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){u}_{i}\right\}=\theta $$

and

$${Maximum}_{1\le i\le m}\left\{\sum\limits_{j=1}^{n}\left(\frac{3\left({\left({b}_{ij}\right)}_{l\rho }+{\left({b}_{ij}\right)}_{l\sigma }+{\left({b}_{ij}\right)}_{l\tau }\right)+{\left({b}_{ij}\right)}_{r\rho }+{\left({b}_{ij}\right)}_{r\sigma }+{\left({b}_{ij}\right)}_{r\tau }}{12}\right){v}_{j}\right\}=\phi $$

to find an optimal solution of the CLPPs (A-13) and (A-14) is equivalent to find an optimal solution of the CLPPs (P41) and (P42) respectively.

Appendix B

It is obvious from Sect. 6.2 that the base of the second proposed modified approach is the CLPPs (P43) and (P44). These CLPPs are obtained as follows:

Step 1: Using the expressions (23) and (24) (named as (F.1) and (F.2) respectively) proved in Sect. 5.2, to find an optimal solution of the crisp NLPPs (P31) and (P32) is equivalent to find an optimal solution of the crisp NLPPs (B-1) and (B-2), respectively.

$${\Pi }_{\alpha }\left(\sum\limits_{j=1}^{n}\left(\sum\limits_{i=1}^{m}\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{l}_{ij},{m}_{ij},{n}_{ij}\Bigg\rangle {u}_{i}\right){v}_{j}\right)=$$

$$\sum\limits_{j=1}^{n}\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{i=1}^{m}{a}_{ij1}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij2}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij3}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij4}{u}_{i}\right);\\ {minimum}_{1\le i\le m}\left\{{l}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{m}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{j}}\right){v}_{j}$$

(F.1)

$${\Pi }_{\alpha }\left(\sum\limits_{i=1}^{m}\left(\sum\limits_{j=1}^{n}\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{l}_{ij},{m}_{ij},{n}_{ij}\Bigg\rangle {v}_{j}\right){u}_{i}\right)=$$

$$\sum\limits_{i=1}^{m}\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{j=1}^{n}{a}_{ij1}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij2}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij3}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij4}{v}_{j}\right);\\ {minimum}_{1\le j\le n}\left\{{l}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{m}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{i}}\right){u}_{i}$$

(F.2)

where,

${\gamma }_{i}=\left[\alpha {\left({minimum}_{1\le j\le n}\left\{{l}_{ij}\right\}\right)}^{2}+\left(1-\alpha \right){\left(1-{maximum}_{1\le j\le n}\left\{{m}_{ij}\right\}\right)}^{2}+\left(1-\alpha \right){\left(1-{maximum}_{1\le j\le n}\left\{{n}_{ij}\right\}\right)}^{2}\right],$ ${\gamma }_{j}=\left[\alpha {\left({minimum}_{1\le i\le m}\left\{{l}_{ij}\right\}\right)}^{2}+\left(1-\alpha \right){\left(1-{maximum}_{1\le i\le m}\left\{{m}_{ij}\right\}\right)}^{2}+\left(1-\alpha \right){\left(1-{maximum}_{1\le i\le m}\left\{{n}_{ij}\right\}\right)}^{2}\right]$ and

$$\gamma =\left[\alpha {\left({minimum}_{\begin{array}{c} 1\le i\le m\\ 1\le j\le n\end{array}}\left\{{l}_{ij}\right\}\right)}^{2}+\left(1-\alpha \right){\left(1-{maximum}_{ \begin{array}{c} 1\le i\le m\\ 1\le j\le n\end{array}}\left\{{m}_{ij}\right\}\right)}^{2}+\left(1-\alpha \right){\left(1-{maximum}_{ \begin{array}{c} 1\le i\le m\\ 1\le j\le n\end{array}}\left\{{n}_{ij}\right\}\right)}^{2}\right]$$

Problem (B-1)

$$Maximize \left\{\begin{array}{c}Minimize\left\{\sum\limits_{j=1}^{n}\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{i=1}^{m}{a}_{ij1}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij2}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij3}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij4}{u}_{i}\right);\\ {minimum}_{1\le i\le m}\left\{{l}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{m}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{j}}\right){v}_{j}\right\}\\ {\text{Subject to}}\\ \sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n\end{array}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (B-2)

$$Minimize \left\{\begin{array}{c}Maximize\left\{\sum\limits_{i=1}^{m}\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{j=1}^{n}{a}_{ij1}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij2}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij3}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij4}{v}_{j}\right);\\ {minimum}_{1\le j\le n}\left\{{l}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{m}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{i}}\right){u}_{i}\right\}\\ {\text{Subject to}}\\ \sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m\end{array}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 2: Since,

$$\sum\limits_{j=1}^{n}\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{i=1}^{m}{a}_{ij1}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij2}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij3}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij4}{u}_{i}\right);\\ {minimum}_{1\le i\le m}\left\{{l}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{m}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{j}}\right){v}_{j}$$

is a convex linear combination of

$$\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{i=1}^{m}{a}_{ij1}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij2}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij3}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij4}{u}_{i}\right);\\ {minimum}_{1\le i\le m}\left\{{l}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{m}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{j}}\right), j=\mathrm{1,2},\dots ,n$$

and

$$\sum\limits_{i=1}^{m}\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{j=1}^{n}{a}_{ij1}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij2}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij3}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij4}{v}_{j}\right);\\ {minimum}_{1\le j\le n}\left\{{l}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{m}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{i}}\right){u}_{i}$$

is a convex linear combination of

$$\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{j=1}^{n}{a}_{ij1}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij2}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij3}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij4}{v}_{j}\right);\\ {minimum}_{1\le j\le n}\left\{{l}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{m}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{i}}\right), i=\mathrm{1,2},\dots ,m$$

. So, to find an optimal solution of the crisp NLPPs (B-1) and (B-2) is equivalent to find an optimal solution of the crisp NLPPs (B-3) and (B-4), respectively.

Problem (B-3)

$$Maximize \left\{{Minimum}_{1\le j\le n}\left\{\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{i=1}^{m}{a}_{ij1}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij2}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij3}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij4}{u}_{i}\right);\\ {minimum}_{1\le i\le m}\left\{{l}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{m}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{j}}\right)\right\}\right\}$$

Subject to

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (B-4)

$$Minimize \left\{{Maximum}_{1\le i\le m}\left\{\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{j=1}^{n}{a}_{ij1}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij2}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij3}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij4}{v}_{j}\right);\\ {minimum}_{1\le j\le n}\left\{{l}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{m}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{i}}\right)\right\}\right\}$$

Subject to

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 3: Assuming,

$${Minimum}_{1\le j\le n}\left\{\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{i=1}^{m}{a}_{ij1}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij2}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij3}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij4}{u}_{i}\right);\\ {minimum}_{1\le i\le m}\left\{{l}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{m}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{j}}\right)\right\}={\Omega }_{1}$$

and

$${Maximum}_{1\le i\le m}\left\{\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{j=1}^{n}{a}_{ij1}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij2}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij3}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij4}{v}_{j}\right);\\ {minimum}_{1\le j\le n}\left\{{l}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{m}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{i}}\right)\right\}={\Omega }_{2}$$

to find an optimal solution of the crisp NLPPs (B-3) and (B-4) is equivalent to find an optimal solution of the CLPPs (B-5) and (B-6), respectively.

Problem (B-5)

$$Maximize \left\{{\Omega }_{1}\right\}$$

Subject to

$$\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{i=1}^{m}{a}_{ij1}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij2}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij3}{u}_{i},\sum\limits_{i=1}^{m}{a}_{ij4}{u}_{i}\right);\\ {minimum}_{1\le i\le m}\left\{{l}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{m}_{ij}\right\},{maximum}_{1\le i\le m}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{j}}\right)\ge {\Omega }_{1}, j=\mathrm{1,2},\dots ,n,$$

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (B-6)

$$Minimize \left\{{\Omega }_{2}\right\}$$

Subject to

$$\gamma \left(\frac{{\Pi }_{\alpha }\left(\Bigg\langle \begin{array}{c}\left(\sum\limits_{j=1}^{n}{a}_{ij1}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij2}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij3}{v}_{j},\sum\limits_{j=1}^{n}{a}_{ij4}{v}_{j}\right);\\ {minimum}_{1\le j\le n}\left\{{l}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{m}_{ij}\right\},{maximum}_{1\le j\le n}\left\{{n}_{ij}\right\}\end{array}\Bigg\rangle \right)}{{\gamma }_{i}}\right)\le {\Omega }_{2}, i=\mathrm{1,2},\dots ,m,$$

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 4: To find an optimal solution of the crisp NLPPs (B-5) and (B-6) is equivalent to find an optimal solution of the CLPPs (B-7) and (B-8) respectively.

Problem (B-7)

$$Maximize \left\{{\Omega }_{1}\right\}$$

Subject to

$$\gamma \left(\frac{{\Pi }_{\alpha }\left(\sum\limits_{i=1}^{m}\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{l}_{ij},{m}_{ij},{n}_{ij}\Bigg\rangle \right){u}_{i}}{{\gamma }_{j}}\right)\ge {\Omega }_{1}, j=\mathrm{1,2},\dots ,n,$$

$$\sum\limits_{i=1}^{m}{u}_{i}=1, {u}_{i}\ge 0, i=\mathrm{1,2},\dots ,m.$$

Problem (B-8)

$$Minimize \left\{{\Omega }_{2}\right\}$$

Subject to

$$\gamma \left(\frac{{\Pi }_{\alpha }\left(\sum\limits_{j=1}^{n}\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{l}_{ij},{m}_{ij},{n}_{ij}\Bigg\rangle \right){v}_{j}}{{\gamma }_{i}}\right)\le {\Omega }_{2}, i=\mathrm{1,2},\dots ,m,$$

$$\sum\limits_{j=1}^{n}{v}_{j}=1, {v}_{j}\ge 0, j=\mathrm{1,2},\dots ,n.$$

Step 5: Using the expressions (F.3) and (F.4), to find an optimal solution of the crisp NLPPs (B-7) and (B-8) is equivalent to find an optimal solution of the CLPPs (B-43) and (B-44) respectively.

$$\frac{{\Pi }_{\alpha }\left(\sum\limits_{i=1}^{m}\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{l}_{ij},{m}_{ij},{n}_{ij}\Bigg\rangle \right){u}_{i}}{{\gamma }_{j}}=\sum\limits_{i=1}^{m}\left(\frac{{a}_{ij1}+2\left({a}_{ij2}+{a}_{ij3}\right)+{a}_{ij4}}{6}\right){u}_{i}$$

(F.3)

$$\frac{{\Pi }_{\alpha }\left(\sum\limits_{j=1}^{n}\Bigg\langle \left({a}_{ij1},{a}_{ij2},{a}_{ij3},{a}_{ij4}\right);{l}_{ij},{m}_{ij},{n}_{ij}\Bigg\rangle \right){v}_{j}}{{\gamma }_{i}}=\sum\limits_{j=1}^{n}\left(\frac{{a}_{ij1}+2\left({a}_{ij2}+{a}_{ij3}\right)+{a}_{ij4}}{6}\right){v}_{j}$$

(F.4)