A note on “Optimal solution of neutrosophic linear fractional programming problems with mixed constraints”

A note on “Optimal solution of neutrosophic linear fractional programming problems with mixed constraints” | Research Square window.SnipcartSettings = { analytics: { enabled: false } }; (function() { var accessVector = localStorage.getItem('access_vector') || ''; window.dataLayer = window.dataLayer || []; if (accessVector) { window.dataLayer.push({ user: { profile: { profileInfo: { snid: accessVector } } } }); } })(); (function(w,d,s,l,i){w[l]=w[l]||[];w[l].push({'gtm.start':new Date().getTime(),event:'gtm.js'});var f=d.getElementsByTagName(s)[0],j=d.createElement(s),dl=l!='dataLayer'?'&l='+l:'';j.async=true;j.src='https://www.googletagmanager.com/gtm.js?id='+i+dl;f.parentNode.insertBefore(j,f);})(window,document,'script','dataLayer','GTM-K279D39R'); Browse Preprints In Review Journals COVID-19 Preprints AJE Video Bytes Research Tools Research Promotion AJE Professional Editing AJE Rubriq About Preprint Platform In Review Editorial Policies Our Team Help Center Sign In Submit a Preprint Cite Share Download PDF Research Article A note on “Optimal solution of neutrosophic linear fractional programming problems with mixed constraints” Parul Tomar, Amit Kumar This is a preprint; it has not been peer reviewed by a journal. https://doi.org/ 10.21203/rs.3.rs-2250652/v1 This work is licensed under a CC BY 4.0 License Status: Posted Version 1 posted You are reading this latest preprint version Abstract Das and Edalatpanah (Soft Comput 26 (2022) 8699–8707) proposed an approach to find an optimal solution of neutrosophic linear fractional programming problems with mixed constraints (linear programming problems with mixed constraints in which each decision variable is represented by a non-negative real number and each other parameter is represented by a triangular neutrosophic number). In this paper, it is pointed out that a mathematical incorrect result is considered in Das and Edalatpanah’s approach. Hence, it is inappropriate to use Das and Edalatpanah’s approach. Also, Das and Edalatpanah’s approach is modified to resolve its inappropriateness. Neutrosophic linear fractional programming Crisp linear fractional programming Triangular neutrosophic numbers Ranking function 1. Introduction Abdel-basset et al. ( 2019 ) proposed an approach to find an optimal solution of neutrosophic linear fractional programming problem (P1). Problem (P1) $$Max \left(or Min\right) \left(\frac{\sum _{j=1}^{n}\left({c}_{j1}, { c}_{j2},{ c}_{j3};{ \mu }_{{\tilde{c}}_{j}},{ \nu }_{{\tilde{c}}_{j}},{ w}_{{\tilde{c}}_{j}}\right){ x}_{j} + \left({p}_{1},{ p}_{2},{ p}_{3};{ \mu }_{\tilde{p}},{ \nu }_{\tilde{p}},{ w}_{\tilde{p}}\right)}{\sum _{j=1}^{n}\left({d}_{j1},{ d}_{j2},{ d}_{j3};{ \mu }_{{\tilde{d}}_{j}},{ \nu }_{{\tilde{d}}_{j}},{ w}_{{\tilde{d}}_{j}}\right){ x}_{j} + \left({q}_{1},{ q}_{2},{ q}_{3};{ \mu }_{\tilde{q}},{ \nu }_{\tilde{q}},{ w}_{\tilde{q}}\right)}\right)$$ Subject to $$\sum _{j=1}^{n}\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right){x}_{j}\le \left({b}_{i1},{b}_{i2},{b}_{i3};{\mu }_{{\tilde{b}}_{i}},{\nu }_{{\tilde{b}}_{i}},{w}_{{\tilde{b}}_{i}}\right), i=\text{1,2},\dots ,m$$ , $$\sum _{j=1}^{n}\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right){x}_{j}=\left(\text{1,1},1;\text{1,0},0\right), i=\text{1,2},\dots ,m$$ , $${x}_{j}\ge 0, j=\text{1,2},\dots ,n.$$ Das and Edalatpanah ( 2022 ) proposed an alternative approach to find an optimal solution of the neutrosophic linear fractional programming problem (P1). Das and Edalatpanah ( 2022 ) also pointed out that it is better to use their proposed approach as compared to Abdel-basset et al. ( 2019 )’s approach due to the following reasons: Much computational efforts are required to find an optimal solution of the neutrosophic linear fractional programming problem (P1) by Abdel-basset et al. ( 2019 )’s approach. While, less computational efforts are required to find an optimal solution of the neutrosophic linear fractional programming problem (P1) by their proposed approach. If the same neutrosophic linear fractional programming problem is solved by their proposed approach and Abdel-basset et al. ( 2019 )’s approach. Then, the optimal solution obtained by their proposed approach is better than the optimal solution obtained by Abdel-basset et al. ( 2019 )’s approach. To validate this claim, Das and Edalatpanah ( 2022 , Section 6, p. 8703) solved the neutrosophic linear fractional programming problem (P2) by their proposed approach and pointed out that the obtained maximum value is 1.4. While, on solving the same problem by Abdel-basset et al. ( 2019 )’s approach, the obtained maximum value is 1. Problem (P2) $$Max \left(\frac{\left(\text{7,8},9;\text{0.5,0.8,0.3}\right){x}_{1}+\left(\text{6,7},8;\text{0.2,0.6,0.5}\right){x}_{2}+\left(\text{8,9},10;\text{0.8,0.1,0.4}\right){x}_{3}}{\left(\text{7,8},9;\text{0.5,0.8,0.3}\right){x}_{1}+\left(\text{8,9},10;\text{0.8,0.1,0.4}\right){x}_{2}+\left(\text{4,6},8;\text{0.75,0.25,0.1}\right){x}_{3}+\left(\text{1,1.5,2};\text{0.75,0.5,0.25}\right)}\right)$$ Subject to $$\left(\text{3,4},5;\text{0.4,0.6,0.5}\right){x}_{1}+\left(\text{2,3},4;\text{1,0.25,0.3}\right){x}_{2}+\left(\text{4,5},6;\text{0.3,0.4,0.8}\right){x}_{3}\le$$ $$\left(\text{25,28,30};\text{0.4,0.25,0.6}\right)$$ , $$\left(\text{4,5},6;\text{0.3,0.4,0.8}\right){x}_{1}+\left(\text{2,3},4;\text{1,0.25,0.3}\right){x}_{2}+\left(\text{2,3},4;\text{1,0.25,0.3}\right){x}_{3}\le$$ $$\left(\text{18,20,22};\text{0.9,0.2,0.6}\right)$$ , $${x}_{1},{x}_{2},{x}_{3}\ge 0$$ . In this paper, it is pointed out that Das and Edalatpanah ( 2022 ) have considered a mathematical incorrect result in their proposed approach. Hence, it is inappropriate to use Das and Edalatpanah ( 2022 )’s approach. Also, Das and Edalatpanah ( 2022 )’s approach is modified to resolve its inappropriateness. Furthermore, a correct optimal solution and the corresponding maximum value of the neutrosophic linear fractional programming problem (P2) are obtained by the modified approach. 2. Das And Edalatpanah’s Approach Das and Edalatpanah ( 2022 ) proposed the following approach to find an optimal solution of the neutrosophic linear fractional programming problem (P1). Step 1 Transform the neutrosophic linear fractional programming problem (P1) into its equivalent crisp linear fractional programming problem (P3). Problem (P3) $$Max \left(or Min\right) \left(\frac{\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, { c}_{j2},{ c}_{j3};{ \mu }_{{\tilde{c}}_{j}},{ \nu }_{{\tilde{c}}_{j}},{ w}_{{\tilde{c}}_{j}}\right){ x}_{j} + \left({p}_{1},{ p}_{2},{ p}_{3};{ \mu }_{\tilde{p}},{ \nu }_{\tilde{p}},{ w}_{\tilde{p}}\right)\right)}{\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{ d}_{j2},{ d}_{j3};{ \mu }_{{\tilde{d}}_{j}},{ \nu }_{{\tilde{d}}_{j}},{ w}_{{\tilde{d}}_{j}}\right){ x}_{j} + \left({q}_{1},{ q}_{2},{ q}_{3};{ \mu }_{\tilde{q}},{ \nu }_{\tilde{q}},{ w}_{\tilde{q}}\right)\right)}\right)$$ Subject to $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right){x}_{j}\right)\le \mathfrak{R}\left({b}_{i1},{b}_{i2},{b}_{i3};{\mu }_{{\tilde{b}}_{i}},{\nu }_{{\tilde{b}}_{i}},{w}_{{\tilde{b}}_{i}}\right), i=\text{1,2},\dots ,m$$ , $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right){x}_{j}\right)=\mathfrak{R}\left(\text{1,1},1;\text{1,0},0\right), i=\text{1,2},\dots ,m$$ , $${x}_{j}\ge 0, j=\text{1,2},\dots ,n$$ , where, $$\mathfrak{R}\left(\sum _{i=1}^{n}\left({a}_{i},{b}_{i},{c}_{i};{\mu }_{i},{\nu }_{i},{w}_{i}\right)\right)=\frac{\left(2+\underset{1\le i\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{i}\right\}-\underset{1\le i\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{i}\right\}-\underset{1\le i\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{i}\right\}\right)}{9}\sum _{i=1}^{n}\left({a}_{i}+{b}_{i}+{c}_{i}\right)$$ . Step 2 Transform the crisp linear fractional programming problem (P3) into its equivalent crisp linear fractional programming problem (P4). Problem (P4) $$Max \left(or Min\right) \left(\frac{\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, { c}_{j2},{ c}_{j3};{ \mu }_{{\tilde{c}}_{j}},{ \nu }_{{\tilde{c}}_{j}},{ w}_{{\tilde{c}}_{j}}\right){ x}_{j} \right)+\mathfrak{ }\mathfrak{R}\left({p}_{1},{ p}_{2},{ p}_{3};{ \mu }_{\tilde{p}},{ \nu }_{\tilde{p}},{ w}_{\tilde{p}}\right)}{\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{ d}_{j2},{ d}_{j3};{ \mu }_{{\tilde{d}}_{j}},{ \nu }_{{\tilde{d}}_{j}},{ w}_{{\tilde{d}}_{j}}\right){ x}_{j}\right)\mathfrak{ }+\mathfrak{ }\mathfrak{R}\left({q}_{1},{ q}_{2},{ q}_{3};{ \mu }_{\tilde{q}},{ \nu }_{\tilde{q}},{ w}_{\tilde{q}}\right)}\right)$$ Subject to $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right){x}_{j}\right)\le \mathfrak{R}\left({b}_{i1},{b}_{i2},{b}_{i3};{\mu }_{{\tilde{b}}_{i}},{\nu }_{{\tilde{b}}_{i}},{w}_{{\tilde{b}}_{i}}\right), i=\text{1,2},\dots ,m$$ , $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right){x}_{j}\right)=\mathfrak{R}\left(\text{1,1},1;\text{1,0},0\right), i=\text{1,2},\dots ,m$$ , $${x}_{j}\ge 0, j=\text{1,2},\dots ,n$$ , where, $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}\right)=\left[\left(\frac{2+\underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{c}}_{j}}\right\} -\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{c}}_{j}}\right\}-\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{c}}_{j}}\right\}}{9}\right)\sum _{j=1}^{n}\left({c}_{j1}+{c}_{j2}+{c}_{j3}\right){x}_{j}\right]$$ (i) , $$\mathfrak{R}\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)= \left[\left(\frac{2+{\mu }_{\tilde{p}}-{\nu }_{\tilde{p}}-{w}_{\tilde{p}}}{9}\right)\left({p}_{1}+{p}_{2}+{p}_{3}\right)\right]$$ (ii) , $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}\right)=\left[\left(\frac{2+\underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{d}_{j}}\right\}-\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{d}}_{j}}\right\}-\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{d}}_{j}}\right\}}{9}\right)\sum _{j=1}^{n}\left({d}_{j1}+{d}_{j2}+{d}_{j3}\right){x}_{j}\right]$$ (iii) , $$\mathfrak{R}\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)=\left[\left(\frac{2 + {\mu }_{\tilde{q}} - {\nu }_{\tilde{q}} - {w}_{\tilde{q}}}{9}\right)\left({q}_{1}+{q}_{2}+{q}_{3}\right)\right]$$ (iv) , $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right){x}_{j}\right)=\left[\left(\frac{2+\underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{a}}_{ij}}\right\}-\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{a}}_{ij}}\right\}-\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{a}}_{ij}}\right\}}{9}\right)\sum _{j=1}^{n}\left({a}_{ij1}+{a}_{ij2}+{a}_{ij3}\right)\right]{x}_{j}$$ (v) , $$\mathfrak{R}\left({b}_{i1},{b}_{i2},{b}_{i3};{\mu }_{{\tilde{b}}_{i}},{\nu }_{{\tilde{b}}_{i}},{w}_{{\tilde{b}}_{i}}\right)=\left[\left(\frac{2+{\mu }_{{\tilde{b}}_{i}}-{\nu }_{{\tilde{b}}_{i}}-{w}_{{\tilde{b}}_{i}}}{9}\right)\left({b}_{i1}+{b}_{i2}+{b}_{i3}\right)\right]$$ (vi) , $$\mathfrak{R}\left(\text{1,1},1;\text{1,0},\right)=1$$ (vii) . Step 3 Find an optimal solution of the crisp linear programming problem (P4) and the corresponding optimal value. The obtained optimal solution and the obtained optimal value represent an optimal solution and the optimal value of the neutrosophic linear fractional programming problem (P1). 3. Inappropriateness Of Das And Edalatpanah’s Approach It is obvious from Section 2 that the following mathematical results are considered to transform the crisp linear fractional programming problem (P3) into linear fractional programming problem (P4). $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}+\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)\right)=\mathfrak{ }\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}\right)+\mathfrak{ }\mathfrak{R}\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)$$ (i) $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}+\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)\right)=\mathfrak{ }\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}\right)\mathfrak{ }+\mathfrak{ }\mathfrak{R}\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)$$ (ii) However, the following examples clearly indicate that in actual case $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}+\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)\right)\ne \mathfrak{ }\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}\right)+\mathfrak{ }\mathfrak{R}\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)$$ (i) $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}+\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)\right)\ne \mathfrak{ }\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}\right)\mathfrak{ }+\mathfrak{ }\mathfrak{R}\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)$$ (ii) Example 1 Let and. Then, $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}+\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)\right)$$ $$=\mathfrak{R}\left(\left(\text{4,8},10;\text{0.5,0.3,0.6}\right)+\left(\text{3,7},11;\text{0.4,0.5,0.6}\right)\right)=\mathfrak{ }\mathfrak{R}\left(\text{7,15,21};\text{0.4,0.5,0.6}\right)$$ $$=\frac{\left(7+15+21\right)}{9}\left(2+0.4-0.5-0.6\right)$$ $$=\frac{43}{9}\left(1.3\right)=6.21$$ 1 $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}\right)=\mathfrak{R}\left(\text{4,8},10;\text{0.5,0.3,0.6}\right)$$ $$=\frac{\left(4+8+10\right)}{9}\left(2+0.5-0.3-0.6\right)$$ $$=\frac{22}{9}\left(1.6\right)=3.$$ 2 $$\mathfrak{R}\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)=\mathfrak{R}\left(\text{3,7},11;\text{0.4,0.5,0.6}\right)$$ $$=\frac{\left(3+7+11\right)}{9}\left(2+0.4-0.5-0.6\right)$$ $$=\frac{21}{9}\left(1.3\right)=3.03$$ 3 Using (2) and (3), $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}\right)+\mathfrak{R}\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)$$ $$=3.91+3.03=6.94$$ 4 It is obvious from (1) and (4) that $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}+\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)\right)\ne \mathfrak{ }\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}\right)+\mathfrak{ }\mathfrak{R}\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)$$ . Example 2 Let and. Then, $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}+\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)\right)=\mathfrak{R}\left(\left(\text{5,7},9;\text{0.4,0.3,0.7}\right)+\left(\text{6,8},10;\text{0.6,0.4,0.2}\right)\right)$$ $$=\mathfrak{ }\mathfrak{R}\left(\left(\text{11,15,19}\right);\text{0.4,0.4,0.7}\right)$$ $$=\frac{\left(11+15+19\right)}{9}\left(2+0.4-0.4-0.7\right)$$ $$=\frac{45}{9}\left(1.3\right)=6.5$$ 5 $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}\right)=\mathfrak{R}\left(\text{5,7},9;\text{0.4,0.3,0.7}\right)$$ $$=\frac{\left(5+7+9\right)}{9}\left(0.4+\left(1-0.3\right)+\left(1-0.7\right)\right)$$ $$=\frac{21}{9}\left(1.4\right)=3.27$$ 6 $$\mathfrak{R}\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)=\mathfrak{R}\left(\text{6,8},10;\text{0.6,0.4,0.2}\right)$$ $$=\frac{\left(6+8+10\right)}{9}\left(2+0.6-0.4-0.2\right)$$ $$=\frac{24}{9}\left(2\right)=5.33$$ 7 Using (6) and (7), $$\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}\right)+\mathfrak{R}\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)$$ $$=3.27+5.33=8.6$$ 8 It is obvious from (5) and (8) that \(\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}+\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)\right)\ne \mathfrak{ }\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}\right)\mathfrak{ }+\mathfrak{ }\mathfrak{R}\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)\) . Hence, it is inappropriate to use Das and Edalatpanah ( 2022 )’s approach to solve the neutrosophic linear fractional programming problem (P1). 4. Modified Approach In this section, Das and Edalatpanah ( 2022 )’s approach is modified to resolve its inappropriateness. The steps of the modified approach are as follows: Step 1 Using the existing scalar multiplication of a non-negative number with a triangular neutrosophic number i.e., (Das and Edalatpanah 2022 , Definition 4, p. 8701), the neutrosophic linear fractional programming problem (P1) can be transformed into its equivalent neutrosophic linear fractional programming problem (P5). Problem (P5) $$Max \left(or Min\right) \left(\frac{\sum _{j=1}^{n}\left({c}_{j1}{x}_{j}, { c}_{j2}{x}_{j},{ c}_{j3}{x}_{j};{ \mu }_{{\tilde{c}}_{j}}, { \nu }_{{\tilde{c}}_{j}},{ w}_{{\tilde{c}}_{j}}\right)+\left({p}_{1},{ p}_{2},{ p}_{3};{ \mu }_{\tilde{p}},{ \nu }_{\tilde{p}},{ w}_{\tilde{p}}\right)}{\sum _{j=1}^{n}\left({d}_{j1}{x}_{j},{ d}_{j2}{x}_{j},{ d}_{j3}{x}_{j};{ \mu }_{{\tilde{d}}_{j}},{ \nu }_{{\tilde{d}}_{j}},{ w}_{{\tilde{d}}_{j}}\right)+\left({q}_{1},{ q}_{2},{ q}_{3};{ \mu }_{\tilde{q}},{ \nu }_{\tilde{q}},{ w}_{\tilde{q}}\right)}\right)$$ Subject to $$\sum _{j=1}^{n}\left({a}_{ij1}{x}_{j},{a}_{ij2}{x}_{j},{a}_{ij3}{x}_{j};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right)\le \left({b}_{i1},{b}_{i2},{b}_{i3};{\mu }_{{\tilde{b}}_{i}},{\nu }_{{\tilde{b}}_{i}},{w}_{{\tilde{b}}_{i}}\right), i=\text{1,2},\dots ,m$$ , $$\sum _{j=1}^{n}\left({a}_{ij1}{x}_{j},{a}_{ij2}{x}_{j},{a}_{ij3}{x}_{j};{\mu }_{{\tilde{a}}_{ij}},{\nu }_{{\tilde{a}}_{ij}},{w}_{{\tilde{a}}_{ij}}\right)=\left(\text{1,1},1;\text{1,0},0\right), i=\text{1,2},\dots ,m$$ , $${x}_{j}\ge 0, j=\text{1,2},\dots ,n.$$ Step 2 Using the existing addition of triangular neutrosophic numbers i.e., (Das and Edalatpanah 2022 , Definition 4, p. 8701), the neutrosophic linear fractional programming problem (P5) can be transformed into its equivalent neutrosophic linear fractional programming problem (P6). Problem (P6) $$Max \left(or Min\right)\left(\frac{\left(\sum _{j=1}^{n}{c}_{j1}{x}_{j} + {p}_{1}, \sum _{j=1}^{n}{c}_{j2}{x}_{j} + {p}_{2}, \sum _{j=1}^{n}{c}_{j3}{x}_{j} + {p}_{3}; \underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{c}}_{j}},{ \mu }_{\tilde{p}}\right\}, \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{c}}_{j}},{ \nu }_{\tilde{p}}\right\}, \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{c}}_{j}},{ w}_{\tilde{p}}\right\}\right)}{\left(\sum _{j=1}^{n}{d}_{j1}{x}_{j} + {q}_{1}, \sum _{j=1}^{n}{d}_{j2}{x}_{j} + {q}_{2}, \sum _{j=1}^{n}{d}_{j3}{x}_{j} + {q}_{3}; \underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{d}}_{j}},{ \mu }_{\tilde{q}}\right\}, \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{d}}_{j}},{ \nu }_{\tilde{q}}\right\}, \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{d}}_{j}},{ w}_{\tilde{q}}\right\}\right)}\right)$$ Subject to $$\left(\sum _{j=1}^{n}{a}_{ij1}{x}_{j}, \sum _{j=1}^{n}{a}_{ij2}{x}_{j}, \sum _{j=1}^{n}{a}_{ij3}{x}_{j};\underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{a}}_{ij}}\right\},\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{a}}_{ij}}\right\},\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{a}}_{ij}}\right\}\right)\le \left({b}_{i1},{b}_{i2},{b}_{i3};{\mu }_{{\tilde{b}}_{i}},{\nu }_{{\tilde{b}}_{i}},{w}_{{\tilde{b}}_{i}}\right), i=\text{1,2},\dots ,m$$ , $$\left(\sum _{j=1}^{n}{a}_{ij1}{x}_{j}, \sum _{j=1}^{n}{a}_{ij2}{x}_{j}, \sum _{j=1}^{n}{a}_{ij3}{x}_{j};\underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{a}}_{ij}}\right\},\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{a}}_{ij}}\right\},\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{a}}_{ij}}\right\}\right)=\left(\text{1,1},1;\text{1,0},0\right), i=\text{1,2},\dots ,m$$ , $${x}_{j}\ge 0, j=\text{1,2},\dots ,n.$$ Step 3 Transform the neutrosophic linear fractional programming problem (P6) into its equivalent crisp linear fractional programming problem (P7). Problem (P7) $$Max \left(or Min\right) \left(\frac{\mathfrak{R}\left(\sum _{j=1}^{n}{c}_{j1}{x}_{j} + {p}_{1}, \sum _{j=1}^{n}{c}_{j2}{x}_{j} + {p}_{2}, \sum _{j=1}^{n}{c}_{j3}{x}_{j} + {p}_{3}; \underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{c}}_{j}},{ \mu }_{\tilde{p}}\right\}, \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{c}}_{j}},{ \nu }_{\tilde{p}}\right\}, \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{c}}_{j}},{ w}_{\tilde{p}}\right\}\right)}{\mathfrak{R}\left(\sum _{j=1}^{n}{d}_{j1}{x}_{j} + {q}_{1}, \sum _{j=1}^{n}{d}_{j2}{x}_{j} + {q}_{2}, \sum _{j=1}^{n}{d}_{j3}{x}_{j} + {q}_{3}; \underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{d}}_{j}},{ \mu }_{\tilde{q}}\right\}, \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{d}}_{j}},{ \nu }_{\tilde{q}}\right\}, \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{d}}_{j}},{ w}_{\tilde{q}}\right\}\right)}\right)$$ Subject to $$\mathfrak{R}\left(\sum _{j=1}^{n}{a}_{ij1}{x}_{j}, \sum _{j=1}^{n}{a}_{ij2}{x}_{j}, \sum _{j=1}^{n}{a}_{ij3}{x}_{j};\underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{a}}_{ij}}\right\},\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{a}}_{ij}}\right\},\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{a}}_{ij}}\right\}\right)\le \mathfrak{R}\left({b}_{i1},{b}_{i2},{b}_{i3};{\mu }_{{\tilde{b}}_{i}},{\nu }_{{\tilde{b}}_{i}},{w}_{{\tilde{b}}_{i}}\right), i=\text{1,2},\dots ,m$$ , $$\mathfrak{R}\left(\sum _{j=1}^{n}{a}_{ij1}{x}_{j}, \sum _{j=1}^{n}{a}_{ij2}{x}_{j}, \sum _{j=1}^{n}{a}_{ij3}{x}_{j};\underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{a}}_{ij}}\right\},\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{a}}_{ij}}\right\},\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{a}}_{ij}}\right\}\right)=\mathfrak{R}\left(\text{1,1},1;\text{1,0},0\right), i=\text{1,2},\dots ,m$$ , $${x}_{j}\ge 0, j=\text{1,2},\dots ,n.$$ Step 4 Using the existing expression (Das and Edalatpanah 2022 , Definition 6, p. 8701), the crisp linear fractional programming problem (P7) can be transformed into its equivalent crisp linear fractional programming problem (P8). Problem (P8) $$Max \left(or Min\right)\left( \frac{\left[\left(\frac{2 + \underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{c}}_{j}},{ \mu }_{\tilde{p}}\right\} - \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{c}}_{j}},{ \nu }_{\tilde{p}}\right\} - \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{c}}_{j}},{ w}_{\tilde{p}}\right\}}{9}\right)\left(\sum _{j=1}^{n}{c}_{j1}{x}_{j} + {p}_{1}+ \sum _{j=1}^{n}{c}_{j2}{x}_{j} + {p}_{2} + \sum _{j=1}^{n}{c}_{j3}{x}_{j} + {p}_{3}\right)\right]}{\left[\left(\frac{2 + \underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{d}}_{j}},{ \mu }_{\tilde{q}}\right\} - \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{d}}_{j}},{ \nu }_{\tilde{q}}\right\} - \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{d}}_{j}},{ w}_{\tilde{q}}\right\}}{9}\right)\left(\sum _{j=1}^{n}{d}_{j1}{x}_{j} + {q}_{1} + \sum _{j=1}^{n}{d}_{j2}{x}_{j} + {q}_{2} + \sum _{j=1}^{n}{d}_{j3}{x}_{j} + {q}_{3}\right)\right]}\right)$$ Subject to $$\left[\left(\frac{2 + \underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{a}}_{ij}}\right\} - \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{a}}_{ij}}\right\} - \underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{a}}_{ij}}\right\}}{9}\right)\left(\sum _{j=1}^{n}{a}_{ij1}{x}_{j}+\sum _{j=1}^{n}{a}_{ij2}{x}_{j}+\sum _{j=1}^{n}{a}_{ij3}{x}_{j}\right)\right]\le \left[\left(\frac{2 + {\mu }_{{\tilde{b}}_{i}} - {\nu }_{{\tilde{b}}_{i}}- {w}_{{\tilde{b}}_{i}}}{9}\right)\left({b}_{i1}+{b}_{i2}+{b}_{i3}\right)\right], i=\text{1,2},\dots ,m$$ , $$\left[\left(\frac{2 +\underset{1\le j\le n}{\text{m}\text{i}\text{n}}\left\{{\mu }_{{\tilde{a}}_{ij}}\right\}-\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{\nu }_{{\tilde{a}}_{ij}}\right\}-\underset{1\le j\le n}{\text{m}\text{a}\text{x}}\left\{{w}_{{\tilde{a}}_{ij}}\right\}}{9}\right)\left(\sum _{j=1}^{n}{a}_{ij1}{x}_{j}+\sum _{j=1}^{n}{a}_{ij2}{x}_{j}+\sum _{j=1}^{n}{a}_{ij3}{x}_{j}\right)\right]=1, i=1, 2,\dots ,m$$ , $${x}_{j}\ge 0, j=\text{1,2},\dots ,n.$$ Step 5 Find an optimal solution of the crisp linear programming problem (P9) and the corresponding optimal value. The obtained optimal solution and the obtained optimal value represent an optimal solution and the optimal value of the neutrosophic linear fractional programming problem (P1). 5. Exact Optimal Solution Of Existing Neutrosophic Linear Fractional Programming Problem Das and Edalatpanah ( 2022 ) solved the neutrosophic linear fractional programming problem (P2) to illustrate their proposed approach. However as discussed in Section 2, Das and Edalatpanah ( 2022 )’s approach is not appropriate as they have considered an incorrect mathematical result in their proposed approach. Therefore, the optimal solution of neutrosophic linear fractional programming problem (P2), obtained by Das and Edalatpanah ( 2022 ), is not correct. In this section, an exact optimal solution of neutrosophic linear fractional programming problem (P2) is obtained by the modified approach. Using the modified approach, an optimal solution of the neutrosophic linear fractional programming problem (P2) can be obtained as follows: Step 1 Using Step 1 of modified method, the neutrosophic linear programming problem (P2) can be transformed into its equivalent neutrosophic linear fractional programming problem (P9). Problem (P9) $$Max \left(\frac{\left(7{x}_{1},8{x}_{1},9{x}_{1};\text{0.5,0.8,0.3}\right)+\left(6{x}_{2},7{x}_{2},8{x}_{2};\text{0.2,0.6,0.5}\right)+\left(8{x}_{3},9{x}_{3},10{x}_{3};\text{0.8,0.1,0.4}\right)}{\left(7{x}_{1},8{x}_{1},9{x}_{1};\text{0.5,0.8,0.3}\right)+\left(8{x}_{2},9{x}_{2},10{x}_{2};\text{0.8,0.1,0.4}\right)+\left(4{x}_{3},6{x}_{3},8{x}_{3};\text{0.75,0.25,0.1}\right)+\left(\text{1,1.5,2};\text{0.75,0.5,0.25}\right)}\right)$$ Subject to $$\left(3{x}_{1},4{x}_{1},5{x}_{1};\text{0.4,0.6,0.5}\right)+\left(2{x}_{2},3{x}_{2},4{x}_{2};\text{1,0.25,0.3}\right)+\left(4{x}_{3},5{x}_{3},6{x}_{3};\text{0.3,0.4,0.8}\right)\le$$ $$\left(\text{25,28,30};\text{0.4,0.25,0.6}\right)$$ , $$\left(4{x}_{1},5{x}_{1},6{x}_{1};\text{0.3,0.4,0.8}\right)+\left(2{x}_{2},3{x}_{2},4{x}_{2};\text{1,0.25,0.3}\right)+\left(2{x}_{3},3{x}_{3},4{x}_{3};\text{1,0.25,0.3}\right)\le$$ $$\left(\text{18,20,22};\text{0.9,0.2,0.6}\right)$$ , $${x}_{1},{x}_{2},{x}_{3}\ge 0$$ . Step 2 Using Step 2 of modified method, the neutrosophic linear programming problem (P9) can be transformed into its equivalent neutrosophic linear fractional programming problem (P10). Problem (P10) $$Max \left(\frac{\left(7{x}_{1}+6{x}_{2}+8{x}_{3},8{x}_{1}+7{x}_{2}+9{x}_{3},9{x}_{1}+8{x}_{2}+10{x}_{3};\text{0.2,0.8,0.5}\right)}{\left(7{x}_{1}+8{x}_{2}+4{x}_{3}+\text{1,8}{x}_{1}+9{x}_{2}+6{x}_{3}+\text{1.5,9}{x}_{1}+10{x}_{2}+8{x}_{3}+2;\text{0.5,0.8,0.4}\right)}\right)$$ Subject to $$\left(3{x}_{1}+2{x}_{2}+4{x}_{3},4{x}_{1}+3{x}_{2}+5{x}_{3},5{x}_{1}+4{x}_{2}+6{x}_{3};\text{0.3,0.6,0.8}\right)\le$$ $$\left(\text{25,28,30};\text{0.4,0.25,0.6}\right)$$ , $$\left(4{x}_{1}+2{x}_{2}+2{x}_{3},5{x}_{1}+3{x}_{2}+3{x}_{3},6{x}_{1}+4{x}_{2}+4{x}_{3};\text{0.3,0.4,0.8}\right)\le$$ $$\left(\text{18,20,22};\text{0.9,0.2,0.6}\right)$$ , $${x}_{1},{x}_{2},{x}_{3}\ge 0$$ . Step 3 Using Step 3 of modified method, the neutrosophic linear programming problem (P10) can be transformed into its equivalent crisp linear fractional programming problem (P11). Problem (P11) $$Max \left(\frac{\mathfrak{R}\left(7{x}_{1}+6{x}_{2}+8{x}_{3},8{x}_{1}+7{x}_{2}+9{x}_{3},9{x}_{1}+8{x}_{2}+10{x}_{3};\text{0.2,0.8,0.5}\right)}{\mathfrak{R}\left(7{x}_{1}+8{x}_{2}+4{x}_{3}+\text{1,8}{x}_{1}+9{x}_{2}+6{x}_{3}+\text{1.5,9}{x}_{1}+10{x}_{2}+8{x}_{3}+2;\text{0.5,0.8,0.4}\right)}\right)$$ Subject to $$\mathfrak{R}\left(3{x}_{1}+2{x}_{2}+4{x}_{3},4{x}_{1}+3{x}_{2}+5{x}_{3},5{x}_{1}+4{x}_{2}+6{x}_{3};\text{0.3,0.6,0.8}\right)\le$$ $$\mathfrak{R}\left(\text{25,28,30};\text{0.4,0.25,0.6}\right)$$ , $$\mathfrak{R}\left(4{x}_{1}+2{x}_{2}+2{x}_{3},5{x}_{1}+3{x}_{2}+3{x}_{3},6{x}_{1}+4{x}_{2}+4{x}_{3};\text{0.3,0.4,0.8}\right)\le$$ $$\mathfrak{R}\left(\text{18,20,22};\text{0.9,0.2,0.6}\right)$$ , $${x}_{1},{x}_{2},{x}_{3}\ge 0$$ . Step 4 Using Step 4 of modified method, the crisp linear programming problem (P11) can be transformed into its equivalent crisp linear fractional programming problem (P12) or its equivalent crisp linear fractional programming problem (P13). Problem (P12) $$Max \left(\frac{⌈\left(\frac{2+0.2 -0.8-0.5}{9}\right)\left(7{x}_{1}+6{x}_{2}+8{x}_{3}+8{x}_{1}+7{x}_{2}+9{x}_{3}+9{x}_{1}+8{x}_{2}+10{x}_{3}\right)⌉}{⌈\left(\frac{2+0.5 -0.8-0.4}{9}\right)\left(7{x}_{1}+8{x}_{2}+4{x}_{3}+1+8{x}_{1}+9{x}_{2}+6{x}_{3}+1.5+9{x}_{1}+10{x}_{2}+8{x}_{3}+2\right)⌉}\right)$$ Subject to $$\left[\left(\frac{2+0.3-0.6-0.8}{9}\right)\left(3{x}_{1}+2{x}_{2}+4{x}_{3}+4{x}_{1}+3{x}_{2}+5{x}_{3}+5{x}_{1}+4{x}_{2}+6{x}_{3}\right)\right]\le$$ $$\left[\left(\frac{2+0.4-0.25-0.6}{9}\right)\left(25+28+30\right) \right]$$ , $$\left[\left(\frac{2+0.3-0.4-0.8}{9}\right)\left(4{x}_{1}+2{x}_{2}+2{x}_{3}+5{x}_{1}+3{x}_{2}+3{x}_{3}+6{x}_{1}+4{x}_{2}+4{x}_{3}\right)\right]\le$$ $$\left[\left(\frac{2+0.9-0.2-0.6}{9}\right)\left(18+20+22\right)\right]$$ , $${x}_{1},{x}_{2},{x}_{3}\ge 0$$ . Problem (P13) $$Max \left(\frac{21.6{x}_{1}+18.9{x}_{2}+24.3{x}_{3}}{31.2{x}_{1}+35.1{x}_{2}+23.4{x}_{3}+5.85}\right)$$ Subject to $$10.8{x}_{1}+8.1{x}_{2}+13.5{x}_{3}\le 128.65,$$ $$16.5{x}_{1}+9.9{x}_{2}+9.9{x}_{3}\le 126,$$ $${x}_{1},{x}_{2},{x}_{3}\ge 0$$ . Step 5 According to Step 5 of the modified method, there is need to find an optimal solution of the crisp linear fractional problem (P13). On solving the problem (P13), the following optimal solution and the corresponding optimal value is obtained $${x}_{1}=0$$ , $${x}_{2}=0$$ , $${x}_{3}=9.53$$ , Optimal value is \(1.01\) . 6. Conclusions It is pointed that Das and Edalatpanah ( 2022 )’s approach is not valid as in this approach a mathematical incorrect assumption is considered. Also, a modified method is proposed to find an optimal solution of the neutrosophic linear fractional programming problems. Finally, an exact optimal solution an existing neutrosophic linear fractional programming problem is obtained by the modified method. Declarations Compliance with ethical standards Conflict of interest The authors declare that they have no conflict of interest Ethical approval This article does not contain any studies with human participants or animals performed by any of the authors. References Abdel-basset M, Mohamed M, Smarandache F (2019) Linear fractional programming based on triangular neutrosophic numbers. Int J Appl Manag Sci 11:1–20 Das S, Edalatpanah SA (2022) Optimal solution of neutrosophic linear fractional programming problems with mixed constraints. Soft Comput 26(17):8699–8707 Cite Share Download PDF Status: Posted Version 1 posted You are reading this latest preprint version Research Square lets you share your work early, gain feedback from the community, and start making changes to your manuscript prior to peer review in a journal. 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Also discoverable on Platform About Our Team In Review Editorial Policies Help Center Resources Author Services Accessibility API Access RSS feed Manage Cookie Preferences © Research Square 2026 | ISSN 2693-5015 (online) Privacy Policy Terms of Service Do Not Sell My Personal Information {"props":{"pageProps":{"initialData":{"identity":"rs-2250652","acceptedTermsAndConditions":true,"allowDirectSubmit":true,"archivedVersions":[],"articleType":"Research Article","associatedPublications":[],"authors":[{"id":155341299,"identity":"1f28c21f-4d2e-4a48-aa3f-f1b5b62a99ed","order_by":0,"name":"Parul Tomar","email":"","orcid":"","institution":"Thapar University: Thapar Institute of Engineering and Technology","correspondingAuthor":false,"submittingAuthor":false,"prefix":"","firstName":"Parul","middleName":"","lastName":"Tomar","suffix":""},{"id":155341300,"identity":"b40b3864-cc16-443a-a4f8-5e06538b0554","order_by":1,"name":"Amit Kumar","email":"data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAZAAAAAyAQMAAABI0h/eAAAABlBMVEX///8AAABVwtN+AAAACXBIWXMAAA7EAAAOxAGVKw4bAAAAz0lEQVRIiWNgGAWjYFACxgbGBjYGBn4QO6GAOC2NjSAtkg0gLQZEWgPWYnAAxCZGC/+0w+0PZ5Rtkzc+vzrxwwMDBnl+sQP4tUjcTmxs3HDutuG2G283SwAdZjhzdgIBa0BaHrbdZtx24+wGkJYEg9sEtMhDtdhvnnF28w+itBiAtGxsu524gb93G3G2GAK1zJxx7nbyjBu82ywSDCQI+0XudvqDjz1lt237+89uvvmjwkaeX5qAFgSQAKuUIFY5CPAfIEX1KBgFo2AUjCQAAHBtT8wn0lrBAAAAAElFTkSuQmCC","orcid":"","institution":"Thapar University: Thapar Institute of Engineering and Technology","correspondingAuthor":true,"submittingAuthor":false,"prefix":"","firstName":"Amit","middleName":"","lastName":"Kumar","suffix":""}],"badges":[],"createdAt":"2022-11-08 10:48:10","currentVersionCode":1,"declarations":"","doi":"10.21203/rs.3.rs-2250652/v1","doiUrl":"https://doi.org/10.21203/rs.3.rs-2250652/v1","draftVersion":[],"editorialEvents":[],"editorialNote":"","failedWorkflow":false,"files":[{"id":32038754,"identity":"e225fa3c-5549-46e3-8a12-fdb3fe1a83c0","added_by":"auto","created_at":"2023-01-25 11:17:57","extension":"pdf","order_by":0,"title":"","display":"","copyAsset":false,"role":"manuscript-pdf","size":313744,"visible":true,"origin":"","legend":"","description":"","filename":"manuscript.pdf","url":"https://assets-eu.researchsquare.com/files/rs-2250652/v1/c4d93d33-7597-4eb0-95d7-c936f16b0ebe.pdf"}],"financialInterests":"","formattedTitle":"A note on “Optimal solution of neutrosophic linear fractional programming problems with mixed constraints”","fulltext":[{"header":"1. Introduction","content":"\u003cp\u003eAbdel-basset et al. (\u003cspan citationid=\"CR1\" class=\"CitationRef\"\u003e2019\u003c/span\u003e) proposed an approach to find an optimal solution of neutrosophic linear fractional programming problem (P1).\u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P1)\u003c/b\u003e \u003cdiv id=\"Equa\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equa\" name=\"EquationSource\"\u003e\n$$Max \\left(or Min\\right) \\left(\\frac{\\sum _{j=1}^{n}\\left({c}_{j1}, { c}_{j2},{ c}_{j3};{ \\mu }_{{\\tilde{c}}_{j}},{ \\nu }_{{\\tilde{c}}_{j}},{ w}_{{\\tilde{c}}_{j}}\\right){ x}_{j} + \\left({p}_{1},{ p}_{2},{ p}_{3};{ \\mu }_{\\tilde{p}},{ \\nu }_{\\tilde{p}},{ w}_{\\tilde{p}}\\right)}{\\sum _{j=1}^{n}\\left({d}_{j1},{ d}_{j2},{ d}_{j3};{ \\mu }_{{\\tilde{d}}_{j}},{ \\nu }_{{\\tilde{d}}_{j}},{ w}_{{\\tilde{d}}_{j}}\\right){ x}_{j} + \\left({q}_{1},{ q}_{2},{ q}_{3};{ \\mu }_{\\tilde{q}},{ \\nu }_{\\tilde{q}},{ w}_{\\tilde{q}}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equb\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equb\" name=\"EquationSource\"\u003e\n$$\\sum _{j=1}^{n}\\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right){x}_{j}\\le \\left({b}_{i1},{b}_{i2},{b}_{i3};{\\mu }_{{\\tilde{b}}_{i}},{\\nu }_{{\\tilde{b}}_{i}},{w}_{{\\tilde{b}}_{i}}\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equc\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equc\" name=\"EquationSource\"\u003e\n$$\\sum _{j=1}^{n}\\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right){x}_{j}=\\left(\\text{1,1},1;\\text{1,0},0\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equd\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equd\" name=\"EquationSource\"\u003e\n$${x}_{j}\\ge 0, j=\\text{1,2},\\dots ,n.$$\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003cp\u003eDas and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e) proposed an alternative approach to find an optimal solution of the neutrosophic linear fractional programming problem (P1). Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e) also pointed out that it is better to use their proposed approach as compared to Abdel-basset et al. (\u003cspan citationid=\"CR1\" class=\"CitationRef\"\u003e2019\u003c/span\u003e)\u0026rsquo;s approach due to the following reasons:\u003c/p\u003e \u003cp\u003e \u003col\u003e \u003cspan\u003e \u003cli\u003e \u003cp\u003eMuch computational efforts are required to find an optimal solution of the neutrosophic linear fractional programming problem (P1) by Abdel-basset et al. (\u003cspan citationid=\"CR1\" class=\"CitationRef\"\u003e2019\u003c/span\u003e)\u0026rsquo;s approach. While, less computational efforts are required to find an optimal solution of the neutrosophic linear fractional programming problem (P1) by their proposed approach.\u003c/p\u003e \u003c/li\u003e \u003c/span\u003e \u003cspan\u003e \u003cli\u003e \u003cp\u003eIf the same neutrosophic linear fractional programming problem is solved by their proposed approach and Abdel-basset et al. (\u003cspan citationid=\"CR1\" class=\"CitationRef\"\u003e2019\u003c/span\u003e)\u0026rsquo;s approach. Then, the optimal solution obtained by their proposed approach is better than the optimal solution obtained by Abdel-basset et al. (\u003cspan citationid=\"CR1\" class=\"CitationRef\"\u003e2019\u003c/span\u003e)\u0026rsquo;s approach. To validate this claim, Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e, Section 6, p. 8703) solved the neutrosophic linear fractional programming problem (P2) by their proposed approach and pointed out that the obtained maximum value is 1.4. While, on solving the same problem by Abdel-basset et al. (\u003cspan citationid=\"CR1\" class=\"CitationRef\"\u003e2019\u003c/span\u003e)\u0026rsquo;s approach, the obtained maximum value is 1.\u003c/p\u003e \u003c/li\u003e \u003c/span\u003e \u003c/ol\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P2)\u003c/b\u003e \u003cdiv id=\"Eque\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Eque\" name=\"EquationSource\"\u003e\n$$Max \\left(\\frac{\\left(\\text{7,8},9;\\text{0.5,0.8,0.3}\\right){x}_{1}+\\left(\\text{6,7},8;\\text{0.2,0.6,0.5}\\right){x}_{2}+\\left(\\text{8,9},10;\\text{0.8,0.1,0.4}\\right){x}_{3}}{\\left(\\text{7,8},9;\\text{0.5,0.8,0.3}\\right){x}_{1}+\\left(\\text{8,9},10;\\text{0.8,0.1,0.4}\\right){x}_{2}+\\left(\\text{4,6},8;\\text{0.75,0.25,0.1}\\right){x}_{3}+\\left(\\text{1,1.5,2};\\text{0.75,0.5,0.25}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equf\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equf\" name=\"EquationSource\"\u003e\n$$\\left(\\text{3,4},5;\\text{0.4,0.6,0.5}\\right){x}_{1}+\\left(\\text{2,3},4;\\text{1,0.25,0.3}\\right){x}_{2}+\\left(\\text{4,5},6;\\text{0.3,0.4,0.8}\\right){x}_{3}\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equg\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equg\" name=\"EquationSource\"\u003e\n$$\\left(\\text{25,28,30};\\text{0.4,0.25,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equh\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equh\" name=\"EquationSource\"\u003e\n$$\\left(\\text{4,5},6;\\text{0.3,0.4,0.8}\\right){x}_{1}+\\left(\\text{2,3},4;\\text{1,0.25,0.3}\\right){x}_{2}+\\left(\\text{2,3},4;\\text{1,0.25,0.3}\\right){x}_{3}\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equi\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equi\" name=\"EquationSource\"\u003e\n$$\\left(\\text{18,20,22};\\text{0.9,0.2,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equj\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equj\" name=\"EquationSource\"\u003e\n$${x}_{1},{x}_{2},{x}_{3}\\ge 0$$\u003c/div\u003e\u003c/div\u003e.\u003c/p\u003e \u003cp\u003eIn this paper, it is pointed out that Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e) have considered a mathematical incorrect result in their proposed approach. Hence, it is inappropriate to use Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e)\u0026rsquo;s approach. Also, Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e)\u0026rsquo;s approach is modified to resolve its inappropriateness. Furthermore, a correct optimal solution and the corresponding maximum value of the neutrosophic linear fractional programming problem (P2) are obtained by the modified approach.\u003c/p\u003e"},{"header":"2. Das And Edalatpanah’s Approach","content":"\u003cp\u003e \u003cdiv class=\"BlockQuote\"\u003e \u003cp\u003eDas and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e) proposed the following approach to find an optimal solution of the neutrosophic linear fractional programming problem (P1).\u003c/p\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 1\u003c/strong\u003e \u003cp\u003eTransform the neutrosophic linear fractional programming problem (P1) into its equivalent crisp linear fractional programming problem (P3).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P3)\u003c/b\u003e \u003cdiv id=\"Equk\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equk\" name=\"EquationSource\"\u003e\n$$Max \\left(or Min\\right) \\left(\\frac{\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, { c}_{j2},{ c}_{j3};{ \\mu }_{{\\tilde{c}}_{j}},{ \\nu }_{{\\tilde{c}}_{j}},{ w}_{{\\tilde{c}}_{j}}\\right){ x}_{j} + \\left({p}_{1},{ p}_{2},{ p}_{3};{ \\mu }_{\\tilde{p}},{ \\nu }_{\\tilde{p}},{ w}_{\\tilde{p}}\\right)\\right)}{\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{ d}_{j2},{ d}_{j3};{ \\mu }_{{\\tilde{d}}_{j}},{ \\nu }_{{\\tilde{d}}_{j}},{ w}_{{\\tilde{d}}_{j}}\\right){ x}_{j} + \\left({q}_{1},{ q}_{2},{ q}_{3};{ \\mu }_{\\tilde{q}},{ \\nu }_{\\tilde{q}},{ w}_{\\tilde{q}}\\right)\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equl\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equl\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right){x}_{j}\\right)\\le \\mathfrak{R}\\left({b}_{i1},{b}_{i2},{b}_{i3};{\\mu }_{{\\tilde{b}}_{i}},{\\nu }_{{\\tilde{b}}_{i}},{w}_{{\\tilde{b}}_{i}}\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equm\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equm\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right){x}_{j}\\right)=\\mathfrak{R}\\left(\\text{1,1},1;\\text{1,0},0\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equn\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equn\" name=\"EquationSource\"\u003e\n$${x}_{j}\\ge 0, j=\\text{1,2},\\dots ,n$$\u003c/div\u003e\u003c/div\u003e,\u003c/p\u003e \u003cp\u003ewhere,\u003cdiv id=\"Equo\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equo\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{i=1}^{n}\\left({a}_{i},{b}_{i},{c}_{i};{\\mu }_{i},{\\nu }_{i},{w}_{i}\\right)\\right)=\\frac{\\left(2+\\underset{1\\le i\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{i}\\right\\}-\\underset{1\\le i\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{i}\\right\\}-\\underset{1\\le i\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{i}\\right\\}\\right)}{9}\\sum _{i=1}^{n}\\left({a}_{i}+{b}_{i}+{c}_{i}\\right)$$\u003c/div\u003e\u003c/div\u003e.\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 2\u003c/strong\u003e \u003cp\u003eTransform the crisp linear fractional programming problem (P3) into its equivalent crisp linear fractional programming problem (P4).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P4)\u003c/b\u003e \u003cdiv id=\"Equp\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equp\" name=\"EquationSource\"\u003e\n$$Max \\left(or Min\\right) \\left(\\frac{\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, { c}_{j2},{ c}_{j3};{ \\mu }_{{\\tilde{c}}_{j}},{ \\nu }_{{\\tilde{c}}_{j}},{ w}_{{\\tilde{c}}_{j}}\\right){ x}_{j} \\right)+\\mathfrak{ }\\mathfrak{R}\\left({p}_{1},{ p}_{2},{ p}_{3};{ \\mu }_{\\tilde{p}},{ \\nu }_{\\tilde{p}},{ w}_{\\tilde{p}}\\right)}{\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{ d}_{j2},{ d}_{j3};{ \\mu }_{{\\tilde{d}}_{j}},{ \\nu }_{{\\tilde{d}}_{j}},{ w}_{{\\tilde{d}}_{j}}\\right){ x}_{j}\\right)\\mathfrak{ }+\\mathfrak{ }\\mathfrak{R}\\left({q}_{1},{ q}_{2},{ q}_{3};{ \\mu }_{\\tilde{q}},{ \\nu }_{\\tilde{q}},{ w}_{\\tilde{q}}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equq\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equq\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right){x}_{j}\\right)\\le \\mathfrak{R}\\left({b}_{i1},{b}_{i2},{b}_{i3};{\\mu }_{{\\tilde{b}}_{i}},{\\nu }_{{\\tilde{b}}_{i}},{w}_{{\\tilde{b}}_{i}}\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equr\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equr\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right){x}_{j}\\right)=\\mathfrak{R}\\left(\\text{1,1},1;\\text{1,0},0\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equs\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equs\" name=\"EquationSource\"\u003e\n$${x}_{j}\\ge 0, j=\\text{1,2},\\dots ,n$$\u003c/div\u003e\u003c/div\u003e,\u003c/p\u003e \u003cp\u003ewhere,\u003cdiv id=\"Equt\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equt\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}\\right)=\\left[\\left(\\frac{2+\\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{c}}_{j}}\\right\\} -\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{c}}_{j}}\\right\\}-\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{c}}_{j}}\\right\\}}{9}\\right)\\sum _{j=1}^{n}\\left({c}_{j1}+{c}_{j2}+{c}_{j3}\\right){x}_{j}\\right]$$\u003c/div\u003e\u003c/div\u003e(i) ,\u003cdiv id=\"Equu\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equu\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)= \\left[\\left(\\frac{2+{\\mu }_{\\tilde{p}}-{\\nu }_{\\tilde{p}}-{w}_{\\tilde{p}}}{9}\\right)\\left({p}_{1}+{p}_{2}+{p}_{3}\\right)\\right]$$\u003c/div\u003e\u003c/div\u003e(ii) ,\u003cdiv id=\"Equv\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equv\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}\\right)=\\left[\\left(\\frac{2+\\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{d}_{j}}\\right\\}-\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{d}}_{j}}\\right\\}-\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{d}}_{j}}\\right\\}}{9}\\right)\\sum _{j=1}^{n}\\left({d}_{j1}+{d}_{j2}+{d}_{j3}\\right){x}_{j}\\right]$$\u003c/div\u003e\u003c/div\u003e(iii) ,\u003cdiv id=\"Equw\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equw\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)=\\left[\\left(\\frac{2 + {\\mu }_{\\tilde{q}} - {\\nu }_{\\tilde{q}} - {w}_{\\tilde{q}}}{9}\\right)\\left({q}_{1}+{q}_{2}+{q}_{3}\\right)\\right]$$\u003c/div\u003e\u003c/div\u003e(iv) ,\u003cdiv id=\"Equx\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equx\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({a}_{ij1},{a}_{ij2},{a}_{ij3};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right){x}_{j}\\right)=\\left[\\left(\\frac{2+\\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{a}}_{ij}}\\right\\}-\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{a}}_{ij}}\\right\\}-\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{a}}_{ij}}\\right\\}}{9}\\right)\\sum _{j=1}^{n}\\left({a}_{ij1}+{a}_{ij2}+{a}_{ij3}\\right)\\right]{x}_{j}$$\u003c/div\u003e\u003c/div\u003e(v) ,\u003cdiv id=\"Equy\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equy\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left({b}_{i1},{b}_{i2},{b}_{i3};{\\mu }_{{\\tilde{b}}_{i}},{\\nu }_{{\\tilde{b}}_{i}},{w}_{{\\tilde{b}}_{i}}\\right)=\\left[\\left(\\frac{2+{\\mu }_{{\\tilde{b}}_{i}}-{\\nu }_{{\\tilde{b}}_{i}}-{w}_{{\\tilde{b}}_{i}}}{9}\\right)\\left({b}_{i1}+{b}_{i2}+{b}_{i3}\\right)\\right]$$\u003c/div\u003e\u003c/div\u003e(vi) ,\u003cdiv id=\"Equz\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equz\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\text{1,1},1;\\text{1,0},\\right)=1$$\u003c/div\u003e\u003c/div\u003e(vii) .\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 3\u003c/strong\u003e \u003cp\u003eFind an optimal solution of the crisp linear programming problem (P4) and the corresponding optimal value. The obtained optimal solution and the obtained optimal value represent an optimal solution and the optimal value of the neutrosophic linear fractional programming problem (P1).\u003c/p\u003e \u003c/p\u003e"},{"header":"3. Inappropriateness Of Das And Edalatpanah’s Approach","content":"\u003cp\u003e \u003cdiv class=\"BlockQuote\"\u003e \u003cp\u003eIt is obvious from Section 2 that the following mathematical results are considered to transform the crisp linear fractional programming problem (P3) into linear fractional programming problem (P4).\u003c/p\u003e \u003c/div\u003e \u003cdiv id=\"Equaa\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equaa\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}+\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)\\right)=\\mathfrak{ }\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}\\right)+\\mathfrak{ }\\mathfrak{R}\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)$$\u003c/div\u003e \u003c/div\u003e(i) \u003cdiv id=\"Equab\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equab\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}+\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)\\right)=\\mathfrak{ }\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}\\right)\\mathfrak{ }+\\mathfrak{ }\\mathfrak{R}\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)$$\u003c/div\u003e\u003c/div\u003e(ii) \u003c/p\u003e \u003cp\u003eHowever, the following examples clearly indicate that in actual case\u003cdiv id=\"Equac\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equac\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}+\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)\\right)\\ne \\mathfrak{ }\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}\\right)+\\mathfrak{ }\\mathfrak{R}\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)$$\u003c/div\u003e\u003c/div\u003e(i) \u003cdiv id=\"Equad\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equad\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}+\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)\\right)\\ne \\mathfrak{ }\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}\\right)\\mathfrak{ }+\\mathfrak{ }\\mathfrak{R}\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)$$\u003c/div\u003e\u003c/div\u003e(ii) \u003c/p\u003e \u003cp\u003e \u003cstrong\u003eExample 1\u003c/strong\u003e \u003cp\u003eLet and. Then,\u003cdiv id=\"Equae\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equae\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}+\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equaf\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equaf\" name=\"EquationSource\"\u003e\n$$=\\mathfrak{R}\\left(\\left(\\text{4,8},10;\\text{0.5,0.3,0.6}\\right)+\\left(\\text{3,7},11;\\text{0.4,0.5,0.6}\\right)\\right)=\\mathfrak{ }\\mathfrak{R}\\left(\\text{7,15,21};\\text{0.4,0.5,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equag\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equag\" name=\"EquationSource\"\u003e\n$$=\\frac{\\left(7+15+21\\right)}{9}\\left(2+0.4-0.5-0.6\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equ1\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equ1\" name=\"EquationSource\"\u003e\n$$=\\frac{43}{9}\\left(1.3\\right)=6.21$$\u003c/div\u003e\u003cdiv class=\"EquationNumber\"\u003e1\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equah\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equah\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}\\right)=\\mathfrak{R}\\left(\\text{4,8},10;\\text{0.5,0.3,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equai\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equai\" name=\"EquationSource\"\u003e\n$$=\\frac{\\left(4+8+10\\right)}{9}\\left(2+0.5-0.3-0.6\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equ2\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equ2\" name=\"EquationSource\"\u003e\n$$=\\frac{22}{9}\\left(1.6\\right)=3.$$\u003c/div\u003e\u003cdiv class=\"EquationNumber\"\u003e2\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equaj\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equaj\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)=\\mathfrak{R}\\left(\\text{3,7},11;\\text{0.4,0.5,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equak\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equak\" name=\"EquationSource\"\u003e\n$$=\\frac{\\left(3+7+11\\right)}{9}\\left(2+0.4-0.5-0.6\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equ3\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equ3\" name=\"EquationSource\"\u003e\n$$=\\frac{21}{9}\\left(1.3\\right)=3.03$$\u003c/div\u003e\u003cdiv class=\"EquationNumber\"\u003e3\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003c/p\u003e \u003cp\u003eUsing (2) and (3),\u003cdiv id=\"Equal\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equal\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}\\right)+\\mathfrak{R}\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equ4\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equ4\" name=\"EquationSource\"\u003e\n$$=3.91+3.03=6.94$$\u003c/div\u003e\u003cdiv class=\"EquationNumber\"\u003e4\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003cp\u003eIt is obvious from (1) and (4) that\u003cdiv id=\"Equam\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equam\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}+\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)\\right)\\ne \\mathfrak{ }\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({c}_{j1}, {c}_{j2},{c}_{j3};{\\mu }_{{\\tilde{c}}_{j}},{\\nu }_{{\\tilde{c}}_{j}},{w}_{{\\tilde{c}}_{j}}\\right){x}_{j}\\right)+\\mathfrak{ }\\mathfrak{R}\\left({p}_{1},{p}_{2},{p}_{3};{\\mu }_{\\tilde{p}},{\\nu }_{\\tilde{p}},{w}_{\\tilde{p}}\\right)$$\u003c/div\u003e\u003c/div\u003e.\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eExample 2\u003c/strong\u003e \u003cp\u003eLet and. Then,\u003cdiv id=\"Equan\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equan\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}+\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)\\right)=\\mathfrak{R}\\left(\\left(\\text{5,7},9;\\text{0.4,0.3,0.7}\\right)+\\left(\\text{6,8},10;\\text{0.6,0.4,0.2}\\right)\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equao\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equao\" name=\"EquationSource\"\u003e\n$$=\\mathfrak{ }\\mathfrak{R}\\left(\\left(\\text{11,15,19}\\right);\\text{0.4,0.4,0.7}\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equap\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equap\" name=\"EquationSource\"\u003e\n$$=\\frac{\\left(11+15+19\\right)}{9}\\left(2+0.4-0.4-0.7\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equ5\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equ5\" name=\"EquationSource\"\u003e\n$$=\\frac{45}{9}\\left(1.3\\right)=6.5$$\u003c/div\u003e\u003cdiv class=\"EquationNumber\"\u003e5\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equaq\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equaq\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}\\right)=\\mathfrak{R}\\left(\\text{5,7},9;\\text{0.4,0.3,0.7}\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equar\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equar\" name=\"EquationSource\"\u003e\n$$=\\frac{\\left(5+7+9\\right)}{9}\\left(0.4+\\left(1-0.3\\right)+\\left(1-0.7\\right)\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equ6\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equ6\" name=\"EquationSource\"\u003e\n$$=\\frac{21}{9}\\left(1.4\\right)=3.27$$\u003c/div\u003e\u003cdiv class=\"EquationNumber\"\u003e6\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equas\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equas\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)=\\mathfrak{R}\\left(\\text{6,8},10;\\text{0.6,0.4,0.2}\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equat\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equat\" name=\"EquationSource\"\u003e\n$$=\\frac{\\left(6+8+10\\right)}{9}\\left(2+0.6-0.4-0.2\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equ7\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equ7\" name=\"EquationSource\"\u003e\n$$=\\frac{24}{9}\\left(2\\right)=5.33$$\u003c/div\u003e\u003cdiv class=\"EquationNumber\"\u003e7\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003c/p\u003e \u003cp\u003eUsing (6) and (7),\u003cdiv id=\"Equau\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equau\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}\\right)+\\mathfrak{R}\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equ8\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equ8\" name=\"EquationSource\"\u003e\n$$=3.27+5.33=8.6$$\u003c/div\u003e\u003cdiv class=\"EquationNumber\"\u003e8\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003cp\u003eIt is obvious from (5) and (8) that \u003cspan class=\"InlineEquation\"\u003e\u003cspan class=\"mathinline\"\u003e\\(\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}+\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)\\right)\\ne \\mathfrak{ }\\mathfrak{R}\\left(\\sum _{j=1}^{n}\\left({d}_{j1},{d}_{j2},{d}_{j3};{\\mu }_{{\\tilde{d}}_{j}},{\\nu }_{{\\tilde{d}}_{j}},{w}_{{\\tilde{d}}_{j}}\\right){x}_{j}\\right)\\mathfrak{ }+\\mathfrak{ }\\mathfrak{R}\\left({q}_{1},{q}_{2},{q}_{3};{\\mu }_{\\tilde{q}},{\\nu }_{\\tilde{q}},{w}_{\\tilde{q}}\\right)\\)\u003c/span\u003e\u003c/span\u003e.\u003c/p\u003e \u003cp\u003eHence, it is inappropriate to use Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e)\u0026rsquo;s approach to solve the neutrosophic linear fractional programming problem (P1).\u003c/p\u003e"},{"header":"4. Modified Approach","content":"\u003cp\u003eIn this section, Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e)\u0026rsquo;s approach is modified to resolve its inappropriateness.\u003c/p\u003e \u003cp\u003eThe steps of the modified approach are as follows:\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 1\u003c/strong\u003e \u003cp\u003eUsing the existing scalar multiplication of a non-negative number with a triangular neutrosophic number i.e., (Das and Edalatpanah \u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e, Definition 4, p. 8701), the neutrosophic linear fractional programming problem (P1) can be transformed into its equivalent neutrosophic linear fractional programming problem (P5).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P5)\u003c/b\u003e \u003cdiv id=\"Equav\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equav\" name=\"EquationSource\"\u003e\n$$Max \\left(or Min\\right) \\left(\\frac{\\sum _{j=1}^{n}\\left({c}_{j1}{x}_{j}, { c}_{j2}{x}_{j},{ c}_{j3}{x}_{j};{ \\mu }_{{\\tilde{c}}_{j}}, { \\nu }_{{\\tilde{c}}_{j}},{ w}_{{\\tilde{c}}_{j}}\\right)+\\left({p}_{1},{ p}_{2},{ p}_{3};{ \\mu }_{\\tilde{p}},{ \\nu }_{\\tilde{p}},{ w}_{\\tilde{p}}\\right)}{\\sum _{j=1}^{n}\\left({d}_{j1}{x}_{j},{ d}_{j2}{x}_{j},{ d}_{j3}{x}_{j};{ \\mu }_{{\\tilde{d}}_{j}},{ \\nu }_{{\\tilde{d}}_{j}},{ w}_{{\\tilde{d}}_{j}}\\right)+\\left({q}_{1},{ q}_{2},{ q}_{3};{ \\mu }_{\\tilde{q}},{ \\nu }_{\\tilde{q}},{ w}_{\\tilde{q}}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equaw\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equaw\" name=\"EquationSource\"\u003e\n$$\\sum _{j=1}^{n}\\left({a}_{ij1}{x}_{j},{a}_{ij2}{x}_{j},{a}_{ij3}{x}_{j};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right)\\le \\left({b}_{i1},{b}_{i2},{b}_{i3};{\\mu }_{{\\tilde{b}}_{i}},{\\nu }_{{\\tilde{b}}_{i}},{w}_{{\\tilde{b}}_{i}}\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equax\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equax\" name=\"EquationSource\"\u003e\n$$\\sum _{j=1}^{n}\\left({a}_{ij1}{x}_{j},{a}_{ij2}{x}_{j},{a}_{ij3}{x}_{j};{\\mu }_{{\\tilde{a}}_{ij}},{\\nu }_{{\\tilde{a}}_{ij}},{w}_{{\\tilde{a}}_{ij}}\\right)=\\left(\\text{1,1},1;\\text{1,0},0\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equay\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equay\" name=\"EquationSource\"\u003e\n$${x}_{j}\\ge 0, j=\\text{1,2},\\dots ,n.$$\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 2\u003c/strong\u003e \u003cp\u003eUsing the existing addition of triangular neutrosophic numbers i.e., (Das and Edalatpanah \u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e, Definition 4, p. 8701), the neutrosophic linear fractional programming problem (P5) can be transformed into its equivalent neutrosophic linear fractional programming problem (P6).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P6)\u003c/b\u003e \u003cdiv id=\"Equaz\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equaz\" name=\"EquationSource\"\u003e\n$$Max \\left(or Min\\right)\\left(\\frac{\\left(\\sum _{j=1}^{n}{c}_{j1}{x}_{j} + {p}_{1}, \\sum _{j=1}^{n}{c}_{j2}{x}_{j} + {p}_{2}, \\sum _{j=1}^{n}{c}_{j3}{x}_{j} + {p}_{3}; \\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{c}}_{j}},{ \\mu }_{\\tilde{p}}\\right\\}, \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{c}}_{j}},{ \\nu }_{\\tilde{p}}\\right\\}, \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{c}}_{j}},{ w}_{\\tilde{p}}\\right\\}\\right)}{\\left(\\sum _{j=1}^{n}{d}_{j1}{x}_{j} + {q}_{1}, \\sum _{j=1}^{n}{d}_{j2}{x}_{j} + {q}_{2}, \\sum _{j=1}^{n}{d}_{j3}{x}_{j} + {q}_{3}; \\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{d}}_{j}},{ \\mu }_{\\tilde{q}}\\right\\}, \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{d}}_{j}},{ \\nu }_{\\tilde{q}}\\right\\}, \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{d}}_{j}},{ w}_{\\tilde{q}}\\right\\}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equba\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equba\" name=\"EquationSource\"\u003e\n$$\\left(\\sum _{j=1}^{n}{a}_{ij1}{x}_{j}, \\sum _{j=1}^{n}{a}_{ij2}{x}_{j}, \\sum _{j=1}^{n}{a}_{ij3}{x}_{j};\\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{a}}_{ij}}\\right\\},\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{a}}_{ij}}\\right\\},\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{a}}_{ij}}\\right\\}\\right)\\le \\left({b}_{i1},{b}_{i2},{b}_{i3};{\\mu }_{{\\tilde{b}}_{i}},{\\nu }_{{\\tilde{b}}_{i}},{w}_{{\\tilde{b}}_{i}}\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbb\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbb\" name=\"EquationSource\"\u003e\n$$\\left(\\sum _{j=1}^{n}{a}_{ij1}{x}_{j}, \\sum _{j=1}^{n}{a}_{ij2}{x}_{j}, \\sum _{j=1}^{n}{a}_{ij3}{x}_{j};\\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{a}}_{ij}}\\right\\},\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{a}}_{ij}}\\right\\},\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{a}}_{ij}}\\right\\}\\right)=\\left(\\text{1,1},1;\\text{1,0},0\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbc\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbc\" name=\"EquationSource\"\u003e\n$${x}_{j}\\ge 0, j=\\text{1,2},\\dots ,n.$$\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 3\u003c/strong\u003e \u003cp\u003eTransform the neutrosophic linear fractional programming problem (P6) into its equivalent crisp linear fractional programming problem (P7).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P7)\u003c/b\u003e \u003cdiv id=\"Equbd\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbd\" name=\"EquationSource\"\u003e\n$$Max \\left(or Min\\right) \\left(\\frac{\\mathfrak{R}\\left(\\sum _{j=1}^{n}{c}_{j1}{x}_{j} + {p}_{1}, \\sum _{j=1}^{n}{c}_{j2}{x}_{j} + {p}_{2}, \\sum _{j=1}^{n}{c}_{j3}{x}_{j} + {p}_{3}; \\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{c}}_{j}},{ \\mu }_{\\tilde{p}}\\right\\}, \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{c}}_{j}},{ \\nu }_{\\tilde{p}}\\right\\}, \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{c}}_{j}},{ w}_{\\tilde{p}}\\right\\}\\right)}{\\mathfrak{R}\\left(\\sum _{j=1}^{n}{d}_{j1}{x}_{j} + {q}_{1}, \\sum _{j=1}^{n}{d}_{j2}{x}_{j} + {q}_{2}, \\sum _{j=1}^{n}{d}_{j3}{x}_{j} + {q}_{3}; \\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{d}}_{j}},{ \\mu }_{\\tilde{q}}\\right\\}, \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{d}}_{j}},{ \\nu }_{\\tilde{q}}\\right\\}, \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{d}}_{j}},{ w}_{\\tilde{q}}\\right\\}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Eqube\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Eqube\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}{a}_{ij1}{x}_{j}, \\sum _{j=1}^{n}{a}_{ij2}{x}_{j}, \\sum _{j=1}^{n}{a}_{ij3}{x}_{j};\\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{a}}_{ij}}\\right\\},\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{a}}_{ij}}\\right\\},\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{a}}_{ij}}\\right\\}\\right)\\le \\mathfrak{R}\\left({b}_{i1},{b}_{i2},{b}_{i3};{\\mu }_{{\\tilde{b}}_{i}},{\\nu }_{{\\tilde{b}}_{i}},{w}_{{\\tilde{b}}_{i}}\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbf\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbf\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\sum _{j=1}^{n}{a}_{ij1}{x}_{j}, \\sum _{j=1}^{n}{a}_{ij2}{x}_{j}, \\sum _{j=1}^{n}{a}_{ij3}{x}_{j};\\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{a}}_{ij}}\\right\\},\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{a}}_{ij}}\\right\\},\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{a}}_{ij}}\\right\\}\\right)=\\mathfrak{R}\\left(\\text{1,1},1;\\text{1,0},0\\right), i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbg\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbg\" name=\"EquationSource\"\u003e\n$${x}_{j}\\ge 0, j=\\text{1,2},\\dots ,n.$$\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 4\u003c/strong\u003e \u003cp\u003eUsing the existing expression (Das and Edalatpanah \u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e, Definition 6, p. 8701), the crisp linear fractional programming problem (P7) can be transformed into its equivalent crisp linear fractional programming problem (P8).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P8)\u003c/b\u003e \u003cdiv id=\"Equbh\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbh\" name=\"EquationSource\"\u003e\n$$Max \\left(or Min\\right)\\left( \\frac{\\left[\\left(\\frac{2 + \\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{c}}_{j}},{ \\mu }_{\\tilde{p}}\\right\\} - \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{c}}_{j}},{ \\nu }_{\\tilde{p}}\\right\\} - \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{c}}_{j}},{ w}_{\\tilde{p}}\\right\\}}{9}\\right)\\left(\\sum _{j=1}^{n}{c}_{j1}{x}_{j} + {p}_{1}+ \\sum _{j=1}^{n}{c}_{j2}{x}_{j} + {p}_{2} + \\sum _{j=1}^{n}{c}_{j3}{x}_{j} + {p}_{3}\\right)\\right]}{\\left[\\left(\\frac{2 + \\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{d}}_{j}},{ \\mu }_{\\tilde{q}}\\right\\} - \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{d}}_{j}},{ \\nu }_{\\tilde{q}}\\right\\} - \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{d}}_{j}},{ w}_{\\tilde{q}}\\right\\}}{9}\\right)\\left(\\sum _{j=1}^{n}{d}_{j1}{x}_{j} + {q}_{1} + \\sum _{j=1}^{n}{d}_{j2}{x}_{j} + {q}_{2} + \\sum _{j=1}^{n}{d}_{j3}{x}_{j} + {q}_{3}\\right)\\right]}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equbi\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbi\" name=\"EquationSource\"\u003e\n$$\\left[\\left(\\frac{2 + \\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{a}}_{ij}}\\right\\} - \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{a}}_{ij}}\\right\\} - \\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{a}}_{ij}}\\right\\}}{9}\\right)\\left(\\sum _{j=1}^{n}{a}_{ij1}{x}_{j}+\\sum _{j=1}^{n}{a}_{ij2}{x}_{j}+\\sum _{j=1}^{n}{a}_{ij3}{x}_{j}\\right)\\right]\\le \\left[\\left(\\frac{2 + {\\mu }_{{\\tilde{b}}_{i}} - {\\nu }_{{\\tilde{b}}_{i}}- {w}_{{\\tilde{b}}_{i}}}{9}\\right)\\left({b}_{i1}+{b}_{i2}+{b}_{i3}\\right)\\right], i=\\text{1,2},\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbj\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbj\" name=\"EquationSource\"\u003e\n$$\\left[\\left(\\frac{2 +\\underset{1\\le j\\le n}{\\text{m}\\text{i}\\text{n}}\\left\\{{\\mu }_{{\\tilde{a}}_{ij}}\\right\\}-\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{\\nu }_{{\\tilde{a}}_{ij}}\\right\\}-\\underset{1\\le j\\le n}{\\text{m}\\text{a}\\text{x}}\\left\\{{w}_{{\\tilde{a}}_{ij}}\\right\\}}{9}\\right)\\left(\\sum _{j=1}^{n}{a}_{ij1}{x}_{j}+\\sum _{j=1}^{n}{a}_{ij2}{x}_{j}+\\sum _{j=1}^{n}{a}_{ij3}{x}_{j}\\right)\\right]=1, i=1, 2,\\dots ,m$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbk\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbk\" name=\"EquationSource\"\u003e\n$${x}_{j}\\ge 0, j=\\text{1,2},\\dots ,n.$$\u003c/div\u003e\u003c/div\u003e\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 5\u003c/strong\u003e \u003cp\u003eFind an optimal solution of the crisp linear programming problem (P9) and the corresponding optimal value. The obtained optimal solution and the obtained optimal value represent an optimal solution and the optimal value of the neutrosophic linear fractional programming problem (P1).\u003c/p\u003e \u003c/p\u003e"},{"header":"5. Exact Optimal Solution Of Existing Neutrosophic Linear Fractional Programming Problem","content":"\u003cp\u003eDas and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e) solved the neutrosophic linear fractional programming problem (P2) to illustrate their proposed approach. However as discussed in Section 2, Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e)\u0026rsquo;s approach is not appropriate as they have considered an incorrect mathematical result in their proposed approach. Therefore, the optimal solution of neutrosophic linear fractional programming problem (P2), obtained by Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e), is not correct. In this section, an exact optimal solution of neutrosophic linear fractional programming problem (P2) is obtained by the modified approach.\u003c/p\u003e \u003cp\u003eUsing the modified approach, an optimal solution of the neutrosophic linear fractional programming problem (P2) can be obtained as follows:\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 1\u003c/strong\u003e \u003cp\u003eUsing Step 1 of modified method, the neutrosophic linear programming problem (P2) can be transformed into its equivalent neutrosophic linear fractional programming problem (P9).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P9)\u003c/b\u003e \u003cdiv id=\"Equbl\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbl\" name=\"EquationSource\"\u003e\n$$Max \\left(\\frac{\\left(7{x}_{1},8{x}_{1},9{x}_{1};\\text{0.5,0.8,0.3}\\right)+\\left(6{x}_{2},7{x}_{2},8{x}_{2};\\text{0.2,0.6,0.5}\\right)+\\left(8{x}_{3},9{x}_{3},10{x}_{3};\\text{0.8,0.1,0.4}\\right)}{\\left(7{x}_{1},8{x}_{1},9{x}_{1};\\text{0.5,0.8,0.3}\\right)+\\left(8{x}_{2},9{x}_{2},10{x}_{2};\\text{0.8,0.1,0.4}\\right)+\\left(4{x}_{3},6{x}_{3},8{x}_{3};\\text{0.75,0.25,0.1}\\right)+\\left(\\text{1,1.5,2};\\text{0.75,0.5,0.25}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equbm\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbm\" name=\"EquationSource\"\u003e\n$$\\left(3{x}_{1},4{x}_{1},5{x}_{1};\\text{0.4,0.6,0.5}\\right)+\\left(2{x}_{2},3{x}_{2},4{x}_{2};\\text{1,0.25,0.3}\\right)+\\left(4{x}_{3},5{x}_{3},6{x}_{3};\\text{0.3,0.4,0.8}\\right)\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equbn\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbn\" name=\"EquationSource\"\u003e\n$$\\left(\\text{25,28,30};\\text{0.4,0.25,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbo\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbo\" name=\"EquationSource\"\u003e\n$$\\left(4{x}_{1},5{x}_{1},6{x}_{1};\\text{0.3,0.4,0.8}\\right)+\\left(2{x}_{2},3{x}_{2},4{x}_{2};\\text{1,0.25,0.3}\\right)+\\left(2{x}_{3},3{x}_{3},4{x}_{3};\\text{1,0.25,0.3}\\right)\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equbp\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbp\" name=\"EquationSource\"\u003e\n$$\\left(\\text{18,20,22};\\text{0.9,0.2,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbq\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbq\" name=\"EquationSource\"\u003e\n$${x}_{1},{x}_{2},{x}_{3}\\ge 0$$\u003c/div\u003e\u003c/div\u003e.\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 2\u003c/strong\u003e \u003cp\u003eUsing Step 2 of modified method, the neutrosophic linear programming problem (P9) can be transformed into its equivalent neutrosophic linear fractional programming problem (P10).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P10)\u003c/b\u003e \u003cdiv id=\"Equbr\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbr\" name=\"EquationSource\"\u003e\n$$Max \\left(\\frac{\\left(7{x}_{1}+6{x}_{2}+8{x}_{3},8{x}_{1}+7{x}_{2}+9{x}_{3},9{x}_{1}+8{x}_{2}+10{x}_{3};\\text{0.2,0.8,0.5}\\right)}{\\left(7{x}_{1}+8{x}_{2}+4{x}_{3}+\\text{1,8}{x}_{1}+9{x}_{2}+6{x}_{3}+\\text{1.5,9}{x}_{1}+10{x}_{2}+8{x}_{3}+2;\\text{0.5,0.8,0.4}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equbs\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbs\" name=\"EquationSource\"\u003e\n$$\\left(3{x}_{1}+2{x}_{2}+4{x}_{3},4{x}_{1}+3{x}_{2}+5{x}_{3},5{x}_{1}+4{x}_{2}+6{x}_{3};\\text{0.3,0.6,0.8}\\right)\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equbt\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbt\" name=\"EquationSource\"\u003e\n$$\\left(\\text{25,28,30};\\text{0.4,0.25,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbu\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbu\" name=\"EquationSource\"\u003e\n$$\\left(4{x}_{1}+2{x}_{2}+2{x}_{3},5{x}_{1}+3{x}_{2}+3{x}_{3},6{x}_{1}+4{x}_{2}+4{x}_{3};\\text{0.3,0.4,0.8}\\right)\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equbv\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbv\" name=\"EquationSource\"\u003e\n$$\\left(\\text{18,20,22};\\text{0.9,0.2,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equbw\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbw\" name=\"EquationSource\"\u003e\n$${x}_{1},{x}_{2},{x}_{3}\\ge 0$$\u003c/div\u003e\u003c/div\u003e.\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 3\u003c/strong\u003e \u003cp\u003eUsing Step 3 of modified method, the neutrosophic linear programming problem (P10) can be transformed into its equivalent crisp linear fractional programming problem (P11).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P11)\u003c/b\u003e \u003cdiv id=\"Equbx\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbx\" name=\"EquationSource\"\u003e\n$$Max \\left(\\frac{\\mathfrak{R}\\left(7{x}_{1}+6{x}_{2}+8{x}_{3},8{x}_{1}+7{x}_{2}+9{x}_{3},9{x}_{1}+8{x}_{2}+10{x}_{3};\\text{0.2,0.8,0.5}\\right)}{\\mathfrak{R}\\left(7{x}_{1}+8{x}_{2}+4{x}_{3}+\\text{1,8}{x}_{1}+9{x}_{2}+6{x}_{3}+\\text{1.5,9}{x}_{1}+10{x}_{2}+8{x}_{3}+2;\\text{0.5,0.8,0.4}\\right)}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equby\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equby\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(3{x}_{1}+2{x}_{2}+4{x}_{3},4{x}_{1}+3{x}_{2}+5{x}_{3},5{x}_{1}+4{x}_{2}+6{x}_{3};\\text{0.3,0.6,0.8}\\right)\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equbz\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equbz\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\text{25,28,30};\\text{0.4,0.25,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equca\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equca\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(4{x}_{1}+2{x}_{2}+2{x}_{3},5{x}_{1}+3{x}_{2}+3{x}_{3},6{x}_{1}+4{x}_{2}+4{x}_{3};\\text{0.3,0.4,0.8}\\right)\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equcb\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcb\" name=\"EquationSource\"\u003e\n$$\\mathfrak{R}\\left(\\text{18,20,22};\\text{0.9,0.2,0.6}\\right)$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equcc\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcc\" name=\"EquationSource\"\u003e\n$${x}_{1},{x}_{2},{x}_{3}\\ge 0$$\u003c/div\u003e\u003c/div\u003e.\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 4\u003c/strong\u003e \u003cp\u003eUsing Step 4 of modified method, the crisp linear programming problem (P11) can be transformed into its equivalent crisp linear fractional programming problem (P12) or its equivalent crisp linear fractional programming problem (P13).\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P12)\u003c/b\u003e \u003cdiv id=\"Equcd\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcd\" name=\"EquationSource\"\u003e\n$$Max \\left(\\frac{\u0026lceil;\\left(\\frac{2+0.2 -0.8-0.5}{9}\\right)\\left(7{x}_{1}+6{x}_{2}+8{x}_{3}+8{x}_{1}+7{x}_{2}+9{x}_{3}+9{x}_{1}+8{x}_{2}+10{x}_{3}\\right)\u0026rceil;}{\u0026lceil;\\left(\\frac{2+0.5 -0.8-0.4}{9}\\right)\\left(7{x}_{1}+8{x}_{2}+4{x}_{3}+1+8{x}_{1}+9{x}_{2}+6{x}_{3}+1.5+9{x}_{1}+10{x}_{2}+8{x}_{3}+2\\right)\u0026rceil;}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equce\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equce\" name=\"EquationSource\"\u003e\n$$\\left[\\left(\\frac{2+0.3-0.6-0.8}{9}\\right)\\left(3{x}_{1}+2{x}_{2}+4{x}_{3}+4{x}_{1}+3{x}_{2}+5{x}_{3}+5{x}_{1}+4{x}_{2}+6{x}_{3}\\right)\\right]\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equcf\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcf\" name=\"EquationSource\"\u003e\n$$\\left[\\left(\\frac{2+0.4-0.25-0.6}{9}\\right)\\left(25+28+30\\right) \\right]$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equcg\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcg\" name=\"EquationSource\"\u003e\n$$\\left[\\left(\\frac{2+0.3-0.4-0.8}{9}\\right)\\left(4{x}_{1}+2{x}_{2}+2{x}_{3}+5{x}_{1}+3{x}_{2}+3{x}_{3}+6{x}_{1}+4{x}_{2}+4{x}_{3}\\right)\\right]\\le$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equch\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equch\" name=\"EquationSource\"\u003e\n$$\\left[\\left(\\frac{2+0.9-0.2-0.6}{9}\\right)\\left(18+20+22\\right)\\right]$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equci\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equci\" name=\"EquationSource\"\u003e\n$${x}_{1},{x}_{2},{x}_{3}\\ge 0$$\u003c/div\u003e\u003c/div\u003e.\u003c/p\u003e \u003cp\u003e \u003cb\u003eProblem (P13)\u003c/b\u003e \u003cdiv id=\"Equcj\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcj\" name=\"EquationSource\"\u003e\n$$Max \\left(\\frac{21.6{x}_{1}+18.9{x}_{2}+24.3{x}_{3}}{31.2{x}_{1}+35.1{x}_{2}+23.4{x}_{3}+5.85}\\right)$$\u003c/div\u003e \u003c/div\u003e \u003c/p\u003e \u003cp\u003eSubject to\u003cdiv id=\"Equck\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equck\" name=\"EquationSource\"\u003e\n$$10.8{x}_{1}+8.1{x}_{2}+13.5{x}_{3}\\le 128.65,$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equcl\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcl\" name=\"EquationSource\"\u003e\n$$16.5{x}_{1}+9.9{x}_{2}+9.9{x}_{3}\\le 126,$$\u003c/div\u003e\u003c/div\u003e\u003cdiv id=\"Equcm\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcm\" name=\"EquationSource\"\u003e\n$${x}_{1},{x}_{2},{x}_{3}\\ge 0$$\u003c/div\u003e\u003c/div\u003e.\u003c/p\u003e \u003cp\u003e \u003cstrong\u003eStep 5\u003c/strong\u003e \u003cp\u003eAccording to Step 5 of the modified method, there is need to find an optimal solution of the crisp linear fractional problem (P13). On solving the problem (P13), the following optimal solution and the corresponding optimal value is obtained\u003c/p\u003e \u003c/p\u003e \u003cp\u003e \u003cdiv id=\"Equcn\" class=\"Equation\"\u003e \u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcn\" name=\"EquationSource\"\u003e\n$${x}_{1}=0$$\u003c/div\u003e \u003c/div\u003e,\u003cdiv id=\"Equco\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equco\" name=\"EquationSource\"\u003e\n$${x}_{2}=0$$\u003c/div\u003e\u003c/div\u003e,\u003cdiv id=\"Equcp\" class=\"Equation\"\u003e\u003cdiv format=\"TEX\" class=\"mathdisplay\" id=\"FileID_Equcp\" name=\"EquationSource\"\u003e\n$${x}_{3}=9.53$$\u003c/div\u003e\u003c/div\u003e,\u003c/p\u003e \u003cp\u003eOptimal value is \u003cspan class=\"InlineEquation\"\u003e\u003cspan class=\"mathinline\"\u003e\\(1.01\\)\u003c/span\u003e\u003c/span\u003e.\u003c/p\u003e"},{"header":"6. Conclusions","content":"\u003cp\u003e \u003cdiv class=\"BlockQuote\"\u003e \u003cp\u003eIt is pointed that Das and Edalatpanah (\u003cspan citationid=\"CR2\" class=\"CitationRef\"\u003e2022\u003c/span\u003e)\u0026rsquo;s approach is not valid as in this approach a mathematical incorrect assumption is considered. Also, a modified method is proposed to find an optimal solution of the neutrosophic linear fractional programming problems. Finally, an exact optimal solution an existing neutrosophic linear fractional programming problem is obtained by the modified method.\u003c/p\u003e \u003c/div\u003e \u003c/p\u003e"},{"header":"Declarations","content":"\u003cp\u003e\u003cstrong\u003eCompliance with ethical standards\u003c/strong\u003e\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003eConflict of interest\u003c/strong\u003e The authors declare that they have no conflict of interest\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003eEthical approval\u0026nbsp;\u003c/strong\u003eThis article does not contain any studies with human\u003cstrong\u003e\u0026nbsp;\u003c/strong\u003eparticipants or animals performed by any of the authors.\u003c/p\u003e"},{"header":"References","content":"\u003col\u003e\u003cli\u003e\u003cspan\u003eAbdel-basset M, Mohamed M, Smarandache F (2019) Linear fractional programming based on triangular neutrosophic numbers. Int J Appl Manag Sci 11:1\u0026ndash;20\u003c/span\u003e\u003c/li\u003e \u003cli\u003e\u003cspan\u003eDas S, Edalatpanah SA (2022) Optimal solution of neutrosophic linear fractional programming problems with mixed constraints. Soft Comput 26(17):8699\u0026ndash;8707\u003c/span\u003e\u003c/li\u003e\u003c/ol\u003e"}],"fulltextSource":"","fullText":"","funders":[],"hasAdminPriorityOnWorkflow":false,"hasManuscriptDocX":true,"hasOptedInToPreprint":true,"hasPassedJournalQc":"","hasAnyPriority":false,"hideJournal":true,"highlight":"","institution":"","isAcceptedByJournal":false,"isAuthorSuppliedPdf":false,"isDeskRejected":"","isHiddenFromSearch":false,"isInQc":false,"isInWorkflow":false,"isPdf":false,"isPdfUpToDate":true,"isWithdrawnOrRetracted":false,"journal":{"display":true,"email":"[email protected]","identity":"researchsquare","isNatureJournal":false,"hasQc":true,"allowDirectSubmit":true,"externalIdentity":"","sideBox":"","snPcode":"","submissionUrl":"/submission","title":"Research Square","twitterHandle":"researchsquare","acdcEnabled":true,"dfaEnabled":false,"editorialSystem":"","reportingPortfolio":"","inReviewEnabled":false,"inReviewRevisionsEnabled":true},"keywords":"Neutrosophic linear fractional programming, Crisp linear fractional programming, Triangular neutrosophic numbers, Ranking function","lastPublishedDoi":"10.21203/rs.3.rs-2250652/v1","lastPublishedDoiUrl":"https://doi.org/10.21203/rs.3.rs-2250652/v1","license":{"name":"CC BY 4.0","url":"https://creativecommons.org/licenses/by/4.0/"},"manuscriptAbstract":"\u003cp\u003eDas and Edalatpanah (Soft Comput 26 (2022) 8699\u0026ndash;8707) proposed an approach to find an optimal solution of neutrosophic linear fractional programming problems with mixed constraints (linear programming problems with mixed constraints in which each decision variable is represented by a non-negative real number and each other parameter is represented by a triangular neutrosophic number). In this paper, it is pointed out that a mathematical incorrect result is considered in Das and Edalatpanah\u0026rsquo;s approach. Hence, it is inappropriate to use Das and Edalatpanah\u0026rsquo;s approach. Also, Das and Edalatpanah\u0026rsquo;s approach is modified to resolve its inappropriateness.\u003c/p\u003e","manuscriptTitle":"A note on “Optimal solution of neutrosophic linear fractional programming problems with mixed constraints”","msid":"","msnumber":"","nonDraftVersions":[{"code":1,"date":"2022-11-30 03:58:19","doi":"10.21203/rs.3.rs-2250652/v1","editorialEvents":[{"type":"communityComments","content":0}],"status":"published","journal":{"display":true,"email":"[email protected]","identity":"researchsquare","isNatureJournal":false,"hasQc":true,"allowDirectSubmit":true,"externalIdentity":"","sideBox":"","snPcode":"","submissionUrl":"/submission","title":"Research Square","twitterHandle":"researchsquare","acdcEnabled":true,"dfaEnabled":false,"editorialSystem":"","reportingPortfolio":"","inReviewEnabled":false,"inReviewRevisionsEnabled":true}}],"origin":"","ownerIdentity":"6dd461ab-af08-4fc2-a24c-bc92ad41a9c7","owner":[],"postedDate":"November 30th, 2022","published":true,"recentEditorialEvents":[],"rejectedJournal":[],"revision":"","amendment":"","status":"posted","subjectAreas":[],"tags":[],"updatedAt":"2023-01-25T11:17:46+00:00","versionOfRecord":[],"versionCreatedAt":"2022-11-30 03:58:19","video":"","vorDoi":"","vorDoiUrl":"","workflowStages":[]},"version":"v1","identity":"rs-2250652","journalConfig":"researchsquare"},"__N_SSP":true},"page":"/article/[identity]/[[...version]]","query":{"redirect":"/article/rs-2250652","identity":"rs-2250652","version":["v1"]},"buildId":"eB-D7MK2yqyqIWuf3Ze0-","isFallback":false,"isExperimentalCompile":false,"dynamicIds":[84888],"gssp":true,"scriptLoader":[]}