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),
$$\mathfrak{R}\left(\sum _{j=1}^{n}\left({c}_{j1}, {c}_{j2},{c}_{j3};{\mu }_{{\tilde{c}}_{j}},{\nu }_{{\tilde{c}}_{j}},{w}_{{\tilde{c}}_{j}}\right){x}_{j}\right)+\mathfrak{R}\left({p}_{1},{p}_{2},{p}_{3};{\mu }_{\tilde{p}},{\nu }_{\tilde{p}},{w}_{\tilde{p}}\right)$$
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),
$$\mathfrak{R}\left(\sum _{j=1}^{n}\left({d}_{j1},{d}_{j2},{d}_{j3};{\mu }_{{\tilde{d}}_{j}},{\nu }_{{\tilde{d}}_{j}},{w}_{{\tilde{d}}_{j}}\right){x}_{j}\right)+\mathfrak{R}\left({q}_{1},{q}_{2},{q}_{3};{\mu }_{\tilde{q}},{\nu }_{\tilde{q}},{w}_{\tilde{q}}\right)$$
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).