Table 1 Decision-matrix of probabilistic q-rung orthopair linguistic neutrosophic set taken by D.

From: A novel probabilistic q-rung orthopair linguistic neutrosophic information-based method for rating nanoparticles in various sectors

 

\({\mathcal {S}}_1\)

\({\mathcal {S}}_2\)

\({\mathcal {S}}_3\)

\({\mathcal {S}}_4\)

\({\mathcal {S}}_5\)

\({\mathcal {S}}_6\)

\({\mathcal {N}}_{1}\)

\(\frac{0.8}{\langle s_{5},\{0.8,0.6,0.4\}\rangle }\)

\(\frac{0.9}{\langle s_{3},\{0.8,0.5,0.5\}\rangle }\)

\(\frac{0.5}{\langle s_{4},\{0.7,0.5,0.5\}\rangle }\)

\(\frac{0.6}{\langle s_{4},\{0.9,0.4,0.5\}\rangle }\)

\(\frac{0.4}{\langle s_{3},\{0.4,0.6,0.9\}\rangle }\)

\(\frac{0.7}{\langle s_{2},\{0.2,0.4,0.8\}\rangle }\)

\({\mathcal {N}}_2\)

\(\frac{0.6}{\langle s_{4},\{0.7,0.4,0.4\}\rangle }\)

\(\frac{0.7}{\langle s_{5},\{0.9,0.2,0.3\}\rangle }\)

\(\frac{0.8}{\langle s_{4},\{0.9,0.4,0.4\}\rangle }\)

\(\frac{0.6}{\langle s_{4},\{0.8,0.5,0.4\}\rangle }\)

\(\frac{0.7}{\langle s_{2},\{0.4,0.6,0.9\}\rangle }\)

\(\frac{0.8}{\langle s_{3},\{0.3,0.6,0.9\}\rangle }\)

\({\mathcal {N}}_3\)

\(\frac{0.7}{\langle s_{4},\{0.7,0.6,0.4\}\rangle }\)

\(\frac{0.7}{\langle s_{3},\{0.7,0.4,0.6\}\rangle }\)

\(\frac{0.4}{\langle s_{4},\{0.8,0.4,0.5\}\rangle }\)

\(\frac{0.6}{\langle s_{4},\{0.8,0.3,0.4\}\rangle }\)

\(\frac{0.4}{\langle s_{1},\{0.4,0.6,0.9\}\rangle }\)

\(\frac{0.4}{\langle s_{2},\{0.4,0.6,0.9\}\rangle }\)

\({\mathcal {N}}_4\)

\(\frac{0.8}{\langle s_{3},\{0.7,0.4,0.4\}\rangle }\)

\(\frac{0.6}{\langle s_{5},\{0.8,0.4,0.5\}\rangle }\)

\(\frac{0.7}{\langle s_{4},\{0.7,0.4,0.5\}\rangle }\)

\(\frac{0.6}{\langle s_{3},\{0.6,0.4,0.4\}\rangle }\)

\(\frac{0.6}{\langle s_{1},\{0.5,0.4,0.9\}\rangle }\)

\(\frac{0.5}{\langle s_{2},\{0.2,0.3,0.9\}\rangle }\)

\({\mathcal {N}}_5\)

\(\frac{0.8}{\langle s_{2},\{0.8,0.4,0.4\}\rangle }\)

\(\frac{0.5}{\langle s_{4},\{0.9,0.4,0.5\}\rangle }\)

\(\frac{0.9}{\langle s_{3},\{0.6,0.3,0.2\}\rangle }\)

\(\frac{0.6}{\langle s_{3},\{0.7,0.4,0.4\}\rangle }\)

\(\frac{0.6}{\langle s_{2},\{0.4,0.5,1.0\}\rangle }\)

\(\frac{0.8}{\langle s_{1},\{0.4,0.6,0.9\}\rangle }\)