The formation of algal solutions in the surface water strongly depends on $\mathrm{pH}$ of water, temperature, and oxygen content. T is a set of water temperatures from a lake given by $\mathrm{T}=\{50,55,60\}$ and $\mathrm{O}$ is a set of oxygen content values in water given by $\mathrm{O}=\{1,2,6\}$. The fuzzy sets of T and $\mathrm{O}$ are,

$ \begin{aligned}\\ &\mathrm{T}=\{0.7 / 50+0.8 / 55+0.9 / 60\} \\\\ &\mathrm{O}=\{0.1 / 1+0.6 / 2+0.8 / 6\}\\ \end{aligned}\\ $

Given $\mathrm{I}=\{0.5 / 50+1 / 55+0.7 / 60\}$ and $\mathrm{R}=\mathrm{T} \times \mathrm{O}$, find i. $A=I \circ R$

ii. $\quad \mathrm{B}=\mathrm{I} \bullet \mathrm{R}$

Solution:

$ \begin{aligned} &T=\{(50,0.7),(55,0.8),(60,0.9)\} \\ &O=\{(1,0.1),(2,0.6),(6,0.81\} \\ &I=\{(50,0.5),(55,1),(60,0.7)\} \end{aligned} $

(i) $R=T \times 0$ using cauterian product.

$ \therefore R=50\left[\begin{array}{lll} 1 & 2 & 6 \\ 0.1 & 0.6 & 0.7 \\ 0.1 & 0.6 & 0.8 \\ 00.1 & 0.6 & 0.8 \end{array}\right]_{3 \times 3} $

ii) $S=I_0 R \quad$ max-min here,

I = I * I

$\therefore I=50\left[\begin{array}{lll}50 & 55 & 60 \\ 55 & 0.5 & 0.5 \\ 0.5 & 1 & 0.7 \\ 0.5 & 0.7 & 0.7\end{array}\right]_{3 \times 3}$

Now,

Let, $I O R=\left[\begin{array}{lll}M_{11} & M_{12} & M_{13} \\ M_{21} & M_{22} & M_{23} \\ M_{31} & M_{32} & M_{33}\end{array}\right] 3 \times 3$

$ \begin{aligned} &M_{11}=\operatorname{Max}\{\min (0.5,0.1), \operatorname{Min}(0.5,0.1), \\ &=\operatorname{Max}\{0.1,0.1,0.1\} \\ &=\operatorname{Max}\{0.1,0.1,0.1\} \\ &=0.1 \\ & \end{aligned}\\ $

$ \begin{aligned} M_{13} &=\operatorname{Max}\{0.5,0.5,0.5\} \\ &=0.5 \\ M_2 &=\operatorname{Max}\{0.1,0.1,0.1\} \\ &=0.1 \end{aligned}\\ $

$ \begin{aligned} M_{22} &=\operatorname{Max}\{0.5,0.6,0.6\} \\ &=0.6 \\ M_{41} &=M_{a x}\{0.5,0.7,0.7\} \\ M_{32} &=\operatorname{Max}\{0.5,0.6,0.6\} \\ &=0.6 \\ M_{33} &=\operatorname{Max}\{0.5,8.7,0.7\} \\ &=0.7 \\ M_{23} &=\operatorname{Max}\{0.5,0.8,0.7\} . \\ &=0.8 \end{aligned} $