Dirt = 80% Grease = 90%

X = 80, Y = 90

Find out req. wash time.

According to given input values, they fall in the category of, $M^*$, large dirt a large grease.

The membership function for the respective one,

$M_{MD} \ ^{(x)} = \frac{100 – 80}{50} = \frac{20}{50} = \frac{2}{5}$

$M_{2D} \ ^{(x)} = \frac{80 – 50}{50} = \frac{30}{50} = \frac{3}{5}$

$M_{MG} \ ^{(y)} = \frac{100 – 90}{50} = \frac{1}{5}$

$M_{LG} \ ^{(y)} = \frac{90 – 50}{50} = \frac{4}{5}$

Place the values in the Rule Base table.

$M_{MD} \ ^{(x)} \cap M_{MG} \ ^{(y)} = \frac{1}{5}$

$M_{MD} \ ^{(x)} \cap M_{LG} \ ^{(y)} = \frac{2}{5}$

$M_{LD} \ ^{(x)} \cap M_{MG} \ ^{(y)} = \frac{1}{5}$

$M_{LD} \ ^{(x)} \cap M_{LG} \ ^{(y)} = \frac{3}{5}$

Max of the four values = $\frac{3}{5} = M_{LD} \ ^{(x)} \cap M_{2G} \ ^{(y)}$

The required wash time is VL

$M_{wash \ Time} \ ^{(z)} = \frac{3}{5}$

$M_{VL} \ ^{(z)} = \frac{z – 40}{20}$

$\frac{3}{5} = \frac{z – 40}{20}$

= 52 mins.