Lecture 7 - Neuro- Fuzzy System - Takagi and Sugeno’s approach

While Mamdani (from Lecture 6) outputs fuzzy shapes that need to be geometrically "defuzzified," the Takagi-Sugeno model uses mathematical linear equations to determine the output of each rule. This makes it highly efficient and very popular in machine learning algorithms.

This is a breakdown of the 6-layer ANFIS architecture and the step-by-step numerical example covered in the slides.

The 6-Layer ANFIS Architecture

Unlike the 5-layer Mamdani model, ANFIS uses 6 layers to process the data:

Layer 1 (Input Layer): Passes the crisp inputs forward unchanged using a linear transfer function.
Layer 2 (Fuzzification Layer): Calculates the membership values ( $μ$ ) for the inputs based on their respective linguistic terms (e.g., Low, High).
Layer 3 (Firing Strength): Key Difference from Mamdani. Instead of taking the minimum value, this layer calculates the product (multiplication) of the $μ$ values to determine the firing strength ( $w$ ) of each rule.
Layer 4 (Normalization): Calculates the normalized firing strength ( $\bar{w}$ ) by dividing a rule's strength by the total sum of all firing strengths.
Layer 5 (Rule Output): Applies the first-order Sugeno equation to calculate the actual output ( $y$ ) of each specific rule: $y_{i} = a_{i} I_{1} + b_{i} I_{2} + c_{i}$ (where $a, b, c$ are predefined coefficients). It then multiplies this output by the normalized firing strength from Layer 4.
Layer 6 (Overall Output): Sums up all the weighted outputs from Layer 5 to produce the final single crisp number.

Step-by-Step Numerical Example Walkthrough

We want to find the deviation in prediction for a specific training scenario.

The Setup:

Inputs: $I_{1} = 1.1$ , $I_{2} = 6.0$ .
Target Output: $O = 2.3$ .
Base Widths ( $d_{1}, d_{2}$ ): Using the provided min-max normalization formula on the normalized weights ( $0.3$ and $0.5$ ), the real base widths are calculated as $d_{1} = 1.01$ and $d_{2} = 5.0$ .

Layer 1

The inputs pass directly through.

$1_{O 1} = I_{1} = 1.1$
$1_{O 2} = I_{2} = 6.0$

Layer 2 (Fuzzification)

Based on the modified right-angled triangle graphs (using the calculated $d_{1}$ and $d_{2}$ base widths):

For $I_{1} = 1.1$ :
- $μ_{L W} = 0.900990$
- $μ_{H} = 0.099009$
For $I_{2} = 6.0$ :
- $μ_{S M} = 0.8$
- $μ_{L R} = 0.2$

Layer 3 (Firing Strengths)

We multiply the membership values to find the strength ( $w$ ) of the 4 activated rules:

$w_{1}$ (LW and SM) = $0.900990 \times 0.8 = 0.720792$
$w_{2}$ (LW and LR) = $0.900990 \times 0.2 = 0.180198$
$w_{3}$ (H and SM) = $0.099009 \times 0.8 = 0.079207$
$w_{4}$ (H and LR) = $0.099009 \times 0.2 = 0.019802$

Layer 4 (Normalization)

We normalize the weights by dividing each $w$ by the sum of all $w$ 's.

(Note: Because of how these specific triangles are structured, the sum of $w_{1} + w_{2} + w_{3} + w_{4}$ happens to equal $1.0$ . Therefore, dividing by $1.0$ means the normalized weights ( $\bar{w}$ ) are identical to the raw weights in this specific instance).

${\bar{w}}_{1} = 0.720792$
${\bar{w}}_{2} = 0.180198$
${\bar{w}}_{3} = 0.079207$
${\bar{w}}_{4} = 0.019802$

Layer 5 (Rule Outputs)

First, we use the rule coefficients ( $a_{i}, b_{i}, c_{i}$ ) from the provided table to calculate the mathematical output ( $y$ ) for each rule:

$y_{1} = (0.2 \times 1.1) + (0.3 \times 6.0) + 0.10 = 2.12$
$y_{2} = (0.2 \times 1.1) + (0.4 \times 6.0) + 0.11 = 2.73$
$y_{3} = (0.3 \times 1.1) + (0.3 \times 6.0) + 0.13 = 2.26$
$y_{4} = (0.3 \times 1.1) + (0.4 \times 6.0) + 0.14 = 2.87$

Next, Layer 5 multiplies these rule outputs by their respective normalized firing strengths:

$5_{O 1} = {\bar{w}}_{1} \times y_{1} = 0.720792 \times 2.12 = 1.528079$
$5_{O 2} = {\bar{w}}_{2} \times y_{2} = 0.180198 \times 2.73 = 0.491941$
$5_{O 3} = {\bar{w}}_{3} \times y_{3} = 0.079207 \times 2.26 = 0.179008$
$5_{O 4} = {\bar{w}}_{4} \times y_{4} = 0.019802 \times 2.87 = 0.056832$

Layer 6 (Final Calculation)

The overall predicted output is simply the sum of the Layer 5 values:

6_{O 1} = 1.528079 + 0.491941 + 0.179008 + 0.056832 = 2.255860

Finally, to find the deviation in prediction, we subtract our predicted output from the known target output:

Target Output = $2.3$
Predicted Output = $2.255860$
Deviation: $2.3 - 2.255860 = 0.044140$