# Q3(b) Center of sums defuzzification [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-final/q3b

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-final--q3b/

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

**Question. Q3(b)** [10 marks · 2024-04-17]

For the given membership functions, determine the defuzzified crisp output using the Center of Sums method.

Swipe across the diagram to read its labels ↔

![Source graph redrawn: A1 corners (1,0),(3,0.5),(7,0.5),(8,0); A2 corners (3,0),(4,0.3),(8,0.3),(9,0).](https://deep-learning-study-room.vercel.app/is2/fig/legacy24-centre-sums.6ec254a338.svg)

Diagram: Source graph redrawn: A1 corners (1,0),(3,0.5),(7,0.5),(8,0); A2 corners (3,0),(4,0.3),(8,0.3),(9,0).. Labels: Source graph redrawn: A1 corners (1,0),(3,0.5),(7,0.5),(8,0); A2 corners (3,0),(4,0.3),(8,0.3),(9,0). | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 0 | 0.25 | 0.5 | 0.75 | 1 | A1 h=0.5 | A2 h=0.3 | μ(x)

Source graph redrawn: A1 corners (1,0),(3,0.5),(7,0.5),(8,0); A2 corners (3,0),(4,0.3),(8,0.3),(9,0).

### Answer

#### Given

A1 has height 0.5 and support [1,8], plateau [3,7]; A2 has height 0.3 and support [3,9], plateau [4,8].

#### To find

Center of sums of the two clipped output sets.

#### Formula and why it applies

Center of sums adds each set’s area and moment separately; overlapping portions are counted once per set. This differs from the centroid of the max-aggregated union. Split each trapezoid into triangles and a rectangle.



$$
z_{COS}=\frac{A_1c_1+A_2c_2}{A_1+A_2}=\frac{M_1+M_2}{A_1+A_2}
$$



#### Solution

##### Step 1. Area and moment of A1

| Piece | Area | Centroid x | Area × centroid |
| --- | --- | --- | --- |
| Left triangle [1,3] | ½×2×0.5=0.5 | (1+3+3)/3=7/3 | 7/6 |
| Rectangle [3,7] | 4×0.5=2 | (3+7)/2=5 | 10 |
| Right triangle [7,8] | ½×1×0.5=0.25 | (7+7+8)/3=22/3 | 11/6 |



$$
A_1=0.5+2+0.25=2.75,\quad M_1=7/6+10+11/6=13,\quad c_1=13/2.75=52/11
$$



##### Step 2. Area and moment of A2

| Piece | Area | Centroid x | Moment |
| --- | --- | --- | --- |
| Left triangle [3,4] | 0.15 | 11/3 | 0.55 |
| Rectangle [4,8] | 1.2 | 6 | 7.2 |
| Right triangle [8,9] | 0.15 | 25/3 | 1.25 |



$$
A_2=0.15+1.2+0.15=1.5,\quad M_2=0.55+7.2+1.25=9,\quad c_2=6
$$



##### Step 3. Weight centroids by areas



$$
z_{COS}=\frac{13+9}{2.75+1.5}=\frac{22}{4.25}=\frac{88}{17}\approx5.176471
$$



Do not simply average 52/11 and 6: A1 has more area and must receive more weight. Do not subtract overlap for Center of Sums.

**Final answer.** Center of Sums = 88/17 ≈ 5.176471.