Intelligent Systems II · TEE · papers

Q5(b) Mean of maxima and centroid of trapezoids [10]

Re-examination · 25 Jun 2024

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

Question. Q5(b) [10 marks · 2024-06-25]

For the given membership functions determine the crisp output using (1) Mean of Maxima and (2) Centroid / Center of Area.

Swipe across the diagram to read its labels ↔
First printed membership function: height 0.75, corners x=0,2,3,4.
Diagram: First printed membership function: height 0.75, corners x=0,2,3,4.. Labels: First printed membership function: height0.75, corners x=0,2,3,4. | 0 | 0.7 | 1.4 | 2.1 | 2.8 | 3.5 | 4.2 | 4.9 | 5.6 | 6.3 | 7 | 0 | 0.25 | 0.5 | 0.75 | 1 | First graph | μ(x)
First printed membership function: height 0.75, corners x=0,2,3,4.
Swipe across the diagram to read its labels ↔
Second printed membership function: height 1, corners x=0,3,4,7.
Diagram: Second printed membership function: height 1, corners x=0,3,4,7.. Labels: Second printed membership function: height1, corners x=0,3,4,7. | 0 | 0.7 | 1.4 | 2.1 | 2.8 | 3.5 | 4.2 | 4.9 | 5.6 | 6.3 | 7 | 0 | 0.25 | 0.5 | 0.75 | 1 | Second graph | μ(x)
Second printed membership function: height 1, corners x=0,3,4,7.

Answer

Given

Two separate printed graphs; graph 1 corners(0,0),(2,.75),(3,.75),(4,0); graph 2 corners(0,0),(3,1),(4,1),(7,0).

To find

MOM and centroid for each graph.

Formula and why it applies

MOM averages the x-values where membership reaches its greatest height. For a continuous flat maximum[a, b], MOM=(a+b)/2. Centroid uses all area: divide the sum of area moments by total area.

z_{MOM}=\frac{a+b}{2},\qquad z_{COA}=\frac{\int x\mu(x)dx}{\int\mu(x)dx}=\frac{\sum A_ic_i}{\sum A_i}LaTeX: z_{MOM}=\frac{a+b}{2},\qquad z_{COA}=\frac{\int x\mu(x)dx}{\int\mu(x)dx}=\frac{\sum A_ic_i}{\sum A_i}

Solution

Step 1. MOM of the first graph

Maximum height.75 occurs throughout[2,3].

z_{MOM,1}=(2+3)/2=2.5LaTeX: z_{MOM,1}=(2+3)/2=2.5
Step 2. Centroid of the first graph
PieceAreax centroidMoment
Triangle 0–2½×2×.75=.75(0+2+2)/3=4/31
Rectangle 2–31×.75=.752.51.875
Triangle 3–4½×1×.75=.375(3+3+4)/3=10/31.25
z_{COA,1}=\frac{1+1.875+1.25}{.75+.75+.375}=\frac{4.125}{1.875}=2.2LaTeX: z_{COA,1}=\frac{1+1.875+1.25}{.75+.75+.375}=\frac{4.125}{1.875}=2.2
Step 3. MOM of the second graph
z_{MOM,2}=(3+4)/2=3.5LaTeX: z_{MOM,2}=(3+4)/2=3.5
Step 4. Centroid of the second graph
PieceAreax centroidMoment
Triangle 0–31.523
Rectangle 3–413.53.5
Triangle 4–71.557.5
z_{COA,2}=\frac{3+3.5+7.5}{1.5+1+1.5}=\frac{14}{4}=3.5LaTeX: z_{COA,2}=\frac{3+3.5+7.5}{1.5+1+1.5}=\frac{14}{4}=3.5

The second graph is symmetric about 3.5, so its area centroid and plateau midpoint coincide. The first graph is asymmetric;2.2 and 2.5 need not agree.

Final answer. Graph 1:MOM 2.5, centroid 2.2. Graph 2:MOM 3.5, centroid 3.5.