Let A and B be two fuzzy integers defined as: A = {(1.0.3), (2, 0.6), (3, 1),…
2015
Let A and B be two fuzzy integers defined as:
A = {(1.0.3), (2, 0.6), (3, 1), (4, 0.7), (5, 0.2)}
B = {(10, 0.5), (11, 1), (12, 0.5)}
Using fuzzy arithmetic operation given by
\(\mu_{A+B^{(Z)}} = \underset{x+y=z}{\oplus} (\mu_A (x) \otimes \mu_B(y))\)
\(f(A+B)\) is _____ . \(\left[ \text{Note :} \quad \begin{array}{c} \oplus \equiv \max \\ \otimes \equiv \min \end{array} \right] \)
- A.
{(11, 0.8), (13, 1), (15, 1)}
- B.
{(11, 0.3), (12, 0.5), (13, 1), (14, 1), (15, 1), (16, 0.5), (17, 0.2)}
- C.
{(11, 0.3), (12, 0.5), (13, 0.6), (14, 1), (15, 1), (16, 0.5), (17, 0.2)}
- D.
{(11, 0.3), (12, 0.5), (13, 0.6), (14, 1), (15, 0.7), (16, 0.5), (17, 0.2)}
Attempted by 21 students.
Show answer & explanation
Correct answer: D
Method: Use the max-min convolution. For each sum z, μ(z) = max over all x+y=z of min(μA(x), μB(y)).
For z = 11: only decomposition 1 + 10 → min(0.3, 0.5) = 0.3, so μ(11) = 0.3.
For z = 12: decompositions 2 + 10 → min(0.6, 0.5) = 0.5, and 1 + 11 → min(0.3, 1) = 0.3. Take max: μ(12) = 0.5.
For z = 13: decompositions 3 + 10 → min(1, 0.5) = 0.5, 2 + 11 → min(0.6, 1) = 0.6, 1 + 12 → min(0.3, 0.5) = 0.3. Take max: μ(13) = 0.6.
For z = 14: decompositions 4 + 10 → min(0.7, 0.5) = 0.5, 3 + 11 → min(1, 1) = 1, 2 + 12 → min(0.6, 0.5) = 0.5. Take max: μ(14) = 1.
For z = 15: decompositions 5 + 10 → min(0.2, 0.5) = 0.2, 4 + 11 → min(0.7, 1) = 0.7, 3 + 12 → min(1, 0.5) = 0.5. Take max: μ(15) = 0.7.
For z = 16: decompositions 5 + 11 → min(0.2, 1) = 0.2, 4 + 12 → min(0.7, 0.5) = 0.5. Take max: μ(16) = 0.5.
For z = 17: only decomposition 5 + 12 → min(0.2, 0.5) = 0.2, so μ(17) = 0.2.
Final result: {(11, 0.3), (12, 0.5), (13, 0.6), (14, 1), (15, 0.7), (16, 0.5), (17, 0.2)}