Unit – 5
Fuzzy Arithmetic
A fuzzy number is a generalization of a regular, real number in the sense that it does not refer to one single value but rather to a connected set of possible values, where each possible value has its own weight between 0 and 1. This weight is called the membership function.
1. Set A must be normal Fuzzy set;
2. of A must be closed interval foe every in (0,1]
3. Support and strong of A must be bounded.
Example 1: It is necessary to compare 2 sensors based upon their detection levels & gain settings the tails of gain settings & sensor detection level with a obtained item being monitored providing typical membership values to reprocess the detection levels for each sensor is given in tales.
Find following membership function:
a) b) c)
Solution:
Gain setting | Detection level of sensor 1(D1) | Detection level sensor 2(D2) |
0 | 0 | 0 |
10 | 0.2 | 0.35 |
20 | 0.35 | 0.25 |
30 | 0.65 | 0.8 |
40 | 0.85 | 0.95 |
50 | 1 | 1 |
= max {}
b) = min {}
c) = 1-
Definition: If aA is a fuzzy set defined on universal set X then its fuzzy cardinality is denoted and defined as
Where is membership grade from given fuzzy set A and is a scalar cardinality (total number of elements present in crisp set) of corresponding
Example 1: Find the fuzzy cardinality of A defined as follows:
Solution: Here X= {0,1,2,3,4,5} s and its scalar cardinality are
{0,2,4,5} | ||
{0,1,2,3,4,5} | ||
{2,5} | ||
{0,2,3,4,5} | ||
{2,4,5} | ||
{5} |
Thus,
Note: Fuzzy cardinality of Fuzzy set is a fuzzy set
Example 2: Find fuzzy cardinality of fuzzy set A and B whose membership function are given as follows , B(x)= where x
Solution:
X | 0 | 1 | 2 | 3 | 4 |
, | 0 | 0.5 | 0.67 | 0.75 | 0.8 |
B(x)= | 1 | 0.9 | 0.8 | 0.7 | 0.6 |
{0,1,2,3,4} | ||
{1,2,3,4} | ||
{2,3,4} | ||
{3,4} | ||
{4} |
{0} | ||
{0,1} | ||
{0,1,2} | ||
{0,1,2,3} | ||
{0,1,2,3,4} |
Thus,
Thus,
Moving from intervals we can define arithmetic on fuzzy numbers based on principles of interval Arithmetic.
Let A and B denote fuzzy numbers and let * denote any of the four basic arithmetic operations. Then, we define a fuzzy set on R, A*B by defining its alpha-cut as:
Arithmetic Operations on Intervals: Arithmetic operations on fuzzy intervals satisfy following useful properties:
Properties:
Examples Illustrating interval-valued arithmetic operations:
Applying the extension principle to arithmetic operations, we have
Fuzzy Addition:
Fuzzy Subtraction
Fuzzy Multiplication
Fuzzy Division
Example 1:
Their
Step 1: Find and
Step 2: Use decomposition theorem which is given as and arithmetic operation on closed interval
Step 3: Find membership function.
Example 2: The membership function of two fuzzy numbers A and B are given as follows
A(x) | = | -1 < x |
B(x) | = | |
= | = | ||||
= 0 | = 0 | otherwise |
Find A+B and A-B
Solution:
Step 1: Find and
For -1 < x
By def. of ,
A(x)
For
= [2-1, 3-2 for all
For
By def. of cut,
B(x)
For 3 < x
B(x)
= [2+1, 5-2 for all
Step 2: By decomposition theorem & arithmetic operation on closed interval we find as follows.
Step 3: Now find membership function
Let 4
| Let 8-4 |
Thus, (A+B) (x) = ; 0 < x
= 4
= 0; Otherwise
To find A-B
Step 4: By decomposition theorem and arithmetic operation on closed interval we find follows
= [
= [4 for all
Step 5: Now find membership function to define fuzzy numbers A-B
Let 4
| Let 2-4
|
Thus,
Algorithm:
Step 1: Find
Step 2: Use decomposition theorem which is given as
Let
A+X = B
Taking alpha cut on both side
Step 3: Find membership function
Example 1: The membership function of two fuzzy numbers A and B are
A(x) | = x-1 | 1 < x |
B(x) | = | |
= | = | ||||
= 0 | = 0 | otherwise |
Solve A+X =B
Solution:
Let
A+X = B
Taking alpha cut on both side
Check
0.1 | 12.8 | 56.9 |
0.6 | 26.8 | 46.4 |
0.8 | 32.4 | 42.2 |
Thus, hence solution exists.
Let
| Let
|
Example 2: The membership function of two fuzzy numbers A and B are
A(x) | = x-3 | 3 < x |
B(x) | = | |
= | = | ||||
= 0 | = 0 | otherwise |
Solve A.X =B
Solution:
Let
A.X = B
Taking alpha cut on both side
Check
0.3 | 4.3636 | 6.0426 |
0.7 | 4.7568 | 5.4884 |
0.9 | 4.9231 | 5.1707 |
Thus, hence solution exists.
Let
| Let
|
Reference