Fuzzy Mathematics- SEMA6B
(those who joined in July 2017 and afterwards)
Unit-I: Crisp sets - Fuzzy sets - Basic Types-Basic concepts - characteristics
and significance of paradigm shift.
Unit II: Additional properties of α-cuts - representation of fuzzy sets extension principle for fuzzy sets.
Unit III: Fuzzy sets operation - Fuzzy complement - Fuzzy intersection- t−norms
Fuzzy union- t− conorm-combination of operation- aggregation operation
Unit IV: Fuzzy numbers - Linguistic variable -Arithmetic operation on intervalsArithmetic operation of fuzzy numbers - Lattice of fuzzy numbers fuzzy equations.
Unit V: Fuzzy decision making - Individual decision making - Multi person
decision making - Fuzzy linear programming.
Text Book: George J. Klir and Bo Bo Yuan - Fuzzy sets and Fuzzy logic theory
Applications- Prentice Hall of India, 2002, New Delhi.
Books for Reference:
1. George J. Klir and Tina A. Folger - Fuzzy sets, uncertainty and Information
- Prentice Hall of India, 2002, New Delhi.
Contents
1 Unit-I
1
1.1
Crisp set-An overview . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Fuzzy sets-Basic Types . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.3
Fuzzy sets and Basic Concepts . . . . . . . . . . . . . . . . . . . . . . 11
1.4
Characteristics and Significance of Paradigm Shift . . . . . . . . . . . 14
2 Unit-II
17
2.1
Properties of α−cuts . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2
Representation of fuzzy sets . . . . . . . . . . . . . . . . . . . . . . . 21
2.3
Extension Principle of Fuzzy Sets . . . . . . . . . . . . . . . . . . . . 24
3 Unit-III
29
3.1
Types of Operation and Fuzzy complement . . . . . . . . . . . . . . . 29
3.2
Fuzzy Intersection: t-norms . . . . . . . . . . . . . . . . . . . . . . . 36
3.3
Fuzzy Union: t-conorm . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.4
Combination of Operations . . . . . . . . . . . . . . . . . . . . . . . . 46
3.5
Aggregation Operation . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4 Unit-IV
55
4.1
Fuzzy Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.2
Arithmetic operation on Fuzzy interval . . . . . . . . . . . . . . . . . 57
4.3
Arithmetic Operation on Fuzzy Numbers . . . . . . . . . . . . . . . . 59
4.4
Lattice of fuzzy numbers . . . . . . . . . . . . . . . . . . . . . . . . . 62
4.5
Fuzzy equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
5 Unit-V
71
5.1
Individual decision making problem . . . . . . . . . . . . . . . . . . . 71
5.2
Multi-person decision making problem . . . . . . . . . . . . . . . . . 73
5.3
Fuzzy Linear Programming problem . . . . . . . . . . . . . . . . . . 77
Chapter 1
Unit-I
1.1
Crisp set-An overview
Zadeh introduced the concept of the theory of fuzzy sets. Fuzzy sets are
sets with boundaries that are not precise the membership in a fuzzy sets. It is not
a matter of affirmation or denial but rather a matter of degree.
When A is a fuzzy set and x is a relevant object, the proposition ”x is a member
of A” is not necessarily true or false as required by the two valued logic, but it may
be true only to some degree, that the degree to which x is actually the member of A.
It is most common but not required to express the degree of membership in fuzzy
sets as well as degrees of truth of the associated proposition by number in the closed
interval [0,1]. The extreme values in this interval 0 or 1 represent respectively the
total deneal and affirmation of the membership in a given fuzzy set.
Some fundamental results
1. Z = {. . . . . . , −2, −1, 0, 1, 2, . . . . . . } - set of all integers.
2. N = {1, 2, 3, . . . . . . } - set of all positive integers
3. N0 = {0, 1, 2, 3, . . . . . . }-set of all non negative integers.
4. Nn = {1, 2, 3, . . . . . . , n} - set of all n positive integers.
5. R- set of all real numbers
6. R+ - set of all non negative integers.
There are 3 basic method by which sets can be defined within the given universal
set X.
1
1.1. CRISP SET-AN OVERVIEW
(i) list method
This set is defined by naming all its members. This method can be used only
for finite sets. Set A whose members are a1 , a2 , . . . , an are usually written as
A = {a1 , a2 , . . . , an }.
(ii) Rule Method
This set is defined by a property satisfy by its members. A common notation
expressing this method is A = {x/p(x)} where the symbol / denotes the phrase
”such that” and p(x) designates a proposition of the form : 0 x has the property
p0 . i.e., A is defined by this notation as the set of all elements of x for which
the proposition p(x) is true.
(iii) Characteristic function
A set is defined by a function is usually called a characteristic function that
declares which element of x are members of a set A and which
 are not members.
1 if x ∈ A
Set A is defined by its characteristic function χA (x) =
0 if x ∈
/A
i.e., Characteristic function maps elements of x to elements of a set {0, 1}
which is normally expressed as χA (x) : X → {0, 1}
For each x belongs to X, when χA (x) = 1 it declares x is a member of A.
When χA (x) = 0 it declares x is not a member of A.
Definition 1.1.1. A set whose elements are themselves set is often referred to as
family of sets. It can be defined as the form {Ai : i ∈ Z} where i, Z are called set
index and the index set respectively. Because the index i is used to referred the
set Ai , the family of sets is also called index set. Family of sets is usually denoted
by script capital letters. A = {a1 , a2 , . . . , an }.
Definition 1.1.2. The family of all subsets of a given set A is called power set of
A and is denoted by P (A). The family of all subsets of P (A) is called second order
power set of A and is denoted by P 2 (A) which stands for P (P (A)). Similarly higher
order power sets P 3 (A), P 4 (A), . . . are defined.
The number of members of a finite set of A is called cardinality of A and is
denoted by |A|. When A is finite then |P (A) |= 2|A| , |P 2 (A) |= 22|A| , . . .
2
1.1. CRISP SET-AN OVERVIEW
Definition 1.1.3. The relative complement of a set A with respect to B is the
set containing all the members of B that are not members of A. This can be written
as B ∼ A. Thus B ∼ A = {x/x ∈ B and x ∈
/ A}.
If the set B is the universal set, the complement is absolute and usually denoted
ˉ The absolute complement is always involutive. i.e., taking the complement
by A.
of a complement yields original set. ie., Aˉ = A.
The absolute complement of the empty set is equal to universal set and the
ˉ = φ.
complement of universal set is equal to empty set. i.e., φˉ = X, X
Definition 1.1.4. The union of sets A and B is the set containing all the elements
that belong either to set A alone or to the set B alone or to both the set A and B.
This is denoted by A ∪ B. Thus A ∪ B = {x/x ∈ A or x ∈ B}. The union operation
can be generalised for any number of sets. For a family of sets {Ai /i ∈ I}, this can
be defined as ∪i∈I Ai = {x/x ∈ Ai , for some i ∈ I}.
Definition 1.1.5. The intersection of sets A and B is the set containing all
the elements belong to both the sets A and B and is denoted by A ∩ B. Thus
A ∩ B = {x/x ∈ A and x ∈ B}.
The generalisation of intersection is the
family of sets {Ai /i ∈ I} and defined as ∩i∈I Ai = {x/x ∈ Ai , for all i ∈ I}.
Fundamental Properties of Crisp set Operation (U.Q)
1. Involution:
Aˉ = A
2. Commutativity: A ∪ B = B ∪ A and A ∩ B = B ∩ A
3. Associativity:
A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C
4. Distributivity: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) and
5. Idempotive:
6. Absorption:
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
A ∪ A = A and A ∩ A = A
A ∪ (A ∩ B) = A and A ∩ (A ∪ B) = A
7. Absorption by X and φ : A ∪ X = X, A ∩ φ = φ
8. Identity:
9. Law of contradiction:
A ∪ φ = A, A ∩ X = A
A ∩ Aˉ = φ
10. Law of excluded middle: A ∪ Aˉ = X
ˉ and A ∪ B=Aˉ ∩ B
ˉ
11. Demorgan’s law: A ∩ B=Aˉ ∪ B
3
1.1. CRISP SET-AN OVERVIEW
Definition 1.1.6. Elements of the power set P (x) of a universal set X can be
ordered by the set inclusion c. This ordering which is only partial forms a lattice in
which the V -Join(least upper bound / supremum) and Λ-meet(greatest lower bound
/ infimum) of any pair of sets A, B ∈ P (x) is given by A ∪ B and A ∩ B respectively.
This lattice is distributive and complemented. It is usually called boolean lattice
or boolean algebra.
Remark 1.1.7. The connection between the two formulations of this lattice
(P (x), ⊆) and (P (x), ∪, ∩) is specificated by the following is equivalent A ⊆ B
iff A ∪ B = B or A ∩ B = A for any A, B ∈ P (x).
Definition 1.1.8. Any two sets have no common members are called disjoint.
i.e., every pair of disjoint sets A and B satisfy the equation A ∩ B = φ.
Definition 1.1.9. A family of pair ways disjoint non empty subset of a set A is
called partition on A if the union of these subsets yields the original set A. We
denote the partition on A by the symbol π(A). Formally π(A) = {Ai /i ∈ I and
i ⊆ A} where Ai 6= φ is a partition on A iff Ai ∩ Aj = φ for each pair i, j ∈ I, i 6= j
and ∪i∈I Ai = A.
Definition 1.1.10. Members of a partition π(A) which are subsets of A are usually
referred to as blocks of the partitions. Each members of A belong to one and only
one block of π(A).
Definition 1.1.11. Given two partitions π1 (A) and π2 (A). We say that π1 (A) is
called refinement of π2 (A) iff each block of π1 (A) is included in some block of π2 (A).
The refinement relation on the set of all partitions of A, π(A) which is denoted by
≤ is a partial ordering. The pair (π(A), ≤) is a lattice referred to as the partition
lattice of A.
Let A = {A1 , A2 , . . . An } be the family of sets such that Ai ⊆ Ai+1 for all
i = 1, 2, . . . n − 1. Then A is called nested family and the sets A1 and An is called
inner most set and outer most set respectively.
Definition 1.1.12. The cartesian product of two sets A and B is the set of of all
ordered pairs such that the first element of each pair is a member of A and second
element is a member of B. Formally
4
1.1. CRISP SET-AN OVERVIEW
A × B = {< a, b > /a ∈ A, b ∈ B} where A × B denote the cartesian product.
Clearly if A 6= B and A, B are non empty then A × B 6= B × A.
The cartesian product of family {A1 , A2 , . . . , An } of sets is the set of all n-tuples
(a1 , a2 , . . . , an ) such that ai ∈ Ai , i ∈ I and written as A1 × A2 × ∙ ∙ ∙ × An or
×1≤i≤n Ai . Thus, ×1≤i≤n Ai = {(a1 , a2 , . . . , an )/ai ∈ Ai , i = 1, 2, . . . , n}. Subset of
partition product are called relations.
Definition 1.1.13. A set whose elements can be labelled by positive integers are
called countable sets. If such labelling is not possible then the set is called uncountable set.
For eg., the set {a/0 ≤ a ≤ 1} is uncountable.
0 0
a is real number. Every
uncountable set is infinite. Countable set can be classified into finite and countably
infinite set.
Definition 1.1.14. A set A is called convex iff for every pair of points R =
{ri /i ∈ Nn } and S = {si /i ∈ Nn } in A and every real number λ ∈ [0, 1], the point
t = {λri + (1 − λ)si }, i ∈ Nn is also in A.
Definition 1.1.15. In otherwords, a set A in Rn is convex iff for every pair of points
r and s in A, all points located on the straight line segment connecting r and s are
also in A.
for eg., The convex and non convex sets in R2 are given as below.
Here A1 − A5 are convex sets and A6 − A9 are non convex sets.
5
1.1. CRISP SET-AN OVERVIEW
A1
A2
A4
A3
A6
A5
A7
A9
A8
In R any set is defined by the single interval of real numbers is convex and any
set defined by more than one interval that does not contain some point between the
interval is not convex. For eg., the set A = [0, 2] ∪ [3, 5] is not convex.
Since let r = 1, s = 4, λ = 0.4 then λr + (1 − λ)s = (0.4)1 + (1 − 0.4)4 = 2.8 ∈
/A
Definition 1.1.16. Let A denote the set of all real numbers. If there is a real
number r such that x ≥ r for every x ∈ A then r is called upper bound of A and
we say that A is bounded above by r.
Let B be the set of all real numbers, B ⊆ R. If there is a real number s such
that x ≤ s for every x ∈ B then s is called lower bound of B and we say that B
is bounded below by s.
Definition 1.1.17. For any set of real numbers A that is bounded above, a real
number r is called supremum of A if
(i) r is upper bound of A
(ii) No number less than r is an upper bound of A.
For any set of real numbers B that is bounded below, a real number s is called
infimum of B if
(i) s is lower bound of B
(ii) No number greater than s is lower bound of B.
6
1.2.
1.2
FUZZY SETS-BASIC TYPES
Fuzzy sets-Basic Types
The characteristic functions can be generalised such that the values assigned to the
elements of universal set fall within a specified range and indicate the membership
grade of these elements in the set.
Largest value denote the higher degrees of set membership.
Such a function is called membership function and the set is defined by it is
called fuzzy set.
Note: The most commonly used range of values of membership function is the
unit interval [0,1]. In this case each membership function maps elements of a given
universal set X which is always a crisp set include real numbers [0,1].
Definition 1.2.1. Two distinct notations are commonly used to denote membership
function. One of them is the membership function of fuzzy set A is denoted by
μA . i.e., μA : X → [0, 1]. The other function is denoted by A and has the form
A : X → [0, 1].
For eg: Applying the concept of high temperature in one contest to weather
and another contest to nuclear reaction would necessarily be represented by very
different fuzzy set.
Each of fuzzy set expressed in particular form in general consecution of a class
of real number are closed to 2.
The four fuzzy set as similar in the sense that the following properties are possed
by each Ai , i ∈ N4 .
(i) Ai (2) = 1 and Ai (x) < 1 for all x 6= 2.
(ii) Ai is symmetric w.r.t. x = 2. i.e., Ai (2 + x) = Ai (2 − x), x ∈ 2
(iii) Ai (x) decreases monotonically from 1 to 0 with increasing difference |2 − x|.
These properties are necessarily inorder to properly represent given connection.
Any additional fuzzy set attempting to represent to the same conception we have
to pass them as well.
Examples of membership functions that may be used in different contexts for
characterizing fuzzy sets of real numbers close to 2.
7
1.2.
FUZZY SETS-BASIC TYPES
1
1
0.5
0
0.5
1
2
0
3
1
1
0.5
0.5
0
1
2
3
0
4
1
2
1
3
2
4
3
4
Example 1.2.2. Three fuzzy sets defined within a finite universal set that consist
of seven level of education.
0- no education, 1- elementary school, 3-High school, 3- 2 year college degree,
4- Bachelor degree, 5- Master degree, 6- Doctoral degree.
Membership function of the three fuzzy sets which attempts to capture the
concepts are little educated, highly educated and very highly educated people are
defined by the symbols ◦, •, ? respectively.
A person who has a bachelor but has no higher degree is view according to the
definition as highly educated to the degree of 0.8 and very highly educated to the
degree of 0.5. Several fuzzy set representing the concepts of such as low, medium,
high and so on are often employed to define the states of variables. Such a variable
is called fuzzy variable.
8
1.2.
FUZZY SETS-BASIC TYPES
0.9
0.8
◦
•
•
◦
?
?
0.7
0.6
?
◦
0.5
0.4
•
?
0.3
?
0.2
0.1
?
•
◦
Interval valued fuzzy sets
A membership function does not assign to each element of the universal set one
real number but a closed interval of real numbers between the identified lower and
upper bounds.
Fuzzy sets defined by membership functional of this type are called Interval
valued fuzzy sets.
These sets are defined formally by the functions of the form A : X → E([0,1])
where E([0, 1]) denote the family of all closed intervals of real numbers in [0, 1].
Note: E([0, 1] ⊂ P ([0, 1]))
For each x, A(x) represented by the segment between the curves which express
the identified lower and upper bounds. Thus A(a) = [α1 , α2 ]
9
1.2.
FUZZY SETS-BASIC TYPES
Interval valued fuzzy sets can further be generalised by allowing the intervals to
be fuzzy sets. Each interval now becomes an ordinary fuzzy set defined within the
universal set [0, 1].
The membership grade assigned to elements of the universal set by the generalised fuzzy sets are ordinary fuzzy set. These sets are referred to as fuzzy sets of
Type II.
Their membership function have the form A : X → F ([0, 1]) where F ([0, 1])
denote the set of all ordinary fuzzy sets that can be defined within the universal set
[0, 1] and F ([0, 1]) is also called power set of [0, 1].
α1
β1
β2
β3
β4
α2
α3
α4
a
b
The concept of Type II fuzzy set is illustrated in above figure where fuzzy
intervals assigned to x = a and x = b. It is assumed that membership functions of
all fuzzy intervals involved trapezoidal shapes and conficlantly each of them x fully
defined by four numbers.
For each x these numbers are produced by 4 functions represented in the above
10
1.3. FUZZY SETS AND BASIC CONCEPTS
fig. by four curves. Thus for eg: If x = a we obtain numbers α1 , α2 , α3 , α4 by which
the fuzzy interval assign to a is uniquely determined.
Similarly if x = b we obtain numbers β1 , β2 , β3 , β4 and the assigned fuzzy interval
is shown on RHS.
When we relax the requirement that membership grade must be represented by
numbers in the unit into [0, 1] and allow them to be represented by the symbol of
an arbitrary set A. i.e., atleast partially orders. We obtain fuzzy sets of another
generalised type. They are called L−Fuzzy sets and their membership function have
the form A : X → L.
A different generalisation of an ordinary fuzzy set involves fuzzy set defined
within a universal set whose elements are ordinary fuzzy set. These fuzzy sets are
known as level 2 fuzzy sets. Their membership function has the form A : F [x] → [0, 1]
where F [x] denote fuzzy power set of x.
Level 2 fuzzy sets can be generalised into level 3 fuzzy set by using a universal
set whose elements are level 2 fuzzy sets.
Higher level fuzzy sets can be obtained recursively in the same way. We can also
conceive of fuzzy sets that are of Type II and also of level 2. Their membership
function have the form A : F [x] → F ([0, 1]).
1.3
Fuzzy sets and Basic Concepts
Define (i) α−cut (ii) strong α−cut (iii) Level set (iv) support (v) Height (vi) Normal
and sub normal fuzzy set. (U.Q)
Given a fuzzy set A defined on X and any number α ∈ [0, 1]. The α−cut α A
and strong α−cut
α
α+
A are the crisp sets
A = {x/A(x) ≥ α} and
α+
A = {x/A(x) > α}
Problem 1.3.1. Let the membership function A1 , A2 , A3 representing the concept
of young, middle and old age persons defined on the interval [0,80] as follows. (U.Q)
11
1.3. FUZZY SETS AND BASIC CONCEPTS
A1 (x) =
A3 (x) =


1



when x ≤ 20
35−x
 15
when 20 < x < 35 A2 (x) =


0
when x ≥ 35


0



when x ≤ 45
x−45
0
15

60−x


15




1
when x ≤ 20 or x ≥ 60
when 20 < x < 35
when 45 < x < 60
when 35 ≤ x ≤ 45
when 45 < x < 60
15



1
1


0





 x−20
when x ≥ 60
A1
10
A2
20
30
A3
40
50
60
70
80
Discrete approximation of membership function A2 by the function D2 has the
form D2 : {0, 2, . . . , 80} → [0, 1]
x
D2 (x)
x∈
/ {22, 24, 26, . . . , 58}
0
x ∈ {22, 58}
0.13
x ∈ {26, 54}
0.4
x ∈ {24, 56}
0.27
x ∈ {28, 52}
0.53
x ∈ {32, 48}
0.8
x ∈ {30, 50}
0.67
x ∈ {34, 46}
0.93
x ∈ {36, 42}
12
1
1.3. FUZZY SETS AND BASIC CONCEPTS
◦
◦
◦
A1 = {x ∈ [0, 80]/A1 (x) ≥ 0},
A3 = {x ∈ [0, 80]/A3 (x) ≥ 0}
◦
A2 = {x ∈ [0, 80]/A2 (x) ≥ 0},
A1 = ◦ A2 = ◦ A3 = [0, 80]
α
α
α
A1 = {x ∈ [0, 80]/A1 (x) ≥ α} = {x ∈ [0, 80]/ 35−x
≥ α}
15
= {x ∈ [0, 80]/35 − x ≥ 15α} = {x ∈ [0, 80]/x ≤ 35 − 15α}
A1 = [0, 35 − 15α] = [0, 20]
A2 = {x ∈ [0, 80]/A2 (x) ≥ α} = {x ∈ [0, 80]/ x−20
≥ α, 60−x
≥ α}
15
15
= {x ∈ [0, 80]/x − 20 ≥ 15α, 60 − x ≥ 15α}
α
α
α
= {x ∈ [0, 80]/x ≥ 20 + 15α, x ≤ 60 − 15α}
A2 = [20 + 15α, 60 − 15α]
A3 = {x ∈ [0, 80]/A3 (x) ≥ α} = {x ∈ [0, 80]/ x−45
≥ α}
15
= {x ∈ [0, 80]/x − 45 ≥ 15α} = {x ∈ [0, 80]/x ≥ 45 + 15α}
A3 = [45 + 15α, 80]
α+
α+
α+
1+
A1 = {x ∈ [0, 80]/A1 (x) > α} = (0, 35 − 15α)
A2 = {x ∈ [0, 80]/A2 (x) > α} = (20 + 15α, 60 − 15α)
A3 = {x ∈ [0, 80]/A3 (x) > α} = (45 + 15α, 80)
A1 = 1+ A2 = 1+ A3 = φ.
Definition 1.3.2. The set of all levels α ∈ [0, 1] that represents distinct α−cuts of
a given fuzzy set A is called level set of A. Formally Λ(A) = {α/A(x) = α, for
some x ∈ X} .
where Λ denote the level set of fuzzy set A defined on X.
For eg. we have Λ(A1 ) = Λ(A2 ) = Λ(A3 ) = [0, 1] and
Λ(D2 (x)) = [0, 0.13, 0.27, 0.4, 0.53, 0.67, 0.8, 0.93, 1].
Note: From the definitions of α− cut and strong α− cut is that the total ordering
values of α in [0,1] is inversely preserved by set inclusion of the corresponding α−cut
as well as strong α− cut. i.e., For any fuzzy set A and pair α1 , α2 ∈ [0, 1] of distinct
values such that α1 < α2 we have
α1
A ⊇ α2 A and
This property can also be expressed as
α+
1
A ∩
α+
2
+
+
A = α2 A, α1 A ∪
α+
2
+
α1
A = α1 A.
A ∩
α+
1
+
A ⊇ α2 A.
α2
A = α2 A, α1 A ∪
α2
A = α1 A
Definition 1.3.3. A support of a fuzzy set A within a universal set X is the crisp
set that contains all the elements of X that have non-zero membership grade in A.
i.e., Support of A = {x ∈ A/A(x) 6= 0}. They are denoted by S(A) or Supp(A).
13
1.4. CHARACTERISTICS AND SIGNIFICANCE OF PARADIGM SHIFT
Hence S(A) = 0+ A and 1-cut A is called core of A.
Definition 1.3.4. The height of h(A) of a fuzzy set A is the largest membership
grade obtained by any element in the set. i.e., h(A) = sup A(x), x ∈ X.
Definition 1.3.5. A fuzzy set A is called normal when h(A) = 1. When h(A) < 1
the fuzzy set A is called sub normal.
Theorem 1.3.6. A fuzzy set A on R is convex iff A(λx1 +(1−λ)x2 ) ≥ min[A(x1 ), A(x2 )]
for all x1 , x2 ∈ R and λ ∈ [0, 1] where min denotes the minimum operator. (U.Q)
Proof. Assume that A is convex. Let α = A1 (x) ≥ A2 (x). Then x1 , x2 ∈ α A
Moreover λx1 + (1 − λ)x2 ∈ α A for any λ ∈ [0, 1].
i.e., A(λx1 + (1 − λ)x2 ) ≥ α = A(x1 ) = min[A(x1 ), A(x2 )].
Conversely assume that A(λx1 + (1 − λ)x2 ) ≥ min[A(x1 ), A(x2 )].
To prove A is convex, it is enough to prove α A is convex.
For any x1 , x2 ∈ α A, we have A(x1 ) ≥ α, A(x2 ) ≥ α
For any λ ∈ [0, 1], A(λx1 + (1 − λ)x2 ) ≥ min[A(x1 ), A(x2 )] ≥ min[α, α] = α.
So,[λx1 + (1 − λ)x2 ] ∈ α A ⇒ α A is convex.
Hence A is convex.
1.4
Characteristics and Significance of Paradigm
Shift
1. The new paradigm allows us to express irreducible observation and measurement and certainties in their various manifestation and make these uncertainties intrinsic to emprical data. Such data which are based on graded distinction among stages of their intrinsic uncertainties are processes as well and the
results obtained are more meaningful in both epistemological and pragmatic
term then their counter parts obtained by processing the usual crisp data.
2. For the reasons briefly discussed earlier the new paradigm offers far greater
resources for managing complexity and controlling computational cost. The
general experience is that the more the complex problem involved, the greater
the superiority of fuzzy method.
14
1.4. CHARACTERISTICS AND SIGNIFICANCE OF PARADIGM SHIFT
3. The new paradigm has considerly greater expressive power, consequently it
can effectively deal with the broader class of problems. In particular it has
the capability to capture and deal with meaning of sentences expressed in
natural languages. This capability of the new paradigm allows us to deal in
Mathematical terms with problems that require the use of natural language.
4. The new paradigm has the greater capability to capture human sense, reasoning, decision making and other aspects of human cognition. When employed
in machine design, the resulting machine are human friendlier.
5. The concept of a scientific paradigm was introduced by Thomas Kuln in his
highly influencial book.
6. In this book, Kuln illustrates the notion of paradigm shift by many well
documental examples from the history of science. Some of the most visible
paradigm shift are associated with the names of coppernicus, Newton (Mechanics), Biology and Mathematics.
7. The paradigm shift initiated by the concept of fuzzy set and the idea of Mathematics based upon the fuzzy sets which is currently on going has similar
characteristics to another paradigm shift recognized in the history of science.
It emerged from the need to bridge the gap between Mathematical Model and
their embrical interpretation.
8. While the Mathematician constructs the theory in terms of perfect objects, th
experimental scientists observes object which the properties demanded by the
theory are very nature of measurement be only approximately true. Mathematical deduction is not useful to the physicist if interpret rigorously.
9. When the new paradigm was proposed by Zadeh in 1965, the usual process
of paradigm shift began. The concept of fuzzy set which underline the new
paradigm was initially ignored, rediculed or attacked by many, while it was
supported only a few mostly young and not influenced.
15
1.4. CHARACTERISTICS AND SIGNIFICANCE OF PARADIGM SHIFT
Cut worthy property
Any property generalised from classical set theory into the domain of fuzzy set
theory that is preserved in all α−cuts for α ∈ [0, 1] in the classical sense is called
cut worthy property. If it is preserved in all strong α− cut for α ∈ [0, 1] then it
is called strong cut worthy property.
Note: Convexity of fuzzy sets is an example of cut worthy property and also
strong cut worthy property.
Remark 1.4.1. We have three basic operation of crisp set namely (i) complement
(ii) union (iii) intersection. The standard complement Aˉ of fuzzy set A w.r.t the
ˉ
universal set X is defined for all x ∈ X by the equation A(x)
= 1 − A(x).
ˉ
Elements of x for which A(x) = A(x)
are called equilibrium points of A. For the
standard complements, membership grades of equilibrium points are 0.5
Given 2-fuzzy set A and B. They are standard intersection A ∩ B and stan-
dard union A ∪ B defined for all x ∈ X by (A ∩ B)(x) = min[A(x), B(x)] and
(A ∪ B)(x) = max[A(x), B(x)] where min and max represents minimum and maximum operators respectively.
∗∗∗∗∗∗∗
16
Chapter 2
Unit-II
2.1
Properties of α−cuts
Theorem 2.1.1. Let A, B ∈ f (x). Then following properties hold for all α, β ∈ [0, 1]
(i)
α+
A ⊆ α A (U.Q)
(ii) α ≤ β ⇒ α A ⊇ β A and
α+
A ⊇ β+ A
(iii) α (A ∩ B) = α A ∩ α B and α (A ∪ B) = α A ∪ α B (U.Q)
(iv)α+ (A ∩ B) = α+ A ∩ α+ B and
(v) α Aˉ = (1−α)+ Aˉ
α+
(A ∪ B) = α+ A ∪ α+ B (U.Q)
Proof. (i) Let x ∈ α+ A ⇒ A(x) > α ⇒ A(x) ≥ α ⇒ x ∈ α A
Thus
α+
A ⊆ αA
(ii) α ≤ β. Let x ∈ β A ⇒ A(x) ≥ β ≥ α ⇒ A(x) ≥ α ⇒ x ∈ α A
Hence β A ⊆ α A. i.e., α A ⊇ β A
Similarly let x ∈ β+ A ⇒ A(x) > β > α ⇒ A(x) > α ⇒ x ∈ α+ A
Hence
β+
A ⊆ α+ A. i.e.,
α+
A ⊇ β+ A
(iii) Let x ∈ α (A ∩ B) ⇔ (A ∩ B)(x) ≥ α ⇔ min[A(x), B(x)] ≥ α
⇔ A(x) ≥ α, B(x) ≥ α ⇔ x ∈ α A ∩ α B
Hence α (A ∩ B) = α A ∩ α B
Let x ∈ α (A ∪ B) ⇔ (A ∪ B)(x) ≥ α ⇔ max[A(x), B(x)] ≥ α
⇔ A(x) ≥ α, B(x) ≥ α ⇔ x ∈ α A ∪ α B
Hence α (A ∪ B) = α A ∪ α B
17
2.1. PROPERTIES OF α−CUTS
(iv) Let x ∈ α+ (A ∩ B) ⇔ (A ∩ B)(x) > α ⇔ min[A(x), B(x)] > α
Hence
α+
⇔ A(x) > α, B(x) > α ⇔ x ∈ α+ A ∩ α+ B
(A ∩ B) = α+ A ∩ α+ B
Let x ∈ α+ (A ∪ B) ⇔ (A ∪ B)(x) > α ⇔ max[A(x), B(x)] > α
⇔ A(x) > α, B(x) > α ⇔ x ∈ α+ A ∪ α+ B
(A ∪ B) = α+ A ∪ α+ B
ˉ
≥ α ⇔ 1 − A(x) ≥ α ⇔1-α ≥ A(x)
(v) Let x ∈ α Aˉ ⇔ A(x)
Hence
α+
⇔ A(x) ≤ 1 − α ⇔ x ∈
/ (1−α)+ A ⇔ x ∈ (1−α)+ Aˉ
ˉ
Hence α Aˉ = (1−α)+ A.
Remark 2.1.2.
F Property (i) is proved nearly using the definition of α A and
α+
A
F Property (ii) of the previous theorem means that the set of all sequence
{α A/α ∈ [0, 1]} and {α+ A/α ∈ [0, 1]} are always monotonically decreasing w.r.t
α and consequently they are nested families of sets.
F Property (iii) and (iv) shows that the standard fuzzy union & fuzzy intersection
are both cutworthy & standard cutworthy.
F Property (v) shows that the standard fuzzy complement is neither cutworthy
nor strong cut worthy.
Theorem 2.1.3. Let Ai ∈ f (x), i ∈ I where I is an index set then
(i) a) ∪i∈I α Ai ⊆ α (∪i∈I Ai ) b) ∩i∈I α Ai = α (∩i∈I Ai )
(ii)a) ∪i∈I α+ Ai = α+ (∪i∈I Ai ) b) ∩i∈I α+ Ai ⊆ α+ (∩i∈I Ai )
Proof. a) Let x ∈ ∪i∈I α Ai ⇒ x ∈ α Ai , for some i ∈ I ⇒ Ai (x) ≥ α,for some i ∈ I
⇒ max Ai (x) ≥ α ⇒ (∪i∈I Ai )(x) ≥ α ⇒ x ∈ α (∪i∈I Ai )
Hence ∪i∈I α Ai ⊆ α (∪i∈I Ai ).
b) Let x ∈ ∩i∈I α Ai ⇔ x ∈α Ai , for all i ∈ I ⇔ (Ai )(x) ≥ α, for all i ∈ I.
⇔ min Ai (x) ≥ α ⇔ (∩i∈I Ai )(x) ≥ α ⇔ x ∈ α (∩i∈I Ai )
Hence ∩i∈I α Ai = α (∩i∈I Ai ).
(ii) a) Let x ∈ ∪i∈I α+ Ai ⇔ x ∈ α+ Ai , for some i ∈ I ⇔ Ai (x) > α,for some i ∈ I
⇔ max Ai (x) > α ⇔ (∪i∈I Ai )(x) > α ⇔ x ∈ α+ (∪i∈I Ai )
Hence ∪i∈I α+ Ai = α+ (∪i∈I Ai ).
18
2.1. PROPERTIES OF α−CUTS
b) Let x ∈ ∩i∈I α+ Ai ⇒ x ∈α+ Ai , for all i ∈ I ⇒ (Ai )(x) > α, for all i ∈ I.
⇒ min Ai (x) > α ⇒ (∩i∈I Ai )(x) > α ⇒ x ∈ α+ (∩i∈I Ai )
Hence ∩i∈I α+ Ai ⊆ α+ (∩i∈I Ai ).
Remark 2.1.4. We have proved that the α−cut union of fuzzy set is equal to the
union of α−cuts. This we have proved only for the finite union but which is not true
for infinite union as seen from following example.
Example 2.1.5. Given an arbitrary universal set X.
Let Ai (x) ∈ F (x)/Ai (x) = 1 − 1i , for all x ∈ X, i ∈ N . Then for any x ∈ X
(∪i∈I Ai )(x) = supi∈N Ai (x) = supi∈N (1 − 1i ) = supi∈N {0, 0.5, 0.66, 0.75, . . . }
Let α = 1. Then 1 ∪i∈N Ai = X. i.e., ∪i∈N Ai (x) ≥ 1, x ∈ X
Also for any i ∈ N, 1 Ai (x) = φ, for any x ∈ X.
∪i∈N 1 Ai (x) = ∪i∈N φ = φ 6= X = 1 (∪i∈N Ai ).
Theorem 2.1.6. Let A, B ∈ F (x). Then for all α ∈ [0, 1]
(i) A ⊆ B ⇔ α A ⊆ α B(U.Q)
(ii) A ⊆ B ⇔ α+ A ⊆ α+ B(U.Q)
(iii) A = B ⇔ α A = α B
(iv) A = B ⇔ α+ A = α+ B
Proof. (i) Assume A ⊆ B. To prove: α A ⊆ α B
Suppose α A * α B . Then there exists α0 ∈ [0, 1] such that
⇒ there exists x0 ∈ X such that x0 ∈ α0 A and x0 ∈
/ α0 B
α0
A ⊆ α0 B
⇒ A(x0 ) ≥ α0 and B(x0 ) α0 ⇒ B(x0 ) < α0 ≤ A(x0 )
⇒ B(x0 ) ≤ A(x0 ) ⇒ B ⊆ A which is ⇒⇐
Hence α A ⊆ α B.
Conversely assume that α A ⊆ α B. To Prove: A ⊆ B
Suppose A * B then there exists x0 such that A(x0 ) > B(x0 )
Let α = A(x0 ). Then x0 ∈ α A and x0 ∈
/ αB ⇒ αA * αB
Which is ⇒⇐. Hence A ⊆ B.
(ii) Assume A ⊆ B. To prove:
Suppose
α+
α+
A * α+ B
A ⊆ α+ B
From the theorem we have the standard fuzzy intersection on infinite set is
cutworthy but not strong cutworthy.
While the standard fuzzy union on infinite sets is strong cutworthy but not
cutworthy.
19
2.1. PROPERTIES OF α−CUTS
So, ∩i∈I α+ Ai * α+ (∩i∈I Ai )
Example Given any arbitrary universal set X.Let Ai (x) ∈ F (x) be defined by
Ai (x) =
1
i+2
i.e.,Ai (x) >
+
1
2
1
2
for all x ∈ X.
for all x ∈ X, i ∈ N .
Now take α = 12 . Then Ai (x) > α ⇒ x ∈ α+ Ai
But
α+
Ai = X. Hence ∩α+ Ai = ∩X = X
1
Now, inf {Ai (x)/i ∈ N } = inf { i+2
+ 12 /i ∈ N }
7
= { 13 + 12 , 14 + 12 , 15 + 12 , . . . } = inf { 56 , 68 , 10
,...} =
1
2
=α
So, ∩Ai (x) = α. Hence x ∈
/ α+ (∩Ai ). This is true for each x ∈ X.
Hence
α+
(∩Ai ) = φ. Now
α+
(∩Ai )=φ 6= X = ∩α+ Ai
Thus, ∩i∈I α+ Ai * α+ (∩i∈I Ai ).
Continuation of proof of theorem:
For all α ∈ [0, 1] there exists α0 ∈ [0, 1] such that
α+
0
+
A * α0 B
+
+
i.e., there exists some x0 ∈ X such that x0 ∈ α0 A and x0 ∈
/ α0 B
⇒ A(x0 ) > α0 and B(x0 ) ≤ α0 ⇒ B(x0 ) ≤ α0 < A(x0 )
⇒ B ⊆ A which is ⇒⇐
Conversely assume that
Hence
α+
α+
A ⊆ α+ B.
A ⊆ α+ B. To Prove: A ⊆ B.
Suppose A * B then there exits x0 ∈ X such that A(x0 ) > B(x0 )
Let α be any number between A(x0 ) and B(x0 ). Then A(x0 ) > α > B(x0 ).
So, x0 ∈ α+ A and x0 ∈
/ α+ B ⇒ α+ A * α+ B which is ⇒⇐Hence A ⊆ B.
(iii) To prove: A = B ⇔ α A = α B
A = B ⇔ A ⊆ B, B ⊆ A
⇔ α A ⊆ α B, α B ⊆ α A
⇔ αA = αB
(iv) To prove: A = B ⇔ α+ A = α+ B
A = B ⇔ A ⊆ B, B ⊆ A
⇔ α+ A ⊆ α+ B, α+ B ⊆ α+ A
⇔ α+ A = α+ B
Note: From the above theorem the properties of fuzzy set inclusion and equality
are both cut worthy and strong cut worthy.
20
2.2.
REPRESENTATION OF FUZZY SETS
Theorem 2.1.7. For any A ∈ F (x) the following properties hold.
(i) α A = ∩β<α β A = ∩β<α β+ A
(ii)
α+
A = ∪α<β β A = ∪α<β β+ A
Proof. (i) For any β < α, we have α A ⊆ β A
Since, for x ∈ α A ⇒ A(x) ≥ α > β ⇒ A(x) ≥ β ⇒ x ∈ β A. So, α A ⊆ β A
Now, α A ⊆ β A ⇒ α A ⊆ ∩β<α β A————–(1)
Also, for all x ∈ ∩β<α β A. For any > 0 we have x ∈ α− A ⇒ A(x) ≥ α − Since > 0 is arbitrary, A(x) ≥ α ⇒ x ∈ α A ⇒ ∩β<α β A ⊆ α A———–(2)
From (1) & (2) α A = ∩β<α β A.
(ii) For any α < β,
Since for x ∈ ∪α<β β A ⇒ x ∈ β A for atleast one β > α.
So, A(x) ≥ β > α ⇒ A(x) > α. we have x ∈ α+ A
So, ∪α<β β A ⊆ α+ A———————(3)
Now let x ∈ α+ A ⇒ A(x) > α ⇒ A(x) > α + ⇒ A(x) > β ≥ β
⇒ x ∈ β A ⇒ x ∈ ∪α<β β A. So,
From (3) & (4)
2.2
α+
α+
A = ∪α<β β A.
A ⊆ ∪α<β β A—————————–(4)
Representation of fuzzy sets
Each fuzzy set can be uniquely represented by either family of all its α− cuts or the
family of all its strong α− cuts and these representations allow us to extend various
properties of crisp sets and operation on crisp set to their fuzzy counter point.
To explain the two representation of fuzzy sets by crisp sets we have the following
example.
Consider the fuzzy sets A =
0.2
x1
+
0.4
x2
+
0.6
x3
+
0.8
x4
+
1
x5
The given fuzzy set A is associated with only 5 distinct α− cuts which are defined
by the following characteristic function which are special membership function.
Consider x.2A , x.4A , x.6A , x.8A , x1A
0.2A = {x/A(x) ≥ 0.2} = {0.2, 0.4, 0.6, 0.8, 1} =
0.4A = {x/A(x) ≥ 0.4} = {0.4, 0.6, 0.8, 1}
0.6A = {x/A(x) ≥ 0.6} = {0.6, 0.8, 1}
0.8A = {x/A(x) ≥ 0.8} = {0.8, 1}
=
=
=
21
1
+ x12 + x13 + x14 + x15
x1
0
+ x12 + x13 + x14 + x15
x1
0
+ x02 + x13 + x14 + x15
x1
0
+ x02 + x03 + x14 + x15
x1
2.2.
REPRESENTATION OF FUZZY SETS
1A = {x/A(x) ≥ 1} = {1} =
0
x1
+
0
x2
+
0
x3
+
0
x4
+
1
x5
We now convert each of the α−cuts to the special fuzzy set (αn ) as follows.
Definition 2.2.1. A special fuzzy set αn from each α−cuts as for each x ∈ X =
{x1 , x2 , x3 , x4 , x5 } as αA(x) = α.αA (x).
Now the above α− cuts can be converted into special fuzzy set as follows.
0.2A =
0.4A =
0.6A =
0.8A =
1A =
0.2
+ 0.2
+ 0.2
+ 0.2
+ 0.2
x1
x2
x3
x4
x5
0
0.4
0.4
0.4
0.4
+
+
+
+
x1
x2
x3
x4
x5
0
0
0.6
0.6
0.6
+
+
+
+
x1
x2
x3
x4
x5
0
0
0
0.8
0.8
+ x2 + x3 + x4 + x5
x1
0
+ x02 + x03 + x04 + x15
x1
The standard fuzzy union of these 5 special fuzzy sets is exactly the original
fuzzy set A. i.e., A = 0.2A ∪ 0.4A ∪ 0.6A ∪ 0.8A ∪ 1A .
Note: For union choose maximum membership grade of x ∈ X
For x1 maximum grade is 0.2x1
For x2 maximum grade is 0.4x2
For x3 maximum grade is 0.6x3
For x4 maximum grade is 0.8x4
For x5 maximum grade is 1x5
The representation of a n arbitrary fuzzy set A in terms of special fuzzy set
αA (x) = αA (x) is referred to as decomposition of A.
Here we formulated and prove three decomposition theorems for a fuzzy set.
Theorem 2.2.2. First Decomposition Theorem (U.Q)
For every A ∈ F (x), A = ∪α∈[0,1] αA where αA is defined by αA (x) = α.αA (x)
and ∪ denotes standard fuzzy union.
Proof. For each particular x ∈ X, let a = A(x)
∴ (∪α∈[0,1] αA )(x) = supα∈[0,1] αA(x) = max{supα∈[0,a] αA(x) , supα∈(a,1] αA(x) }
For each α ∈ (a, 1] we have A(x) = a < α i.e., A(x) α
∴ αA(x) = α.0 = 0
(∵ x ∈
/ αA(x) , by characteristic function,αA (x) = 0)
Also for each α ∈ [0, a], we have A(x) = a ≥ α
∴ αA(x) = α.1 = α
(∵ x ∈ αA(x) ,by characteristic function,αA (x) = 1))
∴ (∪α∈[0,1] αA )(x) = supα∈[0,1] αA(x) = supα∈[0,1] α = a = A(x).
22
2.2.
REPRESENTATION OF FUZZY SETS
Application of First Decomposition Theorem
For every A ∈ F (x), A = ∪α∈[0,1] αA . Consider a fuzzy set A with following
membership function of triangular shape.


x − 1 when x ∈ [1, 2]



A(x) = 3 − x when x ∈ [2, 3]



0
otherwise
Solution: For each α ∈ [0, 1], αA is in the case αA = [α + 1, 3 − α]
(∵ x − 1 ≥ α ⇒ x ≥ α + 1 and 3 − x ≥ α ⇒ x ≤ 3 − α)
The special fuzzy set αA is defined by the membership function

α x ∈ [α + 1, 3 − α]
αA (x) =
0 otherwise

α.1 x ∈ [α + 1, 3 − α]
A
χ(α )(x) =
0.α otherwise
Theorem 2.2.3. Second Decomposition Theorem (U.Q)
For every A
∈
F (x), A
=
∪α∈[0,1] α+A where α+A is defined by
α+A (x) = α.α+A (x) and ∪ denotes standard fuzzy union.
Proof. For each particular x ∈ X, let a = A(x)
∴ (∪α∈[0,1] α+A )(x) = supα∈[0,1] α+A = max{supα∈[0,a) α+A(x) , supα∈[a,1] α+A(x) }
For each α ∈ [a, 1] we have A(x) = a ≤ α i.e., A(x) ≯ α
∴ α+A(x) = α.0 = 0
(∵ x ∈
/ α+A(x) , by characteristic function,α+A (x) = 0)
Also for each α ∈ [0, a), we have A(x) = a > α
∴ α+A(x) = α.1 = α
(∵ x ∈ α+A(x) ,by characteristic function,α+A (x) = 1))
∴ (∪α∈[0,1] α+A )(x) = supα∈[0,1] α+A(x) = supα∈[0,1] α = a = A(x).
Theorem 2.2.4. Third Decomposition Theorem
For every A ∈ F (x), A = ∪α∈Λ(A) αA where Λ(A) is the level set of A and ∪ is
defined as before. Λ(A) = {α/A(x) = α, x ∈ X}.
Proof. For each particular x ∈ X, let a = A(x)
∴ (∪α∈Λ αA )(x) = supα∈Λ(A) αA (x) = max{supα=a αA(x) , supα∈Λ−a αA(x) }
∴ (∪α∈[0,1] αA )(x) = supα=a αA(x) = a = A(x).
23
2.3.
EXTENSION PRINCIPLE OF FUZZY SETS
2.3
Extension Principle of Fuzzy Sets
We say that a crisp function f : X → Y is fuzzified when it is extended to act on a
fuzzy set defined on X and Y .
Then fuzzified function has the form f : F (X) → F (Y ) and its inverse is defined
by f −1 : F (Y ) → F (X)
The principle for fuzzified crisp function is called an extension principle.
Let us first discuss a special case in which the extended functions are restricted
to power set P (x) and P (y) and it is established in classical set theory.
f
Any two function f : X → Y and f −1 : Y → X induces two functions
: F (X) → F (Y ) and f −1 : F (Y ) → F (X) which are defined by
[f (A)](y) = supA(x), A ∈ F (X) and [f −1 (B)](x) = B(f (x)), B ∈ F (Y ).
When X and Y are finite then supremum is replaced by maximum.
Theorem 2.3.1. Let f : X → Y be any arbitrary crisp function. Then for any
Ai ∈ F (X) and Bi ∈ F (Y ), i ∈ I. Then the following properties of function obtained
by the extension principle hold.
(i) f (A) = φ if and only if A = φ
(iii) f (∪Ai ) = ∪f (Ai )
(v) If B1 ⊆ B2 then f
−1
(B1 ) ⊆ f
(vii) f −1 (∪Bi ) = ∪f −1 (Bi )
(ix) A ⊆ f −1 (f (A))
−1
(ii) If A1 ⊆ A2 then f (A1 ) ⊆ f (A2 )
(iv) f (∩Ai ) ⊆ ∩f (Ai )
ˉ
(vi) f −1 (B) = f −1 (B)
(B2 )
(viii) f −1 (∩Bi ) = ∩f −1 (Bi )
(x) B ⊇ f (f −1 (B))
Proof. (i) A = φ ⇔ f (A) = f (φ) ⇔ f (A) = φ
(ii) Given A1 ⊆ A2 ⇒ A1 (x) ≤ A2 (x) for all x ∈ X ⇒ supA1 (x) ≤ supA2 (x)
Consider [f (A1 )](y) = supx/y∈F (x) A1 (x) ≤ supx/y=F (x) A2 (x) = [f (A2 )](y)
∴ [f (A1 )](y) ⊆ [f (A2 )](y) ⇒ f (A1 ) ⊆ f (A2 ).
(iii) Consider f (∪Ai )(y) = supx/y∈F (x) {∪i∈I Ai (x)}
= supi∈I {supx/y∈F (x) Ai (x)}
Hence f (∪Ai ) = ∪f (Ai )
= supi∈I {f (Ai )(y) = ∪f (Ai )(y), for all y ∈ Y
(iv) Consider f (∩Ai )(y) = supx/y∈F (x) {∩i∈I Ai (x)}
= supx/y∈F (x) {infi∈I Ai (x)}
≤ infi∈I {supx/y∈F (x) (Ai )(x)}
24
2.3.
EXTENSION PRINCIPLE OF FUZZY SETS
Hence f (∩Ai ) ⊆ ∩f (Ai )
≤ infi∈I {f (Ai )(y)} = ∩f (Ai )(y), for all y ∈ Y
(v) B1 ⊆ B2 ⇒ B1 (x) ≤ B2 (x)
Consider f −1 (B1 )(x) = B1 f (x) ≤ B2 f (x) = f −1 (B2 )(x)
f −1 (B1 )(x) ⊆ f −1 (B2 )(x) ⇒ f −1 (B1 ) ⊆ f −1 (B2 ).
ˉ (x)) = f −1 (B(x))
ˉ
ˉ
(vi) f −1 (B)(x)=B(f
= f −1 (B)(x)
ˉ
Hence f −1 (B) = f −1 (B)
(vii) f −1 (∪Bi )(x) = B(f (∪Bi )(x))
⇔ B(f (sup Bi (x))), x ∈ X
⇔ sup(B(f (Bi (x)))), x ∈ X
⇔ ∪f −1 (Bi (x)), x ∈ X
Hence f −1 (∪Bi ) = ∪f −1 (Bi )
(viii) f −1 (∩Bi )(x) = B(f (∩Bi )(x))
⇔ B(f (inf Bi (x))), x ∈ X
⇔ inf (B(f (Bi (x)))), x ∈ X
⇔ ∩f −1 (Bi (x)), x ∈ X
Hence f −1 (∩Bi ) = ∩f −1 (Bi ).
(ix) Let x ∈ A ⇒ f (x) ∈ f (A) ⇒ x ∈ f −1 (f (A))
Hence A ⊆ f −1 (f (A)).
(x) Let x ∈ f (f −1 (B)) ⇒ f −1 (x) ∈ f −1 (B) ⇒ x ∈ B.
Hence f (f −1 (B)) ⊆ B.
Theorem 2.3.2. Let f : X → Y be arbitrary crisp function. Then for any
A ∈ F (x), α ∈ [01, ]. The following properties of f fuzzified by the extension
principle hold. (i)
α+
[f (A)] = f (α+ A) (ii) α [f (A)] ⊇ f (α A). (U.Q)
Proof. (i) y ∈ α+ [f (A)] ⇔ [f (A)](y) ≥ α ⇔ supx/y=f (x) A(x) > α
⇔ ∃x0 ∈ X, y = f (x0 ), A(x0 ) > α
⇔ ∃x0 ∈ X, y = f (x0 ), x0 ∈ α+ A
Hence
α+
⇔ y ∈ f (α+ A)
[f (A)] = f (α+ A).
(ii) If y ∈ f (α A) then there exists x0 ∈ α A(A(x0 ) ≥ α) such that y = f (x0 ).
Hence [f (A)](y) = supx/y=f (x) A(x) ≥ A(x0 ) ≥ α ⇒ y ∈ α [f (A)]
Thus f (α A) ⊆ α [f (A)].
25
2.3.
EXTENSION PRINCIPLE OF FUZZY SETS
Example 2.3.3. To Show that f (α A) 6= α [f (A)].

a n ≤ 10
Let X = N, Y = {A, B}. Letf : N → Y by f (A) =
b n > 10
and A(n) = 1 − n1 , n ∈ N . Clearly A(n) ≤ 1.
[f (A)](a) = supA(n) = sup{0, 1 − 12 , 1 − 13 , 1 − 14 , . . . , 1 −
9
= sup{0, 12 , 23 , . . . 10
}=
[f (A)](b) = supA(n) = sup{1 −
9
10
1
,1
11
1
1
,1
12
−
Taking α = 1, 1 [f (A)] = {b} and f ( A) = φ
−
1
,...}
13
1
}
10
=1
(∵= φ).
So, 1 [f (A)] 6= f (1 A). Hence f (α A) 6= α [f (A)].
Note:f (α A) = α [f (A)], α ∈ [0, 1] when x is finite.
Theorem 2.3.4. Let f : X → Y be arbitrary function. Then for any A ∈ F (x), f is
fuzzified by the extension principle that satisfies the equation f (A) = ∪f (α+ A).(U.Q)
Proof. By second decomposition theorem we have A = ∪α∈[0,1]α+ A, A ∈ F (x)
Applying this theorem to f (A) which is a fuzzy set on Y ,
f (A) = ∪α∈[0,1]α+ [f (A)] = ∪α∈[0,1] α.α+ [f (A)] = ∪α∈[0,1] α.f (α+ A)
= ∪α∈[0,1] f (α.α+ A) = ∪α∈[0,1] f (α+ A)
Hence f (A) = ∪f (α+ A)
Definition 2.3.5. The fuzzy sets defined on the cartesian product are referred to
as fuzzy relation.
Example 2.3.6. Let A be fuzzy set defined by A =
all the α−cuts and strong α−cuts of A.
Solution:
0.5
0.4
0.7
0.8
1
A = {x1 , x3 , x4 , x5 } =
1
x1
A = {x1 , x2 , x3 , x4 , x5 } =
A = {x3 , x4 , x5 } =
0
x1
+
0
x2
+
1
x1
0
x2
+
+
1
x3
+
1
x2
1
x3
+
+
1
x4
+
1
x3
1
x4
+
+
+
1
x5
0
+ x02 + x03 + x14 + x15
x1
+ x02 + x03 + x04 + x15
+ x13 + x14 + x15
+ x13 + x14 + x15
+ x03 + x14 + x15
A = {x4 , x5 } =
A = {x5 } =
0.5+
A=
0.4+
A=
0.7+
A=
0
x1
1
x1
0
x1
+
+
+
0
x1
0
x2
0
x2
0
x2
1
x4
26
1
x5
+
1
x5
0.5
x1
+
0.4
x2
+
0.7
x3
+
0.8
x4
+
1
.
x5
List
2.3.
EXTENSION PRINCIPLE OF FUZZY SETS
0.8+
1+
A=
A=
0
x1
0
x1
0
x2
+
+
0
x2
+
+
0
x3
0
x3
+
+
0
x4
0
x4
+
+
1
x5
0
x5
Problem 2.3.7. Let A and B be a fuzzy set defined on a universal set X = Z whose
membership functions are given by A(x)
B(x) =
0.5
2
+
1
3
+
0.5
1
0.3
.
5
+
0.5
−1
=
+
1
0
+
0.5
1
+
0.3
2
Let the function f : X × X → X be defined for all
x1 , x2 ∈ X by (i)f (x1 , x2 ) = x1 .x2
(ii)f (x1 , x2 ) = x1 + x2 . Calculate f (A, B).
Solution: Let f (A, B) = C, C(x) = x1 .x2 and C(x) = x1 + x2 .
table-x1 .x2
table-x1 + x2
x1 /x2
2
3
4
5
x1 /x2
2
3
4
5
-1
-2
-3
-4
-5
-1
1
2
3
4
0
0
0
0
0
0
2
3
4
5
1
2
3
4
5
1
3
4
5
6
2
4
6
8
10
2
4
5
6
7
C(−2) = max{min(0.5, 0.5)} = 0.5 C(−3) = max{min(0.5, 1)} = 0.5
C(−4) = max{min(0.5, 0.5)} = 0.5 C(−5) = max{min(0.5, 0.3)} = 0.3
C(0) = max{min(1, 0.5)} = 0.5
C(0) = max{min(1, 0.5)} = 0.5
C(0) = max{min(1, 1)} = 1
C(0) = max{min(1, 0.3)} = 0.3
C(0) = max{0.5, 1, 0.5, 0.3} = 1
C(2) = max{min(0.5, 0.5)} = 0.5
C(3) = max{min(0.5, 1)} = 0.5
C(4) = max{min(0.5, 0.5)} = 0.5
C(5) = max{min(0.5, 0.3)} = 0.3
C(4) = max{min(0.3, 0.5)} = 0.3[c(4) = max{min(0.5, 0.5), min(0.3, 0.5)}]
C(6) = max{min(0.3, 1)} = 0.3
C(8) = max{min(0.3, 0.5)} = 0.3
C(10) = max{min(0.3, 0.3)} = 0.3
f (A, B) =
0.5
−2
+
0.5
−3
+
0.5
−4
and
+
0.3
−5
+ 10 +
0.5
2
+
27
0.5
3
+
0.5
4
+
0.3
5
+
0.3
6
+
0.3
8
+
0.3
10
2.3.
EXTENSION PRINCIPLE OF FUZZY SETS
Similarly
C(1) = max{min(0.5, 0.5)} = 0.5
C(2) = max{min(0.5, 1)} = 0.5
C(3) = max{min(0.5, 0.5)} = 0.5
C(4) = max{min(0.5, 0.3)} = 0.3
C(2) = max{min(1, 0.5)} = 0.5
C(3) = max{min(1, 1)} = 1
C(4) = max{min(1, 0.5)} = 0.5
C(5) = max{min(1, 0.3)} = 0.3
C(3) = max{min(0.5, 0.5)} = 0.5
C(4) = max{min(0.5, 1)} = 0.5
C(5) = max{min(0.5, 0.5)} = 0.5
C(6) = max{min(0.5, 0.3)} = 0.3
C(4) = max{min(0.3, 0.5)} = 0.3
C(5) = max{min(0.3, 1)} = 0.3
C(6) = max{min(0.3, 0.5)} = 0.3
C(7) = max{min(0.3, 0.3)} = 0.3
C(3) = max{min(0.5, 0.5), min(1, 1)} = max{0.5, 1} = 1
C(4) = max{min(0.5, 0.3), min(0.5, 1), min(0.3, 0.5), min(1, 0.5)}
= max{0.5, 0.3} = 0.5
C(5) = max{min(1, 0.3), min(0.5, 0.5), min(0.5, 0.5)} = max{0.3, 0.5} = 0.5
C(6) = max{min(0.5, 0.3), min(0.5, 0.3)} = max{0.3, 0.3} = 0.3
f (A, B) =
0.5
1
+
0.5
2
+
1
3
+
0.5
4
+
0.5
5
+
0.3
6
+
0.3
7
Problem 2.3.8. Find f (p), f (q), f (r) and f (A1 , A2 ) where A1 ⊆ x1 , A2 ⊆ x2 such
that there exists A1 =
A1 /A2 x
y
a
p
p
b
q
r
c
r
p
0.3
a
+
0.9
b
+
0.5
c
and A2 (x) =
0.5
x
+
1
y
f (p) = max{min(0.3, 0.5), min(0.3, 1), min(0.5, 1)} = max{0.3, 0.3, 0.5} = 0.5
f (q) = max{min(0.9, 0.5)} = 0.5
f (r) = max{min(0.9, 1), min(0.5, 0.5)} = max{0.9, 0.5} = 0.9
f (A1 , A2 ) = f (p) + f (q) + f (r) =
0.5
p
+
0.5
q
+
∗∗∗∗∗∗∗
28
0.9
r
Chapter 3
Unit-III
3.1
Types of Operation and Fuzzy complement
ˉ
A(x)
= 1 − A(x)
(A ∩ B)(x) = min[A(x), B(x)]
A ∪ B(x) = max[A(x), B(x)]
These operations are called standard fuzzy operation.
Fuzzy Complement
A be fuzzy set on X. Let cA is defined by a function c : [0, 1] → [0, 1] which
assigns a value c(A(x)) to each membership grade A(x) of any given fuzzy set A
i.e., c(A(x)) = cA(x), x ∈ X.
To produce meaningful fuzzy complement the fuzzy function c must satisfy at
least the following two axioms.
Axiom c1 : c(0) = 1 and c(1) = 0 (boundary conditions)
Axiom c2 : For all a, b ∈ [0, 1] if a ≤ b then c(a) ≥ c(b) (monotonicity)
Axiom c3 : c is a continuous function
Axiom c4 : c(c(a)) = a for each a ∈ [0, 1] then c is involution.
Note: All axioms are not independent
Theorem 3.1.1. Let a function c : [0, 1] → [0, 1] satisfies axioms c2 and c4 . Then c
also satisfies the axiom c1 and c3 . Moreover c must be a bijective function. (U.Q)
Proof. The range of c is [0,1]⇒ c(0) ≤ 1 and c(1) ≥ 0 ————–(1)
29
3.1. TYPES OF OPERATION AND FUZZY COMPLEMENT
By axiom c2 , c(c(0)) ≥ c(1) and by axiom c4 , 0 = c(c(0)) ≥ c(1)—–(2)
From (1)&(2), 0 = c(1). By axiom c4 , 1 = c(c(1)) = c(0)
∴ c satisfies axiom c1
To Prove: c is bijective function.
We observe that for all a ∈ [0, 1] there exists b = c(a) ∈ [0, 1] such that
c(b) = c(c(a)) = a. So c is onto.
Assume that c(a1 ) = c(a2 ) ⇒ a1 = c(c(a1 )) = c(c(a2 )) = a2 ⇒ a1 = a2
∴ c is 1-1. Hence c is bijective.
To Prove: c is continuous function
Since c is bijective and satisfies axiom c2 (a ≤ b ⇒ c(a) ≥ c(b)) it cannot have
any discontinuous point.
Assuming to the contrary i.e., c has a discontinuity at a. We have c(a0 ).
c(a)
b1
b0
c(a0 )
a
a0
∴ c has a discontinuity at a0 . We have b0 = lim c(a) > c(a0 ) and there exist
a→a0
b1 ∈ [0, 1] such that b0 > b1 > c(a0 ) for which there exist a1 ∈ [0, 1] such that
c(a1 ) = b1 which is ⇒⇐.
Since c is bijection in between c(a0 ) and b0 , c(a) is discontinuous and we cannot
find a pre image for b ∈ [0, 1] such that c(a1 ) = b1 . Thus c is continuous function.
Hence c satisfies axiom c3 .
Nested structure of basic classes of fuzzy complement
Example of general fuzzy complement that satisfies only the axiomatic
skeleton

1 f or a ≤ t
and the threshold type complements which are defined by c(a) =
0 f or a > t
30
3.1. TYPES OF OPERATION AND FUZZY COMPLEMENT
where a ∈ [0, 1] and t ∈ [0, 1), t is called threshold of c. This function is illustrated
in the figure.
all fuzzy complements axiom c1 , c2
ˉ
all classical fzy complemnt A(x)
= 1 − A(x)
All continuos fuzzy
complement axiom c1 , c3
all involutive fuzzy complement
axiom c1 − c4
The fuzzy complement which is continuous but not involutive is the function
c(a) = 12 (1 + cosπa)
c(a)
a
c(0.33) = 12 (1 + cos(180 × 0.33)) = 12 (1 + cos(59.4)) = 12 (1 + 0.5090) = 0.7545
c(0.75) = 12 (1 + cos(180 × 0.75)) = 12 (1 + cos(135)) = 12 (1 − 0.7071) = 0.14645
c(c(a)) = c(c(0.33)) = c(0.75) = 0.15 6= 0.33 = a⇒ c(c(a)) 6= a.
So the given function is not involutive.
1−a
where λ ∈ (−1, ∞)
1 + λa
For each value of the parameter value λ we obtain one particular involutive fuzzy
Definition 3.1.2. The sugeno class defined by cλ (a) =
complement.
For λ = 0, c0 (a) =
1−a
1
which is the classical fuzzy complement.
Definition 3.1.3. The yager class of fuzzy complement is defined by
cω (a) = (1 − aω )1/ω , where ω ∈ (0, ∞) which is also involutive.
When ω = 1 this function becomes c(a) = (1 − a) which is classical fuzzy
complement.
Definition 3.1.4. The equilibrium of fuzzy complement c is defined by c(a) = a
for all a.
31
3.1. TYPES OF OPERATION AND FUZZY COMPLEMENT
eg., The equilibrium value for the classical fuzzy complement is 0.5 which is the
solution of the equation 1 − a = a.
Theorem 3.1.5. Every fuzzy complement has at most one equilibrium. (U.Q)
Proof. Let c be any arbitrary fuzzy complement. An equilibrium of c is the solution
of the equation c(a) − a = 0, a ∈ [0, 1].
It is enough to prove that any equation c(a) − a = b where b is a real constant
has atmost one solution.
Suppose a1 , a2 are two different solutions of the equation c(a) − a = b such that
a1 < a2 then c(a1 ) − a1 = b = c(a2 ) − a2 ——————(1)
Now, a1 < a2 ⇒ c(a1 ) > c(a2 ). Also, −a1 > −a2
So, c(a1 ) − a1 > c(a2 ) − a2 which is ⇒⇐ to (1).
Hence c(a) − a = 0 i.e., c(a) = a. Hence c has at most one equilibrium.
Theorem 3.1.6. Assume that a given fuzzy complement c has an equilibrium ec
which is unique, then (i)a ≤ c(a) ⇔ a ≤ ec
(ii)a ≥ c(a) ⇔ a ≥ ec
Proof. (i) Split a ≤ ec as a < ec and a = ec . Assume a < ec .
a < ec ⇒ c(a) ≥ c(ec ) = ec > a (c has equilibrium ec , c(ec ) = ec ))
⇒ c(a) > a. Also if a = ec ⇒ c(a) = c(ec ) = ec = a ⇒ c(a) = a.
Hence a ≤ ec ⇒ a ≤ c(a). Converse part is similar.
(ii) Split a ≥ ec as a > ec and a = ec . Assume a > ec .
a > ec ⇒ c(a) ≤ c(ec ) = ec < a (c has equilibrium ec , c(ec ) = ec ))
⇒ c(a) < a. Also if a = ec ⇒ c(a) = c(ec ) = ec = a ⇒ c(a) = a.
Hence a ≥ ec ⇒ a ≥ c(a). Converse part is similar.
Theorem 3.1.7. If c is a continuous fuzzy complement then prove that c has unique
equilibrium. (U.Q)
Proof. The equilibrium ec of a fuzzy complement c is the solution of the equation
c(a) − a = 0. This is a special case of the more general equation c(a) − a = b where
b ∈ [−1, 1] is constant.
By axiom c1 , c(0) − 0 = 1 − 0 = 1 and c(1) − 1 = 0 − 1 = −1
Since c is continuous complement, by intermediate value theorem for continuous
function, For each b ∈ [−1, 1] there exists atleast one 0 a0 such that c(a) − a = b.
i.e., c is a continuous fuzzy complement which has a unique equilibrium.
32
3.1. TYPES OF OPERATION AND FUZZY COMPLEMENT
Example 3.1.8. The equilibrium
 for each individual fuzzy complement cλ of the
((1 + λ)1/2 − 1)/λ λ 6= 0
sugeno class is given by ecλ =
1/2
λ=0
ecλ is obtained by selecting the positive solution of the equation.
1 − ecλ
= ecλ ⇒ 1 − ecλ = ecλ (1 + λecλ )
1 + λecλ
⇒ λe2cλ + 2ecλ − 1 = 0———–(1)⇒ e2cλ + λ2 ecλ −
=
=
−2
λ
±
q
( −2
)2 − 4.1. −1
λ
λ
2.1
=
1
λ
−2
±
λ
=0
q
2
4
4
+λ
λ2
=
2( −1
±
λ
q
1+λ
)
λ2
2
−1 1 √
−1 1 √
±
+
1+λ=
1 + λ (taking only positive value)
λ λ
λ λ
(1 + λ)1/2 − 1
, λ 6= 0
λ
(1) ⇒ 2ecλ − 1 = 0 ⇒ ecλ = 1/2.
=
If λ = 0,
The dependence of the equilibrium ec is on the parameter λ.
Definition 3.1.9. Given a fuzzy complement c and the membership grade whose
value is represented by a real number a ∈ [0, 1]. Then any membership grade
represented by a real number d a ∈ [0, 1] such that c(d a) − d a = a − c(a) is called a
dual point of A w.r.t c.
Theorem 3.1.10. If a complement of c has an equilibrium ec then dec = ec .(U.Q)
Proof. If a = ec , then by definition of equilibrium c(a) = a and hence a − c(a) = 0.
Since c(a) = c(ec ) = ec = a ⇒ c(a) = a ⇒ a − c(a) = 0——–(1)
Also if d a = ec then c(d a) = c(ec ) = ec = d a ⇒ c(d a) − d a = 0———–(2)
from (1) & (2), c(d a) − d a = a − c(a)
So, c(dec ) − ec = ec − c(ec ) ⇒ dec − ec = ec − ec
dec − ec = 0. Hence dec = ec .
Theorem 3.1.11. For each a ∈ [0, 1], d a = c(a) if and only if c(c(a)) = a. i.e.,
when the complement is involutive.
Proof. Assume that d a = c(a)
33
3.1. TYPES OF OPERATION AND FUZZY COMPLEMENT
By the definition of dual point, c(d a) − d a = a − c(a)———-(1)
which becomes c(c(a)) − c(a) = a − c(a) (∵ d a = c(a)) ⇒ c(c(a)) = a.
Conversely assume that c(c(a)) = a
By (1), c(d a) − d a = a − c(a) = c(c(a)) − c(a) ⇒ d a = c(a).
Theorem 3.1.12. First Characterization Theorem of Fuzzy Complement
Let c be a function from [0, 1] to [0, 1]. Then c is a fuzzy complement (involutive) if
and only if there exists a continuous function g from [0, 1] to R such that g(0) = 0.
g is strictly increasing and c(a) = g −1 [g(1) − g(a)], for all a ∈ [0, 1].
Proof. Let g be a continuous function from [0,1] to R such that g(0) = 0 and g is
strictly increasing.
Then the 
pseudo inverse of g denote by g −1 is a function from R to [0, 1] defined

0
a ∈ (−∞, 0)



by g −1 (a) = g −1 (a) a ∈ [0, g(1)] where g −1 is the ordinary inverse of g.



1
a ∈ (g(1), ∞)
Let c be a function on [0,1] defined by c(a) = g −1 [g(1) − g(a)], for all a ∈ [0, 1]
Now prove that c is a fuzzy complement.
First we show that c satisfies the axiom c2 for any a, b ∈ [0, 1].
if a < b then g(a) < g(b), since g is strictly increasing.
Hence g(1) − g(a) > g(1) − g(b)
Now c(a) = g −1 [g(1) − g(a)] > g −1 [g(1) − g(b)] > c(b)
So, a < b ⇒ c(a) > c(b). Hence c satisfies axiom c2 .
Next we show that c satisfies the axiom c4 .
c(c(a)) = g −1 [g(1) − g(c(a))]
= g −1 [g(1) − g(g −1 (g(1) − g(a)))]
= g −1 [g(1) − gg −1 (g(1) − g(a))]
= g −1 [g(1) − (g(1) − g(a))]
c(c(a)) = g −1 (g(a)) = a.
Hence c satisfies axiom c4 .
By Theorem 3.1.1, c also satisfies axiom c1 and c3 . Thus c is a fuzzy complement.
Conversely let c be a fuzzy complement that satisfies axiom c2 and c4 .
We need to find a continuous, strictly increasing function g that satisfies
34
3.1. TYPES OF OPERATION AND FUZZY COMPLEMENT
c(a) = g −1 [g(1) − g(a)] and g(0) = 0.
By Theorem 3.1.7, c must have a unique equilibrium say ec .
i.e., c(ec ) = ec ,
ec ∈ [0, 1]
Let h : [0, ec ] → [0, b] be any continuous, strictly increasing bijection function
such that h(0) = 0, h(ec ) = b where b is any positive
 real number.
h(a)
a ∈ [0, ec ]
Now we define a function g : [0, 1] → R by g(a) =
2b − h(c(a)) a ∈ (e , 1]
c
Obviously g(0) = h(0) = 0 and g is continuous as well as strictly increasing,
since h is continuous and strictly increasing. 

a





h−1 (a)
−1
The pseudo inverse of g is given by g (a) =


c(h−1 (2b − a))





1
−1
Now we show that g satisfies c(a) = g [g(1) − g(a)]. For any
a ∈ (−∞, 0)
a ∈ [0, b]
a ∈ [b, 2b]
a ∈ (2b, ∞]
a ∈ [0, 1]
If a ∈ [0, ec ] then g −1 [g(1) − g(a)] = g −1 [g(1) − h(a)] = g −1 [2b − h(a)]
= c(h−1 (2b − (2b − h(a)))) = c(a)
If a ∈ (ec , 1] then g −1 [g(1) − g(a)] = g −1 [2b − (2b − h(c(a)))] = g −1 (h(c(a)))
= h−1 (h(c(a))) = c(a)
Thus for any a ∈ [0, 1], c(a) = g −1 [g(1) − g(a)]. Hence the proof.
[g(1) = 2b − h(c(1)) = 2b − h(0) = 2b − 0 = 2b]
Theorem 3.1.13. Second Characterization Theorem of Fuzzy Complement
Let c be a function from [0, 1] to [0, 1]. Then c is a fuzzy complement (involutive) if
and only if there exists a continuous function f from [0, 1] to R such that f (1) = 0.
f is strictly decreasing and c(a) = f −1 [f (0) − f (a)], for all a ∈ [0, 1]. (U.Q)
Proof. By Theorem 3.1.12, Let c is a fuzzy complement iff there exists a strictly
increasing function g such that c(a) = g −1 [g(1) − g(a)], for all a ∈ [0, 1].
Now let f (a) = g(1) − g(a). ————–(1)
Then f (1) = g(1) − g(1) = 0.
Let g(1) − g(a) = a ⇒ g(a) = g(1) − a ⇒ a = g −1 (g(1) − a)
Now, f (a) = g(1) − g(a) ⇒ a = f −1 (g(1) − g(a)) = f −1 (a)
So, f −1 (a) = a = g −1 (g(1) − a). —————-(2)
35
3.2. FUZZY INTERSECTION: T -NORMS
Also, f (0) = g(1) − g(0) = g(1) ⇒ f −1 (a) = g −1 (f (0) − a)
Now, f (f −1 (a)) = g(1) − g(f −1 (a))
(by (1))
−1
= g(1) − g[g (g(1) − a)] (by (2))
= g(1) − g(1 − a) = g(1) − g(1) + a
=a
Also, f −1 (f (a)) = g −1 [g(1) − f (a)]
(by (2))
−1
= g [g(1) − (g(1) − g(a))]
= g −1 (g(a))
=a
Now,
c(a) = g −1 [g(1) − g(a)]
= f −1 (g(a))
(by (2))
= f −1 (g(1) − a)
= f −1 [g(1) − (g(1) − g(a))]
c(a) = f −1 [f (0) − f (a)]
(by (1))
If a decreasing function f is given, such that f (1) = 0 then f is continuous and
c(a) = f −1 [f (0) − f (a)].
Let us take g(a) = f (0) − f (a) ⇒ g(0) = f (0) − f (0) = 0.
∴ g is continuous and increasing.
c(a) = f −1 [f (0) − f (a)] = f −1 (g(a)) = g −1 (g(1) − g(a)). (by (2))
So, c(a) = g −1 (g(1)−g(a)) and hence by Theorem 3.1.12, c is a fuzzy complement.
3.2
Fuzzy Intersection: t-norms
The fuzzy intersection of two fuzzy sets A and B is a function of the form
i : [0, 1] × [0, 1] → [0, 1] such that (A ∩ B)(x) = i[A(x), B(x)].
Then i is known as t− norms or fuzzy intersection.
The fuzzy intersection t− norm is a binary operation on the unit interval
that satisfies the following axioms.
Axiom 1: i(a, 1) = a (boundary condition)
Axiom 2: b ≤ d ⇒ i(a, b) ≤ i(a, d) (monotonicity)
Axiom 3: i(a, b) = i(b, a) (commutativity)
36
3.2. FUZZY INTERSECTION: T -NORMS
Axiom 4: i(a, i(b, d)) = i(i(a, b), d) (associativity)
This set of axiom are called the axiomatic skeleton for fuzzy intersection or
t− norms. Also we have an additional axioms.
Axiom 5: i is a continuous function (continuity)
Axiom 6: i(a, a) ≤ a (sub-idempotency)
Axiom 7: a1 < a2 &b1 < b2 ⇒ i(a1 , b1 ) < i(a2 , b2 ) (strict monotonicity)
Note: Also i(0, 1) = 0, i(1, 1) = 1 and i(1, 0) = i(0, 1) = 0
Definition 3.2.1. A continuous t− norm that satisfies sub-idempotency is called
an archimedian t− norm.
If it also satisfies strict monotonicity it is called strict archimedian t− norm.
Theorem 3.2.2. The standard fuzzy intersection is the only idempotent t− norm.
(U.Q)
Proof. W.K.T. min(a, a) = a for all a ∈ [0, 1].
Suppose that there exists a t− norm such that i(a, a) = a, a∈ [0, 1]
Then for any a, b ∈ [0, 1] if a < b then
a = i(a, a) ≤ i(a, b) (boundary and monotonicity)
≤ i(a, 1) [a ≤ a, b ≤ 1and a ≤ b]
=a
By the condition i(a, b) = a = min(a, b)
If a ≥ b then min(a, b) = b
b = i(b, b) ≤ i(b, a)
≤ i(a, b)
≤ i(1, b)
≤ i(b, 1) = b
i(a, b) = b = min(a, b)
This is true for all a, b ∈ [0, 1]
Hence the standard fuzzy intersection is the only idempotent t− norm.
Remark 3.2.3.
Standard intersection: i(a, b) = min(a, b)
Algebraic product:
i(a, b) = ab
Bounded difference:
i(a, b) = max(0, a + b − 1)
37
3.2. FUZZY INTERSECTION: T -NORMS
Drastic intersection:


a



i(a, b) = b



0
when b = 1
when a = 1
otherwise
Here each of the above is defined for all a ∈ [0, 1]. The drastic intersection is
denoted by imin . Also imin (a, b) ≤ max(0, a + b − 1) ≤ ab ≤ min(a, b).
Theorem 3.2.4. For all a, b ∈ [0, 1], imin (a, b) ≤ i(a, b) ≤ min(a, b) where imin
denote drastic intersection. (U.Q)
Proof.
upper bound By the boundary condition and monotonicity
b ≤ 1 ⇒ i(a, b) ≤ i(a, 1) = a
a ≤ 1 ⇒ i(a, b) = i(b, a) ≤ i(b, 1) = b
i.e., i(a, b) ≤ a and i(a, b) ≤ b ⇒ i(a, b) ≤ min(a, b).
Lower bound By the boundary condition
i(a, b) = a when b = 1 and i(a, b) = b when a = 1.
Since i(a, b) ≤ min(a, b) and i(a, b) ∈ [0, 1], we have i(a, 0) = 0 and i(0, b) = 0
By the monotonicity
i(a, b) ≥ i(a, 0) = i(0, b) = 0


a when b = 1



i.e., i(a, b) = b when a = 1



0 otherwise
By the definition of drastic intersection, imin (a, b) is the lower bound of i(a, b).
Hence imin (a, b) ≤ i(a, b) ≤ min(a, b).
Theorem 3.2.5. Characterization Theorem of t− norms
Let i be a binary operation on the unit interval. Then i is an Archimedian
t− norms if and only if there exists a decreasing generator f such that
i(a, b) = f −1 [f (a) + f (b)], a, b ∈ [0, 1].
Proof. Assume that f : [0, 1] → [0, ∞] is a continuous strictly decreasing function
with f (1) = 0 and f is constructed by i(a, b) = f −1 [f (a) + f (b)].
To prove: i is boundary t− norms.
boundary i1 : i(a, 1) = f −1 [f (a) + f (1)]
= f −1 [f (a) + 0] = f −1 [f (a)]
38
3.2. FUZZY INTERSECTION: T -NORMS
=a
monotonicity i2 : i(a, b) ≤ i(a, d)
f −1 [f (a) + f (b)] ≤ f −1 [f (a) + f (d)]
f −1 [min(f (a) + f (b), f (0))] ≤ f −1 [min(f (a) + f (d), f (0))]
f −1 [min(f (a) + f (b), 1)] ≤ f −1 [min(f (a) + f (d), 1)]
f −1 [f (a) + f (b)] ≤ f −1 [f (a) + f (d)]
[f (a) + f (b)] ≤ [f (a) + f (d)]
f (b) ≤ f (d) ⇒ b ≤ d
i3 : i(a, b) = f −1 [f (a) + f (b)] = f −1 [f (b) + f (a)] = i(b, a)
i4 : i[i(a, b), d] = f −1 [f (i(a, b)) + f (d)]
= f −1 [f (f −1 (f (a) + f (b))) + f (d)]
= f −1 [f (a) + f (b) + f (d)]
= f −1 [(f (a) + f (f −1 (f (b) + f (d)))]
= f −1 [f (a) + f (i(b, d))]
= i[a, i(b, d)]
i satisfies 4 axioms and it also continuous and idempotent i is a t− norm.
Definition
3.2.6. The
yager
w
class
w 1/w
iw (a, b) = 1 − min(1, [(1 − a) + (1 − b) ]
of
t−
norm
is
given
by
).
Theorem 3.2.7. Let iw denote the class of yager t− norms defined by
iw (a, b) = 1 − min(1, [(1 − a)w + (1 − b)w ]1/w ). Then imin (a, b) ≤ iw (a, b) ≤ min(a, b).
Proof. Lower bound:
iw (1, b) = 1 − min(1, [(1 − a)w + (1 − b)w ]1/w )
= 1 − min(1, [0 + (1 − b)w ]1/w )
= 1 − min(1, [(1 − b)w ]1/w )
= 1 − min(1, (1 − b))
= 1 − (1 − b) = b
iw (a, 1) = 1 − min(1, [(1 − a)w + (1 − b)w ]1/w )
= 1 − min(1, [(1 − a)w + 0]1/w )
= 1 − min(1, [(1 − a)w ]1/w )
= 1 − min(1, (1 − a))
= 1 − (1 − a) = a
39
3.2. FUZZY INTERSECTION: T -NORMS
So, iw (1, b) and iw (a, 1) are independent of w.
lim iw (a, b) = lim [1 − min(1, [(1 − a)w + (1 − b)w ]1/w )]
w→0
w→0
lim iw (a, b) = [1 − min(1, ∞)]
w→0
[∵ lim [(1 − a)w + (1 − b)w ]1/w )] = ∞]
w→0
=1−1=0
∴ iw (a, b) represents drastic intersection imin (a, b).
Upper bound:
lim min(1, [(1 − a)w + (1 − b)w ]1/w )] = max(1 − a, 1 − b)
w→∞
[∵ lim ]min(1, [aw + bw ]1/w )] = max(a, b)]
w→∞
∴ i∞ (a, b) = 1 − max{1 − a, 1 − b} = min(a, b)
Hence imin (a, b) ≤ iw (a, b) ≤ min(a, b).
Note: 1 − max(1 − 0.2, 1 − 0.7) = 1 − max(0.8, 0.3) = 1 − 0.8 = 0.2
min(0.2, 0.7) = 0.2
So, 1 − max(1 − a, 1 − b) = min(a, b).
Example 3.2.8. Yager class fuzzy intersection
When w = 1,i1 (a, b) = 1 − min(1, [(1 − a)1 + (1 − b)1 ]1 )
1. When a = 1, b = 0
i1 (a, b) = 1 − min(1, [(1 − 1)1 + (1 − 0)1 ]1 ) = 1 − min(1, 1) = 1 − 1 = 0
2. When a = 1, b = 0.25
i1 (a, b) = 1−min(1, [(1−1)1 +(1−0.25)1 ]1 ) = 1−min(1, 0.75) = 1−0.75 = 0.25
3. When a = 1, b = 0.5
i1 (a, b) = 1 − min(1, [(1 − 1)1 + (1 − 0.5)1 ]1 ) = 1 − min(1, 0.5) = 1 − 0.5 = 0.5
4. When a = 1, b = 0.75
i1 (a, b) = 1−min(1, [(1−1)1 +(1−0.75)1 ]1 ) = 1−min(1, 0.25) = 1−0.25 = 0.75
40
3.2. FUZZY INTERSECTION: T -NORMS
5. When a = 1, b = 1
i1 (a, b) = 1 − min(1, [(1 − 1)1 + (1 − 1)1 ]1 ) = 1 − min(1, 0) = 1 − 0 = 1
6. When a = 0.75, b = 0
i1 (a, b) = 1 − min(1, [(1 − 0.75)1 + (1 − 0)1 ]1 ) = 1 − min(1, 1.25) = 1 − 1 = 0
7. When a = 0.75, b = 0.25
i1 (a, b) = 1 − min(1, [(1 − 0.75)1 + (1 − 0.25)1 ]1 ) = 1 − min(1, 1) = 1 − 1 = 0
8. When a = 0.75, b = 0.5
i1 (a, b) = 1 − min(1, [(1 − 0.75)1 + (1 − 0.5)1 ]1 ) = 1 − min(1, 0.75) = 1 − 0.75 =
0.25
9. When a = 0.75, b = 0.75
i1 (a, b) = 1−min(1, [(1−0.75)1 +(1−0.75)1 ]1 ) = 1−min(1, 0.5) = 1−0.5 = 0.5
10. When a = 0.75, b = 1
i1 (a, b) = 1−min(1, [(1−0.75)1 +(1−1)1 ]1 ) = 1−min(1, 0.25) = 1−0.25 = 0.75
11. When a = 0.5, b = 0
i1 (a, b) = 1 − min(1, [(1 − 0.5)1 + (1 − 0)1 ]1 ) = 1 − min(1, 1.5) = 1 − 1 = 0
12. When a = 0.5, b = 0.25
i1 (a, b) = 1 − min(1, [(1 − 0.5)1 + (1 − 0.25)1 ]1 ) = 1 − min(1, .25) = 1 − 1 = 0
13. When a = 0.5, b = 0.5
i1 (a, b) = 1 − min(1, [(1 − 0.5)1 + (1 − 0.5)1 ]1 ) = 1 − min(1, 1) = 1 − 1 = 0
14. When a = 0.5, b = 0.75
i1 (a, b) = 1 − min(1, [(1 − 0.5)1 + (1 − 0.75)1 ]1 ) = 1 − min(1, 0.75) = 1 − 0.75 =
0.25
15. When a = 0.5, b = 1
i1 (a, b) = 1 − min(1, [(1 − 0.5)1 + (1 − 1)1 ]1 ) = 1 − min(1, 0.5) = 1 − 0.5 = 0.5
41
3.2. FUZZY INTERSECTION: T -NORMS
16. When a = 0.25, b = 0
i1 (a, b) = 1 − min(1, [(1 − 0.25)1 + (1 − 0)1 ]1 ) = 1 − min(1, 1.75) = 1 − 1 = 0
17. When a = 0.25, b = 0.25
i1 (a, b) = 1 − min(1, [(1 − 0.25)1 + (1 − 0.25)1 ]1 ) = 1 − min(1, 1.5) = 1 − 1 = 0
18. When a = 0.25, b = 0.5
i1 (a, b) = 1 − min(1, [(1 − 0.25)1 + (1 − 0.5)1 ]1 ) = 1 − min(1, 1.25) = 1 − 1 = 0
19. When a = 0.25, b = 0.75
i1 (a, b) = 1 − min(1, [(1 − 0.25)1 + (1 − 0.75)1 ]1 ) = 1 − min(1, 1) = 1 − 1 = 0
20. When a = 0.25, b = 1
i1 (a, b) = 1−min(1, [(1−0.25)1 +(1−1)1 ]1 ) = 1−min(1, 0.75) = 1−0.75 = 0.25
21. When a = 0, b = 0
i1 (a, b) = 1 − min(1, [(1 − 0)1 + (1 − 0)1 ]1 ) = 1 − min(1, 2) = 1 − 1 = 0
22. When a = 0, b = 0.25
i1 (a, b) = 1 − min(1, [(1 − 0)1 + (1 − 0.25)1 ]1 ) = 1 − min(1, 1.75) = 1 − 1 = 0
23. When a = 0, b = 0.5
i1 (a, b) = 1 − min(1, [(1 − 0)1 + (1 − 0.5)1 ]1 ) = 1 − min(1, 1.5) = 1 − 1 = 0
24. When a = 0, b = 0.75
i1 (a, b) = 1 − min(1, [(1 − 0)1 + (1 − 0.75)1 ]1 ) = 1 − min(1, 1.25) = 1 − 1 = 0
25. When a = 0, b = 1
i1 (a, b) = 1 − min(1, [(1 − 0)1 + (1 − 1)1 ]1 ) = 1 − min(1, 1) = 1 − 1 = 0.
Theorem 3.2.9. Let i be a t− norm and let g : [0, 1] → [0, 1] be a function such
that g is strictly increasing and continuous in (0, 1) and g(0) = 0, g(1) = 1. Then
the function ig is defined by ig (a, b) = g −1 [i(g(a), g(b))] for all a, b ∈ [0, 1] where g −1
denote the pseudo inverse of g, is also a t− norm.
42
3.2. FUZZY INTERSECTION: T -NORMS
Proof. To Prove: ig is a t− norm. i.e., We need to prove the axiom i1 − i4 .
For any a ∈ [0, 1], ig (a, 1) = g −1 [i(g(a), g(1))]
= g −1 [i(g(a), 1)]
= g −1 [g(a)] = a
Thus i satisfies axiom i1 .
It is also easy to show that ig satisfies axiom i2 and i3 . In the following we mainly
show that ig satisfies axiom i4 .
For any a, b, d ∈ [0, 1]
ig (a, ig (b, d)) = g −1 [i(g(a), g(g −1 [i(g(b), g(d))]))]————(1)
ig (ig (a, b), d) = g −1 [i(g(g −1 [i(g(a), g(b))], g(d)))]————(2)
Now to prove : (1)=(2).
Let a0 = lim g(x) and b0 = lim g(x). Then we consider the following six cases of
x→0
x→1
possible values of i[g(a), g(b)] and i[g(b), g(d)].
case i: i[g(a), g(b)] ∈ [b0 , 1].
Then a & b must be 1. Hence (1)=(2)=d.
case ii: i[g(b), g(d)] ∈ [0, 1]
By same reason as in case i, (1)=(2)=a
case iii: i[g(a), g(b)] ∈ (a0 , b0 ) and i[g(b), g(d)] ∈ (a0 , b0 )
Then (1)=g −1 [i(g(a), i(g(b), g(d))] = g −1 [i(i(g(a), g(b)), g(d))] = (2).
case iv: i[g(a), g(b)] ∈ (a0 , b0 ) and i[g(b), g(d)] ∈ (0, a0 ]
Then (1) = g −1 [i(g(a), 0)] = 0
= g −1 [i(g(b), g(d))]
≥ g −1 [i(g(a), i(g(b), g(d)))]
= g −1 [i(i(g(a), g(b)), g(d))] = (2)
Case v: i[g(a), g(b)] ∈ [0, a0 ] and i[g(b), g(d)] ∈ (a0 , b0 )
By same reason as in case iv and axiom i3 , (1)=(2).
Case vi: i[g(a), g(b)] ∈ [0, a0 ] and i[g(b), g(d)] ∈ [0, a0 ]
Then (1)=g −1 [i(g(a), 0)] = 0 = g −1 [i(0, g(d))] = (2)
Thus in all cases (1)=(2) and so axiom i4 is satisfied.
Hence ig is a t− norm.
43
3.3. FUZZY UNION: T-CONORM
3.3
Fuzzy Union: t-conorm
The general fuzzy union of two fuzzy sets A and B is defined by the function
u : [0, 1] × [0, 1] → [0, 1] by (A ∪ B)(x) = u[A(x), B(x)].
A fuzzy union t− conorm u is a binary operation on a unit interval which
satisfies at least the following axioms for all a, b, d ∈ [0, 1].
Axiom u1 : u(a, 0) = a (boundary condition)
Axiom u2 : b ≤ d ⇒ u(a, b) ≤ u(a, d)(monotonicity)
Axiom u3 : u(a, b) = u(b, a) (commutativity)
Axiom u4 : u(a, u(b, d)) = u(u(a, b), d)
These set of axioms are called axiomatic skeleton for fuzzy union t− conorm.
Axiom u1 − u4 and Axiom i1 − i4 differ only in the boundary condition.
The additional impartant requirements are expressed by the following axioms
Axiom u5 : u is a continuous function.
Axiom u6 : u(a, a) > a
Axiom u7 : a1 < a2 and b1 < b2 ⇒ u(a1 , b1 ) < u(a2 , b2 ).
Definition 3.3.1. Any continuous and super idempotency t− conorm is called
archimedian t− conorm. If it is also strictly monotonic then it is called strictly
archimedian.
Theorem 3.3.2. The standard fuzzy union is the only idempotent t− conorm.(U.Q)
Proof. W.K.T. max(a, a) = a for all a ∈ [0, 1].
Assume that there exist t− conorm such that u(a, a) = a
For any a, b ∈ [0, 1] if a ≤ b then b = u(b, 0) ≤ u(b, a) ≤ u(b, b) = b
∴ u(b, a) = b = max(a, b)
If a ≥ b then a = u(a, 0) ≤ u(a, b) ≤ u(a, a) = a
∴ u(a, b) = a = max(a, b)
Thus standard fuzzy union is only idempotent t− conorm.
Remark 3.3.3.
Standard union: u(a, b) = max(a, b)
Algebraic sum: u(a, b) = a + b − ab
44
3.3. FUZZY UNION: T-CONORM
Bounded sum:
Drastic union:
u(a, b) = 
min(1, a + b)

a when b = 0



u(a, b) = b when a = 0



0 otherwise
max(a, b) ≤ a + b − ab ≤ min(1, a + b) ≤ umax (a, b) for all a, b ∈ [0, 1] where
umax denote the drastic union.
Theorem 3.3.4. For all a, b ∈ [0, 1], max(a, b) ≤ u(a, b) ≤ umax (a, b)
Proof. By defn, u(a, b) = a when b = 0 and u(a, b) = b when a = 0
For a, b ∈ [0, 1], u(a, 1) = 1 and u(1, b) = 1.
So, u(a, b) ≤ u(a, 1) = 1 and u(a, b) ≤ u(1, b) = 1
⇒ u(a, b) ≤ umax (a, b).
Here u(a, b) ≥ u(a, 0) = a and u(a, b) = u(b, a) ≥ u(b, 0) = b
So, u(a, b) ≥ a and u(a, b) ≥ b. Thus u(a, b) ≥ max(a, b).
Hence max(a, b) ≤ u(a, b) ≤ umax (a, b).
Definition
3.3.5. The
yager
class
of
t−
conorm
is
defined
by
uw (a, b) = min(1, (aw + bw )1/w ), w > 0.
Theorem 3.3.6. Let uw denote the class of yager t− conorm defined by
uw (a, b) = min(1, (aw + bw )1/w ), w > 0. Then max(a, b) ≤ uw (a, b) ≤ umax (a, b).
(U.Q)
Proof. Lower bound: To prove: lim min(1, (aw + bw )1/w ) = max(a, b)
w→∞
R.H.S when a = 0, max(a, b) = max(0, b) = b
when b = 0, max(a, b) = max(a, 0) = a
when a = b, max(a, b) = max(a, a) = a
L.H.S when a = 0, min(1, b) = b
when b = 0, min(1, a) = a
when a = b, lim (min(1, (2aw )1/w ) = lim min(1 + 21/w a) = min(1, a) = a
w→∞
w→∞
∴ L.H.S=R.H.S ⇒ max(a, b) = uw (a, b).
When a 6= b,
lim min(1, (aw + bw )1/w ) = lim (aw + bw )1/w
w→∞
w→∞
45
3.4. COMBINATION OF OPERATIONS
To Prove: lim (aw + bw )1/w = max(a, b)
w→∞
Assume that when a < b
Let θ = (aw + bw )1/w ⇒ logθ =
log(aw + bw )
lim logθ = lim
w→∞
w→∞
w
1
(aw
w
+ bw )
By l-hospital rule in R.H.S,
aw loga + bw logb
(a/b)w loga + logb
=
lim
w→∞
w→∞
a w + bw
(a/b)w + 1
lim logθ = lim
w→∞
Since a < b, (a/b) < 1. So, lim (a/b)w = 0
w→∞
∴ lim logθ = logb ⇒ lim θ = b.
w→∞
w→∞
Hence lim θ = lim (aw + bw )1/w = b = max(a, b).
w→∞
w→∞
It is now enough to prove (aw + bw ) ≥ 1.
when lim , the inequality (aw + bw ) ≥ 1 holds when a = 1 or b = 1.
w→∞
So, lim min(1, (aw + bw )1/w ) = 1 = max(a, b),
w→∞
Upper bound:


a



umax (a, b) = b



1
To Prove: uw (a, b) ≤ umax (a, b)
a, b ∈ [0, 1]
when b = 0
when a = 0 which are independent of w.
otherwise
lim (aw + bw )1/w = ∞ if a, b 6= 0
w→0
∴ lim uw (a, b) = lim min(1, (aw + bw )1/w ) = min(1, ∞) = 1
w→0
w→0
So, the upper bound represents umax (a, b).
3.4
Combination of Operations
The operation intersection and union are dual w.r.t. the complement in the
sense that they satisfy Demorgan’s laws.
ˉ and A ∪ B = Aˉ ∩ B
ˉ
i.e., A ∩ B = Aˉ ∪ B
46
3.4. COMBINATION OF OPERATIONS
We say that t− norms i and t− conorms u are dual w.r.t. the fuzzy complement
c if and only if c(i(a, b)) = u(c(a), c(b)) and c(u(a, b)) = i(c(a), c(b)).
These equations described the Demorgan’s law for fuzzy sets.
Let the triple < i, u, c > denote that i and u are the duals w.r.t c and let any
such triple is called dual triple.
Theorem 3.4.1. The triples < min, max, c > and < imin , umax , c > are dual w.r.t
any fuzzy complement c. (U.Q)
Proof. Assume without loss of generality that a ≤ b.
Then c(a) ≥ c(b) for any fuzzy complement. Hence
max(c(a), c(b)) = c(a) = c(min(a, b)) and min(c(a), c(b)) = c(b) = c(max(a, b))
∴< min, max, c >is a

a


Now, imin (a, b) = b



0
triple.
if b = 1
if a = 1
otherwise
case(i): when a = 1, b 6= 0


a if b = 0



umax (a, b) = b if a = 0



0 otherwise
c(a) = 1 − a = 1 − 1 = 0
∴ umax (c(a), c(b)) = umax (0, c(b)) = c(b)
Also, c(imin (a, b)) = c(b). Hence umax (c(a), c(b)) = c(imin (1, b))
Now, imin (c(a), c(b)) = imin (0, c(b)) = 0
And, c(umax (a, b)) = c(a) = 0. Hence c(umax (a, b)) = imin (c(a), c(b)).
case(ii): when a = 1, b = 0
c(a) = 1 − a = 1 − 1 = 0 and c(b) = 1 − b = 1 − 0 = 1
∴ umax (c(a), c(b)) = umax (0, 1) = 1
Also, c(imin (a, b)) = c(b) = 1.
Hence umax (c(a), c(b)) = c(imin (1, b))
Now, imin (c(a), c(b)) = imin (0, 1) = 0
And, c(umax (a, b)) = c(a) = 0.
Hence c(umax (a, b)) = imin (c(a), c(b)).
case(iii): when a = 0, b = 1
c(a) = 1 − a = 1 − 0 = 1 and c(b) = 1 − b = 1 − 1 = 0
47
3.4. COMBINATION OF OPERATIONS
∴ umax (c(a), c(b)) = umax (1, 0) = 1
Also, c(imin (a, b)) = c(a) = 1.
Hence umax (c(a), c(b)) = c(imin (1, b))
Now, imin (c(a), c(b)) = imin (1, 0) = 0
And, c(umax (a, b)) = c(umax (0, 1)) = c(1) = 0.
Hence c(umax (a, b)) = imin (c(a), c(b)).
case(iv): when a 6= 0, 1 ; b 6= 0, 1
w.l.g. assume that a ≤ b. Since a 6= 0, 1; c(a) = 1, 0 and b 6= 0, 1; c(b) = 1, 0
∴ umax (c(a), c(b)) = umax (0, 1) = 1 = c(0)
Also, c(imin (a, b)) = c(imin (1, 0)) = c(0) = 1.
Hence umax (c(a), c(b)) = c(imin (1, b))
Now, imin (c(a), c(b)) = imin (0, 1) = 0
And, c(umax (a, b)) = c(umax (1, 0)) = c(1) = 0.
Hence c(umax (a, b)) = imin (c(a), c(b)).
Thus, < imin , umax , c > is a dual triple.
Theorem 3.4.2. Given a t− norm i and an involutive fuzzy complement c, the
binary operation u on [0,1] defined by u(a, b) = c(i(c(a), c(b))) for all a, b ∈ [0, 1] is
a t− conorm such that < i, u, c > is a dual triple. (U.Q)
Proof. Given u(a, b) = c(i(c(a), c(b)))————–(1)
To prove that u given by (1) is a t− conorm.
We have to show that it satisfies axioms u1 to u4 .
u1 : u(a, 0) = c(i(c(a), c(0))) = c(i(c(a), 1)) = c(c(a)) = a
u2 : For any a, b, d ∈ [0, 1] if b ≤ d then c(b) ≥ c(d) and
i(c(a), c(b)) ≤ i(c(a), c(d)).
So, by (1) u(a, b) = c(i(c(a), c(b))) ≤ c(i(c(a), c(d))) = u(a, d)
u3 : For any a, b ∈ [0, 1]
By (1) u(a, b) = c(i(c(a), c(b))) = c(i(c(b), c(a))) = u(b, a)
u4 : For any a, b, d ∈ [0, 1]
u(a, u(b, d)) = c(i(c(a), c(u, (b, d))))
= c(i(c(a), c(c(i(c(b), c(d))))))
= c(i(c(a), (i(c(b), c(d)))))
48
3.4. COMBINATION OF OPERATIONS
= c(i(i(c(a), c(b)), c(d)))
= c(i(c(u(a, b)), c(d)))
= u(u(a, b), d)
By employing (1) and axiom c4 we can now show that u satisfies demorgan’s law
c(u(a, b)) = c(c(i(c(a), c(b)))) = i(c(a), c(b))
u(c(a), c(b)) = c(i(c(c(a)), i(c(c(b))))) = c(i(a, b)).
Hence i and u are dual w.r.t. c.
Theorem 3.4.3. Given an involutive fuzzy complement c and an increasing generator g of c, the t− norm and t− conorm generated by g are dual w.r.t. c.
Proof. For any a, b ∈ [0, 1] we have
c(a) = g −1 [g(1) − g(a)]
i(a, b) = g −1 [g(a) + g(b) − g(1)]
u(a, b) = g −1 [g(a) + g(b)]
Hence i[c(a), c(b)] = g −1 [g(g −1 (g(1) − g(a))) + g(g −1 (g(1) − g(b))) − g(1)]
= g −1 [g(1) − g(a) + g(1) − g(b) − g(1)]
= g −1 [g(1) − g(a) − g(b)]
c[u(a, b)] = g −1 [g(1) − g(g −1 (g(a) + g(b)))]
= g −1 [g(1) − min(g(1), g(a) + g(b))]
= g −1 [g(1) − g(a) − g(b)]
So, c[u(a, b)] = i[c(a), c(b)].
Hence the t− norm and t− conorm are dual w.r.t. c.
Theorem 3.4.4. Let < i, u, c > be a dual triple generated by c(u(a, b)) = i(c(a), c(b)).
Then the fuzzy operator i, u, c satisfy the law of excluded middle and the law of
contradiction.
Proof. According to c[u(a, b)] = i[c(a), c(b)]
c(a) = g −1 [g(1) − g(a)]
i(a, b) = g −1 [g(a) + g(b) − g(1)]
u(a, b) = g −1 [g(a) + g(b)]
Then u[a, c(a)] = g −1 [g(a), g(c(a))]
= g −1 [g(a) + g[g −1 (g(1) − g(a))]]
= g −1 [g(a) + g(1) − g(a)] = g −1 [g(1)]
49
3.5. AGGREGATION OPERATION
=1
For all a ∈ [0, 1] law of excluded middle is satisfied.
i[a, c(a)] = g −1 [g(a) + g(c(a)) − g(1)]
= g −1 [g(a) + g[g −1 (g(1) − g(a)) − g(1)]]
= g −1 [g(a) + g(1) − g(a) − g(1)] = g −1 [g(0)]
=0
For all a ∈ [0, 1] law of contradiction is satisfied.
Theorem 3.4.5. Let < i, u, c > be a dual triple that satisfies the law of excluded
middle and law of contradiction. Then < i, u, c > does not satisfy the distributive
laws.
Proof. Assume that the distributive law i(a, u(b, d)) = u[i(a, b), i(a, d)] is satisfied
for all a, b, d ∈ [0, 1].
Let e be the equilibrium of c. clearly e 6= 0, 1 since c(0) = 1, c(1) = 0.
By the law of excluded middle and law of contradiction we obtain
u(e, e) = e(e, c(e)) = 1 and i(e, e) = i(e, c(e)) = 0.
Now applying e to the above distributive laws we have i(e, u(e, e)) = u(i(e, e), i(e, e))
substituting for u(e, e) and i(e, e) we obtain i(e, 1) = u(0, 0)
which result in e = 0(by axiom i1 , u1 )
This contradicts the requirements that e 6= 0
Hence the distributive laws does not hold. In an analogous way we can prove
that the dual distributive law does not hold.
3.5
Aggregation Operation
Any aggregation operation on n fuzzy set is defined by the function
h : [0, 1]n → [0, 1]
when applied to fuzzy sets A1 , A2 , . . . , An defined on λ, function h produces an
aggregate fuzzy set A by operating on the membership grades of these sets for each
x ∈ X. Thus A(x) = h[A1 (x), A2 (x), . . . , An (x)]
Remark 3.5.1. In order to qualify as an intutively meaningful aggregation function
h must satisfy at least the following three axiomatic requirements.
50
3.5. AGGREGATION OPERATION
Axiom h1 : h(0, 0, . . . , 0) = h(1, 1, . . . , 1)
Axiom h2 : For any pair (a1 , a2 , . . . , an ) and (b1 , b2 , . . . , bn ) of n tuples such that
ai , bi ∈ [0, 1] if ai ≤ bi then h(a1 , a2 , . . . , an ) ≤ h(b1 , b2 , . . . , bn )
i.e., h is monotonic increasing in all its arguments.
Axiom h3 : h is continuous function
Axiom h4 : h is symmetric function in all its argument.
i.e., h(a1 , a2 , . . . , an ) = h(ap(1) , ap(2) , . . . , ap(n) )
Axiom h5 : h is an idempotent function. i.e., h(a, a, . . . , a) = a
Note: (i) Fuzzy intersection and fuzzy union are not idempotent
(ii) Fuzzy intersection and union qualify as aggregate operations on fuzzy sets.
Any aggregation operation h that satisfies axiom h2 and h5 satisfies also the
inequality min(a1 , a2 , . . . , an ) ≤ h(a1 , a2 , . . . , an ) ≤ max(a1 , a2 , . . . , an ) ——–(1)
for all n tuples (a1 , a2 , . . . , an ) ∈ [0, 1]n .
Proof. Let a? = min(a1 , a2 , . . . , an ) and a? = max(a1 , a2 , . . . , an )
If h satisfies axiom h2 , &h5 then
a? = h(a? , a? , . . . , a? ) ≤ h(a1 , a2 , . . . , an ) ≤ h(a? , a? , . . . , a? ).
Conversely if h satisfies (1) it must satisfy axiom h5 , since
a = min(a, a, a . . . , a) ≤ h(a, a, . . . , a) ≤ max(a, a, . . . , a) = a
i.e., all aggregation operation between standard fuzzy intersection and standard
fuzzy union are idempotent.
Note: The function h that satisfies (1) are only aggregation operation that are
idempotent. These aggregation operations are usually called averaging operations.
Definition 3.5.2. Another class of aggregation operation that covers the entire
interval between the min and max operation is called the class of ordered weighted
averaging operations. It is denoted by OWA. Let w = {w1 , w2 , . . . , wn } be a
P
weighting vector such that wi ∈ [0, 1] for all i ∈ Nn and
wi = 1.
Then an OWA operation associated with w is the function
hw (a1 , a2 , . . . , an ) = w1 b1 + w2 b2 + ∙ ∙ ∙ + wn bn where bi for any i ∈ Nn is the ith
largest element in a1 , a2 , . . . an . That is < b1 , b2 , . . . bn > is permutation of vector
< a1 , a2 , . . . an > in which the elements are ordered bi ≥ bj if i < j for any part
i, j ∈ Nn .
51
3.5. AGGREGATION OPERATION
Example 3.5.3. The OWA operator hw satisfy axiom h1 to h5 and consequently
also the inequality (1). The lower and upper bounds are obtained for
w = (0.3, 0.1, 0.2, 0.4). We have
hw (0.6, 0.9, 0.2, 0.7) = 0.3(0.9) + 0.1(0.7) + 0.2(0.6) + 0.4(0.2) = 0.54.
Theorem 3.5.4. Let h : [0, 1]n → R+ be a function that satisfies axiom h1 &h2 and
the property h(a1 +b1 , a2 +b2 , . . . , an +bn ) = h(a1 , a2 , . . . , an )+h(b1 , b2 , . . . , bn )—–(1)
P
where ai , bi , ai + bi ∈ [0, 1] then h(a1 , a2 , . . . , an ) = ni=1 wi ai ——-(2) for all i ∈ Nn .
Proof. Let hi (ai ) = h(0, 0, . . . , 0, ai , 0, . . . , 0), i ∈ Nn . Then for any a, b, a + b ∈ [0, 1]
hi (a + b) = hi (a) + hi (b). This is a well investigated functional equation referred
to as cauchy’s functional equation. Its solution is hi (ai ) = wi (ai )
∴ h(a1 , a2 , . . . , an ) = h(a1 , 0, . . . , 0) + h(0, a2 , . . . , 0) + ∙ ∙ ∙ + h(0, 0, . . . , an )
= h1 (a1 ) + h2 (a2 ) + ∙ ∙ ∙ + hn (an )
P
= w1 a1 + w2 a2 + . . . wn an = ni=1 wi ai .
Theorem 3.5.5. Let h : [0, 1]n → [0, 1] be a function that satisfies axiom h1 , h3 and
the properties h(max(a1 , b1 ), . . . (an , bn )) = max(h(a1 , a2 , . . . , an ), h(b1 , b2 , . . . , bn ))—
—(1)and h1 (h1 (a1 )) = hi (ai )—–(2) where hi (ai ) = h(0, 0, . . . ai , . . . , 0), for all i ∈
Nn prove h(a1 , a2 , . . . , an ) = max[h1 (a1 ), h2 (a2 ), . . . , hn (an )] and hi (ai ) = min(wi , ai ).
Proof. Observe that h(a1 , a2 , . . . , an ) = h(max(a1 , 0), max(0, a2 ), . . . max(0, an )).
By (1) we obtain h(a1 , a2 , . . . , an ) = max(h(a1 , 0, . . . , 0), h(0, a2 , . . . , 0)). We
can now replace h(0, a2 , a3 , . . . , an ) with max(h(0, a2 , . . . , 0), h(0, 0, a3 , . . . , an )) and
replacing the same replacement w.r.t a3 , a4 , . . . , an . We eventually obtain
h(a1 , a2 , . . . , an ) = max[h(a1 , 0, . . . , 0), h(0, a2 , . . . , 0), . . . , h(0, 0, . . . , an )]
= max[h1 (a1 ), h2 (a2 ) . . . hn (an )]
It remains to prove that hi (ai ) = min(wi , ai ). Clearly hi (a) is continuous, non
decreasing and such that hi (0) = 0 and hi (hi (ai )) = hi (ai ). Let hi (1) = wi then the
range of hi is [0, wi ]. For any ai ∈ [0, wi ] there exists bi such that ai = hi (bi ).
Hence hi (ai ) = hi (hi (bi )) = hi (bi ) = hi (wi ) ≤ hi (ai ) ≤ h(1) = wi
and consequently hi (ai ) = wi = min(wi , ai ).
Theorem 3.5.6. Let h : [0, 1]n → [0, 1] be a function that satisfies axiom h1 , h3 and
the properties h(min(a1 , b1 ), . . . (an , bn )) = min(h(a1 , a2 , . . . , an ), h(b1 , b2 , . . . , bn ))—
——-(1)and hi (ab) = hi (a)hi (b), hi (0) = 0—–(2) where hi (ai ) = h(1, 1, . . . ai , . . . , 1)
52
3.5. AGGREGATION OPERATION
for all i ∈ Nn . Then there exist numbers α1 , α2 , . . . , αn ∈ [0, 1] such that
h(a1 , a2 , . . . , an ) = min(aα1 1 , aα2 2 , . . . , aαnn ).
Definition 3.5.7. A special kind of aggregation operations are binary operator h
on [0,1] that satisfy the properties monotonicity, commutativity and associativity
of t− norm and t− conorm but replace the boundary conditions of t− norm and
t− conorm with weaker boundary conditions h(0, 0) = 0 and h(1, 1) = 1. Let these
aggregation operations are called norm operation.
Remark 3.5.8. 1. Due to their associativity, norm operations can be extended to
any finite number of arguments . when a norm operator also has the h(a, 1) = a
it becomes t− norm and when it has also the property h(a, 0) = a it becomes a
conorm. Otherwise it is an associativity, averaging operation.
Hence norm operation cover the whole range of aggregation operation from imin
to umax .
2. An example of parameterized class of norm operation that are neither t−
norm nor t− conorm
is the class of binary operation on [0, 1] defined by


min(λ, u(a, b)) when a, b ∈ [0, λ]



hλ (a, b) = max(λ, i(a, b)) when a, b ∈ [λ, 1]



λ
otherwise
where λ ∈ [0, 1] is a t− norm and u is a t− conorm. Let these operation be
called λ averages.
Theorem 3.5.9. Let a norm operation
 h be continuous and idempotent. Then there

max(a, b) when a, b ∈ [0, λ]



exist λ ∈ [0, 1] such that h(a, b) = min(a, b) when a, b ∈ [λ, 1]



λ
otherwise
Proof. Suppose that h is continuous and idempotent norm operation. Then h satisfies the property of continuity, monotonicity, commutativity, associativity and weak
boundary operations. Let λ = h(0, 1) ∈ [0, 1]. We show that this λ is what we need
to find. First we prove that h satisfies the following two properties.
p1 : h(0, a) = a, a ∈ [0, λ]
p2 : h(1, a) = a, a ∈ [λ, 1]
53
3.5. AGGREGATION OPERATION
Let f1 be a function defined by f1 (x) = h(0, x) for all x ∈ [0, 1].
Then f1 (0) = 0, f1 (1) = λ
Since f1 is continuous and monotonically increasing for any a ∈ [0, λ] there exist
x0 ∈ [0, 1] such that f1 (x0 ) = a
Then h(0, a) = h(0, f1 (x0 )) = h(0, h(0, x0 )) = h(h(0, 0), x0 )
= h(0, x0 ) = f1 (x0 ) = a
Hence p1 is proved. it is similar to prove p2 by defining f2 (x) = h(x, 1).
Now we prove that h is actually a median as defined in the theorem. If a, b ∈ [0, λ]
then a = h(a, 0) ≤ h(a, b) and b = h(b, 0) ≤ h(a, b) .Thus max(a, b) ≤ h(a, b)
On the other hand h(a, b) ≤ h(max(a, b), b) ≤ h(max(a, b), max(a, b)) = max(a, b)
∴ h(a, b) = max(a, b)
If a, b ∈ [λ, 1] then h(a, b) ≤ h(a, 1) = a and h(a, b) ≤ h(1, b) = b
Thus h(a, b) ≤ min(a, b).
On the other hand min(a, b) = h(min(a, b), min(a, b)) ≤ h(a, b)
∴ h(a, b) = min(a, b)
If a ∈ [0, λ] and b ∈ [λ, 1] then λ = h(a, λ) ≤ h(a, b) ≤ h(λ, b) = λ
∴ h(a, b) = λ
If a ∈ [λ, 1] and b ∈ [λ, 1] then λ = h(λ, b) ≤ h(a, b) ≤ h(a, λ) = λ
∴ h(a, b) = λ


max(a, b) when a, b ∈ [0, λ]



Hence h(a, b) = min(a, b) when a, b ∈ [λ, 1]



λ
otherwise
***********
54
Chapter 4
Unit-IV
4.1
Fuzzy Numbers
A fuzzy set A on R is said to be a fuzzy number if it must posses at least the
following three properties
(i) A must have a normal fuzzy set
(ii) α A must be a closed interval for every α ∈ [0, 1].
(iii) The support of A, (0+ A) must be bounded.
Note: Since α− cut of any fuzzy number are required to be closed interval for
all α ∈ [0, 1] every fuzzy number is a convex set. The converse is not necessary true
since α− cuts of some convex fuzzy set may be open or half open intervals.
Theorem 4.1.1. let A ∈ F (R). Then A is a fuzzy number if and only if there exist
a closed interval
[a, b] 6= φ such that


1
f or x ∈ [a, b]



A(x) = l(x) f or x ∈ (−∞, a) —————–(F)



r(x) f or x ∈ (b, ∞)
where l is a function from (−∞, a) to [0, 1]. i.e., monotonic increasing contin-
uous from the right and such that l(x) = 0 for x ∈ (−∞, w1 ) and r is a function
from (b, ∞) to [0, 1]. i.e., monotonic decreasing continuous from the left and such
that r(x) = 0 for x ∈ (w2 , ∞).
55
4.1. FUZZY NUMBERS
Proof. Necessity: Assume A is a fuzzy number.
Then, α A is a closed interval for all α ∈ [0, 1]
there exist a, b such that 1 A = [a, b]
∴1 A is closed.
[∵ 1 A is normal.]
Take A(x) = l(x) for x ∈ (−∞, a). So, 0 ≤ l(x) < 1.
To Prove: l(x) is monotonically increase, enough to prove x ≤ y ⇒ l(x) ≤ l(y)
Let x ≤ y ≤ a.
A(λx1 + (1 − λ)x2 ) ≥ min{A(x1 ), A(x2 )}; A(λx + (1 − λ)a) ≥ min{A(x), A(a)}
A(y) ≥ A(x) ⇒ l(x) ≤ l(y)
Assume that l(x) is not continuous from the right.
∴ there exist the sequence of numbers such that
xn ≥ x0 for some x0 ∈ (−∞, a) ⇒ lim xn = x0 ————(1)
n→∞
Now x0 ∈ α A ⇒ A(x0 ) ≥ α ———-(2) for any n (∵ by(1),α A is closed)
lim l(xn ) = A(xn ) = α > l(x0 ) = A(x0 ) ⇒ A(x0 ) < α
n→∞
which is ⇒⇐ to (2). Hence l is continuous from right.
Similarly if we take A(x) = r(x) we can say r(x) is a monotonic decreasing
function in (b, ∞) and continuous from left.
Since A is a fuzzy number,
0+
A is bounded.
Then there exist a pair w1 , w2 ∈ R such that A(x) = 0, x ∈ (−∞, w1 ) ∪ (w2 , ∞).
Conversely if A satisfies (F), then A is a normal fuzzy set. Also 0+ A is bounded.
Since we get a pair (w1 , w2 ) ∈ R such that A(x) = 0, x ∈ (−∞, w1 ) ∪ (w2 , ∞).
choose α ∈ [0, 1] and define
xα = inf {x/l(x) ≥ α, x < a} and yα = sup{x/r(x) ≥ α, x < b}
To Prove: α A must be a closed interval for all α ∈ [0, 1].
It is enough to prove α A = [xα , yα ]
Now for any x0 ∈ α A ⇒ A(x0 ) ≥ α
if x0 < a ⇒ l(x0 ) = A(x0 ) ≥ α (∵ x0 ∈ (−∞, a))
⇒ x0 ∈ {x/l(x) ≥ α, x < a}
⇒ x0 ≥ xα
(By defn of xα ) ⇒ r(x0 ) ≥ α
If x0 > b ⇒ x0 ≤ yα (By defn of yα )
x0 ∈ [xα , yα ]. So, α A ⊆ [xα , yα ]
56
4.2.
ARITHMETIC OPERATION ON FUZZY INTERVAL
By definition of xα , there exist a sequence < xn > in {x/l(x) ≥ l} such that
lim xn = xα . Since l is continuous from right,
n→∞
l(xα ) = l( lim xn ) = lim l(xn ) ≥ α ⇒ A(xα ) ≥ α ⇒ xα ∈ α A.
n→∞
n→∞
Similarly yα ∈ α A. So, [xα , yα ] ⊆ α A. Thus, α A = [xα , yα ]
Hence A is a fuzzy number.
4.2
Arithmetic operation on Fuzzy interval
The four arithmetic operations on closed interval are defined as follows.
(i)[a, b] + [d, e] = [a + d, b + e]
(ii) [a, b] − [d, e] = [a − e, b − d]
(iii) [a, b].[d, e] = [min(ad, ae, bd, be), max(ad, ae, bd, be)]
[a, b]
1 1
a a b b
a a b b
(iv)
= [a, b].[ , ] = [min( , , , ), max( , , , )]
[d, e]
e d
d e d e
d e d e
Problem 4.2.1. (U.Q) a) [−1, 2] + [1, 3] = [0, 5]
b) [−2, 4] − [3, 6] = [−8, 1]
c) [−3, 4].[−3, 4] = [min(9, −12, 16), max(9, −12, 16)] = [−12, 16]
[−4, 6]
= [min(−4, −2, 6, 3), max(−4, −2, 6, 3)] = [−4, 6]
d)
[1, 2]
Properties:
1. A + B = B + A,
A.B = B.A
Proof. A + B = [a1 + b1 , a2 + b2 ] = [b1 + a1 , b2 + a2 ] = B + A
A.B = [min(a1 b1 , a1 b2 , a2 b1 , a2 b2 ), max(a1 b1 , a1 b2 , a2 b1 , a2 b2 )]
= [min(b1 a1 , b1 a2 , b2 a1 , b2 a2 ), max(b1 a1 , b1 a2 , b2 a1 , b2 a2 )] = B.A
2. (A + B) + C = A + (B + C), (A.B).C = A.(B.C)
Proof. (A + B) + C = [(a1 + b1 ) + c1 , (a2 + b2 ) + c2 ]
= [a1 + (b1 + c1 ), a2 + (b2 + c2 )] = A + (B + C)
Similarly (A.B).C = A.(B.C)
57
4.2.
ARITHMETIC OPERATION ON FUZZY INTERVAL
3. A = 0 + A = A + 0, A = 1.A = A.1
Proof. 0 = [0, 0], 1 = [1, 1], A = [a1 , a2 ].
So, A + 0 = [a1 , a2 ] + [0, 0] = [a1 + 0, a2 + 0] = [a1 , a2 ] = 0 + A = A
1.A = A.1 = [min(a1 , a2 ), max(a1 , a2 )] = [a1 , a2 ] = A
4. A.(B + C) ⊆ A.B + A.C
Proof. A.(B + C) = [min(a1 (b1 + c1 ), a1 (b2 + c2 ), a2 (b1 + c1 ), a2 (b2 + c2 ),
max(a1 (b1 + c1 ), a1 (b2 + c2 ), a2 (b1 + c1 ), a2 (b2 + c2 )]
= [min(a1 b1 + a1 c1 + a1 b2 + a1 c2 , a2 b1 + a2 c1 + a2 b2 + a2 c2 ),
max(a1 b1 + a1 c1 + a1 b2 + a1 c2 , a2 b1 + a2 c1 + a2 b2 + a2 c2 )]
⊆ [min(a1 b1 , a1 b2 , a2 b1 , a2 b2 ), max((a1 b1 , a1 b2 , a2 b1 , a2 b2 ))] +
[min(a1 c1 , a1 c2 , a2 c1 , a2 c2 ), max((a1 c1 , a1 c2 , a2 c1 , a2 c2 ))]
A.(B + C) ⊆ A.B + A.C
The distributive laws does not hold for the fuzzy arithmetic w.r.t closed interval.
for. eg. let A = [0, 1], B = [1, 2], C = [−2, −1]. Then
A.B = [0, 2], A.C = [−2, 0], B + C = [−1, 1]
A.(B + C) = [−1, 1] and A.B + A.C = [−2, 2]
∴ A.(B + C) ⊆ A.B + A.C
5. If b.c ≥ 0 for every b ∈ B, c ∈ C
(i) If a1 ≥ 0 then
A.(B+C) = [a1 .(b1 +c1 ), a2 .(b2 +c2 )] = [a1 .b1 , a2 .b2 ]+[a1 .c1 , a2 .c2 ] = A.B+A.C
(ii) If a1 < 0, a2 ≤ 0 then −a2 > 0 and (−A) = [−a2 , −a1 ]
(−A).(B + C) = (−A).B + (−A).C = −(A.B + A.C)
So, A.(B + C) = A.B + A.C
(iii) A.(B + C) = [a1 , a2 ].[b1 + c1 , b2 + c2 ] = [a1 (b1 + c1 ), a2 (b2 + c2 )]
58
4.3. ARITHMETIC OPERATION ON FUZZY NUMBERS
= [a1 .b2 , a2 .b2 ] + [a1 .c2 , a2 .c2 ] = [a1 , a2 ].[b1 .b2 ] + [a1 , a2 ].[c1 .c2 ]
then A.(B + C) = A.B + A.C.
6. 0 ∈ A − A and 1 ∈ A/A
Let A = [a1 , a2 ] . Then A − A = [a1 − a2 , a2 − a1 ] ⇒ 0 ∈ A − A
A/A = min{( aa11 , aa12 , aa21 , aa22 ), max( aa11 , aa12 , aa21 , aa22 )} = [ aa12 , aa21 ] ⇒ 1 ∈ A
7. If A ⊆ E, B ⊆ F then A + B ⊆ E + F, A − B ⊆ E − F, A.B ⊆ E.F,
A
E
⊆ .
B
F
4.3
Arithmetic Operation on Fuzzy Numbers
Let A and B denote fuzzy numbers and ∗ denote any of four basic arithmetic oper-
ations. Then we define a fuzzy set A ∗ B on R by defining its α−cut as
α
(A ∗ B) = α A ∗ α B for α ∈ [0, 1].
when ∗ = / clearly we have to require that 0 ∈
/ αB
By first decomposition theorem, A ∗ B can be expressed as A ∗ B =
S
α(A∗B)
Since α(A∗B) is a closed interval and A, B are fuzzy numbers A ∗ B is also a
fuzzy number.
Problem 4.3.1.Let A, B be two fuzzy numbers whose membership

x + 2/2

when
−
2
<
x
≤
0
x − 1/2






given by A(x) = (2 − x)/2 when 0 < x ≤ 2
B(x) = (6 − x)/2






0
0
otherwise
functions are
when 2 < x ≤ 4
when 0 < x ≤ 6
otherwise
Calculate fuzzy numbers A + B, A − B, B − A, A/B, min(A, B), max(A, B)
Solution: We have α A = [2α − 2, 2 − 2α] and α B = [2α + 2, 6 − 2α]
α
α
(A + B) = α A + α B = [4α, 8 − 4α] and
(A − B) = α A − α B = [4α − 8, −4α] . Now,
4α = x ⇒ α =
x
4
8 − 4α = x ⇒ α =
8−x
4
For, α = x4 , α = 0 ⇒ x = 0
and
For, α =
α=1⇒x=4
8−x
,α
4
=0⇒x=8
x+8
4
− x4
4α − 8 = x ⇒ α =
−4α = x ⇒ α =
For, α =
and
For, α =
59
x+8
,α
4
and
= 0 ⇒ x = −8
α = 1 ⇒ x = −4
−x
,α
4
=0⇒x=0
4.3. ARITHMETIC OPERATION ON FUZZY NUMBERS
α=1⇒x=4
and
α = 1 ⇒ x = −4
and




0
x ≤ 0 and x > 8
0
x > 0 and x ≤ −8






(A+B)(x) = x/4
(A−B)(x) = (x + 8)/4 −8 < x ≤ −4
0<x≤4






(8 − x)/4 4 < x ≤ 8
(−x)/4
−4 < x ≤ 0
α
α
α
(A.B) =αA.α B = [2α − 2, 2 − 2α] .[2α + 2, 6 − 2α]
(A.B) = min{4α2 − 4, 16α − 4α2 − 12, 4 − 4α2 , 4α2 − 16α + 12},
max{4α2 − 4, 16α − 4α2 − 12, 4 − 4α2 , 4α2 − 16α + 12}
(A.B) = [4α2 + 16α − 12, 4α2 − 16α + 12], α ∈ (0, 0.5]
2
= [−4α2 + 16α − 12, 4α√
− 16α + 12], α ∈ [0.5, 1]
4α2 − 16α + 12 = x ⇒ α =
16±
256−4×4×(12−x)
8
4+(4−x)1/2
2
=
4+(4+x)1/2
2
−4α2 + 16α − 12 = x ⇒ α =

 4+(4−x)1/2 −12 ≤ x ≤ 0
2
(A.B)(x) =
 4+(4+x)1/2 0 ≤ x ≤ 12
2
α
A
[2α − 2, 2 − 2α]
A
)= α =
Now, α ( B
[2α + 2, 6 − 2α]
B
α A
(B )
2α−2 2−2α
2α−2 2−2α
= [min{ 2α+2
, 2α+2 }, max{ 2α−2
, 2α+2 }]
α A
(B )
2α−2 2−2α
= [ 2α+2
, 2α+2 ], α ∈ (0, 0.5]
2α−2
2α+2
= [ 2α−2
, 2−2α ], α ∈ (0.5, 1]
2α+2 2α+2
=x⇒α=
A
(B
)(x) =

 1+x
1−x

x−1
−1−x
1+x
1−x
and
2−2α
2α+2
−1 < x ≤ 0
=x⇒α=
x−1
−1−x
.
0<x≤1
Theorem 4.3.2. Let 0 ∗0 ∈ {+, −, ., /} and let à and B̃ denote continuous fuzzy
numbers then the fuzzy set defined by (Ã ∗ B̃)(z) = sup min[Ã(x), B̃(y)] is a conz=x∗y
tinuous fuzzy numbers. (U.Q)
Proof. We shall first prove that α (Ã ∗ B̃) = α Ã ∗ α B̃ by showing that α (Ã ∗ B̃) is a
closed interval for every α ∈ (0, 1]
60
4.3. ARITHMETIC OPERATION ON FUZZY NUMBERS
Let for any z ∈ α Ã ∗ α B̃ there exist some x0 ∈ α Ã and y0 ∈ α B̃ such that
z = x 0 ∗ y0
Thus (Ã ∗ B̃)(z) = sup min[Ã(x), B̃(y)] ≥ min[Ã(x0 ), B̃(y0 )] ≥ α
z=x∗y
α
α
(∵ x0 ∈ Ã, y0 ∈ B̃)
So, (Ã ∗ B̃)(z) ≥ α
Hence, z ∈ α (Ã ∗ B̃) and consequently, α Ã ∗ α B̃ ⊆ α (Ã ∗ B̃)————-(1)
Again let z ∈ α (Ã ∗ B̃). we have
(Ã ∗ B̃)(z) = sup min[Ã(x), B̃(y)] ≥ α
Now for any
z=x∗y
n > [ α1 ] + 1, [ α1 ]
[∵ (Ã ∗ B̃)(z) ≥ α]
is the greatest integer that is less than or equal to
1
α
i.e., [ α1 ] ≤ α1 . Then there exist xn and yn such that z = xn ∗ yn and
min[Ã(xn ), B̃(yn )] > α −
1
n
1
1
⇒ xn ∈ (α− n ) Ã, yn ∈ (α− n ) B̃.
We consider two sequences < xn >, < yn > and since α −
1
)
(α− n+1
1
à ⊆ (α− n ) à and
1
(α− n+1
)
1
B̃ ⊆ (α− n ) B̃
Hence < xn >, < yn > will be contained into some
(α−1/n)
1
n
≤α−
à and
1
n+1
(α−1/n)
B̃.
Since < xn > and < yn > are closed interval and are thus bounded sequence and
Hence there exist convergent subsequence say xni and yni converging to x0 y0 .
i.e., xni → x0 and yni → y0
Corresponding to xni and yni there exist convergent subsequence xnj and ynj
converging to x0 , y0 .
i.e., xni,j → x0 and yni,j → y0 and z = xni,j ∗ yni,j
Since ∗ is continuous, z = lim (xni,j ∗ yni,j )
j→∞
= ( lim xni,j ) ∗ ( lim yni,j )
j→∞
z = x0 ∗ y 0
Also, since Ã(xni,j ) > α −
1
; B̃(xni,j )
ni,j
j→∞
>α−
1
ni,j
Then Ã(x0 ) = Ã( lim xni,j ) = lim Ã(xni,j ) ≥ lim (α −
j→∞
j→∞
j→∞
i.e., Ã(x0 ) ≥ α. Similarly B̃(x0 ) ≥ α
∴ x0 ∈ α Ã, y0 ∈ α B̃ ⇒ x0 ∗ y0 ∈ α Ã ∗ α B̃
z ∈ α Ã ∗ α B̃ ⇒α (Ã ∗ B̃) ⊆ α Ã ∗ α B̃———-(2)
from (1) & (2), α (Ã ∗ B̃) = α Ã ∗ α B̃
61
1
)=α
ni,j
4.4. LATTICE OF FUZZY NUMBERS
Now to show that à ∗ B̃ must be continuous.
Let us assume that à ∗ B̃ is not continuous at z0
i.e., lim (Ã ∗ B̃)(z) < (Ã ∗ B̃)(z0 ) = sup min[Ã(x), B̃(y)]
z→z0
z0 =x∗y
There exist x0 , y0 such that z0 = x0 ∗ y0 then
lim (Ã ∗ B̃)(z) < min[Ã(x0 ), B̃(y0 )]———(3)
z→z0
Since the operation 0 ∗0 ∈ {+, −, ., /} is monotonic w.r.t the first and second
arguments respectively. So we can always find a sequence < xn > and < yn > such
that xn → x0 and yn → y0 as n → ∞ and xn ∗ yn < z0 for any n.
Let zn = xn ∗ yn . then zn → z0 as n → ∞
Thus lim (Ã ∗ B̃)(z) = lim (Ã ∗ B̃)(zn ) =
z→z0
n→∞
sup min[Ã(xn ), B̃(yn )]
zn =xn ∗yn
lim (Ã ∗ B̃)(z) ≥ lim min[Ã(xn ), B̃(yn )]
z→z0
n→∞
= min[Ã( lim xn ), B̃( lim yn )]
n→∞
n→∞
= min[Ã(x0 ), B̃(y0 )]
lim (Ã ∗ B̃)(z) ≥ min[Ã(x0 ), B̃(y0 )]———–(4)
z→z0
from (3) & (4) we get a contradiction.
Hence à ∗ B̃ must be a continuous fuzzy number.
4.4
Lattice of fuzzy numbers
The set R of real numbers is linearly ordered for every pair of real numbers x and
y. We can say that either x ≤ y or y ≤ x. The pair (R, ≤) is a lattice which can
also be expressed
in terms of two lattice operation
x if x ≤ y
y if x ≤ y
min(x, y) =
max(x, y) =
y if y ≤ x
x if y ≤ x
Note: For any pair (x, y) ∈ R operation on fuzzy numbers min and max for any
two fuzzy numbers a, b we define
MIN(A, B)(z) =
sup
min[A(x), B(y)] Max(A, B)(z) =
z=min(x,y)
sup
z=max(x,y)
62
min[A(x), B(y)]
4.4. LATTICE OF FUZZY NUMBERS
Example 4.4.1. It is important to realize that the operation MIN and MAX are
totally different from the standard fuzzy intersection and union, min and max.
This difference is illustrated below. Let R denote set of all fuzzy numbers. Then
operations MIN and MAX are clearly functions of form R × R → R.
Problem 4.4.2. A(x) =


0



x+2
3



 4−x
3
x < −2, x > 4
−2 ≤ x ≤ 1
1≤x≤4


0





 x+2


0
x < 1, x > 3



B(x) = x − 1 1 ≤ x ≤ 2



3 − x 2 ≤ x ≤ 3


x < −2, x > 3
0
x < 1, x > 4





x − 1 1 ≤ x ≤ 2
−2 ≤ x ≤ 1
3
Solution:min (A, B)(x) =
max (A, B)(x) =


4−x

3 − x 2 < x ≤ 2.5
1
≤
x
≤
2.5


3







 4−x
3 − x 2.5 ≤ x ≤ 3
2.5 ≤ x ≤ 4
3


x+2


−2 ≤ x ≤ 0
2≤x≤4

 x−2


 2
 2
Problem 4.4.3. A(x) = 2−x
0 ≤ x ≤ 2 B(x) = 6−x
0≤x≤6
2
2






0
0
otherwise
otherwise
Solution:min (A, B)(x) =

x+2



 2
2−x
2



0
−2 ≤ x ≤ 0
0≤x≤2
max (A, B)(x) =
otherwise

6−x



 2
x−2
2



0
0≤x≤2
2<x≤6
otherwise
Theorem 4.4.4. Let MIN and MAX be binary operations on R defined by
MIN(A, B)(z) =
sup
min[A(x), B(y)] MAX(A, B)(z) =
z=min(x,y)
Then for any A, B, C ⊆ R, the following properties holds.
sup
min[A(x), B(y)]
z=max(x,y)
a) M IN (A, B) = M IN (B, A) and MAX(A,B)=MAX(B,A)(commutative)(U.Q)
b) M IN [M IN (A, B), C] = M IN [A, M IN (B, C)]
M AX[M AX(A, B), C] = M AX [A, M AX(B, C)] (associative)
c) M IN (A, A) = A, M AX(A, A) = A (idempotent)
d) M IN [A, M AX(A, B)] = A and M AX[A, M IN (A, B)] = A (absorption)
e) M IN [A, M AX(B, C)] = M AX[M IN (A, B), M IN (A, C)]
M AX[A, M IN (B, C)] = M IN [M AX(A, B), M AX(A, C)] (distributive)(U.Q)
63
4.4. LATTICE OF FUZZY NUMBERS
Proof. a) MIN(A, B)(z) =
sup
min[A(x), B(y)]
z=min(x,y)
=
min[B(y), A(x)] = M IN (B, A)(z)
sup
z=min(y,x)
MIN(A, B) = M IN (B, A). Similarly M AX(A, B) = M AX(B, A)
b) For all z ∈ R
M IN [A, M IN (B, C)](z) =
sup
min[A(x), M IN (B, C)(y)]
z=min(x,y)
=
min[A(x),
sup
z=min(x,y)
=
sup
min[B(u), C(v)]]
y=min(u,v)
sup
sup
min[A(x), B(u), C(v)]
z=min(x,y)y=min(u,v)
=
min[A(x), B(u), C(v)]
sup
z=min(x,u,v)
=
sup
sup
min[A(x), B(u), C(v)]
z=min(s,v)s=min(x,u)
=
min[
sup
z=min(s,v)
=
sup
sup
min[A(x), B(u)], C(v)]
s=min(x,u)
min[M IN (A, B)(s), C(v)]
z=min(s,v)
= M IN [M IN (A, B), C](z)
Similarly M AX[M AX(A, B), C] = M AX [A, M AX(B, C)]
c) MIN(A, A)(z) =
sup
min[A(x), A(x)] = sup min[A(x), A(x)]
z=x
z=min(x,x)
M IN (A, A)(z) = A(z) ⇒ M IN (A, A) = A
Similarly M AX(A, A) = A.
d) M IN [A, M AX(A, B)](z) =
sup
min[A(x), M AX(A, B)(y)]
z=min(x,y)
=
min[A(x),
sup
z=min(x,y)
=
sup
sup
min(A(u), B(v))]
y=max(u,v)
min[A(x), A(u), B(v)]
z=min(x,max(u,v))
Let M denote the RHS of last equation. Since B is a fuzzy number there exist
v0 ∈ R such that B(v0 ) = 1.
By z = min[z, max(z, v0 )] we have M ≥ min[A(z), A(z), B(v0 )] = A(z)
On the other hand since z = min[x, max(u, v)] we have
min(x, u) ≤ z ≤ x ≤ max(x, u)
By the convexity of fuzzy numbers
A(z) ≥ min[A[min(x, u)], A[max(x, u)]]
= min[A(x), A(u)]
≥ min[A(x), A(u), B(v)] ⇒ A(z) ≥ M
Thus M = A(z) and consequently M IN [A, M AX(B, C)] = A
64
4.4. LATTICE OF FUZZY NUMBERS
Similarly we can prove M AX[A, M IN (A, B)] = A.
e) For any z ∈ R it is easy to see that
M IN [A, M AX(B, C)](z) =
sup
min[A(x), B(u), c(v)]————(1)
z=min(x,max(u,v))
M AX[M IN (A, B), M IN (A, C)](z) =
sup
min[A(m), B(n), A(s), C(t)]
z=max[min(m,n),min(s,t)]
————(2)
To prove that (1) & (2) are equal we first show that E ⊆ F where
E = {min[A(x), B(u), c(v)]/min[x, max(u, v)] = z}
F = {min[A(m), B(n), A(s), C(t)]/max[min(m, n), min(s, t)] = z}
For every a = min[A(x), B(u), C(v)] (a ∈ E) such that min[x, max(u, v)] = z
there exist m = s = x, n = u, t = v such that
max[min(m, n), min(s, t)] = max[min(x, u), min(x, v)]
= min[x, max(u, v)] = z
∴ a = min[A(x), B(u), A(x), C(v)] = min[A(m), B(n), A(s), C(t)] ∈ F
So, E ⊆ F . i.e., F ⊇ E
∴ M AX[M IN (A, B), M IN (A, C)] ≥ M IN [A, M AX(B, C)]
Now for any number b in F there exist a number a in E such that b ≤ a
For any b ∈ F , ∃ m, n, s, t such that max[min(m, n), min(s, t)] = z
b = min[A(m), B(n), A(s), C(t)]
We have
z = min[max(s, m), max(s, n), max(t, m), max(t, n)]
Let x = min[max(s, m), max(s, n), max(t, m)], u = n, v = t
z = min[x, max(u, v)]
On the other hand min(s, m) ≤ x ≤ max(s, m). By convexity of A,
A(x) ≥ min[A(min(s, m)), A(max(s, m))] = min[A(s), A(m)]
∃ a = min[A(x), B(u), C(v)] with min[x, max(u, v)] = z
i.e., a ∈ F, & a = min[A(x), B(u), C(v)] ≥ min[A(s), A(m), B(n), C(t)] = b
i.e., For any b ∈ F, ∃ a ∈ F such that b ≤ a ⇒ sup F ≤ sup E.
i..e, M AX[M IN (A, B), M IN (A, C)] ≤ M IN [A, M AX(B, C)]
Hence M IN [A, M AX(B, C)] = M AX[M IN (A, B), M IN (A, C)]
65
4.5. FUZZY EQUATIONS
4.5
Fuzzy equations
Explain about Fuzzy Equations. (U.Q)
Equations in which the coefficients and unknowns are called fuzzy numbers.
Then the equation is said to be fuzzy equations. These are the two basic fuzzy
equation
A + X = B,———(1)
A . X = B————(2)
Equation A + X = B
Let A = [a1 , a2 ], B = [b1 , b2 ], X = [x1 , x2 ]
Then A + B = [a1 + b1 , a2 + b2 ],
B − A = [b1 − a2 , b2 − a1 ]
A + B − A = [b1 − a2 , b2 − a1 ] + [a1 , b1 ] = [a1 + b1 − a2 , b1 + b2 − a1 ] 6= B
So, X = B − A is not a solution of the equation.
By (1), [a1 , a2 ] + [x1 , x2 ] = [b1 , b2 ]
[a1 + x1 , a2 + x2 ] = [b1 , b2 ]
a1 + x1 = b1 ,
a2 + x2 = b2 ⇒ x1 = b1 − a1 ; x2 = b2 − a2
Thus X = [b1 − a1 , b2 − a2 ] if and only if b1 − a1 ≤ b2 − a2 .
Since every fuzzy number can be represented by an α− cut we can extend the
above definition for α− cut also.
For any α ∈ [0, 1] let α A = [α a1 , α a2 ] and α B = [α b1 , α b2 ], α X = [α x1 , α x2 ].
Then the equation α A + α X = α B has solution X = [α b1 − α a1 , α b2 − α a2 ] iff
(i) α b1 − α a1 ≤ α b2 − α a2
(ii) α ≤ β implies α b1 − α a1 ≤ β b1 − β a1 ≤ β b2 − β a2 ≤ α b2 − α a2 .
Example 4.5.1. Let A and B be following fuzzy numbers
A=
1
0.6
0.8
0.9
0.5
0.1
0.2
+
+
+
+ +
+
[0, 1) [1, 2) [2, 3) [3, 4) 4 (4, 5] (5, 6]
B=
0.1
0.2
0.6
0.7
0.8
0.9 1 0.5
0.4
0.2
0.1
+
+
+
+
+
+ +
+
+
+
[0, 1) [1, 2) [2, 3) [3, 4) [4, 5) [5, 6) 6 (6, 7] (7, 8] (8, 9] (9, 10]
66
4.5. FUZZY EQUATIONS
α
α
A
α
B
α
X
0.1 [0, 6]
[0, 10]
[0, 4]
0.2 [0, 5]
[1, 9]
[1, 4]
0.3 [1, 5]
[2, 8]
[1, 3]
0.4 [1, 5]
[2, 8]
[1, 3]
0.5 [1, 5]
[2, 7]
[1, 2]
0.6 [1, 4]
[2, 6]
[1, 2]
0.7 [2, 4]
[3, 6]
[1, 2]
0.8 [2, 4]
[4, 6]
[2, 2]
0.9 [3, 4]
[5, 6]
[2, 2]
1.0 [4, 4]
[6, 6]
[2, 2]
The solution of equation is the fuzzy number
S
0.1
1
0.7
0.4
0.2
+
+ +
+
.
X = α∈(0,1] X =
[0, 1) [1, 2) 2 (2, 3] (3, 4]
Equation A.X = B
Let A, B are fuzzy numbers on R+ . It is easy to show that X = B/A is not a
solution of the equation A.X = B
For any α ∈ [0, 1] let α A = [α a1 , α a2 ] and α B = [α b1 , α b2 ], α X = [α x1 , α x2 ].
Then the equation α A.α X = α B has solution iff
(i) α b1 /α a1 ≤ α b2 /α a2
(ii) α ≤ β implies α b1 /α a1 ≤ β b1 /β a1 ≤ β b2 /β a2 ≤ α b2 /α a2 .




0
x ≤ 3, x > 5
0






Example 4.5.2. A(x) = x − 3 3 < x ≤ 4 B(x) = x−12
8






5 − x 4 < x ≤ 5
 32−x
12
Find X such that A.X = B.
We have α A = [3 + α, 5 − α] and α B = [8α + 12, 32 − 12α]
x ≤ 12, x > 32
12 < x ≤ 20
20 < x ≤ 32
8α + 12 32 − 12α
32 − 12α
8α + 12
α
≤
. We have X =
,
Clearly
3+α
5−α
3+α
5−α
8α + 12
= x ⇒ 8α + 12 = 3x + αx ⇒ 8α − αx + 12 − 3x = 0
3+α
67
4.5. FUZZY EQUATIONS
12 − 3x
α(x − 8) = 12 − 3x ⇒ α =
x−8
32 − 12α
= x ⇒ 32 − 12α = 5x − αx ⇒ 32 + αx − 12α − 5x = 0
5−α
α(x − 12) = 5x − 32 ⇒ α = 32−5x
12−x
12 − 3x
12 − 3x
= 0 ⇒ x = 4 and
=1⇒x=5
x−8
x−8
32
32 − 5x
32 − 5x
=0⇒x=
and
=1⇒x=5
12 − x
5  12 − x

0
x ≤ 4, x > 32/5



S
Thus, X = α∈(0,1] X = 12−3x
4<x≤5
x−8



 32−5x 5 < x ≤ 32/5
12−x
Problem 4.5.3. A(x) =


0



x+1
2



 3−x
2
Find A.B, A/B. (U.Q)
x ≤ −1, x > 3
−1 < x ≤ 1
B(x) =
1<x≤3


0



x−1
2



 5−x
Solution: α A = [2α − 1, 3 − 2α], α B = [2α + 1, 5 − 2α]
α
2
x ≤ 1, x > 5
1<x≤3
3<x≤5
α
(A + B) = [4α,
4α]
 8 − 4α], (A − B) = [4α − 6, 2 −


0
x ≤ 0, x > 8
0
x ≤ −6, x > 2






(A + B)(x) = x4
0 < x ≤ 4 (A − B)(x) = x+6
−6 < x ≤ −2
4






 8−x 4 < x ≤ 8
 2−x −2 < x ≤ 2
4
4
α
(A.B) = [−4α2 + 12α − 5, 4α2 − 16α + 15] or [4α2 − 1, 4α2 − 16α + 15]
−4α2 + 12α − 5 = x ⇒ −4α2 − 12α − (5 + x) = 0
α=
12±
√
3+ 4−x
2
√
√
144−4×4×(5+x)
8
=
√
3± 4−x
2
√
3± 4−x
2
=0
√
4 − x = −3
x = −5
4α2 − 1 = x ⇒ α =
q
= 0.5
4 − x = −2
x=0
x+1
4
68
4.5. FUZZY EQUATIONS
q
√
x+1
4
q
= 0.5
√
x+1=1
x=0
x+1
4
=1
x+1=2
x=3
2
2
4α − 16α + 15 = x ⇒ 4α − 16α + 15 − x = 0
α=
16±
√
256−4×4×(15−x)
8

(A.B)(x) =

0



3+(4−x)1/2
2



 4+(1−x)1/2
2
α
(A/B) = [
2α−1
2α+1
=
√
4+ 1+x
2
x < −5, x > 15
a−5<x≤0
0≤x≤3
2α − 1 3 − 2α
2α − 1 3 − 2α
,
] or [
,
]
2α + 1 2α + 1
5 − 2α 2α + 1
= x ⇒ 2α − 1 = x(2α + 1) ⇒ α =
x+1
2(1−x)
x+1
2(1−x)
=0
= 0.5
x = −1
x=0
5x+1
2(1+x)
5x+1
2(1+x)
2α−1
5−2α
= x ⇒ 2α − 1 = 5x − 2αx ⇒ α =
= 0.5,
x=0
3−2α
2α+1
5x+1
2(1+x)
=1
x = 1/3
= x ⇒ 3 − 2α = 2αx = x ⇒ α =
3−x
2(1+x)
x+1
2(1−x)
3−x
2(1+x)
=0
x=3
(A/B)(x) =


0





 x+1
2−2x

5x+1


2+2x



 3−x
2+2x
3−x
2(1+x)
=1
x = 1/3
x < −1, x ≥ 3
−1 ≤ x ≤ 0
0 ≤ x ≤ 1/3
1/3 ≤ x ≤ 3
69
4.5. FUZZY EQUATIONS
Problem 4.5.4. A(x) =
C(x) =

x−6



 2
10−x
2



0

x+2



 2
2−x
2



0
−2 < x ≤ 0
B(x) =
0<x<2
otherwise
6<x≤8
8 < x ≤ 10

x−2



 2
6−x
2



0
Find X if A + X = B
2<x≤4
0<x≤6
otherwise
(ii)B.X = C
otherwise
Solution. (i) α A = [2α − 2, 2 − 2α], α B = [2α + 2, 6 − 2α], α C = [2α + 6, 10 − 2α]
α
α
A + α B = [4α, 8 − 4α],
α
A − α B = [4α − 8, 4α]
A + X = α B ⇒ [2α − 2, 2 − 2α] + [x1 , x2 ] = [2α + 2, 6 − 2α]
⇒ [2α − 2 + x1 , 2 − 2α + x2 ] = [2α + 2, 6 − 2α]
2α − 2 + x1 = 2α + 2
⇒ x1 = 4
α
α
2 − 2α + x2 = 6 − 2α
x2 = 4. So, X = [4, 4]
α
B. X = C ⇒ [2α + 2, 6 − 2α].[x1 , x2 ] = [2α + 6, 10 − 2α]
[x1 , x2 ] = [
x=
2α + 6 10 − 2α
,
]
2α + 2 6 − 2α
2α + 6
3−x
⇒α=
2α + 2
x−1
and
3−x
3−x
= 0,
=1
x−1
x−1
Hence X(x) =
10 − 2α
3x − 5
⇒α=
6 − 2α
x−1
3x − 5
3x − 5
= 0,
=1
x−1
x−1
x = 3, x = 2
2≤x<3
x=
x = 5/3, x = 2

5−3x


 1−x

3−x
x−1



0
5/3 ≤ x < 2
5/3 ≤ x < 2
2≤x<3
otherwise
********
70
Chapter 5
Unit-V
5.1
Individual decision making problem
There are several fuzzy model decision making in which relevant goods and constraints are expressed in terms of fuzzy sets. Also we can determine a decision by
an appropriate aggregation of these fuzzy sets. Therefore a decision situation can
be characterised by the following components.
(i) a set A of possible events / actions.
(ii) a set of goals Gi each of which is expressed in terms of fuzzy set defined on
A
(iii) a set of constraints cj each of which is expressed by a fuzzy set defined on A
0
0
Let Gi and Cj be fuzzy sets defined on set Xi and Yj respectively. Also assume
that these fuzzy sets represent goals and constraints expressed by the decision makers./ Then for each i, j ∈ N the meanings of action in a set A in terms of sets Xi
and Yj can be defined by functions
gi : A → Xi , cj = A → Yi
0
and express goals Gi and constraints Cj by the compositions of gi with Gi and
0
the compositions of cj and Cj
0
0
i.e., Gi (a) = Gi (gi (a)) and Cj (a) = Cj (cj (a)) for each a ∈ A.
Definition 5.1.1. Given a decision situation characterized by fuzzy sets A, Gi
(i ∈ Nn ) and Cj (j ∈ Nm ) a fuzzy decision D is conceived as a fuzzy set on A that
simultaneously satisfies the given goals Gi and constraints Cj
71
5.1. INDIVIDUAL DECISION MAKING PROBLEM
i.e., D(a) = min[inf Gi (a), inf Cj (a)], a ∈ A provided that the standard operator
i∈N
j∈N
of fuzzy intersection is employed.
Use of weighted coefficients
The fuzzy model can be extended to accommodate the relative importance of
the various goals and constraints by the use of weighted coefficients. then we define
the decision D as follows.
P
P
D(a) = ni=1 ui Gi (a) + nj=1 vj Cj (a), a ∈ ×———–(1)
where ui , vj are non negative weights attached to each fuzzy goals Gi (∈ N ) and
P
P
each fuzzy constraints cj (j ∈ N ) respectively such that
ui + vj = 1
Remark 5.1.2.
(i) Formula (1) can also be written as D(a) =
Pn
i=1
Gi ui (a)+
Pn
j=1 cj vj (a), a
∈ Ã
(ii) In 1970, Bellman and Zadeh suggest a fuzzy model of decision making in
which relevant goals and constraints are expressed in terms of fuzzy sets.
(iii) Once a fuzzy decision has been arrived at it may be necessary to choose the
best single crisp alternative to the fuzzy set.
Problem 5.1.3. Suppose that an individual needs to decide which of 4 possible jobs
a1 , a2 , a3 , a4 to choose. Assume that the salary of each job is given by the assignment
g(a1 ) = $40, 000; g(a2 ) = $45, 000; g(a3 ) = $50, 000; g(a4 ) = $60, 000. Assume that
the individual assigns to the 4 jobs in A. The following grades is the fuzzy set of
interesting job. c1 =
0.4
a1
+ 0.6
+ 0.2
+ 0.2
. The driving distance of the 4 jobs are given
a2
a3
a4
by c2 (a1 ) = 27;2 (a2 ) = 7.5; c2 (a3 ) = 12; c2 (a4 ) = 2.5 miles. Choose a job that offers
high salary, interesting and within close driving distance.
Solution:
The given data can be tabulated as follows.
Criteria
a1
a2
a3
a4
salary C1
40,000
45,0000
50,000
60,000
Distance C2
27
7.5
12
2.5
Interest C3
??
???
?
?
We define fuzzy set for the goal high salary [35, 000, 55000] and constraints interesting job and close driving distance [5, 30].
72
5.2. MULTI-PERSON DECISION MAKING PROBLEM
The table of fuzzy set is given as follows
Criteria
a1
a2
a3
a4
salary C1
0.11
0.3
0.48
0.8
Distance C2
0.1
0.9
0.7
1
Interest C3 0.4 0.6 0.2 0.2
0
Composing now, the functions g and G we obtain the fuzzy set G =
0.48
a3
+
0.8
a4
0.11
a1
which expresses the goal in terms of the available jobs in set A.
0
By composition the functions c2 and C2 we obtain the fuzzy set c2 =
0.7
a3
+
1
a4
+ 0.3
+
a2
0.1
a1
+
0.9
a2
+
which expresses the constraints in terms of set A.
D(a1 ) = min[G(a1 ), min[c1 (a1 ), c2 (a1 )]] = min[0.11, min[0.1, 0.4]] = 0.1
D(a2 ) = min[G(a) , min[c1 (a2 ), c2 (a2 )]] = min[0.3, min[0.9, 0.6]] = 0.3
D(a3 ) = min[G(a3 ), min[c1 (a3 ), c2 (a3 )]] = min[0.48, min[0.7, 0.2]] = 0.2
D(a4 ) = min[G(a4 ), min[c1 (a4 ), c2 (a4 )]] = min[0.8, min[1, 0.2]] = 0.2
So, D =
0.1
a1
+
0.3
a2
+
0.2
a3
+
0.2
a4
represents the fuzzy characterisation of concept of
desirable job.
a2 is the most desirable job among 4 available jobs.
5.2
Multi-person decision making problem
(U.Q) Let us assume that each member of a group has n individuals decision
makers in reflexive, antisymmetric and transitive preference ordering Pk : k ∈ N
which is totally or partially ordered set X of alternatives. We must found a social
choice function which gives the individual preference ordering, produces the most
acceptable overall group preference ordering.
To deal with the multiplicity of opinion, evidenced in the group the social
preference S may be defined as follows. S : X × X → [0, 1]
which gives the membership grade S(xi , xj ) indicating the degree of group
preference of alternatives xi over xj .
Methods for finding S(xi , xj )
(i) The simple method computes the relative popularity xi over xj by dividing
the number of persons preferring xi to xj denoted by N (xi , xj ) by the total number
73
5.2. MULTI-PERSON DECISION MAKING PROBLEM
of decision makers n.
∴ S(xi , xj ) =
N (xi ,xj )
n
(ii) Let > represents the preference ordering one individual k who exercises complete control over the group decision. Then a dictarial situation can be modeled by
the group preference
relation S for which

1 xi > xk f orsome individual k
S(xi , xj ) =
0 otherwise
(iii) When we have defined the fuzzy relationship S then the final non fuzzy
S
group preference can be determined by using the following formula S = α α S
where α S is the crisp relation comprising the α− cuts of fuzzy relation S scaled by
α. Further α represents the level of agreement between the individual concerning
the particular crisp ordering α S.
Working Aid
(i) The procedure that maximize the final agrement level consist of intersecting
the classes of crisp total ordering that are compatible with the pairs in the α− cuts
α
S for increasingly smaller values of α until a single crisp total ordering is obtained.
(ii) Removed any pair (xi , xj ) if it leads to an intransitivity.
(iii) The largest value of α for which the unique compatible ordering on X × X is
found represents the maximised agreement level of the group and the crisp ordering
itself represents the group decision.
Problem 5.2.1. Assume that each individual of a group of 8 decision makers has the
total preference ordering Pi (i
X = {w, x, y, z} as follows.
=
1, 2, . . . , 8) on a set of alternatives
P1 = {w, x, y, z}; P2 = P5 = {z, y, x, w};
P3 = P7 = {x, w, y, z}; P4 = P8 = {w, z, x, y}; P6 = {z, w, x, y}. Use the fuzzy
multiperson decision making to determine the group decision. (U.Q)
Solution: WKT S(xi , xj ) =
N (xi ,xj )
n
where N (xi , xj ) means xi followed by xj
N (w, x) = {P1 , P4 , P8 , P6 } ⇒ S(w, x) = 4/8 = 0.5
N (w, y) = {P1 , P3 , P7 , P4 , P8 , P6 } ⇒ S(w, y) = 6/8 = 0.75
N (w, z) = {P1 , P3 , P7 , P4 , P8 } ⇒ S(w, z) = 5/8 = 0.625
N (x, w) = {P2 , P5 , P3 , P7 } ⇒ S(x, w) = 4/8 = 0.5
N (x, y) = {P1 , P3 , P7 , P4 , P8 , P6 } ⇒ S(x, y) = 6/8 = 0.75
74
5.2. MULTI-PERSON DECISION MAKING PROBLEM
N (x, z) = {P1 , P3 , P7 } ⇒ S(x, z) = 3/8 = 0.375
N (y, w) = {P2 , P5 } ⇒ S(y, w) = 2/8 = 0.25
N (y, x) = {P2 , P5 } ⇒ S(y, x) = 2/8 = 0.25
N (y, z) = {P1 , P3 , P7 } ⇒ S(y, z) = 3/8 = 0.375
N (z, w) = {P2 , P5 , P6 } ⇒ S(z, w) = 3/8 = 0.375
N (z, x) = {P2 , P5 , P4 , P8 , P6 } ⇒ S(z, x) = 5/8 = 0.625
N (z, y) = {P2 , P5 , P4 , P8 , P6 } ⇒ S(z, y) = 5/8 = 0.625
x
y
z 
 w
w
0
0.5
0.75 0.625


x
0.5
0
0.75 0.375


S=

y 0.25 0.25
0
0.375

z
0.375 0.625 0.625
0
α− cuts:
1
S = N (xi , xj) ≥ 1 = φ,
0.625
0.5
0.75
S = N (xi , xj) ≥ 0.75 = {(w, y), (x, y)}
S = N (xi , xj) ≥ 0.625 = {(w, y), (x, y), (w, z), (z, x), (z, y)}
S = N (xi , xj) ≥ 0.5 = {(w, y), (x, y), (w, z), (z, x), (z, y), (w, x), (x, w)}
0.375
0.25
S = {(w, y), (x, y), (w, z), (z, x), (z, y), (w, x), (x, w), (x, z), (y, z), (z, w)}
S = {(w, y), (x, y), (w, z), (z, x), (z, y), (w, x), (x, w), (x, z), (y, z), (z, w), (y, w), (y, x)}
X × X = {(w, x, y, z), (w, x, z, y), (w, y, x, z), (w, y, z, x), (w, z, x, y), (w, z, y, x)
(x, w, y, z), (x, w, z, y), (x, y, w, z), (x, y, z, w), (x, z, w, y), (x, z, y, w)}
(y, w, x, z), (y, w, z, x), (y, x, w, z), (y, x, z, w), (y, z, x, w), (y, z, w, x)
(z, w, x, y), (z, w, y, x), (z, x, w, y), (z, x, y, w), (z, y, w, x), (z, y, x, w)
To determine the group choice we are going to find the crisp ordering satisfies
above α−cuts.
1
O =X ×X
0.75
O = {(w, x, y, z), (w, x, z, y), (w, z, x, y), (z, w, x, y), (z, x, w, y), (x, w, y, z), (x, w, z, y),
(x, z, w, y)}
0.625
O = {(w, z, x, y)}
The best choice =1 O(n)0.75 ∩ O(n) ∩ 0.625 O = (w, z, x, y).
Group level argument value =0.625.
Problem 5.2.2. Assume that each individual of a group of the 5 judges has a total
preference ordering Pi , i ∈ N5 on four figure skaters X = {a, b, c, d}. The orderings
75
5.2. MULTI-PERSON DECISION MAKING PROBLEM
are
P1 = (a, b, d, c); P2 = (a, c, d, b); P3 = (b, a, c, d) = P5 ; P4 = (a, d, b, c). Use fuzzy
multi person decision making to determine the group decision.
Solution: By given data, we have,
b
c
d
a
a
0 0.6 1
1



b 
0.4
0
0.8
0.6


S=

c 
0
0.2
0
0.6


d
0 0.4 0.4 0
1
S = {(a, c), (a, d)}
0.8
0.6
0.4
S = {(b, c), (a, c), (a, d)}
S = {(b, c), (a, c), (a, d), (a, b), (b, d), (c, d)}
S = {(a, b), (c, d), (b, d), (b, c), (a, c), (a, d), (a, b), (b, d), (c, d), (c, b)}
X × X = {(a, b, c, d), (a, b, d, c), (a, c, b, d), (a, c, d, b), (a, d, c, b), (a, d, b, c),
(b, a, c, d), (b, a, d, c), (b, c, a, d), (b, c, d, a), (b, d, a, c), (b, d, c, a)
(c, a, b, d), (c, a, d, b), (c, b, a, d), (c, b, d, a), (c, d, a, b), (c, d, b, a)
(d, a, b, c), (d, a, c, b), (d, b, a, c), (d, b, c, a), (d, c, a, b), (d, c, b, a)}
To determine the group choice we are going to find the crisp ordering satisfies
the above α− cuts.
1
O = {(a, b, c, d), (a, b, d, c), (a, c, b, d), (a, c, d, b), (a, d, c, b), (a, d, b, c), (b, c, a, d), (b, a, d, c)}
0.8
0.6
O = {(a, b, c, d), (a, b, d, c), (a, d, b, c), (b, a, c, d), (b, a, d, c)}
O = {(a, b, c, d)}
The best choice =1 O(n)0.8 ∩ O(n) ∩ 0.6 O = (a, b, c, d)
Group level argument value =0.6.
Problem 5.2.3. Consider 5 total packages a1 , a2 , a3 , a4 , a5 from which we want to
choose one. These costs are $1000, $3000, $10, 000, $5000, $7000. Their travel time
in hours are 15,10,28,10 and 15. Assume that they are viewed as interesting with
the degrees 0.4,0.3,1,0.6,0.5. Define your own fuzzy set of acceptable costs and your
own fuzzy set of acceptable travel time. Then determine the fuzzy set of interesting
travel packages whose costs and travel time are acceptable and with this set to
choose one of the 5 travel packages.
76
5.3.
FUZZY LINEAR PROGRAMMING PROBLEM
Solution: Let G be the salary and c1 , c2 be the travel time and interest about
job.


1
x ≥ 15000



0.625 0.7 0.875 0.75 0.8
x
,
,
,
,
G = 1 + 0.025( 1000
− 15) 0 ≤ x ≤ 15000 =

a1
a2
a3
a 4 a5


0
x<0
c1 = 0.025(30 − x) =
0.375 0.5 0.05 0.5 0.375
,
,
,
,
a1
a2 a 3 a 4
a5
x
1000
3000
10000
5000
7000
G
0.625
0.7
0.875
0.75
0.8
x
15
10
28
10
15
c1
0.375
0.5
0.05
0.5
0.375
From the above table, we see that package a1 is best choice.
5.3
Fuzzy Linear Programming problem
In this section, we discuss two special cases of fuzzy linear programming.
Define Fuzzy Linear Programming Problem. (U.Q)
Case(i): Fuzzy linear programming problem in which only RHS numbers Bi are
P
P
fuzzy numbers. The general form is max nj=1 cj xj such that nj=1 aj xj ≤ Bi , xi ≥ 0
Case(ii): Fuzzy linear programming problem in which RHS Bi and the coeffi-
cients Aij of the constraint matrix are fuzzy numbers.
Pn
Pn
j=1 cj xj such that
j=1 Aij xj ≤ Bi , xj ≥ 0.
In case (i), fuzzy numbers Bi typically have the form Bi (x) =


1



bi +pi −x
pi



0
x ≤ bi
bi < x < bi + pi
bi + pi ≤ x
The lower bound of the optimal values zl is obtained by solving the standard
P
LPP max z = cx such that nj=1 aij xj ≤ bi , xj ≥ 0.
77
5.3.
FUZZY LINEAR PROGRAMMING PROBLEM
The upper bound of the optimal values zu is obtained by similar LPP in which
bi is replaced with bi + pi .
max z = cx such that
Pn
j=1
aij xj ≤ bi + pi , xj ≥ 0.
Now the problem becomes following classical optimization problem.
P
max λ such that λ(zu − zl ) − cx ≤ −zl and λpi + ni=1 aij xj ≤ bi + pi & λ, xj ≥ 0.
Problem 5.3.1. Assume that a company makes two products. Product P1 has
a $0.40 per unit profit and product P2 has $0.30 per unit profit. Each unit of
product P1 requires twice as many labour hours as each product P2 . The total
available labour hours are at least 500 hours per day and may possibly be extended
to 600 hours per day due to special arrangements for overtime work. The supply
of material is at least sufficient for 400 units of both products P1 and P2 per day
but may possibly be extended to 500 unit per day according to previous experience.
The problem is how many units of products P1 and P2 should be made per day to
maximize the total profit
Solution: Let x1 , x2 denote the number of units of products P1 , P2 made in one
day respectively. Then the problem can be formulated as following LPP
max z = 0.4x1 + 0.3x2 (profit) subject to x1 + x2 ≤ B1 (material) and
2x1 + x2 ≤ B2 (labour) x1 , x2 ≥ 0 where B1 , B2 is defined
by



1
x ≤ 400
1
x ≤ 500






B1 (x) = 500−x
B2 (x) = 600−x
400 < x ≤ 500
500 < x ≤ 600
100
100






0
0
500 < x
600 < x
First we need to calculate the lower and upper bounds of the objective function
by solving the following two classical LPP
P1 : max zl = 0.4x1 + 0.3x2 subject to P2 : max zu = 0.4x1 + 0.3x2 subject to
x1 + x2 ≤ 400 —–(1) and
x1 + x2 ≤ 500———(3) and
2x1 + x2 ≤ 500 ———-(2)
2x1 + x2 ≤ 600——–(4)
(0,400),(400,0) are points on (1) and (0,500), (250,0) are points on (2).
The point of intersection of (1) and (2) is (100,300).
optimum value of zl = 130.
from (3) & (4) proceeding as above we get point of intersection (100, 400).
optimum value of zu = 160.
78
5.3.
FUZZY LINEAR PROGRAMMING PROBLEM
The fuzzy LPP becomes maximize λ such thatλ(zu − zl ) − cx ≤ −zl and λpi +
Pn
i=1 aij xj ≤ bi + pi & λ, xj ≥ 0.
⇒λ(160−130)−(0.4x1 +0.3x2 ) ≤ −130 such that 100λ+x1 +x2 ≤ 500———(5)
100λ + 2x1 + x2 ≤ 600—–(6), x1 , x2 ≥ 0.
We have 30λ − 0.4x1 − 0.3x2 = −130———(7).
6
5
5
4 A
B
4
(100, 300)
3
3
2
2
1
1
O
(100, 400)
C
1
2
3
4
1
5
2
3
4
5
6
To obtain maximum value of λ we solve (5),(6) & (7).
(6)-(5)⇒ x1 = 100; (7)+(0.3 × (5)) ⇒ λ = 0.5. From (6), x2 = 350.
The solution is x1 = 100, x2 = 350. Max profit = z = 145.
Problem 5.3.2. Solve the fuzzy linear programming problem min z = x1 − 2x2
subject to 3x1 − x2 ≥ 1, 2x1 + x2 ≤ 6,
0 ≤ x2 ≤ 2, 0 ≤ x1 .(U.Q)
Solution: Given min z = x1 − 2x2 subject to 3x1 − x2 ≥ 1, 2x1 + x2 ≤ 6,
0 ≤ x2 ≤ 2, 0 ≤ x1
Draw the constraints as straight line 3x1 − x2 = 1————(1)
Put x1 = 0, x2 = −1. The point is (0, −1)
Put x2 = 0, x1 = 1/3. The point is (1/3, 0)
Consider 2x1 + x2 = 6——————(2)
Put x1 = 0, x2 = 6. The point is (0, 6)
Put x2 = 0, x1 = 3. The point is (3, 0)
Also, 0 ≤ x2 ≤ 2 ⇒ point is (0,2) and x1 ≥ 0 ⇒ point is (0,0)
Since the first constraint inequality has ≥ sign, we have to take the area below
straight line (1). Since the second constraint inequality has ≤ sign we have to take
area under straight line (2). ABCD is the common region for all these constraints.
79
5.3.
FUZZY LINEAR PROGRAMMING PROBLEM
To find feasible region.
At (1,2), z = x1 − 2x2 ⇒ z = −3
At (2,2) z = x1 − 2x2 = −2
At (1/3,0) z = 1/3
At (3,0) z = 3.
The solution is x1 = 3, x2 = 0 and maximum profit is z = 3.
7
6
(1)
(2)
5
4
3
2
A
B
1
D
C
1
2
3
4
5
6
7
Problem 5.3.3. Solve the following fuzzy linear programming problem z = 0.5x1 +
0.2x2 subjectto x1 + x2 ≤ B1 , 2x1 + x2 ≤ B2 ,  x1 , x2 ≥ 0 where


1
x ≤ 300
1
x ≤ 400






B1 (x) = 400−x
300 < x ≤ 400 B2 (x) = 500−x
400 < x ≤ 500
100
100







0
0
400 < x
500 < x
(U.Q)
Solution: z = 0.5x1 +0.2x2 subject to x1 +x2 ≤ 300 —(1) & 2x1 +x2 ≤ 400—-(2)
Put x1 = 0 in (1), x2 = 300 (0,300)
Put x2 = 0 in (1), x1 = 300 (300,0)
Put x1 = 0 in (2), x2 = 400 (0,400)
Put x2 = 0 in (2), x1 = 200 (200,0)
At (0,0) z = 0.5(0) + 0.2(0) = 0
At (0,300), z = 0.5(0) + 0.2(300) = 60
At (100,200), z = 0.5(100) + 0.2(200) = 90
At (200,0), z = 0.5(200) + 0.2(0) = 100
80
5.3.
FUZZY LINEAR PROGRAMMING PROBLEM
Hence max zl = 100. Now,
z = 0.5x1 + 0.2x2 subject to x1 + x2 ≤ 400 —(1) & 2x1 + x2 ≤ 500—-(2)
Put x1 = 0 in (1), x2 = 400 (0,400)
Put x2 = 0 in (1), x1 = 400 (400,0)
Put x1 = 0 in (2), x2 = 500 (0,500)
Put x2 = 0 in (2), x1 = 250 (250,0)
5
5
4
4
(100,300)
3
3
2
(100,200)
2
1
1
1
2
3
4
1
5
2
3
4
5
At (0,0) z = 0.5(0) + 0.2(0) = 0
At (0,400), z = 0.5(0) + 0.2(400) = 80
At (100,300), z = 0.5(100) + 0.2(300) = 110
At (250,0), z = 0.5(250) + 0.2(0) = 125
Hence max zu = 125
Then the fuzzy l.p.p becomes λ(125 − 100) − cx ≤ −100
⇒ 25λ − (0.5x1 + 0.2x2 ) ≤ −100—————–(3)
100λ + x1 + x2 ≤ 400——-(4) and 100λ + 2x1 + x2 ≤ 500——–(5)
From (5) − (4), x1 = 100 ; (3) − (0.25 × (5)) ⇒ x2 = 277.78
(3)⇒ 25λ − (0.5(100) + 0.2(277.78)) = −100 ⇒ 25λ − 105.56 = −100
⇒ 25λ = 5.56 ⇒ λ = 0.22.
The optimum value is x1 = 100, x2 = 277.78, z = 105.56.
Problem 5.3.4. Solve the following LPP maxz = 5x1 + 4x2 such that h4, 2, 1ix1 +
h5, 3, 1ix2 ≤ h24, 5, 8i and h4, 1, 2ix1 + h1, 0.5, 1ix2 ≤ h12, 6, 3i, x1 , x2 ≥ 0. (U.Q)
Solution: The above LPP can be written as
max z = 5x1 + 4x2 subject to
81
5.3.
FUZZY LINEAR PROGRAMMING PROBLEM
4x1 + 5x2 ≤ 24;
4x1 + 5x2 = 24————(1)
4x1 + x2 ≤ 12;
4x1 + x2 = 12————(2)
(4 − 1)x1 + (1 − 0.5)x2 ≤ 12 − 6 ⇒ 3x1 + 0.5x2 ≤ 6
3x1 + 0.5x2 = 6———(4)
(4 − 2)x1 + (5 − 3)x2 ≤ 24 − 5 ⇒ 2x1 + 2x2 ≤ 19
2x1 + 2x2 = 19———-(3)
(4 + 1)x1 + (5 + 1)x2 ≤ 24 + 8 ⇒ 5x1 + 6x2 ≤ 32
5x1 + 6x2 = 32———–(5)
(4 + 2)x1 + (1 + 1)x2 ≤ 12 + 3 ⇒ 6x1 + 2x2 ≤ 15
6x1 + 2x2 = 15———-(6)
(0,24/5) and (6,0) are points on (1)
(0,12) and (2,0) are points on (4)
(0,12) and (3,0) are points on (2)
(0,16/5) and (32/5,0) are points on (5)
(0,19/2) and (19/2,0) are points on (3) (0,15/2) and (5/2,0) are points on (6)
12
11
10
9
8
7
6
(6)
5
4
3
(4)
(5)
2
(2)
(3)
1
0.5
1
1.5
2
2.5
3
(1)
3.5
4
4.5
5
5.5
To find B we solve (1) & (4). The point is (1.38,3.7)
At (0,0), z = 5(0) + 4(0) = 0
At (2,0) z = 5(2) + 4(0) = 10
At (0,24/5) z = 5(0) + 4(24/5) = 19.2
At (1.38,3.7) z = 5(1.38) + 4(3.7) = 21.7
The optimum value is x1 = 1.38, x2 = 3.7, z = 21.7.
***********
82
6
6.5
7
7.5 8
8.5
9
9.5