Cmjv05i03p0312 by Journal of Computer and Mathematical Sciences

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ISSN 0976-5727 (Print) ISSN 2319-8133 (Online) Abbr:J.Comp.&Math.Sci. 2014, Vol.5(3): Pg.312-320

Ranking Algorithm for Symmetric Octagonal Fuzzy Numbers Dhanalakshmi V. and Felbin C. Kennedy Research Scholar, Stella Maris College (Autonomous), Chennai, INDIA. Research Guide, Associate Professor, Stella Maris College (Autonomous), Chennai, INDIA. (Received on: June 27, 2014) ABSTRACT Ranking of fuzzy numbers play an important role in decision making, optimization, forecasting etc. Fuzzy numbers must be ranked before an action is taken by a decision maker. In this paper, we propose a ranking algorithm for symmetric octagonal fuzzy numbers, which results in equality only when the numbers coincide. Also the ranking algorithm is verified for some reasonable properties of ordering of fuzzy numbers Keywords: Symmetric Octagonal fuzzy number, ranking, mode, deviation, spread.

1. INTRODUCTION Ranking fuzzy numbers is an important task in analyzing fuzzy information in optimization, data mining, decision making and related areas. Unlike real numbers, fuzzy numbers have no natural order; also no ordering technique is conducive to all. Thus significant contributions have been made in ranking fuzzy numbers1–7,10,12,13. All the ranking ~ ~ procedures in literature defines A f B , ~ ~ ~ ~ ~ ~ A p B and A ≈ B , here A ≈ B implies the

two fuzzy numbers are equivalent, not necessarily be equal. That is, two different fuzzy numbers depending on the ranking method are infered as equivalent. But in many applications the two fuzzy numbers need to be shown different. In this paper, we have proposed a ranking algorithm for symmetric octagonal fuzzy number, in which we obtain the equality only when the two symmetric octagonal fuzzy numbers coincide. Some of the reasonable properties for ranking fuzzy numbers proposed in11 are verified for this algorithm.

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After the introduction, this paper is arranged as follows. In Section 2, the concept of fuzzy number and octagonal fuzzy numbers are recalled. Section 3 introduces the ranking algorithm for symmetric octagonal fuzzy numbers and the properties of the proposed algorithm are discussed in Section 4. The concluding remarks are presented in Section 5. 2. SYMMETRIC OCTAGONAL FUZZY NUMBERS

~ ii. A is normal i.e. ∃x0 ∈ Rsuch that

µ A~ ( x0 ) = 1 iii. µ A~ is piecewise continuous

~ Definition 2.3[9]A fuzzy number A is said to be a generalized octagonal fuzzy number denoted by

~ A = (a1, a2 , a3 , a4 , a5 , a6 , a7 , a8 ; k , w) where a1, a2 , a3 , a4 , a5 , a6 , a7 , a8 , k , w are

real numbers such that For the sake of completeness we recall the required definitions and results. Definition 2.1 [8]The characteristic function µ A~ of a crisp set A ⊆ X assigns a value either 0 or 1 to each member in X . This function can be generalized to a function µ A~ such that the value assigned to the element of the universal set X fall within a specified range i.e. µ A~ : X → [0,1]. The assigned value indicates the membership grade of the element in the set A . The function µ A~ is called the membership function and the set

~ A = {( x, µ A~ ( x)) : x ∈ X } is called a fuzzy

set.

a1 ≤ a2 ≤ a3 ≤ a4 ≤ a5 ≤ a6 ≤ a7 ≤ a8 and 0 < k < w and its membership function µ ~ is given by A

0   x − a1 k  a2 − a1  k   k ( a 4 − x ) + w ( x − a3 )  a4 − a3  w µ A~ ( x) =   k ( x − a5 ) + w(a6 − x)  a6 − a5  k   x − a8 k  a8 − a7   0

x ≤ a1 a1 ≤ x ≤ a2 a2 ≤ x ≤ a3 a3 ≤ x ≤ a4 a 4 ≤ x ≤ a5 a5 ≤ x ≤ a6 a6 ≤ x ≤ a7 a7 ≤ x ≤ a8 x ≥ a8

~ Definition 2.2 [8]A fuzzy set A ,defined on the universal set of real numberR, is said to be a fuzzy number if its membership function has the following characteristics: ~ i. A is convex i.e.

Definition 2.5A fuzzy number A is said to be a symmetric octagonal fuzzy number denoted

~ A = (a L , aU , r, s, t; k , w)

where a L ≤ aU and r, s, t are real numbers such that its representation as generalized octagonal fuzzy number is

µ A~ ( λ x1 + (1 − λ ) x 2 ) ≥ min ( µ A~ ( x1 ), µ A~ ( x 2 ))

( a L − r − s − t , a L − s − t , a L − t , a L , aU , aU + t ,

∀ x1 , x 2 ∈ R , ∀ λ ∈ [ 0,1]

aU + s + t , aU + r + s + t ; k , w ).

Dhanalakshmi V. V., et al., J. Comp. & Math. Sci. Vol.5 (3), 312-320 (2014))

Figure 1: Graphical representation of the symmetric octagonal fuzzy number

Definition 2.6 The α -cut cut of the symmetric octagonal fuzzy number

~ A = (a L , aU , r, s, t; k , w) given in

 ( A L ) , ( A R ) ] α ∈ [0, k ] ~ A = [ AαL , AαR ] =  αL 1 αR 1 ( Aα ) 2 , ( Aα ) 2 ] α ∈ (k ,1] where k −α , k k −α ( AαR )1 = aU + s + t + r , k

w −α , w−k w −α ( AαR ) 2 = aU + t w− k

( AαL ) 2 = a L − t

Definition 2.7 Let

3. RANKING OF SYMMETRIC OCTAGONAL FUZZY NUMBERS Definition 3.1 For any symmetric octagonal fuzzy number A = (a L , aU , r , s, t ; k , w) mode, divergence & spread at w w and k level are defined as : i.

~ w − Divergence( A) = w(aaU − a L + 2t )

iii.

~ Spread k ,w ( A ) = w t

iv.

~ k (a L + aU ) k − Mode( A ) = 2

symmetric octagonal fuzzy numbers, then their sum is defined as

vi.

~ ~ A + B = (aL + bL , aU + bU , r1 + r2 , s1 + s2 , t1 + t2 ;

w(a L + aU ) ~ w − Mode( A) = 2

ii.

~ A = (aL , aU , r1 , s1 , t1; k1 , w1 ) and ~ B = (bL , bU , r2 , s2 , t2 ; k2 , w2 ) be two

min(k1 , k2 ), min(w1 , w2 ))

1 (2,5,1,2,3; ,1) 2

Definition 2.5 is

( AαL )1 = a L − s − t − r

314

vii.

k (2aaU + 2t + s ) ~ k − rightmode( A) = 2 k (2aa L − 2t − s ) ~ k − leftmode( A) = 2 ~ 0 − Divergence( A)

= k (aU − aL + 2r + 2s + 2t )

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~ w − Divergence(B ) then go to Step 3

~ Spread 0, k ( A) = k r

viii.

else

Given any two symmetric octagonal fuzzy numbers we present here an algorithm to compare them.

Case(i): If w − Divergence( A) < ~ ~ ~ w − Divergence(B ) then A f B

ALGORITHM

Case(ii) : If w − Divergence( A) > ~ ~ ~ w − Divergence(B ) then A p B

Let A = (aL , aU , r1 , s1 , t1 ; k1 , w1 ) and

Case(iii): If w − Divergence( A) =

~ B = (bL , bU , r2 , s2 , t2 ; k2 , w2 ) be two

symmetric octagonal fuzzy numbers, then ~ use the following steps to compare A ~ and B

Step1: Find w − Mode(A) and

~ w − Mode(B)

~ w − Mode( A) ~ ~ Case(i): If ~ then A f B > w − Mode( B )

~ w − Divergence(B ) then go to Step 3 ~ Step 3: Find Spread k ,w ( A) and ~ Spreadk , w ( B ) ~ If w − Mode( A) ≥ 0 then ~ ~ Case(i): If Spread k ,w ( A) < Spreadk , w ( B ) ~ ~ then A f B

Case(ii): If w − Mode( A) < w − Mode( B ) ~ ~ then A p B

Case(ii): If Spread k ,w ( A) > ~ ~ ~ Spreadk , w ( B ) then A p B

~ w − Mode( A) Case(iii): If ~ then go to = w − Mode( B )

Case(iii): If Spread k , w ( A) =

~ Spreadk , w ( B ) then go to Step 4

else

Step 2

~ w − Divergence(A) ~ w − Divergence(B ) ~ If w − Mode( A) ≥ 0 then ~ Case(i): If w − Divergence( A) > ~ ~ ~ w − Divergence(B ) then A f B ~ Case(ii): If w − Divergence( A) < ~ ~ ~ w − Divergence(B ) then A p B ~ Case(iii): If w − Divergence( A) = Step2:

Find

and

Case(i): If Spread k ,w ( A) > Spreadk , w ( B ) ~ ~ then A f B Case(ii): If Spread k ,w ( A) < Spreadk , w ( B ) ~ ~ then A p B

Case(iii): If Spread k , w ( A) =

~ Spreadk , w ( B ) then go to Step 4

Step 4: Find w1 and w2

~ ~ Case (i): If w1 > w2 then A f B ~ ~ Case (ii): If w1 < w2 then A p B

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Case (iii): If w1 = w2 then go to Step 6

Step 5: Find k − Mode(A) and

~ k − Mode(B )

~ 0 − Divergence( A) and ~ 0 − Divergence( B ) ~ Case(i): If 0 − Divergence( A) > ~ ~ ~ 0 − Divergence( B ) then A f B ~ Case(ii): If 0 − Divergence( A) < ~ ~ ~ 0 − Divergence( B ) then A p B ~ Case(iii): If 0 − Divergence( A) = ~ 0 − Divergence( B ) then go to Step 9

Step 8: Find

Case(i): If k − Mode( A) > k − Mode( B ) ~ ~ then A f B Case(ii): If k − Mode( A) < k − Mode( B ) ~ ~ then A p B Case(iii): If k − Mode( A) = k − Mode( B ) then go to Step 6

Step 6: Find k − rightmode( A) and

~ k − rightmode(B )

~ k − rightmode( A) ~ ~ ~ then A f B > k − rightmode( B ) ~ k − rightmode( A) ~ ~ Case(ii): If ~ then A p B < k − rightmode( B ) ~ k − rightmode( A) Case(iii): If ~ then go to = k − rightmode( B )

Step 9: Find k1 and k2

~ ~ Case (i): If k1 > k 2 then A f B ~ ~ Case (ii): If k1 < k2 then A p B ~ ~ Case (iii): If k1 = k 2 then A ≈ B

Case(i): If

Step 7

Let A = (aL , aU , r1 , s1 , t1 ; k1 , w1 ) and

~ k − leftmode(B )

~ k − leftmode( A) ~ ~ Case(i): If ~ then A f B > k − leftmode( B ) ~ k − leftmode( A) ~ ~ Case(ii): If ~ then A p B < k − leftmode( B )

Step 8

Proposition 4.1: If equality holds in all the nine steps of the above algorithm, then the two symmetric octagonal fuzzy numbers are equal. Proof:

Step7: Find k − leftmode(A) and

~ k − leftmode( A) Case(iii): If ~ then = k − leftmode( B )

4. PROPERTIES OF THE RANKING ALGORITHM

~ B = (bL , bU , r2 , s2 , t2 ; k2 , w2 ) be two

symmetric octagonal fuzzy numbers, such that all the nine steps of the above algorithm

leads equality. To prove that A and B are equal, i.e. to prove that aL = bL , aU = bU , r1 = r2 , s1 = s2 , t1 = t 2 ,

go to

k1 = k 2 , w1 = w2 .

From step 4,

w1 = w2

(1)

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From step 9,

k1 = k 2

(2)

From step 3 and equation (1),

t1 = t2

Proposition 4.3: Let (3)

From step 1 and equation (1), ~ ~ w − Mode( A) = w − Mode( B ) w1 (a L + aU ) w2 (bL + bU ) = 2 2 ⇒ a L + aU = bL + bU

⇒

(4)

From step 2 and equation (1) and (3), ~ ~ w − Divergence( A) = w − Divergence( B ) ⇒ w1 (aU − aL + 2t1 ) = w2 (bU − bL + 2t 2 ) ⇒ aU − a L = bU − bL

(5)

(6)

From step 6 and equations (3) and (6), ~ ~ k − rightmode ( A ) = k − rightmode( B ) k1 (2aU + 2t1 + s1 ) k 2 (2bU + 2t 2 + s 2 ) = 2 2 ⇒ s1 = s 2

~ A = (aL , aU , r1, s1, t1; k1, w1 ) , ~ B = (bL , bU , r2 , s2 , t2 ; k2 , w2 ) and ~ C = (cL , cU , r3 , s3 , t3; k3 , w3 ) be three

symmetric octagonal fuzzy numbers such ~ ~ ~ ~ ~ ~ that A f B and B f C , then A f C . Proof: Case (i): Let the order be determined in step 1, i.e ~ ~ ~ ~ A f B by w − Mode( A) > w − Mode( B ) ~ ~ and B f C by

~ ~ w − Mode( B) > w − Mode(C ) ,

From equations (4) and (5),

aU = bU , aL = bL

of symmetric octagonal fuzzy numbers is obtained by the proposed algorithm.

⇒

(7)

then ~ ~ ~ w − Mode( A) > w − Mode( B) > w − Mode(C ) ~ ~ ⇒ w − Mode( A) > w − Mode(C ) ~ ~ ⇒ AfC Case (ii): ~ ~ ~ ~ Suppose A f B by step 1 and B f C in step i say i=2,3,...,9

From step 8 and equations (3), (6) and (7), ~ ~ 0 − Divergence( A) = 0 − Divergence( B ) ⇒ k1 (aU − a L + 2r1 + 2s1 + 2t1 ) = k 2 (bU − bL + 2r2 + 2s2 + 2t 2 ) ⇒ r1 = r2

Hence the proof. Remark 4.2: In the above theorem, we have proved that equality, not just the equivalence

i.e. w − Mode( A) > w − Mode( B ) and

~ ~ w − Mode( B) = w − Mode(C ) and the

inequality happens in the later steps, then ~ ~ ~ w − Mode( A) > w − Mode( B) = w − Mode(C ) ~ ~ ⇒ w − Mode( A) > w − Mode(C )

~ ~ ⇒ AfC ~ ~ ~ ~ w − Mode( A) = w − Mode( B ) and

Similarly, A f C if

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~ ~ w − Mode( B) > w − Mode(C ) Case (iii): Similar to case (ii), it can be proved that ~ ~ ~ ~ ~ ~ A f C if A f B by step i and B f C in ~ ~ step j where j=i+1,...,9or B f C by step i ~ ~ and A f B in step j where j=i+1,...,9. ~ ~ ~ ~ Thus in all possibilities, A f B and B f C ~ ~ ⇒ AfC. Hence the proof.

Proposition 4.4: Let A and B be two symmetric octagonal fuzzy numbers with same height and ~ ~ ~ ~ inf supp( A ) > sup supp( B ), then A f B . Proof:

~ ~ ⇒ w − Mode( A) > w − Mode( B ) ~ ~ ⇒ A f B. Hence the proof.

Proof: Let A = (aL , aU , r1, s1, t1; k1, w1 ) ,

~ B = (bL , bU , r2 , s2 , t2 ; k2 , w2 ) and ~ C = (cL , cU , r3 , s3 , t3; k3 , w3 ) be the

three given symmetric octagonal fuzzy numbers. By hypothesis, k1 = k 2 = k 3 ~ ~ and w1 = w2 = w3 Suppose A f B happens in step 1, then

Let A = (aL , aU , r1 , s1 , t1 ; k1 , w1 ) and

~ ~ w − Mode( A) > w − Mode( B )

be the two given symmetric octagonal fuzzy numbers.

⇒

~ B = (bL , bU , r2 , s2 , t2 ; k2 , w2 )

Since they have the same height,

w1 = w2

⇒ a L > a L − r1 − s1 − t1 > bU + r2 + s 2 + t 2 > bU

(8)

~ ⇒ aU > a L > bU > bL , by definition of A

aU + a L > bL + bU

min( w1 , w 3 )( a L + aU + c L + cU ) 2 min( w 2 , w 3 )( b L + bU + c L + cU ) > 2 ~ ~ ~ ~ ⇒ w − Mode ( A + C ) > w − Mode ( B + C ) ~ ~ ~ ~ ⇒ A + C f B + C.

~ ~ Suppose A f B happens in step 2, then

and B

Adding (8) and (9),

⇒ a L + aU + c L + cU > b L + bU + c L + cU ,

⇒

⇒ a L − r1 − s1 − t1 > bU + r2 + s 2 + t 2

⇒ aU > bL

w1 ( a L + aU ) w 2 ( b L + bU ) > 2 2 ⇒ a L + aU > b L + bU adding c L + cU both sides

~ ~ inf supp( A ) > sup supp( B )

⇒ aL > bU

Proposition 4.5: Let A , B and C be three symmetric octagonal fuzzy numbers with ~ ~ same k and w then A f B ~ ~ ~ ~ ⇒ A + C f B + C.

(9)

~ ~ w − Mode( A) = w − Mode( B ) and ~ ~ w − Divergence( A) > w − Divergence( B ) ~ ~ w − Mode( A) = w − Mode( B )

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w1 (a L + aU ) w2 (bL + bU ) = 2 2 ⇒ a L + aU = bL + bU

5. CONCLUSION

⇒

⇒ a L + aU + cL + cU = bL + bU + cL + cU ~ ~ ~ ~ ⇒ w − Mode( A + C ) = w − Mode( B + C ) and ~ ~ w − Divergence( A) > w − Divergence( B ) ⇒ a U − a L + 2 t 1 > bU − b L + 2 t 2 ⇒ a U − a L + 2 t1 + c U − c L + 2 t 3 > bU − b L + 2 t 2 + c U − c L + 2 t 3 ~ ~ ⇒ w − Divergence ( A + C ) ~ ~ > w − Divergence ( B + C ) ~ ~ ~ ~ ⇒ A + C f B + C.

In a similar manner, the result can be proved for all cases. Remark 4.6: The above Proposition may not hold for arbitrary k and w, for

1 2 1 1 ~ B = (2,4,1,1,1; , ) and C~ = (1,2,1,1,1; 1 , 1 ) , 4 2 4 2 ~ ~ then A f B as ~

Let A = (1,4,1,1,1; ,1) ,

~ 5 5 ~ w − Mode( A) = > = w − Mode( B ) , 2 4

1 1 4 2 1 1 ~ ~ B + C = (3,6,2,2,2; , ) 4 2 ~

but A + C = (2,6,2,2,2; , ) and

~ ~ 9 ~ ~ ⇒ w − Mode( A + C ) = 2 < = w − Mode( B + C ) 4 ~ ~ ~ ~ ⇒ A + C p B + C.

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