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On OWA Linear Operators for Decision Making
On OWA Linear Operators for Decision Making
Cables Pérez, E. H.; Lamata, M. T.; Pelta, D.; Verdegay, J. L.
2018-01-02 00:00:00
FUZZY INFORMATION AND ENGINEERING 2018, VOL. 10, NO. 1, 80–90 https://doi.org/10.1080/16168658.2018.1509521 a b b b E. H. Cables Pérez ,M.T.Lamata ,D.Pelta and J. L. Verdegay a b Facultad de Ingeniería de Sistemas, Universidad Antonio Nariño, Bogotá, Colombia; Departamento de CCIA, Universidad de Granada, Granada, Spain ABSTRACT ARTICLE HISTORY Received 3 January 2018 When we deal with problems of decision making where we need Revised 1 February 2018 to give an order of alternatives and the rewards (number of points Accepted 11 February 2018 earned in a race) are not associated with the criteria, but with posi- tions, it makes sense to deal with ordered weighted averaging (OWA) KEYWORDS operators. Within the class of OWA operators we can consider those OWA operators; linear that are based on a linear function. This paper is related to a study functions; decision making of the properties that verify the linear functions set. Taking these properties into account we shall see how the final classification for drivers in Formula One changes when the number of points earned in a race is given by means of a linear function as opposed to the current assignments. 1. Introduction Decision making consists in choosing the best alternative from a given set of alterna- tives, options, candidates, etc., or in obtaining the ranking among them. A decision-making problem takes into account criteria, alternatives and weights. Consider a set of candidates such as Formula One drivers and let us see what happens if instead of assigning the following points {25, 18, 15, 12, 10, 8, 6, 4, 2, 1},wemakealinear assignment: {10, 9, 8, 7, 6, 5, 4, 3, 2, 1}. Our work is based on making a study of those linear functions f = ai + b that give ori- gin to ordered weighted averagings (OWAs) operators. Among those that have a linear behaviour the Borda–Kendal law can be considered as an example: 2(n + 1 − i) −2 2 w = = i + n(n + 1) (1) n(n + 1) n(n + 1) n + 1 or 2(n − i + 1) −2 2n + 1 w = = i + .(2) 2 2 2 n n n For whatever n is selected, it is satisfied that all w are aligned and is satisfied that w = c, i 1 w = w − c, ∀ i. This situation served as the basis to study the behaviour of other linear i i−1 functions which constitute a straight line and contain Equations (5) and (2), with the only CONTACT E. H. Cables Pérez ehcables@uan.edu.co © 2018 The Author(s). Published by Taylor & Francis Group on behalf of the Fuzzy Information and Engineering Branch of the Operations Research Society of China & Operations Research Society of Guangdong Province. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. FUZZY INFORMATION AND ENGINEERING 81 condition that its behaviour is linear. In this way, we may obtain a function class that con- tains a set of infinite linear functions F that allows the w to be obtained, thus completing L i the properties of the Ordered Weighted Averaging (OWA). The idea is to assign values to each of the positions depending on the OWA operator obtained. This paper is organised as follows. Section 2 provides a brief summary about the OWA operator. Section 3 shows the obtaining of the weight vector and its properties associated with an OWA linear operator. In Section 4, the Formula One results are evaluated according to different OWA operators and, finally, the 25 conclusions are presented. 2. OWA Operator: Definition and Properties An OWA operator [1] of dimension n is given by a function F : R → R that has an associated weighting vector W = [w , w , ... , w ] such that w ∈ [0, 1] and w = 1. 1 2 n i i Furthermore, F(a , a , ... , a ) = w b where b is the largest jth of the a .Inour case 1 2 n j j j i b is the element that represents the best finishing position in the race. In particular, a weight w is not associated with a specific argument but with an ordered position in the aggregation. A number of properties can be associated with these operators. It is first noted that the OWA aggregation is commutative, that is, the aggregation is indifferent to the initial indexing of the argument. A second characteristic associated with these operators is that it is monotonic. Thus, if a ˆ ≥ a ∀ i, then F(a ˆ , a ˆ , ... , a ˆ ) ≥ F(a , a , a ). Another character- i i 1 2 n 1 2 n istic associated with these operators is the idempotency. In particular, if ∀ i, a = a then F(a , a , ... , a ) = a. 1 2 n The satisfaction of these three conditions, as noted in [2], ensures that these operators belong to the class of operators called mean operators; the fact that an OWA operator is a mean operator can then be shown. It is known that • The weight vector (1, 0, ... ,0) enables the Max that emphasises the largest element in the argument bag to be obtained. • The weight vector (0, 0, ... ,1) enables the Min that emphasises solely the smallest element in the bag to be obtained. • The weight vector (1/n,1/n, ... ,1/n) enables the simple average to be obtained. Additionally, for any OWA operator F,Min [a ] ≤ F(a , a , ... , a ) ≤ Max [a ]. In [3]a i i 1 2 n i i number of families of these operators are discussed. The various different mean opera- tors are implemented by appropriate selection of the weights in the associated weighting vector. 3. Obtaining the Weight Vector of an OWA with Linear behaviour To calculate the weights vector W = [w ], ∀ i there are several functions, some of which have a linear behaviour [4–6] and others are non-linear [7,8]. This paper deals with a study of the properties that verify the linear functions set F and F as shown in the results obtained by Cables and Lamata [4]. Because of that they identify the linear function that delimits the 82 E. H. CABLES PÉREZ ET AL. Figure 1. Region of the linear functions set FL that veriﬁes the OWAs properties. functions set, which is defined by the region F : −2 2 f (x) = x +,(3) sup n(n − 1) n − 1 f (x) =.(4) inf And obtaining the region represented in Figure 1. We can observe that the bundle of lines contains the point P ((1 + n)/1, 1/n) and all functions have negative slopes, therefore all those w obtained starting from any lin- ear function belonging to this region will be organised in descending order (w > w > 1 2 ··· > w ). Moreover, it is verified that the linear functions set that allows to obtain w verifies the properties of the OWA, has a slope a ∈]−2/n(n − 1), 0[, that depends on the value of n [4]. In the study carried out, another region is also identified, F (see Figure 2) of linear functions that allows to obtain w in decreasing order and to verify the OWAs properties. In addition, the linear functions set also contains the point P ((1 + n)/2, 1/n), which is delimited by the following functions (see Figure 2): 2 2 f (x) = x −,(5) sup n(n − 1) n − 1 f (x) =.(6) inf n FUZZY INFORMATION AND ENGINEERING 83 Figure 2. Region of the linear functions set F that veriﬁes the OWAs’ properties. Figure 3. Region of the linear functions set F and F . Finally, the authors conclude that the general form of the linear functions belong to these regions, and that verify the OWAs properties are given by (see Figure 3): 1 1 + n 1 + n 1 f (x) = ax + − a = a x − +.(7) n 2 2 2 84 E. H. CABLES PÉREZ ET AL. With regard to the slope and the orness measure, we present theorems, the relation between them in the case of the OWA operators obtained by means of linear functions. Theorem 3.1: Given the function f (x) = 1/n, that contends the point P and slope a = 0, inf m it represents an OWA and their α = 0.5. Proof: n n 1 n • w = = = 1, i=1 i=1 n n n n n 1 1 n − 1 1 • α = (n − i)f (i) = − (i) = → α = . inf i=1 i=1 i=1 n − 1 n(n − 1) 2(n − 1) 2 n n because (n) = n and (i) = n(n + 1)/2. i=1 i=1 Theorem 3.2: Let the linear functions set be f (x)giveninEquation (7), with slope a → 0. Then the satisfaction degree α → 0. Proof: By definition, it is known that n n 1 1 1 1 + n α = (n − i)f (i) = (n − i) ai + − a . n − 1 n − 1 n 2 i=1 i=1 n n Taking into account that (i) = n(n + 1)/2and (i ) = n(n + 1)(2n + 1)/6, it is i=1 i=1 obtained that −an(n − 1) + 6n − 6 −an(n − 1)(n + 1) 6(n − 1) −an(n + 1) 1 α = = + = + . 12(n − 1) 12(n − 1) 12(n − 1) 12 2 And when it is calculate −an(n + 1) 1 1 lim + = , n→0 12 2 2 then it is verified that α = 0.5. Theorem 3.3: Let the linear functions set be f (x), with a value of slope a < 0. Then the satisfaction degree is α> 0.5. Proof: Starting from the result obtained in Theorem 3.2, it is known that the biggest slope in the linear functions that constitutes the OWA is a =−2/(n(n − 1)) then, substituting in the value of α, the following is obtained: 1 −2 1 + n 1 2n(n + 1) 1 α = α = (n − i) (i − ) + = + n − 1 n(n − 1) 2 n 12n(n − 1) 2 i=1 (n + 1) 1 (n + 1) + 3(n − 1) (2n − 1) (2n − 1) = + = = ⇒ α = . 6(n − 1) 2 6(n − 1) 6(n − 1) 3(n − 1) And then we calculate (2n − 1) 2 lim = , n→∞ 3(n − 1) 3 then it is verified that α> 0.5. FUZZY INFORMATION AND ENGINEERING 85 Theorem 3.4: Let the linear functions set be f (x), which represents an OWA and if the value of the slope is a > 0. Then the satisfaction degree is α< 0.5. Proof: As in the previous theorem, a = 2/(n(n − 1)) is substituted in f (x) and it is 2 g obtained 1 2 1 + n 1 α = (n − i) i − + n − 1 n(n − 1) 2 n i=1 −2n(n + 1) 1 −(n + 1) 1 −n − 1 + 3(n − 2) 2n − 4 = + = + = = 12n(n − 1) 2 6(n + 1) 2 6(n − 1) 6(n − 1) xn − 2 ⇒ α = . 3(n − 1) And then we calculate n − 2 1 lim = , n→∞ 3(n − 1) 3 then it is verified that α< 0.5. In Figure 4 the behavior of α can be seen. Theorem 3.5: Let the linear functions set be f (x) which represents an OWA with the slope a ∈]−2/(n(n − 1)),0[. Then the satisfaction degree is α ∈ [ , (2n − 1)/(3n − 3),]. Proof: It is known that if a = 0 then the function f (x) is obtained, and by Theorem 3.1 inf it is verified that α = ,and if a =−2/(n(n − 1)) the function f (x) is obtained and sup by the results of Theorem 3.3 it is verified that α = (2n − 1)/(3n − 3). Therefore, it is demonstrated that the value of α ∈ [ , (2n − 1)/(3n − 3)]. Figure 4. Behaviour of the α value. 86 E. H. CABLES PÉREZ ET AL. Table 1. The α intervals for diﬀerent values of n, depending on the positive slope or negative slope. Values of n Negative slope Positive slope 1 1 2[ ,0] [0, ] 2 2 1 5 1 1 3[ ,][ , ] 2 6 6 2 1 7 2 1 4[ ,][ , ] 2 9 9 2 . . . . . . . . . 1 19 8 1 10 [ ,][ , ] 2 27 27 2 Theorem 3.6: Let the linear functions set be f (x) if the slope a ∈]0, 2/(n(n − 1))[. Then the satisfaction degree of α ∈ [(n − 2)/(3n − 1), ]. Proof: It is known that if a = 0 then the function f (x) is obtained and by Theorem 3.1, it inf is verified that α = , also, if a = 2/(n(n − 1)) the function f (x) is obtained and by the sup results of Theorem 3.4 it is verified that α = (n − 2)/(3n − 3). Therefore, it is demonstrated that the value of α ∈ [(n − 2)/(3n − 3), ]. Starting from the results obtained in Theorems 3.5 and 3.6, we may determine the interval of the value of α for a given n. Next, the interval of the value of α is shown for n = 2, ... , 10, for negative and positive slope functions (see Table 1). Theorem 3.7: All the linear functions that verify the OWAs properties, such that the values of 1 1 the satisfaction degree α ∈ [ , (2n − 1)/(3n − 3)] or α ∈ [(n − 2)/(3n − 3), ] have a slope 2 2 a = (6 − 12α)/(n(n + 1)) and vice versa. Proof: It is known by Theorem 3.2 that −an(n + 1) 1 α = + , 12 2 then 1 −an(n + 1) 1 α − = ⇔ 12 α − =−an(n + 1) n 12 2 12α − 6 6 − 12α ⇔−a = ⇔ a = . n(n + 1) n(n + 1) Case 1. α ∈ [ , (2n − 1)/(3n − 3)], then −ifα = the value of a = 0, 18n−18−12(2n−1) 2n−1 6 − 12( ) 2n − 1 3n−3 3n−3 −ifα = , then a = ⇔ a = 3n − 3 n(n + 1) n(n + 1) 18n − 18 − 24n + 12) −6n − 6) a = ⇔ a = 3(n − 1)n(n + 1) 3(n − 1)n(n + 1) −6(n + 1) −2 ⇔ a = ⇔ a = . 3(n − 1)n(n + 1) n(n − 1) FUZZY INFORMATION AND ENGINEERING 87 Therefore, it is verified that the value of the slope a ∈]−2/(n(n + 1)),0[ Case 2 α ∈ [(n − 2)/(3n − 3), ], then −if α = the value of a = 0, 18n−18−12(n−2) n−1 6 − 12( ) n − 2 3n−3 3n−3 −if α = then a = ⇔ a = , 3n − 3 n(n + 1) n(n + 1) 18n − 18 − 12n + 24) 6(n + 1) 2 a = ⇔ a = ⇔ a = . 3(n − 1)n(n + 1) 3(n − 1)n(n + 1) n(n − 1) Therefore, it is verified that the value of the slope of a ∈]0, 2/n(n + 1)[. The procedure to obtain the linear functions are: • Select the quantity of weights n to calculate. 1 1 • Select the value of α,suchthat α ∈ [ , (2n − 1)/(3n − 3),] or α ∈ [(n − 2)/(3n − 1), ]. 2 2 • The slope of the linear function is calculated through the formula a = (6 − 12α)/n(n + 1). • Substitute a in the general function of the linear functions class f (x) ( Equation (7)) which is an OWA. Remark 3.1: On the other hand, by Theorem 3.7 we have that the slope of the straight line canbeexpressedonthe value of α. According to the last theorem we can use the following expression: 6 − 12α a = .(8) n(n + 1) And the general expression for f (x) in function of α is given by 6 − 12α 1 + n 1 f (x) = x − +,(9) n(n + 1) 2 n where the value of α must belong to the following intervals: α ∈ [ , (2n − 1)/(3n − 3)] for negative slopes (decreasing function) and α ∈ [(n − 2)/(3n − 3), ] for positive slopes (increasing function). Remark 3.2: Consider the official scores {25, 18, 15, 12, 10, 8, 6, 4, 2, 1} assigned to the first 10 10 finishing positions in Formula One. Table 2 shows the values in the corresponding weights vector with α = 0, 721672167. 2(10−1) 1 1 19 But for n = 10, ⇒ α ∈ [ , ]and α = 0, 721672167 ∈ / [ , ] and it is therefore not 2 3(10−3) 2 27 possible to obtain a linear function instead of an OWA operator. Table 2. Weights according to F1, the Borda law α = 2/3 and for and α = 0.7. 123456789 10 25 18 15 12 10 8 6 4 2 1 F1 101 101 101 101 101 101 101 101 101 101 10 9 8 7 6 5 4 3 2 1 BORDA 55 55 55 55 55 55 55 55 55 55 109 97 85 73 61 49 37 25 13 1 α = 0.7 550 550 550 550 550 550 550 550 550 550 88 E. H. CABLES PÉREZ ET AL. Thus, we will take the values nearest to = 0.7037 which will be α = 0, 70. Therefore, it is possible to obtain the weights for n = 10 and α = 0.7. Starting from: 6 − 12α 1 + n 1 w = i − + , (10) n(n + 1) 2 n 6 − 12α 1 + n 1 −2.4 −9 1 109 w = 1 − + = + = , n(n + 1) 2 n 110 2 10 550 6 − 12α 1 + n 1 −2.4 −7 1 97 w = 2 − + = + = , n(n + 1) 2 n 110 2 10 550 and so on. The complete results can be seen in Table 2. 4. Results In a Formula One competition only the first 10 drivers who cross the finish line obtain points. These scores are reflected in the first row of Table 3, which correspond to the weights of the first row of Table 2. The following lines of Table 3 (Table 2) represent the scores or weights Table 3. Scores according to F1, the Borda law of α = 2/3, and for an α = 0.7. 1 2345678 910 F1 25 18 15 12 10864 2 1 BORDA 109876543 2 1 α = 0.7 109 97 85 73 61 49 37 25 13 1 Table 4. Thescoresofthe20F1races. F1 1234567 8 9 10 c 18 25 15 12 – 10 6 1 4 – c 2518 8101215 216 – c 18 25 15 12 10 – 6 1 – 2 c 12182515 – 10 8614 c 25 18 – – 15 – 12 10 6 8 c 62512181510 – – 8 – c 25 12 18 6 15 – 10 8 – 4 c 10 12 18 – 25 – – 8 4 – c 12 18 25 10 15 – 6 4 – – c 25 618151012 24 – 8 c 12 25 15 18 – 10 4 2 6 – c 25 18 10 12 15 – – 2 1 8 c 25 15 18 10 12 1 2 8 – – c 25 – 15 – 18 – 10 1 12 – c 18 12 10 – 15 25 8 1 – – c 25 – 12101518 68 – – c 25 18 10 15 – 12 4 8 6 – c 2121815 – 25 6 10 – – c 12 25 18 15 8 10 2 – – 1 c 18 15 25 12 – 10 6 6 – 8 SUM 363 317 305 205 200 168 100 87 54 43 FUZZY INFORMATION AND ENGINEERING 89 Table 5. The scores of the 20 races according to the Borda law. F1 12345 6 7 8 9 10 c 910 8 7 – 6413 – c 10 9 5 6 7 8214 – c 9 10876 – 4 1 – 2 c 7 910 8 – 65413 c 10 9 – – 8 – 7645 c 4 10798 6 – – 5 – c 10 7948 – 6 5 – 3 c 6 7 9 – 10– – 5 3– c 79 10 68 – 4 3 – – c 10 4986 7 2 3 – 5 c 710 8 9 – 6324 – c 10 9678 – – 2 1 5 c 10 8967 1 2 5 – – c 10 –8 –9 – 6 1 7 – c 976 – 8 10 5 1 – – c 10 – 7 6 8 945 – – c 10 9 6 8 – 7354 – c 2798 – 10 4 6 – – c 7 10985 6 2 – – 1 c 9 810 7 – 643 – 5 SUM 166 152 153 114 106 88 67 59 36 29 Table 6. Final scores for F1, Borda law and α = 0.7. SUM 12345 6 7 8 9 10 F1 363 317 305 205 200 168 100 87 54 43 BORDA 166 152 153 114 106 88 67 59 36 29 α = 0.7 1772 1626 1627 1192 1118 913 617 510 322 260 that the pilots would have received if the allocation had been considered by linear OWAs with values of α = (Borda) and an α = 0.70, respectively (Tables 4 and 5). If we consider the results (see Table 6) associated with an α = 0.7, the results are totally analogous to those of Borda’s law. 5. Conclusion Linear functions giving rise to OWA operators have been described and studied in this paper. It can be seen that there are infinite functions from which weights can be calculated with different slopes and orness. The relationships between the slope of the straight line and the orness measure of the α aggregation are shown to be interesting properties. Finally, taking the classification of F1 in the year 2017, the number of points to be assigned to the first 10 finishing positions has been calculated for α = and α = 0.70 and an α = 0.70 using the proposed method and it was observed that in all three cases the first place has not changed, but that when the scores are linear, the second and third positions have been inverted. Disclosure statement No potential conflict of interest was reported by the authors. 90 E. H. CABLES PÉREZ ET AL. Funding Research partially funded by projects TIN2014-55024-P, TIN2017-86647-P and P11-TIC-8001 (all of them including FEDER funds). Also the support provided by the Antonio Nariño University (Colombia) is also acknowledged. References [1] Yager RR. On ordered weighted averaging aggregation operators in multicriteria decisionmaking. IEEE Trans Syst Man Cybern. 1988;18:183–190. [2] Dubois D, Prade H. A review of fuzzy sets aggregation connectives. Inf Sci. 1985;36(1):85–121. [3] Yager RR. Families of OWA operators. Fuzzy Sets Syst. 1993;59(2):125–148. [4] Cables E, Lamata M. Owa weights determination by means of linear functions. Mathware Soft Comput. 2009;16:107–122. [5] Liu X. On the properties of equidifferent OWA operator. Int J Approx Reason. 2006;43(1):90–107. [6] Ahn B. On the properties of OWA operator weights functions with constant level of orness. IEEE Trans Fuzzy Syst. 2006;14(4):511–515. [7] O’Hagan M. Aggregating template or rule antecedents in real-time expert systems with fuzzy set. Proceedings of the 22nd Annual IEEE Asilomar Conference on Signals, Systems, and Computers; 1988, Pacific Grove, California. [8] Filev D, Yager RR. Analytic properties of maximum entropy OWA operators. Inf Sci. 1995;85(1–3): 11–27.
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