for each l
h
∈ L, where (l
h
, . . . , l
h
) and (l
k
, . . . , l
k
) are
the vectors of m replicas of l
h
and l
k
, respectively.
Thus, L(x
j
) consists of those labels that minimize
the distance to the vector of individual assessments.
Notice that L(x
j
) = {l
r
, . . . , l
r+s
} is always an inter-
val, because it contains all the terms from l
r
to l
r+s
,
where r ∈ {1, . . . , g} and 0 ≤ s ≤ g−r. Two different
cases are possible:
1. If s = 0, then L(x
j
) contains a single label, which
will automatically be the collective assessment
l(x
j
) of the alternative x
j
.
2. If s > 0, then L(x
j
) has more than one label. In
order to select the most suitable label of L(x
j
),
we now introduce L
∗
(x
j
), the set of all the labels
l
k
∈ L(x
j
) that fulfill
¯
d
p
((l
k
, . . . , l
k
), (l
r
, . . . , l
r+s
))≤
¯
d
p
((l
h
, . . . , l
h
), (l
r
, . . . , l
r+s
)),
for all l
h
∈ L(x
j
), where (l
k
, . . . , l
k
) and
(l
h
, . . . , l
h
) are the vectors of s + 1 replicas of l
k
and l
h
, respectively.
(a) If the cardinality of L(x
j
) is odd, then L
∗
(x
j
)
has a unique label, the median term, that will
be the collective assessment l(x
j
).
(b) If the cardinality of L(x
j
) is even, then L
∗
(x
j
)
has two different labels, the two median terms.
In this case, similarly to the proposal of (Balin-
ski and Laraki, 2007a), we consider the low-
est label in L
∗
(x
j
) as the collective assessment
l(x
j
).
It is worth pointing out two different cases when
we are using induced Minkowski distances.
1. If p = 1, we obtain the same collective assess-
ments that those given by MJ, the median
4
of the
individual assessments. However, the final results
are not necessarily the same that in MJ because we
use a different tie-breaking process, as is shown
later.
2. If p = 2, each collective assessment is the clos-
est label to the “mean” of the individual assess-
ments
5
, which is the one chosen in the Range Vot-
ing (RV) method
6
(see (Smith, 2007)).
4
It is more precise to speak about the interval of medi-
ans, because if the assessments’ vector has an even number
of components, then there are more than one median. See
(Monjardet, 2008).
5
The chosen label is not exactly the arithmetic mean of
the individual assessments, because we are working with a
discrete spectrum of linguistic terms and not in the continu-
ous one of the set of real numbers.
6
RV works with a finite scale given by equidistant real
numbers, and it ranks the alternatives according to the arith-
metic mean of the individual assessments.
Notice that when we choose p ∈ (1, 2), we find
situations where the collective assessment is located
between the median and the “mean”. This allows us
to avoid some of the problems associated with MJ and
RV.
3.3 Tie-breaking Method
Usually there exist more alternatives than linguistic
terms, so it is very common to find several alterna-
tives sharing the same collective assessment. But irre-
spectively of the number of alternatives, it is clear that
some of them may share the same collective assess-
ment, even when the individual assessments are very
different. For these reasons it is necessary to intro-
duce a tie-breaking method that takes into account not
only the number of individual assessments above or
below the obtained collective assessment (as in MJ),
but the positions of these individual assessments in
the ordered scale associated with L.
As mentioned above, we will calculate the differ-
ence between two distances: one between l(x
j
) and
the assessments higher than l(x
j
) and another one be-
tween l(x
j
) and the assessments lower than the l(x
j
).
Let v
+
j
and v
−
j
the vectors composed by the assess-
ments v
i
j
from
v
1
j
, . . . , v
m
j
higher and lower than
the term l(x
j
), respectively. First we calculate the two
following distances:
D
+
(x
j
) =
¯
d
p
v
+
j
, (l(x
j
), . . . , l(x
j
))
,
D
−
(x
j
) =
¯
d
p
v
−
j
, (l(x
j
), . . . , l(x
j
))
,
where the number of components of (l(x
j
), . . . ,l(x
j
))
is the same that in v
+
j
and in v
−
j
, respectively (obvi-
ously, the number of components of v
+
j
and v
−
j
can
be different).
Once these distances have been determined, a new
index D(x
j
) ∈ R is calculated for each alternative
x
j
∈ X: the difference between the two previous dis-
tances:
D(x
j
) = D
+
(x
j
) − D
−
(x
j
).
By means of this index, we provide a kind of com-
pensation between the individual assessments that are
bigger and smaller than the collectiveassessment, tak-
ing into account the position of each assessment in the
ordered scale associated with L.
For introducing our tie-breaking process, we fi-
nally need the distance between the individual assess-
ments and the collective one:
E(x
j
) =
¯
d
p
(v
1
j
, . . . , v
m
j
), (l(x
j
), . . . , l(x
j
))
.
A LINGUISTIC GROUP DECISION MAKING METHOD BASED ON DISTANCES
461