Creating an acceptable consensus ranking for group decision making

Abstract

This paper examines the problem of combining a set of ordinal rankings to form an acceptable consensus ranking. The objective of traditional group decision making problem is to determine the Minimum Violation Ranking. Motived by the applications of adjusted consensus in recent years, we study this problem from a new perspective, for obtaining an acceptable consensus ranking for group decision making. In this paper, every voter ranks a set of alternatives respectively, and we know the acceptability index, which represents the minimum adjustments that are allowed for each voter. The problem is to find the Minimum Acceptable Violation Ranking (MAVR) which minimizes the sum of voter’s unacceptable violations. Besides, we develop a branch and bound ranking algorithm to solve this problem. The suggested improvement include: (1) analysing the ranking preference by two ways: pairwise preference and ranking-based preference; (2) constructing the lower bound and upper bound, which exclude at most half of the feasible solutions in each iteration process. Furthermore, the effectiveness and efficiency of this algorithm are verified with an example and numerical experiments. Finally, we discuss two extensions of the basic MAVR problem: the Minimum Weighted Acceptable Violation problem, whose voters are accompanied with a set of weights or multiples, and the Minimum Hierarchy Acceptable Violation problem, which uses hierarchical acceptability indexes. In addition, our results can be applied to other ranking and subset selection problems in which provide consensus rankings over the alternatives.

Appendices

Appendix 1: The proof of equivalency of the two linear models

To prove the equivalency, we just need to prove that the two models have the same feasible solution set and the feasible solutions of these two models are the same.

The first linear model is given by

$$\begin{aligned} \min V(C)=\sum _{l=1}^{n} d_{APD}(B^{l},C) =\sum _{l=1}^{n}\sum _{i=1}^{m}\sum _{j=1}^{m} \left| b_{ij}^{l}-c_{ij} \right| \end{aligned}$$

s.t.

$$\begin{aligned} \sum _{i}^{m}\sum _{j}^{m} \left| a_{ij}^{l} - b_{ij}^{l} \right| \le \alpha \ \ \ \ l= & {} 1,\ldots ,n, \\ c_{ij} \ \text {satisfies transitivity property} \ \ \ i= & {} 1,\ldots ,m, j=1,\ldots ,m . \end{aligned}$$

The second linear model can be represented as below.

$$\begin{aligned} \min V_{0}(C) =\sum _{l=1}^{n} d_{APD}(A^{l},B^{l}) =\sum _{l=1}^{n}\sum _{i=1}^{m}\sum _{j=1}^{m} \left| b_{ij}^{l}-a_{ij}^{l} \right| \end{aligned}$$

s.t.

$$\begin{aligned} \sum _{i}^{m}\sum _{j}^{m} \left| c_{ij} - b_{ij}^{l} \right| \le \alpha \ \ \ \ l= & {} 1,\ldots ,n, \\ c_{ij} \ \text {satisfies transitivity property} \ \ \ i= & {} 1,\ldots ,m, j=1,\ldots ,m . \end{aligned}$$

where $A^{l}=(a_{ij}^{l})$ is the initial preference matrix of voter $e_{l}, B^{l}=(b_{ij}^{l})$ is the adjusted preference matrix of voter $e_{l}$ after $k^{l}$ adjustments $( k^{l}\leqslant \alpha ), C=(c_{ij} )$ is a consensus preference decision matrix.

It is obvious to know that the feasible solutions are the set of all possible rankings, so the feasible solutions of these two models are the same. For example, there are 3 alternatives(or candidates) a, b and c in an election, the feasible solution set can be represented as: $\{a, b, c\}, \{a, c, b\}, \{b, a, c\}, \{b, c, a\}, \{c, a, b\}, \{c, b, a\}$.

For a given preference decision matrix C (a feasible solution), we will know the distance between each voter $A^{l}$ and C .

In the first model, because $d(A^{l},C)= d(A^{l},B^{l})+d(A^{l},C)$, to minimizing the distance between adjusted preference matrix $B^{l}$ and C , each voter should adjuste their preference as more as possible (at most $\alpha$), nearing to C . If the distance between adjusted preference matrix $A^{l}$ and C is greater than $\alpha$, $d_{APD}(A^{l},B^{l}) = \alpha$, then $d_{APD}(B^{l},C) = d_{APD}(A^{l},C) - \alpha$. If the distance between adjusted preference matrix $A^{l}$ and C is not greater than $\alpha$, $d_{APD}(A^{l},B^{l}) = d_{APD}(A^{l},C)$, then $d_{APD}(B^{l},C) = 0$.

In the second model, because $d(A^{l},C)= d(A^{l},B^{l})+d(A^{l},C)$, to minimizing the adjusted distance between adjusted preference matrix $B^{l}$ and $A^{l}$, the distance between $B^{l}$ and C is as great as it could be (at most $\alpha$). If the distance between adjusted preference matrix $A^{l}$ and C is greater than $\alpha$, $d_{APD}(B^{l},C) = \alpha$, then $d_{APD}(A^{l},B^{l}) = d_{APD}(A^{l},C) - \alpha$. If the distance between adjusted preference matrix $A^{l}$ and C is not greater than $\alpha$, $d_{APD}(B^{l},C) = d_{APD}(A^{l},C)$, then $d_{APD}(A^{l},B^{l}) = 0$.

Eventually, we choose the minimum distance among all feasible solutions C as the final decision. Because the objectives of these two linear models are same for every feasible solution, so the final decisions of these two models are the same.

From the above we can draw a conclusion that these two two linear models are equivalent.

Appendix 2: The supplemental proof of the upper bound

Because the objective is to minimize the violations, any one of the feasible solutions is no less than the final solution. To prove the upper bound, we just need to prove that the upper bound we represents, is a feasible solution.

If $x_{i},x_{j}\in X_{2}$: The worst case is that we choose all the pairs with a higher aggregated violation $max\left\{ s_{ij},s_{ji}\right\}$ between $\left( x_{i}\prec x_{j}\right)$ and $\left( x_{j}\succ x_{i}\right)$. We have $\sum _{i}\sum _{j} max\left\{ s_{ij},s_{ji}\right\} \ge \sum _{l}\sum _{i}\sum _{j}w_{ij}^{l}$.

Fixing (or choosing) two different pairs of preferences with the highest aggregated violation does not violate the rule of transitivity, they still can be part of some feasible solutions, because the rule of transitivity only exists in more than or equal to three different pairs. For example, $\left\{ a\succ b, b\succ c\right\}$ is not contradict with the transitivity property, however, $\left\{ a\succ b, b\succ c, c\succ a \right\}$ is contradict with the rule of transitivity.

Among these alternatives which we have not yet assigned to the prefix order ($x_{i},x_{j}\in X_{2}$), we choose two different pairs of preferences with the highest aggregated violation as part of solution. For these two pairs , the violation is $2*\left( min^{1}\left\{ s_{ij}\right\} +min^{2}\left\{ s_{ij}\right\} \right)$. For the other pairs, the violation is $\sum _{i}\sum _{j} max\left\{ s_{ij},s_{ji}\right\} -2*\left( max^{1}\left\{ s_{ij}\right\} +max^{2}\left\{ s_{ij}\right\} \right)$. Because the objective is to minimize the violations, any one of the feasible solutions is no less than the final solution. We have $\sum _{i}\sum _{j} max\left\{ s_{ij},s_{ji}\right\} -2*\left( max^{1}\left\{ s_{ij}\right\} +max^{2}\left\{ s_{ij}\right\} \right) +2*\left( min^{1}\left\{ s_{ij}\right\} +min^{2}\left\{ s_{ij}\right\} \right) \ge \sum _{l}\sum _{i}\sum _{j}w_{ij}^{l}$.

From the above and the previous proof in Lemma 2 we can draw a conclusion that $M_{0}+M_{2}$ is an upper bound.