Bi-objective optimization problems with two decision makers: refining Pareto-optimal front for equilibrium solution

Abstract

The Pareto-optimality concept in multi-objective optimization theory is different from the Nash equilibrium concept in noncooperative game theory. When the objective holders are independent decision makers, i.e., human entities or organizations, any solution on the Pareto-optimal front is not necessarily an equilibrium point, hence not a valid solution. The solution has to be a Pareto-optimal-equilibrium (POE) point. In this paper, we convert a bi-objective optimization problem into a two-player game problem by introducing “induced games,” and we propose a new refinement method to find a POE point. We prove that at least one such POE point exists for a class of linear bi-objective optimization problems, and we develop an algorithm to find it. We discuss that the innovative approach considered in this paper is of real future interest to some industrial and social applications. One such example is also presented.

Appendix

1.1 A. Aumann’s polyhedron characterization for 4-vertices payoff space

In this section, we describe the characterization of vertices of the Aumann’s polyhedron for $2\times 2$ bi-matrix games. In fact, we use an expanded form of the characterization proposed in Calvo-Armengo (2006).

We begin by the following Lemma which will be used for converting a $2\times 2$ bi-matrix game to a more tractable form.

Lemma A.1

(Peeters and Potters 1999) The set of correlated equilibria of a bi-matrix game (A, B) does not change if theA-matrix (B-matrix)is multiplied by a positive factor or a fixed row (column) vector is added to all rows (columns) ofA (B).

According to Lemma A.1, we can replace A and B by $A'$ and $B'$, respectively, where

$$\begin{aligned} A' = \begin{bmatrix} a_{1} &{} a_{2} \\ 0 &{} 0 \end{bmatrix}, \quad B' = \begin{bmatrix} b_{1} &{} 0 \\ b_{2} &{} 0 \end{bmatrix} \end{aligned}$$

(A.1)

and $a_{j} = a_{1j}-a_{2j}$, $j=1,2$ and $b_{i} = b_{i1}-b_{i2}$, $i=1,2$. By ${\mathcal {A}}(a_1,a_2,b_1,b_2)$ we denote the correlated equilibria subspace for the bi-matrix game defining by $A'$ and $B'$.

If $a_2,b_2\ne 0$, then by multiplying $A'$ by $1/|a_2|$ and $B'$ by $1/|b_2|$ the set of correlated equilibria does not change, thus any correlated equilibria subspace can be characterized by the following polyhedra

$$\begin{aligned} {\mathcal {A}}(a,1,b,1), {\mathcal {A}}(a,-1,b,1), {\mathcal {A}}(a,1,b,-1), {\mathcal {A}}(a,-1,b,-1). \end{aligned}$$

Two polyhedra ${\mathcal {A}}(a,-1,b,1)$ and ${\mathcal {A}}(a,1,b,-1)$ are the same if we exchange the role of row-player and column-player. Thus, three polyhedra are remained to be considered, as in Lemma A.2. If we have $a_2=0$ or $b_2=0$ in $A'$ and $B'$, then, changing the strategies of player 1 or player 2 puts the problem in the form we considered in Lemma A.2.

Lemma A.2

For an Aumann’s polyhedron${\mathcal {A}}(a,s,b,t)$with$|s|=|t|=1$, we have the following facts:

• (i) The vertices of${\mathcal {A}}(a,1,b,1)$could be characterized by

  $$\begin{aligned} \begin{array}{c|lccc} \hbox { Conditions on}\ a ,b &{} \hbox { Vertices of}\ {\mathcal {A}}(a,1,b,1)\\ \hline a,b\le 0 &{} \frac{1}{(a-1)(b-1)} \begin{bmatrix}1 &{} -a &{} -b &{} ab \end{bmatrix}, \; \frac{1}{a+b-ab} \begin{bmatrix}0&a&b&-ab \end{bmatrix}, \; \frac{1}{a+b-1} \begin{bmatrix}-1 &{} a &{} b &{} 0 \end{bmatrix}, \\ &{} \begin{bmatrix}0 &{} 1 &{} 0 &{} 0 \end{bmatrix}, \; \begin{bmatrix}0 &{} 0 &{} 1 &{} 0 \end{bmatrix}\\ a\le 0< b &{} \begin{bmatrix}0 &{} 0 &{} 1 &{} 0\end{bmatrix}\\ b\le 0<a &{} \begin{bmatrix}0 &{} 1 &{} 0 &{} 0\end{bmatrix}\\ a, b>0 &{} \begin{bmatrix}1 &{} 0 &{} 0 &{} 0 \end{bmatrix}\end{array} \end{aligned}$$

  That is, for$a\le 0$and$b\le 0$, polyhedron${\mathcal {A}}(a,1,b,1)$may have five different vertices. For other cases, only one point is known.

• (ii) The vertices of${\mathcal {A}}(a,-1,b,1)$could be characterized by

  $$\begin{aligned} \begin{array}{c|lccc} \hbox { Conditions on}\ a ,b &{} \hbox { Vertices of}\ {\mathcal {A}}(a,-1,b,1)\\ \hline a<0, b \in {\mathbb {R}}&{} \begin{bmatrix}0 &{} 0 &{} 1 &{} 0 \end{bmatrix}\\ b< 0 = a &{} \frac{1}{1-b} \begin{bmatrix}1 &{} 0 &{} -b &{} 0 \end{bmatrix},\; \begin{bmatrix}0 &{} 0 &{} 1 &{} 0 \end{bmatrix}\\ b\ge 0 = a &{} \begin{bmatrix}0 &{} 0 &{} 1 &{} 0 \end{bmatrix}, \; \begin{bmatrix}1 &{} 0 &{} 0 &{} 0 \end{bmatrix}\\ b<0<a &{} \frac{1}{(1+a)(b-1)} \begin{bmatrix}-1 &{} -a &{} b &{} ab \end{bmatrix}\\ a>0=b &{} \frac{1}{1+a} \begin{bmatrix}1 &{} a &{} 0 &{} 0 \end{bmatrix}, \; \begin{bmatrix}1 &{} 0 &{} 0 &{} 0 \end{bmatrix}\\ a, b>0 &{} \begin{bmatrix}1 &{} 0 &{} 0 &{} 0 \end{bmatrix}\end{array} \end{aligned}$$

• (iii) The vertices of${\mathcal {A}}(a,-1,b,-1)$could be characterized by

  $$\begin{aligned} \begin{array}{c|lccc} \hbox { Conditions on}\ a ,b &{} \hbox { Vertices of}\ {\mathcal {A}}(a,-1,b,-1)\\ \hline a,b\ge 0 &{} \frac{1}{(a+1)(b+1)} \begin{bmatrix}1 &{} a &{} b &{} ab \end{bmatrix}, \; \frac{1}{1+b+ab} \begin{bmatrix}1&0&b&ab \end{bmatrix}, \; \frac{1}{1+a+ab} \begin{bmatrix}1 &{} a &{} 0 &{} ab \end{bmatrix}, \\ &{} \begin{bmatrix}0 &{} 0 &{} 0 &{} 1 \end{bmatrix}\\ a< 0, b\in {\mathbb {R}}&{} \begin{bmatrix}0 &{} 0 &{} 0 &{} 1\end{bmatrix}\\ b < 0 \le a &{} \begin{bmatrix}0 &{} 0 &{} 0 &{} 1\end{bmatrix}\end{array} \end{aligned}$$

Proof

See (Calvo-Armengo 2006). □