Equilibria and Compromises in Two-Person Zero-Sum Multicriteria Games

Abstract

The problem is to formalize the solution of a two-person zero-sum multicriteria (MC) game, providing a payoff to both players with respect to their MC best guaranteed results. As the basic concept of the solution of the MC game, the Shapley equilibrium is selected. It is parametrized by the inverse logical convolution based on scalarization in the Germeyer sense, i.e., on the weighted maximin approach. We investigate the relation between the equilibrium value of the game and its one-sided values defined for each player as the best guaranteed result that does not depend on the order of the moves. We describe the possibilities of a compromise in zero-sum MC games. For such a finite game in mixed strategies, we introduce the notions of the compromise and negotiation values and establish their relation to the equilibrium and one-sided values of the game. We consider a special interpretation of the MC averaging of the result for players oriented on using the inverse logical convolution. For such a case, the nonemptiness of the negotiation set is analyzed. The obtained conclusions are demonstrated on a prototype example.

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Correspondence to I. I. Pospelova.

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Translated by A. Muravnik

APPENDIX

APPENDIX

Computations for figures. Let us find solutions in mixed strategies for the finite prototype two-criteria game (4.3) between two persons using the ILC of their partial criteria. Let \(p\;\mathop = \limits^{{\text{def}}} \;p(x = 1)\) be the probability of the choice of the first row of (4.3) by player 1 and \(q\;\mathop = \limits^{{\text{def}}} \;q(y = 2)\) be the probability of the choice of the first column by player 2, \(p,q \in [0,1]\). If there are two strategies, then the vector of mixed strategies is uniquely determined by its first component; that is why, within the Appendix, p and q denote only the first components of the vectors of mixed strategies but not the vectors themselves. Due to the same reason, we assume that player 1 applies the ILC with parameter \(\mu \;\mathop = \limits^{{\text{def}}} \;{{\mu }_{1}}\), while player 2 applies the ILC with parameter \(\nu \;\mathop = \limits^{{\text{def}}} \;{{\nu }_{1}}\) (respectively, the sets I(μ) and I(ν) are changed depending on whether μ or ν is equal to 0 or 1). The averaged ILC for player 1 and player 2 are represented as follows:

$$\begin{gathered} \forall \mu \not \in {\text{\{ }}0,1{\text{\} }}\,\,{{{\bar {G}}}_{1}}(p,q)[\mu ] = q\min\left( {\frac{{p \cdot 1 + (1 - p) \cdot 0}}{\mu };\;{\kern 1pt} \frac{{p \cdot 0 + (1 - p) \cdot 0.5}}{{1 - \mu }}} \right) \\ \, + (1 - q)\min\left( {\frac{{p \cdot 0.5 + (1 - p) \cdot 0}}{\mu };\;\frac{{p \cdot 0 + (1 - p) \cdot 1}}{{1 - \mu }}} \right) = q\min\left( {\frac{p}{\mu };\;\frac{{1 - p}}{{2(1 - \mu )}}} \right) + (1 - q)\min\left( {\frac{p}{{2\mu }};\;\frac{{1 - p}}{{1 - \mu }}} \right); \\ \end{gathered} $$
$$\begin{gathered} \forall \nu \not \in \{ 0,1\} \,\,{{{\bar {G}}}_{2}}(p,q)[\nu ] = p\max\left( {\frac{{q \cdot 1 + (1 - q) \cdot 0.5}}{\nu };\;\frac{{q \cdot 0 + (1 - q) \cdot 0}}{{1 - \nu }}} \right) \\ \, + (1 - p)\max\left( {\frac{{q \cdot 0 + (1 - q) \cdot 0}}{\nu };\;\frac{{q \cdot 0.5 + (1 - q) \cdot 1}}{{1 - \nu }}} \right) = p\frac{{1 + q}}{{2\nu }} + (1 - p)\frac{{2 - q}}{{2(1 - \nu )}}, \\ \end{gathered} $$
$${{\bar {G}}_{1}}(p,q)[0] = q\frac{{1 - p}}{2} + (1 - q)(1 - p) = (1 - p)\left( {1 - \frac{q}{2}} \right) = (1 - p)\frac{{2 - q}}{2} = {{\bar {G}}_{2}}(p,q)[0],$$

and

$${{\bar {G}}_{1}}(p,q)[1] = qp + (1 - q)\frac{p}{2} = p\frac{{1 + q}}{2} = {{\bar {G}}_{2}}(p,q)[1].$$

To solve the parametric games

$$\bar {\Gamma }[\mu ,\nu ]\;\mathop = \limits^{{\text{def}}} \;\left\langle {[0,1],[0,1],{{{\bar {G}}}_{1}}(p,q)[\mu ], - {{{\bar {G}}}_{2}}(p,q)[\nu ]} \right\rangle ,\quad \mu ,\nu \in [0,1],$$

obtained for this example, we start with the ssearching the optimum strategies \(p{\kern 1pt} {\text{*}}(q,\mu )\) and \(q{\kern 1pt} {\text{*}}(p,\nu )\) for players S and T depending on their mutual actions. The strategies \(p{\kern 1pt} {\text{*}}(q,0) = 0\) and \(p{\kern 1pt} {\text{*}}(q,1) = 1\) maximize the payoff of player 1 for μ = 0 and 1, respectively, while the payoffs themselves are equal to

$$\mathop {\max}\limits_{p \in [0,1]} {{\bar {G}}_{1}}(p,q)[0] = {{\bar {G}}_{1}}(0,q)[0] = 1 - \frac{q}{2}\quad {\text{and}}\quad \mathop {\max}\limits_{p \in [0,1]} {{\bar {G}}_{1}}(p,q)[1] = {{\bar {G}}_{1}}(1,q)[1] = \frac{{1 + q}}{2}.$$

Strategies \(q{\kern 1pt} {\text{*}}(1,0)\) and \(q{\kern 1pt} {\text{*}}(0,1)\) minimizing the loss of player 2 for \(\nu \in \{ 0,1\} \) are arbitrary, \(q{\kern 1pt} {\text{*}}(p,0) = 1\) for p < 1, and \(q{\kern 1pt} {\text{*}}(p,1) = 0\) for p > 0. They lead to the results

$$\mathop {\min}\limits_{q \in [0,1]} {{\bar {G}}_{2}}(p,q)[0] = {{\bar {G}}_{2}}(p,1)[0] = \frac{{1 - p}}{2};\quad {\text{and}}\quad \mathop {\min}\limits_{q \in [0,1]} {{\bar {G}}_{2}}(p,q)[1] = {{\bar {G}}_{2}}(p,0)[1] = \frac{p}{2}.$$

Now, let \(\mu ,\nu \not \in \{ 0,1\} \). Then, \(\forall q \in [0,1]\), we have the value

$$\mathop {\max}\limits_{p \in [0,1]} {{\bar {G}}_{1}}(p,q)[\mu ] = \mathop {\max}\limits_{p \in [0,1]} \left\{ {\begin{array}{*{20}{l}} {\frac{p}{\mu }\frac{{1 + q}}{2},\quad 0 \leqslant p \leqslant \frac{\mu }{{2 - \mu }},} \\ {\frac{{p(1 - q - \mu ) + q\mu }}{{2\mu (1 - \mu )}},\quad \frac{\mu }{{2 - \mu }} \leqslant p \leqslant \frac{{2\mu }}{{1 + \mu }},} \\ {\frac{{1 - p}}{{1 - \mu }}\frac{{2 - q}}{2},\quad \frac{{2\mu }}{{1 + \mu }} \leqslant p \leqslant 1,} \end{array}} \right.$$

which is equal to

$$\left\{ {\begin{array}{*{20}{l}} {\frac{{1 + q}}{{2(2 - \mu )}},\quad q > 1 - \mu ,} \\ {\frac{1}{2},\quad q = 1 - \mu ,} \\ {\frac{{2 - q}}{{2(1 + \mu )}},\quad q < 1 - \mu ,} \end{array}} \right.$$

while the corresponding optimum strategy of player 1 is as follows:

$$p{\kern 1pt} {\text{*}}(q,\mu ) = \left\{ {\begin{array}{*{20}{l}} {\frac{\mu }{{2 - \mu }},\quad q > 1 - \mu ,} \\ {\forall p \in \left[ {\frac{\mu }{{2 - \mu }},\;{\kern 1pt} \frac{{2\mu }}{{1 + \mu }}} \right],\quad q = 1 - \mu ,} \\ {\frac{{2\mu }}{{1 + \mu }},\quad q < 1 - \mu .} \end{array}} \right.$$

For player 2, we have

$$\mathop {\min}\limits_{q \in [0,1]} {{\bar {G}}_{2}}(p,q)[\nu ] = \left\{ {\begin{array}{*{20}{l}} {\frac{3}{2},\quad p = \nu ,} \\ {\frac{p}{{2\nu }} + \frac{{1 - p}}{{1 - \nu }},\quad p > \nu ,} \\ {\frac{p}{\nu } + \frac{{1 - p}}{{2(1 - \nu )}},\quad p < \nu ,} \end{array}} \right.\quad q\text{*}(p,\nu ) = \left\{ {\begin{array}{*{20}{l}} {\forall ,\quad p = \nu ,} \\ {0,\quad p > \nu ,} \\ {1,\quad p < \nu .} \end{array}} \right.$$

Thus, if \(\mu ,\nu \not \in \{ 0,1\} \), then the Nash-equilibrium mixed strategies in game (4.3) in the ILC are the following pairs for the statement \(\bar {\Gamma }_{G}^{0}\) (i.e., for the family of games \(\bar {\Gamma }[\mu ,\nu ]\)):

$$(p\text{*},q\text{*})[\mu ,\nu ] = \left\{ {\begin{array}{*{20}{l}} {(\nu ,1 - \mu ),\quad \nu \in \left( {\frac{\mu }{{2 - \mu }},\;\,\frac{{2\mu }}{{1 + \mu }}} \right),} \\ {\left( {\frac{{2\mu }}{{1 + \mu }},0} \right),\quad \frac{{2\mu }}{{1 + \mu }} > \nu ,} \\ {\left( {\frac{\mu }{{2 - \mu }},1} \right),\quad \frac{\mu }{{2 - \mu }} < \nu ,} \\ {\left( {\frac{{2\mu }}{{1 + \mu }},\;\,q \in \;[0,\;1 - \mu ]} \right),\quad \nu = \frac{{2\mu }}{{1 + \mu }},} \\ {\left( {\frac{\mu }{{2 - \mu }},\;\,q \in [1 - \mu ,1]} \right),\quad \nu = \frac{\mu }{{2 - \mu }}.} \end{array}} \right.$$

For the other μ and ν, the equilibrium pairs are \((p\text{*},q\text{*})[0,1] = (0,\forall )\); \((p\text{*},q\text{*})[0,\nu ] = (0,1)\) \(\forall \nu \ne 1\); \((p\text{*},q\text{*})[1,0] = (1,\forall )\); \((p\text{*},q\text{*})[1,\nu ] = (1,0)\) \(\forall \nu \ne 0\);

$$(p\text{*},q\text{*})[\mu ,0] = \left( {\frac{\mu }{{2 - \mu }},1} \right),\quad (p\text{*},q\text{*})[\mu ,1] = \left( {\frac{{2\mu }}{{1 + \mu }},0} \right)\quad \forall \mu \not \in \{ 0,1\} .$$

It turns out that the set of solutions for the MC mixed extension \(\bar {\Gamma }_{G}^{0}\;\mathop = \limits^{{\text{def}}} \;\{ \bar {\Gamma }[\mu ,\nu ]\,{\text{|}}\,\mu ,\nu \in [0,1]\} \) (as well as for the standard mixed extension \(\bar {\Gamma }_{S}^{0}\)) is not selective in example (4.3): it contains all \(p,q \in [0,1]\) (for certain \(\mu ,\nu \in [0,1]\)).

Construct the sets \(\bar {Z}_{1}^{*}\) and \(\bar {Z}_{2}^{*}\) defined in the criterial space by (4.4). To parametrize \(\bar {Z}_{1}^{*}\), we start with computing the mean values of the ILC for all equilibrium pairs:

$$\bar {G}_{1}^{*}\left( {\frac{{2\mu }}{{1 + \mu }},0} \right) = \frac{1}{{1 + \mu }};\quad \bar {G}_{1}^{*}\left( {\frac{\mu }{{2 - \mu }},1} \right) = \frac{1}{{2 - \mu }};\quad \bar {G}_{1}^{*}\left( {\frac{{2\mu }}{{1 + \mu }},q} \right) = \frac{{2 - q}}{{2(1 + \mu )}}\quad \forall q \in [0,\;{\kern 1pt} 1 - \mu ]$$
$${\text{for}}\quad \nu = \frac{{2\mu }}{{1 + \mu }};\quad \bar {G}_{1}^{*}\left( {\frac{\mu }{{2 - \mu }},q} \right) = \frac{{1 + q}}{{2(2 - \mu )}}\quad \forall q \in [1 - \mu ,1]\quad {\text{for}}\quad \nu = \frac{\mu }{{2 - \mu }};$$
$$\bar {G}_{1}^{*}(\nu ,1 - \mu ) = (1 - \mu )\min\left( {\frac{\nu }{\mu },\;{\kern 1pt} \frac{{1 - \nu }}{{2(1 - \mu )}}} \right) + \mu \min\left( {\frac{\nu }{{2\mu }},\;{\kern 1pt} \frac{{1 - \nu }}{{1 - \mu }}} \right) = \frac{1}{2}\quad \forall \nu \in \left[ {\frac{\mu }{{2 - \mu }},\;{\kern 1pt} \frac{{2\mu }}{{1 + \mu }}} \right].$$

Then we unite the points obtained by multiplying these values for \(\mu \) by the vector \((\mu ,1 - \mu )\) of the coefficients of the convolution, \(\mu \in [0,1]\). Note that the first and second expressions multiplied by the coefficients of the convolution form the segments of the lines in the criterial space. Indeed, denoting the axes by z1 and z2, we have

$${{z}_{1}} = \frac{\mu }{{1 + \mu }},\quad {{z}_{2}} = \frac{{1 - \mu }}{{1 + \mu }} \Rightarrow \mu = \frac{{1 - {{z}_{2}}}}{{1 + {{z}_{2}}}},\quad {{z}_{2}} = 1 - 2{{z}_{1}},$$
$${{z}_{1}} = \frac{\mu }{{2 - \mu }},\quad {{z}_{2}} = \frac{{1 - \mu }}{{2 - \mu }} \Rightarrow \mu = \frac{{1 - 2{{z}_{2}}}}{{1 - {{z}_{2}}}},\quad {{z}_{2}} = \frac{{1 - {{z}_{1}}}}{2}.$$

Two expressions depending on q construct the junctions of the specified segments with the bold line obtained from value 1/2 (see the left-hand part on Fig. 1). The equilibrium pairs corresponding to \(\mu \in \{ 0,1\} \) yield the points on the axes. For variants with an arbitrary strategy of player 2, they yield segments from (0.5, 0) to (1, 0) and from (0, 0.5) to (0, 1).

In the same way, we construct the set \(\bar {Z}_{2}^{*}\). The mean values of the ILC of player 2 in the equilibrium situations \(\forall \nu \not \in \{ 0,1\} \) are as follows:

$$\bar {G}_{2}^{*}(\nu ,1 - \mu ) = \tfrac{3}{2}\quad {\text{for}}\quad \mu \not \in \{ 0,1\} ,\quad \nu \in \left[ {\tfrac{\mu }{{2 - \mu }},\;{\kern 1pt} \tfrac{{2\mu }}{{1 + \mu }}} \right];$$
$$\bar {G}_{2}^{*}\left( {\frac{{2\mu }}{{1 + \mu }},0} \right) = \frac{1}{{1 + \mu }}\left( {\frac{\mu }{\nu } + \frac{{1 - \mu }}{{1 - \nu }}} \right)\quad \forall \nu < \frac{{2\mu }}{{1 + \mu }};\quad \bar {G}_{2}^{*}\left( {\frac{\mu }{{2 - \mu }},1} \right) = \frac{1}{{2 - \mu }}\left( {\frac{\mu }{\nu } + \frac{{1 - \mu }}{{1 - \nu }}} \right)\quad \forall \nu > \frac{\mu }{{2 - \mu }};$$
$$\bar {G}_{2}^{*}\left( {\frac{{2\mu }}{{1 + \mu }},q} \right) = \frac{3}{2}\quad \forall q \in [0,1 - \mu ],\quad \nu = \frac{{2\mu }}{{1 + \mu }};\quad \bar {G}_{2}^{*}\left( {\frac{\mu }{{2 - \mu }},q} \right) = \frac{3}{2}\quad \forall q \in [1 - \mu ,1],\quad \nu = \frac{\mu }{{2 - \mu }}.$$

The equilibrium pairs obtained for \(\nu \in \{ 0,1\} \) add the value \(\bar {G}_{2}^{*}(0,q) = 0\) when ν = 1 and μ = 0, the value \(\bar {G}_{2}^{*}(1,q) = 0\) for ν = 0 and μ = 1, and the value \(\bar {G}_{2}^{*}(0,1) = \bar {G}_{2}^{*}(1,0) = 0.5\) for \(\mu = \nu = 0\) or \(\mu = \nu = 1\). In the criterial space, it is possible that, multiplied by \((\nu ,1 - \nu )\), they do not yield points on the axes, but, according to (4.4), lead to components of C(=1). In particular, we have point (0.5, 1) for \(\mu = \nu = 1\) and point (0, 1) for μ = 0 and ν = 1.

To compute the BGR, we find the best guaranteed mean values of the ILC:

$$\mathop {\min}\limits_{q \in [0,1]} \mathop {\max}\limits_{p \in [0,1]} {{\bar {G}}_{1}}(p,q)[\mu ] = \mathop {\min}\limits_{q \in [0,1]} \left\{ {\begin{array}{*{20}{l}} {\frac{{1 + q}}{{2(2 - \mu )}},\quad q > 1 - \mu ,} \\ {\frac{1}{2},\quad q = 1 - \mu ,} \\ {\frac{{2 - q}}{{2(1 + \mu )}},\quad q < 1 - \mu ,} \end{array}} \right.$$

which is equal to 1/2 (for player S) and is equal to

$$\mathop {\max}\limits_{p \in [0,1]} \mathop {\min}\limits_{q \in [0,1]} {{\bar {G}}_{1}}(p,q)[\mu ]\quad \forall \mu \in [0,1];$$
$$\mathop {\max}\limits_{p \in [0,1]} \mathop {\min}\limits_{q \in [0,1]} {{\bar {G}}_{2}}(p,q)[\nu ] = \mathop {\max}\limits_{p \in [0,1]} \left\{ {\begin{array}{*{20}{l}} {\frac{3}{2},\quad p = \nu ,} \\ {\frac{p}{{2\nu }} + \frac{{1 - p}}{{1 - \nu }},\quad p > \nu ,} \\ {\frac{p}{\nu } + \frac{{1 - p}}{{2(1 - \nu )}},\quad p < \nu ,} \end{array}} \right.$$

which is equal to 3/2 (for player T) and is also equal to

$$\mathop {\min}\limits_{q \in [0,1]} \mathop {\max}\limits_{p \in [0,1]} {{\bar {G}}_{2}}(p,q)[\nu ]\quad \forall \nu \in (0,1).$$

In the considered example, they do not depend on (μ, ν). After the multiplying by the coefficient vectors of the ILC, this yields the segments from (0, 0.5) to (0.5, 0) and from (1, 0.5) to (0.5, 1) in the criterial space (see Fig. 1) for the BGR of players S and T, respectively (the cut-off with C = 1 is taken into account).

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Kreines, E.M., Novikova, N.M. & Pospelova, I.I. Equilibria and Compromises in Two-Person Zero-Sum Multicriteria Games. J. Comput. Syst. Sci. Int. 59, 871–893 (2020). https://doi.org/10.1134/S1064230720060088

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