Dynamic multicriteria games with asymmetric players

Abstract

In this paper a new approach to construct cooperative behavior in dynamic multicriteria games with asymmetric players is presented. To obtain noncooperative and cooperative equilibria the ideas of multi-objective optimization and game theory are combined. To construct a multicriteria Nash equilibrium the bargaining solution is adopted. To design a multicriteria cooperative equilibrium a compromise scheme that guarantees the fulfillment of rationality conditions is applied. The concept of dynamic stability is adopted for dynamic multicriteria games, and the time-consistent payoff distribution procedure is presented. A dynamic bi-criteria bioresource management problem with different discount factors is considered. The strategies and payoffs of players are obtained under cooperative and noncooperative behavior.

Appendices

A Proof of rationality

Consider problem (5) with the rationality conditions. To construct cooperative behavior we should maximize the sum of the Nash products or equally minimize

$$\begin{aligned} H^c=(-V_1^{1c}+J_1^{1N})(V_1^{2c}-J_1^{2N}) +(-V_2^{1c}+J_2^{1N})(V_2^{2c}-J_2^{2N})\rightarrow \min _{u_{1t},u_{2t}} \end{aligned}$$

subject to

$$\begin{aligned}&J_1^{1N} - V_{1}^{1c} \le 0, \\&J_1^{2N} - V_{1}^{2c} \le 0, \\&J_2^{1N} - V_{2}^{1c} \le 0, \\&J_2^{2N} - V_{2}^{2c} \le 0, \\&\quad -(\varepsilon x_t -u_{1t}-u_{2t}) \le 0,\\&u_{1t} \ge 0,\,\,\, u_{2t} \ge 0. \end{aligned}$$

For simplicity, we give the proof for the case in which the players’ criteria depend on their own strategies only, i.e., $V_i^{jc}(u_{1t}^c,u_{2t}^c)=V_i^{jc}(u_{it})$, $i,j=1,2$.

The Kuhn–Tucker (KT) conditions are applicable, and the Lagrangian takes the form

$$\begin{aligned} L= & {} (-V_1^{1c}+J_1^{1N})(V_1^{2c}-J_1^{2N}) +(-V_2^{1c}+J_2^{1N})(V_2^{2c}-J_2^{2N})+\\&+\lambda _{11}(J_1^{1N} - V_{1}^{1c})+\lambda _{12}(J_1^{2N} - V_{1}^{2c})+\lambda _{21}(J_2^{1N} - V_{2}^{1c})+\lambda _{22}(J_2^{2N} - V_{2}^{2c})-\\&- \lambda _3 (\varepsilon x_t -u_{1t} -u_{2t}). \end{aligned}$$

First, we can eliminate the multiplier $\lambda _3$ as the KT conditions for it are

$$\begin{aligned}&\varepsilon x_ -u_{1t}-u_{2t} \ge 0 ,\\&\quad \lambda _3 (\varepsilon x_t -u_{1t}-u_{2t})=0. \end{aligned}$$

If we assume that $\lambda _3 > 0,$ then $\varepsilon x_t -u_{1t}-u_{2t}=0$ and $V_{i}^{jc}(u_{i\tau })=0$ for $\tau =t,\ldots ,m$, $i,j=1,2$, which obviously contradicts the rationality conditions.

Thus, we consider the KT conditions for the four Lagrange multipliers and one time instant $t=1,\ldots ,m.$ Hence, t can be omitted:

$$\begin{aligned}&-(V_1^{1c})'(V_1^{2c}-V_1^{2N}+\lambda _{11})+(V_1^{2c})'(-V_1^{1c}+V_1^{1N}-\lambda _{12}) \ge 0, \nonumber \\&\quad u_1 \Bigl [ -(V_1^{1c})'(V_1^{2c}-V_1^{2N}+\lambda _{11})+(V_1^{2c})'(-V_1^{1c}+V_1^{1N}-\lambda _{12}) \Bigr ] =0 , \end{aligned}$$

(14)

$$\begin{aligned}&-(V_2^{1c})'(V_2^{2c}-V_2^{2N}+\lambda _{21})+(V_2^{2c})'(-V_2^{1c}+V_2^{1N}-\lambda _{22}) \ge 0, \nonumber \\&\quad u_2 \Bigl [ -(V_2^{1c})'(V_2^{2c}-V_2^{2N}+\lambda _{21})+(V_2^{2c})'(-V_2^{1c}+V_2^{1N}-\lambda _{22}) \Bigr ] =0 , \end{aligned}$$

(15)

$$\begin{aligned}&V_1^{1c} - J_1^{1N} \ge 0 , \end{aligned}$$

(16)

$$\begin{aligned}&\lambda _{11}(V_1^{1c} - J_1^{1N})=0, \end{aligned}$$

(17)

$$\begin{aligned}&V_1^{2c} - J_1^{2N} \ge 0 , \end{aligned}$$

(18)

$$\begin{aligned}&\lambda _{12}(V_1^{2c} - J_1^{2N})=0, \end{aligned}$$

(19)

$$\begin{aligned}&V_2^{1c} - J_2^{1N} \ge 0 , \end{aligned}$$

(20)

$$\begin{aligned}&\lambda _{21}(V_2^{1c} - J_2^{1N})=0, \end{aligned}$$

(21)

$$\begin{aligned}&V_2^{2c} - J_2^{2N} \ge 0 , \end{aligned}$$

(22)

$$\begin{aligned}&\lambda _{22}(V_2^{2c} - J_2^{2N})=0, \nonumber \\&u_1 \ge 0,\,\, u_2 \ge 0, \,\, \lambda _{ij} \ge 0,\,\, i,j=1,2. \end{aligned}$$

(23)

• 1. Consider the case $\lambda _{ij}>0$, $i,j=1.2.$ From (17), (19), (21), and (23) it follows that

  $$\begin{aligned} V_1^{1c} - J_{1}^{1N}=0, \,\, V_1^{2c} - J_{1}^{2N}=0, \,\,V_2^{1c} - J_{2}^{1N}=0, \,\,V_2^{2c} - J_{2}^{2N}=0. \end{aligned}$$

  If any of the strategies $u_i$, $i=1,2$, is equal to zero, then conditions (16), (18) or (20), (22) are not satisfied. Hence, $u_i>0$. In this case, the cooperative behavior coincides with the noncooperative one, $u_1=u_1^N$ and $u_2=u_2^N,$ and the goal function $H^c$ is equal to zero.

• 2. Consider the case $\lambda _{11}=0$, $\lambda _{12} >0$, $\lambda _{21} >0$, $\lambda _{22} >0.$ From (19), (21), and (23) it follows that

  $$\begin{aligned} V_1^{2c} - J_{1}^{2N}=0, \,\,V_2^{1c} - J_{2}^{1N}=0, \,\,V_2^{2c} - J_{2}^{2N}=0. \end{aligned}$$

  By analogy, $u_i>0$, $i=1,2,$ and from (14) and (15) it follows that

  $$\begin{aligned} \left\{ \begin{array}{l} (V_1^{2c})'(-V_1^{1c}+V_1^{1N}-\lambda _{12})=0,\\ -(V_2^{1c})'\lambda _{21}-(V_2^{2c})'\lambda _{22}=0. \end{array} \right. \end{aligned}$$

  Consequently,

  $$\begin{aligned} -V_1^{1c}+V_1^{1N}=\lambda _{12}>0, \end{aligned}$$

  which obviously contradicts condition (16).

  Similarly, in all the cases where one of the Lagrange multipliers is equal to zero, we will naturally arrive in contradiction.

• 3. Consider the case $\lambda _{11}=0$, $\lambda _{12}=0$, $\lambda _{21} >0$, $\lambda _{22} >0.$ From (21) and (23) it follows that

  $$\begin{aligned} V_2^{1c} - J_{2}^{1N}=0, \,\,V_2^{2c} - J_{2}^{2N}=0. \end{aligned}$$

  Hence, the cooperative behavior of player 2 coincides with the noncooperative one, $u_2=u_2^N,$ and the strategy of player 1 can be obtained from the equation

  $$\begin{aligned} -(V_1^{1c})'(V_1^{2c}-V_1^{2N})+(V_1^{2c})'(-V_1^{1c}+V_1^{1N})=0. \end{aligned}$$

  (24)

  Here, the goal function becomes

  $$\begin{aligned} H^c=(-V_1^{1c}+V_1^{1N})(V_1^{2c}-V_1^{2N}). \end{aligned}$$

  (25)

• 4. Similarly, if $\lambda _{11}>0$, $\lambda _{12}>0$, $\lambda _{21}=0$, and $\lambda _{22}=0$, then the cooperative behavior of player 1 coincides with the noncooperative one, $u_1=u_1^N,$ and the strategy of player 2 can be obtained from the equation

  $$\begin{aligned} -(V_2^{1c})'(V_2^{2c}-V_2^{2N})+(V_2^{2c})'(-V_2^{1c}+V_2^{1N})=0. \end{aligned}$$

  (26)

  Here, the goal function becomes

  $$\begin{aligned} H^c=(-V_2^{1c}+V_2^{1N})(V_2^{2c}-V_2^{2N}). \end{aligned}$$

  (27)

• 5. Consider the case $\lambda _{11}=0$, $\lambda _{12} >0$, $\lambda _{21} =0$, $\lambda _{22} >0.$ From (19) and (23) it follows that

  $$\begin{aligned} V_1^{2c} - J_{1}^{2N}=0, \,\,V_2^{2c} - J_{2}^{2N}=0. \end{aligned}$$

  Similarly, $u_i>0$, $i=1,2$, and from (14) and (15) we obtain

  $$\begin{aligned} \left\{ \begin{array}{c} (V_1^{2c})'(-V_1^{1c}+V_1^{1N}-\lambda _{12})=0,\\ (V_2^{2c})'(-V_2^{1c}+V_2^{1N}-\lambda _{22})=0. \end{array} \right. \end{aligned}$$

  Consequently,

  $$\begin{aligned} -V_1^{1c}+V_1^{1N}=\lambda _{12}>0, \,\, -V_2^{1c}+V_2^{1N}=\lambda _{22}>0, \end{aligned}$$

  which obviously contradict conditions (16), (20).

  By analogy, if $\lambda _{11}>0$, $\lambda _{12}=0$, $\lambda _{21}>0$, and $\lambda _{22}=0,$ we will arrive in contradiction.

• 6. Finally, consider the case $\lambda _{ij}=0$, $i,j=1,2$.Similarly, it is easy to check that $u_i>0$, $i=1,2$, the minimum is achieved at an interior point and can be found via the first-order optimality conditions (24) and (26).

  Here, the goal function becomes

  $$\begin{aligned} H^c=(-V_1^{1c}+V_1^{1N})(V_1^{2c}-V_1^{2N})+(-V_2^{1c}+V_2^{1N})(V_2^{2c}-V_2^{2N}), \end{aligned}$$

  which is less than zero, (25) and (27).

Thus, the presented scheme gives a solution if the rationality conditions are satisfied.

If all the criteria depend on the strategies of both players, the outline of the proof is the same but with more complicated equations.

B Proof of Proposition 2

First, consider problem (12) for the one-stage game. Let the initial size of the population be x.

As usual, we find the players’ strategies in the linear class $u_{i1}^c= \gamma _{i1} x$, $i=1,2$. As a result, problem (12) takes the form

$$\begin{aligned} p_1(\gamma _{11}^c-N_1)(-(\gamma _{11}^c)^2+M_1)+p_2(\gamma _{21}^c-N_2)(-(\gamma _{21}^c)^2+M_2) \rightarrow \max _{\gamma _{11}^c,\gamma _{21}^c}. \end{aligned}$$

From the first-order optimality conditions we obtain

$$\begin{aligned} \gamma _{11}^c=\frac{N_1+((N_1)^2+3M_1)^{1/2}}{3}, \,\, \gamma _{21}^c=\frac{N_2+((N_2)^2+3M_2)^{1/2}}{3}. \end{aligned}$$

Now, we consider problem (12) for the two-stage game. To determine cooperative strategies we solve the following problem:

$$\begin{aligned}&p_1(\gamma _{12}^c+\delta _1\gamma _{11}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)-N_1)(-(\gamma _{12}^c)^2-(\gamma _{11}^c)^2(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)+M_1)\\\,&\quad +p_2(\gamma _{22}^c+\delta _2\gamma _{21}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)-N_2)\cdot \\&\quad \cdot (-(\gamma _{22}^c)^2-(\gamma _{21}^c)^2(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)+M_2). \rightarrow \max _{\gamma _{11}^c,\gamma _{21}^c,\gamma _{12}^c,\gamma _{22}^c}. \end{aligned}$$

The first-order optimality conditions are given by

$$\begin{aligned}&p_1(D_2-2\gamma _{12}^cD_1)\nonumber \\&\quad -p_1\delta _1\gamma _{11}^c(D_2-2\gamma _{11}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)D_1) -p_2\delta _2\gamma _{21}^c(D_4-2\gamma _{21}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)D_3), \end{aligned}$$

(28)

$$\begin{aligned}&p_2(D_4-2\gamma _{22}^cD_3)\nonumber \\&\quad -p_2\delta _2\gamma _{21}^c(D_4-2\gamma _{21}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)D_3) -p_1\delta _1\gamma _{11}^c(D_2-2\gamma _{11}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)D_1), \end{aligned}$$

(29)

$$\begin{aligned}&D_2=2\gamma _{11}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)D_1, \end{aligned}$$

(30)

$$\begin{aligned}&D_4=2\gamma _{21}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)D_3, \end{aligned}$$

(31)

where $D_1=\gamma _{12}^c+\delta _1\gamma _{11}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)-N_1$, $D_2=-(\gamma _{12}^c)^2-(\gamma _{11}^c)^2(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)+M_1$, $D_3=\gamma _{22}^c+\delta _2\gamma _{21}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)-N_2$, $D_4-(\gamma _{22}^c)^2-(\gamma _{21}^c)^2(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)+M_2)$.

Substituting (30) and (31) into (28) and (29) yields

$$\begin{aligned} D_2= & {} 2\gamma _{12}^cD_1,\\ D_4= & {} 2\gamma _{22}^cD_3. \end{aligned}$$

In view of (30) and (31), these imply

$$\begin{aligned} \gamma _{11}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)=\gamma _{12}^c,\\ \gamma _{21}^c(\varepsilon -\gamma _{12}^c-\gamma _{22}^c)=\gamma _{22}^c, \end{aligned}$$

and consequently

$$\begin{aligned} \gamma _{12}^c=\frac{\varepsilon \gamma _{11}^c}{1+\gamma _{11}^c+\gamma _{21}^c}, \,\,\gamma _{22}^c=\frac{\varepsilon \gamma _{21}^c}{1+\gamma _{11}^c+\gamma _{21}^c}. \end{aligned}$$

Thus, we have expressed all the parameters via the players’ strategies at the last stage; to determine them we need to solve one of equations (28)–(31). From (29) we obtain

$$\begin{aligned} \gamma _{21}^c=(1+\gamma _{11}^c)\frac{-M_2-\varepsilon N_2+\varepsilon (N_2^2+3(1+\delta _2))^{1/2}}{M_2+2\varepsilon N_2-3\varepsilon ^2(1+\delta _2)}, \end{aligned}$$

and $\gamma _{11}^c$ is found from the equation

$$\begin{aligned} (\gamma _{11}^c)^2(M_1+2\varepsilon N_1-3\varepsilon ^(1+\delta _1))+\gamma _{11}^c(M_1+\varepsilon N_1)(1+\gamma _{21}^c)+M_1(1+\gamma _{21}^c)^2=0. \end{aligned}$$

The process can be repeated for the m-stage game, and we will finally arrive in the cooperative strategies as specified by Proposition 2.