The Interaction of Economic Agents in Cournot Duopoly Models under Ecological Conditions: A Comparison of Organizational Modes

Abstract

This paper presents a comparative analysis of the efficiency of organizational modes (information structures) for the interaction of economic agents in static and dynamic Cournot duopoly models. We compare the independent behavior of equal players, their cooperation, and the hierarchy formalized as Germeier games. The efficiency of individual players and the entire society is quantitatively assessed using the private and social relative efficiency indices. The ecological safety conditions of the system are investigated. An organizational and economic interpretation of the results is proposed.

APPENDIX

Let us elucidate the data from Table 1. To find a Nash equilibrium in model (1)–(2), we solve the system  $\frac{{\partial {{g}_{i}}}}{{\partial {{u}_{i}}}}$ = 0, i = 1, 2. As a result, $\left\{ \begin{gathered} 1{\text{/}}2 - 2{{u}_{1}} - {{u}_{2}} = 0 \hfill \\ 1{\text{/}}2 - {{u}_{1}} - 2{{u}_{2}} = 0, \hfill \\ \end{gathered} \right.$ u₁ = u₂ = 1/6.

The Hessian matrix for this system, $\left\| {\begin{array}{*{20}{c}} { - 2 - 1} \\ { - 1 - 2} \end{array}} \right\|$, is negative definite. Therefore, $u_{1}^{{NE}}$ = $u_{2}^{{NE}}$ = 1/6, $g_{1}^{{NE}}$ = $g_{2}^{{NE}}$ = 1/36. Under cooperation, the players jointly maximize the function g($\bar {u}$) = (1/2 – $\bar {u}$)$\bar {u}$, where $\bar {u}$ = u₁ + u₂. We have $\frac{{\partial g}}{{\partial \bar {u}}}$ = 1/2 – 2$\bar {u}$ = 0, $\bar {u}$ = 1/4, and $\frac{{{{\partial }^{2}}g}}{{\partial {{{\bar {u}}}^{2}}}}$ = –2 < 0. Therefore, the set of Pareto-optimal cooperative solutions is the singleton ${{\bar {u}}^{C}}$ = 1/4; the corresponding equal imputation is $u_{1}^{C}$ = $u_{2}^{C}$ = 1/8 and the payoffs are $g_{1}^{C}$ = $g_{2}^{C}$ = 1/32. Assume that player 1 is a Leader in the Stackelberg sense. From the condition $\frac{{\partial {{g}_{2}}}}{{\partial {{u}_{2}}}}$ = 0 the optimal response of player 2 has the form u₂(u₁) = 1/4 – u₁/2. Substituting it into g₁ gives  g₁(u₁, u₂(u₁)) = (1/4 – u₁/2)u₁. The condition  $\frac{{\partial {{g}_{1}}}}{{\partial {{u}_{1}}}}$ = 0 yields u₁ = 1/4. Since $\frac{{{{\partial }^{2}}{{g}_{1}}}}{{\partial u_{1}^{2}}}$ = –1 < 0, we obtain $u_{1}^{{S{{T}_{1}}}}$ = 1/4, $u_{2}^{{S{{T}_{1}}}}$ = u₂($u_{1}^{{S{{T}_{1}}}}$) = 1/8, $g_{1}^{{S{{T}_{1}}}}$ = 1/32, and $g_{2}^{{S{{T}_{1}}}}$ = 1/64.

Finally, let us solve the game (1)–(2) as the Germeier game Γ₂ [25]. We have

$$u_{1}^{D}({{u}_{2}}) = \mathop {\operatorname{Arg} \max }\limits_{0 \leqslant {{u}_{1}} \leqslant 1/2} {{g}_{1}}({{u}_{1}},{{u}_{2}}) = {\text{1/4}} - {{u}_{2}}{\text{/}}2,$$

$$u_{1}^{P}({{u}_{2}}) = \mathop {\operatorname{Arg} \min }\limits_{0 \leqslant {{u}_{1}} \leqslant 1/2} {{g}_{2}}({{u}_{1}},{{u}_{2}}) \equiv {\text{1/2}}{\text{,}}$$

$${{L}_{2}} = \mathop {\max }\limits_{0 \leqslant {{u}_{2}} \leqslant 1/2} \left( {u_{1}^{P}({{u}_{2}}),{{u}_{2}}} \right) = \mathop {\max }\limits_{0 \leqslant {{u}_{2}} \leqslant 1/2} \left( { - u_{2}^{2}} \right) = 0,$$

$${{E}_{2}} = \left\{ {{{u}_{2}} \in {{U}_{2}}:{{g}_{2}}\left( {u_{1}^{P}({{u}_{2}}),{{u}_{2}}} \right) = {{L}_{2}}} \right\} = \{ 0\} ,$$

$${{D}_{2}} = \{ ({{u}_{1}},{{u}_{2}}):{{g}_{2}}({{u}_{1}},{{u}_{2}}) > 0\} ,$$

$${{K}_{2}} = \mathop {\min }\limits_{{{u}_{2}} \in {{E}_{2}}} \mathop {\max }\limits_{0 \leqslant {{u}_{1}} \leqslant 1/2} {{g}_{1}}({{u}_{1}},{{u}_{2}}) = \mathop {\max }\limits_{0 \leqslant {{u}_{1}} \leqslant 1/2} (1 - {{u}_{1}}){{u}_{1}} = 1{\text{/}}16.$$

To find the values K₁ = $\mathop {\sup }\limits_{{{D}_{2}}} {{g}_{1}}({{u}_{1}},{{u}_{2}})$, it is necessary to solve the optimization problem (1/2 – u₁ – u₂)u₁ → max subject to the constraints (1/2 – u₁ – u₂)u₂ > 0 and 0 ≤ u_i ≤ 1/2. Obviously, $u_{2}^{\varepsilon } = \varepsilon$ and $u_{1}^{\varepsilon }$ = 1/4. Then K₁ = 1/16 – ε/4 < K₂ and, therefore, the ε-optimal strategy of the Leader is $\tilde {u}_{1}^{\varepsilon }({{u}_{2}})$ = $\left\{ \begin{gathered} {\text{1/4}}\;{\text{if}}\;{{u}_{2}} = 0 \hfill \\ {\text{1/2}}\;{\text{otherwise}}{\text{.}} \hfill \\ \end{gathered} \right.$ In this case, $g_{1}^{{IS{{T}_{1}}}}$ = 1/16 and $g_{2}^{{IS{{T}_{1}}}}$ = 0.

Note  that ${{\bar {u}}^{{NE}}}$ = 1/3, ${{\bar {u}}^{C}}$ = 1/4, ${{\bar {u}}^{{ST}}}$ = 3/8, and ${{\bar {u}}^{{IST}}}$ = 1/4. Thus, we have arrived at the data from Table 3.