Game-Theoretic Analysis of the Interaction of Economic Agents in the Cournot Oligopoly with Consideration of the Linear Structure, the Green Effect, and Fairness Concern

Abstract

A comparative analysis of the efficiency of various ways of organization of economic agents is carried out, taking into account the structure and regulations of their interaction in the models of the Cournot oligopoly. The Cournot oligopoly models in the form of a supply chain are constructed and analytically investigated, taking into account the green effect and concern for fairness. For symmetric models of the Cournot oligopoly with different ways of organization of economic agents, the respective structures of social and individual preferences are analytically obtained. A numerical study of the Cournot oligopoly models in various forms with asymmetrical agents has been carried out, and the corresponding structures of social and individual preferences have been obtained.

Appendices

APPENDIX A

Proof of Theorem 2.1.

$$\frac{{\partial u_{i}^{G}}}{{\partial {{x}_{i}}}} = a - c - 2{{x}_{i}} - \sum\limits_{j \ne i}^{} {{{x}_{j}}} + \sum\limits_{i = 1}^n {{{g}_{j}}} = 0,\;\;i = 1,\,\, \ldots ,\,\,n;$$

$$\frac{{\partial u_{i}^{G}}}{{\partial {{g}_{i}}}} = {{x}_{i}} - 2{{g}_{i}} = 0,\quad i = 1,\,\, \ldots ,\,\,n;$$

$${{x}_{i}} = 2{{g}_{i}},\;\;i = 1,\,\, \ldots ,\,\,n;$$

$$a - c - 4{{g}_{i}} - 2\sum\limits_{j \ne i} {{{g}_{j}}} + \sum\limits_{j = 1}^n {{{g}_{j}}} = 0;\;\;a - c - 3{{g}_{i}} - \sum\limits_{j \ne i} {{{g}_{j}}} = 0;$$

$$3{{g}_{i}} + \sum\limits_{j \ne i} {{{g}_{j}}} = a - c,\;\;i = 1,\,\, \ldots ,\,\,n;$$

$${{g}_{i}} = g,\;\;i = 1,\,\, \ldots ,\,\,n;\;\;3g + (n - 1)g = a - c;$$

$$g = \frac{{a - c}}{{n + 2}},\;\;x = \frac{{2(a - c)}}{{n + 2}}$$

the value u is found by an immediate substitution.

Proof of Theorem 2.3.

$$\frac{{\partial \bar {u}}}{{\partial {{x}_{i}}}} = - \bar {x} + a - c - \bar {x} + \bar {g} = a - c + \bar {g} - 2\bar {x} = 0,\;\;i = 1,\,\, \ldots ,\,\,n;\;\;\frac{{\partial \bar {u}}}{{\partial {{g}_{i}}}} = \bar {x} - 2{{g}_{i}} = 0,\;\;i = 1,\,\, \ldots ,\,\,n;$$

$${{g}_{i}} = g,\;\;{{x}_{i}} = x,\;\;i = 1,\,\, \ldots ,\,\,n;\;\;\bar {x} = nx,\;\;\bar {g} = ng,$$

$$\bar {x} = 2g;\;\;a - c + ng - 4g = 0;$$

$$g = \left\{ {\begin{array}{*{20}{l}} {\frac{{a - c}}{{4 - n}},\;\;}&{n < 4} \\ {0,}&{{\text{otherwise}}{\text{.}}} \end{array}} \right.$$

If g = 0 then x = x^C is an optimal solution of the cooperative problem without “green” effect. Then

$$x = \left\{ {\begin{array}{*{20}{l}} {\frac{{2(a - c)}}{{n(4 - n)}},\;\;}&{n < 4} \\ {{{x}^{C}},}&{{\text{otherwise}}{\text{,}}} \end{array}} \right.\;\;\bar {u} = \left\{ {\begin{array}{*{20}{l}} {\frac{{{{{(a - c)}}^{2}}}}{{4 - n}},\;\;}&{n < 4} \\ {{{u}^{C}},}&{{\text{otherwise}}{\text{,}}} \end{array}} \right.$$

Proof of Theorem 2.7. A best response of the agents to the Principal’s strategy (g₁, …, g_n), or a Nash equilibrium in the game of agents, is determined by solution of the system

$$\frac{{\partial u_{i}^{{ST}}}}{{\partial {{x}_{i}}}} = a - c - 2{{x}_{i}} - \sum\limits_{j \ne i} {{{x}_{j}} + \bar {g}} = 0,\;\;i = 1, \ldots ,n;$$

and then

$$2{{x}_{i}} + \sum\limits_{j \ne i} {{{x}_{j}}} = a - c + \bar {g},\;\;i = 1, \ldots ,n;$$

$${{x}_{i}} = {{x}^{{ST}}},\;\;i = 1, \ldots ,n;$$

$${{x}^{{ST}}} = \frac{{a - c + \bar {g}}}{{n + 1}};\;\;{{\bar {x}}^{{ST}}} = \frac{{n(a - c + \bar {g})}}{{n + 1}}.$$

A substitution of the found best response in the Principal’s payoff function gives

$$\bar {u}(g,{{x}^{{ST}}}) = (a - c - {{\bar {x}}^{{ST}}} + \bar {g}){{\bar {x}}^{{ST}}} - \sum\limits_{i = 1}^n {g_{i}^{2}} $$

$$ = \left( {a - c + \bar {g} - \frac{{n(a - c + \bar {g})}}{{n + 1}}} \right)\frac{{n(a - c + \bar {g})}}{{n + 1}} - \sum\limits_{i = 1}^n {g_{i}^{2}} = \frac{{n{{{(a - c + \bar {g})}}^{2}}}}{{{{{(n + 1)}}^{2}}}} - \sum\limits_{i = 1}^n {g_{i}^{2}.} $$

Then

$$\frac{{\partial \bar {u}}}{{\partial {{g}_{i}}}} = \frac{{2n(a - c + \bar {g})}}{{{{{(n + 1)}}^{2}}}} - 2{{g}_{i}} = 0,\;\;i = 1,\,\, \ldots ,\,\,n;$$

$${{g}_{i}} = {{g}^{{ST}}},\;\;i = 1,\,\, \ldots ,\,\,n;\;\;{{\bar {g}}^{{ST}}} = n{{g}^{{ST}}};$$

$$n(a - c + n{{g}^{{ST}}}) - {{(n + 1)}^{2}}{{g}^{{ST}}} = 0;$$

$${{g}^{{ST}}} = \frac{{n(a - c)}}{{2n + 1}}$$

and we find x^ST = $\frac{{(n + 1)(a - c)}}{{2n + 1}}$, u^ST = $\frac{{{{{(a - c)}}^{2}}}}{{2n + 1}}$.

Proof of Theorem 2.9. The best response of the agents is still determined by the expression (when n = 2)

$${{x}^{{STFC}}} = \frac{{a - c + \bar {g}}}{3};\;\;{{\bar {x}}^{{STFC}}} = \frac{{2(a - c + \bar {g})}}{3}.$$

$${{x}^{{ST}}} = \frac{{a - c + \bar {g}}}{{n + 1}};\;\;{{\bar {x}}^{{ST}}} = \frac{{n(a - c + \bar {g})}}{{n + 1}}.$$

The Principal’s payoff function is

$${{U}^{{FC}}}(g,x) = (a - c + \bar {g} - \bar {x})\bar {x} - \delta ({{u}_{1}} - {{u}_{2}}) - g_{1}^{2} - g_{2}^{2}.$$

Notice that

$${{u}_{1}}(g,{{x}^{{STFC}}}) - {{u}_{2}}(g,{{x}^{{STFC}}}) = Q(g,{{x}^{{STFC}}})({{x}^{{STFC}}} - {{x}^{{STFC}}}) - g_{1}^{2} + g_{2}^{2} = g_{2}^{2} - g_{1}^{2}.$$

Then a substitution of x^STFC in U^FC gives

$$\begin{gathered} {{U}^{{FC}}}(g,{{x}^{{STFC}}}) = \left( {a - c + \bar {g} - \frac{2}{3}(a - c + \bar {g})} \right)\frac{2}{3}(a - c + \bar {g}) - \delta (g_{2}^{2} - g_{1}^{2}) - g_{1}^{2} - g_{2}^{2} \\ = \frac{2}{9}{{(a - c + \bar {g})}^{2}} - (1 - \delta )g_{1}^{2} - (1 + \delta )g_{2}^{2}. \\ \end{gathered} $$

Then we solve the system of equations

$$\frac{{\partial {{U}^{{FC}}}}}{{\partial {{g}_{1}}}} = \frac{4}{9}(a - c + \bar {g}) - 2(1 - \delta ){{g}_{1}} = 0,$$

$$\frac{{\partial {{U}^{{FC}}}}}{{\partial {{g}_{2}}}} = \frac{4}{9}(a - c + \bar {g}) - 2(1 - \delta ){{g}_{2}} = 0$$

and

$${{g}_{1}} = \frac{{2(1 + \delta )(a - c)}}{{5 - 9{{\delta }^{2}}}},\;\;{{g}_{2}} = \frac{{2(1 - \delta )(a - c)}}{{5 - 9{{\delta }^{2}}}}.$$

These are positive when δ < $\frac{{\sqrt 5 }}{3}$. The pair ($g_{1}^{{STFC}}$, $g_{2}^{{STFC}}$) really maximizes U^FC under the conditions $\frac{{{{\partial }^{2}}{{U}^{{FC}}}}}{{\partial g_{1}^{2}}}$ = $\frac{4}{9}$ – 2(1 – δ) < 0, or δ < $\frac{7}{9}$ and |H| = $\left| {\begin{array}{*{20}{c}} {\frac{4}{9} - 2(1 - \delta )}&{\frac{4}{9}} \\ {\frac{4}{9}}&{\frac{4}{9} - 2(1 + \delta )} \end{array}} \right|$ > 0, or δ < $\sqrt {\frac{{41}}{{81}}}$. A comparison of the three received inequalities gives the final condition δ < $\sqrt {\frac{{41}}{{81}}}$. Immediate calculations give

$${{x}_{1}} = {{x}_{2}} = {{x}^{{STFC}}} = \frac{{3(1 - {{\delta }^{2}})(a - c)}}{{{{{(5 - 9{{\delta }^{2}})}}^{2}}}};$$

$$u_{1}^{{STFC}} = \frac{{{{{(1 + \delta )}}^{2}}(5 - 18\delta + {{\delta }^{2}}){{{(a - c)}}^{2}}}}{{{{{(5 - 9{{\delta }^{2}})}}^{2}}}},\;\;u_{2}^{{STFC}} = \frac{{{{{(1 - \delta )}}^{2}}(5 - 18\delta + {{\delta }^{2}}){{{(a - c)}}^{2}}}}{{{{{(5 - 9{{\delta }^{2}})}}^{2}}}}.$$

APPENDIX B

Proof of Theorem 4.1.

$$\frac{{\partial u_{i}^{G}}}{{\partial {{x}_{i}}}} = - {{x}_{i}} + a - {{c}_{i}} - \sum\limits_{j = 1}^n {{{x}_{j}}} + \alpha \sum\limits_{i = 1}^n {{{g}_{j}}} = 0,\;\;i = 1,\,\, \ldots ,\,\,n;$$

$$\frac{{\partial u_{i}^{G}}}{{\partial {{g}_{i}}}} = \alpha {{x}_{i}} - 2{{\beta }_{i}}{{g}_{i}} = 0,\;\;i = 1,\,\, \ldots ,\,\,n.$$

From the latter expression

$${{g}_{i}} = \frac{{\alpha {{x}_{i}}}}{{2{{\beta }_{i}}}},\;\;i = 1,\,\, \ldots ,\,\,n.$$

A substitution of the received dependency into the first equality gives

$${{x}_{i}} = a - {{c}_{i}} + \sum\limits_{j = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{j}}}} - 1} \right){{x}_{j}}} ,\;\;i = 1,\,\, \ldots ,\,\,n.$$

Multiply both sides of the equality by the expression in brackets:

$$\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right){{x}_{i}} = \left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right)(a - {{c}_{i}}) + \left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right)\sum\limits_{j = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{j}}}} - 1} \right){{x}_{j}}} .$$

The summation of the received equality by i gives

$$\sum\limits_{i = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right){{x}_{i}}} = \sum\limits_{i = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right)(a - {{c}_{i}})} + \sum\limits_{i = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right)\sum\limits_{j = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{j}}}} - 1} \right){{x}_{j}}.} } $$

Now calculate $\sum\limits_{i = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right){{x}_{i}}}$:

$$\sum\limits_{i = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right)x_{i}^{{GNE}}} = \frac{{\sum\limits_{i = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right)(a - {{c}_{i}})} }}{{1 - \sum\limits_{i = 1}^n {\left( {\frac{{{{\alpha }^{2}}}}{{2{{\beta }_{i}}}} - 1} \right)} }}.$$

Proof of Theorem 4.3.

$$\frac{{\partial \bar {u}}}{{\partial {{x}_{i}}}} = - 2\bar {x} + a - {{c}_{i}} + \alpha \bar {g} = 0,\;\;i = 1,\,\, \ldots ,\,\,n;\;\;\frac{{\partial \bar {u}}}{{\partial {{g}_{i}}}} = \alpha \bar {x} - 2{{\beta }_{i}}{{g}_{i}} = 0,\;\;i = 1,\,\, \ldots ,\,\,n.$$

From the latter relation

$${{g}_{i}} = \frac{{\alpha \sum\limits_{i = 1}^n {{{x}_{i}}} }}{{2{{\beta }_{i}}}},\;\;i = 1,\,\, \ldots ,\,\,n.$$

A substitution of the received dependency into the first equality gives

$$\sum\limits_{i = 1}^n {{{x}_{i}}} = \frac{{a - {{c}_{i}}}}{{2 - \frac{{{{\alpha }^{2}}}}{2}\sum\limits_{j = 1}^n {\frac{1}{{{{\beta }_{j}}}}} }},\;\;i = 1,\,\, \ldots ,\,\,n.$$

A substitution of the received result into the objective function of the cooperative problem shows that it is increases when c_i decreases, and then

$$\sum\limits_{i = 1}^n {{{x}_{i}}} = \frac{{a - {{c}_{k}}}}{{2 - \frac{{{{\alpha }^{2}}}}{2}\sum\limits_{j = 1}^n {\frac{1}{{{{\beta }_{j}}}}} }},$$

where k is an index of the agent with a minimal c_k. Then

$$\begin{gathered} {{x}_{k}} = \frac{{a - {{c}_{k}}}}{{2 - \frac{{{{\alpha }^{2}}}}{2}\sum\limits_{j = 1}^n {\frac{1}{{{{\beta }_{j}}}}} }}, \\ {{x}_{{i \ne k}}} = 0. \\ \end{gathered} $$

Proof of Theorem 4.7. A best response of the agents to the Principal’s strategy (g₁, …, g_n), or a Nash equilibrium in the game of agents, is determined by the solution of the system

$$\frac{{\partial u_{i}^{{ST}}}}{{\partial {{x}_{i}}}} = a - {{c}_{i}} - {{x}_{i}} - \sum\limits_{j = 1}^n {{{x}_{j}}} + \alpha \bar {g}$$

and

$${{x}_{i}} = a - {{c}_{i}} - \sum\limits_{j = 1}^n {{{x}_{j}}} + \alpha \sum\limits_{j = 1}^n {{{g}_{j}}.} $$

Sum both sides of the inequality:

$$\sum\limits_{i = 1}^n {{{x}_{i}}} = na - \sum\limits_{i = 1}^n {{{c}_{i}}} - n\sum\limits_{j = 1}^n {{{x}_{j}}} + \alpha n\sum\limits_{j = 1}^n {{{g}_{j}}.} $$

The calculation of $\sum\limits_{i = 1}^n {{{x}_{i}}}$ gives:

$$\sum\limits_{i = 1}^n {{{x}_{i}}} = \frac{{na - \sum\limits_{i = 1}^n {{{c}_{i}}} + \alpha n\sum\limits_{j = 1}^n {{{g}_{j}}} }}{{1 + n}}.$$

A substitution of the received sum into the expression for x_i gives

$$x_{i}^{{ST}} = \frac{{a - (1 + n){{c}_{i}} + n\alpha \bar {g}}}{{n + 1}}.$$

A substitution of the found best response into the Principal’s payoff function gives

$$\bar {u}(g,{{x}^{{ST}}}) = \frac{{n{{a}^{2}} - a\sum\limits_{i = 1}^n {{{c}_{i}}} + \alpha an\sum\limits_{j = 1}^n {{{g}_{j}}} }}{{1 + n}}$$

$$ - \;\frac{{\sum\limits_{i = 1}^n {\left( {{{c}_{i}}a - (1 - n)c_{i}^{2} + {{c}_{i}}\sum\limits_{i = 1}^n {{{c}_{i}}} + \alpha {{c}_{i}}\sum\limits_{j = 1}^n {{{g}_{j}}} } \right)} }}{{1 + n}} + \alpha \sum\limits_{i = 1}^n {{{g}_{j}}} \left( {\frac{{na - \sum\limits_{i = 1}^n {{{c}_{i}}} + \alpha n\sum\limits_{j = 1}^n {{{g}_{j}}} }}{{1 + n}}} \right)$$

$$ - \;{{\left( {\frac{{na - \sum\limits_{i = 1}^n {{{c}_{i}}} + \alpha n\sum\limits_{j = 1}^n {{{g}_{j}}} }}{{1 + n}}} \right)}^{2}} - 2\sum\limits_{i = 1}^n {{{\beta }_{i}}g_{i}^{2}.} $$

Then

$$\frac{{\partial \bar {u}}}{{\partial {{g}_{i}}}} = \frac{{a\alpha n}}{{1 + n}} - \frac{{\alpha \sum\limits_{i = 1}^n {{{c}_{i}}} }}{{1 + n}} + \alpha \frac{{na - \sum\limits_{i = 1}^n {{{c}_{i}}} + \alpha n\sum\limits_{i = 1}^n {{{g}_{j}}} }}{{1 + n}} + {{\alpha }^{2}}\frac{{n\sum\limits_{i = 1}^n {{{g}_{j}}} }}{{1 + n}}$$

$$ - \;\frac{{2n\alpha }}{{{{{(n + 1)}}^{2}}}}\left( {na - \sum\limits_{i = 1}^n {{{c}_{i}}} + \alpha n\sum\limits_{j = 1}^n {{{g}_{j}}} } \right) - 2{{\beta }_{i}}{{g}_{i}} = 0,\;\;i = 1, \ldots ,n$$

and

$$g_{i}^{{ST}} = \frac{{\alpha \left( {na - \sum\limits_{i = 1}^n {{{c}_{i}}} + \alpha n\sum\limits_{j = 1}^n {{{g}_{j}}} } \right)}}{{{{\beta }_{i}}{{{(n + 1)}}^{2}}}}$$

and the summation by all indices gives

$$\sum\limits_{i = 1}^n {g_{i}^{{ST}}} = \frac{{\alpha \left( {na - \sum\limits_{i = 1}^n {{{c}_{i}}} } \right)\sum\limits_{i = 1}^n {\frac{1}{{{{\beta }_{i}}}}} }}{{{{{(n + 1)}}^{2}} - n{{\alpha }^{2}}\sum\limits_{i = 1}^n {\frac{1}{{{{\beta }_{i}}}}} }}.$$

Proof of Theorem 4.9. The best response of the agents is still determined by the expression (when n = 2)

$$x_{1}^{{STFC}} = \frac{{a - 2{{c}_{1}} + {{c}_{2}} + \alpha \bar {g}}}{3};\;\;x_{2}^{{STFC}} = \frac{{a - 2{{c}_{2}} + {{c}_{1}} + \alpha \bar {g}}}{3}.$$

The Principal’s payoff function is

$$\begin{gathered} {{U}^{{FC}}}(g,x) = (1 - \delta )(a - {{c}_{1}} - {{x}_{1}} - {{x}_{2}} + \alpha \bar {g}){{x}_{1}} - (1 - \delta ){{\beta }_{1}}g_{1}^{2} \\ + \;(1 + \delta )(a - {{c}_{2}} - {{x}_{1}} - {{x}_{2}} + \alpha \bar {g}){{x}_{2}} - (1 + \delta ){{\beta }_{2}}g_{2}^{2}. \\ \end{gathered} $$

A substitution of x^STFC into U^FC gives

$${{U}^{{FC}}}(g,{{x}^{{STFC}}}) = \frac{{1 - \delta }}{9}{{(a - 2{{c}_{1}} + {{c}_{2}} + \alpha \bar {g})}^{2}} - (1 - \delta ){{\beta }_{1}}g_{1}^{2}$$

$$ + \;\frac{{1 + \delta }}{9}{{(a - 2{{c}_{2}} + {{c}_{1}} + \alpha \bar {g})}^{2}} - (1 + \delta ){{\beta }_{2}}g_{2}^{2}.$$

The first order conditions determine the system

$$\frac{{\partial {{U}^{{FC}}}}}{{\partial {{g}_{1}}}} = \frac{{2\alpha (1 - \delta )}}{9}(a - 2{{c}_{1}} + {{c}_{2}} + \alpha \bar {g}){{x}_{2}} - 2(1 - \delta ){{\beta }_{1}}{{g}_{1}}$$

$$ + \;\frac{{2\alpha (1 + \delta )}}{9}(a - 2{{c}_{2}} + {{c}_{1}} + \alpha \bar {g}){{x}_{2}} = 0,$$

$${{g}_{2}} = \frac{{(1 - \delta ){{\beta }_{1}}}}{{(1 + \delta ){{\beta }_{2}}}}{{g}_{1}},$$

and

$$g_{1}^{{STFC}} = g_{2}^{{STFC}} = \frac{{2a - (1 - 3\delta ){{c}_{1}} - (1 + 3\delta ){{c}_{2}}}}{{\frac{{9(1 - \delta ){{\beta }_{1}}}}{\alpha } - 2\alpha \left( {\frac{{(1 - \delta ){{\beta }_{1}}}}{{(1 + \delta ){{\beta }_{2}}}} + 1} \right)}}$$

and all other values are found by substitution of the found ones.

APPENDIX C

Table C1. Solution of the Cournot oligopoly with nine asymmetrical agents (when a = 3000)

APPENDIX D

Table D1. Solution of the Cournot oligopoly with two asymmetrical agents (when a = 10), c₁ = 1, d₁ = 1, β₁ = 100, c₂ = 5, d₂ = 1, β₂ = 100

Table D2. Solution of the Cournot oligopoly with three asymmetrical agents and small cost coefficients and c₁ = 1, d₁ = 1, c₂ = 3, d₂ = 1, c₃ = 5, d₃ = 1 without consideration of the “green” effect (when a = 100)

Table D3. Solution of the Cournot oligopoly with three asymmetrical agents and medium cost coefficients and c₁ = 1, d₁ = 3, c₂ = 3, d₂ = 3, c₃ = 5, d₃ = 3 (when a = 100)

Table D4. Solution of the Cournot oligopoly with three asymmetrical agents and big cost coefficients and c₁ = 1, d₁ = 5, c₂ = 3, d₂ = 5, c₃ = 5, d₃ = 50

Table D5. Solution of the Cournot oligopoly with three asymmetrical agents (when a = 100) and n = 3, c₁ = 1, d₁ = 1, β₁ = 100, c₂ = 3, d₂ = 1, β₂ = 100, c₃ = 5, d₃ = 5, β₃ = 100

Table D6. Solution of the GNE Cournot oligopoly with three asymmetrical agents (when a = 100) and n = 3, c₁ = 1, d₁ = 1, c₂ = 3, d₂ = 1, c₃ = 5, d₃ = 5 and different coefficients β_i