Anti-Conservationist Effects of the Conservationist Oil Cartel

Abstract

A resource monopoly behaves as a “conservationist” in models of exhaustible resources. It extracts less today and keeps more resource under the ground than the competitive market does. In this chapter, we consider a model of economically recoverable resources and show that, in the presence of a high-cost competitive fringe, the cartel’s “conservationism” turns into a too early and too intensive resource extraction by the fringe. The “anti-conservationism” of the equilibrium extraction path selected by the fringe proves to be a robust property of the model. As a result, the cartel acts as if it was a strategic player planning its long-term moves to accelerate the depletion of competitors’ resources and to aggravate their competitive disadvantages.

Appendices

Appendices

1.1 A.1 Solution (11.19), (11.20)

The characteristic equation for system (11.17), (11.18) is

$$ \left|\begin{array}{cc}r-\lambda & -r{c}_1/2\\ {}-1/\beta & -\lambda \end{array}\right|={\lambda}^2- r\lambda -r{c}_1/2\beta =0 $$

(11.47)

and the negative root is

$$ \lambda =\left(r-\sqrt{r^2+2r{c}_1/\beta}\right)/2=-\delta . $$

Let us check that time paths (11.19), (11.20) satisfy differential Eqs. (11.17), (11.18). On the one hand, inserting (11.19), (11.20) into (11.17) implies

$$ \dot{P}=0.5r\left(2P-\alpha -{c}_0-{c}_1X\right)=0.5r\left(2\alpha -2\beta \delta {X}^{\ast }{e}^{-\delta t}-\alpha -{c}_0-{c}_1{X}^{\ast}\left(1-{e}^{-\delta t}\right)\right)=0.5r\left(\alpha -{c}_0-{c}_1{X}^{\ast }-2\beta \delta {X}^{\ast }{e}^{-\delta t}+{c}_1{X}^{\ast }{e}^{-\delta t}\right)=0.5r\left({c}_1-2\beta \delta \right){X}^{\ast }{e}^{-\delta t}. $$

On the other hand, differentiating (11.19) yields

$$ \dot{P}=\beta {\delta}^2{X}^{\ast }{e}^{-\delta t}. $$

These two equations are identical if

$$ \beta {\delta}^2=0.5r\left({c}_1-2\beta \delta \right). $$

This is the case for δ =  − λ satisfying the characteristic Eq. (11.47).

1.2 A.2 Solution (11.29), (11.30)

The characteristic equation for system (11.27), (11.28) is

$$ \left|\begin{array}{cc}r-\lambda & -r{c}_{1a}\\ {}-\varphi /\beta & -\lambda \end{array}\right|={\lambda}^2- r\lambda - r\varphi {c}_{1a}/\beta =0. $$

(11.48)

The negative root is $\lambda =\left(r-\sqrt{r^2+4 r\varphi {c}_{1a}/\beta}\right)/2=-\theta$. Solution (11.29), (11.30) satisfies (11.28):

$$ {\dot{Q}}_a=\theta {Q}_a^{\ast }{e}^{-\theta t}=\theta \varphi \left({Q}_a^{\ast }+{Q}_b^{\ast}\right){e}^{-\theta t}=\varphi \left(\alpha -p\right)/\beta, $$

since c_0a = c_0b and ${Q}_a^{\ast }=\varphi \left({Q}_a^{\ast }+{Q}_b^{\ast}\right)$ due to condition (11.26). Insert Eqs. (11.29), (11.30) into (11.27):

$$ {\displaystyle \begin{array}{l}\dot{p}=r\left(p-{c}_{0a}-{c}_{1a}{Q}_a\ \right)=r\left(\alpha -\beta \theta \left({Q}_a^{\ast }+{Q}_b^{\ast}\right){e}^{-\theta t}-{c}_{0a}-{c}_{1a}{Q}_a\right)=\\ {}r\left(\alpha -{c}_{0a}-{c}_{1a}{Q}_a-\beta \theta \left({Q}_a^{\ast }+{Q}_b^{\ast}\right){e}^{-\theta t}\right)=r\left({c}_{1a}\left({Q}_a^{\ast }-{Q}_a\right)-\beta \theta \left({Q}_a^{\ast }+{Q}_b^{\ast}\right){e}^{-\theta t}\right)=\\ {}r\left({c}_{1a}{Q}_a^{\ast }{e}^{-\theta t}-\beta \theta \left({Q}_a^{\ast }+{Q}_b^{\ast}\right){e}^{-\theta t}\right)=r\left({c}_{1a}-\left(\beta \theta /\varphi \right)\right){Q}_a^{\ast }{e}^{-\theta t}.\end{array}} $$

The time derivative of Eq. (11.29) is: $\dot{p}=\beta {\theta}^2\left({Q}_a^{\ast }+{Q}_b^{\ast}\right){e}^{-\theta t}=\left(\beta {\theta}^2/\varphi \right){Q}_a^{\ast }{e}^{-\theta t}$. This coincides with (11.27) for λ =  − θ satisfying (11.48).

1.3 A.3 Equation (11.36)

Equalization of time derivatives of price for Eqs. (11.34), (11.35) implies:

$$ \frac{r}{2}\left(P-\beta {x}_a-{c}_{0a}-{c}_{1a}{X}_a\right)=r\left(P-{c}_{0b}-{c}_{1b}{X}_b\ \right). $$

Rearrange the terms in this equation and take into account (11.33):

$$ P+\beta {x}_a+{c}_{0a}+{c}_{1a}{X}_a=2{c}_{0b}+2{c}_{1b}{X}_b $$

$$ \alpha -\beta \left({x}_a+{x}_b\right)+\beta {x}_a+{c}_{0a}+{c}_{1a}{X}_a=2{c}_{0b}+{c}_{1b}{X}_b $$

$$ \alpha -{c}_{0a}-\beta {x}_b+{c}_{1a}{X}_a=2{c}_{0b}-2{c}_{0a}+{c}_{1b}{X}_b $$

$$ \beta {x}_b={c}_{1a}\left({X}_a+{X}_a^{\ast}\right)-2\Delta {c}_0-{c}_{1b}{X}_b, $$

because $\alpha -{c}_{0a}={c}_{1a}{X}_a^{\ast }$ from Eq. (11.40).

1.4 A.4 Characteristic Equation (11.41)

The characteristic equation for system (11.37)–(11.39) is

$$ {\displaystyle \begin{array}{c}\left|\begin{array}{ccc}r-\lambda & 0& -{rc}_{1b}\\ {}-\frac{1}{\beta }& -\frac{c_{1a}}{\beta }-\lambda & \frac{2{c}_{1b}}{\beta}\\ {}0& \frac{c_{1a}}{\beta }& -\frac{2{c}_{1b}}{\beta }-\lambda \end{array}\right|=\left(r-\lambda \right)\left|\begin{array}{cc}-\frac{c_{1a}}{\beta }-\lambda & \frac{2{c}_{1b}}{\beta}\\ {}\frac{c_{1a}}{\beta }& -\frac{2{c}_{1b}}{\beta }-\lambda \end{array}\right|+\frac{1}{\beta}\cdot \frac{rc_{1a}{c}_{1b}}{\beta}\\ {}=\left(r-\lambda \right)\left({\lambda}^2+\frac{c_{1a}+2{c}_{1b}}{\beta}\lambda \right)+\frac{rc_{1a}{c}_{1b}}{\beta^2}=\\ {}-{\lambda}^3+\left(r-\frac{c_{1a}+2{c}_{1b}}{\beta}\right){\lambda}^2+r\frac{c_{1a}+2{c}_{1b}}{\beta}\lambda +\frac{rc_{1a}{c}_{1b}}{\beta^2}=0,\end{array}} $$

which implies:

$$ -\beta {\lambda}^3+\left( r\beta -\left({c}_{1a}+2{c}_{1b}\right)\right){\lambda}^2+r\left({c}_{1a}+2{c}_{1b}\right)\lambda +r{c}_{1a}{c}_{1b}/\beta =0. $$

1.5 A.5 Formulae (11.45), (11.46)

• (a) Consider Eq. (11.39) for t = 0:

  $$ \beta {\dot{X}}_b(0)={c}_{1a}{X}_a^{\ast }-2\Delta {c}_0 $$

since X_a(0) = X_b(0) = 0. From (11.43):

$$ {\dot{X}}_b(0)={X}_b^{\ast}\left({\sigma}_b{\theta}_1+\left(1-{\sigma}_b\right){\theta}_2\right) $$

hence

$$ {\sigma}_b{\theta}_1+\left(1-{\sigma}_b\right){\theta}_2=\frac{c_{1a}{X}_a^{\ast }-2\Delta {c}_0}{\beta {X}_b^{\ast }} $$

(11.49)

and

$$ {\sigma}_b=\frac{\omega_b-{\theta}_2}{\theta_1-{\theta}_2}=\frac{\theta_2-{\omega}_b}{\theta_2-{\theta}_1} $$

where

$$ {\omega}_b=\frac{c_{1a}{X}_a^{\ast }-2\Delta {c}_0}{\beta {X}_b^{\ast }} $$

which yields (11.46).

• (b) Consider Eq. (11.37) for t = 0:

$$ \dot{P}(0)=r\left(P(0)-{c}_{0b}\ \right) $$

(11.50)

From Eq. (11.44), the time derivative of the cartel price is:

$$ \dot{P}(t)=\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)\left[\xi {\theta}_1^2{e}^{-{\theta}_1t}+\left(1-\xi \right){\theta}_2^2{e}^{-{\theta}_2t}\right], $$

implying that

$$ \dot{P}(0)=\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)\left[\xi {\theta}_1^2+\left(1-\xi \right){\theta}_2^2\right] $$

From Eq. (11.44), we also have it that

$$ P(0)=\alpha -\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)\left[\xi {\theta}_1+\left(1-\xi \right){\theta}_2\right] $$

Combining the last two equations with (11.50) implies:

$$ \beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)\left[\xi {\theta}_1^2+\left(1-\xi \right){\theta}_2^2\right]=r\left\{\alpha -{c}_{0b}-\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)\left[\xi {\theta}_1+\left(1-\xi \right){\theta}_2\right]\right\} $$

Rearrange the terms:

$$ \xi {\theta}_1^2+\left(1-\xi \right){\theta}_2^2+r\left(\xi {\theta}_1+\left(1-\xi \right){\theta}_2\right)=\frac{r\left(\alpha -{c}_{0b}\right)}{\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)} $$

$$ \left({\theta}_1^2+r{\theta}_1-{\theta}_2^2-r{\theta}_2\right)\xi +{\theta}_2^2+r{\theta}_2=\frac{r{c}_{1b}{X}_b^{\ast }}{\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)}=\frac{r{c}_{1b}\left(1-\psi \right)}{\beta }, $$

since $\alpha -{c}_{0b}={c}_{1b}{X}_b^{\ast }$ from (11.40). We have:

$$ \xi =\frac{\omega_a-\left({\theta}_2^2+r{\theta}_2\right)}{\theta_1^2+r{\theta}_1-\left({\theta}_2^2+r{\theta}_2\right)} $$

where ω_a = rc_1b(1 − ψ)/β, implying that

$$ {\sigma}_a={\psi}^{-1}\left(\frac{\theta_2^2+r{\theta}_2-{\omega}_a}{\theta_2^2+r{\theta}_2-\left({\theta}_1^2+r{\theta}_1\right)}-\left(1-\psi \right){\sigma}_b\right) $$

since ξ = ψσ_a + (1 − ψ)σ_b.

• (c) Let us check that Eq. (11.38) is fulfilled as identity for t = 0:

$$ \beta {\dot{X}}_a(0)=\alpha -P(0)-{c}_{1a}{X}_a^{\ast }+2\Delta {c}_0 $$

(11.51)

From Eqs. (11.43), (11.44), we have it that

$$ \beta {\dot{X}}_a(0)=\beta {X}_a^{\ast}\left({\sigma}_a{\theta}_1+\left(1-{\sigma}_a\right){\theta}_2\right), $$

$$ P(0)=\alpha -\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)\left[\xi {\theta}_1+\left(1-\xi \right){\theta}_2\right] $$

Insert these two equations into (11.51):

$$ \beta {X}_a^{\ast}\left({\sigma}_a{\theta}_1+\left(1-{\sigma}_a\right){\theta}_2\right)=\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)\left[\xi {\theta}_1+\left(1-\xi \right){\theta}_2\right]-{c}_{1a}{X}_a^{\ast }+2\Delta {c}_0 $$

Divide both sides by $\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)$ and rearrange:

$$ {\displaystyle \begin{array}{l}\psi \left({\sigma}_a{\theta}_1+\left(1-{\sigma}_a\right){\theta}_2\right)={\xi \theta}_1+\left(1-\xi \right){\theta}_2-\frac{c_{1a}{X}_a^{\ast }-2\varDelta {c}_0}{\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)}=\psi \left({\sigma}_a{\theta}_1+\left(1-{\sigma}_a\right){\theta}_2\right)\\ {}+\left(1-\psi \right)\left({\sigma}_b{\theta}_1+\left(1-{\sigma}_b\right){\theta}_2\right)-\frac{c_{1a}{X}_a^{\ast }-2\varDelta {c}_0}{\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)}=\psi \left({\sigma}_a{\theta}_1+\left(1-{\sigma}_a\right){\theta}_2\right)\end{array}} $$

because, due to (11.49):

$$ \left(1-\psi \right)\left({\sigma}_b{\theta}_1+\left(1-{\sigma}_b\right){\theta}_2\right)=\left(1-\psi \right)\frac{c_{1a}{X}_a^{\ast }-2\Delta {c}_0}{\beta {X}_b^{\ast }}=\frac{c_{1a}{X}_a^{\ast }-2\Delta {c}_0}{\beta \left({X}_a^{\ast }+{X}_b^{\ast}\right)} $$

Consequently, (11.51) holds as identity.