Prices and Rents of Economically Recoverable Resources

Abstract

Models in this chapter are based on the premise of limited resource availability with no explicit resource constraints. A potential resource in the ground is supposed to be inexhaustible but depleting in the sense that the marginal extraction cost increases with cumulative extraction. An economically recoverable resource is determined in the base model for long-term equilibrium as the ultimate cumulative extraction. In the reservoir model, which is relevant to the oil industry, producing reserves are developed through drilling new wells and play a dual role: as reservoirs containing oil inventory and as a factor of production. Finally, we analyse a model with extraction and investment in addition of producing reserves.

Appendices

Appendices

1.1 A.1 Decreasing Marginal Resource Rent w(t) for σ < 1

This statement is a special case of Farzin’s (1992, p. 817) more general result. Integrating (3.11) by parts yields, after some manipulation,

$$ w(t)=\underset{t}{\overset{\infty }{\int }}{e}^{-r\left(\tau -t\right)}{c}^{\prime}\left(Q\left(\tau \right)\right)q\left(\tau \right) d\tau ={r}^{-1}{c}^{\prime}\left(Q(t)\right)q(t)+{r}^{-1}\underset{t}{\overset{\infty }{\int }}{e}^{-r\left(\tau -t\right)}\frac{d\left[{c}^{\prime}\left(Q\left(\tau \right)\right)q\left(\tau \right)\right]}{d\tau} d\tau . $$

Substituting rw(t) into (3.5) implies

$$ \dot{w}(t)=\underset{t}{\overset{\infty }{\int }}{e}^{-r\left(\tau -t\right)}\frac{d\left[{c}^{\prime}\left(Q\left(\tau \right)\right)q\left(\tau \right)\right]}{d\tau} d\tau . $$

We have it that $d\left({c}^{\prime }(Q)q\right)/ d\tau ={c}^{\prime \prime }(Q)\dot{Q}q+{c}^{\prime }(Q)\dot{q}={c}^{\prime \prime }(Q){q}^2+{c}^{\prime }(Q)\dot{q}\le 0$ because c^′′(Q) ≤ 0 for the concave function and $\dot{q}<0$ for the equilibrium extraction path (we have it for system (3.7)–(3.8) that $\dot{p}(t)>0$ for all t ≥ 0). Consequently, $\dot{w}(t)\le 0$.

1.2 A.2 The Solution (3.16), (3.17)

On the one hand, inserting (3.16), (3.17) into (3.7^′) we have:

$$ \dot{p}=r\left(p-{c}_0-{c}_1Q\right)=r\left(\alpha -\beta \theta {Q}^{\ast }{e}^{-\theta t}-{c}_0-{c}_1{Q}^{\ast }+{c}_1{Q}^{\ast }{e}^{-\theta t}\right)=r\left({c}_1{Q}^{\ast }{e}^{-\theta t}-\beta \theta {Q}^{\ast }{e}^{-\theta t}\right)=r\left({c}_1-\beta \theta \right){Q}^{\ast }{e}^{-\theta t}, $$

because Q^∗ = (α − c₀)/c₁. On the other hand, differentiating (3.16) with respect to time yields

$$ \dot{p}=\beta {\theta}^2{Q}^{\ast }{e}^{-\theta t}. $$

These two equations on price growth coincide if

$$ \beta {\theta}^2=r\left({c}_1-\beta \theta \right), $$

which is fulfilled, because θ =  − λ satisfies the characteristic eq. (3.15).

From (3.18), condition θβ/c₁ < 1 is equivalent to:

$$ \sqrt{r^2+4r{c}_1/\beta }<\frac{2{c}_1}{\beta }+r. $$

Squaring both sides implies

$$ {r}^2+4r{c}_1/\beta <{r}^2+4r{c}_1/\beta +4{\left({c}_1/\beta \right)}^2. $$

1.3 A.3 System (3.40)–(3.41)

In the linear case, eq. (3.38) transforms to

$$ r\rho \dot{X}=-\gamma \beta \dot{q}-\gamma r\left(\alpha -\beta q-\xi -\rho X\right). $$

Inserting here q = γS, ξ = rρX/γ from (3.37) and $\dot{S}=\dot{X}-q=\dot{X}-\gamma S$ and rearranging terms implies

$$ \left( r\rho +{\gamma}^2\beta \right)\dot{X}={\gamma}^2\beta \left(\gamma +r\right)S+ r\rho \left(\gamma +r\right)X-\alpha \gamma r, $$

which yields (3.41).

The characteristic equation for system (3.40), (3.41) is

$$ \left|\begin{array}{cc}A-\gamma -\lambda & B\\ {}A& B-\lambda \end{array}\right|={\lambda}^2-\left(A+B-\gamma \right)\lambda -\gamma B={\lambda}^2- r\lambda -\gamma B=0, $$

since A + B = r + γ. The negative root is $\lambda =\frac{1}{2}\left(r-\sqrt{r^2+4\gamma B}\right)=-\theta$.