Is Recycling a Threat or an Opportunity for the Extractor of an Exhaustible Resource?

Abstract

We examine the best extraction strategies for the provider of an exhaustible resource that can be recycled. In a two-period model of resource extraction, the extractor faces prospective entry by a recycler that incurs a fixed cost to produce a perfect substitute of the virgin resource. Its entry is an opportunity or a threat for the extractor, depending on whether it maximizes social welfare or its own revenue. Our results highlight how prospective recycling modifies the Hotelling rule. We characterize various entry possibilities. The benevolent extractor may accommodate or promote recycling, while the self-interested extractor may accommodate or deter recycling.

Appendix

1.1 Appendix 1: Proof of Lemma 1

In the first step, we show that the solution $\widehat{r }(q,\mu )$ of $W_{r}^{n}(q,r)+\mu \pi _{r}(q,r)=0$ is a local maximum.

Let the Lagrangian be $L(q,r)=W^{n}(q,r)+\mu \left[ P_{2}(s-q+r)r-c(r)-F \right]$. If $L_{r}(q,r)=0$ has an interior solution ${\widehat{r}}(q,\mu )$, then we have

$$\begin{aligned} {\widehat{r}}(q,\mu )=-\frac{\mu +\delta }{\mu }\frac{P_{2}(s-q+r)-c^{\prime }(r)}{P_{2}^{\prime }(s-q+r)}. \end{aligned}$$

(39)

We now check that the second-order condition for ${\widehat{r}}(q,\mu )$ to be a local maximum of L(q, r) is satisfied. By substituting ${\widehat{r}} (q,\mu )$ for r in

$$\begin{aligned} L_{rr}\left( q,r\right) =&\delta \left[ P_{2}^{\prime }(s-q+r)-c^{\prime \prime }(r)\right] +\mu \left[ 2P_{2}^{\prime }(s-q+r)\right. \nonumber \\&\left. +P_{2}^{\prime \prime }(s-q+r)r-c^{\prime \prime }(r)\right] , \end{aligned}$$

(40)

we obtain

$$\begin{aligned}&L_{rr}(q,{\widehat{r}}(q,\mu )) \nonumber \\&\quad =(\delta +\mu )\left[ P_{2}^{\prime }(s-q+{\widehat{r}}(q,\mu ))-c^{\prime \prime }({\widehat{r}}(q,\mu ))\right. \nonumber \\&\qquad \left. -\,\frac{P_{2}(s-q+{\widehat{r}}(q,\mu ))-c^{\prime }({\widehat{r}}(q,\mu ))}{P_{2}^{\prime }(s-q+{\widehat{r}}(q,\mu ))} P_{2}^{\prime \prime }(s-q+{\widehat{r}}(q,\mu ))\right] \nonumber \\&\qquad +\,\mu P_{2}^{\prime }(s-q+{\widehat{r}}(q,\mu )). \end{aligned}$$

(41)

Given that $P_{2}(.)$ is log-concave by assumption $\left( A2\right)$, we know that $P_{2}^{\prime \prime }(.)<\frac{P_{2}^{\prime }(.)^{2}}{P_{2}(.)}$. Thus, we have

$$\begin{aligned} L_{rr}(q,{\widehat{r}}(q,\mu ))<\mu P_{2}^{\prime }(.)-\left( \delta +\mu \right) c^{\prime \prime }(.)+\frac{\left( \delta +\mu \right) c^{\prime }(.) }{P_{2}(.)}P_{2}^{\prime }(.). \end{aligned}$$

(42)

Under assumption $\left( A1\right)$, the right-hand side of (42) is negative, and so ${\widehat{r}}(q,\mu )$ is a local maximum.

In the second step, we show that $0<{\widehat{r}}_{q}(q,\mu )<1$.

Differentiating $L_{r}(q,{\widehat{r}}(q,\mu ))=0$ yields the slope of ${\widehat{r}}_{q}(q,\mu )$ with respect to q:

$$\begin{aligned} {\widehat{r}}_{q}(q,\mu )=-\frac{L_{rq}}{L_{rr}},=\frac{\delta P_{2}^{\prime }(.)+\mu \left[ P_{2}^{\prime }(.)+P_{2}^{\prime \prime }(.)r\right] }{ \delta \left[ P_{2}^{\prime }(.)-c^{\prime \prime }(.)\right] +\mu \left[ 2P_{2}^{\prime }(.)+P_{2}^{\prime \prime }(.)r-c^{\prime \prime }(.)\right] } . \end{aligned}$$

(43)

where $-L_{rq}=\delta P_{2}^{\prime }(.)+\mu \left[ P_{2}^{\prime }(.)+P_{2}^{\prime \prime }(.)r\right]$. As $L_{rr}<0$, we have sign $\left( {\widehat{r}}_{q}\right) =$ sign $\left( L_{rq}\right)$. After substitution of ${\widehat{r}}(q,\mu )$ for r, we obtain

$$\begin{aligned} -L_{rq}=(\delta +\mu )\left[ P_{2}^{\prime }(.)-\left( P_{2}(.)-c^{\prime }(.)\right) P_{2}^{\prime \prime }(.)/P_{2}^{\prime }(.)\right] . \end{aligned}$$

(44)

Again using the log-concavity of $P_{2}(.)$, we have

$$\begin{aligned} -L_{rq}<(\delta +\mu )\left[ P_{2}^{\prime }(.)-\left( P_{2}(.)-c^{\prime }(.)\right) \frac{P_{2}^{\prime }(.)}{P_{2}(.)}\right] . \end{aligned}$$

(45)

The right-hand side of (45) reduces to $\left( \delta +\mu \right) c^{\prime }(.)\frac{P_{2}^{\prime }(.)}{P_{2}(.)}<0$. It follows that $-L_{rq}<0$ and, therefore, ${\widehat{r}}_{q}(q,\mu )>0$.

Further calculations yield

$$\begin{aligned} {\widehat{r}}_{q}(q,\mu )-1=\frac{\delta c^{\prime \prime }(.)+\mu \left[ -P_{2}^{\prime }(.)+c^{\prime \prime }(.)\right] }{L_{rr}}. \end{aligned}$$

(46)

The numerator of the fraction given on the right-hand side of (46) is strictly positive, whereas the denominator $L_{rr}$ is negative. We can conclude that ${\widehat{r}}_{q}(q,\mu )-1<0$.

In the third step, we show that ${\widehat{r}}(q,\mu )$ is an interior solution under assumption $\left( A3\right)$, which falls short of q.

Given that recycling cannot emerge ex nihilo, a minimum amount ${{{\underline{q}}}}>0$ of the virgin resource is needed to produce the first unit of recycled output; that is, ${\widehat{r}}({{{\underline{q}}}},\mu )=0$. As ${{{\underline{q}}}}>0$, we have ${\widehat{r}}(0,\mu )<{\widehat{r}}({{{\underline{q}}}},\mu )=0$ since ${\widehat{r}}\left( q,\mu \right)$ is upward sloping with respect to q. This inequality is equivalent to $L_{r}(0,0)=-\left( \mu +\delta \right) \left[ P_{2}(s)-c^{\prime }(0)\right] \le 0$, which holds under assumption $\left( A3\right)$. Moreover, $L_{r}(s,0)=\left( \mu +\delta \right) \left[ P_{2}(0)-c^{\prime }(0)\right]$ is positive under assumption $\left( A3\right)$. Thus, ${\widehat{r}}({{{\underline{q}}}},\mu )=0< {\widehat{r}}(s,\mu )$, and so ${{{\underline{q}}}}<s$. Finally, for all q higher than ${{{\underline{q}}}}$, we have ${\widehat{r}}(q,\mu )-q<{\widehat{r}}({{\underline{q }}},\mu )-{{{\underline{q}}}}$ since ${\widehat{r}}_{q}(q,\mu )-1<0$, and thus ${\widehat{r}}(q,\mu )<q$: the virgin resource is never fully recycled.

1.2 Appendix 2: Proof of Proposition 1

The social planner maximizes $W^{n}(q,r)$ under the break-even constraint (2).

Assuming an interior solution, the first-order conditions for this problem are

$$\begin{aligned} P_{1}(q)-\delta P_{2}(s-q+r)= & {} \mu P_{2}^{\prime }(s-q+r)r, \end{aligned}$$

(47)

$$\begin{aligned} P_{2}(s-q+r)-c^{\prime }(r)= & {} -\frac{\mu }{\left( \delta +\mu \right) } P_{2}^{\prime }(s-q+r)r, \end{aligned}$$

(48)

$$\begin{aligned} \mu \left[ P_{2}(s-q+r)r-c(r)-F\right]= & {} 0. \end{aligned}$$

(49)

If $\mu =0$, (48) immediately shows that the recycling output must be sold at marginal cost. If $\mu$ is positive, then the constraint (2) is binding and the social planner must use the proceeds from sales of the recycled product to finance the recycling costs. Using $\varepsilon$, the first-order condition (48) yields the standard Ramsey formula

$$\begin{aligned} \frac{P_{2}(s-q+r)-c^{\prime }(r)}{P_{2}(s-q+r)}=\frac{\mu }{\left( \delta +\mu \right) }\frac{1}{\varepsilon }. \end{aligned}$$

(50)

Note that (50) can be rewritten as follows:

$$\begin{aligned} p_{2}\left( 1-\frac{\mu }{\left( \delta +\mu \right) \varepsilon }\right) =c^{\prime }(r). \end{aligned}$$

(51)

Thus, we must have $\varepsilon >\frac{\mu }{\delta +\mu }$ for an interior solution. If this were not the case, the left-hand side of (51) would be negative and, hence, could not be equal to marginal cost.

Furthermore, setting $\varepsilon$ in (47) yields:

$$\begin{aligned} \frac{p_{2}}{p_{1}}=\frac{1}{\delta }+\frac{\mu }{\delta \varepsilon }\frac{ p_{2}}{p_{1}}. \end{aligned}$$

(52)

By substituting $\delta =\frac{1}{1+\rho }$ into (52), we obtain the variant of the Hotelling rule (8). Moreover, we can write (52) as

$$\begin{aligned} \frac{p_{2}}{p_{1}}\left( 1-\frac{\mu }{\delta \varepsilon }\right) =\frac{1 }{\delta }, \end{aligned}$$

(53)

which shows that $\varepsilon$ must be higher than $\frac{\mu }{\delta }$ to get an interior solution. If so, then the condition $\varepsilon >\frac{ \mu }{\delta +\mu }$ needed for the existence of an interior solution of Eq. (51) is also satisfied.

1.3 Appendix 3: Proof of Corollary 2

With linear demand and cost functions, assumption $\left( A4\right)$ can be written as¹⁶

$$\begin{aligned} s<\frac{a+c}{3}. \end{aligned}$$

(54)

By substituting $r^{*}(q)$ for r in (24), we obtain the reduced-form function

$$\begin{aligned} {\mathbf {V}}(q)=\left\{ \begin{array}{ll} aq-q^{2}/2+a(a-c+s-q)/2-(a-c+s-q)^{2}/8&{}\quad \text { if }{\widetilde{q}}\le q\le s,\\ \\ -q^{2}+sq+as-s^{2}/2&{}\quad \text { otherwise.} \end{array} \right. \end{aligned}$$

(55)

Assumption (29) boils down to $W\left( q_{0},0\right) <{\mathbf {V}}(s)$, which requires that s belongs to the interval $\left( a-c,\sqrt{\frac{\left( a-c\right) \left( 3a+c\right) }{2}} \right)$.

Substituting (28) into (25 ) yields

$$\begin{aligned} r\left( q_{a}\right) =\frac{4a-3c-2s}{5}. \end{aligned}$$

(56)

From (28), we see that $q_{a}\le s$ provided that $c\ge 3a-4s$. Otherwise, we have ${\widetilde{q}}\le s<q_{a}$ when $F\le {\overline{F}}$, and thus ${\mathbf {V}}(q)$ achieves a global maximum at s from (29). Moreover, $r\left( q_{a}\right) \ge 0$ requires that $c\le \frac{2}{3}\left( 2a-s\right)$, which holds under (29). It turns out that $F_{a}<{\overline{F}}$ when $c\ge 3a-4s$ holds. One can also check that, under assumption $\left( A4\right)$ and $c<a$, we have $q_{0}<q_{a}$.

There are three possibilities:

• (i) ${\widetilde{q}}<q_{0}<q_{a}<s$. The inequality ${\widetilde{q}} <q_{0}$ requires that $F<F_{d}$, where $F_{d}=\frac{(2\left( a-c\right) -s)^{2}}{16}$. Under (29), we have $F_{d}<F_{a}$ because this requires that $c<\frac{26a-13s}{22}$ and we have $\sqrt{\frac{\left( a-c\right) \left( 3a+c\right) }{2}}<\frac{26a-13s}{22}$. Then, ${\mathbf {V}}(q)$ achieves a unique maximum at $q_{a}$.

• (ii) $q_{0}\le {\widetilde{q}}\le q_{a}<s$. This requires that $F_{d}\le F\le F_{a}$. Under (29), $\mathbf {V }(q)$ achieves a global maximum at $q_{a}$ because $W\left( q_{0},0\right)< {\mathbf {V}}(s)<{\mathbf {V}}(q_{a})$.

• (iii) $q_{0}\le q_{a}<{\widetilde{q}}<s$. This requires that $F_{a}<F< {\overline{F}}$. Under (29), ${\mathbf {V}}(q)$ achieves a global maximum at ${\widetilde{q}}$ because $W\left( q_{0},0\right)<{\mathbf {V}}(s)<{\mathbf {V}}({\widetilde{q}})$.

1.4 Appendix 4: Proof of Corollary 3

By substituting $r^{*}(q)$ for r in (34) within the framework of linear demand and cost functions, we obtain the reduced-form function

$$\begin{aligned} {\mathbf {V}}(q)=\left\{ \begin{array}{ll} (a-q)q+(a-s+q+c)(s-q)/2&{}\quad \text { if }{\widetilde{q}}\le q\le s, \\ \\ (a-q)q+(a-s+q)(s-q)&{}\quad \text { otherwise.} \end{array} \right. \end{aligned}$$

(57)

Assumption (35) requires that s belongs to the interval $\left( a-c,\frac{2+\sqrt{3}}{2}\left( a-c\right) \right)$ and assumption $\left( A4\right)$ requires that $s<\mathbf {\ }\frac{a+c}{3}$. Under these assumptions, the fixed costs thresholds are ranked in the following order

$$\begin{aligned} F_{A}<F_{d}<{\overline{F}}. \end{aligned}$$

(58)

By solving the equation $R\left( q,0\right) =R\left( q_{a},r\left( q_{a}\right) \right)$ for q, we obtain an explicit formula for $\overline{ q}$:

$$\begin{aligned} {\overline{q}}=\frac{s}{2}-\frac{\sqrt{3}}{12}\sqrt{4s(2\left( a-c\right) -s)-\left( a-c\right) ^{2}}. \end{aligned}$$

(59)

One can easily check that $4s(2\left( a-c\right) -s)-\left( a-c\right) ^{2}>0$ for all $s\in \left( a\!-\!c,\frac{2+\sqrt{3}}{2}\left( a\!-\!c\right) \right)$. Thus, ${\overline{q}}$ does exist for all $s<\frac{2+\sqrt{3}}{2} \left( a-c\right)$, and we have ${\overline{q}}<q_{0}$. Further calculations show that ${\widetilde{q}}>{\overline{q}}$ holds only if $F\ge F_{A}$.

Depending on the level of the recycler’s fixed costs, which determines the location of ${\widetilde{q}}$, there are three possibilities:

• (i) ${\widetilde{q}}<{\overline{q}}<q_{0}$. In the case where $\widetilde{ q}<{\overline{q}}$, fixed costs are so small that $F<F_{A}$. Then, ${\mathbf {V}} (q)$ has two local maxima at ${\widetilde{q}}$ and $q_{a}$, with a global maximum at $q_{a}$ because ${\widetilde{q}}<{\overline{q}}$ implies that $R\left( {\widetilde{q}},0\right) <R\left( q_{a},r\left( q_{a}\right) \right) = {\mathbf {V}}(q_{a})$.

• (ii) ${\overline{q}}\le {\widetilde{q}}<q_{0}$. In this case, fixed costs are higher, so that $F_{A}\le F<F_{d}$. As ${\overline{q}}\le {\widetilde{q}}$, we have $R\left( q_{a},r\left( q_{a}\right) \right) \le$$R\left( {\widetilde{q}},0\right)$, which rules out $q_{a}$ as a possible equilibrium. Moreover, ${\widetilde{q}}<q_{0}$ implies that $F<F_{d}$, and one can check that $F_{d}<{\overline{F}}$ when $s<\frac{2+\sqrt{3}}{2}\left( a-c\right)$. Since ${\mathbf {V}}(q)$ is increasing on $\left[ 0,{\widetilde{q}} \right]$, ${\mathbf {V}}(q)$ is maximized at ${\widetilde{q}}$.

• (iii) $q_{0}\le {\widetilde{q}}\le s$. Now, fixed costs are even higher so that $F_{d}\le F\le {\overline{F}}$. As $q_{a}<q_{0}$, ${\mathbf {V}} (q)$ is decreasing on $({\widetilde{q}},s]$, therefore ${\mathbf {V}}(q)$ achieves a unique maximum at $q_{0}$.