Optimal Waste Disposal Fees When Product Durability is Endogenous: Accounting for Planned Obsolescence

Abstract

Considering the Coase conjecture (J Law Econ 15:143–149, 1972), we explore the optimal disposal fee of solid waste through the choice of product durability by a producer. We find that introducing a disposal fee curbs planned obsolescence and increases product durability. This increased durability decreases the social cost of waste and increases the service flow from the durable product. Therefore, the optimal disposal fee is higher than the Pigouvian level in the closed-loop.

Appendices

Appendix A: The derivations of (10), (12), (13), and (16)

Here, we solve the multi-stage game. In stage 6, the household discards $G_2$. At this time, the SWM sector collects and disposes them. In stage 5, the household maximizes $u_2$ subject to (8) by x and $q_2$. The Lagrangian for this problem is

$$\begin{aligned} {\mathscr {L}}_2 = x_2 + (Dq_1+q_2) \left( a - \frac{Dq_1+q_2}{2} \right) + \lambda \left\{ RS - x_2- p_2q_2-(Dq_1 +q_2)f \right\} , \end{aligned}$$

where $\lambda$ is the multiplier. The first-order conditions for this problem are given by

$$\begin{aligned} \frac{\partial {\mathscr {L}}_2}{\partial x_2}= & {} 1 - \lambda = 0, \end{aligned}$$

(A.1)

$$\begin{aligned} \frac{\partial {\mathscr {L}}_2}{\partial q_2}= & {} a-Dq_1-q_2-\lambda p_2 -\lambda f = 0 \;\;\;\; \text {and} \end{aligned}$$

(A.2)

$$\begin{aligned} \frac{\partial {\mathscr {L}}_2}{\partial \lambda }= & {} RS -x_2-p_2q_2 - (Dq_1 +q_2)f =0. \end{aligned}$$

(A.3)

From (A.1) and (A.2), we have the second-period inverse demand function (10). In stage 4, the producer chooses $q_2$, which maximizes $\pi _2=p_2q_2 -C_2$, given the first-period decisions $q_1$ and D. The first-order condition for the producer’s profit maximization in period 2 is given by (11). Then, we have (12).

Next, we consider the first-period decisions. In stage 3, the household discards $G_1$. At this time, the SWM sector collects and disposes them. In stage 2, the household maximizes (6) subject to (9) by choosing $x_1$ and $q_1$. The second-period price and quantity are given for the representative household. The Lagrangian for this problem is

$$\begin{aligned} {\mathscr {L}}_1= & {} x_1 + q_1 \left( a - \frac{q_1}{2} \right) \\&+\, \delta \left\{ x_2 + (Dq_1+q_2) \left( a - \frac{Dq_1+q_2}{2} \right) \right\} \\&+ \,\mu \left\{ I - x_1 - p_1q_1- (1-D)q_1f -\frac{1}{R}x_2-\frac{1}{R}p_2q_2 -\frac{1}{R}(Dq_1 +q_2 )f \right\} , \end{aligned}$$

where $\mu$ is the multiplier. The first-order conditions for this problem are given by

$$\begin{aligned} \frac{\partial {\mathscr {L}}_1}{\partial x_1}= & {} 1 - \mu = 0, \end{aligned}$$

(A.4)

$$\begin{aligned} \frac{\partial {\mathscr {L}}_1}{\partial q_1}= & {} a - q_1+ \delta D(a-Dq_1-q_2) -\mu p_1-\mu (1-D)f \nonumber \\&-\, \mu \delta Df = 0 \end{aligned}$$

(A.5)

and

$$\begin{aligned} \frac{\partial {\mathscr {L}}_1}{\partial \mu }= & {} I - x_1 - p_1q_1 -(1-D)q_1 f- \frac{1}{R}x_2 \nonumber \\&-\, \frac{1}{R}p_2q_2 -\frac{1}{R} (Dq_1+q_2) f =0. \end{aligned}$$

(A.6)

From (A.1), (A.2), (A.4), and (A.5) and considering $R^{-1}=\delta$, we have the first-period inverse demand function (13).

In stage 1, the profit given (10), (13), and $q_2(q_1,D)$ defined in (12) is

$$\begin{aligned} \Pi (q_1,D)= & {} \{ p_1(q_1, D, q_2(q_1,D)) - (1+\delta D)c \} q_1 \nonumber \\&+\,\delta \{ p_2(q_1, D, q_2(q_1,D)) -c \} q_2(q_1,D). \end{aligned}$$

(A.7)

The producer chooses not only $q_1$ but also D, which maximizes (A.7), given (10), (12), and (13). The first-order conditions for the producer’s overall profit maximization are given by (14) and (15). From (12), (14), and (15), we have the equilibrium (16).

Appendix B: Optimal disposal fee in the closed-loop equilibrium without planned obsolescence

First, we examine the market equilibrium where the monopolist does not practice planned obsolescence. If ${\tilde{f}} \equiv \frac{\delta }{4+\delta } (a-c) \le f$, durability reaches the upper bound. By substituting $D=1$ into (12) and (14), we have

$$\begin{aligned} D^{CL2}=1, \;\; q_1^{CL2} = \frac{2}{4+\delta }(a-c) \;\; \text {and} \;\; q_2^{CL2} = \frac{2+\delta }{2(4+\delta )}(a-c)-\frac{1}{2}f, \end{aligned}$$

(A.8)

where the superscript CL2 denotes the closed-loop equilibrium without planned obsolescence. We restrict our analysis to the interior solution of $q_2$. Otherwise, the case is $q_2=0$, that is, the producer can commit to the future production schedule. In other words, the case contradicts the assumption of the closed-loop strategy. The condition for $q_2>0$ is given as $f < {\bar{f}} \equiv \frac{2+\delta }{4+\delta }(a-c)$.

Next, we derive the optimal disposal fee in the closed-loop equilibrium when ${\tilde{f}} \le f < {\bar{f}}$. Without planned obsolescence, social welfare is

$$\begin{aligned} W^{CL2}= & {} \left\{ a - \frac{q_1^{CL2}}{2} - c \right\} q_1^{CL2} + \delta \left( a- \frac{q_1^{CL2}+q_2^{CL2} }{2}-c \right) (q_1^{CL2}+q_2^{CL2}) \nonumber \\&-\, \delta (\tau +k)(q_1^{CL2}+q_2^{CL2}). \end{aligned}$$

(A.9)

The first-order condition of welfare maximization yields

$$\begin{aligned} \frac{dW^{CL2}}{df}= & {} \delta \{ a-(q_1^{CL2}+q_2^{CL2}) -c -(\tau +k) \} \frac{dq_2^{CL2}}{df} \nonumber \\= & {} \delta \left[ a- \left\{ \frac{(6+\delta )(a-c)}{2(4+\delta )}-\frac{1}{2}f \right\} -c-(\tau + k) \right] \frac{dq_2^{CL2}}{df} =0. \end{aligned}$$

(A.10)

From (A.10), the optimal disposal fee without planned obsolescence is

$$\begin{aligned} f^{CL2} = 2(\tau + k) -\frac{2+\delta }{4+\delta } (a-c). \end{aligned}$$

(A.11)

The condition that makes $f^{CL2}$ an interior solution is given by ${\tilde{f}} \le f^{CL2} < {\bar{f}}$; this is rewritten as

$$\begin{aligned} \frac{1+\delta }{4+\delta } \equiv {\widehat{K}} \le \tau + k < \frac{2+\delta }{4+\delta }(a-c) \equiv {\bar{K}}. \end{aligned}$$

(A.12)

Under condition (A.12), the optimal disposal fee is lower than the Pigouvian level since

$$\begin{aligned} f^{CL2} - (\tau +k) = (\tau +k) -\frac{2+\delta }{4+\delta }(a-c) <0. \end{aligned}$$

(A.13)

Finally, we consider the globally optimal disposal fee that includes cases both with and without planned obsolescence. From (22) and (A.12), we have the following relationship:

$$\begin{aligned} {\widetilde{K}}< {\widehat{K}} < {\bar{K}}. \end{aligned}$$

(A.14)

Moreover, we know that $f^{CL}$ is optimal if $\tau + k < {\widetilde{K}}$ and $f^{CL2}$ is optimal if ${\widehat{K}} \le \tau +k < {\bar{K}}$. Thus, our remaining problem is to investigate the optimal disposal fee when

$$\begin{aligned} {\widetilde{K}} \le \tau +k < {\widehat{K}}. \end{aligned}$$

(A.15)

(A.15) is rewritten as $f^{CL2}<{\tilde{f}} \le f^{CL}$. This condition implies that neither $f^{CL}$ nor $f^{CL2}$ is a solution. From (20) and (A.15), we have

$$\begin{aligned} \left. \frac{dW^{CL}}{df} \right| _{f={\tilde{f}}} = \frac{1}{2\delta (4+\delta )} \{ (4+\delta )(4-\delta +\delta ^2)(\tau +k)-\delta ^3(a-c) \} <0. \end{aligned}$$

(A.16)

From (A.10) and (A.15), we have

$$\begin{aligned} \left. \frac{dW^{CL2}}{df} \right| _{f={\tilde{f}}} = -\frac{\delta }{2(4+\delta )} \{ (1+\delta )(a-c)-(4+\delta )(\tau +k) \} >0. \end{aligned}$$

(A.17)

In addition, from (16), (19), and (A.9), the social welfare function is continuous at $f={\tilde{f}}$,

$$\begin{aligned}&\left. W^{CL} \right| _{f={\tilde{f}}} = \left. W^{CL2} \right| _{f={\tilde{f}}} \nonumber \\&\quad = \frac{a-c}{2(4+\delta )^2} \{ (12+19\delta +6\delta ^2)(a-c)-6\delta (4+\delta )(\tau +k) \}. \end{aligned}$$

(A.18)

Because the social welfare functions are single-peaked, (A.16), (A.17), and (A.18) imply that the social welfare function is continuous, monotonically increasing when $f<{\tilde{f}}$ and monotonically decreasing when ${\tilde{f}} <f$. Therefore, we conclude that ${\tilde{f}}$ is optimal if (A.15) holds.

Summarizing above results, in our closed-loop model the globally optimal disposal fee $f^*$ is the following:

$$\begin{aligned} f^* = \left\{ \begin{array}{ll} f^{CL} &{} if \;\; \tau +k< {\widetilde{K}} \\ {\tilde{f}} &{} if \;\; {\tilde{K}} \le \tau + k< {\widehat{K}} \\ f^{CL2} &{} if \;\; {\widehat{K}} \le \tau + k < {\bar{K}}. \\ \end{array} \right. \end{aligned}$$

(A.19)

Appendix C: The derivations of (24) and (25)

We consider the open-loop maximization problem of the monopolistic producer. The first-order derivatives of (3) yield

$$\begin{aligned} \frac{\partial \Pi }{\partial q_1}= & {} p_1 -(1+\delta D)c +\frac{\partial p_1}{\partial q_1}q_1 + \delta \frac{\partial p_2}{\partial q_1} q_2 \nonumber \\= & {} a -2q_1 +\delta D(a-2Dq_1 -q_2)-(1-D)f-\delta Df \nonumber \\&-\,(1+\delta D)c- \delta Dq_2 \end{aligned}$$

(A.20)

$$\begin{aligned} \frac{\partial \Pi }{\partial q_2}= & {} \frac{\partial p_1}{\partial q_2} q_1 + \delta \left( p_2 -c+ \frac{\partial p_2}{\partial q_2} q_2 \right) \nonumber \\= & {} -\,\delta Dq_1 + \delta (a-Dq_1 -2q_2 -f)- \delta c \end{aligned}$$

(A.21)

$$\begin{aligned} \frac{\partial \Pi }{\partial D}= & {} \left( \frac{\partial p_1}{\partial D} - \delta c \right) q_1 + \delta \frac{\partial p_2}{\partial D} q_2 \nonumber \\= & {} \delta (a-2Dq_1-q_2)q_1 + (1-\delta )f q_1 -\delta cq_1 -\delta q_1 q_2. \end{aligned}$$

(A.22)

In order that the producer provides a positive amount of durable products in the first period, $\frac{\partial \Pi }{\partial q_1}=0$ must be valid. If $f=0$, however, from (A.21) and (A.22), we have $\frac{\partial \Pi }{\partial q_2}=\frac{\partial \Pi }{\partial D}$. Thus, we classify this into two cases: $f>0$ and $f=0$.

Case I: :

$f > 0$. This case is classified into three sub-cases, namely $\frac{\partial \Pi }{\partial D} = 0$, $\frac{\partial \Pi }{\partial D} < 0$, and $\frac{\partial \Pi }{\partial D} > 0$, which respectively correspond to $D^{OL} \in (0,1)$, $D^{OL}=0$, and $D^{OL}=1$.

Case I-1: :

$f > 0$ and $\frac{\partial \Pi }{\partial D} = 0$. From (A.21), (A.22), and $\frac{\partial \Pi }{\partial D} = 0$, we have $\frac{\partial \Pi }{\partial q_2} <0$. This condition leads to a corner solution $q_2 =0$. Thus, the conditions $\frac{\partial \Pi }{\partial q_1}=0$, $\frac{\partial \Pi }{\partial D} = 0$, and $q_2 =0$ lead to $q_1 = \frac{1}{2} (a-c-f)$ and $D=1+ \frac{f}{\delta (a-c-f)} >1$. However, $D>1$ contradicts the case requirement, $D^{OL} \in (0,1)$; this implies that case I-1 never holds in equilibrium.

Case I-2: :

$f > 0$ and $\frac{\partial \Pi }{\partial D} < 0$. From (A.21) and (A.22), if $\frac{\partial \Pi }{\partial D} < 0$, then $\frac{\partial \Pi }{\partial q_2} < 0$. The conditions $\frac{\partial \Pi }{\partial D} < 0$ and $\frac{\partial \Pi }{\partial q_2} < 0$ lead to the corner solutions $D^{OL} =0$ and $q_2^{OL}=0$. Because $Q_2=Dq_1 + q_2$, we have $Q_2=0$. This implies that the market vanishes in the second period despite the existence of market demand (10). Therefore, this case does not hold in equilibrium.

Case I-3: :

$f > 0$ and $\frac{\partial \Pi }{\partial D} > 0$. This case implies that $D^{OL}=1$. From (A.20) and $\frac{\partial \Pi }{\partial q_1}=0$, we have

$$\begin{aligned} a-2q_1 = \frac{2\delta }{1+\delta }q_2 + c + \frac{\delta }{1+\delta } f. \end{aligned}$$

(A.23)

By substituting (A.23) into (A.21), we have $\frac{\partial \Pi }{\partial q_2} = - \frac{2 \delta }{1+\delta } q_2 - \frac{\delta }{1+\delta } f <0$. This leads to the corner solution $q_2^{OL}=0$. By substituting $q_2^{OL}=0$ into (A.23), we have (24). From the above three sub-cases, we find that (24) is the only equilibrium if $f>0$.

Case II: :

$f = 0$. As mentioned above, $\frac{\partial \Pi }{\partial q_2} = \frac{\partial \Pi }{\partial D}$ is valid. From $\frac{\partial \Pi }{\partial q_1}=0$, $\frac{\partial \Pi }{\partial q_2} = \frac{\partial \Pi }{\partial D}=0$, (A.20), and (A.21), we have (25).