How supply chain coordination affects the environment: a carbon footprint perspective

Abstract

Environmental responsibility has become an important part of doing business. Government regulations and customers’ increased awareness of environmental issues are pushing supply chain entities to reduce the negative influence of their operations on the environment. In today’s world, companies must assume joint responsibility with their suppliers for the environmental impact of their actions. In this paper, we study coordination between a buyer and a vendor under the existence of two emission regulation policies: cap-and-trade and tax. We investigate the impact of decentralized and centralized replenishment decisions on total carbon emissions. The buyer in this system faces a deterministic and constant demand rate for a single product in the infinite horizon. The vendor produces at a finite rate and makes deliveries to the buyer on a lot-for-lot basis. Both the buyer and the vendor aim to minimize their average annual costs resulting from replenishment set-ups and inventory holding. We provide decentralized and centralized models for the buyer and the vendor to determine their ordering/production lot sizes under each policy. We compare the solutions due to independent and joint decision-making both analytically and numerically. Finally, we arrive at coordination mechanisms for this system to increase its profitability. However, we show that even though such coordination mechanisms help the buyer and the vendor decrease their costs without violating emission regulations, the cost minimizing solution may result in increased carbon emission under certain circumstances.

Appendix

1.1 Proof of Lemma 1

For any order quantity $Q$, the amount of traded carbon credits by the buyer is $X_b(Q)=C_b-\frac{f_bD}{Q}-\frac{g_bQ}{2}-e_bD$. Observe that $\hat{Q}_d$ minimizes $\frac{f_bD}{Q}+\frac{g_bQ}{2}$ with a minimum function value $\sqrt{2f_bg_bD}$. That is,

$$\begin{aligned} \frac{f_bD}{Q}+\frac{g_bQ}{2}\ge \sqrt{2f_bg_bD} \end{aligned}$$

for all $Q\ge 0$. This implies

$$\begin{aligned} X_b(Q)\le C_b-e_bD-\sqrt{2f_bg_bD}. \end{aligned}$$

Given that $(C_b-e_bD)\le \sqrt{2g_bf_bD}$, it turns out that $X_b(Q)\le 0$ for all $Q\ge 0$. That is, the retailer does not sell carbon credits at any order quantity. In this case, Expression (1) implies that the retailer’s inventory replenishment problem reduces to minimizing $BC_1(Q,X_b(Q))$ over $Q\ge 0$. As given by Expression (15), $Q^*_{d1}$ is the optimal solution of this problem. $\square$

1.2 Development of the other results for the Proof of Theorem 1

Lemma 2

The buyer sells carbon credits (i.e., $X_b(Q)>0$) only when $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $Q_1<Q<Q_2$.

Proof

From Lemma 1, we know that if $(C_b-e_bD)\le \sqrt{2g_bf_bD}$, then the buyer does not sell carbon credits. Therefore, selling carbon credits is possible only when $(C_b-e_bD)> \sqrt{2g_bf_bD}$. Furthermore, under this condition, $X_b(Q)>0$ should be satisfied. $X_b(Q)=C_b-\frac{f_bD}{Q}-\frac{g_bQ}{2}-e_bD>0$ holds for order quantities $Q$ such that $Q_1<Q<Q_2$. Note that, as $(C_b-e_bD)> \sqrt{2g_bf_bD}$, both $Q_1$ and $Q_2$ are defined and $Q_1<Q_2$. $\square$

Lemma 2 implies that in addition to the case of $(C_b-e_bD)\le \sqrt{2g_bf_bD}$ suggested by Lemma 1, there are two cases in which the retailer does not sell carbon credits: if $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $Q\le Q_1$, or if $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $Q\ge Q_2$.

Lemma 3

Depending on how $f_bh_b$ compares to $K_bg_b$, the following ordinal relations exist between $Q^*_{d1}$ and $Q^*_{d2}$:

• If $f_bh_b>K_bg_b$, then $Q^*_{d1}>Q^*_{d2}$.

• If $f_bh_b=K_bg_b$, then $Q^*_{d1}=Q^*_{d2}$.

• If $f_bh_b<K_bg_b$, then $Q^*_{d1}<Q^*_{d2}$.

Proof

We will prove the first part of the lemma. The proofs of the remaining two parts are similar.

Since $p_c^b\ge p_c^s$, $f_bh_b>K_bg_b$ implies that $(p_c^b-p_c^s)f_bh_b>(p_c^b-p_c^s)K_bg_b$. Adding $K_bh_b+p_c^bp_c^sf_bg_b$ to both sides of this inequality, and after some rearrangement of terms, we have

$$\begin{aligned} (K_b+p_c^bf_b)(h_b+p_c^sg_b)>(K_b+p_c^sf_b)(h_b+p_c^bg_b). \end{aligned}$$

The above expression can be rewritten as

$$\begin{aligned} \frac{(K_b+p_c^bf_b)}{(h_b+p_c^bg_b)}>\frac{(K_b+p_c^sf_b)}{(h_b+p_c^sg_b)}, \end{aligned}$$

which further implies

$$\begin{aligned} \sqrt{\frac{2(K_b+p_c^bf_b)D}{(h_b+p_c^bg_b)}}>\sqrt{\frac{2(K_b+p_c^sf_b)D}{(h_b+p_c^sg_b)}}. \end{aligned}$$

Observe that the left-hand side of the above inequality is $Q^*_{d1}$ and the right-hand side is $Q^*_{d2}$, and therefore, $Q^*_{d1}>Q^*_{d2}$. $\square$

In the next lemma, we present further properties of the retailer’s problem in the case of $(C_b-e_bD)> \sqrt{2g_bf_bD}$.

Lemma 4

When $(C_b-e_bD)> \sqrt{2g_bf_bD}$, the following cases cannot be observed.

• $Q_1<Q_2\le Q^*_{d2}\le Q^*_{d1}$.

• $Q^*_{d1}\le Q^*_{d2} \le Q_1<Q_2$.

Proof

Let us start with the first part of the lemma. Using Expression (17) and Expression (19), $Q_2\le Q^*_{d2}$ implies that

$$\begin{aligned} \frac{C_b-e_bD+\sqrt{(C_b-e_bD)^2-2g_bf_bD}}{g_b}\le \sqrt{\frac{2(K_b+p^s_cf_b)D}{h_b+p^s_cg_b}}. \end{aligned}$$

Since $(C_b-e_bD)>\sqrt{2g_bf_bD}$, the left-hand side is positive. Therefore, taking the square of both sides leads to

$$\begin{aligned} \frac{(C_b-e_bD)^2+(C_b-e_bD)\sqrt{(C_b-e_bD)^2-2g_bf_bD}-g_bf_bD}{g_b}\le \frac{(K_bg_b+p_c^sf_bg_b)D}{h_b+p_c^sg_b}. \end{aligned}$$

Due to Lemma 3, we know that having $Q^*_{d2}\le Q^*_{d1}$ is possible only when $f_bh_b\ge K_bg_b$, which implies

$$\begin{aligned} \frac{(f_bh_b+p_c^sf_bg_b)D}{h_b+p_c^sg_b}\ge \frac{(K_bg_b+p_c^sf_bg_b)D}{h_b+p_c^sg_b}. \end{aligned}$$

Combining the last two inequalities, we obtain

$$\begin{aligned}&\frac{(C_b-e_bD)^2+(C_b-e_bD)\sqrt{(C_b-e_bD)^2-2g_bf_bD}-g_bf_bD}{g_b}\\&\quad \le \frac{(f_bh_b+p_c^sf_bg_b)D}{h_b+p_c^sg_b}=f_bD. \end{aligned}$$

Multiplying both sides of the above expression by $g_b$ and after some rearrangement of terms, it follows that

$$\begin{aligned} (C_b-e_bD)^2-2g_bf_bD\le -(C_b-e_bD)\sqrt{(C_b-e_bD)^2-2g_bf_bD}. \end{aligned}$$

Recall that, $Q_1$ and $Q_2$ were formed by considering the positive square root of the discriminant in $X_b(0)$, and $Q_2$ was defined as the larger root. Since $(C_b-e_bD)>\sqrt{2g_bf_bD}$, the above inequality cannot hold for the positive square root of $(C_b-e_bD)^2-2g_bf_bD$. Therefore, we cannot have $Q_1<Q_2\le Q^*_{d2}\le Q^*_{d1}$.

Now, let us continue with the second part of the lemma. Using Expression (17) and Expression (18), $Q^*_{d2}\le Q_1$ implies that

$$\begin{aligned} \sqrt{\frac{2(K_b+p^s_cf_b)D}{h_b+p^s_cg_b}} \le \frac{C_b-e_bD-\sqrt{(C_b-e_bD)^2-2g_bf_bD}}{g_b}. \end{aligned}$$

Taking the square of both sides of this inequality leads to

$$\begin{aligned} \frac{(K_b+p^s_cf_b)D}{h_b+p^s_cg_b} \le \frac{(C_b-e_bD)^2-(C_b-e_bD)\sqrt{(C_b-e_bD)^2-2g_bf_bD}-g_bf_bD}{(g_b)^2}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{(K_bg_b+p^s_cf_bg_b)D}{h_b+p^s_cg_b} \le \frac{(C_b-e_bD)^2-(C_b-e_bD)\sqrt{(C_b-e_bD)^2-2g_bf_bD}-g_bf_bD}{g_b}. \end{aligned}$$

Based on Lemma 3, having $Q^*_{d2}\ge Q^*_{d1}$ suggests that $f_bh_b\le K_bg_b$, which implies

$$\begin{aligned} \frac{(f_bh_b+p^s_cf_bg_b)D}{h_b+p^s_cg_b} \le \frac{(C_b-e_bD)^2-(C_b-e_bD)\sqrt{(C_b-e_bD)^2-2g_bf_bD}-g_bf_bD}{g_b}. \end{aligned}$$

Observe that the left-hand side of the above inequality reduces to $f_bD$. Therefore, after some rearrangement of terms, it can be rewritten as

$$\begin{aligned} (C_b-e_bD)^2-2g_bf_bD\ge (C_b-e_bD)\sqrt{(C_b-e_bD)^2-2g_bf_bD}. \end{aligned}$$

Again, the above inequality cannot hold for the positive square root of $(C_b-e_bD)^2-2g_bf_bD$. Therefore, we cannot have $Q^*_{d1}\le Q^*_{d2}\le Q_1<Q_2$. $\square$

The first part of Lemma 4 implies that when $(C_b-e_bD)> \sqrt{2g_bf_bD}$, the case of $Q_1<Q_2\le Q^*_{d2}=Q^*_{d1}$ cannot occur. Likewise, the second part implies that when $(C_b-e_bD)> \sqrt{2g_bf_bD}$, the case of $Q^*_{d1}= Q^*_{d2} \le Q_1<Q_2$ cannot take place. Combining this result with Lemma 3 further leads to the following implication: If $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b=K_bg_b$, the only possible ordering of $Q_1$, $Q_2$, $Q^*_{d1}$ and $Q^*_{d2}$ is $Q_1<Q^*_{d1}= Q^*_{d2}<Q_2$, because having $(C_b-e_bD)> \sqrt{2g_bf_bD}$ implies $Q_2>Q_1$, and it follows due to Lemma 3 and the fact that $f_bh_b=K_bg_b$ that $Q^*_{d1}= Q^*_{d2}$. Under these conditions, excluding the cases covered in Lemma 4 from further consideration, the only possible ordering that remains is $Q_1<Q^*_{d1}= Q^*_{d2}<Q_2$.

Lemma 5

If $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b=K_bg_b$, then $Q_d^*=Q^*_{d1}= Q^*_{d2}$.

Proof

Under the conditions of the lemma, the only possible ordering of $Q_1$, $Q_2$, $Q^*_{d1}$, and $Q^*_{d2}$ is $Q_1<Q^*_{d1}= Q^*_{d2}<Q_2$. To prove the lemma, we will consider three regions of $Q$ separately: $Q\le Q_1$, $Q_1<Q<Q_2$, and $Q\ge Q_2$. Expression (1) and Lemma 2 together imply that if $(C_b-e_bD)> \sqrt{2g_bf_bD}$, for order quantities $Q$ such that $Q_1<Q<Q_2$, we have $BC\left( Q,X_b(Q)\right) =BC_2\left( Q,X_b(Q)\right)$; for order quantities $Q$ such that $Q\le Q_1$, we have $BC\left( Q,X_b(Q)\right) =BC_1\left( Q,X_b(Q)\right)$; for order quantities $Q$ such that $Q\ge Q_2$, we have $BC\left( Q,X_b(Q)\right) =BC_1\left( Q,X_b(Q)\right)$.

Let us start with $Q$ such that $Q_1<Q<Q_2$ and $Q\ne Q^*_{d2}$. Since $Q^*_{d2}$ is the unique minimizer of $BC_2(Q,X_b(Q))$ and $BC\left( Q,X_b(Q)\right) =BC_2\left( Q,X_b(Q)\right)$, it follows that

$$\begin{aligned} BC\left( Q,X_b(Q)\right)= & {} BC_2\left( Q,X_b(Q)\right) >BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) =BC\left( Q^*_{d2},X_b(Q^*_{d2})\right) ,\nonumber \\&\forall Q \text{ s.t. } Q_1<Q<Q_2 \text{ and } Q\ne Q^*_{d2}. \end{aligned}$$

(33)

Now, let us continue with $Q\le Q_1$. Recall that at $Q_1$, we have $BC_1\left( Q_1,X_b(Q_1)\right) =BC_2\left( Q_1,X_b(Q_1)\right)$. Since $BC_1\left( Q,X_b(Q)\right)$ is a strictly convex function with a unique minimizer $Q^*_{d1}$, and $Q\le Q_1<Q^*_{d1}$, it follows that

$$\begin{aligned} BC_1\left( Q,X_b(Q)\right) \ge BC_1\left( Q_1,X_b(Q_1)\right) =BC_2\left( Q_1,X_b(Q_1)\right) . \end{aligned}$$

Using the fact that $BC_2\left( Q,X_b(Q)\right)$ is a strictly convex function with a unique minimizer $Q^*_{d2}$, and $Q_1\ne Q^*_{d2}$, we further have

$$\begin{aligned} BC_2\left( Q_1,X_b(Q_1)\right) >BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) . \end{aligned}$$

Combining the last two inequalities leads to

$$\begin{aligned} BC_1\left( Q,X_b(Q)\right) >BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) , \end{aligned}$$

which is equivalent to

$$\begin{aligned} BC\left( Q,X_b(Q)\right) >BC\left( Q^*_{d2},X_b(Q^*_{d2})\right) ,\quad \quad \quad \quad \quad \forall Q \text{ s.t. } Q\le Q_1. \end{aligned}$$

(34)

Finally, let us consider order quantities $Q$ such that $Q\ge Q_2$. Recall that at $Q_2$, we have $BC_1\left( Q_2,X_b(Q_2)\right) =BC_2\left( Q_2,X_b(Q_2)\right)$. Since $BC_1\left( Q,X_b(Q)\right)$ is a strictly convex function with a unique minimizer $Q^*_{d1}$, and $Q^*_{d1}<Q_2\le Q$, it follows that

$$\begin{aligned} BC_1\left( Q,X_b(Q)\right) \ge BC_1\left( Q_2,X_b(Q_2)\right) =BC_2\left( Q_2,X_b(Q_2)\right) . \end{aligned}$$

Using the fact that $BC_2\left( Q,X_b(Q)\right)$ is a strictly convex function with a unique minimizer $Q^*_{d2}$, and $Q_2\ne Q^*_{d2}$, we further have

$$\begin{aligned} BC_2\left( Q_2,X_b(Q_2)\right) >BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) . \end{aligned}$$

Combining the last two inequalities leads to

$$\begin{aligned} BC_1\left( Q,X_b(Q)\right) >BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) , \end{aligned}$$

which, is also equivalent to

$$\begin{aligned} BC\left( Q,X_b(Q)\right) >BC\left( Q^*_{d2},X_b(Q^*_{d2})\right) ,\quad \quad \quad \quad \quad \forall Q \text{ s.t. } Q\ge Q_2. \end{aligned}$$

(35)

Based on Expressions (33), (34), and (35), we conclude that $Q^*=Q^*_{d2}$. $\square$

Lemma 1 and Lemma 5 constitute parts of our solution algorithm for the retailer’s decentralized replenishment problem. Lemma 1 suggests the solution in the case of $(C_b-e_bD)\le \sqrt{2g_bf_bD}$, and Lemma 5 provides the solution in the case of $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b=K_bg_b$. At this point, there is one more case to be considered, that is, $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b\ne K_bg_b$. Before proceeding with a detailed analysis of this case, let us present another result that applies to the case of $(C_b-e_bD)> \sqrt{2g_bf_bD}$ in general.

Lemma 6

When $(C_b-e_bD)> \sqrt{2g_bf_bD}$, we have $BC_1\left( Q,X_b(Q)\right) \le BC_2\left( Q,X_b(Q)\right)$ for all $Q$ such that $Q_1\le Q\le Q_2$, and $BC_1\left( Q,X_b(Q)\right) > BC_2\left( Q,X_b(Q)\right)$ for all $Q$ such that $Q<Q_1$ or $Q>Q_2$.

Proof

Recall that $X_b(Q)=C_b-\frac{f_bD}{Q}-\frac{g_bQ}{2}-e_bD$, and $X_b(Q)=0$ when $Q=Q_1$ and $Q=Q_2$. Furthermore, we have $X_b(Q)> 0$ for all $Q$ s.t. $Q_1< Q< Q_2$, and we have $X_b(Q)< 0$ for all $Q$ s.t. $Q<Q_1$ and for all $Q$ s.t. $Q>Q_2$. We will show that $BC_1\left( Q,X_b(Q)\right) \le BC_2\left( Q,X_b(Q)\right)$ if $Q\in [Q_1,Q_2]$. The proofs of the other parts of the lemma, which are omitted, follow in a similar fashion.

Since $p_c^b\ge p_c^s$, it follows that

$$\begin{aligned} (p_c^b-p_c^s)\left( C_b-\frac{f_bD}{Q}-\frac{g_bQ}{2}-e_bD\right) \ge 0. \end{aligned}$$

After adding $\frac{K_bD}{Q}+\frac{h_bQ}{2}+cD$ to both sides of the above inequality and rearranging the terms, we have

$$\begin{aligned}&\frac{K_bD}{Q}+\frac{h_bQ}{2}+cD-p_c^b\left( C_b-\frac{f_bD}{Q}-\frac{g_bQ}{2}-e_bD\right) \\&\quad \le \frac{K_bD}{Q}+\frac{h_bQ}{2}+cD-p_c^s\left( C_b-\frac{f_bD}{Q}-\frac{g_bQ}{2}-e_bD\right) , \end{aligned}$$

which implies $BC_1\left( Q,X_b(Q)\right) \le BC_2\left( Q,X_b(Q)\right)$. $\square$

The above lemma will be used in the proofs of the next two results.

Lemma 7

When $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b< K_bg_b$, the following orderings among $Q_1$, $Q_2$, $Q^*_{d1}$, and $Q^*_{d2}$ cannot take place:

• $Q^*_{d1}\le Q_1 <Q_2\le Q^*_{d2}$,

• $Q^*_{d1}\le Q_1 < Q^*_{d2}<Q_2$, and

• $Q^*_{d1}< Q^*_{d2} \le Q_1 < Q_2$.

Proof

We will prove the first two parts of the lemma. Note that the third part is a special case of $Q^*_{d1}\le Q^*_{d2}\le Q_1<Q_2$ and is covered in Lemma 4.

Due to the strict convexity of $BC_2\left( Q,X_b(Q)\right)$ and the fact that $Q^*_{d2}$ is its minimizer, having $Q_1<Q_2\le Q^*_{d2}$ implies

$$\begin{aligned} BC_2\left( Q_1,X_b(Q_1)\right) >BC_2\left( Q_2,X_b(Q_2)\right) . \end{aligned}$$

At $Q=Q_1$ and $Q=Q_2$, we have $BC_1\left( Q,X_b(Q)\right) =BC_2\left( Q,X_b(Q)\right)$. Therefore, the above inequality is equivalent to the following:

$$\begin{aligned} BC_1\left( Q_1,X_b(Q_1)\right) >BC_1\left( Q_2,X_b(Q_2)\right) . \end{aligned}$$

(36)

However, due to the strict convexity of $BC_1\left( Q,X_b(Q)\right)$ and $Q^*_{d1}$ being its unique minimizer, having $Q^*_{d1}\le Q_1 <Q_2$ would imply

$$\begin{aligned} BC_1\left( Q_1,X_b(Q_1)\right) <BC_1\left( Q_2,X_b(Q_2)\right) . \end{aligned}$$

(37)

Expression (36) and (37) contradict, therefore, it is not possible to have $Q^*_{d1}\le Q_1 <Q_2\le Q^*_{d2}$.

Now, let us continue with the proof of the second part. Note that Lemma 6 and its proof imply

$$\begin{aligned} BC_1\left( Q^*_{d2},X_b(Q^*_{d2})\right) <BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) \end{aligned}$$

in case of $Q_1<Q^*_{d2}<Q_2$. Furthermore, having $Q_1\!<\!Q^*_{d2}$ leads to $BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) <BC_2\left( Q_1,X_b(Q_1)\right)$ due to the strict convexity of $BC_2\left( Q,X_b(Q)\right)$ and the fact that $Q^*_{d2}$ is its minimizer. Combining this with the above inequality implies

$$\begin{aligned} BC_1\left( Q^*_{d2},X_b(Q^*_{d2})\right) <BC_2\left( Q_1,X_b(Q_1)\right) . \end{aligned}$$

At $Q=Q_1$, we have $BC_2\left( Q_1,X_b(Q_1)\right) =BC_1\left( Q_1,X_b(Q_1)\right)$. Therefore, the above expression is equivalent to

$$\begin{aligned} BC_1\left( Q^*_{d2},X_b(Q^*_{d2})\right) <BC_1\left( Q_1,X_b(Q_1)\right) . \end{aligned}$$

(38)

However, due to the strict convexity of $BC_1\left( Q,X_b(Q)\right)$ and $Q^*_{d1}$ being its unique minimizer, having $Q^*_{d1}\le Q_1 <Q^*_{d2}$ would imply

$$\begin{aligned} BC_1\left( Q^*_{d2},X_b(Q^*_{d2})\right) >BC_1\left( Q_1,X_b(Q_1)\right) . \end{aligned}$$

(39)

As Expressions (38) and (39) contradict, it is not possible to have $Q^*_{d1}\le Q_1 < Q^*_{d2}$ $<Q_2$.$\square$

Notice that, since $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b< K_bg_b$ are the two conditions of Lemma 7, two common properties of the cases considered are $Q_1<Q_2$ and $Q^*_{d1}<Q^*_{d2}$. Lemma 7 further leads to the result in Corollary 4.

Corollary 4

When $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b<K_bg_b$, the following orderings are possible:

• $Q_1<Q_2\le Q^*_{d1}<Q^*_{d2}$,

• $Q_1<Q^*_{d1}<Q^*_{d2}\le Q_2$, and

• $Q_1<Q^*_{d1}<Q_2<Q^*_{d2}$.

Numerical instances to illustrate the cases in Corollary 4 (and Corollary 5) are presented in Table 10. The first three examples of Table 10 correspond to the different cases of the corollary in the order they are presented.

Table 10 Numerical illustrations of Corollaries 4 and 5 given $D=50, c=12$ and $g_b=0.5$

In the next lemma, we provide a similar result to Lemma 7, now for the case of $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b> K_bg_b$.

Lemma 8

When $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b> K_bg_b$, the following orderings among $Q_1$, $Q_2$, $Q^*_{d1}$, and $Q^*_{d2}$ cannot take place:

• $Q^*_{d2}< Q_1 <Q_2\le Q^*_{d1}$,

• $Q_1\le Q^*_{d2}<Q_2\le Q^*_{d1}$, and

• $Q_1 < Q_2 \le Q^*_{d2}<Q^*_{d1}$.

Proof

Similar to the proof of Lemma 7, we will prove the first two parts of the lemma. The third part is a special case of $Q_1<Q_2\le Q^*_{d2}\le Q^*_{d1}$ and is covered in Lemma 4.

Let us assume that the ordering in the first part of the lemma takes place. Due to Lemma 6, having $Q_2\le Q^*_{d1}$ implies $BC_1\left( Q^*_{d1},X_b(Q^*_{d1})\right) \ge BC_2\left( Q^*_{d1},X_b(Q^*_{d1})\right)$. Furthermore, it follows from the strict convexity of $BC_2\left( Q,X_b(Q)\right)$ that having $Q^*_{d2}<Q_1<Q_2\le Q^*_{d1}$ leads to

$$\begin{aligned} BC_2\left( Q^*_{d1},X_b(Q^*_{d1})\right) \ge BC_2\left( Q_2,X_b(Q_{2})\right) >BC_2\left( Q_1,X_b(Q_{1})\right) , \end{aligned}$$

and hence, $BC_1\left( Q^*_{d1},X_b(Q^*_{d1})\right) >BC_2\left( Q_1,X_b(Q_{1})\right)$. At $Q=Q_1$, we have $BC_2\left( Q,X_b\right.$ $\left. (Q)\right) =BC_1\left( Q,X_b(Q)\right)$. Therefore, if the ordering is true, as it is assumed, it would follow that $BC_1\left( Q^*_{d1},X_b(Q^*_{d1})\right) >BC_1\left( Q_1,X_b(Q_{1})\right)$, which contradicts with $Q^*_{d1}$ being the minimizer of $BC_1\left( Q,X_b(Q)\right)$. Therefore, it is not possible to have $Q^*_{d2}< Q_1 <Q_2\le Q^*_{d1}$.

Let us continue with the proof of the second part by assuming that there exists an instance with this ordering. Due to Lemma 6, having $Q_1\le Q^*_{d2}<Q_2$ implies $BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) \ge BC_1\left( Q^*_{d2},X_b(Q^*_{d2})\right)$. Furthermore, it follows from the strict convexity of $BC_1\left( Q,X_b(Q)\right)$ that having $Q^*_{d2}<Q_2\le Q^*_{d1}$ leads to

$$\begin{aligned} BC_1\left( Q^*_{d2},X_b(Q^*_{d2})\right) > BC_1\left( Q_2,X_b(Q_{2})\right) \ge BC_1\left( Q^*_{d1},X_b(Q^*_{d1})\right) , \end{aligned}$$

and hence, $BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) > BC_1\left( Q^*_{d1},X_b(Q^*_{d1})\right)$. Using Lemma 6 once again and the fact that $Q^*_{d1}\ge Q_2$, we must have $BC_1\left( Q^*_{d1},X_b(Q^*_{d1})\right) \ge BC_2\left( Q^*_{d1},X_b(Q^*_{d1})\right)$, which would imply $BC_2\left( Q^*_{d2},X_b(Q^*_{d2})\right) >BC_2\left( Q^*_{d1},X_b(Q^*_{d1})\right)$. However, this contradicts with the fact that $Q^*_{d2}$ is the minimizer of $BC_2\left( Q,X_b(Q)\right)$. Therefore, it is not possible to have $Q_1\le Q^*_{d2}<Q_2\le Q^*_{d1}$. $\square$

Note that under the two conditions of Lemma 8, two common properties of the cases considered are $Q_1<Q_2$ and $Q^*_{d1}>Q^*_{d2}$. Lemma 8 further leads to the result in the next corollary.

Corollary 5

When $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b>K_bg_b$, the following orderings are possible:

• $Q_1\le Q^*_{d2}< Q^*_{d1}<Q_2$,

• $Q^*_{d2}<Q_1<Q^*_{d1}<Q_2$, and

• $Q^*_{d2}<Q^*_{d1}\le Q_1<Q_2$.

Numerical instances to illustrate the cases in Corollary 5 are also presented in Table 10. The last three examples of Table 10 correspond to the different cases of the corollary in the order they are presented.

1.3 Proof of Theorem 1

The proof will follow based on considering the cases presented in Lemma 1, Lemma 5, Corollary 4, and Corollary 5.

Case 1: $(C_b-e_bD)\le \sqrt{2g_bf_bD}$

It follows due to Lemma 1 that in this case $Q_d^*=Q^*_{d1}$.

Case 2: $(C_b-e_bD)> \sqrt{2g_bf_bD}$

We have the following three subcases ($f_bh_b = K_bg_b$, $f_bh_b < K_bg_b$, and $f_bh_b > K_bg_b$ ):

Case 2.1: $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b = K_bg_b$

It follows due to Lemma 5 that in this case $Q_d^*=Q^*_{d2}$.

Case 2.2: $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b < K_bg_b$

Corollary 4 implies the following three subcases: $Q_1<Q_2\le Q^*_{d1}<Q^*_{d2}$, $Q_1<Q^*_{d1}<Q^*_{d2}\le Q_2$, and $Q_1<Q^*_{d1}<Q_2<Q^*_{d2}$. We present a detailed proof for the first subcase. Since the proofs of the other subcases are similar, we present sketches of the proofs for those.

• Case 2.2.1: $Q_1<Q_2\le Q^*_{d1}<Q^*_{d2}$ Note that the subcase of $Q_1<Q_2\le Q^*_{d1}<Q^*_{d2}$ is distinguished from the other two by the fact that $Q_2\le Q^*_{d1}$. The proof will follow by considering three different regions of $Q$ ($Q>Q_2$, $Q_1\le Q\le Q_2$, $Q<Q_1$), and in each case by showing that $BC\left( Q^*_{d1},X_b(Q^*_{d1})\right) \le BC\left( Q,X_b(Q)\right)$. Let us start with $Q$ values such that $Q>Q_2$. Expression (1) and Lemma 2 imply that $BC\left( Q,X_b(Q)\right) =BC_1\left( Q,X_b(Q)\right)$. By definition, $Q^*_{d1}$ is the minimizer of $BC_1\left( Q,X_b(Q)\right)$, therefore, $BC_1\left( Q,X_b(Q)\right) \ge BC_1\left( Q^*_{d1},X_b(Q^*_{d1})\right)$. Since $Q^*_{d1}$ is also in the region of $Q$ values considered (i.e., $Q^*_{d1}\ge Q_2$), this, in turn, is equivalent to $BC\left( Q,X_b(Q)\right) \ge BC\left( Q^*_{d1},X_b(Q^*_{d1})\right)$. Now, let us consider $Q$ values such that $Q_1\le Q\le Q_2$. Expression (1) and Lemma 2 imply that $BC\left( Q,X_b(Q)\right) =BC_2\left( Q,X_b(Q)\right)$. Since $BC_2\left( Q,X_b(Q)\right)$ is a strictly convex function with a unique minimizer $Q^*_{d2}$ and $Q<Q^*_{d2}$, $BC_2\left( Q,X_b(Q)\right)$, and hence $BC\left( Q,X_b(Q)\right)$, is decreasing in this region. Therefore, $BC\left( Q,X_b(Q)\right) \ge BC\left( Q_2,X_b(Q_2)\right)$ for all $Q$ such that $Q_1\le Q\le Q_2$. Furthermore, we have $BC\left( Q_2,X_b(Q_2)\right) =BC_2\left( Q_2,X_b(Q_2)\right) =BC_1\left( Q_2,X_b(Q_2)\right)$ and $BC_1\left( Q_2,X_b(Q_2)\right) \ge BC_1\left( Q^*_{d1},X_b(Q^*_{d1})\right) =BC\left( Q^*_{d1},X_b(Q^*_{d1})\right)$. Hence, $BC\left( Q,X_b(Q)\right) \ge BC\left( Q^*_{d1},X_b(Q^*_{d1})\right)$. Finally, let us consider $Q$ values such that $Q<Q_1$. Again, due to Expression (1) and Lemma 2, we know that $BC\left( Q,X_b(Q)\right) =BC_1\left( Q,X_b(Q)\right)$. Since $BC_1\left( Q,X_b(Q)\right)$ is a strictly convex function with a unique minimizer $Q^*_{d1}$ and $Q<Q^*_{d1}$, $BC_1\left( Q,X_b(Q)\right)$, and hence $BC\left( Q,X_b(Q)\right)$, is decreasing in this region. Therefore, $BC\left( Q,X_b(Q)\right) > BC\left( Q_1,X_b(Q_1)\right)$ for all $Q$ such that $Q<Q_1$. We have discussed above that $BC\left( Q,X_b(Q)\right)$ is decreasing over $Q_1\le Q \le Q_2$, hence $BC\left( Q_1,X_b(Q_1)\right) >BC\left( Q_2,X_b(Q_2)\right)$. Combining the last two results implies $BC\left( Q,X_b(Q)\right) >BC\left( Q_2,X_b(Q_2)\right)$. We have also argued above that $BC\left( Q_2,X_b(Q_2)\right) \ge BC\left( Q^*_{d1},X_b(Q^*_{d1})\right)$. Therefore, we conclude $BC\left( Q,X_b(Q)\right) >BC\left( Q^*_{d1},X_b(Q^*_{d1})\right)$.

• Case 2.2.2: $Q_1<Q^*_{d1}<Q^*_{d2}\le Q_2$ We have $BC\left( Q,X_b(Q)\right) \ge BC\left( Q^*_{d2},X_b(Q^*_{d2})\right)$ for all $Q\in [Q_1,Q_2]$, because, $Q_1<Q^*_{d2}\le Q_2$ and $BC\left( Q,X_b(Q)\right) =BC_2\left( Q,X_b(Q)\right)$ in this region of $Q$ values. Next, we use the facts that $BC\left( Q,X_b(Q)\right) =BC_1\left( Q,X_b(Q)\right)$ for all $Q\in (Q_2,\infty )$, $BC_1\left( Q,X_b(Q)\right)$ is increasing in this region, and $BC_1\left( Q_2,X_b(Q_2)\right) =BC_2\left( Q_2,X_b(Q_2)\right)$ to conclude that $BC\left( Q,X_b(Q)\right) >BC\left( Q_2,X_b(Q_2)\right)$. This further implies $BC\left( Q,X_b(Q)\right) >BC\left( Q^*_{d2},X_b(Q^*_{d2})\right)$ for all $Q\in (Q_2,\infty )$. Finally, using the facts that $BC\left( Q,X_b(Q)\right) =BC_1\left( Q,X_b(Q)\right)$ for all $Q$ such that $Q<Q_1$, $BC_1\left( Q,X_b(Q)\right)$ is decreasing in this region, and $BC_1\left( Q_1,X_b(Q_1)\right) =BC_2\left( Q_1,X_b(Q_1)\right)$ to conclude that $BC\left( Q,X_b(Q)\right) >BC\left( Q_1,X_b(Q_1)\right)$. This further implies $BC\left( Q,X_b(Q)\right) >BC\left( Q^*_{d2},X_b(Q^*_{d2})\right)$ for all $Q$ such that $Q<Q_1$.

• Case 2.2.3: $Q_1<Q^*_{d1}<Q_2<Q^*_{d2}$ We have $BC\left( Q,X_b(Q)\right) \ge BC\left( Q_{2},X_b(Q_{2})\right)$ for all $Q\in [Q_1,Q_2]$, because, $BC\left( Q,X_b(Q)\right) =BC_2\left( Q,X_b(Q)\right)$ and $Q_1<Q_2<Q^*_{d2}$ (implying that $BC_2\left( Q,X_b(Q)\right)$ is decreasing in this region of $Q$ values). Next, we use the facts that $BC\left( Q,X_b(Q)\right) =BC_1\left( Q,X_b(Q)\right)$ for all $Q\in (Q_2,\infty )$ and $Q^*_{d1}<Q_2$ (implying that $BC_1\left( Q,X_b(Q)\right)$ is increasing in this region) to conclude that $BC\left( Q,X_b(Q)\right) >BC\left( Q_2,X_b(Q_2)\right)$. Finally, using the facts that $BC\left( Q,X_b(Q)\right) =BC_1\left( Q,X_b(Q)\right)$ for all $Q$ such that $Q<Q_1$, and $Q_1<Q^*_{d1}$ (implying that $BC_1\left( Q,X_b(Q)\right)$ is decreasing in this region), we conclude that $BC\left( Q,X_b(Q)\right) >BC\left( Q_1,X_b(Q_1)\right)$. Combining this with the fact that $BC\left( Q_1,X_b(Q_1)\right) >BC\left( Q_2,X_b(Q_2)\right)$ further leads to $BC\left( Q,X_b(Q)\right) >BC\left( Q_2,X_b(Q_2)\right)$ for all $Q$ such that $Q<Q_1$.

Case 2.2: $(C_b-e_bD)> \sqrt{2g_bf_bD}$ and $f_bh_b >K_bg_b$

Corollary 5 implies the following three subcases: $Q^*_{d2} < Q_1 < Q_2\le Q^*_{d1}$, $Q_1 \le Q^*_{d2} < Q_2 \le Q^*_{d1}$, $Q_1 < Q_2 \le Q^*_{d2}<Q^*_{d1}$. A detailed proof will be omitted for this case as it follows by analyzing the different subcases, as in the proof of Case 2.1. $\square$

1.4 Proof of Proposition 1

Suppose $C_b-e_bD\le \sqrt{2g_bf_bD}$. Using Expression (1) and Lemma 1, $BC(Q,X_b(Q))=BC_1(Q,X_b(Q))$ for all $Q$. Since $BC_1(Q,X_b(Q))$ is a strictly convex function of $Q$, $BC(Q,X_b(Q))$ is also a strictly convex function of $Q$.

Now, suppose that $C_b-e_bD>\sqrt{2g_bf_bD}$. The proof will follow by showing that $BC(\alpha Q_a+(1-\alpha )Q_b, X_b(\alpha Q_a+(1-\alpha )Q_b))< \alpha BC(Q_a,X_b(Q_a))+(1-\alpha )BC(Q_b,X_b$ $(Q_b))$ for all $Q_a\geqslant {0}$, $Q_b\geqslant {0}$ ($Q_a\ne Q_b$), and $\alpha \epsilon (0,1)$. First, observe from Expression (1), Lemma 2, and Lemma 6 that $BC(Q,X_b(Q))=\max \{BC_1(Q,X_b(Q)),BC_2(Q,X_b(Q))\}$ when $C_b-e_bD>\sqrt{2g_bf_bD}$. Since $BC_1(Q,X_b(Q))$ and $BC_2(Q,X_b(Q))$ are strictly convex functions of $Q$, it follows that $BC_1(\alpha Q_a+(1-\alpha )Q_b,X_b(\alpha Q_a+(1-\alpha )Q_b))< \alpha BC_1(Q_a,X_b(Q_a))+(1-\alpha )BC_1(Q_b,X_b(Q_b))$ and $BC_2(\alpha Q_a+(1-\alpha )Q_b,X_b(\alpha Q_a+(1-\alpha )Q_b))< \alpha BC_2(Q_a,X_b(Q_a))+(1-\alpha )BC_2(Q_b,X_b(Q_b))$ for all $Q_a\geqslant {0}$, $Q_b\geqslant {0}$ ($Q_a\ne Q_b$) and $\alpha \epsilon (0,1)$. Combining this with $BC(Q,X_b(Q))=\max \{BC_1(Q,X_b(Q)),BC_2(Q,X_b(Q))\}$ leads to $BC_1(\alpha Q_a+(1-\alpha )Q_b, X_b(\alpha Q_a+(1-\alpha )Q_b))<{\alpha BC(Q_a,X_b(Q_a))+(1-\alpha )BC(Q_b,X_b(Q_b))}$ and $BC_2(\alpha Q_a+(1-\alpha )Q_b,X_b(\alpha Q_a+(1-\alpha )Q_b))<{\alpha BC(Q_a,X_b(Q_a))+(1-\alpha )BC(Q_b,X_b(Q_b))}$. Hence, $\max \{BC_1(\alpha Q_a+(1-\alpha )Q_b,X_b(\alpha Q_a+(1-\alpha )Q_b)),BC_2(\alpha Q_a+(1-\alpha )Q_b,X_b(\alpha Q_a+(1-\alpha )Q_b))\}< \alpha BC(Q_a,X_b(Q_a))+(1-\alpha )BC(Q_b,X_b(Q_b))$. Note that the left-hand side of this inequality is $BC(Q,X_b(Q))$. Thus, $BC(Q,X_b(Q))$ is also a strictly convex function of $Q$ if $C_b-e_bD>\sqrt{2g_bf_bD}$. $\square$

1.5 Proof of Proposition 3

Using Equations (28) and (31), we have $Q^*_d\leqslant {Q^*_c}$ if and only if

$$\begin{aligned} \sqrt{\frac{2(K_b+t_bf_b)D}{h_b+t_bg_b}} \leqslant {\sqrt{\frac{2(K_b+K_v+t_bf_b+t_vf_v)D}{h_b+t_bg_b+(h_v+t_vg_v)\frac{D}{P}}}}. \end{aligned}$$

Taking the square of both sides leads to

$$\begin{aligned} \frac{2(K_b+t_bf_b)D}{h_b+t_bg_b} \leqslant {\frac{2(K_b+K_v+t_bf_b+t_vf_v)D}{h_b+t_bg_b+(h_v+t_vg_v)\frac{D}{P}}}. \end{aligned}$$

This, in turn, implies

$$\begin{aligned} \begin{array}{ll} K_bh_b+K_bt_bg_b+K_bh_v\frac{D}{P}+K_bt_vg_v\frac{D}{P}+t_bf_bh_b+t^2_bf_bg_b+h_vt_bf_b\frac{D}{P}+t_bf_bt_vg_v\frac{D}{P} \\ \leqslant {K_bh_b+K_vh_b+h_bt_bf_b+h_bt_vf_v+K_bt_bg_b+K_vt_bg_b+t^2_bf_bg_b+t_bg_bt_vg_v}. \end{array} \end{aligned}$$

After some cancellations and rearrangement of terms, we get

$$\begin{aligned} (K_b+t_bf_b)(h_v+t_vg_v)\frac{D}{P} \leqslant {(K_v+t_vf_v)(h_b+t_bg_b)}. \end{aligned}$$

This results in

$$\begin{aligned} \frac{K_b+t_bf_b}{h_b+t_bg_b}\leqslant {\frac{K_v+t_vf_v}{h_v+t_vg_v}\frac{P}{D}}. \end{aligned}$$

$\square$

1.6 Proof of Proposition 4

• (i) It follows from Proposition 3 that $\frac{K_b+t_bf_b}{h_b+t_bg_b}\leqslant {\frac{K_v+t_vf_v}{h_v+t_vg_v}\frac{P}{D}}$ is equivalent to $Q^*_d\le Q^*_c$. Multiplying both sides of the inequality $\frac{t_bf_b+t_vf_v}{t_bg_b+\frac{t_vg_vD}{P}}\leqslant {\frac{K_b+t_bf_b}{h_b+t_bg_b}}$ with $2D$ and taking the square root of both sides, we obtain

  $$\begin{aligned} \sqrt{\frac{2(t_bf_b+t_vf_v)D}{t_bg_b+\frac{t_vg_vD}{P}}}\leqslant {\sqrt{\frac{2(K_b+t_bf_b)D}{h_b+t_bg_b}}}, \end{aligned}$$

  which implies $Q^t_c\le Q^*_d$. Combining this result with the fact that $Q^*_d\le Q^*_c$ implies that $Q^t_c\le Q^*_d \le Q^*_c$. Since $TT(Q)$ is a strictly convex function, this implies that $TT(Q^*_c)\ge TT(Q^*_d)$. That is, in the centralized solution, the government collects at least the same amount of taxes as it collects in the decentralized solution.

• (ii) Again, due to Proposition 3, we know that $\frac{K_b+t_bf_b}{h_b+t_bg_b}\leqslant {\frac{K_v+t_vf_v}{h_v+t_vg_v}\frac{P}{D}}$ implies $Q^*_d\le Q^*_c$. Multiplying both sides of the inequality $\frac{t_bf_b+t_vf_v}{t_bg_b+\frac{t_vg_vD}{P}}\geqslant {\frac{K_b+K_v+t_bf_b+t_vf_v}{h_b+t_bg_b+(h_v+t_vg_v)\frac{D}{P}}}$ with $2D$ and taking the square root of both sides, we obtain

  $$\begin{aligned} \sqrt{\frac{2(t_bf_b+t_vf_v)D}{t_bg_b+\frac{t_vg_vD}{P}}}\geqslant {\sqrt{\frac{2(K_b+K_v+t_bf_b+t_vf_v)D}{h_b+t_bg_b+(h_v+t_vg_v)\frac{D}{P}}}}, \end{aligned}$$

  which is equivalent to $Q^t_c\ge Q^*_c$. Combining this result with the fact that $Q^*_d\le Q^*_c$ implies $Q^t_c\ge Q^*_c \ge Q^*_d$. Since $TT(Q)$ is a strictly convex function, this implies $TT(Q^*_d)\ge TT(Q^*_c)$. That is, in the decentralized solution, the government collects at least the same amount of taxes as it collects in the centralized solution.

$\square$