Subsidizing strategies in a sustainable supply chain

Abstract

This paper inspects the effects of government subsidies on a supply chain that involves a supplier, a manufacturer, and consumers with electric vehicles (EVs). In practice, the government usually subsidizes different players (the supplier, the manufacturer, or consumers) in the supply chain using two subsidy schemes, namely the linear subsidy model (Model L) and the fixed subsidy model (Model F). We examine the equilibria of the EV supply chain that incorporates the influence from the government who is an external player in each subsidy model. To reach the target adoption of EVs, subsidizing the supplier / consumer presents the least costs to the government in Model L / Model F. We further compare all subsidizing policies in Model L / Model F to show that offering fixed subsidies to the consumer is the optimal subsidizing policy. We also conduct numerical experiments to show the robustness of the optimal subsidizing policy with varying demand uncertainty and target adoption. We extend the model to explore the effects of different subsidizing policies on consumer surplus and social welfare.

Appendix

Proof of Proposition 1

In the benchmark model, the manufacturer’s expected profit is $\pi _M(p,q)=p {\mathbf{\mathsf{E}}}min[q, \epsilon ap^{-b}] - wq$. By setting $z=\frac{q}{ap^{-b}}$, manufacturer’s problem turns to $\pi _M(p,z)=ap^{-b} [p A(z) - wz]$, where $A(z) = {\mathbf{\mathsf{E}}}min[z, \epsilon ] = \int _{L}^{z} f(x)x\hbox{d}x + \bar{F}(z)z$. $\partial \pi _M(p,z)/\partial p= ap^{-b-1}\{bzw-(b-1)pA(z)\}$. For $b>1$, optimal p is uniquely determined by z. $\partial \pi _M(p,z)/\partial p=0$ implies $p^*(z)=\frac{bwz}{(b-1)A(z)}$. Since $p^*(z)$ is the unique maximizer of ${\mathbf{\mathsf{E}}}\pi _M(p, z)$, next we will show $z_{0}$ optimizes $\pi _M[p^*(z),z]$.

$$\begin{aligned} \frac{d \pi _M[p^*(z),z]}{dz}=&\, \frac{\partial \pi _M[p^*(z),z]}{\partial z} + \frac{\partial \pi _M[p^*(z),z]}{\partial p} \cdot \frac{\hbox{d}p^*(z)}{\hbox{d}z} \\=&\, \frac{\partial \pi _M[p^*(z),z]}{\partial z} \\=&\, \frac{awp^*(z)^{-b}}{(b-1)A(z)} \cdot G(z) \quad with\\ G(z)=&\, bz\bar{F}(z)-(b-1)A(z), \end{aligned}$$

where we have used the fact that $\frac{\partial \pi _M[p^*(z),z]}{\partial p}=0$ due to the optimality of $p^*(z)$. Since $b>1$, first-order condition requires the optimal $z_{0}$ satisfies $G(z_0)=0$. Now we should show such a $z_{0}$ always exists and is uniquely determined. We can easily check that $G(L)=L>0$ and $G(H)=-(b-1)<0$. Since G(z) is continuous, we have $G'(z)=\bar{F}(z)[1-\frac{zf(z)b}{\bar{F}(z)}]$. Increasing General Failure Rate means $zf(z)/\bar{F}(z)$ is increasing, and then $G'(z)$ is decreasing in z and G(z) is a concave function. This, in conjunction with $G(L)>0$ and $G(H)<0$, guarantees the uniqueness of $z_{0} \in [L, H]$. Therefore, $z_{0}$ is the unique solution of equation $(b-1)A(z)=bz\bar{F}(z)$.

Anticipating the manufacturer’s strategy, supplier chooses w to maximize her profit $\pi _{S}=(w-c)q$, where $q=z_{0}a{p^*}^{-b}$ and $p^*=p^*(z_{0})$. It can be easily solved that $w^*=bc/(b-1)$. Based on this, the equilibrium of the game has been solved. $p^* = \frac{b^2 z_{0}c}{(b-1)^2 A(z_{0})}$, $q^* = a z_{0}Tc^{-b}$, $\pi_{S}^* = \frac{az_{0}T}{b-1} c^{1-b}$, $\pi _{M}^{*} = \frac{abz_{0}T}{(b-1)^2} c^{1-b}$, ${ES^*} = aA(z_{0})Tc^{-b}$.

Proofs of Propositions 2 and 3 are similar to the proof of Proposition 1 and are omitted here.□

Proof of Proposition 4

With linear subsidy $r_3$ offered to consumers, the manufacturer’s profit expected profit is

$$\begin{aligned} \pi _M(p,q)=&\, p\cdot {\mathbf{\mathsf{E}}}\min \{q, \epsilon a[(1-r_3)p]^{-b}\} - wq\\=&\, \frac{1}{1-r_3} \{p(1-r_3) {\mathbf{\mathsf{E}}}\min \{q, \epsilon a[(1-r_3)p]^{-b}\} - (1-r_3)wq\}\\=&\, \frac{1}{1-r_3} \{p' {\mathbf{\mathsf{E}}}\min [q, \epsilon a(p')^{-b}] - w'q\}\quad where\quad p'=p(1-r_3),\quad w' = (1-r_3)w. \end{aligned}$$

Then, we can see the problem is almost the same with the benchmark model. We omit the analysis and results here since it can be solved similarly. In addition, Proposition 2 and Proposition 3 can also be converted into the benchmark model and we can prove them similarly.□

Proof of Proposition 5 and Proposition 6

With the fixed subsidy s ₂ to the manufacturer, his expected profit is

$$\begin{aligned} \pi _M(p, q)=&\, p {\mathbf{\mathsf{E}}}{\text{min}}[q, \epsilon ap^{-b}] - (w-s_2)q\\=&\, p {\mathbf{\mathsf{E}}}{\text{min}}[q, \epsilon ap^{-b}] - w'q\quad where\quad w'=(w-s_2). \end{aligned}$$

The only difference between this problem and the benchmark for the manufacturer is his unit cost. We can directly infer $p^*=\frac{bz_{0}(w-s_2)}{(b-1)A(z_{0})}$ and $q=az_{0}T(w-s_2)^{-b}$. Then, the supplier’s problem is

$$\begin{aligned} \max _{w}\,\pi _{S}(w)=&\, (w-c)q \\=&\, (w-c) az_{0}T(w-s_2)^{-b}. \end{aligned}$$

First-order condition requires $w^{FM}=\frac{bc-s_2}{b-1}$. Then, $w^{FM}-s_2=\frac{b(c-s_2)}{b-1}$ is the manufacturer’s actual unit cost, and $p^{FM}=\frac{b^2 z_{0}(c-s_2)}{(b-1)^2 A(z_{0})}$.

When the fixed subsidy s ₁ are offered to the supplier, her profit is $\pi _{w}=(w-c+s_1)q$ as her unit cost turns into $c-s_1$ from c. The manufacturer’s problem keeps the same. Then, we can directly infer from the benchmark that $w^{FS}=\frac{b(c-s_1)}{b-1}$ and $p^{FS}=\frac{b^2 z_{0}(c-s_1)}{(b-1)^2 A(z_{0})}$. Under these two policies, if unit subsidy is always s, we can see the manufacturer’s problem does not change since his cost and demand both keep the same under these two policies. Therefore, their equilibria are almost the same.□

Proof of Proposition 7

Under P _FC , the manufacturer’s expected profit is

$$\begin{aligned} \pi _M (p, q)=&\, p {\mathbf{\mathsf{E}}}{\text{min}}[q, \epsilon a (p-s_3)^{-b}] - wq\\=&\, a(p-s_3)^{-b} [p {\mathbf{\mathsf{E}}}\cdot {\text{min}}(z, \epsilon )- wz]\quad {\text{where}}\quad z=\frac{q}{a(p-s_3)^{-b}}\\ \pi _M(p, z)=&\, a(p-s_3)^{-b}[pA(z)-wz] \end{aligned}$$

$\partial \pi _M(p,z)/\partial p=a(p-s_3)^{-b-1}\{ bzw-A(z)s_3-(b-1)A(z)p\}$. Since $p>c>s_3$, first-order condition requires $p^*(z)=\frac{bzw-A(z)s_3}{(b-1)A(z)}$. Given z, $p^*(z)$ is uniquely determined. Next we will find $z^{FC}$ that optimizes $\pi _M[p^*(z),z]$.

$$\begin{aligned} \frac{d \pi _M[p^*(z),z]}{dz}=&\, \frac{\partial \pi _M[p^*(z),z]}{\partial z} + \frac{\partial \pi _M[p^*(z),z]}{\partial p} \cdot \frac{\hbox{d}p^*(z)}{\hbox{d}z} \\=&\, \frac{\partial \pi _M[p^*(z),z]}{\partial z} \\=&\, \frac{a[p^*(z)-s_3]^{-b}}{(b-1)A(z)} \cdot K(z) \quad with\\ K(z)=&\, [bz\bar{F}(z)-(b-1)A(z)]w-A(z)\bar{F}(z)s_3. \end{aligned}$$

Since $p>s_3$ and $b>1$, first-order condition requires $K(z)=0$. We can easily check that $K(L)=L(w-s_3)>0$ and $K(H)=-(b-1)w<0$. Since K(z) is continuous, optimality of $z^{FC}$ requires $K(z^{FC})=0$. Since $A(z^{FC})\bar{F} (z^{FC})s_3>0$, this indicates $G(z^{FC})=bz^{FC}\bar{F}(z^{FC})-(b-1)A(z^{FC})>0$. In the proofs of benchmark model, we have show G(z) is decreasing and concave in [L, H]. This, in conjunction with $G(L)>0$, $G(H)<0$ and $G(z_{0})=0$, guarantees $z^{FC}<z_0$. When $bz/A(z)-(b-1)/\bar{F}(z)$ is unimodal, we can easily obtain the optimal $z^{FC}$ is uniquely determined by w and $z^{FC}$ increases as w increases but decreases as s ₃ increases. Since $p^{FC}=\frac{w}{\bar{F}(z^{FC})}$, we can easily get $p^{FC}$ increases as w increases and decreases as s ₃ increases.□

Proofs of Proposition 8 and Lemma 1

From Propositions 2, 3 and 4, we observed $ES^{LC}=ES^{LM}=ES^{LS}$ when $r_1=r_2=r_3$. Meanwhile, $EX^{LC}=\frac{b}{b-1} EX^{LM}$ and $EX^{LM}=\frac{b}{b-1} EX^{LS}$. With $b>1$, we can easily obtain $EX^{LC}>EX^{LM}>EX^{LS}$. Additionally, we can observe $(1-r) \pi _M^{LC} = \pi _M^{LM} = \pi _M^{LS}$ and $\pi _{S}^{LC}=\pi _{S}^{LM}=\pi _{S}^{LS}/(1-r)$. Since $r>0$, we obtain $\pi _M^{LC} > \pi _M^{LM} = \pi _M^{LS}$ and $\pi _{S}^{LC}=\pi _{S}^{LM}>\pi _{S}^{LS}$. We thus complete the proof of Lemma 1.□

Proofs of Propositions 9, 10 and 13 are straightforward and omitted here.

Proofs of Proposition 11

Suppose unit subsidy under P _FC is s and unit subsidy under P _FS is r, then $EX^{FC}=\tau s$ and $EX_{FS}=\frac{z_0}{A(z_0)}\tau r$. Thus, we just need to compare s and $\frac{z_0}{A(z_0)} r$. By solving the chain’s problem while keeping the constraint in Eq. (11) tight, and denoting $z^{FC}=z^*$, we have $s=\left\{ \frac{bz^*}{A(z^*)}-\frac{b-1}{F(z^*)} \right\} c$ and $\frac{z_0}{A(z_0)}r=\left\{ \frac{z_0}{A(z_0)}-(b-1)\root b \of {\frac{A(z_0)}{A(z^*)}} \left(\frac{1}{\bar{F}(z^*)}-\frac{z^*}{A(z^*)}\right) \right\} c = \left\{ \frac{z_0}{A(z_0)}+(b-1)\root b \of {\frac{A(z_0)}{A(z^*)}} \frac{z^*}{A(z^*)} - \root b \of {\frac{A(z_0)}{A(z^*)}} \frac{b-1}{\bar{F}(z^*)} \right\} c$. Since we can show $z^*<z_0$, we have $\root b \of {\frac{A(z_0)}{A(z^*)}}>1$. If $b>\frac{z \bar{F}(z)}{\int _{L}^{z}xf(x)\hbox{d}x}$ for any $z\in [L, z_0]$, we can conclude $\frac{z}{A(z) \root b \of {A(z)}}$ increases in z within $[L, z_0]$, which indicates $\frac{z_0}{z^*} \left[\frac{A(z^*)}{A(z_0)}\right]^{(1+1/b)}>1$. Then, $\frac{z_0}{A(z_0)}=\frac{z_0}{A(z_0)} \root b \of {\frac{A(z_0)}{A(z^*)}} \root b \of {\frac{A(z^*)}{A(z_0)}} \frac{z^*}{A(z^*)} \frac{A(z^*)}{z^*} >\root b \of {\frac{A(z_0)}{A(z^*)}} \frac{z^*}{A(z^*)}$. Therefore, $\frac{z_0 r}{A(z_0)c}>b \root b \of {\frac{A(z_0)}{A(z^*)}} \frac{z^*}{A(z^*)} - \root b \of {\frac{A(z_0)}{A(z^*)}} \frac{b-1}{\bar{F}(z^*)}= \root b \of {\frac{A(z_0)}{A(z^*)}} \left[\frac{bz^*}{A(z^*)}-\frac{b-1}{F(z^*)}\right]> \frac{bz^*}{A(z^*)}-\frac{b-1}{F(z^*)} =\frac{s}{c}$. Then, we have $\frac{z_0}{A(z_0)}r>c$, $EX^{FS}>EX^{FC}$, and $P_{FC}\succ P_{FS}$. $P_{FS}\succ P_{LC}$ is straightforward and we omit it.□

Proof of Proposition 12

The proofs of the equalities are straightforward and now we focus on the inequality. $z_0>z^{FC}$ indicates $\frac{z^{FC}}{A(z^{FC})}<\frac{z_0}{A(z_0)}$ because $\frac{z}{A(z)}$ increases in z. Since $ES^{FS}=A(z_0)a(p^{FS})^{-b}=A(z^{FC})a(p^{FC}-s^{FC})^{-b}=\tau$ and $A(z_0)>A(z^{FC})$, we can easily get $p^{FS}>p^{FC}-s^{FC}$. Meanwhile, $\tau =\frac{A(z^{FC})}{z^{FC}} q^{FC}=\frac{A(z_0)}{z_0} q^{FS}=\frac{A(z_0)}{z_0} q^{FM}$ indicates $q^{FC}<q^{FS}=q^{FM}$.□