Optimal subsidy in promoting distributed renewable energy generation based on policy benefit

Abstract

Distributed renewable energy generation via micro-grid plays a strategic role in defining energy policy for mitigating the pressure of global climate changes and energy reservation. As the initial installation of the renewable generation equipment is costly, it is necessary that the government provides incentive schemes to private investors aiming at mobilizing private capital to support distributed renewable energy generation. This paper brings forward optimal subsidy to stimulate private investment and focuses more on the government’s expected policy benefit. We formulate principal–agent model in which the private investor’s preference toward renewable generation is described as asymmetric information. We analyze the optimal subsidy with the purpose of maximizing the expected policy benefit; besides, this paper reveals benefit conflicts between the policymakers and the private investors, and examines the parameters’ effect on the government’s purpose under the condition of asymmetric information. Finally, a numerical example is presented to test the effectiveness of the model. The results shed new light on the role of investor’s preference in determining the share of renewable energy generation; moreover, it has important implication for policymakers: the results suggest that radically innovative systems will get down the cost curve and may display higher long-term potentials, but in the short run, the government should eliminate asymmetric information as far as possible and improve the investors’ environment-friendly awareness.

Appendix

Proof of Proposition 1

Proposition 1 including analysis of incentive-compatibility constraint and participation constrains. First, let L(x, y) = s(y)q(y) + v(α, y) − C(β, ρ, x, q(y)) and L(x, y) mean the utility function of private investor toward preference x but choosing the pair (s(x),q(x)), where x, y ∊ [0, b] and x ≠ y. Thus, for any given x, the incentive-compatibility constraint can be written as

$L(x,x) \ge L(x,y),\forall y \in \left[ {0,b} \right]$, L(x, y) satisfies the first-order condition $\frac{\partial L(x,y)}{\partial y}|_{y = x} = 0$ and the second-order condition $\frac{{\partial^{2} L(x,y)}}{{\partial y^{2} }}|_{y = x} < 0$. It follows from the first-order condition that

$$s(x)\frac{{{\text{d}}q(x)}}{{{\text{d}}x}} + q(x)\frac{{{\text{d}}s(x)}}{{{\text{d}}x}} + \frac{{{\text{d}}v(\alpha ,x)}}{{{\text{d}}x}} - \frac{\partial C}{\partial q}\frac{{{\text{d}}q(x)}}{{{\text{d}}x}} = 0, \, \forall x \in \left[ {0,b} \right]$$

(P1)

Differentiating (P1) with respect to x,

$$\begin{aligned} s(x)\frac{{{\text{d}}^{2} q(x)}}{{{\text{d}}x^{2} }} + & q(x)\frac{{{\text{d}}^{2} s(x)}}{{{\text{d}}x^{2} }} + 2\frac{{{\text{d}}q(x)}}{{{\text{d}}x}}\frac{{{\text{d}}s(x)}}{{{\text{d}}x}} + \frac{{{\text{d}}^{2} v(\alpha ,x)}}{{{\text{d}}x^{2} }} - \frac{{{\text{d}}^{2} C}}{{{\text{d}}q^{2} }}\left(\frac{{{\text{d}}q(x)}}{{{\text{d}}x}}\right)^{2} \\ \, - & \frac{{{\text{d}}^{2} C}}{{{\text{d}}q{\text{d}}x}}\frac{{{\text{d}}q(x)}}{{{\text{d}}x}} - \frac{{{\text{d}}C}}{{{\text{d}}q}}\frac{{{\text{d}}^{2} q(x)}}{{{\text{d}}x^{2} }} = 0, \, \quad \forall x \in \left[ {0,b} \right] \\ \end{aligned}$$

(P2)

It follows from the second-order condition that

$$\begin{aligned} s(x)\frac{{{\text{d}}^{2} q(x)}}{{{\text{d}}x^{2} }} + & q(x)\frac{{{\text{d}}^{2} s(x)}}{{{\text{d}}x^{2} }} + 2\frac{{{\text{d}}q(x)}}{{{\text{d}}x}}\frac{{{\text{d}}s(x)}}{{{\text{d}}x}} + \frac{{{\text{d}}^{2} v(\alpha ,x)}}{{{\text{d}}x^{2} }} \\ \, - & \frac{{{\text{d}}^{2} C}}{{{\text{d}}q^{2} }}(\frac{{{\text{d}}q(x)}}{{{\text{d}}x}})^{2} - \frac{{{\text{d}}C}}{{{\text{d}}q}}\frac{{{\text{d}}^{2} q(x)}}{{{\text{d}}x^{2} }} < 0,\quad \, \forall x \in \left[ {0,b} \right] \\ \end{aligned}$$

(P3)

Applying (P2) to (P3) yields, $\frac{{{\text{d}}^{2} C}}{{{\text{d}}q{\text{d}}x}}\frac{{{\text{d}}q(x)}}{{{\text{d}}x}} < 0,\quad \forall x \in \left[ {0,b} \right]$. Because we have assumed $\frac{{\partial^{2} C}}{\partial x\partial q} \le 0$, thus we derive $\frac{{{\text{d}}q(x)}}{{{\text{d}}x}} > 0$. The proof of part (i) is completed.

On the other hand, when x > y, by $\frac{dq(x)}{dx} > 0$, $\frac{{\partial^{2} r(q(x),x)}}{\partial x\partial q} \le 0$ and integrating (P1) yields

$$\begin{aligned} [s(x)q(x) + v(\alpha ,x)] - [s(y)q(y) + v(\alpha ,y)] = & \int_{y}^{x} {\frac{\partial C(\beta ,\rho ,s,q(s))}{\partial q}} \frac{{{\text{d}}q(s)}}{{{\text{d}}s}}{\text{d}}s \\ \, \ge & \int_{y}^{x} {\frac{\partial C(\beta ,\rho ,x,q(s))}{\partial q}} \frac{{{\text{d}}q(s)}}{{{\text{d}}s}}{\text{d}}s = C(\beta ,\rho ,x,q(x)) - C(\beta ,\rho ,x,q(y)) \\ \end{aligned}$$

and when x < y,

$$\begin{aligned} [s(x)q(x) + v(\alpha ,x)] - [s(y)q(y) + v(\alpha ,y)] = & - \int_{x}^{y} {\frac{C(\beta ,\rho ,s,q(s))}{\partial q}} \frac{{{\text{d}}q(s)}}{ds}{\text{d}}s \\ \, \ge & \int_{x}^{y} {\frac{\partial C(\beta ,\rho ,x,q(s))}{\partial q}} \frac{{{\text{d}}q(s)}}{{{\text{d}}s}}{\text{d}}s = C(\beta ,\rho ,x,q(x)) - C(\beta ,\rho ,x,q(y)) \\ \end{aligned}$$

Therefore, no matter what is the relationship between x and y, we can always obtain incentive constraints (1). That is, we can get (1) from equation (i) and (ii).

For equation (iii), differentiating $\prod (s(x),q(x),x) = s(x)q(x) + v(\alpha ,x) - C(\beta ,\rho ,x,q(x))$ with respect to x: $\frac{\partial \prod }{\partial x} = q(x)\frac{{{\text{d}}s(x)}}{{{\text{d}}x}} + s(x)\frac{{{\text{d}}q(x)}}{{{\text{d}}x}} + \frac{{{\text{d}}v(\alpha ,x)}}{{{\text{d}}x}} - \frac{\partial C}{\partial q(x)}\frac{{{\text{d}}q(x)}}{{{\text{d}}x}} - \frac{\partial C}{\partial x}$ Substituting (P1) into $\frac{\pi \prod }{\partial x}$, there will be $\frac{\partial \pi }{\partial x} = - \frac{\partial C(\beta ,\rho ,x,q(x))}{\partial x} \ge 0$ which means П is increasing with respect to x. Thus, we can rewrite participant constraint as $\prod (s(0),q(0),0) = s(0)q(0) + v(a,0) - C(\beta ,\rho ,0,q(0)) = C_{0},$ which represents the minimum of private investor’s profits.

Proof of Proposition 2

Since $\frac{\partial \prod }{\partial x} = - \frac{\partial C(\beta ,\rho ,x,q(x))}{\partial x}$ , that is equivalent to $\prod (s(x),q(x),x) - \prod (s(0),q(0),0) = \int_{0}^{x} { - \frac{C(\beta ,\rho ,x,q(x))}{\partial s}} {\text{d}}s,$ according to equation (iii), therefore, we can yield

$$s(x)q(x) = C(\beta ,\rho ,x,q(x)) - v(\alpha ,x) + C_{0} - \int_{0}^{x} {\frac{\partial C(\beta ,\rho ,s,q(s))}{\partial s}} {\text{d}}s$$

(P4)

then substituting (P4) into the objective function yields

$$\begin{aligned} E[U(q(x) + R(x) - s(x)q(x)] \hfill \\ = \int_{0}^{b} {\left[ {U(q(x) + R(x) - C(\beta ,\rho ,x,q(x)) + v(\alpha ,x) - C_{0} + \int_{0}^{x} {\frac{{\partial C(\beta ,\rho ,s,q(s))}}{{\partial s}}} {\text{d}}s} \right]f(x){\text{d}}x} \hfill \\ = \int_{0}^{b} {\left[ {U(q(x)) + R(x) - C(\beta ,\rho ,x,q(x)) + v(\alpha ,x) - C_{0} + \frac{{1 - F(x)}}{{f(x)}}\frac{{\partial C(\beta ,\rho ,x,q(x))}}{{\partial x}}} \right]f(x){\text{d}}x} \hfill \\ \end{aligned}$$

(P5)

Remarks

The decision vector (s(x), q(x)) in objective function is simplified as q(x) which indicates that (P5) is only decided by q(x) and is irrelevant with s(x).

Proof of Proposition 3

Let $M(x) = U(q(x)) + R(x) - C(\beta ,\rho ,x,q(x)) + v(\alpha ,x) - C_{0} + \frac{1 - F(x)}{f(x)}\frac{\partial C(\beta ,\rho ,x,q(x))}{\partial x},$ then integrating assumption (iii) and the second-order variation of (P5) with respect to q, we can derive $\int_{0}^{b} {\frac{{\partial^{2} M(q,x)}}{{\partial q^{2} }}(\delta q)^{2} dF(x)} < 0$. Hence, the optimal consumption of electricity under the preference satisfies ∫ ^b₀ M(q(x), x)dF(x) = 0, i.e.,

$$\frac{{\partial U(q^{**} (x))}}{{\partial q^{**} }} - \frac{{\partial C(\beta ,\rho ,x,q^{**} (x))}}{{\partial q^{**} }} + \frac{1 - F(x)}{f(x)}\frac{{\partial^{2} C(\beta ,\rho ,x,q^{**} (x))}}{{\partial x\partial q^{**} }} = 0$$

(P6)

Under the same circumstance, the optimal subsidy s**(x) can be obtained from equation (P4):

$$s^{**} (x) = \frac{{C(\beta ,\rho ,x,q^{**} (x)) - v(\alpha ,x) + C_{0} - \int_{0}^{x} {\frac{{\partial C(\beta ,\rho ,s,q^{**} (s))}}{\partial s}} {\text{d}}s}}{{q^{**} (x)}}$$

(P7)

Let $G(x) = \frac{1 - F(x)}{f(x)}$, the derivation of (P6) with respect to x can be written as,$\begin{aligned}&\left[\frac{{\partial^{2} U(q^{**} (x))}}{{\partial q^{**2} }} -\frac{{\partial^{2} C(\beta ,\rho ,s,q^{**} (s))}}{{\partial q^{**2}}} + G(x)\frac{{\partial^{3} C(\beta ,\rho ,s,q^{**}(s))}}{{\partial x\partial q^{**2} }}\right]\frac{{dq^{**}(x)}}{{{\text{d}}x}}=\frac{{\partial^{2} C(\beta ,\rho ,s,q^{**} (s))}}{{\partial x\partial q^{**} }} - \frac{\partial G}{\partial x}\frac{{\partial^{2} C(\beta ,\rho ,s,q^{**} (s))}}{{\partial x\partial q^{**} }} - G(x)\frac{{\partial^{3} C(\beta ,\rho,s,q^{**} (s))}}{{\partial x^{2} \partial q^{**} }} \end{aligned}$, according to the assumption part (ii) and (iii) $\frac{\partial G}{\partial x} < 0$, we can obtain that $\frac{{dq^{**} (x)}}{{{\text{d}}x}} > 0$. Consequently, the optimal and feasible solutions in model (3) is (s**(x), q**(x)). Therefore, all the proofs are completed.