Promoting the Development of Renewable Energy Under Uncertainty

Abstract

Foreign direct investment in renewable energy projects, particularly where biomass is used as input, has been attracting increasing attention, partly due to the Clean Development Mechanism (CDM) of the Kyoto Protocol. In these situations, there may be an information gap between a host country’s government and the foreign firm that will invest: while the firm can collect information regarding the project, for example through a feasibility study, it will be difficult for the government to know whether the foreign firm is undertaking the project efficiently. It is supposed that the government will offer the foreign firm some remunerations, consisting of feed-in premiums (FIPs) and capital subsidies, to encourage investment in such a project. The purpose of this chapter is to determine an optimal combination of FIPs and capital subsidies that encourages investment in a CDM project by a foreign firm, while minimizing the cost of FIPs and capital subsidies. To this end, we develop a microeconomic model that accounts for this information gap. It is shown that a single combination of remunerations is determined to be optimal regardless of the existence of such an information gap. The model developed in this chapter may be considered an extension of the model developed in Chap. 5, this time applied to an investigation that accounts for uncertainty.

Appendix 1: The Solution to the Minimization Problem in the Case of Asymmetric Information

To begin, note that for a firm of type G, the participation constraint (8.28) may be eliminated because

$$ \begin{aligned}\Pi _{G} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right) + s_{G} - F & \ge \pi_{G} \left( {x_{G}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F \\ & > \pi_{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F. \\ & \ge 0 \\ \end{aligned} $$

(8.33)

The first inequality in (8.33) follows from (8.26), and the third inequality follows from (8.29). The second inequality follows from $C_{G}^{{\prime }} (x) < C_{B}^{{\prime }} (x),\,p + r = C_{G}^{{\prime }} \left( {x_{G}^{*} (p)} \right)$, and $p + r = C_{B}^{{\prime }} \left( {x_{B}^{*} (p)} \right)$.

The Lagrangian $\widetilde{L}$ is

$$\begin{aligned} \tilde{L} = & t\left( {p_{G} x_{G}^{*} \left( {p_{G} } \right) + s_{G} } \right) + (1 - t)\left( {p_{B} x_{B}^{*} \left( {p_{B} } \right) + s_{B} } \right) \\ & - \mu _{1} \left[ {\left( {\pi _{G} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right) + s_{G} - F} \right) - \left( {\pi _{G} \left( {x_{G}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F} \right)} \right] \\ & - \mu _{2} \left[ {\left( {\pi _{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F} \right) - \left( {\pi _{B} \left( {x_{B}^{*} \left( {p_{G} } \right)} \right) + s_{G} - F} \right)} \right], \\ & - \mu _{3} \left[ {\pi _{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F} \right] \\ & - \mu _{4} p_{G} - \mu _{5} p_{B} - \mu _{6} s_{G} - \mu _{7} s_{B} \\ \end{aligned} $$

(8.34)

where $\mu_{i} \,(i = 1, \ldots ,7)$ is a Lagrange multiplier.

Then, we write out the complete set of first order conditions:

$$ \frac{{\partial \widetilde{L}}}{{\partial p_{G} }} = tx_{G}^{*} \left( {p_{G} } \right) + \frac{{tp_{G} }}{{C_{G}^{{\prime \prime }} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right)}} - \mu_{1} x_{G}^{*} \left( {p_{G} } \right) + \mu_{2} x_{B}^{*} \left( {p_{G} } \right) - \mu_{4} = 0, $$

(8.35)

$$ \begin{aligned} \frac{{\partial \tilde{L}}}{{\partial p_{B} }} = & (1 - t)x_{B}^{*} \left( {p_{B} } \right) + \frac{{(1 - t)p_{B} }}{{C_{B}^{{\prime \prime }} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right)}} + \mu _{1} x_{G}^{*} \left( {p_{B} } \right) \\ & - \mu _{2} x_{B}^{*} \left( {p_{B} } \right) - \mu _{3} x_{B}^{*} \left( {p_{B} } \right) - \mu _{5} = 0, \\ \end{aligned} $$

(8.36)

$$ \frac{{\partial \widetilde{L}}}{{\partial s_{G} }} = t - \mu_{1} + \mu_{2} - \mu_{6} = 0, $$

(8.37)

$$ \frac{{\partial \widetilde{L}}}{{\partial s_{B} }} = (1 - t) + \mu_{1} - \mu_{2} - \mu_{3} - \mu_{7} = 0, $$

(8.38)

$$ \mu_{1} \left[ {\left\{ {\pi_{G} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right) + s_{G} - F} \right\} - \left\{ {\pi_{G} \left( {x_{G}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F} \right\}} \right] = 0, $$

(8.39)

$$ \mu_{2} \left[ {\left\{ {\pi_{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F} \right\} - \left\{ {\pi_{B} \left( {x_{B}^{*} \left( {p_{G} } \right)} \right) + s_{G} - F} \right\}} \right] = 0, $$

(8.40)

$$ \mu_{3} \left\{ {\pi_{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F} \right\} = 0, $$

(8.41)

$$ \mu_{4} p_{G} = 0, $$

(8.42)

$$ \mu_{5} p_{B} = 0, $$

(8.43)

$$ \mu_{6} s_{G} = 0, $$

(8.44)

$$ \mu_{7} s_{B} = 0, $$

(8.45)

$$ \left\{ {\pi_{G} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right) + s_{G} - F} \right\} - \left\{ {\pi_{G} \left( {x_{G}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F} \right\} \ge 0, $$

(8.46)

$$ \left\{ {\pi_{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F} \right\} - \left\{ {\pi_{B} \left( {x_{B}^{*} \left( {p_{G} } \right)} \right) + s_{G} - F} \right\} \ge 0, $$

(8.47)

$$ \pi_{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right) + s_{B} - F \ge 0, $$

(8.48)

$$ p_{G} \ge 0,\,p_{B} \ge 0,\,s_{G} \ge 0,\,s_{B} \ge 0, $$

(8.49)

$$ {\text{and}}\,\mu_{1} ,\mu_{2} ,\mu_{3} ,\mu_{4} ,\mu_{5} ,\mu_{6} ,\mu_{7} \ge 0. $$

(8.50)

First, suppose $\mu_{1} = 0$. From Eq. (8.37), $\mu_{6} = t + \mu_{2} > 0$. Thus, $s_{G} = 0$ from Eq. (8.44). On the other hand, $\mu_{4} > 0$ from Eq. (8.35). Hence, $p_{G} = 0$ from Eq. (8.42). However, $p_{G} = 0,\,s_{G} = 0$ does not satisfy (8.33). Therefore, $\mu_{1} > 0$.

Next, we will show $p_{G} = p_{B}$. If $p_{G} < p_{B} ,\,p_{B} > 0$ follows from (8.49). Thus, $\mu_{5} = 0$ from Eq. (8.43). Because $\mu_{2} + \mu_{3} = (1 - t) + \mu_{1} - \mu_{7}$ from Eq. (8.38), the left hand side of Eq. (8.36) is

$$ \begin{aligned} \frac{{\partial \tilde{L}}}{{\partial p_{B} }} & = (1 - t)x_{B}^{*} \left( {p_{B} } \right) + \frac{{(1 - t)p_{B} }}{{C_{B}^{{\prime \prime }} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right)}} + \mu _{1} x_{G}^{*} \left( {p_{B} } \right) - \left[ {(1 - t) + \mu _{1} - \mu _{7} } \right]x_{B}^{*} \left( {p_{B} } \right) \\ & = \frac{{(1 - t)p_{B} }}{{C_{B}^{{\prime \prime }} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right)}} + \mu _{1} \left( {x_{G}^{*} \left( {p_{B} } \right) - x_{B}^{*} \left( {p_{B} } \right)} \right) + \mu _{7} x_{B}^{*} \left( {p_{B} } \right) \\ & > 0. \\ \end{aligned} $$

(8.51)

However, this contradicts Eq. (8.36). Hence, $p_{G} \ge p_{B}$.

If $p_{G} > p_{B} ,\,p_{G} > 0$ follows from (8.49). Then, $\mu_{4} = 0$ from Eq. (8.42). Furthermore, $\mu_{2} = 0$: if $\mu_{2} > 0$,

$$ \pi_{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right) + s_{B} = \pi_{B} \left( {x_{B}^{*} \left( {p_{G} } \right)} \right) + s_{G} $$

(8.52)

holds from Eq. (8.40). On the other hand, because $\mu_{1} > 0$,

$$ \pi_{G} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right) + s_{G} = \pi_{G} \left( {x_{G}^{*} \left( {p_{B} } \right)} \right) + s_{B} $$

(8.53)

holds from Eq. (8.39). Taken together, Eqs. (8.52) and (8.53) yield

$$ \pi_{G} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right) - \pi_{B} \left( {x_{B}^{*} \left( {p_{G} } \right)} \right) = \pi_{G} \left( {x_{G}^{*} \left( {p_{B} } \right)} \right) - \pi_{B} \left( {x_{B}^{*} \left( {p_{B} } \right)} \right). $$

(8.54)

However, this is not the case: the left hand side must be larger than the right hand side because $p_{B} > p_{B}$. Hence, $\mu_{2} = 0$.

Then, because $\mu_{1} = t - \mu_{6}$ from Eq. (8.37), the left hand side of Eq. (8.35) is

$$\begin{aligned} \frac{{\partial \tilde{L}}}{{\partial p_{G} }} & = tx_{G}^{*} \left( {p_{G} } \right) + \frac{{tp_{G} }}{{C_{G}^{{\prime \prime }} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right)}} - \left( {t - \mu _{6} } \right)x_{G}^{*} \left( {p_{G} } \right) \\ & = \mu _{6} x_{G}^{*} \left( {p_{G} } \right) + \frac{{tp_{G} }}{{C_{G}^{{\prime \prime }} \left( {x_{G}^{*} \left( {p_{G} } \right)} \right)}} \\ & > 0. \\ \end{aligned} $$

(8.55)

This contradicts Eq. (8.35). Hence, $p_{G} = p_{B}$ holds. Accordingly, $s_{G} = s_{B}$ follows from (8.53).

Therefore, the minimization problem is reduced to the following minimization problem with respect to p, s:

$$ {\text{minimize }}t\left( {px_{G}^{*} (p) + s} \right) + (1 - t)\left( {px_{B}^{*} (p) + s} \right), $$

(8.56)

$$ {\text{subject to}}\,\pi_{B} \left( {x_{B}^{*} (p)} \right) + s - F \ge 0, $$

(8.57)

$$ p \ge 0, $$

(8.58)

$$ {\text{and}}\,s \ge 0. $$

(8.59)

The Lagrangian $\widehat{L}$ is

$$ \widehat{L} = t\left( {px_{G}^{*} (p) + s} \right) + (1 - t)\left( {px_{B}^{*} (p) + s} \right) - \xi_{1} \left[ {\pi_{B} \left( {x_{B}^{*} (p)} \right) + s - F} \right] - \xi_{2} p - \xi_{3} s, $$

(8.60)

where $\xi_{1} ,\xi_{2}$, and $\xi_{3}$ are Lagrange multipliers.

We now obtain the set of first-order conditions:

$$ \frac{{\partial \widehat{L}}}{\partial p} = tx_{G}^{*} (p) + \frac{tp}{{C_{G}^{{\prime \prime }} \left( {x_{G}^{*} (p)} \right)}} + (1 - t)x_{B}^{*} (p) + \frac{(1 - t)p}{{C_{B}^{{\prime \prime }} \left( {x_{B}^{*} (p)} \right)}} - \xi_{1} x_{B}^{*} (p) - \xi_{2} = 0, $$

(8.61)

$$ \frac{{\partial \widehat{L}}}{\partial s} = 1 - \xi_{1} - \xi_{3} = 0, $$

(8.62)

$$ \xi_{1} \left[ {\pi \left( {x_{B}^{*} (p)} \right) + s - F} \right] = 0, $$

(8.63)

$$ \xi_{2} p = 0, $$

(8.64)

$$ \xi_{3} s = 0, $$

(8.65)

$$ \pi \left( {x_{B}^{*} (p)} \right) + s - F \ge 0, $$

(8.66)

$$ p \ge 0,\,s \ge 0, $$

(8.67)

$$ {\text{and}}\,\xi_{1} ,\xi_{2} ,\xi_{3} \ge 0. $$

(8.68)

First, suppose $\xi_{1} = 0$. Then, $\xi_{3} = 1 > 0$ from Eq. (8.62). Hence, $s = 0$ from Eq. (8.65). Furthermore, $\xi_{2} > 0$ from Eq. (8.61). Thus, $p = 0$ from Eq. (8.64). However, $p = 0,\,s = 0$ does not satisfy (8.66) due to assumption (8.6). Therefore, $\xi_{1} > 0$ holds.

Accordingly, from Eq. (8.63),

$$ \pi \left( {x_{B}^{*} (p)} \right) + s - F = 0 $$

(8.69)

holds.

Next, suppose $\xi_{2} = 0$. Then, because $\xi_{1} = 1 - \xi_{3}$ from Eq. (8.62), the left hand side of Eq. (8.61) is

$$ \begin{aligned} \frac{{\partial \hat{L}}}{{\partial p}} & = tx_{G}^{*} (p) + \frac{{tp}}{{C_{G}^{{\prime \prime }} \left( {x_{G}^{*} (p)} \right)}} + (1 - t)x_{B}^{*} (p) + \frac{{(1 - t)p}}{{C_{B}^{{\prime \prime }} \left( {x_{B}^{*} (p)} \right)}} - \left( {1 - \xi _{3} } \right)x_{B}^{*} (p) \\ & = t\left( {x_{G}^{*} (p) - x_{B}^{*} (p)} \right) + \xi _{3} x_{B}^{*} (p) + \frac{{tp}}{{C_{G}^{{\prime \prime }} \left( {x_{G}^{*} (p)} \right)}} + \frac{{(1 - t)p}}{{C_{B}^{{\prime \prime }} \left( {x_{B}^{*} (p)} \right)}} \\ & > 0. \\ \end{aligned}$$

(8.70)

This contradicts Eq. (8.61). Hence, $\xi_{2} > 0$.

Accordingly, $p = 0$ from Eq. (8.64). Then, Eq. (8.69) yields $s = F - \pi \left( {x_{B}^{*} (0)} \right)$.