Welfare optimal reliability and reserve provision in electricity markets with increasing shares of renewable energy sources

Abstract

We develop an analytical model to derive the competitive market equilibrium for electricity spot and reserve markets under uncertain demand and renewable electricity generation. The first-best market equilibrium of the model resembles the result of a centralized co-optimization of the markets involved. We then derive the welfare-optimal provision of reserves. At first-best, cost of reserve capacity is balanced against expected cost of outages. Comparative statics and a numerical application based on German secondary reserve provision imply an increase of reserve provision with a growing share of renewable generation. Furthermore, a growing share of renewable generation decreases the level of reliability as measured in energy not served since required reserves to balance higher expected deviations will be more expensive, resulting in a trade-off between higher reserve costs and costs of energy not served.

Appendix: Proofs

1.1 Proposition 1

Proof

Evaluating the derivative (3.39) at the lowest possible value of positive reserves, i.e. zero, we get:

$$\begin{aligned} \left. \frac{\partial E\left[ S\right] }{\partial r^{+}}\right| _{r^{+}=0}= & {} \left( v-c\left( \overline{D}\right) \right) \left( 1-F_{D}\left( \overline{D}\right) \right) \nonumber \\&+ \beta \int _{\underline{D}}^{\overline{D}}\left[ c\left( \overline{D}\right) -c\left( K_{r}^{-}\right) \right] dF_{d}\nonumber \\&+\beta \left[ c\left( \overline{D}\right) -c\left( k_{0}^{-}\right) \right] F_{D}\left( \underline{D}\right) >0 \end{aligned}$$

(.1)

where the inequality follows from $v>c\left( \overline{D}\right)$, $c^{\prime }{>0 }$, $K_{r}^{-}\le \overline{D}$ and $k_{0}^{-}\le \overline{D}$.

Evaluating the derivative at the largest possible value of positive reserves,³⁷$r_{max}^{+}=\frac{\alpha }{1-\alpha }\overline{D}$ we get

$$\begin{aligned} \left. \frac{\partial E\left[ S\right] }{\partial r^{+}}\right| _{r^{+}=r_{max}^{+}}= & {} \left( v-c\left( \frac{\overline{D}}{1-\alpha }\right) \right) \left( 1-F_{d}\left( \frac{\overline{D}}{1-\alpha }\right) \right) \nonumber \\&- \left( 1-\alpha \right) \int _{\overline{D}}^{\frac{\overline{D}}{1-\alpha }}\left[ c\left( \frac{\overline{D}}{1-\alpha }\right) -c\left( K_{r}^{+}\right) \right] dF_{D}\nonumber \\&- \int _{\underline{D}}^{\overline{D}}\left[ \left( 1-\alpha -\beta \right) c\left( \frac{\overline{D}}{1-\alpha }\right) +\beta c\left( K_{r}^{-}\right) -\left( 1-\alpha \right) c\left( 0\right) \right] dF_{D}\nonumber \\&- \left[ \left( 1-\alpha -\beta \right) c\left( \frac{\overline{D}}{1-\alpha }\right) +\beta c\left( \overline{D}-r^{-}\right) -\left( 1-\alpha \right) c\left( 0\right) \right] F_{D}\left( \underline{D}\right) .\nonumber \\ \end{aligned}$$

(.2)

The last three terms are easily seen to be negative, since c is increasing. The first term is positive, but for reasonable distributions of D it is very small. In what follows we assume $\left. \frac{\partial E\left[ S\right] }{\partial r^{+}}\right| _{r^{+}=r_{max}^{+}}<0$, so that there is at least one solution to the first-order condition (3.40).

As for the second-order condition for maximum expected welfare, since $r^{-}$ is set and fixed at its maximum value, it suffices to show that the second derivative with respect to $r^{+}$ is negative. After some simplification we get

$$\begin{aligned} \frac{\partial ^{2}E\left[ S\right] }{\left( \partial r^{+}\right) ^{2}}= & {} -c'\left( k_{m}\right) \left( 1-F\left( k_{m}\right) \right) -\left( v-c\left( k_{m}\right) \right) f\left( k_{m}\right) \nonumber \\&- \left( 1-\alpha \right) \int _{\overline{D}}^{k_{m}}\left[ c'\left( k_{m}\right) +\frac{1-\alpha }{\alpha }c'\left( K_{r}^{+}\right) \right] dF\nonumber \\&- \int _{\underline{D}}^{\overline{D}}\left[ \left( 1-\alpha -\beta \right) c'\left( k_{m}\right) +\beta c'\left( K_{r}^{-}\right) +\frac{\left( 1-\alpha \right) ^{2}}{\alpha }c'\left( k_{0}^{+}\right) \right] dF\nonumber \\&- \left[ \left( 1-\alpha -\beta \right) c'\left( k_{m}\right) +\beta c'\left( k_{0}^{-}\right) +\frac{\left( 1-\alpha \right) ^{2}}{\alpha }c'\left( k_{0}^{+}\right) \right] F\left( \underline{D}\right) , \end{aligned}$$

(.3)

where we omit the subscript D on the distribution function $F_{D}$ and probability density $f_{D}$ in order to reduce clutter. Since $c^{\prime }{>0, v>c\left( k_{m}\right) }$ and $1-\alpha -\beta >0$, we get $\frac{\partial ^{2}E\left[ S\right] }{\left( \partial r^{+}\right) ^{2}}<0$. $\square$

1.2 Proposition 2

Proof

First, if $\tau$ is a parameter (e.g., v, $\alpha$, $\beta$), then, differentiating through the first-order condition (3.40) with respect to that parameter and rearranging yields:

$$\begin{aligned} \frac{d\hat{r}^{+}}{d\tau }=\frac{\frac{\partial ^{2}E\left[ S\right] }{\partial \tau \partial r^{+}}}{-\frac{\partial ^{2}E\left[ S\right] }{\left( \partial r^{+}\right) ^{2}}}, \end{aligned}$$

(.4)

where the right-hand side is evaluated at $\hat{r}^{+}$. Since the denominator is positive (by virtue of the second-order condition for maximum), the sign of the numerator will also provide the sign of the derivative of $\hat{r}^{+}$ with respect to the parameter.

In the remainder of this proof, to reduce clutter, we shall omit the ‘hat’ on the optimal value of positive reserves and simply write $r^{+}$ for the optimal value.

First, differentiating (3.39) with respect to v we get

$$\begin{aligned} \frac{\partial ^{2}E\left[ S\right] }{\partial v\partial r^{+}}=1-F\left( k_{m}\right) >0, \end{aligned}$$

(.5)

which establishes (a).

To prove (b), we differentiate (3.39) with respect to $\alpha$ and get

$$\begin{aligned} \frac{\partial ^{2}E\left[ S\right] }{\partial \alpha \partial r^{+}}= & {} -\int _{\overline{D}}^{k_{m}}\left[ -c\left( k_{m}\right) +c\left( K_{r}^{+}\right) -c'\left( K_{r}^{+}\right) \left( -\frac{1}{\alpha ^{2}}\left\{ D-k_{m}\right\} \right) \right] dF\nonumber \\&- \int _{\underline{D}}^{\overline{D}}\left[ -c\left( k_{m}\right) +c\left( k_{0}^{+}\right) -\left( 1-\alpha \right) c'\left( k_{0}^{+}\right) \frac{1}{\alpha ^{2}}r^{+}\right] dF\nonumber \\&- \left[ -c\left( k_{m}\right) +c\left( k_{0}^{+}\right) -\left( 1-\alpha \right) c'\left( k_{0}^{+}\right) \frac{1}{\alpha ^{2}}r^{+}\right] F\left( \underline{D}\right) , \end{aligned}$$

(.6)

where we have used $\frac{dK_{r}^{+}}{d\alpha }=-\frac{1}{\alpha ^{2}}\left[ D-k_{m}\right]$, $\frac{dk_{0}^{+}}{d\alpha }=\frac{1}{\alpha ^{2}}r^{+}$, $\frac{dK_{r}^{-}}{d\alpha }=0$ and $\frac{dk_{0}^{-}}{d\alpha }=0$. Since $c'>0$, $k_{m}\ge K_{r}^{+}$, $k_{m}\ge k_{0}^{+}$ all the integrands are found to be negative. Thus, the right-hand side as a whole is positive and (b) is confirmed.

To see that (c) holds we differentiate (3.39) with respect to $\beta$ and get

$$\begin{aligned} \frac{\partial ^{2}E\left[ S\right] }{\partial \beta \partial r^{+}}= & {} \int _{\underline{D}}^{\overline{D}}\left[ c\left( k_{m}\right) -c\left( K_{r}^{-}\right) -c'\left( K_{r}^{-}\right) \frac{1}{\beta }\left( \overline{D}-D\right) \right] dF\nonumber \\&+ \left[ c\left( k_{m}\right) -c'\left( k_{0}^{-}\right) -c'\left( k_{0}^{-}\right) \frac{1}{\beta }r^{-}\right] F\left( \underline{D}\right) , \end{aligned}$$

(.7)

where we have used $\frac{dK_{r}^{+}}{d\beta }=0$, $\frac{dK_{r}^{-}}{d\beta }=\frac{1}{\beta }\left[ \overline{D}-D\right]$ and $\frac{dk_{0}^{-}}{d\beta }=\frac{1}{\beta }r^{-}$. Note that both expressions inside the square brackets are of the form

$$\begin{aligned} \left\{ c\left( D_{1}\right) -c\left( D_{0}\right) \right\} -c'\left( D_{0}\right) \left\{ D_{1}-D_{0}\right\} \end{aligned}$$

(.8)

with $D_{0}<D_{1}$. This expression is strictly positive if c is strictly convex, zero if c is linear, and negative if c is concave. Hence, (c) is established.

As for (d), assume, for simplification of presentation, but without loss of generality, that the variance of D is equal to 1. Introduce a mean-preserving perturbation of the variance by introducing a new distribution, viz.

$$\begin{aligned} F_{\sigma }\left( D\right) =F_{D}\left( \frac{D-\overline{D}}{\sigma }+\overline{D}\right) \end{aligned}$$

(.9)

Here it is crucial that we assume D to be an unbiased prediction of d so $E\left[ D\right] =\overline{D}$ under the original distribution $F_{D}$.

Under $F_{\sigma }$ the expectation of D is still $\overline{D}$, but its variance is $\sigma ^{2}$. Obviously, the probability density of $F_{\sigma }$ is $f_{\sigma }\left( D\right) =\frac{1}{\sigma }f_{d}\left( \frac{D-\overline{D}}{\sigma }+\overline{D}\right)$ and its derivative is $f_{\sigma }'\left( D\right) =\frac{1}{\sigma ^{2}}f_{d}'\left( \frac{D-\overline{D}}{\sigma }+\overline{D}\right)$.

The derivatives of $F_{\sigma }$ and $f_{\sigma }$ with respect to $\sigma$ are given by

$$\begin{aligned} \frac{\partial }{\partial \sigma }F_{\sigma }\left( D\right)= & {} -\frac{D-\overline{D}}{\sigma }f_{\sigma }\left( D\right) \end{aligned}$$

(.10)

$$\begin{aligned} \frac{\partial }{\partial \sigma }f_{\sigma }\left( D\right)= & {} -\frac{1}{\sigma }f_{\sigma }\left( D\right) -\frac{D-\overline{D}}{\sigma }f_{\sigma }'\left( D\right) . \end{aligned}$$

(.11)

A lengthy calculation, employing integration by parts of integrals involving $f_{\sigma }'\left( D\right)$, leads, after much simplification, to the expression

$$\begin{aligned} \frac{\partial ^{2}E\left[ S\right] }{\partial \sigma \partial r^{+}}= & {} \left[ v-c\left( k_{m}\right) \right] \frac{r^{+}}{\sigma }f_{\sigma }\left( k_{m}\right) \nonumber \\&+ \frac{1-\alpha }{\alpha \sigma }\int _{\overline{D}}^{k_{m}}c'\left( k_{r}^{+}\right) \left( D-\overline{D}\right) dF_{\sigma }\nonumber \\&+ \frac{1}{\sigma }\int _{\underline{D}}^{\overline{D}}c'\left( K_{r}^{-}\right) \left( D-\overline{D}\right) dF_{\sigma }. \end{aligned}$$

(.12)

All three terms of the right-hand side are positive and (d) follows.

To see (e) introduce a new marginal cost function

$$\begin{aligned} c_{a}=a+c\left( D\right) \end{aligned}$$

(.13)

where a is a non-negative constant. Clearly, $c_{a}$ satisfies all conditions imposed on c. Moreover, a drops out of all but the first term of (3.39) so differentiating with respect to a yields

$$\begin{aligned} \frac{\partial ^{2}E\left[ S\right] }{\partial a\partial r^{+}}=-\left[ 1-F\left( k_{m}\right) \right] <0, \end{aligned}$$

(.14)

which implies that (e) holds.

Finally, introduce a multiplicative shift of c by defining

$$\begin{aligned} c_{b}\left( D\right) =c\left( 0\right) +b\left[ c\left( D\right) -c\left( 0\right) \right] . \end{aligned}$$

(.15)

Employing $c_{b}$ as a new marginal cost function, differentiating (3.39) with respect to b and using the first-order condition (3.40) for the optimal value of positive reserves (this is the only place in this proof that we rely on the fact that we evaluate the derivatives calculated at that value) we get

$$\begin{aligned} \frac{\partial ^{2}E\left[ S\right] }{\partial b\partial r^{+}}=-v\left[ 1-F\left( k_{m}\right) \right] <0, \end{aligned}$$

(.16)

which establishes (f). $\square$