Optimal control of electricity input given an uncertain demand

Abstract

We consider the problem of determining an optimal strategy for electricity injection that faces an uncertain power demand stream. This demand stream is modeled via an Ornstein–Uhlenbeck process with an additional jump component, whereas the power flow is represented by the linear transport equation. We analytically determine the optimal amount of power supply for different levels of available information and compare the results to each other. For numerical purposes, we reformulate the original problem in terms of the cost function such that classical optimization solvers can be directly applied. The computational results are illustrated for different scenarios.

6 Appendix

The detailed calculation of the second moment of the JDP used in Sect. 4.1 is as follows:

$$\begin{aligned} {\mathbb {E}}\left[ Y_t^2\right]&= {\mathbb {E}}\left[ \left( e^{-\kappa t}y_0 + \sigma \int _{0}^{t} e^{-\kappa (t-s)} dW_s + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds + \sum _{i=1}^{N_t} \gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2\right] \nonumber \\&= {{\mathbb {E}}\left[ \underbrace{\left( e^{-\kappa t}y_0 + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds\right) ^2}_{=A} + \underbrace{\left( \sigma \int _{0}^{t} e^{-\kappa (t-s)} dW_s + \sum _{i=1}^{N_t} \gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2}_{=B}\right. }\nonumber \\&\quad {\left. +\,\underbrace{2\left( e^{-\kappa t}y_0 + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds\right) \cdot \left( \sigma \int _{0}^{t} e^{-\kappa (t-s)} dW_s + \sum _{i=1}^{N_t} \gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) }_{=C} \right] } \end{aligned}$$

For a better readability, we evaluate A, B, and C in expectation separately:

$$\begin{aligned} {\bar{A} = {\mathbb {E}}[A]}&= \left( e^{-\kappa t}y_0 + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds\right) ^2\\&=y_0^2e^{-2\kappa t} + 2y_0e^{-\kappa t} \int _{0}^{t}e^{-\kappa (t-s)}\kappa \mu (s) ds + \left( \kappa \int _{0}^{t}e^{-\kappa (t-s)} \mu (s) ds\right) ^2. \end{aligned}$$

For the diffusion- and jump-driven quadratic term B, we calculate

$$\begin{aligned} {\bar{B} = {\mathbb {E}}[B]}&= {\mathbb {E}}\left[ \left( \sigma \int _{0}^{t} e^{-\kappa (t-s)} dW_s\right) ^2\right] +{\mathbb {E}}\left[ \left( \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2\right] \\&\quad +\,2{\mathbb {E}}\left[ \sigma \int _{0}^{t} e^{-\kappa (t-s)} dW_s\sum _{i=1}^{N_t} \gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right] \\&= {\mathbb {E}}\left[ \sigma ^2 \int _{0}^{t} e^{-2 \kappa (t-s)} ds\right] +{\mathbb {E}}\left[ \left( \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2\right] \\&\quad +\,2{\mathbb {E}}\left[ \sigma \int _{0}^{t} e^{-\kappa (t-s)} dW_s\right] {\mathbb {E}}\left[ \sum _{i=1}^{N_t} \gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right] \\&= \sigma ^2 \left[ \frac{e^{-2 \kappa (t-s)}}{2\kappa }\right] _{s=0}^{s=t} + {\mathbb {E}}\left[ \left( \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2\right] \\&= \frac{\sigma ^2}{2\kappa }\left( 1-e^{-2 \kappa t}\right) + {\mathbb {E}}\left[ \left( \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2\right] . \end{aligned}$$

Note that the expectation of the integral with respect to the Brownian motion is zero. It remains to calculate the expectation of the mixed summand C:

$$\begin{aligned} {\bar{C} = {\mathbb {E}}[C]}&= 2\left( e^{-\kappa t}y_0 + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds\right) \cdot {\mathbb {E}}\left[ \sigma \int _{0}^{t} e^{-\kappa (t-s)} dW_s + \sum _{i=1}^{N_t} \gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right] \\&= 2\left( e^{-\kappa t}y_0 + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds\right) \cdot {\mathbb {E}}\left[ \sigma \int _{0}^{t} e^{-\kappa (t-s)} dW_s\right] \\&\quad +\,2\left( e^{-\kappa t}y_0 + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds\right) \cdot {\mathbb {E}}\left[ \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right] \\&= 2\left( e^{-\kappa t}y_0 + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds\right) \cdot {\mathbb {E}}\left[ \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right] \\&= 2\left( e^{-\kappa t}y_0 + \kappa \int _{0}^{t} e^{-\kappa (t-s)}\mu (s) ds\right) \cdot {\bar{\gamma }} \frac{\nu }{\kappa }\left( 1-e^{-\kappa t}\right) . \end{aligned}$$

Again the expectation of the integral with respect to the Brownian motion vanishes. Note that the second moment of the time-dependent OUP is obtained by setting $\gamma _{\tau _i} \equiv 0$ for all jump times $\tau _i$.

Thus, it remains to calculate ${\mathbb {E}}\left[ \left( \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2\right]$. Note that ${\mathbb {E}}\left[ \gamma \right] ^2=\bar{\gamma }^2$.

$$\begin{aligned} {\mathbb {E}}\left[ \left( \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2\right]&= {\mathbb {E}}\left[ \sum _{i=1}^{N_t}\gamma _{\tau _i}^2 e^{-2\kappa (t-\tau _i)} + \sum _{i\ne j}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\gamma _{\tau _j} e^{-\kappa (t-\tau _j)}\right] \\&= {\mathbb {E}}\left[ \sum _{i=1}^{N_t} e^{-2\kappa (t-\tau _i)}\right] \bar{\gamma }_2+ {\mathbb {E}}\left[ \sum _{i\ne j}^{N_t} e^{-\kappa (t-\tau _i)}e^{-\kappa (t-\tau _j)}\right] \bar{\gamma }^2\\&= {\mathbb {E}}\left[ \sum _{i=1}^{N_t} e^{-2\kappa (t-\tau _i)}\right] \bar{\gamma }_2+ 2\cdot {\mathbb {E}}\left[ {\mathbb {E}}\left[ \sum _{i=2}^{N_t} \sum _{j=1}^{i-1} e^{-\kappa (t-\tau _i)}e^{-\kappa (t-\tau _j)}|N_t\right] \right] \bar{\gamma }^2. \end{aligned}$$

We know from (Mikosch 2009, Prop. 2. 1. 16) that $\tau _i \sim {\mathcal {U}}[0,t]$. Thus, we calculate

$$\begin{aligned} {\mathbb {E}}\left[ e^{-2\kappa (t-\tau _i)}\right]&= \int _{0}^{t}e^{-2\kappa (t-s)}\frac{1}{t}ds = \frac{1-e^{-2\kappa t}}{2\kappa t}. \end{aligned}$$

We can then deduce

$$\begin{aligned} {\mathbb {E}}\left[ \sum _{i=1}^{N_t} e^{-2\kappa (t-\tau _i)}\right]&= \sum _{i=1}^{\infty } i \cdot e^{-\nu t} \frac{(\nu t)^{i}}{i !} \frac{1-e^{-2\kappa t}}{2\kappa t} = \nu \cdot \frac{(1-e^{-2\kappa t})}{2\kappa }. \end{aligned}$$

It remains to compute the mixed-term expectation.

$$\begin{aligned} {\mathbb {E}}\left[ e^{-\kappa (t-\tau _i)}e^{-\kappa (t-\tau _j)}\right| \tau _j<\tau _i, N_t=n]&= \int _{0}^{t}e^{-\kappa (t-s)} \frac{1}{t} \int _{0}^{s} e^{-\kappa (t-u)}\frac{2}{t} du ds \\&= \int _{0}^{t} \frac{2\cdot (e^{-2\kappa (t-s)}-e^{-\kappa (2t-s)})}{\kappa t^2} ds \\&= \frac{1+e^{-2\kappa t}-2e^{-\kappa t}}{\kappa ^2 t^2}. \end{aligned}$$

Thus, we have

$$\begin{aligned} {\mathbb {E}}\left[ \sum _{i=2}^{N_t} \sum _{j=1}^{i-1} e^{-\kappa (t-\tau _i)}e^{-\kappa (t-\tau _j)}|N_t=n\right]&= \sum _{i=2}^{n} \sum _{j=1}^{i-1} {\mathbb {E}}\left[ e^{-\kappa (t-\tau _i)}e^{-\kappa (t-\tau _j)}|N_t=n\right] \\&= \sum _{i=1}^{n-1} i \cdot {\mathbb {E}}\left[ e^{-\kappa (t-\tau _{i+1})}e^{-\kappa (t-t_1)}|N_t=n\right] \\&= {\mathbb {E}}\left[ e^{-\kappa (t-t_{2})}e^{-\kappa (t-t_1)}|N_t=n\right] \cdot \frac{n \cdot (n-1)}{2} \\&= \frac{1+e^{-2\kappa t}-2e^{-\kappa t}}{\kappa ^2 t^2} \cdot \frac{n \cdot (n-1)}{2}.\\ {\mathbb {E}}\left[ {\mathbb {E}}\left[ \sum _{i=2}^{N_t} \sum _{j=1}^{i-1} e^{-\kappa (t-\tau _i)}e^{-\kappa (t-\tau _j)}|N_t\right] \right]&= {\mathbb {E}}\left[ N_t^2-N_t\right] \cdot \frac{1+e^{-2\kappa t}-2e^{-\kappa t}}{2 \kappa ^2 t^2} \\&= \nu ^2 \cdot \frac{1+e^{-2\kappa t}-2e^{-\kappa t}}{2 \kappa ^2}. \end{aligned}$$

Finally, the closed-form expression is

$$\begin{aligned} {\mathbb {E}}\left[ \left( \sum _{i=1}^{N_t}\gamma _{\tau _i} e^{-\kappa (t-\tau _i)}\right) ^2\right]&= \nu \cdot \frac{(1-e^{-2\kappa t})}{2\kappa } \cdot \bar{\gamma }_2+ \nu ^2 \cdot \frac{1+e^{-2\kappa t}-2e^{-\kappa t}}{\kappa ^2} \cdot \bar{\gamma }^2. \end{aligned}$$