Decentralized Charging Coordination of Large-Population PEVs Under a Hierarchical Structure

Abstract

Centralized or decentralized charging schedules of large-scale PEVs coordinated by a system operator usually require significant management, computation and communication capabilities on the system. Alternatively in this chapter, it constructs a hierarchical model for the PEV charging coordination problems where a collection of agents are introduced between the system operator and individual vehicles, and proposes an off-line decentralized method for the constructed hierarchical optimization problems. Under the decentralized method, each PEV implements its best behavior with respect to a given local charging price curve set by its agent. Each agent submits the collected aggregated charging behaviors under this agent to the system operator who then updates the electricity generation price and broadcasts it to PEVs via agents. To reimburse the transaction operation costs on agents, the charging price on the PEVs under an agent comprises both the generation price broadcasted from the system operator and the operation price set by this agent. It is shown that, under certain conditions, the proposed dynamical procedure converges to the efficient (or socially optimal) solution. The proposed method under the hierarchical structure presents the advantage of the autonomy of the individual PEVs and the low computation and communication capability requirements on the system.

4.6    Appendices

4.1.1 4.6.1    Proof of Lemma 4.1

To solve the optimization problem, a Lagrangian function is defined as follow $L_{nm}\left( \varvec{u}_n^m;\varvec{\rho }_m\right) \triangleq J_{nm}\left( \varvec{u}_n^m;\varvec{\rho }_m\right) + A_m\left( \Gamma _{nm}-\displaystyle \sum _{t\in \mathscr {T}}\varvec{u}_n^m\right)$, where $A_m$ is Lagrange multiplier. Since $J_{nm}\left( \varvec{u}_n^m;\varvec{\rho }_m\right)$ is convex with respect to $\varvec{u}_n^m$ due to the convexity of $f_{nm}$, the charging control that minimizes $J_{nm}\left( \varvec{u}_n^m;\varvec{\rho }_m\right)$, subject to $\varvec{u}_n^m\in \mathscr {U}_n^m$, must satisfy,

• (I). $\displaystyle \frac{\partial {L_{nm}}}{\partial {A}_m} = 0$,

• (II). $\displaystyle \frac{\partial {L_{nm}}}{\partial {u_{nt}^m}}\le {0}, u_{nt}^m\ge {0}$, with complementary slackness.

Condition (I) contains the constraint $\displaystyle \sum _{t\in \mathscr {T}}\left( \varvec{u}_n^m\right) =\Gamma _{nm}$. It can be derived from condition (II) that

$$\begin{aligned} \left[ f_{nm}'\right] ^{-1}\left( u_{nt}^m\right) + \rho _{mt} - A_m \left\{ \begin{array}{ll} = 0, &{} \text {in case}\,\,u_{nt}^m > 0, \\ < 0, &{} { otherwise}, \end{array} \right. \end{aligned}$$

which is equivalent to (4.8).

The dependence of $u_{nt}^m\left( \varvec{\rho }_{m}, A_m\right)$ on $A_m$ expressed in (4.8) ensures that, for any fixed $\varvec{\rho }_{m}$,

• There exists an $A_{m-}$ such that for $A_m \le A_{m-}$, $\displaystyle \sum _{t\in \mathscr {T}}u_{nt}^m\left( \varvec{\rho }_{m},A_m\right) = 0$.

• For $A_m > A_{m-}$, $\displaystyle \sum _{t\in \mathscr {T}}u_{nt}^m\left( \varvec{\rho }_{m},A_m\right)$ is strictly increasing with $A_m$, with the relationship continuous, though not smooth. Hence, $\displaystyle \sum _{t\in \mathscr {T}}u_{nt}^m\left( \varvec{\rho }_{m},A_m\right)$ is invertible over this domain.

Therefore a constraint $\displaystyle \sum _{t\in \mathscr {T}}u_{nt}^m\left( \varvec{\rho }_{m},A_m\right) = K > 0$ defines a unique $A_m > A_{m-}$ for each fixed $\varvec{\rho }_m$, which may be written as $A_m\left( \varvec{\rho }_m\right)$. The particular value of $A_m$ that ensures satisfaction of the constraint $\displaystyle \sum _{t\in \mathscr {T}}u_{nt}^m = \Gamma _{nm}$ shall be denoted by $A_m^{*}\left( \varvec{\rho }_m\right)$. The resulting control trajectory can be written $\varvec{u}_n^m\left( \varvec{\rho }_m,A_m^{*}\left( \varvec{\rho }_m\right) \right) = \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right)$.

Since $J_{nm}\left( \varvec{u}_n^m;\varvec{\rho }_m\right)$ is convex with respect to $\varvec{u}_n^m$, the minimizing control defined by (4.8) must be unique.    $\blacksquare$

4.1.2 4.6.2    Proof of Lemma 4.2

Recall that $\varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) \triangleq \underset{\varvec{u}_n^m\in \,\mathscr {U}_n^m}{\text {argmin}}\left\{ J_{nm}\left( \varvec{u}_n^m;\varvec{\rho }_m\right) \right\}$, it means that $\varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right)$ represents the charging behavior that minimizes the individual cost function (4.6) with respect to a given price curve $\varvec{\rho }_m$.

By the same token, $\varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) \in \mathscr {U}_n^m$ represents the best response of PEV n with respect to $\widehat{\varvec{\rho }}_m$, such that $\varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) \triangleq \underset{\varvec{u}_n \in \,\mathscr {U}_n}{\text {min}} \, J_{n}^m\left( \varvec{u}_n^m; \widehat{\varvec{\rho }}_m\right)$. Then, by Lemma 4.1, it obtains that $\varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) = \varvec{u}_{n}^m\left( \varvec{\rho }_m, A_m\left( \varvec{\rho }_m\right) \right)$ and $\varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) = \varvec{u}_{n}^m\left( \widehat{\varvec{\rho }}_m, A_m\left( \widehat{\varvec{\rho }}_m\right) \right)$.

Another behavior $\varvec{v}_{n}^m\left( \varvec{\rho }_m,\widehat{\varvec{\rho }}_m\right)$ is defined as follow

$$\begin{aligned}&v_{nt}^m\left( \varvec{\rho }_m,\widehat{\varvec{\rho }}_m\right) = \left\{ \begin{array}{ll} u_{nt}^m\left( \widehat{\varvec{\rho }}_m,A_m^{*}\left( \varvec{\rho }_m\right) \right) , &{} \text {in case}\,\,t\in \mathscr {T}, \\ 0, &{} { otherwise} \end{array} \right. \end{aligned}$$

(4.21)

Remark: As specified in (4.21), the charging behavior $v_{nt}^m\left( \varvec{\rho }_m,\widehat{\varvec{\rho }}_m\right)$ describes an individual charging control satisfying (4.8) with respect to $\widehat{\varvec{\rho }}_m$ and $A_m^{*}\left( \varvec{\rho }_m\right)$. There is no guarantee that $v_{nt}^m\left( \varvec{\rho }_m,\widehat{\varvec{\rho }}_m\right)$ is an admissible charging behavior, such that $\sum v_{nt}^m\left( \varvec{\rho }_m,\widehat{\varvec{\rho }}_m\right) = \Gamma _{nm}$.

The inequality (4.14) will be testified in (I) and (II) below.

• (I) It will be show that:

  $$\begin{aligned}&\left| u_{nt}^{m,*}\left( \varvec{\rho }_m\right) - v_{nt}^m\left( \varvec{\rho }_m,\widehat{\varvec{\rho }}_m\right) \right| \nonumber \\ \le&\Big |\left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \rho _{mt}\right) - \left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \widehat{\rho }_{mt}\right) \Big |. \end{aligned}$$

  (4.22)

  Obviously, (4.22) holds for $t\not \in \mathscr {T}_n^m$, since $u_{nt}^{m,*}\left( \varvec{\rho }_m\right) = v_{nt}^m\left( \varvec{\rho }_m, \widehat{\varvec{\rho }}_m\right) = 0$ It is shown that (4.22) holds for all $t \in \mathscr {T}_n^m$. Note: For notational simplicity, $\varvec{v}_n^m\equiv \varvec{v}_{n}^m\left( \varvec{\rho }_m,\widehat{\varvec{\rho }}_m\right)$ will be adopted throughout the following proof.

  • (I.a)   $v_{nt}^m = u_{nt}^{m,*} = 0$. It follows immediately that $v_{nt}^m - u_{nt}^{m,*} =0$

  • (I.b)   $v_{nt}^m > 0$ and $u_{nt}^{m,*} = 0$. By (4.8), $v_{nt}^m > 0$ implies $\left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \widehat{\rho }_{mt}\right) > 0$, and $u_{nt}^{m,*} = 0$ implies $\left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \rho _{mt}\right) \le 0$. Together these give:

    $$\begin{aligned} 0&< v_{nt}^m - u_{nt}^{m,*}\\&\le \left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \widehat{\rho }_{mt}\right) - \left[ f_{nm}'\right] ^{-1}\left( A^{*}\left( \varvec{\rho }_m\right) - \rho _{mt}\right) . \end{aligned}$$

  • (I.c)   $v_{nt}^m = 0$ and $u_{nt}^{m,*} > 0$. Similar to (I.b),

    $$\begin{aligned} 0&< u_{nt}^{m,*} - v_{nt}^m\\&\le \left[ f_{nm}'\right] ^{-1}\left( A^{*}\left( \varvec{\rho }_m\right) - \rho _{mt}\right) - \left[ f_{nm}'\right] ^{-1}\left( A^{*}\left( \varvec{\rho }_m\right) - \widehat{\rho }_{mt}\right) . \end{aligned}$$

  • (I.d)   $v_{nt}^m > 0$ and $u_{nt}^{m,*} > 0$. By (4.8) the following holds:

    $$\begin{aligned}&\left| v_{nt}^m - n_{nt}^{m,*}\right| \\ =&\Big |\min \left\{ \gamma _{nm}, \left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \widehat{\rho }_{mt}\right) \right\} \\&\quad - \min \left\{ \gamma _{nm}, \left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \rho _{mt}\right) \right\} \Big | \\ \le&\Big |\left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \widehat{\rho }_{mt}\right) - \left[ f_{nm}'\right] ^{-1}\left( A_m^{*}\left( \varvec{\rho }_m\right) - \rho _{mt}\right) \Big | \end{aligned}$$

• (II)   The following inequality below will be verified:

  $$\begin{aligned}&\left\| \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) - \varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) \right\| _{1} \le 2 \left\| \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) - \varvec{v}_{n}^m\left( \varvec{\rho }_m,\widehat{\varvec{\rho }}_m\right) \right\| _{1}. \end{aligned}$$

  (4.23)

  For notational simplicity, $\sum \varvec{u}_n^m \equiv \displaystyle \sum _{t\in \mathscr {T}}u_{nt}^m$ is applied in the following (II.a) to (II.c).

  • (II.a) $\sum \varvec{v}_n^m\ = \sum \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right)$. This equality ensures $\varvec{v}_n^m \in \mathscr {U}_n^m$. Also, charging control $\varvec{v}_n^m$ has the form (4.8) with $A_m = A_m\left( \varvec{\rho }_m\right)$. Therefor, by Lemma 4.1, $\varvec{v}_n^m$ is the local optimal control with respect to $\widehat{\varvec{\rho }}_m$. Hence, $\varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) = \varvec{v}_n^m$. It follows that:

    $$\begin{aligned}&\left\| \varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) - \varvec{u}_n^m\left( \varvec{\rho }_m\right) \right\| _{1} = \left\| \varvec{v}_n^m - \varvec{u}_n^m\left( \varvec{\rho }_m\right) \right\| _{1}\le 2\left\| \varvec{v}_n^m - \varvec{u}_n^m\left( \varvec{\rho }_m\right) \right\| _{1}. \end{aligned}$$

  • (II.b) $\sum \varvec{v}_n^m <\sum \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right)$. By (4.1) and the definition of $\varvec{u}_n^m\left( \widehat{\varvec{\rho }}_m\right)$, $\sum \varvec{u}_n^m\left( \varvec{\rho }_m\right) = \sum \varvec{u}_n^m\left( \widehat{\varvec{\rho }}_m\right) = \Gamma _{nm}$. Therefor $\sum \varvec{v}_n^m < \sum \varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right)$, together with (4.8) and the definition of $\varvec{v}_n^m$ and $\varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right)$ implies,

    $$\begin{aligned} A_m^{*}\left( \widehat{\varvec{\rho }}_m\right)> A_m^{*}\left( \varvec{\rho }_m\right) , u_{nt}^{m,*}\left( \widehat{\varvec{\rho }}_m\right) > v_{nt}^m, \text { for all } t, \end{aligned}$$

    Hence,

    $$\begin{aligned} 0&\le \left\| \varvec{v}_n^m - \varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) \right\| _{1}\\&= \sum \varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) - \sum \varvec{v}_n^m = \sum \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) - \sum \varvec{v}_n^m \\&\le \left\| \varvec{v}_n^{m} - \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) \right\| _{1} \end{aligned}$$

    where the last inequality is a consequence of the triangle inequality for norms, taking into account that $\sum (\cdot ) = \Vert \cdot \Vert _{1}$ for all admissible non-negative control trajectories; then:

    $$\begin{aligned}&\left\| \varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) - \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) \right\| _{1} \\ \le&\left\| \varvec{v}_n^{m} - \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) \right\| _{1} + \left\| \varvec{v}_n^{m} - \varvec{u}_n^{m,*}\left( \widehat{\varvec{\rho }}_m\right) \right\| _{1} \\ \le&2 \left\| \varvec{v}_n^{m} - \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right) \right\| _{1} \end{aligned}$$

  • (II.c) $\sum \varvec{v}_n^m > \sum \varvec{u}_n^{m,*}\left( \varvec{\rho }_m\right)$. A similar argument to (II.b) can be used to show the inequality above holds in this case.

Together with (I) and (II), the conclusion of (4.14) in Lemma 4.2 can be obtained.   $\blacksquare$