Intervene in advance or passively? Analysis and application on congestion control of smart grid

Abstract

This paper models a frequently encountered problem regarding optimal control and queuing. When the arriving and leaving of queuer are predicted, congestion is forecasted, the timing of intervene and control can really improve the effectiveness. A practical case of its application in industry is for the congestion management of smart grid, designed for which, modeling and solving the relevant optimal stopping problem refine the management to be even smarter, especially for the decision of whether intervene in advance or until the congestion happens, towards an ultimate reduction of congestion time. This model has no assumption expect letting the arriving time with a period time follows a uniform distribution, and has no parameters expect the congestion threshold. This conciseness makes this study applicable for very general cases and even for similar situations in other topics, and the findings illustrate deep wisdom of management especially for queueing control.

Appendices

Appendix A: Expression of function H

For the function H defined by (3.1) that

$$\begin{aligned} H^n(t,i,j):=\mathop {\hbox \mathrm{IE}}\nolimits \left[ \int _t^{t+T/n}\mathbf{1}_{(N_s\ge K)ds}\,|\,N_t=i,F_t=j\right] , \end{aligned}$$

we have its expression as follows,

• (1) $i<K-1$: $H^n(t,i,j)=0$.

• (2)

  • a) $i=K-1$: $H^n(t,K-1,n-K+1)=0$;

  • b) $j<n-k+1$:

    $$\begin{aligned} H^n(t,k-1,j)=\frac{(n-k+1-j)(\exp (-A_1T/n)+A_1T/n-1)}{(T-t)A_1^2}, \end{aligned}$$

    where $A_1:=(K-1)/t+(n-K+1-j)/(T-t)$.

• (3) $i=K$:

  $$\begin{aligned} H^n(t,K,j)=&\frac{T}{n}\left( e^{-B_1{T}/{n}}+\frac{(1-\exp (-B_1T/n))(n-K-j)}{(T-t)T/n}\right) \\&+\frac{(1-\exp (-B_1T/n)(B_1T/n+1))K}{tB_1^2}, \end{aligned}$$

  where $B_1:=K/t+(n-K-j)/(T-t)$.

• (4) $i\le K+1$: $H^n(t,i,j)=t+O({1}/{n^2})$.

Appendix B: Expression of function G

Here we proceed to prove the Proposition 3.2. Given $r\in [0,T]$, $(i,j)\in {{\mathcal {M}}}_2$, $N_r=i$, $F_r=j$, denote by p(r, i, j, s, m) the probability of the event $N_s=m$ for any $s\in (r,T]$ and $m\in {{\mathcal {M}}}$. Hence $P(N_s\ge K\,|\,N_r=i,F_r=j)=\sum _{m=K}^{n-j}p(r,i,j,s,m)$ and

$$\begin{aligned}&p(r,i,j,s,m)=P(N_s=m\,|\,N_r=i,F_r=j)\nonumber \\&\quad =\sum _{\begin{array}{c} d_1\ge 0\vee i-m\\ d_1\le i\wedge (n-j-m) \end{array}} \left( {\begin{array}{c}d_1\\ i\end{array}}\right) \frac{(s-r\vee L)^{d_1}(r-s+r\vee L)^{i-d_1}}{r^i}\nonumber \\&\qquad \left( {\begin{array}{c}d_2\\ n-i-j\end{array}}\right) \frac{(L\vee (s-r))^{d_2}(T-r-L\vee (s-r))^{n-i-j-d_2}}{(T-r)^{n-i-j}}\,\Big |_{d_2=d_1+m-i}\nonumber \\&\quad =\sum _{\begin{array}{c} d_1\ge 0\vee i-m\\ d_1\le i\wedge (n-j-m) \end{array}} \left( {\begin{array}{c}d_1\\ i\end{array}}\right) \frac{(s-r\vee L)^{d_1}(r-s+r\vee L)^{i-d_1}}{r^i}\nonumber \\&\qquad \left( {\begin{array}{c}d_1+m-i\\ n-i-j\end{array}}\right) \frac{(L\vee (s-r))^{d_1+m-i}(T-r-L\vee (s-r))^{n-j-d_1-m}}{(T-r)^{n-i-j}}. \end{aligned}$$

(5.1)

By definition (2.4) of G, and use the notation that $\hat{t}=(t+\eta )\wedge T$, we express the function G as follows,

$$\begin{aligned} G(t,i,j)&=\mathop {\hbox \mathrm{IE}}\nolimits \left[ {{\hat{t}}}-t+\int _{{\hat{t}}}^T\mathbf{1}_{(N^*_s(\hat{t}, N_t-F_{\hat{t}}+F_t, F_{\hat{t}})\ge K)}ds\,\Big |\,N_t=i,F_t=j\right] \\&\quad ={{\hat{t}}}-t+\sum \limits _{r=0}^{(n-j)\wedge i}\mathop {\hbox \mathrm{IE}}\nolimits \left[ \int _{{\hat{t}}}^T\mathbf{1}_{(N^*_s(\hat{t}, i-r, j+r)\ge K)}ds\right] \left( {\begin{array}{c}i\\ r\end{array}}\right) \\&\qquad \frac{({{\hat{t}}}-t\vee L)^r(t-{{\hat{t}}}+t\vee L)^{i-r}}{t^r}\\&\quad ={{\hat{t}}}-t+\sum \limits _{r=0}^{(n-j)\wedge i}\int _{{\hat{t}}}^TP(N_s\ge K\,|\,N_{{\hat{t}}}=i-r,F_{{\hat{t}}}=j+r)ds\left( {\begin{array}{c}i\\ r\end{array}}\right) \\&\qquad \frac{({{\hat{t}}}-t\vee L)^r(t-{{\hat{t}}}+t\vee L)^{i-r}}{t^r}\\&\quad ={{\hat{t}}}-t+\sum \limits _{r=0}^{(n-j)\wedge i}\int _{{\hat{t}}}^T\sum _{m=K}^{n-j}p({{\hat{t}}},i-r,j+r,s,m)ds\\&\qquad \left( {\begin{array}{c}i\\ r\end{array}}\right) \frac{({{\hat{t}}}-t\vee L)^r(t-{{\hat{t}}}+t\vee L)^{i-r}}{t^r}. \end{aligned}$$

Hence we complete the proof of Proposition 3.2.

Appendix C: Pseudo code of the algorithm