A general storage model with applications to energy systems

Abstract

We consider a general storage system with applications to energy systems. Energy inflow, energy demand and costs interact with an environmental process. In each period, a limited amount of energy can be purchased from a supplementary source of energy at a cost per unit lower than a penalty cost for unmet demand. Sufficient conditions are found for the optimality of a zone-based release rule with critical numbers depending on the current state of the environment. There one buffer is needed to build up a storage level in the usual sense and a second one to protect against possible losses due to the variability in the randomly varying costs for unmet demand. The great versatility of our model is shown within the context of reservoir, hybrid energy and natural gas storage systems.

Appendix

Lemma 11

Let $v\in \mathfrak B_0$. Then, for all $i\in I$,

• (i) Uv(s, i) is nonnegative and decreasing in s.

• (ii) Jv(s, i, a) is convex in s for all a.

• (iii) There exists a minimizer $f^*\in F$ of Uv (i.e., $U_{f^*}v=Uv$) fulfilling

  $$\begin{aligned} f^*(s,i)\le f^*(s+1,i)\le f^*(s,i)+1,\quad s\in S. \end{aligned}$$

• (iv) Uv(s, i) is convex in s.

Proof

• (i) Since c and v are nonnegative, also Uv is nonnegative. Let $f\in F$ such that $Uv=U_{f}v$. Fix $(s,i)\in S\times I$. Set $a=f(s,i)$. Note that $a\in D(s+1,i)$, too. Then, since v is decreasing in s, using $T(s,a,x):=\min \{s-a+x,C\}$,

  $$\begin{aligned}&Uv(s+1,i)-Uv(s,i)\\&\quad \le Lv(s+1,i,a)-Lv(s,i,a)\\&\quad =\beta (i)\sum _{x=0}^\infty q_i(x)\int W_{i,x}(dj)\left[ v(T(s+1,a,x),j)-v(T(s,a,x),j)\right] \\&\quad \le 0, \end{aligned}$$

  which implies that Uv(s, i), $i\in I$, is decreasing in s.

• (ii) By assumption, $v(\cdot ,i)$, $i\in I$, is convex and decreasing in s. Further, for all a and x, $\min \{s-a+x,C\}$ is concave in s. By considering the three cases (a) $s+2-a+x\le C$, (b) $s+1-a+x=C$, (c) $s-a+x\ge C$, the composition $v(\min \{s-a+x,C\},j)$ is easily verified to be convex in s (where in the second case also use has to be made of the monotonicity of v). Finally, Jv(s, i, a) is easily verified to be convex in s.

• (iii) We first show that $f^*(s,i)\le f^*(s+1,i)$. Let $a^*=f^*(s,i)$ and $a\in D(s,i)$, $a<a^*$. Then $Lv(s,i,a)-Lv(s,i,a^*)\ge 0$. If $a^*=\min \{0,d(i)-K(i)\}$, then $f^*(s,i)\le f^*(s+1,i)$ holds trivially. Therefore let $a^*>\min \{0,d(i)-K(i)\}$. Then, since c(i, a) is independent of s and Jv(s, i, a) is convex in s, we have

  $$\begin{aligned}&Lv(s+1,i,a)-Lv(s+1,i,a^*)\\&\quad =Lv(s,i,a)-Lv(s,i,a^*)+\Delta Jv(s,i,a)-\Delta Jv(s,i,a^*)\\&\quad \ge \sum _{\tilde{a}=a}^{a^*-1}[\Delta Jv(s,i,\tilde{a})-\Delta Jv(s,i,\tilde{a}+1)]\\&\quad =\sum _{\tilde{a}=a}^{a^*-1}[\Delta Jv(s+1,i,\tilde{a}+1)-\Delta Jv(s,i,\tilde{a}+1)]\\&\quad \ge 0. \end{aligned}$$

  Thus, $f^*(s,i)\le f^*(s+1,i)$. We next show that $f^*(s+1,i)\le f^*(s,i)+1$. Let $a\in D(s,i)$, $a>a^*$. Then $Lv(s,i,a)-Lv(s,i,a^*)\ge 0$ by definition of $a^*$. If $a+1\not \in D(s+1,i)$, then the assertion follows immediately. Thus, let $a+1\in D(s+1,i)$. Then also $a^*+1\in D(s+1,i)$ and we obtain

  $$\begin{aligned}&Lv(s+1,i,a+1)-Lv(s+1,i,a^*+1)\\&\quad =c(i,a+1)-c(i,a^*+1)+ Jv(s+1,i,a+1)-Jv(s+1,i,a^*+1)\\&\quad =c(i,a+1)-c(i,a^*+1)+Jv(s,i,a)-Jv(s,i,a^*)\\&\quad =c(i,a+1)-c(i,a)-c(i,a^*+1)+c(i,a^*)+Lv(s,i,a)-Lv(s,i,a^*)\\&\quad \ge (c(i,a+1)-c(i,a))-(c(i,a^*+1)-c(i,a^*))\\&\quad \ge 0, \end{aligned}$$

  where the last inequality follows from the convexity of c(i, a) in a. Hence, $f^*(s+1,i)\le f^*(s,i)+1$.

• (iv) To prove convexity of $Uv(\cdot ,i)$, using (ii) and (iii) we have to consider the following three cases (a)–(c).

  • (a) $f^*(s,i)=f^*(s+1,i)=f^*(s+2,i)$.

    $$\begin{aligned}&\Delta Uv(s+1,i)-\Delta Uv(s,i)=\Delta Jv(s+1,i,f^*(s,i))\\ {}&\quad -\Delta Jv(s,i,f^*(s,i))\ge 0. \end{aligned}$$

  • (b) $f^*(s,i)\le f^*(s+1,i)\le f^*(s+2,i)=f^*(s,i)+1$.

    $$\begin{aligned}&\Delta Uv(s+1,i)-\Delta Uv(s,i)\\&\quad \ge \Delta Lv(s+1,i,f^*(s,i)+1)-\Delta Lv(s,i,f^*(s,i))\\&\quad =\Delta Jv(s+1,i,f^*(s,i)+1)-\Delta Jv(s,i,f^*(s,i))\\&\quad =\Delta Jv(s,i,f^*(s,i))-\Delta Jv(s,i,f^*(s,i))\\&\quad =0. \end{aligned}$$

  • (c) $f^*(s,i)< f^*(s+1,i)<f^*(s+2,i)$. Set $a^*=f^*(s,i)$. Then we have $f^*(s+\varsigma ,i)= a^*+\varsigma$ and $Jv(s+\varsigma ,i, f^*(s+\varsigma ,i))=Jv(s,i,a^*)$ for $\varsigma \in \{1,2\}$ and, together with the convexity of c(i, a) in a,

    $$\begin{aligned} \Delta Uv(s+1,i)-\Delta Uv(s,i)=c(i,a^*+2)-2c(i,a^*+1)+c(i,a^*) \ge 0, \end{aligned}$$

  which completes the proof. $\square$

Let $i\in I$, $v\in \mathfrak B_0$. Together with Lemma 11(ii) it follows that both $a\rightarrow \rho (i) a+Jv(a,i,0)$ and $a\rightarrow \ell (i) a+Jv(a,i,0)$ are convex in a and tend to infinity as $a\rightarrow \infty$. Thus, there exists $z(i)\in \mathbb N_0$, for which $a\rightarrow \rho (i)a+Jv(a,i,0)$ becomes minimal on $\mathbb N_0$. Further, there exists $Z(i)\in \mathbb N_0$ for which $a\rightarrow \ell (i)a+Jv(a,i,0)$ becomes minimal on $\mathbb N_0$. Let $a^*=Z(i)$. Then, by definition of $a^*$ and $\ell (i)\le \rho (i)$,

$$\begin{aligned} Jv(a^*,i,0)\le \ell (i)+Jv(a^*+1,i,0)\le \rho (i)+Jv(a^*+1,i,0), \end{aligned}$$

which implies $z(i)\le Z(i)$.

Lemma 12

Let $v\in \mathfrak B_0$. Then there exists $f\in F$ such that $Uv=U_fv$ with the following properties:

• (i) $f(s,i)=s-\min \{s,z(i)\}$ for all $0\le s\le d(i)-K(i)+z(i)$, where z(i) is the largest minimum point of the convex function $a\rightarrow \rho (i) a+Jv(a,i,0)$ on $\mathbb N_0$.

• (ii) Let $\Delta v(s,i)\ge -\rho (i)$ for all $(s,i)\in S\times I$. If ($A_\rho$) is fulfilled, then $\Delta Uv(s,i)\ge -\rho (i)$, $(s,i)\in S\times I$, and (i) holds with $z(i)=0$ for all $i\in I$.

Proof

Let $i\in I$ and $K(i)\le d(i)$. Set $\tilde{s}:=z(i)$, $d:=d(i)$, $K:=K(i)$, and $\rho :=\rho (i)$. By definition of $\tilde{s}$,

$$\begin{aligned} \rho \tilde{s}+Jv(\tilde{s},i,0)-\rho (\tilde{s}+1) -Jv(\tilde{s}+1,i,0)=-(\rho +\Delta Jv(\tilde{s},i,0))<0. \end{aligned}$$

If $\tilde{s}>0$, we additionally have

$$\begin{aligned} \rho \tilde{s}+Jv(\tilde{s},i,0)-\rho (\tilde{s}-1) -Jv(\tilde{s}-1,i,0)=\rho +\Delta Jv(\tilde{s}-1,i,0))\le 0. \end{aligned}$$

Next we show that $\tilde{s}$ is the largest $s\in S$, for which action $a=0$ is optimal. The proof follows in two steps

• (1) For $s=1,\ldots ,\tilde{s}$, using $f(0,i)=0$ and Lemma 11(iii),

  $$\begin{aligned}&\rho (d-K)+Jv(s,i,0)-\rho (d-K-1)-Jv(s-1,i,0)\\&\quad =\rho +\Delta Jv(s-1,i,0))\le \rho +\Delta Jv(\tilde{s}-1,i,0)\le 0. \end{aligned}$$

• (2) For $s=\tilde{s}+1,\ldots ,\tilde{s}+d-K$, using (1) and Lemma 11(iii),

  $$\begin{aligned}&\rho (d-K-(s-\tilde{s}))+Jv(s-(s-\tilde{s}),i,0)\\&\quad -\rho (d-K-(s-\tilde{s}-1))-Jv(s-(s-\tilde{s}-1)),i,0)\\&\quad = -(\rho +\Delta Jv(\tilde{s},i,0))<0 \end{aligned}$$

  The inequalities in step (2) additionally prove that $f(s,i)=s-\tilde{s}$ for $s=\tilde{s}+1,\ldots ,\tilde{s}+d-K$. Thus, (i) holds.

Let $v\in \mathfrak B_0^\rho :=\{v\in \mathfrak B_0~|~\Delta v(s,i)\ge -\rho (i)\}$. Suppose $v\in \mathfrak B_0^\rho$. To show $Uv\in \mathfrak B_0^\rho$, fix $(s,i)\in S\times I$. If $f(s+1,i)=s+1$, then

$$\begin{aligned} \Delta Uv(s,i)\ge Lv(s+1,i,s+1)-Lv(s,i,s)= c(i,s+1)-c(i,s)\ge - \rho (i). \end{aligned}$$

Otherwise, i.e., $a:=f(s+1,i)<s+1$, we get, exploiting $v\in \mathfrak B_0^\rho$,

$$\begin{aligned} \Delta Uv(s,i)\ge Lv(s+1,i,a)-Lv(s,i,a)=\Delta Jv(s,i,a)\ge -\rho (i), \end{aligned}$$

where the last inequality follows from $v\in \mathfrak B_0^\rho$ and assumption ($A_\rho$). Thus, $Uv\in \mathfrak B_0^\rho$ and, in particular,

$$\begin{aligned} Jv(0,i,0)-\rho (i)- Jv(1,i,0)= -\rho (i)-\Delta Jv(0,i,0)\le 0, \end{aligned}$$

which results in $z(i)=0$ and completes the proof. $\square$

Lemma 13

Let $v\in \mathfrak B_0$. For $i\in I$, define z(i) as in Lemma 12 in case of $K(i)\le d(i)$ and $z(i)=0$ in case of $K(i)>d(i)$. Then there exists a minimizer $f\in F$ of Uv such that for all $i\in I$

• (i) $f(s,i)=\max \{s-Z(i), d(i)-K(i)\}$,   $d(i)-K(i)+z(i)<s\le d(i)$,

• (ii) $f(s,i)=\min \{d(i),\max \{s-Z(i), d(i)-K(i)\}\}$,   $s> d(i)$,

where Z(i) is the largest minimum point of the convex function $a\rightarrow \ell (i) a+Jv(a,i,0)$ on $\mathbb N_0$.

Proof

(i) Fix $i\in I$. Set $d:=d(i)$, $K:=K(i)$, $\ell :=\ell (i)$. First, let us consider $s\le d-K+z(i)$, which is of relevance in case of $d-K\ge 0$ only. By Lemma 12(i), $f(s,i)=s-\min \{s,z(i)\}$. In particular, $f(d-K+z(i),i)=d-K$. Now, let $d-K+z(i)<s\le d$. Then, together with Lemma 11(iii), we already know that $f(s,i)\ge d-K$. Thus,

$$\begin{aligned} Uv(s,i)=\min _{d-K\le a\le s}\{\ell \cdot (d-a)+Jv(s,i,a)\}. \end{aligned}$$

(7)

Note that $Jv(s,i,a)=Jv(s-a,i,0)$. Thus, using $a'=s-a$, (7) can be rewritten as

$$\begin{aligned} Uv(s,i)=\ell \cdot (d-s)+\min _{0\le a'\le s-d+K}\{\ell \cdot a'+Jv(a',i,0)\}. \end{aligned}$$

(8)

The minimum is achieved at $\min \{Z(i),s-d+K\}\in \mathbb N_0$. By retransforming action a, (i) then follows. Finally, let $s> d$. Then, similar to (i), we obtain together with $a'=s-a$

$$\begin{aligned} Uv(s,i)= & {} \min _{d-K\le a\le d}\{\ell \cdot (d-a)+Jv(s,i,a)\}\\= & {} \ell \cdot (d-s)+\min _{s-d\le a'\le s-d+K}\{\ell \cdot a'+Jv(a',i,0)\}. \end{aligned}$$

If $Z(i)\in \{s-d,\ldots ,s-d+K\}$, it is admissible. Otherwise the minimum is taken on in one of the endpoints $s-d$ and $s-d+K$, respectively. By retransforming action a, we finally arrive at (ii). $\square$

Proof of Theorem 2

Let $v_0=0$. Then $\Delta v_1(s,i)=\min _{a\in D(s+1,i)}\{c(i,a)\}-\min _{a\in D(s,i)}\{c(i,a)\}\le 0=\Delta v_0(s,i)$. Now the proof follows by induction on n. First, for $n=1$, (i) holds by assumption. To show (ii)–(iv), fix $(s,i)\in S\times I$. Let $a>f_1(s,i)=:a^*$. Then, using (i),

$$\begin{aligned}&Lv_1(s,i,a)-Lv_1(s,i,a^*)\\&\quad =c(i,a)-c(i,a^*)+\sum _{a'=a^*+1}^{a}[Jv_1(s,i,a')-Jv_1(s+1,i,a')]\\&\quad \ge c(i,a)-c(i,a^*)+\sum _{a'=a^*+1}^{a}[Jv_0(s,i,a')-Jv_0(s+1,i,a')]\\&\quad \ge 0. \end{aligned}$$

Hence, $f_2(s,i)\le f_1(s,i)$. If $a^*:=z_1(i)>0$, then

$$\begin{aligned}&\rho (i)(a^*-1)+Jv_1(a^*-1,i,0)-\rho (i)a^*-Jv_1(a^*,i,0)\\&\quad =-\rho (i)-J\Delta v_1(a^*-1,i,0)\\&\quad \ge -\rho (i)-J\Delta v_0(a^*-1,i,0)\\&\quad \ge 0. \end{aligned}$$

Thus, $z_2(i)\ge z_1(i)$. The inequality $Z_2(i)\ge Z_1(i)$ follows analogously.

Next we prove (i) for $n+1=2$, i.e., verify $\Delta v_2(s,i)\le \Delta v_1(s,i)$. Using (ii) for $n=1$ and Lemma 11(iii), we have to consider the following three cases:

• (a) $f_{2}(s,i)=f_{1}(s+1,i)$.

  $$\begin{aligned} \Delta v_{2}(s,i)=\Delta Jv_1(s,i,f_1(s,i))\le \Delta Jv_{0}(s,i,f_1(s,i))=\Delta v_{1}(s,i). \end{aligned}$$

• (b) $f_{2}(s,i)\le f_{1}(s,i)<f_{1}(s+1,i)$. Then we have $f_{2}(s,i)+1\in D(s+1,i)$ and

  $$\begin{aligned}&\Delta v_{2}(s,i)\le c(i,f_{2}(s,i)+1)-c(i,f_{2}(s,i))\\&\quad \le c(i,f_{1}(s,i)+1)-c(i,f_{1}(s,i))\\&\quad =\Delta v_{1}(s,i). \end{aligned}$$

• (c) $f_{2}(s,i)< f_{2}(s+1,i)\le f_{1}(s+1,i)$. Then we have $f_{1}(s+1,i)-1\in D(s,i)$ and (i) follows in analogy to (b), using $c(i,f_{2}(s+1,i))-c(i,f_{2}(s+1,i)-1)\le c(i,f_{1}(s+1,i))-c(i,f_{1}(s+1,i)-1)$.

Exactly the same arguments can be used to prove (i)–(iv) for an arbitrary $n\in \mathbb N$. Finally, the comparison with $\Delta V(s,i)$, $f^*(s,i)$, $z^*(i)$ and $Z^*(i)$ follows by letting $n\rightarrow \infty$. $\square$

Proof of Proposition 4

Fix $(s,i)\in S\times I$ such that $a^*:=f^*(s,i)>\max \{a^+(i),d(i)\}$. Then we have $LV(s,i,a^*)<LV(s,i,a^*-1)$ by assumption. Further, since $c(i,a^*)=0$, we get

$$\begin{aligned}&LV(s+1,i,a^*+1)\\&\quad =(a^*+1-a^+(i))\varrho _a^+(i)+\tilde{c}_s(i,s+1)+JV(s+1,i,a^*+1)\\&\quad =\varrho _a^+(i)+\tilde{c}_s(i,s+1)-\tilde{c}_s(i,s)+LV(s,i,a^*)\\&\quad <\varrho _a^+(i)+\tilde{c}_s(i,s+1)-\tilde{c}_s(i,s)+LV(s,i,a^*-1)\\&\quad =\varrho _a^+(i)+(a^*-1-a^+(i))\varrho _a^ +(i)+\tilde{c}_s(i,s+1)+JV(s,i,a^*-1)\\&\quad =LV(s+1,i,a^*) \end{aligned}$$

and thus $f^*(s+1,i)= a^*+1$, since $f^*(s+1,i)\le f^*(s,i)+1$ by Theorem 3(ii). $\square$