A Proof of Theorems 1 and 2
A.1 Lemmata for DNFR Approximation Results
We present a series of lemmata, which will be needed later in the proofs to Theorems 1 and 2.
Lemma 2
(Family of dwell time block sets) The family of dwell time block interval sets \(\{{\mathscr {J}}_b\}_{b\in [n_b]}\) as defined in Definition 5 is a partition of the set of all interval indices [N].
Proof
This follows directly from the definition of dwell time block interval sets. \(\square \)
Lemma 3
(Block accumulated control deviation properties) For all \(b\in [n_b]\) we have that
$$\begin{aligned} \sum \limits _{i \in [{n_\omega }]} \varGamma _{i,b} = \mathscr {L}_b, \qquad \qquad \sum \limits _{i \in [{n_\omega }]} \varTheta _{i,b} = 0. \end{aligned}$$
Proof
Let us derive an auxiliary result for \(j\in [N]\):
$$\begin{aligned} \sum \limits _{i \in [{n_\omega }]} \theta _{i,j} = \sum \limits _{i \in [{n_\omega }]} \sum \limits _{l \in [j]} (a_{i,l}-w_{i,l})\varDelta _l {\mathop {=}\limits ^{\text {(Conv)}}} \sum \limits _{l \in [j]} \varDelta _l - \sum \limits _{l \in [j]} \varDelta _l = 0. \end{aligned}$$
(A.1)
We use this and rearrange the sums in order to proof the first assertion:
$$\begin{aligned} \sum \limits _{i \in [{n_\omega }]}\varGamma _{i,b}=\sum \limits _{i \in [{n_\omega }]} \left( \theta _{i,l_{b-1}}+\sum \limits _{j\in {\mathscr {J}}_b}a_{i,j}\varDelta _j \right)&= 0 + \sum \limits _{i \in [{n_\omega }]} \sum \limits _{j\in {\mathscr {J}}_b}a_{i,j}\varDelta _j \\&= \sum \limits _{j\in {\mathscr {J}}_b} \sum \limits _{i \in [{n_\omega }]} a_{i,j}\varDelta _j {\mathop {=}\limits ^{\text {(Conv)}}} \sum \limits _{j\in {\mathscr {J}}_b} \varDelta _j \\&= \mathscr {L}_b. \end{aligned}$$
The auxiliary result is also useful for the second statement:
$$\begin{aligned} \sum \limits _{i \in [{n_\omega }]}\varTheta _{i,b}=\sum \limits _{i \in [{n_\omega }]} \theta _{i,l_{b-1}}{\mathop {=}\limits ^{(A.1)}} 0. \end{aligned}$$
\(\square \)
Lemma 4
(Accumulated difference of \(\varGamma \)and \(\varTheta \)over active controls) Let \(b_1,b_2\in [n_b]\) and we define \(S_{b_1,b_2}\) as the set of active controls between \(b_1\) and \(b_2\):
$$\begin{aligned} S_{b_1,b_2}: = \{i\in [{n_\omega }] \ \mid \ \exists b: \ b_1< b < b_2 \text { with } w_{i,j}=1, \ \forall j\in {\mathscr {J}}_b \}. \end{aligned}$$
Then, we have
$$\begin{aligned} \sum \limits _{i\in S_{b_1,b_2}} \left( \varGamma _{i,b_2} - \varTheta _{i,b_1}\right) \le \overline{\mathscr {L}}. \end{aligned}$$
Proof
Using Definition 7 of \(\varGamma ,\varTheta \) and rearranging sums yields
$$\begin{aligned} \sum \limits _{i\in S_{b_1,b_2}} \left( \varGamma _{i,b_2} - \varTheta _{i,b_1}\right)&= \sum \limits _{i\in S_{b_1,b_2}} \left( \sum \limits _{b=b_1+1}^{b_2} \sum \limits _{j\in {\mathscr {J}}_b} a_{i,j} \varDelta _j - \sum \limits _{b=b_1+1}^{b_2-1}\sum \limits _{j\in {\mathscr {J}}_b} w_{i,j} \varDelta _j \right) \\&= \sum \limits _{b=b_1+1}^{b_2} \sum \limits _{j\in {\mathscr {J}}_b} \varDelta _j \underbrace{\sum \limits _{i\in S_{b_1,b_2}}a_{i,j}}_{\le 1} - \sum \limits _{b=b_1+1}^{b_2-1} \sum \limits _{j\in {\mathscr {J}}_b} \varDelta _j \underbrace{\sum \limits _{i\in S_{b_1,b_2}}w_{i,j}}_{=1} \\&\le \sum \limits _{b=b_1+1}^{b_2} \sum \limits _{j\in {\mathscr {J}}_b} \varDelta _j - \sum \limits _{b=b_1+1}^{b_2-1} \sum \limits _{j\in {\mathscr {J}}_b} \varDelta _j \\&= \mathscr {L}_{b_2} \le \overline{\mathscr {L}}. \end{aligned}$$
\(\square \)
Note that \(S_{b_1,b_2}\) is trivially the empty set, if \(b_2 \le b_1 + 1 \), but the result remains true in this case. We employ the concept of \(S_{b_1,b_2}\) for a contradiction in the proofs of Theorems 1 and 2.
Lemma 5
(Control with negative \(\varGamma \)value has not been future forced) Let \((C_1,C_2,\chi _D)\) be given and assume that the forward control deviation of a control \(i\in [{n_\omega }]\) and a block \(b_2 \ge 2\) after executing Algorithm 1 satisfies:
$$\begin{aligned} \varGamma _{i,b_2}\le C_2\overline{\mathscr {L}} - \overline{\mathscr {L}}, \qquad \text {and} \qquad \varGamma _{i,b_2}< 0. \end{aligned}$$
Then, there is an earlier activation of i on some block \(b_1<b_2\) and this activation has not been \(b_2\)-future forced on \(b_1\).
Proof
Note that \(\varGamma _{i,b}\) is monotonically increasing in b for deactivated controls i. We conclude from this and \(\varGamma _{i,b_2}< 0\) that there is an earlier activation of i on some block \(b_1<b_2\). We take a closer look on the forward control deviation on block \(b_2\):
$$\begin{aligned} C_2\overline{\mathscr {L}} - \overline{\mathscr {L}}\ge & {} \varGamma _{i,b_2} = \sum \limits _{k=1}^{b_2} \sum \limits _{j\in {\mathscr {J}}_{k}} a_{i,j} \varDelta _j - \sum \limits _{k=1}^{b_1} \sum \limits _{j\in {\mathscr {J}}_k} w_{i,j} \varDelta _j\\= & {} \sum \limits _{k=1}^{b_2} \sum \limits _{j\in {\mathscr {J}}_k} a_{i,j} \varDelta _j - \sum \limits _{k=1}^{b_1-1} \sum \limits _{j\in {\mathscr {J}}_k} w_{i,j} \varDelta _j - \mathscr {L}_{b_1}, \end{aligned}$$
and rearranging terms implies
$$\begin{aligned} \qquad \sum \limits _{k=1}^{b_2} \sum \limits _{j\in {\mathscr {J}}_k} a_{i,j} \varDelta _j - \sum \limits _{k=1}^{b_1-1} \sum \limits _{j\in {\mathscr {J}}_k} w_{i,j} \varDelta _j \le C_2\overline{\mathscr {L}} - \overline{\mathscr {L}} + \mathscr {L}_{b_1} \le C_2\overline{\mathscr {L}}. \end{aligned}$$
The last inequality shows us that i has been not \(b_2\)-future forced on \(b_1\). \(\square \)
A.2 Proof of Theorem 1
Proof
We proceed via induction.
Base case: We consider the first b on which a down time forbidden control \(i_D\in [{n_\omega }]\) appears and let us assume
$$\begin{aligned} \varGamma _{i_D,b} {\mathop {>}\limits ^{}} \tfrac{3}{2}\overline{\mathscr {L}} \end{aligned}$$
(A.2)
holds and we prove that it results in a contradiction. It follows from Lemma 3
$$\begin{aligned} \tfrac{3}{2} \overline{\mathscr {L}}\ge \mathscr {L}_{b} = \sum \limits _{i\ne i_D} \varGamma _{i,b} + \varGamma _{i_D,b}, \end{aligned}$$
so that there must be a control \(i_1\ne i_D\) with negative forward control deviation on b:
$$\begin{aligned} \exists \ i_1 \ne i_D: \ \varGamma _{i_1,b} < 0. \end{aligned}$$
We apply Lemma 5 to the last inequality: \(i_1\) has not been b-future forced on its last activation and we denote the block of this activation with \(b_1\). In other words, we know that there is at least one block \(b_1\) and one control \(i_1\) that was not b-future forced on \(b_1\) and still was activated on \(b_1\). Now, we denote by \(i_1\) the control of this property with the last activation before b. By this definition, we observe that all controls being activated after \(b_1\) would become forced until b. We notate
$$\begin{aligned} F_{b_1,b} : = \{i\in [{n_\omega }] \ \mid \ \exists k(i):\ b_1< k(i) \le b \text { on which } i \text { is forced or } b\text {-future forced} \}. \end{aligned}$$
In particular, we have \(i_D\in F_{b_1,b}\). For \(i\in F_{b_1,b}\backslash \{i_D\}\) we conclude
$$\begin{aligned} \varGamma _{i,b}= \varTheta _{i,b-1} + \sum \limits _{j\in {\mathscr {J}}_{b}} a_{i,j}\varDelta _j = \sum \limits _{k=1}^{b} \sum \limits _{j\in {\mathscr {J}}_k} a_{i,j}\varDelta _j - \sum \limits _{k=1}^{k(i)-1} \sum \limits _{j\in {\mathscr {J}}_k} w_{i,j}\varDelta _j {\mathop {>}\limits ^{}} \tfrac{3}{2} \overline{\mathscr {L}}, \end{aligned}$$
and therefore
$$\begin{aligned} \varGamma _{i,b} > \tfrac{3}{2} \overline{\mathscr {L}} - \mathscr {L}_{k(i)}, \qquad i\in F_{b_1,b}\backslash \{i_D\}. \end{aligned}$$
(A.3)
The last inequality holds, since control i was last activated at block k(i). For block \(b_1\) we know that \(i_1\) has been chosen, despite not being b-future forced. We use this observation and our assumption of \(b>b_1\) being the first block with a down time forbidden control to conclude that all controls out of \(F_{b_1,b}\) have been inadmissible on \(b_1\). Hence, it results for \(i\in F_{b_1,b}\)
$$\begin{aligned} \varGamma _{i,b_1}< - \tfrac{3}{2} \overline{\mathscr {L}} + \mathscr {L}_{b_1}, \qquad \Rightarrow \qquad \varTheta _{i,b_1} < - \tfrac{3}{2} \overline{\mathscr {L}} + \mathscr {L}_{b_1}. \end{aligned}$$
(A.4)
We sum up the inequalities (A.2) and (A.3) over \( F_{b_1,b}\) and similarly for (A.4), which yields
where we set \(b_2:=\arg \min \{\mathscr {L}_{k} \; \mid \; b_1< k \le b \}\) and notate with \(| F_{b_1,b}|\) the cardinality of \(F_{b_1,b}\). Subtracting (A.6) from (A.5) results in
$$\begin{aligned} \sum \limits _{i\in F_{b_1,b}} \left( \varGamma _{i,b} - \varTheta _{i,b_1} \right) \>&\ \tfrac{3}{2} \overline{\mathscr {L}} + (| F_{b_1,b}| -1 ) \left( \tfrac{3}{2} \overline{\mathscr {L}} - \mathscr {L}_{b_2} \right) - | F_{b_1,b}| (- \tfrac{3}{2} \overline{\mathscr {L}} + \mathscr {L}_{b_1}) \nonumber \\ =&\tfrac{3}{2} \overline{\mathscr {L}} + (2 | F_{b_1,b}| -1) \tfrac{3}{2} \overline{\mathscr {L}} - (| F_{b_1,b}| -1 ) \mathscr {L}_{b_2} - |F_{b_1,b}| \mathscr {L}_{b_1}\nonumber \\ \overset{}{\ge }&\tfrac{3}{2} \overline{\mathscr {L}} + (2 | F_{b_1,b}| -1) \tfrac{1}{2} \overline{\mathscr {L}} \nonumber \\ >&\overline{\mathscr {L}}. \end{aligned}$$
(A.7)
We used \(\mathscr {L}_{b_1},\mathscr {L}_{b_2}\le \overline{\mathscr {L}}\) in the second inequality. We finish our calculations by considering the property of \(F_{b_1,b}\) comprising all control activations between time blocks \(b_1+1\) and \(b-1\), therefore we can apply Lemma 4 with \(F_{b_1,b}=S_{b_1,b}\) and obtain
$$\begin{aligned} \sum \limits _{i\in F_{b_1,b}} \left( \varGamma _{i,b} - \varTheta _{i,b_1} \right) \le \overline{\mathscr {L}}, \qquad \qquad \lightning \end{aligned}$$
(A.8)
which is a contradiction to inequality (A.7).
Step case: Let the assertion hold until any block \(b-1\in [n_b]\) and we prove that then the statement holds for b. Again, we consider \(i_D\in [{n_\omega }]\) and assume
$$\begin{aligned} \varGamma _{i_D,b} {\mathop {>}\limits ^{}} \tfrac{3}{2}\overline{\mathscr {L}} \end{aligned}$$
(A.9)
holds and we prove that it results in a contradiction. With a similar argumentation as in the base case we deduce that there is a control \(i_1\) that has not been b-future forced on block \(b_1<b\) and reuse the definition of \(F_{b_1,b}\). Thus, inequality (A.3) still holds. Now, we distinguish between two cases why the controls out of \(F_{b_1,b}\) have not been activated on \(b_1\). If all controls \(i\in F_{b_1,b}\) have been inadmissible on \(b_1\), we can argue as in the base case. Hence, we focus on the other case: there is an \(i_2\in F_{b_1,b}\), which was down time forbidden on \(b_1\) and all other controls \(i\in F_{b_1,b}\backslash \{i_2 \}\) were inadmissible. By the induction hypothesis and the previously derived inequality (A.4) we have
$$\begin{aligned} \varTheta _{i_2,b_1} \le \tfrac{3}{2} \overline{\mathscr {L}}, \qquad \qquad \varTheta _{i,b_1} < - \tfrac{3}{2} \overline{\mathscr {L}} + \mathscr {L}_{b_1}, \quad i\in F_{b_1,b}\backslash \{i_2 \}. \end{aligned}$$
Summing up these inequalities over \(F_{b_1,b}\) results therefore in
$$\begin{aligned} \sum \limits _{i\in F_{b_1,b}} \varTheta _{i,b_1}\ < \ \tfrac{3}{2} \overline{\mathscr {L}} +| F_{b_1,b}-1| \left( -\tfrac{3}{2} \overline{\mathscr {L}} + \mathscr {L}_{b_1}\right) . \end{aligned}$$
(A.10)
Next, we argue that \(F_{b_1,b}\) contains at least two controls: the case \(b_1=b-1\) is not possible, since \(i_D\) is via assumption forced and down time forbidden on b, hence admissible on \(b-1\). Therefore, \(b_1 \le b-2\) and there is a control \(i\ne i_D, \ i \in F_{b_1,b}\), which is activated on \(b-1\). Altogether we have \(|F_{b_1,b}|\ge 2\). With this observation we subtract inequality (A.10) from (A.5):
$$\begin{aligned} \sum \limits _{i\in F_{b_1,b}} \left( \varGamma _{i,b} - \varTheta _{i,b_1} \right) \ >&\ \tfrac{3}{2} \overline{\mathscr {L}} + (| F_{b_1,b}| -1 ) \left( \tfrac{3}{2} \overline{\mathscr {L}} - \mathscr {L}_{b_2} \right) \nonumber \\&\ - \left( \tfrac{3}{2} \overline{\mathscr {L}} + (| F_{b_1,b}|-1) (- \tfrac{3}{2} \overline{\mathscr {L}} + \mathscr {L}_{b_1}) \right) \nonumber \\ =&3(| F_{b_1,b}| -1) \overline{\mathscr {L}} - (| F_{b_1,b}| -1 ) \mathscr {L}_{b_2} - |F_{b_1,b}-1| \mathscr {L}_{b_1}\\ \overset{}{\ge }&(| F_{b_1,b}| -1) \overline{\mathscr {L}}\\ \ge&\overline{\mathscr {L}}. \end{aligned}$$
Notice that \(|F_{b_1,b}|\ge 2\) is used in the last inequality. Finally, we build again on Lemma 4, where it is justified to set \(F_{b_1,b}=S_{b_1,b}\). Thus, the above inequality is a contradiction to the inequality from the lemma and we have shown that the assertion holds for all \(b\in [n_b]\) on which a down time forbidden control exists. \(\square \)
A.3 Proof of Theorem 2
Proof
The assertion can be shown for the parameter choices I. and II. in a very similar way, which is why we prove both cases in parallel. Since the algorithm activates for each block \(b\in [{n_\omega }]\) either a forced, or future forced, or admissible control and the family of blocks is a partition [N] by Lemma 2, exactly one control is activated on each interval \(j\in [N]\) and therefore the (Conv) constraint satisfied. Hence, DNFR guarantees feasibility of \(\mathbf {w}^{\text {DNFR}}\). If down time forbidden controls are neglected, i.e., \(\chi _D=0\), \(\mathbf {w}^{\text {DNFR}}\) yields an objective value with at most the claimed upper bound by the definition of admissible and forced activation. The same holds for the choice \(\chi _D=1\), since the control deviation does not become greater than the claimed upper bound during a MD time phase by Theorem 1. Therefore, we need only to prove that DNFR always provides a solution. For this, we show that for each interval there is (1.) at least one admissible control and (2.) at most one forced control.
-
(1.)
We prove by contradiction that there exists at least one admissible control: Let us assume there is no admissible activation for block \(b\in [n_b]\) and distinguish between the cases:
-
I.
With \(C_2=\frac{2n_{\omega }-3}{2n_{\omega }-2}\) we assume
$$\begin{aligned} \varGamma _{i,b} {\mathop {<}\limits ^{}} - \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + \mathscr {L}_b, \qquad i\in [n_{\omega }], \end{aligned}$$
and we prove that it results in a contradiction. It follows from summing up all controls and from Lemma 3:
$$\begin{aligned} \mathscr {L}_b = \sum \limits _{i \in [{n_\omega }]} \varGamma _{i,b} < n_{\omega } \left( - \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + \mathscr {L}_b\right) =-n_{\omega } \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + {n_\omega }\mathscr {L}_b, \end{aligned}$$
and subtracting \( {n_\omega }\mathscr {L}_b\) from the right hand side yields
$$\begin{aligned} (1- n_{\omega }) \overline{\mathscr {L}}\le & {} (1-{n_\omega }) \mathscr {L}_b < -n_{\omega } \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} \\= & {} - n_{\omega } \overline{\mathscr {L}} + \frac{n_{\omega }}{2(n_{\omega }-1)} \overline{\mathscr {L}} \, \, {\mathop {\le }\limits ^{{n_\omega }\ge 2}} (1- n_{\omega }) \overline{\mathscr {L}}. \qquad \lightning \end{aligned}$$
-
II.
If there is no down time forbidden control on b, we can proceed as in I. Otherwise, we may have one control \(i_D\) that is down time forbidden. We assume all other controls are inadmissible, i.e.,
$$\begin{aligned} \varGamma _{i,b} {\mathop {<}\limits ^{}} - \tfrac{3}{2} \overline{\mathscr {L}} + \mathscr {L}_b, \qquad i\in [n_{\omega }], \ i \ne i_D, \end{aligned}$$
and we prove that it results in a contradiction. Hence, again by Lemma 3
$$\begin{aligned} \mathscr {L}_b = \sum \limits _{i \in [{n_\omega }]} \varGamma _{i,b} = \sum \limits _{i \ne i_D} \varGamma _{i,b} + \varGamma _{i_D,b} < (n_{\omega }-1)(- \tfrac{3}{2} \overline{\mathscr {L}} + \mathscr {L}_b)+ \varGamma _{i_D,b}, \end{aligned}$$
and therefore
$$\begin{aligned} \tfrac{3}{2} (n_{\omega }-1) \overline{\mathscr {L}} - (n_{\omega }-2)\mathscr {L}_b \le \tfrac{3}{2} \overline{\mathscr {L}} < \varGamma _{i_D,b}. \qquad \lightning \end{aligned}$$
The last inequality is a contradiction to Theorem 1.
We conclude that there must be an admissible activation for all blocks and thereby for all intervals.
-
(2.)
If there were more than one forced control at a time step, the algorithm would be ambiguous at line 3–4. Moreover, DNFR would provide, in this case, a solution that does not satisfy the upper bound on the objective. Therefore, we prove that this case is impossible and we do so again by contradiction. Assume that there \(b\in [n_b]\) is the block with smallest index on which at least two controls \(i_1,i_2\) are being forced, i.e.,
$$\begin{aligned} I. \quad \varGamma _{i_1,b}, \varGamma _{i_2,b}> \tfrac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}}, \qquad II. \quad \varGamma _{i_1,b}, \varGamma _{i_2,b} > \tfrac{3}{2} \overline{\mathscr {L}}. \end{aligned}$$
(A.11)
In the proof of Theorem 1 we have already shown how to obtain a contradiction with only one forward control deviation \(\varGamma _{i,b}\) greater than the rounding threshold. So, case II. is settled with this theorem and we focus on case I. for which we proceed very similarly as in the proof of Theorem 1. Let us first apply Lemma 3:
$$\begin{aligned} \overline{\mathscr {L}} \ge \mathscr {L}_b = \sum \limits _{i\in [n_{\omega }]} \varGamma _{i,b} = \sum \limits _{\begin{array}{c} i\in [n_{\omega }],\\ i \ne i_1,i_2 \end{array}} \varGamma _{i,b} + \sum \limits _{i=i_1,i_2} \varGamma _{i,b} > \sum \limits _{\begin{array}{c} i\in [n_{\omega }],\\ i \ne i_1,i_2 \end{array}} \varGamma _{i,b} + 2 \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}}. \end{aligned}$$
Hence, we have
$$\begin{aligned} \sum \limits _{i\in [n_{\omega }], i \ne i_1,i_2} \varGamma _{i,b} < \overline{\mathscr {L}} - 2\ \dfrac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} = - \dfrac{2n_{\omega }-4}{2n_{\omega }-2} \overline{\mathscr {L}}, \end{aligned}$$
which implies that there is at least one control \(i_3\) such that
$$\begin{aligned} \varGamma _{i_3,b} < - \dfrac{1}{n_{\omega }-2} \dfrac{2n_{\omega }-4}{2n_{\omega }-2} \overline{\mathscr {L}} = - \dfrac{2}{2n_{\omega }-2} \overline{\mathscr {L}}. \end{aligned}$$
Then, by Lemma 5, there is an earlier activation of \(i_3\) on some block \(b_3<b\) and this activation has not been b-future forced on \(b_3\). Let \(i_3\) denote the control of this property with the last activation before b. This definition implies that all controls that are active between \(b_3\) and b become forced until b. We use again the notation
$$\begin{aligned}&F_{b_3,b} : = \{i\in [{n_\omega }] \ \mid \ \exists k(i):\ b_3< k(i) \le b\\&\quad \text { on which } i \text { is forced or } b\text {-future forced.} \}. \end{aligned}$$
In particular, we find \(i_1,i_2\in F_{b_3,b}\). For \(i\in F_{b_3,b}\backslash \{i_1,i_2\}\), we apply the definition of \(F_{b_3,b}\) and \(\varGamma \):
$$\begin{aligned} \varGamma _{i,b} = \varTheta _{i,b-1} + \sum \limits _{j\in {\mathscr {J}}_b} a_{i,j}\varDelta _j = \sum \limits _{k=1}^{b} \sum \limits _{j\in {\mathscr {J}}_k} a_{i,j}\varDelta _j - \sum \limits _{k=1}^{k(i)} \sum \limits _{j\in {\mathscr {J}}_k} w_{i,j}\varDelta _j. \end{aligned}$$
Since control i was last activated on block k(i) and b- future forced on k(i), we have
$$\begin{aligned} \varGamma _{i,b} > \dfrac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} - \mathscr {L}_{k(i)}, \qquad i\in F_{j_3,j}\backslash \{i_1,i_2\}. \end{aligned}$$
(A.12)
For block \(b_3\) we know that \(i_3\) has been chosen, even though not being b-future forced. That implies \(i_3\) was selected on \(b_3\) because all controls out of \(F_{b_3,b}\) were not admissible at this block. Hence, it results for \(i\in F_{b_3,b}\)
$$\begin{aligned} \varGamma _{i,b_3}< - \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + \mathscr {L}_{b_3}, \qquad \Rightarrow \qquad \varTheta _{i,b_3} < -\frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + \mathscr {L}_{b_3}.\nonumber \\ \end{aligned}$$
(A.13)
Now, we consider the sum of inequalities (A.11), (A.12) and sum up (A.13) over \(F_{b_3,b}\), which yields
$$\begin{aligned} \sum \limits _{i\in F_{b_3,b}} \varGamma _{i,b}&> \ 2 \dfrac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + (| F_{b_3,b}| -2 ) \left( \dfrac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} - \mathscr {L}_{b_2} \right) , \end{aligned}$$
(A.14)
$$\begin{aligned} \sum \limits _{i\in F_{b_3,b_3}} \varTheta _{i,b_3}&< \ | F_{b_3,b}| \left( - \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + \mathscr {L}_{b_3}\right) , \end{aligned}$$
(A.15)
where \(b_2:=\arg \min \{\mathscr {L}_{k} \; \mid \; b_3< k \le b \}\). Substracting (A.15) from (A.14), we obtain
$$\begin{aligned} \sum \limits _{i\in F_{b_3,b}} \left( \varGamma _{i,b} - \varTheta _{i,b_3} \right)&> \ 2 \dfrac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + (| F_{b_3,b}| -2 ) \left( \dfrac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} - \mathscr {L}_{b_2} \right) \nonumber \\&\ \ \ - | F_{b_3,b}| \left( - \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} + \mathscr {L}_{b_3}\right) \nonumber \\&= \ 2 | F_{b_3,b}| \frac{2n_{\omega }-3}{2n_{\omega }-2} \overline{\mathscr {L}} - (| F_{b_3,b}| -2 ) \mathscr {L}_{b_2} - | F_{b_3,b}| \ \mathscr {L}_{b_3} \nonumber \\&\overset{}{\ge } \overline{\mathscr {L}} \left( 2 | F_{b_3,b}| \frac{2n_{\omega }-3}{2n_{\omega }-2} - 2 | F_{b_3,b}| +2 \right) \end{aligned}$$
(A.16)
$$\begin{aligned}&= \overline{\mathscr {L}} \left( 2- \dfrac{| F_{b_3,b}|}{n_{\omega }-1} \right) \nonumber \\&\overset{}{\ge } \overline{\mathscr {L}}. \end{aligned}$$
(A.17)
In (A.16) we used \(\mathscr {L}_{b_2},\mathscr {L}_{b_3}\le \overline{\mathscr {L}}\), while the last inequality holds due to \(| F_{b_3,b}| \le n_{\omega }-1\). As in the proof of Theorem 1, we invoke now Lemma 4 with \(F_{b_3,b}=S_{b_1,b}\) in order to raise a contradiction to inequality (A.17). Overall, we have shown that there is at most one forced activation per block and thereby per interval. This completes the proof. \(\square \)
Remark 7
The proceeding in the proof of Theorem 2 shows us, on closer inspection, that DNFR provides a solution with control deviation bounded by \(C_2 \overline{\mathscr {L}}\) for the absence of MD time constraints, i.e., \(\chi _D=0\), and any chosen rounding threshold \(C_2\ge \frac{2{n_\omega }- 3}{2{n_\omega }- 2}\) and any block length parameter \(C_1\ge 0\). This implies the previously known result by NFR
[21], \(\theta (\mathbf {w}^{\text {NFR}}) \le \bar{\varDelta }\), evolves as special case of DNFR with \(C_1=0, \) and \(C_2=1\).
B Discussion on the Tightness of the obtained Minimum Down Time Bounds
Proposition 5
(Tightness of the bound for (CIA-D)) Let us assume the MD time constraint is active, i.e., \(C_D> \underline{\varDelta }\) is given. Then, the following is true:
-
1.
The bound for (CIA-D) stated in Proposition 3 can not be improved by the DNFR scheme with \(\chi _D=1\) for \({n_\omega }\ge 3\).
-
2.
The bound for (CIA-D) is tight up to at most the constant \(\frac{1}{4}C_D+ \bar{\varDelta }\).
Proof
The assumption of an active MD time constraint ensures that the bound cannot be improved by the bound for MU times from Proposition 2. We use again the concept of a minimal \(C_1\)-overlapping grid, here with \(C_1 = C_D/2\).
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1.
We want to prove that the DNFR scheme with \(\chi _D=1\) and \(C_2 < \tfrac{3}{2}\) may provide solutions with a (CIA-D) objective greater than \(C_2 \overline{\mathscr {L}}\). Let us consider first \(C_2 \le \tfrac{3}{2} - \varepsilon _1,\) with \(\ 0 < \varepsilon _1 \le 0.5\). We present example values for \(\mathbf {a}\in A\) with a time horizon length of at least \( 12 C_1\), so that at least six blocks with length \(\overline{\mathscr {L}}\) exist by Lemma 1. Let \(0< \varepsilon _2 < \varepsilon _1\) be small and let the relaxed control values \(\mathbf {a}\) be given as
$$\begin{aligned} (a_{i,b})_{i\in [{n_\omega }],b\in [n_b]} = \left( \begin{matrix} 1 &{} 0.5-\varepsilon _1 + \varepsilon _2 &{} 1-\varepsilon _2 &{} 2\varepsilon _1 - 2\varepsilon _2 &{} 0.5 &{} 0.5 &{} \cdots &{} 0.5 \\ 0 &{} 0.5+\varepsilon _1 - \varepsilon _2 &{} 0 &{} 1- 2\varepsilon _1 + 2\varepsilon _2 &{} 0.5 &{} 0.5 &{} \cdots &{} 0.5\\ 0 &{} 0 &{} \varepsilon _2 &{} 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots &{} 0 \end{matrix} \right) . \end{aligned}$$
With this example values we discuss the thereby constructed DNFR solution as well its objective quality.
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First block: \(i_1\) is 2-future forced and activated.
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Second block: Both \(i_1\) and \(i_2\) are 4-future forced. The DNFR algorithm breaks ties arbitrarily, so that activating \(i_2\) is legitimate.
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Third block: \(i_1\) is down time forbidden, while \(i_2\) is not admissible. DNFR activates therefore \(i_3\).
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Fourth block: \(i_1\) is activated, since it is forced.
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Fifth block: We have in the meantime \(\varTheta _{i_1,4}=(0.5+\varepsilon _1 -2 \varepsilon _2)\overline{\mathscr {L}}\) and \(\varTheta _{i_2,4}=(0.5-\varepsilon _1 +\varepsilon _2) \overline{\mathscr {L}}\). Since \(\varepsilon _2\) satisfies \(\varepsilon _2 < \varepsilon _1\), both controls are 6-future forced on the fifth block. Let DNFR activate \(i_2\).
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Sixth block: \(i_1\) is still down time forbidden and can not be active, which implies
$$\begin{aligned} \varTheta _{i_1,6} = (0.5+\varepsilon _1 -2 \varepsilon _2 +1) \overline{\mathscr {L}} >( \tfrac{3}{2} - \varepsilon _1) \overline{\mathscr {L}} = C_2 \overline{\mathscr {L}}, \end{aligned}$$
so that the proposed control deviation bound is not fulfilled.
Finally, if \(\varepsilon _1 > 0.5\), thus \(C_2 < 1\), we can construct a similar example for which the control \(i_1\) is already forced on the first block and the control deviation is again greater than \(C_2 \overline{\mathscr {L}}\).
-
2.
The MD time constraints are equivalent to MU time constraints with \(C_U= C_D\) for a problem with only two controls \({n_\omega }=2\). Proposition 2 provides an example for this case, where \(\theta \ge \frac{1}{2}( C_U + \bar{\varDelta })\) holds. This example can be also applied for more than two controls by setting the relaxed control values \(a_{i,b}\) to zero, for \(i>2\). Then, the difference to the upper bound \(\frac{3}{4} C_D + \frac{3}{2} \bar{\varDelta }\) from Proposition 3 is the one stated in the assertion.
\(\square \)
Proposition 5 tells us the DNFR scheme with \(C_2=\tfrac{3}{2}\) and \(\chi _D=1\) can not be improved. The following example motivates why we have chosen the set of the down time forbidden control in such a way that the active control can only be changed after a duration of \(C_D/2\) at the earliest. If it is possible to switch already after one interval \(\varDelta _j\), DNFR may construct greedy solutions with a large control deviation at long MD times \(C_D\). The following example shows the reason for this.
Example 1
Let \({n_\omega }=3\) and a rounding threshold \(C_2\ge \frac{2{n_\omega }-3}{2{n_\omega }-2}\) be given. We alter DNFR in the following way: instead over blocks we iterate forward over all intervals. We keep for a given MD time \(C_1=C_D\) the threshold \(C_2 \overline{\mathscr {L}}\) for forced, future forced and admissible activation. In order to construct feasible solutions for (CIA-D), we extend the definition of \(\mathscr {I}_b^D\) by letting all controls to be down time forbidden that are inactive and were active in the previous period of length \(C_D\). Next, we are going to construct exemplary relaxed values for this modified DNFR scheme with large control deviation. We first introduce recursively the indices
$$\begin{aligned} \quad j_i:=\min \left\{ j\in [N] \, \mid \, \sum \nolimits _{l=j_{i-1}}^j \varDelta _l >C_2 \overline{\mathscr {L}} \right\} , \quad i=1,2,3, \end{aligned}$$
where \(j_{0}:=1\). Let the relaxed values be given as follows
$$\begin{aligned} (a_{i,j})_{i\in [{n_\omega }],j\in [N]}= \left( \begin{array}{cccccccccc} 1 &{} \cdots &{} \overbrace{1}^{j_1} &{} 0 &{} \cdots &{} \overbrace{0}^{j_2} &{} 0 &{} \cdots &{} \overbrace{0}^{j_3} &{} \cdots \\ 0 &{} \cdots &{} 0 &{} 1 &{} \cdots &{} 1 &{} 0 &{} \cdots &{} 0 &{} \cdots \\ 0 &{} \cdots &{} 0 &{} 0 &{} \cdots &{} 0 &{} 1 &{} \cdots &{} 1 &{} \cdots \end{array} \right) . \end{aligned}$$
Then, the modified DNFR may construct the following solution:
$$\begin{aligned} (w_{i,j})_{i\in [{n_\omega }],j\in [N]}= \left( \begin{array}{cccccc} 1 &{} 0 &{} 0 &{} \cdots &{} \overbrace{0}^{j_4} &{} \cdots \\ 0 &{} 1 &{} 0 &{} \cdots &{} 0 &{} \cdots \\ 0 &{} 0 &{} 1 &{} \cdots &{} 1 &{} \cdots \end{array} \right) , \end{aligned}$$
where \( j_4:=\min \{j\in [N] \, \mid \, \sum \nolimits _{l\in [j]} \varDelta _l < C_D+\varDelta _1 \}\) is the last index before the MD phase of \(i_1\) ends. At first, \(i_1\) is earliest future forced, but not anymore after being active on \(j=1\). Then, \(i_2\) is activated on the second interval before \(i_3\) is the earliest future forced control and needs to stay active until \(j_4\), since the other controls are down time forbidden. We notice that with small \(\varDelta _1,\varDelta _2\) it can result \(j_4\le j_2\) and therefore
$$\begin{aligned} |\theta _{i_3,j_4}|= |\sum \limits _{l=3}^{j_4} (0-1)\varDelta _l | \le | C_D+\bar{\varDelta } - \varDelta _2 | \le C_D+\bar{\varDelta } -\underline{\varDelta } . \end{aligned}$$
If we compare the term on the right inequality side with the bound from Proposition 3, i.e., \(\frac{2{n_\omega }-3}{2{n_\omega }-2}(C_D+\bar{\varDelta })=\frac{3}{4}(C_D+\bar{\varDelta })\), we conclude that the latter is less only if \(C_D< 4\underline{\varDelta } - \bar{\varDelta }\). Since \(\underline{\varDelta }\) can be arbitrarily small and we assumed \(C_D\) big compared with the grid length, the modified DNFR scheme would construct no improving bounds. Similar “greedy” examples can be constructed for \({n_\omega }> 3\) and block lengths greater than \(\underline{\varDelta }\).
Some comments on these tightness properties are in order.
Remark 8
(Quality of bound for (CIA-D)) The MD time configuration of DNFR, i.e., \(\chi _D=1\), yields smaller upper bounds compared with the DNFR algorithm with MU time configuration, i.e. \(\chi _D=0\) and \(C_1=C_D\), only for instances with more than three controls and a large MD time \(C_D\) compared with the grid length \(\bar{\varDelta }\). In fact, we conjecture the upper bound for any \({n_\omega }\) to be \(\theta ^{\max } = \frac{1}{2} C_D + \bar{\varDelta }\), therefore only slightly greater than the one for \({n_\omega }=2\). With this threshold, there would be no forced control until the first down time forbidden control appears and we postulate that active controls that become forced without activation during the next \(C_D\) time duration may stay active without other controls becoming forced. Of course, this argumentation justifies no proof - Proposition 5 together with Example 1 states that a generic solution fulfilling this bound can not be found via the DNFR scheme and it is presumably hard, if not even impossible, to construct it by another polynomial time algorithm.
Remark 9
(Quality of bound for (CIA-UD)) The integrality gap bound for (CIA-UD) as stated in Proposition 3 is tight for \(C_U \ge C_D\) by the result of Proposition 2. For \(C_U < C_D\), the bound is not necessarily tight, but it is again difficult to prove tight bounds due to the problem’s combinatorial structure.
Remark 10
If we deal with an MDT \(C_1\) that begins and ends exactly on the grid points, the upper bounds become \(\ \frac{2{n_\omega }- 3}{2{n_\omega }- 2} C_U \) for (CIA-U), \(\ \frac{3}{4} C_D\) for (CIA-D), and accordingly reduced for (CIA-UD).
C Proof of Proposition 4
Proof
Using (6.1) and the definition of \(M_D\), we obtain \( M_D \ge 2({n_\omega }-1)\). Notice that even if \(C_D/\bar{\varDelta } \notin \mathbb {N}\), we still find for the cardinality of the dwell time index sets \(|{\mathscr {J}}_{k}^{\text {SUR}}(C_D)| = M_D \in \mathbb {N}\) for \(k\le N-M_D\) because we deal with an equidistant grid. Hence, we calculate the forward control deviation of the currently activated control in the DSUR algorithm (line 3) on the next \(M_D\) intervals.
We prove the claim by proving the following claims: For any \({n_\omega }\ge 2\), \(C_D\) and N fulfilling (6.1) and (6.2), there is an \(\mathbf {a}\in A\) with
$$\begin{aligned} a_{i,j} = 0, \qquad \text { for } \ \ i=2,\ldots ,{n_\omega }, \ j= 1,\ldots ,i-1, \end{aligned}$$
(C.1)
resulting in a constructed \(\mathbf {w}^{\text {DSUR}}\) with
$$\begin{aligned} w^{\text {DSUR}}_{i,j} = 0, \qquad \text { for } \ \ i=2,\ldots ,{n_\omega }, \ j= 1,\ldots ,i-1, \end{aligned}$$
(C.2)
and
$$\begin{aligned} \theta _{2,j} = \left( \frac{M_D}{2} + ({n_\omega }-2)\right) \bar{\varDelta }, \quad j = ({n_\omega }-1)(1+M_D) + \lceil M_D/2 \rceil - M_D. \end{aligned}$$
(C.3)
This implies the Claim 6.3 by the definition of the objective value of (CIA-D). We proceed via induction.
Base case:
\({n_\omega }= 2\): By assumption we have \(M_D \ge 2\bar{\varDelta }\) and thus a nontrivial MD time. We construct an \(\mathbf {a}\in A\) on \(N= (1+M_D)+\lceil M_D/2 \rceil -1\) intervals. If the Claim C.3 is true for this \(\mathbf {a}\), it does also hold for \(N\ge (1+M_D)+\lceil M_D/2 \rceil -1\) because we can extend \(\mathbf {a}\) by inserting arbitrary unit vector columns after the last column without affecting Claim C.3. We consider
Because of \(a_{2,1}=0\), C.1 is true. The DSUR algorithm activates the first control on interval \(j=1\). Then, \(i=1\) is the currently activated control. Assuming \(i=1\) is active until \(\, 2 \le k-1\le M_D/2 \,\) for \(M_D\) being even, respectively \(2 \le k-1\le \lceil M_D/2 \rceil \) for \(M_D\) being odd, its dwell time block index set is \({\mathscr {J}}_{k}^{\text {SUR}}(C_D)=\{k,\ldots ,k+M_D-1\}\) and its forward control deviation on interval k as given in line 4 of DSUR amounts to
$$\begin{aligned} \theta _{1,k-1}+\sum \limits _{l\in {\mathscr {J}}_{k}^{\text {SUR}}(C_D)}a_{1,l}\bar{\varDelta }=- (k-2)\bar{\varDelta }+ \left( M_D/2 + (k-2) \right) \bar{\varDelta } = \frac{M_D}{2} \bar{\varDelta }. \end{aligned}$$
On the other hand, the forward control deviation for \(i=2\) on these intervals k amounts to
$$\begin{aligned} \gamma _{2,k} = \theta _{2,k-1}+ a_{2,k}\bar{\varDelta }= {\left\{ \begin{array}{ll} (k-2)\bar{\varDelta }+ 0.5 \bar{\varDelta } = M_D/2, &{} \quad \text { if } M_D \text { odd and } k-1=\lceil M_D/2 \rceil , \\ (k-2)\bar{\varDelta }+ 1 \bar{\varDelta }=(k-1) \bar{\varDelta }\le M_D/2, &{} \quad \text { else.} \end{array}\right. } \end{aligned}$$
We observe that the forward control deviation for control \(i=1\) for all these intervals k is greater or equal to the one of \(i=2\) and we let DSUR deliberately choose \(i=1\) to be active in case of equality. Hence, \(w_{1,j}^{\text {DSUR}}=1\), for \(j\in [N]\). This implies the control \(i=2\) stays inactive and in particular (C.2) is true. Combining this with the above forward control deviation for \(i=2\) yields
$$\begin{aligned} \theta _{2,1+\lceil M_D/2 \rceil } = \frac{M_D}{2} \bar{\varDelta }, \end{aligned}$$
which settles the Claim (C.3) for \({n_\omega }=2\).
Inductive step: We show that, if the claim holds for \({n_\omega }-1\), then it is also true for \({n_\omega }\).
Let \(\mathbf {a}^{({n_\omega }-1)}\in [0,1]^{({n_\omega }-1) \times (({n_\omega }-2)(1+M_D) + \lceil M_D/2 \rceil -1)}\) denote a matrix for which DSUR constructs a \(\mathbf {w}^{\text {DSUR}}\) that satisfies the Claims (C.1)–(C.3) for \({n_\omega }-1\). We construct an \(\mathbf {a}\in A\) on \(N= ({n_\omega }-1)(1+M_D) + \lceil M_D/2 \rceil -1\) intervals and with \({n_\omega }\) controls. We can argue similarly to the base case that we can neglect the case \(N> ({n_\omega }-1)(1+M_D) + \lceil M_D/2 \rceil -1\). Let \({\mathbf {I}}_{k} \) denote the identity matrix of dimension \(k\times k\) and let \({\mathbf {0}}_{k}\) denote the zero matrix of dimension \(k\times n\), where n is specified by the dimension of the block matrix below the zero matrix. We consider the following matrix
where the third block of columns may be nonexistent, if \(2{n_\omega }> M_D +1\). The first two blocks of columns, however, are well-defined due to \(M_D \ge 2({n_\omega }-1)\) by (6.1) and thus \(2{n_\omega }-1 \le M_D+1\). The above matrix is defined on N intervals, with N chosen as above, since we add \(M_D + 1\) intervals to the existing \(({n_\omega }-2)(1+M_D) + \lceil M_D/2 \rceil -1\) intervals from \(\mathbf {a}^{({n_\omega }-1)}\). At first, we see that (C.1) is satisfied by \(\mathbf {a}\). Second, we claim that DSUR constructs the following \(\mathbf {w}^{\text {DSUR}}\in W\):
where \(\mathbf {w}^{\text {DSUR}, ({n_\omega }-1)}\) denotes the obtained solution of DSUR for \(\mathbf {a}^{({n_\omega }-1)}\). We first justify this value for the intervals \(k=1,\ldots ,{n_\omega }\):
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\(k=1\): DSUR selects the control \(i=1\) because of \(a_{1,1}=1\).
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\(k=2\): The control \(i=1\) is currently activated with a forward control deviation of \(\bar{\varDelta }\), calculated on the next \(M_D\) intervals. The forward control deviation for control \(i=2\) amounts to \(\gamma _{2,2}=\theta _{2,1}+a_{2,2}\bar{\varDelta }=0+\bar{\varDelta }\). Therefore, DSUR may set the control \(i=2\) to be active.
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\(k=3\): We use the induction hypothesis for \(\mathbf {a}^{({n_\omega }-1)}\) and Claim (C.1) that yields \(a_{2,M_D+2}^{({n_\omega }-1)}=0\). Thus, the forward control deviation of control \(i=2\) is \(\bar{\varDelta }\), which is the same for \(i=3\). We let DSUR deliberately set the control \(i=3\) to be active.
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\(k=4,\ldots ,{n_\omega }\): We argue analogously to the case \(k=3\).
Hence, (C.2) is established. After control \(i={n_\omega }\) has been activated on interval \(k={n_\omega }\), all other controls are down time forbidden until interval \(M_D+1\). Thus, control \(i={n_\omega }\) stays to be active up to and including interval \(M_D+1\). Because the controls \(i=1,\ldots ,{n_\omega }-1\) are only once active until interval \(M_D+1\), but \(\sum _{k\in [M_D+1]}a_{i,k}\bar{\varDelta }=2\bar{\varDelta }\), we conclude \(\theta _{i,M_D+1}=\bar{\varDelta }\). This justifies why DSUR constructs \(\mathbf {w}^{\text {DSUR}, ({n_\omega }-1)}\) after interval \(M_D+1\):
-
The controls \(i=2,\ldots ,{n_\omega }-1\) are down time forbidden on the intervals \(k=(M_D+1) +1,\ldots ,(M_D+1)+ i-1\), but are not activated in \(\mathbf {w}^{\text {DSUR}, ({n_\omega }-1)}\) on these intervals according to the induction hypothesis (C.2) anyway.
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The control deviation for control \({n_\omega }\) is negative, i.e., \(\theta _{{n_\omega },k}=-(M_D+1-{n_\omega })\bar{\varDelta }\) for \(k\ge M_D+1\) so that control \({n_\omega }\) is not activated after interval \(M_D+1\).
-
All other controls \(1,\ldots ,{n_\omega }-1\) start with the same control deviation \(\theta _{i,M_D+1}=\bar{\varDelta }\), when DSUR iterates on interval \(M_D+2\). Thus, DSUR constructs the same \(\mathbf {w}\) from \(\mathbf {a}^{({n_\omega }-1)}\) as it would construct from \(\mathbf {a}^{({n_\omega }-1)}\) starting with the first interval and \(\theta _{i,0}=0\). This implies by the induction hypothesis DSUR generates \(\mathbf {w}^{\text {DSUR}, ({n_\omega }-1)}\).
The induction hypothesis regarding (C.3) implies for \(\mathbf {w}^{\text {DSUR}, ({n_\omega }-1)}\)
$$\begin{aligned} \theta _{2,j}= & {} \left( \frac{M_D}{2} + (({n_\omega }-1)-2)\right) \bar{\varDelta }, \\ j= & {} (({n_\omega }-1)-1)(1+M_D) + \lceil M_D/2 \rceil - M_D. \end{aligned}$$
We argued that this control deviation value is increased in \(\mathbf {w}^{\text {DSUR}}\) by \(\bar{\varDelta }\) and before the choice \(\mathbf {w}^{\text {DSUR}, ({n_\omega }-1)}\) there exist \(M_D+1\) columns in \(\mathbf {w}^{\text {DSUR}}\). So, (C.3) is also true for \({n_\omega }\). \(\square \)