## Appendix A The Strong Duality  
**Theorem A1**. : _The strong duality of the prime optimization problem OPT-2 and its Lagrange dual problem holds, i.e., the optimal values of these two problems are identical._  
Proof. : The proof of this theorem is a little bit complicated. Before we move forward, we firstly rewrite OPT-2 as follows:  
\[ \begin{array}{ll}\min&g(\vec{d})\\ \text{s.t.}&\mathbf{A}^{T}\vec{d}-L=0\\ &g_{i}(\vec{d})\leqslant 0,i=0,\ldots,2n+1\end{array} \]  
where \(\vec{d}\) is a column vector, which is \(\{d_{0},d_{1},\ldots,d_{n}\}^{T}\), and \(\mathbf{A}\) is a column vector of which the entries are all one. The objective function \(g(\vec{d})\) stands for \(\Phi_{2f}\sum\limits_{i=0}^{n}d_{i}^{k}\). For \(0\leqslant i\leqslant n\), the inequality constraints stand for \(d_{i}-R\leqslant 0\). For \(n+1\leqslant i\leqslant 2n+1\), \(g_{i}(\vec{d})\) stands for \(-d_{i-(n+1)}\leqslant 0\). First of all, we denote the feasible solution set to OPT-2 as \(\mathcal{D}\). Then, we have the following facts:  
Figure 9: The energy consumption time curve of the node with the number 53 in Layer A. * It is obvious that \(\mathcal{D}\) is not empty, that is it has interior points. Additionally, we have certain \(\vec{d}\in\text{int}\mathcal{D}\), such that \(g_{i}(\vec{d})<0\) and \(\mathbf{A}^{T}\vec{d}=L\) (int\(\mathcal{D}\) stands for the set of interior points of \(\mathcal{D}\)). * The function \(g(\vec{d})\) and \(g_{i}(\vec{d})\) are all convex functions for \(\vec{d}\in\text{int}\mathcal{D}\), and the rank of \(\mathbf{A}\) is one. We denote the optimal value of OPT-2 as \(p^{\star}\), and we denote the set \(\mathcal{A}\) as: \[\mathcal{A}=\left\{g_{0}(\vec{d}),g_{1}(\vec{d}),\ldots,g_{2n+1}(\vec{d}), \mathbf{A}^{T}\vec{d},g(\vec{d}))\right\}\] (1) Then, we have: \[p^{\star}=\inf\left\{t\mid(u,w,t)\in\mathcal{A},u\leqslant 0,w=0\right\}\] (2) where \(u=\left\{g_{0}(\vec{d}),g_{1}(\vec{d}),\ldots,g_{2n+1}(\vec{d})\right\}\), \(w=\mathbf{A}^{T}\vec{d}\) and \(t=g(\vec{d})\). The Lagrangian dual function of OPT-2 can be written as: \[G(\lambda,v)=G(\lambda_{1,i},\lambda_{2,i},v)=\inf\left\{(\lambda,v,1)^{T}(u,w,1)\mid(u,w,t)\in\mathcal{A}\right\}\] (3) Now, we define the \"epigraph\" (the word \"epigraph\" means \"upper graph\") of set \(\mathcal{A}\) as: \[\mathcal{A}_{E}=\mathcal{A}+(\mathcal{R}_{+}^{2n+2}\times\{0\}\times\mathcal{ R}_{+})\] (4) The convexity of the set \(\mathcal{A}_{E}\) can be justified easily through the definition of convex sets. We define another convex set: \[\mathcal{B}=\left\{(\vec{0},0,s)\in\mathcal{R}^{2n+2}\times\mathcal{R}\times \mathcal{R}\mid s<p^{\star}\right\}\] (5) Then, we can prove that the two convex sets, \(\mathcal{A}_{E}\) and \(\mathcal{B}\), do not intersect. Suppose that \(\mathcal{A}_{E}\cap B\
eq\emptyset\); then, we have at least one element \((u,w,t)\in\mathcal{A}_{E}\cap\mathcal{B}\). Therefore, we have \(u=0\),\(w=0\) and \(t<p^{\star}\). Since \((u,w,t)\in\mathcal{A}_{E}\), then we have at least one \(\vec{d}\), such that \(g_{i}(\vec{d})\leqslant 0\), \(\mathbf{A}^{T}\vec{d}=0\) and \(g(\vec{d})\leqslant t<p^{\star}\), which is impossible, since \(p^{\star}\) is already the optimal solution to OPT-2. Hence, \(\mathcal{A}_{E}\cap\mathcal{B}=\emptyset\). Since \(\mathcal{A}_{E}\) and \(\mathcal{B}\) are both convex sets that do not intersect, there must exist a separating hyperplane between them. Therefore, there exists a non-zero vector \((\vec{\lambda},\vec{0},\mu)\) and a non-negative number \(\alpha\), such that: \[\forall(u,w,t)\in\mathcal{A}_{E}\Rightarrow(\vec{\lambda},\vec{0},\mu)^{T}(u,w,t)\geqslant\alpha\] (6) \[\forall(u,w,t)\in\mathcal{B}\Rightarrow(\vec{\lambda},\vec{0},\mu)^{T}(u,w,t)\leqslant\alpha\] (7)  
From Equation (6), we have \(\vec{\lambda}\geq 0\) and \(\mu\geqslant 0\), otherwise \((\vec{\lambda},\vec{0},\mu)^{T}(u,w,t)\) is unbounded below. From Equation (7), we have \(\mu t\leqslant\alpha\) for any \(t<p^{\star}\).