# Optimal Resource Allocation Policies for Multi-User Backscatter Communication Systems
## 4 Optimal Resource Allocation of Passive Tags
\begin{array}{ll}\mathbf{(P3)}&\max_{\mathbf{P},\mathbf{\hat{r}}^{a},\mathbf{n}} &\sum_{i=1}^{N}G_{i}(P_{i},t_{i}^{a},n_{i})\\ \text{s.t.}&P_{i}=P_{i}\mathbb{1}\,(P_{i}\geq P_{i}^{\text{th}}),&\forall i,\\ &P_{i}\leq P^{\text{max}},\quad 0\leq t_{i}^{a}\leq T,&\forall i,\\ &n_{i}^{\min}\mathbb{1}\,(P_{i}\geq P_{i}^{\text{th}})\leq n_{i}\leq 1(P_{i} \geq P_{i}^{\text{th}}),&\forall i,\\ &P^{\text{c}}t_{i}^{a}\leq\eta P_{i}h_{i}(T-t_{i}^{a})+\eta(1-n_{i})P_{i}h_{i} t_{i}^{a},&\forall i\\ &\frac{1}{N}\sum_{i=1}^{N}P_{i}\leq P^{\text{ave}},\end{array} \]

where \(\mathbf{t}^{a}=[t_{1}^{a},\cdots,t_{N}^{a}]\). Problem P3 is also a non-convex problem. Different from the case of passive tags, Problem P3 is more complicated. In the following section, both the priority of the allocated reader transmission power and the biconvex property are used to solve Problem P3. ## 6 The Optimal Resource Allocation of Semi-Passive Tags

Like the case of passive tags, the allocated reader transmission power controls the decision to \"activate or not\". Hence, there are also at most \(N\) permutations for activated semi-passive tags. Based on the analysis in Section 4, Problem P3 is also decomposed into \(N\) sub-problems. The formulation of the \(i\)-th sub-problem is given by

\[ \begin{array}{ll}\mathbf{(P4)}&\max_{\mathbf{\hat{p}},\mathbf{\hat{r}}^{a}, \mathbf{n}}&\sum_{m=1}^{M}t_{m}^{a}\left[\frac{1}{2}+\frac{1}{2} \text{erf}(\sqrt{a_{m}P_{m}n_{m}})\right]\\ &\text{s.t.}&P_{m}^{\text{th}}\leq P_{m},\qquad P_{m}\leq p^{\text{max}},& \forall m,\\ &0\leq t_{m}^{a}\leq T,\quad n_{m}^{\min}\leq n_{m}\leq 1,&\forall m,\\ &P^{\text{c}}t_{m}^{a}\leq\eta P_{m}h_{m}(T-t_{m}^{a})+\eta(1-n_{m})P_{m}h_{m} t_{m}^{a},&\forall m,\\ &\sum_{m=1}^{M}P_{m}\leq NP^{\text{ave}},\end{array} \]

where \(\mathbf{\hat{t}}^{a}=[t_{1}^{a},...,t_{M}^{a}]\) and \(\mathbf{\hat{n}}=[n_{1},...,n_{M}]\). The feasible condition of Problem P4 is also given as \(P_{m}^{\text{th}}\leq p^{\text{max}},\forall m\) and \(\sum_{m=1}^{M}P_{m}^{\text{th}}\leq NP^{\text{ave}}\). Assume Problem P4 is feasible. Before solving Problem P4, a useful structure for the optimal solution is given in the following lemma. Lemma 2: _The optimal \(P_{m}^{*}\), \(t_{m}^{a*}\), and \(n_{m}^{*}\) for solving Problem P4 satisfy:_

\[P^{\text{c}}t_{m}^{a*}=\eta P_{m}^{*}h_{m}(T-t_{m}^{a*})+\eta(1-n_{m}^{*})P_{m }^{*}h_{m}t_{m}^{a*},\quad\forall m.\]

Proof: See Appendix C.

To solve Problem P4, two sub-cases mentioned in Section 4 are considered. If \(NP^{\text{ave}}\geq MP^{\text{max}}\), let \(P_{m}^{*}=p^{\text{max}},\forall m\). Lemma 2 indicates that extending the active time by full use of harvested power leads to the maximum total goodput. With Lemma 2 and \(P_{m}^{*}=P^{\text{max}},\forall m\), \(n_{m}\) is rewritten as \(n_{m}=\frac{T}{t_{m}^{a}}-\frac{p^{\text{max}}}{\eta P^{\text{max}}h_{m}}\). Combining with the constraints \(n_{m}^{\min}\leq n_{m}\leq 1\) and \(0\leq t_{m}^{a}\leq T\) leads to the new constraint \(\frac{\eta P^{\text{max}}h_{m}T}{P^{\text{c}}+p^{\text{max}}h_{m}}\leq t_{m}^{a }\leq\hat{T}\), where \(\hat{T}_{m}=\min\{\frac{\eta P^{\text{max}}h_{m}T}{P^{\text{c}}+\eta P^{\text{ min}}_{m}\text{max}h_{m}},T\}\). Based on the above analysis, Problem P4 can be reduced to Problem P5 as follows. \[(\mathbf{P5}) \max_{\mathbf{i}\epsilon} \sum_{m=1}^{M}t_{m}^{a}\left[\frac{1}{2}+\frac{1}{2}\mathrm{erf} \left(\sqrt{\frac{\gamma_{m}}{t_{m}^{a}}-b_{m}}\right)\right]\] s.t. \[\frac{\eta P^{\max}h_{m}T}{P^{c}+\eta P^{\max}h_{m}}\leq t_{m}^{ a}\leq\hat{T}_{m},\quad\forall m,\]

where \(\gamma_{m}=c_{m}P^{\max}\). It is easy to prove that Problem P5 is a convex problem. Exploiting the Lagrange method and KKT conditions, the optimal active time and reflection ratio with \(P_{m}^{*}=P^{\max},\forall m\) are given in the following theorem.