# Optimal Resource Allocation Policies for Multi-User Backscatter Communication Systems
## 4 Optimal Resource Allocation of Passive Tags  
In this section, the optimal solution for Problem P1 is derived. First, Problem P1 is decomposed into \(N\) convex sub-problems. Then, the optimal solutions for these feasible sub-problems are derived by convex optimization techniques. Last, the solution for the sub-problem with the maximum total goodput also solves Problem P1. \(G_{i}(P_{i},n_{i})\) is an increasing function with respect to \(P_{i}\). To maximize the total goodput, these tags with better channel conditions will be allocated power (or be activated) with priority. Hence, there are at most \(N\) permutations for activated passive tags. Without loss of generality, assuming the channel gains of \(N\) passive tags satisfy \(h_{1}\geq h_{2}\geq...\geq h_{N}\), these \(N\) permutations for the indices of activated passive tags are given as \([1,...,N]\), \([1,...,N-1]\),..., \([1]\). Based on the above permutations, Problem P1 can be decomposed into \(N\) sub-problems. That is to say, each permutation corresponds to a sub-problem. For the \(i\)-th sub-problem, the number of activated passive tags is \(M\), where \(M=N-i+1\). \(G_{i}(P_{i},n_{i})\) is also an increasing function with respect to \(n_{i}\), so that \(n_{i}\) can keep as \(n_{i}=n_{i}^{\text{max}}\) over the whole duration in order to maximize the power of backscattered signals. The \(i\)-th sub-problem is given by \[(\mathbf{P2}) \max_{\hat{\mathbf{P}}} \sum_{m=1}^{M}\hat{G}_{m}(P_{m})\] s.t. \[P_{m}^{\text{th}}\leq P_{m},\quad P_{m}\leq P^{\max},\quad\forall m,\] \[\sum_{m=1}^{M}P_{m}\leq NP^{\text{ave}},\]  
where \(\hat{G}_{m}(P_{m})=T\left[\frac{1}{2}+\frac{1}{2}\text{erf}\left(\sqrt{a_{m}P_{ m}-b_{m}}\right)\right]\), \(b_{m}=\frac{P^{\text{ave}}g_{m}}{\sigma^{2}\eta}\), and \(\hat{\mathbf{P}}=[P_{1},...,P_{M}]\). It can prove that \(\hat{G}_{m}(P_{m})\) is a concave function by deriving its second derivative with respect to \(P_{m}\). It follows that \(\sum_{m=1}^{M}\hat{G}_{m}(P_{m})\) is also a concave function, since it is the summation of \(\hat{G}_{m}(P_{m})\). Hence, Problem P2 is a convex problem [26]. Before solving the \(i\)-th sub-problem, we first check its feasibility. The feasibility of Problem P2 is given as \(P_{m}^{\text{th}}\leq P^{\max}\), \(\forall m\) and \(\sum_{m=1}^{M}P_{m}^{\text{th}}\leq NP^{\text{ave}}\). If Problem P2 is feasible, the optimal solution for Problem P2 can be derived by considering two sub-cases. First, check whether \(NP^{\text{ave}}\geq MP^{\max}\). If so, the optimal solution for Problem P2 can be easily derived below. Theorem 4.1: _The optimal power allocation of Problem P2 under the condition \(NP^{\text{ave}}\geq MP^{\max}\) is given by_  
\[P_{m}^{*}=P^{\max},\qquad\forall m.\]  
In order to derive the optimal solution for Problem P1, all feasible sub-problems need to be solved sequentially. To reduce the complexity, a useful conclusion is given in the following lemma. Lemma 1: _If \(NP^{\text{ave}}\geq MP^{\max}\) for the \(i\)-th sub-problem, then the optimal solutions for \(i+1\)-th,..., \(N\)-th sub-problem are not the optimal solution for Problem P1._  
Proof: See Appendix A. Based on Lemma 1, it is not necessary to solve the \(i+1\)-th,..., \(N\)-th sub-problem if \(NP^{\text{ave}}\geq MP^{\max}\), since the optimal solution for Problem P1 can be found from the first \(i\) sub-problems. If \(NP^{\text{ave}}<MP^{\max}\), Problem P2 is solved by Lagrange method. The optimal power allocation is given in the following theorem.