A characterization of max–min SIR-balanced power allocation with applications

Abstract

We consider a power-controlled wireless network with an established network topology in which the communication links (transmitter–receiver pairs) are corrupted by the co-channel interference and background noise. We have fairly general power constraints since the vector of transmit powers is confined to belong to an arbitrary convex polytope. The interference is completely determined by a so-called gain matrix. Assuming irreducibility of this gain matrix, we provide an elegant characterization of the max–min SIR-balanced power allocation under such general power constraints. This characterization gives rise to two types of algorithms for computing the max–min SIR-balanced power allocation. One of the algorithms is a utility-based power control algorithm to maximize a weighted sum of the utilities of the link SIRs. Our results show how to choose the weight vector and utility function so that the utility-based solution is equal to the solution of the max–min SIR-balancing problem. The algorithm is not amenable to distributed implementation as the weights are global variables. In order to mitigate the problem of computing the weight vector in distributed wireless networks, we point out a saddle point characterization of the Perron root of some extended gain matrices and discuss how this characterization can be used in the design of algorithms in which each link iteratively updates its weight vector in parallel to the power control recursion. Finally, the paper provides a basis for the development of distributed power control and beamforming algorithms to find a global solution of the max–min SIR-balancing problem.

Appendix 1

1.1 1.1 Auxiliary results

For completeness and reader’s convenience, we restate here a useful result after [45, Theorem A.35], [42, Theorem 2.1], which we refer to in some proofs below.

Theorem 6

LetXbe an arbitrary non-negative matrix, and let α > 0 be any scalar. A necessary and sufficient condition for a solutionp ≥ 0, p ≠ 0, to

$$ (\alpha{\bf I}-{\bf X}){\bf p}={\bf b} $$

(25)

to exist for anyb > 0 is that α > r = ρ(X). In this case, there is only one solutionp, which is strictly positive and given byp = (αI − X)⁻¹b.

1.2 1.2 Proof of Lemma 1

By Definition 1 and (3), we have $t'\leq\min_{k\in{\mathcal{K}}}(\hbox{SIR}_k({\bf p})/\gamma_k),{\bf p}\in\hbox{P}$, if and only if p is a max–min SIR-balanced power vector. Therefore, in a matrix form, we have ${\varvec{\Upgamma}}{\bf z}\leq(\frac{1}{t^{\prime}}{\bf I}-{\varvec{\Upgamma}}{\bf V}){\bf p},\,{\bf p}\in\hbox{P}$ if and only if p is a solution to (2). Since a solution exists and is positive, it follows from Theorem 6 that

$$ \rho({\varvec{\Upgamma}}{\bf V})<1/t^{\prime} , $$

(26)

where ρ(·) is used to denote the spectral radius. So, by Theorem 6 we know that there exists a unique positive vector $\bar{\bf p}^{\prime}\in\hbox{P}$ given by (6) and this vector is a solution to the max–min SIR-balancing problem (2). Writing (6) as a system of K SIR equations we get $\hbox{SIR}_k(\bar{{\bf p}}')/\gamma_k=t'$ for each $k\in{\mathcal{K}}$ which completes the proof.

1.3 1.3 Proof of Lemma 2

By the definition of F in (5), for any q ∈ F, there exists p(q) ∈ P₊ given by

$$ {\bf p}({\bf q})=({\bf I}-{\bf G}({\bf q}){\varvec{\Upgamma}}{\bf V})^{-1}{\bf G}({\bf q}){\varvec{\Upgamma}}{\bf z},\;\quad{\bf q}\in\hbox{F} $$

(27)

where ${\bf G}({\bf q}):=\hbox{diag}(g(q_1),\ldots,g(q_K))$ with g(x) defined by (A.3). From Theorem 6 we know that $\rho({\bf G}({\bf q}){\varvec{\Upgamma}}{\bf V})<1$ and p(q) is unique so that p(q) defines a function (map) from F into P₊. Furthermore, the Neumann series $\sum_{l=0}^\infty({\bf G}({\bf q}){\varvec{\Upgamma}}{\bf V})^{l}$ exists and converges (see for instance [29, p. 618] and [45, Theorem A.11]) so that

$$ {\bf p}({\bf q})=\sum_{l=0}^\infty({\bf G}({\bf q}){\varvec{\Upgamma}}{\bf V})^{l}{\bf G}({\bf q}){\varvec{\Upgamma}}{\bf z}. $$

(28)

Now, considering strict monotonicity and continuity of $g:\hbox{Q}\to{{\mathbb{R}}_{++}}$ shows that for every p ∈ P, there is exactly one q ∈ F such that p = p(q). Combining this with the above let us conclude that there is a one-to-one correspondence between F and P. This map is continuous and defined by (27). So, by [40, Theorem 4.22], F is a connected set. Moreover, as g is strictly increasing, the right-hand side of (28) shows that if q ∈ F and q′ ≤ q, q′ ∈ Q^K, then we must have q′ ∈ F, which implies downward comprehensivity. Finally, with (A.3), the convexity property follows from [7] or [45, Sect. 5.3].

1.4 1.4 Proof of Proposition 1

Let (A.3) be satisfied, and let $\bar{\bf p}\in{\hbox{P}_{+}}$ be any solution to (2) or, equivalently, (4). Let $\bar{q}_k=\phi(\hbox{SIR}_k(\bar{\bf p})/\gamma_k),k\in{\mathcal{K}}$ . By Lemma 2, F is a convex downward comprehensive set and, by (A.2), (4) and (5), $\bar{\bf q}=(\bar{q}_1,\ldots,\bar{q}_K)\in\partial\hbox{F}$ is its boundary point since at least one power constraint is active in the optimum (see Lemma 4). Thus, by irreducibility of V and Lemma 3, there exists w > 0 such that ${\bf w}^T(\bar{{\bf q}}-{\bf u})\geq 0$ for all u ∈ F. Due to positivity of w, this implies that for any ${\bf u}\in\hbox{F},{\bf u}\neq\bar{{\bf q}}$, there is at least one index $i=i(\bar{{\bf q}}, {\bf u})\in{\mathcal{K}}$ such that $\bar{q}_{i}>u_{i}$. One particular solution to (2) is $\bar{{\bf p}}^{\prime}$ given by (6), therefore for any ${\bf u}\in\hbox{F},{\bf u}\neq\bar{\bf q}^{\prime}$ there is $i=i(\bar{\bf q}^{\prime},{\bf u})\in{\mathcal{K}}$ such that $\bar{q}_{i}^{\prime}>u_i$ where $\bar{q}^{\prime}_k=\phi(\hbox{SIR}_k(\bar{\bf p}^{\prime})/\gamma_k),k\in{\mathcal{K}}$. On the other hand, however, we have $\bar{\bf q}^{\prime}\leq\bar{\bf q}$, since by Observation 1 and Definition 1 there holds $\bar{q}^{\prime}_1=\cdots=\bar{q}^{\prime}_K= \min_{k\in{\mathcal{K}}}\phi(\hbox{SIR}_k(\bar{{\bf p}})/\gamma_k)$. Combining both inequalities shows that $\bar{{\bf q}}=\bar{\bf q}^{\prime}$, and hence, by bijectivity, we obtain $\bar{\bf p}=\bar{\bf p}'$, which is unique by Theorem 6.

1.5 1.5 Proof of Lemma 4

Part (1) should be obvious since if we had $g_n(\bar{{\bf p}})<1$ for all $n\in{\mathcal{N}}$, then it would be possible to increase $\min_{k\in{\mathcal{K}}}\hbox{SIR}_k(\bar{{\bf p}})/\gamma_k$ by allocating the power vector $c\bar{{\bf p}}\in{\hbox{P}_{+}}$ with $c=1/\max_{n\in{\mathcal{N}}}g_n(\bar{{\bf p}})>1$. In order to show part (ii), note that if (A.2) and (A.3) hold, then, by Lemma 2, F is a convex downward comprehensive set. Moreover, $\bar{{\bf p}}\in{\hbox{P}_{+}}$ given by (4) corresponds to a point $\bar{{\bf q}}\in\partial\hbox{F}$, with $\bar{q}_k=\phi(\hbox{SIR}_k(\bar{{\bf p}})/\gamma_k),k\in{\mathcal{K}}$. Thus, by irreducibility of V, it follows from Lemma 3 that $\bar{{\bf q}}$ is a maximal point of F, and hence $\bar{{\bf q}}\leq{\bf q}$ for any q ∈ F implies that ${\bf q}=\bar{{\bf q}}$ [10]. That is, there is no vector in F that is larger in all components than $\bar{{\bf q}}$. On the other hand, by the discussion in Sect. 3, $\bar{{\bf q}}$ is a point where the hyperplane in the direction of the vector $(1/K,\ldots,1/K)$ intersects the boundary of F. As a result, $\bar{q}_1=\cdots=\bar{q}_K$, which together with the maximality property and strict monotonicity of ϕ, shows that $\hbox{SIR}_k(\bar{{\bf p}})/\gamma_k=\beta$ for each $k\in{\mathcal{K}}$ where β is positive due to (i). If V is irreducible, the uniqueness of $\bar{{\bf p}}$ follows from Proposition 1.

1.6 1.6 Proof of Lemma 6

Let $n\in{\mathcal{N}}$ be arbitrary. First we prove part (i). Since $1/P_n{\bf z}{\bf c}_n^T\geq 0$ and V is irreducible, we can conclude from (15) that B ⁽ⁿ⁾ ≥ 0 is irreducible as well. Thus, by the Perron–Frobenius theorem for irreducible matrices [29, 23], there exists a positive vector p which is an eigenvector of B ⁽ⁿ⁾ associated with ρ(B ⁽ⁿ⁾), and there are no nonnegative eigenvectors of B ⁽ⁿ⁾ associated with ρ(B ⁽ⁿ⁾) other than p and its positive multiples. Among all the positive eigenvectors, there is exactly one eigenvector p > 0 such that g _n (p) = c ₁. This proves part (i). In order to prove (ii), note that if A ⁽ⁿ⁾ was irreducible, then we could invoke the Perron–Frobenius theorem and proceed essentially as in part (i) to conclude (ii) (with the uniqueness property resulting from the normalization of the eigenvector so that its last component is equal to c ₂ > 0). In order to show that A ⁽ⁿ⁾ is irreducible, let ${\mathcal{G}}({\bf A}^{(n)})$ be the associated directed graph of $\{1,\ldots,K+1\}$ nodes [29]. Since ${\varvec{\Upgamma}}{\bf V}$ is irreducible, it follows that the subgraph ${\mathcal{G}}({\varvec{\Upgamma}}{\bf V})$ is strongly connected [29]. Furthermore, as the vector ${\varvec{\Upgamma}}{\bf z}$ is positive, we can conclude from (13) that there is a directed edge leading from node K + 1 to each node n < K + 1 belonging to the subgraph ${\mathcal{G}}({\varvec{\Upgamma}}{\bf V})$. Finally, note that as ${\varvec{\Upgamma}}{\bf V}$ is irreducible, each row of ${\varvec{\Upgamma}}{\bf V}$ has at least one positive entry. Hence, the vector $1/P_n{\bf c}_n^T{\varvec{\Upgamma}}{\bf V}$ has at least one positive entry as well, from which and (13) it follows that there is a directed edge leading from a node belonging to ${\mathcal{G}}({\varvec{\Upgamma}}{\bf V})$ to node K + 1. So, ${\mathcal{G}}({\bf A}^{(n)})$ is strongly connected, and thus A ⁽ⁿ⁾ is irreducible.

1.7 1.7 Proof of Theorem 1

(i)→(ii): By Lemma 5, $\bar{{\bf p}}\in{\hbox{P}_{+}}$ satisfies (14) for some β > 0. Thus, by Lemma 6, part (i) implies part (ii). (ii)→(iii): Given any $n\in{\mathcal{N}}_0(\bar{{\bf p}})$, it follows from (14) that $\rho({\bf B}^{(n)})\bar{{\bf p}}={\bf B}^{(n)}\bar{{\bf p}}$ with $g_n(\bar{{\bf p}})=1$ is equivalent to $\rho({\bf B}^{(n)})\bar{{\bf p}}={\varvec{\Upgamma}}{\bf V}\bar{{\bf p}}+{\varvec{\Upgamma}}{\bf z}$, which in turn can be rewritten to give (12) with $\beta=\rho({\bf B}^{(n)})$ and $\bar{\underline{\bf p}}=(\bar{{\bf p}},1)$. Since $\bar{{\bf p}}$ is positive, so is also $\bar{\underline{\bf p}}$. Thus, $\bar{\underline{\bf p}}$ with $\bar{\underline{p}}_{K+1}=1$ is a positive right eigenvector of A ⁽ⁿ⁾ and the associated eigenvalue is equal to ρ(B ⁽ⁿ⁾) > 0. So, considering part (ii) of Lemma 6, $\bar{\underline{\bf p}}$ is unique and therefore we can conclude that (iii) follows from (ii). (iii)→(i): The solution to the problem (8) always exists. By Lemma 5 and Proposition 1, the irreducibility of V implies that the unique solution $\bar{{\bf p}}$ satisfies (12) with $\bar{\underline{\bf p}}=(\bar{{\bf p}},1)$ for each $n\in{\mathcal{N}}_0(\bar{{\bf p}})$. On the other hand, the normalization $\bar{\underline{p}}_{K+1}=1$ guarantees, by Lemma 6, that for each $n\in{\mathcal{N}}_0(\bar{{\bf p}})$, there exists exactly one positive vector $\bar{\underline{\bf p}}$ with $\bar{\underline{p}}_{K+1}=1$ such that (12) is satisfied. It corresponds thus to the unique solution of (8) which proves the last missing implication.

1.8 1.8 Proof of Lemma 7

By (5) with (A.2), we have q ∈ F if and only if there is p ∈ P such that ϕ$(\hbox{SIR}_k({\bf p})/\gamma_k)$ ≥ q _k for each $k\in{\mathcal{K}}$. Thus, q ∈ F if and only if $1/\lambda:=\max_{{\bf p}\in\hbox{P}}\min_{k\in{\mathcal{K}}}(\hbox{SIR}_k({\bf p})/\gamma_kg(q_k))\geq 1$ where the maximum always exists. Comparing the left hand side of the inequality above with (2) shows that the only difference to the original problem formulation is that γ _k is substituted by γ _k g(q _k ) or, equivalently, ${\varvec{\Upgamma}}$ by ${\bf G}({\bf q}){\varvec{\Upgamma}}$, which is positive definite as well. Thus, by (9), Theorem 1 and Theorem 2, we have q ∈ F if and only if $\lambda=\max_{n\in{\mathcal{N}}}\lambda_n({\bf q})\leq 1$. Moreover, p(q) given by (22) is the unique power vector such that $q_k = \phi(\text{SIR}_k(\mathbf{p}(\mathbf{q}))/\gamma_k) \text{ for each } k\in{\mathcal{K}}$. Since the Neumann series converges for any q ∈ F, we have ${\bf p}({\bf q})=\sum_{l=0}^\infty({\bf G}({\bf q}){\varvec{\Upgamma}}{\bf V})^l{\bf G}({\bf q}){\varvec{\Upgamma}}{\bf z}$. Now as ${\bf G}({\bf q}){\varvec{\Upgamma}}{\bf z}$ is positive and ${\bf G}({\bf q}){\varvec{\Upgamma}}{\bf V}$ is irreducible, we can conclude from [45, Lemma A.22] and (A.2) that each entry of p(q) is strictly increasing in each entry of q. Thus, as F is downward comprehensive and ${\bf q}\notin\partial\hbox{F}$ holds if and only if all power constraints are inactive, for every q ∈ int(F), there is $\tilde{{\bf q}}\in\partial\hbox{F}$ such that $\tilde{{\bf q}}={\bf q}+{\bf u}$ for some u > 0. By irreducibility of B ⁽ⁿ⁾, this implies that $\lambda_n({\bf q})<\lambda_n({\bf q}+{\bf u})=\lambda_n(\tilde{{\bf q}})\leq 1$ for each $n\in{\mathcal{N}}$ So, if $\max_{n\in{\mathcal{N}}}\lambda_n({\bf q})=1$, then q ∈ ∂F. Conversely, if q ∈ ∂F, we must have $\max_{n\in{\mathcal{N}}}\lambda_n({\bf q})=1$ since otherwise there would exist $\tilde{{\bf q}}\notin\hbox{F}$ such that $\max_{n\in{\mathcal{N}}}\lambda_n(\tilde{{\bf q}})=1$, which would contradict Theorem 2. This completes the proof.

1.9 1.9 Proof of Theorem 3

Let $\tilde{{\bf q}}\in\partial\hbox{F}$ and $n_0=\arg\max_{n\in{\mathcal{N}}}\lambda_n({\bf q})$ be arbitrary and note that F is a convex set. So, by Lemma 3, there is w > 0 such that $\tilde{{\bf q}}$ maximizes ${\bf q}\mapsto{\bf w}^T{\bf q}$ over F. Lemma 7 implies that this convex problem can be stated as $\tilde{{\bf q}}=\arg\max_{{\bf q}}{\bf w}^T{\bf q}$ subject to $\lambda_{n_0}({\bf q})=1,{\bf q}\in\hbox{Q}^K$. Due to (A.2), the spectral radius is continuously differentiable on Q^K. Thus, the Karush–Kuhn–Tucker conditions [10], which are necessary and sufficient for optimality (due to the convexity property), imply that w is parallel with $\nabla\lambda_{n_0}({\bf q})$. Now, by [14], we have $\frac{\partial\lambda_{n_0}({\bf q})}{\partial q_k}= y_kg^{\prime}(q_k)\sum_{l\in{\mathcal{K}}} b_{k,\,l}^{(n_0)}x_l =\frac{g^{\prime}(q_k)}{g(q_k)}y_k\sum_{l\in{\mathcal{K}}} g(q_k)b_{k,\,l}^{(n_0)}x_l =\lambda_{n_0}({\bf q})\frac{g^{\prime}(q_k)} {g(q_k)}y_k x_k =\frac{g^{\prime}(q_k)}{g(q_k)}y_k x_k$ for each $k\in{\mathcal{K}}$ where y and x are left and right positive eigenvectors of ${\bf G}({\bf q}){\bf B}^{(n_0)}$ associated with $\lambda_{n_0}({\bf q})$, which, by irreducibility, are unique up to positive multiples.

1.10 1.10 Proof of Corollary 1

As V is irreducible, Observation 1 and Proposition 1 (see also the proof) imply that $\bar{{\bf p}}$ corresponds to a point $\bar{{\bf q}}\in\partial\hbox{F}$. Since ${\bf q}^\ast({\bf w})\in\partial\hbox{F}$ for any w > 0, it follows from Theorem 3 that $\bar{{\bf q}}={\bf q}^\ast({\bf w})$ if w is has the form (23). Now by Observation 1, we have $\bar{q}_1=\cdots=\bar{q}_K$. Thus, as both g and its derivative g′ are strictly monotonic (by (A.2) and (A.3)), we must have ${\bf u}(\bar{{\bf q}})=a{\bf 1},a>0$ and ${\bf G}(\bar{\bf q})=1/\rho({\bf B}^{(n_0)}){\bf I}$. Thus, the corollary follows from Theorem 3.