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8. Suppose that 1) and 2) are fulfilled. Also, p ≥ 0 fulfills γk Ik (p ) = C(Γ)pk , ∀k, and there exists an index k0 such that pk0 = 0, then p has at least two zero components. 28 Axiomatic SIR-Balancing Theory Proof. Suppose that pk0 = 0 is the only zero component. Because of 1) and 2), we have Ik0 (p ) > 0, which leads to the contradiction 0 < γk0 Ik0 (p ) = C(Γ)pk0 = 0. 9. Suppose that 1) and 2) are fulfilled. For K = 2, 3, there exists exactly one vector p ≥ 0, p = 0, such that C(Γ)pk = γk Ik (p), ∀k, and this vector fulfills p > 0.

Similarly, we have (n) K2 ≤ Ck · p − p(n) ∞ . Thus, |Ik (p) − Ik (p(n) )| ≤ Ck · p − p(n) ∞ . Since p(n) → p, we can conclude that lim Ik (p(n) ) = Ik (p) , n→∞ which means that Ik is continuous. g. 3, where an iterative algorithm is derived. 4 Interference-coupling and irreducibility The interference coupling can be modeled by a directed graph, where each user is represented by a node. The nodes are connected by directed edges which are given by the positive entries of the link gain matrix Ψ. If Ψij > 0 then there is a connection from node i to node j.

2), is not achieved, but approached by maxk γk /SIRk (p) arbitrarily close, so all quantities γk /SIRk are asymptotically balanced at the common level C(Γ). In this sense, the expression ‘SIR balancing’ is justified. These effects will be illustrated by the following example. 5. (no min-max optimizer exists) Consider the simple interference function I(p) = Ψp, where Ψ= Ψ(1) 0 V Ψ(2) , with blocks Ψ(1) and Ψ(2) along the diagonal, and an off-diagonal Block V > 0. The first sub-system Ψ(1) receives no interference, because its off-diagonal block is zero.

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Answers Sydsaeter & Hammond - Mathematics for Economic Analysis


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