Sentence examples for be stated as max from inspiring English sources

Exact(1)

Using (7), (10) and the power constraint T ̄ ≤ T, the power-limited average rate maximization problem can be stated as max g ̃ i, ĝ i, T i 2 ∑ i = 2 N P i R i + ∑ i = 1 N Q i R i subject to 2 ∑ i = 2 N P ′ i T i + ∑ i = 1 N Q ′ i T i ≤ T (13).

Similar(59)

Then, the optimization problem can be stated as solving max t j j = 1, …, m Φ. (2).

In this perspective, for a power constraint T, the power-limited throughput optimization problem can be stated as R ̄ max ( 1, …, N ) = max ∀ g ~ n, m, ĝ n, m, T n, m ∑ n = 1 N ∑ m = 1 M n φ n, m β n, m log 1 + ĝ n, m T n, m s.t.

The dual function g can now be stated as g = max L ( μ k, λ ) s.t.

The maximum of (1) is then found as the supremum of the finite set {G(0),…,G(M−1)}, generated from (6) by varying c(m) in M. The optimal coding mode assignment c m opt can be stated as follows m opt = arg max m ∈ { 0, …, M − 1 } G ( m ) (7).

We first consider the optimization of the time allocation vector θ - = [ θ 1 θ 2... θ N - 1 ] T for a given rate ratio vector σ - = [ σ 1 σ 2... σ N ] T. Considering the conditions in (24), the optimization can be stated as follows: θ - * = arg max θ - R (25).

Virtual SINR maximizing beamforming scheme proposes that each BS solve a virtual SINR maximization problem and it can be stated as {mathbf{w}_{k}} = mathop{arg max }limits_{||mathbf{w}|{|^{2}} = 1} frac{{{rho_{k}}|{mathbf{h}_{kk}}{mathbf{w}_{k}}{|^{2}}}}{{{sigma^{2}} + sumlimits_{j ne k}^{M} {{rho_{kj}}|{mathbf{h}_{kj}}{mathbf{w}_{k}}{|^{2}}} }}.

Using the revelation principle, the monopolist's problem under incomplete information can be stated as choosing a pair (q that solves: max q - C ( q - G ( q p d θ, (11).

To maximize the achievable sum rate, a rate maximization problem with power constraint can be stated as follows begin{array}{*{20}l} mathop{text{max}} limits_{{p}}&quadquadsum_{k=1}^{K}{r}_{k}left({p}right) end{array} (8a).

Note that the condition (1.4), which is stated as ρ * > max { ( δ σ B 1 * ) δ, ( δ σ B 1 * ) 1 σ }. is crucial to guarantee that R > 1 > r0, and in the proof of Theorem 1.3 we require R > 1 > r0 because the exponents in inequalities of (A3) is different.

Formally, the optimization is stated as σ * = arg max σ - R = arg max σ - 1 a + ∑ j = 1 N σ j C ( γ j R ) - 1 (13).

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