If an upper bound u of A is in A, then u is called a greatest element (or a maximum element) of A. The collection of all greatest elements (maximum elements) of A is denoted by maxA, that is, max A = { y ∈ A : y ≽ x, for all x ∈ A }.
If A is a subset of P, then an element u of P is called an upper bound of A if x ≼ u for each x ∈ A ; if u ∈ A, then u is called a greatest element (or a maximum element) of A. The collection of all greatest elements (maximum elements) of A is denoted by maxA.
Given an initial guess (u^{(0)}), for (k=0,1,ldots ) , until ({u^{ k)}}) converges, compute begin{aligned} (omega I+C+iI u^{ k+1)}= omega I-D+C-T u^{ k)}+b, end{alI-D+C-T u^{ khere (omega =frac{d_{mI-D+C-T u^{ k}+b}) is thend{aligned (d_{min }) and (d_{max }), where (d_{min }) and (d_{max }) are the minimum and maximum elements of the diagonal matrix D.
Let Dmin and Dmax be the minimum and maximum elements of D, respectively.
where |·| and ∥·∥2 denote the absolute value and Euclidean norm, respectively, and w max denotes the maximum element of w.
Identify the maximum element of each column, and corresponding row index k' denotes the assignment of that user to the i th sub-carrier.
Step 2: Find the maximum element of |g (i |, then get its column index as ( j=underset{nin left{1,2,cdots, Nright}}{ arg max}left|{boldsymbol{g}}^{(i)}(n right| ).
The ≽ i -downward set of the inverse image { z i ∈ S i : P i ( z i, x − i ) is a maximum element of f i ( S i, x − i ) } is an inductive subset of S i ; g2′.
Without loss of generality, assuming that x m is the maximum element of { x j : j = 1, 2, 3, …, m }, that is, x j ⪯ x m, for j = 1, 2, 3, …, m. (4).
Compute the maximum element of the vector ΔI j, i, max{ΔI j, i }, and remove a bit from the sub-carrier identified by the corresponding column index i.
Since cl ( { x α } ) is totally ordered, and { x j : j = 1, 2, 3, …, m } ⊆ cl ( { x α } ), there is a maximum element of the finite subset { x j : j = 1, 2, 3, …, m }.
