12 be a good choice for practical implementations. However, in order to reduce the complexity and DoD of the power control phase, we have developed the following less complex centralized scheme. In this scheme, firstly the subcarriers are allocated in eachcell l using the heuristic criterion defined in (11). The allocation of each subcarrier is followe d by the power allocation phase (based on equalization) as mentioned in the initial allocation phase of Algorithm 2 (i.e., Steps 1 to 5). Once the subcarrier allocations are finalized, we thencompute GP based powers for the allocated users at any arbitrarily selected subcarrier n in all cells. Setting the equalization based powers p$_{n,k,l}$$_{eq}$ as the upper bound on p$_{n,k,l}$ and considering a high SINR regime, we now define the following less complex GP problem with the objective to maximize the throughput at the n th subcarrier: C. Centralized Scheme B minimize p$_{n,k,l}$ log$_{2}$ L ∏ l =1 ( σ 2 + I$_{n,l}$ p$_{n,k,l}$h$_{n,k,l}$ ) α$_{n,k,l}$ subject to p$_{n,k,l}$ ≤ p$_{n,k,l}$$_{eq}$, ∀ l (14) Clearly, the resulting GP based power of each competing user at subcarrier n in the different cells may not succeed in achieving the upper bound, due to the ICI effect. We call this power as left-over power. The left-over power can then be distributed equally among the remaining allocated subcarriers of the user. The procedure is detailed in Algorithm 3. Since at the end of the initial allocation phase, the subcarrier allocations become fixed and the total power is distributed equally among the allocated subcarriers of a user, we cannot set an upper bound which depicts higher power than the previously allocated power. If we do so, this may cause power reduction or even no power at some other allocated subcarrier of that user in order to maintain the total power constraint. Thus, this may results in an invalid subcarrier allocation. Next follows an example which demonstrates the significanceof GP as well as centralized scheme B over equal power allocation. Consider H$_{1}$ , H$_{2}$ = [0 . 30 0 . 25; 0 . 04 0 . 15] × 10 − $^{9}$, DRAFT March 19, 2018