$W_m(\emptyset)=0$. For example, consider Case II where $M=1$. We have $\mathcal{S}_l^0 = \mathcal{S}_l^1 = \mathcal{U}_l$, $W_0(\mathcal{U}_l)=G_l^0$, and $W_1(\mathcal{U}_l)=G_l^1$.

The power allocation algorithm for Case III is presented in Table \[tab:Case3Algo\]. The algorithm iteratively picks users from the [*eligible*]{} subset $\mathcal{S}_l^m$ and assigns them to the [*allocated*]{} subset $\mathcal{U}_l^m$. In each step $l$, $\Psi$ is the subset of FBS’s that will transmit the level $l$ packet; the complementary set $\overline{\Psi}$ is the subset of FBS’s that will not transmit the level $l$ packet. The expected increment in total power for each partition is computed, and the partition with the smallest expected increment will be chosen. $\Delta_l^m$ is the power of BS $m$ for transmitting the level $l$ data packet. In Steps $6$–$15$, the MBS and FBS combination $\Psi$ is determined for transmitting the level $l$ packet, with the lowest power $\Delta_0$. In Steps $16$–$30$, elements in $\mathcal{S}_l^m$ are assigned to $\mathcal{U}_l^m$ according to $\Psi$. In Steps $31$–$35$, power sums $Q_l^m$ and the corresponding power allocations $P_l^m$ are calculated in the reverse order from the known $\mathcal{U}_l^m$’s.

The complexity of the algorithm is $\mathcal{O}(ML)$.

  ----- ------------------------------------------------------------------------------------------------------------------------
  1:    Initialize: $c_l^m=0$ and $Q_{L+1}^m=0$, for all $l$, $m$;
  2:    FOR $l=1$ TO $L$
  3:    $\;\;\;$ 