Consider a -receiver parallel Gaussian broadcast channel as shown in Figure 1

where is the transmitted signal, represents a complex noise with variance , and corresponds to the channel seen by receiver on tone . The maximum common information rate that can be supported by the channel is given by

with the variance of the input signal on channel , the power allocation among all subchannels, the total power constraint, and the SNR gap which measures the loss with respect to theoretically optimum performance [22]. To achieve the maximum common information rate, the common message codebook cannot be broken into different codebooks for each channel, that is, joint encoding and joint decoding must be performed across all subchannels [23]. This transmission scheme is referred to as "single codebook, variable power" transmission [24].

The expression in (2) is the maximization of the minimum of a set of sums of concave functions of . Since the sum and the minimum operations preserve concavity, the objective is concave, and maximizing a concave function yields a convex optimization problem. This max-min optimization problem can be efficiently solved by the approach based on minimax hypothesis testing given in [14, 17, 18]. For two receivers, the optimal power allocation algorithm is given by three steps

Step 1.

Find given by

with

If , then the optimal power allocation is and finish.

Step 2.

Find given by

If then the optimal power allocation is and finish.

Step 3.

For a given set of weights corresponding to the index with , find given by

Search over all to find that satisfies , then the optimal power allocation is and finish.

First consider the optimization problem of Steps 1 and 2. As the objective function is concave, the power allocation can be derived by the standard Karush-Kuhn-Tucker (KKT) conditions [25]. The modified Lagrangian function for Steps 1 and 2 is given by

with the Lagrange multiplier associated with the total power constraint. By taking the derivative of the modified Lagrangian function with respect to , we can solve the KKT system of the optimization problem. The derivative with respect to is given by

Nulling the derivative gives

The optimal power allocation corresponds to Gallager's water-filling strategy for single-user parallel Gaussian channels [11].

Step 1.

Step 2.

We now consider the optimization problem of Step 3. As the objective is a weighted sum of concave functions, the power allocation can also be derived by the standard KKT conditions. The modified Lagrangian function for Step 3 is given by

with the Lagrange multiplier associated with the total power constraint. By taking the derivative of the modified Lagrangian function with respect to , we can solve the KKT system of the optimization problem. The derivative with respect to is given by

Nulling the derivative gives

The quadratic equation to be solved is

The discriminant is given by

The power allocation is given by the positive root

In this formula, the optimal power allocation takes into account the difference between the water-fill functions and the weights of the different receivers.

For more than two receivers, the optimal power allocation algorithm is driven by the solutions of higher-degree polynomials and involves more steps under multiple hypothesis testing. For instance, with three receivers , the optimal power allocation algorithm is given by seven steps involving the solutions of three linear equations, three quadratic equations and a cubic equation [26]. Therefore, for three receivers and four receivers , the optimal power allocation is a type of water-filling strategy given by the solutions up to a cubic and a quartic equation, respectively. The optimal power allocation can also be found analytically (the solution is not given in this paper due to space limitations). With , the optimal power allocation is given by the solutions of polynomial equations up to degree from the formula

In general, the roots for polynomials with a higher degree than four cannot be expressed analytically but can be solved numerically. Note that to reduce the complexity in a practical algorithm, the weights are taken from a given data set in interval with samples, leading to a possible exhaustive search over possibilities. Therefore, to satisfy the conditions requiring the rates of the different receivers to be equal, the optimal value should minimize the dispersion of the rates

Figure 2 shows the power control for a single tactical radio network. An inner loop determines the power allocation maximizing the common rate subject to a total power constraint. Then, an outer loop minimizes the power such that a common rate constraint is achieved. Algorithm 1 provides the proposed power allocation for power minimization subject to a common rate constraint. The inner loop and the outer loop correspond to lines 13–21 and 6–30, respectively. Note that if all the steps in the multiple hypothesis testing are needed, the complexity of the algorithm increases exponentially with the number of receivers .

**Algorithm 1:** Minimization of the power subject to a common rate constraint.

(1) init

(2) init

(3) init

(4) init

(5) while

(6) for all steps

(7) init according to step

(8) init

(9) init

(10) init

(11) init

(12) while

(13) Calculate according to the roots of(18)

(14) if

(15)

(16)

(17)

(18) end if

(19)

(20) end while

(21) If condition satisfied on exit step loop

(22) end for

(23) if

(24)

(25)

(26)

(27) end if

(28)

(29) end while