We search for the optimal coordinated power allocation by approaching the following optimization problem:
{\mathbf{P}}_{i}^{*}=\left({P}_{i,1}^{*},...,{P}_{i,N}^{*}\right)=\text{arg}\phantom{\rule{1em}{0ex}}\underset{{\mathbf{P}}_{i}\in {\Omega}^{N}}{\text{min}}\phantom{\rule{2.77695pt}{0ex}}{P}_{\text{overall}},\phantom{\rule{1em}{0ex}}i=1,2,
(3)
where {\Omega}^{N}=\left\{{\mathbf{P}}_{i}\left\forall j\in \left\{1,...,N\right\},0\le {P}_{i,j}\le {P}_{0},{\sum}_{\text{j}=1}^{\text{N}}{P}_{i,j}\le {P}_{0}\right\right\} is the feasible set. Since Ω^{N}is a closed and bounded set, and R : Ω^{N}→ ℝ is continuous, function (1) has a solution [[16], Theorem 0.3].
For the sake of mathematical derivation, we denote {\gamma}_{i,j}=\left({P}_{0}{\left{H}_{i,j}\right}^{2}\right)/{\sigma}_{j}^{2} and x_{
i,j
}= P_{
i,j
}/P_{0}. It is indicated that γ_{
i,j
}is the SNR associated with CTP i over the j th subchannel when assuming the entire power P_{0} is allocated to the j th subchannel, and x_{
i,j
}represents the power allocation ratio. Since the logarithm is monotonically increasing function, the objective (2) combined with the constraints can be described as
\begin{array}{c}\text{minimize}\phantom{\rule{1em}{0ex}}Z={P}_{\text{overall}}/{P}_{0}=\sum _{j=1}^{N}{x}_{1,j}+\sum _{j=1}^{N}{x}_{2,j}\\ \text{subject}\phantom{\rule{2.77695pt}{0ex}}\text{to}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\prod _{j=1}^{N}\left(1+\sum _{i=1}^{2}{\gamma}_{i,j}{x}_{i,j}\right)={2}^{{R}_{S}/\left(\frac{B}{N}\right)}={2}^{{R}_{S}^{*}},\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\sum _{j=1}^{N}{x}_{i,j}\le 1,\phantom{\rule{1em}{0ex}}i=1,2,\end{array}
(4)
where Z ≤ 2 represents the total power consumption ratio.
In order to obtain a closedform expression for the required power allocation, we divide the constrained problem into two different cases:

a)
The power constraint is assumed to be large enough, so that solving the primal objective is equivalent to solving an unconstrained problem. In this case, the power constraint at each CTP (i.e., {\sum}_{j=1}^{N}{x}_{i,j}\le 1 for i = 1, 2) is always satisfied.

b)
The power constraint is not large enough, so that one of the CTPs may exceed the power constraint if the distribution of the users is not even. In this situation, the solution in (a) leads to {\sum}_{j=1}^{N}{x}_{1,j}>1 or {\sum}_{j=1}^{N}{x}_{2,j}>1. In this situation the exceeded CTP should provide its full power for transmission, while the other CTP should increase its transmit power until it meets the throughput requirement for the system. In mathematic terms, the constraint in the primal objective should be changes to {\sum}_{j=1}^{N}{x}_{1,j}=1 and {\sum}_{j=1}^{N}{x}_{2,j}\le 1, or to {\sum}_{j=1}^{N}{x}_{2,j}=1 and {\sum}_{j=1}^{N}{x}_{1,j}\le 1. Actually, this case is more meaningful in practical systems.
3.1. Optimal solution for the unconstrained case
It is noted that for arbitrary γ_{
i,j
}, the likelihood of having γ_{1,j}= γ_{2,j}for j ∈ {1, 2} in an actual system is almost zero. Without loss of generality, we divide the N subchannels into two parts: the first part contains M subchannels that satisfy γ_{1,m}> γ_{2,m}for m ∈ {1, 2,..., M}, while the second part contains K subchannels that satisfy γ_{1,k}< γ_{2,k}for k ∈ {1,2,..., K}. According to this model, N = M + K.
We first present the following lemma:
Lemma 1: When ignoring the power constraint {\sum}_{j=1}^{N}{x}_{1,j}\le 1;{\sum}_{j=1}^{N}{x}_{2,j}\le 1, the objective problem (4) can be degenerated into the unconstrained function
\begin{array}{c}\text{minimize}\phantom{\rule{1em}{0ex}}Z=\sum _{m=1}^{M}{x}_{1,m}+\sum _{k=1}^{K}{x}_{2,k}\\ \text{subject}\phantom{\rule{2.77695pt}{0ex}}\text{to}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\prod _{m=1}^{M}\left(1+{\gamma}_{1,m}{x}_{1,m}\right)\cdot \prod _{k=1}^{K}\left(1+{\gamma}_{2,k}{x}_{2,k}\right)={2}^{{R}_{S}^{*}},\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}M+K=N.\end{array}
(5)
Proof: To prove this lemma we use reductio ad absurdum. Assume that vectors {{\mathbf{P}}_{1}}^{*}=\left({{x}_{1,m}}^{*},\dots ,{{x}_{1,k}}^{*}\ne 0,\dots \right) and {{\mathbf{P}}_{2}}^{*}=\left({{x}_{2,m}}^{*}\ne 0,\dots ,{{x}_{2,k}}^{*},\dots \right) are optimal solutions for m ∈ {1,..., M} and k ∈ {1,..., K}, which satisfy the throughput requirement and achieve the minimum Z*. However, we could find another set of vectors {{\mathbf{P}}_{1}}^{**}=\left({{x}^{**}}_{1,m}={{x}^{*}}_{1,m}+\left({\gamma}_{2,m}/{\gamma}_{1,m}\right){{x}^{*}}_{2,m},\dots ,{{x}_{1,k}}^{**}=0\dots \right) and {{\mathbf{P}}_{2}}^{**}=\left({{x}_{2,m}}^{**}=0,\dots ,{{x}^{**}}_{2,k}={{x}^{*}}_{2,k}+\left({\gamma}_{1,k}/{\gamma}_{2,k}\right){{x}^{*}}_{1,k},\dots \right) which also satisfy the throughput requirement, and achieve a smaller Z**, due to γ_{1,m}> γ_{2,m}, γ_{1,k}< γ_{2,k}and
\begin{array}{ll}\hfill {Z}^{**}& =\sum _{m=1}^{M}\left({{x}^{*}}_{1,m}+\frac{{\gamma}_{2,m}}{{\gamma}_{1,m}}{{x}^{*}}_{2,m}\right)+\sum _{k=1}^{K}\left({{x}^{*}}_{2,k}+\frac{{\gamma}_{1,k}}{{\gamma}_{2,k}}{{x}^{*}}_{1,k}\right)\phantom{\rule{2em}{0ex}}\\ <{Z}^{*}=\sum _{m=1}^{M}\left({{x}^{*}}_{1,m}+{{x}^{*}}_{2,m}\right)+\sum _{k=1}^{K}\left({{x}^{*}}_{2,k}+{{x}^{*}}_{1,k}\right).\phantom{\rule{2em}{0ex}}\end{array}
(6)
Therefore, the objective (4) could be equivalently transformed into (5). The degeneration implies that a group of M subchannels (that verify γ_{1,m}> γ_{2,m}) should receive power allocation from CTP 1, while the other K = N  M subchannels (that verify γ_{1,k}< γ_{2,k}) should receive power allocation from CTP 2. In other words, to achieve the goal of minimizing the total transmit power in the system, each CTP should select the better subchannels (according to the channel quality) to allocate power on them.
The degraded objective function (5) is strictly convex. Then, by Lagrange dual function we have that
Z=\sum _{m=1}^{M}{x}_{1,m}+\sum _{k=1}^{K}{x}_{2,k}\lambda \left[\prod _{m=1}^{M}\left(1+{\gamma}_{1,m}{x}_{1,m}\right)\cdot \prod _{k=1}^{K}\left(1+{\gamma}_{2,k}{x}_{2,k}\right){2}^{{R}_{S}^{*}}\right].
(7)
First, we derive λ as
\lambda =\frac{1}{{2}^{{R}_{S}^{*}}}{\left(\frac{{2}^{{R}_{S}^{*}}}{\prod _{m=1}^{M}{\gamma}_{1,m}\cdot \prod _{k=1}^{K}{\gamma}_{2,k}}\right)}^{\frac{1}{N}}.
(8)
Then, the closedform solution for the power allocation ratio at each CTP is obtained as follows:
{x}_{1,m}=\lambda {2}^{{R}_{S}^{*}}\frac{1}{{\gamma}_{1,m}}\phantom{\rule{1em}{0ex}}m\in \left\{1,\dots ,M\right\},\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{x}_{2,k}=\lambda {2}^{{R}_{S}^{*}}\frac{1}{{\gamma}_{2,k}}\phantom{\rule{1em}{0ex}}k\in \left\{1,\dots ,K\right\}.
(9)
Finally, the minimum total power consumption ratio is given by the following expression:
Z=N{\left(\frac{{2}^{{R}_{S}^{*}}}{\prod _{m=1}^{M}{\gamma}_{1,m}\cdot \prod _{k=1}^{K}{\gamma}_{2,k}}\right)}^{\frac{1}{N}}\sum _{m=1}^{M}\frac{1}{{\gamma}_{1,m}}\sum _{k=1}^{K}\frac{1}{{\gamma}_{2,k}}.
(10)
Note: As mentioned before, formula (9) only represents the optimal solution for the primal objective when the power constraint {\sum}_{j=1}^{N}{x}_{i,j}\le 1 for i = 1,2 holds. However, when the power constraint is not large enough and the user density is not evenly distributed, one of the CTPs may exceed its individual power constraint. In this situation, the solution in (9) leads to {\sum}_{j=1}^{N}{x}_{1,j}>1 or {\sum}_{j=1}^{N}{x}_{2,j}>1. Therefore, the exceeded CTP should provide its full power for transmission, and the power allocation for the other user should be recalculated to meet the throughput requirement for the system. The analyisis for this case is specified in the following section.
3.2. Optimal solution for the constrained case
Without loss of generality, we assume that CTP 1 is the overloaded transmission point. Thus, it is straightforward to know that CTP 1 should provide its full power for transmission (i.e., {\sum}_{j=1}^{N}{x}_{1,j}=1). Then the objective (4) can be redefined as
\begin{array}{c}\text{minimize}\phantom{\rule{1em}{0ex}}Z=1+\sum _{j=1}^{N}{x}_{2,j}\\ \text{subject}\phantom{\rule{2.77695pt}{0ex}}\text{to}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\prod _{j=1}^{N}\left(1+\sum _{i=1}^{2}{\gamma}_{i,j}{x}_{i,j}\right)={2}^{{R}_{S}^{*}};\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\sum _{j=1}^{N}{x}_{1,j}=1,\sum _{j=1}^{N}{x}_{2,j}\le 1.\end{array}
(11)
Before going forward in the analysis, we first need to present the following lemma.
Lemma 2: In order to minimize the overall power consumption, the optimal power allocation strategy consists in transmitting simultaneously from both CTPs in only one of the Nsubchannels. The rest of the subchannels should be divided in two groups, one per CTP. The selection of the subchannels for each group should be carried out according to the individual subchannel gains from each CTP.
Proof: By Lagrange dual function we have that
\begin{array}{l}\Gamma =1+{x}_{2,1}+\cdots +{x}_{2,N}\lambda \{(1+{\gamma}_{1,1}{x}_{1,1}+{\gamma}_{2,1}{x}_{2,1})\cdots \\ [1+{\gamma}_{1,N}(1{x}_{1,1}\cdots {x}_{1,N1})+{\gamma}_{2,N}{x}_{2,N}]{2}^{{R}_{S}^{*}}\}.\end{array}
(12)
For the sake of mathematic simplification, we denote A_{
j
}= (1 +γ_{1,j}x_{1,j}+ γ_{2,j}x_{2,j}) and obtain the Hessian matrix ∇^{2}f as follows:
{\nabla}^{2}\Gamma =\left[\begin{array}{cccc}\hfill \frac{{\partial}^{2}\Gamma}{\partial {{x}_{1,1}}^{2}}\hfill & \hfill \frac{{\partial}^{2}\Gamma}{\partial {x}_{1,1}\partial {x}_{2,1}}\hfill & \hfill \cdots \phantom{\rule{0.3em}{0ex}}\hfill & \hfill \frac{{\partial}^{2}\Gamma}{\partial {x}_{1,1}\partial {x}_{2,N}}\hfill \\ \hfill \frac{{\partial}^{2}\Gamma}{\partial {x}_{2,1}\partial {x}_{1,1}}\hfill & \hfill \frac{{\partial}^{2}\Gamma}{\partial {{x}_{2,1}}^{2}}\hfill & \hfill \cdots \phantom{\rule{0.3em}{0ex}}\hfill \\ \hfill \vdots \hfill & \hfill \ddots \hfill \\ \hfill \frac{{\partial}^{2}\Gamma}{\partial {x}_{2,N}\partial {x}_{1,1}}\hfill & \hfill \cdots \phantom{\rule{0.3em}{0ex}}\hfill & \hfill \frac{{\partial}^{2}\Gamma}{\partial {{x}_{2,N}}^{2}}\hfill \end{array}\right].
(13)
The firstorder principal minor determinant of ∇^{2}Γ is given by {\partial}^{2}\Gamma /\partial {{x}_{1,1}}^{2}=2{\gamma}_{1,1}{\gamma}_{1,N}{A}_{2}{A}_{3}\dots {A}_{N1}<0, while the secondorder principal minor determinant attains the form
\frac{{\partial}^{2}\Gamma}{\partial {{x}_{1,1}}^{2}}\cdot \frac{{\partial}^{2}\Gamma}{\partial {{x}_{2,1}}^{2}}\frac{{\partial}^{2}\Gamma}{\partial {x}_{1,1}\partial {x}_{2,1}}\cdot \frac{{\partial}^{2}\Gamma}{\partial {x}_{2,1}\partial {x}_{1,1}}={\left(\lambda {\gamma}_{2,1}{\gamma}_{1,N}{A}_{2}{A}_{3}\dots {A}_{N1}\right)}^{2}<0.
(14)
Therefore, the Hessian ∇^{2}Γ is indefinite since both, the firstorder and the secondorder principal minor determinant are negative. This means that function (12) has no extreme points, and as a consequence, the minimum value must be achieved on the domain boundary [17]. In other words, one of the x_{i,1}for i ∈ {1,2} equals to 0; i.e., A_{1} = (1 + γ_{1,1}x_{1,1} + γ_{2,1}x_{2,1}) should be degenerated to {A}_{1}^{\Delta}=\left(1+{\gamma}_{i,1}{x}_{i,1}\right) for i ∈ {1, 2}.
Similarly, for arbitrary A_{
j
}= (1 + γ_{1,j}x_{1,j}+ γ_{2,j}x_{2,j}), we still have the following equation
\frac{{\partial}^{2}\Gamma}{\partial {{x}_{1,j}}^{2}}\frac{{\partial}^{2}\Gamma}{\partial {{x}_{2,j}}^{2}}\frac{{\partial}^{2}\Gamma}{\partial {x}_{1,j}\partial {x}_{2,j}}\frac{{\partial}^{2}\Gamma}{\partial {x}_{2,j}\partial {x}_{1,j}}={\left(\lambda {\gamma}_{2,j}{\gamma}_{1,N}{A}_{1}\dots {A}_{j1}{A}_{j+1}\dots {A}_{N1}\right)}^{2}<0.
(15)
Hence, in order to achieve the minimum, all the A_{
j
}= (1+γ_{1j}x_{1j}+γ_{2j}x_{2j}), j ∈ {1, ..., N 1} ought to be degenerated to {A}_{j}^{\Delta}=\left(1+{\gamma}_{i,j}{x}_{i,j}\right) for i ∈ {1, 2}. At the end, due to the constraint {x}_{1,n}=1{\sum}_{j=1,j\ne n}^{N}{x}_{1,j}, a certain A_{
n
}for n∈ {1...N} is left with two additive variables (1 + γ_{1,n}x_{1,n}+ γ_{2,n}x_{2,n}), and the other A_{
j
}for j ∈ {1... N} (j ≠ n) are degenerated to the form of {A}_{j}^{\Delta}=\left(1+{\gamma}_{i,j}{x}_{i,j}\right) for i = 1 or 2. Thus, function (12) has been degenerated into strictly convex.
Based on the above analysis, the optimal power allocation mode (as depicted in Figure 2) is that only one of the Nsubchannels (i.e., a certain A_{
n
}) should be transmitted together by the two CTPs; the others, according to the different subchannel gains, should be separated into two parts (i.e., x_{1,j}or x_{2,j}) and transmitted by an individual CTP.
A specific 2subchannel overloaded case is presented here to illustrate Lemma 2 by numerical simulation (see Figure 3). As an example of function (11), we assume {2}^{{R}_{S}^{*}}=10. Considering the power constraint, path loss attenuation, shadowing, and noise, we randomly generate γ_{
i,j
}as follows: γ_{1,1} = 3.205, γ_{2,1} = 2.311, γ_{1,2} = 4.108, γ_{2,2} = 3.406. Figure 3 shows the minimum power consumption Z versus ergodic power allocation ratio. Based on this figure it is possible to see that, the total power consumption reaches its minimum value (i.e., 1.229) when (x_{1,2} = 0.44, x_{2,1} = 0) and (x_{1,1} = 0.56, x_{2,2} = 0.229). Note that this set of values verifies Lemma 2, that stated that the optimal power allocation is obtained on the domain boundary, when only one of the available subchannels is being used simultaneously by the two CTPs.
Now we are ready to solve the degraded convex problem, and obtain a closedform solution for optimal power allocation. Without loss of generality, we consider that the n th subchannel represents the common channel which is used simultaneously by both CTPs. Then, we divide the N  1 subchannels into two groups: the first group of M subchannels is used exclusively by CTP 1, and the second group of K = NM1 subchannels is used solely by CTP 2. Thus, the objective function (12) can be expressed as
\begin{array}{l}\Gamma =1+{x}_{2,n}+{x}_{2,1}+\dots {x}_{2,K}\lambda \{{\displaystyle {\prod}_{m=1}^{M}(1+{\gamma}_{1,m}{x}_{1,m})\cdot {\displaystyle {\prod}_{k=1}^{K}(1+{\gamma}_{2,k}{x}_{2,k})}}\\ \cdot \left[1+{\gamma}_{1,n}\left(1{\displaystyle \sum _{m=1}^{M}{x}_{1,m}}\right){x}_{1,n}+{\gamma}_{2,n}{x}_{2,n}\right]{2}^{{R}_{S}^{*}}\}.\end{array}
(16)
By partial derivation on (16), it is possible to see that
\lambda =\frac{1}{{2}^{{R}_{S}^{*}}}{\left(\frac{{\left({\gamma}_{1,n}/{\gamma}_{2,n}\right)}^{M}{2}^{{R}_{S}^{*}}}{{\gamma}_{2,n}\cdot \prod _{m=1}^{M}{\gamma}_{1,m}\cdot \prod _{k=1}^{K}{\gamma}_{1,k}}\right)}^{\frac{1}{N}}.
(17)
Then, the optimal solution is obtained as follows:
\left\{\begin{array}{c}\hfill {x}_{1,m}=\frac{{\gamma}_{2,n}}{{\gamma}_{1,n}}\lambda {2}^{{R}_{S}^{*}}\frac{1}{{\gamma}_{1,m}},\phantom{\rule{1em}{0ex}}{x}_{1,n}=1\sum _{m=1}^{M}{x}_{1,m},\hfill \\ \hfill {x}_{2,k}=\lambda {2}^{{R}_{S}^{*}}\frac{1}{{\gamma}_{2,k}},\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{x}_{2,n}=\lambda {2}^{{R}_{S}^{*}}\frac{1+{\gamma}_{1,n}{x}_{1,n}}{{\gamma}_{2,n}}.\hfill \end{array}\right.
(18)
Remark: It is seen that the solutions (9) and (18) turn out to take the form of traditional WF solution, but having a coordinated feature. The coordinated feature here means that the optimal power allocation ratio x_{1,m}correlates not only with γ_{1,m}for m ∈ (1,...,M) and γ_{2,k}for k ∈ (1,..., K), but also depends on the relationship γ_{2,n}/γ_{1,n}in (18). The optimal solution can be only obtained through perfect coordination between both CTPs. According to the derived solution, both CTPs have to determine jointly the power ratio that they should allocate on each of their active subchannel, to achieve an effective minimization of the overall power consumption of the system.
Finally, by γ_{
i,j
}= (P_{0}H_{
i,j
}^{2})/N_{0}ΔB and x_{
i,j
}= P_{
ij
}/P_{0}, the optimal power allocation is given by
\begin{array}{l}{{\mathbf{P}}_{1}}^{*}=\left({{P}_{1,1}}^{*},\dots ,{{P}_{1,M}}^{*},{{P}_{1,n}}^{*},\underset{K}{\underset{\u23df}{0,\dots ,0}}\right),\phantom{\rule{2em}{0ex}}\\ {{\mathbf{P}}_{2}}^{*}=\left(\underset{M}{\underset{\u23df}{0,\dots ,0}},{{P}_{2,n}}^{*},{{P}_{2,1}}^{*},\dots ,{{P}_{2,NM1}}^{*}\right),\phantom{\rule{2em}{0ex}}\\ M=\underset{M\in \left\{1,\dots ,N1\right\}}{\text{arg}}\text{min}{P}_{\text{overall}}.\phantom{\rule{2em}{0ex}}\end{array}
(19)
The intuition behind the optimal power allocation in (9) and (18) is also to take advantage of good channel conditions: when subchannel condition is good, more power (and a higher data) rate is sent over the subchannel. On the other hand, as the subchannel quality degrades, less power (and a lower data rate) is sent over the channel. It is emphasized that the solution should be satisfied by P_{
i,j
}≥ 0; i.e., if the channel SNR falls below the cutoff threshold, the channel is not used and
{{P}_{i,j}}^{*}={\left[{P}_{i,j}\right]}^{+}=\text{max}\left\{0,{P}_{i,j}\right\}
(20)
results. Hence, an iterative JMPCPA algorithm is presented (as depicted in Figure 4), not only to determine the optimal power allocation vectors {{\mathbf{P}}_{1}}^{*} and {{\mathbf{P}}_{2}}^{*}, but also to evaluate the solutions numerically.