The EE optimization model and its solution of this study are addressed in this section. As discussed before, EE is defined as the achievable sum rate divided by the consumed energy. Hence, in line with prior study [20], the global EE optimization problem of this modified system (within all time intervals of all BS areas), while integrated taking the antenna, RF chain, circuit, and dynamic and constant power into consideration, can be described as
$$ \underset{P_{i,j}^{k}, P_{i,j}^{\text{RF}}}{\text{max}}\quad {\int_{0}^{T}}\sum\limits_{i=1}^{B}\sum\limits_{j=1}^{K}\frac{R_{i, j}^{k}}{{P}_{\text{clu}}}dt, $$
(5)
$$\begin{array}{*{20}l} \text{subject to:} \\ C1: & \int_{0}^{T}\sum\limits_{i=1}^{B} \sum\limits_{j=1}^{K} \left(\frac{1}{\eta}{P}_{i,j}^{k} +\! {P}_{i,j}^{c} + {P}_{i,j}^{\text{RF}}\right)dt \leq P_{T}^{\text{total}}, \end{array} $$
(6a)
$$\begin{array}{*{20}l} C2: & \int_{0}^{T}\sum\limits_{i=1}^{B} {P}_{b}^{\mathrm{dyn,BS}}dt \leq P_{\text{bs}}^{\text{total}}, \end{array} $$
(6b)
$$\begin{array}{*{20}l} C3\!:\! &\int_{0}^{T}\!\! \sum\limits_{i=1}^{B} \sum\limits_{j=1}^{K} \!\left(\!\!\Big\lceil\! \frac{R_{i,j}^{k}}{ \log_2 \!\left(1+\rho\frac{M-K}{K}\!\right)}\! \Big\rceil\!\!\right)\!dt\! \leq\! N_{\text{CCs}}^{\text{total}}, \end{array} $$
(6c)
$$\begin{array}{*{20}l} C4: & \int_{0}^{T}\sum_{i=1}^{B} \sum_{j=1}^{K} N_{i,j}^{a}dt \leq N_{\text{bs}}^{a,\text{total}}, \end{array} $$
(6d)
$$\begin{array}{*{20}l} C5: & \int_{0}^{T}\sum\limits_{i=1}^{B} \sum\limits_{j=1}^{K} N_{i,j}^{\text{UE}}dt \leq N_{bs}^{rf, \text{total}}, \end{array} $$
(6e)
$$\begin{array}{*{20}l} C6: & {P}_{i,j}^{k}, {P}_{i,j}^{c}, {P}_{i,j}^{\text{RF}}, {P}_{b}^{\mathrm{dyn,BS}} \geq 0, \forall k, \end{array} $$
(6f)
where \(P_{i,j}^{k}, {P}_{i,j}^{c}, {P}_{i,j}^{\text {RF}}\) are of the same value as \({P_{j}^{k}}, {P_{j}^{c}}\), \(P_{j}^{\text {RF}}\). Here \(P_{T}^{\text {total}}, P_{bs}^{\text {total}}, N_{CCs}^{\text {total}}\), \(N_{bs}^{a, \text {total}}, N_{bs}^{rf, \text {total}}\) yield the total transmission power, total BS power, the number of CC, the number of antenna, and the number of RF chain, respectively. Further, \(N_{i,j}^{a}\), \( N_{i,j}^{\text {UE}}, N_{\text {bs}}^{\mathrm {rf, total}}\) denote the number of antenna allocated to each UE, number of UE indicator, and the total number of RF chain within one cluster, respectively. Note that here C1, C2 specify that in one cluster, the total transmission power should be no greater than the allocated power, and the total cellular power consumption should be no greater than the total BS power, respectively. In addition, C3, C4, C6 denote that in one cluster, the number constraints of CC, antenna, as well as the power constraints, should be satisfied. Whereas C5 yields the UE number cannot exceed the RF chain number. This is because of the assumption that one active RF chain could only serve one active UE, in line with the prior MIMO structure, for the sake of compactness.
The EE optimization problem, although, has been addressed so far and can be solved by the concave optimization method, but it will cost more resources and calculation time due to the constraint C3 comparing with the linear searching method. Thus, another purpose of this paper is to find an alternative method to simplify the processing. Fortunately, in line with prior study [21], it is noted that if C
k
was taken as the used CCs of the previous UEs, a minimum CC selection method can be obtained beforehand, whereas I
m
(t) is defined as
$$ I_{m}(t) = \left \{ \begin{array}{rl} 1, \quad &~C_{k} \in \Omega_{m}, ~C_{k} = \text{arg}\underset{C_{k} \in \Omega_{m}}{\text{min}}{P_{j}^{k}}, \\ 0, \quad &\text{otherwise}, \\ \end{array} \right. $$
(7)
with C
k
,Ω
m
denote the optimal CC for kth UE and the set of CCs. For the left optimization problem, we dig more and find the following Corollary existing.
Corollary.
According to Lemma 2, the binary value of power allocation \(P_{i,j}^{k}\) and its time distribution τ
on both obey the identically Poisson distribution in every time interval (t(i−1),t(i)]. While linearly adding every individual power allocation within each time interval, it turns out to be the optimal solution of the total time intervals. This can be proved as follows.
Proof.
In τ
on time period, whenever there is a transmission request, power allocation is triggered with a constant value of \(P_{i,j}^{k}\) (proof of Lemma 1 in [22]). Otherwise, in τ
off period, no power is allocated. This is because τ
on and \(P_{i,j}^{k}\) display the same performances like the active UEs’ existences with their on–off binary values. Thus, it is clear that the binary values of \(\tau, P_{i,j}^{k}\) are of the same distribution performances just like UEs. □
With the aforementioned Lemma 1, Lemma 2, and Corollary 1 in hand, the global EE optimization solution can be obtained while summing up the single time interval EE optimization problems other than the aforementioned integration to the total time intervals. In addition, the single time interval EE optimization problem is the same among different time periods on condition that among each of them, the UEs are performing the same distribution. Thus once the optimal solution is obtained in a time interval, the global optimization is achieved. In this case, while removing the integration to the total time intervals, the problem can be further described as follows
$$ \underset{P_{i,j}^{k}, P_{i,j}^{\text{RF}}}{\text{max}}\quad \sum\limits_{i=1}^{B} \sum_{j=1}^{K}\frac{ R_{i,j}^{k}}{P_{\text{clu}}}, $$
(8)
$$\begin{array}{*{20}l} &\text{subject to} \\ C1: &\sum\limits_{i=1}^{B} \sum\limits_{j=1}^{K} (\frac{1}{\eta}P_{i,j}^{k} + P_{i,j}^{c} + P_{i,j}^{\text{RF}}) \leq P_{T}^{\text{total}}, \end{array} $$
(9a)
$$\begin{array}{*{20}l} C2: &\sum\limits_{i=1}^{B} {P}_{b}^{\mathrm{dyn,BS}} \leq P_{\text{bs}}^{\text{total}}, \end{array} $$
(9b)
$$\begin{array}{*{20}l} C3: &\sum\limits_{i=1}^{B} \sum\limits_{j=1}^{K} \left(\Big\lceil \frac{R_{i,j}^{k}}{ \log_{2} (1+\rho\frac{M-K}{K})} \Big\rceil\right) \leq N_{\text{CCs}}^{\text{total}}, \end{array} $$
(9c)
$$\begin{array}{*{20}l} C4: &\sum\limits_{i=1}^{B} \sum\limits_{j=1}^{K} N_{i,j}^{a} \leq N_{\text{bs}}^{a, \mathrm total}, \end{array} $$
(9d)
$$\begin{array}{*{20}l} C5: &\sum\limits_{i=1}^{B} \sum\limits_{j=1}^{K} N_{i,j}^{\text{UE}} \leq N_{\text{bs}}^{rf, \mathrm total}, \end{array} $$
(9e)
$$\begin{array}{*{20}l} C6: &{P}_{i,j}^{k}, {P}_{i,j}^{c}, {P}_{i,j}^{\text{RF}}, {P}_{b}^{\mathrm{dyn,BS}} \geq 0, \forall k. \end{array} $$
(9f)
Note that the purpose of C3 is to find the optimal CC solution under CC number constraint, which has less relationship with the power allocation. In this case, it can be solved via an offline method. That is, whenever request coming, select the least CC number for its transmission under CC constraint in (7). The remaining part after this selection turns out to be a component selection and power allocation problem with respect to the achievable transmission rate. In addition, without loss of generality, it is further assumed that each antenna consumes the same power once activated, as well as the circuit, RF chain.
In addition, similar to Lemma 2, while adding up all users’ location performances within every BS areas, it is the UEs’ location performances of the whole cluster. Moreover, the PPP distribution of UE within one cluster, with large time range, can be taken as a sum of separate PPP distribution of each BS area. Whereas in each area, the UE displays the same PPP distribution but with a smaller range. Thus, the optimal solution can be obtained by optimizing the EE problem of one BS area within one time interval. Which is, the remaining problem can be decomposed as a optimization problem with
$$ \begin{aligned} \underset{{P_{j}^{k}}, P_{j}^{\text{RF}}}{\text{max}}\quad \sum\limits_{k=1}^{K}\frac{R_{i,j}^{k}}{{P_{j}^{k}} + P_{j}^{\text{RF}}}, \end{aligned} $$
(10)
$$\begin{array}{*{20}l} \text{subject to} \\ C1: &\sum\limits_{j=1}^{K} \left(\frac{1}{\eta}{P_{j}^{k}} + {P_{j}^{c}} + P_{j}^{\text{RF}}\right) \leq P_{T}, \end{array} $$
(11a)
$$\begin{array}{*{20}l} C2: &\sum\limits_{j=1}^{K} {N_{j}^{a}} \leq N_{\text{bs}}^{a}, \forall b, \end{array} $$
(11b)
$$\begin{array}{*{20}l} C3: &\sum\limits_{j=1}^{K} N_{j}^{\text{UE}} \leq N_{\text{bs}}^{\text{rf}}, \forall t, \end{array} $$
(11c)
$$\begin{array}{*{20}l} C4: &{P_{j}^{k}}, {P_{j}^{c}}, P_{j}^{\text{RF}} \geq 0, \forall k, \end{array} $$
(11d)
whereas the global optimization solution is the summarization of each of this single-time-interval-single-BS-area optimization problem. To this optimization problem, it is proved by Theorem 1 in [23] that, if we define a maximum weighted solution with \(S^{*} = \text {max}({P_{j}^{k}}, P_{j}^{RF})= \sum _{k=1}^{K} R_{j}^{k*}/({P_{J}^{k}} + P_{j}^{RF})^{*}\), and further suppose the problem has an optimal solution S
∗. Optimal solution of the EE problem in (10), if any, should satisfy the following constraint
$$ \begin{aligned} &\text{max}\left({P_{j}^{k}}, P_{j}^{\text{RF}}\right) - S^{*} \left({P_{j}^{k}} + P_{j}^{\text{RF}}\right)^{*} \\ &= \sum\limits_{k=1}^{K} R_{j}^{k*} - S^{*} \left({P_{j}^{k}} + P_{j}^{\text{RF}}\right)^{*} = 0, \end{aligned} $$
(12)
and its optimal solution with \(\text {max}\left ({P_{j}^{k}}, P_{j}^{\text {RF}}\right)\) must be equal to S
∗ if any. Note that RF chain power consumption has nothing to do with the achievable sum rate although it is needed for the transmission. Under this circumstance, while searching for the optimal solution, it will converge to zero. To avoid this, in line with [24], we assume that in each search step, the number of RF chain is equal to the number of antenna in order to satisfy the transmission requirement.
After this, the extreme point existence is verified by the Lagrange method as shown in (13) and (14). As we can see, it always has an extreme point on condition that (13) is equal to zero and (14) lower than zero. Thus, at least one optimal solution of our EE optimization problem exists. In addition, this optimal solution can be straightforward obtained while comparing all of those extreme points and keeping the one with best EE performance.
$$ \begin{aligned} &\frac{\partial \text{max}\left({P_{j}^{k}}, P_{j}^{\text{RF}}\right)}{\partial {P_{j}^{k}}} =\frac{\partial \sum\limits_{j=1}^{K}\frac{R_{j}}{{P_{j}^{k}} + P_{j}^{\text{RF}}}}{\partial {P_{j}^{k}}}\\&= \sum\limits_{j=1}^{K}\frac{B_{\text{CCs}}\left(\frac{M-K} {\mathrm{ln2} [{KN}_{0}+{P_{j}^{k}}(M-K)]} -\text{log}_{2}\frac{{KN}_{0} + {P_{j}^{k}}(M-K)}{K N_{0}}\right) }{\left({P_{j}^{k}} + P_{j}^{\text{RF}}\right)^{2}}, \end{aligned} $$
(13)
$$ \begin{aligned} &\frac{\partial^{2} \text{max}\left({P_{j}^{k}}, P_{j}^{\text{RF}}\right)}{\partial ({P_{j}^{k}})^{2}} = \frac{\partial^{2} \sum\limits_{j=1}^{K}\frac{{R_{j}^{k}}}{{P_{j}^{k}} + P_{j}^{\text{RF}}}}{\partial ({P_{j}^{k}})^{2}}= \\& -\sum\limits_{j=1}^{K}\frac{B_{\text{CCs}} \left(\frac{(M-K)^{2}}{\mathrm{ln2}[{KN}_{0} + {P_{j}^{k}}(M-K)]^{2}} + \frac{M-K}{\mathrm{ln2}[{KN}_{0} + {P_{j}^{k}}(M-K)]} \right) }{({P_{j}^{k}} + P_{j}^{\text{RF}})^{2}} \\ &- \sum\limits_{j=1}^{K}\frac{2B_{\text{CCs}}\!\left(\frac{M-K} { \mathrm{ln2}[{KN}_{0}+{P_{j}^{k}}(M-K)]} \,-\, \text{log}_{2}\frac{{KN}_{0} +{P_{j}^{k}}(M-K)}{{KN}_{0}}\right)}{\left({P_{j}^{k}} + P_{j}^{\text{RF}}\right)^{3}}. \end{aligned} $$
(14)
As all constraints are in linear functions now, one can use the linear searching method to obtain the optimal solution of (10) with constraints (11) plus an offline searching method of CC. That is, in each step, search the optimal solution with every possible combination of the constraints, compare them, and only keep the best one. As the linear constraints, one can obtain the optimal solution within a much shorter time. Whereas the searching can be achieved by IBM Cplex (refer to IBM ILOG CPLEX Optimization Studio V12.60 documents) or MATLAB-based linear programming (LP) [25].