# Information-Theoretic Study of Time-Domain Energy-Saving Techniques in Radio
Access

François Rottenberg François Rottenberg is with ESAT-DRAMCO, Ghent Technology
Campus, KU Leuven, 9000 Ghent, Belgium (e-mail:
francois.rottenberg@kuleuven.be).

###### Abstract

Reduction of wireless network energy consumption is becoming increasingly
important to reduce environmental footprint and operational costs. A key
concept to achieve it is the use of lean transmission techniques that
dynamically (de)activate hardware resources as a function of the load. In this
paper, we propose a pioneering information-theoretic study of time-domain
energy-saving techniques, relying on a practical hardware power consumption
model of sleep and active modes. By minimizing the power consumption under a
quality of service constraint (rate, latency), we propose simple yet powerful
techniques to allocate power and choose which resources to activate or to put
in sleep mode. Power consumption scaling regimes are identified. We show that
a “rush-to-sleep” approach (maximal power in fewest symbols followed by sleep)
is only optimal in a high noise regime. It is shown how consumption can be
made linear with the load and achieve massive energy reduction (factor of 10)
at low-to-medium load. The trade-off between energy efficiency (EE) and
spectral efficiency (SE) is also characterized, followed by a multi-user study
based on time division multiple access (TDMA).

###### Index Terms:

Energy consumption, radio access technologies, physical layer, channel
capacity.

##  I Introduction

###  I-A Motivation

As of 2022, yearly data volume has gone up to more than 3 Zettabytes
($10^{21}$ bytes) and the traffic continues to rise at a rate of about
25%/year [1]. Energy efficiency has improved over the years but not fast
enough, which results in an annual energy consumption growth of 2.5% for the
ICT sector [1]. It is becoming increasingly important to reduce energy
consumption of wireless communication networks to reach climate ambitions and
reduce operational expenses, in other words, “break the energy curve” [2, 3].
Most energy of wireless networks is consumed by the radio access network (RAN)
and more specifically at base stations [4, 5]. Moreover, the traffic load at a
base station is highly varying across the day and most often lightly loaded,
with traffic at night being about 10 times lower than during the day [6]. This
opens a big potential for energy reduction through the use of lean
transmission techniques that dynamically activate or deactivate resources as a
function of the load, letting the system dynamically switch from fully active
to deep sleep mode.

###  I-B State of the Art

Energy-saving techniques are a popular topic. A large effort has been made to
integrate these techniques into industrial products and standards. The 5G
standard was for instance designed with a lean paradigm in mind which resulted
in, e.g., less reference signaling to increase sleep duration [7, 8, 9]. At
low-to-medium load, a popular scheduling technique is a “rush-to-sleep”
approach which compacts transmission in as few symbols as possible. These
symbols are transmitted at maximal power, leaving the remaining symbols in the
frame free, so that the sleep duration is maximized. Many studies have been
performed to evaluate the gains of such techniques based on standardized power
models [5], system-level evaluations [10, 11, 12, 13, 14] and aided by actual
measurements [15]. The use of machine learning was also identified as an
interesting tool to predict traffic and/or to optimize energy-saving features
[16, 17]. We refer to [8] for a review of these techniques.

Despite much work in the domain, there remains a fundamental gap to be filled
by establishing an information-theoretic study of time-domain energy-saving
techniques, even for basic systems such as single input single output (SISO)
transceivers. Going back to the underlying physics of energy consumption of
base stations and with a proper mathematical formulation of the communication
link, a lot of additional understandings and improvements can be obtained:
optimization of algorithms, finding optimal power scaling regimes as a
function of load, guarantees of optimality for energy-saving features and/or
finding the gap from it with existing techniques… As an example, it is not
clear if or when a rush-to-sleep approach is optimal or not. We should mention
that many recent works have performed this kind of energy-saving studies but
they have focused on the spatial domain and more specifically the optimal
operation of massive MIMO systems (number of active antennas, served users,
power allocation) as a function of the load [18, 19, 20, 21, 22]. On the other
hand, information-theoretic time-domain studies are lacking. The fundamental
studies on energy-efficient communications have mainly focused on a stationary
transmission of symbols in time at a constant rate and average transmit power
$P_{\mathrm{T}}$ [23, 24, 25, 26, 27]. Considering an ideal consumption model
and a quasi-static channel, this choice seems intuitive. To clarify it, let us
formalize the problem. The channel capacity of a complex discrete memoryless
additive white gaussian noise (AWGN) channel, under an average transmit power
constraint $P_{\mathrm{T}}$, is

$\displaystyle R$
$\displaystyle=\log_{2}\left(1+\frac{P_{\mathrm{T}}}{L\sigma_{n}^{2}}\right)=\log_{2}\left(1+\frac{P_{\mathrm{T}}}{{\sigma}^{2}}\right)\
\text{[bits/channel use]},$

where $\sigma^{2}=\sigma_{n}^{2}L$ is the noise power at the receiver
$\sigma_{n}^{2}$ normalized by the path loss $L$. If the transmission is
divided in frames of $N$ symbols, the average rate and transmit power are

$\displaystyle R$
$\displaystyle=\frac{1}{N}\sum_{n=0}^{N-1}\log_{2}\left(1+\frac{p_{n}}{\sigma^{2}}\right),\
P_{\mathrm{T}}=\frac{1}{N}\sum_{n=0}^{N-1}p_{n}$

where $p_{n}$ is the transmit power of the $n$-th symbol. Considering a rate
constraint $R$, let us find the power allocation that minimizes the consumed
power $P_{\mathrm{cons}}$. Under an ideal consumption model, we have
$P_{\mathrm{cons}}^{\mathrm{ideal}}=P_{\mathrm{T}}$ and the problem can be
written as111This problem can be seen as a conventional waterfilling problem
where water/noise levels are the same at each time slot.

$\displaystyle\min_{p_{0},...,p_{N-1}}\frac{1}{N}\sum_{n=0}^{N-1}p_{n}\
\text{s.t.}\
\frac{1}{N}\sum_{n=0}^{N-1}\log_{2}\left(1+\frac{p_{n}}{\sigma^{2}}\right)=R.$

Given the concavity of the $\log(.)$ function, we can write using the Jensen’s
inequality

$\displaystyle R$
$\displaystyle\leq\log_{2}\left(1+\frac{1}{\sigma^{2}}\frac{1}{N}\sum_{n=0}^{N-1}p_{n}\right)\leftrightarrow(2^{R}-1)\sigma^{2}\leq\frac{1}{N}\sum_{n=0}^{N-1}p_{n}=P_{\mathrm{T}}$

and the bound is tight if uniform power allocation is used, i.e.,
$p_{n}=P_{\mathrm{T}}=(2^{R}-1)\sigma^{2}$, for $n=0,...,N-1$. Intuitively,
the log dependence of the rate implies diminishing returns. Starting from a
non-uniform allocation, it can always be improved by reallocating some power
from the time interval with the highest allocated power to the one with the
lowest power.

In practice however, the consumed power $P_{\mathrm{cons}}$ is far from being
equal or even linearly proportional to the transmit power $P_{\mathrm{T}}$.
This is due to two main reasons, namely: i) as soon as a given time slot is
active, a static load-independent power consumption is present due to
activation of hardware components such as radio-frequency chains and baseband
processing units; ii) the load-dependent power consumption, i.e., the
dependence of $P_{\mathrm{cons}}$ in $p_{n}$, is typically concave as power
amplifiers (PAs) are more energy-efficient close to their saturation.
Intuitively, this implies that the “cost” of using more power decreases when a
large output power is transmitted. These two effects will counterbalance the
log penalty and push towards using a reduced number of active time slots,
especially in low-to-medium load scenarios.

###  I-C Contributions

This paper presents a pioneering information-theoretic study of time-domain
energy-saving techniques, using a realistic power consumption model. The
transmission model considers single-antenna base stations and users. Even for
such a basic system, a comprehensive study of energy-saving features is
lacking, which is the gap this paper is aiming to fill. The investigated
techniques provide drastic energy reduction by dictating how to dynamically
(de)activate hardware resources as a function of the load. The optimization
problems are formalized as the minimization of $P_{\mathrm{cons}}$ for a given
rate. More specifically, the structure of our paper and our contributions are
structured as follows. Section II presents the hardware power consumption
model used in this work, with two distinct contributions: active and sleep
energy consumption. The active power consumption model is shown to address a
large variety of PA classes. Section III considers the optimal allocation of
time resources in a single-user scenario. The solution is approached step by
step through lemmas to get more insight on its nature. Linear and exponential
scaling regimes of $P_{\mathrm{cons}}$ as a function of the load $R$ are
identified. Asymptotic results for large $N$ are provided that greatly
simplify the analysis while having negligible performance penalty. We prove
that a rush-to-sleep approach is optimal in a noise limited regime but not
otherwise. The optimal trade-off EE-SE is also derived from previous results
and we show that a maximal SE does not always provide a maximal EE. Section IV
then extends previous results by considering successive sleep modes, resulting
in drastic energy reductions. Section V considers the extension to a multi-
user scenario where users are multiplexed using TDMA. The optimal allocation
is provided for the most promising regime in terms of energy-savings, i.e.,
the low-to-medium-load scenario where $P_{\mathrm{cons}}$ linearly scales with
the rate of each user and the system is not fully active. Finally, Section VI
concludes the paper.

Notations: The operators $\lceil.\rceil$, $\lfloor.\rfloor$ and
$\left[.\right]$ are the ceil, floor and round operators, respectively. The
operator $\lfloor x\rceil$ which we refer to as the ceil-floor operator
selects among the upper and lower bounding integers of $x$ the one that
optimizes the cost function. The function $W(z)$ is the Lambert W function,
i.e., the solution of $z=W(z)e^{W(z)}$. We use the notation $f(x)=O(g(x))$, as
$x\rightarrow a$, if there exist positive numbers $\delta$ and $\lambda$ such
that $|f(x)|\leq\lambda g(x)$ when $0<|x-a|<\delta$.

##  II Power Consumption Model

As described in the introduction, we consider the transmission of $N$ symbols,
each of duration $T$ [s]. The full frame has thus a duration $NT$ [s]. Out of
the $N$ intervals, a number $N_{\mathrm{a}}$ are active and actually
transmitting information while the remaining $N-N_{\mathrm{a}}$ are inactive
and in sleep mode. This implies that $0\leq N_{\mathrm{a}}\leq N$. For
minimizing latency and maximizing the sleep duration which allows entering a
deeper sleep mode [28], the $N_{\mathrm{a}}$ active intervals are grouped
together at the beginning of the transmission. We define as
$E_{\mathrm{active}}$ the energy consumed during active time slots, which is
assumed to depend on the transmit power at each time interval, i.e.,
$p_{0},...,p_{N_{\mathrm{a}}-1}$. On the other hand, $E_{\mathrm{sleep}}$
represents the energy consumed during sleep modes, which is non zero as all
hardware components cannot be switched-off. It is assumed to depend on
$(N-N_{\mathrm{a}})T$ as a longer sleep duration allows to enter a deeper
sleep mode [28]. The average consumed power over the frame duration is thus
given by

$\displaystyle P_{\mathrm{cons}}$
$\displaystyle=\frac{E_{\mathrm{active}}(p_{0},...,p_{N_{\mathrm{a}}-1})+E_{\mathrm{sleep}}((N-N_{\mathrm{a}})T)}{NT}.$
(1)

In the following, we detail the models of the sleep and active energy
consumption. As a benchmark, we also introduce the ideal consumption model

$\displaystyle P_{\mathrm{cons}}^{\mathrm{ideal}}$
$\displaystyle=\frac{1}{N}\sum_{n=0}^{N_{\mathrm{a}}-1}p_{n},$ (2)

implying that the consumed power is equal to the transmit power. In other
words, no losses are present.

###  II-A Active Energy Consumption

Using the well-established model from [29], the active power consumption can
be modelled as

$\displaystyle P_{\mathrm{active}}$
$\displaystyle=\frac{P_{\mathrm{PA}}+P_{\mathrm{RF}}+P_{\mathrm{BB}}}{(1-\sigma_{\mathrm{DC}})(1-\sigma_{\mathrm{MS}})(1-\sigma_{\mathrm{cool}})}$

where $P_{\mathrm{PA}}$, $P_{\mathrm{RF}}$ and $P_{\mathrm{BB}}$ are the
powers consumed by the PAs, the radio-frequency chains and the baseband unit
respectively. The coefficients $\sigma_{\mathrm{DC}}$, $\sigma_{\mathrm{MS}}$
and $\sigma_{\mathrm{cool}}$ are the loss factors related to DC-DC power
supply, mains supply and active cooling respectively.

TABLE I: Values of loss factors and efficiency used in evaluations [29]. DC-DC | $\sigma_{\mathrm{DC}}$ | 7.5%
---|---|---
Mains supply | $\sigma_{\mathrm{MS}}$ | 9.0%
Cooling | $\sigma_{\mathrm{cool}}$ | 10.0%
Efficiency | $\eta=(1-\sigma_{\mathrm{DC}})(1-\sigma_{\mathrm{MS}})(1-\sigma_{\mathrm{cool}})$ | 75.8%

We are interested in modelling the dependence of the active consumed power
$P_{\mathrm{active}}$ in the output power at each active time slot
$p_{0},...,p_{N_{\mathrm{a}}-1}$. The base station (BS) power consumption
analysis of [29] showed that mainly the PA consumed power $P_{\mathrm{PA}}$
scales with the output power. The other terms are thus considered load-
independent. The active energy consumed across the frame duration can thus be
written as

$\displaystyle E_{\mathrm{active}}(p_{0},...,p_{N_{\mathrm{a}}-1})$
$\displaystyle=\frac{T}{\eta}\left(N_{\mathrm{a}}\tilde{P}_{0}+\sum_{n=0}^{N_{\mathrm{a}-1}}P_{\mathrm{PA}}(p_{n})\right)$

where $\tilde{P}_{0}=P_{\mathrm{BB}}+P_{\mathrm{RF}}$ and
$\eta=(1-\sigma_{\mathrm{DC}})(1-\sigma_{\mathrm{MS}})(1-\sigma_{\mathrm{cool}})$.
Values of loss factors and efficiencies used in evaluations are shown in Table
I. To model the PA consumption, we use the following model

$\displaystyle P_{\mathrm{PA}}(p)$ $\displaystyle=P_{\mathrm{PA},0}+\beta
p^{\alpha},\ 0\leq p\leq P_{\mathrm{max}}$ (3)

with $\alpha\in]0,1]$ and $\beta\geq 0$. The first term $P_{\mathrm{PA},0}$
represents the load-independent consumption while the second is load-
dependent. This load dependency does not typically scale linearly with $p$.
The fact that $\alpha\in]0,1]$ implies concavity of $P_{\mathrm{PA}}(p)$. This
concavity comes from the fact that a typical PA efficiency is improved when
moving closer to saturation [30]. The constant $P_{\mathrm{max}}$ denotes the
maximal transmit power.In practice, $P_{\mathrm{max}}$ is (much) lower than
the PA saturation power, that we denote by $P_{\mathrm{sat}}$. The use of the
so-called back-off $P_{\mathrm{max}}/P_{\mathrm{sat}}$ is required as recent
technologies, e.g., orthogonal frequency division multiplexing (OFDM), have
high peak-to-average power ratio (PAPR). A back-off (typically from -12 dB to
-6 dB) prevents the PA to enter the saturation region, which would otherwise
create nonlinear distortion impacting the signal quality and creating out-of-
band emissions. The authors of [31] have justified in details the use of a
similar model as (3) through their own measurements and a literature review
[32, 33]. Model (3) is more general as it also includes a load-independent
component, which is useful for particular PA architectures.

####  II-A1 Ideal Power Amplifier

PA consumed power is linearly proportional to the output power giving
$P_{\mathrm{PA},0}=0$, $\beta=\alpha=1$ and

$\displaystyle P_{\mathrm{PA}}^{\mathrm{ideal}}(p)=p.$

####  II-A2 Class A Power Amplifier

PA consumed power is independent of the load and has a maximal efficiency of
$1/2$ giving $P_{\mathrm{PA},0}=2P_{\mathrm{sat}}$, $\beta=0$ and

$\displaystyle P_{\mathrm{PA}}^{\mathrm{A}}(p)=2P_{\mathrm{sat}}$

where $P_{\mathrm{sat}}$ is the saturation power of the PA.

####  II-A3 Class B Power Amplifier

The PA consumed power has a load dependence that scales with the square root
of the output power giving $P_{\mathrm{PA},0}=0$,
$\beta=\frac{4}{\pi}\sqrt{P_{\mathrm{sat}}}$ and $\alpha=1/2$

$\displaystyle
P_{\mathrm{PA}}^{\mathrm{B}}(p)=\frac{4}{\pi}\sqrt{P_{\mathrm{sat}}}\sqrt{p}$

This model has much relevance for typical base stations working with a
significant back-off from saturation [31]. Some authors sometimes call it the
“traditional” PA model [34]. Therefore, by default, we will use it in
following evaluations, with a 8 dB back-off.

####  II-A4 Envelope Tracking Power Amplifier

According to the curve fitted model proposed in [35], PA consumption was shown
to be modelled as

$\displaystyle
P_{\mathrm{PA}}^{\mathrm{ET}}(p)\approx\frac{aP_{\mathrm{sat}}}{(1+a)\eta_{\mathrm{max}}}+\frac{1}{(1+a)\eta_{\mathrm{max}}}p,$

where $a=0.0082$. This model can again be seen as a special case of (3).

####  II-A5 Doherty Power Amplifier

The $\ell$-way Doherty PA consumed power is given by [25]

$\displaystyle
P_{\mathrm{PA}}^{\mathrm{Doherty}}(p)=\frac{4P_{\mathrm{sat}}}{\ell\pi}\begin{cases}\sqrt{\xi}&0<\xi\leq\frac{1}{\ell^{2}}\\\
(\ell+1)\sqrt{\xi}-1&\frac{1}{\ell^{2}}<\xi\leq 1\end{cases}$

where $\xi=p/P_{\mathrm{sat}}$. The class B model is obtained as a special
case when $\ell=1$. Except in such particular cases, the model proposed in (3)
cannot exactly represent such PAs but can provide an approximation depending
on the operating range of the Doherty amplifier.

######  Remark 1.

We previously described how model (3) can address typical theoretical PA
models. However, it can also be fitted for practical PAs based on measurements
and/or datasheets. The PA was identified as the main load-dependent
contribution. However, more generally, the model proposed in (3) can take into
account other load-dependent terms.

In the light of this remark, we formalize the underlying assumption about the
load-dependent active energy consumption model used throughout this work.

$\mathbf{(As1)}$: The active energy consumed across the frame is

$\displaystyle E_{\mathrm{active}}(p_{0},...,p_{N_{\mathrm{a}}-1})$
$\displaystyle=T\left(N_{\mathrm{a}}P_{0}+\gamma\sum_{n=0}^{N_{\mathrm{a}-1}}p_{n}^{\alpha}\right)$
(4)

where $\gamma=\frac{\beta}{\eta}\geq 0$ and
$P_{0}=\frac{P_{\mathrm{BB}}+P_{\mathrm{RF}}+P_{\mathrm{PA},0}}{\eta}\geq 0$,
$\alpha\in]0,1]$, $0\leq p_{n}\leq P_{\mathrm{max}}$ for
$n=0,\ldots,N_{\mathrm{a}}-1$, $0\leq N_{\mathrm{a}}\leq N$ and $p_{n}=0$ for
$n=N_{\mathrm{a}},\ldots,N-1$. The averaged consumed power is then

$\displaystyle P_{\mathrm{cons}}$
$\displaystyle=\frac{N_{\mathrm{a}}}{N}P_{0}+\frac{\gamma}{N}\sum_{n=0}^{N_{\mathrm{a}-1}}p_{n}^{\alpha}+\frac{E_{\mathrm{sleep}}((N-N_{\mathrm{a}})T)}{NT}.$
(5)

###  II-B Sleep Energy Consumption

In the ideal consumption model (2), the sleep energy consumption is exactly
null. In practice, not all hardware can be switched off as always-on reference
signals are required to allow users to access the network. Moreover, different
hardware components have different activation/reactivation latencies. Hence,
depending on the sleep duration, more or less components can be switched off.
Therefore, power models have been proposed that consider successive sleep
modes as a function of the sleep depth. We define the power consumption in
sleep mode as $P_{\mathrm{sleep}}(t)$, where $t=0$ is used a reference for the
system entering sleep and $t$ is the sleep duration so that
$E_{\mathrm{sleep}}(0)=0$. Given the fact that increasing sleep duration
allows to switch off more hardware components, we introduce the following
assumption.

$\mathbf{(As2)}$: $P_{\mathrm{sleep}}(t)$ is monotonically non-increasing.

######  Proposition 1.

Under $\mathbf{(As2)}$, the sleep energy consumption $E_{\mathrm{sleep}}(t)$
is a concave function of the sleep duration $t$.

###### Proof.

Directly follows from $\mathbf{(As2)}$.∎

######  Remark 2.

$\mathbf{(As2)}$ does not imply continuity of $P_{\mathrm{sleep}}(t)$, which
can have jump discontinuities. Switching-off components might lead to a non-
continuous drop of $P_{\mathrm{sleep}}(t)$, as shown in Fig. 1.

$10^{0}$$10^{1}$$10^{2}$$10^{3}$$0$$20$$40$$60$Time [ms]$\displaystyle
P_{\mathrm{sleep}}(t)$ [W]Piecewise constant sleep $\mathbf{(As3)}$Constant
sleep $\mathbf{(As4)}$

Figure 1: Two classical sleep power models are shown: constant or successive
sleep modes based on values in Table II. TABLE II: Numerical values of power
and power models used in evaluations.

Maximal transmit power |  $P_{\mathrm{max}}=20$ W
---|---
Deep sleep power |  $P_{3}=1$ W
Load-independent active power | $P_{0}=110P_{3}$
Successive sleep power model $\mathbf{(As3)}$ [5] | $P_{\mathrm{sleep}}(t)$
Micro sleep | Light sleep | Deep sleep | Hibernating sleep
$T_{1}=0$ |  $T_{2}=6$ ms |  $T_{3}=50$ ms |  $T_{4}=1$ s
$P_{1}=50P_{3}$ | $P_{2}=25P_{3}$ | $P_{3}$ | $P_{4}=0.1P_{3}$
Constant sleep power model $\mathbf{(As4)}$ | $P_{\mathrm{sleep}}(t)=P_{\mathrm{sleep}}=P_{1}$

Popular models for $P_{\mathrm{sleep}}(t)$ include the use of different sleep
modes. The model in [28] has been used as a qualitative and quantitative
reference for several years by companies [36]. More recently, 3GPP has
introduced an improved model, expressed in relative units with respect to the
deep sleep mode, that better reflects current trends [5, 37]. The model
contains four sleep modes and numerical values are given in Table II for one
configuration described in [5]. These values will be used as an example in
evaluations.

We can formalize this model mathematically. Let us define as $S$ the number of
sleep modes, starting from mode 0 (no sleep) to mode $S$ (deepest sleep mode).
The start and end of each sleep mode are denoted by $T_{s}$ and $T_{s+1}$,
with $T_{0}=0$ and $T_{S+1}=+\infty$. The sleep power consumption during mode
$s$ is denoted by $P_{s}$. This notation is consistent with the definition of
$P_{0}$ which denotes the load-independent active power consumption. This
corresponds to “sleep mode zero”, taking place until $T_{1}$, where no
hardware components are actually switched off.

$\mathbf{(As3)}$: the sleep power consumption is piecewise constant implying
that $P_{\mathrm{sleep}}(t)=P_{s}$ and

$\displaystyle
E_{\mathrm{sleep}}(t)=\int_{0}^{t}P_{\mathrm{sleep}}(t^{\prime})dt^{\prime}=\sum_{s^{\prime}=0}^{s-1}P_{s^{\prime}}(T_{s^{\prime}+1}-T_{s^{\prime}})+(t-T_{s})P_{s}$

where $s$ is the index such that $T_{s}<t\leq T_{s+1}$ and $P_{s+1}\leq P_{s}$
according to $\mathbf{(As2)}$.

We also introduce another popular sleep power model widely used in the
literature and shown to be accurate to characterize 4G long term evolution
(LTE) macro base stations [29].

$\mathbf{(As4)}$: the sleep power consumption is constant implying that
$P_{\mathrm{sleep}}(t)=P_{\mathrm{sleep}}$ and
$E_{\mathrm{sleep}}(t)=P_{\mathrm{sleep}}t$ with $P_{0}\geq
P_{\mathrm{sleep}}$. The averaged consumed power is then

$\displaystyle P_{\mathrm{cons}}$
$\displaystyle=\frac{N_{\mathrm{a}}}{N}P_{0}+\frac{\gamma}{N}\sum_{n=0}^{N_{\mathrm{a}-1}}p_{n}^{\alpha}+\frac{N-N_{\mathrm{a}}}{N}P_{\mathrm{sleep}}.$
(6)

######  Remark 3.

$\mathbf{(As4)}$ is a particular case of $\mathbf{(As3)}$ with a single sleep
mode: $S=1$, $P_{\mathrm{sleep}}=P_{1}$ and $T_{1}=0$. In the following, we
use the term “$\mathbf{(As3)}-\mathbf{(As4)}$”, when both assumptions hold.

##  III Optimal Allocation of Time Resources

This section considers the solution of minimizing the average consumed power
under $(\mathbf{As1})$ and a rate constraint

$\displaystyle\min_{N_{\mathrm{a}},p_{0},...,p_{N_{\mathrm{a}}-1}}P_{\mathrm{cons}}\quad\text{s.t.
}\frac{1}{N}\sum_{n=0}^{N_{\mathrm{a}}-1}\log_{2}\left(1+\frac{p_{n}}{\sigma^{2}}\right)=R.$
(7)

We define a constant that will be useful throughout this section. Under
$(\mathbf{As1})-(\mathbf{As4})$, $R_{\mathrm{a}}$ is the constant that
minimizes the convex problem

$\displaystyle R_{\mathrm{a}}$ $\displaystyle=\arg\min_{x\geq
0}\frac{P_{0}-P_{\mathrm{sleep}}+\gamma\sigma^{2\alpha}\left(2^{x}-1\right)^{\alpha}}{x},$

where $P_{\mathrm{sleep}}$ is the constant sleep power consumption defined in
$(\mathbf{As4})$. When $P_{0}-P_{\mathrm{sleep}}=0$, it is given by

$\displaystyle
R_{\mathrm{a}}=(W(-\alpha^{-1}e^{-\alpha^{-1}})+\alpha^{-1})/\log(2)$ (8)

where $W(z)$ is the Lambert W function, i.e., the solution of
$z=W(z)e^{W(z)}$. In the following, the general solution of (7) is approached
step by step, introducing several lemmas, which provide insight on its form
and scaling as a function of $R$. Finally, the trade-off SE versus EE will be
characterized.

###  III-A Optimal Allocation for Load-Dependent Consumed Power

The following lemma provides the power allocation that minimizes the load-
dependent part of the average consumed power under a rate constraint and
without a maximal per-time slot power constraint. Under $\mathbf{(As1)}$, the
load-dependent part of $P_{\mathrm{cons}}$ can be identified as

$\displaystyle P_{\mathrm{ld}}(p_{0},...,p_{N-1})$
$\displaystyle=\frac{\gamma}{N}\sum_{n=0}^{N-1}p_{n}^{\alpha},$ (9)

which equals $P_{\mathrm{cons}}$ for $P_{0}=0$ and $E_{\mathrm{sleep}}(t)=0$.

######  Lemma 1.

Under $\mathbf{(As1)}$, for $P_{0}=E_{\mathrm{sleep}}(t)=0$, $\gamma>0$ and
$P_{\mathrm{max}}\to+\infty$, the minimum of (7), is achieved by uniformly
allocating power among $N_{\mathrm{a}}$ time slots

$\displaystyle N_{\mathrm{a}}$
$\displaystyle=\lfloor\min(NR/R_{\mathrm{a}},N)\rceil,$ $\displaystyle p_{n}$
$\displaystyle=\begin{cases}\left(2^{R\frac{N}{N_{\mathrm{a}}}}-1\right)\sigma^{2}&\text{if
}n=0,...,N_{\mathrm{a}}-1\\\ 0&\text{otherwise}\end{cases}.$

###### Proof.

See Appendix VII-A. ∎

$0$$2$$4$$6$$8$$0$$2$$4$$6$$8$$\displaystyle p_{0}$$\displaystyle
p_{1}$$\displaystyle
R=\frac{1}{2}(\log_{2}(1+p_{0})+\log_{2}(1+p_{1}))$$\frac{1}{2}(p_{0}+p_{1})=3$$\frac{1}{2}(\sqrt{p_{0}}+\sqrt{p_{1}})=0.75$$\frac{1}{2}(\sqrt{p_{0}}+\sqrt{p_{1}})=2$

(a)

$0$$0.2$$0.4$$0.6$$0.8$$1$$0$$2$$4$$6$$8$$10$$\delta$$\displaystyle\alpha$$\displaystyle
R_{\mathrm{a}}$$\delta=0$$\delta=1$$\delta=5$

(b)

Figure 2: (a) Contour plot of i) rate constraint and ii) load-dependent power
consumption ($\gamma=1$) for a frame of $N=2$ time slots and two load-
dependent power exponents $\alpha$: $1$ and $1/2$. (b) Constant
$R_{\mathrm{a}}$ as a function of the load-dependent power exponent $\alpha$
and the ratio $\delta=(P_{0}-P_{\mathrm{sleep}})/(\gamma\sigma^{2\alpha})$.

######  Remark 4 (Frame of $N=2$ symbols).

To illustrate Lemma 1, Fig. 2a considers the particular case $N=2$,
$\gamma=1$. It shows contour plots of the constraint $R$ and the cost function
$P_{\mathrm{ld}}$, as a function of $p_{0}$ and $p_{1}$. In the case
$\alpha=1$, implying linearity of consumed and transmit power, the objective
function curve $(p_{0}+p_{1})/2=3$ is a straight line. As shown in the
introduction, utilization of the two time slots is always optimal
($N_{\mathrm{a}}=2$), with uniform power allocation. However, for
$\alpha=1/2$, this is not anymore the case. For low values of $R$, it is
better to only use $N_{\mathrm{a}}=1$ slot while for large $R$,
$N_{\mathrm{a}}=2$ is optimal, again with uniform power.

######  Remark 5 (Scaling of $N_{\mathrm{a}}$ for general $N$ and $\alpha$).

For an arbitrary value of $N$ and $\alpha\in]0,1]$, the number of activated
time slots $N_{\mathrm{a}}$ scales approximately linearly with the rate $R$ up
to the point where the maximal number $N$ is allocated, i.e., when
$R>R_{\mathrm{a}}$. When $R\leq R_{\mathrm{a}}$, the rate per activated time
slot is approximately equal to $R_{\mathrm{a}}$. The “approximate” nature
comes from the rounding operation. This error disappears for large $N$, as
will be formalized properly in the following. The constant $R_{\mathrm{a}}$ is
independent of $R$ and is a function of $\alpha$. As shown in Fig. 2b for the
case $\delta=0$ ($P_{0}=P_{\mathrm{sleep}}=0$), $R_{\mathrm{a}}$ monotonically
decreases as a function of $\alpha$, implying that more time slots are
activated for a fixed value of $R$. In the asymptotic cases of $\alpha$
approaching 0 or 1, a single ($N_{\mathrm{a}}=1$) or all time slots
($N_{\mathrm{a}}=N$) are allocated, respectively.

###  III-B Optimal Allocation with no Maximal Power Constraint

The following lemma gives the power allocation that minimizes the averaged
consumed power under a rate constraint and without a maximal per-time slot
power constraint. Removing this constraint provides the solution when it is
not binding, i.e., when the user experiences a good channel (low normalized
noise variance $\sigma^{2}=L\sigma_{n}^{2}$) and its target rate is not too
high. The general case will be addressed in next subsection.

######  Lemma 2.

Under $\mathbf{(As1)}$-$\mathbf{(As2)}$ and for $P_{\mathrm{max}}\to+\infty$,
the minimum of the problem (7) is achieved by uniformly allocating power among
$N_{\mathrm{a}}$ time slots

$\displaystyle p_{n}$
$\displaystyle=\begin{cases}\left(2^{R\frac{N}{N_{\mathrm{a}}}}-1\right)\sigma^{2}&\text{if
}n=0,...,N_{\mathrm{a}}-1\\\ 0&\text{otherwise}\end{cases}$ (10)

where $N_{\mathrm{a}}$ is the argument that minimizes

$\displaystyle\min_{0\leq N_{\mathrm{a}}\leq
N}\frac{N_{\mathrm{a}}}{N}\left(P_{0}+\gamma\sigma^{2\alpha}\left(2^{R\frac{N}{N_{\mathrm{a}}}}-1\right)^{\alpha}\right)+\frac{E_{\mathrm{sleep}}((N-N_{\mathrm{a}})T)}{NT}.$
(11)

Under $\mathbf{(As3)}-\mathbf{(As4)}$, the solution is

$\displaystyle N_{\mathrm{a}}=\lfloor\min(NR/R_{\mathrm{a}},N)\rceil.$

###### Proof.

See Appendix VII-A. ∎

######  Remark 6 (Relation with Lemma 1).

Lemma 2 extends the result of Lemma 1 by considering a non-zero sleep power
$P_{\mathrm{sleep}}$ and load-independent active power consumption $P_{0}$.
Under a constant sleep power model $\mathbf{(As4)}$, the problem has a similar
solution.

######  Remark 7 (Scaling of $R_{\mathrm{a}}$).

As shown in Fig. 2b, $R_{\mathrm{a}}$ increases with
$\delta=(P_{0}-P_{\mathrm{sleep}})/(\gamma\sigma^{2\alpha})$. This implies
that less time slots should be activated if normalized noise power
$\sigma^{2}$, $P_{\mathrm{sleep}}$ and $\gamma$ are small and if the active
load-independent power consumption $P_{0}$ is high. A major difference with
Lemma 1 is that $R_{\mathrm{a}}$ does not go to zero as $\alpha$ approaches 1
when $\delta>0$. This implies that, given nonzero static power consumption,
not all time slots should always be activated. The particular load-independent
case $\gamma=0$ implies that $R_{\mathrm{a}}=+\infty$ and $N_{\mathrm{a}}=1$.
This makes sense as $P_{\mathrm{cons}}$ then only depends on $N_{\mathrm{a}}$.
As shown in next subsection, considering a finite maximal transmit power per
time slot will change this result.

###  III-C Optimal Allocation with Maximal Power Constraint

We now consider the additional constraint of a finite maximal per-time slot
power $P_{\mathrm{max}}$. This constraint may render the problem unfeasible.
Therefore, we introduce the following assumption.

Algorithm 1 Iterative resource allocation

$\sigma^{2},R,N,P_{\mathrm{max}},R_{\mathrm{max}},R_{\mathrm{a}}$

$N_{\mathrm{a}}\leftarrow(\ref{eq:theor_no_max_problem})$$\triangleright$
Init. by sol. of Theor. 2

$p_{0},...,p_{N-1}\leftarrow(\ref{eq:opt_p_n})$

$N_{\mathrm{max}}\leftarrow 0$

$\hat{N}\leftarrow N$

while $p_{N_{\mathrm{a}}-1}>P_{\mathrm{max}}$ do $\triangleright$ Check max
power constraint

$N_{\mathrm{max}}\leftarrow N_{\mathrm{max}}+1$ $\triangleright$ Set one more
time slot to max power

$R\leftarrow(NR-R_{\mathrm{max}})/({N}-1)$ $\triangleright$ Adapt rate
constraint

$N\leftarrow N-1$

$N_{\mathrm{a}}\leftarrow(\ref{eq:theor_no_max_problem})$$\triangleright$
Update with sol. of Theor. 2

$p_{0},...,p_{N-1}\leftarrow(\ref{eq:opt_p_n})$

end while

$p_{N_{\mathrm{a}}},...,p_{N_{\mathrm{a}}+N_{\mathrm{max}}-1}\leftarrow
P_{\mathrm{max}}$

$p_{N_{\mathrm{a}}+N_{\mathrm{max}}},...,p_{\hat{N}-1}\leftarrow 0$

(As5): Problem (7) is feasible, i.e.,

$\displaystyle R\leq
R_{\mathrm{max}}=\log_{2}\left(1+\frac{P_{\max}}{\sigma^{2}}\right).$

The exact solution of problem (7) generally requires an iterative solution.

######  Proposition 2.

Under $\mathbf{(As1)}-\mathbf{(As2)},\mathbf{(As5)}$, the solution of the
problem (7) can be obtained by using Algorithm 1.

###### Proof.

See Appendix VII-B. ∎

To avoid the need of an iterative solution and the ceil-floor operator, we use
the fact that the problem greatly simplifies by considering the asymptotic
case of a large $N$. Then, the ratio $N_{\mathrm{a}}/N$ can be considered
asymptotically continuous instead of only taking discrete values.

######  Remark 8 (Large $N$ assumption).

The large $N$ assumption is realistic in practice as frames are typically made
of many symbols. Moreover, the assumption of having a sufficiently large $N$
is central and implicit in this work. This article investigates the gain of
activating only a share of transmission time slots. To be able to do this,
some flexibility in the number of activated resources should be available,
implying a sufficiently large $N$.

To provide the closed-form asymptotic solution, we define

$\displaystyle\tilde{R}=\min(R_{\mathrm{a}},R_{\mathrm{max}}),P_{\mathrm{a}}=(2^{R_{\mathrm{a}}}-1)\sigma^{2},\tilde{P}=\min(P_{\mathrm{a}},P_{\mathrm{max}}).$

######  Theorem 1.

Under $\mathbf{(As1)}-\mathbf{(As5)}$, the solution of problem (7) and two
scaling regimes of $P_{\mathrm{cons}}$ as a function of $R$ can be found:

$\square$ (Linear) If $R\leq\tilde{R}$, as $N\rightarrow+\infty$, the
allocation

$\displaystyle N_{\mathrm{a}}$ $\displaystyle=\left[NR/\tilde{R}\right],\
p_{n}=\begin{cases}\tilde{P}&\text{if }n=0,...,N_{\mathrm{a}}-1\\\
0&\text{otherwise}\end{cases}$

is asymptotically optimal and achieves a consumed power

$\displaystyle P_{\mathrm{cons}}$
$\displaystyle=P_{\mathrm{sleep}}+R\frac{P_{0}-P_{\mathrm{sleep}}+\gamma\tilde{P}^{\alpha}}{\tilde{R}}+\epsilon$

where $\epsilon$ is the gap from the optimum which asymptotically vanishes:
$|\epsilon|=O(1/N)$.

$\square$ (Exponential) If $R>\tilde{R}$:

$\displaystyle N_{\mathrm{a}}$ $\displaystyle=N,\
p_{n}=\left(2^{R}-1\right)\sigma^{2}\quad\text{for }n=0,...,N-1,$
$\displaystyle P_{\mathrm{cons}}$
$\displaystyle=P_{0}+\gamma\sigma^{2\alpha}\left(2^{R}-1\right)^{\alpha}.$

$0$$2$$4$$6$$8$$10$$0$$50$$100$$150$$200$$R_{\mathrm{max}}$$R_{\mathrm{a}}$$P_{\mathrm{cons,max}}$$P_{0}$$P_{\mathrm{sleep}}$LinearExponential$\displaystyle
R$ [bits/channel use]$\displaystyle P_{\mathrm{cons}}$ [W]$\displaystyle N=10$
symbols, $\displaystyle\sigma^{2}=10$ mWUniformAsympt. opt.Opt.Rush-to-sleep

(a)

$0$$0.5$$1$$1.5$$2$$2.5$$0$$50$$100$$150$$200$$R_{\mathrm{max}}$$R_{\mathrm{a}}$$P_{\mathrm{cons,max}}$$P_{0}$$P_{\mathrm{sleep}}$Linear$\displaystyle
R$ [bits/channel use]$\displaystyle P_{\mathrm{cons}}$ [W]$\displaystyle N=50$
symbols, $\displaystyle\sigma^{2}=5$ WUniform alloc.Asympt. opt. alloc.Opt.
alloc.Rush-to-sleep

(b)

Figure 3: Power consumption versus load using optimal/uniform power allocation
with constant sleep mode $\mathbf{(As4)}$.

###### Proof.

See Appendix VII-C. ∎

######  Remark 9.

For large $N$, the solution has a simple form: the ceil-floor operator is
replaced by a rounding operator. The power and number of bits per-activated
time slot ($\tilde{P}$ and $\tilde{R}$) are constant in the linear regime. As
shown in Fig. 3a and 3b, the approximation error can barely be seen and is
already negligible for small values of $N$ such as 10 or 50.

######  Remark 10 (Scaling regimes).

Lemma 1 puts forward two scaling regimes: linear and exponential. They can
easily be identified in Fig. 3a. In Fig. 3b, a higher noise regime is
considered so that only the linear regime is present
($\tilde{R}=R_{\mathrm{max}}$).

######  Remark 11 (Gain with respect to uniform allocation).

Fig. 3a and 3b also plot the gain with respect to a uniform allocation, i.e.,
$p_{n}=\left(2^{R}-1\right)\sigma^{2}$ for $n=0,...,N-1$. As expected, the
gain is larger at low load (rate) where the optimal allocation only activates
few resources.

######  Remark 12 (Rush-to-sleep).

If $R_{\mathrm{max}}\leq R_{\mathrm{a}}$ (high noise regime, Fig. 3b), a rush-
to-sleep approach is optimal: active time slots transmitting at full power
$P_{\mathrm{max}}$ and rate $R_{\mathrm{max}}$. This minimizes
$N_{\mathrm{a}}$ and maximize sleep duration. If
$R_{\mathrm{a}}<R_{\mathrm{max}}$ (low noise regime, Fig. 3a), reduced
transmit power should be used instead and not even using sleep for
$R>R_{\mathrm{a}}$. It is then better to fully activate the system with a
uniform allocation.

######  Remark 13 (Converse problem).

We here consider the minimization of $P_{\mathrm{cons}}$ for a fixed $R$. The
maximization of $R$ for a fixed $P_{\mathrm{cons}}$ can also be considered. It
could occur if power is available and should directly be used, e.g., a solar
panel or wind turbine without battery and/or not connected to the grid. The
solution can be found using same methodology or directly by “reverting” the
result of Theor. 1. Indeed, the allocation that minimizes $P_{\mathrm{cons}}$
for a fixed rate $R$ is also the allocation that maximizes $R$ for the minimal
value of $P_{\mathrm{cons}}$ of the inital problem. Scaling regimes of $R$ as
a function of $P_{\mathrm{cons}}$ will be linear and logarithmic. It is here
omitted due to space constraints.

###  III-D Trade-Off: Spectral Efficiency versus Energy Efficiency

Considering that the transmission occupies a bandwidth
$B=(1+\alpha_{\mathrm{rol}})/T$, where $\alpha_{\mathrm{rol}}\in[0,1]$ is the
roll-off factor, so that the SE and the EE are

$\displaystyle\mathrm{SE}$
$\displaystyle=\frac{R}{TB}=\frac{R}{1+\alpha_{\mathrm{rol}}}\quad\text{[bits/s/Hz]},\
\mathrm{EE}=\frac{B\mathrm{SE}}{P_{\mathrm{cons}}}=\frac{R}{TP_{\mathrm{cons}}}\quad\text{[bits/Joule]}.$
(12)

Given these relationships, the trade-off SE-EE can be easily identified from
previous results.

######  Corollary 1.

Under $\mathbf{(As1)}-\mathbf{(As5)}$, the optimal EE for a given SE is:

$\square$ If $\mathrm{SE}\leq\tilde{R}/(1+\alpha_{\mathrm{rol}})$ (implying
$R\leq\tilde{R}$), as $N\rightarrow+\infty$,

$\displaystyle\mathrm{EE}$
$\displaystyle=\frac{B\mathrm{SE}}{P_{\mathrm{sleep}}+\mathrm{SE}\frac{1+\alpha_{\mathrm{rol}}}{\tilde{R}}\left(P_{0}-P_{\mathrm{sleep}}+\gamma\tilde{P}^{\alpha}\right)}+O(1/N).$

$\square$ If $\mathrm{SE}>\tilde{R}/(1+\alpha_{\mathrm{rol}})$ (implying
$R>\tilde{R}$):

$\displaystyle\mathrm{EE}$
$\displaystyle=\frac{B\mathrm{SE}}{P_{0}+\gamma\sigma^{2\alpha}\left(2^{\mathrm{SE}(1+\alpha_{\mathrm{rol}})}-1\right)^{\alpha}}.$

###### Proof.

From the definition of the EE in (12), it is clear that its maximization is
equivalent to the minimization of $P_{\mathrm{cons}}$. As a result, the
optimal allocations and the results of Theor. 1 can directly be used, which
lead to the above results. The two cases correspond to the linear and
exponential scaling regimes of Theor. 1, respectively. ∎

######  Corollary 2.

Under $\mathbf{(As1)}-\mathbf{(As5)}$, the SE that maximizes the optimal EE
given in Corollary 1 is:

$\square$ If $R_{\mathrm{a}}\geq R_{\mathrm{max}}$:

$\displaystyle\bar{\mathrm{SE}}=\mathrm{SE}_{\mathrm{max}}$
$\displaystyle=\frac{R_{\mathrm{max}}}{1+\alpha_{\mathrm{rol}}},\
\mathrm{EE}_{\mathrm{max}}=\frac{B\mathrm{SE}_{\mathrm{max}}}{P_{0}+\gamma{P}_{\mathrm{max}}^{\alpha}}.$

$\square$ If $R_{\mathrm{a}}<R_{\mathrm{max}}$:

$\displaystyle\bar{\mathrm{SE}}$
$\displaystyle=\frac{\bar{R}}{1+\alpha_{\mathrm{rol}}},\
\mathrm{EE}_{\mathrm{max}}=\frac{B\bar{\mathrm{SE}}}{P_{0}+\gamma\sigma^{2\alpha}\left(2^{\bar{R}}-1\right)^{\alpha}}$

where $\bar{R}$ is the constant that minimizes the convex problem

$\displaystyle\min_{\bar{R}\in[R_{\mathrm{a}},R_{\mathrm{max}}]}\frac{P_{0}+\gamma\sigma^{2\alpha}\left(2^{\bar{R}}-1\right)^{\alpha}}{\bar{R}}.$

###### Proof.

See Appendix VII-D. ∎

$0$$2$$4$$6$$8$$0$$200$$400$$600$$800$$1{,}000$$\mathrm{EE}_{\mathrm{max}}$$\bar{\mathrm{SE}}$$\displaystyle\mathrm{SE}$
[bits/s/Hz]$\displaystyle\mathrm{EE}$
[kbits/Joule]$\displaystyle\sigma^{2}=10$ mW, $\displaystyle B=20$ MHz,
$\displaystyle\alpha_{\mathrm{rol}}=0.1$UniformOptimalRush-to-sleep

(a)

$0$$0.5$$1$$1.5$$2$$0$$50$$100$$150$$200$$\mathrm{EE}_{\mathrm{max}}$$\bar{\mathrm{SE}}=\mathrm{SE}_{\mathrm{max}}$$\displaystyle\mathrm{SE}$
[bits/s/Hz]$\displaystyle\mathrm{EE}$ [kbits/Joule]$\displaystyle\sigma^{2}=5$
W, $\displaystyle B=20$ MHz,
$\displaystyle\alpha_{\mathrm{rol}}=0.1$UniformOptimalRush-to-sleep

(b)

Figure 4: Energy efficiency versus spectral efficiency for optimal/uniform
allocation with constant sleep mode $\mathbf{(As4)}$.

######  Remark 14 (Scaling of EE-SE).

Fig. 4a and 4b plot the EE as a function of the SE for the optimal and uniform
allocation. Most gain in terms of EE (not absolute energy) is obtained at
medium SE. The plots are obtained by varying the rate $R$, all other
parameters being fixed.

######  Remark 15 (Optimal EE).

As shown in Fig. 4a, for a low normalized noise, the maximal EE is not
obtained at maximal SE while it is for a higher noise, as shown in Fig. 4b.

##  IV Optimal Allocation for Piecewise Constant Successive Sleep Modes

Fig. 3a and 3b have shown promising gains to reduce consumed power at low
load. However, they are still limited by the relatively high sleep power
consumption $P_{\mathrm{sleep}}$. Similarly, Fig. 4a and 4b can seem limited
as one would hope to obtain an EE that is approximately flat as a function of
the SE. The reason is the same: a too high $P_{\mathrm{sleep}}$.

To drastically reduce energy consumption at low-to-medium load, it is of
paramount importance to implement successive sleep modes and use a frame
duration long enough so that the system can enter these sleep modes. To find
the optimal allocation, the iterative algorithm of Prop. 2 can be used, which
requires to solve problem (11) at each iteration. This is an integer
programming problem which can have a significant complexity. If the problem is
relaxed by considering $N_{\mathrm{a}}$ continuous, it remains challenging to
solve as it implies the minimization of the concave function
$E_{\mathrm{sleep}}(t)$ (Prop. 1).

If the sleep power consumption is assumed piecewise constant, according to
$\mathbf{(As3)}$, a simple allocation can still be found. This sleep power
model was detailed in Section II-B and Fig. 1. For a given $N_{\mathrm{a}}$,
only sleep modes such that $T_{s}\leq(N-N_{\mathrm{a}})T$ can be entered. We
define $N_{\mathrm{a},s}^{+}=N-\lfloor T_{s}/T\rfloor$ so that sleep mode $s$
can be used if $N_{\mathrm{a}}\leq N_{\mathrm{a},s}^{+}$. Depending on the
target rate $R$, it might be unfeasible to use a given sleep mode because of
the maximal power constraint per time slot. A minimum of active time slots
$\frac{NR}{R_{\mathrm{max}}}$ is required to satisfy it. Otherwise, the power
per time slot would have to be higher than $P_{\mathrm{max}}$ to satisfy the
rate constraint. Mode $s$ is thus feasible only if
$\frac{NR}{R_{\mathrm{max}}}\leq N_{\mathrm{a},s}^{+}$. As a result, deepest
sleep modes are only possible for low values of $R$, which intuitively makes
sense. For a target rate $R$, we define the set of feasible sleep modes as
$\mathcal{S}_{R}=\left\\{s\ |\ s\in\\{0,\cdots,S-1\\}\text{ and
}\frac{NR}{R_{\mathrm{max}}}\leq N_{\mathrm{a},s}^{+}\right\\}$. Moreover, we
generalize the definition of $R_{\mathrm{a}}$ per sleep mode as the constant
$R_{\mathrm{a}}(s)$ that minimizes the convex problem

$\displaystyle R_{\mathrm{a}}(s)=\arg\min_{x\geq
0}\frac{P_{0}-P_{s}+\gamma\sigma^{2\alpha}\left(2^{x}-1\right)^{\alpha}}{x}.$

We also define $\tilde{R}_{s}=\min(R_{\mathrm{a}}(s),R_{\mathrm{max}})$.

######  Theorem 2.

Under $\mathbf{(As1)}-\mathbf{(As3)},\mathbf{(As5)}$, as
$N\rightarrow+\infty$, the solution of problem (7) is found by computing for
all feasible sleep modes $s\in\mathcal{S}_{R}$

$\displaystyle N_{\mathrm{a},s}$
$\displaystyle=\left[\min\left(NR/\tilde{R}_{s},N_{\mathrm{a},s}^{+}\right)\right],\
p_{n,s}=\begin{cases}(2^{\frac{RN}{N_{\mathrm{a},s}}}-1)\sigma^{2}&\text{if
}n=0,...,N_{\mathrm{a},s}-1\\\ 0&\text{otherwise}\end{cases}$ $\displaystyle
P_{\mathrm{cons},s}$
$\displaystyle=\frac{N_{\mathrm{a},s}}{N}\left(P_{0}+\gamma\sigma^{2\alpha}\left(2^{\frac{RN}{N_{\mathrm{a},s}}}-1\right)^{\alpha}\right)+\frac{E_{\mathrm{sleep}}((N-N_{\mathrm{a,s}})T)}{NT}+O({1}/{N})$

and choosing among these modes the one that has the minimal
$P_{\mathrm{cons},s}$.

###### Proof.

See Appendix VII-E. ∎

$0$$2$$4$$6$$8$$10$$0$$50$$100$$150$$200$$R_{\mathrm{max}}$$P_{\mathrm{cons,max}}$$\displaystyle
R$ [bits/channel use]$\displaystyle P_{\mathrm{cons}}$
[W]$\displaystyle\sigma^{2}=10$ mW, $\displaystyle NT=200$ msUniformAsympt.
opt.Opt.Rush-to-sleep

(a)

$0$$0.5$$1$$1.5$$2$$2.5$$0$$50$$100$$150$$200$$R_{\mathrm{max}}$$P_{\mathrm{cons,max}}$$\displaystyle
R$ [bits/channel use]$\displaystyle P_{\mathrm{cons}}$
[W]$\displaystyle\sigma^{2}=5$ W, $\displaystyle NT=200$ msUniformAsympt.
opt.Opt.Rush-to-sleep

(b)

Figure 5: Power consumption versus load with successive sleep power model
$\mathbf{(As3)}$.

######  Remark 16.

Fig. 5a and Fig. 5b are plotted based on the same simulation parameters as
Fig. 3a and 3b, but with a different sleep model. Using successive power modes
has a drastic impact on the consumed energy at low load. A key parameter to
allow drastic savings is to have a large enough frame duration so that deepest
sleep modes can be used. For these figures, it was fixed to 200 ms. Hence,
deep sleep power mode can be entered but not hibernating sleep power mode.

$0$$2$$4$$6$$8$$0$$200$$400$$600$$800$$1{,}000$$\displaystyle\mathrm{SE}$
[bits/s/Hz]$\displaystyle\mathrm{EE}$
[kbits/Joule]$\displaystyle\sigma^{2}=10$ mW, $\displaystyle B=20$ MHz,
$\displaystyle\alpha_{\mathrm{rol}}=0.1$, $\displaystyle NT=200$
msUniformOptimalRush-to-sleep

(a)

$0$$0.5$$1$$1.5$$2$$0$$50$$100$$150$$200$$\displaystyle\mathrm{SE}$
[bits/s/Hz]$\displaystyle\mathrm{EE}$ [kbits/Joule]$\displaystyle\sigma^{2}=5$
W, $\displaystyle B=20$ MHz, $\displaystyle\alpha_{\mathrm{rol}}=0.1$,
$\displaystyle NT=200$ msUniformOptimalRush-to-sleep

(b)

Figure 6: Energy efficiency versus spectral efficiency with successive sleep
power model $\mathbf{(As3)}$.

######  Remark 17.

Similarly, Fig. 6a and 6b can be compared to Fig. 4a and 4b, where only the
sleep model differs. As ideally expected, the EE quickly reaches a plateau. To
still improve this behaviour, a longer frame duration can be used.

##  V Optimal Allocation for TDMA System

We now extend previous results by considering a downlink transmission from the
BS to $K$ users. We consider the constant power sleep model so that
$(\textbf{As1})-(\textbf{As4})$ hold. The users are multiplexed using TDMA. In
the frame of $N$ symbols, each symbol is allocated to the transmission towards
at most one user so that no inter-user interference is present. Out of the $N$
symbols, $N_{k}$ symbols are allocated to user $k$, for $k=0,...,K-1$. The
power associated to the $n$-th symbol transmitted to user $k$ is denoted by
$p_{k,n}$, where $n=0,...,N_{k}-1$ and $k=0,...,K-1$. The normalized noise
variance $\sigma_{k}^{2}=\sigma_{n}^{2}L_{k}$ is considered different at each
user, as each can have a specific path loss $L_{k}$. The consumed power is
then

$\displaystyle P_{\mathrm{cons}}^{\mathrm{TDMA}}$
$\displaystyle=\frac{N_{\mathrm{a}}}{N}P_{0}+\frac{\gamma}{N}\sum_{k=0}^{K-1}\sum_{n=0}^{N_{k-1}}p_{k,n}^{\alpha}+\frac{N-N_{\mathrm{a}}}{N}P_{\mathrm{sleep}}$

where $N_{\mathrm{a}}=\sum_{k=0}^{K-1}N_{k}$ so that $0\leq N_{\mathrm{a}}\leq
N$. We consider the generalization of problem (7) of minimizing the power
consumption, under $(\textbf{As1})-(\textbf{As4})$ and per-user rate
constraints $R_{k}$. The problem can be formulated as

$\displaystyle\min_{\begin{subarray}{c}N_{k},p_{k,n}\\\ n=0,...,N_{k}-1\\\
k=0,...,K-1\end{subarray}}P_{\mathrm{cons}}^{\mathrm{TDMA}}\ \text{s.t. }$
$\displaystyle\frac{1}{N}\sum_{n=0}^{N_{k}-1}\log_{2}\left(1+\frac{p_{k,n}}{\sigma_{k}^{2}}\right)=R_{k}\
\forall k,\ \sum_{k=0}^{K-1}N_{k}\leq N.$ (13)

We assume in the following that the problem has a feasible solution, which
generalizes (As5).

(As6): Problem (13) is feasible. Defining the maximal per-user rate
$R_{k,\mathrm{max}}=\log_{2}\left(1+\frac{P_{\max}}{\sigma_{k}^{2}}\right)$,
it implies that

$\displaystyle\sum_{k=0}^{K-1}\lceil\frac{NR_{k}}{R_{k,\mathrm{max}}}\rceil\leq
N.$

Moreover, we define the constant $R_{k,\mathrm{a}}$ that minimizes the convex
problem

$\displaystyle R_{k,\mathrm{a}}$ $\displaystyle=\arg\min_{x\geq
0}\frac{P_{0}-P_{\mathrm{sleep}}+\gamma\sigma^{2\alpha}_{k}\left(2^{x}-1\right)^{\alpha}}{x}.$

We also define $\hat{R}_{k}=\min(R_{k,\mathrm{a}},R_{k,\mathrm{max}})$ and
$\hat{P}_{k}=\min(P_{k,\mathrm{a}},P_{\mathrm{max}})$. Two regimes can be
considered for Problem (13), depending if the constraint
$\sum_{k=0}^{K-1}N_{k}\leq N$ is binding or not.

In terms of power savings, the most promising case is the low-to-medium load
regime where the constraint is not binding. In that case, the problem fully
decouples per-user and the linear regime solution of Theor. 1 can directly be
used on a per-user basis.

######  Theorem 3 (TDMA - linear regime).

Under $\mathbf{(As1)}-\mathbf{(As4)},\mathbf{(As6)}$, as
$N\rightarrow+\infty$, if $\sum_{k=0}^{K-1}\frac{R_{k}}{\hat{R}_{k}}\leq
1+O(1/N)$, the allocation

$\displaystyle N_{k}$ $\displaystyle=\left[NR_{k}/\hat{R}_{k}\right],\
p_{k,n}=\hat{P}_{k}+O(1/N)\quad\text{for }n=0,...,N_{k}-1$

for user $k=0,...,K-1$ is an asymptotic solution of Problem (13) and

$\displaystyle P_{\mathrm{cons}}$
$\displaystyle=P_{\mathrm{sleep}}+\sum_{k=0}^{K-1}R_{k}\frac{P_{0}-P_{\mathrm{sleep}}+\gamma\hat{P}_{k}^{\alpha}}{\hat{R}_{k}}+O(1/N).$

###### Proof.

See Appendix VII-F. ∎

(a)  $\sigma_{0}^{2}=\sigma_{1}^{2}=10$ mW

(b)  $\sigma_{0}^{2}=\sigma_{1}^{2}=5$ W

Figure 7: Power consumption versus load in a $K=2$ user TDMA system using
optimal/uniform power allocation with constant sleep mode $\mathbf{(As4)}$.
The optimal allocation is valid in the asymptotic linear regime, i.e., when
$\frac{R_{0}}{\hat{R}_{0}}+\frac{R_{1}}{\hat{R}_{1}}\leq 1$.

######  Remark 18 ($K=2$ \- TDMA).

For $K=2$, Fig. 7a and 7b plot $P_{\mathrm{cons}}$ as a function of $R_{1}$
and $R_{2}$. The optimal solution of Theor. 3 is plotted where it is valid,
i.e., the asymptotic linear regime
$\frac{R_{0}}{\hat{R}_{0}}+\frac{R_{1}}{\hat{R}_{1}}\leq 1$. As already
observed in the single-user case (Fig. 3b), this regime can cover the whole
feasible rate region, as shown in Fig. 7b, characterized by a relatively
higher noise power implying that $\hat{R}_{k}=R_{k,\mathrm{max}}$, $\forall
k$. As a benchmark, a fully active uniform allocation was plotted where a
share $N_{k}/N=R_{k}/(R_{0}+R_{1})$ of time slots was allocated to each user.

######  Remark 19 (rush-to-sleep - TDMA).

If $\sum_{k=0}^{K-1}\frac{R_{k}}{\hat{R}_{k}}\leq 1$, a rush-to-sleep
allocation is asymptotically optimal for each user such that
$R_{k,\mathrm{max}}\leq R_{k,\mathrm{a}}$. On the other hand, if
$\sum_{k=0}^{K-1}\frac{R_{k}}{\hat{R}_{k}}>1$, it is optimal to use a fully
active system and no sleep.

If the constraint $\sum_{k=0}^{K-1}N_{k}=N$ is binding, Problem (13) is
coupled between users and challenging to solve. As the sleep duration is zero,
the power consumption becomes

$\displaystyle P_{\mathrm{cons}}^{\mathrm{TDMA}}$
$\displaystyle=P_{0}+\frac{\gamma}{N}\sum_{k=0}^{K-1}\sum_{n=0}^{N_{k-1}}p_{k,n}^{\alpha}.$

As this regime does not allow switching-off components, relatively small
energy reduction potentials are expected. Given space constraints, we do not
provide a more detailed solution. One possibility is to reduce the target
rates of the users to make the constraint non-binding and then use Theor. 3.
Another possibility is to use conventional scheduling policies, that are not
aware of sleep capabilities, which makes sense as the system is fully active.

##  VI Conclusion

In this work, we have proposed a fundamental study of time-domain energy-
saving techniques in radio access. The results provide key novel insights from
an information-theoretic perspective. Considering equal gain parallel
communication channels, conventional information-theoretic results state that
all channels (time slots in this study) should be equally used to minimize
transmit (not consumed) power under a rate constraint. On the contrary,
popular energy-saving techniques steer towards an extreme opposite “rush-to-
sleep” approach: compact transmission in as few time slots as possible, at
maximal transmit power, to maximize sleep duration.

Using a realistic power consumption model, our information-theoretic study
bridges the gap between these two extremes. Simple allocations are provided
that allow drastic energy savings reaching factors of 10 at low load. At low-
to-medium load, the optimal number of active time slots is linearly
proportional to the rate, resulting in a power consumption which linearly
scales with the rate. At a higher load, all time slots become allocated. The
rush-to-sleep approach is shown to be optimal in a high-noise regime but not
otherwise. In a low-noise regime, it might be better to use a fully active
system. Moreover, the fundamental trade-off between EE and SE is revisited
leveraging the time-domain hardware sleep capabilities. Transmitting at
maximal SE maximizes the SE in a high-noise regime while, in a low-noise
regime, a reduced SE maximizes the EE. Considering a sleep model with
increasing depth complicates the study but also greatly increases the energy-
saving gains. For a piecewise constant model, simple allocations can still be
found. Finally, for a multi-user TDMA system, single-user results are
applicable on a per-user basis in the low-to-medium load regime, where the
system should not be fully active and where sleep-aware energy-saving gains
can be achieved.

##  VII Appendix

We start by introducing two lemmas that will be useful in the following.

######  Lemma 3.

An optimal solution of the following problem

$\displaystyle\min_{p_{0},...,p_{N-1}}\frac{\gamma}{N}\sum_{n=0}^{N-1}p_{n}^{\alpha}\quad\text{s.t.
}\frac{1}{N}\sum_{n=0}^{N-1}\log_{2}\left(1+\frac{p_{n}}{\sigma^{2}}\right)=R$
(14)

must have a uniform allocation among active time slots: $\forall
n,n^{\prime}$, if $p_{n}>0$, $p_{n^{\prime}}>0$ then $p_{n}=p_{n^{\prime}}$.

###### Proof.

The case $\alpha=1$ was already treated in the introduction.For $\alpha=0$,
the cost function only depends on the number of active time slots so that a
single time slot must be allocated power. In the following, we will use a
proof by contradiction for the case $0<\alpha<1$. Any non-uniform power
allocation has at least two time slots, say $n=0$ and $n=1$ (potentially using
a re-indexing), that are such that $p_{0}>0$, $p_{1}>0$ and $p_{0}\neq p_{1}$.
We consider the power allocated to the other time slots as fixed and optimize
the cost function with respect to $p_{0}$ and $p_{1}$ only. For the sake of
clarity, we define $\rho_{n}=p_{n}/\sigma^{2}$. The reduced problem is

$\displaystyle\min_{\rho_{0},\rho_{1}}\tilde{f}(\rho_{0},\rho_{1})=\sum_{n=0}^{1}\rho_{n}^{\alpha}\
\text{s.t. }\sum_{n=0}^{1}\log\left(1+\rho_{n}\right)=Z$

where $Z=NR\log 2-\sum_{n=2}^{N-1}\log\left(1+\rho_{n}\right)$. The constraint
implies that

$\displaystyle\rho_{1}$ $\displaystyle=\frac{e^{Z}}{1+{\rho_{0}}}-1$ (15)

so that $\rho_{0}$ and $\rho_{1}$ take values only in the domain $[0,e^{Z}-1]$
and have a one-to-one relationship. As $\rho_{0}\rightarrow 0$,
$\rho_{1}\rightarrow e^{Z}-1$ and vice versa. Moreover, the problem is
symmetrical so that
$\tilde{f}(\rho_{0},\rho_{1})=\tilde{f}(\rho_{1},\rho_{0})$. Using (15), the
problem can be rewritten as a monovariable unconstrained problem

$\displaystyle\min_{\rho_{0}}\tilde{f}(\rho_{0})=\rho_{0}^{\alpha}+\rho_{1}^{\alpha}=\rho_{0}^{\alpha}+\left(\frac{e^{Z}}{1+{\rho_{0}}}-1\right)^{\alpha}.$

The derivative of $\tilde{f}(\rho_{0})$ and its limit at the bounds of its
domain are given by

$\displaystyle\tilde{f}^{\prime}(\rho_{0})$
$\displaystyle=\frac{\alpha}{\rho_{0}^{1-\alpha}}-\frac{e^{Z}}{(1+\rho_{0})^{2}}\frac{\alpha}{\rho_{1}^{1-\alpha}},\
\lim_{\rho_{0}\rightarrow 0}\tilde{f}^{\prime}(\rho_{0})=+\infty,\
\lim_{\rho_{0}\rightarrow e^{Z}-1}\tilde{f}^{\prime}(\rho_{0})=-\infty.$ (16)

To find the critical points, we set $f^{\prime}(\rho_{0})=0$ and combining
with (15), we find the condition

$\displaystyle 0$
$\displaystyle=\rho_{1}^{1-\alpha}(1+\rho_{0})-\rho_{0}^{1-\alpha}(1+\rho_{1})$
(17)

which shows that there is always one critical point in
$\rho_{0}=\rho_{1}=e^{Z/2}-1$. Moreover, the fact that the problem is
symmetric implies an odd number of critical points: if there is a critical
point in $\tilde{\rho}_{0}$, there is one in
$\frac{e^{Z}}{1+{\tilde{\rho}_{0}}}-1$. Using again (15), we can rewrite the
condition (17) as

$\displaystyle 0$
$\displaystyle=(e^{Z}-1-\rho_{0})^{1-\alpha}(1+\rho_{0})^{1+\alpha}-\rho_{0}^{1-\alpha}e^{Z}.$

The point $\rho_{0}=0$ is not a critical point. Hence, we can restrict to
$\rho_{0}>0$ and divide by $\rho_{0}^{1-\alpha}$

$\displaystyle 0$
$\displaystyle=\underbrace{(e^{Z}-1-\rho_{0})^{1-\alpha}(1+\rho_{0})^{1+\alpha}\rho_{0}^{\alpha-1}-e^{Z}}_{\tilde{g}(\rho_{0})}.$

The function $\tilde{g}(\rho_{0})$ is infinite in $\rho_{0}=0$ and $-e^{Z}$
for $\rho_{0}=e^{Z}-1$. The number of roots of $\tilde{g}(\rho_{0})$ and thus
critical points of $\tilde{f}(\rho_{0})$ is at most equal to one plus the
number of critical points/alternations of $\tilde{g}(\rho_{0})$. Setting
$\tilde{g}^{\prime}(\rho_{0})=0$ gives the condition

$\displaystyle
0=-(1-\alpha)(1+\rho_{0})\rho_{0}+(e^{Z}-1-\rho_{0})(1+\alpha)\rho_{0}+(e^{Z}-1-\rho_{0})(1+\rho_{0})(\alpha-1)$

which is a quadratic equation in $\rho_{0}$. It has thus at most two
solutions. As a result, $\tilde{g}(\rho_{0})$ has max two alternations and
$\tilde{f}(\rho_{0})$ has at most 3 critical points. The second order
derivative of $\tilde{f}(\rho_{0})$ and its limit at the bounds of its domain
are given by

$\displaystyle\tilde{f}^{\prime\prime}(\rho_{0})$
$\displaystyle=-\alpha(1-\alpha)\frac{1}{\rho_{0}^{2-\alpha}}+\frac{1}{\rho_{1}^{2-\alpha}}\frac{\alpha
e^{Z}}{(1+\rho_{0})^{3}}\left(-(1-\alpha)e^{Z}\frac{1}{1+\rho_{0}}+2\rho_{1}\right)$
$\displaystyle\lim_{\rho_{0}\rightarrow 0}\tilde{f}^{\prime\prime}(\rho_{0})$
$\displaystyle=-\infty,\ \lim_{\rho_{0}\rightarrow
e^{Z}-1}\tilde{f}^{\prime\prime}(\rho_{0})=-\infty.$ (18)

Domain | $[0,e^{Z}-1]$
---|---
Intercept | $\tilde{f}(0)=(e^{Z}-1)^{\alpha}$
Symmetry | $\tilde{f}(\rho_{0})=\tilde{f}\left(\frac{e^{Z}}{1+{\rho_{0}}}-1\right)$
Derivatives |  $\tilde{f}^{\prime}(0)=+\infty$, $\tilde{f}^{\prime}(e^{Z}-1)=-\infty$
Concavity | $\tilde{f}^{\prime\prime}(0)=\tilde{f}^{\prime\prime}(e^{Z}-1)=-\infty$
Critical points | Always 1 in $\rho_{0}=\rho_{1}=e^{Z/2}-1$
| Potentially 2 others symmetrical

Figure 8: Sketch of function $\tilde{f}(\rho_{0})$.

As shown in Fig. 8, three cases can be distinguished. In case (a), there is a
single critical point in $\rho_{0}=\rho_{1}=e^{Z/2}-1$ which is a maximum. In
cases (b) and (c), there are three critical points: the middle one in
$\rho_{0}=\rho_{1}=e^{Z/2}-1$ will now be a minimum (local in (b), global in
(c)) while the two on its sides are maxima. Hence, global minima can only be
obtained for either $\rho_{0}=0,\rho_{1}=e^{Z}-1$ or
$\rho_{1}=0,\rho_{0}=e^{Z}-1$ or $\rho_{0}=\rho_{1}=e^{Z/2}-1$. Hence, it is
impossible to find an optimal allocation such that $p_{0}>0$, $p_{1}>0$ and
$p_{0}\neq p_{1}$. ∎

######  Lemma 4.

Under $\mathbf{(As1)}-\mathbf{(As4)}$, the following function is convex for
$x>0$

$\displaystyle f(x)$
$\displaystyle=\frac{P_{0}-P_{\mathrm{sleep}}+\gamma\sigma^{2\alpha}\left(2^{x}-1\right)^{\alpha}}{x}.$

###### Proof.

Under $\mathbf{\mathbf{(As4)}}$, $P_{0}-P_{\mathrm{sleep}}\geq 0$ and thus
$(P_{0}-P_{\mathrm{sleep}})/x$ is convex. Given that the sum of two convex
functions is convex, it is sufficient to show that

$\displaystyle\frac{\gamma\sigma^{2\alpha}\left(2^{x}-1\right)^{\alpha}}{x}\text{
or equivalently }g(y)=\frac{\left(e^{y}-1\right)^{\alpha}}{y}$

is convex for $y>0$ (using $y=x\log 2$). Its second derivative is

$g^{\prime\prime}(y)=\frac{(e^{y}-1)^{\alpha-2}(e^{2y}(\alpha^{2}y^{2}-2\alpha
y+2)-e^{y}(\alpha y^{2}-2\alpha y+4)+2)}{y^{3}}.$

Given that $y>0$, we have directly that $y^{3}>0$, $(e^{y}-1)^{\alpha-2}\geq
0$ and it is sufficient to show that

$h(y)=e^{2y}(\alpha^{2}y^{2}-2\alpha y+2)-e^{y}(\alpha y^{2}-2\alpha
y+4)+2\geq 0.$

Using the Taylor series expansion
$e^{y}=\sum_{r=0}^{+\infty}\frac{y^{r}}{r!}$, which converges for all $y$, we
find

$\displaystyle h(y)$
$\displaystyle=\sum_{r=0}^{+\infty}\frac{(2x)^{r}}{r!}(\alpha^{2}y^{2}-2\alpha
y+2)-\sum_{r=0}^{+\infty}\frac{y^{r}}{r!}(\alpha y^{2}-2\alpha y+4)+2$
$\displaystyle=\sum_{r=2}^{+\infty}y^{r}\left(\alpha\frac{\alpha
2^{r-2}-1}{(r-2)!}+\alpha\frac{2-2^{r}}{(r-1)!}+\frac{2^{r+1}-4}{r!}\right).$

To show that $h(y)\geq 0$ for $y>0$ and $\alpha\in[0,1]$, it is sufficient to
show that for all $r\geq 2$

$\displaystyle\alpha\frac{2^{r-2}\alpha-1}{(r-2)!}+\alpha\frac{2-2^{r}}{(r-1)!}+\frac{2^{r+1}-4}{r!}\geq
0$ $\displaystyle\leftrightarrow\alpha^{2}2^{r-2}r(r-1)+\alpha
r(3-2^{r}-r)+2^{r+1}-4$ $\displaystyle\geq 0.$

For $r=2$, this is verified as $2\alpha^{2}-6\alpha+4=2(\alpha-1)(\alpha-2)$
is always positive for $\alpha\in[0,1]$. For $r=3$, this is also verified as
$\alpha^{2}12-\alpha 24+12=12(\alpha-1)^{2}$ is again positive for
$\alpha\in[0,1]$. For $r\geq 4$, we have $r(r-1)\geq r^{2}/2$,
$3-2^{r}-r\geq-2^{r+1}$ and $-2^{-r+4}\geq-1$ so that

$\displaystyle\alpha^{2}2^{r-2}r(r-1)+\alpha r(3-2^{r}-r)+2^{r+1}-4$
$\displaystyle\geq 2^{r-2}(\alpha^{2}r^{2}/2-\alpha r8+8-2^{-r+4})$
$\displaystyle\geq 2^{r-2}(\alpha^{2}r^{2}/2-\alpha r8+7)$

which roots are in $8r\pm\sqrt{50}r$. Given that $8r-\sqrt{50}r\geq 0.92r>1$
for $r\geq 2$, both roots are strictly larger than 1 and the term is positive
for $\alpha\in[0,1]$, which concludes the proof. ∎

###  VII-A Proof of Lemmas 1 and 2

One can first note that Lemma 1 is a particularization of Lemma 2 when
$P_{0}=E_{\mathrm{sleep}}(t)=0$. Hence, the result will be found as a specific
case in the following. Under $\mathbf{(As1)}$-$\mathbf{(As2)}$, Problem (7)
can be rewritten as

$\displaystyle\min_{N_{\mathrm{a}}}\frac{N_{\mathrm{a}}}{N}P_{0}+\left[\min_{p_{0},...,p_{N_{\mathrm{a}}-1}}\frac{\gamma}{N}\sum_{n=0}^{N_{\mathrm{a}-1}}p_{n}^{\alpha}\right]+\frac{E_{\mathrm{sleep}}((N-N_{\mathrm{a}})T)}{NT}$
(19)

which shows that only the second term depends on the power allocation, i.e.,
the term defined as $P_{\mathrm{ld}}$. From Lemma 3, an optimal allocation
needs to be uniform in the number of activated time slots. For a given
$N_{\mathrm{a}}$, the rate constraint fixes the transmit power per active time
slot

$\displaystyle p_{n}$
$\displaystyle=\left(2^{R\frac{N}{N_{\mathrm{a}}}}-1\right)\sigma^{2}\text{ if
}n=0,...,N_{\mathrm{a}}-1.$

Hence, the problem can be reformulated as finding the optimal number of active
slots $N_{\mathrm{a}}$ that minimizes the consumed power, i.e., Problem (11)
in Lemma 2. Moreover, under $\mathbf{(As3)}-\mathbf{(As4)}$, the problem
becomes

$\displaystyle\min_{N_{\mathrm{a}}}P_{\mathrm{sleep}}+\frac{N_{\mathrm{a}}}{N}\left(P_{0}-P_{\mathrm{sleep}}+\gamma\sigma^{2\alpha}\left(2^{\frac{NR}{N_{\mathrm{a}}}}-1\right)^{\alpha}\right).$

We define $x=\frac{RN}{N_{\mathrm{a}}}$ and relax the problem by considering
$x$ as continuous

$\displaystyle\min_{x}f(x)$
$\displaystyle=\frac{P_{0}-P_{\mathrm{sleep}}+\gamma\sigma^{2\alpha}\left(2^{x}-1\right)^{\alpha}}{x}.$

From Lemma 4, we know that $f(x)$ is convex. Given the definition of $x$ and
the integer nature of $N_{\mathrm{a}}$, $x$ can only take discrete values in
practice. Given the fact that $f(x)$ is convex, it is guaranteed that one of
the neighboring possible values of $R_{\mathrm{a}}=\arg\min f(x)$ is optimal.
As a result, the solution is given by either
$N_{\mathrm{a}}=\lceil\frac{RN}{R_{\mathrm{a}}}\rceil$,
$N_{\mathrm{a}}=\lfloor\frac{RN}{R_{\mathrm{a}}}\rfloor$ or $N$ if
$\frac{RN}{R_{\mathrm{a}}}>N$. Using the ceil-floor notation concludes the
proof of Lemma 2. The above result can also be particularized to the problem
of Lemma 1 by setting $P_{0}=E_{\mathrm{sleep}}(t)=0$ and the problem
simplifies to

$\displaystyle\min_{x}f(x)$
$\displaystyle=\frac{\left(2^{x}-1\right)^{\alpha}}{x}\leftrightarrow\min_{y}g(y)=\frac{\left(e^{y}-1\right)^{\alpha}}{y}.$

where $y=x\log 2$. Setting its derivative to zero, we find

$\displaystyle\alpha e^{y}y$ $\displaystyle=e^{y}-1\leftrightarrow
y=W\left(-\alpha^{-1}e^{-\alpha^{-1}}\right)+\alpha^{-1}\leftrightarrow
R_{\mathrm{a}}=W\left(-\alpha^{-1}e^{-\alpha^{-1}}\right)+\alpha^{-1}/\log 2,$

which concludes the proof of Lemma 1.

###  VII-B Proof of Proposition 2

The algorithm is initialized by the solution of the relaxed problem assuming
that no max power constraints are binding. The solution is then given by
Theor. 2. If the solution is such that $p_{n}\leq P_{\mathrm{max}},\forall n$,
the problem is solved. On the other hand, if, for at least one time slot
$p_{n}>P_{\mathrm{max}}$, at least one of the constraints must be binding.
Hence, the algorithm allocates the maximal power $P_{\mathrm{max}}$ to one
additional time slot. The rate constraint on the remaining time slots is then
adapted. The power allocation is re-computed assuming that no max power
constraint is binding on the remaining time slots not yet set to
$P_{\mathrm{max}}$. Again, the solution is given by Theor. 2. Again, the
algorithm checks if the max constraint is verified. If yes, the algorithm has
converged. If not, it enters a novel iteration and so on until convergence.

###  VII-C Proof of Theorem 1

Let us consider one by one four different cases. On the one hand, the
exponential regime mentioned in the theorem where $R>\tilde{R}$ and i)
$\tilde{R}=R_{\mathrm{max}}$ or ii) $\tilde{R}=R_{\mathrm{a}}$. On the other
hand, the linear regime mentioned in the theorem where $R\leq\tilde{R}$ and
iii) $\tilde{R}=R_{\mathrm{a}}$ or iv) $\tilde{R}=R_{\mathrm{max}}$.

Case i): This case is not applicable according to $\mathbf{(As5)}$ as it is
unfeasible to have $R>R_{\mathrm{max}}$.

Case ii): The case together with $\mathbf{(As5)}$ implies
$R_{\mathrm{max}}\leq R>R_{\mathrm{a}}$. From Lemma 2, we can find that
$N_{\mathrm{a}}=N$ and the corresponding power allocation, which is well
feasible as it does not violate the $P_{\mathrm{max}}$ constraint. The
exponential regime result of Theor. 1 is then found.

Case iii): This case implies $R\leq R_{\mathrm{a}}\leq R_{\mathrm{max}}$. Let
us first consider that the maximal power constraint per time slot is not
active such that we can use the result of Lemma 2. If $R\leq R_{\mathrm{a}}$,
the optimal number and ratio of active time slots are

$\displaystyle N_{\mathrm{a}}$
$\displaystyle=\lfloor{RN}/{R_{\mathrm{a}}}\rceil=\left[\frac{RN}{R_{\mathrm{a}}}\right]+\epsilon_{1}=\frac{RN}{R_{\mathrm{a}}}+\epsilon_{2}$
$\displaystyle\frac{N_{\mathrm{a}}}{N}$
$\displaystyle=\left[\frac{RN}{R_{\mathrm{a}}}\right]/N+\frac{\epsilon_{1}}{N}=\frac{R}{R_{\mathrm{a}}}+\frac{\epsilon_{2}}{N}.$

where $|\epsilon_{1}|<1$ and $|\epsilon_{2}|<1$. The optimal ratio
${N_{\mathrm{a}}}/{N}$ asymptotically converges to $R/R_{\mathrm{a}}$ and the
same occurs if the optimal number of time slots $N_{\mathrm{a}}$ is
approximated using a rounding operator instead of the ceil-floor operator.
Using this result, as $N\rightarrow+\infty$, the optimal power allocation per
active time slot of Lemma 2 can be rewritten as

$\displaystyle p_{n}$
$\displaystyle=\left(2^{R\frac{N}{N_{\mathrm{a}}}}-1\right)\sigma^{2}=P_{\mathrm{a}}+O(1/N)$
(20)

where $P_{\mathrm{a}}=\left(2^{R_{\mathrm{a}}}-1\right)\sigma^{2}$ and
$N_{\mathrm{a}}$ can be the ideal value $\lfloor{RN}/{R_{\mathrm{a}}}\rceil$
or its approximation using the rounding operator. Given that
$R_{\mathrm{a}}\leq R_{\mathrm{max}}$, this allocation does not violate the
$P_{\mathrm{max}}$ constraint and the result is feasible. The power
consumption becomes

$\displaystyle P_{\mathrm{cons}}$
$\displaystyle=P_{\mathrm{sleep}}+\frac{N_{\mathrm{a}}}{N}\left(P_{0}-P_{\mathrm{sleep}}+\gamma
P_{\mathrm{a}}^{\alpha}+O(1/N)\right)=P_{\mathrm{sleep}}+R\frac{P_{0}-P_{\mathrm{sleep}}+\gamma
P_{\mathrm{a}}^{\alpha}}{R_{\mathrm{a}}}+O(1/N).$

Case iv): This case implies $R\leq R_{\mathrm{max}}\leq R_{\mathrm{a}}$ and
thus $P_{\mathrm{a}}>P_{\mathrm{max}}$ such that allocation (20) is not
feasible. As an alternative, the iterative Algorithm 1 can be used and
simplified in the asymptotic regime. Indeed, as $N\rightarrow+\infty$, at each
iteration, the algorithm allocates a constant power $P_{\mathrm{a}}$
(independent of $R$) to active time slots, not yet set to $P_{\mathrm{max}}$.
At the convergence of the algorithm, the allocation will have approximately
${RN}/{R_{\mathrm{max}}}$ active time slots with maximal power
$P_{\mathrm{max}}$ and rate $R_{\mathrm{max}}$. As a result, as
$N\rightarrow+\infty$, at the optimum,
$N_{\mathrm{a}}/N=\left[{R}/{R_{\mathrm{max}}}\right]+O(1/N)={{R}/{R_{\mathrm{max}}}}+O(1/N)$
and the allocation

$\displaystyle N_{\mathrm{a}}$
$\displaystyle=\left[\frac{RN}{R_{\mathrm{max}}}\right],\
p_{n}=P_{\mathrm{max}}\text{ for $n=0,...,N_{\mathrm{max}}$ }$

is asymptotically optimal and achieves a consumed power

$\displaystyle P_{\mathrm{cons}}$
$\displaystyle=P_{\mathrm{sleep}}+R\frac{P_{0}-P_{\mathrm{sleep}}+\gamma
P_{\mathrm{a}}^{\alpha}}{R_{\mathrm{max}}}+O(1/N).$

Cases iii) and iv) can be written more compactly using the definitions of
$\tilde{R}$ and $\tilde{P}$, giving the linear regime result of Theor. 1.

###  VII-D Proof of Corollary 2

The results of Corol. 2 can be found by minimizing the EE expression in the
two regimes of Corol. 1. In the regime where
$\mathrm{SE}\leq\tilde{R}/(1+\alpha_{\mathrm{rol}})$, it is clear that the EE
is maximized for the largest SE, i.e., when
$\mathrm{SE}=\tilde{R}/(1+\alpha_{\mathrm{rol}})$. Moreover, if
$R_{\mathrm{a}}\geq R_{\mathrm{max}}$, we have $\tilde{R}=R_{\mathrm{max}}$
and the second regime of Corol. 1 is not feasible. The optimal SE corresponds
to the maximal SE,
$\mathrm{SE}_{\mathrm{max}}=R_{\mathrm{max}}/(1+\alpha_{\mathrm{rol}})$. On
the other hand, if $R_{\mathrm{a}}<R_{\mathrm{max}}$,
$\tilde{R}=R_{\mathrm{a}}$ and the regime
$\mathrm{SE}>\tilde{R}/(1+\alpha_{\mathrm{rol}})$ can be entered. The
optimization over $\bar{R}$ then provides the optimum and can only improve the
optimum as $\bar{R}$ is allowed to take value $R_{\mathrm{a}}$.

###  VII-E Proof of Theorem 2

It is direct to see that one should choose the optimal allocation among
feasible sleep modes. Under $\mathbf{(As3)}$, the sleep energy consumption at
time $t$ if sleep mode $s$ is used is
$E_{\mathrm{sleep},s}(t)=E_{\mathrm{sleep}}(T_{s})+(t-T_{s})P_{s}$ where
$E_{\mathrm{sleep}}(T_{s})=\sum_{s^{\prime}=0}^{s-1}P_{s^{\prime}}(T_{s^{\prime}+1}-T_{s^{\prime}})$.222No
deeper sleep mode than $s$ is considered even if $T_{s+1}<t$, which could
decrease sleep energy consumption. Still, this does not affect the
optimization result a deeper sleep mode will perform better and will be chosen
instead. The consumed power using sleep mode $s$ can then be written as

$\displaystyle P_{\mathrm{cons},s}$
$\displaystyle=\frac{\tilde{E}_{s}}{NT}+\frac{N_{\mathrm{a}}}{N}P_{0}+\frac{\gamma}{N}\sum_{n=0}^{N_{\mathrm{a}-1}}p_{n}^{\alpha}+\frac{N-N_{\mathrm{a}}}{N}P_{s}$

where $\tilde{E}_{s}=E_{\mathrm{sleep}}(T_{s})-T_{s}P_{s}$. This form is
similar to the one given in (6), under $\mathbf{(As4)}$. The sole differences
are the presence of $P_{s}$ instead of $P_{\mathrm{sleep}}$ and the constant
$\tilde{E}_{s}/(NT)$, which affects the cost function but does not impact the
optimization. The result of Theorem 1 can then be used: uniform allocation
among $N_{\mathrm{a},s}$ active mode is optimal. The only difference is the
fact that the maximal value of $N_{\mathrm{a},s}$ is $N_{\mathrm{a},s}^{+}$
instead of $N$, so that the sleep duration is sufficient to enter mode $s$. As
a result, we find
$N_{\mathrm{a},s}=\left[\min\left(NR/\tilde{R}_{s},N_{\mathrm{a},s}^{+}\right)\right]$.

###  VII-F Proof of Theorem 3

If the constraint $\sum_{k=0}^{K-1}N_{k}\leq N$ is not binding, Problem (13)
is fully decoupled between users and can be solved by solving for
$k=0,...,K-1$ an independent per-user problem

$\displaystyle\min_{\begin{subarray}{c}N_{k},p_{k,n}\\\
n=0,...,N_{k}-1\end{subarray}}$
$\displaystyle\frac{N_{k}}{N}P_{0}+\frac{\gamma}{N}\sum_{n=0}^{N_{k-1}}p_{k,n}^{\alpha}+\frac{N-N_{k}}{N}P_{\mathrm{sleep}}\text{
s.t.
}\frac{1}{N}\sum_{n=0}^{N_{k}-1}\log_{2}\left(1+\frac{p_{k,n}}{\sigma_{k}^{2}}\right)=R_{k}$

so that the asymptotic solution of Theorem 1 can be used giving
$N_{k}=\left[NR_{k}/\hat{R}_{k}\right]$. If the constraint
$\sum_{k=0}^{K-1}N_{k}\leq N$ is not violated, this is the asymptotic solution
of Problem (13). From Section VII-C, we know that, as $N\rightarrow+\infty$,
$N_{k}/N=R_{k}/\hat{R}_{k}+O(1/N)$ and the constraint can thus be equivalently
written as $\sum_{k=0}^{K-1}R_{k}/\hat{R}_{k}\leq 1+O(1/N)$.

## Acknowledgment

The author would like to thank his colleagues from the DRAMCO-KU Leuven lab
and Dr. Pål Frenger for many fruitful discussions and valuable comments.

## References

  * [1] J. Malmodin, “The power consumption of mobile and fixed network data services - The case of streaming video and downloading large files,” _Electronics Goes Green 2020+_ , p. 10, 2020.
  * [2] E. Ekudden, “Breaking the energy curve,” Ericsson, Tech. Rep., Mar. 2020.
  * [3] C. Andersson, J. Bengtsson, G. Byström, P. Frenger, Y. Jading, and M. Nordenström, “Improving energy performance in 5G networks and beyond,” _Ericsson Technology Review_ , no. 8, pp. 2–11, August 2022.
  * [4] M. Gruber, O. Blume _et al._ , “EARTH — Energy Aware Radio and Network Technologies,” in _2009 IEEE 20th International Symposium on Personal, Indoor and Mobile Radio Communications_ , Sep. 2009, pp. 1–5.
  * [5] 3GPP, “3rd Generation Partnership Project; Technical Specification Group Radio Access Network; Study on network energy savings for NR (Release 18),” 3rd Generation Partnership Project (3GPP), Tech. Rep. 38.864, Dec. 2022, version 18.0.0.
  * [6] “Green 5G White Paper,” Huawei, Tech. Rep., Oct. 2021.
  * [7] E. Dahlman, S. Parkvall, and J. Sköld, _5G NR: The next generation wireless access technology_. Academic Press, 2020.
  * [8] D. López-Pérez, A. De Domenico _et al._ , “A Survey on 5G Radio Access Network Energy Efficiency: Massive MIMO, Lean Carrier Design, Sleep Modes, and Machine Learning,” _IEEE Comm. Surv. & Tut._, vol. 24, no. 1, pp. 653–697, 2022.
  * [9] S. Zhang, Q. Wu, S. Xu, and G. Y. Li, “Fundamental Green Tradeoffs: Progresses, Challenges, and Impacts on 5G Networks,” _IEEE Communications Surveys & Tutorials_, vol. 19, no. 1, pp. 33–56, 2017.
  * [10] S. Tombaz, P. Frenger, F. Athley, E. Semaan, C. Tidestav, and A. Furuskar, “Energy Performance of 5G-NX Wireless Access Utilizing Massive Beamforming and an Ultra-Lean System Design,” in _2015 IEEE Global Communications Conference (GLOBECOM)_ , Dec. 2015, pp. 1–7.
  * [11] P. Lahdekorpi, M. Hronec, P. Jolma, and J. Moilanen, “Energy efficiency of 5G mobile networks with base station sleep modes,” in _2017 IEEE Conference on Standards for Communications and Networking_ , Helsinki, Sep. 2017, pp. 163–168.
  * [12] M. Matalatala, M. Deruyck _et al._ , “Simulations of beamforming performance and energy efficiency for 5G mm-wave cellular networks,” in _2018 IEEE Wireless Communications and Networking Conference (WCNC)_ , Apr. 2018, pp. 1–6.
  * [13] P. Frenger and K. W. Helmersson, “Energy Efficient 5G NR Street-Macro Deployment in a Dense Urban Scenario,” in _2019 IEEE Global Communications Conference (GLOBECOM)_ , Dec. 2019, pp. 1–6.
  * [14] P. Frenger and R. Tano, “More Capacity and Less Power: How 5G NR Can Reduce Network Energy Consumption,” in _2019 IEEE 89th Vehicular Technology Conference (VTC2019-Spring)_ , Apr. 2019, pp. 1–5.
  * [15] L. Golard, J. Louveaux, and D. Bol, “Evaluation and projection of 4G and 5G RAN energy footprints: the case of Belgium for 2020–2025,” _Annals of Telecommunications_ , Nov. 2022.
  * [16] F. E. Salem, T. Chahed, Z. Altman, and A. Gati, “Traffic-aware Advanced Sleep Modes management in 5G networks,” in _2019 IEEE Wireless Communications and Networking Conference (WCNC)_ , Marrakesh, Apr. 2019, pp. 1–6.
  * [17] N. Piovesan, A. De Domenico, M. Bernabe _et al._ , “Forecasting Mobile Traffic to Achieve Greener 5G Networks: When Machine Learning is Key,” in _2021 IEEE 22nd International Workshop on Signal Processing Advances in Wireless Communications (SPAWC)_ , 2021, pp. 276–280.
  * [18] H. V. Cheng, D. Persson, E. Bjornson, and E. G. Larsson, “Massive MIMO at night: On the operation of massive MIMO in low traffic scenarios,” in _2015 IEEE International Conference on Communications (ICC)_ , London, 2015, pp. 1697–1702.
  * [19] K. Senel, E. Björnson, and E. G. Larsson, “Joint Transmit and Circuit Power Minimization in Massive MIMO With Downlink SINR Constraints: When to Turn on Massive MIMO?” _IEEE Trans. Wireless Commun._ , vol. 18, no. 3, pp. 1834–1846, Mar. 2019.
  * [20] E. Bjornson, L. Sanguinetti, J. Hoydis, and M. Debbah, “Optimal Design of Energy-Efficient Multi-User MIMO Systems: Is Massive MIMO the Answer?” _IEEE Trans. Wireless Commun._ , vol. 14, no. 6, pp. 3059–3075, Jun. 2015.
  * [21] J. C. Marinello, T. Abrao, A. Amiri, E. de Carvalho, and P. Popovski, “Antenna Selection for Improving Energy Efficiency in XL-MIMO Systems,” _IEEE Transactions on Vehicular Technology_ , vol. 69, no. 11, pp. 13 305–13 318, Nov. 2020.
  * [22] E. Peschiera and F. Rottenberg, “Linear Precoder Design in Massive MIMO under Realistic Power Amplifier Consumption Constraint,” in _2022 Joint WIC/IEEE Symposium on Information Theory and Signal Processing in the Benelux_ , 2022, p. 5.
  * [23] Ł. Budzisz, F. Ganji, Rizzo _et al._ , “Dynamic Resource Provisioning for Energy Efficiency in Wireless Access Networks: A Survey and an Outlook,” _IEEE Communications Surveys & Tutorials_, vol. 16, no. 4, pp. 2259–2285, 2014.
  * [24] Y. Chen, S. Zhang, S. Xu, and G. Y. Li, “Fundamental trade-offs on green wireless networks,” _IEEE Communications Magazine_ , vol. 49, no. 6, pp. 30–37, Jun. 2011.
  * [25] Jingon Joung, Chin Keong Ho, and Sumei Sun, “Spectral Efficiency and Energy Efficiency of OFDM Systems: Impact of Power Amplifiers and Countermeasures,” _IEEE J. Sel. Areas Commun._ , vol. 32, no. 2, pp. 208–220, Feb. 2014.
  * [26] J. Wu, S. Rangan, and H. Zhang, Eds., _Green Communications: Theoretical Fundamentals, Algorithms, and Applications_. CRC Press, Apr. 2016\.
  * [27] Q. Wu, G. Y. Li, W. Chen, D. W. K. Ng, and R. Schober, “An Overview of Sustainable Green 5G Networks,” _IEEE Wireless Commun._ , vol. 24, no. 4, pp. 72–80, Aug. 2017.
  * [28] B. Debaillie, C. Desset, and F. Louagie, “A Flexible and Future-Proof Power Model for Cellular Base Stations,” in _2015 IEEE 81st Vehicular Technology Conference (VTC Spring)_ , May 2015, pp. 1–7.
  * [29] G. Auer, V. Giannini, C. Desset, I. Godor, P. Skillermark, M. Olsson _et al._ , “How much energy is needed to run a wireless network?” _IEEE Wireless Communications_ , vol. 18, no. 5, pp. 40–49, Oct. 2011.
  * [30] S. C. Cripps, _RF power amplifiers for wireless communications_ , 2nd ed., ser. Artech House microwave library. Boston, Mass.: Artech House, 2006.
  * [31] D. Persson, T. Eriksson, and E. G. Larsson, “Amplifier-Aware Multiple-Input Multiple-Output Power Allocation,” _IEEE Communications Letters_ , vol. 17, no. 6, pp. 1112–1115, Jun. 2013.
  * [32] A. Grebennikov, _RF and microwave power amplifier design_. McGraw-Hill Education, 2015.
  * [33] S. Mikami, T. Takeuchi _et al._ , “An Efficiency Degradation Model of Power Amplifier and the Impact against Transmission Power Control for Wireless Sensor Networks,” in _2007 IEEE Radio and Wireless Symposium_ , Long Beach, CA, USA, 2007, pp. 447–450.
  * [34] M. M. A. Hossain, C. Cavdar, E. Bjornson, and R. Jantti, “Energy Saving Game for Massive MIMO: Coping With Daily Load Variation,” _IEEE Transactions on Vehicular Technology_ , vol. 67, no. 3, pp. 2301–2313, Mar. 2018.
  * [35] M. M. A. Hossain and R. Jantti, “Impact of efficient power amplifiers in wireless access,” in _2011 IEEE Online Conference on Green Communications_ , Sep. 2011, pp. 36–40.
  * [36] Ericsson, “Modeling and evaluation methodology for network energy saving,” Tech. Rep. 3GPP TSG RAN WG1 #109-e, May 2022, R1-2204881.
  * [37] T. Islam, D. Lee, and S. S. Lim, “Enabling Network Power Savings in 5G-Advanced and Beyond,” _IEEE Journal on Selected Areas in Communications_ , vol. 41, no. 6, pp. 1888–1899, Jun. 2023.