1

Power-and Rate-Adaptation Improves the
Effective Capacity of C-RAN for Nakagami-m
Fading Channels
arXiv:1704.01924v4 [] 20 Feb 2018

Hong Ren, Nan Liu, Cunhua Pan, Maged Elkashlan, Arumugam Nallanathan,
Fellow, IEEE, Xiaohu You, Fellow, IEEE and Lajos Hanzo, Fellow, IEEE

Abstract
We propose a power-and rate-adaptation scheme for cloud radio access networks (C-RANs), where
each radio remote head (RRH) is connected to the baseband unit (BBU) pool through optical links.
The RRHs jointly support the users by efficiently exploiting the enhanced spatial degrees of freedom.
Our proposed scheme aims for maximizing the effective capacity (EC) of the user subject to both perRRH average-and peak-power constraints, where the EC is defined as the maximum arrival rate that
can be supported by the C-RAN under the statistical delay requirement. We first transform the EC
maximization problem into an equivalent convex optimization problem. By using the Lagrange dual
decomposition method and solving the Karush-Kuhn-Tucker (KKT) equations, the optimal transmission power of each RRH can be obtained in closed-form. Furthermore, an online tracking method is
provided for approximating the average power of each RRH for the sake of updating the Lagrange
dual variables. For the special case of two RRHs, the expression of the average power of each RRH
can be calculated in explicit form. Hence, the Lagrange dual variables can be computed in advance
in this special case. Furthermore, we derive the power allocation for two important extreme cases: 1)
no delay constraint; 2) extremely stringent delay-requirements. Our simulation results show that the
proposed scheme significantly outperforms the conventional algorithm without considering the delay
H. Ren, N. Liu and X. You are with the Southeast University, Nanjing, China. (e-mail:{renhong, nanliu, xhyu}@seu.edu.cn).
C. Pan, M.Elkashlan and Arumugam Nallanathan are with the Queen Mary University of London, London E1 4NS, U.K.
(Email:{c.pan, maged.elkashlan, a.nallanathan}@qmul.ac.uk). L. Hanzo is with the School of Electronics and Computer Science,
University of Southampton, Southampton, SO17 1BJ, U.K. (e-mail:lh@ecs.soton.ac.uk).
This work is partially supported by the National Natural Science Foundation of China under Grants 61571123 and 61521061,
the National 863 Program (2014AA01A702). L.Hanzo would like to acknowledge the financial support of the European Research
Council, Advanced Fellow grant Beam-Me-Up.

2

requirements. Furthermore, when appropriately tuning the value of the delay exponent, our proposed
algorithm is capable of guaranteeing a delay outage probability below 10−9 when the maximum tolerable
delay is 1 ms. This is suitable for the future ultra-reliable low latency communications (URLLC).

I. I NTRODUCTION
The fifth-generation (5G) wireless system to be deployed by 2020 is expected to offer a
substantially increased capacity, despite reducing the energy consumption of the fourth-generation
(4G) system [1]. To achieve this ambitious goal, the cloud radio access network (C-RAN) concept
has been regarded as one of the most promising solutions, which relied on the techniques of
network function virtualization (NFV), software-defined networks (SDN) and cloud computing
[1]. In particular, C-RAN is composed of three key components: 1) a pool of baseband units
(BBUs) centrally located at a cloud data center; 2) low-cost, low-power distributed radio remote
heads (RRHs) deployed in the network; 3) high-bandwidth low latency fronthaul links that
connect the RRHs to the BBU pool. Under the C-RAN architecture, most of the baseband signal
processing of conventional base stations has been shifted to the BBU pool and the RRHs are
only responsible for simple transmission and reception functions, some of the hitherto centralized
signal processing operations can be relegated to the BBU pool, such as cooperative transmission,
user scheduling, etc. Hence, the network capacity can be significantly improved.
Recently, the performance of C-RAN has been extensively studied [2]–[7], albeit these papers
have been focused on the physical layer issues, which giving no cognizance to the delay of the
upper layer’s. However, most of the multimedia services, such as video conferencing, interactive
gaming and mobile TV have stringent delay requirements. Due to the time-varying characteristics
of fading channels, it is impossible to impose a deterministic delay-bound guarantee for wireless
communications. For the sake of analyzing the statistical QoS performance, Wu et al. [8] introduced the notion of effective capacity (EC), which can be interpreted as the maximum constant
packet arrival rate that can be supported by the system, whilst satisfying a maximum bufferviolation probability constraint. Since then, substantial research attention has been dedicated to
studying the EC maximization problem by accounting for diverse QoS requirements.
However, due to the complex expression of EC, most of the existing papers focus on the
EC maximization problem for the simple scenario, where there is only a single transmitter [9]–
[15]. Specifically, a QoS-driven power-and rate-adaptation scheme was proposed for single-input
single-output (SISO) systems communicating over flat-fading channels in [9], with the objective

3

of maximizing the system throughput subject to both delay-QoS and average power constraints,
which is characterized by the QoS exponent θ. A smaller θ corresponds to a looser QoS guarantee,
while a higher value of θ represents a more stringent QoS requirement. The power allocation
scheme developed in [9] can be adapted to both the time-variant channel conditions and the QoS
requirements. The results of [9] showed that in the extreme case of very loose QoS requirements
(θ → 0), the power allocation reduces to the conventional water-filling solution. By contrast,
when the QoS requirements are extremely stringent (θ → ∞), the optimal power allocation
becomes the channel inversion scheme, where the system operates under a fixed transmission
rate. The problem of maximizing the EC of a frequency-selective fading channel was studied
in [10], where the authors showed that independently optimizing the power of each channel by
using the method of [9] yields a poor performance. Hence, jointly optimizing the power allocation
over all channels was derived in [10], which can achieve both high throughput and a low delay
at the same time. The EC maximization problem was studied in [11] in the context of cognitive
radio networks, where the power constraints of [9], [10] were replaced by the maximum tolerable
interference-power at the primary user. Closed-form expressions of both the power allocation and
the EC were derived for the secondary user. Michail et al. [12] provided a detailed EC analysis of
both Nakagami-m, as well as of Rician and generalized-K multiple-input single-output (MISO)
fading channels by using random matrix theory. The power minimization problem subject to EC
constraints was considered in [13] for three different scenarios, namely for a point-to-point link,
a multihop amplify-and-forward relay network and for a multiuser downlink cellular network.
Significant power savings can be achieved by using the power allocation scheme of [13]. To a
further advance, both subcarrier and power allocation were investigated in [14] for a one-way
relay network wherein the optimal subcarrier and power allocation was derived by adopting
the Lagrangian dual decomposition method. Most recently, Wenchi et al. [15] considered both
the average-and peak-power constraints when maximizing the EC, and provided the specific
conditions, when the peak power constraints can be removed. Some other related contributions
that aimed for maximizing the effective energy efficiency (EE) of single-transmitter systems are
[16]–[20], where the effective EE is defined as the ratio of EC to the average power consumption.
To avoid any traffic congestion and attain a good C-RAN performance, the adaptive power
allocation scheme should take the diverse QoS requirements into account to guarantee the
satisfaction of users. In this paper, we aim for jointly optimizing the power allocation of each
RRH in order to maximize the EC of a user of the C-RAN, where both the average power and

4

peak power constraints of each RRH are considered. This user is jointly served by all RRHs
of the C-RAN. Unfortunately, the power allocation schemes developed in the aforementioned
papers for single-transmitter scenarios [9]–[20] cannot be directly applied to C-RAN’s relying on
multiple RRHs for serving the user and to simultaneously exploit the spatial degrees of freedom.
The reason can be explained as follows. In these papers, there is only a single transmitter and
only a sum-power constraint is imposed. The Lagrange method can be used to find the optimal
power allocation, which is in the form of a water-filling-like solution in general. However, for
the C-RAN, all RRHs have their individual power constraints and the power cannot be shared
among the RRHs. Yu et al. [21] provided a detailed reason as to why the classic Lagrange method
cannot be readily applied in C-RAN. Hence, new methods have to be developed. Specifically,
the contributions of this paper can be summarized as follows:
1) In this paper, we derive the optimal power allocation for each RRH by resorting to the
Lagrangian dual decomposition and the Karush-Kuhn-Tucker (KKT) conditions. The power
allocation solutions depend both on the user’s QoS requirements and on the joint channel
conditions of the RRHs.
2) The Lagrangian dual decomposition requires us to calculate the subgradient, where the
average power of each iteration should be obtained. However, it is numerically challenging
to derive the expression of average power for each RRH. To tackle this issue, we provide
an online training method for tracking the average power of each RRH. For the special
case of a single RRH, the closed-form expression of average power can be obtained. For
the more complex case of two RRHs, we also provide the expression of the average power
for each RRH in explicit form, which can be numerically evaluated.
3) We provide the closed-form power allocation solutions for two extremely important cases:
1) When the QoS exponent θ approaches zero, which corresponds to the conventional
ergodic capacity maximization problem; 2) When the QoS exponent θ tends to infinity,
which represents the strictly delay-limited case.
4) Our simulation results will show that the proposed algorithms significantly outperforms
the existing algorithms. Additionally, by appropriately choosing the QoS exponent, our
proposed delay-aware algorithm can guarantee a delay-outage probability bellow 10−9 ,
when the maximum tolerable delay is 1 ms, which satisfies the strict delay requirements
of the future ultra-reliable low latency communications (URLLC).

5

The rest of this paper is organized as follows. In Section II, the C-RAN system is introduced
along with the concept of EC and our problem formulation. In Section III, we provide the
optimal power allocation for the general case of any number of RRHs. In Section IV, we
derive the integral expressions’s closed-form solution concerning the average transmit power
of each RRH for the sake of updating the Lagrangian dual variables. In Section V, we obtain
the power allocation for two extreme cases, namely for the delay-tolerant scenario and for
extremely stringent delay constraints. Numerical results are also provided in Section VI. Finally,
our conclusions are drawn in Section VII.
II.

SYSTEM MODEL

Consider a downlink C-RAN consisting of I RRHs and a single user 1 , where each RRH
and the user have a single antenna, as depicted in Fig. 1. The set of RRHs is denoted as
I = {1, 2, · · · , I}. All the RRHs are assumed to be connected to the BBU pool through the
fronthaul links relying on high-speed fiber-optic cables. In Fig. 2, the upper layer packets are
first buffered in first-in-first-out (FIFO) queue, which will be transmitted to the physical layer
of the RRHs. At the data-link layer, the upper-layer packets are partitioned into frames and then
each frame will be mapped into bit-streams at the physical layer. The channel is assumed to
obey the stationary block fading model, implying that they are fixed during each time frame of
length of Tf , while it is switched independently over different time frames.
To elaborate, we consider a Nakagami-m block-fading channel, which is very general and
includes most of the practical wireless communication channels as special cases [22]. The
parameter m represents the severeness of the channel, where the fading channel fluctuations are
reduced with m. The probability density function (PDF) of the Nakagami-m channel spanning
from the i-th RRH to the user is given by:
 m


m
αim−1 m
exp − αi , αi ≥ 0,
(1)
f (αi ) =
Γ(m) ᾱi
ᾱi
R∞
where Γ(m) = 0 wm−1 e−w dw is the Gamma function, αi denotes the instantaneous channelpower-to-noise ratio (CPNR) from the ith RRH to the user, and ᾱi is the average received CPNR
at the user from the ith RRH, denoted as P Li /σ 2 , where P Li is the large-scale fading channel
1

The method developed in this paper can also be applied to the multi-user scenario, where all users apply the classical

orthogonal frequency division multiplexing access (OFDMA) technique to remove the multi-user interference.

6

Transmitted signals
BBU pool
User Data

Fronthual links

User
P , arrival rate

Delay QoS
Requirement

RRH1

RRH I

BBU pool
Coherent
Receiver

AMC

..
RRH 2

FIFO
Queue

Receiver

FIFO
Queue

Transmitter

User

Feedback
channel

Adaptive
Power Control

Channel
Estimation

RRH 3
RRH1

..
.

RRH I

Fig. 1. Topology of a C-RAN with I RRHs.

Discrete-Time
Wireless Fading Channel

r[t ] log 2 1  SNR[t ]

Fig. 2. Cross-layer transmission model.

gain spanning from the ith RRH to the user that includes the path loss and shadowing effect,
and σ 2 is the noise power.
Let us define α = [α1 , α2 , · · · , αI ]T . Since α1 , · · · , αI are independent, the joint PDF of α
is given by
f (α) = f (α1 )f (α2 ) · · · f (αI ).

(2)

A. Effective Capacity (EC)
The EC is defined as the maximum constant frame arrival rate that a given service process
can support, while obeying the delay requirement indicated by the QoS exponent θ that will be
detailed later. Let the sequence R[k], k = 1, 2, · · · represent the data service-rate, which follows
a discrete-time stationary and ergodic stochastic process. The parameter k is the time frame
P
index. Let us denote by S(t) , tk=1 R[k] the partial sum of the service process over the time
sequence spanning from k = 1 to k = t. Let us furthermore assume that the Gartner-Ellis limit
of S[t], which is denoted by ΛC (θ) = limt→∞ (1/t) log(E{eθS[t] }), is a convex differentiable
function for all real-value of θ [9]. Then, the EC of the service process specified by θ is
EC(θ) = −

1
ΛC (−θ)
= − log(E{e−θR[k] })
θ
θ

(3)

where E{·} denotes the expectation operator.
Let us assume that a data source enters a queue of infinite buffer size at a constant data rate
µ. The probability that the delay exceeds a maximum delay bound of Dmax satisfies [8]
out
Pdelay
= Pr {Delay ≥ Dmax } ≈ εe−θµDmax ,

(4)

where f (x) ≈ g(x) indicates that lim [f (x)/g(x)] = 1, and ε is the probability that the buffer
x→∞

is non-empty, which can be calculated by using the method in [8]. The parameter θ can be
found by letting the constant arrival rate to be equal to the EC, i.e. setting it to µ = EC(θ).

7

out
Hence, if the delay-bound violation probability is required to be below Pdelay
, one should limit

its incoming data rate to a maximum of µ = EC(θ).
It is seen from (4) that θ is an important parameter, representing the decay rate of the delay
violation probability. A smaller θ corresponds to a slower decay rate, which indicates that the
delay requirement is loose, while a larger θ corresponds to a faster decay rate, which implies that
the system is capable of supporting a more stringent delay requirement. In other words, when
θ → 0, an arbitrarily long delay can be tolerated by the system, which corresponds to the capacity
studied in Shannon information theory. On the other hand, when θ → ∞, this implies that no
delay is allowed by the system, which corresponds to the very stringent statistical delay-bound
QoS constraint of allowing no delay at all.
B. Problem formulation
For the user, we assume that the maximal ratio combining (MRC) is adopted at the receiver
[23]. Then, the instantaneous service rate of a single frame, denoted by R(ν), can be expressed
as follows [24], [25]:
R(ν) = Tf B log2 (1 +

X
i∈I

pi (ν)αi ),

(5)

∆

where B is the system’s bandwidth, ν = (α, θ) represents the network condition that includes
both the channel’s power gains and the EC exponent requirement, and pi (ν) represents the
power allocation for RRH i that depends on the network’s condition ν. In this paper, we aim
for optimizing the transmit power in order to maximize the EC for the user under two different
types of power limitations for each RRH: under an average power constraint and a peak power
constraint. The first one is related to the long-term power budget, while the second guarantees
that the instantaneous transmit power is below the linear range of practical power amplifiers.
Mathematically, this optimization problem can be formulated as
 h
i
P
1
max
− log Eα e−θTf B log2 (1+ i∈I pi (ν)αi )
{pi (ν),i∈I}
θ

(6a)

s.t. Eα [pi (ν)] ≤ Piavg , ∀i ∈ I,

(6b)

0 ≤ pi (ν) ≤ Pipeak , ∀i ∈ I,

(6c)

where Eα {·} denotes the expectation over α, while Piavg and Pipeak denote the ith RRH’s
maximum average transmit power constraint and peak transmit power constraint, respectively.

8

III. O PTIMAL P OWER A LLOCATION M ETHOD
By exploiting the fact that log(·) is a monotonically increasing function, Problem (6) can be
equivalently simplified as
min



{pi (ν),∀i∈I}

where we have ε (θ) =

θTf B
.
ln 2

Eα

1+

X
i∈I

pi (ν) αi

−ε(θ) 

(7a)

s.t. Eα [pi (ν)] ≤ Piavg , ∀i ∈ I,

(7b)

0 ≤ pi (ν) ≤ Pipeak , ∀i ∈ I,

(7c)

In Appendix A, we prove that Problem (7) is a convex optimization

problem. Hence, the Lagrangian duality method can be used to solve Problem (7) with zero
optimality gap.
Note that in Problem (7), only the average power constraints are ergodic, while the others are
instantaneous power constraints. Similar to [26], we should first introduce the dual variables
associated with the average transmit power constraints. Then, the original problem can be
decomposed into several independent subproblems, where each one corresponds to one fading
state. In addition, the instantaneous power constraints are enforced for each fading state. Let
λ = [λ1 , · · · , λI ]T represent the nonnegative dual variables associated with the average power
constraints. The Lagrangian function of Problem (7) can be written as
X
X
λi (Eα [pi (ν)] − Piavg ),
pi (ν)αi )−ε(θ) ] +
L(P (ν) , λ) = Eα [(1 +
i∈I

i∈I

(8)

where P (ν) = [p1 (ν), · · · , pI (ν)]T . Let us now define P = {P (ν) |(7c)}. The Lagrange dual
function is then given by
g (λ) = min L(P (ν) , λ).
P(ν)∈P

(9)

The dual problem is defined as
max g (λ) ,

λi ≥0,∀i

(10)

As proven in Appendix A, Problem (7) is a convex optimization problem, which implies that
there is no duality gap between the dual problem and the original problem. Thus, solving the
dual problem is equivalent to solving the original problem.
To solve the dual problem in (10), we should solve the problem in (9) for a fixed λ, then
update the Lagrangian dual variables λ by solving the dual problem in (10). Iterate the above
two steps until convergence is reached.
1) Solving the dual function in (9): For a given λ, we should find the dual function g(λ),
which can be rewritten as

9

g(λ) = E [g̃ (λ)] −

X

λi Piavg ,

(11)

i∈I

where we have:
g̃ (λ) = min



1+

X

P(ν)∈P

i∈I

pi (ν) αi

−ε(θ)

+

X
i∈I

λi pi (ν).

(12)

Note that g̃ (λ) can be decoupled into multiple independent subproblems, each corresponding to
a specific fading state. Those subproblems have the same structure for each fading state. Hence,
to simplify the derivation, ν is omitted in the following. Each subproblem can be expressed as:

−ε(θ) X
X
min 1 +
pi αi
+
λi pi
(13a)
i∈I

P

i∈I

s.t. 0 ≤ pi ≤ Pipeak , ∀i ∈ I.

(13b)

The above problem is convex. In the following, we obtain the closed-form solution by solving
the KKT conditions of Problem (13).
First, we introduce the nonnegative dual variables of µi , ∀i, and δi , ∀i for the associated
constraints in (13b). The KKT conditions for Problem (13) can be expressed as
X
−ε(θ)(1 +
p∗i αi )−ε(θ)−1 αi + λi + µ∗i − δi∗ = 0, ∀i
i∈I

(14a)

µ∗i (p∗i − Pipeak ) = 0, ∀i

(14b)

δi∗ p∗i = 0, ∀i

(14c)

p∗i ≤ Pipeak , ∀i

(14d)

p∗i ≥ 0, ∀i,

(14e)

with p∗i ≥ 0, δi∗ ≥ 0, and µ∗i ≥ 0, ∀i. Then, we have the following lemma.
Lemma 1: For any two arbitrary RRHs i and j, if p∗i > 0 and p∗j = 0, then the following
relationship must hold

λj
λi
≤ .
αi
αj

(15)

Proof : Please see Appendix B.
Lemma 1 shows that the RRHs associated with a smaller λi /αi should be assigned a non-zero
power, while the RRHs having a larger value should remain silent. Let us hence introduce π as
a permutation over I, so that we have

λπ(i)
απ(i)

≤

λπ(j)
,
απ(j)

when i < j, i, j ∈ I. Let I 0 ⊆ I be the

set of RRHs that transmit at a non-zero power. Then, according to Lemma 1, it can be readily
verified that we have I 0 = {π(1), · · · , π(|I 0 |)}.

10

Lemma 2: There is at most one RRH associated with 0 < p∗i < Pipeak , and the RRH index is
i = π(|I 0 |).
Proof : Please see Appendix C.
Next, we derive the optimal power allocation of Problem (13) shown as in Theorem 1.
Theorem 1: The optimal solution of Problem (13) is shown as follows,

peak


,
 Pπ(a)(

#)
"

0 |−1
1

 λ
−ε(θ)+1
|IP
peak
peak
π(|I 0 |)
∗
1
−
Pπ(a) απ(a) − 1
pπ(a)= min Pπ(|I 0 |) , α 0
ε(θ)απ(|I 0 |)
π(|I |)

a=1




 0,

a < |I 0 |;
a = |I 0 |;

(16)

a > |I 0 |;

where |I 0 | is the largest value of integer x, so that we have
#−ε(θ)−1
" x−1
X peak
λπ(x)
.
<
Pπ(b) απ(b) + 1
ε(θ)απ(x)
b=1

(17)

Proof : Please refer to Appendix D.
Corollary 1: When the number of RRHs is equal to one, i.e. I = 1, the optimal transmit
power is given
by



0,




1

−
ε(θ)
1
∗
λ
p1 =
1+ε(θ) α 1+ε(θ)
1
( ε(θ) )
1



peak

P
,


 1

if α1 <
1
,
α1

if α1 ≥

λ1
;
ε(θ)
λ1
and
ε(θ)

1

(
if α1 ≥

λ1
ε(θ)

λ1
ε(θ)

)

ε(θ)
1
1+ε(θ) α 1+ε(θ)
1

1

and
λ1
( ε(θ)
)

ε(θ)
1
1+ε(θ) α 1+ε(θ)
1

−
−

1
α1

< P1peak ;

1
α1

≥ P1peak .

(18)

Proof : It is can be readily derived using Theorem 1.
Note that the above result is consistent with the point-to-point result obtained in [15].
2) Solving the dual problem (10): To solve the dual problem (10), we invoke the subgradient
method, which is a simple method of optimizing non-differentiable objective function [27]. The
(k)

(k)

subgradient is required by the subgradient of g(·) at λ(k) = [λ1 , · · · , λI ]T in the kth iteration2 .
Theorem 2: The subgradient of g(·) at λ(k) in the kth iteration is given by


d(k) = Eα P∗λ(k) (ν) − Pavg ,
(19)
h
iT
where P∗λ(k) (ν) = p∗(1,λ(k) ) (ν), · · · , p∗(I,λ(k) ) (ν) is the optimal solution of Problem (9) when
we have λ = λ(k) , and Pavg = [P1avg , · · · , PIavg ]T .
Proof : Please see Appendix E.
2

According to [27], a vector d is a subgradient of g(λ) at λ(k) , if for all λ, g(λ) ≤ g(λ(k) ) + dT (λ − λ(k) ) holds.

11

Based on Theorem 2, the Lagrangian dual variables can be updated as
λ(k+1) = λ(k) + ζ (k) d(k) ,

(20)

where ζ (k) is the step in the k th iteration. The subgradient method is guaranteed to converge if
P
(k)
ζ (k) satisfies limk→∞ ζ (k) = 0 and ∞
= ∞ [28]. The step size in the simulation section
k=1 ζ
is set as ζ (k) = a/k, where a is the constant step-size parameter.
In summary, the solution of Problem (6) is given in Algorithm 1.
Algorithm 1 Solving Problem (6)
Initialize:
Iteration number k = 0, λ(0) = [λ1 (0) , · · · , λI (0) ];
Repeat
1. Compute P(k) with fixed λ(k) using (16);
2. Compute the subgradient d(k) , using (19);
3. Update λ(k+1) using (20), increase k by 1;
Until convergence

To execute Algorithm 1, there is another issue that has to be tackled, namely how to calculate
the average power for each RRH in order to obtain the subgradient d(k) . Given the dual variables
λ(k) , the expression of optimal power at each RRH depends on the generations of channel gains
α. Different generations of α will lead to different orders of λi /αi , i = 1, · · · , I, and thus
different power expressions. Even for a fixed problem-order the power allocation expressions
require multiple integrations, which imposes a high computational complexity. As a result, it
is a challenge to obtain the expression of average power for each RRH in closed-form for any
given λ(k) . In fact, even for the simple case of two RRHs, the average powers generally do not
have simple closed-form solutions, as shown in the next section.
To resolve the above issue, we propose an online calculation method for tracking the average
power required for each RRH. The main idea is to replace the expectation operator by averaging the power allocations for all samples of channel generations during the fading process.
(k−1)

Specifically, let us define P̄i

as
k−1

(k−1)
P̄i


1 X ∗
p(i,λ(j) ) α(j) , θ , i = 1, · · · , I,
=
k − 1 j=1

(21)

12

h x

h x

h x

0

x

(a) Case 1) hc x t 0

0

.x

*

x

(b) Case 2) hc 0 d 0, h x t 0
*

0

. . .
x

*

x

(c) Case 3) hc 0 d 0, h x* d 0

Fig. 3. The properties of the function h(x) = (1 + ax)b − cx.


where p∗(i,λ(j) ) α(j) , θ denotes the optimal power allocation of the ith RRH for the jth channel
generation, and λ(j) represents the corresponding dual variables. Then, the expectation over
p∗(i,λ(k) ) (ν) can be approximated as

(k−1)
h
i
p∗(i,λ(k) ) α(k) , θ + (k − 1)P̄i
(k)
∗
Eα p(i,λ(k) ) (ν) ≈ P̄i =
, ∀i ∈ I,
k
(k−1)

which can be recursively obtained based on the previous P̄i

(22)

, ∀i. It is worth noting that this

algorithm can be readily applied to the case when the fading statistics are unknown.
Fortunately, for the case of a single RRH3 , the average power can be obtained in closed-form,
which is given in Appendix F. The average power expression for the case of I = 2 is even more
complex, which is extensively studied in the following section.
IV. S PECIAL C ASE : I = 2
To show the difficulty of obtaining the closed-form solution for each RRH, we only consider
the simplest scenario of two RRHs, i.e., I = 2. To this end, we first introduce the following
lemma, which will be used in the ensuing derivations.
Lemma 3: Let us define the function h(x) = (1 + ax)b − cx, where a > 0, c > 0, b > 1. Then,
in the region of x ≥ 0, there are only three possible curves for the function h(x), which are
shown in Fig. 3. The conditions for each case are given as follows:
1) Case 1: ab − c ≥ 0;
2) Case 2: ab − c < 0 and
3) Case 3: ab − c < 0 and


c 1/(b−1)
(1
ab

1/(b−1)
c
(1
ab

− b) + b ≥ 0;
− b) + b < 0;

Proof : Please see Appendix G.
3

Note that [15] did not provide the closed-form expression for the average power for the case of one RRH.

13

Based on Lemma 3, we commence by deriving the expression of the average transmit power.
According to Theorem 1, there are three different possible cases for I = 2: 1) |I 0 | = 0; 2)
|I 0 | = 1; 3) |I 0 | = 2. Obviously, the average power contributed by the first case is zero, hence
we only consider the latter two cases, which are discussed in the following two subsections.
Without loss of generality (w.l.o.g), we assume that
λ1
λ2
≤
α1
α2
holds. The case for

λ1
α1

>

λ2
α2

(23)

can be similarly derived.

A. Scenario 1: There is only one RRH that transmits at a non-zero power, i.e., |I 0 | = 1
Since we have assumed that the inequality (23) holds, only RRH 1 is transmitting at a nonzero power and RRH 2 remains silent in this case. According to Theorem 1, the conditions for
|I 0 | = 1 are given by

−ε(θ)−1
λ2
≥ P1peak α1 + 1
,
ε (θ) α2
λ1
< 1,
ε (θ) α1
and only RRH 1 transmits at non-zero power, which is given by
#)
(
"
1
− ε(θ)+1
1
λ
1
peak
p∗1 = min P1 ,
−1
.
α1
ε (θ) α1

(24)
(25)

(26)

It may be readily seen that, there are two possible values of p∗1 , and the conditions for each
value will be discussed as follows.
1) p∗1 = P1peak : In this case, the following condition should be satisfied:
"
#
1
− ε(θ)+1
1
λ1
− 1 ≥ P1peak ,
α1
ε (θ) α1

(27)

which is equivalent to


1 + P1peak α1

1+ε(θ)

−

ε (θ)
α1 ≤ 0.
λ1

(28)

Notice that the left hand side of (28) is in the form of h(x) defined in Lemma 3, with x = α1 ,
a, b, c given by
a = P1peak , b = 1 + ε (θ) , c = ε (θ)/λ1 .

(29)

To guarantee that there exists a positive α1 , the function h(x) should be reminiscent of Fig. 3(c), and the conditions for Case 3 in Lemma 3 should be satisfied. Otherwise, p∗1 is not equal to

14

P1peak . In the following, we assume that the conditions are satisfied. Let us denote the solutions
of h(α1 ) = 0 as α1l and α1u , where α1l < α1u . These solutions can be readily obtained by using
the classic bisection method. Then, condition (28) is equivalent to
α1l ≤ α1 ≤ α1u .

(30)

By combining (23) and (24), we obtain the feasible region of α2 in the form of:

1+ε(θ) 
1+ε(θ)
λ2 
λ2 
λ2
peak
0 ≤ α2 ≤ min
=
α1 ,
1 + P1 α1
1 + P1peak α1
,
λ1
ε (θ)
ε (θ)

(31)

where the last equality holds by using (28). Similarly, by combining (25) and (30), we obtain
the feasible region of α1 as:


λ1
l
max α1 ,
≤ α1 ≤ α1u .
ε (θ)

(32)

In this context, we prove that λ1 /ε (θ) < α1l in Appendix H. Hence, the feasible region of α1 is
given by α1l ≤ α1 ≤ α1u .
Based on the above discussions, we obtain the conditions for p∗1 = P1peak , p∗2 = 0 as follows:
 c 1/(b−1)
(1 − b) + b < 0,
(33)
C1 :ab − c < 0,
ab
where a, b, c are given in (29).
If Condition C1 in (33) is satisfied, the average power assigned to RRH 1 in this case is given
by
∆
C1
TRRH
=
1

Z

αu
1

Z

αl1

λ2
ε(θ)

(1+P1peak α1 )

1+ε(θ)

P1peak f (α1 ) f (α2 ) dα2 dα1 ,

(34)

0

By substituting the PDF of α2 in (2) into (34), (34) can be simplified to

Z u
1+ε(θ) 
P1peak α1
mλ2 
peak
C1
TRRH1 =
f (α1 )γ m,
1 + P1 α1
dα1 .
Γ(m) αl1
ᾱ2 ε (θ)

(35)

C1
Unfortunately, the closed-form expression of TRRH
cannot be obtained even for the special
1
C1
case of m = 1. However, the value of TRRH
can be obtained at a good accuracy by using the
1

numerical integration
function of Matlab.


1
− ε(θ)+1
λ1
1
∗
− 1 : In this case, the following condition should be satisfied:
2) p1 = α1
ε(θ)α1
1
α1

"

λ1
ε (θ) α1

1
− ε(θ)+1

#
− 1 < P1peak ,

(36)

which leads to h(α1 ) > 0 with a, b, c defined in (29). As seen from Fig. 3, when the first two
conditions in Lemma 3 are satisfied, the inequality (36) holds for any α1 ≥ 0. When the third

15

condition in Lemma 3 is satisfied, the inequality (36) holds when 0 ≤ α1 ≤ α1l and α1 ≥ α1u ,
where α1l and α1u (α1l < α1u ) are the solutions of h(α1 ) = 0 with a, b, c defined in (29).
Now, we first assume that the first two conditions in Lemma 3 are satisfied
 c 1/(b−1)
C2 :ab − c ≥ 0, or ab − c < 0 and
(1 − b) + b ≥ 0,
ab

(37)

where a, b, c are given in (29). Then, the condition in (36) can be neglected. According to (23)
and (24), the feasible region of α2 is given by

1+ε(θ)  λ
λ2
λ2 
2
peak
0 ≤ α2 ≤ min
α1 ,
1 + P1 α 1
= α1 ,
λ1
ε (θ)
λ1
where (36) is used in the last equality. From (25), the feasible region of α1 is given by α1 ≥

(38)
λ1
.
ε(θ)

As a result, the average power of RRH 1 contributed by the case when condition C2 is satisfied
is given by
C2
TRRH
1

Z

∞

λ2
α
λ1 1

Z

=
λ1
ε(θ)

1
=
Γ (m)

p∗1 f (α1 ) f (α2 ) dα2 dα1 ,

0

Z

∞
λ1
ε(θ)

1
α1

"

λ1
ε (θ) α1

1
− ε(θ)+1

# 

mλ2 α1
− 1 γ m,
f (α1 ) dα1 .
ᾱ2 λ1

(39)

C2
Fortunately, when m is an integer, the closed-form expression of TRRH
can be obtained. Let
1

us define
1
(m/ᾱ1 )m
mλ2
m
λ1
∆ mλ2
,V ,
,W =
, Z=
,Y =
+ .
U,
ε (θ)
1 + ε (θ)
ᾱ2 λ1
(m − 1)!
ᾱ2 λ1 ᾱ1

(40)

If m = 1, the Nakagami-m channel reduces to the Rayleigh channel, and the average power of
RRH 1 in (39) can be simplified to:
C2
TRRH
1



1
1
ᾱ1V −1 Γ (V, U /ᾱ1 )
U
Γ (V, U Y )
+ Ei −
− Ei (−U Y ) .
=
−
V
V
V
U
ᾱ1
ᾱ1
ᾱ1 Y U
ᾱ1

(41)

If m is an integer that is larger than one, i.e., m ≥ 2, we can obtain the closed-form expression
C2
of TRRH
:
1
C2
TRRH
= J1 + J2 − J3 − J4 ,
1

where J1 , J2 , J3 and J4 are respectively given by


m−1
ᾱ1V −1 Γ V + m − 1, mU
X
ᾱ1
Wl
J1 =
, J2 = Z
Γ (l + m − 1, Y U ),
(m − 1)!mV −1 U V
l!Y l+m−1
l=0


m−1
Z X
Wl
m
mU
J3 = V
Γ (l + V + m − 1, Y U ), J4 =
Γ m − 1,
.
U l=0 l!Y l+V +m−1
ᾱ1 (m − 1)!
ᾱ1

(42)

(43)

(44)

16

The details of the above derivations can be found in Appendix I.
Next, we consider the case, where Condition 3) in Lemma 3 is satisfied. According to the
condition in (36), the feasible region of α1 is 0 ≤ α1 ≤ α1l and α1 ≥ α1u . Additionally, from
(25), α1 ≥ U should hold. According to Appendix H, U ≤ α1l always holds. Hence, the overall
feasible region is α1l ≥ α1 ≥ U and α1 ≥ α1u . Furthermore, the feasible region of α2 is the same
as in (38). As a result, the expression of the average power for RRH 1 contributed in this case
can be similarly obtained as in (39), except for the different integration intervals for α1 . The
expression for the special case, when m is an integer can be similarly obtained.
Let us now define the following function


Z ∞
1  α1 V
1
− 1 γ (m, W α1 ) f (α1 )dα1 .
F (x) =
Γ (m) x α1
U
C2
C2
Then, TRRH
given in (39) is equal to TRRH
= F (U ).
1
1

If the following condition is satisfied:
C3 : ab − c < 0 and

 c 1/(b−1)
(1 − b) + b < 0,
ab

(45)

with a, b, c given in (29), the average power for RRH 1 contributed under this condition is

C3
l
TRRH
=
F
(U
)
−
F
α
+ F (α1u ) .
1
1

(46)

Note that Condition C3 is the same as Condition C1, but the power allocation for RRH 1 is
different.
B. Scenario 2: Both RRHs are transmitting at a non-zero power, i.e., |I 0 | = 2
In this case, RRH 1 will transmit with full power, i.e., p∗1 = P1peak , and RRH 2 will transmit
at a non-zero power. According to Theorem 1, the following condition should be satisfied:

−ε(θ)−1
λ2
peak
< 1 + P1 α1
(47)
ε (θ) α2
and the transmit power of RRH 2 is given by
(
"
#)
1
− 1+ε(θ)
1
λ
2
peak
peak
p∗2 = min P2 ,
− P1 α 1 − 1
.
α2
ε (θ) α2
The conditions for each value of p∗2 will be discussed as follows.

(48)

17

1) p∗2 = P2peak : In this case, the following condition should be satisfied:
#
"
1
− 1+ε(θ)
1
λ
2
P2peak ≤
− P1peak α1 − 1 .
α2
ε (θ) α2

(49)

Combining conditions (23), (47) and (49), we can obtain the feasible region of α1 as follows
#
"
 1
λ1
1
ε (θ) α2 1+ε(θ)
α2 ≤ α1 < peak
− P2peak α2 − 1 , A.
(50)
λ2
λ
P1
2
To guarantee that we have a nonempty set of α1 , the following condition should be satisfied:

1+ε(θ)
λ1 peak
ε (θ)
peak
1 + P2 α2 + P1 α2
<
α2 ,
(51)
λ2
λ2
which is in the form of function h(x) defined in Lemma 3 with x = α2 , and a, b, c given by
a = P2peak +

λ1 peak
ε (θ)
P1 , b = 1 + ε (θ) , c =
.
λ2
λ2

(52)

To guarantee the existence of a positive α2 , the graphical curve of the function h(α2 ) should be
similar to that in Fig. 3-c, and the conditions are given by
 c 1/(b−1)
C4 :ab − c < 0, and
(1 − b) + b < 0,
ab

(53)

where a, b, c are given in (52). Let us denote the solutions of h(α2 ) = 0 as α2l and α2u , where
we have α2l < α2u . Then, the feasible region of α2 is given by α2l < α2 < α2u .
Under Condition C4, the average powers of RRH 1 and RRH 2 are respectively calculated as
 



Z u
P1peak α2
mA
mλ1 α2
C4
TRRH1 =
f (α2 ) γ m,
− γ m,
dα2 ,
(54)
Γ (m) αl2
ᾱ1
ᾱ1 λ2
C4
TRRH
2

C4
TRRH
P peak
1 2

,
(55)
P1peak
where A is defined

 in (50).1
− 1+ε(θ)
peak
λ2
1
∗
2) p2 = α2
− P1 α1 − 1 : In this case, the following condition should be
ε(θ)α2
=

satisfied:
P2peak

1
>
α2

"

λ2
ε (θ) α2

1
− 1+ε(θ)

#
− P1peak α1 − 1 .

By combining (23), (47) and (56), the feasible region of α1 can be obtained as follows:
"
#

 1

1
ε (θ) α2 1+ε(θ)
λ1
B , peak
−1 > α1 ≥ max
α2 , A ,
λ2
λ2
P1

(56)

(57)

where A is defined in (50). Note that B > A always holds. Hence, to guarantee that there exists
a feasible α1 , the following condition should be satisfied:

1+ε(θ)
λ1 peak
ε (θ) α2
0 ≥ 1 + P1 α 2
−
.
λ2
λ2

(58)

18

Again, the right hand side of (58) is in the form of the function h(x) defined in Lemma 3, with
x = α2 , and a, b, c are given by
a=

λ1 peak
ε (θ)
P1 , b = 1 + ε (θ) , c =
.
λ2
λ2

(59)

To guarantee that there exists a positive α2 , the third condition in Lemma 3 should be satisfied:
 c 1/(b−1)
(1 − b) + b < 0,
(60)
CX :ab − c < 0 and
ab
where a, b, c are given in (59). When a, b, c are given in (59), we can denote the solutions
of h (α2 ) = 0 as α̃2l and α̃2u with α̃2l < α̃2u . Hence, the feasible region of α2 is given by
α̃2l < α2 < α̃2u .
The remaining task is to determine the lower bound of α1 as shown in (57).
Case I: If the following condition is satisfied:
λ1
α2 ≥ A,
λ2

(61)

the feasible region of α1 is given by
B > α1 ≥

λ1
α2 ,
λ2

(62)

where B is defined in (57). The inequality (61) can be rewritten as
!1+ε(θ)
ε (θ) α2
λ1 P1peak
peak
1 + P2 α2 +
α2
≥
,
λ2
λ2

(63)

which is equivalent to h(α2 ) ≥ 0 with a, b and c given in (52).
As seen from Fig. 3, when the first two conditions in Lemma 3 are satisfied, the inequality
(63) holds for any α2 > 0. Combining this with the condition (58), the feasible region of α2 is
given by α̃2l < α2 < α̃2u , where α̃2l and α̃2u are the solutions of h(α2 ) = 0 with a, b, c defined in
(59). Define the following condition with a, b, c defined in (52)4 :
 c 1/(b−1)
(1 − b) +b ≥ 0,
C5 : CX and ab − c ≥ 0, or CX , ab − c < 0 and
ab
and C as
1
C,
α2
4

Under Condition C5, a, b, c are defined in (59).

"

ε (θ) α2
λ2

1
 1+ε(θ)

(64)

#
−1 .

(65)

19

Under Condition C5, the average power of RRH 1 and RRH 2 are respectively given by



Z u 
mλ1
P1peak α̃2
mB
C5
− γ m,
α2 f (α2 )dα2 ,
(66)
TRRH1 =
γ m,
Γ (m) α̃l2
ᾱ1
ᾱ1 λ2
C5
TRRH
= J1 − J2
2

with

(67)




 
mλ1
mB
−
γ
m,
α
f (α2 ) dα2
C
γ
m,
2
α̃l2
  ᾱ1
 ᾱ1 λ2

peak R u
α̃2 1
ᾱ1 P1
mλ1
mB
J2 = mΓ(m) α̃l α2 γ m + 1, ᾱ1 − γ m + 1, ᾱ1 λ2 α2 f (α2 )dα2 ,

J1 =

1
Γ(m)

R α̃u2

2

where B is defined in (57), α̃2l and α̃2u are the solutions of h(α2 ) = 0 with a, b, c defined in (59).
On the other hand, when the third condition in Lemma 3 is satisfied, inequality (63) holds for
0 ≤ α2 ≤ α2l or α2 ≥ α2u , where α2l and α2u are the solutions of h(α2 ) = 0 with a, b, c defined
in (52). Additionally, it is easy to verify that the curve of h(α2 ) associated with a, b, c defined
in (52) is above the curve of h(α2 ) with a, b, c defined in (59). Hence, the following relations
hold:
α̃2l < α2l < α2u < α̃2u .

(68)

Combining (68) with condition (58), the feasible region of α2 is given by
α̃2l < α2 < α2l , α2u < α2 < α̃2u .
As a result, when the following conditions are satisfied with a, b, c defined in (52):
 c 1/(b−1)
(1 − b) +b < 0,
C6 : CX , ab − c < 0 and
ab

(69)

(70)

C6
C6
the average power required for RRH 1 and RRH 2, denoted as TRRH
and TRRH
respectively, is
1
2
C5
C5
similar to the expressions of TRRH
and TRRH
except that the integration interval of α2 becomes
1
2

(69).
Case II: If the following condition is satisfied:
λ1
α2 < A,
λ2

(71)

the feasible region of α1 is B > α1 > A, where B is defined in (57). The above condition is
equivalent to h(α2 ) < 0 with a, b, c defined in (52). To guarantee the existence of a positive α2 ,
the third condition in Lemma 3 should be satisfied. Assuming that this condition is satisfied, the
feasible region of α2 is given by α2l < α2 < α2u , where α2l and α2u are the solutions of h(α2 ) = 0
with a, b, c defined in (52). Again, by using the relations in (68) and the condition (58), the
feasible region of α2 is given by α2l < α2 < α2u .

20

TABLE I
M AIN RESULTS WHEN I = 2,
Conditions

p∗1
p∗1 =

C2

p∗1 =

C3

λ2
α2

1
α1
1
α1




=

P1peak
− 1
ε(θ)+1

λ1
ε(θ)α1
λ1
ε(θ)α1

−

1
ε(θ)+1

Average Power
C1
TRRH
1


−1

−1

C2
TRRH
1
C3
TRRH
1

p∗1 = P1peak , p∗2 = P2peak


− 1
1+ε(θ)
peak
λ2
α
−
1
p∗1 = P1peak , p∗2 = α12
−
P
1
1
ε(θ)α2


− 1
1+ε(θ)
peak
peak
λ2
∗
∗
1
− P1 α1 − 1
p 1 = P 1 , p 2 = α2
ε(θ)α2


− 1
1+ε(θ)
peak
λ2
p∗1 = P1peak , p∗2 = α12
α
−
1
−
P
1
1
ε(θ)α2

C4
|I 0 | = 2

≤

Power Allocation

C1
|I 0 | = 1

λ

AND α1
1

C5
C6
C7

C4
C4
TRRH
, TRRH
1
2
C5
C5
TRRH
, TRRH
1
2
C6
C6
TRRH
, TRRH
1
2
C7
C7
TRRH
, TRRH
1
2

Hence, if the following conditions are satisfied with a, b, c defined in (52):
 c 1/(b−1)
C7 : CX , ab − c < 0 and
(1 − b) +b < 0,
ab
that the average powers required for RRH 1 and RRH 2 are respectively given by



Z u 
mB
mA
P1peak α2
C7
γ m,
− γ m,
f (α2 )dα2 ,
TRRH1 =
Γ (m) αl2
ᾱ1
ᾱ1
C7
TRRH
= J˜1 − J˜2 ,
2

(72)

(73)
(74)

where J˜1 and J˜2 are given by


R αu2   mB 
1
mA
˜
J1 = Γ(m) αl C γ m, ᾱ1 − γ m, ᾱ1
f (α2 ) dα2 ,
2
 



peak R u
α2 1
ᾱ1 P1
mB
mA
J˜2 = mΓ(m)
γ
m
+
1,
−
γ
m
+
1,
f (α2 )dα2 .
ᾱ1
ᾱ1
αl α2
2

C. Discussion of the results
In this subsection, we summarize the results discussed in the above two subsections in Table
I. The average power required for each RRH contributed under condition (23) is given by
λ1

≤

λ2

α1
α2
PRRH
=
1

7
X

λ1

≤

λ2

α1
α2
Ci
TRRH
ε(Ci), PRRH
=
1
2

i=1

7
X

Ci
TRRH
ε(Ci),
2

(75)

i=4

where ε (·) is an indicator function, defined as

 1, if Condition Ci holds,
ε (Ci) =
 0, otherwise.

(76)

Then, the average power required for each RRH is given by
PRRH1 = P

λ1
λ
≤ α2
α1
2
RRH1

+P

λ1
λ
> α2
α1
2
RRH1

, PRRH2 = P

λ1
λ
≤ α2
α1
2
RRH2

+P

λ1
λ
> α2
α1
2
RRH2

,

(77)

21

λ1

>

λ2

λ1

>

λ2

α1
α2
α1
α2
where PRRH
and PRRH
denotes the average transmit power under condition
1
2

RRH 1 and RRH 2, which can be calculated as the condition of

λ1
α1

≤

λ1
α1

>

λ2
α2

for

λ2
.
α2

To provide more insights concerning the joint power allocation results for the case of I = 2,
in Fig. 4 we plot the regions corresponding to different cases of the dual variables. For clarity,
we only consider the case of

λ1
α1

>

λ2
.
α2

Fig. 4-(a) corresponds to the case of λ1 = λ2 = 1. From this figure, we can see that our
proposed joint power allocation algorithm divides the region of

λ1
α1

>

λ2
α2

into two exclusive

regions by the solid lines. If (α1 , α2 ) falls into the region T1 , both RRHs remian silent. On the
other hand, if (α1 , α2 ) falls into the region T2 , Condition C2 is satisfied and only RRH 1 will
transmit with a non-zero power, but less than the peak power.
Fig. 4-(b) corresponds to the case of λ1 = 1/5, λ2 = 1. In this case, the region of

λ1
α1

>

λ2
α2

is divided into five exclusive regions. Similarly, in region T1 , none of the RRHs transmit. In
region T2 and T4 , Condition C3 is satisfied and only RRH 1 is assigned non-zero power for
data transmission, but less than the peak power. If (α1 , α2 ) falls into region T3 , Condition C1 is
satisfied, and RRH 1 will transmit at peak power and RRH 2 still remains silent. However, if
(α1 , α2 ) falls into the region T5 , Condition C5 holds, RRH 1 will transmit at peak power and
RRH 2 is allocated positive power that is lower than the peak power.
Fig. 4-(c) corresponds to the case of λ1 = λ2 = 1/10. In this case, the region is partitioned
into as many as eight regions. Regions T1 -T4 are the same as the case in Fig. 4-(b). If (α1 , α2 )
falls into regions T6 and T8 , Conditions C6 is satisfied, hence RRH 1 will transmit at its peak
power and RRH 2 is assigned non-zero power for its transmission, but its power is less than the
peak power. If (α1 , α2 ) falls into region T7 , Condition C7 is satisfied. Similarly, RRH 1 transmits
at its maximum power and RRH 2 transmits at a non-zero power below its peak power. On the
other hand, if (α1 , α2 ) falls into region T7 , Condition C4 is satisfied and both RRHs will transmit
at their peak power.
It is interesting to find that with the reduction of the dual variables, more RRHs will transmit
at non-zero power or even the peak power. This can be explained as follows. The dual variables
can be regarded as a pricing factor, where a lower dual variable will encourage the RRHs to be
involved in transmission due to the low cost.

22

D2

D2

D2

D2

O2
D
O1 1

O2
D
O1 1

D2

D 2u

.

D2
T8

D 2u

T5

T4
T5

T2

T1

0

.

T1

D1

O1
H T

0

. .D

T3

T2

O1
H T

l
1

a

D2

.D

O2
1 P1peakD1
H T

D

1H T

D 2l

0

D1

u
1

T1

b

Fig. 4. The partitions of the region of

λ1
α1

>

λ2
α2

l
2

O2
D
O1 1

.
.

D2

O2
1  P1peakD1
H T

D1

1
ª
º
1 «§ H T D 2 ·1H T
 P2peakD 2  1»
¨
¸
»
P1peak «© O2 ¹
¬«
¼»

T7
T6

.
O D

T3

T2
1

H T

l
1

.
D

1H T

T4
u
1

D1

c

under different dual variables: (a) λ1 = λ2 = 1; (b) λ1 = 1/5, λ2 = 1;

(c) λ1 = λ2 = 1/10. The system parameters are set as [15]: Time frame of length Tf = 0.2 ms, the system bandwidth
B = 100 KHz, delay QoS exponent θ = (ln2)/20 so that ε(θ) = 1, peak power constraint Pipeak = 1 W, ∀i = 1, 2.

V. P OWER A LLOCATION IN E XTREME C ASES
In this section, we discuss the power allocation for two extreme cases: 1) when θ → 0,
representing no delay requirement; 2) when θ → ∞ and Pipeak → ∞, ∀i ∈ I, representing the
extremely strict zero delay requirement.
A. θ → 0 (No delay requirements)
When θ → 0, the system is delay-tolerant. The EC maximization problem is equivalent to the
ergodic capacity maximization problem, which is given by


P
max Eα log2 1 + i∈I pi (α) αi
{pi (α),∀i}

s.t.

(78)

Eα [pi (α)] ≤ Piavg , ∀i
0 ≤ pi (α) ≤ Pipeak , ∀i.

By using a similar method as in Section III, the optimal transmit power at each RAU is given
by

p∗π(a)


peak


Pπ(a)
,


(


peak
1
=
min Pπ(|I
0 |) , α
π (|I 0 |)





 0,

a < |I 0 | ;
1
ln 2

·

απ(|I 0 |)
λπ(|I 0 |)

−

0 |−1
|IP

a=1

!)
peak
Pπ(a)
απ(a) − 1

, a = |I 0 | ;

(79)

a > |I 0 | ;

where |I 0 | is the largest value of x such that
x−1

X peak
1 απ(x)
·
>1+
Pπ(a) απ(a)
In2 λπ(x)
a=1
and π is a permutation over I so that

λπ(i)
απ(i)

≤

λπ(j)
,
απ(j)

when i < j, i, j ∈ I.

(80)

23

When there is only one RRH, the optimal power allocation reduces to the conventional waterfilling solution that is given by


1
1
p∗1 =
−
λ1 ln 2 α1

P1peak
,

(81)

0

where [x]y0 represents that the value is zero if x < 0, x if 0 < x < y, and y if x > y.
B. θ → ∞ and Pipeak → ∞, ∀i ∈ I (Extremely strict delay requirement)
For this case, the system cannot tolerate any delay, and the EC is equal to the zero-outage
capacity. Hence, the optimal power allocation for each RRH reduces to the channel inversion
associated with a fixed data rate [29] given by
p∗i =

βi
, ∀i ∈ I,
αi

(82)

where βi is specifically selected to satisfy the average power constraints. For the Nakagami-m
channel defined in (1), we have

Eαi

1
αi


=




m
,m
(m−1)ᾱi

 0,

>1

(83)

1 ≥ m.

Then, βi can be calculated as
βi =




(m−1)ᾱi Piavg
,m
m

 0,

>1

(84)

1 ≥ m.

The EC can be obtained as

 T Blog (1 + P β ), m > 1
f
2
i∈I i
EC =
 0,
1 ≥ m.

(85)

Note that when m = 1, the EC is equal to zero, which is consistent with the result in [29], namely
that the zero-outage capacity for Rayleigh fading is equal to zero. However, when m > 1, the
EC increases with m. This coincides with the intuition that with the increase of m, the channel
variations reduce and the channel becomes more stable. As a result, the queue-length variations
are reduced and hence higher delay requirements can be satisfied.

24

VI. S IMULATIONS
In this section, we characterize the performance of our proposed algorithm. We consider a
C-RAN network covering a square area of 2 km × 2 km. The user is located at the center of this
region and the RRHs are independently and uniformly allocated in this area. Unless otherwise
stated, we adopt the same simulation parameters as in [15]: Time frame of length Tf = 0.2 ms;
system bandwidth of B = 100 KHz; average power constraint of Piavg = 0.5 W, ∀i; peak power
constraint of Pipeak = 1 W, ∀i; QoS exponent of θ = 0.05; Nakagami fading parameter m = 2.
The path-loss is modeled as P Li = 148.1+37.6log10 di (dB) [30], where di is the distance between
the ith RRH and the user measured in km. The noise density power is -174 dBm/Hz [30].
We compare our algorithm to the following algorithms:
1) Nearest RRH serving algorithm: In this algorithm, the user is only served by its nearest
RRH, and the optimization method proposed in [15] for point-to-point systems is adopted
to solve the power allocation problem in this setting. This algorithm is proposed to show
the benefits of cooperative transmission in C-RAN.
2) Constant power allocation algorithm: In this algorithm, the transmit power is set to be
equal to the average power constraint for any time slots. Since the peak power limit is
higher than the average power limit, both the average power constraints and peak power
constraints can be satisfied. This approach was assumed to illustrate the benefits of dynamic
power allocation proposed in this paper.
3) Independent power allocation algorithm: In this algorithm, each RRH i independently
optimizes its power allocation without considering the joint channel conditions of the
other RRHs. The optimization problem is given by


1
max − log Eα e−θTf B log2 (1+pi (ν)αi )
pi (ν)
θ

(86a)

s.t. Eα [pi (ν)] ≤ Piavg ,

(86b)

0 ≤ pi (ν) ≤ Pipeak .

(86c)

The optimal power allocation solution is given by (18). This algorithm is invoked for
showing the benefits of jointly optimizing the power allocation according to the joint
channel conditions of all RRHs.
4) Ergodic capacity maximization algorithm: This algorithm aims for maximizing the ergodic
capacity without considering the delay requirement. The optimal power allocation solution

25

1000

500

θ=0.005,a=0.4

0.4
RRH

0.2
0

0

User

-500

RRH

1

θ=0.005,a=0.6

0.4
0.2

RRH

20

40

RRH

1

60

80

2

100

θ=0.005,a=1

1

RRH1

RRH2

0.5

-1000
-1000

-500

0

500

X coordinate (m)

1000

0

RRH

0.2

2

200 400 600 800 1000

0.6

0

θ=0.05,a=0.6

0.6

0.4

2

1

Average Power (W)

Y coordinate (m)

RRH

0.6

0

Average Power (W)

RRH

RRH

1

500

1000

2

1500

θ=0.05,a=1

0.6
0.4
0.2

RRH1

0

20

40

60

RRH2

80

100

θ=0.05,a=1.6

1

RRH1

RRH2

0.5
5

10

15

Number of iterations

20

0

5

10

15

Number of iterations

20

Fig. 5. Random network topology with two RRHs. The user

Fig. 6. Convergence behaviour of the proposed algorithm for

is located at the center of the C-RAN. The coordinates of

the special case of two RRHs under different step parameters.

RRH 1 and RRH 2 are given by [−600, 800] and [900, 946],

Left three subplots correspond to the QoS exponent θ = 0.005,

respectively. The CPNRs are ᾱ1 = 3.89 and ᾱ2 = 1.43.

and the right three correspond to θ = 0.05.

is given by (79).
5) Channel inversion algorithm: This algorithm imposes an extremely strict delay requirement.
The power allocation is the channel inversion given by (82).
In the following, we first consider the case of I = 2, where the average power required for
each RRH can be numerically obtained according to the results of Section IV. Then, we study
the more general case associated with more than two RRHs.
Fig. 5 shows a randomly generated network topology relying on two RRHs. The average
CPNRs related to RRH 1 and RRH 2 are then given by ᾱ1 = 3.89 and ᾱ2 = 1.43. Figs. 6-8
are based on this network topology. In this setting, the user is only served by RRH 1 due to its
shorter distance compared to RRH 2, when using the nearest RRH algorithm.
In Fig. 6, we study the convergence behaviour of the proposed algorithm under different delay
exponents and step-size parameters. The left three subplots correspond to the QoS exponent
θ = 0.005, while the right three correspond to θ = 0.05. It can be observed from this figure that
the convergence speed mainly depends on step-size parameter a. For both cases of θ, a smaller
step-size parameter leads to slower convergence and larger one leads to faster convergence speed.
For example, for the case of θ = 0.005, nearly 700 iterations are required for convergence when
a = 0.4, while only 20 iterations are needed when a = 1. However, since the subgradient method
is not an ascent method, a larger value of a may lead to a fluctuation of the power value. A
small value of a yields a smooth but slow convergence. Hence, the step-size parameter should
be carefully chosen to strike a tradeoff between the convergence speed and the smoothness of

26

100

1.8

1.6

1.4

Pout
delay

Normalized EC (bit/s/Hz)

2

Proposed allo. alg.
Constant power allo. alg.
Nearest RRH serving alg.
Independent power allo. alg.
Ergodic capacity maxi.
Channel inversion

Proposed alg., D

10

10-2

QoS Exponent θ

max

=2ms

Proposed alg., Dmax=1ms

1.2

1

10-5

10-1

-10

Proposed alg., D

10-2

max

=0.5ms

QoS Exponent θ

10-1

Fig. 7. Normalized EC for various algorithms vs QoS exponent

Fig. 8. Delay-outage probability versus QoS exponent θ for

θ.

various values of Dmax for our proposed algorithm.

the power curve. It should be emphasized that the main advantage for the case of two RRHs is
that the Lagrange dual variables can be calculated in an off-line manner by using the results of
Section IV, and stored in the memory.
Let us now study the impact of the delay requirements on the EC performance for various
algorithms in Fig. 7. The normalized EC performance (which is defined as the EC divided
by B and Tf with the unit of ’bits/s/Hz’) is considered. As expected, when the delay-QoS
requirement becomes more stringent, i.e., the value of QoS exponent θ increases, the normalized
EC achieved by all algorithms is reduced. By employing two RRHs for jointly serving the user to
exploit higher spatial degrees of freedom, the proposed algorithm significantly outperforms the
nearest RRH based algorithm, where only one RRH is invoked for transmission. For example,
for θ = 0.1 the normalized EC provided by the former algorithm is roughly 50% higher than
that of the latter algorithm. It is observed that the performance of the latter algorithm is even
inferior to that of the naive constant power allocation algorithm. Since our proposed algorithm
adapts its power allocation to delay-QoS requirement, whilst additionally taking into account
the channel conditions, it achieves a higher normalized EC than the constant power allocation
algorithm, and the performance gain attained is about 20% when θ = 0.1. By optimizing the
power allocation based on the joint channel conditions, the proposed algorithm performs better
than the independent one, where the joint relationship of the channel conditions is ignored. The
performance gain is more prominent when the delay-QoS requirement is very stringent, which
can be up to 18.6% in this example. The proposed algorithm has similar performance to that of
the ergodic capacity maximization algorithm for small θ, while the performance gain provided by

27

1000

Average Power(W)

500
RRH1
User

0

-500

-1000
-1000

0

500

0

1000

X coordinate (m)

20

Fig. 9. Random network topology with five RRHs.

RRH

3

RRH2

RRH

1

40

2

RRH3

RRH

60

RRH

4

RRH

5

80

40
60
θ=0.5,a=1
RRH

20

2

40
60
θ=0.1,a=5
RRH1

1

0

RRH

1

20

1

0.5

RRH5

-500

0

0.5

RRH3
RRH2

RRH

0.5

RRH4

Y coordinate (m)

θ=0.05,a=40

1

RRH4

100
RRH5

80
3

Number of iterations

RRH

4

80

100
RRH

5

100

Fig. 10. Convergence behaviour of the online tracking method.

the proposed algorithm becomes higher for a large θ. For a high θ, the normalized EC achieved
by the ergodic capacity maximization algorithm is even lower than that of the constant power
allocation algorithm. It is seen from Fig.7 that the performance of the channel inversion based
power allocation method is much worse than that of our proposed algorithm, since it aims for
an extremely strict delay budget rather than adaptively adjusting the power allocation according
to the delay requirements.
Fig. 8 shows the delay-outage probability versus QoS exponent θ for various values of Dmax
for our proposed algorithm. The maximum incoming data rate is set to be equal to the EC, i.e. to
µ = EC(θ). The method in [8] is used for calculating the delay-outage probability. As expected,
the delay outage probability increases with the reduction of Dmax . By appropriately choosing
the QoS exponent θ, the delay outage probability can be reduced below 10−9 . For example, for
the case of Dmax = 1 ms, the delay outage probability is as low as 4 × 10−10 when θ = 0.1259,
which satisfies the stringent delay requirement of the ultra-reliable low latency communications
out
(URLLC) in 5G [31]. For a given target Pdelay
, the corresponding value of θ obtained from

Fig. 8 can be used in Fig. 7 to find the achievable EC. For example, when the delay outage
out
probability requirement is Pdelay
= 10−6 for the case of Dmax = 1 ms, the corresponding θ is

given by 0.0794. The resultant normalized EC becomes 1.81 bit/s/Hz according to Fig. 7.
We now consider a more general case in Fig. 9, where five RRHs are randomly located in a
square. The average CPNRs received from these RRHs are then ᾱ1 = 64.3, ᾱ2 = 5.3, ᾱ3 = 63.1,
ᾱ4 = 3.8 and ᾱ5 = 5.1.
Based on Fig. 9, the convergence behaviour of the online tracking method is shown in Fig. 10,
where different QoS requirements are tested. As shown in Fig. 10, the online tracking method

28

converges promptly for all the values of θ considered which is achieved by properly choosing the
step parameter a. These observations demonstrate the efficiency of the online tracking method.
VII. C ONCLUSIONS
We considered joint power allocation for the EC maximization of C-RAN, where the user has to
guarantee a specific delay-QoS requirement to declare successful transmission. Both the per-RRH
average and peak power constraints were considered. We first showed that the EC maximization
problem can be equivalently transformed into a convex optimization problem, which was solved
by using the Lagrange dual decomposition method and by studying the KKT conditions. The
online tracking method was proposed for calculating the average power for each RRH. For the
special case of two RRHs, the expression of average power for each RRH can be obtained in
closed-form. We also provided the power allocation solutions for the extreme cases of θ → ∞
and θ → 0. The simulation results showed that our proposed algorithm converges promptly and
performs much better than the existing algorithms, including the conventional ergodic capacity
maximization method operating without delay requirements. A delay outage probability of 10−9
can be achieved at Dmax = 1 ms by our proposed algorithm through controlling the value of θ,
which is very promising for the applications in future URLLC in 5G.
A PPENDIX A
P ROOF OF CONVEXITY OF P ROBLEM (7)
Since the expectation operator is a linear additive operator, we only have to consider the objective function for each new channel fading generation. For simplicity, we omit the dependency
of pi (ν) on ν and the objective function can be expressed as

−ε(θ)
X
∆
y(P) = 1 +
pi αi
,

(A.1)

i∈I

where P denotes the collection of power allocations. The second partial derivatives of y(P) can
be calculated as

−ε(θ)−2
X
∂y 2 (P)
2
=
α
ε(θ)
(ε(θ)
+
1)
1
+
p
α
i
i
i
i∈I
∂p2i

(A.2)


−ε(θ)−2
X
∂y 2 (P)
= αi αj ε(θ) (ε(θ) + 1) 1 +
pi αi
.
i∈I
∂pi ∂pj

(A.3)

and

Hence, the Hessian matrix of the function y(P) is

X
2
∇ y(P) = ε(θ) (ε(θ) + 1) 1 +

i∈I

pi α i

−ε(θ)−2

ααT ,

29

which is a positive semidefinite matrix. Obviously, the constraints in Problem (7) are linear.
Hence, the proof is completed.
A PPENDIX B
P ROOF OF LEMMA 1
As p∗i > 0, p∗j = 0, from (14b) and (14c) we have µ∗i ≥ 0, δi∗ = 0 and µ∗j = 0, δj∗ ≥ 0,
respectively. Then, from (14a) we have
X
µ∗i = ε(θ)(1 +
p∗m γ̃m )−ε(θ)−1 αi − λi ≥ 0,
m∈I
X
δj∗ = −ε(θ)(1 +
p∗m αm )−ε(θ)−1 αj + λj ≥ 0.
m∈I

(B.1)
(B.2)

Hence, it follows that
X
λj
λi
≤ ε(θ)(1 +
p∗m αm )−ε(θ)−1 ≤ .
m∈I
αi
αj

(B.3)

A PPENDIX C
P ROOF OF LEMMA 2
We first prove the first part of Lemma 2. Assume that there are two RRHs i and j that
0 < p∗i < ppeak
and 0 < p∗j < Pjpeak for i 6= j. Then, according to (14b) and (14c), we have
i
u∗i = 0, δi∗ = 0, u∗j = 0, δj∗ = 0. Substituting them into (14a) yields
X
p∗m αm )−ε(θ)−1 αi + λi = 0,
−ε(θ)(1 +
m∈I
X
−ε(θ)(1 +
p∗m αm )−ε(θ)−1 αj + λj = 0.
m∈I

(C.1)
(C.2)

Hence, we have
λj
λi
= .
αj
αi

(C.3)

Since αi is independent of αj , λi and λj are fixed, it is concluded that the above equality holds
with a zero probability. Thus, there is at most one RRH i with 0 < p∗i < Pipeak .
Let us also assume that there are two users i, k ∈ I 0 with 0 < p∗i < Pipeak and p∗k = Pkpeak .
By using (14b) and (14c), we have u∗i = 0, δi∗ = 0, u∗k ≥ 0, and δk∗ = 0. According to (14a), it
follows that
X
λk
λi
= ε(θ)(1 +
p∗m αm )−ε(θ)−1 ≥
.
m∈I
αi
αk
Hence, we conclude that i = π(|I 0 |).

(C.4)

30

A PPENDIX D
P ROOF OF THEOREM 1
Lemma 2 implies that the optimal solution must be one of the following two cases:
•
•

Case I: p∗π(a) = Papeak , a = 1, · · · , |I 0 |;
Case II: p∗π(a) = Papeak , a = 1, · · · , |I 0 | − 1, p∗π(I 0 ) can be calculated as:


1

−ε(θ)+1
|I 0 |−1
X
λπ(|I 0 |)
1 
peak
Pπ(a)
απ(a) − 1 .
−
p∗π(I 0 ) =
απ(|I 0 |)
ε(θ)απ(|I 0 |)
a=1

Then, we have to prove that |I 0 | is the largest value of x that satisfies the following condition
" x−1
#−ε(θ)−1
X peak
λπ(x)
<
Pπ(b) απ(b) + 1
.
(D.1)
ε(θ)απ(x)
b=1
First, we show that in both Case I and Case II, for any user π(a) ∈ I 0 , the above inequality
holds. For Case I, as p∗π(|I 0 |) = P|Ipeak
0 | , by using (14b) and (14c), it follows that µ|I 0 | ≥ 0 and
δ|I 0 | = 0. Then, substituting them into (14a) yields
−ε(θ)−1
−ε(θ)−1  0
 0
|I |−1
|I |
X
X
λπ(|I 0 |)
peak
peak
.
απ(b) + 1
<
Pπ(b)
απ(b) + 1
≤
Pπ(b)
ε(θ)απ(|I 0 |)
b=1
b=1
Hence, the above inequality holds for a = |I 0 |. From Lemma 1, the left hand side (LHS) of
(D.1) increases as x increases, while the right hand side of (RHS) (D.1) decreases as x increases.
Hence, the inequality (D.1) holds for x = 1, · · · , |I 0 | − 1. As a result, for x = 1, · · · , |I 0 |, the
inequality (D.1) holds. A similar proof can be extended to Case II, the details of which are
omitted.
Finally, we show that for any RRH π(j), j ∈ {|I 0 | + 1, · · · , I}, (D.1) does not hold. Due to
the fact that the LHS of (D.1) increases as x increases while the RHS of (D.1) decreases as x
increases, it is sufficient to prove that for RRH π(|I 0 | + 1), (D.1) does not hold. According to
∗
∗
(14b) and (14c), we have δπ(|I
0 |+1) ≥ 0 and µπ(|I 0 |+1) = 0. Then, from (14a), it follows that
 0
−ε(θ)−1  0
−ε(θ)−1
|I |
|I |
X
X
λπ(|I 0 |+1)
peak
≥
p∗πb απ(b) + 1
≥
Pπ(b)
απ(b) + 1
,
ε(θ)απ(|I 0 |+1)
b=1
b=1

which does not satisfy inequality (D.1). Therefore, the proof is completed.

31

g D1
g D

g D1
P1peak

*
1

g D1*
P1peak

O1
H T

O1 D1L D1* D1U
H T
(b)

D1

D1*
(a)

D1

Fig. 11. Two cases for computing the average power.

A PPENDIX
E
.
P ROOF OF T HEOREM 2
h
iT
For any given λ̃, denote the optimal solution of Problem (9) by P∗λ̃ (ν) = p∗(1,λ̃) (ν), · · · , p∗(I,λ̃) (ν)
when λ = λ̃. Then, we have

X
g(λ̃) = min Eα [(1 +

i∈I

P∈P

pi (ν)αi )

−ε(θ)

]+

X
i∈I

λ̃i (Eα [pi (ν)] −

Piavg )



X
λ̃i (Eα [p∗(i,λ(k) ) (ν)] − Piavg )
p∗(i,λ(k) ) (ν)αi )−ε(θ) ] +
i∈I
i∈I

X 
(k)
λ̃i − λi (Eα [p∗(i,λ(k) ) (ν)] − Piavg )
= g(λ(k) ) +
i∈I

T
= g(λ(k) ) + λ̃ − λ(k) (Eα [P∗λ(k) (ν)] − P̃avg ),
≤ Eα [(1 +

X

(E.1)
(E.2)
(E.3)
(E.4)

where (E.1) uses the definition of g(λ̃) in (9), (E.2) follows since P∗λ(k) (ν) is not the optimal
solution of Problem (E.1) when λ = λ̃.
A PPENDIX F
T HE AVERAGE P OWER FOR THE C ASE OF I = 1
Let us now define the following function
g(α1 ) = 

λ1
ε(θ)

1
1
 1+ε(θ)

ε(θ)
1+ε(θ)

−

α1

1
.
α1

(F.1)

Taking the first-order derivative of g(α1 ) with respect to α1 and setting it to zero yields:
g 0 (α1 ) = − 

λ1
ε(θ)

1
1
 1+ε(θ)

ε(θ)
1+ 1+ε(θ)

α1

By solving (F.2), we can obtain the solution α1∗ =

ε(θ)
1
+ 2 = 0.
1 + ε(θ) α1

λ1
ε(θ)



1+ε(θ)
ε(θ)

that the function g(α1 ) first increases in the region of α1 ∈

h

1+ε(θ)

λ1
, α1∗
ε(θ)

i

(F.2)

. It can be easily verified
and then decreases when

32

α1 ∈ [α1∗ , ∞]. Hence, we can obtain the maximum value of the function g(α1 ) by substituting
α1∗ into (F.1), which yields
g(α1∗ )


1+ε(θ)
ε(θ)
1
=
.
λ1 1 + ε(θ)

(F.3)

It can be readily verified that when α1 → ∞, g(α1 ) → 0. In addition, g



λ1
ε(θ)



= 0. Hence,

depending on the comparative value between g(α1∗ ) and P1peak , two cases should be considered,
when computing the average power as illustrated in Fig. 11. The details are given as follows:
Case I: g(α1∗ ) ≤ P1peak . In this case, the transmit power is always lower than the peak power
as shown in Fig. 11. Hence, the peak power constraints are redundant and can be removed. Thus,
the average power can be expressed as
Z
Eα1 {p1 } =

∞
λ1
ε(θ)

g(α1 )f (α1 )dα1 ,

which can be expanded as

#

 1   ε(θ) "
ε(θ)
m− 1+ε(θ)




ε(θ) 1+ε(θ)
m 1+ε(θ)
λ1 m

−

λ1
ᾱ1
m
m
1

− ᾱλ11ε(θ)
e ᾱ1 ε(θ) + Γ m + 1+ε(θ)
, ᾱλ11ε(θ)

ε(θ)

Γ(m)
m−
(
)

1+ε(θ)

 
m−1 λ1 m


Eα1 {p1 } =
− ᾱ ε(θ)
λ
m
λ1 m
m
− (m−1)Γ(m)ᾱ1 − ᾱ1 ε(θ)
e 1 + Γ m, ᾱ11ε(θ) , when m 6= 1




1
   1+ε(θ)




ε(θ)

− 1+ε(θ)

λ1
λ1
1
1
 ε(θ)
+
,
when m = 1,
ᾱ
,
Ei
−
Γ
1
λ1
1+ε(θ) ᾱ1 ε(θ)
ᾱ1
ᾱ1 ε(θ)
(F.4)
R ∞ −t
R ∞ s−1 −t
where Ei(x) = − −x (e /t) dt is the exponential integral function, while Γ (s, x) = x t e dt
is the upper incomplete gamma function.
Case II: g(α1∗ ) > P1peak . In this case, there must exist two solutions that satisfy g(α1 ) = P1peak
as shown in Fig. 11. Denote these two solutions as α1L and α1U with α1L < α1U . These two solutions
can be numerically obtained by existing algorithms, such as the classic bisection search method.
h
i


λ1
, α1L and α1U , ∞ ,
As seen from Fig. 11, the transmit power is equal to g(α1 ) when α1 ∈ ε(θ)


equal to P1peak when α1 ∈ α1L , α1U . Hence, the average power can be expressed as:
Z αU1
Z ∞
Z αL1
peak
g(α1 )f (α1 )dα1 + P1
f (α1 )dα1 +
g(α1 )f (α1 )dα1 .
(F.5)
Eα1 {p1 } =
λ1
L
U
α
α
1
{z
} | 1
{z
}
| ε(θ)
{z
} |
O1

O2

O3

Note that O1 and O3 can be derived similarly as the average power in Case I, which are omitted
for simplicity. We only provide the expression of O2 as follows:
 



P1peak
m U
m L
γ m, α1 − γ m, α1
,
O2 =
Γ(m)
ᾱ1
ᾱ1

33

where γ (s, x) =

Rx
0

ts−1 e−t dt is the lower incomplete gamma function.

For the case of a single RRH, there is only one dual variable λ1 , which can be obtained by
solving equation Eα1 {p1 } = P1avg with one-dimension bisection search method.
A PPENDIX G
P ROOF OF L EMMA 3
For x ≥ 0, the second-order derivative of h(x) with respect to (w.r.t.) x is given by h00 (x) =
a2 b(b − 1)(1 + ax)b−2 , which is positive since a > 0, b > 1. Hence, h(x) is a convex function,
when x ≥ 0. Combining with the fact that h(0) = 1, there are only three possible curves for
h(x) when x ≥ 0, as shown in Fig. 3. Next, we derive the conditions for each case.
1) Case 1: In this case, the first-order derivative of h(x) is positive when x = 0, i.e., h0 (0) ≥ 0,
which yields the condition for Case 1 in Lemma 3;
2) Case 2: The first-order derivative of h(x) is negative for x = 0, i.e., h0 (0) < 0, which
yields the condition of ab − c < 0. The minimum point of h(x) is achieved by solving
  1

c b−1
1
0
∗
− 1 . Then, according
equation h (x) = 0, and the solution is given by x = a ab
to Fig. 3-b, h(x∗ ) ≥ 0 should hold, which leads to the second part of the conditions in
Case 2 in Lemma 3;
3) Case 3: The proof is similar to that in Case 2, and is omitted for simplicity.
A PPENDIX H
P ROOF OF

λ1
ε(θ)

< α1l

It is easy to verify that h (λ1 /ε (θ)) > 0 when a, b, c are given in (29). Hence, when the
third condition of Lemma 3 given in Condition C1 is satisfied, according to Fig. 3-c, λ1 /ε (θ)
must fall into two regions: 1) 0 < λ1 /ε (θ) < α1l ; 2) λ1 /ε (θ) > α1u . In the following, we prove
that the probability of λ1 /ε (θ) falling into the second region is zero by using the method of
contradiction.
Suppose that λ1 /ε (θ) > α1u holds. Then, according to Fig. 3-c, λ1 /ε (θ) must be larger than
the optimal point, i.e., λ1 /ε (θ) > x∗ , where x∗ is given in Appendix G. By substituting a, b, c
in (29) into this inequality and after some further simplifications, we have
!ε(θ)
P1peak λ1
ε (θ)
.
+1
> peak
ε (θ)
P1 λ1 (1 + ε (θ))

(H.1)

34

Additionally, by inserting a, b, c in (29) into Condition C1 and after some further simplifications,
we have
P1peak λ1


<

ε (θ)
1 + ε (θ)

1+ε(θ)
< 1.

(H.2)

By substituting (H.2) into the right hand side of (H.1) and with some simple further operations,
one obtains λ1 P1peak > 1, which contradicts with (H.2) that λ1 P1peak < 1. Hence, the assumption
that λ1 /ε (θ) > α1u cannot hold, which completes the proof.
A PPENDIX I
C2
D ERIVATIONS OF TRRH
1

WHEN

m IS A POSITIVE INTEGER

C2
The closed-form expression of TRRH
for the case of m = 1 can be easily obtained by inserting
1

m = 1 into (39). In the following, we only focus on the case when m ≥ 2.
When m ≥ 2, function γ (m, x) can be expanded as [8.352.1, Page 899, [32]]
!
m−1
X xl
.
γ (m, x) = (m − 1)! 1 − e−x
l!
l=0

(I.1)

By plugging the above expression with x = W α1 into (39), we obtain the average power for
RRH 1 as follows:
C2
TRRH
1

Z
= V
U
|

Z

∞

m

− α
α1V +m−2 e ᾱ1 1 dα1

+Z

U

{z

}

J1

m−1
X
l=0

Wl
l!

Z

∞

α1l+m−2 e−Y α1 dα1

U

|

{z

}

J2

Z
Z ∞
m−1
Z X W l ∞ l+V +m−2 −Y α1
−mα
α1
e
− V
dα1 − Z
α1m−2 e ᾱ1 1 dα1 .
U l=0 l! U
U
{z
}
|
{z
} |
J
4

J3

With some simple variable substitutions, the closed-form expressions of J1 , J2 , J3 and J4 can
be easily calculated as in (43) and (44), respectively.

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