# Sum-Rate Maximization of Uplink Rate Splitting Multiple Access (RSMA)
Communication

Zhaohui Yang, Mingzhe Chen, Walid Saad, , Wei Xu, , and Mohammad Shikh-Bahaei,
A preliminary version of this work was submitted to IEEE Globecom 2019 [1].Z.
Yang and M. Shikh-Bahaei are with the Centre for Telecommunications Research,
Department of Informatics, King’s College London, WC2B 4BG, UK, Emails:
<EMAIL_ADDRESS><EMAIL_ADDRESS>Chen is with the Chinese
University of Hong Kong, Shenzhen, China, and also with the Electrical
Engineering Department of Princeton University, USA, Email:
<EMAIL_ADDRESS>Saad is with Wireless@VT, Bradley Department of
Electrical and Computer Engineering, Virginia Tech, Blacksburg, VA, USA,
Email<EMAIL_ADDRESS>W. Xu is with the National Mobile Communications
Research Laboratory, Southeast University, Nanjing 210096, China, Email:
<EMAIL_ADDRESS>
###### Abstract

In this paper, the problem of maximizing the wireless users’ sum-rate for
uplink rate splitting multiple access (RSMA) communications is studied. In the
considered model, each user transmits a superposition of two messages to a
base station (BS) with separate transmit power and the BS uses a successive
decoding technique to decode the received messages. To maximize each user’s
transmission rate, the users must adjust their transmit power and the BS must
determine the decoding order of the messages transmitted from the users to the
BS. This problem is formulated as a sum-rate maximization problem with
proportional rate constraints by adjusting the users’ transmit power and the
BS’s decoding order. However, since the decoding order variable in the
optimization problem is discrete, the original maximization problem with
transmit power and decoding order variables can be transformed into a problem
with only the rate splitting variable. Then, the optimal rate splitting of
each user is determined. Given the optimal rate splitting of each user and a
decoding order, the optimal transmit power of each user is calculated. Next,
the optimal decoding order is determined by an exhaustive search method. To
further reduce the complexity of the optimization algorithm used for sum-rate
maximization in RSMA, a user pairing based algorithm is introduced, which
enables two users to use RSMA in each pair and also enables the users in
different pairs to be allocated with orthogonal frequency. For comparisons,
the optimal sum-rate maximizing solutions with proportional rate constraints
are obtained in closed form for non-orthogonal multiple access (NOMA),
frequency division multiple access (FDMA), and time division multiple access
(TDMA). Simulation results show that RSMA can achieve up to 10.0%, 22.2%, and
83.7% gains in terms of sum-rate compared to NOMA, FDMA, and TDMA.

###### Index Terms:

Rate splitting multiple access (RSMA), decoding order, power management,
resource allocation.

##  I Introduction

Driven by the rapid development of advanced multimedia applications, next-
generation wireless networks [2] must support high spectral efficiency and
massive connectivity. In consequence, rate splitting multiple access (RSMA)
has been recently proposed as an effective approach to provide more general
and robust transmission framework compared to non-orthogonal multiple access
(NOMA) [3, 4, 5, 6, 7, 8] and space-division multiple access (SDMA). However,
implementing RSMA in wireless networks also faces several challenges [9] such
as decoding order design and resource management for message transmission.

Recently, a number of existing works such as in [9, 10, 11, 12, 13, 14, 15,
16, 17, 18, 19, 20] have studied a number of problems related to the
implementation of RSMA in wireless networks. In [9], the authors outlined the
opportunities and challenges of using RSMA for multiple input multiple output
(MIMO) based wireless networks. The authors in [10] developed a rate splitting
algorithm for the maximization of users’ data rates. The authors in [12]
developed an algorithm to optimize the users’ sum-rate in downlink RSMA under
imperfect channel state information (CSI). The authors in [12] optimized
users’ sum-rate in downlink multi-user multiple input single output (MISO)
systems under imperfect CSI. The work in [13] showed that RSMA can achieve
better performance than NOMA and SDMA. In [14], the application of linearly-
precoded rate splitting is studied for multiple input single output (MISO)
simultaneous wireless information and power transfer (SWIPT) broadcast channel
systems. The authors in [15] investigated the rate splitting-based robust
transceiver design problem in a multi-antenna interference channel with SWIPT
under the norm-bounded errors of CSI. The work in [16] developed a
transmission scheme that combines rate splitting, common message decoding,
clustering and coordinated beamforming so as to maximize the weighted sum-rate
of users. In [17], the energy efficiency of the RSMA and NOMA schemes is
studied in a downlink millimeter wave transmission scenario. The authors in
[18] used RSMA for a downlink multiuser MISO system with bounded errors of
CIST. The data rate of using RSMA for two-receiver MISO broadcast channel with
finite rate feedback is studied in [19]. Our prior work in [20] investigated
the power management and rate splitting scheme to maximize the sum-rate of the
users. However, most of the existing works such as in [9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20] studied the use of RSMA for the downlink rather than
in the uplink. In fact, using RSMA for uplink data transmission can
theoretically achieve the optimal rate region [21]. Moreover, none of the
existing works in [9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20] jointly
considered the optimization of power management and message decoding order for
uplink RSMA. In practical RSMA deployments, the message decoding order will
affect the transmission rate of the uplink users and, thus, it must be
optimized.

The main contribution of this paper is a novel framework for optimizing power
allocation and message decoding for uplink RSMA transmissions. Our key
contributions include:

  * •

We consider the uplink of a wireless network that uses RSMA, in which each
user transmits a superposition of two messages with different power levels and
the base station (BS) uses a successive interference cancellation (SIC)
technique to decode the received messages. The power allocation and decoding
order problem is formulated as an optimization problem whose goal is to
maximize the sum-rate of all users under proportional rate constraints.

  * •

The non-convex sum-rate maximization problem with discrete decoding variable
and transmit power variable is first transformed into an equivalent problem
with only the rate splitting variable. Then, the optimal solution of the rate
splitting is obtained in closed form. Based on the optimal rate splitting of
each user, the optimal transmit power can be derived under a given decoding
order. Finally, the optimal decoding order is determined by exhaustive search.
To reduce the computational complexity, a low-complexity RSMA scheme based on
user pairing is proposed to show near sum-rate performance of RSMA without
user pairing.

  * •

We provide closed-form optimal solutions for sum-rate maximization problems in
uplink NOMA, frequency division multiple access (FDMA), and time division
multiple access (TDMA). Simulation results show that RSMA achieves better
performance than NOMA, FDMA, and TDMA in terms of sum-rate.

The rest of this paper is organized as follows. The system model and problem
formulation are described in Section II. The optimal solution is presented in
Section III. Section IV presents a low-complexity sum-rate maximization
scheme. The optimal solutions of sum-rate maximization for NOMA, FDMA and TDMA
are provided in Section V. Simulation results are analyzed in Section VI.
Conclusions are drawn in Section VII.

##  II System Model and Problem Formulation

Consider a single cell uplink network with one BS serving a set $\mathcal{K}$
of $K$ users using RSMA. In uplink RSMA, each user first transmits a
superposition code of two messages to the BS. Then, the BS uses a SIC
technique to decode the messages of all users [21].

The transmitted message $s_{k}$ of user $k\\!\in\\!\mathcal{K}$ is given by:

$s_{k}=\sum_{j=1}^{2}\sqrt{p_{kj}}s_{kj},\quad\forall k\in\mathcal{K},$ (1)

where $p_{kj}$ is the transmit power of message $s_{kj}$ from user $k$.

The total received message $s_{0}$ at the BS can be given by:

$s_{0}=\sum_{k=1}^{K}\sqrt{h_{k}}s_{k}+n=\sum_{k=1}^{K}\sum_{j=1}^{2}\sqrt{h_{k}p_{kj}}s_{kj}+n,$
(2)

where $h_{k}$ is the channel gain between user $k$ and the BS and $n$ is the
additive white Gaussian noise. Each user $k$ has a maximum transmission power
limit $P_{k}$, i.e., $\sum_{j=1}^{2}p_{kj}\leq P_{k}$.

To decode all messages $s_{kj}$ from the received message $s_{0}$, the BS will
use SIC. The decoding order at the BS is denoted by a permutation $\bm{\pi}$.
The permutation $\bm{\pi}$ belongs to set $\Pi$ defined as the set of all
possible decoding orders of all $2K$ messages from $K$ users. The decoding
order of message $s_{kj}$ from user $k$ is ${\pi}_{kj}$. The achievable rate
of decoding message $s_{kj}$ is:

$r_{kj}=B\log_{2}\left(1+\frac{h_{k}p_{kj}}{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}>{\pi}_{kj}\\}}h_{l}p_{lm}+\sigma^{2}B}\right),$
(3)

where $B$ is the bandwidth of the BS, $\sigma^{2}$ is the power spectral
density of the Gaussian noise. The set
$\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}>{\pi}_{kj}\\}$ in (3)
represents the messages $s_{lm}$ that are decoded after message $s_{kj}$.

Since the transmitted message of user $k$ includes messages $s_{k1}$ and
$s_{k2}$, the achievable rate of user $k$ is given by:

$r_{k}=\sum_{j=1}^{2}r_{kj}.$ (4)

Our objective is to maximize the sum-rate of all users with proportional rate
constraints. Mathematically, the sum-rate maximization problem can be formally
posed as follows:

$\displaystyle\mathop{\max}_{\bm{\pi},\bm{p}}\quad$
$\displaystyle\sum_{k=1}^{K}r_{k},$ (5) s.t. $\displaystyle
r_{1}:r_{2}:\cdots:r_{K}=D_{1}:D_{2}:\cdots D_{K},$ (5a)
$\displaystyle\sum_{j=1}^{2}p_{kj}\leq P_{k},\quad\forall k\in\mathcal{K},$
(5b) $\displaystyle\bm{\pi}\in\Pi,p_{kj}\geq 0,\quad\forall
k\in\mathcal{K},j\in\mathcal{J},$ (5c)

where $\bm{p}\\!=\\![p_{11},p_{12},\cdots,p_{K1},p_{K2}]^{T}$, $r_{k}$ is
defined in (4), and $\mathcal{J}=\\{1,2\\}$. $D_{1},\cdots,D_{K}$ is a set of
predetermined nonnegative values that are used to ensure proportional fairness
among users. The fairness index is defined as

$\frac{\left(\sum_{k=1}^{K}D_{k}\right)^{2}}{K\sum_{k=1}^{K}D_{k}^{2}}$ (6)

with the maximum value of 1 to be the greatest fairness case in which all
users would achieve the same data rate [22]. With proper unitization, we set

$\sum_{k=1}^{K}D_{k}=1.$ (7)

Although it was stated in [21] that RSMA can reach the optimal rate region, no
practical algorithm was proposed to compute the decoding order and power
allocation. It is therefore necessary to quantify the uplink performance gains
that RSMA can obtain compared to conventional multiple access schemes.

Due to the non-linear equality constraint (5a) and discrete variable
$\bm{\pi}$, problem (5) is a non-convex mixed integer problem. Hence, it is
generally hard to solve problem (5). Despite the non-convexity and discrete
variable, we will next develop a novel algorithm to obtain the globally
optimal solution to problem (5).

##  III Optimal Power Allocation and Decoding Order

In this section, an effective algorithm is proposed to obtain the optimal
power allocation and decoding order of sum-rate maximization problem (5).

###  III-A Optimal Sum-Rate Maximization

Let $\tau$ be the sum-rate of all $K$ users. Given this new variable $\tau$,
problem (5) can be rewritten as:

$\displaystyle\mathop{\max}_{\tau,\bm{\pi},\bm{p}}\quad$ $\displaystyle\tau,$
(8) s.t. $\displaystyle r_{k}={D_{k}}\tau,\quad\forall k\in\mathcal{K},$ (8a)
$\displaystyle\sum_{j=1}^{2}p_{kj}\leq P_{k},\quad\forall k\in\mathcal{K},$
(8b) $\displaystyle\bm{\pi}\in\Pi,p_{kj}\geq 0,\quad\forall
k\in\mathcal{K},j\in\mathcal{J},$ (8c)

where $\tau$ is the sum-rate of all users since
$\tau=\sum_{k=1}^{K}D_{k}\tau=\sum_{k=1}^{K}r_{k}$ according to (7) and (8a).

Problem (8) is challenging to solve due to the decoding order variable
$\bm{\pi}$ with discrete value space. To handle this difficulty, we provide
the following lemma, which can be used for transforming problem (8) into an
equivalent problem without decoding order variable $\bm{\pi}$.

###### Lemma 1

In RSMA, under a proper decoding power order $\bm{\pi}$ and splitting power
allocation $\bm{p}$, the optimal rate region can be fully achieved, i.e.,

$\sum_{k\in\mathcal{K}^{\prime}}r_{k}\leq
B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\right),\quad\forall\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset,$
(9)

where $\emptyset$ is an empty set and
$\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset$ means that
$\mathcal{K}^{\prime}$ is a non-empty subset of $\mathcal{K}$.

Lemma 1 follows directly from [21, Theorem 1]. Based on Lemma 1, we can use
the rate variable to replace the power and decoding variables. In consequence,
problem (8) can be equivalently transformed to

$\displaystyle\mathop{\max}_{\tau,\bm{r}}\quad$ $\displaystyle\tau,$ (10) s.t.
$\displaystyle{r_{k}}={D_{k}}\tau,\quad\forall k\in\mathcal{K},$ (10a)
$\displaystyle\sum_{k\in\mathcal{K}^{\prime}}r_{k}\leq
B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\right),\quad\forall\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset,$
(10b)

where $\bm{r}=[r_{1},r_{2},\cdots,r_{K}]^{T}$. In problem (10), the dimension
of the variable is smaller than that in problem (8). Moreover, the discrete
decoding order variable is replaced by rate variable in problem (10).
Regarding the optimal solution of problem (10), we have the following lemma.

###### Lemma 2

For the optimal solution $(\tau^{*},\bm{r}^{*})$ of problem (10), there exists
at least one $\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset$ such
that
$\sum_{k\in\mathcal{K}^{\prime}}r_{k}^{*}=B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\right)$.

Proof: See Appendix A. $\Box$

###### Theorem 1

The optimal solution of problem (10) is

$\tau^{*}=\min_{\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset}\frac{B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\right)}{\sum_{k\in\mathcal{K}^{\prime}}D_{k}},$
(11)

and

$r_{k}^{*}={D_{k}}{\min_{\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset}\frac{B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\right)}{\sum_{k\in\mathcal{K}^{\prime}}D_{k}}},\quad\forall
k\in\mathcal{K}.$ (12)

Proof: According to Lemma 2, there exists at least one
$\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset$ such that

$\sum_{k\in\mathcal{K}^{\prime}}r_{k}^{*}=B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\right).$
(13)

To ensure the feasibility of (10b), the optimal $\tau^{*}$ is given by (11).
Then, according to (10a), the optimal $r_{k}^{*}$ is determined as in (12).
$\blacksquare$

From (11), one can directly obtain the optimal sum-rate of problem (10) in
closed form, which can be helpful in characterizing the rate performance of
RSMA.

Having obtained the optimal solution $(\tau^{*},\bm{r}^{*})$ of problem (10),
we still need to calculate the optimal $(\bm{\pi}^{*},\bm{p}^{*})$ of the
original problem (8). Next, we introduce a new algorithm to obtain the optimal
$(\bm{\pi}^{*},\bm{p}^{*})$ of problem (8).

Substituting the optimal solution $(\tau^{*},\bm{r}^{*})$ of problem (10) into
problem (8), we can obtain the following feasibility problem:

find $\displaystyle\bm{\pi},\bm{p},$ (14) s.t.
$\displaystyle\sum_{j=1}^{2}\\!B\log_{2}\\!\left(\\!\\!1\\!+\\!\frac{h_{k}p_{kj}}{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}\\!>\\!{\pi}_{kj}\\}}\\!h_{l}p_{lm}\\!+\\!\sigma^{2}B}\\!\\!\right)\\!\\!=r_{k}^{*},\quad\forall
k\in\mathcal{K},$ (14a) $\displaystyle\sum_{j=1}^{2}p_{kj}\leq
P_{k},\quad\forall k\in\mathcal{K},$ (14b)
$\displaystyle\bm{\pi}\in\Pi,p_{kj}\geq 0,\quad\forall
k\in\mathcal{K},j\in\mathcal{J}.$ (14c)

Due to the decoding order constraint (14c), it is challenging to find the
optimal solution of problem (14). To solve this problem, we first fix the
decoding order $\bm{\pi}$ to obtain the power allocation and then exhaustively
search $\bm{\pi}$. Given decoding order $\bm{\pi}$, problem (14) can be
simplified as:

find $\displaystyle\bm{p},$ (15) s.t.
$\displaystyle\sum_{j=1}^{2}\\!B\log_{2}\\!\left(\\!\\!1\\!+\\!\frac{h_{k}p_{kj}}{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}\\!>\\!{\pi}_{kj}\\}}\\!h_{l}p_{lm}\\!+\\!\sigma^{2}B}\\!\\!\right)\\!\\!\geq
r_{k}^{*},\quad\forall k\in\mathcal{K},$ (15a)
$\displaystyle\sum_{j=1}^{2}p_{kj}\leq P_{k},\quad\forall k\in\mathcal{K},$
(15b) $\displaystyle p_{kj}\geq 0,\quad\forall
k\in\mathcal{K},j\in\mathcal{J}.$ (15c)

Note that the equality in (14a) is replaced by the inequality in (15a). The
reason is that any feasible solution to problem (14) is also feasible to
problem (15). Meanwhile, for a feasible solution to problem (15), we can
always construct a feasible solution to problem (14).

To verify the feasibility of problem (15), we can construct the following
problem by introducing a new variable $\alpha$:

$\displaystyle\max_{\alpha,\bm{p}}\quad$ $\displaystyle\alpha,$ (16) s.t.
$\displaystyle\sum_{j=1}^{2}\\!B\log_{2}\\!\left(\\!\\!1\\!+\\!\frac{h_{k}p_{kj}}{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}\\!>\\!{\pi}_{kj}\\}}\\!h_{l}p_{lm}\\!+\\!\sigma^{2}B}\\!\\!\right)\\!\\!\geq\alpha
r_{k}^{*},\quad\forall k\in\mathcal{K},$ (16a)
$\displaystyle\sum_{j=1}^{2}p_{kj}\leq P_{k},\quad\forall k\in\mathcal{K},$
(16b) $\displaystyle p_{kj},\alpha\geq 0,\quad\forall
k\in\mathcal{K},j\in\mathcal{J}.$ (16c)

To show the equivalence between problems (15) and (16), we provide the
following lemma.

###### Proposition 1

Problem (15) is feasible if and only if the optimal objective value
$\alpha^{*}$ of problem (16) is equal to or larger than 1.

Proof: On one side, if $\bm{p}$ is a feasible solution of problem (15), we can
show that $(\alpha=1,\bm{p})$ is a feasible solution of problem (16), which
indicates that the optimal objective value of problem (16) should be equal to
or larger than 1.

On the other side, if the optimal solution $(\alpha^{*},\bm{p}^{*})$ of
problem (16) satisfies $\alpha^{*}\geq 1$, we can show that $\bm{p}^{*}$ is a
feasible solution of problem (15). $\Box$

Problem (16) is non-convex due to constraints (16a). To handle the non-
convexity of (16), we adopt the difference of two convex function (DC) method,
using which a non-convex problem can be solved suboptimally by converting a
non-convex problem into convex subproblems. In order to obtain a near globally
optimal solution of problem (16), we can try multiple initial points
$(\alpha,\bm{p})$, which can lead to multiple locally optimal solutions. Thus,
a near globally optimal solution can be obtained by choosing the locally
optimal solution with the highest objective value among all locally optimal
solutions. To construct an initial feasible point, we first arbitrarily
generate $\bm{p}$ that satisfies linear constraints (16b)-(16c), and then we
set:

$\alpha=\min_{k\in\mathcal{K}}\frac{\sum_{j=1}^{2}B\log_{2}\left(1+\frac{h_{k}p_{kj}}{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}>{\pi}_{kj}\\}}h_{l}p_{lm}+\sigma^{2}B}\right)}{r_{k}^{*}}.$
(17)

By using the DC method, the left hand side of (16a) satisfies:

$\displaystyle\begin{aligned}
\>\>&\sum_{j=1}^{2}B\log_{2}\left(1+\frac{h_{k}p_{kj}}{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}>{\pi}_{kj}\\}}h_{l}p_{lm}+\sigma^{2}B}\right)\\\
=&\sum_{j=1}^{2}B\log_{2}\\!\left(\\!{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}\geq{\pi}_{kj}\\}}h_{l}p_{lm}\\!+\\!\sigma^{2}B}\right)-\\!\sum_{j=1}^{2}B\log_{2}\\!\left(\\!{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}>{\pi}_{kj}\\}}h_{l}p_{lm}\\!+\\!\sigma^{2}B}\\!\right)\end{aligned}$
$\displaystyle\begin{aligned}
\geq&\sum_{j=1}^{2}B\log_{2}\\!\left(\\!{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}\geq{\pi}_{kj}\\}}h_{l}p_{lm}\\!+\\!\sigma^{2}B}\right)-\\!\sum_{j=1}^{2}B\log_{2}\\!\left(\\!{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}>{\pi}_{kj}\\}}h_{l}p_{lm}^{(n)}\\!+\\!\sigma^{2}B}\\!\right)\\\
&-\sum_{j=1}^{2}B\frac{\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}>{\pi}_{kj}\\}}h_{l}(p_{lm}-p_{lm}^{(n)})}{(\ln
2)\sum_{\\{(l\in\mathcal{K},m\in\mathcal{J})|{\pi}_{lm}>{\pi}_{kj}\\}}h_{l}p_{lm}^{(n)}+\sigma^{2}B}\\\
\triangleq&r_{k,\text{lb}}(\bm{p},\bm{p}^{(n)}),\end{aligned}$

where $p_{lm}^{(n)}$ represents the value of $p_{lm}$ at iteration $n$, and
the inequality follows from the fact that $\log_{2}(x)$ is a concave function
and a concave function is always no greater than its first-order
approximation. By substituting the left term of constraints (16a) with the
concave function $r_{k,\text{lb}}(\bm{p},\bm{p}^{(n)})$, problem (16) becomes
convex, and can be effectively solved by the interior point method [23].

Algorithm 1 Optimal Sum-Rate Maximization for RSMA

1: Obtain the optimal solution $(\tau^{*},\bm{r}^{*})$ of problem (10)
according to Theorem 1.

2: for $\bm{\pi}\in\Pi$ do

3: for $1:1:N$ do

4: Arbitrarily generate a feasible solution $(\alpha^{(0)},\bm{p}^{(0)})$ of
problem (16), and set $n=0$.

5: repeat

6: Obtain the optimal solution $(\alpha^{(n+1)},\bm{p}^{(n+1)})$ of convex
problem (16) by replacing the left term of constraints (16a) with
$r_{k,\text{lb}}(\bm{p},\bm{p}^{(n)})$.

7: Set $n=n+1$.

8: until the objective value (16a) converges.

9: end for

10: Obtain the optimal solution ($\alpha^{*},\bm{p}^{*}$) of problem (16) with
the highest objective value.

11: If $\alpha^{*}\geq 1$, break and jump to step 13.

12: end for

13: Obtain the optimal decoding order $\bm{\pi}^{*}=\bm{\pi}$ and power
allocation $\bm{p}^{*}$ of problem (14).

The optimal sum-rate maximization algorithm for RSMA is provided in Algorithm
1, where $N$ is the number of initial points to obtain a near globally optimal
solution of non-convex problem (16).

###  III-B Complexity Analysis

In Algorithm 1, the major complexity lies in solving problem (10) and problem
(14). To solve (10), from Theorem 1, the complexity is $\mathcal{O}(2^{K}-1)$
since the set $\mathcal{K}$ has $2^{K}-1$ non-empty subsets. According to
steps 2-12, a near globally optimal solution of problem (14) is obtained via
solving a series of convex problems with different initial points and decoding
order strategies. Considering that the dimension of variables in problem (16)
is $1+2K$, the complexity of solving convex problem in step 6 by using the
standard interior point method is $\mathcal{O}(K^{3})$ [23, Pages 487, 569].
Since the network consists of $K$ users and each user transmits a
superposition two messages (there are $2K$ messages in total), the decoding
order set $\Pi$ consists of $(2K)!/2^{K}$ elements. Given $N$ initial points,
the total complexity of solving problem (14) is
$\mathcal{O}(NK^{3}(2K)!/2^{K})$. As a result, the total complexity of
Algorithm 1 is $\mathcal{O}(2^{K}+NK^{3}(2K)!/2^{K})$.

In practice, we consider small $K$ to reduce the SIC complexity, the
computational complexity of Algorithm 1 can be practical. To deal with a large
number of users, the users can be classified into different groups with small
number of users in each group. The users in different groups occupy different
frequency bands and users in the same group are allocated to the same
frequency band using RSMA [24, 25]. For the special case with $K=2$, we can
show that the optimal optimal decoding order and power allocation of problem
(14) can be obtained in closed form.

###  III-C RSMA with Two Users

Based on Lemma 1, the rate region of RSMA with two users can be expressed by:

$\\{(r_{1},r_{2})|0\leq r_{1}\leq R_{1},0\leq r_{2}\leq R_{2},r_{1}+r_{2}\leq
R_{\max}\\},$ (18)

where

$R_{1}=B\log_{2}\left(1+\frac{h_{1}P_{1}}{\sigma^{2}B}\right),R_{2}=B\log_{2}\left(1+\frac{h_{2}P_{2}}{\sigma^{2}B}\right),R_{\max}=B\log_{2}\left(1+\frac{h_{1}P_{1}+h_{2}P_{2}}{\sigma^{2}B}\right).$
(19)

According to Algorithm 1, the computational complexity needed to obtain the
boundary point (as shown in Lemma 2 the optimal point to minimize always lies
in the boundary point) of the rate region for RSMA is high. In the following,
we introduce a low-complexity method to obtain all boundary points of the rate
region in two-user RSMA.

In two-user RSMA, only one user needs to transmit a superposition code of two
messages and the other user transmits one message. Without loss of generality,
user 1 only transmits one message $s_{11}$, i.e., the transmit power for
message $s_{12}$ is always 0.

###### Lemma 3

For two-user RSMA, the optimal decoding order is $s_{21}$, $s_{11}$ and
$s_{22}$. For the boundary rate $(r_{1},r_{2})$, we consider the following
three cases.

Case (1) $r_{1}=R_{1}$, $0\leq r_{2}\leq R_{\max}-R_{1}$: the optimal power
allocation is

$p_{11}=P_{1},p_{12}=0,p_{21}=\frac{1}{h_{2}}\left(2^{\frac{r_{2}}{B}}-1\right)(h_{1}P_{1}+\sigma^{2}B),p_{22}=0.$
(20)

Case (2) $r_{2}=R_{2}$, $0\leq r_{1}\leq R_{\max}-R_{2}$: the optimal power
allocation is

$p_{11}=\frac{1}{h_{1}}{\left(2^{\frac{r_{1}}{B}}-1\right)(h_{2}P_{2}+\sigma^{2}B)},p_{12}=0,p_{21}=0,p_{22}=P_{2}.$
(21)

Case (3) $r_{1}+r_{2}=R_{\max}$, $0\leq r_{1}\leq R_{1}$, $0\leq r_{2}\leq
R_{2}$: the optimal power allocation is

$p_{11}=P_{1},p_{12}=0,p_{21}=P_{2}-\frac{h_{1}P_{1}}{h_{2}\left(2^{\frac{r_{1}}{B}}-1\right)}+\frac{\sigma^{2}B}{h_{2}},p_{22}=\frac{h_{1}P_{1}}{h_{2}\left(2^{\frac{r_{1}}{B}}-1\right)}-\frac{\sigma^{2}B}{h_{2}}.$
(22)

Proof: See Appendix B. $\Box$

##  IV Low-Complexity Sum-Rate Maximization

According to Section III-B, the computational complexity of sum-rate
maximization for RSMA is extremely high. In this section, we propose a low-
complexity scheme for RSMA, where users are classified into different
pairs111In this paper, we assume that the user pairing is given, which can be
obtained according to matching theory [24] or the order of channel gains [25].
and each pair consists of two users. RSMA is used in each pair and different
pairs are allocated with different frequency bands. Assume that $K$ users are
classified into $M$ pairs, i.e., $K=2M$. The set of all pairs is denoted by
$\mathcal{M}$.

For pair $m$, the allocated fraction of bandwidth is denoted by $f_{m}$. Let
$c_{mj}$ denote the data rate of user $j$ in pair $m$. According to Lemma 1,
we have:

$c_{mj}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{mj}P_{mj}}{\sigma^{2}Bf_{m}}\right),\quad\forall
m\in\mathcal{M},j\in\mathcal{J},$ (23) $c_{m1}+c_{m2}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{m1}P_{m1}+h_{m2}P_{m2}}{\sigma^{2}Bf_{m}}\right),\quad\forall
m\in\mathcal{M},$ (24)

where $h_{mj}$ denotes the channel gain between user $j$ in pair $m$ and the
BS, and $P_{mj}$ is the maximal transmission power of user $j$ in pair $m$.

Similar to (5), the sum-rate maximization problem for RSMA with user pairing
can be formulated as:

$\displaystyle\mathop{\max}_{\bm{f},\bm{c}}\quad$
$\displaystyle\sum_{m=1}^{M}\sum_{j=1}^{2}c_{mj},$ (25) s.t. $\displaystyle
c_{11}:c_{12}:\cdots:c_{M2}=D_{11}:D_{12}:\cdots:D_{M2}$ (25a)
$\displaystyle\sum_{m=1}^{M}f_{m}=1,$ (25b) $\displaystyle c_{mj}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{mj}P_{mj}}{\sigma^{2}Bf_{m}}\right),\quad\forall
m\in\mathcal{M},j\in\mathcal{J},$ (25c) $\displaystyle c_{m1}+c_{m2}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{m1}P_{m1}+h_{m2}P_{m2}}{\sigma^{2}Bf_{m}}\right),\quad\forall
m\in\mathcal{M},$ (25d) $\displaystyle f_{m},c_{m1},c_{m2}\geq 0,\quad\forall
m\in\mathcal{M},$ (25e)

where $\bm{f}=[f_{1},f_{2},\cdots,f_{M}]^{T}$,
$\bm{c}=[c_{11},c_{12},\cdots,c_{M1},c_{M2}]^{T}$, and
$D_{11},D_{12},\cdots,D_{M1},D_{M2}$ is a set of predetermined nonnegative
values that are used to ensure proportional fairness among users with
$\sum_{m=1}^{M}\sum_{j=1}^{2}D_{mj}=1$.

Introducing a new variable $\tau$, problem (25) can be rewritten as:

$\displaystyle\mathop{\max}_{\tau,\bm{f},\bm{c}}\quad$ $\displaystyle\tau,$
(26) s.t. $\displaystyle c_{mj}={D_{mj}}\tau,\quad\forall
m\in\mathcal{M},j\in\mathcal{J},$ (26a) $\displaystyle\sum_{m=1}^{M}f_{m}=1,$
(26b) $\displaystyle c_{mj}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{mj}P_{mj}}{\sigma^{2}Bf_{m}}\right),\quad\forall
m\in\mathcal{M},j\in\mathcal{J},$ (26c) $\displaystyle c_{m1}+c_{m2}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{m1}P_{m1}+h_{m2}P_{m2}}{\sigma^{2}Bf_{m}}\right),\quad\forall
m\in\mathcal{M},$ (26d) $\displaystyle f_{m},c_{m1},c_{m2}\geq 0,\quad\forall
m\in\mathcal{M}.$ (26e)

To solve problem (26), we can use the bisection method to obtain the optimal
solution. Denote the optimal objective value of problem (26) by $\tau^{*}$.

Figure 1: An illustration of the bisection method.

We can conclude that problem (26) is always feasible with $\tau<\tau^{*}$ and
infeasible with $\tau>\tau^{*}$. This motivates us to use the bisection method
to find the optimal $\tau^{*}$, as shown in Fig. 1, where $\tau^{(n)}$ is the
value of $\tau$ in the $n$-th iteration and $[\tau_{\min},\tau_{\max}]$ is the
initial value interval of $\tau$. To show the feasibility of problem (26) for
each given $\tau$, we solve a feasibility problem with constraints
(26a)-(26e). With given $\tau$, the feasibility problem of (26) becomes

find $\displaystyle{\bm{f},\bm{c}},$ (27) s.t. $\displaystyle
c_{mj}={D_{mj}}\tau,\quad\forall m\in\mathcal{M},j\in\mathcal{J},$ (27a)
$\displaystyle\sum_{m=1}^{M}f_{m}=1,$ (27b) $\displaystyle c_{mj}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{mj}P_{mj}}{\sigma^{2}Bf_{m}}\right),\quad\forall
m\in\mathcal{M},j\in\mathcal{J},$ (27c) $\displaystyle c_{m1}+c_{m2}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{m1}P_{m1}+h_{m2}P_{m2}}{\sigma^{2}Bf_{m}}\right),\quad\forall
m\in\mathcal{M},$ (27d) $\displaystyle f_{m},c_{m1},c_{m2}\geq 0,\quad\forall
m\in\mathcal{M}.$ (27e)

Substituting (27a) into (27c) and (27d), we have:

${D_{mj}}{\tau}\leq
Bf_{m}\log_{2}\left(1+\frac{h_{mj}P_{mj}}{\sigma^{2}Bf_{m}}\right),\quad
j\in\mathcal{J},$ (28)
$({D_{m1}\\!+\\!D_{m2}}){\tau}\\!\leq\\!Bf_{m}\log_{2}\\!\left(\\!1\\!+\\!\frac{h_{m1}P_{m1}\\!+\\!h_{m2}P_{m2}}{\sigma^{2}Bf_{m}}\\!\right).\\!\\!$
(29)

It can be proved that $g(x)=x\ln\left(1+\frac{1}{x}\right)$ is a monotonically
increasing function. Thus, to satisfy (28) and (29), bandwidth fraction
$f_{m}$ should satisfy:

$f_{m}\geq\max\\{f_{m1},f_{m2},f_{m3}\\},$ (30)

where

$f_{mk}=-\frac{(\ln 2)D_{mk}h_{mk}P_{mk}}{Bh_{mk}P_{mk}\tau W\left(-\frac{(\ln
2)D_{mk}\sigma^{2}}{h_{mk}P_{mk}\tau}\text{e}^{-\frac{(\ln
2)D_{mk}\sigma^{2}}{h_{mk}P_{mk}\tau}}\right)+{(\ln
2)D_{mk}\sigma^{2}B}},\quad k=1,2,$ (31) $f_{m3}=-\frac{(\ln
2)(D_{m1}+D_{m2})(h_{m1}P_{m1}+h_{m2}P_{m2})}{\beta+{(\ln
2)(D_{m1}+D_{m2})\sigma^{2}B}},$ (32) $\beta=B(h_{m1}P_{m1}+h_{m2}P_{m2})\tau
W\left(-\frac{(\ln
2)(D_{m1}+D_{m2})\sigma^{2}}{(h_{m1}P_{m1}+h_{m2}P_{m2})\tau}\text{e}^{-\frac{(\ln
2)(D_{m1}+D_{m2})\sigma^{2}}{(h_{m1}P_{m1}+h_{m2}P_{m2})\tau}}\right),$ (33)

and $W(\cdot)$ is the Lambert-W function.

Based on (30) and (27b), we have:

$\sum_{m=1}^{M}\max\\{f_{m1},f_{m2},f_{m3}\\}\leq 1.$ (34)

According to (28)-(34), problem (27) has a feasible solution if and only if
(34) is satisfied. As a result, the algorithm for obtaining the maximum sum-
rate of problem (27) is summarized in Algorithm 2, where $\tau^{*}$ is the
optimal sum-rate of problem (10).

Algorithm 2 : Low-Complexity Sum-Rate Maximization

1: Initialize $\tau_{\min}=0$, $\tau_{\max}=\tau^{*}$, and the tolerance
$\epsilon$.

2: Set $\tau=\frac{\tau_{\min}+\tau_{\max}}{2}$, and calculate $f_{m1}$,
$f_{m2}$ and $f_{m3}$ according to (31) and (32), respectively.

3: Check the feasibility condition (34). If problem (27) is feasible, set
$\tau_{\min}=\tau$. Otherwise, set $\tau=\tau_{\max}$.

4: If $(\tau_{\max}-\tau_{\min})/\tau_{\max}\leq\epsilon$, terminate.
Otherwise, go to step 2.

The complexity of the proposed Algorithm 2 in each step lies in checking the
feasibility of problem (27), which involves the complexity of $\mathcal{O}(M)$
according to (31)-(32). As a result, the total complexity of the proposed
Algorithm 2 is $\mathcal{O}(M\log_{2}(1/\epsilon))$, where
$\mathcal{O}(\log_{2}(1/\epsilon))$ is the complexity of the bisection method
with accuracy $\epsilon$.

##  V  Sum-Rate Maximization for Uplink NOMA/FDMA/TDMA

To evaluate the performance gain of the RSMA scheme proposed in Sections II-
IV, and for comparison purposes, we will solve the sum-rate maximization
problems for uplink NOMA, FDMA and TDMA schemes.

###  V-A NOMA

Without loss of generality, the channel gains are sorted in descending order,
i.e., $h_{1}\geq h_{2}\geq\cdots\geq h_{K}$. In NOMA, the BS first decodes the
messages of users with high channel gains and then decodes the messages of
users with low channel gains by subtracting the interference from decoded
strong user. The achievable rate of user $k$ with NOMA is calculated as [26]:

$r_{k}^{\text{NOMA}}=B\log_{2}\left(1+\frac{h_{k}q_{k}}{\sum_{j=k+1}^{K}h_{j}q_{j}+\sigma^{2}B}\right),$
(35)

where $q_{k}$ is the transmit power of user $k$. The transmission power
$q_{k}$ has a maximum transmit power limit $P_{k}$, i.e., we have $q_{k}\leq
P_{k}$, $\forall k\in\mathcal{K}$.

Similar to (8), the sum-rate maximization problem for uplink NOMA can be given
by:

$\displaystyle\mathop{\max}_{\tau,\bm{q}}\quad$ $\displaystyle\tau,$ (36) s.t.
$\displaystyle
B\log_{2}\left(1+\frac{h_{k}q_{k}}{\sum_{j=k+1}^{K}h_{j}q_{j}+\sigma^{2}B}\right)={D_{k}}\tau,\quad\forall
k\in\mathcal{K},$ (36a) $\displaystyle 0\leq q_{k}\leq P_{k},\quad\forall
k\in\mathcal{K},$ (36b)

where $\bm{q}=[q_{1},q_{2},\cdots,q_{K}]^{T}$. To obtain the optimal solution
of problem (36), we provide the following theorem.

###### Theorem 2

The optimal solution of problem (36) is

$\tau^{*}=\min_{k\in\mathcal{K}}\tau_{k},$ (37)

and

$q_{k}^{*}=\frac{1}{h_{k}}{\left(2^{\frac{D_{k}\tau^{*}}{B}}-1\right)}\sum_{j=k+1}^{K}2^{\frac{\sum_{l=k+1}^{j-1}D_{l}\tau^{*}}{B}}\left({2^{\frac{D_{j}\tau^{*}}{B}}}-1\right)\sigma^{2}B+\frac{1}{h_{k}}\left({2^{\frac{D_{k}\tau^{*}}{B}}}-1\right)\sigma^{2}B,\quad\forall
k\in\mathcal{K},$ (38)

where $\tau_{k}$ is the solution to

$\displaystyle
P_{k}=\frac{1}{h_{k}}{\left(2^{\frac{D_{k}\tau_{k}}{B}}-1\right)}\sum_{j=k+1}^{K}2^{\frac{\sum_{l=k+1}^{j-1}D_{l}\tau_{k}}{B}}\left({2^{\frac{D_{j}\tau_{k}}{B}}}-1\right)\sigma^{2}B+\frac{1}{h_{k}}\left({2^{\frac{D_{k}\tau_{k}}{B}}}-1\right)\sigma^{2}B,\quad\forall
k\in\mathcal{K}.$ (39)

Proof: See Appendix C. $\Box$

Since the right hand side of (39) monotonically increases with $\tau_{k}$, the
solution of $\tau_{k}$ to (39) can be effectively obtained by the bisection
method.

###  V-B FDMA

In FDMA, each user will be allocated a fraction of the BS bandwidth. Let
$b_{k}$ denote the fraction of bandwidth allocated to user $k$. Then the data
rate of user $k$ is:

$r_{k}^{\text{FDMA}}=Bb_{k}\log_{2}\left(1+\frac{h_{k}P_{k}}{\sigma^{2}Bb_{k}}\right).$
(40)

Note that user $k$ transmits with maximum power in (40) since there is no
inter-user interference and large power leads to high data rate. Due to
limited bandwidth, we have $\sum_{k=1}^{K}b_{k}=1$.

Similar to (8), the sum-rate maximization problem for uplink FDMA can be given
by:

$\displaystyle\mathop{\max}_{\tau,\bm{b}}\quad$ $\displaystyle\tau,$ (41) s.t.
$\displaystyle
Bb_{k}\log_{2}\left(1+\frac{h_{k}P_{k}}{\sigma^{2}Bb_{k}}\right)=D_{k}\tau,\quad\forall
k\in\mathcal{K},$ (41a) $\displaystyle\sum_{k=1}^{K}b_{k}=1,$ (41b)
$\displaystyle b_{k}\geq 0,\quad\forall k\in\mathcal{K},$ (41c)

where $\bm{b}=[b_{1},b_{2},\cdots,b_{K}]^{T}$. Regarding the optimal solution
of problem (41), we provide the following theorem.

###### Theorem 3

The optimal solution of problem (41) is $(\tau^{*},\bm{b}^{*})$, where
$\tau^{*}$ is the solution of

$-\sum_{k=1}^{K}\frac{(\ln
2)D_{k}h_{k}P_{k}\tau}{Bh_{k}P_{k}W\left(-\frac{(\ln
2)D_{k}\tau\sigma^{2}}{h_{k}P_{k}}\text{e}^{-\frac{(\ln
2)D_{k}\sigma^{2}\tau}{h_{k}P_{k}}}\right)+{(\ln 2)D_{k}\sigma^{2}B\tau}}=1$
(42)

and

$b_{k}^{*}=-\frac{(\ln
2)D_{k}h_{k}P_{k}\tau^{*}}{Bh_{k}P_{k}W\left(-\frac{(\ln
2)D_{k}\sigma^{2}\tau^{*}}{h_{k}P_{k}}\text{e}^{-\frac{(\ln
2)D_{k}\sigma^{2}\tau^{*}}{h_{k}P_{k}}}\right)+{(\ln
2)D_{k}\sigma^{2}B\tau^{*}}}.$ (43)

Proof: See Appendix D. $\Box$

###  V-C TDMA

In TDMA, each user will be assigned a fraction of time to use the whole BS
bandwidth. Let $a_{k}$ be the fraction of time allocated to user $k$. The data
rate of user $k$ is:

$r_{k}^{\text{TDMA}}=Ba_{k}\log_{2}\left(1+\frac{h_{k}P_{k}}{\sigma^{2}B}\right)$
(44)

with $\sum_{k=1}^{K}a_{k}=1$.

Similar to (8), the sum-rate maximization problem for uplink TDMA can be given
by:

$\displaystyle\mathop{\max}_{\tau,\bm{a}}\quad$ $\displaystyle\tau,$ (45) s.t.
$\displaystyle
Ba_{k}\log_{2}\left(1+\frac{h_{k}P_{k}}{\sigma^{2}B}\right)=D_{k}\tau,\quad\forall
k\in\mathcal{K},$ (45a) $\displaystyle\sum_{k=1}^{K}a_{k}=1,$ (45b)
$\displaystyle a_{k}\geq 0,\quad\forall k\in\mathcal{K},$ (45c)

where $\bm{a}=[a_{1},a_{2},\cdots,a_{K}]^{T}$. For problem (45), the optimal
solution is given by the following theorem.

###### Theorem 4

The optimal solution of problem (45) is:

$\tau^{*}=\frac{1}{\sum_{k=1}^{K}\frac{D_{k}}{B\log_{2}\left(1+\frac{h_{k}P_{k}}{\sigma^{2}B}\right)}},$
(46)

and

$a_{k}^{*}=\frac{D_{k}\tau^{*}}{B\log_{2}\left(1+\frac{h_{k}P_{k}}{\sigma^{2}B}\right)},\quad\forall
k\in\mathcal{K}.$ (47)

Proof: See Appendix E. $\Box$

###  V-D Analysis and Discussion

We define the optimal uplink sum-rates for RSMA, NOMA, FDMA and TDMA as
$\tau^{\text{RSMA}}$, $\tau^{\text{NOMA}}$, $\tau^{\text{FDMA}}$ and
$\tau^{\text{TDMA}}$, respectively. For the optimal sum-rate with various
uplink multiple access schemes, we can state the following lemma.

###### Lemma 4

$\tau^{\text{RSMA}}\geq\tau^{\text{NOMA}}$ and
$\tau^{\text{RSMA}}\geq\tau^{\text{FDMA}}\geq\tau^{\text{TDMA}}$.

Lemma 4 can be easily proved by the fact that for any feasible solution to
NOMA/FDMA scheme, we can construct a feasible solution to RSMA with the same
or better objective value and for any feasible solution to TDMA scheme, we can
construct a feasible solution to FDMA with the same or better objective value.
In order to illustrate the sum-rate performance of different multiple access
schemes, we consider the special case with two users, i.e., $K=2$.

Denote $\mathcal{R}_{2}^{\text{X}}$ as the rate region of two users with
multiple access $\text{X}\in\\{\text{RSMA},\text{NOMA},\text{FDMA},$
$\text{TDMA}\\}$. Based on Lemma 2, the rate region of RSMA with two users can
be expressed by:

$\mathcal{R}_{2}^{\text{RSMA}}=\\{(r_{1},r_{2})|0\leq r_{1}\leq R_{1},0\leq
r_{2}\leq R_{2},r_{1}+r_{2}\leq R_{\max}\\},$ (48)

where $R_{1}$, $R_{2}$ and $R_{\max}$ are defined in (19).

###### Lemma 5

For two-user NOMA/FDMA/TDMA, we have

$\mathcal{R}_{2}^{\text{NOMA}}=\left\\{(r_{1},r_{2})|0\leq r_{1}\leq
R_{1},0\leq r_{2}\leq
R_{2},r_{2}+B\log_{2}\left({2^{\frac{r_{1}}{B}}-1}\right)\leq
B\log_{2}\left(\frac{h_{1}P_{1}}{\sigma^{2}B}\right)\right\\},$ (49)

$\mathcal{R}_{2}^{\text{FDMA}}=\left\\{(r_{1},r_{2})|0\leq r_{1}\leq
R_{1},0\leq r_{2}\leq R_{2},f_{1}(r_{1})+f_{2}(r_{2})\leq 1,\right\\},$ (50)

and

$\mathcal{R}_{2}^{\text{TDMA}}=\left\\{(r_{1},r_{2})|0\leq r_{1}\leq
R_{1},0\leq r_{2}\leq R_{2},\frac{r_{1}}{R_{1}}+\frac{r_{2}}{R_{2}}\leq
1\right\\},$ (51)

where

$f_{k}(r_{k})=-\frac{(\ln 2)h_{k}P_{k}r_{k}}{Bh_{k}P_{k}W\left(-\frac{(\ln
2)\sigma^{2}r_{k}}{h_{k}P_{k}}\text{e}^{-\frac{(\ln
2)\sigma^{2}r_{k}}{h_{k}P_{k}}}\right)+{(\ln 2)\sigma^{2}Br_{k}}},\quad
k=1,2.$ (52)

Proof: See Appendix F. $\Box$

Figure 2: An example of rate region for RSMA, NOMA, FDMA and TDMA with $K=2$
users, $h_{1}=9.45\times 10^{-9}$, $h_{2}=6.17\times 10^{-9}$, $P_{1}=P_{2}=1$
dBm, $\sigma^{2}$= -174 dBm/Hz, and $B=1$ MHz.

Based on Lemmas 3 and 5, we provide an example of the rate region for multiple
access schemes, as shown in Fig. 2. From Fig. 2, it is observed that RSMA has
the largest rate region, while TDMA has the smallest rate region. According to
the proportional rate constraints (5a), we have
$\frac{r_{1}}{r_{2}}=\frac{D_{1}}{D_{2}}$, which is a linear function through
the original point in the rate coordinate $(r_{1},r_{2})$. Thus, the optimal
rate point can be obtained by finding the intersection between the rate region
and the line $r_{2}=\frac{D_{2}}{D_{1}}r_{1}$. Thus, it is shown from Fig. 2
that RSMA has the best sum-rate performance and TDMA has the worst sum-rate
performance, which verifies the theoretical findings in Lemma 5.

##  VI Numerical Results

TABLE I: System Parameters Parameter | Value
---|---
Bandwidth of the BS $B$ | 1 MHz
Noise power spectral density $\sigma^{2}$ | -174 dBm/Hz
Path loss model |  $128.1+37.6\log_{10}d$ ($d$ is in km)
Standard deviation of shadow fading |  $8$ dB
Maximum transmit power $P$ | 1 dBm

For our simulations, we deploy $K$ users uniformly in a square area of size
$500$ m $\times$ $500$ m with the BS located at its center. The path loss
model is $128.1+37.6\log_{10}d$ ($d$ is in km) and the standard deviation of
shadow fading is $8$ dB. In addition, and the noise power spectral density is
$\sigma^{2}=-174$ dBm/Hz. Unless specified otherwise, we choose an equal
maximum transmit power $P_{1}=\cdots=P_{K}=1$ dBm, and a bandwidth $B=1$ MHz.
The main system parameters are summarized in Table I. All statistical results
are averaged over a large number of independent runs.

We compare the sum-rate performance of RSMA, NOMA, FDMA, and TDMA. Fig. 3
shows how the sum-rate changes as the maximum transmit power of each user
varies for a network having two users. We can see that the sum-rate of all
multiple access schemes linearly increases with the logarithmic maximum
transmission power of each user. This is because the sum-rate is a logarithmic
function of the maximum power of the users. It is found that RSMA achieves the
best performance among all multiple access schemes. From Fig. 3, RSMA can
increase up to 4.1%, 10.2% and 28.8% sum-rate compared to NOMA, FDMA and TDMA,
respectively. This is because that RSMA can achieve the largest rate region
and users with RSMA can achieve higher rate than other multiple access
schemes. Fig. 3 also shows that TDMA achieves the worst sum-rate performance,
which corroborates the theoretical findings in Lemma 5.

Figure 3: Sum-rate versus maximum transmit power of each user ($K=2$ users,
$D_{1}=0.5$, and $D_{2}=0.5$). Figure 4: Sum-rate versus proportional rate
parameter $D_{2}$ ($K=2$ users and $D_{1}=1-D_{2}$).

Fig. 4 shows the sum-rate versus the proportional rate parameter $D_{2}$ of
user 2. From this figure, we can observe that RSMA achieves the best
performance in terms of sum-rate, while TDMA has the worst sum-rate
performance. This is because RSMA exhibits a better spectrum efficiency
compared to FDMA, and RSMA uses an optimized power allocation among two
messages for each user to achieve the optimal rate region, while each user
only transmits one message in NOMA. In addition, both FDMA and NOMA achieve
better sum-rate performance than TDMA due to the superiority of spectrum
efficiency. Fig. 4 shows that the sum-rate of NOMA is lower than that of FDMA
for small proportional rate parameter of user 2 ($D_{2}<0.2$) and the sum-rate
of NOMA is higher than that of FDMA for proportional rate parameter of user 2
($D_{2}>0.2$). According to the proportional rate constraints (5a)
$\frac{r_{1}}{r_{2}}=\frac{D_{1}}{D_{2}}$, the optimal rate point can be
obtained by finding the intersection between the rate region and the line
$r_{2}=\frac{D_{2}}{D_{1}}r_{1}$. For a small proportional rate parameter of
user 2, the slope $\frac{D_{2}}{D_{1}}$ is low and user rate value of the
intersection between the rate region (Fig. 2) of NOMA and line
$r_{2}=\frac{D_{2}}{D_{1}}r_{1}$ is smaller than that of the intersection
between the rate region of FDMA and line $r_{2}=\frac{D_{2}}{D_{1}}r_{1}$.
Consequently, the sum-rate of FDMA is higher than NOMA. WBy using a similar,
the sum-rate performance of NOMA is better than FDMA for large proportional
rate parameter of user 2.

Figure 5: Sum-rate versus bandwidth of the BS ($K=2$ users, $D_{1}=0.5$, and
$D_{2}=0.5$). Figure 6: Sum-rate versus maximum transmit power of each user
under different user pairing metods ($K=10$ users, $D_{1}=\cdots=D_{10}=0.1$).

Fig. 5 shows the sum-rate versus the bandwidth of the BS. From this figure, we
can see that RSMA always achieves a better performance than NOMA, FDMA, and
TDMA. Fig. 5 demonstrates that the sum-rate increases rapidly for a small
bandwidth, however, this increase becomes slower for a larger bandwidth. This
is because a high bandwidth leads to high noise power, which consequently
decreases the slope of increase of the sum-rate for all multiple access
schemes. Fig. 5 also demonstrates that the sum-rate resulting from RSMA is
greater than the one achieved by all other multiple access schemes,
particularly when the bandwidth is large.

Figure 7: CDF of sum-rate ($K=10$ users, $D_{1}=\cdots=D_{10}=0.1$).

For low-complexity RSMA with user pairing (labeled as ‘RSMA-UP’), we study the
influence of user pairing by considering three different user-pairing methods
[25]. For strong-weak (SW) pair selection, the user with the strongest channel
condition is paired with the user with the weakest in one pair, and the user
with the second strongest is paired with one with the second weakest in one
pair, and so on. For strong-middle (SM) pair selection, the user with the
strongest channel condition is paired with the user with the middle strongest
user in one pair, and so on. For strong-strong (SS) pair selection, the user
with the strongest channel condition is paired with the one with the second
strongest in one pair, and so on.

In Fig. 6, we show how the sum-rate changes as the maximum transmit power of
each user varies for a network having ten users. From this figure, we observe
that RSMA always achieves the best performance. For RSMA-UP with different
user-pairing methods, SW outperforms the other two methods in terms of sum-
rate for RSMA-UP. To maximize the sum-rate, it tends to pair users with
distinctive gains for sum-rate maximization. Due to the superiority of SW, we
choose SW for pair selection of RSMA-UP in the following simulations.

Fig. 7 presents the cumulative distribution function (CDF) of sum-rate
resulting from RSMA, NOMA, FDMA, and RSMA-UP-SW for a network with $K=10$
users. From Fig. 7, the CDFs for RSMA, RSMA-UP-SW, and NOMA all improve
significantly compared FDMA, particularly when the sum-rate is high, which
shows that RSMA, RSMA-UP-SW, and NOMA are suitable for high sum-rate
transmission. Moreover, we can observe that RSMA outperforms NOMA. This is
because RSMA can adjust the splitting power of two messages for each user so
as to control the interference decoding thus optimizing the sum-rate of all
users, while there is no power splitting for each user in NOMA. Moreover,
RSMA-UP-SW can achieve a similar performance to RSMA. However, the complexity
of RSMA-UP-SW is much lower compared to RSMA according to Section III-B and
Section IV, which shows the effectiveness of RSMA-UP-SW.

Figure 8: Sum-rate versus number of users ($D_{1}=\cdots=D_{K}=1/K$).

In Fig. 8, we plot the sum-rate versus the number of users is given. Clearly,
the proposed RSMA or RSMA-UP-SW will always achieve a better performance
compared to NOMA, FDMA, and TDMA especially when the number of users is large.
In particular, RSMA can achieve sum-rate gains of up to 10.0%, 22.2%, and
83.7% compared to NOMA, FDMA, and TDMA, respectively, while RSMA-UP-SW can
improve the sum-rate by up to 4.1%, 11.6% and 66.8% compared to NOMA, FDMA,
and TDMA, respectively. When the number of users is large, the multiuser gain
is more pronounced for the proposed RSMA compared to conventional NOMA, FDMA,
and TDMA. This is due to the fact hat RSMA can effectively determine the power
splitting of each user to achieve the theoretically maximal rate region, while
there is no power splitting in NOMA and the allocated bandwidth/time of each
user is low for FDMA/TDMA when the number of users is large. However, RSMA
achieves a better performance compared to NOMA, FDMA, and TDMA at the cost of
additional computational complexity according to Section III-B. Fig. 8 also
shows that RSMA-UP-SW achieves a better sum-rate performance compared to NOMA,
FDMA, and TDMA but with low complexity according to Section IV, which shows
that RSMA-UP-SW is promising solution that strikes a desirable tradeoff
between performance gain and computational complexity.

##  VII Conclusion

In this paper, we have investigated the decoder order and power optimization
in an uplink RSMA system. We have formulated the problem as a sum-rate
maximization problem. To solve this problem, we have transformed it into an
equivalent problem with only rate splitting variables, which has closed-form
optimal solution. Given the optimal rate requirement of each user, the optimal
transmit power of each user is obtained under given the decoding order and the
optimal decoding order is found by an exhaustive search method. To reduce the
computational complexity, we have proposed a low-complexity RSMA with user
pairing. Simulation results show that RSMA achieves higher sum-rate than NOMA,
FDMA, and TDMA.

##  Appendix A Proof of Lemma 2

Assume that for the optimal solution $(\tau^{*},\bm{r}^{*})$ of problem (10),
we have

$\sum_{k\in\mathcal{K}^{\prime}}r_{k}^{*}<B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\right),\quad\forall\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset.$
(A.1)

In this case, we can construct a new rate solution
$\bm{r}^{\prime}=[r_{1}^{\prime},\cdots,r_{K}^{\prime}]$ with
$r_{k}^{\prime}=\epsilon r_{k}^{*}$ and

$\epsilon\\!=\\!\min_{\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset}\\!\frac{B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\\!\right)}{\sum_{k\in\mathcal{K}^{\prime}}r_{k}^{*}}>1.$
(A.2)

According to (A.2), we can show that

$\sum_{k\in\mathcal{K}^{\prime}}r_{k}^{\prime}\leq
B\log_{2}\left(1+\frac{\sum_{k\in\mathcal{K}^{\prime}}h_{k}P_{k}}{\sigma^{2}B}\right),\>\forall\mathcal{K}^{\prime}\subseteq\mathcal{K}\setminus\emptyset,$
(A.3)

which ensures that $\bm{r}^{\prime}$ satisfies constraints (10b).

Based on (10a), we have $\tau^{*}=\frac{r_{k}^{*}}{D_{k}},\forall
k\in\mathcal{K}$. We set $\tau^{\prime}$ as

$\tau^{\prime}=\frac{r_{k}^{\prime}}{D_{k}}=\frac{\epsilon
r_{k}^{*}}{D_{k}}>\tau^{*}.$ (A.4)

According to (A.3) and (A.4), we can see that new solution
$(\tau^{\prime},\bm{r}^{\prime})$ is feasible and the objective value (10) of
new solution is better than that of solution $(\tau^{*},\bm{r}^{*})$, which
contradicts the fact that $(\tau^{*},\bm{r}^{*})$ is the optimal solution.
Lemma 2 is proved.

##  Appendix B Proof of Lemma 3

Given decoding orders $s_{21}$, $s_{11}$ and $s_{22}$, we have

$r_{1}=B\log_{2}\left(1+\frac{h_{1}p_{11}}{h_{2}p_{22}+\sigma^{2}B}\right)$
(B.1)

and

$r_{2}\\!=\\!B\log_{2}\\!\left(\\!1\\!+\\!\frac{h_{2}p_{21}}{h_{1}p_{11}\\!+\\!h_{2}p_{22}\\!+\\!\sigma^{2}B}\\!\right)\\!+\\!B\log_{2}\\!\left(\\!1\\!+\\!\frac{h_{2}p_{22}}{\sigma^{2}B}\\!\right).$
(B.2)

For case (1), since user 1 reaches its maximum rate point, we have
$p_{11}=P_{1}$ and $p_{22}=0$. Transmission power $p_{21}$ can be calculated
according to (B.2).

For case (2), since user 2 reaches its maximum rate point, we have $p_{21}=0$
and $p_{22}=P_{2}$. Transmission power $p_{11}$ can be calculated according to
(B.1).

For case (3), since $r_{1}+r_{2}=R_{\max}$, the sum-rate of users 1 and 2 only
reaches its maximum point when both users $u_{1}$ and $u_{2}$ transmit maximum
power, i.e., $p_{11}=P_{1}$ and $p_{21}+p_{22}=P_{2}$. According to (B.1), we
can obtain power $p_{22}$. With $p_{22}$, we can calculate
$p_{21}=P_{2}-p_{22}$.

##  Appendix C Proof of Theorem 2

Denote

$z_{k}=\sum_{j=k}^{K}h_{j}q_{j},\quad\forall k\in\mathcal{K}.$ (C.1)

According to (36a), we can obtain:

$B\log_{2}\left(\frac{z_{k}+\sigma^{2}B}{z_{k+1}+\sigma^{2}B}\right)={D_{k}}{\tau}.$
(C.2)

Based on (C.2), we have:

$z_{k}={2^{\frac{D_{k}\tau}{B}}}z_{k+1}+\left({2^{\frac{D_{k}\tau}{B}}}-1\right)\sigma^{2}B.$
(C.3)

Using the recursive formulation (C.3) and $z_{K+1}=\sum_{k=K+1}^{K}q_{k}=0$,
we can calculate:

$z_{k}=\sum_{j=k}^{K}2^{\frac{\sum_{l=k}^{j-1}D_{l}\tau}{B}}\left({2^{\frac{D_{j}\tau}{B}}}-1\right)\sigma^{2}B,$
(C.4)

where we set $\sum_{l=k}^{k-1}D_{l}=0$. Based on (C.1), we have:

$q_{k}=\frac{z_{k}-z_{k+1}}{h_{k}},\quad\forall k\in\mathcal{K}.$ (C.5)

Combining (C.4) and (C.5) yields:

$q_{k}=\frac{1}{h_{k}}{\left(2^{\frac{D_{k}\tau}{B}}-1\right)}\sum_{j=k+1}^{K}2^{\frac{\sum_{l=k+1}^{j-1}D_{l}\tau}{B}}\left({2^{\frac{D_{j}\tau}{B}}}-1\right)\sigma^{2}B+\frac{1}{h_{k}}\left({2^{\frac{D_{k}\tau}{B}}}-1\right)\sigma^{2}B,\quad\forall
k\in\mathcal{K},$ (C.6)

which monotonically increases with $\tau$. Considering the maximum uplink
transmission power constraints (36b), we can obtain that

$\tau\leq\tau_{k},\quad\forall k\in\mathcal{K},$ (C.7)

where $\tau_{k}$ is the solution to (39). To maximize sum-rate $\tau$, the
optimal $\tau^{*}$ is given by (37). Consequently, the optimal transmission
power is provided by (38).

##  Appendix D Proof of Theorem 3

Solving (41a) yields

$b_{k}=-\frac{(\ln 2)D_{k}h_{k}P_{k}\tau}{Bh_{k}P_{k}W\left(-\frac{(\ln
2)D_{k}\sigma^{2}\tau}{h_{k}P_{k}}\text{e}^{-\frac{(\ln
2)D_{k}\sigma^{2}\tau}{h_{k}P_{k}}}\right)+{(\ln 2)D_{k}\sigma^{2}B\tau}}.$
(D.1)

Substituting (D.1) to constraint (41b), the optimal $\tau^{*}$ is calculated
as (42), and the optimal bandwidth allocation is accordingly given by (43).

##  Appendix E Proof of Theorem 4

Solving (45a) yields

$a_{k}=\frac{D_{k}\tau}{B\log_{2}\left(1+\frac{h_{k}P_{k}}{\sigma^{2}B}\right)}.$
(E.1)

Combining (E.1) and constraint (45b), the optimal sum-rate is given by (46).
Then, the optimal time sharing is (47).

##  Appendix F Proof of Lemma 5

For two-user NOMA, according to (35), we can obtain:

$r_{1}=B\log_{2}\left(1+\frac{h_{1}q_{1}}{h_{2}q_{2}+\sigma^{2}B}\right),r_{2}=B\log_{2}\left(1+\frac{h_{2}q_{2}}{\sigma^{2}B}\right),0\leq
q_{1}\leq P_{1},0\leq q_{2}\leq P_{2}.$ (F.1)

Based on (F.1), we have:

$r_{2}=B\log_{2}\left(1+\frac{h_{2}q_{2}}{\sigma^{2}B}\right)\leq
B\log_{2}\left(1+\frac{h_{2}P_{2}}{\sigma^{2}B}\right)=R_{2},$ (F.2)

and

$h_{2}q_{2}+\sigma^{2}B=\frac{h_{1}q_{1}}{2^{\frac{r_{1}}{B}}-1}.$ (F.3)

Substituting (F.3) into (F.2) yields:

$\displaystyle r_{2}$
$\displaystyle=B\log_{2}\left(1+\frac{h_{2}q_{2}}{\sigma^{2}B}\right)=B\log_{2}\left(\frac{h_{1}q_{1}}{\left(2^{\frac{r_{1}}{B}}-1\right)\sigma^{2}B}\right)$
$\displaystyle=B\log_{2}\left(\frac{h_{1}q_{1}}{\sigma^{2}B}\right)-B\log_{2}\left({2^{\frac{r_{1}}{B}}-1}\right)\leq
B\log_{2}\left(\frac{h_{1}P_{1}}{\sigma^{2}B}\right)-B\log_{2}\left({2^{\frac{r_{1}}{B}}-1}\right),$
(F.4)

where the inequality follows from the fact that $0\leq q_{1}\leq P_{1}$.
Combining (F.1) to (F), two-user NOMA rate region is given in (49).

For two-user FDMA, according to (40), we have:

$r_{1}=Bb_{1}\log_{2}\left(1+\frac{h_{1}P_{1}}{\sigma^{2}Bb_{1}}\right),r_{2}=Bb_{1}\log_{2}\left(1+\frac{h_{2}P_{2}}{\sigma^{2}Bb_{2}}\right),b_{1}+b_{2}=1,b_{1},b_{2}\geq
0.$ (F.5)

Based on (F.5), we have

$b_{k}=f_{k}(r_{k}),\quad k=1,2,$ (F.6)

where $f_{k}(r_{k})$ is defined in (52). Combining (F.5) and (F.6), two-user
FDMA rate region is given in (50).

For two-user TDMA, according to (44), we can calculate:

$r_{1}=a_{1}R_{1},r_{2}=a_{2}R_{2},a_{1}+a_{2}=1,a_{1},a_{2}\geq 0.$ (F.7)

According to (F.7), two-user TDMA rate region is given in (51).

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