# Resource Allocation for URLLC and eMBB Traffic in Uplink Wireless Networks

Duan-Shin Lee, Cheng-Shang Chang, Ruhui Zhang and Mao-Pin Lee Duan-Shin Lee is
with the Department of Computer Science and the Institute of Communications
Engineering, Cheng-Shang Chang is with the Institute of Communications
Engineering, Ruhui Zhang is with the Department of Computer Science, and Mao-
Pin Lee is with the Institute of Communications Engineering, National Tsing
Hua University, Hsinchu 300, Taiwan, R.O.C<EMAIL_ADDRESS>cschang@ee.nthu.edu.tw,huibrana@gapp.nthu.edu.tw,teddy1998mb@gmail.com.)This
research was supported in part by the Ministry of Science and Technology,
Taiwan, R.O.C., under Contract 109-2221-E-007-093-MY2.

###### Abstract

In this paper we consider two resource allocation problems of URLLC traffic
and eMBB traffic in uplink 5G networks. We propose to divide frequencies into
a common region and a grant-based region. Frequencies in the grant-based
region can only be used by eMBB traffic, while frequencies in the common
region can be used by eMBB traffic as well as URLLC traffic. In the first
resource allocation problem we propose a two-player game to address the size
of the grant-based region and the size of the common region. We show that this
game has specific pure Nash equilibria. In the second resource allocation
problem we determine the number of packets that each eMBB user can transmit in
a request-grant cycle. We propose a constrained optimization problem to
minimize the variance of the number of packets granted to the eMBB users. We
show that a water-filling algorithm solves this constrained optimization
problem. From simulation, we show that our scheme, consisting of resource
allocation according to Nash equilibria of a game, persistent random
retransmission of URLLC packets and allocation of eMBB packets by a water-
filling algorithm, works better than four other heuristic methods.

keywords: wireless networks, resource allocation, game theory, constrained
optimization, water-filling algorithm

##  1 Introduction

The fifth-generation networks (5G) and beyond aim to cover three generic
connectivity types: (i) enhanced mobile broadband (eMBB), (ii) ultra-reliable
low-latency communications (URLLC), and (iii) massive machine-type
communications (mMTC) (see, e.g., [1, 2, 3] and references therein). The
reliability defined in 3GPP for supporting URLLC services, such as autonomous
driving, drones, and augmented/virtual reality, requires a $1-10^{-5}$ success
probability of transmitting a layer 2 packet of length 32 bytes within 1
millisecond. mMTC services are characterized by a large number of simple
devices. Motivated by these emerging needs in 5G, research communities in
wireless networking commonly believe that grant-based communications continue
to be feasible for eMBB services, but multiple access schemes in a grant-free
manner are more suitable for URLLC and mMTC services. Many multiple access
schemes have been proposed in the literature recently. We refer the readers to
[4, 5] for references. In this paper we propose a method to deliver URLLC
packets reliably with their latency bounds, and to schedule eMBB packets in a
fair manner.

Providing satisfactory services to ultra-reliable low-latency communications
(URLLC) traffic and enhanced mobile broadband (eMBB) traffic simultaneously is
an interesting and challenging research problem. These two traffic types are
characterized by very different service requirements. URLLC traffic demands a
latency as low as 1 millisecond and a packet loss probability as low as
$10^{-5}$ [6]. On the other hand, an eMBB user cares very much about its
throughput and also the fairness in the throughput among all eMBB users. eMBB
services can be efficiently supported by a request and grant paradigm.
However, a grant-based service is not likely to meet the stringent latency
requirement of URLLC services. A grant-free paradigm is more likely to satisfy
the latency requirement of URLLC users. Non-Orthogonal Multiple Access (NOMA)
offers a good solution to the problem above [7]. The whole frequency band can
be used for both grant-based eMBB services and grant-free URLLC services [8,
9, 10, 11, 12]. Using NOMA techniques, one can divide the frequency band into
a grant-based region and a common region in uplink 5G networks. Using wireless
resource blocks in the grant-based region, eMBB users make requests, receive
grants, and transmit packets according to the grants. Properly designed
resource allocation algorithms can ensure that eMBB users receive their
services in a fair manner. Using wireless resource blocks in the common
region, URLLC users sporadically transmit packets in a grant-free manner. eMBB
users also can be granted packets in the common region. If so, eMBB users
transmit their granted packets at a smaller power in the common region. If
eMBB packets collide with URLLC packets in the common region, the eMBB packets
hopefully can be recovered by the interference cancellation technique [5].

In this paper we propose to divide wireless resource blocks into a common
region and a grant-based region. URLLC packets are transmitted sporadically in
a grant-free manner in the common region. We propose a persistent
retransmission scheme for the URLLC packets to cope with their stringent
latency requirements. One of the contributions of this paper is that we
analyze the probability that an URLLC packet fails to meet its latency
requirement. In this paper we study two resource allocation problems. In the
first problem, we determine the size of the common region and the size of the
grant-based region. We propose to solve this problem by formulating a two-
player game. The two players of the game agent URLLC and agent eMBB, who
negotiate resource blocks on behalf of all URLLC users and eMBB users,
respectively. The payoff of the URLLC agent depends on the probability that an
URLLC packet fails to meet its latency requirement. The second contribution of
this paper is that we analyze the Nash equilibria of this game. eMBB users can
use resource blocks in the grant-based region as well as in the common region
to transmit their packets. The transmissions are carried out in a request and
grant manner. In the second resource allocation problem, we study the
allocation of the overall bits that can be transmitted in a request-grant
cycle to individual eMBB users. We solve this problem by formulating a
constraint optimization problem to minimize the variance of the number of bits
transmitted by the eMBB users. The third contribution of this paper is that we
show a water-filling algorithm, solves the nonlinear program. From simulation,
we show that our scheme, consisting of resource allocation according to Nash
equilibria of a game, persistent random retransmission of URLLC packets and
allocation of eMBB packets by a water-filling algorithm, works better than
four other heuristic methods.

The outline of this paper is as follows. In Section 2 we present the frame
structure of the wireless uplink transmission system. In Section 3 we present
a random persistent retransmission scheme for the URLLC packets. We present an
analysis of the probability that an URLLC packet fails to be transmitted
successfully in this section. In Section 4 we describe a resource block
assignment game to allocate the number of resource blocks in the common region
and that in the grant-based region. In Section 5 we describe a constrained
optimization problem of the eMBB users. We present numerical and simulation
results in Section 6. Finally, we conclude this paper in Section 7.

##  2 Resource blocks and frame structure

In this paper we propose that the frequency band is divided into a common
region and a grant-based (GB) region in Fig. 1. We assume that URLLC packets
can only be transmitted in the common region in a grant-free manner. The eMBB
users make transmission requests. The eMBB users can use resource blocks in
both the grant-based region and the common region to transmit their granted
packets. Since URLLC packets are transmitted in the common region in a grant-
free manner, it is possible that URLLC packets and eMBB packets are collided
in the common region. We assume that eMBB packets are transmitted at a smaller
power, so that URLLC packets are likely to be received successfully. Then,
successive interference cancellation techniques are applied to recover the
collided eMBB packets.

Figure 1: Resource blocks are divided into a common region and a GB region. In
part (a), resource blocks are exactly divided into the two regions. In part
(b), there are resource blocks that are assigned into neither of the two
regions. In part (c), the two regions are overlapped. In other words, there
are resource blocks that belong to the two regions.

In this paper we assume that time is discrete and is divided into time slots.
Multiple time slots are grouped into a time frame. A time frame is divided
into two periods, including a control period and a transmission period. In
control periods, eMBB users make transmission requests. In transmission
periods, eMBB users transmit their data packets that are granted by a base
station. In the control period, the base station also announces how resource
blocks are divided into a common region and a grant-based region. In each time
slot, one eMBB packet can be transmitted. It has been widely recognized that
conventional automatic repeat request (ARQ) mechanisms such as acknowledgments
or timeouts cannot solve the reliability issue of URLLC traffic due to its
stringent delay requirement. To cope with the ultra stringent latency
requirement, it has been proposed that time slots are divided into mini time
slots [13, 14, 15]. An URLLC packet can be transmitted in one mini time slot.
We assume that a time slot is divided into $\tau$ mini time slots. A typical
length of a time slot is 1 millisecond (ms), and a typical value for $\tau$ is
$8$. We show time frames, time slots and mini time slots in Fig. 2.

Figure 2: Schematic description of time slots, mini time slots, time frames
and resource blocks. Two URLLC users may transmit packets in the same mini
time slot. The two packets are shown in rectangles and circles, respectively.
The two packets collide and are retransmitted persistently and randomly.

We summarize in Table I some symbols used in this paper.

Symbol |  Definition
---|---
$\rho$ |  URLLC packet arrival rate
$\tau$ |  Delay requirement of an URLLC packet
$p$ |  URLLC packet retransmission probability
$N$ |  Total number of resource blocks
$\epsilon$ |  Upper bound of the URLLC packet loss probability
$E_{k}$ |  Event that a tagged URLLC packet cannot be successfully transmitted given that the common region has $k$ resource blocks
$b$ |  Cost of resource blocks
$m$ |  Number of eMBB users
$c$ |  Number of bits that the eMBB users can transmit using one resource block in the grant-based region
TABLE I: Definition of parameters used in this paper.

##  3 Retransmission scheme for URLLC traffic

URLLC traffic is characterized by its ultra stringent delay requirement and
extremely high reliability requirement. An URLLC packet needs to be delivered
to its destination within 1 millisecond with a success probability higher than
$1-10^{-5}$. The ultra low latency requirement makes grant-based transmission
mode infeasible for URLLC services. It is commonly believed that grant-free
transmission is suitable for URLLC services [16].

The traditional semi-persistent scheduling (SPS) scheme [6] pre-schedules
resources before sessions begin. SPS would be suitable for URLLC traffic if
the traffic were periodic and predictable. However, URLLC traffic is expected
to be sporadic. SPS could be very inefficient in resource utilization. It has
been widely recognized that conventional automatic repeat request (ARQ)
mechanisms such as acknowledgments or timeouts cannot solve the reliability
issue of URLLC traffic due to its stringent delay requirement [13, 14, 15].
Packets must be retransmitted persistently without waiting for their negative
acknowledgments or expiration of timeout clocks. That is, each URLLC packet
will be transmitted multiple number of times. An URLLC packet is successfully
received if at least one copy is received successfully. This approach has been
adopted by the 3GPP standard [14]. However, the multiple access nature of the
uplink communication makes it possible that multiple URLLC users attempt to
transmit their packets using the same resource block in the same mini time
slot. A collision thus can happen. A persistent transmission scheme would
imply that all subsequent retransmissions are also collided. In this paper we
propose a randomized persistent transmission scheme for URLLC traffic. Suppose
that up to $\tau$ copies of each URLLC packet are transmitted. The first copy
is transmitted at the time when it arrives. Instead of repeating the
transmission of the remaining $\tau-1$ copies, one in each of the $\tau-1$
subsequent mini time slots, each copy is transmitted with probability $p$.
With probability $1-p$, the URLLC user does not transmit a copy. This
mechanism can reduce the number of subsequent collisions and let the collided
URLLC users have a chance to successfully transmit their packets. Fig. 2
contains a graphical illustration. A time slot is divided into several mini
time slots. On the upper side of Fig. 2, the two URLLC users attempt to
transmit packets. Original packet transmissions are shown by colored shapes,
and retransmissions are shown by uncolored shapes. Both of the two URLLC users
transmit in the first mini time slot of a specific time slot using the first
resource block. They persistently and randomly retransmit. The first user
(shown in rectangle) successfully retransmits its packet in the second
retransmission attempt. The second user (shown in circle) successfully
retransmits its packet in the third retransmission attempt.

We now present an analysis of the probability that a tagged URLLC packet
cannot be successfully transmitted. Suppose that there are $n_{1}$ resource
blocks in the common region. Assume that there are $n$ URLLC users. Assume
that URLLC packets arrive according to a Poisson process with rate $\rho$
packets per mini time slot. Assume that each URLLC user randomly chooses a
resource block to transmit its packet. It follows that the URLLC packet
arrivals to a specific resource block, say block $j$, are Poisson with rate
$\rho/n_{1}$ packets per mini time slot. Let $\tilde{\rho}=\rho/n_{1}$ be the
traffic intensity per resource block. We consider a specific packet called
tagged packet. We label the mini time slot that the tagged URLLC packet
arrives as mini time slot $0$. The sender of this tagged packet continuously
attempts to retransmit the packet with probability $p$ in each of the next
$\tau-1$ mini time slots. The sender does not retransmit the packet with
probability $1-p$ in each of the next $\tau-1$ mini time slots. Let
$E_{n_{1}}$ denote the event that the tagged URLLC packet cannot be
successfully transmitted given that the common region has $n_{1}$ resource
blocks.

We now analyze ${\bf\sf P}(E_{n_{1}})$. Recall that we denote the URLLC packet
arrival rate to a resource block by $\tilde{\rho}=\rho/n_{1}$. We also simply
denote the event that the tagged packet fails to be successfully transmitted
by $E$, rather than $E_{n_{1}}$. Let $G_{i}$ be the event that the tagged
packet is not successfully transmitted in mini time slot $i$ for
$i=0,1,\ldots,\tau-1$. Then,

$E=G_{0}\cap G_{1}\cap\cdots\cap G_{\tau-1}.$

Let $X_{j}$ be the number of arrivals in mini time slot $j$ not including the
tagged packet for $j=-(\tau-1),-(\tau-1)+1,\ldots,-1,0,1,\ldots,\tau-1$. Let
$A$ denote the event that $X_{j}=x_{j}$ for $-(\tau-1)\leq j\leq\tau-1$. That
is,

$A=\\{X_{j}=x_{j},-(\tau-1)\leq j\leq\tau-1\\}.$

Conditioning on event $A$, events $G_{0},G_{1},\ldots,G_{\tau-1}$ are
independent. By the law of total probability and conditional independence, one
has

$\displaystyle{\bf\sf P}(E)$ $\displaystyle=\sum_{{x_{j}=0}\atop{-(\tau-1)\leq
j\leq\tau-1}}^{\infty}{\bf\sf
P}\left(\mathop{\bigcap}_{i=0}^{\tau-1}G_{i}\;\Biggl{|}\;A\right)\cdot{\bf\sf
P}(A)$ $\displaystyle=\sum_{{x_{j}=0}\atop{-(\tau-1)\leq
j\leq\tau-1}}^{\infty}\prod_{i=0}^{\tau-1}{\bf\sf P}(G_{i}\;|\;A)\cdot{\bf\sf
P}(A),$ (1)

where

$\displaystyle{\bf\sf P}(A)$ $\displaystyle={\bf\sf
P}(X_{j}=x_{j},-(\tau-1)\leq j\leq\tau-1)$
$\displaystyle=\prod_{j=-(\tau-1)}^{\tau-1}\frac{e^{-\tilde{\rho}}\tilde{\rho}^{x_{j}}}{x_{j}!}.$
(2)

Now we derive the conditional probabilities in (1). If there are new arrivals
in mini time slot $0$, the tagged packet cannot be successfully transmitted.
Assume that there are no new arrivals in mini time slot $0$, the tagged packet
still cannot be successfully transmitted if there is at least one arrival in
mini time slots $-(\tau-1),-(\tau-1)+1,\ldots,-1$ that is retransmitted in
mini time slot $0$. Combining these possibilities, we have

$\displaystyle{\bf\sf P}(G_{0}\;|\;A)$ $\displaystyle={1_{\left\\{x_{0}\geq
1\right\\}}}+{1_{\left\\{x_{0}=0\right\\}}}\left(1-(1-p)^{\sum_{i=-(\tau-1)}^{-1}x_{i}}\right)$
$\displaystyle=1-{1_{\left\\{x_{0}=0\right\\}}}(1-p)^{\sum_{i=-(\tau-1)}^{-1}x_{i}},$
(3)

where notation ${1_{\left\\{A\right\\}}}$ is an indicator function of event
$A$. The value of ${1_{\left\\{A\right\\}}}$ is equal to $1$ if event $A$ is
true, and is equal to $0$ otherwise. For $i=1,2,\ldots,\tau-1$, we have

$\displaystyle{\bf\sf P}(G_{i}\;|\;A)$ $\displaystyle=(1-p)$
$\displaystyle+p\cdot\left[{1_{\left\\{x_{i}\geq
1\right\\}}}+{1_{\left\\{x_{i}=0\right\\}}}\left(1-(1-p)^{\sum_{j=i-(\tau-1)}^{i-1}x_{j}}\right)\right]$
$\displaystyle=1-p\cdot{1_{\left\\{x_{i}=0\right\\}}}(1-p)^{\sum_{j=i-(\tau-1)}^{i-1}x_{j}}.$
(4)

In (3) and (4), notation ${1_{\left\\{A\right\\}}}$ is an indicator function
of event $A$. The value of ${1_{\left\\{A\right\\}}}$ is equal to $1$ if event
$A$ is true, and is equal to $0$ otherwise. Substituting (3) and (4) into (1),
we obtain an expression for ${\bf\sf P}(E)$. For general values of $\tau$, we
are not able to obtain a closed form expression for ${\bf\sf P}(E)$. However,
we show in the next proposition that ${\bf\sf P}(E)$ is an increasing function
of $\tilde{\rho}$. This result is used to study the Nash equilibria of a
resource allocation game in Section 4. The proof is presented in Appendix A.

###### Proposition 1

For any $p$ and $\tau$, ${\bf\sf P}(E)$ is increasing with respect to
$\tilde{\rho}$.

To proceed further, we consider a specific case, in which $\tau=3$. That is,
each new arrival has two opportunities to be retransmitted. Substituting Eqs.
(3) and (4) into (1), we have

$\displaystyle{\bf\sf P}(G_{0}\;|\;A){\bf\sf P}(G_{1}\;|\;A){\bf\sf
P}(G_{2}\;|\;A)$
$\displaystyle\quad=\left(1-{1_{\left\\{x_{0}=0\right\\}}}(1-p)^{x_{-2}+x_{-1}}\right)$
$\displaystyle\quad\cdot\left(1-p\cdot{1_{\left\\{x_{1}=0\right\\}}}(1-p)^{x_{-1}+x_{0}}\right)$
$\displaystyle\quad\cdot\left(1-p\cdot{1_{\left\\{x_{2}=0\right\\}}}(1-p)^{x_{0}+x_{1}}\right).$
(5)

In addition, from the independent increment property of Poisson process, we
have

${\bf\sf
P}(A)=\prod_{j=-2}^{2}\frac{e^{-\tilde{\rho}}\tilde{\rho}^{x_{j}}}{x_{j}!}.$
(6)

Substituting (5) and (6) into (1) and simplifying, we obtain

$\displaystyle{\bf\sf P}(E)$
$\displaystyle=1-(1+2p)\cdot\exp(-3\tilde{\rho}+2(1-p)\tilde{\rho})$
$\displaystyle\quad+p(1+p)\cdot\exp(-4\tilde{\rho}+(1-p)\tilde{\rho}+(1-p)^{2}\tilde{\rho})$
$\displaystyle\quad+p\cdot\exp(-5\tilde{\rho}+3(1-p)\tilde{\rho})$
$\displaystyle\quad-p^{2}\cdot\exp(-5\tilde{\rho}+(1-p)\tilde{\rho}+(1-p)^{2}\tilde{\rho}).$
(7)

A Taylor expansion of (7) around $\tilde{\rho}=0$ is

$\displaystyle{\bf\sf P}(E)$
$\displaystyle\approx(1-p^{2}+p^{3})\tilde{\rho}+o(\tilde{\rho}^{2}).$ (8)

For general values of $\tau$, it is difficult to evaluate (1). Since URLLC
traffic requires a very low packet loss probability of the order $10^{-5}$,
the system must operate in a light traffic condition. We now propose a light
traffic approximation for ${\bf\sf P}(E)$ with general values of $\tau$. It is
interesting to point out that the order of this light traffic approximation
with $\tau=3$ agrees with the Taylor expansion in (8).

We consider a light traffic model, in which excluding the tagged packet in
mini time slot $0$, in mini time slots in range $[-(\tau-1),\tau-1]$, there is
either exactly one packet arrival, or there are no packet arrivals. With
probability $1-(2\tau-1)\tilde{\rho}$ there are no additional arrivals. With
probability $(2\tau-1)\tilde{\rho}$, there is exactly one arrival. If there is
exactly one arrival, this packet arrives in any one of the $2\tau-1$ mini time
slots in range $[-(\tau-1),\tau-1]$ with an equal probability. We now derive
${\bf\sf P}(E)$ for this light traffic model. Let $H_{i}$ be the event that a
packet arrives in mini time slot $i$, where $-(\tau-1)\leq i\leq\tau-1$. If
$1\leq i\leq\tau-1$,

${\bf\sf P}(E\;|\;H_{i})=0.$ (9)

If $-(\tau-1)\leq i\leq-1$,

${\bf\sf P}(E\;|\;H_{i})=p(1-p+p\cdot p)^{\tau+i-1}\cdot(1-p)^{-i}.$ (10)

If $i=0$,

${\bf\sf P}(E\;|\;H_{0})=(1-p+p\cdot p)^{\tau-1}.$ (11)

For $i$ in range $[-(\tau-1),\tau-1]$,

${\bf\sf P}(H_{i})=\frac{(2\tau-1)\tilde{\rho}}{2\tau-1}=\tilde{\rho}.$ (12)

Let $H^{c}$ denote the event that there are no additional packet arrivals.
Then,

$\displaystyle{\bf\sf P}(E\;|\;H^{c})$ $\displaystyle=0$ $\displaystyle{\bf\sf
P}(H^{c})$ $\displaystyle=1-(2\tau-1)\tilde{\rho}.$ (13)

By the law of total probability and Eqs. (9) - (13), we have

$\displaystyle{\bf\sf P}(E)$ $\displaystyle=\sum_{i=-(\tau-1)}^{-1}{\bf\sf
P}(E\;|\;H_{i}){\bf\sf P}(H_{i})$ $\displaystyle+{\bf\sf
P}(E\;|\;H_{0}){\bf\sf P}(H_{0})+{\bf\sf P}(E\;|\;H^{c}){\bf\sf P}(H^{c})$
$\displaystyle=\Bigg{(}\sum_{i=-(\tau-1)}^{-1}p(1-p+p^{2})^{\tau+i-1}(1-p)^{-i}$
$\displaystyle+(1-p+p^{2})^{\tau-1}\Bigg{)}\tilde{\rho}.$ (14)

In a special case where $\tau=3$, (14) agrees with (8).

##  4 A two-player game

In this section we propose a two-step procedure to the resource block
allocation to URLLC users and eMBB users. We first apply the Nash equilibria
of a two-player game to determine the number of resource blocks in the common
region and that in the grant-based region. We then apply the solution of a
constrained nonlinear optimization problem to allocate the number of bits
granted to the eMBB users. We explain the two-player game in this section. We
present the constrained nonlinear optimization problem in Section 5.

We now explain a two-player resource allocation game. The two players are an
URLLC agent and an eMBB agent, negotiating resource blocks on behalf of URLLC
users and eMBB users, respectively. The URLLC agent negotiates the number of
resource blocks in the common region. The eMBB agent negotiates the number of
resource blocks in the grant-based region. Assume that there are $n$ URLLC
users and $m$ eMBB users. Assume that there are totally $N$ resource blocks to
be allocated. The set of actions for the URLLC agent is $\\{0,1,\ldots,N\\}$.
The eMBB agent has the same set of actions. When the URLLC agent takes action
$n_{1}$ and the eMBB agent takes action $n_{2}$, where $0\leq n_{1},n_{2}\leq
N$, the payoff to the URLLC agent is denoted by $P_{URLLC}(n_{1},n_{2})$ and
the payoff to the eMBB agent is denoted by $P_{eMBB}(n_{1},n_{2})$. Recall
that $E_{k}$ denotes the event that a tagged URLLC packet cannot be
successfully transmitted given that the common region has $k$ resource blocks.
From the light traffic approximation of (14), we have

$\displaystyle{\bf\sf P}(E_{k})=$
$\displaystyle\quad\left\\{\begin{array}[]{ll}\Bigg{(}\displaystyle\sum_{i=-(\tau-1)}^{-1}p(1-p+p^{2})^{\tau+i-1}(1-p)^{-i}\\\
+(1-p+p^{2})^{\tau-1}\Bigg{)}\cdot\rho/k&\mbox{if $k\geq 1$}\\\ 1&\mbox{if
$k=0$.}\end{array}\right.$ (18)

We assume that the URLLC agent’s payoff is

$P_{URLLC}(n_{1},n_{2})={1_{\left\\{{\bf\sf
P}(E_{\min(n_{1},N-n_{2})})\leq\epsilon\right\\}}}\cdot{1_{\left\\{n_{1}+n_{2}\leq
N\right\\}}}-\frac{n_{1}}{N}\cdot b,$ (19)

where $\epsilon$ is an upper bound of the URLLC packet loss probability, and
$b$ is a cost of using $N$ resource blocks. We assume that

$0<b<1.$ (20)

Suppose that an eMBB user can transmit $c$ bits using one resource block in
the grant-based region in a time frame. If action profile $(n_{1},n_{2})$ is
taken by the two players, the grant-based region has $\min(n_{2},N-n_{1})$
resource blocks that are not overlapped with resource blocks in the common
region. Thus, the eMBB users can transmit $\min(n_{2},N-n_{1})\cdot c$ bits in
this region. On the other hand, the common region has $\min(n_{1},N-n_{2})$
resource blocks that are not overlapped with resource blocks in the grant-
based region. We assume that the eMBB users transmit at a smaller power, and
their packets are recovered by interference cancellation if they are collided
with URLLC packets. Thus, the number of bits that the eMBB users can transmit
in this region is $a\cdot\min(n_{1},N-n_{2})\cdot c$, where $0<a<1$. Parameter
$a$ reflects the fact that an eMBB user can transmit less data using resource
blocks in the common region. We refer the readers to Fig. 1 for an
illustration of these regions. The total number of bits that eMBB users can
potentially transmit in a time frame given that action profile $(n_{1},n_{2})$
is taken is $T_{n_{1},n_{2}}$, where

$\displaystyle
T_{n_{1},n_{2}}=[\min(n_{2},N-n_{1})+a\cdot\min(n_{1},N-n_{2})]\cdot c.$ (21)

Define

$\displaystyle n_{2}^{*}(n_{1})=$
$\displaystyle\quad\left\\{\begin{array}[]{ll}\min\\{n_{2}:T_{n_{1},n_{2}}\geq
r\\}&\mbox{if $\max_{0\leq n_{2}\leq N}T_{n_{1},n_{2}}\geq r$}\\\
N&\mbox{otherwise,}\end{array}\right.$ (24)

where $r$ is the number of bits requested by eMBB users collectively in a time
frame. Quantity $n_{2}^{*}(n_{1})$ is the minimum number of resource blocks
for the common region in order for eMBB users to transmit $r$ bits. We assume
that the eMBB agent’s payoff is

$\displaystyle P_{eMBB}(n_{1},n_{2})=$
$\displaystyle\quad\left\\{\begin{array}[]{ll}\frac{n_{2}^{*}(n_{1})}{N}\cdot{1_{\left\\{n_{1}+n_{2}\leq
N\right\\}}}-\frac{n_{2}}{N}\cdot b&\mbox{if $n_{2}>n_{2}^{*}(n_{1})$}\\\
\frac{n_{2}}{N}\cdot{1_{\left\\{n_{1}+n_{2}\leq
N\right\\}}}-\frac{n_{2}}{N}\cdot b&\mbox{if $n_{2}\leq
n_{2}^{*}(n_{1})$.}\end{array}\right.$ (27)

If $\min_{0\leq n_{1}\leq N}{\bf\sf P}(E_{n_{1}})\leq\epsilon$, define

$n_{1}^{*}=\min\\{n_{1}:{\bf\sf P}(E_{n_{1}})\leq\epsilon\\}.$ (28)

Clearly, $1\leq n_{1}^{*}\leq N$.

The following theorem characterizes the Nash equilibria of the two-player
resource allocation game. Its proof is presented in Appendix B.

###### Theorem 2

If $\min_{0\leq i\leq N}{\bf\sf P}(E_{i})\leq\epsilon$, there are three cases
as follows.

  1. 1.

If

$n_{2}^{*}(0)\leq N-n_{1}^{*}\ \mbox{and}\ n_{2}^{*}(n_{1}^{*})\leq
N-n_{1}^{*}.$ (29)

action profile $(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ is the only one pure Nash
equilibrium of the two-player resource allocation game.

  2. 2.

If

$n_{2}^{*}(0)>N-n_{1}^{*}\ \mbox{and}\ n_{2}^{*}(n_{1}^{*})\leq N-n_{1}^{*}.$
(30)

holds, action profiles $(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ and
$(0,n_{2}^{*}(0))$ are the only pure Nash equilibria of the two-player
resource allocation game. In addition,

    1. (a)

if $n_{1}^{*}<N$ or $n_{2}^{*}(0)-n_{2}^{*}(n_{1}^{*})<N$,
$(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ is socially better than the other
equilibrium.

    2. (b)

If $n_{1}^{*}=N$ and $n_{2}^{*}(0)-n_{2}^{*}(n_{1}^{*})=N$, both
$(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ and $(0,n_{2}^{*}(0))$ are socially
optimal.

  3. 3.

If

$n_{2}^{*}(0)>N-n_{1}^{*}\ \mbox{and}\ n_{2}^{*}(n_{1}^{*})>N-n_{1}^{*}.$ (31)

holds, action profiles $(n_{1}^{*},N-n_{1}^{*})$ and $(0,n_{2}^{*}(0))$ are
the only pure Nash equilibria of the two-player resource allocation game. In
addition,

    1. (a)

if $n_{1}^{*}<N$ or $n_{2}^{*}(0)<N$, $(n_{1}^{*},N-n_{1}^{*})$ is socially
better than the other equilibrium.

    2. (b)

If $n_{1}^{*}=N$ and $n_{2}^{*}(0)=N$, both $(n_{1}^{*},N-n_{1}^{*})$ and
$(0,n_{2}^{*}(0))$ are socially optimal.

If $\min_{0\leq i\leq N}{\bf\sf P}(E_{i})>\epsilon$, action profile
$(0,n_{2}^{*}(0))$ is the only one pure Nash equilibrium of the two-player
resource allocation game.

Remark. We remark that Theorem 2 contains only three cases (i.e. (29), (30)
and (31)) when $\min_{0\leq i\leq N}{\bf\sf P}(E_{i})\leq\epsilon$. We now
show that $n_{2}^{*}(0)\leq N-n_{1}^{*}$ and $n_{2}^{*}(n_{1}^{*})\leq
N-n_{1}^{*}$ can not occur at the same time. To see this, note from the
definition of $n_{1}^{*}$ in (28) that

$\displaystyle T_{n_{1}^{*},N-n_{1}^{*}}$
$\displaystyle=[\min(N-n_{1}^{*},N-n_{1}^{*})+a\cdot\min(n_{1}^{*},n_{1}^{*})]\cdot
c$ $\displaystyle=(N-n_{1}^{*}+a\cdot n_{1}^{*})\cdot c$
$\displaystyle>[\min(N-n_{1}^{*},N)+a\cdot\min(0,n_{1}^{*})]\cdot c$
$\displaystyle=T_{0,N-n_{1}^{*}}.$ (32)

Suppose that $n_{2}^{*}(0)\leq N-n_{1}^{*}$. From (32), we have
$T_{n_{1}^{*},N-n_{1}^{*}}>T_{0,N-n_{1}^{*}}\geq r$, Thus,
$n_{2}^{*}(n_{1}^{*})\leq N-n_{1}^{*}$.

##  5 A constrained optimization problem

In this section we present a constrained nonlinear program to assign the
number of bits to be granted to the eMBB users. As before, we assume that
there are $m$ eMBB users. eMBB user $j$, where $1\leq j\leq m$, has
cumulatively been granted $z_{j}(t)$ bits up to time frame $t$. Also assume
that user $j$ requests to transmit $r_{j}$ bits in time frame $t+1$. We assume
that $r_{j}>0$ for all $1\leq j\leq m$. Suppose that user $j$ is granted to
transmit $x_{j}$ bits in time frame $t+1$. It follows that

$z_{j}(t+1)=z_{j}(t)+x_{j}.$ (33)

We aim to find $\\{x_{1},x_{2},\ldots,x_{m}\\}$ such that the sample variance
of $\\{z_{1}(t+1),z_{2}(t+1),\ldots,z_{m}(t+1)\\}$ is minimized. That is, we
attempt to minimize

$\frac{1}{m-1}\sum_{j=1}^{m}\left(z_{j}(t+1)-\frac{1}{m}\sum_{k=1}^{m}z_{k}(t+1)\right)^{2}.$
(34)

Since the units of $\\{r_{j},1\leq j\leq m\\}$, $\\{z_{j},1\leq j\leq m\\}$,
and $\\{x_{j},1\leq j\leq m\\}$ are bits, they tend to be large numbers. It is
reasonable to ignore their integer constraints and approximate them by real
numbers.

Suppose that the common region has $n_{1}$ resource blocks and the grant-based
region has $n_{2}$ resource blocks. We propose to use the two-player game
presented in Section 4 to determine $n_{1}$ and $n_{2}$. Specifically, we
assume that $(n_{1},n_{2})$ is the only one pure strategy Nash equilibrium if
the game has only one pure strategy Nash equilibrium, and it is the socially
optimal Nash equilibrium otherwise. In both cases, $n_{1}+n_{2}\leq N$.
According to (21), at most

$\displaystyle L=\min\left((n_{2}+a\cdot n_{1})\cdot
c,\sum_{k=1}^{m}r_{k}\right)$ (35)

eMBB bits can be transmitted in time frame $t+1$, where $r_{k}$ is the number
of bits that user $k$ requests to transmit. We rewrite (34) using $L$, i.e.

$\displaystyle z_{j}(t+1)-\frac{1}{m}\sum_{k=1}^{m}z_{k}(t+1)$
$\displaystyle=z_{j}(t)+x_{j}-\frac{1}{m}\left(\sum_{k=1}^{m}z_{k}(t)+\sum_{k=1}^{m}x_{k}\right)$
$\displaystyle=z_{j}(t)+x_{j}-\frac{1}{m}\left(\sum_{k=1}^{m}z_{k}(t)+L\right)$
$\displaystyle=x_{j}-\eta_{j},$ (36)

where

$\eta_{j}=\frac{1}{m}\left(\sum_{k=1}^{m}z_{k}(t)+L\right)-z_{j}(t)$ (37)

and we have set

$\sum_{k=1}^{m}x_{k}=L.$

For notational simplicity, we shall use notation $z_{j}$ without $t$ in the
rest of this paper. Define

$F_{j}(x)=\frac{(x-\eta_{j})^{2}}{m-1}.$

We have reached the following separable convex nonlinear program with linear
constraints.

$\min_{x_{1},x_{2},\ldots,x_{m}}\sum_{j=1}^{m}F_{j}(x_{j})$ (38)

subject to

$\displaystyle\sum_{j=1}^{m}x_{j}=L$ (39) $\displaystyle 0\leq x_{j}\leq
r_{j}\ \ \mbox{for all $j$}.$ (40)

We use bold face letters to denote vectors. Let
${\mbox{\boldmath$x$}}=(x_{1},x_{2},\ldots,\\\ x_{m})$,
${\mbox{\boldmath$r$}}=(r_{1},r_{2},\ldots,r_{m})$,
${\mbox{\boldmath$e$}}=(1,1,\ldots,1)$ and
${\mbox{\boldmath$0$}}=(0,0,\ldots,0)$. Expressing (38), (39) and (40) in
terms of vectors, we have

$\min_{{\mbox{\boldmath\scriptsize$x$}}}F({\mbox{\boldmath$x$}})$ (41)

subject to

$\displaystyle h({\mbox{\boldmath$x$}})$ $\displaystyle=0$ (42)
$\displaystyle{\mbox{\boldmath$g$}}_{1}({\mbox{\boldmath$x$}})$
$\displaystyle\leq{\mbox{\boldmath$0$}}$ (43)
$\displaystyle{\mbox{\boldmath$g$}}_{2}({\mbox{\boldmath$x$}})$
$\displaystyle\leq{\mbox{\boldmath$0$}},$ (44)

where

$\displaystyle F({\mbox{\boldmath$x$}})$
$\displaystyle\stackrel{{\scriptstyle\scriptstyle\rm
def}}{{=}}\sum_{i=1}^{m}F_{i}(x_{i})$ (45) $\displaystyle
h({\mbox{\boldmath$x$}})$ $\displaystyle\stackrel{{\scriptstyle\scriptstyle\rm
def}}{{=}}\sum_{j=1}^{m}x_{j}-L$ (46)
$\displaystyle{\mbox{\boldmath$g$}}_{1}({\mbox{\boldmath$x$}})$
$\displaystyle\stackrel{{\scriptstyle\scriptstyle\rm
def}}{{=}}-{\mbox{\boldmath$x$}}$ (47)
$\displaystyle{\mbox{\boldmath$g$}}_{2}({\mbox{\boldmath$x$}})$
$\displaystyle\stackrel{{\scriptstyle\scriptstyle\rm
def}}{{=}}{\mbox{\boldmath$x$}}-{\mbox{\boldmath$r$}}.$ (48)

Suppose that ${\mbox{\boldmath$x$}}^{*}$ is a regular point that satisfies
constraints (42), (43) and (44). Then by the Karush-Kuhn-Tucker theorem [17,
p. 352], a necessary condition for ${\mbox{\boldmath$x$}}^{*}$ to be a
relative minimum point for the problem in (41) is that there exist a real
number $\lambda$ and column vectors $\mu$ and $\omega$ in $R^{m}$ with
${\mbox{\boldmath$\mu$}}\geq{\mbox{\boldmath$0$}}$ and
${\mbox{\boldmath$\omega$}}\geq{\mbox{\boldmath$0$}}$ such that

$\displaystyle\nabla F({\mbox{\boldmath$x$}}^{*})+\lambda\nabla
h({\mbox{\boldmath$x$}}^{*})+{\mbox{\boldmath$\mu$}}^{T}\nabla{\mbox{\boldmath$g$}}_{1}({\mbox{\boldmath$x$}}^{*})+{\mbox{\boldmath$\omega$}}^{T}\nabla{\mbox{\boldmath$g$}}_{2}({\mbox{\boldmath$x$}}^{*})$
$\displaystyle={\mbox{\boldmath$0$}}$ (49)
$\displaystyle{\mbox{\boldmath$\mu$}}^{T}{\mbox{\boldmath$g$}}_{1}({\mbox{\boldmath$x$}}^{*})$
$\displaystyle={\mbox{\boldmath$0$}}$ (50)
$\displaystyle{\mbox{\boldmath$\omega$}}^{T}{\mbox{\boldmath$g$}}_{2}({\mbox{\boldmath$x$}}^{*})$
$\displaystyle={\mbox{\boldmath$0$}},$ (51)

where ${\mbox{\boldmath$\mu$}}^{T}$ denotes the transpose of $\mu$. We remark
that $\nabla F({\mbox{\boldmath$x$}})$ and $\nabla h({\mbox{\boldmath$x$}})$
are $1\times m$ row vectors.
$\nabla{\mbox{\boldmath$g$}}_{i}({\mbox{\boldmath$x$}}^{*})$, $i=1,2$, are
$m\times m$ matrices. An inequality constraint such as (43) is said to be
active at a feasible point $x$ if
$({\mbox{\boldmath$g$}}_{1})_{i}({\mbox{\boldmath$x$}})=0$ for some $1\leq
i\leq m$. Otherwise, the inequality constraint is said to be inactive. An
entry of $\mu$ or $\omega$ may be non-zero only if the corresponding
constraint is active. If one knew in advance which constraints were active,
the solution of (49) can easily be solved. Specifically, let sets $S$, $S_{1}$
and $S_{2}$ be defined as

$\displaystyle S$ $\displaystyle=\\{j:\mbox{(\ref{ineq-constraint-1}) and
(\ref{ineq-constraint-2}) are both inactive}\\}$ $\displaystyle S_{1}$
$\displaystyle=\\{j:\mbox{(\ref{ineq-constraint-1}) is active and (\ref{ineq-
constraint-2}) is inactive}\\}$ $\displaystyle S_{2}$
$\displaystyle=\\{j:\mbox{(\ref{ineq-constraint-1}) is inactive and
(\ref{ineq-constraint-2}) is active}\\}.$

Note that it is impossible that both inequality constraints are active at the
same time. It follows that

$S\cup S_{1}\cup S_{2}=\\{1,2,\ldots,m\\}$

and the three sets are mutually exclusive. From the definition in (45),
$\nabla F({\mbox{\boldmath$x$}})$ is a $1\times m$ row vector whose $i$-th
entry is

$\left(\nabla
F({\mbox{\boldmath$x$}})\right)_{i}=\frac{2(x_{i}-\eta_{i})}{m-1}.$ (52)

From the definition in (46), $\nabla h({\mbox{\boldmath$x$}})$ is a $1\times
m$ row vector whose entries are all one, i.e.

$\left(\nabla h({\mbox{\boldmath$x$}})\right)_{i}=1.$ (53)

From the definitions in (47) and (48), both
$\nabla{\mbox{\boldmath$g$}}_{1}({\mbox{\boldmath$x$}})$ and
$\nabla{\mbox{\boldmath$g$}}_{2}({\mbox{\boldmath$x$}})$ are $m\times m$
matrices. It is easy to verify that

$\nabla{\mbox{\boldmath$g$}}_{1}({\mbox{\boldmath$x$}})=-{\mbox{\boldmath$I$}},\quad\nabla{\mbox{\boldmath$g$}}_{2}({\mbox{\boldmath$x$}})={\mbox{\boldmath$I$}},$
(54)

where $I$ denotes the $m\times m$ identity matrix. From (52), (53) and (54),
one can simplify (49) to obtain

$x_{i}^{*}=\eta_{i}-\frac{m-1}{2}(\lambda-\mu_{i}+\omega_{i}),$ (55)

where $\mu_{i}$ and $\omega_{i}$ are the $i$-th entry of $\mu$ and $\omega$,
respectively.

Now we derive $\lambda$, $\mu$ and $\omega$. One can rewrite constraint (39)
as follows.

$\displaystyle L=\sum_{j=1}^{m}x_{j}^{*}$ $\displaystyle=\sum_{j\in
S}x_{j}^{*}+\sum_{j\in S_{1}}x_{j}^{*}+\sum_{j\in S_{2}}x_{j}^{*}$
$\displaystyle=\sum_{j\in S}x_{j}^{*}+0+\sum_{j\in S_{2}}r_{j}$
$\displaystyle=\sum_{j\in S}\eta_{j}-\frac{m-1}{2}\lambda\cdot|S|+\sum_{j\in
S_{2}}r_{j},$ (56)

where the last equality is due to (55) and $|S|$ denotes the number of
elements in set $S$. From (56), we have

$\lambda=\frac{\sum_{j\in S}\eta_{j}+\sum_{j\in S_{2}}r_{j}-L}{(m-1)|S|/2}.$
(57)

Now we determine $\mu$ and $\omega$. For index $i$, if

$0<\eta_{i}-(m-1)\lambda/2<r_{i},$

both inequality constraints are satisfied and are inactive. In this case,
$\mu_{i}=0$ and $\omega_{i}=0$. If $\eta_{i}-(m-1)\lambda/2\leq 0$, constraint
(43) is violated and constraint (44) is satisfied. In this case,

$\displaystyle\omega_{i}$ $\displaystyle=0$ $\displaystyle\mu_{i}$
$\displaystyle=-\frac{2\eta_{i}}{m-1}+\lambda.$ (58)

Clearly, $\mu_{i}>0$. Finally, if $\eta_{i}-(m-1)\lambda/2\geq r_{i}$,
constraint (43) is satisfied and constraint (44) is violated. In this case,

$\displaystyle\mu_{i}$ $\displaystyle=0$ $\displaystyle\omega_{i}$
$\displaystyle=\frac{2(\eta_{i}-r_{i})}{m-1}-\lambda.$ (59)

Clearly, $\omega_{i}>0$.

In the following proposition, we present some properties of the optimal point
given in (55). The proof of the proposition is presented in Appendix C. This
proposition suggests a method to find the optimal solution in (55). This
method is called water-filling algorithm and will be presented in the next
section.

###### Proposition 3

Let ${\mbox{\boldmath$x$}}^{*}$ be the optimal solution of (41).

  1. 1.

Suppose that $z_{i}+r_{i}\geq z_{j}+r_{j}$, then

$x_{i}^{*}=r_{i}\ \mbox{implies that}\ x_{j}^{*}=r_{j}.$ (60)

  2. 2.

Suppose that $z_{i}\geq z_{j}$, then

$x_{j}^{*}=0\ \mbox{implies that}\ x_{i}^{*}=0.$ (61)

  3. 3.

Suppose that $0<x_{i}^{*}<r_{i}$ and $0<x_{j}^{*}<r_{j}$, then

$x_{i}^{*}+z_{i}=x_{j}^{*}+z_{j}.$ (62)

###  5.1 A water-filling algorithm

In this section we propose an algorithm to find the optimal point
${\mbox{\boldmath$x$}}^{*}$ and sets $S,S_{1}$ and $S_{2}$. Proposition 3 at
the end of the last section provides a view to the optimization problem (41).
One can view bits as water, requests as buckets, the number of cumulatively
granted bits as the height of bins on the top of which buckets sit, and the
total number of bits that can be transmitted as the amount of water. In this
view, the solution of problem (41) is a method of filling water into buckets
of finite sizes, sitting on tops of bins of various heights. Specifically, let
there be $m$ buckets. The sizes of the buckets are $r_{1},r_{2},\ldots,r_{m}$.
The $j$-th bucket sits on the top of a bin of height $z_{j}$. There are $L$
units of water. The goal is to fill water into the buckets such that the
height of water surfaces in buckets are as even as possible. See Fig. 3 for a
graphical illustration. Statements (1) and (2) of Proposition 3 imply that
water starts to fill buckets on lower bins first, and buckets with higher tops
are full only after buckets with lower tops are full. Statement (3) says that
partially filled buckets have the same water level. We refer the reader to
Fig. 3 for an example. Finding the optimal solution of (41) is equivalent to
filling water into buckets simultaneously starting from buckets on lowest
bins. Once a bucket is full, it stops receiving water, and water is filled
into other buckets until it is exhausted. This water-filling problem can
easily be solved by a recursive algorithm shown in Algorithm 1. This algorithm
works by raising the water level from one possible level to the next higher
possible level. There are three possible water levels in the algorithm. The
possible water levels are

  1. 1.

bottoms of one of the buckets;

  2. 2.

tops of one of the buckets;

  3. 3.

water level at a partially filled bucket.

Let $P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$ denote the recursive
function in Algorithm 1. The output of this function is the optimal point of
the optimization problem (41), i.e.

${\mbox{\boldmath$x$}}^{*}=P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L).$

Let

$j=\mathop{\rm argmin}_{i}\\{z_{i}:r_{i}>0,1\leq i\leq m\\}.$

User $j$ is said to be least granted.

Figure 3: Illustration of a water-filling problem. There are six buckets
sitting on six bins of heights $z_{1},z_{2},\ldots,z_{6}$. The bucket sizes
are $r_{1},r_{2},\ldots,r_{6}$. Fill $L$ units of water into the buckets
starting from the buckets whose bottoms are the lowest until the water is
exhausted. That is, $x_{1}+x_{2}+\ldots+x_{6}=L$. In this example, buckets 1
and 2 are completely full. Buckets 3 and 4 are partially filled. Bucket 5 and
bucket 6 are completely empty.

Let ${\cal N}$ be the set of users whose requests are non-zero, i.e.

${\cal N}=\\{i:r_{i}>0,1\leq i\leq m\\}.$

Then, sort $z_{i},i\in{\cal N}$, into an ascending sequence. Label user’s
identities such that

$z_{1}=z_{2}=\ldots=z_{k}<z_{k+1}\leq\ldots\leq z_{|{\cal N}|}$ (63)

for some $k$, where $1\leq k\leq|{\cal N}|$, and such that

$r_{1}\leq r_{2}\leq\ldots\leq r_{k}.$ (64)

There are $k$ eMBB users that are least granted in the number of bits. Let

$\tilde{m}=\min\\{r_{1},z_{k+1}-z_{k}\\}.$ (65)

If $L\leq\tilde{m}k$, we will show in the proof of Theorem 4 to be presented
below that the optimal solution is

$x_{j}=\left\\{\begin{array}[]{ll}L/k&1\leq j\leq k\\\ 0&k+1\leq j\leq|{\cal
N}|.\end{array}\right.$

If $L>\tilde{m}k$, we raise the water level by $\tilde{m}$. We introduce a new
set of parameters ${\mbox{\boldmath$z$}}^{\prime}$,
${\mbox{\boldmath$r$}}^{\prime}$, $L^{\prime}$, and compute

${\mbox{\boldmath$y$}}=P({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},m,L^{\prime}),$
(66)

where

$\displaystyle z_{\ell}^{\prime}$
$\displaystyle=\left\\{\begin{array}[]{ll}z_{\ell}+\tilde{m}&1\leq\ell\leq
k\\\ z_{\ell}&k+1\leq\ell\leq|{\cal N}|\end{array}\right.$ (69) $\displaystyle
r_{\ell}^{\prime}$
$\displaystyle=\left\\{\begin{array}[]{ll}r_{\ell}-\tilde{m}&1\leq\ell\leq
k\\\ r_{\ell}&k+1\leq\ell\leq|{\cal N}|\end{array}\right.$ (72) $\displaystyle
L^{\prime}$ $\displaystyle=L-\tilde{m}k.$ (73)

The optimal point $x$ of the original problem is related to $y$. There are two
cases depending on how the minimum in (65) is achieved. In the first case,
$r_{1}<z_{k+1}-z_{k}$. We assume that there are $j$ users whose requests are
the same as $r_{1}$, i.e.

$\tilde{m}=r_{1}=r_{2}=\ldots=r_{j}$

for some $1\leq j\leq k$. In this case, the optimal point of the original
problem $x$ is related to $y$ through

$x_{\ell}=\left\\{\begin{array}[]{ll}\tilde{m}&\ell=1,2,\ldots,j\\\
y_{\ell}+\tilde{m}&\ell=j+1,j+2,\ldots,k\\\ y_{\ell}&k+1\leq\ell\leq|{\cal
N}|.\end{array}\right.$ (74)

In the second case, $r_{1}\geq z_{k+1}-z_{k}$ and

$\tilde{m}=z_{k+1}-z_{k}.$

In this case, the optimal point of the original problem $x$ is related to $y$
through

$x_{\ell}=\left\\{\begin{array}[]{ll}y_{\ell}+\tilde{m}&1\leq\ell\leq k\\\
y_{\ell}&k+1\leq\ell\leq|{\cal N}|.\end{array}\right.$ (75)

The following theorem states that the water-filling algorithm solves the
constrained optimization problem in (38). Its proof is presented in Appendix
C.

###### Theorem 4

If $L\leq\sum_{j=1}^{m}r_{j}$, the water-filling algorithm finds the optimal
point ${\mbox{\boldmath$x$}}^{*}$ that satisfies (55), (57), (58) and (59).

We remark that the idea of viewing resources as water and allocating resources
as if filling water in buckets have been used in several resource allocation
problems [18, 19]. In fact, several algorithms that solve resource allocation
problems are also called water-filling algorithms. There are two types of
water-filling problems in the literature. We refer the reader to Fig. 1 in
[19]. The first type, shown in Fig. 1(a) of the reference, arises in a power
control problem of Gaussian interference channels [20, 21]. However, the
corresponding optimization problem has a logarithmic objective function. In
addition, the assigned power has no upper limit. The corresponding water-
filling problem can be viewed as a special case of our problem, in which the
buckets have infinite capacities. The water-filling problem in Fig. 1(b) of
[19] arises in a max-min fairness allocation problem. The water-filling
problem is also a special case of ours, in which the heights of bins
$\\{z_{j}:1\leq j\leq m\\}$ are all zero. We refer the reader to [18] for more
information on max-min fairness allocation.

Algorithm 1 Water filling algorithm

Function:
${\mbox{\boldmath$x$}}=P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$
Inputs: ${\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m$, and $L$
Outputs: $x$

1:Let ${\cal N}=\\{i:r_{i}>0\\}$.

2:Label the users in set ${\cal N}$ and let $k$ be such that
$z_{1}=z_{2}=\ldots=z_{k}<z_{k+1}\leq\ldots\leq z_{|{\cal N}|}.$

3:In addition, label users $1,2,\ldots,k$ such that  $r_{1}\leq
r_{2}\leq\ldots\leq r_{k}.$

4:Let $\tilde{m}=\min\\{r_{1},z_{k+1}-z_{k}\\}$.

5:if $L\leq\tilde{m}k$ then

6: $x_{1}=x_{2}=\ldots=x_{k}=L/k$

7: $x_{j}=0,\ j=k+1,k+2,\ldots,|{\cal N}|$

8: return $x$

9:end if

10:Evaluate ${\mbox{\boldmath$z$}}^{\prime}$ according to (69).

11:Evaluate ${\mbox{\boldmath$r$}}^{\prime}$ according to (72).

12:Evaluate $L^{\prime}$ according to (73).

13:Call
${\mbox{\boldmath$y$}}=P({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},m,L^{\prime})$.

14:if $r_{1}<z_{k+1}-z_{k}$ then

15: Compute $x$ according to (74).

16:else if $z_{k+1}-z_{k}\leq r_{1}$ then

17: Compute $x$ according to (75).

18:end if

19:return $x$

##  6 Numerical and simulation results

In this section we present numerical and simulation results. We present the
results in the following two sub-sections. In Section 6.1 we examine the light
traffic analysis of ${\bf\sf P}(E_{i})$ by comparing numerical results with
simulation results. In Section 6.2, we study the Nash equilibria of the two-
player resource allocation games and the performance of the water-filling
algorithm.

###  6.1 Light traffic approximation

In this section we first examine the light traffic analysis of ${\bf\sf
P}(E_{i})$ by comparing numerical results with simulation results. Then, we
study the effect of $p$ and $\tau$ to the packet loss probability ${\bf\sf
P}(E_{i})$.

In Fig. 4 we compare the light traffic approximation with $\tau=3$ in (18) and
its exact result in (7). We assume that the retransmission probability $p$ for
URLLC packets is 0.3. We consider three sizes for the common region: $i=24$
for a small common region, $i=48$ for a medium sized common region and $i=100$
for a large common region. From Fig. 4 we see that the light traffic
approximation and the exact result are very close to each other. Next we show
${\bf\sf P}(E_{i})$ versus $p$ in Fig. 5. As expected, ${\bf\sf P}(E_{i})$
decreases with $p$ initially. Later it increases with $p$. There is an optimal
value of $p$ at which ${\bf\sf P}(E_{i})$ achieves a minimal value.

Figure 4: ${\bf\sf P}(E_{n_{1}})$ versus $\rho$.

Figure 5: ${\bf\sf P}(E_{n_{1}})$ versus $p$.

###  6.2 Resource allocation

In this section we study the two resource allocation algorithms presented in
Section 4 and Section 5, respectively. We simulate the system shown in Fig. 6.
The sizes of common region and grant-based region are determined by a resource
allocation algorithm. Each eMBB user has a buffer which can store $B$ eMBB
bits. eMBB packet arrivals in frame $t$ are stored in the corresponding
buffer. If the residual space of a buffer is not enough to store new arrivals,
eMBB packets can be lost. At the beginning of frame $t+1$, eMBB users make a
request to transmit all bits in their buffers. A resource allocation algorithm
determines the number of bits that each user can transmit. Packets that can
not be transmitted stay in the buffers. We will compare the water-filling
algorithm with a few other resource allocation algorithms.

First, we study the packet loss probability of eMBB traffic when wireless
resources are assigned according to the socially optimal Nash equilibrium. We
assume that eMBB bits are scheduled using the water-filling algorithm in
Algorithm 1. We assume that the number of bits to each eMBB buffer in a time
frame is an independent discrete uniform random variable over $[0,3\times
10^{5}]$. Values of other parameters are shown in Table II. We compare the
performance with that of a few other resource allocation strategies.
Specifically, we compare the performance of socially optimal Nash equilibrium
$(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ with the non-optimal Nash equilibrium
$(0,n_{2}^{*}(0))$, profiles $(n_{1}^{*}+1,N-n_{1}^{*}-1)$ and $(N-1,1)$, and
a random profile. In time slot $t$, the random profile assigns $N_{1,t}$
resource blocks to the common region, and assigns $N_{2,t}$ resource blocks to
the grant-based region. Discrete random variable $N_{1,t}$ is uniform over the
interval $[0,N]$. Conditioning on $N_{1,t}$, discrete random variable
$N_{2,t}$ is uniform over the interval $[0,N-N_{1,t}]$. The packet loss
probability of eMBB traffic is shown in Fig. 7. The packet loss probability of
URLLC traffic is shown in Fig. 8. From these two figures, we see that the
resource allocation based on socially optimal profile
$(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ achieves a good balance between the two
loss probabilities. We have also simulated the social payoff, which is defined
as the sum of payoffs of two game players. The result is shown in Fig. 9. As
expected, the socially optimal Nash equilibrium has the largest social
payoffs.

Figure 6: System architecture. Parameter | Value | Parameter | Value
---|---|---|---
$\rho$ | $6.5\times 10^{-4}$ | $\tau$ | 8
| packets per |  |
| mini-slot |  |
$p$ | 0.3 | $N$ | 60
$\epsilon$ | $10^{-5}$ | $n_{1}^{*}$ | 30
$b$ | 0.8 | $c$ |  $3.2\times 10^{4}$ bits
$B$ |  $3.8\times 10^{5}$ bits | $m$ | 8
TABLE II: Parameter values.

Figure 7: Packet loss probability of eMBB traffic versus $a$.

Figure 8: Packet loss probability of URLLC traffic versus $a$.

Figure 9: Social payoffs versus $a$.

Finally, we compare the water-filling algorithm with a few other algorithms.
In this study, we assume that resource blocks are partitioned into common
region and grant-based region according to the profile
$(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$. Besides the water-filling algorithm, other
resource allocation algorithms that we consider are smallest request first,
largest request first, random order allocation, two-step averages, and max-min
fairness allocation. We now briefly explain these allocation methods. The
smallest request first and the largest request first methods are
straightforward. Sort the requests $\\{r_{j}:1\leq j\leq m\\}$ into an
ascending list. In the case of smallest request first, grant requests in the
ascending order. On the other hand, in the case of largest request first,
grant requests in the descending order. In either case, grant requests until
the allowance $L$ is exhausted or all requests have been satisfied. The random
order allocation is similar. Arrange requests according to a randomly selected
permutation and grant requests according to the order of this permutation
until $L$ bits are exhausted or all requests are satisfied. The two-step
average method works as follows. There are two steps in this method. In the
fist step, compute and grant requests according to the average $v=L/m$. The
number of bits granted to user $i$ in step 1 is $x_{i}^{(1)}=\min(r_{i},v)$.
The residual request size of user $i$ is $r_{i}^{(1)}=r_{i}-x_{i}^{(1)}$. The
original grant allowance is $L$ and the residual allowance after the first
step is $L^{(1)}=L-\sum_{i=1}^{m}x_{i}^{(1)}$. In the second step, sort
$\\{r_{i}^{(1)}:1\leq i\leq m\\}$ into a descending list. Grant requests in
the order of this list until $L^{(1)}$ is exhausted or all requests have been
satisfied. Finally, we explain the max-min fairness method. This allocation
method was proposed to allocate bandwidth [18]. The max-min fairness
allocation can be solved by a water-filling problem. This problem is a special
case of $P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$, in which $z$ is
set to a zero vector, i.e. ${\mbox{\boldmath$z$}}={\mbox{\boldmath$0$}}$.
Although the max-min allocation can be solved by a water-filling algorithm,
there is a simpler algorithm to solve it. First, sort $\\{r_{i}:1\leq i\leq
m\\}$ into an ascending sequence. Relabeling indexes if necessary, we can
assume that the sequence $\\{r_{i}:1\leq i\leq m\\}$ is ascending. For
convenience, let $r_{i}^{(0)}=r_{i}$ for $i=1,2,\ldots,m$, and let
$L^{(0)}=L$. In step $j$, compute the average number of bits that can be
granted to $j$ users, i.e.

$v^{(j)}=L^{(j-1)}/(m-j+1).$

In step $j$, allocate

$x_{i}^{(j)}=\min(v^{(j)},r_{i}^{(j-1)})$

to user $i$. Calculate the residual request sizes for the next step, i.e.

$r_{i}^{(j)}=r_{i}^{(j-1)}-x_{i}^{(j-1)},\qquad 1\leq i\leq m.$

If $r_{i}^{(j)}=0$ for all $i$, the algorithm stops. Otherwise, the algorithm
continues and we compute the residual allowance

$L^{(j)}=L^{(j-1)}-\sum_{i=1}^{m}x_{i}^{(j)}.$

We simulate the water-filling algorithm and the five other resource allocation
algorithms for $T=10^{6}$ frames. The sample variance of a sequence
$\\{z_{i}(t+1):1\leq i\leq m\\}$ is defined in (34). Another well known
measure of fairness is the Jain’s fairness index defined as

$\frac{\left(\sum_{i=1}^{m}z_{i}(t+1)\right)^{2}}{m\cdot\sum_{i=1}^{m}z_{i}(t+1)^{2}}.$

The eMBB packet loss probability, sample variance (defined in (34)) and Jain’s
fairness index in Figs. 10, 11 and 12, respectively. From Fig. 10, the
smallest request first method has the smallest eMBB packet loss probability.
The water-filling algorithm and the max-min fairness method have very similar
results. From Figs. 11 and 12, the water-filling algorithm has the best
performance in both sample variance and the Jain’s fairness index.

Figure 10: eMBB packet loss probability versus $m$.

Figure 11: Sample variance versus $m$.

Figure 12: Jain’s fairness index versus $m$.

##  7 Conclusions

In this paper we studied a wireless resource allocation problem and a packet
scheduling problem of the URLLC traffic and eMBB traffic in an uplink wireless
network. We proposed to divide frequencies into a grant-based region for eMBB
traffic only, and a common region for both URLLC traffic and eMBB traffic. To
cope with the ultra stringent latency and reliability requirement of URLLC
packets, we proposed a persistent random retransmission scheme for URLLC
packets. We derived the probability that a randomly selected URLLC packet
fails to meet its latency requirement. We proposed a two-player game to
determine the sizes of the common region and the grant-based region. We
studied the Nash equilibria of this game. For the request-and-grant allocation
of eMBB packets, we proposed a constraint nonlinear program which minimizes
the variance of the number of bits granted to the eMBB users. We show that a
water-filling algorithm can recursively solve the nonlinear program. From
simulation, we show that our scheme, consisting of resource allocation
according to Nash equilibria of a game, persistent random transmission of
URLLC packets and scheduling eMBB packets by a water-filling algorithm, works
better than four other heuristic methods.

Appendix A

Proof of Proposition 1. Let $X_{j}$ denote the number of Poisson arrivals in
mini time slot $j$, $-(\tau-1)\leq j\leq\tau-1$, not including the tagged
arrival in mini time slot $0$. We use bold face letters to denote vectors. Let
$x$ be a $1\times(2\tau-1)$ row vector, i.e.

${\mbox{\boldmath$x$}}=(x_{-(\tau-1)},\ldots,x_{-1},x_{0},x_{1},\ldots,x_{\tau-1}).$

Define function $f_{i}({\mbox{\boldmath$x$}})$ according to (3) and (4)

$\displaystyle f_{i}({\mbox{\boldmath$x$}})=$
$\displaystyle\left\\{\begin{array}[]{ll}1-{1_{\left\\{x_{0}=0\right\\}}}(1-p)^{\sum_{i=-(\tau-1)}^{-1}x_{i}}&\mbox{if
$i=0$}\\\
1-p\cdot{1_{\left\\{x_{i}=0\right\\}}}(1-p)^{\sum_{j=i-(\tau-1)}^{i-1}x_{j}}&\mbox{if
$1\leq i\leq\tau-1$.}\end{array}\right.$ (78)

Also define function

$f({\mbox{\boldmath$x$}})=\prod_{i=0}^{\tau-1}f_{i}({\mbox{\boldmath$x$}}).$

It is easy to verify that $f_{i}({\mbox{\boldmath$x$}})$ is non-decreasing
with respect to $x_{j}$ for any $j=-(\tau-1),\ldots,\tau-1$. Thus,
$f({\mbox{\boldmath$x$}})$ is also non-decreasing. Let
${\mbox{\boldmath$X$}}=(X_{-(\tau-1)},\ldots,X_{-1},X_{0},X_{1},\ldots,X_{\tau-1})$,
where $X_{i}$ is the number of Poisson arrivals in mini time slot $i$. One can
rewrite (1) as follows,

$\displaystyle{\bf\sf E}[{1_{\left\\{E\right\\}}}\;|\;{\mbox{\boldmath$X$}}]$
$\displaystyle=\prod_{i=0}^{\tau-1}f_{i}({\mbox{\boldmath$X$}})=f({\mbox{\boldmath$X$}})$
$\displaystyle{\bf\sf P}(E)$ $\displaystyle={\bf\sf E}[{\bf\sf
E}[{1_{\left\\{E\right\\}}}\;|\;X]]={\bf\sf E}[f({\mbox{\boldmath$X$}})].$

Let ${\mbox{\boldmath$Y$}}=\\{Y_{j}:-(\tau-1)\leq j\leq\tau-1\\}$ be a
sequence of independent Poisson random variables with mean $\rho^{\prime}$,
where $\rho^{\prime}\geq\rho$. It is well known that Poisson random variables
are stochastically increasing in means (see Example 9.2(B) on page 411 of Ross
[22]). Thus,

$Y_{j}\geq_{\rm st}X_{j}$

for all $j=-(\tau-1),\ldots,\tau-1$. By Example 9.2(A) of Ross [22, p. 410],

${\bf\sf E}[f({\mbox{\boldmath$Y$}})]\geq{\bf\sf
E}[f({\mbox{\boldmath$X$}})].$

It follows that ${\bf\sf P}(E)$ increases with $\tilde{\rho}$.

Appendix B

Proof of Theorem 2. Suppose that $(n_{1},n_{2})$ is an action profile of the
two-player resource allocation game. Recall that profile $(n_{1},n_{2})$ is
called a pure Nash equilibrium if action $n_{1}$ and action $n_{2}$ are best
responses to each other.

Now we consider the case, in which $\min_{0\leq i\leq N}{\bf\sf
P}(E_{i})\leq\epsilon$. There are three sub-cases. Since their proofs are
similar, we only present the proof of the second sub-case. Proofs of the other
sub-cases are omitted. We also omit the proof of the case, in which
$\min_{0\leq i\leq N}{\bf\sf P}(E_{i})>\epsilon$. Suppose that (30) holds. We
shall show that profiles $(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ and
$(0,n_{2}^{*}(0))$ are pure Nash equilibria. We also need to show that profile
$(i,j)$ is not a Nash equilibrium if
$(i,j)\neq(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ and $(i,j)\neq(0,n_{2}^{*}(0))$.
In addition, we shall discuss the social optimality of these two equilibria.

To begin, we show that profile $(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ is a pure
Nash equilibrium. We show that action $n_{1}^{*}$ of the URLLC agent is a best
response to action $n_{2}^{*}(n_{1}^{*})$ of the eMBB agent. For
$i<n_{1}^{*}$,

$P_{URLLC}(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))=1-\frac{n_{1}^{*}}{N}b$ (79)

and

$P_{URLLC}(i,n_{2}^{*}(n_{1}^{*}))=-\frac{i}{N}b.$

Since $0<b<1$, it follows that

$P_{URLLC}(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))>P_{URLLC}(i,n_{2}^{*}(n_{1}^{*}))$
(80)

for all $i<n_{1}^{*}$. Now consider $i$, such that $i>n_{1}^{*}$. We have

$\displaystyle P_{URLLC}(i,n_{2}^{*}(n_{1}^{*}))$
$\displaystyle=1\cdot{1_{\left\\{i+n_{2}^{*}(n_{1}^{*})\leq
N\right\\}}}-\frac{ib}{N}$
$\displaystyle=\left\\{\begin{array}[]{ll}-\frac{ib}{N}&\mbox{if
$i>N-n_{2}^{*}(n_{1}^{*})$}\\\ 1-\frac{ib}{N}&\mbox{if $i\leq
N-n_{2}^{*}(n_{1}^{*})$}\end{array}\right.$ (83)

Comparing (79) with (83), we obtain

$P_{URLLC}(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))>P_{URLLC}(i,n_{2}^{*}(n_{1}^{*}))$
(84)

for all $i>n_{1}^{*}$. (80) and (84) establish that action $n_{1}^{*}$ is a
best response to action $n_{2}^{*}(n_{1}^{*})$. We also need to show that
action $n_{2}^{*}(n_{1}^{*})$ is a best response to action $n_{1}^{*}$. The
payoff of the eMBB agent when profile $(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ is
taken is

$\displaystyle P_{eMBB}(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$
$\displaystyle=\frac{n_{2}^{*}(n_{1}^{*})}{N}{1_{\left\\{n_{1}^{*}+n_{2}^{*}(n_{1}^{*})\leq
N\right\\}}}-\frac{n_{2}^{*}(n_{1}^{*})b}{N}$
$\displaystyle=\frac{n_{2}^{*}(n_{1}^{*})(1-b)}{N}.$ (85)

The payoff of the eMBB agent for an arbitrary action $j$ is

$P_{eMBB}(n_{1}^{*},j)=\left\\{\begin{array}[]{ll}\frac{n_{2}^{*}(n_{1}^{*})}{N}{1_{\left\\{n_{1}^{*}+j\leq
N\right\\}}}-\frac{jb}{N}&\mbox{if $j>n_{2}^{*}(n_{1}^{*})$}\\\
\frac{j}{N}{1_{\left\\{n_{1}^{*}+j\leq N\right\\}}}-\frac{jb}{N}&\mbox{if
$j\leq n_{2}^{*}(n_{1}^{*})$.}\end{array}\right.$

Since $n_{1}^{*}+n_{2}^{*}(n_{1}^{*})\leq N$, the preceding payoff is equal to

$P_{eMBB}(n_{1}^{*},j)=\left\\{\begin{array}[]{ll}\frac{n_{2}^{*}(n_{1}^{*})}{N}{1_{\left\\{n_{1}^{*}+j\leq
N\right\\}}}-\frac{jb}{N}&\mbox{if $j>n_{2}^{*}(n_{1}^{*})$}\\\
\frac{j}{N}-\frac{jb}{N}&\mbox{if $j\leq
n_{2}^{*}(n_{1}^{*})$.}\end{array}\right.$ (86)

From (85), (86) and assumption $b<1$, we have

$P_{eMBB}(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))\geq P_{eMBB}(n_{1}^{*},j).$

Hence, action $n_{2}^{*}(n_{1}^{*})$ is a best response to action $n_{1}^{*}$.

Next, we show that profile $(0,n_{2}^{*}(0))$ is a pure Nash equilibrium.
First, we show that action $0$ of the URLLC agent is a best response to action
$n_{2}^{*}(0)$ of the eMBB agent. Since

${\bf\sf P}(E_{\min(0,N-n_{2}^{*}(0))})={\bf\sf P}(E_{0})=1>\epsilon$

and $0+n_{2}^{*}(0)\leq N$, it follows that

$P_{URLLC}(0,n_{2}^{*}(0))=0.$ (87)

For arbitrary $i$,

$P_{URLLC}(i,n_{2}^{*}(0))={1_{\left\\{{\bf\sf
P}(E_{\min(i,N-n_{2}^{*}(0))})\leq\epsilon\right\\}}}{1_{\left\\{i+n_{2}^{*}(0)\leq
N\right\\}}}-\frac{ib}{N}.$ (88)

We know that $n_{1}^{*}+n_{2}^{*}(0)>N$. For $1\leq i<n_{1}^{*}$, the first
indicator function is $0$. For $i\geq n_{1}^{*}$, the second indicator
function is $0$. Thus, (88) is reduced to

$P_{URLLC}(i,n_{2}^{*}(0))=-\frac{ib}{N}<P_{URLLC}(0,n_{2}^{*}(0))$

for the both cases. Hence, action $0$ is a best response to action
$n_{2}^{*}(0)$. We also need to show that action $n_{2}^{*}(0)$ is a best
response to action $0$. Note that

$\displaystyle P_{eMBB}(0,n_{2}^{*}(0))$
$\displaystyle=\frac{n_{2}^{*}(0)}{N}{1_{\left\\{0+n_{2}^{*}(0)\leq
N\right\\}}}-\frac{n_{2}^{*}(0)b}{N}$
$\displaystyle=\frac{n_{2}^{*}(0)(1-b)}{N},$ (89)

since $n_{2}^{*}(0)\in[0,N]$. On the other hand, the payoff of the eMBB agent
for an arbitrary action $j$ is

$\displaystyle P_{eMBB}(0,j)$
$\displaystyle=\left\\{\begin{array}[]{ll}\frac{n_{2}^{*}(0)}{N}{1_{\left\\{0+j\leq
N\right\\}}}-\frac{jb}{N}&\mbox{if $j>n_{2}^{*}(0)$}\\\
\frac{j}{N}{1_{\left\\{0+j\leq N\right\\}}}-\frac{jb}{N}&\mbox{if $j\leq
n_{2}^{*}(0)$}\end{array}\right.$
$\displaystyle=\left\\{\begin{array}[]{ll}\frac{n_{2}^{*}(0)}{N}-\frac{jb}{N}&\mbox{if
$j>n_{2}^{*}(0)$}\\\ \frac{j(1-b)}{N}&\mbox{if $j\leq
n_{2}^{*}(0)$.}\end{array}\right.$

Since $b<1$, it follows that

$P_{eMBB}(0,n_{2}^{*}(0))\geq P_{eMBB}(0,j)$

This implies that action $n_{2}^{*}(0)$ is a best response to action $0$.

Now we show that any profile $(i,j)$ such that
$(i,j)\neq(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ and $(i,j)\neq(0,n_{2}^{*}(0))$,
cannot be a Nash equilibrium. For each profile $(i,j)$ that is not an
equilibrium, we identify a better response $i^{\prime}$ to action $j$, or a
better response $j^{\prime}$ to action $i$. In the process of proving that
$(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ is a pure Nash equilibrium, for instance,
we have shown that actions $n_{1}^{*}$ and $n_{2}^{*}(n_{1}^{*})$ are best
responses to each other. Thus, to show that profile $(i,j)$ is not a Nash
equilibrium, it is sufficient to assume that

$\begin{array}[]{l}i\neq n_{1}^{*},\ \ i\neq 0\\\ j\neq n_{2}^{*}(n_{1}^{*}),\
\ j\neq n_{2}^{*}(0).\end{array}$ (90)

For any profile $(i,j)$ such that (90) holds, we show that

  1. 1.

if $j\leq N-n_{1}^{*}$, then action $n_{1}^{*}$ is a better response to action
$j$ than action $i$;

  2. 2.

if $j>N-n_{1}^{*}$, then action $0$ is a better response to action $j$ than
action $i$.

Indeed, suppose that $j\leq N-n_{1}^{*}$, we have

$\displaystyle P_{URLLC}(n_{1}^{*},j)$
$\displaystyle=1\cdot{1_{\left\\{n_{1}^{*}+j\leq
N\right\\}}}-\frac{n_{1}^{*}b}{N}$ $\displaystyle=1-\frac{n_{1}^{*}b}{N}$
$\displaystyle P_{URLLC}(i,j)$ $\displaystyle={1_{\left\\{{\bf\sf
P}(E_{\min(i,N-j)}\leq\epsilon\right\\}}}{1_{\left\\{i+j\leq
N\right\\}}}-\frac{ib}{N}$ $\displaystyle\ \
\left\\{\begin{array}[]{ll}=-\frac{ib}{N}&\mbox{if $i<n_{1}^{*}$}\\\ \leq
1-\frac{ib}{N}&\mbox{if $i>n_{1}^{*}$.}\end{array}\right.$

The preceding equations imply that

$P_{URLLC}(n_{1}^{*},j)\geq P_{URLLC}(i,j)$

for $i\neq n_{1}^{*}$ and $j\leq N-n_{1}^{*}$. Now assume that
$j>N-n_{1}^{*}$, we have

$\displaystyle P_{URLLC}(0,j)$ $\displaystyle=0-\frac{0\cdot b}{N}=0$
$\displaystyle P_{URLLC}(i,j)$
$\displaystyle=\left\\{\begin{array}[]{ll}{1_{\left\\{{\bf\sf
P}(E_{\min(i,N-j)})\leq\epsilon\right\\}}}\cdot 0-\frac{ib}{N}&\mbox{if
$i>n_{1}^{*}$}\\\ -\frac{ib}{N}&\mbox{if $1\leq i\leq
n_{1}^{*}$}\end{array}\right.$ $\displaystyle=-\frac{ib}{N}.$

From the preceding equations, clearly $P_{URLLC}(0,j)>P_{URLLC}(i,j)$. Thus,
we have proven that action $0$ is a better response than action $i$.

Finally, we prove the social optimality of these two equilibria. Let
$P_{sum}(i,j)$ denote the sum of payoffs of the two players when action
profile $(i,j)$ is used. That is,

$P_{sum}(i,j)=P_{URLLC}(i,j)+P_{eMBB}(i,j).$

Now

$\displaystyle P_{sum}(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$
$\displaystyle=1-\frac{n_{1}^{*}}{N}\cdot
b+\frac{n_{2}^{*}(n_{1}^{*})}{N}\cdot(1-b)$ $\displaystyle
P_{sum}(0,n_{2}^{*}(0))$ $\displaystyle=\frac{n_{2}^{*}(0)}{N}\cdot(1-b).$

It follows that

$\displaystyle
P_{sum}(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))-P_{sum}(0,n_{2}^{*}(0))=$
$\displaystyle\qquad\frac{N-n_{1}^{*}b+(n_{2}^{*}(n_{1}^{*})-n_{2}^{*}(0))(1-b)}{N}.$
(91)

Clearly, $n_{1}^{*}\leq N$. From (30), we have

$0\leq n_{2}^{*}(0)-n_{2}^{*}(n_{1}^{*})\leq N.$

Thus,

$P_{sum}(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))-P_{sum}(0,n_{2}^{*}(0))\geq 0.$

If $n_{1}^{*}<N$ or $n_{2}^{*}(0)-n_{2}^{*}(n_{1}^{*})<N$, strategy profile
$(n_{1}^{*},n_{2}^{*}(n_{1}^{*}))$ is strictly socially better than profile
$(0,n_{2}^{*}(0))$. On the other hand, if $n_{1}^{*}=N$ and
$n_{2}^{*}(0)-n_{2}^{*}(n_{1}^{*})=N$, the two profiles are equally good
socially.

We have completed the proof.

Appendix C

In this appendix we prove Proposition 3 and then Theorem 4.

Proof of Proposition 3. We first prove part (1) of Proposition 3. Since
$x_{i}^{*}=r_{i}$, we have

$\eta_{i}-(m-1)\lambda/2\geq r_{i}.$

From (37) we have

$\frac{1}{m}\left(\sum_{k=1}^{m}z_{k}+L\right)-(m-1)\lambda/2\geq
r_{i}+z_{i}\geq z_{j}+r_{j}.$

Hence,

$\eta_{j}-(m-1)\lambda/2\geq r_{j},$

which implies that $x_{j}^{*}=r_{j}$. We have completed the proof of part (1).
The proof of part (2) is similar and we omit the proof. We now consider part
(3). Since $0<x_{i}^{*}<r_{i}$ and $0<x_{j}^{*}<r_{j}$, it follows that
$\mu_{i}=\omega_{i}=\mu_{j}=\omega_{j}=0$. Thus,

$\displaystyle x_{i}^{*}+z_{i}$
$\displaystyle=\frac{1}{m}\left(\sum_{k=1}^{m}z_{k}+L\right)-z_{i}-\frac{(m-1)\lambda}{2}+z_{i}$
$\displaystyle=\frac{1}{m}\left(\sum_{k=1}^{m}z_{k}+L\right)-z_{j}-\frac{(m-1)\lambda}{2}+z_{j}$
$\displaystyle=x_{j}^{*}+z_{j}.$

The proof is completed.

Proof of Theorem 4. Now we prove Theorem 4. We note that if
$\sum_{j=1}^{m}r_{j}<L$, the feasible region of the optimization problem in
(38) is empty. If $\sum_{j=1}^{m}r_{j}=L$, it is clear that the feasible
region contains only one point. Thus, the solution is $x_{j}=r_{j}$ for all
$1\leq j\leq m$. Thus, in the proof we assume that $\sum_{j=1}^{m}r_{j}>L$.

Suppose that $L\leq\tilde{m}k$. In this case, the optimal solution of (41) is
stated in Claim 1 below.

  * •

Claim 1 If $L\leq\tilde{m}k$, the optimal solution is $x_{j}=L/k$ for $1\leq
j\leq k$, and $x_{j}=0$ for $k+1\leq j\leq|{\cal N}|$.

On the other hand, if $\tilde{m}k<L<\sum_{j=1}^{m}r_{j}$, the proof needs the
following claims.

  * •

Claim 2 The objective function of
$P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$ and that of
$P({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},m,L^{\prime})$
are the same.

  * •

Claim 3 The objective function is convex. There is a unique optimal point.

  * •

Claim 4 Assume that $L\geq\tilde{m}k$. The feasible region $\Omega^{\prime}$
of
$P({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},m,L^{\prime})$
is a subset of the feasible region $\Omega$ of
$P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$. In addition, the optimal
point of $P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$ is in
$\Omega^{\prime}$.

  * •

Claim 5 Since $L<\sum_{j=1}^{m}r_{j}$ and each call to $P$ strictly decreases
$L$, the recursive function terminates in a finite number of calls.

If $L>\tilde{m}k$, the water-filling algorithm in Algorithm 1 raises the water
level by $\tilde{m}$ and call
$P({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},m,L^{\prime})$.
From Claim 2, the original objective function and the new objective function
are the same. From Claim 3, the constrained nonlinear program has a unique
solution. From Claim 4, this unique solution is in the feasible region of
$P({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},m,L^{\prime})$.
The water-filling algorithm successively reduces the feasible region. In this
process, $L$ is strictly reduced. In a finite number of steps, the condition
$L\leq\tilde{m}k$ will hold and the algorithm will stop.

Now we prove Claims 1, 2, 4. Claims 3 and 5 are obvious. We omit their proofs.
Proof of Claim 1. First, if $L=0$, it is obvious that the feasible region has
only one point and $x_{i}=0$ for all $i$ gives the optimal solution. We assume
that $0<L\leq\tilde{m}k$. Let $x$ denote the optimal solution. If $x_{1}=0$,
by the second statement in Proposition 3 we have $x_{i}=0$ for all $2\leq
i\leq|{\cal N}|$. It follows that $\sum_{i=1}^{m}x_{i}=0$, which contradicts
the assumption that $L>0$. If $x_{1}=r_{1}$, then

$\frac{1}{m}\left(\sum_{i=1}^{m}z_{i}+L\right)-z_{1}-\frac{m-1}{2}\lambda>r_{1}.$

Since $z_{j}=z_{1}$ for all $j=2,3,\ldots,k$, it follows that

$\frac{1}{m}\left(\sum_{i=1}^{m}z_{i}+L\right)-z_{j}-\frac{m-1}{2}\lambda>r_{1}.$

Thus, $x_{j}>r_{1}$. We have

$L=\sum_{i=1}^{m}x_{i}\geq\sum_{i=1}^{k}x_{i}>r_{1}k\geq k\tilde{m},$

which is a contradiction. Thus,

$0<x_{1}<r_{1}.$ (92)

Since $z_{1}=z_{2}=\ldots=z_{k}$ and $r_{1}\leq r_{2}\leq\ldots\leq r_{k}$, it
follows from the definition of $\eta_{i}$ in (37) that
$x_{1}=x_{2}=\ldots=x_{k}$. We now show that $x_{j}=0$ for $k+1\leq
j\leq|{\cal N}|$. If $z_{k+1}-z_{k}\geq r_{1}$, i.e. $\tilde{m}=r_{1}$, we
have

$\displaystyle\eta_{k+1}-\frac{m-1}{2}\lambda$
$\displaystyle=\frac{1}{m}\left(\sum_{i=1}^{m}z_{i}+L\right)-z_{1}+z_{1}-z_{k+1}-\frac{m-1}{2}\lambda$
$\displaystyle=x_{1}+z_{1}-z_{k+1}$ $\displaystyle<r_{1}-(z_{k+1}-z_{k})\leq
0.$

Thus, $x_{j}=0$ for $k+1\leq j\leq|{\cal N}|$. On the other hand, consider the
case where $z_{k+1}-z_{k}<r_{1}$, i.e. $\tilde{m}=z_{k+1}-z_{k}$. Assume that
$x_{j}>0$ for some $j=k+1,\ldots,|{\cal N}|$. We look for a contradiction.
Since $x_{j}>0$, we have

$\frac{1}{m}\left(\sum_{i=1}^{m}z_{i}+L\right)-z_{j}-\frac{m-1}{2}\lambda>0.$

From the preceding inequality and the definition of $\eta_{1}$ in (37), we
have

$\eta_{1}-\frac{m-1}{2}\lambda>z_{j}-z_{1}.$

This implies that $x_{1}>z_{j}-z_{1}$. Thus,

$L=\sum_{i=1}^{m}x_{i}\geq\sum_{i=1}^{k}x_{i}>k(z_{j}-z_{1})\geq k\tilde{m}.$

This is a contradiction.

Proof of Claim 2. We assume that $L>\tilde{m}k$. We use notation
$\eta_{j}({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},L)$ to explicitly show
the dependency with ${\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},$ and $L$.
From (37) and ${\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime}$,
and $L^{\prime}$ defined in (69), (72) and (73), we have

$\displaystyle\eta_{j}({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},L^{\prime})$
(93)
$\displaystyle=\frac{1}{m}\left(\sum_{\ell=1}^{m}z_{\ell}^{\prime}(t)+L^{\prime}\right)-z_{j}^{\prime}(t)$
$\displaystyle=\frac{1}{m}\left(\sum_{\ell=1}^{k}z_{\ell}^{\prime}(t)+\sum_{\ell=k+1}^{m}z_{\ell}^{\prime}(t)+L^{\prime}\right)-z_{j}^{\prime}(t)$
$\displaystyle=\frac{1}{m}\left(\sum_{\ell=1}^{k}(z_{\ell}(t)+\tilde{m})+\sum_{\ell=k+1}^{m}z_{\ell}(t)+L-\tilde{m}k\right)$
$\displaystyle\quad-z_{j}^{\prime}(t)$
$\displaystyle=\eta_{j}({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},L)-\tilde{m}{1_{\left\\{1\leq
j\leq k\right\\}}},$ (94)

where ${1_{\left\\{A\right\\}}}$ is equal to $1$ if $A$ is true, and is equal
to $0$ otherwise. It follows from (94) that one can rewrite

$x_{j}-\eta_{j}({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},L)=y_{j}-\eta_{j}({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},L^{\prime})$

through a change of variables

$y_{j}=x_{j}-\tilde{m}{1_{\left\\{1\leq j\leq k\right\\}}}.$

Thus, the objective functions in
$P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$ and
$P({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},m,L^{\prime})$
are identical.

Proof of Claim 4. Let $x$ and $y$ denote the solution of problems
$P({\mbox{\boldmath$x$}},{\mbox{\boldmath$r$}},m,L)$ and
$P({\mbox{\boldmath$z$}}^{\prime},{\mbox{\boldmath$r$}}^{\prime},m,L^{\prime})$,
respectively, where $x$ and $y$ are related by (74) or (75) depending on
whether $r_{1}<z_{k+1}-z_{k}$ or $r_{1}\geq z_{k+1}-z_{k}$. We assume that
indexes are labeled such that (63) and (64) hold. Recall that
${\mbox{\boldmath$z$}}^{\prime}$, ${\mbox{\boldmath$r$}}^{\prime}$ and
$L^{\prime}$ are defined in (69), (72) and (73), respectively. The feasible
region of problem $P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$ is

$\Omega=\left\\{{\mbox{\boldmath$x$}}:\sum_{i=1}^{m}x_{i}=L,0\leq x_{i}\leq
r_{i},1\leq i\leq m\right\\}.$ (95)

In terms of $y$, feasible region $\Omega^{\prime}$ is

$\displaystyle\Omega^{\prime}$
$\displaystyle=\Biggl{\\{}{\mbox{\boldmath$y$}}:\sum_{i=1}^{m}y_{i}=L^{\prime},0\leq
y_{i}\leq r_{i}^{\prime},1\leq i\leq m\Biggr{\\}}.$

If $r_{1}<z_{k+1}-z_{k}$, one can apply a change of variables using (74) to
express $\Omega^{\prime}$ as

$\displaystyle\Omega^{\prime}$
$\displaystyle=\Biggl{\\{}{\mbox{\boldmath$x$}}:\sum_{i=1}^{j}0+\sum_{i=j+1}^{k}(x_{i}-\tilde{m})+\sum_{i=k+1}^{m}x_{i}=L^{\prime},$
$\displaystyle\quad x_{i}=\tilde{m},1\leq i\leq j,$ $\displaystyle\quad 0\leq
x_{\ell}-\tilde{m}\leq r_{\ell}-\tilde{m},j+1\leq\ell\leq k,$
$\displaystyle\quad 0\leq x_{\nu}\leq r_{\nu},k+1\leq\nu\leq m\Biggr{\\}}.$

With a little manipulation, we obtain from the preceding

$\displaystyle\Omega^{\prime}$
$\displaystyle=\Biggl{\\{}{\mbox{\boldmath$x$}}:\sum_{i=1}^{m}x_{i}=L,x_{i}=r_{i},1\leq
i\leq j,$ $\displaystyle\quad 0\leq x_{\ell}\leq r_{\ell},j+1\leq\ell\leq
m\Biggr{\\}}.$ (96)

From (95) and (96), it is clear that $\Omega^{\prime}\subseteq\Omega$. The
proof for the case, where $r_{1}\geq z_{k+1}-z_{k}$ is similar. We omit the
details.

We now show that the optimal point of problem
$P({\mbox{\boldmath$z$}},{\mbox{\boldmath$r$}},m,L)$ actually lies in region
$\Omega^{\prime}$. We prove the case where

$r_{1}<z_{k+1}-z_{k}.$ (97)

From (96), we need to show that the optimal solution satisfies $x_{i}=r_{i}$
for $i=1,2,\ldots,j$. Assume that $x_{1}<r_{1}$. From (55), since
$x_{1}<r_{1}$, it follows that $\mu_{1}=\omega_{1}=0$ and

$\frac{1}{m}\left(\sum_{\ell=1}^{m}z_{\ell}+L\right)-z_{1}-\frac{m-1}{2}\lambda<r_{1}.$
(98)

Since $z_{1}=z_{i}$ for $i=2,3,\ldots,k$, it follows that

$x_{i}=x_{1}<r_{1}.$

In addition, for $i=k+1,\ldots,m$, we have

$\displaystyle\frac{1}{m}\left(\sum_{\ell=1}^{m}z_{\ell}+L\right)-z_{i}-\frac{m-1}{2}\lambda$
$\displaystyle=\frac{1}{m}\left(\sum_{\ell=1}^{m}z_{\ell}+L\right)-z_{1}-\frac{m-1}{2}\lambda+z_{1}-z_{i}.$
(99)

From (98) and (99), we have

$\displaystyle\frac{1}{m}\left(\sum_{\ell=1}^{m}z_{\ell}+L\right)-z_{i}-\frac{m-1}{2}\lambda$
$\displaystyle<r_{1}+z_{1}-z_{i}$ $\displaystyle\leq 0,$

where the last inequality is due to the assumption (97). Thus, $x_{i}=0$ for
$i=k+1,\ldots,m$. It follows that

$L=\sum_{\ell=1}^{m}x_{\ell}=\sum_{\ell=1}^{k}x_{\ell}+\sum_{\ell=k+1}^{m}x_{\ell}<\tilde{m}k,$

which contradicts the assumption that $L>\tilde{m}k$. The proof for the other
case where $r_{1}\geq z_{k+1}-z_{k}$ is similar. We omit the details. The
proof is complete.

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