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TV Broadcast Efficiency in 5G Networks from Subscriber Prospective

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TV Broadcast Efficiency in 5G Networks from
Subscriber Prospective
Chun Pong Lau and Basem Shihada
Computer, Electrical and Mathematical Science Engineering Division
King Abdullah University of Science and Technology (KAUST)
{lau.pong, basem.shihada}@kaust.edu.sa
Abstract—The flexibility of radio access network facilitated by
5G networks opens the gateway of deploying dynamic strategies
for broadcasting TV channels in an efficient way. Currently,
spectrum efficiency and bandwidth efficiency are the two common
metrics measuring the efficiency of a system. These metrics assess
the net bitrate for a unit of spectrum bandwidth. However, there
is a lack of measurement, quantifying the effectiveness of a
broadcasting strategy from the user perspective. In this paper,
we introduce a novel measurement approach, called broadcast
efficiency which considers the mobile user as a main reference.
Broadcast efficiency is calculated as the number of served
audiences per unit of radio resource. From numerical analysis,
we show that broadcasting unpopular TV channels dramatically
decreases the broadcast efficiency. This finding is evaluated by
employing multiple distributions on the size of audience among
TV channels. Furthermore, by conducting a real-life simulation,
we discover that a high broadcast efficiency may result in a low
percentage of served audiences if the audiences of TV channels
are quite evenly distributed.
Index Terms—TV, broadcast, efficiency, 5G.
I. INTRODUCTION
After the successful operation of the fourth-generation (4G)
networks, researchers are now enthusiastically designing the
next generation mobile networks with even greater capability.
With the support of Evolved MBMS (eMBMS) in Long-Term
Evolution (LTE) networks, service providers are able to offer
broadcast and multicast services to mobile subscribers [1].
Currently, mobile service providers are launching multimedia
services to their subscribers. For example, in Saudi Arabia,
the major operator Mobily has launched a new broadcasting
service known as ‘mView’ to deliver live television (TV) and
video on demand services since 2014 [2]. In US, Verizon
utilizes the 4G LTE networks to provide wireless TV [3]. In
2014, Nokia successfully launched a trial TV broadcasting
service on LTE network in Munich, Germany [4]. When
shaping the concept of the future 5G networks, researchers
expect at least more than a thousand times of system capacity
and ten times more spectral efficiency than the 4G networks
[5]. By utilizing these capabilities, broadcasting TV channels
on 5G networks becomes attractive and profitable in the short
coming future.
In the current mobile network TV broadcasting service,
service providers offer tens of TV channels for audiences to
select from. The bitrate of each TV channel requires several
hundreds to thousands kilobits per second. It is a challenge
for mobile operators to simultaneously broadcast all of the TV
channels to audiences within the limited amount of available
bandwidth in each cell. Reported in [6], in order to broadcast
60 TV channels in suburban area, a spectrum of more than
400 MHz is required to deliver quality service in the worst
scenario. It is almost impossible, as normally mobile service
providers are not licensed with sufficient bandwidth to fulfill
this requirement. Besides, researchers proposed cross-layer
systems to reduce bandwidth consumption, such as superpo-
sition coded multicasting (SCM). In SCM, the base station
(BS) superimposes two video streams, both good and fair
quality streams of a video, in one radio signal to broadcast
[7], [8]. Although this reduces the radio resource consumption
for broadcasting multiple copies of the same video stream and
ensure reliable receptions for users in various channel quali-
ties, it does not reduce bandwidth for broadcasting multiple
TV channels [9].
It is well known in literature that spectrum or bandwidth
efficiency is the common approach used to gauge the rate
that information is transmitted with a given unit of radio
resources. However, in a broadcasting scenarios, this is no
longer sufficient to quantify the effectiveness of a broadcasting
strategy from the user perspective since the broadcasting
strategies will influence the system performance when the size
of audience is varying. In [10], Ye et al. introduced a metric
to measure the effectiveness of broadcasting in vehicular ad-
hoc network. However, their measurement is restricted to
measure the number of nodes per unit time, which does
not take the transmission bitrate into account. In [11], the
broadcast efficiency is defined as the number of successfully
received packets in a unit duration regardless of the senders.
On the other hand, in [12], the broadcast efficiency is defined
as the number of newly covered nodes reached per unit
transmit power in wireless ad-hoc networks. Both definitions
are considered inappropriate for video broadcasting in cellular
networks since the transmission power is not the highest
priority concern of the BS and each node might not be in
the position to receive packets from different sources. To the
best of our knowledge, a measurement that takes both user
perspective and transmission bitrate into account for video
broadcast has not been proposed. Therefore, it urges us to
redefine broadcast efficiency to evaluate broadcasting strategy
in cellular network.
978-1-4799-5952-5/15/$31.00 ©2015 IEEE
In this paper, we first propose a new definition of broadcast
efficiency. It is the ratio between the number of served audi-
ences and the consumed radio resources, which is the bitrate.
Second, we analyze the growth rate of the number of served
audiences and the required bitrate. Then, we investigate the
effect of broadcasting various TV channels on broadcast effi-
ciency. By the analysis, we found that the broadcast efficiency
decreases when less popular TV channels are broadcasted.
Finally, we validate our findings by numerical evaluation and
a real-life simulation.
The rest of this paper is organized as follows. Section II
describes the formulation of the model in detail. Numerical
analysis is presented in Section III. The evaluation is demon-
strated in Section IV. Finally, conclusions are made in Section
V.
II. MODEL FORMULATION
In this section, the model formulation of the system is
presented.
A. TV channel properties
In the system, we assume that each cell independently
decides on choosing qpopular TV channels out of a set of
TV channels Cfor broadcasting. The total number of TV
channels in Cis defined as |C|. The number of audiences
of each channel cCis defined as n(c). The total number of
audiences in the cell is N. The TV channels are sorted by the
popularity, which is the number of audiences of each channel,
in descending order. This is defined as follows,
Definition 1. n(c1)n(c2)when c1<c
2;c1,c
2C.
n(c)is a decreasing discrete function, which implies that the
most popular TV channel in the set Chas a larger or equal
number of audiences than the second popular TV channel, and
so on.
The required bitrate of each TV channel for broadcasting is
defined as r(c). For simplicity, and without loss of generality,
we assume all of the channels have the same bitrate where,
r(c1)=r(c2),c1,c
2C. (1)
B. Broadcast efficiency
The broadcast efficiency E(q)is a measure for the number
of users are served per unit throughput. The unit is defined
as users per bit per second. The numerator of the broadcast
efficiency, which is the total number of audiences for broad-
casting qchannels, is formulated as follows,
fn(q)=
q
i=1
n(i),(2)
where qCis the number of popular channels that are
selected to be broadcasted. The denominator, which is the total
required bitrates for broadcasting qchannel, is formulated as
follows,
fr(q)=
q
i=1
r(i),(3)
Finally, the formulation of broadcast efficiency is defined as
follows,
Definition 2.
E(q)=fn(q)
fr(q),
where qCis the number of popular channels that are
selected to be broadcasted.
A larger value of E(q)indicates a higher broadcast effi-
ciency. This indicates that either a larger number of audiences
are being served or a lower bitrate is required for broadcasting.
In other words, the broadcast efficiency increases when the
number of watching audiences of the broadcasting channels
increase or the required bitrate for broadcasting decrease.
III. ANALYSIS
In this section, we first analyze the growth rate of the
numerator and denominator of E(q)in Definition 2, which
are the total number of TV channels audiences and the
total required bitrate. Then, we investigate how the broadcast
efficiency E(q)is varied with q, which is the number of
channels that are selected to be broadcasted.
Lemma 1. fn(q)is an increasing function and the upper
bound of the growth rate of the function is O(q).
Proof. First, we analyze the extreme case that all of the
numbers of audiences of each TV channel are equal.
Case 1. If n(c1)=n(c2),then fn(q)=q·n(1) = Θ(q).
In this case, the total number of audiences increases linearly
when the number of TV channels being broadcasted increases.
It implies that the growth rate of fn(q)is strictly linear. Since
n(c1)n(c2), the upper bound of growth rate of fn(q)is
O(q).
Then, we analyze the general case that more popular chan-
nels have more audiences than less popular channels.
Case 2. If n(c1)n(c2),then fn(q)q·n(1) = O(q)
Since n(c)is a decreasing discrete function, the numbers of
audiences of a more popular TV channels are always greater
or equal to the less popular TV channels. In this case, fn(q)is
increasing and the growth rate is slower than a linear function.
Finally, we analyze the growth rate if the values of n(c) is
equal to zero.
Case 3. If n(c1)=0,then n(c2)=0,c2>c
1;
fn(q)fn(c1)=O(1), where fn(c1)is a constant.
Since n(c)is a decreasing discrete function, once there is a
TV channel c1has no audience, the channels which are less
popular than it, have no audience as well. Therefore, the value
of fn(q)will not increase after q=c1. In this case, the growth
rate of fn(q)is upper bounded by a constant.
By summarizing the above cases, fn(q)=O(q)and Lemma
1 is proven.
Lemma 2. fr(q)is an increasing function and the growth
rate of the function is Θ(q).
Proof. From (1), the bitrates of TV channels are equal. The
summation of the bitrates increases linearly with respect to q.
It implies that the growth rate of fr(q)is strictly linear. It is
upper bounded and lower bounded by linear growth functions.
Therefore, fr(q)=Θ(q).
Lemma 3. E(q)is a decreasing function and the value of
E(q)is lower bounded by N
|C|r(1) .
Proof. E(q)is composited by the numerator with fn(q)and
the denominator with fr(q). From Lemma 1 and 2, we prove
that fn(q)=O(q)and fr(q)=Θ(q). Hence, fn(q)=
O(fr(q)). It implies that the denominator fr(q)growth faster
than or equal to the numerator fn(q). Therefore, E(q)is a
decreasing function.
Since qC, the maximum value of qis |C|. The maximum
value of fn(|C|)which is the total number of audience N.
The maximum bitrate for broadcasting is fr(|C|). The fastest
growth rate of the numerator is linear. When it is divided
by the linear-growth denominator, the decrease rate of E(q)
will become slowest as a constant function and become the
lower bound of E(q). The lower bound value is fn(|C|)
fr(|C|)=
N
|C|r(1)
In summary, according to Lemma 3, we prove that the
broadcast efficiency decreases when broadcasting more and
more unpopular TV channels. Furthermore, the broadcast
efficiency is lower bounded by N
|C|r(1) .
IV. EVAL U AT I O N
In this section, we first numerically evaluate the analysis by
applying known distributions of popularity for TV channels.
Then, we conducted a real-life simulation to further evaluate
the broadcast efficiency.
A. Numerical evaluation
We numerically evaluate the lemmas by applying different
cases and distributions for n(c), which are different distri-
butions for the size of audience of TV channels. Reported
in [13] and [14], the popularity of TV channels following
Zipf-like distribution. We use Zipf distribution with parameter
s=1in our evaluation. In addition, we include another
two distributions, Zipf-Mandelbrow and Zeta, with parameter
s=2and s=3, for comparison. Furthermore, two extreme
cases are added in the evaluation. One case is all of the
TV channels having the same size of audience, which is an
uniform distribution. It is related to Case 1 in Lemma 1.
Another case is when all of the audiences watching only the
most popular TV channel, which is denoted as n(2) = 0 in
the figures and related to Case 2 in Lemma 1. For the other
settings, we assume there are 2000 audiences in the cell and
10 channels are being selected to be broadcasted. The required
bitrate for each TV channel is 1 Mb/s.
Figure 1 shows the number of audiences for each channel.
The channels in the x-axis are sorted in a descending order
0 2 4 6 8 10
0
500
1000
1500
2000
c−th popular channel
number of audiences
Zipf, s=1
Zipf−M., s=2
Zeta, s=3
n(2)=0
uniform
Fig. 1. Number of audience of TV channels, c is sorted by popularity
0 2 4 6 8 10
0
500
1000
1500
2000
2500
q
fn(q) (users)
Zipf, s=1
Zipf−M., s=2
Zeta, s=3
n(2)=0
uniform
Fig. 2. fn(q)
according to the popularity, which is the size of audience.
All of the cases follows Definition 1 that the less popular
channels have fewer or equal number of audiences. In Zipf
distribution, the most popular TV channel has 683 audiences.
The numbers of audiences drop gently to 68 audiences in
the least popular channel. In Zip-Mandelbrow distribution, it
starts with 1291 audiences in the most popular channel, drop
quickly and negative exponentially to only 13 audiences in
the least popular channel. In Zeta distribution, it starts with
1664 audiences and has a fastest drop among these three
distribution to having less than 10 audiences in the last five
TV channels each. In the special case that n(2) = 0,allof
the 2000 audiences watch the most popular TV channel and
no audiences for the remains. For the uniform case, all of the
TV channels has 200 audiences.
Figure 2 shows the function fn(q), which is the total number
of audiences when broadcasting qpopular TV channels. In
Zipf distribution, fn(q)has a low starting number of audiences
and grows quickly when qincreases. In Zipf-Mandelbrow and
Zeta distribution, both of them start with a higher number
of audiences and grow slower than the Zipf distribution. In
n(2) = 0, the first TV channel has all of the 2000 audiences
therefore, it remains constant and not growing in fn(q).In
0246810
0
2
4
6
8
10
q
fr(q) (Mbps)
Fig. 3. fr(q)
0 2 4 6 8 10
0
500
1000
1500
2000
q
Broadcast Efficiency (users/Mbps)
Zipf, s=1
Zipf−M., s=2
Zeta, s=3
n(2)=0
uniform
Fig. 4. Broadcast Efficiency
the uniform distribution, it grows linearly and has the fastest
growth rate compare to the other cases. This validates Lemma
1 that fn(q)is an increasing function and the fastest growth
rate of the function is linear.
Figure 3 shows the function fr(q), which is the total require
bitrate for broadcasting qTV channels. From the assumption
in (1), all of the TV channels have the same bitrate. Therefore,
fr(q)grows exactly linearly. This validates Lemma 2 that
fr(q)is a strictly linear growth function.
Figure 4 shows the broadcast efficiency for each case. It
shows that E(q)is a decreasing function. The case n(2) =
0has the fastest decreasing rate since broadcasting more
channels in this case does not serve any more audience. It
is a waste of resources if broadcasting more than one TV
channel when there is no audience in the remaining 9 TV
channels. Zipf distribution has a slower decreasing rate than
the other two distributions because more audiences in the
unpopular channels. It leads to a conclusion that a more
even distribution of audiences on the TV channels helps to
slow down the decreasing rate of broadcast efficiency when
broadcasting more TV channels. In the uniform distribution,
broadcast efficiency remains constant since both the numerator
and denominator of E(q)are linear. It is also the lower
Fig. 5. BS assignment for each house and building grouped and colored by
assigned BS, with the cell boundaries
1 2 3 4 5 6 7
0
100
200
300
400
number of channels
Cumulative Number of Audiences
5918
5915
5914
5913
5912
Fig. 6. Average cumulative number of audiences for TV channels in each
BS coverage area
bound of the broadcast efficiency. The lower bound value is
200 users/Mb/s which is equal to N
|C|r(1) described in Lemma
3.
In summary, the numerical evaluation validates the lemmas
presented in the previous section. We further show the effect of
broadcasting different numbers of TV channels on broadcast
efficiency in the following subsection by a real-life simulation.
B. Real-life simulation
A real-life simulation is conducted to evaluate the broadcast
efficiency with various number of broadcasting channels.
1) Simulation setup: We obtained a real-life dataset from
statistical data and actual geographical location for evaluating
our work. The statistical data are extracted from the surveys
of TV audience behavior in Saudi Arabia reported by Dubai
Press Club and Pan Arab Research Center in 2007, 2011
and 2012 [15]–[17]. The daily average people using television
(PUT) level presented in [17] is adopted to estimate the overall
size of audience. The seven most popular TV channels are
considered for simplicity. The audience sizes of the channels
are approximated from [16]. We assumed that the bitrates of
the TV channels are 700 kb/s.
0600 1100 1600 2100 0200
0
20
40
60
80
100
time segment grouped by 15 minutes
number of audiences
5918
MBC 1
Al Arabiya
MBC 4
Al Jazeera
MBC 2
MBC Action
MBC Drama
Fig. 7. Number of audiences in BS 5918
0600 1100 1600 2100 0200
0
200
400
600
800
time segment grouped by 15 minutes
number of audiences
5915
MBC 1
Al Arabiya
MBC 4
Al Jazeera
MBC 2
MBC Action
MBC Drama
Fig. 8. Number of audiences in BS 5915
For the geographical location, we considered King Abdullah
University of Science and Technology (KAUST), located
in Saudi Arabia, as our study location. KAUST is a mid-
size living compound with about 2000 town houses and 80
apartment buildings. We assumed that there are two sets of
watching equipment, such as, TV sets, tablets, or smartphones,
in each town house, and twenty sets of watching equipment in
each apartment building, in a total of 5490 sets. We assumed a
static BS assignment for each house and building by selecting
the strongest reception signal from the BS. Furthermore, there
are five BSs operated by the mobile network service provider.
These are shown as stars with the BS name in Figure 5.
Each BS has the same transmission power and coverage area.
The receivers under the cell coverage have similar channel
conditions to receive fair video quality. The assignment is
colored in Figure 5 according to the assigned BS. Figure 6
shows the average cumulative number of audiences for sorted
TV channels, in each BS coverage area. BS 5915 has the most
number of audiences in average and BS 5918 has the least
number of audiences.
2) Simulation result: Two BSs which have the most and the
least number of audiences, BS 5915 and BS 5918, respectively,
are selected to be discussed in this paper.
0600 1100 1600 2100 0200
0
20
40
60
80
5918
time segment
Broadcast efficiency (users/Mbps)
1 channel
2 channels
3 channels
4 channels
5 channels
6 channels
7 channels
Fig. 9. Broadcast efficiency in BS 5918
0600 1100 1600 2100 0200
0
100
200
300
400
500
600
700
5915
time segment
Broadcast efficiency (users/Mbps)
1 channel
2 channels
3 channels
4 channels
5 channels
6 channels
7 channels
Fig. 10. Broadcast efficiency in BS 5915
In Figure 7 and 8, the number of audiences of each channel
in a day from 6AM to 2AM are shown. The data are stacked
in the bar charts and grouped by 15 minutes. These two BSs
shows a big difference in the audience distribution. BS 5918
has more audiences for Al Arabiya across the day and more
audiences for MBC 4 in the evening. BS 5915 is mainly
dominated by the audiences for MBC 1 across the day.
Figure 9 and 10 show the broadcasting efficiency of these
two BSs. The figures have seven cases which show the
broadcasting efficiency for broadcasting different numbers of
TV channels. This validates our finding that broadcasting more
channels decreases the broadcast efficiency. Furthermore, by
comparing these two figures, it clearly shows that the differ-
ence of audience distributions affect the broadcast efficiency.
First, the average value of broadcast efficiency of BS 5918
is lower than BS 5915 due to the lower number of average
total audiences. It shows that the total number of audiences is
directly proportional to the broadcast efficiency. Second, the
differences of broadcast efficiency between broadcasting more
channels in BS 5915 is larger than the differences in BS 5918.
For example, at 21:30, the difference of broadcasting 1 channel
and 2 channels in BS 5918 drops 18.7%, but in BS 5915, it
drops 28.2%. This is because BS 5915 has a larger portion
0600 1100 1600 2100 0200
20
40
60
80
100
5918
time segment
Percentage of served users (%)
1 channel
2 channels
3 channels
4 channels
5 channels
6 channels
7 channels
Fig. 11. Percentage of served audiences in BS 5918
0600 1100 1600 2100 0200
30
40
50
60
70
80
90
100
5915
time segment
Percentage of served users (%)
1 channel
2 channels
3 channels
4 channels
5 channels
6 channels
7 channels
Fig. 12. Percentage of served audiences in BS 5915
of audiences watching the most popular TV channel than in
BS 5918, where the audiences are more evenly distributed.
By this observation, we summarize that broadcasting more
unpopular channels in audience uneven distributed cells drops
the broadcast efficiency faster.
However, there is a trade-off for high broadcast efficiency.
Figure 11 and 12 show the percentage of served audiences
for different numbers of broadcasting channels. The fewer TV
channels are broadcasted, the higher broadcast efficiency, but a
lower number of served audiences. In BS 5918, since the dis-
tribution of audiences between different TV channels is more
even, broadcasting only one popular channel can serve only
42.9% of audience in average. However, in average, 59.8%
of total audiences are served in BS 5915 when broadcasting
only the most popular channel. In order to guarantee that more
than 90% of audiences are served at any time, BS 5918 needs
to broadcast at least 6 TV channels to fulfill the requirement.
In BS 5915, it only requires 5 TV channels to serve more
than 90% of the audiences. Broadcasting 3 TV channels can
mostly serve 80% of the requirement. In summary, more
audiences can be served when broadcasting fewer TV channels
if the audience distribution is not evenly distributed, which
indicates that the difference of audiences between popular and
unpopular TV channels should be as large as possible.
V. C ONCLUSION
In this paper, we defined a new metric known as the
user broadcast efficiency. It measures the size of audience
served when a cell is broadcasting TV channels in terms of
consumed bitrates. Numerical analysis and evaluation proved
that broadcast efficiency decreases when broadcasting less
popular TV channels. By conducting a real-life simulation, we
demonstrated that high broadcast efficiency comes with a low
percentage of served audiences as a trade-off. In the future, we
will investigate the effects of broadcast efficiency by varying
TV channel bitrates. Furthermore, we will balance between
broadcast efficiency and the number of served audiences as
an essential target for future research.
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