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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 c∈Cis 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

2∈C.

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

2∈C. (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 q∈Cis 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 q∈Cis 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 q∈C, 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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