0
17kviews
Trunking - Theory
5
459views

When number of users increase in the system, the demand for radio channels also increases. It gives rise to congestion and quality of service given to the users is also affected. Hence, trunking theory came into prominence. The fundamentals of trunking theory were developed by Erlang, a Danish mathematician who, in the late 19th century, studied how a large population could be accommodated by a limited number of servers. It is a theory which proposes strategies to accommodate maximum number of users in limited frequency spectrum.

The concept of trunking allows a large number of users to share the relatively small number of channels in a cell by providing access to each user, on demand, from a pool of available channels. In a trunked radio system, each user is allocated a channel on a per call basis, and upon termination of the call, the previously occupied channel is immediately returned to the pool of available channels.

To design trunked radio systems that can handle a specific capacity at a specific “grade of service,” it is essential to understand trunking theory and queuing theory.

Definition of Common Terms used in Trunking Theory

• Set-up Time : The time required to allocate a trunked radio channel to a requesting user.
• Blocked Call: A call which can not be completed at the time of request, due to congestion. Also referred as a lost call.
• Holding Time: Average duration of a typical call. Denoted by H (in seconds).
• Traffic Intensity: Measure of channel time utilization, which is the average channel occupancy measured in Erlangs. This is dimensionless quantity and may be used to measure the time utilization of single or multiple channels. Denoted by A.
• Load: Traffic intensity across the entire trunked radio system, measured in Erlangs.
• Grade of Service (GOS) : A measure of congestion which is specified as the probability of a call being blocked ( for Erlang B), or the probability of a call being delayed beyond a certain amount of time ( for Erlang C).
• Request Rate: The average number of call requests per unit time. Denoted by $\lambda$ $second^{-1}$

One Erlang represents the amount of traffic intensity carried by a channel that is completely occupied (i.e.one call-hour per hour or one call-minute per minute). For example, a radio channel that is occupied for thirty minutes during an hour carries 0.5 Erlangs of traffic.

The grade of service (GOS) is a measure of the ability of a user to access a trunked system during the busiest hour. The busy hour is based upon customer demand at the busiest hour during a week, month, or year. It is the wireless designer’s job to estimate the maximum required capacity and to allocate the proper number of channels in order to meet the GOS. GOS is typically given as the likelihood that a call is blocked, or the likelihood of a call experiencing a delay greater than a certain queuing time. Lesser the GOS, better is the system.

Traffic Intensity per Channel

If H is the average duration of a call and $\lambda$ is the average number of call requests per unit time, each user generates a traffic intensity of $A_u$ Erlangs,

$A_u = \lambda \times H$

For a system containing U users, the total offered traffic A is

$A = U \times A_u$

In a C channel trunked system, if the traffic is equally distributed among the channels, then the traffic intensity per channel

$A_c = UA_u/C$

The maximum possible carried traffic is the total number of channels C, in Erlangs. The AMPS cellular system is designed for a GOS of $2 \%$ blocking. This implies that the channel allocations for cell sites are designed so that 2 out of 100 calls will be blocked due to channel occupancy during the busiest hour.

Two types of trunked systems are used in cellular systems: Figure 18: Classification of trunked system

Blocked Cleared Systems (BCC)

In this type of system, if no channels are available, the requesting user is blocked without access and is free to try again later. It is a memoryless system. To estimate GOS, traffic intensity of a BCC system, we use the Erlang B chart.

1) This system does not offer queuing for call request.

2) For every user who requests service, it is assumed there is no setup time and the user is given immediate access to a channel if one is available. If no channels are available, the requesting user is blocked without access and is free to try again later.

3) Calls arrive as determined by poisson distribution. It is assumed that there are an infinite number of users. This system is based on following assumptions.

$\quad$ a. Call requests are memoryless implying that all users including blocked users may request for channel at any time.

$\quad$ b. The probability of users occupying a channel is exponentially distributed. That means longer calls are less likely to happen.

$\quad$ c. There are a finite number of channels available in pool (group of channels).

$\quad$ d. In arrival times all call requests are independent of each other.

4) This system uses M/M/m queue and leads to the derivation of the Erlang B formula. The Erlang B formula determines the probability that a call is blocked and is a measure of the GOS for a trunked system which provides no queuing for blocked calls.

5) The probability of blocking or the GOS is gven by the formula,

$Pr_{blocking} = \frac{A^c/C!}{\sum_{k=0}^C \frac{A^k}{k!}} = GOS$

Where,

A = Traffic intensity

C = Total number of channels

k =$k^{th}$ channel

Erlang B formula gives a conservative estimate of GOS as the finite users always predict smaller likelihood of blocking.

Blocked Calls Delayed System (BCD)

1) In a BCD system , queue is provided to hold calls which are blocked. If a channel is not available immediately, the call request is delayed until a channel becomes available. This type of trunking system is called as Blocked Calls Delayed or Erlang C system.

2) It's measure of GOS is defined as the probability that a call is blocked after waiting a specific length of time in the queue. To find the GOS it is necessary to find the likelihood that a call is initially denied access to the system.

3) The likelihood of a call not having immediate access to a channel is determined by the Erlang C formula

$Pr_{(delay \gt 0)} = \frac{A^c}{A^c + C!(1-\frac{A}{c}) \sum_{k=0}^{c-1} \frac{A^k}{k!}}$

A = Traffic intensity

C = Total number of channels

K = $k^{th}$ channel

If no channels are immediately available, the call is delayed and the probability that the delayed call is forced to wait more than t seconds is given by the probability that a call is delayed multiplied by the conditional probability that the delay is greater than t seconds. The GOS of a trunked system where blocked calls are delayed is hence given by

$Pb_{(delay \gt t)} = Pb_{(delay \gt 0)} \times Pb_{(delay \gt t) /(delay \gt 0)} = Pb_{(delay \gt 0)}e^{(-(C-A)t/H}$

H= Holding Time.

The average delay D for all calls in a queued system is given by,

$D=Pr_{delay \gt 0} \times \frac{H}{C-A}$

The Erlang B and Erlang C chart are as shown: Figure 19: The Erlang B chart showing the probability of blocking as function of the number of channels and traffic intensity in Erlangs

The X axis is traffic intensity in Erlangs which is in log scale. The Y axis is the probability of blocking or the GOS and is also in log scale. As can be seen by the reader, there are some curves which intersect the horizontal and vertical lines of the chart. Each curve corresponds to a channel number. The curves shown are from channel number 1 to 100. If two parameters are given, the third can be found from this chart.

For example, if traffic intensity is 1 Erlang and GOS is 0.003, first it is to be found where these two lines are intersecting each other. We then have to follow the curve on or near this intersecting point and read the channel number. It is seen that the curve corresponding to channel number 5 lies on the intersecting point.

If the GOS and number of trunked channels are given, then their point of intersection has to be found . This value has to be traced vertically down and the corresponding value on the lower X axis has to read. For eg: For channel number 5 and GOS 0.04, traffic intensity read from the chart is 2 Erlangs. To ease out the process of reading Erlang charts, Erlang calculator are widely available online which can be used during calculations.

The Erlang C is as shown in figure 19. The GOS is expressed as probability of delay. The same procedure as described for Erlang B chart has to be followed for reading this chart. Figure 20: The Erlang C chart showing the probability of delay as function of the number of channels and traffic intensity in Erlangs

Numericals

1) If 20 MHz of total spectrum is allocated for a duplex wireless cellular system and each simplex channel has 25 KHz RF bandwidth, find:

(a) The number of duplex channels.

(b) The total number of channels per cell site, if N = 4 cell reuse is used.

Solution:

a) Given: A simplex channel’s Bandwidth=25 KHz

Total bandwidth allotted=20 MHz

Number of duplex channels = $\frac{20 MHz}{25 KHz \times 2} = 400$ $Channels$

b) Cluster size N=4

Total number of channels per cell= 100 channels

2) A cellular service provider decides to use a digital TDMA scheme which can tolerate a signal to-interference ratio of 15 dB in the worst case. Find the optimal value of N for

(a) omnidirectional antennas,

(b) $120^°$ sectoring, and

(c) $60^°$ sectoring. Should sectoring be used? If so, which case ($60^°$ or $120°$) should be used?

(Assume a path loss exponent of n = 4 and consider trunking efficiency.)

Solution:

Worst case S/I tolerated by the service provider:

$\frac{S}{I} = 15dB = 10 \ log\frac{S}{I}$

$\frac{S}{I} = 31.62$

a) For $360^0$ sectoring , $i_0=6$

$\frac{S}{I} = \frac{\sqrt{3N}^4}{i_0}$

$\quad$ Cluster size N= 7

b) For $120^0$ sectoring , $i_0=2$

$\quad$ Cluster size N= 3

c) For $60^0$ sectoring , $i_0=1$

N = 1.87. Rounding of this value to the next standard cluster size value, N=3

In both, $120^o$ and $60^o$ sectoring,capacity is increased by 2.33$\%$. However, in $60^0$ sectoring, there is a problem of more number of handoffs.

3) If an intensive propagation measurement campaign showed that the mobile radio channel provided a propagation path loss exponent of n = 3 instead of four, how would your design decision change.

Solution:

Given:

Signal to interference, $\frac{S}{I}=31.62$

Path loss exponent, n=3

a) For $360^0$ sectoring, $i_0=6$

$\frac{S}{I} = \frac{\sqrt{(3N)}^3)}{i_0}$

Calculating , we get ,N= 11

Therefore standard cluster size , N= 12

Reuse factor, $n=\frac{1}{12} = 0.083$

b) For $120^0$ sectoring, $i_0=2$

N= 5.29

Therefore standard cluster size , N= 7

Reuse factor, $n=\frac{1}{7} = 0.142$

c) For $60^0$ sectoring, $i_0=1$

N= 3.3

Therefore standard cluster size , N= 4

Reuse factor, $n=\frac{1}{4} = 0.25$

In this case when n=3,the capacity is increased in $60^o$ sectoring by a factor of $\frac{12}{4}$ i:e 3 times. $120^o$ in this case offers an increase in capacity by a factor of $\frac{12}{7}$ i:e 2.33 times. Hence,$60^o$ sectoring is done.

4) A total of 24 MHz of bandwidth is allocated to a particular FDD cellular telephone system that uses two 30 kHz simplex channels to provide full duplex voice and control channels.Assume each cell phone user generates 0.1 Erlangs of traffic. Assume Erlang B is used.

(a) Find the number of channels in each cell for a four-cell reuse system.

(b) If each cell is to offer capacity that is 90% of perfect scheduling, find the maximum number of users that can be supported per cell where omnidirectional antennas are used at each base station.

(c) What is the blocking probability of the system in (b) when the maximum number of users are available in the user pool?

(d) If each new cell now uses $120^°$ sectoring instead of omnidirectional for each base station, what is the new total number of users that can be supported per cell for the same blocking probability as in (c)?

(e) If each cell covers five square kilometers, then how many subscribers could be supported in an urban market that is 50 km for the case of omnidirectional base station antennas?

(f) If each cell covers five square kilometers, then how many subscribers could be supported in an urban market that is 50 km for the case of $120^o$ sectored antennas?

Solution:

Given data:

No of cells = 4

Total spectrum = 24 MHz

Therefore, a duplex channel's bandwidth = 60 Hz

(a) Number of channels in each cell = $\frac{24 \times 10^6}{60 \times 10^6 \times 4} =100$ Channels/cell

(b) A perfect scheduled system is a fully utilized system in which every channel carries 1 Erlang of traffic. (If there are 100 channels and the system is perfectly scheduled system, then it carries 100 Erlangs of traffic.)

Since there are 100 channels per cell & the system is 90 $\%$ perfect scheduling.

$A=\frac{90}{100} \times 100 = 90$ Erlangs.

No of users=$\frac{A}{A_u}=\frac{90}{0.1}=900$ users.

(c) Reading for 100 channels and 90 Erlangs of traffic, From Erlang B chart, GOS = 0.03

(d) Assuming that number of channels remains constant even after sectoring,

Number of channels/sector =100/3=33.33

GOS=0.03

Reading from Erlang B channel $A_{sec}=26$ Erlang ( C=33, GOS= 0.03 )

No. of users per sector, $U_{sec}=\frac{26}{0.01} = 260$

Total no of users/ cell = $260 \times 3=780$ users.

Please note: Number of users before sectoring was employed=900. Number of users after 120 degree sectoring= 780 users

(It is observed that after sectoring , the total number of users have decreased in this case. This is because, we have assumed number of channels/cell to be same even after sectoring. In practice , after sectoring, the cluster size decreases. Hence number of channels per cell increases which results in more number of users per cell.)

(e) Each cell area = 5 sq. km

Total area=$50 \times 50$ sq. km

Therefore, total number of cells = 500 cells.

Total number of users in omnidirectional antenna scenario= 4,50,000 users

(f) Each cell area = 5 sq. km

Total area=$50 \times 50$ sq. km

Therefore, total number of cells = 500 cells.

Total number of users in omnidirectional antenna scenario= 390,000 users

5) A hexagonal cell has a cell radius of 1.387 Km. It is a 4 cell reuse system. The total number of channels in the entire system is 60. If the load per user is 0.029 Erlangs, and $\lambda=1$ call/hour , compute the following for an Erlang C system that has a 5 $\%$ probability of a delayed call :

$\quad$ a. How many users per square km will this system support?

$\quad$ b. What is the probability that a delayed call will have to wait for more than 10 seconds?

$\quad$ c. What is the probability that a call will be delayed for more than 10 seconds?

Solution:

Given:

Cluster size = 4

Cell radius (R) = 1.387 km

Total no of channels used for entire system = 60

Load per user ($A_u$) = 0.029 Erlangs

$\lambda$=1 call/hour

GOS= 0.05

Total no of channels per cell (c)=$\frac{60}{4} = 15$ Channels.

From Erlang C with GOS = 0.05

Traffic Intensity = 9 Erlangs

$\text {Therefore ,Total number of user} = \frac{\text{Total Traffic intensity}}{\text{Traffic per user}} = \frac{9}{0.029} = 310 \text {users}$

$\text{Number of users per Sq.Km} = \frac{\text{Number of Users}}{\text{Area covered per cell}} = \frac{310}{2.598 \times (1.387)^2} = \frac{310}{5} =62 \text{Users/Sq. km}$

$\lambda=1$ call / hour

$A_u = \lambda \times H$

Therefore,

$H =\frac{A_u}{\lambda} = \frac{0.029}{1} = 0.029 \ hours \ \times 3600 =104.4 \ seconds$.

The Probability $(P_r)$ that a delayed call will have to wait longer than 10 second i.e

$Pr(delay \gt 10 second/delay) = e^{−(C−A)E⁄H}$

$\quad$ $\quad$ $\quad$ $\quad$ = $e^{−\{(15−9)10⁄104.4\}}$

$\quad$ $\quad$ $\quad$ $\quad$ = 56.29 $\%$

Therefore,

Probability that a call is delayed more than 10 seconds

$P_r(delay \gt 10 second) = P_r(delay \gt 10 second/delay)\times P_r(delay)$

$\quad$ $\quad$ $\quad$ = $0.05 \times0.5629$

$\quad$ $\quad$ $\quad$ =$2.81 \%$

6) Exercises in trunking (queueing) theory:

(a) What is the maximum system capacity (total and per channel) in Erlangs when providing a 2 $\%$ blocking probability with four channels, with 20 channels, with 40 channels?

(b) How many users can be supported with 40 channels at 2$\%$ blocking? Assume H = 105 s, $\lambda$= 1 call/hour.

(c) Using the traffic intensity calculated in part (a), find the grade of service in a lost call delayed system for the case of delays being greater than 20 seconds. Assume that H = 105 s, and determine the GOS for four channels, for 20 channels, for 40 channels.

(d) Comparing part (a) and part (c), does a lost call delayed system with a 20 second queue perform better than a system that clears blocked calls?

Solution:

Given,

GOS =0.02

For N= 4 Channels

(a) As read from the Erlang B chart,

$A_{total}$=1.1 Erlang

$A_{per \ channel} =\frac{1.1}{4} = 0.275$ Erlang

• For N=20 channels

$A_{total}$ =14 Erlang

$A_{per \ channel} =\frac{14}{20} = 0.7$ Erlang

• For N=40 channels

$A_{total}$ = 31 Erlang

$A_{per \ channel} =\frac{31}{40} = 0.775$ Erlang

(b) G=0.02

H=105 second

$\lambda$=1 call/hour

Number of channels =40

$A_u=\lambda \times H =\frac{1 \times 105}{60 \times 60} = 0.024$

$U = \frac{A}{A_u} = \frac{31}{0.029} =1068$ user

(c) For C = 4 channel

$A_{total}$=1.1 Erlang

H=105 second

$P_r(delay \gt 20 second)= P_r(delay \gt 0) \times e^{-[(C-A)tH]}$

= $0.03 \times e^{−[(4-1.1)\frac{20}{105}]} = 0.017$

For c = 20 channel

$A_{total}$=14 Erlang

$P_r(delay \gt 0) =0.06$

$P_r(delay \gt 20 second)= P_r(delay \gt 0) \times e^{-[(C-A)tH]}$

= $0.06 \times e^{−[(20-14)\frac{20}{105}]}= 0.019$

For c = 40 channel

$A_{total}$= 31 Erlang

$P_r(delay \gt 0) =0.07$

$P_r(delay \gt 20 second)= P_r(delay \gt 0) \times e^{-[(C-A)tH]}$

= $0.07 \times e^{−[(40-31)\frac{20}{105}]} = 0.013$