Cellular concept 2

Cellular Concept Cellular Concept Part IIPart II

Prepared byPrepared by

Assoc. Prof. Dr. Jafri B DinAssoc. Prof. Dr. Jafri B Din

WCC FKE UTMWCC FKE UTM

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Trunking and Grade of ServiceTrunking and Grade of ServiceTrunkingTrunking : :

to accommodate a large number of to accommodate a large number of users in a limited radio spectrum.users in a limited radio spectrum.

- allows a large # of users to share the relatively - allows a large # of users to share the relatively small # of chsmall # of chss in a cell by providing access to each in a cell by providing access to each user, on demand, from a pool of available chuser, on demand, from a pool of available chss..

- each user is allocated a channel on a per call basis, - each user is allocated a channel on a per call basis, and upon termination of the call, the previously and upon termination of the call, the previously occupied channel is immediately returned to the occupied channel is immediately returned to the pool of available chpool of available chss..

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Trunking and Grade of ServiceTrunking and Grade of Service

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The Grade of Service (GOS)The Grade of Service (GOS)- Is a measure of the ability of a user to access a Is a measure of the ability of a user to access a

trunked system during the busiest hour.trunked system during the busiest hour.

- The busy hour is based upon customer demand at The busy hour is based upon customer demand at the busiest hour during a week, month, or year.the busiest hour during a week, month, or year.

- It is the wireless engineer’s job to estimate the It is the wireless engineer’s job to estimate the maximum required capacity and to allocate the maximum required capacity and to allocate the proper # of chproper # of chss in order to meet the GOS. in order to meet the GOS.

- GOS is typically given as the likelihood that a call GOS is typically given as the likelihood that a call is blocked, or the likelihood of a call experiencing is blocked, or the likelihood of a call experiencing a delay greater than a certain queuing time.a delay greater than a certain queuing time.

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The The traffic intensitytraffic intensity offered by each user is equal to the call offered by each user is equal to the call request rate multiplied by the holding time. Each user generates request rate multiplied by the holding time. Each user generates a traffic intensity of Aa traffic intensity of A

uu Erlangs Erlangs

AAu u = = H HWhere Where HH : average duration of a call : average duration of a call

: average number of a call request per unit time : average number of a call request per unit time for each userfor each user

For a system containing For a system containing UU users and an unspesified # of chs, the users and an unspesified # of chs, the total offered traffic intensitytotal offered traffic intensity, A, A

A = U AA = U Auu

In a C channel trunked system, the traffic intensity per In a C channel trunked system, the traffic intensity per channel, channel, AAcc

AAcc = = U AU Au u

CC

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GOS of 2 % :GOS of 2 % : the channel allocations for cell sites are designed so that 2 out the channel allocations for cell sites are designed so that 2 out of 100 calls will be blocked due to channel occupancy during of 100 calls will be blocked due to channel occupancy during the busiest hour.the busiest hour.

Two types of trunked systems:Two types of trunked systems:

• No queuing for call requests (No queuing for call requests (Blocked Call ClearedBlocked Call Cleared))(Erlang B)(Erlang B)

• Queuing is provided to hold calls which are blocked Queuing is provided to hold calls which are blocked ((Blocked Calls DelayedBlocked Calls Delayed))(Erlang C)(Erlang C)

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Erlang BErlang B No queuing for call request.No queuing for call request. If no ch are available, the requesting user is blocked without If no ch are available, the requesting user is blocked without

access and is free to try again later.access and is free to try again later. It assumes that calls arrive as determined by a It assumes that calls arrive as determined by a Poisson Poisson

distributiondistribution, and there are an infinite # of users, also, and there are an infinite # of users, also

• There are memoryless arrivals of requests ( all users, There are memoryless arrivals of requests ( all users, including blocked users, may request a channel at any including blocked users, may request a channel at any time)time)

• The probability of a user occupying a channel is The probability of a user occupying a channel is exponentially distributedexponentially distributed (longer calls are less likely to (longer calls are less likely to occur as described by an exponential distribution)occur as described by an exponential distribution)

• There are finite # of channels available in the trunking There are finite # of channels available in the trunking pool.pool.

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Erlang B Trunking GOSErlang B Trunking GOS

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Example 3Example 3::How many users can be supported for 0.5% blocking probability How many users can be supported for 0.5% blocking probability for the following number of trunked channels in a blocked calls for the following number of trunked channels in a blocked calls cleared system?cleared system?a) 5a) 5 b) 10b) 10 c) 20c) 20 d) 100d) 100Assume each user generates 0.1 Erlangs of traffic.Assume each user generates 0.1 Erlangs of traffic. {0.1 Erlang : {0.1 Erlang : = 2 calls per hour ; H = av of 3 minutes each call = 2 calls per hour ; H = av of 3 minutes each call

Au = 2/60 * 3 = 0.1 Erlangs.}Au = 2/60 * 3 = 0.1 Erlangs.}SolutionSolutiona) C = 5, GOS = 0.5 %, Au = 0.1 ; From graph, A = 1.13a) C = 5, GOS = 0.5 %, Au = 0.1 ; From graph, A = 1.13

U = A/Au = 1.13/0.1 U = A/Au = 1.13/0.1 ≈≈ 11 users 11 usersb) C = 10, GOS = 0.5 %, Au = 0.1 ; From graph, A = 3.98b) C = 10, GOS = 0.5 %, Au = 0.1 ; From graph, A = 3.98

U = A/Au = 3.98/0.1 U = A/Au = 3.98/0.1 ≈≈ 40 users 40 usersc) c) U = A/Au = 13/0.1 U = A/Au = 13/0.1 ≈≈ 130 users 130 usersd) d) U = A/Au = 85/0.1 U = A/Au = 85/0.1 ≈≈ 850 users 850 users

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Erlang BErlang B

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Erlang CErlang C

Queue is provided to hold calls which are blockedQueue is provided to hold calls which are blocked If a ch is not available immediately, the call request If a ch is not available immediately, the call request

may be delayed until a ch becomes available.may be delayed until a ch becomes available. GOS is defined as the probability that a call is GOS is defined as the probability that a call is

blocked after waiting a specific length of time in blocked after waiting a specific length of time in queue.queue.

The likelihood of a call not having immediate access The likelihood of a call not having immediate access to a ch is determined by the Erlang C.to a ch is determined by the Erlang C.

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Erlang CErlang C

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Example 4Example 4

An urban area has a population of two million residents. Three An urban area has a population of two million residents. Three competing trunked mobile networks (Systems A, B and C) competing trunked mobile networks (Systems A, B and C) provide cellular services in this area. System A has 394 cells provide cellular services in this area. System A has 394 cells with 19 channels each, System B has 98 cells with 57 channels with 19 channels each, System B has 98 cells with 57 channels each, and system C has 49 cells, each with 100 channels.each, and system C has 49 cells, each with 100 channels.

i) Find the number of users that can be supported at 2% blocking i) Find the number of users that can be supported at 2% blocking if each user averages two calls per hour at an average call if each user averages two calls per hour at an average call duration of three minutes.duration of three minutes.

ii) Assuming that all three trunked systems are operated at ii) Assuming that all three trunked systems are operated at maximum capacity, compute the percentage market maximum capacity, compute the percentage market penetration of each cellular provider.penetration of each cellular provider.

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SolutionSolutionSystem ASystem Ai) i) Probability of blocking = 2% = 0.02Probability of blocking = 2% = 0.02

# of Channels, # of Channels, CC = 19 = 19

Traffic Intensity per user, Traffic Intensity per user, AAu u = = H = H = 2/60 x 32/60 x 3 = = 0.1 Erlangs0.1 Erlangs

From the graph, for GOS = 0.02, C = 19, get From the graph, for GOS = 0.02, C = 19, get AA = 12 Erlangs = 12 Erlangs

Total # of users per cell, Total # of users per cell, UU = = A/AA/Auu = 12/0.1 = 120 = 12/0.1 = 120

Total # of subscribers = 120 x 394 cells = 47,280.Total # of subscribers = 120 x 394 cells = 47,280.System BSystem Bi) i) Probability of blocking = 2% = 0.02Probability of blocking = 2% = 0.02

# of Channels, # of Channels, C C = 57= 57

Traffic Intensity per user, Traffic Intensity per user, AAu u = = H = H = 2/60 x 32/60 x 3 = = 0.1 Erlangs0.1 Erlangs

From the graph, for GOS = 0.02, C = 57, get From the graph, for GOS = 0.02, C = 57, get AA = 45 Erlangs = 45 Erlangs

Total # of users per cell, Total # of users per cell, U = A/AU = A/Auu = 45/0.1 = 450 = 45/0.1 = 450

Total # of subscribers = 450 x 98 cells = 44,100.Total # of subscribers = 450 x 98 cells = 44,100.

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System CSystem Ci) i) Probability of blocking = 2% = 0.02Probability of blocking = 2% = 0.02

# of Channels, C = 100# of Channels, C = 100

Traffic Intensity per user, Traffic Intensity per user, AAu u = = H = H = 2/60 x 32/60 x 3 = = 0.1 Erlangs0.1 Erlangs

From the graph, for GOS = 0.02, From the graph, for GOS = 0.02, CC = 100, get = 100, get AA = 88 Erlangs = 88 Erlangs

Total # of users per cell, Total # of users per cell, U = A/AU = A/Auu = 88/0.1 = 880 = 88/0.1 = 880

Total # of subscribers = 880 x 49 cells = 43,120.Total # of subscribers = 880 x 49 cells = 43,120.ii) Total cellular subscribersii) Total cellular subscribers = 47,280 + 44,100 + 43,120= 47,280 + 44,100 + 43,120

= 134,500.= 134,500.Total population = 2,000,000Total population = 2,000,000Therefore % market penetration for :Therefore % market penetration for :

System A : 47,280/2,000,000 = 2.364 %System A : 47,280/2,000,000 = 2.364 %System B : 44,100/2,000,000 = 2.205 %System B : 44,100/2,000,000 = 2.205 %System C : 43,120/2,000,000 = 2.156 %System C : 43,120/2,000,000 = 2.156 %

Total market penetration = 6.725 %Total market penetration = 6.725 %

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Example 5Example 5A certain city has an area of 1,300 square miles and is covered by A certain city has an area of 1,300 square miles and is covered by a cellular system using a seven-cell reuse pattern. Each cell has a a cellular system using a seven-cell reuse pattern. Each cell has a radius of four miles and the city is allocated 40 MHz of radius of four miles and the city is allocated 40 MHz of spectrum with a full duplex channel bandwidth of 60kHz. spectrum with a full duplex channel bandwidth of 60kHz. Assume a GOS of 2% for an Erlang B system is specified. If the Assume a GOS of 2% for an Erlang B system is specified. If the offered traffic per user is 0.03 Erlangs, compute offered traffic per user is 0.03 Erlangs, compute

a) The number of cells in the service areaa) The number of cells in the service area

b) The number of channels per cellb) The number of channels per cell

c) Traffic intensity of each cellc) Traffic intensity of each cell

d) The maximum carried trafficd) The maximum carried traffic

e) The total number of users that can be served for 2% GOSe) The total number of users that can be served for 2% GOS

f) The number of mobiles per unique channelf) The number of mobiles per unique channel

g) The teoretical max # of users could be served at one timeg) The teoretical max # of users could be served at one time

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SolutionSolutionArea = 1,300 milesArea = 1,300 miles22 N = 7N = 7 R = 4 milesR = 4 miles

GOS = 2 %GOS = 2 % BW = 40 MHzBW = 40 MHz BWBWchch = 60 kHz = 60 kHz

AAuu = 0.03 Erlangs = 0.03 Erlangs

a) # of cell = 1,300/(2.5981*4a) # of cell = 1,300/(2.5981*422) = 31.27 cells = 31 cells.) = 31.27 cells = 31 cells.

b) # of ch per cell b) # of ch per cell = BW / (BW= BW / (BWchch * 7) * 7) = 40 x 10= 40 x 1066 / 60 x 10 / 60 x 1033 * 7 * 7= 95.238 = 95 channels/cell= 95.238 = 95 channels/cell

c) From graph ; C = 95, GOS = 0.02,c) From graph ; C = 95, GOS = 0.02, A = 85 ErlangsA = 85 Erlangsd) Max carried traffic d) Max carried traffic = # of cell * A= # of cell * A

= 31 * 85 = 2635 Erlangs= 31 * 85 = 2635 Erlangs

e) Ae) Auu = 0.03 E, A= 0.03 E, Amaxmax = 2635 E ; = 2635 E ; From A = U AFrom A = U Auu

U U = A / A= A / Au u = 2635 / 0.03= 2635 / 0.03= 87,833 users.= 87,833 users.

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Cellular Concept (Handover)Cellular Concept (Handover)

12

34

5

7

6

Handover allows established calls to continue by switching them to another radio resource.

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Cellular concept (Cluster)Cellular concept (Cluster)

Network capacity can be increased by reusing the frequencies.

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Cellular concept (Macrocell)Cellular concept (Macrocell)

Macrocells are used for remote or sparsely populated. The cells typically have a radius of 10-35km.

10km.

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f) # of mobiles per ch f) # of mobiles per ch = # of users / # of ch= # of users / # of ch

= 87,833/(40 x 10= 87,833/(40 x 1066/60 x 10/60 x 1033) =87,833/666) =87,833/666

= 132 mobiles per ch= 132 mobiles per ch

g) Max # of user (theoretical) = # of the available ch in the systemg) Max # of user (theoretical) = # of the available ch in the system

( all channels occupied)( all channels occupied)

= C * Nc= C * Nc

= 95 * 31= 95 * 31

= 2,945 users= 2,945 users

( 3.4 % of the customer base )( 3.4 % of the customer base )

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Quiz 1Quiz 1

i) Show that S/I (co-channel interference) for a mobile may i) Show that S/I (co-channel interference) for a mobile may be expressed asbe expressed as

Where Where n : propogation power lawn : propogation power law

N : number of cell per clusterN : number of cell per cluster

iioo : number of co-channel interfering cells : number of co-channel interfering cells

ii) State 2 assumptions to be made for the above expressions.ii) State 2 assumptions to be made for the above expressions.

o

n

i

N

I

S )3(

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Exercise 2Exercise 2

If a signal-to-interference ratio of 15 dB is If a signal-to-interference ratio of 15 dB is required for satisfactory for a cellular system, required for satisfactory for a cellular system, what is the frequency reuse factor and cluster size what is the frequency reuse factor and cluster size that should be used for maximum capacity if the that should be used for maximum capacity if the path loss exponent is (a) n=4, (b) n = 3?path loss exponent is (a) n=4, (b) n = 3?

Assume that there are six co-channel cell in the Assume that there are six co-channel cell in the first tier, and all of them are at the same distance first tier, and all of them are at the same distance from the mobile. Use suitable approximations.from the mobile. Use suitable approximations.