# Chapter No.3 : The cellular concept system design fundamentals¶

## Example 3.1 Page No.61¶

In [14]:
from math import ceil
# To compute the number of channels available per cell for a)four-cell reuse system a)seven-cell reuse system a)12-cell reuse system

# Given data
B=33*10**6#                                                  # Total bandwidth allocated to particular FDD system in Hz
Bc=25*10**3#                                                 # Bandwidth per channel in Hz
Nc=2#                                                       # Number of simplex channels
Bc=Bc*Nc#                                                   # Channel bandwidth in Hz

Ntotal=B/Bc#                                                # Total number of channels

#a) To compute the number of channels available per cell for four-cell reuse system
N=4#                                                        # frequency reuse factor
chpercell=Ntotal/N#                                         # number of channels available per cell for four-cell reuse system

# Displaying the result in command window
print "\n The number of channels available per cell for 4-cell reuse system = %0.0f channels"%(chpercell)
print "\n One control channel and 160 voice channels would be assigned to each cell."

# b) To compute the number of channels available per cell for seven-cell reuse system
N=7#                                                        # frequency reuse factor
chpercell=ceil(Ntotal/N)#                                         # number of channels available per cell for seven-cell reuse system

# Answer is varrying due to round-off error

# Displaying the result in command window
print "\n \n The number of channels available per cell for 7-cell reuse system = %0.0f channels"%(chpercell)
print "\n Each cell would have one control channel, four cells would have 90 voice channels and three cells would have 91 voice channels."

# c) To compute the number of channels available per cell for 12-cell reuse system
N=12#                                                        # frequency reuse factor
chpercell=Ntotal/N#                                          # number of channels available per cell for seven-cell reuse system

# Displaying the result in command window
print "\n \n The number of channels available per cell for 12-cell reuse system = %0.0f channels"%(chpercell)
print "\n Each cell would have one control channel, eight cells would have 53 voice channels and four cells would have 54 voice channels."

 The number of channels available per cell for 4-cell reuse system = 165 channels

One control channel and 160 voice channels would be assigned to each cell.

The number of channels available per cell for 7-cell reuse system = 95 channels

Each cell would have one control channel, four cells would have 90 voice channels and three cells would have 91 voice channels.

The number of channels available per cell for 12-cell reuse system = 55 channels

Each cell would have one control channel, eight cells would have 53 voice channels and four cells would have 54 voice channels.


## Example 3.2 Page No.72¶

In [15]:
from math import sqrt, log10
from __future__ import division
# To find frequency reuse factor for path loss exponent (n) a)n=4 b)n=3

# Given data
SIdB=15#                     # Signal to interference(dB)
io=6#                        # Number of cochannel cell

# For n=4
n1=4#                        # Path loss exponent
N1=7#                        # First consideration: frequency reuse factor N=7
DR1=sqrt(3*N1)#              # Co-channel reuse ratio
si1=(1/io)*(DR1)**n1#         # Signal to interference
sidB1=10*log10(si1)#         # Signal to interference(dB)

# For n=3
n2=3#                        # Path loss exmponent
si=(1/io)*(DR1)**n2#          # Signal to interference for first consideration: frequency reuse factor N=7
sidB=10*log10(si)#           # Signal to interference(dB)

N2=12#                       # second consideration : frequency reuse factor N=12 since sidB<SIdB
DR2=sqrt(3*N2)#              # Co-channel reuse ratio
si2=(1/io)*(DR2)**n2#         # Signal to interference
sidB2=10*log10(si2)#         # Signal to interference(dB)

# Displaying the result in command window
print "\n Signal to noise ratio for n=4 with frequency reuse factor N=7 = %0.2f dB"%(sidB1)
print "\n Signal to noise ratio for n=3 with frequency reuse factor N=7 = %0.2f dB"%(sidB)
print "\n Signal to noise ratio for n=3 with frequency reuse factor N=12 = %0.2f dB"%(sidB2)
print "\n Since SIR is for n=3 with frequency reuse factor N=7 greater than the minimum required, so N=12 is used."

 Signal to noise ratio for n=4 with frequency reuse factor N=7 = 18.66 dB

Signal to noise ratio for n=3 with frequency reuse factor N=7 = 12.05 dB

Signal to noise ratio for n=3 with frequency reuse factor N=12 = 15.56 dB

Since SIR is for n=3 with frequency reuse factor N=7 greater than the minimum required, so N=12 is used.


## Example 3.4 Page No.80¶

In [1]:
# To find number of users for Number of channels (C) a)C=1 b)C=5 c)C=10 d)C=20 e)C=100

# Given data
Au=0.1#                # Traffic intensity per user

# a)To find number of users for C=1
C1=1#                  # Number of channels
A1=0.005#              # Total traffic intensity from Erlangs B chart
U1=(A1/Au)#            # Number of users
U1=1#                  # Since one user could be supported on one channel

# b)To find number of users for C=5
C2=5#                  # Number of channels
A2=1.13#               # Total traffic intensity from Erlangs B chart
U2=round(A2/Au)#       # Number of users

# c)To find number of users for C=10
C3=10#                 # Number of channels
A3=3.96#               # Total traffic intensity from Erlangs B chart
U3=round(A3/Au)#       # Number of users

# Answer is varrying due to round off error

# d)To find number of users for C=20
C4=20#                 # Number of channels
A4=11.10#              # Total traffic intensity from Erlangs B chart
U4=round(A4/Au)#       # Number of users

# Answer is varrying due to round off error

# e)To find number of users for C=100
C5=100#                # Number of channels
A5=80.9#               # Total traffic intensity from Erlangs B chart
U5=round(A5/Au)#       # Number of users

# Displaying the result in command window
print "\n Total number of users for 1 channel = %0.0f"%(U1)
print "\n Total number of users for 5 channel = %0.0f"%(U2)
print "\n Total number of users for 10 channel = %0.0f"%(U3)
print "\n Total number of users for 20 channel = %0.0f"%(U4)
print "\n Total number of users for 100 channel = %0.0f"%(U5)

 Total number of users for 1 channel = 1

Total number of users for 5 channel = 11

Total number of users for 10 channel = 40

Total number of users for 20 channel = 111

Total number of users for 100 channel = 809


## Example 3.5 Page No.83¶

In [2]:
from __future__ import division
# To find number of users for a)system A b)system B c)system C

# Given data
GOS=0.02#                           # Grade of Service (Probability of bloacking)
lamda=2#                            # Average calls per hour
H=(3/60)#                           # Call duration in seconds

Au=lamda*H#                         # Traffic intensity per user

# a)To find number of users for System A
C1=19#                             # Number of channels used
A1=12#                             # Traffic intensity from Erlang B chart
U1=round(A1/Au)#                   # Number of users per cell
cells1=394
TU1=U1*cells1#                     # Total number of users
MP1=TU1/(2*10**6)*100#              # Market penetration percentage

# b)To find number of users for System B
C2=57#                            # No. of channels used
A2=45#                            # Traffic intensity from Erlang B chart
U2=round(A2/Au)#                  # Number of users per cell
cells2=98
TU2=U2*cells2#                    # Total no. of users
MP2=TU2/(2*10**6)*100#             # Market penetration percentage

# c)To find number of users for System C
C3=100#                          # Number of channels used
A3=88#                           # traffic intensity from Erlang B chart
U3=round(A3/Au)#                 # Number of users per cell
cells3=49
TU3=U3*cells3#                   # Total no. of users
MP3=TU3/(2*10**6)*100#            # Market penetration percentage

TU=TU1+TU2+TU3#                  # Total number of users in all 3 systems
MP=TU/(2*10**6)*100#              # Combined Market penetration percentage

# Displaying the result in command window
print "\n Total number of users in system A = %0.0f"%(TU1)
print "\n The percentage market penetration of system A  = %0.2f"%(MP1)
print "\n \n Total number of users in system B = %0.0f"%(TU2)
print "\n The percentage market penetration of system B  = %0.3f"%(MP2)
print "\n \n Total number of users in system C = %0.0f"%(TU3)
print "\n The percentage market penetration of system C  = %0.3f"%(MP3)
print "\n \n Total number of users in all 3 systems = %0.0f"%(TU)
print "\n The combined Market penetration percentage of all systems = %0.3f"%(MP)

 Total number of users in system A = 47280

The percentage market penetration of system A  = 2.36

Total number of users in system B = 44100

The percentage market penetration of system B  = 2.205

Total number of users in system C = 43120

The percentage market penetration of system C  = 2.156

Total number of users in all 3 systems = 134500

The combined Market penetration percentage of all systems = 6.725


## Example 3.6 Page No.84¶

In [2]:
# To find a)Number of cells in given area b)Number of channels/cell c)Traffic intensity per cell d)Maximum carried traffic e)Total number of users for 2% GOS f) Number of mobiles per unique channel   g)Maximum number of users could be served at one time

# Given data
Area=1300#                          # Total coverage area in m**2
R=4#                                # Radius of cell in m
N=7#                                # Frequecy reuse factor
S=40*10**6#                          # Allocated spectrum in Hz
Ch=60*10**3#                         # Channel width in Hz

# a)Number of cells
CA=2.5981*R**2#                      # Area of hexagonal cell in m**2
Nc=round(Area/CA)#                  # Number of cells

# Displaying the result in command window
print "\n Number of cells in given system = %0.0f cells"%(Nc)

# b)Number of channels/cell
C1=round(S/(Ch*N))#                  # Number of channels

# Displaying the result in command window
print "\n \n Number of channels per cell in given system = %0.0f channels/cell"%(C1)

# c) Traffic intensity per cell
C1=95#                               # Number of channels from b)
A=84#                                # Traffic intensity from Erlang B chart

# Displaying the result in command window
print "\n \n Traffic intensity in given system = %0.0f Erlangs/cell"%(A)

# d)Maximum carried traffic
traffic=Nc*A#                       # Maximum carried traffic

# Displaying the result in command window
print "\n \n Maximum carried traffic in given system = %0.0f Erlangs"%(traffic)

# e)Total number of users for 2% GOS
trafficperuser=0.03#                 # Given traffic per user
U=traffic/trafficperuser#            # Total number of users

# Displaying the result in command window
print "\n \n Total number of users = %0.0f users"%(U)

# f) Number of mobiles per unique channel
C=666#                               # Number of channels
mobilesperchannel=round(U/C)#        # Number of mobiles per unique channel

# Displaying the result in command window
print "\n \n Number of mobiles per unique channel = %0.0f mobiles/channel"%(mobilesperchannel)

# g)Maximum number of users could be served at one time
print "\n \n Theoretically maximum number of served mobiles is the number of available channels in the system."
C=C1*Nc#                              # Maximum number of users could be served at one time

# Displaying the result in command window
print "\n Theoretical Maximum number of users could be served at one time = %0.0f users"%(C)
print "It is 3.4% of customer base."

 Number of cells in given system = 31 cells

Number of channels per cell in given system = 95 channels/cell

Traffic intensity in given system = 84 Erlangs/cell

Maximum carried traffic in given system = 2604 Erlangs

Total number of users = 86800 users

Number of mobiles per unique channel = 130 mobiles/channel

Theoretically maximum number of served mobiles is the number of available channels in the system.

Theoretical Maximum number of users could be served at one time = 2945 users
It is 3.4% of customer base.


## Example 3.7 Page 85¶

In [2]:
from math import exp
# To find a)number of users per square km b)probability that delayed call have to wait longer than t=10sec c)probability that call is delayed more than 10 sec

# Given data
R=1.387#                                  # Radius of cell in m
Area=2.598*R**2#                           # Area of hexagonal cell in m**2
cellpercluster=4#                         # Number of cells/cluster
channels=60#                              # Number of channels

channelspercell=channels/cellpercluster#  # Number of channels per cell

# a)To find number of users per square km
A=0.029#                                  # Traffic intensity per user
traffic=9#                                # Traffic intensity from Erlang chart C
U1=traffic/A#                             # Total number of users in 5sq.km.
U=round(U1/Area)#                         # Number of users per square km

# Displaying the result in command window
print "\n Number of users per square km in given system = %0.0f users/sq km"%(U)

# b)To find the probability that delayed call have to wait longer than t=10sec
lamda=1#                                 # Holding time
H1=A/lamda#                              # Duration of call
H=H1*3600#                                # Duration of call in second
t=10
Pr=exp(-(channelspercell-traffic)*t/H)*100#         # probability that delayed call have to wait longer than t=10sec.

# Displaying the result in command window
print "\n \n Percentage of probability that delayed call have to wait longer than t=10 sec = %0.2f percent"%(Pr)

# c)To find the probability that call is delayed more than 10 sec
Pr10=delayprob*Pr#                        # probability that call is delayed more than 10 sec

# Displaying the result in command window
print "\n \n Percentage of probability that call is delayed more than 10 sec = %0.2f percent"%(Pr10)

 Number of users per square km in given system = 62 users/sq km

Percentage of probability that delayed call have to wait longer than t=10 sec = 56.29 percent

Percentage of probability that call is delayed more than 10 sec = 2.81 percent


## Example 3.8 Page 89¶

In [3]:
# To find number of channels in 3 km by 3 km square centered around A in Figure 3.9 for a)without use of microcell b)with the use of lettered microcells c)all base stations are replaced by microcells

# Given data
R=1#                                                                      # Cell radius in km
r=0.5#                                                                    # Micro-cell radius in km
Nc=60#                                                                    # Number of channels in base station

# a)To find number of channels without use of microcell
Nb1=5#                                                                     # Number of base stations in given area
N1=Nb1*Nc#                                                                 # Number of channels without use of microcell

# b)To find number of channels with the use of lettered microcells
Nb2=6#                                                                     # Number of lettered microcells
Nb2=Nb1+Nb2#                                                               # Total number of base stations in given area
N2=Nb2*Nc#                                                                 # Number of channels with the use of lettered microcells

# c)To find number of channels if all base stations are replaced by microcells
Nb3=12#                                                                    # Number of all the microcells
Nb3=Nb1+Nb3#                                                               # Total number of base stations in given area
N3=Nb3*Nc#                                                                 # Number of channels if all base stations are replaced by microcells

# Displaying the result in command window
print "\n Number of channels without use of microcell = %0.0f channels"%(N1)
print "\n \n Number of channels with the use of lettered microcells = %0.0f channels"%(N2)
print "\n \n Number of channels if all base stations are replaced by microcells = %0.0f channels"%(N3)

 Number of channels without use of microcell = 300 channels

Number of channels with the use of lettered microcells = 660 channels

Number of channels if all base stations are replaced by microcells = 1020 channels


## Example 3.9 Page 92¶

In [7]:
from __future__ import division
# To analyze trunking efficiency capacity of sectoring and unsectoring

# Given data
H=2/60#                                                       # Average call duration in hour
GOS=0.01#                                                     # Probability of blocking

# Unsectored system
C1=57#                                                         # Number of traffic channels per cell in unsectored system
A=44.2#                                                        # Carried traffic in unsectored system
calls1=1326#                                                   # Number of calls per hour in unsectored system from Erlangs B table

# 120 degree sectored system
C2=C1/3#                                                      # Number of traffic channels per antenna sector in 120 degree sectored system
calls2=336#                                                   # Number of calls per hour in 120 degree sectored system from Erlangs B table
Ns1=3#                                                        # Number of sectors
capacity=Ns1*calls2#                                          # Cell capacity or number of calls handled by system per hour

dif=calls1-capacity#                                          # decrease in cell capacity in 120 degree sectored system
percentdif=(dif/calls1)*100#                                  # decrease in cell capacity in 120 degree sectored system in percentage

# Displaying the result in command window
print "\n Cell capacity of unsectored system = %0.0f calls/hour"%(calls1)
print "\n \n Cell capacity of 120 degree sectored system = %0.0f calls/hour"%(capacity)
print "\n \n Decrease in cell capacity in 120 degree sectored system = %0.2f percent"%(percentdif)

 Cell capacity of unsectored system = 1326 calls/hour

Cell capacity of 120 degree sectored system = 1008 calls/hour

Decrease in cell capacity in 120 degree sectored system = 23.98 percent