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."
```

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."
```

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
GOS=0.005# #G rade of Service
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)
```

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)
```

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)
GOS=0.02# # Grade of service
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."
```

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
delayprob=0.05# # Grade of service
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)
```

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)
```

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)
```