MRN – Exercises – part A
Transcript of MRN – Exercises – part A
Politecnico di Milano Facoltà di Ingegneria dell’Informazione
MRN – Exercises – part A
Mobile Radio Networks Prof. Antonio Capone
Exercise 1
o Let us consider a cellular network based on multicarrier-TDMA with 24 carriers, and 3 channels per carrier a) Using the cluster model, dimension the cluster so as to
guarantee a minimum SIR equal to 14 dB assuming a propagation factor equal to 3
b) In the cells of the cluster the offered traffic is equal to 5 Erlang. What is the blocking probability?
c) Assuming the offered traffic is 80% of new calls and 20% of handover requests, calculated the blocking probability and the handover failure probability
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Exercise 1 (sol)
a) the cluster dimensioning formula is:
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Kmin ≅ 9.44 so we select K = 12
η2
minmin 3
)6( SIRK ⋅=
12.251014 1014
≅=dB
Not all values are admissible: K∈ {3,4,7,9,12,13,...}
Replacing the SIRmin and η numerical values in the formula we get:
SIRmin is equal to
3
Exercise 1 (sol)
In our case B(6,5) = 0.2
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612243
=⋅
=N
∑=
=N
k
k
N
kANA
ANB
0 !
!),(
b) We can use the Erlang-B formula B(N,A). The number N of channels is:
4
Exercise 1 (sol)
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Exercise 1 (sol)
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c) Since we don’t have guard channels, the system does not make any difference between new calls and handover requests Therefore the handover failure probability Phf and call blocking probability Pb are the same
2.0== bhf PP
6
Exercise 2
o Let us consider a cellular network based on multicarrier-TDMA using carriers with 5 channels each a) Assuming an offered traffic per cell equal to 5 Erlang,
calculate the minimum number of carriers necessary for guaranteeing a blocking probability of 2%
b) Assuming a minimum required SIR of 12 dB and a propagation factor of 2.8, calculate the total number of carriers necessary for the system
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Exercise 2 (sol) a) From the Erlang-B graph we see that the number of channels necessary is 10.
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5
0.02
With 5 channels per carriers, we then need 2 carriers per cell
8
Exercise 2 (sol)
b)
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8.285.151012 10
12
min
=
≅==
η
dBSIR
9
63.83
)6(2
minmin
=
=⋅
=
K
SIRKη
Since we need 2 carriers per cell and we have 9 cells per cluster, the system needs in total 18 carriers
9
Exercise 3
o The statistical counter of a base station indicates that a cell with 10 channels serve a traffic equal to 7 Erlang. Calculate: a) The offered traffic b) The blocking probability c) The channels utilization factor
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Exercise 3 (sol)
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8
Pb = B(8,10) = 0.12 Ao = As/(1-Pb)= = 7/0.88 = 8 ρ = As/N = 0.7
11
Exercise 4 o Let us consider a cellular network based on multicarrier-
TDMA with 18 carriers, and 4 channels per carrier a) Using the cluster model, dimension the cluster so as to
guarantee a minimum SIR equal to 13.5 dB with a propagation factor of 3.
b) Assuming the offered traffic has a uniform distribution in the area with an intensity of 15 Erlang per km2, dimension the cell radius so that the blocking probability is equal to 2%.
c) After one year the statistical counter indicates that the cell serves 4,65 Erlang. Calculate the offered traffic and the blocking probability.
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Exercise 4 (sol)
a) Using the formula with the numerical values: We get and then K=9
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( )3
6 /2
min
ηSIRK ⋅=
74,8min =K
339.2210 5.13 10
5.13
min
=
≅==
η
dBSIR
13
Exercise 4 (sol) b) With K = 9 each cell has 2 carriers and 8 channels. With a maximum probability of 2% we get from the Erlang-B graph that the maximum offered traffic is approx. 3.5 Erlang With a maximum traffic of 3.5 Erlang and a traffic density of 15 Erlang/ km2, the area of the cell must be limited to 3.5/15 = 0.233 km2. The area of an hexagon is: And the the radius
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2332rA =
m 299=r
14
Exercise 4 (sol)
c) From the Erlang-B formula we get: B(5,8)=0.07 As=Ao(1-B(5,8))=4.65
And therefore the offered traffic is 5 Erlang and the blocking probability 7%.
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Exercise 5 o a) Derive a model for calculating the blocking probability
in a CDMA based mobile radio network with different traffic classes characterized by different information rates (and therefore different processing gain), considering the uplink only and assuming ideal power control.
o b) Define the blocking states for each traffic class assuming: n Thermal noise negligible n (Eb/No)min equal to 10 n 3 traffic classes with processing gain equal to 50, 25 and
20 respectively.
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Exercise 5 (sol)
a) In the case of CDMA with single cell, single service, uplink direction, the capacity is obtained calculating the SIR:
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1111)1(
+≅++≤
≥+−
=
bPbn
bPnPSIR
ηη
17
Exercise 5 (sol)
o Parameter b is given by:
o Where Gp is the processing gain:
o B is the bandwidth at radio frequency and R is the information rate.
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pGSIRb min=
RBGp =
18
Exercise 5 (sol)
o In the case of the multiservice system, information rate has multiple values. For simplicity we consider this values R as multiple of a base rate r:
o It is easy to observe that for a constant SIR (SIRmin) increasing the rate from r to dir we need to increase the power from P to diP.
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{ },,...,, 21 rdrdrdR m=
19
Exercise 5 (sol)
o Therefor the SIR constraints becomes:
o Where br is the value of b corresponding to the rate r and nj is the number of users with rate djr.
o Therefore we have:
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r
jjj
bPPdn
PSIR ≥−
=∑
Cb
dnrj
jj =+≤∑ 11
20
Exercise 5 (sol)
o We can then apply the analysis of the extended Erlang-B formula in the case of Poisson arrival process and exponentially distributed service times
o … (see lecture slides)
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Exercise 5 (sol)
o b) In the considered case we have:
o We need to define a base rate: o And then:
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20/25/50/
3
2
1
BRBRBR
=
=
=
100/Br =
R1 = 2rR2 = 4rR3 = 5r
22
Exercise 5 (sol)
o The capacity C with the rate r is equal to: C = 11 o The blocking states are:
– States for R1: (5,0,0);(3,1,0);(3,0,1);(1,2,0);(1,1,1);(0,0,2) – States for R2: Those of R1 + (4,0,0);(2,1,0);(2,0,1);(0,1,1);(0;2;0) – States for R3: Those of R1 + those of R2 + (1,0,1)
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rRrRrR
542
3
2
1
=
=
=
23
Exercise 6 o a) Derive a model for calculating the call blocking probability
and the call dropping probability due to handover failure for a mobile radio system with n channels per cell of which g reserved for handover requests only.
o b) Calculate the average handover frequency λh , the average channel holding time τc and the call dropping probability Pd in the case of: – Users moving at 30 km/h and random direction; – Square cells with edge l=300 m – New call frequency λi equal to 3.5 calls/min; – Average call duration τ =1/µ equal to 2 min; – n=15, g=0;
o The channel holding time and the call duration are exp distributed random variables and arrival processes (new calls and handover) are Poisson processes.
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Exercise 6 (sol)
o a) see lecture slides
o b)
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38.05.012.2
111=
+=
+==
ηµµτ
cc
η =V LπS
= 2.12 cells changes/min
2m 90000 m 1200
m/min 500
=
=
=
SLV
25
Exercise 6 (sol)
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ν =ττ h
=ηµ=2.120.5
= 4.24
λh =Ph 1−Pb( )
1−Ph 1−Phf( )"# $%λi ≅
Ph1−Ph( )
λi =ηµλi =νλi =
= 4.24 ⋅3.5=14.84 handover/min
26
Exercise 6 (sol)
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A = λi +λhµc
=18.34 ⋅0.38 ≅ 7
Pb = B(A,N ) = B(7,15) ≅ 0.003Pd ≅νPb ≅ 0.0127
27
Exercise 7
o Consider a CDMA cellular system with 2 base stations and 4 active terminals. Attenuations aij (=1/gij) between each base station i and terminal j are reported in matrix A.
o A power control mechanism based on a closed loop control keeps the Eb/No at a constant value of 10 dB. The processing gain Gp is 80 and noise power N is 0.25.
o Calculate emitted powers pi of terminals (uplink) assuming terminals are connected to the base station with the minimum attenuation.
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{ } ⎥⎦
⎤⎢⎣
⎡==
42888844
ijaA
28
Exercise 7 (sol)
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⎪⎪⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪⎪
⎨
⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
∑
∑
∑
∑
≠
≠
≠
≠
ptar
b
iii
ptar
b
iii
ptar
b
iii
ptar
b
iii
GNE
Ngpgp
GNE
Ngpgp
GNE
Ngpgp
GNE
Ngpgp
1
1
1
1
04
2
244
03
2
233
02
1
122
01
1
111
⎪⎪
⎩
⎪⎪
⎨
⎧
≅
≅
≅
≅
166.0083.0160.0160.0
1
3
2
1
pppp
29
Exercise 8
o Consider a cell with new calls traffic equal to 2 Erlang and handover traffic of 1 Erlang. The cell has 4 channel of which one is guard channel for handover.
o Calculate the call blocking probability and the handover failure probability
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Exercise 8 (sol)
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In our case we have:
c = 3
g = 1
( )
( ) ( )
( )
( ) ( )∑ ∑
∑ ∑
∑∑
=
+
+=
−+
=
+
+=
−
+
=
−
+
=
+++
++
==
+++
+==
c
k
gc
ck
ckhc
ih
kih
ghc
ih
gchf
c
k
gc
ck
ckhc
ih
kih
gc
ck
ckhc
ihgc
ckkb
kk
gcP
kk
kP
0 1
0 1
!!
)!(
!!
!
ρρρ
ρρ
ρρρ
π
ρρρ
ρρ
ρρρ
π
12
=
=
h
i
ρ
ρ
31
Exercise 8 (sol)
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We have:
π0 ≅ 0.071
π1 ≅ 0.212 π2 ≅ 0.318 π3 ≅ 0.319 π4 ≅ 0.080
399,0 08,0 == bhf PP