Cellular architecture

Introduction Mobile computing
Informal Deﬁnition of Mobile Computing
Ability to work from a non-ﬁxed location.
Mobile computing: anytime, anywhere computing
Basic requirements are:
- Portability.
- Wireless network.
R. K. Ghosh Mobile Computing CS634 4 / 116

Portability Related Issues
Scarcity resource:
- Low power CPU
- Limited memory
- Low batter life
Less reliable:
- Lost or physical damaged.
- Administered by novice users.

Wireless Network Related Issues
Low bandwidth and high error rates
Frequent network outage
- Disconnection due to limited coverage
- Voluntary disconnection for saving battery.
Single hop connectivity and asymmetry in connection.
High tariff.

System Related Issues
Established basis for system design may not hold:
time
battery life
size
resources Research trends based on time
traditional
mobile
adding :

System Related Issues
Heterogeneity and variability increases in mobile computing
Location becomes non-static.
Connectivity becomes variable.
Bandwidth becomes variable.
Device interface becomes variable.
Environment related inﬂuence increases.
Security and vulnerability increases.

Consequences of System Related Issues
Must deal with resources variation.
Must support heterogeneity.
Must adapt to environmental condition.
Must handle intermittent connectivity.
Must handle mobility across domains.
Must handle scalability.

Introduction Mobile/distributed computing
Mobile Versus Distributed Computing
Distributed systems
No resource scarcity.
All computers have
comparable resources.
Symmetry in computation.
All computers have ﬁxed
location.
Failure model is simple.
Mobile systems
Suffer from resource scarcity.
Location search is required.
Involuntary disconnection is
not necessarily failure.
Asymmetry in computation as
end nodes are resource poor.

Introduction Radio communication
Capacity & Coverage
Solution: reuse frequencies, but without interferences.
- spectral congestion can be eliminated by developing an
architecture which allows spatial multiplexing.

Cellular Architecture
Radio Connectivity
Coverage is the main focus of radio systems.
– A high power antenna mounted on a large tower can provide better
coverage
– But spectrum for private communication is limited (25MHz).
– Abour 30kHz required for voice communication.
– So, about 25MHz/30kHz = 833 pair of connections possible
How to increase both capacity & coverage?

Capacity & Coverage
Solution: reuse frequencies, but without interferences.
- spectral congestion can be eliminated by developing an
architecture which allows spatial multiplexing.

Cellular architecture
Cellular architecture: is the turning point in frequency reuse
based wireless communication.
It allows spatial multiplexing which not only eliminates
interferences but also provide continuous uninterrupted
coverage.
It is equivalent to partitioning of a large coverage area using small
cells each serviced by one transceiver.

Could be viewed as equivalent to marking out counties in a
districtby using colors.
A subtle difference exists between map coloring and spatial
multiplexing of spectrum:
- To avoid co-channel and adjacent channel interferences while
planning frequency reuse.
Frequency reuse plan needs an engineering solution.

Theoretical Abstraction
Coverage area of an antenna can be considered as a circle.
Continuous coverage of an area means: packing of the desired
area by equal sized circles such that
- No uncovered gaps should exist.
- Possible if the circles overlap minimally (guard band).

Approximated as hexagons which completely packs an area.

Cellular architecture Frequency planning
Frequency Planning

Frequency Planning
The group of channels allocated to one cell should be different
from those assigned its geographically adjacent cells.
- C: total number of duplex channels
- Cx: number allocated to a cell x
- N: number of cells among which C channels equally divided
- I.e., C = N × Cx.
The group of N cells is called a cluster.
- Rc: number of replicated clusters in a system.
- The system capacity: K = Rc × C.
So, the size of a cluster determines system capacity.

Capacity Enhancements
Suppose a cluster is replicated 4 times.
Effective frequency: 25MHz×4=100MHz.
Let a cluster size be 7:
- Each cell gets a spectrum of 25MHz/7 = 3.57MHz.
- And can support 3.57MHz/30kHz = 119 users.
The number of users in each cluster = 119×7 = 833.
With four clusters the total number of users = 3332.

System Design
The maximum number of calls/hour/cell is decided on the basis of
trafﬁc conditions in each cell.

System Design
The maximum number of calls/hour/cell is decided on the basis of
trafﬁc conditions in each cell.
Example 1
Let Q denote the number of calls per hour during busy period. Let the
values of Q for 10 cells be 2000, 3000, 500, 1000, 1200, 1800, 2500,
2800, 900, 1500. Assume 60% of the car phones will be used during
this period, and one call is made per phone.
Total number of calls i Qi = 17200 calls per hour.
Since 17200 is 60% of capacity, the maximum number of possible
calls is 17200/0.6 = 28667.

Geometry of a Cell
R
( 3/2)R

Geometry of a Cell
R
( 3/2)R
Area of the triangle = (
√
3/4)R2.

Geometry of a Cell
R
( 3/2)R
Area of the hexagon = 6(
√
3/4)R2.

Reuse Distance
Lemma
Let the coordinate of a cell C be (0,0) and that of its co-channel cell C
be (i, j). Then distance CC is equal to R 3(i2 + ij + j2), where R
denotes the length of a side of the regular hexagon representing a cell.

Reuse Distance
3
π/3
jR sin
3jRcosπ/3
3
iR
3jR
D
C
(0,0)
C’
(i,j)
B
A
π/3

Reuse Distance
3
π/3
jR sin
3jRcosπ/3
3
iR
3jR
D
C
(0,0)
C’
(i,j)
B
A
π/3
Axes at 60◦ to each other.

Reuse Distance
3
π/3
jR sin
3jRcosπ/3
3
iR
3jR
D
C
(0,0)
C’
(i,j)
B
A
π/3
Coordinates are represented by integers.

Reuse Distance
Proof.
C can be reached from C by traversing i cells in one axial
direction then turning 60◦ in clockwise and then hopping j cells in
the other axial direction.
AB =
√
3jR cos π/3 =
√
3
2 jR, and
BC =
√
3jR sin π/3 = 3
2iR.

Reuse Distance
Proof.
D2
=
√
3iR +
√
3
2
jR
2
+
3
2
jR
2
= R2
3i2
+ 3i.j +
3
4
j2
+
9
4
j2
= R2
(3i2
+ 3ij + 3j2
)
= 3R2
(i2
+ ij + j2
)

Cells Per Cluster
The number of cells N per cluster can be found by ﬁnding
proportion of the cell area to the cluster area.
The area of a cluster can be found by analyzing the cell geometry.

Replicated Adjacent Clusters
C C’
A
B

Cells Per Cluster
D/2
D/
√
3
Area = 6D2
/4
√
3
An adjacent pair of small triangles are congruent.

Cells Per Cluster
Lemma
The number of cells in a cluster is equal D2
3R2 = i2 + ij + j2
Proof.
The area A (of a cluster) is equal to 6 D2
4
√
3
and the area Ac (of a
cell) = 6R2
√
3/4. Therefore, the number of cells in a cluster:
N =
A
Ac
=
6D2
4
√
3
/
6R2
√
3
4
=
D
R
√
3
2
=
3R2(i2 + ij + j2)
3R2
= i2
+ ij + j2

Frequency Planning
For example, if (2, 2) is the closest co-channel cell of a cell at position
(0, 0), then
The cluster size is N = 22 + 2.2 + 22 = 12.
1
2
5
6
1
8
9
10
10 8
6
9
8
2
7
12
9 5
10
11
6
5
4
5
6 4
10
11
4
3
9
2
5
2
37
8
11
37
12
12
12
1
1
1
1
1
3
9
4
7
211
11

Cellular architecture Splitting and sectoring
Splitting & Sectoring
Increased reuse of frequency increases chance of interference.
How to keep interference low and increase the capacity?
Service area has been planned and infrastructure is in place, any
incremental change is difﬁcult to execute.
To address this two simple ideas
1 Cell splitting
2 Cell sectoring

Cell Splitting
Creates smaller cells out of a congested cell.
– By reducing both antenna size and transmitter power.
So increased spatial multiplexing happens with smaller cells.
Smaller cells are placed in or between large cells.
If cell radius becomes R/2, then D also becomes D/2
Frequency reuse plan is preserved by keeping Q = D/R
unchanged.
– Reducing D would imply increase in interference.
But micro cells require more frequent handoffs.

Cell Splitting
in−splitting between−splitting
Areas with high traffic load
Two possible splittings.

Cell Splitting
Let the transmit power of the base station in the original cell be Po
and that of micro-cell be Pm.
The received power Pr at cell boundaries of the two cells are:
Pr[original cell] ∝ PoR−n
Pr[micro-cell] ∝ Pm(R/2)−n
If n = 4, then transmit power of micro cells should be reduced by
1/16, i.e. Pt2 = Pt1/16.

Effect of Splitting
It is not necessary to split all the cells.
Sometimes it becomes difﬁcult to exactly identify the coverage
area that would require cell splitting.
So in practice different cell sizes may co-exist.
Therefore, a careful ﬁne-tuning of power outputs by transceivers is
needed to keep co-channel intereference at minimum level.
The channel assignment becomes quite complicated.

Cell Sectoring
Cell sectoring another technique.
Transmit power of a channel is concentrated into a ﬁnite sector of
the cell.
The sectoring causes co-channel interference and transmission
only within a speciﬁed region of the cell.
So, it leads to greater reuse of frequencies.
Normally a cell is partitioned into three/six sectors.

Cell sectoring
120
o
60o

Cell sectoring
A
B
C
D
affect D
does not affect D

Signal measurement & interference
Signal Measurements
DeciBel (dB): measurement unit for relative strengths of radio
signals.
10 DeciBel equals one Bel representing power ratio 1:10.
Power ratio 1:100 equals 2 Bels or 20 deciBels.
Similarly, power ratio 1:1000 is 3 Bels or 30 deciBels.
Power gain due to amplification is measured by relative power
strengths of input power P1 and amplified power P2.
With log scale, log10 (P2/P1) measures relative power strength
due to amplification in Bels.
Eg., if an amplifier outputs 100watt with an input of 100 milliwatts,
then power gain is log10(100/0.1) = log10 1000 = 3 Bels or
30 deciBels.

Trafﬁc Measurement
Example 2
Suppose, a micro-wave system uses a 10 watt transmitter. The
transmitter is connected by a cable with 0.7 dB loss to a 13 dB
antenna. Let atmospheric loss be 137 dB on transmission. The
receiver antenna with 11 dB gain connected to cable with 1.4 dB loss
to the receiver. Then the what is the power at the receiver?

Trafﬁc Measurement
Solution:
10 watts = 10000 mW.
10log10(10000/1) = 40 dB power output of transmitter.
The relative strength of power at the receiver end = (40 - 0.7 + 13
- 137) dB = -84.7 dB.
The loss at receiver side (11 - 1.4) dB.
So the net power at the receiver = (-84.7 + 9.6) dB = -75.1 dB.

Signal to Interference Ratio
Deﬁnition
The quality of received signals from the current BS affected by
interference from the signals of its nearby BS using the same
frequency.

Signal to Interference Ratio
Deﬁnition
The quality of received signals from the current BS affected by
interference from the signals of its nearby BS using the same
frequency.
Deﬁnition
Co-channel interference is measured by Signal to Interference Ratio
(SIR) at mobile terminals. This ratio is
S/I = S/
i0
i=1
Ii ,
where Ii is the interfering signal received from co-channel i, and i0 is
the number of co-channel cells nearby.

Signal Attenuation
In free space, average signal strength decays according a power law
involving distances between the transmitter and the receiver.
d: is the distance between the transmitter and the receiver.
P0: is the power measured at a reference point which is at
distance d0 from the transmitter.
Then the average received power Pr at the receiver from the
transmitting antenna is given by:
Pr = P0
d
d0
−n
,
where n is the path loss exponent.
In reality Pr will be proportional to the expression in RHS.

Signal Propagation
The relation between power strengths at the transmitter and the
receiver in log scale can be expressed as
log10 Pr = log10 P0 − n log10
d
d0
.
In terms of deciBel (dB) units, it is
Pr(dB) = P0(dB) − 10n log10
d
d0
.
Note: n is in Bel so 10n is deciBel equivalent.

SIR for Co-channel Interference
Di: the distance of a mobile terminal (MT) from ith co-channel
cell.
R: is the radio range of current BS.
Signal attenuation from the co-channel cell is proportional to D−n
i .
The strength of signal received from the current BS is proportional
to R−n.

SIR for Co-channel Interference
If MT located at the center of the current cell, then all interfering
co-channel BSs at equal distance from MT.
That is Di = D, ∀ i, SIR (in dB) is:
S/I = 10 log10 R−n
/
i0
i=1
D−n
i
= 10 log10(D/R)−n
/i0
= 10 log10
√
3N
n
/i0 .

Co-channel Interference
Assume that 18dB is the minimum SIR for good voice quality.
Let n = 4 be path loss exponent.
Let N = 7 be the cluster size.
S/I = 10 log10
√
3N
4
/i0
= 10 log10
√
21
4
/6
= 10 log10 73.5
= 10 × 1.866
= 18.66

Topological Consideration for SIR
In the worst case scenario, an MT may be located at the edge of a cell.
D
DD−R
D−R
D+R
D+R
R
MT
X1
X4
X3
X2X6
X5

The distances between MT and the BSs of different co-channel
cells will be in the range {D − R, D, D + R}.
– Two co-channel cells at a distance D − R
– Two at a distance D and
– Two others at a distance D + R.

Thus, the ratio of power strengths of current BS and the other
interfering BSs, is
S/I =
R−4
2(D − R)−4 + 2D−4 + 2(D + R)−4
=
1
2(
√
21 − 1)−4 + 2(
√
21)−4 + 2(
√
21 + 1)−4
= 49.56
So, the value of SIR = 10 log10 49.56 = 17 dB.
Implying the voice quality will not be good.

SIR with Cell Sectorization
With 120o sectors, the number of co-channel cells is reduced from
6 to 2 for N = 7.
Therefore, the SIR is:
10 log10(S/I) = 10 log (
√
3 × 7)4
/2
= 10 log 220.5 = 23.43dB
Which is substantial improvement from 18.66dB in case of
omni-directional antenna.
60o sectorization reduces interference from co-channel cells to 1.

Cell Sectorization
Example 3
Suppose each cell uses 60 channels irrespective of size. Original cell
radius is 1km and micro cell radius is 0.5km Find the number of
channels in a square with center at A in ﬁgure below.
C
D
E
B C
F
F E
FDG E B
GF
D C
D B
GE
A

Cell Sectorization
Solution:
The sides of a larger hexagon are 1km long.
To cover 3km×3km area around A, we need to walk 1.5km (1.5
times of a hexagon) on NEWS.
It covers 5 BS (in red) with 300 channels before splitting.
A is surrounded by 6 micro BS B, C, D, E, F, G.
If A is replaced then the total number of BS = 5+6 = 11.
So, the number of channels = 11× 60 = 660 (2.2 times).
If all original BSs are replaced by micro cells in the square area
then 17 micro BSs will be required.
So the total number becomes 17× 60 = 1020 channels.

Signal measurement & interference Traffic modeling
Traffic Intensity
Traffic intensity varies over the day.
Grade of Service (GoS) is directly related to traffic intensity.
TI is measured in a unit called Erlang.
One Erlang: traffic volume for one hour.
Example 4
If 40 calls/hour serviced with each of average call duration of 5
minutes, then the traffic in Erlang:
Traffic in hour = (40 × 5)/60
= 3.33 Erlangs

Erlang B Model
In a lossy system, GoS is computed by Erlang B trafﬁc model.
λ: arrival rate, and µ: service rate.
1/λ: average time between arrival of two consecutive requests
1/µ: average service time.
Eg., if average duration of connection is 3 minutes, then
1/µ = 3/60 = 0.05 hour, so, µ = 20.

Erlang B Model
dd1 a d23 a4 d3 a5 4 d5a2a1
Depicts connection requests and servicing requests for 5 users.
Interval Ii = ai+1 − ai represent the inter-arrival time.
Duration of service represented intervals S1 = d1 − a1,
S2 = d2 − d1, S3 = d3 − d2, S4 = d4 − d3, S5 = d5 − d4.
The arrival rate and service rate are given by expressions 1/E(Ii)
and 1/E(Si).

Erlang B Model
The inter-arrival times for connection requests is modeled by
Poisson distribution.
The rate λ of a Poisson process is the average number of number
events per unit time over a long period.
The probability of n call requests arriving during an interval of time
[0, t) under Poisson process is,
Prn[t] =
(λt)n
n!
e−λt
, for n = 0, 1, . . . .

Erlang B Model
Under Poisson arrivals, call requests arriving during two
non-overlapping intervals are independent.
I.e., Prn[t2 − t1] and Prn[t4 − t3] are independent,
Let t be an arbitrary starting point in time.
Suppose T1 is the time that has elapsed until arrival of next call
request, then
Pr[T1 > t] = Pr0[t] = e−λt

Erlang B Model
Thus, the probability of inter-arrival time between call requests
being less than t is
FT1(t) = Pr[T1 ≤ t] = 1 − e−λt
So, probability distribution function of T1 is
fa(t) = λe−λt
That is, T1 is distributed exponentially with mean λ.

Erlang B Model
For every t ≥ 0, and δ ≥ 0:
Pr[nt+δ − nt = 0] = 1 − λδ + O(δ)
Pr[nt+δ − nt = 1] = λδ + O(δ)
Pr[nt+δ − nt ≥ 2] = O(δ)
O(δ): probability of more than one call request arriving, and it is
such that limδ→0 O O(δ)
δ = 0

Erlang B Model
Every successful call requires some service time, with mean
service rate µ, the mean service time is 1/µ.
Probability that the holding time of nth call will be less than some
time t is given by
Pr[cn < t] = 1 − e−µt
, t > 0
and the probability density function of service time is
fs(cn) = µe−µt

Markov Chain for Channel Occupancy
We can use Markov chain to represent channel occupancy.
The number of channels is C can service C requests concurrently.
Therefore, it is M/M/C/C queuing system with following
parameters:
– Arrival process is Poisson with arrival rate λ.
– The service time is exponential with servicing rate µ.
– The number of servers or the channels for serving the connection
requests is C.
– The capacity (number clients) which may be in the queue is C.

1
C
2
E(x) = 1/
Pb
b(1−P )λ
λ
µ
n(t)
limited
number
of lines

Suppose 0 channels being used by the system.
Over a small interval, system may continue in 0 state is 1-λδ.
The probability there will be change to 1 state (1 channel in use) is
λδ.
But, if one channel is already in use, then the transition to 0 will be
with probability µδ.
The system will continue in state 1 with 1-λδ − µδ.
The sum of probabilities of all transitions out of a state will be 1.
.......
..............
.......
C0 1 2
λδ
µδ
λδ λδ
1−λδ−µδ
2µδ
1−λδ−2µδ C
µδC
1−λδ− µδ

Over a long period of time, system reaches steady state.
At steady state, the global balance equation is
λδPn−1 = nµδPn, n ≤ C
λPn−1 = nµPn
P1 = (λP0)/µ
Further, we have C
0 Pn = 1, i.e., P0 = 1 − C
n=1 Pn
Solving this equation we have
Pn = P0
λ
µ
n
1
n!
and
P0 =
µ
λ
n
n!Pn = 1 −
C
i=1
Pi

Substituting for Pis in terms of P0 and simplifying we get
P0 =
1
C
n=0
λ
µ
n
1
n!
We already know that PC = P0
λ
µ
C
1
C! .
So,
PC =
λ
µ
C
1
C!
C
n=0
λ
µ
n
1
n!

Total trafﬁc is A = λ(1/µ)
So,
PC =
AC 1
C!
C
n=0 An 1
n!
Above equation is called Erlang B formula.

Example of Erlang B
Example 5
Suppose there are 200 connection requests per hour in peak time.
Average call duration be 3 minutes. If the system has 25 channels then
ﬁnd out the probability of call dropping.
Solution:
Arrival arrival rate is λ = 200.
Time per call 0.05 hour, or the service rate µ = 20.
The average number of requests per hour λ, and the average call
duration is 1/µ.
The product λ × 1
µ = A is called the busy hour trafﬁc (BHT).

Example of Erlang B
Solution (contd):
It gives BHT A = 200/20 = 10.
If there are 25 channels then the probability of call dropping is
2.927×10−5.

Channel assignment
Frequency and Channel Numbers
FVC/RVC
FCC/RCC
FVC/RVC: Forward and reverse voice channels.
FCC/RCC: Forward and reverse control channels.

Channel assignment
Example 6
Suppose 33MHz and 1MHz are allocated for trafﬁc channels and
control channel respectively for a coverage area. Suppose BW for one
simple channel (RX or TX) = 25kHZ, so duplex (RX+TX) 50kHz. If
cluster size of 7 is used then ﬁnd out an near equitable distribution of
channels in a cell.
Solution:
33000/50 = 660 channels.
1MHz for control = 1000/50 = 20 CTRL channels
So, 660-20=640 Voice channels.

Channel assignment
Solution (contd):
For N = 7 case, one possible allocation:
– Five cells get 92 voice channels each and two remaining get 90
channels each.
– Out of 20 CTRL channels, six cells get 3, and remaining 2 for one
cell.
Other possible allocation:
– 4 cells get 91 channels 3 cells get 92 channels each.
– Distribution of CTRL channels remain same.

Channel assignment
Original AMPS Resource Allocation
825-845 MHz 870-890 MHz
25 MHz GAP
1 12 23 3666 666666
Downplink channels Uplink channels
312 voice and 21 control channels in each spectrum block
Initial allocation two bands 825-845MHz and 870-890MHz.
Each duplex channel is 60kHz.
21 CTRL channels in each band (channel # 313-354).

Channel assignment
Extension of AMPS Channels
5GHz added to spectrum later for 166 extra channels.
– 1MHz (CTRL) at the begining and 4MHz (Voice) at the end two
bands.
Extra 83 channels of A partitioned as 33 and 50.
But 83 channels of B located at the end of the band.
1MHz CTRL channels are numbered from 991 to 1023.
825-894MHz voice channels numbered from 1 to 799.

Channel assignment
Extension of AMPS Channels
824-849 MHz
869-894 MHz
33
33 312
312312
312
21+21
21+21
50
50
83
83
A
A
B
BB
BA+B
A+B
A
A
A
A
Downplink channels
Uplink channels
Summary of extended AMPS channels.

Channel assignment
Multiplexing channel
Partitioning the spectrum along frequency, time, or code are used
for this.
FDMA: partitions spectrum allocating distinct frequency bands.
TDMA: achieves channel separation by disjoint time intervals
called slots,
CDMA: ensures channel separation by using different modulation
codes.
Combination of different channel separation schemes is also
possible.
Eg., TDMA and FDMA can be combined to divide frequency band
into time slots for logical cannels.

Channel assignment
Fixed Channel Assignment
Channel assignment is classified either as fixed or dynamic.
Each cell is allocated a fixed number of channels in FCA scheme.
An active communication gets terminated if a connected MT
moves from a cell to a cell that has no free channel.
An active connection can be maintained by handoff.
We discuss about handoff later.

Channel assignment
Dynamic Channel Assignment
No channel is permanently allocated to any cell.
Each time a channel is required, it is allocated. by mobile
switching center (MSC).
MSC use some sophisticated algorithms taking care of: future call
blocking, inter-cell and intra-cell handoffs, and co-channel
interferences among other things.
The effectiveness of DCA depends on collection of real-time data
on channel occupancy, trafﬁc distribution and received signal
strength indication (RSSI) of all channels on a regular basis.
So, dynamic channel allocation increases both storage and
computational load on the MSCs.

Channel assignment
Channel Assignment
DCA schemes can be implemented either in centralized or in
distributed fashion.
In centralized assignment channels are assigned by a central
controller.
In distributed assignment channels are assigned either by local
cell or from the cell where the call originated.
In a cell based control, base station is responsible for keeping
track of available channels in its vicinity.
The channel status is updated on a regular basis by exchange of
information among BSs.
In mobile device managed allocation, mobile chooses a channel
based on the SIR ratio involving co-channel cells.

Channel assignment
Channel Assignment
General approach: use graph abstraction for representing cellular
system, and transform it into graph coloring.
However, in most general setting, it can be posed as constraint
satisfaction problem.
An n × n symmetric matrix C = {cij}, known as compatibility
matrix is deﬁned.
cij represents the minimum frequency separation required
between cells i and j.
Since frequency bands are evenly spaced, they can be identiﬁed
by integers.
The number of channels required for each cell is represented by a
requirement vector M = {mi}, i = 1, . . . , n.

Channel assignment
Channel Assignment
The frequency assignment vector F = {Fi} is such that Fi is a
finite subset of the positive integers which defines the frequencies
assigned to cell i. F is admissible provided it satisfies the
following constraints:
Fi = mi, for i = 1, . . . n
|f − f | ≥ cij, where f ∈ Fi and f ∈ Fj

Channel assignment
Channel Assignment
The largest integer contained in F is called the span of the
frequency assignment.
Note that the largest integer represents the minimum number of
channels required for the frequency assignment.
So, F with the minimum span constitutes the solution to the
problem channel assignment.
The problem is known to be NP hard.

Channel assignment
Interference Graph

Channel assignment
Interference Graph
Each vertex represent a cell or BS.
Edge (u, v) is associated with a weight W(u, v) proportional to
strength of intereference.
Every node v is associated with a non-negative integer Tv for
channel requirement.
To avoid interference, two channel a, b used in different cells u and
v, must differ by |a − b| ≤ W(u, v)
Also a system wide number W used for setting minimum
difference between two channels used in same cell, i.e.,
|a − b| ≤ W, if a and b are used in same cell.

Channel assignment Fixed Channel Assignment
Used when distribution of trafﬁc load is uniform.
The set of available channels is partitioned into N disjoint sets.
N = 1
3 × D
R
2
Where R is range and D is reuse distance.
The overall average call blocking probability will be same as call
blocking probability in a cell.

FCA: Borrow From Richest
A common sense driven approach is to borrow a free channel
when no free channel is found.
The borrower is called acceptor
Lender is known as donor.
Free channel should be selected in such way that:
1 It does not affect donor cell.
2 Does not introduce interferences on existing connections.
By selecting the cell with largest number of free channels (BFR)
as donor both conditions can be met.

C2
X Y
C3
C4
C1
1
2
34
6
5
donor
C’
C
acceptor
Borrowed channel is blocked in the co-channel cells (of the
donor) which are within reuse distance.

Borrowing becomes possible if the borrowed channel c is
simultaneously free in three nearby co-channel cells.
So, c should blocked C1 and C2.
If planned carefully, c may concurrently serve as a borrowed
channel in different acceptor cells.
– Eg., if C borrowed c from C then, X (a neighbor of C2) can not
borrow c, though C and X could use a channel c without
interference as they are three cells apart.
– Also as channel c is locked in C, C1 and C2, cell Y can not borrow
it from cell C3, because this borrowing is permitted if c is free in C1
and C2.
– Here again, Y and C are three cells apart.

FCA: Borrow the First Available
Any sophisticated borrowing method will incur penalties for
complex searchings.
A simpler option is to use BFA channel.
But, for implementing BFA, the initial channels assignment should
be different from direct assignment of channels to cells.
The set of channels is ﬁrst divided into sets and each set is
assigned to cells at a reuse distance D.
Then channel ordering is used for borrowing.

FCA: Borrowing with Channel Ordering
Channels with highest priority used for call locally.
Channel with lowest priority used for borrowing in neighboring cell.
It dynamically adjusts the ratio of the channels used in cell and
those lent to neighboring cells.
After borrowing channel is locked in co-channel cells within reuse
distance.

FCA: Borrowing with Directional Channel Locking
BDCL compares favorably with the system that performs
exhaustive complex searches, yet computationally less expensive.
Consider the ﬁgure in slide # 73, c was locked in all directions in
the cells C, C1, and C2
However, locking of c in C2 should sufﬁce only in directions 2, 3, 4.
It leaves c unlocked in directions 1, 5, 6 in C2.
Channel is locked in direction i by cell to prevent the ith neighbor
to borrow the channel.

FCA: BDCL
X being in direction 1 from C2, could borrow c.
X and C are not within reuse distance D, thus, the concurrent use
of c is possible.
Of course, whether or not X can actually borrow c, depends on its
locking status cells C3 and C4.
Note: in C, c is blocked in all directions, and in C1 it is locked in
directions 3, 4, 5, 6

FCA: BDCL
Borrowed channel should be returned to the donor cell.
The question is: when a borrowed channel should be returned
donor cell?
Answer depends on how it could inﬂuence systems performance.
Performance here concerns:
1 Inability of the system to satisfy a new connection request.
2 The number of channel switchings for ongoing connections.
Channel switching is not only costly but irritating.

FCA: Channel Reallocation
3 51 2 6 7
switch
4
Higher order nominal channel is released then an existing call on lower
order channel switched.

15
31 2 4
11 12 13 14 19181716
switch
5 6 7 9 108
20
A nominal channel is released then an existing call on borrowed channel
then release borrowed channel switching call to nominal channel.

15
31 2
11 12 13 14 19181716
5 6 7 9 108
20
switch
4
A call on borrowed channel terminates but a call on lower order bor-
rowed channel exits then release higher order borrowed channel switch-
ing call to lower order channel.

15
31 2 4
11 12 13 14 19 20181716
switch
5 6 7 9 108
completely unlocked
A channel is completely unlocked by termination of call in interfering
cell, existing call on borrowed channel or a higher order channel is
switched to this channel.

Under heavy trafﬁc condition channel borrowing could create
domino effect.
Domino effect may require a comprehensive channel reallocation
strategy.
Simple FCA sometimes may provide better performance than FCA
with channel borrowing.

Channel assignment Dynamic Channel Assignment Policies
DCA: Parameters
Key idea behind DCA scheme is to evolve evaluation for allocation
of candidate channels.
The cost consists of:
– number of future call blocking,
– channel occupancy under the current trafﬁc conditions,
– co-channel/adjacent channel interferences,
– acceptable average call blocking and other QoS related to radio
measurements.

DCA: Centralized Algorithms
First available (FA): assigns the ﬁrst channel ensuring channel
reuse constraints.
Locally optimized dynamic assignment (LODA): assigns the
channel by minimizing the future call block possibilities in cells in
vicinity.
Channel reuse optimization: tries to optimize reuse distance.
Maximizes utilization of every channel (by shorter reuse distance).
Maximum use in reuse ring: channel for allocation is selected by
ﬁnding the one that is used in most cells in co-channel set.
MSQ: selects channel that minimizes mean square of distance
among the cells using same channel.
NN: NN strategy selects the available channel occupied in the
nearest cell in distance ≥ D
R

DCA: Centralized Algorithms
Most of these try employ local optimizations. The 1-clique scheme
attempts a global optimization scheme.
Builds a graph for each channel where each vertex represents a
cell, and two vertices in this graph are connected by an edge if
and only if the cells corresponding to the end vertices do not have
co-channel interference.
So, each graph presents channel allocations possibilities.
Actual channel assignment is done from several possibilities so
that as many vertices as possible, still remain available for
allocation.

DCA: Distributed Algorithms
Each cell keeps track of free channels, the information stored in
an augmented channel occupancy (ACO) matrix
It is an (M + 1) × (ki + 1) matrix, where M is the number of
channels in the system and ki is the number of neigboring cells
within the co-channel interference distance from cell i.
Last column gives number of free channels in the cell
corresponding to the row.

DCA: Distributed Algorithms
BS No. Channel number assignable channels
1 2 3 4 . . . M
i x x . . . 0
i1 x x . . . 0
i2 x x . . . 3
...
...
...
...
...
...
...
...
iki
x x . . . 5
The contents of ACO matrix is updated by collecting channel
occupancy information from interfering cells.

DDCA
The cell ﬁnds an empty column and assigns the channel
corresponding to ﬁrst empty column.
– A non-zero entry in last column imply existence of empty column.
If no empty column exist, column having 1 occupancy is
considered. If cell occupying this channel has assignable
channels then:
– That cell is requested to shift to some other channel.
– The channel then becomes free for assignment.

DCA
It is possible to address the issue of adjoint channel interference
(ACI) adding extra restriction on the channel selection from ACO
matrix in DDCA mentioned above.
ACI effects are negligible if the minimum channel separation of
Nadj is maintained.
At the time of assigning a new channel c to cell i, the algorithms
ensure that the channels corresponding to columns Nadj − 1 to the
left or right of column c in ACO matrix do not have entries for row i.

Channel Assignment and Mutual Exclusion
Channel is a resource.
Neighboring cells can not share this resource simultaneously.
So, it is similar to mutual exclusion problem.
There are differences:
– In ME no two processes can share a resource simultaneously.
– But in CA, channel can be used by two cell provided minimum
reuse contraint is preserved.
– In CA, a collection of resources (channels) is to be shared.
However, techniques of ME could lead to a solution for CA.

Channel Assignment and Mutual Exclusion
Consider it as relaxed ME (RME). Certain pair of cell can not use while
certain other pair can use the same channel simultaneously. The
problems are:
How to implement RME on single resource.
Resolving deadlocks.
Extending RME to multiple resources.
Designing information structures.
Implementing efﬁcient channel selection strategy.

Handoff
Service with Mobility
Provisioning continuity of service despite users’ mobilities is a
challenging.
Interestingly, solution based on a simple idea of the game of
football!
The continuity can achieved by handoffs (or handovers).
Handoff process is induced either by cell crossing, or when the
quality of channel deteriorates.

Handoff
Service Degradation
The deterioration of service is due two reasons:
1 Signal quality deterioration.
2 Trafﬁc load.

Handoff
Keeping Connection Active
If the link to new BS is formed before or almost immediately as the
link to old BS goes down.
Thus, a handoff is the transition of signal transmission from one
BS to another.
Frequency switching may also be required when MT is moving
inside cell. Eg., intra-cell handoffs discussed in channel allocation
schemes.
Our focus is on inter cell handoffs.

Handoff
Keeping Connection Active
Cells overlap: it means MT is within the range of multiple BSs at
the boundary of a cell.
The N/W decides which BS will handle the transmission to/from
MT. The decision could be
– With assistance of MT, or
– Without assistanced of MT.
The critical part of the handoff: the detection of the handoff
condition.
Once an active connection is completely severed nothing can be
done.

Handoff
Hystersis
Threshold: signal level slightly stronger than minimum.
Hystersis: the margin between the threshold and the minimum
usable signal.
Hystersis can be deﬁned by value
∆ = Shandoff − Smin,
∆ should not be too small or too large.

Handoff
Hystersis
t0
movement of mobile terminal
2BSBS1
RSS from BS1 RSS from BS2
t1
2t
hystersis
A B C D
At t0: MT receives signal only from BS1.

Handoff
Hystersis
t0
2BSBS1
t1
2t
hystersis
A B C D
At t1: RSSI from BS1 and BS2 become comparable.

Handoff
Hystersis
t0
2BSBS1
t1
2t
hystersis
A B C D
The handoff must begin after A and completed before C.

Handoff
Hystersis
The value of ∆ depends on:
Environment.
Speed of mobile.
Time required to perform handoff.

Handoff Handoff policies
Channel Allocation
Prioritize the channel assignment for handoff before the new call.
Pre-allocate a certain number of handoff channels called guard
channels.
If the guard channels are not available then the handoff will be
serviced by other channels but a handoff would compete with new
call.
– Increases the probability of a dropped call.

Channel Allocation
channels.
call.
– Guard channels may remain under utilized.

Channel Allocation
channels.
call.
– Reserving channels may be suitable for DCA scheme.

Handoff Handoff protocols
Entities
Entities involved are:
1 User’s mobile handset (MH),
2 BS to which MH is currently connected and BSs in the
neighborhood of MH’s movements, and
3 MSCs controlling the above group of BSs.
Both network entities (BSs and MSCs) and MH may initiate and
control a handoff.

Handoff Classes
Depending on controlling entity or the entities, the handoff classiﬁed
as:
1 Network controlled.
2 Mobile assisted.
3 Mobile controlled.

Handoff Classes
In N/W controlled protocol, handoff decision is based on
measurements of RSSs of MH adjoining BSs
The process includes measurements, channel switching, takes
approximately around 100-200ms.
In MH assisted handoff, MH measures RSSs it receives from BSs
and the decision for handoff is made N/W. Takes about 1 second.
In mobile controlled handoff, MH measures RSS of neighboring
BSs, and interference levels of all channels, and initiates handoff.

Goals of Handoff Protocol
1 Should be performed quickly.
2 Interruption in connection should be imperceptible to users.
3 Should be performed infrequently.
4 Should be performed successfully.

Generic Procedure
MH BSMSCBSold new
Handoff request ack
Link establishment
Handoff access
Handoff complete
Flush complete
Flush command
Handoff required
Handoff request
Handoff command
Reports measurements
Handoff command
1
2
1 Handoff decision 2 Resource allocation

Cellular architecture

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