Introduction To Wireless and Mobile Systems 4th Edition Agrawal Solutions Manual Download
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Chapter 8: Traffic Channel Allocation
P8.1. What are the specific advantages of static channel allocation over dynamic
channel allocation strategies? Explain clearly.
[Solution]
FCA is simple. It needs low computational effort and is easy to implement.
Therefore, it has a lower call setup delay.
From the view of performance, it has better performance under heavy traffic and
maximum channel reusability. And it’s suitable for large cell environment.
P8.2. Are there collisions present in traffic or information channels in a cellular system?
[Solution]
No collisions are possible in traffic or information channels because the BS
controls all the mobile stations and the channel assignment is collision free.
[Solution]
In FDMA/TDMA system, the number of frequencies and time slot is limited.
Channel allocation problem in FDMA/TDMA is how to efficiently allocate the
frequency or time slot to support more users. While in CDMA system, the number
of possible codes indicates the number of users that can be supported. Hence, the
allocation problem in CDMA is converted to how to control the transmission
power so as to minimize the interference.
P8.4. If you do not sector the cells, can you still borrow channels from adjacent cells?
Explain clearly.
[Solution]
You can still borrow channel from adjacent cells. However, you must carefully
choose the channel to make sure that the same channels cannot be used in
adjacent cells at the same time and there is no interference with the cells of other
clusters, which are within reuse distance of it.
63
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
[Solution]
(a) Choose a free channel, which is not being used by the other cells of this
cluster.
(b) Choose a free channel, which could minimize the possible interference with
the cells of the other clusters, which are within reuse distance of the center
cell.
P8.6. Which cell(s) may borrow channels and which could be an appropriate donor(s) in
Problem 5.11?
[Solution]
The cell with 4000 calls per hour may borrow channels. The cells with 900 and
1000 calls per hour could be appropriate donors.
P8.7. What are the advantages of cell sectoring? How do you compare this with
SDMA?
[Solution]
Cell sectoring minimizes the interference by reducing possible number of co-
channels that could cause interference within each cell, SDMA facilitates
simultaneous multiple connections using the same frequency. Both these
techniques are useful in optimizing the reuse of RF resources.
P8.8. In a cellular system, with 7-cell clusters, has the following average number pf
calls at a given time:
Cell number Average number of calls/unit time
1 900
2 2000
3 2500
4 1100
5 1200
6 1800
7 1000
If the system is assigned 49 traffic channels, how would you distribute the
channel if
(a) Static allocation is used based on traffic load.
(b) A FCA Simple borrowing scheme is used (no traffic load considered).
(c) A dynamic channel allocation scheme is used.
[Solution]
(a) Distribute the channels according the traffic load of each cell.
Cell number 1 2 3 4 5 6 7
Channel number 4 9 12 5 6 8 5
64
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
(b) Allocate seven channels to each cell and borrow channels from adjacent
cells with low traffic.
(c) Channels are allocated dynamically as new call arrival in the system and is
achieved by keep all free channels in a central pool.
P8.9. Each cell is sectored in a slightly different way into three sectors as follows:
What will be the impact of such sectoring on channel borrowing and what will be
its effect on co-channel interference? Explain carefully.
[Solution]
If sector x borrows channels from sector a of A1, clusters A5, A6, and A7 satisfy
the reuse distance. The directions of a sectors of A2, A3, and A4 make them not
interfere with sector x.
Solution 8.9
65
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
P8.10. Each cell of a wireless system, is partitioned in 6-sector as shown in Figure 8.13.
66
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
(b) Similar with (a). Co-channel interference is the same as the analysis done in
Chapter 5. There are only one co-channel interference sector for each sector.
P8.11. In a cellular system, a cluster of 7-cell, is assigned 48 traffic channels. Show the
assignment of channels to each cell if
(a) Omni-directional antennas are used.
(b) 3-sector directional antennas are used.
(c) 6-sector directional antennas are used.
[Solution]
(a) 6 channels each cell
(b) 2 channels each sector
(c) 1 channel each sector
67
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
[Solution]
(a) This figure shows the 3-way cell sectoring of a system with 4-cell as its basic
building block. For example, sector x of A4 borrows channels from sector c of
A0. A1, A2, and A6 satisfy the reuse distance and the directions of c sectors of
A3, A4, and A5 will not cause interference with sector x.
Solution 12 (a)
68
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
Solution 12 (b)
P8.13. How do you compare hybrid with flexible channel allocation? Which one would
you prefer and why?
[Solution]
A hybrid channel allocation scheme is a combination of fixed and dynamic
channel allocation schemes. Hybrid scheme is complicated and different and its
performance analysis by an analytical model is difficult. Generally, hybrid
allocation has a better service than FCA at low and moderate traffic load.
P8.14. For a wireless network with integrated services, e.g., including both voice and
data applications, there are two basic channel allocation schemes: Complete
Sharing (CS) and Complete Partitioning (CP). The CS policy allows all users
equal access to the channels available at all times. The CP policy, on the other
hand, divides up the available bandwidth into separate sub-pools according to user
type. Compare both the advantages and disadvantages of these two schemes.
69
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
[Solution]
The advantages of CS for integrated services are: efficient bandwidth utilization
and reduce the total call blocking probability. The disadvantage of CS for
integrated services is that when data traffic is heavy, the blocking probability of
voice calls will be increased. The advantage of CP for integrated service is that
the QoS for voice can be easily supported, as data traffic does not influence the
blocking probability of voice calls. But, the unused bandwidth in each category
will be wasted.
P8.15. What kind of technique(s) you could possibly use to serve a new call if all the
channels in the current cell have been occupied and no channel can be borrowed
from neighboring cells.
[Solution]
Preempt lower priorities calls or borrow some bandwidth from other lower
priority calls.
[Solution]
R
The radius of each microcell is 7.69 km (see Figure 8.6). The size is 154
1.5 3
km2. Because we use smaller cell, we could use low signal power to cover the
whole cell.
D
q 3 N 21 4.5825,
R
4.5,
1
CIR
2 q 1 2q q 1
106.
70
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
P8.17. Providing cellular service along a freeway is a tough job and such a scenario is
illustrated in the following. A typical road-width varies from 200 m to 400 m. If
you select 1000 m as the radius of each cell, then, you require one cell for each
km, while the radius of a conventional normal cell is about 20 km. From the
freeway usage point of view, very small segment of each cell is useful. Do you
have any suggestions for alternative designs? What are the tradeoffs? Do you
suggest the use of SDMA technique?
[Solution]
It is better to use smaller size cells. In this way, we could use low power BS and
improve the utility of each cell. But the MS will experience more frequent
handoff.
P8.18. In a cellular system with 4 channels, one channel is reserved for handoff calls.
(a) What is the value of BO and BH , given λO = λH = 0.001 and μ = 0.0003?
(b) What are the values of probabilities P(0), P(1), P(2), P(3) and P(4).
(c) What is the average number of occupied channels in this Problem?
[Solution]
μ = 0.0003, S = 4, Sc = 3, λO = 0.001, λH = 0.001
Using Equations (8.11), (8.12), and (8.13), we can compute
P(0) = 0.01, P(1) = 0.06, P(2) = 0.18, P(3) = 0.41, P(4) = 0.34
BO = P(3) + P(4) = 0.75
BH = P(4) = 0.34
The average number of occupied channels
N 1 P 1 2 P 2 3 P 3 4 P 4 3
P8.19. Repeat Problem P8.18 for the case that the number of channels is increased to ten.
71
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8: Traffic Channel Allocation
[Solution]
Use the same method as Problem P8.18.
P8.20. In a cellular system, the total number of channels per cell, is given as 6 and 2
channels are reserved exclusively for handoff calls. What are the blocking
probabilities for originating and if handoff request rate is 0.0001, the originating
call rate is 0.001, and the service rate μ = 0.0003?
[Solution]
Similar to Problem P8.18.
μ = 0.0003, S = 6, Sc = 4, λO = 0.001, λH = 0.0001
P(0) = 0.04, P(4) = 0.27, P(5) = 0.018, P(6) = 0.0010
BO = P(4) + P(5) + P(6) = 0.29
BH = P (6) = 0.0012
P8.21. What is the impact on the answer for Problem P8.20 if the number of reserved
channels is changed to
[Solution]
(a) Similar to Problem P8.18.
μ = 0.0003, S = 6, Sc = 5, λO = 0.001, λH = 0.0001
P(0) = 0.04, P(5) = 0.17, P(6) = 0.01
BO = P(5) + P(6) = 0.18
BH = P(6) = 0.01
BO decreases, but BH increases.
(b) P(0) = 0.05
BO = 0.45, BH = 0.000012
BH decreases, but BO increases
72
©2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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