Optimization of a wireless access network
K. Kraimeche, B. Kraimeche, K. Chiang
Abstract— N base stations (BSs) at fixed locations are to be location updating, handovers, and call routing to a roaming
connected to a subset of M possible sites for mobile switching subscriber.
centers (MSCs) in a cellular access network. The MSCs are, in To switch calls from/to local mobile users to/from remote
turn, to be connected to a number of local exchanges of the users, MSCs are connected by land-cables to nearby LEs of
public switched telephone network. Given the BS-MSC the PSTN. The potential locations of MSCs are judiciously
connection cost matrix, the MSC-LE connection vector, the determined with respect to the BS locations and to the LEs
MSC equipment cost vector, and the MSC capacity constraint
in the region. Typically, the locations of the LEs are fixed,
vector, we formulate the network design problem as a matrix
optimization problem with constraints. We introduce a simple and a single LE serves an area with many BS’s and multiple
exhaustive search algorithm to find the optimum solution MSCs.
matrix. We compare our algorithm to a heuristic algorithm
from the literature. Both algorithms have been implemented in
Matlab, and their computational efficiency compared on
The wireless access network of a cellular telephone system
consists primarily of base stations, mobile switching centers,
and local exchanges of the public switched telephone
network (PSTN). Each cell in the hexagonal cell grid
contains one base station (BS). A set of BS’s are physically
connected to and served by a mobile switching center
(MSC). In turn, a set of MSC’s are physically connected to
and served by a local exchange (LE). Fig. 1 depicts the
general configuration of a cellular access network. Each BS
is typically assigned a group of radio channels (frequency
carriers) to support a number of mobile stations (MSs) in its
Fig. 1: A cellular access network
cell. BS’s at adjacent cells are assigned different sets of
frequencies. The antennas of a BS are designed to achieve
Our interest in this paper is in the topological design of the
coverage only within the particular cell. By limiting
network connecting the BSs to the MSCs and the MSCs to
coverage of a BS to its cell area, the set of frequencies
the LEs in a typical region of the cellular system. From the
assigned to this BS can be reused at other BS’s that are
preceding discussion, it’s clear that this access network has a
distant enough to keep co-channel interference within
centralized tree topology. That is, a single LE facility
controls a set of MSCs, and a single MSC controls, in turn, a
set of BS’s. Finally, a BS supports a group of mobile
A BS contains the radio transceivers defined for its cell, and
stations (MSs) through wireless connections.
handles the radio-link protocols with the user’s wireless
device (cell phone). In addition, it may house a controller
A tree topology of the access network, consisting of one LE,
that handles radio-channel setup, frequency hopping and
5 MSCs and 19 BSs, is shown in Fig. 2. The central node is
handovers. In a large metro area, a potentially large number
the LE switch, denoted by S0, the MSC sites are denoted by
of BS’s are deployed at pre-determined locations. The BS
S1, S2, …,S5, and the BS towers are denoted by T1, T2, …,
controllers are connected by land-wires to nearby MSC’s in
T19. Here, towers T1, T2 and T19 are homed to the central LE
S0 (assumed to have MSC functionalities), towers T3, T4, T5
and T6 are homed to an MSC at S1; etc.; and towers T16, T17,
The MSC provides all the functionality needed to handle a
and T18 are homed to the last MSC at S5.
mobile subscriber, such as registration, authentication,
K. Kraimeche is a graduate student in the ISAT department at James
The topological design of access networks has been an
Madison University, Harrisonburg, VA 22807 (firstname.lastname@example.org). important part of cellular network research in recent years.
B. Kraimeche is a Professor in the ISAT and CS departments, James Recent studies are given in references [1, 2, 3]. Generally,
Madison University (email@example.com). the design problem is formulated as a constrained
K. Chiang is an undergraduate student in the ISAT department, James
optimization problem, where the goal is to find a network
Madison University (firstname.lastname@example.org).
topology such that an objective function is optimized, transmission cabling, interfacing, maintenance, leasing).
subject to a set of constraints. The objective function may Note that a tower may be located at the site of an MSC, in
be the total cost, or some performance measure like which case the corresponding Cij cost is zero.
utilization, call blocking or throughput. The constraints may These cost elements Cij can be written in the form of an
be bounds on link capacities, cost elements, or some Nx(M+1) matrix, as follows:
network performance measure.
C10 C11 C12 C1M
C20 C21 C23 C2M
T1 T19 C30 C31 C32 C3M (1)
T3 CN 0 CN1 CN 2 ...CNM
S2 S3 S4
If an MSC at site Sj is utilized, the MSC capital cost and its
connection cost to the LE are also incurred. Let Fj be the
cost of connecting an MSC at Sj to the central LE S0, and Bj
the capital cost of the MSC at Sj . We can write these 2 costs
as row vectors, as follows:
F (F0 , F1 , F2 , FM )
B (B0 , B1 , B2 , BM )
Fig. 2: Centralized access network
It’s assumed that the capital cost of the central LE is not
Often, however, the optimization is too complex, or it’s
counted, that is, B0=0, and clearly F0=0.
computationally impractical to search for the optimal
solution. So, one usually resorts to heuristic methods that
Similarly, we can write the MSC constraints as the
enable one to determine a near-optimum network topology
following row vector:
In this paper, we consider the simple design of an access P (P0 , P , P2 ,
1 PM ) (3)
network with one single LE, as shown in Fig. 2. The
objective function is the total cost of connecting the BSs to Recall that Pj is the maximum number of BSs that MSC at
the MSCs, and connecting the MSCs to the LE. Let N be the site Sj can handle, j=1, 2, …, M, with P0=N (i.e., the central
number of BSs, T1, T2, …, TN. The locations of the N LE can handle all the N towers).
terminals are assumed known and fixed. Let M be the
number of potential sites, S1, S2, …, SM, where up to M A network design can be defined by the following Nx(M+1)
MSCs can be placed. In one extreme situation, none of the matrix variable:
M sites is used, and all the N BSs are linked directly to the
central LE, S0. In the other extreme, all the M MSC sites are X 10 X 11 X 12 X 1M
used, each serving a subset of BS’s.
X 20 X 21 X 23 X 2M
The principal constraint is that the MSC at site Sj can handle X 30 X 31 X 32 X 3M (4)
up to a maximum of Pj BSs, j=1, 2, …, M. This can be a
hardware limitation, or a capacity constraint of the land-
cable connecting the MSC to the LE. The central site is
assumed to have no such constraint (i.e., we set P0 =N). X N0 X N1 X N 2 ...X NM
II. OPTIMIZATION PROBLEM
In the above matrix, the element variable Xij, i=1, 2, …, N,
and j=0, 1, 2, …, M, is defined as:
We want to formulate the network design problem as an
optimization problem. Let Cij be the cost of connecting 1 if Ti is connected to S j
tower Ti, i=1, 2, …, N, to MSC site Sj, j=1, 2, …, M or to X ij (5)
the central site S0 0 if Ti is not connected to S j
The cost Cij is measured in some unit (e.g., dollar/month),
and represents the overall BS-MSC connection cost (e.g.,
Note that since a BS may be connected to at most one of the matrix X must be less than or equal to Pj , j = 0, 1, 2, …,
M MSC sites or to the central LE site, there must be only M. In the matrix inequality of Eq. (11), the inequality
one “1” in each row of matrix X. In addition, note that the relation is defined element by element.
number of 1’s in column j is the number of BSs connected to
the MSC at site Sj, j=0, 1, …, M. Thus, an all-zero column III. SOLUTION ALGORITHMS
of matrix X corresponds to an MSC site that is not used
The preceding formulation of the network design problem is
From matrix X, we extract the following MSC-usage vector: a matrix linear programming problem with constraints. The
decision variable is matrix X . Since the matrix variable is
Y (Y0 , Y1 , , YM ) (6) quite sparse, we introduce a simple exhaustive search
algorithm for finding the optimum solution. For comparison,
Where the element variable, Yj, j=0, 2, …, M, is defined as: we also adapt a heuristic algorithm from . We’ve
implemented both algorithms in Matlab.
1 if S j used , i.e., if X ij 0 A. AN EXHAUSTIVE SEARCH ALGORITHM
i 1 (7)
0 if S j not used , i.e., if X ij 0 The algorithm consists of generating all the possible
Nx(M+1) X matrices and searches for the matrix that yields
the minimum cost. The matrices are such that the number of
The cost of a network design (defined by matrix X and 1’s in each row is exactly one while the number of 1’s in
vector Y ) is thus expressed as follows: each column does not exceed the corresponding MSC
capacity constraint. The number of such matrices is bounded
T by (M+1)N. The algorithm follows the steps:
Z sumdiag (C X ) F Y T B YT (8)
1. Generate matrix X with exactly one 1 in each row
2. If the number of 1’s in each column satisfies the
In the above expression, the superscript T means transpose associated constraint, the matrix is valid; then
of matrix or vector, and sumdiag(A) is a function that sums
up the diagonal elements of matrix A. generate vector Y from X and compute the cost.
3. If the cost if smaller than currently smallest cost,
The first term in the cost function Z is the cost of connecting then update smallest cost and store matrix X . Go
the N BSs to the MSCs used or to the central LE, the second to step 1.
term is the cost of connecting the MSCs to the LE, and the 4. Stop when all matrices have been tried.
third term is the hardware cost of the MSCs used.
The optimum solution is the matrix X with the smallest
The network design problem can thus be stated as the
following optimization problem: Given the following:
a) N BSs at known locations, T1, T2, …, TN, B. A HEURISTIC ALGORITHM
b) M possible MSC sites, S1, S2, …, SM,
c) the BS-connection cost matrix, C The algorithm is described in . It consists of an
d) the MSC cost vectors F and B, and initialization step, and M iterations.
e) the mux capacity constraint vector P
Initialization: Connect all N BSs to the LE. Compute cost.
Find the matrix X (thus the vector Y ) that minimizes the
network cost Z: Iteration 1: Start with initial topology from iteration 0. Add
Z sumdiag (C X )
F Y T
B Y T
(9) the M MSC sites, one at a time; move BSs from S0 to the
new MSC so as to reduce cost while checking the P
constraint; Select the topology with the smallest cost.
Subject to the following 2 constraints:
X E E (10) Iteration 2: start with best topology from iteration 1. Add a
X P (11) 2nd MSC from the remaining (M-1) MSCs, one at a time;
move BSs to the new MSC so as to reduce cost while
In the preceding equations, E is the column vector of all 1’s. checking the P constraint. Select the topology with the
The first constraint indicates that the sum of the elements in smallest cost
row i of matrix X must be 1, i=1, …, N. The second Iteration i: start with best topology from iteration (i-1).
constraint indicates that the sum of elements in column j of Add an ith MSC from the remaining (M-i+1) MSCs, one at
a time; move BSs to the new MSC so as to reduce cost while
checking the P constraint. Select the topology with the Iteration 0: Connect all N=6 BSs to S0. The resulting cost
smallest cost. is: Z 4 5 8 9 10 7 43
Iteration 1: Add S1.
Iteration M: start with best topology from iteration (M-1). The best topology is: (T1 , T5 , T6 ) S0 ; (T2 , T3 , T4 ) S1 .
Add the Mth MSC; move BSs to the new MSC so as to
reduce cost while checking the P constraint. Select the The cost is: Z 3 4 6 4 10 7 4 1 39
topology with the smallest cost. Add S2.
The best topology is (T1 , T2 , T6 ) S0 ; (T3 , T4 , T4 ) S2 .
The best solution is the topology with the smallest cost The cost is: Z 4 5 7 2 7 6 6 1 38
across all the iterations.
The best topology is:
(T2 , T4 , T6 ) S0 ; (T1 , T3 , T5 ) S3 . The cost is:
We’ve implemented both the heuristic algorithm and the
exhaustive search algorithm in Matlab, and we’ve tested the Z 5 9 7 2 3 7 9 1 43
algorithms on several examples. The best topology so far is the one with S2
We first consider a network of N=6 BSs and M=3 MSC
sites. The vector parameters are: F = ( 0, 4, 6, 9), B = (0, 1, Iteration 2: Add S1.
1, 1), and P = (6, 3, 3, 3), and the cost matrix is: The best topology is:
(T1 , T6 ) S0 ; (T2 , T4 ) S1; (T3 , T5 ) S2 .
4 5 3 2 The cost is:
5 3 4 5 Z 4 7 3 6 2 6 4 1 6 1 40
8 4 2 3 Add S3.
C The best topology is:
9 6 7 8
(T3 , T4 , T5 ) S 2 ; (T1 , T2 , T6 ) S3 .
10 9 6 7
The cost is:
7 8 10 6
Z 2 7 6 6 1 2 5 6 9 1 45
The best topology so far is the one with S2 with Z=38
Both the heuristic algorithm and exhaustive search algorithm
generated the same matrix, given below, for the optimum Iteration 3: Add S3 to topology with S1 and S2. No cost
design at a cost of 38 units: reduction.
1 0 0 0 A second example consists of N=10 BSs and M=4 MSC
1 0 0 0 sites. The vector parameters are: F = (0, 1, 3, 1, 1), B = (0, 2,
0 0 1 0 2, 1, 5), and P = ( 10, 3, 3, 3, 3), and the cost matrix is:
X , Z 38
0 0 1 0
3 2 4 1 3
0 0 1 0
4 1 5 2 2
1 0 0 0
2 3 6 4 1
The optimum network topology is given in Fig. 3. 1 4 2 7 6
6 2 4 6 7
S0 8 4 1 8 4
4 5 3 1 8
T6 5 6 7 3 4
7 8 8 2 5
2 4 3 5 1
The heuristic algorithm generated a sub-optimum design
with matrix X h given below and a cost of 28 units. The
Fig. 3: Final topology exhaustive search algorithm generated matrix X o for the
optimum design with a cost of 26 units. The exhaustive
In this case, the heuristic found the optimum solution. The search algorithm, however, took more time to arrive at the
computation details of the heuristic are as follows: solution, compared to the heuristic. This highlights the
computational value of the heuristic in finding a sub-
optimum solution in comparison to the exhaustive-search in V. CONCLUSION
finding the optimum solution. The problem of connecting a set of base stations to a set of
potential mobile switching centers, and connecting these
0 0 0 1 0 1 0 0 0 0 latter to a central exchange in a cellular network is
0 1 0 0 0 0 1 0 0 0 formulated in this paper as a matrix optimization problem.
1 0 0 0 0 1 0 0 0 0 An exhaustive-search algorithm and a heuristic algorithm
1 0 0 0 0 1 0 0 0 0 are used to solve the problem. Work will be pursued to
0 1 0 0 0
compare the computational performance of the 2 algorithms
0 1 0 0 0
Xh , Xo on more examples, and to extend them to handle multiple
1 0 0 0 0 0 1 0 0 0 exchanges.
0 0 0 1 0 0 0 0 1 0
1 0 0 0 0 0 0 0 1 0 ACKNOWLEDGEMENT
0 0 0 1 0 0 0 0 1 0 The authors acknowledge the contribution of Henry Olschofka,
0 1 0 0 0 1 0 0 0 0 undergraduate student in the CS department at JMU, in the Matlab
implementation of the heuristic algorithm.
The sub-optimal and optimal networks are given in Figs. 4 REFERENCES
and 5, respectively.  Mathar, R., Niessen, T., “Optimum positioning of base stations for
cellular radio networks”, Wireless Networks, N0. 6, 2000
 Raisane, L., Whitaker, R., Hurley, S., “A comparison of randomized
and evolutionary approaches for optimizing base station site
selection”, ACM Symposium on Applied Computing, 2004
 Galota, M., Glaber, C., Reith, S. Vollmer, H., “A Polynomial-time
scheme for base station positioning in UMTS networks”, ACM, 2001
 Schwartz, M., “Computer-communication Networks”, Prentice-Hall,
Fig. 4: Optimum topology
Fig. 5: Sub-optimum topology