1. CSE 550 Computer Network Design
Dr. Mohammed H. Sqalli
COE, KFUPM
Spring 2012 (Term 112)

2. Outline Topology Design for Centralized Networks
Multipoint Line Topology
Terminal Assignment
Concentrator Location
CSE-550-T112
Lecture Notes - 4

3. Centralized Network Design
Centralized network: is where all communication is to and from a single central site
The “central site” is capable of making routing decisions
→ Tree topology provides only one path through the center (For reliability, lines between other sites can be included)
Center
Concentrator
Terminal
High speed lines
Low speed lines
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4. Centralized Network Design Problems
Multipoint line topology: selection of links connecting terminals to concentrators or directly to the center
Terminal assignment: association of terminals with specific concentrators
Concentrator location: deciding where to place concentrators, and whether or not to use them at all
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5. Multipoint Line Topology

6. The Greedy Algorithm
At each stage, select the shortest edge possible (myopic or near-sighted):
Start with empty solution s
While elements exist
Find e, the best element not yet considered
If adding e to s is feasible, add it; if not, discard it
May not find a feasible solution when one exists
Efficient and simple to implement → widely used
Basis of other more complex and effective algorithms
A Minimum Spanning Tree (MST) is a tree of minimum total cost
In the case of MST, the greedy algorithm guarantees both optimality and reasonable computational complexity
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7. Constrained/Capacitated MST (CMST)
CMST Problem:
Given:
A central node N0
A set of other nodes (N1, N2, …, Nn)
A set of weights (W1, W2, …, Wn) for each node
The capacity of each link Wmax
A cost matrix Cij = Cost(i,j)
Find a set of trees T1, T2, …, Tk such that:
Each Ni belongs to exactly one Tj, and
Each Tj contains N0
is minimized
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8. CMST
Objective: Find a tree of minimum cost and which satisfies a number of constraints such as:
Flow over a link
Number of ports
The CMST problem is NP-hard (i.e., cannot be solved in polynomial time)
→ Resort to heuristics (approximate algorithms)
These heuristics will attempt to find a good feasible solution, not necessarily the best, that:
Minimizes the cost
Satisfies all the constraints
Well-known heuristics:
Kruskal
Prim
Esau-Williams
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9. Kruskal’s Algorithm for CMST
Example: Given a network with five nodes, labelled 1 to 5, and characterized by the following cost matrix
Node 1 is the central backbone node
fmax=5, f1=0, f2=2, f3=3, f4=2, f5=1
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10. Prim’s Algorithm for CMST
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11. Esau-Williams Algorithm for CMST
Node 1 is the central node.
tij: is the tradeoff of connecting i to j or i directly to the root
If (tij < 0) → better to connect i to j
If (tij ≥ 0) → better to connect i directly to the root
Algorithm:
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12. Terminal Assignment

13. Problem Statement
Terminal Assignment: Association of terminals with specific concentrators
Given:
T terminals (stations) i = 1, 2, …, T
C concentrators (hubs/switches) j = 1, 2, …, C
Cij: cost of connecting terminal i to concentrator j
Wj: capacity of concentrator j
Assume that terminal i requires Wi units of a concentrator capacity
Assume that the cost of all concentrators is the same
xij = 1; if terminal i is assigned to concentrator j
xij = 0; otherwise
Objective:
Minimize:
Subject to:
i = 1, 2, …, T (Each terminal associated with one Concentrator)
j = 1, 2, …, C (Capacity of concentrators is not exceeded)
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14. Assignment Problem Given a cost matrix: Assume that:
One column per concentrator
One row per terminal
Assume that:
Weight of each terminal is 1 (i.e., each terminal consumes exactly one unit of concentrator capacity)
A concentrator has a capacity of W terminals (e.g., number of ports)
A feasible solution exists iff T ≤ W * C C1 C2 T1 T2 T3
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Lecture Notes - 4

15. Augmenting Path Algorithm
It is based on the following observations:
Ideally, every terminal is assigned to the nearest concentrator
Terminals on concentrators that are full are moved only to make room for another terminal that would cause a higher overall cost if assigned to another concentrator
An optimal partial solution with k+1 terminals can be found by finding the least expensive way of adding the (k+1)th terminal to the k terminal solution
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16. Augmenting Path Algorithm
Initially, try to associate each terminal to its nearest concentrator
If successful in assigning all terminals without violating capacity constraints, then stop (i.e., an optimal solution is found)
Else,
Repeat
Build a compressed auxiliary graph
Find an optimal augmentation
Until all terminals are assigned
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17. Building a Compressed Auxiliary Graph
U: set of unassociated terminals
T(Y): set of terminals associated with Y
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18. Building a Compressed Auxiliary Graph
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19. Example Cost Matrix: W = 2 (capacity of each concentrator)
Solution: (a, H) (b, I), (c, H), (d, G), (e, I), (f, G)
Cost = 15 G H I a 6 3 8 b 2 9 4 c 1 d 5 e f 7
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20. Concentrator Location

21. Problem Statement
Concentrator location: deciding where to place concentrators, and whether or not to use them at all
Given:
T terminals (stations) i = 1, 2, …, T
C concentrators (hubs/switches) j = 1, 2, …, C
Cij: cost of connecting terminal i to concentrator j
dj: cost of placing a concentrator at location j (i.e., cost of opening a location j)
Kj: maximum capacity (of terminals) that can be handled at possible location j
Assume that terminal i requires Wi units of a concentrator capacity
xij = 1; if terminal i is assigned to concentrator j; 0, otherwise
yj = 1; if a concentrator is decided to be located at site j; 0, otherwise
Objective:
Minimize:
Subject to:
i = 1, 2, …, T (Each terminal associated with one Concentrator)
j = 1, 2, …, C (Capacity of concentrators is not exceeded)
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22. Add Algorithm Greedy Algorithm
Start with all terminals connected directly to the center
Evaluate the savings obtainable by adding a concentrator at each site
Greedily select the concentrator which saves the most money
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23. Add Algorithm Initialization: For , do where
M: set of locations
Select an initial location m
Assume all terminals are connected to m
Set L0 = {m}
Set k = 0 (iteration count)
c'i = cim, i= 1, 2, …, N
Compute:
For , do where
Determine a new m such that:
If there is no such m, go to step 4.
Update: and for
Set and k:= k+1, and go to Step 1.
No more improvement possible; stop.
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24. Example
Cost Matrix:
Kj = 3, j = 1, 2, 3, 4 (capacity of each concentrator)
d1 = 0, d2 = d3 = d4 = 2
Solution: S2 (1, 2, 3), S1 (4), S4 (5, 6)
Cost = 9 S1 S2 S3 S4 1 2 4 3 5 6
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25. References
A. Kershenbaum, “Telecommunications Network Design Algorithms”, McGraw-Hill,1993
M. Pióro and D. Medhi, “Routing, Flow, and Capacity Design in Communication and Computer Networks”, Morgan Kaufmann Publishers, Inc., 2004
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