Network analysis

Engineering Management 12/19/2013
Network Analysis
MD. MOUDUD HASAN
LECTURER,AIE,HSTU.
Network Analysis
 Network Analysis.- OR technique(s) for the solution of
problems based on the visual and conceptual representation
of the interrelationships between the components of a
system.
 A network consists of a set of points and a set of lines
connecting certain points.
The points are called nodes.
The lines are called arcs (edges, branches).
 The arcs may have certain flow.
 If the flow can go in only one direction in the arc, the arc is
said to be directed arc,
 if not the arc is undirected (or links).
•
M M Hasan, Lecturer, AIE, HSTU,
Dinajpur. 1

Examples of Network Applications
Network Terminology
 A network that has only directed arcs is said to
be a directed network. If all the arcs are
undirected, the network is undirected.
 A path between two nodes is a sequence of
arcs connecting them. A directed path from
node i to node j allows flow from node i to j. An
undirected path from i to j allows flow in either
direction.
Dinajpur. 2

Network Terminology
 A path that begins and ends at the same node is called
a cycle. A cycle can be directed or undirected if the
path is directed or undirected.
 Two nodes are connected if the network contains at
least one undirected path between them
 A connected network is a network where every pair of
nodes is connected.
 A tree is a connected network with no cycles. A
spanning tree is a tree connecting all the nodes of a
network.
 Arc capacity is the maximum amount of flow that can
be carried on a directed arc.
 In a supply node the flow out exceeds the flow in. The
opposite occurs in a demand node. In a transshipment
node, flow in = flow out.
Directed and Undirected Arcs
 Directed arcs- used to represent cases when flow is
allowed only one direction. Examples: traffic in one-way
streets, water flow in channels, etc. They are graphically
represented with arrows.
 Undirected Arcs (link)- used to represent cases when flow
is allowed in both directions. Examples: traffic in two-way
streets, pipes allowing flow in both direction, etc. They
are graphically represented with a simple straight line.
Dinajpur. 3

Directed and Undirected Networks
 Directed Network
1
2
3
4
5
 Undirected Network
6
7
8
1
2
3
4
6
5
7
Paths
1
2
3
4
6
7
5
Directed
Path
8
Undirected
Path
Directed
Cycle
Undirected
Cycle
Dinajpur. 4

Network Problems
 Some problems typically solved by using
network methods:
1. Minimum Spanning Tree
2. Shortest Route
3. Maximum Flow
4. Critical Path
Examples of Network Problems
 Minimum Spanning Tree- The objective is to connect all of the
nodes of a network with links that minimize the total “cost”
(“cost” could be time, distance, etc.)
• Common Examples:
 Design of computer networks
 Design of a network of roads
 Shortest Route- The objective is to minimize the “cost” to go
from a Node A (the origin) to a Node B (the destination)
 Selection of highways to go from City A to City B such that the total
traveled distance is minimized.
 Deciding what branches of a pipeline to build to connect City A to City B
such that the total cost is minimized.
Dinajpur. 5

Examples of Network Problems
 Maximum Flow- The objective is to maximize the
flow that can be sent from one node (the source) to
another (the sink)
 Traffic management
 Logistics systems
 Critical Path- The objective is to minimize the time it
takes to complete the total project
• Common Example:
 Project Management (CPM & PERT Networks)
Minimum Spanning Tree
 A minimum spanning tree is the set of the arcs of a
network that have the shortest total length while
providing a route (path) between each pair of nodes.
• For a spanning tree to be a minimum spanning tree, it has to
satisfy the path optimality conditions
Dinajpur. 6

Applications?
 Constructing highways or railroads spanning
several cities
 Laying pipelines connecting offshore drilling sites,
refineries, and consumer markets
 Designing computer networks
 Others?
Algorithm for Minimum Spanning
Tree Problem
 Procedure:
1. Select any node arbitrarily and then connect it
to the nearest distinct node.
2. Identify the unconnected node that is the closest
to a connected node, and then connect these two
nodes (ties may be broken arbitrarily). Repeat this
until all nodes have been connected.
Dinajpur. 7

Example1
1
3
8
2
6
7
8
7
7
8
6
2
6
4
7
Select any node
Connect the closest node
Select the closest node to
any of the connected nodes
V13, V37, V27, V67, V68,=1 w(T) = 7+6+2+6+4=25
Select the closest node to
any of the connected nodes
Example2
 Apply the procedure to solve the minimum
spanning tree for the graph shown below:
3
8
8
1
6 8
5
9
2
2 7
Dinajpur. 8

Kruskal’s Algorithm
(a more refined greedy algorithm)
 A more efficient algorithm (and also optimal) for obtaining
the minimum spanning tree
 Sort all the arcs in non-decreasing order of cost
 Define a set, LIST, that is the set of arcs chosen as part of
a minimum spanning tree
 Initially LIST is empty
 Examine the arcs in the sorted order one by one and
check whether adding the arc we are currently examining
to LIST creates a cycle with the arcs already in LIST. If
not, add arc to LIST, otherwise discard it.
 Terminate procedure when the cardinality of LIST equals
the number of vertices less one. The result is minimum
spanning tree T*
Example3
1
3
8
2
6
7
8
7
7
8
6
2
6
4
7
V27
From the sorted list, select the least cost arc
V27 ,V68
V27 ,V68 ,V37
V27 ,V68 ,V37 ,V67
From the sorted list select the least cost arc
Selecting this arc, will create a
cyclewe skip it
V27 ,V68 ,V37 ,V67 ,V13
Since the number of arcs is n-1 (number of nodes –1), we stop
Dinajpur. 9

Shortest Route Problem
 Objective: To determine the shortest route (path) from the source node to
a destination node.
 Alternatively, we could define the shortest path problem as determining
how to send 1 unit of flow as cheaply as possible from the source node to
another node in an incapacitated network.
 Corresponding to each arc ( i,j), there is a non-negative number dij called
the distance from i to j (dij =  if we cannot get from i to j directly).
 Example: Find the shortest route between Nodes 1 and 8.
1
3
1 2
2
4
5
4
6
7
8
2
1
1
2
5
3
3
7
6
3
5
2
6
Applications
 Want to send a truck from a DC to a customer.
 Want to send a message across a
telecommunications network.
 What does “shortest” mean in these cases?
Dinajpur. 10

Dijkstra's Algorithm
 Dijkstra’s algorithm is an efficient method to find
the shortest path between two nodes.
 Some of the characteristics of the algorithm:
• Assigns a temporary (unsolved) or permanent
(solved) label to each node in the network.
• The temporary (unsolved) label is an upper bound
on the shortest distance from the source (origin)
node to that node.
• The shortest route from the source to a node is
given by the permanent label.
Steps of the Shortest Path (Dikjstra’s) Algorithm
 Objective of the nth iteration: Find the nth nearest node to the
origin.
 Input for the nth iteration: n-1 nearest nodes to the origin
(solved at the previous iteration), including their shortest path
and the distance from the origin (call these solved nodes).
 Candidates for the nth nearest node: Each solved node that is
directly connected by a link to one or more unsolved nodes
provides one candidate - the unsolved node with the shortest
connecting link (ties provide additional candidates).
 Calculation of the nth nearest node: For each such solved
node and its candidate, add the distance between them and
the distance of the shortest path from the origin to the solved
node. The candidate with the smallest such total distance is
the nth nearest node (ties provide additional solved nodes),
and its shortest path is the one generating this distance.
Dinajpur. 11

EXAMPLE:
The Park management needs to find the shortest path
from the park entrance (node O) to the scenic wonder
(node T) through the road system shown in Fig.
A
2 2
5
4 1
4 T
3
4
7
D
1
5
7
o
C
B
E
n
Solved Nodes
Directly
Connected to
Unsolved Nodes
Closest
Connected
Unsolved
Node
Total
Distance
Involved
nth
Nearest
Node
Minimum
Distance
Last
connection
1 O A 2 A 2 OA
2
O C 4 C 4 OC
A B 2+2=4 B 4 AB
3
A D 2+7=9
B E 4+3=7 E 7 BE
C E 4+4=8
A D 2+7=9
B D 4+4=8 D 8 BD
E D 7+1=8 D 8 ED
D T 8+5=13 T 13 DT
E T 7+7=14
Possible Shortest paths are
i) OA(2) --- AB(2)---BD(4)-----DT(5)=13
ii) OA(2) --- AB(2)---BE(3)--ED(1)-----DT(5)=13
Dinajpur. 12

Problem:
Find the shortest path from the node O to the node T
through the road system shown in Fig.
7
1
3
4
4
3
1
E
6
1
B
C
1
2
A
4
o
D
F
T
5
Maximum Flow :
Flow Network Definition
 It is a weighted directed graph G=(V,E) where every
edge has a non-negative weight called capacity
and receives a flow on each edge.
 It has two distinguished vertices s and t (source
and sink) such that s has no incoming edges and t
has no outgoing edges.
Application:
Flow networks can be used to model traffic in roads,
fluids flow in pipes, current flow in electrical
systems, etc
Dinajpur. 13

2. Flow definition
It is a function f : V x V → R that satisfies
o Capacity constraint
o Flow conservation : for u, v Є V- {s,t}
 The value of the flow, denoted |f|, is
the total flow from the source s; and in
the same way, to the sink t.
Max Flow Problem
Given a network G, a flow is said to
be maximum if |f| is the largest of
all flows in G.
The Max Flow Problem consists of
finding the maximum feasible flow
from s to t without violating any
capacity constraints.
| f | max = 4 + 3 + 3 = 10
Dinajpur. 14

• Given a flow network and a flow, the residual
network consists of edges that can admit more net
flow.
• G=(V,E) --a flow network with source s and sink t
• f: a flow in G.
• The amount of additional net flow from u to v
before exceeding the capacity c(u,v) is the residual
capacity of (u,v), given by: cf(u,v)=c(u,v)-f(u,v)
in the other direction: cf(v, u)=c(v, u)+f(u, v).
29
Residual networks:
20
4/4
30
s
Example of residual network
20
7
12
16 v1 v3
9
10 4
v2 v4
t
13
4
14
4/12
4/16 v1 v3
4/9
s
10 4
v2 v4
t
13
7
4/14
(a)
Dinajpur. 15

31
8
Example of Residual network (continued)
12 v1 v3
5
s
10 4
4
4
10
v2 v4
t
13
20
4
7
4
4
(b)
• Residual capacity of p : the maximum amount of
net flow that we can ship along the edges of an
augmenting path p, i.e., cf(p)=min{cf(u,v):(u,v) is
on p}.
32
Augmenting paths:
• Given a flow network G=(V,E) and a flow f, an
augmenting path is a simple path from s to t in
the residual network Gf.
2 3 1
The residual capacity is 1.
Dinajpur. 16

33
8
Example of an augment path (bold edges)
12 v1 v3
5
s
10 4
4
4
10
v2 v4
t
13
20
4
7
4
4
(b)
DDeetteerrmmiinnee tthhee mmaaxxiimmuumm ffllooww iinn tthhee nneettwwoorrkk ooff
this figure. Source 1 to sink 5.
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Dinajpur. 17

Steps:
• Step 1: start from source node (S1)
• Step 2: find connection with that node where flow is possible. e.g – t2,
t3, t4
• Step3: identify maximum flow capacity among these node. e.g- t3
• Step 4: connect source node (S1) to maximum capacity node (t3) .
• Step 5: now find connection with t3 where flow is possible. e.g. – t4 , t5
• Step 6: identify maximum flow capacity among these node. e.g- t5
• Step 6: connect source node (t3) to maximum capacity node (t5) .
• Step 7: continue this until reach at sink node.
• Step 8: find minimum flow capacity along the augmenting path
and that is the assign flow fi.
• Step 9: draw residual network.
• Step 10: repeat step 1 to 9 until there are no more augmenting paths.
• Step 11: find maximum flow capacity, = Σ
Step-1
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Dinajpur. 18

Step-2
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Step-3
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Dinajpur. 19

Step-4
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Step-5
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Dinajpur. 20

Step-6
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Step-7
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Dinajpur. 21

Step-8
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
Step-9
10
1
5
0
0
5
0
0
20
20
0
0
0
2
4
10
3
30
20
30
40
f1=20
Dinajpur. 22

Iteration-1
Step-10
Cr13=(30-20)=10
Cr35=(20-20)=0
10
1
Augmenting path: 1-3-5
f1=20
5
20
0
20
0
20
0
0
0
2
4
10
3
5
10
20
0
30
40
Iteration-2
Step-10
10
1
Augmenting path: 1-2-3-4-5
f2=10
10
15
20
10
20
0
10
10
0
2
4
0
3
5
10
10
0
30
30
Dinajpur. 23

Iteration-3
Step-10
10
1
Augmenting path: 1-2-5
10
f3=10
15
20
10
20
0
10
20
0
2
4
0
3
5
10
0
10
20
30
Iteration-4
Step-10
10
1
Augmenting path: 1-3-2-5
10
f4=10
15
30
0
20
0
10
20
0
2
4
0
3
5
0
0
20
10
40
Dinajpur. 24

Iteration-5
Step-10
1
Augmenting path: 1-3-2-5
0
f5=10
15
30
0
20
0
20
20
10
0
2
4
0
3
5
0
0
20
10
40
Iteration-6
Step-10
1
Augmenting path: no more path
0
15
30
0
20
0
20
20
10
0
2
4
0
3
5
0
0
20
10
40
Dinajpur. 25

Step:11
Iteration Augmenting path Flow amount
1 1 -ᗒ3 -ᗒ5 20
2 1 -ᗒ2 -ᗒ3 -ᗒ4 -ᗒ5 10
3 1 -ᗒ2 -ᗒ5 10
4 1 -ᗒ3 -ᗒ2-ᗒ5 10
5 1 -ᗒ4 -ᗒ5 10
Maximum flow: 60
Dinajpur. 26

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